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$\cos \left(\frac{2\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{4\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{6\pi }{7}\right)^{\frac{1}{3}} =?$ I have been trying to solve the following question for a long time:
Find $a,b,c,d$ such that: ($a,b,c,d$ are primes)
$\cos \left(\frac{2\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{4\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{6\pi }{7}\right)^{\frac{1}{3}} = \left(\frac{a-b\sqrt[3]{c}}{d}\right)^{\frac{1}{3}} $
What I have tried so far is that:
Let $x=\cos \left(\frac{2\pi }{7}\right), y=\cos \left(\frac{4\pi }{7}\right),z=\cos \left(\frac{6\pi }{7}\right)$, then
$x+y+z=-1/2$,
$xyz =1/8 $
$xy+yz+zx=-1/2$
So, $x,y,z$ are the roots of $8t^3+4t^2-4t-1=0$ and can be found by Cardan's method and hence $x^{1/3}+y^{1/3}+z^{1/3}$ can be found. However that method was way too long and it seems that the question may have a more elegant solution (given the form).
Also, I tried to cube both sides of the equation in the question and substitute using the above three relations, but that required the value of $x^{2/3}+y^{2/3}+z^{2/3}$, which I coundn't find.
Would someone pls help me out (without using Cardan)?
Thanks in advance!
| Try Ramanujan's cubic polynomial, this polynomial don't need to find 3 roots. If coefficients are satisfied the condition, then sum of cuberoots of zeroes can be evaluated directly from its coefficients.
EDIT: With your results, you established the equation: $8t^3+4t^2-4t-1=0$ which is equivalent $t^3+\frac{1}{2}t^2-\frac{1}{2}t-\frac{1}{8}=0$
The Ramanujan's cubic polynomial said that: If the polynomial $x^3+px^2+qx+r$ has 3 real roots $x_1, x_2, x_3$ and the relation $pr^{\frac{1}{3}}+3r^{\frac{2}{3}}+q=0$ then $x_1^{\frac{1}{3}}+ x_2^{\frac{1}{3}}+ x_3^{\frac{1}{3}}=\left(-p-6r^{\frac{1}{3}}+3(9r-pq)^{\frac{1}{3}}\right)^{\frac{1}{3}}$
Back to our case: $p=\frac{1}{2}, q=-\frac{1}{2}, r=-\frac{1}{8}$ and $pr^{\frac{1}{3}}+3r^{\frac{2}{3}}+q=\left(\frac{1}{2}\right)\left(-\frac{1}{8}\right)^{\frac{1}{3}}+3\left(-\frac{1}{8}\right)^{\frac{2}{3}}-\frac{1}{2}=-\frac{1}{4}+\frac{3}{4}-\frac{1}{2}=0$ then you can calculate the sum of cuberoot of zeroes follow the relation with $x_1, x_2, x_3$. Thanks for reading.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
| You want to divide by $x^3+x = x(x^2+1)$, hence by the Chinese remainder theorem it is enough to check the remainders $\pmod{x}$ (which is obviously zero) and $\pmod{x^2+1}$. This remainder is simply given by setting $x^2\equiv -1$, such that
$$ x^{10}+x^{11}+\ldots+x^{19}+x^{20} \equiv (-1-x+1+x)+(-1-x+1+x)+(-1-x+1+x)\color{red}{-x} $$
and option $(B)$ is apparent now.
| {
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"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $ \frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx$ I recently came across a problem on definite integration and couldn't solve it despite my efforts. It goes as
$$
\frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx
$$
The $2-x^2$ term and the upper limit along with the $2+x^2$ term at the bottom motivated me to substitute $x=2^{1/2}\tan a$. However subsequent steps proceeded to a stage I couldnt simplify.
I tried other substitutions but was unable to proceed
Any help to solve this problem would be appreciated. Please feel free to share your first thoughts and intuition as well.
| After the first step from @ClaudeLeibovici’s answer, you can do a crazy u-sub: $$u=\frac{x^2-1}{x^2+1}$$ so $$x=\sqrt{\frac{1+u}{1-u}}$$ By chain rule, $$\implies dx=\frac{1}{2\sqrt{\frac{1+u}{1-u}}}\cdot\frac{2}{(1-u)^2}du$$$$=\frac{du}{(1+u)^\tfrac12(1-u)^\tfrac32}$$ Also, $$\frac{1}{\sqrt{1+x^4}}= \frac{1}{\sqrt{1+\left(\frac{1+u}{1-u}\right)^2}}=\frac{1-u}{\sqrt{2(1+u^2)}}$$ so the integral simplifies to $$-\int u\cdot\left(\frac{1-u}{\sqrt{2(1+u^2)}}\right)\frac{du}{(1+u)^\tfrac12(1-u)^\tfrac32}$$$$=-\int\frac{udu}{\sqrt{2(1+u^2)} (1+u)^\tfrac12(1-u)^\tfrac12}$$$$=-\int\frac{udu}{\sqrt{2(1-u^4)}}=-\frac{1}{2\sqrt2}\arcsin(u^2)+C$$ Putting appropriate limits, you get $\displaystyle\frac{\pi}{4\sqrt2}$ so the final answer should be $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine the power series representation of the function $f(x)=\sqrt{(4+x)^3}$ and indicate the radius of convergence I want to find a representation of the function mentioned above, so I took into account that:
$$f(x)=\sqrt{(4+x)^3}=8\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$$ and developing the binomial series for $\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$ we have to
$$\left(1+\frac{x}{4}\right)^{\frac{3}{2}}=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdot\cdot\cdot$$
Now, to obtain the terms of the original series simply multiply by 8; now, how can I express the function as a series, since I have only managed to express that as a sum of terms
Any help is appreciated
| Starting from
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\,x^n \tag{1}$$
which is important and should be treated as a fundamental "building block", in my opinion
one gets by termwise integration
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(1-2n)}\,x^n \tag{2}$$
and by integrating again
$$ (1-x)^{3/2} = \sum_{n\geq 0}\frac{3\binom{2n}{n}}{4^n(2n-1)(2n-3)}\,x^n. \tag{3}$$
By replacing $x$ with $-x/4$ we get
$$ \left(1+\frac{x}{4}\right)^{3/2} = \sum_{n\geq 0}\frac{3\binom{2n}{n}(-1)^n}{16^n(2n-1)(2n-3)}\,x^n \tag{4}$$
so
$$\boxed{ \sqrt{(4+x)^3} = \sum_{n\geq 0}\frac{24\binom{2n}{n}(-1)^n}{16^n(2n-1)(2n-3)}\,x^n} \tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503835",
"timestamp": "2023-03-29T00:00:00",
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How to calculate $\int_0^{2\pi}\frac{\cos(\phi)-R}{1-2R\cos(\phi)+R^2}\cos(n\phi)~d\phi$? I wish to calculate
$$I(R)=\int_0^{2\pi}\frac{\cos(\phi)-R}{1-2R\cos(\phi)+R^2}~\cos(n\phi)~d\phi,$$
where $n\in\mathbb{N}$, $R\in[0,1)$.
Based on trial and error from plugging numbers into Wolfram alpha I think the answer is
$$I(R)=\begin{cases}
0, & n=0, \\
\pi R^{n-1}, & n\ge1.
\end{cases}$$
However, I don't know how to show this more rigorously. Do you know how to calculate the above integral, and do you know if my formula is correct?
| Define: $z=e^{i\phi}$
$$\cos(\phi)=\frac{1}2\left(z+\frac{1}z \right),~~~~\cos(n\phi)=\frac{1}2\left(z^n+\frac{1}{z^n} \right),~~~~d\phi=\frac{1}{iz}dz$$
If $n=0$,
$$\begin{align}
I&=\frac{1}{2i}\oint \frac{z^2-2Rz+1}{z(z-R)(1-Rz)}dz=\frac{1}{2i}\cdot 2\pi i\cdot \left(Res[z=0]+Res[z=R]\right)\\
\\
&=\pi\cdot\left( -\frac{1}R+\frac{1-R^2}{R(1-R^2)}\right)=0
\end{align}$$
If $n\ge 1$
$$\begin{align}
I&=\frac{1}{4i}\oint \frac{(z^2-2Rz+1)z^{n-1}}{(z-R)(1-Rz)}dz+\frac{1}{4i}\oint \frac{(z^2-2Rz+1)}{z^{n+1}(z-R)(1-Rz)}dz=I_1+I_2\\
\\
I_1&=\frac{1}{4i}\cdot 2\pi i\cdot Res[z=R]=\frac{\pi}2R^{n-1}\\
\\
I_2&=\frac{1}{4i}\cdot 2\pi i\cdot \left(Res[z=0]+Res[z=R] \right)=\frac{\pi}2 \left( -\frac{1}{R^{n+1}}+R^{n-1}+\frac{1}{R^{n+1}} \right)\\
\\
I&=\pi\cdot R^{n-1}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .
Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{j}(x-1)^j. \end{align*}
Hence \begin{align*} f^{(n)}(1)&=n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2. \end{align*}
Is it correct? How to compute $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$
| A partial answer
Let
$$F=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}$$
So we can write $F=G+H$, where
$$G=\sum_{j=0}^{n+1}{2n+2\choose 2j}{j \choose n}, \quad H=\sum_{j=1}^{n+1} {2n+2 \choose 2j-1}{j-1/2 \choose n}$$
Only 2 terms in $G$ are nonzero, when $j=n,n+1$.
Hence $$G={2n+2 \choose 2n} {n \choose n}+{2n+2 \choose 2n+2}{n+1 \choose n}=2(n+1)^2.$$
Though Mathematica gives $H$ in terms of hypergemetric $_pF_q$, which gives $H=2(n+1)^2$ (numerically) again!
I wish to come back.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Interesting integral $\int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$? Noting that
$\displaystyle d(x \sin x+\cos x)=x \cos xdx,\tag*{} $
we have
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x=-\int_{0}^{\frac{\pi}{4}} \frac{x}{\cos x} d\left(\frac{1}{x \sin x+\cos x}\right)\tag*{} $
Integration by parts gives
$\displaystyle \begin{aligned}I &=-\left[\frac{x}{\cos x(x \sin x+\cos x)}\right]_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}} \frac{\cos x+x \sin x}{\cos ^{2} x} \cdot \frac{1}{x \sin x+\cos x} d x \\&=-\frac{\frac{\pi}{4}}{\frac{1}{\sqrt{2}}\left(\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)}+\int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\&=-\frac{2 \pi}{\pi+4}+[\tan x]_{0}^{\frac{\pi}{4}} \\&=\frac{4-\pi}{4+\pi}\end{aligned}\tag*{} $
Is there any other method?
| Well, we are trying to solve:
$$\mathcal{I}\left(x\right):=\int\underbrace{\left(\frac{x}{x\sin\left(x\right)+\cos\left(x\right)}\right)^2}_{:=\space\mathscr{I}\left(x\right)}\space\text{d}x\tag1$$
Rewrite the integrand using $\sin^2\left(x\right)+\cos^2\left(x\right)=1$:
$$\mathscr{I}\left(x\right)=\left(\frac{x}{x\sin\left(x\right)+\cos\left(x\right)}\right)^2=\frac{x^2\sin^2\left(x\right)+x^2\cos^2\left(x\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}\tag2$$
Look to cancel terms with the denominator by adding and subtracting a term from the numerator:
$$\mathscr{I}\left(x\right)=\frac{x^2\sin^2\left(x\right)+x^2\cos^2\left(x\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}=$$
$$\frac{x^2\cos^2\left(x\right)-x\sin\left(x\right)\cos\left(x\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}+\frac{x^2\sin^2\left(x\right)+x\sin\left(x\right)\cos\left(x\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}\tag3$$
Factor and cancel terms:
$$\mathscr{I}\left(x\right)=\frac{x\cos\left(x\right)\left(x\cos\left(x\right)-\sin\left(x\right)\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}+\frac{x\sin\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}\tag4$$
Integrate the sum term by term:
$$\mathcal{I}\left(x\right)=\int\frac{x\cos\left(x\right)\left(x\cos\left(x\right)-\sin\left(x\right)\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}\space\text{d}x+\int\frac{x\sin\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}\space\text{d}x\tag5$$
Using IBP: $\int\text{f}\space\text{dg}=\text{fg}-\int\text{g}\space\text{df}$, where $f=\sin\left(x\right)-x\cos\left(x\right)$, $\text{dg}=-\frac{x\cos\left(x\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)^2}\space\text{d}x$, $\text{g}=\frac{1}{x\sin\left(x\right)+\cos\left(x\right)}$ and $\text{df}=x\sin\left(x\right)\space\text{d}x$. This leads to:
$$\mathcal{I}\left(x\right)=\frac{\sin\left(x\right)-x\cos\left(x\right)}{\cos\left(x\right)+x\sin\left(x\right)}\underbrace{-\int\frac{x\sin\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}\space\text{d}x+\int\frac{x\sin\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}\space\text{d}x}_{=\space0}\tag6$$
So, we end up with:
$$\mathcal{I}\left(x\right)=\frac{\sin\left(x\right)-x\cos\left(x\right)}{\cos\left(x\right)+x\sin\left(x\right)}+\text{C}\tag7$$
So, your definite integral gives:
$$\mathcal{I}\left(\frac{\pi}{4}\right)-\mathcal{I}\left(0\right)=\frac{8}{4+\pi}-1-0=\frac{8}{4+\pi}-1=\frac{4-\pi}{4+\pi}\tag8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solve the differential equation: $6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$ I have the following differential equation before me:
$6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$
I tried solving it by reducing to Bernoulli form of first order differential equation.
I divided both sides of the equation by $6 \cos^2(x)y^4$ to get:
$\dfrac{1}{y^4} \dfrac{dy}{dx}-\dfrac{1}{y^3} \dfrac{\sin(x)}{6\cos^2(x)}=\dfrac{-\sin^3(x)}{3\cos^2(x)}$
Then I took $\dfrac{1}{y^3}=z$
so that $\dfrac{-3}{y^4}\dfrac{dy}{dx}=\dfrac{dz
}{dx}$
This gave me:
$\dfrac{dz}{dx}+\dfrac{\sin(x)}{2\cos^2(x)}z=\dfrac{\sin^3(x)}{\cos^2(x)}$
This is first order linear differential equation of the type
$ \dfrac{dz}{dx}+P(x)z=Q(x)$
Integrating Factor (IF) is given by:
$IF=e^{\int P(x) \,dx}= e^{\int \dfrac{\sin(x)}{2\cos^2(x)}\,dx}= e^{\dfrac{1}{2\cos(x)}}$
Next, solution is given by:
$z.IF= \int IF.Q(x)$
or
$z.e^{\dfrac{1}{2\cos(x)}}=\int e^{\dfrac{1}{2\cos(x)}}.\dfrac{\sin^3(x)}{\cos^2(x)} \,dx$
This is the step where I falter as I find myself unable to tackle the integral on the RHS of the above equation.
You are requested to help me with the evaluation of this integral or come up with another way of tackling this differential equation. Any help will be highly appreciated.
| I am writing $\frac{dy}{dx}$ as $y'$
Our equation is $$6\operatorname{cos}^2(x)y'=\operatorname{sin}(x)y-2\operatorname{sin}^3(x)y^4$$ $=$
$$y'-\frac{\operatorname{sin}(x)y}{6\operatorname{cos}^2(x)}=-\frac{\operatorname{sin}^3(x)y^4}{3\operatorname{cos}^2(x)}$$ Dividing by $y^4$,
$$\frac{y'}{y^4}-\frac{\operatorname{sin}(x)}{6\operatorname{cos}^2(x)y^3}=-\frac{\operatorname{sin}^3(x)}{3\operatorname{cos}^2(x)}$$
Let $u=\frac{1}{y^3}$ therefore $y=\frac{1}{u^{\frac13}}$ and $u'=\frac{-3y'}{y^4}$ and $y'=\frac{-u'y^4}{3}$
Substituting all the values, our equation becomes $$-\frac{u\operatorname{sin}(x)}{6\operatorname{cos}^2(x)}-\frac{u'}{3}=-\frac{\operatorname{sin}^3(x)}{3\operatorname{cos}^2(x)}$$ which is equivalent to $$\frac{u\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+u'=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ Let $u=tv$ where $t,v$ are two functions $\implies$ $u'=tv'+t'v$ Rewriting the equation in terms of functions and then grouping we get, $$tv'+v\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ Solving the first equation $$\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=0$$ By seeing the work that you showed in your question, I am pretty sure you can calculate this equation easily. So after solving we get $$t=\frac{1}{\sqrt[2\operatorname{cos}(x)]e}$$ Now solving the second equation $$tv'+v\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ at $$t=\frac{1}{\sqrt[2\operatorname{cos}(x)]e}$$ and$$\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=0$$ Therefore the second equation becomes $$\frac{v'}{\sqrt[2\operatorname{cos}(x)]e}=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ After transforming things $$dv=\frac{\sqrt[2\operatorname{cos}(x)]e\operatorname{sin}^3(x)\:\:dx}{\operatorname{cos}^2(x)}$$ Integrating both sides,
$$\int1\cdot dv=\int\frac{\sqrt[2\operatorname{cos}(x)]e\operatorname{sin}^3(x)\:\:dx}{\operatorname{cos}^2(x)}$$ $\implies$
$$v=-\frac{\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)}{2}+\sqrt[2\operatorname{cos}(x)]e\operatorname{cos}(x)+2\cdot\sqrt[2\operatorname{cos}(x)]e+C$$ or
$$u=-\frac{\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)-2\cdot\sqrt[2\operatorname{cos}(x)]e\operatorname{cos}(x)-4\cdot\sqrt[2\operatorname{cos}(x)]e-2C}{2\cdot\sqrt[2\operatorname{cos}(x)]e}$$ $\implies$
$$\frac{1}{y^3}=\frac{C-\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)}{2\cdot\sqrt[2\operatorname{cos}(x)]e}+\operatorname{cos}(x)+2$$ or
$$y=-\frac{\sqrt[3]2\cdot\sqrt[6\operatorname{cos}(x)]e}{\sqrt[3]{\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)-2\cdot\sqrt[2\operatorname{cos}(x)]e\operatorname{cos}(x)-4\cdot\sqrt[2\operatorname{cos}(x)]e-C}}$$ This is the required answer and the solution $y=0$ is achieved when $C=\infty$
| {
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"url": "https://math.stackexchange.com/questions/4510057",
"timestamp": "2023-03-29T00:00:00",
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If $x^2+2(\alpha-1)x-\alpha+7=0$ has distinct negative solutions... Let $\displaystyle{ \alpha }$ be real such that the equation $\displaystyle{ x^2+2(\alpha-1)x-\alpha+7=0 }$ has two different real negative solutions. Then
*
*$ \ \displaystyle{ \alpha<-2 }$ ;
*$ \ \displaystyle{ 3<\alpha<7 }$ ;
*it is impossible ;
*none of (a)-(c).
$$$$
I have done the following :
The value $\alpha$ is real and such that the equation $x^2+2(\alpha-1)x-\alpha+7=0$ has two different real negative solutions.
The solutions of the equation are given from the quadratic formula \begin{align*}x_{1,2}&=\frac{-2(\alpha-1)\pm \sqrt{[2(\alpha-1)]^2-4\cdot 1\cdot (-\alpha+7)}}{2}\\ & =\frac{-2(\alpha-1)\pm \sqrt{4(\alpha^2-2\alpha-1)-4\cdot (-\alpha+7)}}{2}\\ & =-(\alpha-1)\pm \sqrt{2(\alpha^2-2\alpha-1)-2\cdot (-\alpha+7)} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-4\alpha-2+2\alpha-14} \\ & =-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}\end{align*}
So that we have two different solutions the discriminant must be non-zero.
So that we have two negative solutions, it must hold $-(\alpha-1)\pm \sqrt{2\alpha^2-2\alpha-16}<0$.
So that we have real solutions the expression under the square root must be non negative.
The expression under the square root has the sign of the coefficient of $x^2$, i.e. positive, outside the roots.
We have that \begin{equation*}2\alpha^2-2\alpha-16=0 \Rightarrow \alpha_{1,2}=\frac{1}{2}\pm \frac{\sqrt{33}}{2}\end{equation*}
So we have that the expression under the square root if $\alpha<\frac{1}{2}- \frac{\sqrt{33}}{2}$ and if $\alpha>\frac{1}{2}+ \frac{\sqrt{33}}{2}$.
Is my attempt correct so far? Now do we check if the first two intervals of $\alpha$ can hold for all these conditions? Or how do we continue? Or is there a better way to solve that exercise ?
| If $r$ and $s$ are the roots of $x^2+2(\alpha-1)x-\alpha+7$, then$$\left\{\begin{array}{l}r+s=2-2\alpha\\rs=7-\alpha.\end{array}\right.$$Therefore$$\left\{\begin{array}{l}2-2\alpha<0\\7-\alpha>0;\end{array}\right.$$in other words, $\alpha\in(1,7)$. But, in fact, we cannot have $\alpha\in(1,3]$, because $(r-s)^2>0$, and\begin{align}(r-s)^2>0&\iff(r+s)^2-4rs>0\\&\iff(2-2\alpha)^2-4(7-\alpha)>0\\&\iff4\alpha^2-4\alpha-24>0\\&\iff\alpha^2-\alpha+6>0\\&\iff(\alpha+2)(\alpha-3)>0\\&\iff\alpha<-2\vee\alpha>3.\end{align}So, $\alpha\in(3,7)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}$ where $x,y,z>0$ and $x+y+z=3$.
Prove $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x},$$ where $x,y,z>0$ and $x+y+z=3$.
Maybe we can show $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge x^3+y^3+z^3,\tag1$$
then $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\ge \frac{1}{2}\left(\frac{x}{y^2}+x^3+\frac{y}{z^2}+y^3+\frac{z}{x^2}+z^3\right)\ge \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}.$$
But it's also difficlut to show $(1)$.
| Let $z=\max\{x,y,z\}.$
Thus, $$\sum_{cyc}\left(\frac{x}{y^2}-\frac{x^2}{y}\right)=$$
$$=\left(\tfrac{x}{y^2}+\tfrac{y}{z^2}+\tfrac{z}{x^2}-\tfrac{1}{x}-\tfrac{1}{y}-\tfrac{1}{z}\right)-\left(\tfrac{x^2}{y}+\tfrac{y^2}{z}+\tfrac{z^2}{x}-x-y-z\right)+\sum_{cyc}\left(\tfrac{1}{x}-1\right)=$$
$$=\left(\tfrac{x}{y^2}+\tfrac{y}{x^2}-\tfrac{1}{x}-\tfrac{1}{y}+\tfrac{y}{z^2}+\tfrac{z}{x^2}-\tfrac{y}{x^2}-\tfrac{1}{z}\right)-\left(\tfrac{x^2}{y}+\tfrac{y^2}{x}-x-y+\tfrac{y^2}{z}+\tfrac{z^2}{x}-\tfrac{y^2}{x}-z\right)+$$
$$+\frac{1}{3}\left(\tfrac{x}{y}+\tfrac{y}{x}-2+\tfrac{y}{z}+\tfrac{z}{x}-\tfrac{y}{x}-1+\tfrac{x}{y}+\tfrac{y}{x}-2+\tfrac{x}{z}+\tfrac{z}{y}-\tfrac{x}{y}-1\right)=$$
$$=\left(\tfrac{(x-y)^2(x+y)}{x^2y^2}+\tfrac{(z-x)(z-y)(x+z)}{x^2z^2}\right)-\left(\tfrac{(x-y)^2(x+y)}{xy}+\tfrac{(z-x)(z-y)(y+z)}{xz}\right)+$$
$$+\frac{1}{3}\left(\frac{2(x-y)^2}{xy}+(z-x)(z-y)\left(\frac{1}{xz}+\frac{1}{yz}\right)\right)=$$
$$=(x-y)^2\left(\tfrac{x+y}{x^2y^2}-\tfrac{x+y}{xy}+\tfrac{2}{3xy}\right)+(z-x)(z-y)\left(\tfrac{x+z}{x^2z^2}-\tfrac{y+z}{xz}+\tfrac{x+y}{3xyz}\right).
$$
We'll prove that:$$\frac{x+y}{x^2y^2}-\frac{x+y}{xy}+\frac{2}{3xy}\geq0.$$
Indeed, by the River Li's beautiful point since $x+y\leq2$, by AM-GM we obtain:
$$\frac{x+y}{x^2y^2}-\frac{x+y}{xy}+\frac{2}{3xy}>\frac{x+y}{xy\left(\frac{x+y}{2}\right)^2}-\frac{x+y}{xy}\geq0.$$
Id est, it's enough to prove that:
$$\frac{x+z}{x^2z^2}-\frac{y+z}{xz}+\frac{x+y}{3xyz}\geq0$$
or
$$\frac{(x+y+z)(x+z)}{xz}+\frac{x+y}{y}+\frac{9x}{x+y+z}-3(x+y+z)\geq0$$ or
$$\frac{x^2+z^2+y(x+z)}{xz}+\frac{x}{y}+\frac{9x}{x+y+z}\geq6$$ or $f(z)\geq0,$ where $$f(z)=yz^3+(x^2-5xy+2y^2)z^2+(x^3+5x^2y-4xy^2+y^3)z+xy(x+y)^2.$$
But $$f''(z)=6yz+2(x^2-5xy+2y^2)\geq$$
$$\geq3y(x+y)+2(x^2-5xy+2y^2)=2x^2-7xy+7y^2>0,$$
which says $$f'(z)=3yz^2+2(x^2-5xy+2y^2)z+x^3+5x^2y-4xy^2+y^3\geq$$
$$\geq\frac{3y(x+y)^2}{4}+(x+y)(x^2-5xy+2y^2)+x^3+5x^2y-4xy^2+y^3=$$
$$=\frac{8x^3+7x^2y-22xy^2+15y^3}{4}>0,$$
which says $$f(z)\geq f\left(\frac{x+y}{2}\right)=\frac{3(x+y)(2x^3+7x^2y-4xy^2+3y^3)}{8}>0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically? I'm trying to solve the said equation in the thread title algebraically.
$$(x^3+1)=2\sqrt[3]{2x-1}$$
Cubing both sides and simplifying:
$$x^9+3x^6+3x^3-16x+9 = 0$$
Not sure if this can be solved algebraically?
Edit: WA gives $3$ solutions $x=1,\frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(-1+\sqrt{5})$
| Rearrange as $$\underbrace{\frac{x^3+1}{2}}_{f(x)}=\underbrace{\sqrt[3]{2x-1}}_{g(x)}$$
Since $f(x)$ is a bijective function on $\mathbb R$, it must have an inverse. But note that the inverse of $f(x)$ is $g(x)$. Thus, if the two curves intersect, they must intersect ON the line $y=x$ (because $f$ and $g$ are mirror images about this line).
Thus we solve the two curves $y=\dfrac{1+x^3}{2}$ and $y=x$ to get $$x^3-2x+1=0$$ which boils down to $$(x-1)(x^2+x-1)=0$$ which can be readily solved to get solutions $\displaystyle 1, \frac{-1\pm\sqrt5}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Bounding a simple ratio of bivariate functions For any $x,y \ge 0$, define $f(x,y) := x^2+y^2 + 2c xy$, where $c := \sqrt{2/\pi}$. It is clear that
$$
\sqrt{2/\pi}\cdot (x+y)^2 \le f(x,y) \le (x+y)^2.
\tag{1}
$$
This is an immediate consequence of the fact that $c \le 1 \le 1/c$.
Question. What is a good upper-bound for $\alpha := \sup_{x,y}\dfrac{(x+y)^2}{f(x,y)}$ ?
Note that (1) gives the trivial upper-bound $\alpha \le \sqrt{\pi/2}$.
| It is natural to guess that the maximum is achieved when $x=y$, that is:
\begin{equation}
\frac{(x+y)^2}{x^2+y^2+2cxy}\le \frac{2}{1+c},
\end{equation}
which is equivalent to
\begin{equation}
2(x^2+y^2+2cxy) \ge (1+c)(x+y)^2,
\end{equation}
which is true because $LHS-RHS = (1-c)(x-y)^2 \ge 0$.
| {
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The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th term becomes $(n-1) \cdot n$ and so it's summation becomes $\frac{n(n+1) (2n+1)}{6} - \frac{n(n+1)}{2}$ but it's the same summation as above but the results are different what am I doing wrong?
| This reduces the problem to the evaluation of a second derivative.
$$
1\cdot 2 + 2\cdot 3+ 3\cdot 4+ \cdots + n\cdot (n+1)
= \sum_{k=1}^{n} k(k+1) \\
= \left.\frac{d^2}{dx^2}(1+x+x^2+x^3+\cdots+x^{n+1})\right|_{x=1} \\
= \left.\frac{d^2}{dx^2}\frac{x^{n+2}-1}{x-1}\right|_{x=1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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I can't come up with the intended solution to $\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18$ I was going trough some easy algebra problems when I encountered
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18.
$$
As you can see the problem is easily solvable with AM > GM
I fairly quickly came up with this solution:
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab
= \frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2}
$$
and using AM-GM
$$
\frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2} \ge 7\sqrt[7]{\frac{a^{10}b^{10}9^8}{a^{10}b^{10}2^{10}3^3}} \approx 20
$$
(Yes I was able to do that by hand and later check with my calculator)
I am not sure that this is the intended solution though
| Here is the intended solution (this is from a Junior Balkan Mathematical Olympiad TST of Bulgaria).
By repeated use of AM-GM,
$$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab \geq \frac{2ab}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + \frac{9ab}{2} + \frac{9ab}{2} \geq 4 \sqrt[4]{\frac{9^4}{2^4}} = 18$$
For equality to hold, we must have $a=b$, $\frac{9ab}{2} = \frac{81a^2b^2}{4}$ and $ \frac{9ab}{2} = \frac{1}{a^4b^4}$, which is not possible.
I did not know before that there is a simple AM-GM solution which gives a better integer constant than $18$, so nice work!
| {
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Arc length of curve stuck with integration Data from exercise
$$y=\frac{4}{3}x^2+2\\
x\in[-1,1]$$
Formula for length of curve
$$L=\int_a^{b}\sqrt{1+(f(x)')^2}\ dx$$
So far i have
$$y'=\frac{8}{3}x$$
$$\int_{-1}^{1}\sqrt{1+\frac{64}{9}x^2}\ dx$$
Substition
$$t^2=\frac{64}{9}x^2$$
$$t=\frac{8}{3}x$$
$$\frac{3}{8}dt=dx$$
So i have
$$\int_{-\frac{8}{3}}^{\frac{8}{3}}\sqrt{1+t^2}\ dt$$
I this point i dont have a clue how to integrate this
| Call the indefinite integral $I$
$$I\equiv\int\sqrt{1+t^2}\,\mathrm dt$$
Through integration by parts with $u=\sqrt{1+t^2}$ and $v=x$, then
$$I=t\sqrt{1+t^2}-\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt$$
The integrand can also be rewritten as
\begin{align*}
I & =\int\frac {1+t^2}{\sqrt{1+t^2}}\,\mathrm dt\\ & =\int\frac {\mathrm dt}{\sqrt{1+t^2}}+\int\frac {t^2}{\sqrt{1+t^2}}\,\mathrm dt
\end{align*}
Adding the two expressions for $I$ together and dividing both sides by two eliminates one of the integral terms.
\begin{align*}
I & =\frac 12t\sqrt{1+t^2}+\frac 12\int\frac {\mathrm dt}{\sqrt{1+t^2}}
\end{align*}
The last integral is somewhat well-known and can be easily evaluated with an Euler Substitution $x=t+\sqrt{1+t^2}$.
$$\int\frac {\mathrm dt}{\sqrt{1+t^2}}=\log\left(t+\sqrt{1+t^2}\right)$$
Therefore, the indefinite version of the integral becomes
$$\int\sqrt{1+t^2}\,\mathrm dt\color{blue}{=\frac 12t\sqrt{1+t^2}+\frac 12\log\left(t+\sqrt{1+t^2}\right)+C}$$
Now, substitute in the bounds for $t=8/3$ and $t=-8/3$ to arrive at your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches)
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.
I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\min(\sqrt{x^2+y^2})=3$.
I'm wondering is it possible to solve this problem differently?
| Go with Trigonometry.
Let $x=a\sin \alpha ,\, y=a\cos \alpha$, then we have:
$$\begin{aligned}&3a\sin \alpha+4a\cos \alpha=15\\
\implies &15\leq \sqrt {9a^2+16a^2}=5|a|\\
\implies &|a|\ge 3.\end{aligned}$$
$$\sqrt {x^2+y^2}=|a|\ge 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
| As an alternative by half-angle formula $\tan \frac{\theta}2=\frac{1-\cos \theta}{\sin \theta}$ we have (note that $\cos \alpha \neq \sin \alpha$):
$$\tan\left(\dfrac{3\pi}{4}+\alpha\right)
=\frac{1-\cos \left(\frac{3\pi}{2}+2\alpha\right)}{\sin \left(\frac{3\pi}{2}+2\alpha\right) }
=\frac{1-\sin(2\alpha)}{-\cos(2\alpha)}=\frac{(\cos \alpha-\sin \alpha)^2}{-(\cos^2 \alpha-\sin^2 \alpha)}=\frac{\cos \alpha-\sin \alpha}{-(\cos \alpha+\sin \alpha)}$$
and then
$$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\frac{(\cos \alpha-\sin \alpha)^2}{(\cos \alpha+\sin \alpha)^2}=\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$$
| {
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
| I've tried to answer in a way which skips over some of the tedious calculation so as to make it easier to understand what this question means.
$$f(a)=4x^2-2ax+a^2-5a+4>0$$
Consider, for the moment, $x$ to be a constant. Then we have a quadratic expression in $a$ with roots at (details omitted):
$$a_1=- \sqrt{-3x^2+5x+\tfrac94}+x+\tfrac52$$
$$a_2= \sqrt{-3x^2+5x+\tfrac94}+x+\tfrac52$$
In the range of interest of $x$, it turns out that the expression under the square root is always positive, so we always have two real roots. We can plot the roots as follows:
We want $f(a)>0$ for all $x \in (0,2)$. It is easy to see that this requires that $a$ is less than the minimum value of $a_1$ or $a$ is greater than the maximum value of $a_2$.
We therefore differentiate $a_1$ and $a_2$ with respect to $x$ (details omitted) to find the turning points which are, for $a_1$:
$$x=\tfrac16(5-\sqrt{13}), \;\; a=-\tfrac23(\sqrt{13}-5)$$
and for $a_2$:
$$x=\tfrac16(5+\sqrt{13}), \;\; a=\tfrac23(\sqrt{13}+5)$$
And our full solution is therefore:
$$a<-\tfrac23(\sqrt{13}-5) \text{ and } a>\tfrac23(\sqrt{13}+5) $$
| {
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The number of four-digit numbers that have distinct digits and are divisible by $99$ We try to find the number of four-digit numbers that have distinct digits and are divisible by $99$.
Let a number be $N = abcd$, then we have $9| N$ and $11|N$.
Thus $9| a+b+c+d$ and $a+c \equiv b+d \mod 11$.
I listed all the four-digit numbers that are divisible by $99$:
From here we can find the numbers with distinct digits. But this method is not elegant.
Is there an elegant method of doing it?
Update: $9| a+b+c+d$ and $a+c \equiv b+d \mod 11$ implies that $9 | a+c$.
Thus $a+c$ is either $9$ or $18$. We can't have $18$ since $a \neq c$.
Hence we have $(a,c) = (1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)$.
| Note that:
For a number to be divisible by $99$, the sum of doublets should be divisible by $99$.
Proof:
$abcdef=ab(10^4)+cd(10^2)+ef$
and $10^{2n} \mod 99 =1$
So, $abcdef \mod 99 = ab + cd + ef$
So, $ab+cd+ef$ should be divisible by $99$.
Now let's solve your problem:
Now for a $4$-digit number ($abcd$) to be divisible by $99$ we know that:
$ab+cd=99k$ where $k\in\Bbb Z$
Now we can easily see that $k$ must be $1$ as $k=0$ would mean $0000$, invalid and $k=2$ means $9999$, again invalid.
So, we can write $ab+cd=99k$ as $10(a+c) + b+ d = 99$
This means that $a+c=9$ and also, $b+d=9$.
Now moving on to cases:
Case $1$: Excluding $0$ (which also means excluding $9$):
We have now $8$ numbers left: $1,2,3,4,5,6,7,8$
Now $a$ can assume any of the $8$ values in $\binom 81$ ways corresponding to which $c$ will automatically assume it's complement.
Since we need all $4$ digits to be distinct so $b$ can now assume any value out of the remaining $6$ digits in $\binom 61$ ways and $d$ will assume it's complement.
So, in total, we get, $\binom 81 \binom 61 = 48$ ways for forming a $4$ digit number in required format.
Case $2$: Including $0$ (and thus $9$):
Now, out of $a$ and $c$, only $c$ can assume $0$ and $a$, $9$. Then $b$ can take any value out of the remaining $8$ digits in $\binom 81$ ways and $d$ will assume it's complement.
Now, out of $b$ and $d$ anyone can assume $0$, in $\binom 21$ ways and then $a$ can take any value from the remaining digits in $\binom 81$ ways, correspondingly $c$ will assume the complement.
So, in total, we get, $\binom 81 + \binom 21\binom 81 = 24$ ways of forming a $4$ digit number in the required format.
Total: (Case $1+$ Case $2$):
Thus, we get $48+24=\bf 72$ as the number of four-digit numbers that have distinct digits and are divisible by $99$.
Checked my result from your image by manual counting too ;) .
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Proving Rate of Convergence I am investigating the following coupled sequence:
\begin{align*}
y_0 &= 1\\
x_{n+1} &= \sqrt{1 + \frac{1}{y_n}}\\
y_{n+1} &= \sqrt{1 - \frac{1}{x_{n+1}}}\\
\end{align*}
I am trying to show
\begin{align*}
\lim_{n \to \infty} \frac{\left |x_{n+1} - \varphi\right |}{\left |x_n - \varphi\right |} &= \frac{1}{4}\\
\lim_{n \to \infty} \frac{\left |y_{n+1} - \varphi^{-1}\right |}{\left |y_n - \varphi^{-1}\right |} &= \frac{1}{4}
\end{align*}
where $\varphi = \frac{-1 + \sqrt{5}}{2}$.
So far, I've shown:
\begin{align*}
x_{n+1}^2 - \varphi^2 &= 1 + \frac{1}{y_n} - \varphi^2\\
&= \sqrt{1 + \frac{1}{x_n - 1}} - \varphi\\
&= \frac{1 + \frac{1}{x_n - 1} - \varphi^2}{\sqrt{1 + \frac{1}{x_n - 1}} + \varphi}\\
&= \frac{\frac{1}{x_n - 1} - \frac{1}{\varphi - 1}}{\sqrt{1 + \frac{1}{x_n - 1}} + \varphi}\\
&= \frac{\varphi - x_n}{\left (\varphi - 1\right )\left (\sqrt{x_n^2 - x_n} + \varphi\left(x_n - 1\right ) \right )}\\
\frac{x_{n+1} - \varphi}{\varphi - x_n} &= \frac{\varphi}{\left (x_{n+1} + \varphi\right )\left (\sqrt{x_n^2 - x_n} + \varphi\left(x_n - 1\right )\right )}\\
\end{align*}
This result (which is true when the limits are substituted in) could be promising but I've come up on to a dead-end it seems.
Any help would be great.
| After the edit, it should be simple. You have:
$$x_{n+1} = f(x_n): \quad f(x) = \sqrt{1 + \dfrac{1}{\sqrt{1-\frac 1x}}},\,\, x_1 =\sqrt{2}.$$
Some, but not too terrible, calculus will show that $1 <x<f(x)$ if $x\in (1,\varphi).$ This tells us that $x_n$ is increasing. In fact, after removing square roots $f(x)>x$ is equivalent to:
$$(x-1)^3(x+1)^2-x = (x^2-x-1)(x^3-x+1) < 0$$
which is easy to verify on $(1,\varphi).$
It's also trivially bounded too since because $f$ is a decreasing function:
$$f(x_n)\leq f(x_1) = \sqrt{1 + \dfrac{1}{\sqrt{1-\frac{1}{\sqrt{2}}}}}.$$
So you have your convergence and the only thing left to do is to solve the equation:
$$x = f(x)\iff (x^2-x-1)(x^3-x+1) = 0.$$
The unique positive solution is $\varphi$ and the problem follows.
EDIT:
The part of the above work that says $x_n$ is increasing is incorrect, as I mistakenly computed $f(x_1) < \varphi.$ But in reality, $f(x_1)\approx 1.68...$ and indeed the sequence oscillates. However, this makes it possible for a simpler proof using OP's work. One simply needs to have a lower bound on $x_n$ and in fact $x_n > 1$ will work since:
$$\frac{x_{n+1} - \varphi}{\varphi - x_n} = \frac{\varphi}{\left (x_{n+1} + \varphi\right )\left (\sqrt{x_n^2 - x_n} + \varphi\left(x_n - 1\right )\right )} < \dfrac{\varphi}{1+\varphi} = \varphi^{-1}<1.$$
This proves $x_n\to\varphi$ and in return OP's equation above proves the convergence rate is $\dfrac 14.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Compute the area of Quadrilateral $ABCD$ As title suggests, the question is to solve for the area of the given convex quadrilateral, with two equal sides, a side length of 2 units and some angles:
I have solved the problem with a synthetic geometric approach involving some angle chasing. However, I believe my solution (which I will post in a comment below since I don't want to clutter the question) is a little messy and not efficient. Are there any better ways to do this? Geometric and/or trigonometric approaches are all welcomed!
EDIT: I have posted my solution below!
| Alternative approach:
Area$(\triangle ADC) = \dfrac{1}{2} \times \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)].$
Area$(\triangle ABC) = \dfrac{1}{2} \times \overline{BC} \times [ ~\overline{AC} \sin(75^\circ)].$
Therefore, $~\text{Area}(\triangle ADC) ~=~ \text{Area}(\triangle ABC).$
Therefore
$$\text{Area(quadrilateral)} ~=~ \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)]. \tag1 $$
Plan of Attack:
*
*Let $X$ denote the intersection of the two diagonals.
*Use the
Law of Sines
and the
Law of Cosines to compute
$~\overline{AX}, ~\overline{BX}, ~\overline{CX}, ~\overline{DX}.$
*Compute $~\overline{AD}.$
*Apply the formula in (1) above.
$\displaystyle \frac{\overline{DX}}{\sin(105^\circ)} =
\frac{\overline{AD}}{\sin(45^\circ)} =
\frac{\overline{BC}}{\sin(45^\circ)} =
\frac{\overline{BX}}{\sin(75^\circ)} \implies $
$$\overline{DX} = \overline{BX}. \tag2 $$
$\displaystyle \frac{\overline{AX}}{\sin(30^\circ)} =
\frac{\overline{AD}}{\sin(45^\circ)} =
\frac{\overline{BC}}{\sin(45^\circ)} =
\frac{\overline{CX}}{\sin(60^\circ)} \implies $
$\displaystyle \frac{1}{2} \times \overline{CX} = \frac{\sqrt{3}}{{2}} \times \overline{AX} \implies $
$$\overline{CX} = \sqrt{3} \times \overline{AX}. \tag3 $$
$$\overline{DX}^2 + \overline{AX}^2
- \left[2\overline{AX}~~\overline{DX} \times \frac{1}{\sqrt{2}}\right]$$
$$= ~\overline{CX}^2 + \overline{BX}^2
- \left[2\overline{CX}~~\overline{BX} \times \frac{1}{\sqrt{2}}\right] \implies $$
$$\overline{DX}^2 + \overline{AX}^2
- \left[\sqrt{2} ~\overline{AX}~~\overline{DX}\right]$$
$$= ~3\overline{AX}^2 + \overline{DX}^2
- \left[\sqrt{2} ~\left(\sqrt{3} ~\overline{AX}\right)~~\overline{DX}\right] \implies $$
$$= ~2\overline{AX}^2 + \left[ ~\sqrt{2} ~\overline{AX}~~\overline{DX} ~~\left(1 - \sqrt{3}\right) ~\right] = 0 \implies $$
$$= ~2\overline{AX} + \left[ ~\sqrt{2} ~~\overline{DX} ~~\left(1 - \sqrt{3}\right) ~\right] = 0 \implies $$
$$\overline{AX} ~=~ \overline{DX} ~~\left[\frac{\sqrt{3} - 1}{\sqrt{2}}\right]
~=~ \overline{BX} ~~\left[\frac{\sqrt{3} - 1}{\sqrt{2}}\right]. \tag4 $$
Note $\displaystyle ~: ~\cos(135^\circ) = \frac{-1}{\sqrt{2}}.$
\begin{alignat*}{2}
4 ~ &=~ \overline{AX}^2 + \overline{BX}^2 + \sqrt{2} ~\overline{AX} ~~\overline{BX} \\ \\
&=~ \overline{BX}^2\left(\frac{\sqrt{3} - 1}{\sqrt{2}}\right)^2
~+~ \overline{BX}^2
~+~ \sqrt{2} ~\left(\frac{\sqrt{3} - 1}{\sqrt{2}}\right) ~\overline{BX}^2 \\ \\
&=~ \overline{BX}^2 ~\left[ ~\left( ~\frac{4 - 2\sqrt{3}}{2} ~\right) ~+~ 1 ~+~ \sqrt{3} - 1 ~\right] \\ \\
&=~ \overline{BX}^2 ~\left[ ~\frac{4 - 2\sqrt{3}}{2} ~+~ \frac{2\sqrt{3}}{2} ~\right] \\ \\
&=~ 2 ~\overline{BX}^2 \implies
\end{alignat*}
$$\overline{BX} = \sqrt{2}. \tag5 $$
Therefore
*
*$~\overline{AX} = \sqrt{3} - 1.$
*$~\overline{BX} = \sqrt{2}.$
*$~\overline{CX} = 3 - \sqrt{3}.$
*$~\overline{DX} = \sqrt{2}.$
$\underline{\text{Computation of} ~\overline{AD}}$
\begin{alignat*}{2}
\overline{AD}^2 ~ &=~ \overline{AX}^2 + \overline{DX}^2 - \sqrt{2} ~\overline{AX} ~~\overline{DX} \\ \\
&=~ 4 - 2\sqrt{3} + 2 - 2\sqrt{3} + 2 \\ \\
&= 8 - 4\sqrt{3} \implies
\end{alignat*}
$$\overline{AD} = \sqrt{8 - 4\sqrt{3}}. \tag6 $$
$\underline{\text{Final Computations}}$
In general,
$\cos(2\theta) = 2\cos^2(\theta) - 1 \implies $
$\displaystyle \frac{\sqrt{3}}{2} = \cos(30^\circ) = 2\cos^2\left(15^\circ\right) - 1 \implies $
$\displaystyle \cos\left(15^\circ\right) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{3}} =
\sin\left(75^\circ\right) = \sin\left(105^\circ\right).$
$\overline{AC} = \overline{AX} + \overline{CX} = 2.$
Therefore, using (1) above,
$$\text{Area(quadrilateral)} ~=~
\sqrt{8 - 4\sqrt{3}} \times 2 \times
\frac{1}{2} \sqrt{2 + \sqrt{3}}$$
$$=~ 2 \times \sqrt{2 - \sqrt{3}} \times \sqrt{2 + \sqrt{3}} = 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
From a question I asked before this, I have trouble actually with the numbers manipulating part.
Using trigo identity, $\sin^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{12} = 1$ so , $\cos^2 \frac{\pi}{12} = 1- \sin^2 \frac{\pi}{12}$
To find $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$
$\sin^2 \frac{\pi}{12} = (\frac{\sqrt{3} -1}{2 \sqrt{2}})^2 = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2} = \frac{2- \sqrt{3}}{4}$
$\cos \frac{\pi}{12} = \sqrt{1-(\frac{\sqrt{3} -1}{2 \sqrt{2}})^2} $
$\cos \frac{\pi}{12} = \sqrt{1- \frac{2-\sqrt{3}}{4}}$
$\cos \frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$
What is wrong with my steps?
| You can also use $$\cos(x)+\sin(x)=\sqrt{2}\sin(x+\frac{\pi}4)$$
In this instance with $x=\frac{\pi}{12}$ you get to calculate $\sin(\frac{\pi}3)$ which is known.
The advantage is that you don't get nested root, nor have to rationalize $\sqrt{3}-1$ on denominator, just fraction addition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
SO(2)(K) isomorphic to a $K$-split Torus and its rank Let $K$ be a field and $SO(2)(K):= \{Q \in GL_2(K) \ \vert \ Q^T Q= I \text{ and } \det(Q)=1 \}$ the special orthogonal group.
Is there any characterisation known to decide for which fields $K$ the special orthogonal group $SO(2)(K)$ is toric, ie $SO(2)(K) \cong (K^{\times})^n $ and if that's the case the rank $n$ is always $1$?
My naive approach to define a map $K^{\times} \to SO(2)(K)$ via $$ x \mapsto \begin{pmatrix} x & -\sqrt{1-x^2} \\ \sqrt{1-x^2} & x \end{pmatrix}$$
suggest that the rank of $SO(2)(K)$ is always at least one (rank = maximal $n$ such that $SO(2)(K)$ contains a torus $(K^{\times})^n $ as subgroup).
Can $n$ be bigger than $1$? Under which conditions on $K$ the map above is an isomorphism?
| First as an algebraic variety, $SO_2(K)\simeq V(x^2+y^2=1)$. Indeed, let $\begin{pmatrix} x & a \\ y & b\end{pmatrix}\in SO_2(K)$, we have the two conditions $xa+yb=0$ and $xb-ya=1$ have completely determined $a,b$ since $\det\begin{pmatrix} x & y \\ -y & x\end{pmatrix}=1\not=0$.
On the other hand, $K^{\times}\simeq V(xy=1)$. If $\sqrt{-1}\in K$, then we have the splitting $x^2+y^2=(x+\sqrt{-1}y)(x-\sqrt{-1}y)=1$, which connects the two varieties by solving the equations $x=a+b\sqrt{-1}y, y=a-b\sqrt{-1}y$ assuming $\text{char}(K)\not=2$ to get $$a=\frac{x+y}{2}=\frac{x+\frac{1}{x}}{2}, b=\frac{x-y}{2\sqrt{-1}}=\frac{x-\frac{1}{x}}{2\sqrt{-1}}$$
In other words, $x\in K^{\times}\mapsto \begin{pmatrix} \frac{x+\frac{1}{x}}{2} & -\frac{x-\frac{1}{x}}{2\sqrt{-1}} \\ \frac{x-\frac{1}{x}}{2\sqrt{-1}} & \frac{x+\frac{1}{x}}{2} \end{pmatrix}$ is the desired isomorphism $K^{\times}\simeq SO_2(K)$.
So $\text{char}(K)\not=2$ and $\sqrt{-1}\in K$ are sufficient. Now we show they are also necessary.
As $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^2=-I_2$, if there is a group isomorphism $K^{\times}\simeq SO_2(K)$, $K^{\times}$ must contain $\sqrt{-1}$.
If $\text{char}(K)=2$, then $x^2+y^2=(x+y)^2=1$, $x+y=1$. And from here it's not hard to show $\begin{pmatrix} x & a \\ y & b \end{pmatrix}=\begin{pmatrix} x & 1-x \\ 1-x & x\end{pmatrix}$ must be symmetric, hence its square is just $1$. But in $K^{\times}$, the only element whose square is $1$ is $1$ itself, and $SO_2(K)$ contains at least two elements $I_2, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, hence $SO_2(K)\not\simeq K^{\times}$ for any $K$ with characteristic $2$. Indeed, $x\mapsto\begin{pmatrix} 1-x & x \\ x & 1-x\end{pmatrix}$ defines an isomorphism from $(K, +)$ to $SO_2(K)$.
Note that while we have used a little algebraic geometry to figure out the exact isomorphism, our arguments are purely group theoretical. In any case, $n$ cannot be strictly greater than $1$, since the varieties $V(x^2+y^2=1)$ and $V(xy=1)$ have the same dimension.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus
*
*Verify the following limits
(iii) $\lim\limits_{n\to \infty} \left
[\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$
The solution manual says
$$\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right
]$$
$$=\lim\limits_{n\to \infty} \left [\left
(\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right )+\left
(\sqrt[4]{n}-\sqrt[4]{n+1}\right )\right ]$$
$$=0+0=0$$
(Each of these two limits can be proved in the same way that
$\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was proved in the
text)
How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right ]=0$?
Note that in the main text, $\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was solved by multiplying and dividing by $(\sqrt{n+1}+\sqrt{n})$ to reach
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$
| I want to present you a very general method that does not involve any trick. Basically this is using the taylor expansion of $(1+x)^\alpha$ around $0$ to get en equivalent of the sequence :
\begin{aligned}
(1+x)^{\alpha} &=1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^{2}+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+o\left(x^{n}\right)(x \to 0) \\
\end{aligned}
First let's factorize by the dominant term in each term of the sum :
$$\sqrt[8]{n^2+1}-\sqrt[4]{n+1}= n^{\frac{1}{4}}(1+\frac{1}{n^2})^\frac18 -n^\frac{1}{4} (1+\frac1n)^\frac14 $$
Then we use the Taylor expansion of $(1+x)^\alpha$ around $0$ at order $1$ :
$$ (1+\frac{1}{n^2})^\frac18 = 1+ o(\frac1{n}) $$
$$ (1+\frac1n)^\frac14 = 1+\frac14\frac1n + o(\frac1{n}) $$
Therefore
\begin{align}
\sqrt[8]{n^2+1}-\sqrt[4]{n+1} &= n^{\frac{2}{8}}(1+\frac{1}{n^2})^\frac18 -n^\frac{1}{4} (1+\frac1n)^\frac14 \\
&= n^{\frac{1}{4}}(1+ o(\frac1{n})) - n^\frac{1}{4}(1+\frac14\frac1n +o(\frac1{n})) \\
&= -\frac14 \frac{1}{n^\frac34}+o(\frac{1}{n^\frac34}) \\
&\sim -\frac14 \frac{1}{n^\frac34}
\end{align}
Conclusion : $$\boxed{\sqrt[8]{n^2+1}-\sqrt[4]{n+1} \sim -\frac14 \frac{1}{n^\frac34} }$$
Since $ -\frac14 \frac{1}{n^\frac34} \to 0$ then the $\sqrt[8]{n^2+1}-\sqrt[4]{n+1} \to 0$.
And we have a little bonus, we know that it's approaching $0$ from the negative side since the equivalent is negative. We also know that it converges towards $0$ with a a certain speed given by the equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
Find the total number of roots of $(x^2+x+1)^2+2=(x^2+x+1)(x^2-2x-6)$, belonging to $(-2,4)$.
My Attempt:
On rearranging, I get, $(x^2+x+1)(3x+7)+2=0$
Or, $3x^3+10x^2+10x+9=0$
Derivative of the cubic is $9x^2+20x+10$
It is zero at minus zero point something and minus one point something.
So, even at local minima, the cubic is positive. It means it would cross x-axis only once.
At $x=-2$, cubic is positive and at $-3$, it is negative.
It means the only root is minus two point something.
Is there any other way to solve this question? Something that doesn't involve calculator? Or maybe something that doesn't involve calculus?
| $x^2+x+1 + \dfrac{2}{x^2+x+1} \ge 2\sqrt 2$
Largest value of $x^2 - 2x-6$ in the given interval is $2$ at $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Closed Form Formula for Nonlinear Recurrence $a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$ I'm trying to find a closed form solution to the sequence
$a_{n+1}=\frac{a_{n}}{2} + \frac{5}{a_{n}}$
I tried using a generating function approach in the following way:
Let $$f(x) = \sum_{n=1}^\infty a_n x^n$$
Then multiplying the original equation with $x^n$ and summing over all $n$, we get:
$$\sum_{n=1}^\infty a_{n+1}x^n= \sum_{n=1}^\infty \frac{a_{n}}{2} x^n + \sum_{n=1}^\infty\frac{5}{a_{n}} x^n$$
$$= \frac{f(x) - a_1}{x} = \frac{f(x)}{2} + 5 \sum_{n=1}^\infty \frac{x^n}{a_n}$$
But now I don't know how to simplify the last term and I get stuck. I put it on Wolframalpha and
found the general solution to the recurrence relation as:
$$a_n = -i \sqrt{10} \cot\left(c_1 2^n\right)$$
| Here is an intuition: Rewrite the recurrence relation s
$$ a_{n+1} = \frac{1}{2}\left(a_n + \frac{\ell^2}{a_n}\right), \qquad \ell = \sqrt{10}. $$
This is precisely what we get when we apply the Newton-Raphson method to $x^2 - \ell^2$. So,
*
*for any positive initial value, $(a_n)$ converges to $\ell$, and
*for any negative initial value, $(a_n)$ converges to $-\ell$.
This naturally leads us to study the behavior of the sequences $(x_n^+)$ and $(x_n^-)$ defined by
$$ x_n^{\pm} = a_n \pm \ell. $$
Let us find the recurrence relation for these sequences:
$$ x_{n+1}^{\pm} = a_{n+1} \pm \ell = \frac{1}{2}\left(a_n + \frac{\ell^2}{a_n}\right) \pm \ell = \frac{(a_n \pm \ell)^2}{2a_n} = \frac{(x_n^{\pm})^2}{2a_n}. $$
This gives an intuitive explanation to @NN2's wonderful trick. Indeed, taking ratio of the two sequences,
$$ \frac{x_{n+1}^+}{x_{n+1}^-} = \left( \frac{x_{n}^+}{x_{n}^-} \right)^2,
\qquad\text{i.e.,}\qquad
\frac{a_{n+1} + \ell}{a_{n+1} - \ell} = \left( \frac{a_n + \ell}{a_n - \ell} \right)^2. $$
Iterating this, we get
$$ \frac{a_n + \ell}{a_n - \ell} = \left( \frac{a_0 + \ell}{a_0 - \ell} \right)^{2^n}. $$
Now solving this for $a_n$ gives the desired recurrence relation for $(a_n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Showing that $(x,y)=\left(\frac{c\sin t+d}{a+b\cos t},\frac{e\cos t+f}{a+b\cos t}\right)$ parameterizes an ellipse I want to show that
$$\mathbf{P}(t) = (X, Y) = \left( \dfrac{ c \sin t + d}{a + b \cos t} , \dfrac{ e \cos t + f }{a + b \cos t } \right)$$
is actually an ellipse, given that $ | b | \lt | a | $
How can I prove this?
What I have tried:
Guided by @orangeskid solution to a recent problem, I've defined the vector
$ \mathbf{r}(t) = (x(t),y(t), z(t)) = (c \sin t + d , e \cos t + f , a + b \cos t ) $
which is an ellipse in $3D$ space whose parametric equation is
$ \mathbf{r}(t) = [d, f, a]^T + \cos t \ [ 0, e, b ]^T + \sin t \ [c , 0, 0 ]^T $
Using matrix-vector notation, this can be written as
$ \mathbf{r}(t) = M \mathbf{v}(t) \hspace{50pt}(1) $
where
$ M = \begin{bmatrix} 0 && c && d \\ e && 0 && f \\ b && 0 && a \end{bmatrix} $
and
$ \mathbf{v}(t) = \begin{bmatrix} \cos t \\ \sin t \\ 1 \end{bmatrix} $
Now, we note that
$ \mathbf{v}^T(t) Q_0 \mathbf{v}(t) = 0 \hspace{50pt}(2) $
where
$Q_0 = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && -1 \end{bmatrix} $
From eq. $(1)$,
$ \mathbf{v}(t) = M^{-1} \mathbf{r}(t) \hspace{50pt} $
Substituting this into $(2)$,
$\mathbf{r}^T (t) Q \mathbf{r}(t) = 0 \hspace{50pt}(3) $
where
$ Q = M^{-T} Q_0 M^{-1} $
Thus, $(3)$ becomes
$\begin{bmatrix} x && y && z \end{bmatrix} \begin{bmatrix} Q_{11} && Q_{12} && Q_{13} \\ Q_{12} && Q_{22} && Q_{23} \\ Q_{13} && Q_{23} && Q_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 $
Dividing both sides of the above equation by $z^2$, we get
$\begin{bmatrix} \dfrac{x}{z} && \dfrac{y}{z} && 1 \end{bmatrix} \begin{bmatrix} Q_{11} && Q_{12} && Q_{13} \\ Q_{12} && Q_{22} && Q_{23} \\ Q_{13} && Q_{23} && Q_{33} \end{bmatrix} \begin{bmatrix} \dfrac{x}{z} \\ \dfrac{y}{z} \\ 1 \end{bmatrix} = 0 $
But, $ \dfrac{x}{z} = P_x = X, \dfrac{y}{z} = P_y = Y $, hence,
$\begin{bmatrix} X && Y && 1 \end{bmatrix} \begin{bmatrix} Q_{11} && Q_{12} && Q_{13} \\ Q_{12} && Q_{22} && Q_{23} \\ Q_{13} && Q_{23} && Q_{33} \end{bmatrix} \begin{bmatrix} X \\ Y \\ 1 \end{bmatrix} = 0 $
And this an equation of a conic in $X$ and $Y$. Further investigation is needed to determine if it is indeed an ellipse.
| Try to solve $\cos t$ and $\sin t$ in terms of $x$ and $y$:
\begin{align}
\cos t &= \frac{f-ay}{by-e} \\
\sin t &= \frac{(ae-bf)x+bdy-de}{c(e-by)} \\
\cos^2 t+\sin^2 t &=
\left(
\frac{f-ay}{by-e}
\right)^2+
\left[
\frac{(ae-bf)x+bdy-de}{c(e-by)}
\right]^2 \\
c^2(by-e)^2 &=c^2(f-ay)^2+[(ae-bf)x+bdy-de]^2 \\
\end{align}
Now,
$0=
\begin{pmatrix}
x & y & 1
\end{pmatrix}
\begin{pmatrix}
(ae-bf)^2 & (ae-bf)bd & (bf-ae)de \\
(ae-bf)bd & c^2(a^2-b^2)+b^2d^2 & c^2(be-af)-bd^2e \\
(bf-ae)de & c^2(be-af)-bd^2e & c^2(f^2-e^2)+d^2e^2
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ 1
\end{pmatrix}$
and discriminants
$$\delta = (ae-bf)^2 (a^2-b^2)c^2 > 0$$
$$\Delta = -c^4 (ae-bf)^4 < 0$$
which gives a real ellipse providing $ae\ne bf$ and $c\ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Determine the values of c and d so that E[Y] = c and Var(Y ) = 1 Question
Supppose $X$ is a random PDF $f(x)=\frac{2x}{c^{2}d^{2}}, \quad 0<x<cd$ where $c > 0, \; d > 0$.
Determine the values of $c$ and $d$ so that $\mathbb{E}[X]= c $ and $\mathrm{Var}(x) = 1$
My approach
My approach has been to try and isolate the variables using the given information. I have used $\mathrm{Var}(X) = 1$ as a reference point to start. My work has been as follows:
\begin{align}
1 &= \mathbb{E}(X^2) - \mathbb{E}[X]^2 \\
1 &= \int_{0}^{cd} x^{2} \cdot \frac{2x}{c^{2}d^{2}} - \left(\int_{0}^{cd} x \cdot \frac{2x}{c^{2}d^{2}}\right)^2
\end{align}
Following this, I have approached this result:
\begin{align}
1 = \frac{(cd)^4}{2c^2d^2} - \frac{(cd)^4}{c^4d^4}
\end{align}
Which yields the following:
\begin{align}
1 &= \frac{c^2d^2}{2} \\
2 &= c^2d^2 \\
\end{align}
I have then derived the possible inter solutions of $c$ and $d$.
\begin{align*}
\therefore c &= \pm 1\; ,\pm 2 \\
d &= \pm 1\; , \pm 2
\end{align*}
I have then tried to prove the $\mathbb{E}[X] = c$ through the following case. Suppose $c = 1$ and $d = 2$ thus $0 < x < 2$.
\begin{align}
\mathbb{E}[X] &= \int_{0}^{2} x\cdot \frac{2y}{c^{2}d^{2}} \; \mathrm{d}y \\
&= \int_{0}^{2} \frac{2x^2}{4} \; \mathrm{d}x \\
&= \frac{1}{4}\int_{0}^{2} 2x^2 \; \mathrm{d}x \\
&= \frac{1}{4} \cdot \frac{2x^3}{3} \Biggr|_{0}^{2}
\end{align}
However, the answer does not evalute to the case. I have taken the wrong approach and am lost on how to approach the problem. Any help with the solution would be really appreciated.
Update
Thank you for the suggestions. Errors were made in the integration. The error was found within the 2nd integral.
$$
\frac{(cd)^4}{c^4d^4} \rightarrow \frac{2(cd)^3}{3(c^4d^4)}
$$
Squaring the results would give the following $\frac{4c^2d^2}{9}$
The derivation of the expected value for $c$ is as follows
\begin{align}
c &= \int_{0}^{ac} x \cdot \frac{2x}{c^2d^2} \; \mathrm{d}x \\
c &= \frac{2}{c^2d^2} \int_{0}^{cd} x^2 \\
c &= \frac{2}{c^2d^2} \cdot \frac{x^3}{3} \Biggr|_{0}^{cd} \\
c &= \frac{2x^3}{3c^2d^2} \xrightarrow[]{} \frac{2(cd)^3}{3c^2d^2} \\
c &= \frac{2}{3}cd \\
&\therefore d = \frac{3}{2}
\end{align}
Following that, the answer would be simply substituting it into the corrected equation for variance.
\begin{align}
1 &= \frac{1}{2}c^2\cdot \frac{9}{4} - \frac{4}{9}c^2\cdot\frac{9}{4} \\
1 &= \frac{9}{4}c^2\left(\frac{1}{2}-\frac{4}{9}\right) \\
18 &= \frac{9}{4}c^2 \\
8 &= c^2 \\
c &= 2\sqrt{2}
\end{align}.
| I am sorry, I am having a hard time following what you were trying to do. You are given that the pdf of $X$ is $f(x)=\frac{2x}{c^{2}d^{2}}$ with $c,d>0$ and $0<x<cd$.
The first sanity check is that your pdf has to be non-negative everywhere and integrate to $1$. You have
$$
\int_{-\infty}^\infty f(x) dx
= \int_0^{cd} \frac{2x}{c^{2}d^{2}}\ dx
= \left. \frac{x^2}{c^2 d^2} \right|_0^{cd}
= 1,
$$
so this makes sense. To find the expected value, compute
$$
\mathbb{E}[X]
= \int_0^{cd} xf(x)\ dx = \int_0^{cd} x \frac{2x}{c^2d^2} dx
= \frac{2}{c^2d^2} \int_0^{cd} x^2dx = \frac{2cd}{3}.
$$
However, you are given that $\mathbb{E}[X]= c$, so we must have $d = 3/2$.
Now to find the variance, as you did, note that
$$
\mathbb{Var} X
= \mathbb{E}\left[X^2\right] - \mathbb{E}[X]^2
= \int_0^{cd} x^2 f(x)\ dx - c^2
= \int_0^{cd} x^2 \frac{2x}{c^2d^2} dx - c^2.
$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can we generalize factorisation of $(a+b)^n-(a^n+b^n)$
How can we generalize factorisation of
$$(a+b)^n-(a^n+b^n)\,?$$
where $n$ is an odd positive integer.
I found the following cases:
$$(a+b)^3-a^3-b^3=3ab(a+b)$$
$$(a+b)^5-a^5-b^5=5 a b (a + b) (a^2 + a b + b^2)$$
$$(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$$
Exapansions and factoring seem terrible. Maybe there is a good way for generalization.
The question is, can we generalize these factorisations?
$$(a+b)^n-(a^n
+b^n)$$
for all odd integer $n$?
| There doesn't seem to be any general factored form, but on the few polynomial factors you mention in the question, we can answer exactly when they appear. From symmetry and homogeneity, it is enough to consider the simpler $$p(x)=(1+x)^n-(1+x^n)$$ as with $x = b/a$, we have $a^n p(x)$ is the expression you want to factor.
*
*Now $p(0)=0 \implies x$ is always a factor of $p(x) \implies ab$ is always a factor of the original bivariate expression. From the Binomial theorem, it is also clear that no higher power of $x$ or $(ab)$ will be a factor.
*Noting $p(-1)= -1-(-1)^n \implies x+1$ is a factor when $n$ is odd, or equivalently $(a+b)$ is a factor for odd $n$. From $p'(x) = n(1+x)^{n-1}-nx^{n-1}, p'(-1) \neq 0$ so the multiplicity of this factor is always $1$.
*Let $\omega$ be a non-real cube root of unity, i.e. $\omega^2+\omega+1=0$. Then
$p(\omega) = (-1)^n(\omega)^{2n}-1-\omega^n$.
Case $n=3k$: we have $p(\omega)=(-1)^n$
Case $n=3k+1$: we have $p(\omega)= (-1)^n\omega^2+\omega^2$, which is zero iff $n$ is also odd, i.e. for $n=6k+1$.
Case $n=3k+2$: we have $p(\omega)= (-1)^n\omega+\omega$, which vanishes iff $n$ is also odd, i.e. for $n=6k+5$.
If $p(\omega)=0$ we have $1+x+x^2$ as a factor, or equivalently $a^2+ab+b^2$ is a factor iff $n \equiv \pm1 \pmod 6$. Let us also check multiplicity for these cases using $p'(x) = n(1+x)^{n-1}-nx^{n-1} \implies p'(\omega)=n\left( (-\omega^2)^{n-1} - \omega^{n-1}\right)$:
Case $n=6k+1$: we always have $p'(\omega) = 0$, so always $(a^2+ab+b^2)^2$ is always a factor. Also $p''(\omega) = n(n-1) \neq 0$, so the multiplicity is exactly $2$.
Case $n=6k+5$: we have $p'(\omega) = n\left(\omega^2-\omega \right)\neq 0$ so here the multiplicity is always $1$.
Summarising all of that, for all odd $n$, we have $ab(a+b)$ as a factor, with no higher power for any of the terms. In case $n=1 \pmod 6$, we have additionally $(a^2+ab+b^2)^2$ as a factor (with no higher power possible) and in case $n=5 \pmod 6$ we have just $a^2+ab+b^2$ as a factor. In no other case will $a^2+ab+b^2$ be a factor.
The above few factors with limited multiplicity are clearly not enough to provide required degree for higher order polynomials, so there would necessarily be other polynomial factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $\sqrt{x^2+x+1}+\sqrt{x^2+\frac{3x}{4}}=\sqrt{4x^2+3x}$ Solve the equation $$\sqrt{x^2+x+1}+\sqrt{x^2+\dfrac{3x}{4}}=\sqrt{4x^2+3x}$$
The domain is $$x^2+\dfrac{3x}{4}\ge0,4x^2+3x\ge0$$ as $x^2+x+1>0$ for every $x$. Let's raise both sides to the power of 2: $$x^2+x+1+x^2+\dfrac{3x}{4}+2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=4x^2+3x\\2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}$$ Let's raise both sides to the power of 2 again but this time the roots should also satisfy $A:2x^2+\dfrac54x\ge0$:$$4(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)=(2x^2+\dfrac54x)^2$$ I came at $$x(2x^2+\dfrac{87}{16}x+3)=0$$ I obviously made a mistake as the answer is $x=-4$, but is there an easier approach?
| following your original approach you should have gotten in line 4:
$$2\sqrt{(x^2+x+1)\left(x^2+\dfrac{3x}{4}\right)}=2x^2+\dfrac{5x}{4}-1$$
and, squaring both sides and collecting like terms, yields
$32x^3+151x^2+88x-16=0$
which can be factored as
$(x+4)(32x^2+23x-4)=0$
from which you get $x=-4$ and two extraneous irrational roots
However,as already noted in other answers, this isn't the simplest approach
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove PQ is parallel to BC with angle chase? I've been working on this geometry problem for a long time, but I haven't been able to make any progress. I have tried angle chase but that didn't get me anywhere.
Acute triangle $ABC$ has $AB<AC.$ Let $D, E, F$ be the foots of perpendiculars from $A, B, C \text{ and } H $ is orthocenter. $EF$ and $AD$ intersect at $P.$ The reflection of $AB$ in $AD$ intersects BE at $Q$, making $\angle DAB = \angle DAB' $
I want to prove that line joining $P,Q$ is parallel to $BC.$ How can I prove?
| Let D be the origin and DA be the $y$-axis. Suppose $A(0,a)$, $B(b,0)$, $C(c,0)$, with $a\not=0$ and $b\not=c$.
$AB$: $x/b+y/a=1$
$CH$: $y=b/a(x-c)$
$F$: $(\frac{b(a^2+bc)}{a^2+b^2}, \frac{ab(b-c)}{a^2+b^2})$
$AC$: $x/c+y/a=1$
$BH$: $y=c/a(x-b)$
$E$: $(\frac{c(a^2+bc)}{a^2+c^2}, \frac{ac(c-b)}{a^2+c^2})$
The $y$-coordinate of $P$ is $\frac{\frac{b(a^2+bc)}{a^2+b^2} \frac{ac(c-b)}{a^2+c^2}-\frac{c(a^2+bc)}{a^2+c^2}\frac{ab(b-c)}{a^2+b^2}}{\frac{b(a^2+bc)}{a^2+b^2}-\frac{c(a^2+bc)}{a^2+c^2}}=\frac{-2abc}{a^2-bc}$
$AB'$: $x/(-b)+y/a=1$
$BH$: $y=c/a(x-b)$
The $y$-coordinate of $Q$ is $\frac{\frac{-2cb}a}{1-\frac{bc}{a^2}}=\frac{-2abc}{a^2-bc}$.
Hence $PQ$ is parallel to the $x$-axis, i.e., $BC$.
Because of the generic nature of the coordinates, we have proved the proposition is true for all triangles, including right triangles and obtuse triangles, except when $a^2=bc$, in which case $EF$ is parallel to $AD$ and $BH$ is parallel to $AB'$, both $P$ and $Q$ "at infinity".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
I can't find the solution of $\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$ I can't find the solution of
$$\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}}$$
Computing for $x$ goes to $0$ it gives a $1^\infty$ type of indeterminate form. I tried to solve it by making it similar to $0/0$ type of indeterminate form by taking log of both sides and writing as
$$\frac{\ln\left(1+\frac{x}{(x-1)^2}\right)}{\sqrt{1+x}-1}$$
Then I applied L'Hôpital's rule to find the limit and find it infinity for $0^+$ and $0$ for $0^-$ but my solution did not match with the answer.
Thanks for help!
| $$\begin{align*}
\lim_{x\rightarrow 0} \left(1+\frac{x}{(x-1)^2}\right)^{\frac{1}{\sqrt{1+x}-1}} &= \lim_{x\to0} \exp\left(\ln\left(\left(1+\frac x{(x-1)^2}\right)^{\frac1{\sqrt{x+1}-1}}\right)\right) \tag{1} \\[1ex]
&= \exp\left(\lim_{x\to0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}{\sqrt{x+1}-1}\right) \tag{2} \\[1ex]
&= \exp\left(\lim_{x\rightarrow 0} \frac{\ln\left(1+\frac x{(x-1)^2}\right) \left(\sqrt{1+x}+1\right)}x\right) \tag{3} \\[1ex]
&= \exp\left(2\lim_{x\rightarrow 0} \frac{\ln\left(1+\frac x{(x-1)^2}\right)}x\right) \tag{4} \\[1ex]
&= \exp\left(-2 \lim_{x\to0} \frac{x+1}{(x-1)(x^2-x+1)}\right) \tag{5} \\[1ex]
&= \exp(-2\times-1) = \boxed{e^2}
\end{align*}$$
*
*$(1)$ : $x = e^{\ln(x)}$
*$(2)$ : continuity of $e^x$; $\ln(a^b)=b\ln(a)$
*$(3)$ : introduce the conjugate $\sqrt{1+x}+1$
*$(4)$ : $\lim\limits_{x\to c} f(x)g(x) = \lim\limits_{x\to c} f(x) \lim\limits_{x\to c}g(x)$
*$(5)$ : L'Hôpital's rule
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Isolating $θ$ in $a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$. I have an equation that follows this form:
$a \sin^2 \theta + b \sin \theta + c + d \cos^2 \theta + e \cos \theta + f = 0$ where $a,b,c,d,e,f \in \mathbb{R}$ and $a\neq 0, d\neq 0$. I am trying to isolate $θ.$
How can this be achieved?
| You can convert this expression to a 4-th order polynomial using a wassestrass substitution
$$ \cos \theta = \frac{1-t^2}{1+t^2} $$
$$ \sin \theta = \frac{2 t}{1+t^2} $$
which converts the trig expression into the following polynomial
$$ a \frac{4 t^2}{(1+t^2)^2} +b \frac{2 t}{1+t^2} + c +d \frac{(1-t^2)^2}{(1+t^2)^2} + e \frac{1-t^2}{1+t^2}+f =0 $$
Multiply the whole thing by $(1+t^2)^2$ to get a 4-order polynomial in terms of $t$.
After you solve for $t$, then use $$\theta = 2 \tan^{-1} (t)$$
You can prove the above by using $t = \tan \left( \tfrac{\theta}{2} \right)$ and use trig identities.
I have another solution that involves reducing the order of the expression into a 3rd-order polynomial and then solving that.
*
*Consider the solution as having two parts, $\theta = \alpha + \beta$, and using this in the original expression. Expand all the trig addition terms to get $$\small \begin{gather} \cos^2 \alpha (a - (a-d) \cos^2 \beta) + \underbrace{ \cos \alpha \left( 2 (a-d) \sin \alpha \sin \beta \cos \beta + e \cos \beta + b \sin \beta \right)}_\text{assume = 0} + \\ + \sin^2 \alpha (d + (a-d) \cos \beta^2) + \sin \alpha ( b \cos \beta - e \sin \beta) + c + f = 0 \end{gather} $$
*Since $\alpha$ and $\beta$ are somewhat arbitrary we can find the value of $\alpha$ that makes the underlying part above zero. This way the above expression is "simplified" by removing this term, and the remaining expression is only in terms of $\beta$ $$ \alpha = \sin^{-1} \left( \tfrac{b}{2(d-a)\cos \beta} + \tfrac{e}{2(d-a) \sin \beta} \right)$$
*The remaining expression contains only even powers of $\sin \beta$ and $\cos \beta$ which can all be collected into one variable by substituting $s = \sin^2 \beta$ and derive this 3-rd order polynomial in terms of $s$ $$\small \begin{gathered} \underbrace{ 4 (a-d)^3}_{k_3} \;s^3 + \underbrace{ -4 \left( a^3-a^2 (c+4 d+f)+a d (2 c + 5 d + 2 f)-d^2 (c+2 d+f)\right)}_{k_2}\; s^2 + \\ + \underbrace{-4 a^2 (c+d+f) + a (b^2 +8 c d + 8 d^2 + 8 d f + e^2) - d ( b^2 + 4 c d + 4 d^2 + 4 d f+e^2)}_{k_1}\; s + \underbrace{d e^2-a e^2}_{k_0} = 0 \end{gathered} $$ where the $k_0$, $k_1$, $k_2$ and $k_3$ coefficients are known and defined as above in $k_0 + k_1 s + k_2 s^2 + k_3 s^3 = 0$
*Eliminate the quadratic term with the substitution $s = -\tfrac{k_2}{3 k_3} + \tfrac{1}{k_3^{1/3}} t$
$$ \underbrace{ \left( k_0 - \tfrac{k_1 k_2}{3 k_3} + \tfrac{2 k_2^2}{27 k_3^2} \right) }_{v} + \underbrace{\left( \tfrac{k_1}{k_3^{1/3}} - \tfrac{k_2^2}{3 k_3^{4/3}} \right) }_{u}\,t + t^3 = 0 $$
to get $v + u \,t + t^3 =0 $ with known coefficients $u$ and $v$.
*One more substitution, this time to reduce the 3rd order equation above into a 2rd order quadratic which is solvable. Use $t = z^{1/3} - \tfrac{u}{3 z^{1/3}}$ in the equation above to get $$ \frac{27 z^2 + 27 v\,z - u^3}{27 z} =0 $$ and solve for $z$
The two solutions are $$z = -\tfrac{v}{2} \pm \sqrt{ \tfrac{u^3}{27} + \tfrac{v^2}{4} }$$
*Back substitutions $$ \begin{aligned}
t & = z^{1/3} - \tfrac{u}{3 z^{1/3}} \\
s & = -\tfrac{k_2}{3 k_3} + \tfrac{1}{k_3^{1/3}} t \\
\beta &= \sin^{-1}( \sqrt{s}) \\
\alpha &= \sin^{-1} \left( \tfrac{b}{2(d-a)\cos \beta} + \tfrac{e}{2(d-a) \sin \beta} \right) \\
\theta & = \alpha + \beta
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Which number is larger? Using Binomial Theorem Which expression is larger,
$$
99^{50}+100^{50}\quad\textrm{ or }\quad 101^{50}?
$$
Idea is to use the Binomial Theorem:
The right hand side then becomes
$$
101^{50}=(100+1)^{50}=\sum_{k=0}^{50}\binom{50}{k}1^{50-k}100^k=100^{50}+\sum_{k=0}^{49}\binom{50}{k}100^k
$$
The left hand side reads
$$
99^{50}+100^{50}=(100-1)^{50}+100^{50}=\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k+100^{50}
$$
Thus, since both sides have the summand $100^{50}$, it remains to compare
$$
\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k\quad\textrm{and}\quad \sum_{k=0}^{49}\binom{50}{k}100^k
$$
| It could be amazing to compare
$$(2x)^x +(2 x-1)^x \qquad \text{and} \qquad (2 x+1)^x$$ or, better, their logarithms.
So, consider that you look for the zero of function
$$f(x)=\log \left((2x)^x +(2 x-1)^x\right)-\log \left((2 x+1)^x\right)$$ which has a trivial solution $x=2$.
So
$$x\gt 2 \qquad \implies\qquad (2x)^x +(2 x-1)^x \lt (2 x+1)^x$$
Checking
$$\left(
\begin{array}{cccc}
x &(2x)^x +(2 x-1)^x & (2 x+1)^x &(2x)^x +(2 x-1)^x- (2 x+1)^x\\
1 & 3 & 3 & 0 \\
2 & 25 & 25 & 0 \\
3 & 341 & 343 & -2 \\
4 & 6497 & 6561 & -64 \\
5 & 159049 & 161051 & -2002 \\
6 & 4757545 & 4826809 & -69264 \\
\end{array}
\right)$$
May be, you could use induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What is the solution for $x+22=-6\sqrt{2x+9}$ So, I want to solve for x in the radical equation:
$x + 22 = -6\sqrt{2x+9}$
By Squaring each expression we get:
$(x + 22)^2 = (-6\sqrt{2x+9})^2$
$ x^2 + 44x + 484 = 36\cdot(2x+ 9) $
$ x^2 + 44x + 484 = 72x + 324 $
Now by solving the quadratic equation:
$ x^2 - 28x + 160 = 0 $
$x^2 - 20x - 8x + 160 = 0$
$x\cdot(x-20) -8\cdot(x-20) = 0 $
$ (x-20)\cdot(x-8) = 0 $
$ x = 20 $ or $ x = 8 $
But, none of the values of x satisfies the equation $x + 22 = -6\sqrt{2x+9}$, they satisfy the equation $x + 22 = 6\sqrt{2x+9}$. There should be an extraneous root that satisfies the equation $x + 22 = 6\sqrt{2x+9}$, but boot the roots satisfies this equation and none of them satisfies $x + 22 = -6\sqrt{2x+9}$.
Why is that so? And is there any complex/imaginary solution to the equation?
| Just because we can write a question doesn't mean the question has any solutions. For example I can with a perfectly straight face ask you to solve $|x-3|=-7$ (which is impossible as $|x-3| \ge 0 > -7$) and if you solve by squaring both sides to get $(x-3)^2 = (-7)^2$ and you do the work ($x^2-6x + 9 = 49; x^2-6x-40=(x-10)(x+4)=0$ so $x=10$ or $x=-4$) you get only extraneous solutions.
To show there is no solution we can look at the conditions. For $\sqrt{2x+9}$ to exist we must have $2x +9 \ge $ or $x \ge -4\frac 12$. But we must also have $\sqrt{2x+9}\ge 0$ and so $-6\sqrt{2x+9} \le 0$ and must have $x+22 < 0$ so $x < -22$. A contradiction.
....
What about complex numbers....
The issue isn't about taking the square root of a negative number. It's about defining which of the two square roots $2x+9$ we are going to consider to be the square root. The only way for $x+22 = -6\sqrt{2x+9}$ is if we allow our selves have either square root be valid.
If we do that we get your extraneous roots.
| {
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"url": "https://math.stackexchange.com/questions/4580801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is wrong in the proof of max value of $|iz+z_0|$
If $|z-i|\leq 2,z_0 = 5+3i$ then find max value of $|iz+z_0|.$
Now ,
$|iz+z_0| \leq |iz|+|z_0| \cdots \text{(By Triangle Inequality.)}$
Now consider the diagram in Argand Plane -
Observe that $|z| \leq 3.$
$\Rightarrow |iz+z_0| \leq \boxed{3 + \sqrt{34}}$
But the Answer is $\boxed{7}.$ What is wrong in this proof $?$
| Where you were incorrect is not accounting for the change of $z \to iz$. While its true that $|z| = |iz|$, this does not always apply when adding complex numbers, $|z + z_0| ≠ |iz + z_0|$.
In your method, when you substituted $|iz| = 3$, you assimilated that similar substitution also applies to $|iz + z_0|$ towards, as you would for $|z + z_0|$.
If $z = x + yi$, then $|z + z_0| = |x + 5 + (y + 3)i|$; its easy to see that you would choose a set of possible positive ${x, y}$ within the bound $|z - i| \le 2$, while obtaining the longest length ($|z + z_0|_{max}$).
However, $|iz + z_0| = |xi - y + 3i + 5|$. Notice the $y$ is now negated. This means that using positive $y$ values, will reduce the coeff of the complex number $iz + z_0$ (as $x_{z0} > 0$ and $y_{z0} > 0$) and decrease $|iz + z_0|$, instead of obtaining the maximum.
Instead, you need a set of ${x, -y}$ to obtain $|iz + z_0|_{max}$. [$x$ is unchanged as rotation is only by $\frac{\pi}{2}$, $z\to iz$].
Graphically, your solution lies in 4 quadrant (sector) of the $|z - i| = 2$:
While I did not yet solve it algebraically, I tried trial and error and reached the approximation of $\approx 7$
Finding an exact answer is left as an exercise.
UPDATE
While the the triangle inequality is not used in the method below, consider if it might suit the question better.
From above, we know the solution lies on the arc of the curve, $y = -\sqrt{4 - x^2} + 1, x \in [0, \sqrt{3}]$
Hence, substituting this for $x, y$ in the complex number $iz$, we get:
$$|iz + z_0|$$
$$\to |ix - y + 5 + 3i|$$
$$\to |ix - \big[-\sqrt{4 - x^2} + 1 \big] + 5 + 3i|$$
$$\to |(x + 3)i + (\sqrt{4 - x^2} + 4)|$$
$$\to \sqrt{(x + 3)^2 + \big[ \sqrt{4 - x^2} + 4 \big]^2}$$
$$\to \sqrt{8\sqrt{4 - x^2} + 6x + 29}$$
In order to find the maximum length, we need to find where the length function above reaches a maximum.
Hence,
$$\frac{d}{dx}\sqrt{8\sqrt{4 - x^2} + 6x + 29} = 0$$
$$\frac{d}{dx}\big[8\sqrt{4 - x^2} + 6x + 29\big] \cdot \frac{1}{\sqrt{8\sqrt{4 - x^2} + 6x + 29}} = 0$$
We have numerator = 0,
$$\to 3\sqrt{4 - x^2} + 4x = 0$$
$$\therefore x = \frac{6}{5}, x \in [0, \sqrt{3}]$$
Finally,
$$\to |(\frac{6}{5} + 3)i + (\sqrt{4 - \big(\frac{-6}{5}\big)^2} + 4)|$$
$$ = 7 \text{, as required}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with parameters Let $2x(ac+bd)+2y(ad-bc)+(a^2+b^2)(1+z)+(c^2+d^2)(1-z) ≥ 0$ for any $a, b, c , d \in \mathbb{R}$. Prove that $x^2+y^2+z^2 ≤ 1$.
My attempt:
Take $2$ vectors: ${\vec{u}}=(a;b),{\vec{v}}=(c;d)$.
Then $a c+b d=\left(\vec{u},\vec{v}\right)\nonumber=\,u v\cos\varphi$ is a dot product of these vectors and $a d-b c=[\vec{u},\vec{v}]=u v\sin\varphi$ is a cross product of said vectors.
$\varphi$ is an oriented angle between those vectors measured from $u$ to $v$.
$$a^{2}+b^{2}=u^{2}$$
$$c^{2}+d^{2}=v^{2}$$
As a result, we get: $(1+z)u^{2}+2u v(x\cos\varphi+y\sin\varphi)+(1-z)v^{2}\geq0\forall u,v,\varphi$.
Setting $v=0$ gives $(1+z)\geq0$. Same drill gets us $1-z\geq0$.
Now if $z\ne-1$ the expression is always non-negative iff its discriminant is non-positive i.e.
$$(x\cos\varphi+y\sin\varphi)^{2}+z^{2}\leq1\forall\varphi\in[0;2\pi).$$
Note that $\operatorname*{max}(x\cos\varphi+y\sin\varphi)^{2}=\operatorname*{max}((x;y),(\cos\varphi;\sin\varphi))^{2}=x^{2}+y^{2}$. And thus $x^{2}+y^{2}+z^{2}\leq1$.
The case $z=−1$ gives us:
$$2v(u(x\cos\varphi+y\sin\varphi)+v)\geq0.$$
As we need our equation work for all values of $u,v,\varphi$ we get that
$$\begin{array}{l c r}{{}}&{{}}\\ {{x\cos\varphi+y\sin\varphi\geq-{\frac{v}{u}},u,v\neq0}}.\end{array}$$
As the right side can be made arbitrary close to zero we need to make $x\cos\varphi+y\sin\varphi\geq0\forall\varphi$, which is possible only if $x=y=0$. So in all cases $x^2+y^2+z^2≤1$.
Is my proof correct? Is there any other way to solve this problem?
| Letting $c = 1, d = 0$, the hypothesis yields
$$2ax - 2by + (a^2 + b^2)(1+z) + 1 - z \ge 0, \quad \forall a, b\in \mathbb{R}. \tag{1}$$
From (1), clearly we have $z + 1 \ge 0$.
If $z + 1 = 0$, (1) yields
$$2ax - 2by + 2 \ge 0, \quad \forall a, b \in \mathbb{R}$$
which results in $x = y = 0$.
Thus, $x^2 + y^2 + z^2 \le 1$.
If $z + 1 > 0$, (1) is written as: $\forall a, b \in \mathbb{R}$,
$$(z + 1)\left(a + \frac{x}{z+1}\right)^2
+ (z + 1)\left(b - \frac{y}{z+1}\right)^2 + \frac{1 - x^2 - y^2 -z^2}{z + 1}\ge 0. \tag{2}$$
From (2), we have $1 - x^2 - y^2 - z^2 \ge 0$.
We are done.
| {
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"url": "https://math.stackexchange.com/questions/4582837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(x^2−1) \bmod 8$ $\in \{ 0 , 3 , 7 \}, \forall x \in \mathbb{Z}$. It must be verified that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$. First some definitions. Using the following theorem a definition for $\bmod$ is provided:
Theorem. For all $a \in \mathbb{Z}$ and $d \in \mathbb{N}$ unique integers q, r exist satisfying $a = q \cdot d + r \wedge 0 \leq r <d$
The integers $q$, $r$ correspond to the quotient and remainder, respectively, of $a$, these being defined as:
$a = (a / d) \cdot d + a \bmod d \wedge 0 \leq a \bmod d < d$
With $q = (a/d)$ and $r = a \bmod d$. To verify the claim I use induction. I use a brute force approach where I search for all numbers $x$ such that $x^2 - 1 \bmod 8 = 0$, $x^2 - 1 \bmod 8 = 3$ or $x^2 - 1 \bmod 8 = 7$. First I search for all $x$ such that $x^2 - 1 \bmod 8 = 0$. If $x^2 - 1 \bmod 8 = 0$ then 8 is a divisor of $x^2 - 1$, i.e.: $8 \mid x^2 - 1$. Now I search for some other $x$ such that 8 is a divisor of $x^2 - 1$. Let $f(x) = x^2 - 1$ and $8 \mid f(x)$ then $8 \mid f(x + a)$ for some $a \in \mathbb{N}$. I compute this a:
$8 \mid f(x) = 8 \mid x^2 - 1 \implies 8 \mid (x + a)^2 - 1 = 8 \mid x^2 + a^2 + 2ax - 1 = 8 \mid x^2 - 1 + a^2 + 2ax$
The implication holds for e.g.: $a = 4$ since:
$8 \mid x^2 - 1 + a^2 + 2ax \implies 8 \mid x^2 - 1 + 16 + 8x = 8 \mid f(x) \wedge 8 \mid 16 + 8x = true$
So it can be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4)$. In fact it may be concluded that if $8 \mid f(x)$ then $8 \mid f(x + 4k), \forall x \in \mathbb{Z} \wedge k \in \mathbb{N}$, since:
$8 \mid (x + 4k)^2 - 1 = 8 \mid x^2 - 1 + 16k^2 + 8kx = 8 \mid f(x) \wedge 8 \mid 16k^2 + 8kx$
So it holds that $8 \mid f(x) \implies 8 \mid f(x + 4k)$. By inspection one finds that $8 \mid f(1) = 8 \mid 1^2 - 1 = 8 \mid 0 = true$ and so $f(x) \bmod 8 = 0$ for all $x \in \{1, 5, 9, 13, ...\}$. Since $f(x)$ is symmetric it holds that $f(-x) = f(x)$ and so it follows that $f(x) \bmod 8 = 0$ for all $x \in \{-1, -5, -9, -13, ...\}$. Upon further inspection one finds that $8 \mid f(3) = 8 \mid 3^2 - 1 = 8 \mid 8 = true$ and so it follows that $x^2 - 1 \bmod 8 = 0$ for all $x \in \{3, 7, 11, 15, ...\}$ and all $x \in \{-3, -7, -11, -15, ...\}$
Next I search for all x such that $x^2 - 1 \bmod 8 = 3$, or equivalently $x^2 - 4 \bmod 8 = 0$. Using the same approach I find that $x^2 - 1 \bmod 8 = 3$ for all $x \in \{2, 6, 10, 14, ...\}$ and all $x \in \{-2, -6, -10, -14, ...\}$. Finally $x^2 - 1 \bmod 8 = 7$ for all $x \in \{0, 4, 8, 12, ...\}$ and $x \in \{0, -4, -8, -12, ...\}$. And so one concludes that for all $x \in \mathbb{Z}$ it holds that $x^2 - 1 \bmod{8} \in \{0, 3, 7\}$.
This question comes from a course in discrete mathematics in computer science. I feel that my approach is overkill and that I'm doing something wrong and that there must be some cleverer way of solving the problem. If anyone can help with a better, cleaner, approach, or point out errors, it will be greatly appreciated :).
| Since $x^2-1\bmod8$ only depends on $x\bmod8,$ its possible values are obtained by taking 8 consective values for $x,$ e.g. $x\in\{-3,-2,-1,0,1,2,3,4\}.$
$(\pm3)^2-1\equiv0\bmod8.$
$(\pm2)^2-1\equiv3\bmod8.$
$(\pm1)^2-1\equiv0\bmod8.$
$0^2-1\equiv7\bmod8.$
$4^2-1\equiv7\bmod8.$
Your observation that $(x+4)^2\equiv x^2\bmod8$ was clever but not really worthwile.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a_n$ is a Cauchy sequence, if $|a_{n+1} - a_n| < \frac{1}{n(n+1)}$ Obviously, $\frac{1}{n(n+1)}$ converges to $0$, as $n \to \infty$
But it doesn't say anything to me, since, for example, $a_n = \sqrt{n}$ diverges, but $|a_{n+1} - a_n| \to 0$
So I need to look at this problem from another point of view.
Sequence is Cauchy, if $$(\forall \varepsilon >0)(\exists N \in \mathbb{N})(\forall n \geq N)(\forall m \geq N): |a_n - a_m| < \varepsilon$$
Now considering some inequalities:
*
*$|a_{n} - a_{n+1}| < \frac{1}{(n) (n+1)} = \frac{1}{n} - \frac{1}{n+1}$
*$|a_{n+1} - a_{n+2}| < \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}$
*$|a_{n+2} - a_{n+3}| < \frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$ etc.
Adding them:
$|a_n - a_{n+3}| \leq |a_{n} - a_{n+1}| + |a_{n+1} - a_{n+2}| + |a_{n+2} - a_{n+3}| < \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+2} - \frac{1}{n+3} = \frac{1}{n} - \frac{1}{n+3}$
And that makes me assume, that $|a_n - a_m| < \frac{1}{n \times m}$ (if it is a correct assumption, induction is needed to prove it, I guess)
Hence, in order to make $|a_n - a_m| < \varepsilon$, we need to ensure that $n \times m > \frac{1}{\varepsilon}$
But I would like to have only $n$ on the left side.
So, besides the need to prove that $a_n$ is a Cauchy sequence, I have four additional questions:
*
*Is my reasoning correct?
*If it is, then do we need to get rid of $m$ (since $m$ is not a fixed number).
*If we do, how to do that?
*If we are talking about basic Cauchy sequences, is tag "sequences-and-series" appropriate?
Thank you
Edit:
instead of $|a_n - a_m| < \frac{1}{n \times m}$ there should be $|a_n - a_m| < \frac{1}{n} - \frac{1}{m}$
| For $n<m$, write
$$
a_n-a_m = \sum_{k=n}^{m-1}a_k-a_{k+1},
$$
so that the triangle inequality yields
$$
|a_n-a_m| \leqslant \sum_{k=n}^{m-1}|a_k-a_{k+1}|.
$$
From what you have done, you should find
$$
|a_n-a_m| \leqslant \frac{1}{n}-\frac{1}{m} \leqslant \frac{1}{n},
$$
and the result should come up easily now.
This is the so called trick of telescoping series, and getting rid of $m$ is easier than you thought here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find y from $x^2 y^3 d x+x\left(1+y^2\right) d y=0$ Suppose $\alpha (x,y)=\frac{1}{xy^3}$ is integral factor of equation $$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$
Check $\alpha (x,y)$:
$x^2 y^3 d x+x\left(1+y^2\right) d y=0 \mid \cdot \frac{1}{x y^3} \quad x \neq 0, y \neq 0$
$x d x+y^3\left(1+y^2\right) d y=0 \quad (\square)$
Let $M(x)=x, \quad \text{and} \quad N(x)=y^3\left(1+y^2\right)$ where $
D=\left\{ \left( x;y \right) \in \mathbb{R} ^2\,\,| \begin{matrix}
x>0& \land& y>0\\
\end{matrix} \right\}
$
I notice:
*
*$D$ is connected set
*$M(x),\: N(x)$ are continuous function over $D$ ($C^1$)
Let us check that the partial derivatives are equal:
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
therefore
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
Back to $ (\square)$:
$\int M(x, y) d x+\int N(x, y) d y=\mathbf{C}$
$\int x d x+\int \frac{1+y^2}{y^3} d y=\mathbf{C}$
$\frac{x^2}{2}+\int\left(\frac{1}{y^3}+\frac{1}{y}\right) d y=\mathbf{C}$
$\frac{x^2}{2}+\frac{y^{-3+1}}{-3+1}+\ln (y)=\mathbf{C}$
$\frac{x^2}{2}-\frac{1}{2 y^2}+\ln (y)=\mathbf{C}$
Question:
*
*How find solutions for the variable $y$ like WolframAlpha. I know WolframAlpha use Lambert function, but i don't know find y.
$$
y=-\frac{1}{\sqrt{W\left(e^{x^2-2 C}\right)}}
$$
$$
y=\frac{1}{\sqrt{W\left(e^{x^2-2 C}\right)}}
$$
*What happens if $(0,0) \in R^2$. Is solution differential equation?
*What happens if $ (x, y) \in R^2 - D$ without (0,0)?
*Why D must be connected set?
Thank you in advance.
| $$x^2 y^3 d x+x\left(1+y^2\right) d y=0$$
Separable ODE :
$$xdx=-\left(\frac{1}{y^3}+\frac{1}{y} \right)dy$$
Integrate :
$$\frac12 x^2=\frac{1}{2y^2}+\ln|y|+\text{constant}$$
Solution on the form of inverse function :
$$x(y)=\pm\sqrt{\frac{1}{y^2}+2\ln|y|+c}$$
Explicit solution :
$$\ln|y|=\frac12 x^2+\frac{1}{2y^2}+\text{constant}$$
$$y^2=C\:e^{x^2}e^{1/y^2}$$
$$y^{-2}e^{y^{-2}}=C\:e^{x^2}$$
From the definition of the LambertW function $Ae^A=B\implies A=W(B)$
with $A=y^{-2}$ and $B=C\:e^{x^2}$ we get :
$$y^{-2}=W\left( C\:e^{x^2}\right)$$
Finally the WolframAlpha solution :
$$y(x)=\pm\frac{1}{\sqrt{W\left(C\:e^{x^2} \right)}}=\pm\frac{1}{\sqrt{W\left(\:e^{x^2+2C'} \right)}}$$
$C=e^{2C'}$
| {
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Minimum square sum of inscripted triangle Problem: In a right triangle $ABC$, the hypotenuse is long $4$ and the angle in $B$ is $30°$. Calling $N$ the midpoint on the side $AB$ (the hypotenuse), and $M$ the middle point on the side $CB$, consider a random point on the side $AC$, lets call it $P$ and $AP = x$. Find the value of $x$ such that is minimum the sum of the squares of the sides of triangle $PNM$.
My Solution so far:
I have to minimise $NM^2+NP^2 + PM^2$ and
*
*$NM$ is fixed and equal to $1$, so I should only care about $NP$ and $MP$
*$ABC$ is a right triangle with two angles of one angle of $30°$ then the other angle will be $60°$ and therefore the side $AC$ is half the hypotenuse, therefore $AC=2$, and from this follow that $CB = 2 \sqrt{3}$
*$PM$ is the hypothenuse of the right triangle $CPM$ and therefore its equation is $PM^2 = CM^2 + PC^2 = (\sqrt{3})^2 + (2-x)^2 = x^2 -4x + 7$
But now I'm stuck since I don't know how to correctly parametrise $PN$ as a function of $AP$.
Once found a way to write $PN$ as a function of $AP$ I'm done, since I can write them inside my, equation, take the derivative and find the minimum of the function of $AP$
Final Solution:
Thanks to @mathlove, the triangle $APN$, can be solved by using the Law of Cosines, in our case used on the side $PN$ and with angle $\beta$ it results in:
\begin{equation}\begin{aligned}
PN^2 &= AP^2 + AN^2 - 2\cdot AP\cdot AN\cdot cos(\beta) = \\
&= x^2 + 2^2 - 4xcos(60) =\\
&= x^2 -2x + 4
\end{aligned}\end{equation}
Now I finally have an equation of both $PN^2$ and $PM^2$ both depending on $x$ and therefore I can find the minimum:
\begin{equation}\begin{aligned}
PN^2+PM^2 &= x^2 -2x + 4 + x^2 -4x +7 = \\
& = 2x^2 + -6x + 11
\end{aligned}\end{equation}
Which by deriving in $x$ we obtain $4x -6 = 0$ which implies $x = 3/2$
| You can apply the law of cosines to $\triangle{APN}$ to have
$$PN^2=x^2+2^2-4x\cos 60^\circ=x^2-2x+4$$
Also, note that $CB=2\color{red}{\sqrt 3}$.
| {
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Proving $\,\sin^272^\circ - \sin^260^\circ = \frac18\left(\sqrt5 - 1\right)$ I'm fairly new into proving trigonometric equations, but I believe I'm getting better at it day after day. Yet, I have stumbled upon one such which left me at an impasse. The problem is presented as follows:
$$\sin^272^\circ - \sin^260^\circ = \dfrac{\sqrt5 - 1}8$$
The goal is to prove the truth of the so-presented equation. I first complicated the right-hand side of the equation to $\dfrac{\sqrt5-1}4\cdot\dfrac12$ to then simplify it into $\sin18^\circ\cdot\sin30^\circ$ (knowing, of course, that $\dfrac{\sqrt5-1}4$ is equal to $\sin18^\circ$, and that $\dfrac12$ is equal to $\sin30^\circ$).
I then proceeded to work the proof from the left-hand side, now that both sides of the equation are termed in trigonometric ratios. The left-hand side is easily noticed to be a difference of squares, hence being apt for complication as $(\sin72^\circ - \sin60^\circ)(\sin72^\circ + \sin60^\circ)$. It is only a coincidence to notice that the expressions can be further simplified knowing the sum/difference to product rule (i.e. that $\sin A + \sin B = 2\sin\dfrac{A + B}2\cos\dfrac{A - B}2$, and also that $\sin A - \sin B =2\cos\dfrac{A + B}2\sin\dfrac{A - B}2$).
Executing these simplifications, we get the next result: $2\sin66^\circ\cdot\cos66^\circ\cdot2\sin6^\circ\cdot\cos6^\circ$. It is evident to see how the double angle rules could be of use here (that $2\sin\theta\cos\theta =\sin2\theta$). The penultimate result with this rule applied is thus, so far: $\sin132^\circ\cdot\sin12^\circ$. I can only go as far as to simply right the ratios in terms of angles less that $45$ degrees, which is: $\cos42^\circ\cdot\sin12^\circ$.
For the most of it, I am stuck; and any rules I apply to simplify either expression gets me to a result previously acquired in the run. How am I to finish proving that $\cos42^\circ\cdot\sin12^\circ$ is equal to $\sin18^\circ\cdot\sin30^\circ$, or equivalently $\sin18^\circ\cdot\cos60^\circ$. This problem can be found in S.L Loney's book on plane trigonometry, a really helpful book for getting a good picture on trigonometry. Thank you in advance.
| Now
$$\sin^2 72°=\cos^2(90°-72°)=\cos^2(18°)=\frac{1+\cos36°}{2}=\frac{1+\frac{\sqrt 5+1}{4}}{2}$$
$$=\frac{5+\sqrt 5}{8}$$
Thus
$$\sin^2 60°=\frac 34$$
Hence
$$\color{orange}{\sin^272^\circ - \sin^260^\circ} = \frac{5+\sqrt 5}{8}-\frac 34=\frac{5+\sqrt 5-6}{8}=\color{red}{\frac18\left(\sqrt5 - 1\right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$ I encountered the following binomial equality:
$$\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$$
which I know it's true, but I don't know how to prove it directly. I entered the left-hand-side to Mathematica, and it directly gave me the right-hand-side. So I wonder if anyone knows an elementary proof.
| We seek to show that
$$\sum_{q=0}^K (-1)^q {2n+1-q\choose q}
{2n-2q\choose K-q} = \frac{1}{2} (1+(-1)^K).$$
The LHS is
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\sum_{q=0}^K
(-1)^q \frac{1}{z^{q+1}} (1+z)^{2n+1-q}
\frac{1}{w^{K-q+1}} (1+w)^{2n-2q}
\; dz \; dw
\\ = \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z} (1+z)^{2n+1}
\\ \times
\sum_{q=0}^K
(-1)^q \frac{1}{z^q} (1+z)^{-q}
w^q (1+w)^{-2q}
\; dz \; dw.$$
Here we may extend $q$ beyond $K$ to infinity because the pole at zero
in $w$ is canceled for the extra values. We obtain
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z} (1+z)^{2n+1}
\\ \times
\frac{1}{1+w/(1+w)^2/z/(1+z)}
\; dz \; dw
\\ = \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n+2}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
(1+z)^{2n+2}
\\ \times
\frac{1}{(1+z(1+w))(w+z(1+w))}
\; dz \; dw.$$
The pole at zero in $z$ is gone but a new pole has appeared inside the
contour. Note that when we summed the geometric series we required
$|w/(1+w)^2| \lt |z(1+z)|.$ We have with $\gamma\ll 1$ and
$\varepsilon\ll 1$ that $|w/(1+w)^2| \le \gamma/(1-\gamma)^2 \lt
2\gamma$ and $|z(1+z)| \ge \varepsilon (1-\varepsilon) \gt
\frac{1}{2}\varepsilon.$ Therefore taking $\varepsilon = 4\gamma$ will
work e.g. $\gamma=1/11$ and $\varepsilon = 4/11.$
We have for the first simple pole at $z_0=-1/(1+w)$ that $|-1/(1+w)| \gt
1/(1+\gamma) \gt 4\gamma = \varepsilon.$ This pole is not inside the
contour. The second pole is at $z_1=-w/(1+w)$ and we have $|-w/(1+w)|
\lt \gamma/(1-\gamma) \lt 4\gamma = \varepsilon.$ This pole is inside
the contour. We thus write
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n+1}
\frac{1}{2\pi i} \int_{|z|=\varepsilon}
(1+z)^{2n+2}
\\ \times
\frac{1}{(1+z(1+w))(w/(1+w)+z)}
\; dz \; dw.$$
Evaluating the residue from the simple pole at $z_1$ we find
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} (1+w)^{2n+1}
(1-w/(1+w))^{2n+2} \frac{1}{1-(1+w)w/(1+w)} \; dw
\\ = \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{K+1}} \frac{1}{1-w^2} \; dw.$$
This is
$$[w^K] \frac{1}{1-w^2} = \frac{1}{2}(1+(-1)^K)$$
as claimed.
| {
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} |
To find the sum of this series I need to find the sum of this series $$
\frac{n}{n^2}+\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{n}{n^2+(n-1)^2}
$$
Here general term $=\frac{n}{n^2+r^2}, r=0,1,2, \ldots$
$$
\begin{aligned}
& =\frac{1}{n}\left[\frac{1}{1+\left(\frac{r}{n}\right)^2}\right] \\
& =\frac{d x}{1+x^2}\left[\text { Replace } \frac{1}{n} \text { by } d x \text { and } \frac{r}{n} \text { by } x\right]
\end{aligned}
$$
Required sum $=\int_0^1 \frac{d x}{1+x^2}=\left.\tan ^{-1} x\right|_0 ^1=\frac{\pi}{4}$
In this how to get the lower limit and upper limit values $0$ and $1$ respectively ???
Thanks in advance.
| Upper limit= $\lim_{n\to\infty} \frac{n}{n}$,
Lower limit= $\lim_{n\to\infty} \frac{0}{n}$.
That gives $1$ and $0$.
In general if this type of sum comes up with the lower limit being $a_n$ and upper being $ b_n$. Then Upper limit of integral = $\lim_{n\to\infty} \frac{b_n}{n}$,
Lower limit of integral = $\lim_{n\to\infty} \frac{a_n}{n}$.
| {
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Find the value of ${a_n^2}+{b_n^2}+{c_n^2}-a_nb_n-b_nc_n-c_na_n$
Let $$a_n=\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots$$
$$b_n=\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\cdots$$
$$c_n=\binom{n}{2}+\binom{n}{5}+\binom{n}{8}+\cdots$$
Then find the value of $${a_n^2}+{b_n^2}+{c_n^2}-a_nb_n-b_nc_n-c_na_n$$
I wrote all expressions in a summation form but couldn't find a closed form.
Surprisingly, wolframaplha shows that all of $a_n,b_n$ and $c_n$ converges. And it also gives the closed form, but that is in the form of trigonometric expressions. Like for example,
$$\sum_{i=0}^{\infty}\binom{n}{3i}=\frac13\left(2\cos\left(\frac{n\pi}{3}\right)+2^n\right)$$
Till now I just knew the methods of integration, differentiation, sequences series and putting values to find sum of particular binomial coefficients. How did trigonometry come into play$?$ Also, we need to find the above asked expression and squaring these trigonometric expressions is not a good idea imo so there must be some other method.
Any help is greatly appreciated.
| I saw this question or a similar in this site but it is not easy to find it. With the hints in the comments:
By binomial expansion, we have
$$\begin{array}
.(1+1)^n&=&\binom{n}{0}&+\binom{n}{1}&+\binom{n}{2}&+\binom{n}{3}\cdots\\
(1+w)^n&=&\binom{n}{0}&+\binom{n}{1}w&+\binom{n}{2}w^2&+\binom{n}{3}w^3+\cdots\\
(1+w^2)^n&=&\binom{n}{0}&+\binom{n}{1}w^2&+\binom{n}{2}w^4&+\binom{n}{3}w^6+\cdots
\end{array}$$
Now, by using the identities $1+w+w^2=0$ and $w^3=1$, we can compute the linear combinations below. For example, $3a_n$ is just the sum of the equations above.
$$a_n=\frac{1}{3}\left((1+1)^n+(1+w)^n+(1+w^2)^n\right)=\frac{1}{3}\left(2^n+2\cos(\frac{n\pi}{3})\right)$$
$$b_n=\frac{1}{3}\left((1+1)^n+w^2(1+w)^n+w(1+w^2)^n\right)=\frac{1}{3}\left(2^n+2\cos(\frac{n\pi}{3}+\frac{4\pi}{3})\right)$$
$$c_n=\frac{1}{3}\left((1+1)^n+w(1+w)^n+w^2(1+w^2)^n\right)=\frac{1}{3}\left(2^n+2\cos(\frac{n\pi}{3}+\frac{2\pi}{3})\right)$$
Now you can check these by using the identities $\cos(u+v)=\cos u\cos v-\sin u\sin v$, $\cos(\frac{2\pi}{3})=\cos(\frac{4\pi}{3})=-\frac{1}{2}$, $\sin(\frac{2\pi}{3})=-\sin(\frac{4\pi}{3})=\frac{\sqrt{3}}{2}$ or some better identities you know:
$$\cos(\frac{n\pi}{3})+\cos(\frac{n\pi}{3}+\frac{2\pi}{3})+\cos(\frac{n\pi}{3}+\frac{4\pi}{3})=0,$$
$$\cos^2(\frac{n\pi}{3})+\cos^2(\frac{n\pi}{3}+\frac{2\pi}{3})+\cos^2(\frac{n\pi}{3}+\frac{4\pi}{3})=\frac{3}{2},$$
$$\cos(\frac{n\pi}{3})\cos(\frac{n\pi}{3}+\frac{2\pi}{3})+\cos(\frac{n\pi}{3})\cos(\frac{n\pi}{3}+\frac{4\pi}{3})+\cos(\frac{n\pi}{3}+\frac{2\pi}{3})\cos(\frac{n\pi}{3}+\frac{4\pi}{3})=-\frac{3}{4}.$$
Hence, the sum is $\frac{4}{9}\left(\frac{3}{2}-\left(-\frac{3}{4}\right)\right)=1.$
This also immediately follows from achille hui's hint and these computations are not necessary:
$$x^2+y^2+z^2-(xy+yz+zx)=(x+yw+zw^2)(x+yw^2+w^4)$$
gives
$$a_n^2+b_n^2+c_n^2-(a_nb_n+a_nc_n+b_nc_n)=(a_n+b_nw+c_nw^2)(a_n+b_nw^2+c_nw^4)=(1+w)^n(1+w^2)^n=(1+w+w^2+w^3)^n=w^{3n}=1.$$
| {
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"answer_count": 2,
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Does $\sum\limits_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$ converge? My Attempt
First, check the limit
$$\lim_{n \to \infty} \frac{3^n + 4^n}{2^n + 5^n} =
\lim_{n \to \infty} \frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} = 0.$$
So, we cannot conclude anything.
I used Comparison Test and Ratio Test.
Consider that
$$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{5^n}{2^n + 5^n}.$$
If I can proof the convergence of the series on the right-side, then it's done by comparison test.
I used ratio test, in order to proof the convergence of
$$\sum_{n= 1}^{\infty} \frac{5^n}{2^n + 5^n}.$$
$$\lim_{n \to \infty} \frac{55^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{5^n} = 5 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = 1.$$
I didn't find the way to proof.
Any suggestion? Thanks in advanced.
Solution (@abiessu & @Thomas Andrew)
Consider that
$$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{4^n}{2^n + 5^n}.$$
Proof this series converge.
$$\sum_{n= 1}^{\infty} \frac{4^n}{2^n + 5^n}.$$
Proof (Ratio Test)
$$\lim_{n \to \infty} \frac{44^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{4^n} = 4 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = \frac{4}{5}.$$
The series converge.
Hence
$$\sum_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$$
converge.
| I prefer the Ratio Test. In fact,
$$ \lim_{n\to\infty}\frac{\frac{3^n+4^n}{2^n+5^n}}{(\frac{4}{5})^n}=\lim_{n\to\infty}\frac{3^n+4^n}{4^n}\cdot\frac{5^n}{2^n+5^n}=1>0 $$
which implies that $\sum_{n=1}^\infty\frac{3^n+4^n}{2^n+5^n}$ converges since $\sum_{n=1}^\infty(\frac{4}{5})^n$ converges.
| {
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Why is $2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$ for $x \in \mathbb{N}$? I have found following formula stands:
$2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$
($x \in \mathbb{N}$)
I have wondered why this is true. I have checked its rightness on the graph drawing tools, but I want to know the mathematical proof of this.
Although I can prove it for $x \in \mathbb{N}, x\geq2$, as following:
$2\sqrt{x+1} - \sqrt{x}/2 \geq 2\sqrt{x} - \sqrt{x}/2 = \frac{3}{2}\sqrt{x}\geq\frac{3}{2}\cdot \sqrt{2} \geq2$
and prove the rightness in case $x=1$ by direct calculation,
I want to know whether it is possible to prove this formula without dividing the case of $x=1$ and others.
Thank you.
| You can set $f(x)=2\sqrt{x+1}-2-\frac 12\sqrt{x}$
The derivative $f'(x)=\frac 1{\sqrt{x+1}}-\frac 1{4\sqrt{x}}$
It is zero when $(4\sqrt{x})^2=(\sqrt{x+1})^2\iff 16x=x+1\iff x=\frac 1{15}$
The derivative is $-$ then $+$ so $f$ is $\searrow$ then $\nearrow$ with a minimum in $\frac 1{15}$.
We could calculate $f(\frac 1{15})$ but since we are only interested in naturals let just calculate
$f(1)=2\sqrt{2}-2-\frac 12\sqrt{2}>0$ as you have noticed yourself.
And since it is increasing, it's true for all $x\ge 1$ not only integers.
For sake of completion, $f(\frac 1{15})=\frac 12\sqrt{15}-2<\frac 12\sqrt{16}-2=0$
So there exists a zero of $f$ in the interval $[\frac 1{15},1]$ in addition to $x=0$.
We can calculate it by squaring both sides:
$(2\sqrt{x+1})^2=(2+\frac 12\sqrt{x})^2\iff 4x+4=4+\frac 14x+2\sqrt{x}\iff\frac {15}4x=2\sqrt{x}\iff \\\sqrt{x}=\frac{8}{15}\text{ or }x=0$
So actually the inequality is true for all $x\ge \frac{64}{225}$
Side note: this direct calculation of the zero, shows there are only two zeroes, $x=0$ and $x=\frac{64}{225}$, so you can deduce directly (without derivative) that since $f(1)>0$ and $f$ is continuous then $f$ cannot change sign on $[1,+\infty)$
| {
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Assuming x is small, expand $\frac{\sqrt{1-x}}{\sqrt{1+2x}}$ up to and including the term in $x^{2}$ I have tried this many times but can't quite land on the correct answer.
The correct answer:
$1-\frac{3x}{2}+\frac{15x^{2}}{8}$
These are the steps I took:
*
*Re wrote it as: $\left ( 1-x \right )^{\frac{1}{2}}\left ( \left ( 1+2x \right )^{\frac{1}{2}} \right )^{-1}$
*Using bionmal expansion I expanded $\left ( 1-x \right )^{\frac{1}{2}}$ up to $x^{2}$
*
*For this I get : $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$
*Then to solve the second bracket written in step 1 I did $\left (1+2x \right )^{\frac{1}{2}}$
*
*For this I got $\left (1-x+\frac{x^{2}}{4} \right)$
*Then finally I raised this to the power one so $\left (1-x+\frac{x^{2}}{4} \right)^{-1}$
*For this I got $\left (1+x+\frac{3x^{2}}{4} \right)$
*Then I just expaned those two brakcets so: $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$$\left (1+x+\frac{3x^{2}}{4} \right)$
Once expanded, I then neglected powers bigger than $x^{2}$ (mentioned in question). Then collected like terms. However either my method or the algebra is going wrong, and I just need some help with this.
| The first terms of the Taylor expansion of $\sqrt{1+2x}$ near $0$ are $1+x-\frac{x^2}{\color{red}2}$; therefore, the first terms of the Taylor expansion of $\frac1{\sqrt{1+2x}}$ near $0$ are $1-x+\frac{3x^2}2$. And if you multiply $1-\frac x2-\frac{x^2}8$ with $1-x+\frac{3x^2}2$ and then you eliminate those terms whose degree is greater than $2$, you do get indeed $1-\frac{3 x}2+\frac{15 x^2}8$.
| {
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Verifying $\sqrt{\frac{2}{3}}\cos\left(\frac{1}{3}\arccos\sqrt{\frac{3^3}{2^5}}\,\right)=\cos\frac\pi5$ I can verify the equation using Excel. But is there another derivation?
$$\sqrt{\frac{2}{3}}\cos\left[\frac{1}{3}\arccos\left(\!\!\sqrt{\frac{3^3}{2^5}}\right)\right]=\cos\left(\frac{\pi}{5}\right)$$
Thank you for help.
| Let $u=\sqrt{\frac{2}{3}}\cos\left(\theta\right)$, where $\theta=\frac{1}{3}\arccos\left(\sqrt{\frac{3^3}{2^5}}\right)$.
By the triple angle formula, we have
$$4\cos^3(\theta) -3\cos(\theta)-\frac{3}{4}\sqrt{\frac{3}{2}} = 0.$$
i.e. $u$ is a root of the cubic $$q(x) = 6\sqrt{\frac{3}{2}}x^3-3\sqrt\frac{3}{2}x-\frac{3}{4}\sqrt{\frac{3}{2}}.$$
or, after clearing coefficients,
$$ p(x)=8x^3-4x-1.$$
On the other hand, $\cos\left(\frac{\pi}{5}\right)$ is also a root of $p(x)$, as
\begin{align*}
p\left(\cos\left(\frac{\pi}{5}\right)\right)&=8\cos^3\left(\frac{\pi}{5}\right)-4\cos\left(\frac{\pi}{5}\right)-1\\
&= 2\left(3\cos\left(\frac{\pi}{5}\right)+\cos\left(\frac{3\pi}{5}\right)\right)-4\cos\left(\frac{\pi}{5}\right)-1\\
&= 2\cos\left(\frac{\pi}{5}\right)+2\cos\left(\frac{3\pi}{5}\right)-1\\
&= 0.
\end{align*}
Factoring $p(x)$, we have
$$p(x)=(2x+1)(4x^2-2x-1).$$
We see that $p(x)$ has a unique positive root. Since $u$ and $\cos\left(\frac{\pi}{5}\right)$ are both positive, we must have $u=\cos\left(\frac{\pi}{5}\right)$.
| {
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Summation of $\sum_{r=1}^n \sin(2r-1)\theta$ In a CIE A Level Further Mathematics question paper, the following question appeared:
By considering $$\sum_{r=1}^n z^{2r-1}$$ where $z=\cos\theta +i\sin\theta$, show that if $\sin ≠ 0$,
$$\sum_{r=1}^n \sin(2r-1)\theta =\frac{\sin^2n\theta}{\sin\theta}$$
My attempt at solving this:
$$\sum_{r=1}^n z^{2r-1}=z+z^3+z^5+...z^{2n-1}=\frac{z((z^2)^n-1)}{z^2-1}$$
Then, expanding the brackets and dividing the numerator and denominator by $z$,
$$\frac{z^{2n}-1}{z-z^{-1}}$$
Applying de Moivre's theorem, we get
$$\frac{\cos(2n\theta)+i\sin(2n\theta)-1}{2i\sin\theta}$$
Since we only need the imaginary part,
$$\operatorname{Im}\left(\sum_{r=1}^n z^{2r-1}\right)=\frac{\cos(2n\theta)-1}{2\sin\theta}$$
Lastly, applying the double angle formula,
$$\frac{-\sin^{2}(n\theta)}{\sin\theta}$$
Maybe I'm being completely oblivious to one single elementary error, but what mistake have I made to get a negative sign?
| The evaluation should be
$$
\eqalign{
\Im\left(\frac{\cos\left(2n\theta\right)+i\sin\left(2n\theta\right)-1}{2i\sin\left(\theta\right)}\right) &= \Im\left(\frac{\cos\left(2n\theta\right)}{2i\sin\left(\theta\right)}+\frac{\sin\left(2n\theta\right)}{2\sin\left(\theta\right)}-\frac{1}{2i\sin\left(\theta\right)}\right) \cr
&= \Im\left(\frac{\sin\left(2n\theta\right)}{2\sin\left(\theta\right)}+\frac{i}{2\sin\left(\theta\right)}-\frac{i\cos\left(2n\theta\right)}{2\sin\left(\theta\right)}\right) \cr
&= \frac{1}{2\sin\left(\theta\right)}-\frac{\cos\left(2n\theta\right)}{2\sin\left(\theta\right)} \cr
&= \frac{1-\cos\left(2n\theta\right)}{2\sin\left(\theta\right)} \cr
&= \frac{\sin^{2}\left(n\theta\right)}{\sin\left(\theta\right)}.
}
$$
Maybe there was a missing negative in your attempt.
| {
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How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
| Let us begin with noticing that
\begin{align*}
f(x) & = x^{6} + x^{5} + 4x^{4} - 12x^{3} + 4x^{2} + x + 1\\\\
& = x^{3}\left[\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12\right] \tag{$x\neq 0$}
\end{align*}
Hence, if we make the change of variable $u = x + 1/x$, we get that:
\begin{align*}
\left(x^{3} + \frac{1}{x^{3}}\right) + \left(x^{2} + \frac{1}{x^{2}}\right) + 4\left(x + \frac{1}{x}\right) - 12 & = u(u^{2} - 3) + u^{2} - 2 + 4u -12\\\\
& = u^{3} + u^{2} + u - 14\\\\
& = (u^{3} - 2u^{2}) + (3u^{2} - 6u) + (7u - 14)\\\\
& = u^{2}(u - 2) + 3u(u - 2) + 7(u - 2)\\\\
& = (u^{2} + 3u + 7)(u - 2)
\end{align*}
If $x = 0$, the proposed relation clearly holds. If $x > 0$, then $u\geq 2$. Finally, if $x < 0$, then $u\leq -2$.
Gathering all these results, one concludes the validity of the proposed claim.
Hopefully this contributes!
| {
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Evaluating $\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)$. (WolframAlpha says it doesn't exist; I get $0$.) I have tried to solve limit, which wolfram says that DNE, but according to my calculations it is equal to 0. Limit is given below
$$\begin{align}
\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)
&=\sin(\sqrt{n^2+1}\pi-n\pi+n\pi) \\
&=(-1)^n\sin(\sqrt{n^2+1}\pi-n\pi) \\
&=(-1)^n\sin\left(\frac{(\sqrt{n^2+1}\pi-n\pi)(\sqrt{n^2+1}\pi+n\pi)}{(\sqrt{n^2+1}\pi+n\pi)}\right) \\
&=(-1)^n\sin\left(\frac{n^2\pi^2+\pi^2-n^2\pi^2}{\sqrt{n^2+1}\pi+n\pi}\right) \\
&=(-1)^n\sin\left(\frac{\pi^2}{n\pi(\sqrt{1+\frac{1}{n^2}}+1}\right) \\
&=0
\end{align}$$
It is because denominator of sin goes to infinity so everything inside sin goes to 0. as we know, sin of that would go to 0 too. And we know that $(-1)^n$ is bounded so we got that bounded * 0 has to be equal to 0.
Am I doing some mistake here ?
| The step $\sin(\sqrt{n^2+1}\pi) = (-1)^n\sin(\sqrt{n^2+1}\pi-n\pi)$ holds if $n$ is an integer but does not hold if $n$ is not an integer.
Let's look at this another way.
$\sqrt{n^2+1}$ is unbounded and increasing. When $n$ is large $\sqrt{n^2+1}\approx n.$
Suppose $n\in \mathbb R$
For large values of $n$ there will be values of $\sin(\sqrt{n^2+1}\pi)$ that equal $1, -1$ and $0$ and the limit does not exist.
e.g. $\sin (\sqrt{100.5^2 + 1}\pi) = 0.9999$
If $n\in \mathbb N$ then for large values of $n$
$\sin(\sqrt{n^2+1}\pi)\approx \sin n\pi = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so far, because they are all positive numbers, is to square $x,y,z$ but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that $x<z$ is easy and so is $y<z$, but I'm stuck at $x < y \text{ or } x>y$.
| One way is as follows. It can be proved that $f(x) = \sqrt{x}$ is an increasing function. So we have:
$$x_1\lt x_2 \implies \sqrt{x_1}\lt \sqrt{x_2} \implies \sqrt{x_2} - \sqrt{x_1} \gt 0$$This implies that $x,y\gt 0$. Also $g(x) = x^2$ is an increasing function for $x\ge 0$, then:
$$x_1\lt x_2 \implies x_1^2\lt x_2^2 \ \ \ \ x_1,x_2\in [0,\infty)$$
Now suppose $x\gt y$:
$$x>y \implies x^2\gt y^2 \implies (\sqrt{3} -1)^2>(\sqrt{5} - \sqrt{2})^2 \implies 3+1-2\sqrt{3}\gt 5+2-2\sqrt{10} \implies -3-2(\sqrt{3}-\sqrt{10})\gt 0 \implies -3\gt 2(\sqrt{3}-\sqrt{10}) \implies 3<2(\sqrt{10}-\sqrt{3})\implies \frac{3}{2}\lt \sqrt{10}-\sqrt{3} \implies \frac{9}{4} \lt 10 +3-2\sqrt{30} \implies 2\sqrt{30}\lt 13 - \frac{9}{4} \implies 4\times30\lt \frac{43^2}{4^2} \implies 4^3\times30 \lt 43^2 \implies 1920\lt 1849$$
This is a contradiction and implies $x\lt y$. Other cases can be checked similarly.
| {
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Proving $\left\lfloor(\frac{1+\sqrt{5}}{2})^{4n+2}\right\rfloor-1$ is a perfect square for $n=0,1,2,\ldots$ Let $$S_n = \left \lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1$$ ($n=0, 1, 2, \ldots$).
Prove that $S_n$ is a perfect square.
In Art of Problem Solving website, there is a hint
$$
\begin{align}
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1 & =\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}+\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}-2\\
&=\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n+1}+\left(\frac{1-\sqrt{5}}{2}\right)^{2n+1}\right)^2
\end{align}
$$
I don't know how to get the first equal sign, is it mean
$$
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor = \left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}-\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\} $$
But how to prove the decimal part of $\phi^{4n+2}$
$$ \left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2} \right\} =1-\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}$$
| By considering the binomial expansions of $\left(\frac{1+\sqrt5}2\right)^{4n+2}$ and $\left(\frac{1-\sqrt5}2\right)^{4n+2}$ it can be all non-integer terms cancel out when the two are added together, giving an integer result. Therefore we can say that $$\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)^{4n+2}\in\mathbb{Z}$$
However, since $\left|\frac{1-\sqrt5}2\right|<1$, $\left|\left(\frac{1-\sqrt5}2\right)^{4n+2}\right|<1$. Since the exponent of $\frac{1-\sqrt5}2$ is even, $\left(\frac{1-\sqrt5}2\right)^{4n+2}>0$. Therefore, $$\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)^{4n+2}=\left\lceil\left(\frac{1+\sqrt5}2\right)^{4n+2}\right\rceil$$
If $m\notin\mathbb{Z}, \lceil m\rceil=\lfloor m\rfloor+1$.
$$\therefore\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)^{4n+2}-1=\left\lceil\left(\frac{1+\sqrt5}2\right)^{4n+2}\right\rceil-1=\left\lfloor\left(\frac{1+\sqrt5}2\right)^{4n+2}\right\rfloor,$$
as required.
| {
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Solving $y''(x) + \epsilon y'(x) + 1 = 0$ using power series We are given
\begin{align}
\begin{cases}
y''(x) + \epsilon y'(x) + 1 =0, \ 0 < \epsilon <<1\\
y(0)=0, \ y'(0)=1
\end{cases}
\end{align}
and asked to solve this using the solution form $y(x) = \sum_{n=0}^{\infty}\epsilon^{n}y_n(x)$.
Doing the known method for power series ODEs, we have
\begin{align}
&y(x) = \sum_{n=0}^{\infty}y_n(x) \epsilon^n\\
&y'(x) = \sum_{n=1}^{\infty}ny_n(x) \epsilon^{n-1} = \sum_{n=0}^{\infty}(n+1)y_{n+1}(x)\epsilon^{n}\\
&y''(x) = \sum_{n=2}^{\infty}n(n-1)y_n(x) \epsilon^{n-2} = \sum_{n=0}^{\infty}(n+2)(n+1)y_{n+2}(x)\epsilon^{n}.
\end{align}
So we must plug them into the system and find the general type $y_n$. I know how to do this in general. However, the present systems grinds to a halt, at least to my eyes.
Doing the substitution
\begin{align}
&\sum_{n=0}^{\infty} (n+2)(n+1)y_{n+2}(x)\epsilon^n + \sum_{n=0}^{\infty}(n+1)y_{n+1}(x) \epsilon^{n+1} + 1 =0\\
&1+ \sum_{n=0}^{\infty}(n+1)(n+1)y_{n+2}(x) \epsilon^{n} + \sum_{n=1}^{\infty}ny_n(x)\epsilon^{n} =0\\
&1 + 2y_2(x) + \sum_{n=1}^{\infty}\left\{ (n+2)(n+1)y_{n+2}(x) + ny_{n}(x) \right\}\epsilon^{n} = 0
\end{align}
and this means that
\begin{align}
\begin{cases}
y_2 = -\dfrac{1}{2}\\
y_{n+2} = \dfrac{-n}{(n+2)(n+1)}y_n
\end{cases}
\end{align}
but I think that leads to an algebraic fault, since by plugging $n=0$ at the second one we get $y_2(x) = 0$, but we found out that $y_2(x) = -\dfrac{1}{2}$.
Any thoughts on how to proceed?
EDIT: The answer is hinted to be
\begin{align}
y_{n}(x) = (-1)^{n} \left[ \dfrac{x^{n+1}}{(n+1)!} - \dfrac{x^{n+2}}{(n+2)!} \right]
\end{align}
| You methodology is basically correct but there are some slips in the algebra. I feel more comfortable writing $y_n(x)=a_nx^n.$ Then $y=a_0+\epsilon a_1x+\epsilon^2a_2x^2+...$ Then $y(0)=a_0=0$ and$y^{\prime}(0)=\epsilon a_1$, so $a_1=1/\epsilon$. To find $\epsilon y^{\prime}$, differentiate $y$ termwise, multiply by $\epsilon$ and then make a change of dummy variables $m=n-1.$ Then differentiate $y$ twice termwise and make a change of dummy variables $m=n-2$. The differential equation becomes $$\sum_{m=0}^{\infty}\epsilon^{m+2}a_{m+2}(m+2)(m+1)x^m+\epsilon^{m+2}a_{m+1}(m+1)x^m+\delta_{0,m}x^{m}=0$$ where $\delta_{0,m}$ is the kronecker delta. Putting $m=0,$we solve for $a_2.$ Then for $m>0$, we obtain a recursive relation. Then we can change back to the $y_n(x)$ notation.
| {
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"source": "stackexchange",
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Determining the sum of $\frac{a_{n+1}}{a_n}$ where $a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$
Let $a_0=1, a_1=\frac{1}{2}, a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$, then find $\lim_{n\to\infty} \sum_{k=0}^{n}\frac{a_{k+1}}{a_k}=\dots$
We have $(a_n)$ is strictly decreasing as $a_{n+1}-a_n=\frac{-a_n(a_n+1)}{1+(n+1)a_n}<0, a_n>0$ and $ |{\frac {a_{n+1}}{a_n}|}<1$, then the sum is convergent.
Tried to telescope the sum by defining $\frac{a_{n+1}}{a_n}=\frac{na_n}{1+(n+1)a_n}=b_n$ (say). Eventually there was coming another sequence that was not giving any conclusion. Tried examining the behaviour of the terms of this sequence $$ a_0=1,a_1=\frac{1}{2},a_2=\frac{1}{2^3},a_3=\frac{1}{2^2(2^3+2+1)}\\a_4=\frac{3^2}{2^4(2^3+2^2)(2^3+3)(2^4(2^3+2^2)(2^3+2^2+1)+2^3+2^2+2+1)}$$
There would come up more $2^n$'s in the denominator for larger n's. Tried mimicking the solution as in the Convergence of $\left(a_{n+1}= \cfrac{{a_n}^2}{1+{a_n}} (n\ge 1) , a_1=1 \right)$ (seems a bit relatable), but I am stuck here.
Any hint would be appreciated. Thanks.
| Consider the sequence $b_n=na_n$ and notice that $b_{n}-b_{n+1}=\frac{b_n}{1+\frac{n+1}{n}b_n}\in(0,b_n)$ for $n>0$, so $b_n$ is positive and decreasing, and thereby convergent. Let $b=\lim_{n\rightarrow\infty}b_n$. Combining the convergence with the above we have $0=\lim_{n\rightarrow\infty}(b_{n+1}-b_n)=\frac{b}{1+b}$, which gives $b=0$.
On the other hand, notice that $\frac{a_{n+1}}{a_n}=\frac{b_n^2}{na_n(1+(n+1)a_n)}=b_{n}-b_{n+1}$ and hence $\sum_{k=0}^n\frac{a_{n+1}}{a_n}=\frac{1}{2}+\sum_{k=1}^n(b_{k}-b_{k+1})=b_1+b_1-b_{n+1}=1-b_{n+1}$. Since we have already established that $b=0$, we have $\sum_{k=0}^\infty\frac{a_{n+1}}{a_n}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How fast does $a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}$ decay as $n \to \infty$? Let
\begin{equation}
a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}
\end{equation}
for $n \in \mathbb{N}$.
Then, $a_n$ is clearly a monotone decreasing sequence of positive numbers and by the Dominated Convergence Theorem, $a_n \to 0^+$ as $n \to \infty$.
However, I have some difficulty estimating how fast $a_n$ decays. For example, $\frac{a_n}{f(n)}=O(1)$ as $n \to \infty$ for some polynomial $f$?
Could anyone please provide insight into the decay rate of $a_n$?
| Alternative solution:
With the substitution $x = \sqrt y$, we have
$$a_n = \int_{1/2}^1 \frac{1}{(1/2 + y)^{n+1/2}}\frac{1}{2\sqrt y}\,\mathrm{d}y.$$
It is easy to prove that, for all $y \in [1/2, 1]$,
$$\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2}(y - 1/2) \le \frac{1}{2\sqrt y} \le \frac{\sqrt 2}{2}.$$
We have
$$\int_{1/2}^1 \frac{\frac{\sqrt 2}{2}}{(1/2 + y)^{n+1/2}}\,\mathrm{d}y - \int_{1/2}^1 \frac{\frac{\sqrt 2}{2}(y - 1/2)}{(1/2 + y)^{n+1/2}}\,\mathrm{d}y \le a_n \le \int_{1/2}^1 \frac{\frac{\sqrt 2}{2}}{(1/2 + y)^{n+1/2}}\,\mathrm{d}y$$
which results in
\begin{align*}
&\frac{\sqrt 2}{2n-1} - \frac{\sqrt 2}{2n-1}(2/3)^{n-1/2} - \frac{2\sqrt 2}{4n^2 - 8n + 3} + \frac{(n + 3/2)\sqrt 2}{4n^2 - 8n + 3}(2/3)^{n-1/2}\\
\le{}& a_n \le \frac{\sqrt 2}{2n-1} - \frac{\sqrt 2}{2n-1}(2/3)^{n-1/2}.
\end{align*}
Thus,
$$a_n = \frac{\sqrt 2}{2n} + O\left(\frac{1}{n^2}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Nth Derivative of $\frac{x}{x-1}$ So starting off, rewrite $x*\dfrac1{x-1}$. $d/dx=\dfrac{d}{dx}(x)*\dfrac{1}{x-1}+\dfrac{d}{dx}\dfrac{1}{x-1}*x$.
$\dfrac{d}{dx}(x)=1$. $x^{n}=nx^{n-1}$.
$\dfrac{1}{x-1}=(x-1)^{-1}$.
$\dfrac{d}{dx}(\dfrac{1}{x-1})=(-1)(x-1)^{-2}=\dfrac{-1}{(x-1)^{2}}$.
Combining them both we get $\dfrac{1}{x-1}+\dfrac{-1}{(x-1)^{2}}*x$.
$\dfrac{d}{dx}=\dfrac{1}{x-1}+\dfrac{-x}{(x-1)^{2}}$.
Now to find $\dfrac{d^2y}{d^2x}$, using product rule and sum rule
$\dfrac{d}{dx}\dfrac{1}{x-1}+\dfrac{d}{dx}-x\dfrac{1}{(x-1)^2}+\dfrac{d}{dx}\dfrac{1}{(x-1)^{2}}x$, so anyway doing it we get
$\dfrac{-1}{(x-1)^2}+\dfrac{-1}{(x-1)^2}+\dfrac{-2x}{(x-1)^{3}}$.
Simplifying $\dfrac{d^2y}{d^2x}=\dfrac{-2}{(x-1)^2}+\dfrac{-2x{(x-1)^{3}}$; and finally $\dfrac{d^3y}{d^3x}$. $\dfrac{d}{dx}\dfrac{-2}{(x-1)^2}+\dfrac{d}{dx}-2x\dfrac{1}{(x-1)^{3}}+\dfrac{d}{dx}\dfrac{1} {(x-1)^{3}}*(-2x)$,which gives $\dfrac{4}{(x-1)^3}+\dfrac{-2}{(x-1)^3}+\dfrac{6x{(x-1)^4}$; simplyfing we get
$\dfrac{2}{(x-1)^3}+\dfrac{6x}{(x-1)^4}$; now there is supposed to be a pattern here but i don't see it.
is there something wrong with my answer?
| $\frac {x}{1-x} = \frac {x}{1-x}+1-1 = \frac {1}{1-x} - 1\\
\frac {d}{dx} (1-x)^{-1}-1= (1-x)^{-2}\\
\frac {d^n}{dx^n} (1-x)^{-1}-1= n!(1-x)^{-(n+1)}\\
$
With what you have above, you do change the sign of your expression in the first step. I don't know if that is intentional.
at $\frac {d}{dx} x\frac {1}{x-1} = \frac {1}{x-1} - \frac {x}{(x-1)^2}$ combine the terms.
$\frac {1}{x-1} - \frac {x}{(x-1)^2} = \frac {x-1 - x}{(x-1)^2} = \frac {-1}{(x-1)^2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integration $\int_0^{\pi/2} \frac{dx}{(3 + 5 \cos x)^2}$ I had tried to solve this integral; using the substitution $\tan(x/2) =t$, and $\cos x= \frac{1-t^2}{1+t^2}$. But after making terms in $t$, I am not able to integrate further as numerator contains quadratic and denominator contains biquadratic.
$\int\limits_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}\ dx$.
| An alternative approach is to integrate both sides of
$$\left(\frac{5\sin x}{3 + 5 \cos x}\right)’
= \frac{16}{(3 + 5 \cos x)^2}+\frac3{3 + 5 \cos x}$$
to simply the integral
\begin{align}
\int_0^{\pi/2} \frac{1}{(3 + 5 \cos x)^2}dx
= \frac{5}{48}
-\frac1{16}\int_0^{\pi/2} \frac3{3 + 5 \cos x}dx
\end{align}
| {
"language": "en",
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Finding angle with geometric approach I would like to solve this problem just with an elementary geometric approach. I already solved with trigonometry, e.g. using the Bretschneider formula, finding that the angle $ x = 15° $. Any idea?
I edited showing how I computed the $ x $ value using the Bretschneider formula for the area of the quadrilateral $ ABDE $ and equating to the sum of the triangles' area $ ABE + EFD + BDF $
$$\begin{cases}
BC = a \\
AB = a(1/\tan(2x) - 1) \\
BD = a\sqrt{2} \\
AE = AB/\cos(2x+\pi/6) = a(1/\tan(2x) -1)/\cos(2x+\pi/6) \\
ED = a/\cos(x)
\end{cases}
$$
So I solved this equation with Mathematica, and the only solution that fit the problem is $ x = \pi/12 $
$ a^2/2+(a^2(1/\tan(2x) - 1)(1+\tan(x)))/2 + a^2 \tan(x)/2 =
((a\sqrt{2})^2 + \\
(a(1/\tan(2x)- 1)/\cos(2x+\pi/6))^2 - (a/\cos(x))^2 -(a(1/\tan(2x) - 1))^2)/4 \tan(\pi/2 -2x) $
I guess there is a simpler trigonometric approach, but I just wanted to try with that formula.
| Here I report the original figure, to which I added the segments $DH$ bisecting the angle $\widehat{FDG}$ and $EK$ orthogonal to $AD$.
Let's define
$$
d = AB, \qquad e = EF.
$$
We have, by Pythagorean theorem,
$$
AD = \sqrt{ AC^2 + CD^2 } = \sqrt{ (a+d)^2 + a^2 }.
$$
where $a$ is the known length of the side of the square.
For what follows, it would be useful to set $R=\sqrt{ (a+d)^2 + a^2 }.$
From the similar triangles $DFG$ and $ACD$
$$
\frac{FG}{CD} = \frac{DF}{AC} \quad \implies \quad FG = \frac{a^2}{a+d}.
$$
By Pythagorean theorem
$$
DG = \sqrt{ FG^2 + DF^2 } = \frac{aR}{a+d}.
$$
By the angle bisector theorem
\begin{align}
& \frac{GH}{FH} = \frac{DG}{DF}, \\[2mm]
& \frac{GH+FH}{FH} = \frac{DG+DF}{DF}, \\[2mm]
& \frac{FG}{FH} = \frac{DG+DF}{DF}, \\[2mm]
& FH = R-(a+d).
\end{align}
Note that
$$
EF = FH \quad \implies \quad e = R-(a+d)
$$
By Pythagorean theorem
$$
AE = \sqrt{ AB^2 + BE^2 } = \sqrt{ d^2 + (a+e)^2 }
$$
Being $AEK$ a 30-60-90 triangle, then
\begin{align}
& AK = \frac{\sqrt{3}}{2} AE = \frac{\sqrt{3}}{2} \sqrt{ d^2 + (a+e)^2 } \\
& EK = \frac{1}{2} AE = \frac{1}{2} \sqrt{ d^2 + (a+e)^2 }
\end{align}
We have
$$
DK = AD - AK = R - \frac{\sqrt{3}}{2} \sqrt{ d^2 + (a+e)^2 }
$$
By Pythagorean theorem
$$
DE^2 = EK^2 + DK^2 = \frac{1}{4} [ d^2 + (a+e)^2 ] + \left( R - \frac{\sqrt{3}}{2} \sqrt{ d^2 + (a+e)^2 } \right)^2
$$
but also
$$
DE^2 = DF^2 + EF^2 = a^2 + e^2
$$
Comparing these two expressions for $DE^2$, we have
$$
d^2+(a+e)^2+(a+d)^2-e^2=\sqrt{3}R\sqrt{d^2+(a+e)^2}
$$
Taking into account
\begin{align}
& e^2 = 2(a+d)^2+a^2-2(a+d)R \\
& (a+e)^2 = 2(a+d)^2-2d(R+a)
\end{align}
and squaring both sides
$$
4 \left(aR+d^2\right)^2 = 3 \left[(a+d)^2+a^2\right] \left[2 (a+d)^2+d^2-2 d \left(R+a\right)\right]
$$
by expanding the products and isolating the root $R$
$$
2 d \left(6 a^2+10 a d+3 d^2\right) R=4 a^4+16 a^3 d+32 a^2 d^2+24 a d^3+5 d^4
$$
squaring again both sides and factoring
$$
\left(2 a^2-2 a d-d^2\right) \left(8 a^6+72 a^5 d+188 a^4 d^2+208 a^3 d^3+122 a^2 d^4+50 a d^5+11 d^6\right)=0
$$
The only positive solution of this equation is
$$
d=(\sqrt{3}-1)a
$$
but this means that $AC = \sqrt{3}a$, the triangle $ACD$ is a 30-60-90 triangle, then $\widehat{CAD}=2x=30°$, so $x=15°$.
| {
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How do I justify $\cot 60$ being $\frac{\sqrt{3}}{3}$? Apologies if this is a basic question, my trigonometry skills simply isn't nowhere near up to scratch.
I have a real problem grappling with how $\cot$ functions. At least in my limited understanding, the standard values found in most table simply doesn't reflect how the identity seems to be calculated.
Take for instance $\cot \frac{\pi}{3}$
$\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3}=\frac{1}{2}$
$\tan 60=\frac{\sin 60}{\cos 60}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
I have seen two definitions of $\cot$, $\frac{\cos}{\sin}$ and $\frac{1}{\tan}$. According to both these definitions, the value of $\cot 60$ should be $\frac{1}{\sqrt{3}}$
Instead in most tables the value is claimed to be $\frac{\sqrt{3}}{3}$.
Are my calculations off? Am I reading the table wrong? Or is this actually the same value, written in this way due to some kind of strange convention?
| $\ \sqrt{3}\ $ is defined to be the (unique*) positive real number whose square is $\ 3.\ $ We therefore have:
$$ \frac{1}{\sqrt{3}} = 1\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\left( \sqrt{3} \right)^2} = \frac{\sqrt{3}}{3}. $$
$$$$
Alternatively,
$$\frac{1}{\sqrt{3}} = 1\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{9}} = \frac{\sqrt{3}}{3}. $$
*Uniqueness is not usually a part of the definition, but it is not difficult to show that the number is unique...
| {
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All natural number solutions for the equation $a^2+b^2=2c^2$ $a$, $b$ and $c$ of all Pythagorean triplets can be written in the form
$$
\begin{split}
a &= 2mn\\
b &= m^2-n^2 \\
c &= m^2+n^2
\end{split}
$$
where $m$ and $n$ are natural numbers. For any natural number $m$ and $n$, this set of equations will give a Pythagorean triplet. And all Pythagorean triplets satisfy this set of equations.
Can $a$, $b$ and $c$ of all triplets satisfying the equation $$a^2+b^2=2c^2$$ where $a$, $b$ and $c$ are natural numbers, be written as a set of equations as for the Pythagorean triplets?
So, I need a set of equations that generates triplets that satisfy the equation $a^2+b^2=2c^2$ for any natural numbers I plug into the set of equations. Also, every natural number triplets satisfying the equation $a^2+b^2=2c^2$ must satisfy the set of equations.
I tried to derive the set of equations myself, no attempts have been successful yet.
I would like to have the proof of the set of equations, (otherwise I won't know if every triple will satisfy the set of equations)
Any comments that helps to give an insight into solving the problem are really appreciated.
| Here is a proof on why natural solutions exist for $a, b, c$, as a matter of fact, infinite solutions.
$$a^2 + b^2 \equiv 0 \mod 2 \to (a + b)^2 - 2ab \equiv 0 \mod 2 \\ \quad \\ \implies a + b \equiv 0 \mod 2 $$
More importantly, this suggests $$a \equiv b \mod 2 \implies \text{ both $a - b$ and $a + b$ must be even}$$
Then there are such integers $x, y$ which satisfy $a - b = 2x$ and $a + b = 2y$.
Hence,
$$4x^2 + 4y^2 = (a + b)^2 + (a-b)^2 = 2(2c^2) \\ \quad \\ \implies x^2 + y^2 = c^2 * \qquad \square$$
For better understanding, $a, b$ have to be both even or odd, hence have the same remainder when divided by $2$. Furthermore, $*$ suggests there are infinite solutions to $a^2 + b^2 = c^2$ as there are infinite integer pythagorean triples.
| {
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Probability of a sum being divisible by $4$ Roll $n$ six-sided dice.
What is the probability that the sum of the results is divisible by $4$?
This is an 11th grade problem, and I really don't know -after a lot of searching-what tools do I need to solve this problem using only high school knowledge.
| I'm not quite sure that generating functions and some complex numbers are included in 11th grade (not sure what it corresponds to in Sweden), but you can get a closed solution for each value of $n$ if you know about these. Even though I suspect this is too complicated.
Let the generating function be
\begin{equation}
f_n(x) = (x+x^2+x^3+x^4+x^5+x^6)^n
\end{equation}
For two dices, expanding $f_2(x)$ gives
\begin{equation}
f_2(x) = x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9 +3x^10+2x^{11}+x^{12}
\end{equation}
here, it is the coefficients of $x^4,x^8$ and $x^{12}$ which are interesting, the sum of these corresponds to the number of outcomes which are divisible by four. Summing up, we see that for two dices, there are $3+5+1=9$ ways to get a sum divisible by four.
Now, define the number $\zeta = e^{\pi i/2}$. This has a periodicity of $4$, e.g. $\zeta^1 = \zeta^5$. If you expand $f_n(x)$, you get $f_n(x) = \sum_{k=0}^{6n} c_kx^k$, so here we want to try to extract the coefficients of interest, i.e. $c_4,c_8,c_{12}, ...$.
If you now look at the function $g_k(x)=x^k, k=0,1,2,3,...$, note that
\begin{align}
g_0(\zeta^0) + g_0(\zeta^1) + g_0(\zeta^2) + g_0(\zeta^3) &= 4 \newline
g_1(\zeta^0) + g_1(\zeta^1) + g_1(\zeta^2) + g_1(\zeta^3) &= 0 \newline
g_2(\zeta^0) + g_2(\zeta^1) + g_2(\zeta^2) + g_2(\zeta^3) &= 0 \newline
g_3(\zeta^0) + g_3(\zeta^1) + g_3(\zeta^2) + g_3(\zeta^3) &= 0 \newline
g_4(\zeta^0) + g_4(\zeta^1) + g_4(\zeta^2) + g_4(\zeta^3) &= 4 \newline
\end{align}
and so on. As you can see, we extract the correct coefficients. This means that we have that the number of possible ways $N$ to have a sum divisible by four is given by
\begin{equation}
N = \frac{1}{4} \sum_{k=0}^3 f_n(\zeta^k)
\end{equation}
Calculating gives
\begin{align}
f_n(\zeta^0) &= 6^n \newline
f_n(\zeta^1) &= (-1+i)^n \newline
f_n(\zeta^2) &= 0 \newline
f_n(\zeta^3) &= (-1-i)^n
\end{align}
such that
\begin{equation}
N = \frac{1}{4} \sum_{k=0}^3 f_n(\zeta^k) = \frac{1}{4} \left(6^n + 2^{n/2+1} \cos\left(\frac{3\pi n}{4} \right) \right)
\end{equation}
The first few terms give that the number of total outcomes which are divisible by four are then $N=1,9,55,322,1946$ for $n=1,2,3,4,5$. These are the same I got when expanding the generating functions so I think things match up, even though I'm seldom certain.
Furthermore, since you asked about the probability, the total number of outcomes are $6^n$, the probability for the sum being divisible by four, is given by
\begin{equation}
P_n = \frac{1}{4} \left(1 + \frac{2^{n/2+1}}{6^n} \cos\left(\frac{3\pi n}{4} \right) \right)
\end{equation}
Finally, note that $P_n\rightarrow 1/4$ when $n\rightarrow \infty$.
| {
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Continuity of function of two variable by Epsilon-Delta technique. Given a function
$f(x,y)= \frac{x^3y^2+x}{x^2+y^2}$ when $(x,y) \neq (0,0)$. Show that the function $f(x,y)$ is continuous at $(1,1)$ by Epsilon-Delta definition.
I can show it easily by the fraction of two continuous functions is continuous. But I need to prove it directly by Epsilon-Delta technique. I'm not able to think, how to do it. Please give some hint!
Effort: For any given $\epsilon>0$, we have
$|f(x,y)-1|= |\frac{x^3y^2+x}{x^2+y^2}-1|=|\frac{x^3y^2+x-x^2-y^2}{x^2+y^2}|$
Now, what's next?
| For any $\epsilon > 0$ there exists a $\delta>0$ such that
$\sqrt {(x-1)^2 + (y-1)^2} < \delta \implies \left | \frac {x^3y^2 + x - x^2 - y^2}{x^2+y^2}\right | <\epsilon$
The numerator can be simplified and factored.
$(x^3-1)y^2 - x(x-1) = (x-1)((x^2+x+1)(y^2)-x)$
$(x-1) < \delta$
Let $\delta\le 1$
$|x|< 2, |y| < 2$
$|((x^2+x+1)(y^2)-x)| \le (4 + 2 + 1)(4) + 2 \le 30$
$x^2+y^2\ge 1$
$\left | \frac {((x^2+x+1)y^2-x)(x-1)}{x^2+y^2}\right | < 30\delta \le \epsilon$
$\delta = \min (1,\frac {1}{30}\epsilon)$
| {
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Triple summation having $i,j,k$ as variables
Finding value of
$(1)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk$
$(2)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i\leq j\leq k\leq n}ijk$
For $(1) $ what I have done as
For without any restriction as
$ (S)=\displaystyle \mathop{\sum\sum\sum}_{1\leq i,j,k\leq n}ijk=\bigg(\sum^n_{i=1}i\bigg)^3=\bigg(\frac{n(n+1)}{2}\bigg)^3$
For exactly one equal
$\displaystyle S'=\displaystyle \mathop{\sum\sum\sum}_{1\leq i=j,k\leq n}ijk=\bigg(\sum^n_{j=1}j^2\bigg)\bigg(\sum^n_{k=1}k\bigg)=\bigg(\frac{n(n+1)(2n+1)}{6}\bigg)\bigg(\frac{n(n+1)}{2}\bigg)$
There are 3 such pairs $( i=j,j=k,k=i)$
All three equal
$S''=\displaystyle \mathop{\sum\sum\sum}_{1\leq i=j=k\leq n}ijk=\sum^n_{k=1}k^3=\bigg(\frac{n(n+1)}{2}\bigg)^2$
Now we have
$\displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk=S-3S'+S''$
$\displaystyle =\frac{n^2(n+1)^2}{4}\bigg(\frac{3n^2-n+4}{6}\bigg)$
But this is not correct, please tell me where I am wrong
Also please tell me part $2$
| Your approach is smart. It just needs a little revision. Based on Faulhaber's formula
we use
\begin{align*}
S_n&=\sum_{1\leq i,j,k\leq n}ijk=\left(\sum_{i=1}^ni\right)^3=\frac{1}{8}n^3(n+1)^3\\
S^{\prime}_n&=\sum_{1\leq i=j,k\leq n}ijk=\left(\sum_{j=1}^{n}j^2\right)\left(\sum_{k=1}^nk\right)\\
&=\frac{1}{6}n(n+1)(2n+1)\frac{1}{2}n(n+1)\\
&=\frac{1}{12}n^2(n+1)^2(2n+1)\\
S^{\prime\prime}_n&=\sum_{1\leq i=j=k\leq n}ijk=\sum_{i=1}^ni^3=\frac{1}{4}n^2(n+1)^2\\
\color{blue}{S^{\prime\prime\prime}_n}&\color{blue}{=\sum_{1\leq i<j<k\leq n}ijk}\\
\end{align*}
The identity we use to get $\color{blue}{S^{\prime\prime\prime}_n}$ is
\begin{align*}
\color{blue}{S_n=6\cdot S^{\prime\prime\prime}_n+3\cdot S^{\prime}_n-2\cdot S^{\prime\prime}_n}\tag{1}
\end{align*}
Note, in (1)
*
*we use $6$ times $S^{\prime\prime\prime}_n$ since we have $3!=6$ permutations of $(i,j,k)$ to respect.
*we also have to consider that $1\leq i=j,k\leq n$ also contains the cases $1\leq i=j\color{blue}{=}k\leq n$, which are also counted when we sum over $1\leq i=k,j\leq n$ and $1\leq i, j=k\leq n$. To compensate this overcounting we subtract $S^{\prime\prime}_n$ twice.
We obtain according to (1) and the identities above
\begin{align*}
\color{blue}{S^{\prime\prime\prime}_n}&=\frac{1}{6}S-\frac{1}{2}S^{\prime}+\frac{1}{3}S^{\prime\prime}\\
&=\frac{1}{48}n^3(n+1)^3-\frac{1}{24}n^2(n+1)^2(2n+1)+\frac{1}{12}n^2(n+1)^2\\
&=\frac{1}{48}n^2(n+1)^2\left(n(n+1)-2(2n+1)+4\right)\\
&=\frac{1}{48}n^2(n+1)^2\left(n^2-3n+4\right)\\
&\,\,\color{blue}{=\frac{1}{48}n^2(n+1)^2(n-1)(n-2)}
\end{align*}
in accordance with a comment from @JG.
The other wanted sum (2) can be derived from $S^{\prime\prime\prime}_{n}$. We obtain
\begin{align*}
\color{blue}{\sum_{1\leq i\leq j\leq k\leq n}}\color{blue}{ijk}
&=\sum_{1\leq i<j<k\leq n}ijk+\sum_{1\leq i=j<k\leq n}ijk+\sum_{1\leq i<j=k\leq n}ijk\\
&\qquad\qquad+\sum_{1\leq i=j=k\leq n}ijk\\
&=\sum_{1\leq i<j<k\leq n}ijk+\sum_{1\leq i=j<k\leq n}ijk+\sum_{1\leq k<i=j\leq n}ijk\\
&\qquad\qquad+\sum_{1\leq i=j=k\leq n}ijk\\
&=\sum_{1\leq i<j<k\leq n}ijk+\sum_{1\leq i=j,k\leq n}ijk\\
&\,\,\color{blue}{=S^{\prime\prime\prime}_n+S^{\prime}_n}\\
&=\frac{1}{48}n^2(n+1)^2(n-1)(n-2)+\frac{1}{12}n^2(n+1)^2(2n+1)\\
&=\frac{1}{48}n^2(n+1)^2(n^2+5n+6)\\
&\,\,\color{blue}{=\frac{1}{48}n^2(n+1)^2(n+2)(n+3)}
\end{align*}
in accordance with a comment from @JG.
| {
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Sums with nested radicals related to $1/\pi^2$ Context:
I evaluated some sums related to $1/\pi^{2}$ that for a reason I don't know I can't find in the literature.
$$(1)\hspace{.5cm}{1\over 2\sqrt{2}}+{1\over 2^{3}\sqrt{2+\sqrt{2}}}+{1\over 2^{5}\sqrt{2+\sqrt{2+\sqrt{2}}}}+\cdots=1-{8\over \pi^{2}},$$
$$(2)\hspace{.5cm}{1/4\over 2+\sqrt{3}}+{1/16\over 2+\sqrt{2+\sqrt{3}}}+{1/64\over 2+\sqrt{2+\sqrt{2+\sqrt{3}}}}+\cdots=1-{9\over \pi^{2}},$$
$$(3)\hspace{.5cm}{1/4\over 2+\sqrt{2+\varphi}}+{1/16\over 2+\sqrt{2+\sqrt{2+\varphi}}}+{1/64\over 2+\sqrt{2+\sqrt{2+\sqrt{2+\varphi}}}}+\cdots={1+\varphi}-{25\over \pi^{2}},$$
where $\varphi=\frac{1+\sqrt{5}}{2}.$
Note that this series are related to Viète's formulas.
Question:
Do you know any reference where these formulas are shown? Can you prove them with a general result?
| I will consider here only the third series with $\varphi$ the golden ratio.
I will establish it as an application of the following identity, justifying the presence of a $\frac{1}{\pi^2}$ term :
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
Its proof by different means is given in the answers to this question. Among them, one finds a proof using Viète's formulas...
Indeed, taking $x=\frac{\pi}{5}$ in (1), one gets, in the RHS :
$$\frac{1}{\sin^2 \frac{\pi}{5}}-\frac{25}{\pi^2}=\frac{4}{3-\pi}-\frac{25}{\pi^2}$$
On the LHS, detailing the first terms :
$$\color{red}{\underbrace{\frac{1}{4\cos^2\frac{\pi}{5 \times 2}}}_{= \ \frac{1}{2+\varphi}}}+\frac{1}{4^2\cos^2\frac{\pi}{5 \times 2^2}}+\frac{1}{4^3\cos^2\frac{\pi}{5 \times 2^3}}+\cdots$$
Why is the first term highlighted ? Because it is the single term which is not present in the series you have given. All the other terms are exactly the right ones :
Indeed, one can verify that :
$$4 \cos^2 \left(\frac{\pi}{5.2^k}\right)=\begin{cases}2+\varphi&(k=1)\\ 2+\sqrt{2+\varphi}&(k=2)\\2 + \sqrt{2+\sqrt{2+\varphi}}&(k=3)\\ \cdots \end{cases}$$
due to relationship :
$$\cos\left(\frac{\pi}{5.2^{k+1}}\right)=\frac12\left(1+\cos\left(\frac{\pi}{5.2^{k}}\right)\right) \iff $$
$$4\left(\cos\left(\frac{\pi}{5.2^{k+1}}\right)\right)^2=2+2 \cos\left(\frac{\pi}{5.2^{k}}\right) $$
It remains for completing the proof to subtract to the RHS the "added term" :
$$\frac{4}{3-\pi}-\frac{25}{\pi^2}-\color{red}{\frac{1}{2+\varphi}}=1+\varphi-\frac{25}{\pi^2}$$
as desired (using relationshup $\varphi^2=\varphi+1$, etc.).
Remark : the different special values such as $\sin \frac{\pi}{5}$, etc. can be found for example here.
| {
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Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$
| Without derivatives:
Remark: I used this trick in my answer before.
We want to prove that $2\cos \frac{50\pi}{180} + \sqrt 3 > 3$.
Let $u = \cos \frac{50\pi}{180}$ and $v = \frac{3 - \sqrt 3}{2}$. We need to prove that $u > v$.
We have $u \ge \cos \frac{60\pi}{180} = 1/2$ and $v > \frac{3 - 2}{2} = 1/2$.
Thus, $4u^2 + 4uv + 4v^2 > 3$.
Thus, it suffices to prove that
$$(u - v)[4u^2 + 4uv + 4v^2 - 3] > 0$$
or
$$4u^3 - 3u > 4v^3 - 3v$$
or (using triple angle formula $\cos 3x = 4\cos^3 x - 3\cos x$)
$$\cos \left(3 \times \frac{50\pi}{180}\right) > \frac{45}{2} - \frac{27}{2}\sqrt 3$$
or
$$13\sqrt 3 > \frac{45}{2}$$
which is true.
We are done.
| {
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Find the roots of the equation $(1+\tan^2x)\sin x-\tan^2x+1=0$ which satisfy the inequality $\tan x<0$ Find the roots of the equation $$(1+\tan^2x)\sin x-\tan^2x+1=0$$ which satisfy the inequality $$\tan x<0$$
Shold I solve the equation first and then try to find which of the roots satisfy the inequality? Should I use $\tan x$ in the solution itself and obtain only the needed roots?
I wasn't able to see how the inequality can be used beforehand, so I solved the equation. For $x\ne \dfrac{\pi}{2}(2k+1),k\in\mathbb{Z}$, the equation is $$\dfrac{\sin^2x+\cos^2x}{\cos^2x}\sin x-\left(\dfrac{\sin^2x}{\cos^2x}-\dfrac{\cos^2x}{\cos^2x}\right)=0$$ which is equivalent to $$\sin x+\cos2x=0\\\sin x+1-2\sin^2x=0\\2\sin^2x-\sin x-1=0$$ which gives for the sine $-\dfrac12$ or $1$. Then the solutions are $$\begin{cases}x=-\dfrac{\pi}{6}+2k\pi\\x=\dfrac{7\pi}{6}+2k\pi\end{cases}\cup x=\dfrac{\pi}{2}+2k\pi$$ How do I use $\tan x<0$?
| $tanx < 0 $ essentially means $ x \in \frac{(2n-1)\pi}{2} ,n\pi $ and the by taking intersection of this and the solution that you obtained, the final answer is $x = 2n\pi -\frac{\pi}{6}$
| {
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Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
| Given integral $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Let's factor out $x^2 $ from the denominator.
$\Rightarrow \int \frac{1}{x}\sqrt{\frac{1+x^2}{1-x^2}}dx$
Now multiply and divide the numerator and the denominator by $\sqrt{1+x^2}$
$\Rightarrow \int \frac{1+x^2}{x\sqrt{1-x^4}}dx$
Now we can split the integral into 2 sub-integrals
$\Rightarrow \int \frac{1}{x\sqrt{1-x^4}}dx+\int \frac{x^2}{x\sqrt{1-x^4}}dx$
$\Rightarrow \int \frac{x^3}{x^4\sqrt{1-x^4}}dx+\int \frac{x}{\sqrt{1-x^4}}dx$
For the first integral, use $1-x^4=u^2$ and for the second, use $x^2=v$. The integrals reduce to
$\Rightarrow \frac{-1}{2}\int \frac{du}{1-u^2}+\frac{1}{2}\int \frac{dv}{\sqrt{1-v^2}}$
$\Rightarrow \frac{1}{2}\int \frac{du}{u^2-1}+\frac{1}{2} \sin ^{-1}v$
$\Rightarrow \frac{1}{4}\log |\frac{u-1}{u+1}| +\frac{1}{2}\sin ^{-1}v+c$
$\Rightarrow \frac{1}{4}\log |\frac{\sqrt{1-x^4}-1}{\sqrt{1-x^4}+1}| +\frac{1}{2}\sin ^{-1}(x^2)+c$
Hence, $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx=\frac{1}{4}\log |\frac{\sqrt{1-x^4}-1}{\sqrt{1-x^4}+1}| +\frac{1}{2}\sin ^{-1}(x^2)+c$
| {
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Trying to do integration using residue theorem prove using residue theorem $$\int_{0}^{2\pi}\frac{\cos2\theta}{5+4\cos\theta}d\theta={\pi}/6$$
I tried by using $$z=e^{i\theta}$$
now $$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+z^{-1}}{2}$$
$$dz=izd\theta$$
$$\int_{0}^{2\pi}\frac{\cos2\theta}{5+4\cos\theta}d\theta=\int_{0}^{2\pi}\frac{2\cos^2\theta-1}{5+4\cos\theta}d\theta=\int_{|z|=1}\frac{(z^4+1)dz}{2iz^2(2z+1)(z+2)}$$
Now poles occur at $$z=0, -1/2, -2$$
rejecting $$z=-2,$$as $$ |z|=2>1$$
Now $$Res_{z=-1/2}=\frac{(-1/2)^4+1)}{2i(-1/2)^2(-1/2+2)}=17/12i$$
Pole at z=0 is of order 2
$$Res_{z=0}=\lim_{z \to 0} \frac{d}{dz}(\frac{(z^4+1)}{2i(2z+1)(z+2)})=-5/8i$$
$$\int_{0}^{2\pi}\frac{\cos2\theta}{5+4\cos\theta}d\theta={2\pi}i(Res_{z=-1/2}+Res_{z=0})=19{\pi}/12$$
But the answer is $${\pi}/6$$
I tried many times, but i get the same answer
| $\displaystyle \int_{|z|=1}\frac{(z^4+1)dz}{2iz^2(2z+1)(z+2)}$, should be converted to $\displaystyle \frac{1}{i}\int_{|z|=1} \frac{1+z^4}{4z^2(z+2)(z+1/2)}$. From there, you can see you have poles at $z=-2$, $z=0$ (order $2$), and $z=-\dfrac{1}{2}$. Since we only care about the poles inside the unit circle, we can evaluate the two resides as follows:
$$2\pi i \operatorname{Res}\left(\frac{1+z^4}{4z^2(z+2)(z+1/2)}, z=0\right) = \frac{2\pi i}{(2-1)!}\lim_{z \to 0}\frac{d^{2-1}}{dz^{2-1}}\frac{(z-0)^2\left(1+z^4\right)}{4z^2(z+2)(z+1/2)} = -\frac{10\pi i}{8}$$
and
$$2\pi i\operatorname{Res}\left(\frac{1+z^4}{4z^2(z+2)(z+1/2)}, z=-\dfrac{1}{2}\right) = 2\pi i \lim_{z\to -1/2}\frac{\left(z+\frac{1}{2}\right)\left(1+z^{4}\right)}{4z^{2}\left(z+2\right)\left(z+1/2\right)} = \frac{34\pi i}{24}.$$
Therefore, the integral is
$$\frac{1}{i}\left(\frac{34\pi i}{24}-\frac{10\pi i}{8}\right)=\frac{\pi}{6}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the volume between sphere and hyperboloid I am trying to find the volume between the sphere: $x^2+y^2+z^2=9$ and the hyperboloid $x^2+y^2-z^2=1$. I set the integral as: $$\int_{0}^{2\pi}\int_{-3}^{3}\int_{\sqrt{1+z^2}}^{\sqrt{9-z^2}}rdrdzd\theta$$ but it does not give me the correct answer. Can somebody tell me what is wrong with my calculations?
| Your parametrization of the solid is wrong. Notice that $\sqrt{z^2+1}<\sqrt{9-z^2}$ when $-2<z< 2$, but $\sqrt{z^2+1}>\sqrt{9-z^2}$ when $|z|>2$.
When $-2\leq z\leq 2$, the part of the solid looks like a pipe and it is parametrized as $-2\leq z\leq 2$, $0\leq r\leq\sqrt{z^2+1}$, $0\leq\theta\leq 2\pi$. We can find the volume of this part by cross-section method: $V_1=\int_{-2}^2A(z)dz=2\int_0^2\pi r^2dz=2\pi\int_0^2(z^2+1)dz=\frac{28}3\pi$.
When $2\leq z\leq 3$, the part of the solid is a spherical cap parametrized by $2\leq z\leq\sqrt{9-r^2}$, $0\leq r\leq\sqrt 5$, $0\leq\theta\leq2\pi$. Hence, its volume is $V_2=\int_0^{2\pi}\int_0^{\sqrt5}(\sqrt{9-r^2}-2)rdrd\theta=\frac83\pi$. Bottom cap is also $V_3=\frac83\pi$. The total volume $V=V_1+V_2+V_3=\frac{28+8+8}{3}\pi=\frac{44}3\pi.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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About an integral of the MIT Integration Bee Finals (2023) I would like to solve the first problem of the MIT Integration Bee Finals, which is the following integral :
$$\int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx$$
I tried substitution $u=\tan(x)$, King Property, but nothing leads me to the solution which is apparently $\frac{2\sqrt{3}}{9} \pi$.
If anybody knows how to solve it I would be grateful.
| I want to generalize the integral as
$$
I(a,n)=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(\sin x+\cos x)^{2 n}}=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(1+\sin 2 x)^n} d x
$$
where $0<a<1$ and $n\in N$.
Letting $t=\tan x$ gives
$$
I(a,n)=\int_0^{\infty} \frac{t^{a}}{\left(1+\frac{2 t}{1+t^2}\right)^n} \cdot \frac{d t}{1+t^2}=\int_0^{\infty} \frac{t^{a}\left(1+t^2\right)^{n-1}}{(t+1)^{2 n}} d t
$$
Using Binomial expansion and beta function, we have
$$\begin{aligned}
I(a,n)&=\sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \int_0^{\infty} \frac{t^{2 k+a}}{(t+1)^{2 n}} d t\\&=\sum_{k=0}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right) B\left(2 k+a+1,2n-2k-a-1 \right)\\&= \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \frac{\Gamma\left(2 k+a+1\right) \Gamma\left(2 n-2 k—a-1\right)}{(2 n-1)!}\end{aligned}
$$
In particular,
$$I(\frac{1}{3} ,1)=\Gamma(\frac{4}{3} )\Gamma(\frac{2}{3} )= \frac{1}{3} \Gamma(\frac{1}{3} )\Gamma(\frac{2}{3} ) =\frac{2\pi}{3\sqrt 3} $$
| {
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"source": "stackexchange",
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Hint with floors and ceiling proof Prove that $$\forall n \in \mathbb{Z},\left \lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$$
I decided to approach it by extending floor and ceiling definition and got
$$ \frac{n-1}{2} \leq \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor \leq \frac{n}{2} $$
And now I am stuck with $$\frac{n}{2} = \frac{n-1}{2} $$
which only made it worse.
Any advice or hint on how I can approach this proof differently?
| Let $n = 2m + 1$ for some $m \in \mathbb{Z}$. In this case, $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = m$. Similarly, $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor m + \frac{1}{2} \right\rfloor = m$. Consequently, $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$.
We can similarly consider the case $n = 2m$. For this case, $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor = m$ and $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lceil m - \frac{1}{2} \right\rceil = m$. Once again, we can conclude $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$, giving us the required result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find integral of $\int\frac{x^2+3}{\sqrt{x^2+6}}dx$ from Cambridge IGCSE Additional Mathematics $$\int\frac{x^2+3}{\sqrt{x^2+6}}dx$$
I tried to $$\int\frac{x^2+6-3}{\sqrt{x^2+6}}dx=\int\frac{x^2+6}{\sqrt{x^2+6}}dx-\int\frac{3}{\sqrt{x^2+6}}dx= \int\sqrt{x^2+6}dx - 3\int\frac{1}{\sqrt{x^2+6}}dx$$
Then got stuck
In exam-mate website the answer is $$\frac{1}{2}x\sqrt{x^{2}+6}$$
Thanks for replies in advance
| Since your work $$\int \frac{x^{2}+3}{\sqrt{x^{2}+6}}dx=\int \frac{x^{2}+6-3}{\sqrt{x^{2}+6}}dx=\int \sqrt{x^{2}+6}dx-3\int \frac{1}{\sqrt{x^{2}+6}}dx,$$ which it is correct. Now we are looking for primitives of $G(\sqrt{a^{2}+x^{2}})$, we can try with the substitution $x=a\tan t$.
*
*$\int \sqrt{x^{2}+6}dx\underset{x=\sqrt{6}\tan t}{=}6$ $\int \sec^{3}t\, dt$;
*$\int \frac{1}{\sqrt{x^{2}+6}}dx\underset{x=\sqrt{6}\tan t}{=}$$\int \sec t\, dt$,
Thus,
\begin{align*}\int \frac{x^{2}+3}{\sqrt{x^{2}+6}}\, dx &\underset{x=\sqrt{6}\tan t}{=}-3\ln|\tan t+\sec t|+3\tan t\sec t+3\ln|\tan t+\sec t|+C\\
&=3\tan t\sec t+C\\
&=3\tan(\tan^{-1}\frac{x}{\sqrt{6}})\sec(\tan^{-1}\frac{x}{\sqrt{6}})+C\\
&=3\frac{x}{\sqrt{6}}\sqrt{(\frac{x}{\sqrt{6}})^{2}+1}+C\\
&=3\frac{1}{\sqrt{6}}\frac{1}{\sqrt{6}}\sqrt{x^{2}+6}+C\\
&=\frac{1}{2}x\sqrt{x^{2}+6}+C
\end{align*}
Therefore,
$$\int \frac{x^{2}+3}{\sqrt{x^{2}+6}}dx=\frac{1}{2}x\sqrt{x^{2}+6}+C$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficulties with estimation of epsilon-delta limit proof I have to proof $\lim_{x\to 5}\frac{4x-9}{3x-16}=-11$. I have hard time to evaluate $\frac{1}{|3x-16|}$.
So I start that let $|x-5|<$. I need to show that $|\frac{4x-9}{3x-16}-(-11)|<ε$.
$|\frac{4x-9}{3x-16}-(-11)|$ = $|\frac{37x-185}{3x-16}|$ = $\frac{5|x-5|}{|3x-16|}$ < $ε$.
I choose $:=1$, then $|x-5|<1$, where I get $-1 < x-5 < 1$, then multiply with 3 and subtract -1, so $-4 < 3x-16 < 2$. If I think about it as $\frac{-1}{4} > \frac{1}{3x-16} > \frac{1}{2}$ contradiction.
Is it wrong to continue the way that $ |\frac{1}{3x-16}| < \frac{1}{2}$ ? So I get that $\frac{5|x-5|}{|3x-16|}$< $5 * * \frac{1}{2}$ <= $ε $ if $ = \frac{2}{5}*ε$.
Finaly I got that := min {1, $\frac{2}{5}*ε$}
I would need some help with $\frac{1}{|3x-16|}$.
| If $|x-5|<\frac16$, then $5-\frac16<x<5+\frac16$, and therefore $-\frac32<3x-16<-\frac12$. So, $|3x-16|>\frac12$.
So, if you take $\delta\leqslant\frac16$ and if $|x-5|<\delta$, you have$$\left|\frac{4x-9}{3x-16}-(-11)\right|=\frac{37|x-5|}{|3x-16|}<74|x-5|.$$So, you can take $\delta=\min\left\{\frac\varepsilon{74},\frac16\right\}$, in order that$$\left|\frac{4x-9}{3x-16}-(-11)\right|<\varepsilon.$$
In your answer, you had $-4<3x-16<2$ and you deduced that $-\frac14>\frac1{3x-16}>\frac12$, which is wrong. You have indeed $a<b<c\implies\frac1a>\frac1b>\frac1c$, but only when $a>0$ or $c<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
| $\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.
$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$
$= \int_{n+1}^{2n+1} \frac{1}{x}\ dx \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \int_{n}^{2n} \frac{1}{x}\ dx$
$= \ln(2n)-\ln(n) = \ln(2)$.
So by squeeze theorem we are done.
| {
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"url": "https://math.stackexchange.com/questions/716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
| Take base cases 1, 2, and 3, and then deduce case $n+3$ from case $n$.
| {
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Finding a Function That Approaches Another Function One of my math professors gave me the following challenge. It isn't graded, it's just for fun.
Consider the function:
\begin{equation*}
f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1).
\end{equation*}
I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:
$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$
$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$
Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?
| Yes looks like it must be b)
Consider the following:
There is only one $x$ factor, so this eliminates choices c) and d).
The function is odd, i.e $f(-x) = -f(-x)$, so this eliminates a).
Thus b) must be the answer.
To find the answer by hand you can do the following:
Start with
$$f(x) = x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} $$
Now differentiate
$$f'(x) = 1 + 3x^2 + 5x^4 + \cdots $$
Multiply by $x$
$$f'(x)x = x + 3x^3 + 5x^5 + \cdots $$
Differentiate again
$$(f'(x)x)' = 1 + 3^2 x^2 + 5^2 x^4 + \cdots $$
Multiply by $x$, differentiate again and then multiply by $x$ to give the answer.
| {
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partitions with even number of even parts - partitions with odd number of even parts Let $e(n)$ be the number of partitions of $n$ with even number of even parts and let $o(n)$ denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that $\sum_{n \geq 0} (e(n) - o(n)) x^n = \frac{1}{(1-x) \times (1+x^2) \times (1 - x^3 ) \times (1+x^4) \times ... }$. I have been racking my head over this for the past few hours, and I can't see any light.
I noticed, that $e(n) - o(n) = 2e(n) - p(n)$ where $p(n)$ is the number of partitions of $n$, so the above claim is equivalent to showing $\sum_{ n \geq 0} e(n)x^n = \frac{1}{2} \frac{1}{(1-x)(1-x^3)(1-x^5)...}( \frac{1}{(1-x^2)(1-x^4)....} + \frac{1}{(1+x^2)(1+x^4)........})$, and similarly, it is equivalent to $\sum_{ n \geq 0} o(n)x^n = \frac{1}{2} \frac{1}{(1-x)(1-x^3)(1-x^5)...}( \frac{1}{(1-x^2)(1-x^4)....} - \frac{1}{(1+x^2)(1+x^4)........})$, but these identities appear more difficult than the original one.
Any hints and suggestions appreciated.
| Now,
$$ \frac{1}{(1-x)(1+x^2)(1-x^3)\dots} $$
$$ = (1+x+x^2+\dots)(1-x^2+x^4-\dots)(1+x^3+x^6+\dots)(1-x^4+x^8-\dots)\dots $$
Consider a term which provides a negative coefficient to $x^n$
If we pick $x^{2n_2}$ from the second term, $x^{4n_4}$ from the fourth and so on,
The coefficient of $-1$ is $$(-1)^{n_2 + n_4 + \dots}$$
this will be negative if and only if $n_2 + n_4 + \dots$ is odd which comes from $o(n)$.
The positive ones come from $e(n)$.
Thus we must have that
$$\sum (e(n) - o(n))x^n = \frac{1}{(1-x)(1+x^2)(1-x^3)\dots}$$
| {
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differential equations I am trying to work through an example from Jordan and Smiths book Nonlinear Ordinary Differential Equations. It's example $6.1$ on page $195$. The question reads:
Obtain an approximate solution of the family of equations $x^{\prime\prime} + x = ex^3$
with $x(e, 0) = 1$, $x^{\prime}(e,0) = 0$ and error $O(e^3)$ uniformly on $t \geq 0$, by the method of coordinate pertubation.
This what I have so far: $x_0^{\prime\prime} + x_0 = 0$ so we get $x_0 = A_0\cos t + B_0\sin t$.
From the conditions given: $x_0(0) = 1$ and $x^{\prime}_0(0) = 0$, so we have $x_0 = \cos t$.
The second equation is $x^{\prime\prime}_1 + x_1 = x_0^3$ which becomes $x_1 + x_1 = \cos^3 t$, so $x^{\prime\prime}_1 + x_1 = \frac{3}{4}\cos t + \frac{1}{4}\cos 3t$.
Now I assume $x_1 = A\cos t + B\cos 3t$, so $x^{\prime}_1 = -A\sin t - 3B\sin 3t$, and $\ddot{x}_1 = -A\cos t - 9B\cos 3t$.
Therefore $\ddot{x}_1 + x_1 = -8B\cos 3t = \frac{3}{4}\cos t + \frac{1}{4}\cos 3t$, hence $-8B = \frac{1}{4}$, so $B = \frac{-1}{32}$ which gives me $x_1 = A\cos t - \frac{1}{32}\cos 3t$.
Using this and the condition that $x_0(0) = 0$ gives me $A = \frac{1}{32}$. However the solution in the book contains another term: $\frac{3}{8}t\sin t$. I don't know how to get this term. Could someone please help.
| Well, as you can see in the book, proposing a regular perturbation
$$
x(t) = x_0(t) + \varepsilon x_1(t) + \varepsilon^2 x_2(t) + ...
$$
we have
$$
x_0'' + \varepsilon x_1 '' + \varepsilon^2 x_2 '' + ... + x_0 + \varepsilon x_1 + \varepsilon^2 x_2 + ... = \varepsilon (x_0 + \varepsilon x_1 + ... )^3
$$
and
$$
(x_0 + \varepsilon x_1 + ... )^3 = x_0^3 + 3 \varepsilon x_0^2 x_1 + \varepsilon^2 (3 x_0 x_1^2 + 3 x_0^2 x_2) + ...
$$
then
\begin{align}
O\big(1):& &x_0'' + x_0 &= 0\\
O(\varepsilon):& &x_1'' + x_1 &= x_0^3\\
O(\varepsilon^2):& &x_2'' + x_2 &= 3 x_0^2 x_1\\
\vdots\quad
\end{align}
Finally, the initial condition is $O(1)$, meaning $x_0(0) = 1$, $x_0'(0) = 0$, and $$x_1(0) = x_1'(0) = x_2(0) = x_2'(0) = \ldots = 0.$$
Combining all this information,
\begin{align}
x_0(t) &= \cos(t) \\
x_1(t) &= \tfrac{1}{32} \big(\cos(t) - \cos(3 t)\big) + \tfrac{3}{8} t \sin(t) \\
x_2(t) &= \tfrac{1}{1024}\big(23 \cos(t) - 24 \cos(3 t) + \cos(5 t)\big) + \tfrac{3 t}{256}\big(8 \sin(t) - 3 \sin(3 t)\big) - \tfrac{9 t^2}{128} \cos(t)
\end{align}
and we have constructed the approximation $$x(t) = x_0(t) + \varepsilon x_1(t) + \varepsilon^2 x_2(t) + O(\varepsilon^3).$$
| {
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
| Let $\omega$ be a primitive fifth root of unity. The quadratic residues in $\mathbb{Z}/(5\mathbb{Z})^*$ are $1$ and $4$, hence by setting $q=\omega^1-\omega^2-\omega^3+\omega^4$ such Gauss sum fulfills $q^2=5$. On the other hand
$$\omega^1-\omega^2-\omega^3+\omega^4= 2\cos\frac{2\pi}{5}-2\cos\frac{4\pi}{5} \tag{A}$$
is real an positive, so $q=\sqrt{5}$, and
$$\omega^1+\omega^2+\omega^3+\omega^4=2\cos\frac{2\pi}{5}+2\cos\frac{4\pi}{5}=-1,\tag{B}$$
so by summing $(A)$ and $(B)$ we get $4\cos\frac{2\pi}{5}=\sqrt{5}-1$ as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $ I am stuck while proving this identity I verified it using induction but like the other two (1) (2), I am seeking a more of a general way (algebraic will be much appreciated)
$$\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $$
| HINT:
$$(1+x)^{n} - (1-x)^{n} = 2 \Biggl[ {n \choose 1} x + {n \choose 3} x^{3} + \cdots \Biggr]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/7835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate this efficiently? If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?
I solved it in the following way :
Expanding we get, the coefficient of $ x $ as $ (m-n) = 3 \cdots (1)$ and coefficient of $ x^2 $ as $ \frac{n(n-1)}{2} + \frac{m(m-1)}{2} +- m \cdot n = -6 \cdots (2)$ after substituting and some algebraic treatment I got m = 12 and n = 9.
Now this is correct but I am interested if there exists any short procedure such that the entire problem could be done under a mint.
| For the coefficient of $x$ to be 3, $m-n=3$ as you said, so $m>n$ and $m-n>0$ and we can factor the original product:
$$\begin{align}
(1+x)^m(1-x)^n&=(1+x)^{m-n+n}(1-x)^n
\\
&=(1+x)^{m-n}(1+x)^n(1-x)^n
\\
&=(1+x)^{m-n}\left((1+x)(1-x)\right)^n
\\
&=(1+x)^{m-n}(1-x^2)^n
\\
&=(1+x)^3(1-x^2)^n
\end{align}$$
Now, the coefficient of $x^2$ is the sum of the coefficients of $x^2$ in $(1-x^2)^n=1-nx^2+\cdots$ and $(1+x)^3=1+3x+3x^2+x^3$, so $-6=-n+3$ and $n=9$. Since $m-n=3$, $m=12$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Just as a curiosity, a one-line-real-analytic-proof I found by combining different ideas from this thread and this question:
$$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.\end{eqnarray*}$$
Update. By collecting pieces, I have another nice proof. By Euler's acceleration method or just an iterated trick like my $(1)$ here we get:
$$ \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} = \color{red}{\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}}\tag{A}$$
and the last series converges pretty fast. Then we may notice that the last series comes out from a squared arcsine. That just gives another proof of $ \zeta(2)=\frac{\pi^2}{6}$.
A proof of the identity
$$\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi}{2}\sum_{k\geq 0}\frac{(-1)^k}{2k+1}=\frac{\pi}{2}\cdot\frac{\pi}{4}$$
is also hidden in tired's answer here. For short, the integral
$$ I=\int_{-\infty}^{\infty}e^y\left(\frac{e^y-1}{y^2}-\frac{1}{y}\right)\frac{1}{e^{2y}+1}\,dy $$
is clearly real, so the imaginary part of the sum of residues of the integrand function has to be zero.
Still another way (and a very efficient one) is to exploit the reflection formula for the trigamma function: $$\psi'(1-z)+\psi'(z)=\frac{\pi^2}{\sin^2(\pi z)}$$ immediately leads to: $$\frac{\pi^2}{2}=\psi'\left(\frac{1}{2}\right)=\sum_{n\geq 0}\frac{1}{\left(n+\frac{1}{2}\right)^2}=4\sum_{n\geq 0}\frac{1}{(2n+1)^2}=3\,\zeta(2).$$
2018 update. We may consider that $\mathcal{J}=\int_{0}^{+\infty}\frac{\arctan x}{1+x^2}\,dx = \left[\frac{1}{2}\arctan^2 x\right]_0^{+\infty}=\frac{\pi^2}{8}$.
On the other hand, by Feynman's trick or Fubini's theorem
$$ \mathcal{J}=\int_{0}^{+\infty}\int_{0}^{1}\frac{x}{(1+x^2)(1+a^2 x^2)}\,da\,dx = \int_{0}^{1}\frac{-\log a}{1-a^2}\,da $$
and since $\int_{0}^{1}-\log(x)x^n\,dx = \frac{1}{(n+1)^2}$, by expanding $\frac{1}{1-a^2}$ as a geometric series we have
$$ \frac{\pi^2}{8}=\mathcal{J}=\sum_{n\geq 0}\frac{1}{(2n+1)^2}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$ In this post, David Speyer, actually, gave an expression for $\displaystyle \frac{t}{e^{t}-1}$.
The question is can we sum the given series, using that expression, if not how does one sum this series. $$\sum\limits_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1}=\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$$
| What you require here are the Eisenstein series. In particular the evaluation of
$$E_2(\tau) = 1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}},$$
at $\tau = i. $ Rearrange to get
$$\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau} } = \frac{1}{24}(1 – E_2(i) ).$$
See Lambert series for additional information.
EDIT: The function
$$G_ 2(\tau) = \zeta(2) \left(
1 – 24\sum_{n=1}^\infty \frac{ne^{2\pi i n \tau} }{1 - e^{2\pi i n \tau}} \right)
=\zeta(2)E_2(\tau)$$
satisfies the quasimodular transformation
$$G_ 2\left( \frac{a\tau+b}{c\tau+d} \right) =
(c\tau+d)^2G_ 2(\tau) - \pi i c (c\tau + d).$$
And so with $a=d=0,$ $c=1$ and $b=-1$ we find $G_ 2(i) = \pi/2.$ Therefore
$$E_2(i) = \frac{ G_ 2( i)}{ \zeta(2)} = \frac{\pi}{2}\frac{6}{\pi^2} = \frac{3}{\pi}.$$
Hence we obtain
$$\sum_{n=1}^\infty \frac{n}{e^{2\pi n} – 1} = \frac{1}{24} - \frac{1}{8\pi},$$
as given in the comment to the question by Slowsolver.
EDIT:
There is a very nice generalisation of the sum in the question.
For odd $ m > 1 $ we have
$$\sum_{n=1}^\infty \frac{n^{2m-1} }{ e^{2\pi n} -1 } = \frac{B_{2m}}{4m},$$
where $B_k$ are the Bernoulli numbers defined by
$$\frac{z}{e^z - 1} = \sum_{k=0}^\infty \frac{B_k}{k!} z^k \quad \textrm{ for }
|z| < 2 \pi.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
| This may be an easier method $$\int\frac{1}{\sin{x} \cdot \cos{x}} \ dx = \int\frac{\sec^{2}{x}}{\tan{x}} \ dx$$ by multiplying the numerator and denominator by $\sec^{2}{x}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Height of a tetrahedron How do I calculate the height of a regular tetrahedron having side length $1$ ?
Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table?
| You can also use trig based on the dihedral angle between two faces of the tetrahedron.
Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$
$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.
As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:
$$\begin{eqnarray}
|OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\
s^2 &=& \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\
s^2 &=& \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\
1 &=& \frac{3}{2} - \frac{3}{2} \cos{M} \\
\frac{-1}{2} &=& - \frac{3}{2} \cos{M} \\
\frac{1}{3} &=& \cos{M} \;\;\; (**)\\
\Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&=& \sin{M}
\end{eqnarray}$$
Therefore,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$
(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):
$$\begin{eqnarray}
\cos\left({\text{angle between two sides of a regular triangle}}\right) &=& \frac{1}{2}\\
\cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &=& \frac{1}{3}\\
\cos\left({\text{angle between two facets of a regular n-simplex}}\right) &=& \frac{1}{n}
\end{eqnarray}$$
(Who would've suspected, upon first encountering it, that the "$2$" in "$\cos{60^{\circ}}=\frac{1}{2}$" was actually a reference to the dimension of the triangle?)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Suggest a tricky method for this problem Find the value of:
$$ 1+ \biggl(\frac{1}{10}\biggr)^2 + \frac{1 \cdot 3}{1 \cdot 2} \biggl(\frac{1}{10}\biggr)^4 + \frac{1\cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \biggl(\frac{1}{10}\biggr)^6 + \cdots $$
| This can also be done using Wallis' product.
For instance see a similar problem: Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$
and
Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$
Use
$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \sin^{2n} x \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} = \dfrac{1}{2^n} \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n}$$
Thus
$$\displaystyle \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} = \dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots n} \dfrac{1}{10^{2n}}$$
Thus your sum is
$$\displaystyle \sum_{n=0}^{\infty} \dfrac{2}{\pi} \int_{0}^{\pi/2} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx} = \dfrac{2}{\pi} \int_{0}^{\pi/2} \sum_{n=0}^{\infty} \left(\dfrac{2\sin^2 x}{100}\right)^n \ \text{dx}$$
$$\displaystyle = \dfrac{2}{\pi}\int_{0}^{\pi/2} \dfrac{100}{100 - 2\sin^2x} \ \text{dx} = \dfrac{2}{\pi} \dfrac{5\pi}{7\sqrt{2}} = \dfrac{10}{7\sqrt{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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integral of $\arcsin(\sin(x))$ I'm having trouble with this integral
$$\int\arcsin(\sin x)\,\mathrm dx$$
The problem is with the intervals of definition for each function :/ if someone could dumb it down for me.
*
*$\arcsin\colon [-1, 1] \to [-\pi/2, \pi/2]$.
*$\sin:\mathbb{R}\to [-1, 1]$.
right?
But what about $\arcsin(\sin x)$ ?
| Let's have $f(x)=\arcsin(\sin(x))$ and calculate $\displaystyle F(x)=\int_0^x f(t)dt$
Since $f$ is $2\pi-$periodic and $F(2\pi)=0$ by symetry, we have that $F$ is $2\pi-$periodic too.
The idea to introduce $s(x)=\operatorname{sgn}(\cos(x))$ will be useful to find a closed form.
$F(x)=x\arcsin(\sin(x))-\frac{s(x)}2x^2+C(x)$
In fact his formula is correct on $1$ period interval under the condition that we assign a constant value for $C(x)$ in each of the $3$ intervals described below.
But I'll propose a slightly different form : $F(x)=s(x)(\frac 12f(x)^2)+B(x)$ where $B(x)$ is piecewise constant too, but zero on two of the intervals.
Under this form the periodicity of $F$ is more obvious knowing that $f$ is periodic.
$\displaystyle x\in[0,\frac{\pi}{2}];\quad F(x)=\int_0^x tdt=\frac {x^2}2$
$\displaystyle x\in[\frac{\pi}{2},\frac{3\pi}{2}];\quad F(x)=F(\frac{\pi}2)+\int_{\frac\pi2}^x (\pi-t)dt=\frac {\pi^2}8+\bigg[\pi t-\frac{t^2}2\bigg]_{\frac{\pi}2}^x=\pi x-\frac{x^2}2-\frac{\pi^2}4$
$\displaystyle x\in[\frac{3\pi}{2},2\pi];\quad F(x)=F(\frac{3\pi}2)+\int_{\frac{3\pi}2}^x (t-2\pi)dt=\frac {\pi^2}8+\bigg[\frac{t^2}2-2\pi t\bigg]_{\frac{3\pi}2}^x=\frac{x^2}2-2\pi x+2\pi^2$
$\begin{cases}
x\in[0,\frac{\pi}{2}] & f(x)=x & s(x)=+1 & \frac 12s(x)f(x)^2=\frac {x^2}2 & B(x)=0\\
x\in[\frac{\pi}{2},\frac{3\pi}{2}] & f(x)=\pi-x & s(x)=-1 & \frac 12s(x)f(x)^2=-\frac{\pi^2}2+\pi x-\frac {x^2}2 & B(x)=\frac{\pi^2}4\\
x\in[\frac{3\pi}{2},2\pi] & f(t)=x-2\pi & s(x)=+1 & \frac 12s(x)f(x)^2=\frac{x^2}2-2\pi x+2\pi^2 & B(x)=0\\
\end{cases}$
Now if we consider $\displaystyle \frac{1-s(x)}2=\chi_{[\frac{\pi}{2},\frac{3\pi}{2}]+2k\pi}$ then $B(x)=\big(\frac{1-s(x)}2\big)\frac{\pi^2}4$
Finally a simplified closed form for $F(x)$ can be given by :
$\displaystyle \int_0^x \arcsin(\sin(t))dt=\frac{\pi^2}8+\operatorname{sgn}(\cos(x))\bigg(\frac{\arcsin(\sin(x))^2}2-\frac{\pi^2}8\bigg)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?
| If you want to know if $a^n + b^n$ is divisible by $a+b$ (or by $a-b$, perhaps), you can always try long division, whether explicitly or in your head. I can't figure out a way to do the LaTeX or ASCII art for it here to do it explicitly, so I'll show you how one would do it "in one's head".
For example, for $a^3+b^3$, to divide $a^3+b^3$ by $a+b$, start by writing $a^3+b^3= (a+b)(\cdots)$. Then: we need to multiply $a$ by $a^2$ to get the $a^3$, so we will get $a^3+b^3=(a+b)(a^2+\cdots)$. The $a^2$ makes an unwanted $a^2b$ when multiplied by $b$, so we need ot get rid of it; how? We multiply $a$ by $-ab$. So now we have
$$a^3+b^3 = (a+b)(a^2-ab+\cdots).$$
But now you have an unwanted $-ab^2$ you get when you multiply $b$ by $-ab$; to get rid of that $-ab^2$, you have to "create" an $ab^2$. How? We multiply $a$ by $b^2$. So now we have:
$$a^3 + b^3 = (a+b)(a^2-ab+b^2+\cdots)$$
and then we notice that it comes out exactly, since we do want the $b^3$ that wee get when we multiply $b^2$ by $b$; so
$$a^3 + b^3 = (a+b)(a^2-ab+b^2).$$
If the expression you want is not divisible by what you are trying, you'll run into problems which require a "remainder". For instance, if you tried to do the same thing with $a^4+b^4$, you would start with $a^4+b^4 = (a+b)(a^3+\cdots)$; then to get rid of the extra $a^3b$, we subtract $a^2b$: $a^4+b^4 = (a+b)(a^3 - a^2b+\cdots)$. Now, to get rid of the unwanted $-a^2b^2$, we add $ab^2$, to get $a^4+b^4 = (a+b)(a^3-a^2b+ab^2+\cdots)$. Now, to get rid of the unwanted $ab^3$, we subtract $b^3$, to get
$$a^4+b^4 = (a+b)(a^3-a^2b+ab^2 - b^3+\cdots).$$
At this point we notice that we get an unwanted $-b^4$ (unwanted because we want $+b^4$, not $-b^4$). There is no way to get rid of it with the $a$, so this will be a "remainder". We need to cancel it out (by adding $b^4$) and then add what is still missing (another $b^4$), so
$$a^4 + b^4 = (a+b)(a^3-a^2b+ab^2 -b^3) + 2b^4.$$
(Which, by the way, tells you that $a^4-b^4 = (a+b)(a^3-a^2b+ab^2-b^3)$, by moving the $2b^4$ to the left hand side).
| {
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Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*}
\lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}
&=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
&=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\
&=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\
&=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
&=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
&=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\
&=\frac{1}{16}
\end{align*}$
wolframalpha says it's negative. What am I doing wrong?
| An argument why it should be negative.
When $x>0$, we have $\sqrt{4+x} > 2$ and hence $\frac{1}{\sqrt{4+x}} < \frac{1}{2}$ and hence
$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$
Similarly, When $x<0$, we have $\sqrt{4+x} < 2$ and hence $\frac{1}{\sqrt{4+x}} > \frac{1}{2}$ and hence
$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$
So either way, as $x \rightarrow 0$, the value is negative.
| {
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exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now
| Observe that $(5 - 2\sqrt{6}) = \frac{1}{5+2\sqrt{6}}$.
So, if we set $a = 5+ 2 \sqrt{6}$, we have
\begin{equation}
a^x +a^{-x} +2 = 100.
\end{equation}
The above expression is symmetric in $x$, so if $x = k$ is a solution, $x = -k$ is also a solution. Consider $f(x) = a^x +a^{-x} +2$ for $x>0$. You can show that this is an increasing function for $x > 0$. This implies that there can be at most one value of $x>0$ for which the equality is satisfied. You can check by substitution that $x = 2$ is a solution. So, the only solutions for the equation are $x = +2$ and $x=-2$.
| {
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All positive integers $n$ such that $n \mid 2^{n-1}+1$ How can one find all positive integers $n$ for which $n \mid 2^{n-1}+1$?
| The idea behind this solution is that if $n>1$ then $n$ must be odd, so $2^{n-1}+1$ is a sum of coprime squares, and we get $n\equiv 1 \bmod{4}$, so that $2^{n-1}+1$ is a sum of coprime fourth powers and $n\equiv 1 \bmod{8}$, etc...
We claim that if $p$ is an odd prime such that $p\mid x^{2^k}+1$ for some natural $k$ and integer $x$, then $p\equiv 1 \bmod{2^{k+1}}$. Because $p\mid x^{2^k}+1$ implies that $x^{2^{k+1}}\equiv 1 \bmod{p}$ and $x^{2^k}\equiv -1 \not \equiv 1 \bmod{p}$, then the order of $x$ modulo $p$ divides $2^{k+1}$ and doesn't divide $2^k$. We conclude that it's $2^{k+1}$, and using Fermat's little theorem we get that $2^{k+1}\mid p-1$, as we wanted.
Back to the original problem, we claim that $n=1$ is the only solution. It's clearly a solution so suppose that $n>1$. Then $n$ is necessarily odd, so let $k$ be the greatest integer such that $n\equiv 1 \bmod{2^k}$.
Using the lemma proved before with $x=2^{\frac{n-1}{2^k}}$, we get that all prime factors of $n$ must be $1\bmod{2^{k+1}}$ (because they're odd). This immediately implies that $n\equiv 1 \bmod{2^{k+1}}$, contradicting the maximality of $k$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/31892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Characterization of two-step 2x2 stochastic matrices Show that:
A 2 x 2 stochastic matrix is two-step transition matrix of a Markov chain if and only if the sum of its principal diagonal terms is greater than or equal to $1$.
| Encode the Markov matrix $\begin{pmatrix}1-a & a\\ b & 1-b\end{pmatrix}$ by the couple $(a,b)$ with $a$ and $b$ between $0$ and $1$. The square of this matrix is encoded by the couple
$$
(a(2-a-b),b(2-a-b)),
$$
hence the question is to determine for which $a$ and $b$ between $0$ and $1$, there exists $x$ and $y$ between $0$ and $1$ such that
$$
a=x(2-x-y),\qquad b=y(2-x-y).
$$
If $x$ and $y$ exist, their sum $s$ solves $s(2-s)=a+b$ hence $(s-1)^2=1-a-b$, in particular
$$
a+b\le1.
$$
Assume this condition holds. Then $s=1\pm\sqrt{1-a-b}$, hence
$$
x=a/(2-s)=a/(1\mp\sqrt{1-a-b}),
$$
and
$$y=b/(2-s)=b/(1\mp\sqrt{1-a-b}).
$$
Since $1+\sqrt{1-a-b}$ is between $1$ and $2$, at least the choice $+$ in $\mp$ yields solutions $x$ and $y$ between $0$ and $1$.
Finally the Markov matrix $\begin{pmatrix}1-a & a\\ b & 1-b\end{pmatrix}$ is the square of a Markov matrix if and only if
$$
a+b\le1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/32000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving indetermination in limit I'm trying to solve this limit, but I can't get it out of a $\frac{0}{0}$ indetermination:
$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$
Maybe there is something I'm missing. Thanks a lot in advanced.
| $$\frac{x-4}{5 - \sqrt{x^2 + 9}} = \frac{x-4}{25 - (x^2 + 9)} \times (5 + \sqrt{x^2 + 9}) = \frac{x-4}{16 - x^2} \times (5 + \sqrt{x^2 + 9}) = \frac{5 + \sqrt{x^2 + 9}}{-(x+4)}$$
Hence, $$\lim_{x \rightarrow 4} \frac{x-4}{5 - \sqrt{x^2 + 9}} = \lim_{x \rightarrow 4} \frac{5 + \sqrt{x^2 + 9}}{-(x+4)} = \frac{5+5}{-8} = - \frac{5}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Linear Algebra Question My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system
$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$
is consistent for any values of $a,b$, and $c$ that satisfy that equation.
Thanks,
| Perform Gaussian elimination on the system. For each row with 3 zeros on the left, the corresponding equation on the right is a restriction on $a$, $b$, $c$.
Here are the details:
$$
\begin{array}{rrrl}
2 & 1 & -1 & a \\
1 & -2 & -3 & b \\
-3 & -1 & -2 & c
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & -3 & b \\
2 & 1 & -1 & a \\
-3 & -1 & -2 & c
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & -3 & b \\
0 & 5 & 5 & a-2b \\
0 & -7 & -7 & c+3b
\end{array}
\quad\to\quad
$$
$$
\begin{array}{rrrl}
1 & -2 & -3 & b \\
0 & 35 & 35 & 7a-14b \\
0 & -35 & -35 & 5c+15b
\end{array}
\quad\to\quad
\begin{array}{rrrl}
1 & -2 & -3 & b \\
0 & 35 & 35 & 7a-14b \\
0 & 0& 0 &7a+b+5c
\end{array}
$$
Hence we must have $7a+b+5c=0$ if the original system has a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/39844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Multiplication of coefficients in Dirichlet series This appears to be a relationship:
$\sum\limits_{p\;\text{prime}} \frac{1}{p^s} = \log\zeta (s) - \sum\limits_{n=1}^{\infty}\frac{\sqrt{a_{n}b_{n}}}{n^{s}}$
where $a_{n}$ is a sequence of fractions beginning $\frac{0}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{2}, \frac{1}{1}, \frac{0}{1}, \frac{1}{1}, \frac{1}{3}, \frac{1}{2}, \frac{0}{1}, \frac{1}{1}, \frac{0}{1}...$
and $b_{n}$ is another sequence of fractions $\frac{0}{1}, \frac{0}{1}, \frac{0}{1}, \frac{1}{2}, \frac{0}{1}, \frac{1}{1}, \frac{0}{1}, \frac{1}{3}, \frac{1}{2}, \frac{1}{1}, \frac{0}{1}, \frac{0}{1}...$
for which the Dirichlet series are defined by:
$\sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{s}} = +\frac{1}{1}(\zeta (s)-1)^1 -\frac{1}{2}(\zeta (s)-1)^2 +\frac{1}{3}(\zeta (s)-1)^3 -
\frac{1}{4}(\zeta (s)-1)^4 +\frac{1}{5}(\zeta (s)-1)^5 -...$
$\sum\limits_{n=1}^{\infty}\frac{b_{n}}{n^{s}} = -\frac{0}{1}(\zeta (s)-1)^1 +\frac{1}{2}(\zeta (s)-1)^2 -\frac{2}{3}(\zeta (s)-1)^3 +
\frac{3}{4}(\zeta (s)-1)^4 -\frac{4}{5}(\zeta (s)-1)^5 +...$
The square root of the elementwise multiplication of $a_{n}$ times $b_{n}$,$\;$
$\sqrt{a_{n}b_{n}}$ is then: $\frac{0}{1},\frac{0}{1},\frac{0}{1},\frac{1}{2},\frac{0}{1},\frac{0}{1},\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{0}{1},\frac{0}{1},\frac{0}{1}...$
Is there any literature describing multiplication of coefficients of Dirichlet series?
| After a quick glance at your sequences, it appears that $$a_n =\frac{\Lambda(n)}{\log n}$$ and $$\sqrt{a_n b_n}=\frac{\Lambda(n)}{\log n} -\chi_p (n)$$ where $\chi_p(n)$ is the indicator function for the primes.
The sequence $b_n$ doesn't really matter too much, except the fact that it is $$\frac{\Lambda(n)}{\log n} -\chi_p(n)$$ on prime powers. Since $a_n$ is zero elsewhere, the sequence $b_n$ can be whatever we want elsewhere. (Similarly the sequence $a_n$ can be whatever we want on the primes, and you still have the relation in the first line.)
So, why does this hold? Since $\sqrt{a_nb_n}=\frac{\Lambda(n)}{\log n} -\chi_p (n)$ it follows that $$\sum_{n=1}^\infty \frac{\sqrt{a_n b_n}}{n^s}=\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} n^{-s} +\sum_p p^{-s}.$$ Then the prime zeta functions cancel, and we recall that $$\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} n^{-s}=\log \zeta(s).$$ Hope that helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/42840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute the expression $(a^2 + 4b - 1)(b^2 + 4a - 1)$ without calculating the roots of $x^2 - x - 5 = 0$ Let $a$ and $b$ be the roots of this equation:
$$x^2 - x - 5 = 0$$
Find the value of
$$(a^2 + 4b - 1)(b^2 + 4a - 1)$$
Without calculating the values of a and b.
I saw this on a problems site and tried it but I got 100 and I don't think that's correct.
| This is an exercise in elementary symmetric polynomials. Write $s_1=a+b$ and $s_2=ab$. From the equation
$$
x^2-x-5=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-s_1x+s_2
$$
we can read that $s_1=1$ and $s_2=-5$.
The number that you wanted to know is
$$
(a^2+4b-1)(b^2+4a-1)=a^2b^2+4(a^3+b^3)+16ab-4(a+b)-(a^2+b^2)+1.
$$
We need to express the quantities in parenthesis in terms of $s_1$ and $s_2$. This is not too difficult, because from
$$
s_1^2=a^2+2ab+b^2=(a^2+b^2)+2s_2
$$
we get $a^2+b^2=s_1^2-2s_2=11$. Similarly from
$$
s_1^3=a^3+3a^2b+3ab^2+b^3=(a^3+b^3)+3ab(a+b)
$$
we get that $a^3+b^3=s_1^3-3s_1s_2=16.$
Putting all this together gives
$$
(a^2+4b-1)(b^2+4a-1)=25+64-80-4-11+1=-5.
$$
What makes this tick is that $(a^2+4b-1)(b^2+4a-1)$ is symmetric in the unknowns $a$ and $b$. IOW if you swap the values of $a$ and $b$ nothing will change. Such polynomial functions can always be written in terms of the elementary symmetric polynomials $s_1$ and $s_2$. This result can be generalized to several unknowns. The buzzword "elementary/basic symmetric polynomials" should give you enough material. The power sums such as $a^3+b^3$ are a well-known special case. The buzzword "Newton's identities" helps you there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/42996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Finding $A^n$ for a matrix I have a matrix $$
A =
\left[ {\begin{array}{cc}
1 & c \\
0 & d \\
\end{array} } \right]
$$
with $c$ and $d$ constant. I need to find $A^n$ ($n$ positive) and then need to prove that formula using induction.
I would like to check that the formula I derived is correct:
$$
A^n =
\left[ {\begin{array}{cc}
1 & c^{n-2}(dc + c) \\
0 & d^n \\
\end{array} } \right]
$$
If this is correct, how can I prove this? I suppose I can write $A^{n+1} = A^n A$, which would be
$$
\left[ {\begin{array}{cc}
1 & c^{n-2}(dc + c) \\
0 & d^n \\
\end{array} } \right]
\left[ {\begin{array}{cc}
1 & c \\
0 & d \\
\end{array} } \right]
$$
But then what would I do?
Thanks.
| Mathematical induction is the way to go, but first you want to have a "target." I'm not sure what you did, but I think you confused $c$ and $d$ somewhere along the line in your calculations. Let's see a first few values:
$$\begin{align*}
A&= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\\
A^2 &= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\left(\begin{array}{cc}1&c\\0&d\end{array}\right) = \left(\begin{array}{cc}1 & c+cd\\0&d^2\end{array}\right).\\
A^3 &= \left(\begin{array}{cc}
1&c+cd\\
0 &d^2\end{array}\right) \left(\begin{array}{cc}
1 & c\\
0 & d\end{array}\right) = \left(\begin{array}{cc}
1 & c+cd+cd^2\\
0 & d^3
\end{array}\right).\\
A^4 &= \left(\begin{array}{cc}
1&c+cd+cd^2\\
0 & d^3
\end{array}\right)\left(\begin{array}{cc}
1 & c\\0 & d\end{array}\right) = \left(\begin{array}{cc}
1 & c + cd + cd^2 + cd^3\\
0 & d^4
\end{array}\right).
\end{align*}$$
Okay, that suggests the pattern:
Conjecture. For every positive integer $n$,
$$A^n = \left(\begin{array}{cc}
1 & c(1+d+d^2+\cdots + d^{n-1})\\
0 & d^n
\end{array}\right).$$
To prove the conjecture, we use mathematical induction. Prove the formula is true for $n=1$, and then, assuming the formula holds for $k$, prove it holds for $k+1$. We have:
Basis. For $n=1$, we have
$$A = \left(\begin{array}{cc}1 & c\\0 & d\end{array}\right),$$
so the formula holds.
Inductive step. Show that if the formula holds for $k$, then it also holds for $k+1$.
Induction hypothesis. Assume the result holds for $k$; that is, that
$$A^k = \left(\begin{array}{cc}
1 & c(1+d+\cdots+d^{k-1})\\0 & d^k
\end{array}\right).$$
Now we have:
$$\begin{align*}
A^{k+1} & = A^kA\\
&= \left(\begin{array}{cc}
1 & c(1+d+\cdots + d^{k-1})\\0 & d^k\end{array}\right)
\left(\begin{array}{cc} 1 & c\\0 & d\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & c + cd(1+d+\cdots + d^{k-1})\\
0 & d^{k+1}\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & c(1 + d(1+d+\cdots + d^{k-1}))\\
0 & d^{k+1}\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & c(1+d+d^2+\cdots + d^k)\\
0 & d^{k+1}\end{array}\right).
\end{align*}$$
That is exactly the formula we have evaluated at $k+1$. Therefore, if the formula holds for $k$, then it holds for $k+1$ as well.
So: the formula holds for $n=1$; and if it holds for $n=k$, then it also holds for $n=k+1$. By Mathematical Induction, we conclude that the formula holds for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/44368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Factoring Quantities Question I was doing an exercise and ran into a problem with their use of factoring. Here is the problem specific to where the issue occurs:
$$
\frac{(x^2 + 1)^{1/2} - x^2 (x^2 + 1)^{-1/2}}{x^2 + 1} = \frac{(x^2 + 1)^{-1/2}\left( x^2 + 1 - x^2\right)}{x^2 + 1}
$$
If $(x^2+1)^{-1/2}$ was factored out on the top, shouldn't the amount after factoring be just be $1 - x^2$? I am not seeing where the extra $x^2$ is coming from. Can anyone spot the issue I am having?
| When you factor out a $(x^2+1)^{-\frac{1}{2}}$, the remaining quantity is multiplied by $(x^2+1)^{\frac{1}{2}}$. That is, if we have some expression
$$(x^2+1)^a \cdot f(x)$$
where $f(x)$ is some polynomial and $a$ is any number, when we factor out a $(x^2+1)^{-\frac{1}{2}}$ we get
$$(x^2+1)^{-\frac{1}{2}}\left((x^2+1)^{a+\frac{1}{2}}\cdot f(x)\right).$$
More concretely in the case of your example, because $\frac{1}{2}=-\frac{1}{2}+1$, we have that
$$(x^2+1)^{\frac{1}{2}}=(x^2+1)^{-\frac{1}{2}}(x^2+1)^1,$$
and therefore
$$\begin{align} (x^2+1)^{\frac{1}{2}}-x^2(x^2+1)^{-\frac{1}{2}} =&(x^2+1)^{-\frac{1}{2}}(x^2+1)^1-x^2(x^2+1)^{-\frac{1}{2}}\\ =& (x^2+1)^{-\frac{1}{2}}\left((x^2+1)-x^2\right)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/44613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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