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Problems based on triangles and trigonometry. In an acute triangle $x,y,z$ are the given angles where $\cos x=\tan y$, $\cos y = \tan z$ and $\cos z = \tan x$. Find the sum of sines in the triangle.
Could be done by substituting values of $\sin$ function but in vain. Can anyone please help me?
| The question states that it's given that
in an acute triangle
$x,y,z$ are the given angles where
\begin{align}
\cos x&=\tan y
\tag{1}\label{1}
,\\
\cos y&=\tan z
\tag{2}\label{2}
,\\
\cos z&=\tan x
\tag{3}\label{3}
.
\end{align}
There are many ways to prove that
there is no valid triangle with such properties.
For one is the incomplete answer
which implies that
it follows from \eqref{1}-\eqref{3}
that
\begin{align}
\tan x&=\tan y=\tan z=
\cos x=\cos y=\cos z=
=\sqrt{\tfrac12\,(\sqrt 5-1)}
,
\end{align}
which is absurd.
Another way is:
rewriting \eqref{1}-\eqref{3} as
\begin{align}
\cos x\cos y&=\sin y
\tag{4}\label{4}
,\\
\cos y\cos z&=\sin z
\tag{5}\label{5}
,\\
\cos z\cos x&=\sin x
\tag{6}\label{6}
,
\end{align}
so
\begin{align}
\sin x+\sin y+\sin z
&=
\cos x\cos y+\cos y\cos z+\cos z\cos x
\tag{7}\label{7}
.
\end{align}
Using known identities
\begin{align}
\sin x+\sin y+\sin z&=u
\tag{8}\label{8}
\end{align}
and
\begin{align}
\cos x\cos y+\cos y\cos z+\cos z\cos x
&=\frac{u^2+v^2}4-1
\tag{9}\label{9}
,
\end{align}
where $u=\rho/R$, $v=r/R$
and $\rho,r,R$ are the semiperimeter,
inradius and circumradius of given triangle (if such exists).
From equations \eqref{7}-\eqref{9},
\begin{align}
\frac{u^2+v^2}4-1=v
,\\
u&=2+\sqrt{8-v^2}
,
\end{align}
and this expression for
\begin{align}
u>\tfrac{3\sqrt3}2
&=\max_{v\in[0,1/2]}u(v)
,
\end{align}
that is, there is no a pair $(u,v)$
that simultaneously agree with \eqref{7}-\eqref{9}
and represent a valid triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the radius of the circle given in the picture below. This is the image of the question. I am not able to get how to find the radius. Please help with that.
This is my try. I can't proceed now after it.
Thanks
|
Given: $CD=2,\,BD=3,\,AB=\sqrt{11}$.
Let's define $\angle B=\angle DBE,\, CE=d$.
$2R=\frac{\sqrt{11}}{\cos A}$, $BE=\frac{\sqrt{11}}{\sin A}, \angle DBE=\angle DAE$, thus $CBE\sim CAD$
$$\frac{CB}{CA}=\frac{BE}{AD}=\frac{CE}{CD}$$
$$\frac{5}{d+2R}=\frac{\frac{\sqrt{11}}{\sin A}}{AD}=\frac{d}{2}$$
By applying the cosine rule to both $\Delta CBA$, $\Delta CDA$ we get
$$\begin{cases}
11=25+(2R+d)^2-2\cdot 5\cdot (2R+d)\cos C,\\
AD^2=4+(2R+d)^2-2\cdot 2\cdot (2R+d)\cos C,
\end{cases}$$
By eliminating $\cos C$ we get $$
AD^2=4+(2R+d)^2-\frac{2}{5}\left(25+(2R+d)^2-11\right)=$$
$$\frac{3}{5}(2R+d)^2-\frac{12}{5}=15\left(\frac{2R+d}{5}\right)^2-\frac{12}{5}=\frac{60}{d^2}-\frac{12}{5}$$
Thus we get $$\begin{cases}
10=d(d+2R)\\
2\sqrt{11}=d\sqrt{\frac{60}{d^2}-\frac{60}{25}}\sin A
\end{cases}$$
$$\begin{cases}
10=d(d+\frac{\sqrt{11}}{\cos A})\\
2\sqrt{11}=\frac{\sqrt{60}}{5}\sqrt{25-d^2}\sin A
\end{cases}$$ hence we get $d$ and then $R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Integral solution to a³+3ab²=4c³ Please help if integral solutions to this equation exists or not
A³+3AB²=4C³
Such that A,B,C are disninct
I thought we may be possible to prove solutions exists if and only if A=B=C condition is satisfied.
| Note that
$$(A-B)^3=A^3-3A^2B+3AB^2-B^3 \\
(A+B)^3=A^3+3A^2B+3AB^2+B^3$$
This gives
$$A^3+3AB^2= \frac{1}{2} \left((A+B)^3+(A-B)^3 \right)$$
Therefore, if you have a solution
$$(A-B)^3+(A+B)^3=(2C)^3$$
Use FLT to deduce that all solutions satisfy $A=B$ or $A=-B$ or $A=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.
| If I recall correctly, you make the assumption that your expression takes the form $\sqrt{a}\pm\sqrt{b}$:
$\sqrt{8-4\sqrt{3}} = \sqrt{a}\pm\sqrt{b}$
$8-4\sqrt{3} = 8-\sqrt{48} = a\pm2\sqrt{ab}+b$
We can see that the irrational part must be assigned the negative sign.
Equating rational and irrational parts:
$a+b = 8$
$-2\sqrt{ab} = -\sqrt{48} \implies ab = 12$
Then $a = \frac{12}{b} \implies \frac{12}{b}+b = 8 \implies 12 + b^2 = 8b \implies b^2 - 8b + 12 = (b-6)(b-2) = 0$
Since we know the answer is positive, take b = 2 and a = 6.
$\sqrt{8-4\sqrt{3}} = \sqrt{6}-\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}}$? I need to resolve the convergence of $$\sum_{n = 1}^{\infty} \frac{n^3}{2^n + 5^n}$$ where $n \in \mathbb{N}$.
Mainly, I need advice concerning Cauchy's convergence criterion because I just can't evaluate
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}}.$$
My initial idea was to convert it to $\exp$ but that didn't help a bit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}} = \mathrm{e}^{\lim_{n \to \infty}\frac{1}{n}\ln\frac{n^3}{2^n + 5^n}}$$ but now I have to evaluate the bit
$$\lim_{n \to \infty} \ln \frac{n^3}{2^n + 5^n}.$$
It clearly goes to $0$ since $2^n + 5^n$ grows much faster than $n^3$, but what now? $\ln 0 = -\infty$ and plugging it back will result in $0 \cdot \left(-\infty\right)$ which is undefied.
I didn't give up there. Let's take a different approach.
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}} = \lim_{n \to \infty} \left(\frac{n^3}{2^n + 5^n}\right)^{\frac{1}{n}}$$
this will result in $\left(\frac{\infty}{\infty}\right)^0$, so maybe we can expand it:
$$\lim_{n \to \infty} \left(\frac{n^3}{2^n + 5^n}\right)^{\frac{1}{n}} = \lim_{n \to \infty}\frac{n^{\frac{3}{n}}}{\left(2^n + 5^n\right)^\frac{1}{n}} = \lim_{n \to \infty} \frac{e^{\frac{3}{n}\ln{n}}}{e^{\frac{1}{n} \ln\left(2^n + 5^n\right)}}$$ and it's still in an indetermined form. If I were to continue in this fashion I would, of course, get the same expression as before.
| Just like started, first we switch to the exponent of $e$:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{1/n} = \exp{(\lim_{n\rightarrow\infty}\frac{\ln(\frac{n^3}{2^n+5^n})}{n}})$$
Using l'Hospitals rule we need to evaluate the derivative of the numerator:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{-1}\frac{3n^2(2^n+5^n)-n^3(2^n\ln2+5^n\ln5)}{(2^n+5^n)^2}=$$
$$\lim_{n\rightarrow\infty}\frac{3}{n}-\frac{2^n\ln2+5^n\ln5}{2^n+5^n}=\lim_{n\rightarrow\infty}\frac{3}{n}-\lim_{n\rightarrow\infty}\frac{(\frac{2}{5})^n\ln2+\ln5}{(\frac{2}{5})^n+1}=$$
$$0-\frac{0*\ln2+\ln5}{0+1}=-\ln5$$
Hence the result is:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{1/n} = \exp{(\lim_{n\rightarrow\infty}\frac{\ln(\frac{n^3}{2^n+5^n})}{n}})=\exp(-\ln5)=\frac{1}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve PDE $xzu_x+yzu_y-(x^2+y^2)u_z=0$ using characteristics method I was asked to find the characteristics of the following EDP and then to solve it.
$$xzu_x+yzu_y-(x^2+y^2)u_z=0$$
I've reached the following characteristic system $$x'=-\frac{xz}{x^2+y^2}\qquad y'= -\frac{yz}{x^2+y^2}
$$ but i dont know how to continue from this. How does one solve an PDE using the characteristic method?
| Multiplying the first equation by $x$ and the second one by $y$ gives
$$
\tfrac12 (x^2)' = -\frac{x^2 z}{x^2 + y^2}, \qquad \tfrac12 (y^2)' = -\frac{y^2 z}{x^2 + y^2} .
$$
The sum leads to $(r^2)' = -2z$ where $r^2 = x^2 + y^2$. Multiplying the first equation by $y$ and the second one by $x$ gives
$$
yx' = -\frac{xy z}{x^2 + y^2}, \qquad xy' = -\frac{xy z}{x^2 + y^2} .
$$
The difference leads to $t' = 0$ where $t = y/x$. The additional equations $u' = 0$ and $z' = 1$ yield $u = c_1$ and $z = s + c_2$, respectively. Thus, integrating the differential equations for $r^2$ and $t$, we find $r^2 = -s(s + 2 c_2) + c_3$ and $t = c_4$. The fact that
$z^2 = -r^2+ c_3 + (c_2)^2$
leads to a general solution of the form $u = F(x^2+y^2+z^2, y/x)$ for some arbitrary $F$.
The above steps suggest that the cylindrical coordinates $(x,y,z) \mapsto (r, \theta, z)$ may be relevant here. Indeed, the PDE rewrites as
$zu_r - r u_z = 0$, which general solution is $u = G(r^2 + z^2, \theta)$ for some arbitrary $G$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need to find maximum in order to prove that the function converges uniformly. So, here is the function: $f_n = \sqrt{n}\left(\sqrt{x+\frac{1}{n}}-\sqrt{x}\right)$
I found that the limit for it is $0$. Now I want to show whether it converges uniformly. So first I tried to find a maximum.
$$f'_n(x) = \frac{1}{2}\sqrt{n}\left(\frac{1}{\sqrt{x+\frac{1}{n}}}-\frac{1}{\sqrt{x}}\right) = \\ \frac{1}{2}\sqrt{n}\left(\frac{\sqrt{x}-\sqrt{x+\frac{1}{n}}}{\sqrt{x}\sqrt{x+\frac{1}{n}}}\right)$$
Then I solve for the x at denominator.
$$\sqrt{x}-\sqrt{x+\frac{1}{n}}=0 \Rightarrow x -2\sqrt{x}\sqrt{x+\frac{1}{n}}+x+\frac{1}{n}= 0 \\ 2\sqrt{x}\sqrt{{x+\frac{1}{n}}}=\frac{1}{n}+2x \Rightarrow 2x^2+2x\frac{1}{n}=\frac{1}{n^2}+4x\frac{1}{n}+4x^2 \\ 2x^2+2x\frac{1}{n}+\frac{1}{n^2}=0 \ \ x_1 = \frac{-2\frac{1}{n}+ \sqrt{\frac{4}{n^2}-\frac{8}{n^2}}}{6} = -\frac{1}{3n}+\frac{1}{3n}=0$$
$$x_2 = -\frac{2}{3n}$$
I get very clumsy $x_1$ so I gather I have made some mistake all along. Can someone help?
| Your sequence of functions converges pointwise to $0$ on $\mathbb{R}_+^*$. Thus, it converges uniformly to zero if $||f_n||_{\infty}= \sup_{x}|f_n(x)| \to 0$. But $f_n(x) \to 1$ as $x \to 0$, so $||f_n||_{\infty}\geqslant 1$ for all $n$ and $(f_n)$ does not converge uniformly
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence
$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,
My strategy is to use Stolz's Cesaro theorem for this sequence.
Now, the numerator is given by :
$x_r = n^1+ n^2 +n^3 + \ldots +n^r$, so $x_{n+1} - x_{n} = n^{n+1}$
Similarly for denominator
$y_r = 1^n + 2^n + 3^n +\ldots +r^n$, so $y_{n+1}- y_{n} = (n+1)^n$
Using Stolz Cesaro, this limit is equivalent to
$\displaystyle \lim \dfrac{n^{n+1}}{(n + 1)^n}$, which diverges to $ +\infty$,
However ans given to me is $\dfrac{e-1}{e}$, Can anyone tell where is the error in my solution ?
Thanks.
| Divide nunberator and denominator by $n^n$. So your question consists of two limits, numerator and denominator, we'll deal with them separately.
For the numerator the limit would become $lim_{n \to \infty} 1+\frac{1}{n}+\ldots+\frac{1}{n^n} = 1*\frac{(1/n)^{n+1}-1}{(1/n)-1} = \lim_{h \to 0} \frac{h^{1+1/h} -1}{h-1} = -[e^{(1+1/h)\ln(h)} -1] = -[e^{-\infty}-1]=1$
For the denominator I'll prove a even more general limit, for any constant $k \neq 0$
$$\lim_{n to \infty}\frac{1^{kn}+2^{kn}+\ldots+n^{kn}}{n^{kn}}=\sum_{r=1}^{n} \frac{r^{kn}}{n^{kn}} = \sum_{r=0}^{n-1} \frac{(n-r)^{kn}}{n^{kn}}= \sum_{r=0}^{n-1} (1-r/n)^{kn} = \sum_{r=0}^{n-1} e^{-rk} = \frac{1}{1-e^{-k}} = \frac{e^k}{e^k-1}$$.
Here k=1, so final answer is $\frac{1}{(e/e-1)}=\frac{e-1}{e}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factor a polynomial with complex roots of the form $a+bi$, where $a\neq 0$? (Not just find the root) For example, the quadratic formula reveals that the roots of $x^2 - 4x + 5$ are $x = 2\pm 2i $
But how do we use these roots to actually factor $x^2 - 4x + 5$?
My best guess was that, since $x = 2\pm 2i $, we would have $x^2 - 4x + 5 = (x-2-2i)(x-2+2i)$
But that is incorrect because $(x-2-2i)(x-2+2i) = x^2 - 8i \neq x^2 - 4x + 5 $
So how do we factor $x^2 - 4x + 5$ ?
| You got the wrong roots in the first place. The roots are $2 \pm i$, please check your work (application of the quadratic formula).
Once you correct that, your factorisation will be correct. But you made another mistake in expanding the product of the factors.
$(x-2-i)(x-2+i) = (x-2)^2 - i(x-2)+i(x-2) +(i)(-i) = (x-2)^2 + 1 = x^2 - 4x +4 + 1 = x^2 -4x + 5$ as expected.
There is no factorisation possible purely in the reals.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following equation for $x$:
$$1-x=(1-u)^2$$ $$x=u(2-u)$$
and replacing $u$ by and $x$, and $=$ by $\to$ , which leads to
$$\frac{1}{(1-x)^2}=1+x(2-x)+x^2(2-x)^2+x^3(2-x)^3+... \tag{2}$$
Succumbing to the same 'substitution'approach with $\frac{x}{1+x^2}$, I put
$$\frac{1}{1-x}=\frac{u}{1+u^2}$$ $$x=1-\frac{1}{u}-u$$
so with $x \to 1-\frac{1}{x}-x$
I should have:
$$\frac{x}{1+x^2}=1+\left(1-\frac{1}{x}-x\right)+\left(1-\frac{1}{x}-x\right)^2+\left(1-\frac{1}{x}-x\right)^3+...\tag{3}$$
but this does not seem correct (at least according to my Desmos graph for $|x|<1$).
Can someone please explain what my conceptual errors are?
| Because
$$u^2+1=ux(1-x)$$
$$u^2-ux(1-x)+1=0$$
has no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703913",
"timestamp": "2023-03-29T00:00:00",
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Limit of $\frac{\cos{x}-1-\frac{x^2}{2}}{x^4+y^4}$ as it goes to the origin How can I solve a limit like this one:
$$\lim_{(x,y) \rightarrow (0,0)}\frac{\cos{x}-1-\frac{x^2}{2}}{x^4+y^4}$$
If I do $x=0$ so it aproaches through $y$ or $y=0$ so it approaches through $x$ it doesn't really help because of the division over $0$ in the denominator. As well as $x=y$, it doesn't make things better.
Thanks for the help.
| If $x = 0$ there is no division by zero ! Since when $x = 0$
$$f(x ,y) = \frac{\cos(x) - 1 -x^2/2}{x^4 + y^4} = \frac{1-1 - 0}{y^4} = 0$$
Therefore
$$ \lim_{y \to 0} f(0,y) = \lim_{y \to 0} 0 = 0.$$
If $y = 0$ then
$$ f(x,0) = \frac{\cos(x)-1-x^2/2}{x^4}.$$
Since the numerator and the denominator both converge to $0$ we can use L'Hospital's rule :
$$ \frac{- \sin x - x}{4x^3}$$
Since the numerator and denominator both converge to $0$ we use L'Hospital's rule a second time
$$ \frac{- \cos x - 1}{12x^2}$$
so we have
$$ \lim_{x \to 0} f(x,0) = \lim_{x \to 0} \frac{-\cos(x)-1}{12x^2} = - \infty$$
Since the limit of $(x,y)$ to $(0,0)$ is different along two different paths the limit
$$\lim_{(x,y) \to (0,0)} \frac{\cos(x)-1-x^2/2}{x^4 + y^4}$$
does not exist !
| {
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"timestamp": "2023-03-29T00:00:00",
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For any $n ≥ 5,$ the value of $1+ \frac{1}2 + \frac{1}3+···+\frac{1}{2^n −1}$ lies between
QUESTION: For any $n ≥ 5,$ the value of $$1+ \frac{1}2 + \frac{1}3+···+\frac{1}{2^n −1}$$ lies between
$(A)$ $0$ and $\frac{n}2$
$(B)$ $\frac{n}2$ and $n$
$(C)$ $n$ and $2n$
$(D)$ none of the above.
MY APPROACH: This is what I did-
We do our calculation considering $n=5$.
Call the given series $S$.
First let us compare the series $S$ with the series $A$ where the denominator of every term of $S$ is changed with the next higher power of $2$ as: $$A=\frac{1}2+\frac{1}2+\frac{1}4+\frac{1}4+\frac{1}8+\frac{1}8+\frac{1}8+\frac{1}8+...+\frac{1}{32}$$ which on grouping gives us-$$\frac{5}2$$ therefore, $A=2.5$
Again, I consider the series $B$ is replaced with a lower power of $2$ which goes like- $$1+\frac{1}2+\frac{1}2+\frac{1}4.4+\frac{1}8.8+... +\frac{1}{16}$$ which gives $B=5$.
Now observe that $A<S<B$. So the value of $S$ must lie in $[2.5,5]$.
So the answer should be $\frac{n}2$ to $n$.
Am I correct? How do I prove it rigorously?
Also, since there is no closed formula for the summation of harmonic progressions, how do I calculate the value of the sum (if it was asked for some large value of $n$)
Can anyone please help me out?
Thank you in advance.
| Divide the sum into blocks with denominators running from $2^m$ through $2^{m+1}-1$, where $0\le m<n$:
$$\begin{align*}
\sum_{k=1}^{2^n-1}\frac1k&=\underbrace{1}+\underbrace{\frac12+\frac13}+\underbrace{\frac14+\frac15+\frac16+\frac17}+\ldots+\underbrace{\frac1{2^{n-1}}+\ldots+\frac1{2^n-1}}\\
&=\sum_{m=0}^{n-1}\sum_{k=2^m}^{2^{m+1}-1}\frac1k\;.\end{align*}$$
The inner summation has $2^m$ terms, so we have
$$\frac12=2^m\cdot\frac1{2^{m+1}}<\sum_{k=2^m}^{2^{m+1}-1}\frac1k\le 2^m\cdot\frac1{2^m}=1\;:$$
each term is larger than $\frac1{2^{m+1}}$, and each term is less than or equal to $\frac1{2^m}$. And the outer summation has $n$ terms, so
$$\frac{n}2<\sum_{k=1}^{2^n-1}\frac1k\le n\;.$$
It is possible to get good approximations to the harmonic numbers quite easily. For instance
$$H_n\sim \ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{240n^4}\;,$$
where $\gamma\approx0.5772156649$ is the Euler-Mascheroni constant.
| {
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Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117\, \left( b+c \right) ^{4} \left( b+c-2\,a \right) ^{2}}{1024}}+{\frac {3\,{a}^{2} \left( 40\,{a }^{2}+7\,{b}^{2}+14\,bc+7\,{c}^{2} \right) \left( b-c \right) ^{2}}{ 32}}$$
$$+{\frac {3\, \left( b+c \right) ^{2} \left( 3\,a-2\,b-2\,c \right) ^{2} \left( b-c \right) ^{2}}{32}}+\frac{3}{16}\, \left( a+2\,b+2\,c \right) \left( 4\,a+b+c \right) \left( b-c \right) ^{4}$$
$$+{\frac { \left( 16\,{a}^{2}+24\,ab+24\,ac+11\,{b}^{2}+22\,bc+11\,{c}^{ 2} \right) \left( 4\,a-b-c \right) ^{2} \left( b+c-2\,a \right) ^{2} }{1024}} \geqq 0$$
I think$,$ $uvw$ is the best way here but it's not concordant for student in The Secondary School.
Also$,$ BW helps here, but not is nice, I think.
So I wanna nice solution for it! Thanks for a real lot!
| Yes, SOS helps:
$$(a^2+b^2+c^2)^3-9(a^3+b^3+c^3)=(a^2+b^2+c^2)^3-9abc(a^3+b^3+c^3)=$$
$$=\frac{1}{2}\sum_{cyc}(2a^6+6a^4b^2+6a^4c^2-18a^4bc+4a^2b^2c^2)=$$
$$=\frac{1}{2}\sum_{cyc}(2a^6-a^4b^2-a^4c^2+7a^4b^2+7a^4c^2-14c^4ab-4a^4bc+4a^2b^2c^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)^2(a^2+b^2)+7c^4-2abc(a+b+c))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2(7c^4-2abc^2-2ab(a+b)c+(a+b)^2(a^2+b^2))\geq0,$$
where the last inequality is true by AM-GM:
$$c^4+\frac{1}{8}(a^2+b^2)(a+b)^2\geq c^4+a^2b^2\geq2abc^2$$ and
$$6c^4+\frac{7}{8}(a+b)^2(a^2+b^2)\geq2ab(a+b)c.$$
Can you prove the last inequality by AM-GM?
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$
where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$.
It is evident that $x=1/2$ satisfies the above equation. Please help.
| Seems the fastest way to show this is indeed considering the polynomial left after clearing the denominator and dividing by $2(x-\frac{1}{2})$
$$f(x)=-3 t^4 - 2 t^2 x^2 + 2 t^2 x - t^2 + x^4 - 2 x^3 + x^2$$
Considering $$f'(x)= -2 (2 x - 1) (t^2 - x^2 + x)$$ we see that as $(t^2+x(1-x))>0$ for $0<x<1$ thus the function increases on $\left(0,\frac{1}{2}\right)$ and decreases on $\left(\frac{1}{2},1\right)$ thus we have to check only that $f(0),\,f\left(\frac12\right),\,f(1)$ has the same sign, indeed:
$$f(0)=-3t^4-t^2\le 0$$ $$f(1)=-3t^4-t^2\le 0$$
$$f\left(\frac{1}{2}\right)=-3 t^4 - \frac{t^2}{2} + \frac{1}{16}$$
But the latter expression factors as
$$-\frac{1}{16} (4 t^2 + 1) (12 t^2 - 1)$$
so for $12t^2-1<0$ the statement is not true.
$$$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What does it mean for an ODE to be conservative? What does it mean for an ODE to be conservative?
For example, I already read somewhere that the equation
$$w\cdot y''-y+y^{2k+1}=0,$$
with $w>0$ and $k\in \mathbb{N}$ constants fixeds, is conservative. In practice, what does this mean?
| Given the differential equation
$wy'' - y + y^{2k + 1} = 0, \tag 1$
we may multiply it through by $y'$:
$wy'y'' - yy' + y^{2k + 1}y' = 0, \tag 2$
and observe that
$\left ( \left ( \dfrac{w}{2} y' \right )^2 \right )' = wy''y', \tag 3$
and
$\left ( -\dfrac{y^2}{2} + \dfrac{y^{2k + 2}}{2k + 2} \right )' = - yy' + y^{2k + 1}y'; \tag 4$
then (2) may be written
$\left ( \left ( \dfrac{w}{2} y' \right )^2 \right )' + \left ( -\dfrac{y^2}{2} + \dfrac{y^{2k + 2}}{2k + 2} \right )' = 0, \tag 5$
or
$\left ( \left ( \dfrac{w}{2} y' \right )^2 -\dfrac{y^2}{2} + \dfrac{y^{2k + 2}}{2k + 2} \right )' = 0; \tag 6$
thus,
$\left ( \dfrac{w}{2} y' \right )^2 -\dfrac{y^2}{2} + \dfrac{y^{2k + 2}}{2k + 2} = C, \; \text{a constant} \tag 7$
along the solution curves $(x, y(x))$ of (1). That is, the quantity on the left of (7) is a conserved quantity of the equation(1); hence we deem (1) a conservative ordinary differential equation, since the function
$F(y, y') = \left ( \dfrac{w}{2} y' \right )^2 -\dfrac{y^2}{2} + \dfrac{y^{2k + 2}}{2k + 2} \tag 8$
is invariant in value on the solution curves.
In general, a conservative second-order equation or system is one for which a function such as $F(y, y')$ exists. For these $F(y, y')$, it follows that
$\dfrac{\partial F(y, y')}{\partial y} y' + \dfrac{\partial F(y, y')}{\partial y'}y'' = \dfrac{dF(y, y')}{dx} = 0, \tag 9$
and thus $y(x)$ satisfies the deifferential equation (9). We may in fact invoke these principles to construct differential equations corresponding to many different $F(y, y')$; for example, if
$F(y, y') = \cos y + e^{y'}, \tag{10}$
then we have
$- y' \sin y + e^{y'}y'' = 0, \tag{11}$
or
$y'' = e^{-y'}y'\sin y. \tag{12}$
| {
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$p^2+1=q^2+r^2$. Strange phenomenon of primes Problem:
Find prime solutions to the equation
$p^2+1=q^2+r^2$
I welcome you to post your own solutions as well
I have found a strange solution which I can't understand why it works(or what's the math behind it.) Here it is through examples
Put $r=17$(prime)
Now $17^2-1=16\times 18=288=2 \times 144$ (a particular factorization)
$\frac {2+144}{2}=73$
$\frac {144-2}{2}=71$
Solution pair $(p,q,r)=(73,71,17)$
Put $r=23$,
$23^2-1=22\times 24=8\times 66$
$\frac {8+66}{2}=37$
$\frac {66-8}{2}=29$
Solution pair $(37,29,23)$
It works for each prime except for $2,3,5$ Which generate
$(2,2,1),(3,3,1),5,5,1)$
Please explain me how it's working
| This seems to hold much more broadly than just for primes.
Let$$r^2-1=ab$$where $b>a$, and let $p=\frac{b+a}{2}$ and $q=\frac{b-a}{2}$.
Then the equation $p^2+1=q^2+r^2$ becomes$$\left(\frac{b+a}{2}\right)^2+1=\left(\frac{b-a}{2}\right)^2+ab+1$$and we have$$\frac{b^2+2ab+a^2}{4}+1=\frac{b^2-2ab+a^2+4ab}{4}+1=\frac{b^2+2ab+a^2}{4}+1$$
Addendum: If $r$ is an odd prime, then $p$ and $q$, that is $\frac{b+a}{2}$ and $\frac{b-a}{2}$, can be prime only if both are odd, i.e. only if $b\equiv 2\pmod 4$ and $a\equiv 0\pmod 4$, or vice-versa.
This restricts the number of divisor pairs $a, b$ that need to be considered in determining whether $r^2-1=ab$ for some odd primes $p$ and $q$.
To illustrate, let $r=47$. Then$$r^2-1=ab=2208=2^5\cdot3\cdot23$$Thus the possible $ab$ are $2\cdot 1104$, $6\cdot 368$, and $2^4\cdot 138$.
For $b=1104$, $a=2$, $p=\frac{b+a}{2}=553=7\cdot79$, not prime.
For $b=368$, $a=6$, $p=\frac{b+a}{2}=187=11\cdot17$, not prime.
For $b=138$, $a=2^4$, $p=\frac{b+a}{2}=77=7\cdot11$, not prime.
It seems $47$ is the least prime value of $r>3$ for which$$p^2+1=q^2+r^2$$is true for no primes $p$, $q$.
As @Gerry Myerson shows, $r=193$ is another instance. I've yet to find other primes between these two, but have not looked much beyond $r=97$.
| {
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How to Prove : $ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $ How to Prove :
$$ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $$
I have tried looking at Series definitions of the Polygamma function from which we can obtain $\gamma$ but I'm a little bit lost since the given definitions on Wikipedia are not exactly like this one.
Thank you kindly for your help and time.
|
Lemma:
Let $f(z)=\sum_{n=2}^{\infty} a_nz^n$ be convergent with radius $>1.$ Then:
$$\sum_{n=2}^{\infty} a_n\zeta(n)=\sum_{k=1}^{\infty} f\left(\frac1k\right)$$
Proof:
$$\begin{align}\sum_{n=2}^{\infty} a_n\zeta(n)&=\sum_{n=2}^{\infty} a_n\sum_{k=1}^{\infty} \frac{1}{k^n} \\
&=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}a_n\left(\frac 1k\right)^n\\
&=\sum_{k=1}^{\infty}f\left(\frac1k\right)
\end{align}$$
Now, in your case, $a_n=\frac{(-1)^{n}}{2^{n-1}n}$ gives $$f(z)=2\sum_{n=2} \frac{(-z/2)^n}{n}=z-2\log(1+z/2)$$
Now, $$\sum_{k=1}^{N}f(1/k)=H_N - 2\log\left(\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\right)$$
Now, $H_N-\log N\to \gamma.$ So the limit is equal to the limit $$\gamma -2 \log\left(\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\right)$$ as $N\to\infty.$
Thus, you just need to show:
$$\lim_{N\to\infty}\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}=\frac{2}{\sqrt{\pi}}$$
But: $$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}=\frac{2N+1}{2^{2N}}\binom{2N}{N}$$
And we have that $\binom{2n}{n}\sim \frac{2^{2n}}{\sqrt{\pi n}}$ (see here.)
So we have:
$$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\sim\frac{2N+1}{N\sqrt{\pi}}\sim \frac{2}{\sqrt{\pi}}$$
| {
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The summation $\sum_{n\geqslant1} \frac1n\sum_{d\mid n}\frac{d}{n^2+d}.$
I wish to evaluate $\sum\limits_{n\geqslant1}\frac1n\sum\limits_{d\mid n}\frac{d}{n^2+d}.$
Some observations:
Let $f(n)=\sum\limits_{d\mid n}\frac{d}{n^2+d}$.
Then $f(p)=\frac{p^2+p+2}{(p^2+1)(p+1)},\ f(2)=\frac8{15},\ f(4)=\frac{283}{765}.\ \ f$ isn't multiplicative, there is no way to relate $f(4)$ to $f(2),$ and trying to add the summation for $f$ to itself by considering $\frac{d}{n^2+d}+ \frac{n\mid d}{n^2+n\mid d}= \frac{n^2+d^2n+2d}{(n^2+d)(nd+1)}$ doesn't look promising.
Because none of these approaches worked, I turned to changing the order of summation.
Each summand for $f(n)$ is $\leqslant \frac{n}{n^2+n}=\frac1{n+1}<\frac1n$ so, the sum is $\leqslant\sum\limits_n \frac{d(n)}{n^2}\leqslant\sum\limits_n \frac{2\sqrt{n}}{n^2},$ which converges. Each term is positive, so the sum absolutely converges and we may rearrange the order of summation to sum over $d$ and $n=dk$ for $k\geqslant 1.$ We get $\sum\limits_{d\geqslant1} \sum\limits_{k\geqslant1}\frac1{dk} \frac{d}{(dk)^2+d}=\sum\limits_{d\geqslant 1}\frac1{d}\sum\limits_{k\geqslant 1}\frac1{k(dk^2+1)}=\sum\limits_{d \geqslant1}\frac1d\sum\limits_{k\geqslant1}\left(\frac1k- \frac{dk}{dk^2+1}\right).$ But now we have a difference of $2$ divergent series. How do we proceed when the partial fraction decomposition turns against you? Even the sum for $d=1$ does not appear to have a nice closed form.
Update: The problem has a typo and should have $d^2+n$ in the denominator. I found a solution to the corrected version.
Let $f(n) = \sum\limits_{d|n} \frac{d}{n+d^2}$ so that we seek $\sum\limits_{n \ge 1} \frac{f(n)}{n}.$ Since $\frac{n/d}{n+(n/d)^2} = \frac{d}{n+d^2},$ we have $f(n) \le 2\sum\limits_{d|n \atop d \ge \sqrt{n}} \frac{d}{n+d^2} < 2\sum\limits_{d|n \atop d \ge \sqrt{n}} \frac{1}{d} = \frac{2}{n} \sum\limits_{d|n \atop d \le \sqrt{n}} d \le \frac{d(n)\sqrt{n}}{n} < \frac{1}{n^{1/4}}$ for $n$ sufficiently large since $d(n) = o(n^{\epsilon})$ for any $\epsilon>0$ (here we chose $\epsilon = 1/4$)*. By comparison to $\sum\limits_{n \ge 1} n^{-5/4},$ the sum converges. Each term is positive, so the sum absolutely converges and we may rearrange the order of summation to sum over $d$ and $n = dk$ for $k \ge 1.$ We get $\sum\limits_{d \ge 1} \sum\limits_{k \ge 1} \frac{1}{dk} \frac{d}{dk+d^2} = \sum\limits_{d \ge 1} \frac{1}{d} \sum\limits_{k \ge 1} \frac{1}{k(d+k)} = \sum\limits_{d \ge 1} \frac{1}{d^2} \sum\limits_{k \ge 1} \left(\frac{1}{k} - \frac{1}{d+k}\right) = \sum\limits_{d \ge 1} \frac{H_d}{d^2}.$
Using the integral representation $H_n = \int_0^1 \frac{1-x^n}{1-x} \, dx,$ we get that the sum is $\int_0^1 \frac{g(x)}{1-x} \, dx$ where $g(x) = \sum\limits_{n \ge 1} \frac{1-x^n}{n^2}.$ Since $g(1) = 0$ and $xg'(x) = -\sum\limits_{n \ge 1} \frac{x^n}{n} = \log(1-x),$ we get $g(x) = -\int_x^1 \frac{\log(1-t)}{t} \, dt.$ Using integration by parts we get $\int_0^1 \frac{g(x)}{1-x} \, dx = -\log(1-x)g'(x) \big|_0^1 + \int_0^1 g'(x)\log(1-x) \, dx = \int_0^1 \frac{\log(1-x)^2}{x} \, dx = \int_0^1 \frac{\log(x)^2}{1-x} \, dx.$ Let $I_n = \int_0^1 \log(x)^2 x^n \, dx.$ By expanding $\frac{1}{1-x}$ as a power series and exchanging the order of summation and integration, we get $\int_0^1 \frac{\log(x)^2}{1-x} = \sum\limits_{n \ge 0} I_n.$
We turn our attention to $I_n = \int_0^1 \log(x)^2 x^n \, dx = \int_{-\infty}^0 u^2 e^{u(n+1)} \, du = \frac{1}{(n+1)^3} \int_{-\infty}^0 v^2 e^v \, dv = \frac{1}{(n+1)^3} \int_0^{\infty} v^2 e^{-v} = \frac{2!}{(n+1)^3}.$ Finally, the original sum is nothing but $2 \sum\limits_{n \ge 0} \frac{1}{(n+1)^3} = 2\zeta(3).$
*The elementary bound $d(n) \le 2\sqrt{n}$ is almost but not quite enough (you get the harmonic series), so we must resort to a result of Apostol. If there is a very fast proof of $d(n) = o(n^c)$ for some $c<1/2,$ I'd like to hear it.
| By switching the order of summation and by applying the inverse Laplace transform we have
$$S=\sum_{n\geq 1}\sum_{d\mid n}\frac{d}{n(n^2+d)} = \sum_{d\geq 1}\sum_{k\geq 1}\frac{1}{k(k^2 d^2+d)}=\int_{0}^{+\infty}\frac{ds}{e^s-1}\sum_{k\geq 1}\frac{1-e^{-s/k^2}}{k}\,ds=\sum_{k\geq 1}\frac{H_{1/k^2}}{k} $$
and now we may invoke the Maclaurin series of $H_s$ in a right neighbourhood of the origin
$$ H_s = \zeta(2)s-\zeta(3)s^2+\zeta(4)s^3-\ldots $$
in order to get
$$ S = \frac{1}{2}+\sum_{m\geq 2}(-1)^m\left[\zeta(m)\zeta(2m-1)-1\right] $$
which turns $S$ into an essentially geometric series, whose value is $\approx 1.29534$.
| {
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What kind of inequality is this? What kind of inequality is this?
$|z^2 + 1|$ is greater than or equal to $|z|^2 - 1$
why $-1$? , and why not $-100\,000$?
does $z$ have to be a complex variable?
correction : sorry it was not $|z^2|$, it was $|z|^2$
| In the reals,
$$z^2+1\ge z^2$$ holds (and the modulus is not even necessary).
But in the complex, a counter-example is
$$|i^2+1|=|i^2|-1=0.$$
Let $z^2=a+ib$ and let us look for the minimum of
$$\sqrt{(a+1)^2+b^2}-\sqrt{a^2+b^2}.$$
We cancel the gradient,
$$\begin{cases}\dfrac{a+1}{\sqrt{(a+1)^2+b^2}}=\dfrac{a}{\sqrt{a^2+b^2}},
\\\dfrac{b}{\sqrt{(a+1)^2+b^2}}=\dfrac{b}{\sqrt{a^2+b^2}}\end{cases}$$
and this implies $b=0$, and $z^2$ is real. The function reduces to
$$|a+1|-|a|$$ which has the minimum $-1$. Hence
$$|z^2+1|\ge|z^2|-1$$ is tight.
| {
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Is there a formula for adding say: $A\sin^{-1}C-B\sin^{-1}D$ where A, B, C and D are constants? I'm looking for a formula or a method that can deal with the case when $A\ne B$.
$$A\sin^{-1}C-B\sin^{-1}D$$
There is a formula for when $A=B$:
$$\sin^{-1}(C)-\sin^{-1}(D)=\sin^{-1}(C\sqrt{1-D^2}-D\sqrt{1-C^2})$$
The specific case I have is: $$4\sin^{-1}(38r)-3\sin^{-1}(52r)-\sin^{-1}(-53r)$$
Where $r$ is some constant.
| I don’t know about the general case, but for your specific case you can do the following:
First notice that $$ \sin(4x) = 2\sin 2x\cos 2x \\= 4\sin x \cos x (1-2\sin^2 x) \\= 4\sin x(1-2\sin^2x)\sqrt{1-\sin^2 x} $$
So if $4\sin^{-1} x = y$, then
$$\sin(4\sin^{-1} x) = \sin y = 4x(1-2x^2)\sqrt{1-x^2} \\ \implies y=\sin^{-1} \left(4x(1-2x^2)\sqrt{1-x^2}\right)$$ It can be shown similarly that $$3\sin^{-1}(x)= \sin^{-1} (3x-4x^3) $$
Hence, $$4\sin^{-1}(38r)-3\sin^{-1}(52r)-\sin^{-1}(-53r) \\= \sin^{-1} \left(4(38r)(1-2(38r)^2\sqrt{1-(38r)^2} \right)-\sin^{-1} \left(3(52r)-4(52r)^3\right) +\sin^{-1} (53r) $$ Now you can use the formula you mentioned.
| {
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Why $\int_{ \mathbb{R}^2 } \frac{dx\,dy }{(1+x^4+y^4)} $ converges? Why $\int_{ \mathbb{R}^2 } \frac{dx\,dy }{(1+x^4+y^4)} $ converges?
Apparently this integral is quite similar to the integral
$\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ diverges or converges?
and it converges.
So this is quite remarkable that it does converge.
| Answer. Yes.
Note that
$$
1+x^4+y^4\ge 1+\frac{1}{2}(x^2+y^2)^2
$$
and hence, using polar coordinates ($x=r\cos\theta, \,y=r\sin\theta$), we have
$$
\int_{\mathbb R^2}\frac{dx\,dy}{1+x^4+y^4}\le \int_{\mathbb R^2}\frac{dx\,dy}{1+\frac{1}{2}(x^2+y^2)^2}=
\int_0^{2\pi}\int_0^\infty\frac{r\,dr\,d\theta}{1+\frac{1}{2}r^4}\le
\int_0^{2\pi}\int_0^\infty\frac{2r\,dr\,d\theta}{1+r^4}\\
=2\pi \int_0^\infty\frac{ds}{1+s^2}=2\pi\cdot\frac{\pi}{2}=\pi^2.
$$
Note. If we set
$$
A=\{(x,y): |xy|\le 1\},
$$
then $A$ has infinite area, since $\int_0^\infty\frac{dx}{x}=\infty$. Meanwhile,
$$
\frac{1}{1+x^{10}y^{10}}\ge \frac{1}{2}, \quad \text{for all $(x,y)\in A$}
$$
and hence
$$
\int_{\mathbb R^2}\frac{dx\,dy}{1+x^{10}y^{10}}\ge \int_{A}\frac{dx\,dy}{1+x^{10}y^{10}}\ge \int_{A}\frac{1}{2}\,dx\,dy=\infty.
$$
| {
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How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long integral.So i didn't proceed with partial fraction.I'm unable figure out any other method.I think that there might be some other method for evaluation of this definite integral since its value is
$\frac{π}{2\sqrt{3}}$.
Does anyone have nice way to solve it!
| Taking $x\mapsto \frac{1}{x}$ transforms the integral into $I= \int_0^{\infty} \frac{x^6}{1+x^4+x^8} d x$ and then taking the average of them gives $$
\begin{aligned}
I & =\frac{1}{2} \int_0^{\infty} \frac{x^6+1}{x^8+x^4+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{\left(x^2+1\right)\left(x^4-x+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x+1\right)} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3} \\
& =\frac{1}{2 \sqrt{3}}\left[\tan ^{-1}\left(\frac{\left.x-\frac{1}{x}\right)}{\sqrt{3}}\right)\right]_0^{\infty} \\
& =\frac{\pi}{2 \sqrt{3}}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.
| Let $I=\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$. Divide Numerator and denominator of integrand by $x^2$ to get: $I=\int \frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}} \mathop{dx}$. Substitute $y=2-2x^{-2}+x^{-4}$ so that $dy=4(x^{-3}-x^{-5})dx$ and hence $I=\int \frac{1}{4 \sqrt{y}} \mathop{dy}=\frac{y^{1/2}}{2}+C=\frac{\sqrt{2-2x^{-2}+x^{-4}}}{2}+C$, where $C$ is integration constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplified I got
$$\int\dfrac{2-2\tan^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\tan^2\dfrac{x}{2}+6\tan\dfrac{x}{2}-5}dx$$
substituted $\tan^2\dfrac x2=\sec^2\dfrac x2-1$
$$\int\dfrac{4-2\sec^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\left(\tan\dfrac{x}{2}+\dfrac{3}{5}\right)^2-\dfrac{34}{5}}dx$$
I can't eliminate $\tan\frac x2$ term in numerator. I think I am not in right direction. your help to solve this integral is appreciated. thank in advance
| The trick that I use for solving integrals with $\sin{x}$ and $\cos{x}$ with varying coefficients in the numerator and denominator as follows.
For your integral, consider the two easy integrals:
\begin{align*}
\int \frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x&=x+\mathrm{C}\\
\int \frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x&=\ln{\big | 3\sin{x}+5\cos{x} \big |}+\mathrm{C}\\ \end{align*}
Now, set up a system of equations as the following:
$$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; \mathrm{d}x=\int \mathrm{A}\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +\mathrm{B} \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; \mathrm{d}x$$
Such that,
$$2\cos{x}=5\mathrm{A}\cos{x}+3\mathrm{B}\cos{x}$$
$$-\sin{x}=3\mathrm{A}\sin{x}-5\mathrm{B}\sin{x}$$
$$ \mathrm{A}=\frac{7}{34} \; \text{and} \; \mathrm{B}=\frac{11}{34}$$
$$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\int \frac{7}{34}\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +\frac{11}{34} \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; \mathrm{d}x$$
$$\boxed{\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\frac{7x}{34}+\frac{11\ln{\big | 3\sin{x}+5\cos{x} \big |}}{34}+\mathrm{C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
Is this correct?
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)'=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)=\int \frac{3x^2}{{(2x+1)}^{\frac{3}{2}}} \mathop{dx}$$
| Since the ODE is linear $$(2x+1)y'-3y=3x^2(2x+1)$$
We can proceed without integration factor solving homogeneous equation first:
$\dfrac{y'}{y}=\dfrac{3}{2x+1}\ $ gives $\ y=C\,(2x+1)^{3/2}$
And then find a polynomial of degree $3$ satisfying RHS:
$(2x+1)(3ax^2+2bx+c)-3(ax^3+bx^2+cx+d)=3x^2(2x+1)\iff\begin{cases}3a-6=0\\3a+b-3=0\\2b-c=0\\c-3d=0\end{cases}$
Which gives $$p(x)=2x^3-3x^2-6x-2=(2x+1)(x^2-2x-2)$$
Finally $$y=(2x+1)(C\sqrt{\vphantom{|}2x+1}+x^2-2x-2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Asymptotic analysis of $\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}$ For large positive constant $D$, I want an asymptotic evaluation of the sum
$$\sum_{n=-\infty}^\infty \tan^{-1} \left(\frac{D}{2n+1}\right) \log\left(\frac{D}{|2n+1|}\right) \frac{1}{n+3/4}.$$
Note that the sum is convergent since for large $n$ since $\tan^{-1}(D/(2n+1)) \approx D/(2n+1)$ and therefore the term decays fastly. This question is motivated from a calculation of Feynman diagram in quantum field theory.
| We follow the analogous development in this answer. Let $S(D)$ be given by
$$\begin{align}
S(D)&=\sum_{n=-\infty}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=-\infty}^{-1} \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\
&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+1/4}\tag1
\end{align}$$
We analyze the first series on the right-hand side of $(1)$. We begin by writing
$$\begin{align}
\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\
&+\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}
\end{align}\tag2$$
For the first series on the right-hand side of $(2)$ we have
$$\begin{align}
\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\log(D)\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\
&-\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(2n+1\right)}{n+3/4}\tag3
\end{align}$$
For the first series on the right-hand side of $(3)$ we find using the Euler-McLaurin Summation Formula that
$$\begin{align}
\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac1{n+3/4}-\sum_{2n+1\le D}\frac{\arctan\left(\frac {2n+1}{D}\right)}{n+3/4}\\\\
&=\frac\pi2\left(\log(D)+O(1)\right)-O(1)\\\\
&=\frac\pi2 \log(D)+O(1)\tag4
\end{align}$$
For the second series on the right-hand side of $(3)$ we find using the Euler-McLaurin Summation Formula that
$$\begin{align}
\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log(2n+1)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac{\log(2n+1)}{n+3/4}\\\\
&-\sum_{2n+1\le D}\frac{\arctan\left(\frac{2n+1}{D}\right)\log(2n+1)}{n+3/4}\\\\
&=\frac\pi4 \log^2(D)+O\left(\frac{\log(D)}{D}\right)-O(1)\tag5
\end{align}$$
Using $(4)$ and $(5)$ in $(3)$ reveals
$$\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}=\frac\pi4 \log^2(D)+O(\log(D))$$
Next, we anlayze the second series on the right-hand side of $(2)$. It is evident that
$$\left|\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\right|\le D^2\sum_{2n+1>D}\frac{1}{(2n+1)^2(n+3/4)}=O(1)$$
Putting it all together, we find that for $D\to\infty$
$$S(D)=\frac{\pi}{2}\log^2(D)+O(\log(D))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 17 \text{ (mod $k$)}$.
Knowing the sum of cubes formula we get that ($\frac{(k-5)(k-4)}{2})^2\equiv 17 \text{ (mod $k$)}$.
From here I'm not sure how I should continue. What would be my options?
| If $S_m=\sum_{r=1}^mr^3$
$$2S_{n+4}=\sum_{r=1}^{n+4}r^3+(n+5-r)^3\equiv0\pmod{n+5}$$ as $r^3+(n+5-r)^3\equiv0\pmod{n+5}, 1\le r\le n+4$
$$\implies2S_n\equiv-2[(n+4)^3+(n+3)^3+(n+2)^3+(n+1)^3]\pmod{n+5}$$
$$\equiv-2[(n+5-1)^3+(n+5-2)^3+(n+5-3)^3+(n+5-4)^3]$$
$$\equiv2(1^3+2^3+3^3+4^3)$$
$$\implies2S\equiv200\pmod{n+5}\text{ but we need }2S\equiv34\pmod{n+5}$$
So, the necessary & sufficient condition is
$$200\equiv34\pmod{n+5}\iff (n+5)|(200-34)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
| HINT
Firstly, let
$$\tan \dfrac A2 = x,\quad \tan \dfrac B2 = y,\quad s=(x+y)^2,\quad p=xy,\quad x,y \in(0,1),\tag1$$
then
\begin{align}
&\tan\dfrac C2 = \cot\left(\dfrac A2+\dfrac B2\right) = \dfrac{1-xy}{x+y} = \dfrac{1-p}{\sqrt s}\in(0,1),\\[4pt]
&\cos A = \dfrac{1-x^2}{1+x^2},\quad \cos B = \dfrac{1-y^2}{1+y^2},\quad
\cos C = \dfrac{s-(1-p)^2}{s+(1-p)^2} > 0,\\[4pt]
&\cos A + \cos B = \dfrac{(1-x^2)(1+y^2)+(1+x^2)(1-y^2)}{(1+x^2)(1+y^2)}
= \dfrac{2-2p^2}{s+(1-p)^2},\\[4pt]
&\cos A\cos B = \dfrac{(1-x^2)(1-y^2)}{(1+x^2)(1+y^2)}
= \dfrac{(1+p)^2-s}{s+(1-p)^2},\\[4pt]
&s \ge 2p,\quad \sqrt s>1-p,\quad s+2\sqrt s>2,
\end{align}
\begin{cases}
s\in(0,4)\\[4pt]
p\in\left(|1-\sqrt s|,\dfrac s4\right].\tag2
\end{cases}
Secondly,
\begin{align}
&\cos A + \cos B + \cos C = \dfrac{s+(1-p)(2+2p-1+p)}{s+(1-p)^2}
= 1+\dfrac{4p(1-p)}{s+(1-p)^2},\\[4pt]
&\cos A\cos B\cos C = \dfrac{((1+p)^2-s)(s-(1-p)^2)}{(s+(1-p)^2)^2},\\[4pt]
&\cos A + \cos B + \cos C + 4\cos A \cos B \cos C \\[4pt]
&= 1+\dfrac{4\big(p(1-p)(s+(1-p)^2)+((1+p)^2-s)(s-(1-p)^2)\big)}{(s+(1-p)^2)^2} \ge 2 (?)\\[4pt]
\end{align}
Then should
\begin{align}
&F = \cos^3A+\cos^3B+\cos^3C + 64\cos^3A\cos^3B\cos^3C\\[4pt]
&\ge \dfrac1{16}(\cos A+\cos B+\cos C + 4\cos A\cos B\cos C)^3\ge \dfrac12.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3737759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Doubt about how to compute $\sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$ In the book I am reading, I have encountered the following sum.
$$S = \sum_{n=1}^{2019}{\frac{1+2+3+4...+n}{1^3+2^3+3^3...+2019^3}}$$
From here, I factored the denominator since it does not seem to be dependent on $n,$ and I rewrote the expression as
$$\frac{1}{1^3+2^3+3^3+ \cdots +2019^3} \sum_{n=1}^{2019}{1+2+3+4+ \cdots +n}=\frac{1}{1^3+2^3+3^3 + \cdots +2019^3} \sum_{n=1}^{2019}{\sum_{k=1}^n}k$$
From here, I proceeded to compute the sum since there is an equation for the sum of the first $n$ terms, and then, I solved the sum of this equation. Also, I applied the equation of the sum of the first $n^3$ terms. Finally, my result was wrong. My question is regarding my initial approach to the problem since everything else is very simple. I have thought about other ways to solve this, and I can not find it. The official answer is
$$\frac{2019}{2010}.$$
Give this problem a try, and if you arrive at the solution, I invite you to help me understand it.
| Like you mentioned, the denominator is independent of $n,$ hence we can consider it separately. We have that $D = 1^3 + 2^3 + \cdots + 2019^3 = \sum_{i = 1}^{2019} i^3.$ Observe that $S$ can be written as $$\begin{align*} S = \frac 1 D \sum_{n = 1}^{2019} (1 + 2 + \cdots + n) &= \frac 1 D \sum_{n = 1}^{2019} \sum_{k = 1}^n k \\ \\ &= \frac 1 D \sum_{n = 1}^{2019} \frac{n(n + 1)} 2 \\ \\ &= \frac 1 {2D} \biggl(\sum_{n = 1}^{2019} n^2 + \sum_{n = 1}^{2019} n \biggr) \\ \\ &= \frac 1 {2D} \biggl(\frac {2019 \cdot 2020 \cdot 4039} 6 + \frac {2019 \cdot 2020} 2 \biggr) \end{align*}.$$ One can evaluate $D$ using the sum of cubes formula $\sum_{k = 1}^n k^3 = \frac{n^2 (n + 1)^2} 4$ to find that $S = \frac{2021}{6117570}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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prove formula by induction I need to prove by induction that $\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$.
I've seen some similar questions about the convergence of the infinite series, however none about the finite case, I've tried in several ways to factorize or add things in order the find the case m+1 but without success.
| Note: If
$\sum_{n=2}^{m} \frac{1}{n^2 - 1} = \frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})$
Then $\sum_{n=2}^{m+ 1} \frac{1}{n^2 - 1} = (\sum_{n=2}^{m} \frac{1}{n^2 - 1}) + \frac {1}{(m+1)^2 - 1} =$
$(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{(m^2 + 2m +1)-1}=$
$(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{m(m + 2)}$
Can you finish?
Hint: $\frac 1k - \frac 1{k+a} = \frac {k+a}{k(k+a)} - \frac {k}{k(k+a)} = \frac {(k+a)-k}{k(k+a)} = \frac {a}{k(k+a)}$.
Hint 2: Equality goes both ways.
Hint 3:You did put "telescoping-series" as a tag.....
Hint 4: You know that you've got to end up with $\frac 12(1+\frac 12 -\frac 1{m+1} - \frac 1{m+2})$.... Can you show $(\frac{1}{2}(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1})) + \frac 1{m(m + 2)} = \frac 12(1+\frac 12 -\frac 1{m+1} - \frac 1{m+2})$. If you can you are done. If you can't you're dinked.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to integrate $\int\frac{1}{x^2-12x+35}dx$? How to integrate following
$$\int\frac{1}{x^2-12x+35}dx?$$
What I did is here:
$$\int\frac{dx}{x^2-12x+35}=\int\frac{dx}{(x-6)^2-1}$$
substitute $x-6=t$, $dx=dt$
$$=\int\frac{dt}{t^2-1}$$
partial fraction decomposition,
$$=\int{1\over 2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt$$
$$=\frac12(\ln|1-t|-\ln|1+t|)+c$$
$$=\frac12\left(\ln\left|\frac{1-t}{1+t}\right|\right)+c$$
substitute back to $t$
$$=\frac12\ln\left|\frac{7-x}{x-5}\right|+c$$
I am not sure if my answer correct. Can I integrate this without substitution? Thank you
| You can do that without substitution. Use partial fractions by factorizing denominator:$x^2-12x+35=(x-5)(x-7)$
$$\int \frac{dx}{x^2-12x+35}=\int \frac{dx}{(x-7)(x-5)}$$
$$=\int\frac12\left( \frac{1}{x-7}-\frac{1}{x-5}\right)dx$$
$$=\frac12(\ln\left| x-7\right|-\ln\left| x-5\right|)$$
$$=\frac12\ln\left| \frac{x-7}{x-5}\right|+C$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Improper integral $\int_{-\infty}^\infty \frac{1}{x^n + 1}$ for $n$ integer, especially $n$ odd The exercise is to analyze the integral $\int_{-\infty}^\infty \frac{1}{x^n + 1}$ for $n$ integer.
For $n$ even, the integrand is well defined, and I discovered that the integral converges in this case.
My problem is analysing the $n$ odd case. The problem arises because for $x=-1$ the integrand is not defined. This way, I tried breaking the integral in:
$\int_{-\infty}^{-A} \frac{1}{x^n + 1}$ + $\int_{-A}^{-1} \frac{1}{x^n + 1}$ + $\int_{-1}^B \frac{1}{x^n + 1}$+ + $\int_{B}^{\infty} \frac{1}{x^n + 1}$
I tried analysing these integrals, but was unable to proceed.
When I asked to fiends, I heard things like Cauchy principal value, but this exercise was from my course of calculus for engineers, so I was looking for a simple study of improper integrals like this one.
| Let
\begin{equation}
I=\int\limits_{-\infty}^{+\infty} \frac{1}{x^{n}+1}\,dx
\end{equation}
for some positive integer $n$. For any odd positive integer, the integral diverges, so we will only have to compute the case for even positive integers. Given that $n$ is even, we can write the integral as follows:
\begin{equation}
I=2\int\limits_{0}^{+\infty} \frac{1}{x^{n}+1}\,dx
\end{equation}
This integral can be expressed in terms of the beta function. Let $z=\frac{1}{x^{n}+1}$, which implies that $dx=-\frac{1}{n}\left(\frac{1}{z}-1\right)^{\frac{1}{n}-1}z^{-2}\,dz$. Plugging everything in leaves us with:
\begin{equation}
I=\frac{2}{n}\int\limits_{0}^{+\infty} z\left(\frac{1}{z}-1\right)^{\frac{1}{n}-1}z^{-2}\,dz
\end{equation}
\begin{equation}
I=\frac{2}{n}\int\limits_{0}^{+\infty} \left(\frac{1-z}{z}\right)^{\frac{1}{n}-1}z^{-1}\,dz
\end{equation}
\begin{equation}
I=\frac{2}{n}\int\limits_{0}^{+\infty} z^{-1}\left(1-z\right)^{\frac{1}{n}-1}z^{1-\frac{1}{n}}\,dz
\end{equation}
\begin{equation}
I=\frac{2}{n}\int\limits_{0}^{+\infty} z^{-\frac{1}{n}}\left(1-z\right)^{\frac{1}{n}-1}\,dz
\end{equation}
The last integral is equal to $B\left(1-\frac{1}{n},\frac{1}{n}\right)$, so:
\begin{equation}
I=\frac{2}{n}B\left(1-\frac{1}{n},\frac{1}{n}\right)
\end{equation}
Let $s=1/n$, which implies that:
\begin{equation}
I=\frac{2}{n}B\left(1-s,s\right)
\end{equation}
Using the definition of the beta function in terms of the gamma function:
\begin{equation}
I=\frac{2}{n}B\left(1-s,s\right)=\frac{2}{n}\frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1)} = \frac{2}{n}\Gamma(1-s)\Gamma(s)
\end{equation}
From Euler's reflection formula, we conclude that for even $n$:
\begin{equation}
\boxed{I=\int\limits_{-\infty}^{+\infty} \frac{1}{x^{n}+1}\,dx=\frac{2\pi}{n\sin\left(\frac{\pi}{n}\right)}}
\end{equation}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3.
$f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$
Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?.
| Another possibility is to use Lagrange polynomials, and express $f(x)$ as a function that has the given four values in the given points plus a term that produces the given leading term coefficient, but vanishes in all four points:
\begin{align}
f(x)=\
&y_1\frac{(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}+\\
&y_2\frac{(x-x_1)(x-x_3)(x-x_4)}{(x_2-x_1)(x_2-x_3)(x_2-x_4)}+\\
&y_3\frac{(x-x_1)(x-x_2)(x-x_4)}{(x_3-x_1)(x_3-x_2)(x_3-x_4)}+\\
&y_4\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_1)(x_4-x_2)(x_4-x_3)}+\\
&a(x-x_1)(x-x_2)(x-x_3)(x-x_4)\\
=\
&-\frac{5}{24}(x-3)(x+1)(x+6)+\\
&+\frac{5}{18}(x-2)(x+1)(x+6)+\\
&+\frac{1}{30}(x-2)(x-3)(x+6)+\\
&-\frac{37}{360}(x-2)(x-3)(x+1)+\\
&3(x-2)(x-3)(x+1)(x+6)
\end{align}
| {
"language": "en",
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Is the radius of convergence related to the ratio limit or half of the interval of convergence? I have a series $S$ with general terms $a_n=\frac{(-1)^n(x-1)^n}{(2n-1)2^n}$, $n\ge 1$:
$$S = \sum_{n=1}^\infty \frac{(-1)^n(x-1)^n}{(2n-1)2^n}$$
Finding the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ and then finding the limit of the ratio as $n\to\infty$, I find the limit to be $1$ and the interval to be $-1 \lt x \lt 3$. More declaratively, the interval is $\left|\frac{x−1}{2}\right| \lt 1$ which I've refined to what was said earlier.
I've read conflicting sites that state the radius $R$ of convergence is $\frac{1}{N}$, where $N$ is the limit as found earlier, but also that it's half the interval length.
Here's my work:
$$\begin{align}
\left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{\frac{(-1)^{n+1}(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\frac{(-1)^{n}(x-1)^{n}}{(2n-1)2^{n}}}\right| \\
&= \left|\frac{(-1)^{n+1}(x-1)^{n+1}(2n-1)(2^n)}{(-1)^n(x-1)^n(2(n+1)-1)(2^{n+1})}\right| \\
&= \left|\frac{(-1)(x-1)(2n-1)}{(2n+2-1)(2)}\right| \\
&= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{2n-1}{2n+2}\right|
\end{align}$$
Then, finding the limit $L$:
$$\begin{align}
L &= \lim_{n\to\infty} \left(\left|\frac{-(x-1)}{2}\right| \times \left|\frac{2n-1}{2n+1}\right|\right) \\
&= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{2n-1}{2n+1}\right| \\
&= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{\frac{2n}{n}-\frac{1}{n}}{\frac{2n}{n}+\frac{1}{n}}\right| \\
&= \left|\frac{-(x-1)}{2}\right| \times \lim_{n\to\infty} \left|\frac{2-\frac{1}{n}}{2+\frac{1}{n}}\right| \\
&= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{\lim_{n\to\infty} \left(2-\frac{1}{n}\right)}{\lim_{n\to\infty} \left(2+\frac{1}{n}\right)}\right| \\
&= \left|\frac{-(x-1)}{2}\right| \times \left|\frac{2}{2}\right| \\
&= \left|\frac{-(x-1)}{2}\right| \times 1 \\
&= \left|\frac{-(x-1)}{2}\right|
\end{align}$$
Then I know my interval is $\left|\frac{-(x-1)}{2}\right| \lt 1$:
$$\left|\frac{-(x-1)}{2}\right| \lt 1 \\
-1 \lt \frac{x-1}{2} \lt 1 \\
-2 \lt x-1 \lt 2 \\
-1 \lt x \lt 3$$
If the limit found earlier is $1$, the radius would be $R = \frac{1}{1} = 1$, yet I've found the interval to be $(-1, 3)$, which would imply $R = 2$. Where have I made an error?
| Note that a power series takes the form
$$\sum_{n=0}^\infty a_n(x-x_0)^n$$
In your case you have
$$a_n=\begin{cases}\frac{(-1)^n}{(2n-1)2^n}&n\ne0\\0&n=0\end{cases}\qquad x_0=1$$
If you calculate the limit you call $N$ we get
$$N=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac12$$
So the radius of convergence is $R=1/N=2$ and hence the interval of convergence is
$$x\in(x_0-R,x_0+R)=(-1,3)$$
as expected.
| {
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Find $2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$
Function $f$ $\in \mathbb{R}$ is odd and has a period of $4$. On a $[0,2]$ segment function $f$ is defined as $f(x)= 4x - 2x^2$. Find the set of solutions for the equation: $$2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$$
So, here's my attempt: function has a period of $4$ means that $f(x) = f(x+4) = f(x+8) = f(x+12)$. Also, from the fact that $f$ is odd, $f(x-8)$ just means $-f(x+8) = -f(x)$ (I might be wrong here).
Now let $4x-2x^2 = z$. We get the equation: $$2z\cdot(-z)-3z-2=0$$ $$-2z^2-3z-2=0$$ $$2z^2+3z+2 = 0 $$
$D < 0$, so this doesn't have a solution. What am I doing wrong?
| In your solution, you make the following error
$$f(x) \space \text{is odd} \implies f(x-8) = -f(x+8)$$
This is not true, $f$ being odd means that $f(-x) = -f(-x)$, hence $f(x-8) = -f(-x+8)$
Now, since the period is 4, what we can say is that $f(x-8) = f(x-4) = f(x)$
Hence, replacing $f(x) = z$
$$2z^2 - 3z-2 = 0$$
$$\implies z = \frac{3 \pm 5}{4} = -\frac{1}{2},2$$
Now is where your odd function restriction will come into play
EDIT
Adding as it is not straightforward how to continue. Since $f$ is only defined on $[0,2]$, you extend it to $[-2,2]$ using the odd restriction. Hence $f(x)$ is defined as
$$f(x) = \begin{cases}4x-2x^2 & x\in[0,2] \\ 4x+2x^2 & x \in[-2,0]\end{cases}$$
Now, if you solve on this full domain for the above, you have
$$4x+2x^2 = -\frac{1}{2} \implies x = -1 \pm \frac{\sqrt{3}}{2}$$
$$4x - 2x^2 = 2 \implies x = 1$$
Now, since our full function $f$ is a periodic repetition of this, the roots will get repeated as well, and get shifted either up or down by 4 units each. Hence the roots will be of the form
$$x = \begin{cases}1 + 4k \\ 4k-1 + \frac{\sqrt{3}}{2} \\ 4k - 1 - \frac{\sqrt{3}}{2}\end{cases}$$
| {
"language": "en",
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If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions.
I am also not sure if I should consider $c$ as a coefficient of $x^0$, suppose if I take that $c$ is also odd,
then $$b^2-4ac $$ will be odd. But that $-b$(odd), in the quadratic formula will cancel out the oddness of $\sqrt{b^2-4ac}$, in case if it is a perfect square. If it's not a perfect square then the root is irrational.
If I take $c$ to be even, even then the same argument runs but we noticed that when we take $c$ odd, we get that when discriminant is perfect square, so that means the question asks for $c$ not to be an coefficient.
Final question: Is this right to take $c$ as one of the coefficients of the equation $ax^2+bx=c=0$?
| Take $a=1,b=3,c=2$ to get the rational solutions $-2,-1$. So the statement is false unless $c$ is also required to be odd.
Now consider squares modulo $8$. Any odd number has the form $8n+1$, $8n+3$, $8n+5$, or $8n+7$ (these are abbreviated as $\equiv1,3,5,7\bmod8$). So an odd number squared is
$$1^2=1$$
$$3^2=9=8\cdot1+1\equiv1$$
$$5^2=25=8\cdot3+1\equiv1$$
$$7^2=49=8\cdot6+1\equiv1.$$
And any odd number times $4$ is
$$4\cdot1=4$$
$$4\cdot3=12=8\cdot1+4\equiv4$$
$$4\cdot5=20=8\cdot2+4\equiv4$$
$$4\cdot7=28=8\cdot3+4\equiv4.$$
Therefore, if $a,b,c$ are all odd, then $ac$ is also odd, and
$$b^2-4ac\equiv1-4=-3=8\cdot(-1)+5\equiv5\not\equiv1$$
so $b^2-4ac$ cannot be a square.
| {
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"answer_id": 4
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Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easily solved using the substitution $y=\cos(x)$.
What if the power of sine is even?
| Denote
$${I_n = \int\frac{\sin^{2n}(x)}{1 + \cos^2(x)}dx}$$
Then
$${I_{n}=\int\sin^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=\int(1-\cos^2(x))\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx}$$
If you expand this, you get
$${=I_{n-1} - \int \cos^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=I_{n-1}-\int \sin^{2n-2}(x) - \frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx}$$
Hence
$${I_{n}=2I_{n-1} - \int \sin^{2n-2}(x)dx}$$
Now define ${S_n = \int\sin^{2n}(x)dx}$. Then
$${S_{n}=\int \sin^2(x)\sin^{2n-2}(x)dx=S_{n-1}-\int \cos^2(x)\sin^{2n-2}(x)dx}$$
On the rightmost integral, using integration by parts yields
$${\int\cos^2(x) \sin^{2n-2}(x)dx=\frac{\cos(x)\sin^{2n-1}(x)}{2n-1}+\frac{1}{2n-1}\int \sin^{2n}(x)dx}$$
So overall
$${\Rightarrow S_n = S_{n-1}-\frac{\cos(x)\sin^{2n-1}(x)}{2n-1} - \frac{1}{2n-1}S_n}$$
And so
$${\left(\frac{2n}{2n-1}\right)S_n = S_{n-1} - \frac{\cos(x)\sin^{2n-1}(x)}{2n-1}}$$
$${\Rightarrow S_n = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
Now you have two recursive relations that will help you compute the integral for higher even powers of ${\sin(x)}$:
$${I_n = 2I_{n-1} - S_{n-1}}$$
$${S_{n} = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Converting parametric equation to Cartesian How can I convert these parametric equations into a Cartesian equation (I think that is what its called). Wolfram alpha doesn't seem like it can handle this one.
$$
x=\cos{t}+\cos{-6t}\\
y=\sin{t}+\sin{-6t}
$$
Is there a name for this operation?
| I think you're looking for an implicit equation for this parametric plane curve. The process of turning a system of parametric equations into a system of implicit equations is called implicitization. See $\S3.3$ of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms for more on this.
Let $u = \cos(t)$ and $v = \sin(t)$. Using de Moivre's formula, we have
\begin{align*}
\cos(6t) &= T_6(\cos(t)) = 32u^6 - 48u^4 + 18u^2 - 1\\
\sin(6t) &= \sin(t) \, U_6(\cos(t)) = v(32u^5 - 32u^3 + 6u)
\end{align*}
where $T_6$ and $U_6$ are the Chebyshev polynomials of the first- and second kind, respectively. (Depending on your indexing, the second polynomial might be $U_5$.)
We form the ideal
$$
(u^2 + v^2 - 1, x - (u + 32u^6 - 48u^4 + 18u^2 - 1), y - (v - v(32u^5 - 32u^3 + 6u)))
$$
in $\mathbb{Q}[u,v,x,y]$ and compute a Gröbner basis under the elimination ordering. (This can be done in SageMath.) The last generator in this basis is
\begin{align*}
f(x,y) &:= x^{12} + 6 x^{10} y^{2} - 5 x^{10} + 15 x^{8} y^{4} - 25 x^{8} y^{2} + 5 x^{8} - 2 x^{7} + 20 x^{6} y^{6} - 50 x^{6} y^{4}\\
& \qquad + 20 x^{6} y^{2} + 42 x^{5} y^{2} + 15 x^{4} y^{8} - 50 x^{4} y^{6} + 30 x^{4} y^{4} - 70 x^{3} y^{4} + 6 x^{2} y^{10}\\
& \qquad - 25 x^{2} y^{8} + 20 x^{2} y^{6} + 14 x y^{6} + y^{12} - 5 y^{10} + 5 y^{8}
\end{align*}
so the implicit equation of the curve is $f(x,y) = 0$. Here are the parametric plot and implicit plot for comparison.
| {
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Given $\cos(a) +\cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$ Given $\cos(a) + \cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$.
I have tried using the identity $\cos(a) = \frac{1-t^2}{1+t^2}$. but manipulating this seems to have got me nowhere.
| Hint:
$$\dfrac{1-s^2}{1+s^2}+\dfrac{1-t^2}{1+t^2}=1$$
$$\iff\dfrac{1-s^2}{1+s^2}=1-\dfrac{1-t^2}{1+t^2}$$
$$\iff(1-s^2)(1+t^2)=2t^2(1+s^2)$$
| {
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Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x\leqslant1\;\text{and}-1\leqslant\cos x\leqslant1$$
I think I need to have only one term.
Edit : I have made some more progress
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$
| Another way:
$$y(\sin^2x -6\sin x\cos x+3\cos^2x)=5$$
Divide both sides by $\cos^2x$
$$y\tan^2x-6y\tan x+3y=5(1+\tan^2x)$$
Rearrange to form a quadratic equation in $\tan x$ which is real
So, the discriminant must be $\ge0$
| {
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Find x intercepts of a higher degree polynomial $2x^4+6x^2-8$ I am to factor and then find the x intercepts (roots?) of $2x^4+6x^2-8$
The solutions are provided as 1 and -1 and I am struggling to get to this.
My working:
$2x^4+6x^2-8$ =
$2(x^4+3x^2-4)$
Focus on just the right term $(x^4+3x^2-4)$:
Let $u$ = $x^2$, then:
$u^2+3u-4$ =
master term is 1 * -4 = -4. Factors that give minus 4 and sum to 3 are 4 and -1...
$(u^2-u)+(4u-4)$ =
$u(u-1)+4(u-1)$ =
$(u+4)(u-1)$
I don't know where to go from here. If I write $u$ back into it's original $x^2$ I get:
$(x^2+4)(x^2-1)$
Where do I go from here to arrive at x intercepts of 1 and -1?
| you are almost done
$$(x^2+4)(x^2-1)=0$$
$$x^2=-4, \ x^2=1$$
$$x=\pm2 i, \ x=\pm 1$$
considering the real values, the x-intercepts are $x=-1, y=0$ and $x=1, y=0$
x-intercepts in point form: (-1,0) and (1,0)
| {
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On the diophantine equation $x^{m-1}(x+1)=y^{n-1}(y+1)$ with $x>y$, over integers greater or equal than two I don't know if the following diophantine equation (problem) is in the literature. We consider the diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$ over integers $x\geq 2$ and $y\geq 2$ with $x>y$, and over integers $m\geq 2$ and $n\geq 2$. These are four integral variables $x,y,m$ and $n$. The solutions that I know for the problem $(1)$ are two, the solution $(x,y;m,n)=(3,2;2,3)$ and $(98,21;2,3)$.
Question 1. Do you know if this problem is in the literature? Alternatively, if this problem isn't in the literature can you find more solutions?
If the equation or problem $(1)$ is in the literature please refer it answering this question as a reference request, and I try to search and read the statements for new solutions from the literature. In other case compute more solutions or add upto what uppers limits you got evidence that there aren't more solutions.
Question 2. I would like to know what work can be done with the purpose to know if the problem $(1)$ have finitely many solutions $(x,y;m,n)$. I mean what relevant reasonings or heuristics you can to deduce with the purpose to study if the problem have finitely solutions.
If this second question is in the literature, please refer the literature answering this question as a reference request, and I try to search and read the statements from the literature.
| My approach is to restrict the exponents to $m = 2, n = 3$, because both of your solutions have these values, and due to $m - 1 = 1$, this could makes the question different, as the LHS becomes simply a product of two consecutive numbers.
This approach of mine might not be very useful, as I thought I can come up with something that proves that there are infinitely many solutions for this type of equality, but at the end you see it might be the other way round, your two solutions might be the only ones.
I will use the identity $a^b - 1 = (a - 1)(a^{b - 1} + a^{b - 2} + ... + a^2 + a + 1)$ in my approach. $\tag{*}$
We have $x(x + 1) = y^2(y + 1) = y^3 + y^2$
and to be able to use $(*)$, subtract $2$ from both sides:
$x(x + 1) - 2 = y^3 + y^2 - 2$
$x^2 + x - 2 = y^3 - 1 + y^2 - 1 = (y - 1)(y^2 + y + 1) + (y - 1)(y + 1)$
Now I would like to have an expression on both sides so that I can piecewise make them equal:
$x^2 + x - 2 = (y - 1)(y^2 + 2y + 1) + y - 1$
$x^2 + x - 2 = (y - 1)(y + 1)^2 + (y + 1) - 2$
$x^2 + x = (y - 1)(y + 1)^2 + (y + 1)$
Now we had on both sides pairwise equal terms of $x = (y + 1)$, if $(y - 1) = 1$. So one solution is $y = 2, x = (y + 1) = 3$. This is what you got. But this is not giving your other solution. Something is needed to make it more general.
My idea is that as we already have an expression on the RHS that has the right exponents, we need to get rid of the coefficient $(y - 1)$. Therefore we change it to
$x^2 + x = (y - 1)(y + 1)^2 + (y + 1) = (y + 1)^2 + (y + 1) + (y - 2)(y + 1)^2$
and so we must have $(y - 2)(y + 1)^2 = 0$ to obtain $x = (y + 1)$. But this only give $y = -1$ besides $y = 2$, so nothing useful.
So let us change the delta between $x$ and $y$, and instead of having $x = (y + 1)$, let us change to $x = (y + \alpha)$.
We need the following:
$(y + \alpha) = (y + 1) + (\alpha - 1) \implies (y + 1) = (y + \alpha) - (\alpha - 1)$
$(y + \alpha)^2 = y^2 + 2\alpha y + \alpha^2 = y^2 + 2y + 1 + (2\alpha - 2)y + (\alpha^2 - 1) = (y + 1)^2 + (2\alpha - 2)y + (\alpha^2 - 1) \implies (y + 1)^2 = (y + \alpha)^2 - (2\alpha y - 2y + \alpha^2 - 1)$
Now we have
$x^2 + x = (y + 1)^2 + (y + 1) + (y - 2)(y + 1)^2 = (y + \alpha)^2 + (y + \alpha) - (2\alpha y - 2y + \alpha^2 - 1) - (\alpha - 1) + (y - 2)((y + \alpha)^2 - (2\alpha y - 2y + \alpha^2 - 1))$
and so we must have $ - (2\alpha y - 2y + \alpha^2 - 1) - (\alpha - 1) + (y - 2)((y + \alpha)^2 - (2\alpha y - 2y + \alpha^2 - 1)) = 0$ to obtain $x = (y + \alpha)$.
After a lot of transformations one gets a quadratic for $\alpha$:
$\alpha_{1, 2} = \alpha^2 + (2y + 1)\alpha + y - y^3$
from which $\alpha$ based on the quadratic formula is
$\alpha_{1, 2} = \frac{-(2y + 1) \pm \sqrt{4y^3 + 4y^2 + 1}}{2}$
I thought I can come up with something useful for the expression under the square root, but I could not determine for what values of $y$ it gives a square number. When it does, $\alpha$ is an integer, as the number under the square root is odd, therefore its square root is odd, and $(2y + 1)$ is odd, so the $\pm$ of these are even. For $y = 2$ it gives $\alpha = 1$ so $x = 3$, which is your first solution and my previous one, and for $y = 21$ it gives $\alpha = 77$ so $x = 98$ which is your other solution. I brute-forced the expression under the square root up to one million, but there are no other solutions.
Of course even if it is provable that there are no other solutions for this expression to be a square, other solutions might exist for the original equality, as it only related to the piecewise equality logic applied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\tan^2\phi}{2\tan \phi }=\frac{1}{x^2}$$
$$\sin 2\phi=x^2$$
$$\phi=\frac{\pi}{4}-\frac 12 \cos^{-1}x^2$$
Where am I going wrong?
| Domain of $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ is $x \in (-1,1]$, nothing wrong with that.
But range of the above term(argument of $\arctan$) is $(1,\infty)$ so this means, when you assume it be $\phi$, it is restricted to the interval $\left[ \frac{\pi}{4},\frac{\pi}{2}\right]$
This creates a problem in the last line, because $\sin^{-1}(x^2)$ should be in its principal range and $\frac{\pi}{2}\le 2\phi \le \pi$
Edit: to find the domain of the argument, the easiest way, divide both sides by $\sqrt{1+x^2}$ and then substitute $x^2=\cos(2\theta)$, $\theta \in \left(0,\frac{\pi}{4}\right)$, to get
$$\frac{1+\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}}{1-\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}}$$
and then use the identity, $\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}$
$$\frac{1+\tan \theta}{1-\tan \theta}=\tan \left(\frac{\pi}{4}+ \theta\right)$$
Now, $\theta \in \left(0,\frac{\pi}{4}\right)$ so, $\tan \left(\frac{\pi}{4}+\theta\right) \in [1, \infty)$
Which is ironically your question...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis This particular integral evaluates to,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)-\frac{G}{6}$$
And its been proven here.
But i'd like to know how to evaluate this without complex analysis.
One of the answers uses differentiation under the integral sign directly and partial fraction decomposition on a similar integral, but doing it that way doesnt help me with this case here
I tried to evaluate this way but got stuck,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\int _0^1\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx+\int _1^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx\:\:\:\:\:\: \text{then sub}\:\:x=\frac{1}{t}\:\:\text{for the 2nd integral}$$
$$=\int _0^1\frac{\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt+\int _0^1\frac{t^2\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt-3\int _0^1\frac{t^2\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$
$$=\int _0^1\frac{\ln \left(t^3+1\right)}{t^2+1}\:dt+3G+3\int _0^1\frac{\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$
I managed to evaluate the last integral expanding the denominator but i cant think of a way to evaluate the 1st integral, please help me.
| I am going to prove it by integration by parts and partial fractions.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{\left(1+x^{2}\right)^{2}} d x
&=-\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{2 x} d\left(\frac{1}{1+x^{2}}\right)\\
&=-\left[\frac{\ln \left(1+x^{3}\right)}{2 x\left(1+x^{2}\right)}\right]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{1}{1+x^{2}} \cdot \frac{\frac{3 x^{3}}{1+x^{3}}-\ln \left(1+x^{3}\right)}{x^{2}} d x\\
&=\frac{3}{2} \underbrace{ \int_{0}^{\infty} \frac{x}{\left(1+x^{2}\right)\left(1+x^{3}\right)} d x}_{\frac{1}{36}(8 \sqrt{3}-9) \pi}-\frac{1}{2} \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{x^{2}\left(1+x^{2}\right)} d x\\&= \frac{\sqrt{3} \pi}{3}-\frac{3 \pi}{8} -\frac{1}{2} \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{x^{2}} d x}_{\frac{2 \sqrt{3} \pi}{3}} +\frac{1}{2} \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}}}_{\frac{\pi \ln 2}{4}-\frac{G}{3}+\frac{2 \pi}{3} \ln (2+\sqrt{3})}
\end{aligned}
$$
where the last integral comes from the post.
Sum them up gives the result
$$
I=-\frac{3 \pi}{8}+\frac{\pi \ln 2}{8}-\frac{G}{6}+\frac{\pi}{3} \ln (2+\sqrt{3})
$$
| {
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"url": "https://math.stackexchange.com/questions/3761814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ we will use the substitution: $$ a+bx^k = u^{\frac{1}{n}}$$
Therefore,
$$ u^3 = 1 + x^2 \iff x = \sqrt{u^3 +1} \text{ (Mistake Here. Check the comments) } $$ $$\iff dx = \frac{3u^2}{2\sqrt{u^3}+1}$$
So the new integral is,
$$ \int x^5 (1+x^2)^{\frac{2}{3}} dx = \frac{3}{2} \int u^4 (u^3+1)^{\frac{7}{6}} du$$
Instead of simplifying the integral, the substitution did nothing by keeping it at the same form, with different values on the variables $m,k,n$.
I tried to substitute once again and it doesn't seem to lead in any known paths, anytime soon.
Any ideas on how this could be evaluated?
| HINT:
Let $1+x^2=t^3\implies 2xdx=3t^2dt$ or $xdx=\frac{3}{2}t^2dt$
$$\int x^5(1+x^2)^{2/3}dx=\int (x^2)^2(1+x^2)^{2/3}xdx$$
$$=\int (t^3-1)^2(t^3)^{2/3}\ \frac{3t^2}{2}dt$$
$$=\frac32\int(t^3-1)^2t^4 dt $$
$$=\frac{3}{2}\int (t^{10}-2t^7+t^4)dt$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$
$$ \int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right) dx = $$
$$\int \left(\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx = $$
$$\int 1*\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx $$
That's a dead end.
I also tried other conjugate approaches (only the numerator, only the denominator etc) with no better luck.
Any ideas?
| $$I=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \,dx $$
Rationalize:
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\times\frac{-\sqrt{1+x}-\sqrt{1-x}}{-\sqrt{1+x}-\sqrt{1-x}}$$
$$=\frac{\sqrt{(1+x)(1-x)}+1}{x}$$
$$=\frac{\sqrt{1-x^2}}{x}+\frac 1x$$
Therefore
$$I=\int\left(\frac{\sqrt{1-x^2}}{x}+\frac 1x\right)\,dx=\int\frac{\sqrt{1-x^2}}{x}\,dx+\ln|x|+C_1$$
Now, integrate the second integral by parts:
$$\int \frac{\sqrt{1-x^2}}{x^2}dx=\sqrt{1 - x^2}\left( -\frac{1}{x}\right) + \int \frac{x}{\sqrt{1-x^2}}\left( -\frac{1}{x}\right)\,dx$$
$$=-\frac{\sqrt{1 - x^2}}{x} - \int \frac{1}{\sqrt{1-x^2}}\,dx$$
$$= -\frac{\sqrt{1 - x^2}}{x} - \arcsin(x) +C_2$$
Thus, letting $C=C_1+C_2$ gives
$$I=-\frac{\sqrt{1 - x^2}}{x} - \arcsin(x)+\ln|x|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities? You can't use geometric sums, minimal polynomials, pentagon, and exact values with radicals.
All the five, fifth-roots of unity are :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\right),\left(\cos \left(\frac{4 \pi}{5}\right)+i \sin \left(\frac{4 \pi}{5}\right)\right),\left(\cos \left(\frac{6 \pi}{5}\right)+i \sin \left(\frac{6 \pi}{5}\right)\right), \left(\cos \left(\frac{8 \pi}{3}\right)+i \sin \left(\frac{8 \pi}{5}\right)\right)$
| $$z^5=1$$
$$z^5-1=0$$
$$(z-1)(z^4+z^3+z^2+z+1)=0$$
Note that the parenthesis with 5 terms are the roots but as we know the complex is a filed no zero divisors thus the sum of the roots is zero!
Plug in $e^{\frac{2\pi i }{5}}=z$
$$(e^{\frac{2\pi i }{5}}-1)(e^{\frac{8\pi i }{5}}+ e^{\frac{6\pi i }{5}}+ e^{\frac{4\pi i }{5}}+ e^{\frac{2\pi i }{5}}+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the constant for $\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} \sim C\ln(1-k)$ I encounter a problem for Elliptic integral, to find the exact $C$ for
$$\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} \sim C\ln(1-k)$$
as $k\uparrow1\;(0<k<1)$.
to establish such asymptotic behavior around $k\uparrow1$ is nothing special, we have
$$\begin{aligned}
\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}}
& = \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x)(1+x)(1-kx)(1+kx)(1+(kx)^2)}}} \\
& \le \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x)(1-kx)}}} = \frac{2\operatorname{artanh}(\sqrt{k})}{\sqrt{k}}
\end{aligned}$$
and
$$\begin{aligned}
\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}}
& \ge \frac1{2\sqrt{2}} \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x)(1-kx)}}} \\
& \ge \frac1{2\sqrt{2}} \int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-kx)^2}}} = -\frac1{2\sqrt{2}}\frac{\ln(1-k)}{k}
\end{aligned}$$
notice
$$\frac{\operatorname{artanh}(\sqrt{k})}{\sqrt{k}} \sim -\frac1{2}\frac{\ln(1-k)}{k} \text{ as } k\uparrow1$$
What I know is the original integral as well belonging to Elliptic integral of the first kind, namely
$$\int_{0}^{1} {\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-(kx)^4)}}} = \frac1{\sqrt{1+k^2}}K\left(\frac{2k^2}{1+k^2}\right)$$
as little knowledge I have for Elliptic integral, I can't figure out what exactly $C$ is for such behavior, and I am wondering if there is a relatively fundamental way to find it.
thanks in advance for any suggestion.
| I will propose a creative approach based on Fourier-Legendre expansions. In $L^2(0,1)$ we have$^{(*)}$
$$ K(x)=\sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1),\qquad -\log(1-x)=1+\sum_{n\geq 1}\left(\frac{1}{n}+\frac{1}{n+1}\right)P_n(2x-1) $$
so
$$ K(x)+\frac{1}{2}\log(1-x)=\frac{3}{2}-\sum_{n\geq 1}\frac{P_n(2x-1)}{2n(n+1)(2n+1)} $$
where the RHS is continuous and bounded on $[0,1]$ due to $|P_n(2x-1)|\leq 1$. This implies $K(x)\sim -\frac{1}{2}\log(1-x)$ as $x\to 1^-$.
A straightforward consequence is that
$$\begin{eqnarray*} \int_{0}^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^4 x^4)}}=\frac{1}{\sqrt{1+k^2}}K\left(\frac{2k^2}{1+k^2}\right)&\sim& -\frac{1}{2\sqrt{2}}\log\left(1-\frac{2k^2}{k^2+1}\right)\\&\sim&-\frac{1}{2\sqrt{2}}\log\left(\frac{1+k}{1+k^2}(1-k)\right)\end{eqnarray*} $$
so the wanted constant is $C=-\frac{1}{2\sqrt{2}}$. Keeping track of the first term of the regular part,
$$\boxed{ \int_{0}^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^4 x^4)}} \sim -\frac{1}{2\sqrt{2}}\log(1-k)+\sqrt{2}\log(2)\qquad\text{as }k\to 1^-.}$$
(*)The first identity is easily derived from the generating function for Legendre polynomials; the second identity can be proved by computing $\int_{0}^{1}\log(x) P_n(2x-1)\,dx$ through Rodrigues' formula and integration by parts.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x \in [3,5]$.
Calculate the area between the two functions, $f(x)$, $g(x)$, for $x \in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$
To determine the area between the functions I used the formula $A= \int_a^b|f(x)-g(x)|dx$. Therefore, I have:
\begin{align}
A&=\int_3^5|x^2+3x+7-xe^{x^3+4}|dx \\
&= \int_3^5|x^2|dx+\int_3^5|3x|dx+\int_3^5|7|dx-\int_3^5|e^{x^3+4}|dx\\
&= \left(\frac{x^3}{3}+\frac{3x^2}{2}+7x\right)_3^5 - \int_3^5|xe^{x^3+4}|dx
\end{align}
Here I can use $u$-subsitution and the Incomplete Gamma Function to find the integral of $xe^{x^3+4}$. All in all I get:
\begin{align}
A&=\left(\frac{x^3}{3}+\frac{3x^2}{2}+7x-e^4\frac{1}{2}\left(-\frac{x^2Γ\left(\frac{1}{\frac{3}{2}},\:-\left(x^2\right)^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-\left(x^2\right)^{\frac{3}{2}}}}\right)\right)_3^5\\
&=-\frac{98}{3}-24-14+e^4\frac{1}{2}\left(-\frac{25Γ\left(\frac{1}{\frac{3}{2}},\:-25^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-25^{\frac{3}{2}}}}-\left(-\frac{9Γ\left(\frac{1}{\frac{3}{2}},\:-9^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-9^{\frac{3}{2}}}}\right)\right)
\end{align}
Can this be simplified further or is the given solution enough?
| You can do this:
$$The \ area \ between \ the \ two \ functions = \int_{3}^{5}{xe^{x^3+4}}dx \ - \ \int_3^5{x^2+3x+7}dx$$ Because $\forall x \in [3, 5] \ \ \ xe^{x^3+4} > x^2+3x+7$.
You can look here: https://www.desmos.com/calculator/csh4alwmeu
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$ and $ b< 0$
Given both terms are negative, $ a + b < 0 $
$$
|a+b| = -(a + b) = -a -b = |a| + |b|\\
$$
Case 3) $ a> 0$ and $ b< 0$
Case 3a) $ a + b \geq 0 $
Knowing that $ b < -b $ because $ b < 0 < -b $:
$$
|a+b| = a + b < a - b = |a| + |b|
$$
Case 3b) $ a + b \leq 0 $
Knowing that $ -a < a $ because $ -a < 0 < a $:
$$
|a+b| = -(a + b) = -a - b < a - b = |a| + |b|
$$
Case 4) $ a< 0$ and $ b> 0$
Case 4a) $ a + b \geq 0 $
Knowing that $ a < -a $ because $ a < 0 < -a $:
$$
|a+b| = a + b < -a + b = |a| + |b|
$$
Case 4b) $ a + b \leq 0 $
Knowing that $ -b < b $ because $ -b < 0 < b $:
$$
|a+b| = -(a + b) = -a -b < -a + b = |a| + |b|
$$
Is my proof correct?
I feel like it's somewhat incomplete as I'm not considering when either $a$ or $b$ are equal to zero in the cases 3 and 4.
| Something similar to yours:
The first part was, indeed, easy.
For the second part:
Without loss of generality, take $|a|>|b|~~~~~~~~~(*)$
*
*$a>0, b<0 \overset{(*)}\implies a+b>0$. Then, indeed, $\underbrace{|a+b|}_{=a+b} < \underbrace{|a|}_{=a}+\underbrace{|b|}_{=-b}$ since $b<-b$.
*$a<0, b>0 \overset{(*)}\implies a+b<0$. Then, indeed, $\underbrace{|a+b|}_{=-a-b} < \underbrace{|a|}_{=-a}+\underbrace{|b|}_{=b}$ since $-b<b$.
| {
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Given $c =\sqrt {a^2-b^2}$, $c= ea$, conclude that $\frac {2\pi ab}{(a+c)T} =\frac {2\pi a}T\sqrt {\frac {1-e}{1+e}}$ Given $c =\sqrt {a^2-b^2}$, $c= ea$, conclude that $$\frac {2\pi ab}{(a+c)T} =\frac {2\pi a}T\sqrt {\frac
{1-e}{1+e}}$$
Sorry for the awful formatting of this question. I know this is a pretty simple problem, but for some reason I cannot figure it out. The furthest I have gotten is reducing the b in the numerator to $\sqrt {a^2(1-e)}$, but then I get stuck. I appreciate any help I can get. Thanks in advance.
| Since $\;c =\sqrt {a^2-b^2},\;$ we get $\;b=\sqrt {a^2-c^2}$.
Moreover,
$$\frac {2\pi ab}{(a+c)T} =\frac{2\pi a\sqrt{a^2-c^2}}{T(a+c)}=\frac{2\pi a}T\sqrt{\frac
{(a+c)(a-c)}{(a+c)^2}}=\frac{2\pi a}T\sqrt{\frac
{a-c}{a+c}}=\frac{2\pi a}T\sqrt{\frac
{a-ea}{a+ea}}=\frac{2\pi a}T\sqrt{\frac
{a(1-e)}{a(1+e)}}=\frac{2\pi a}T\sqrt{\frac
{1-e}{1+e}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}
The answer is $x=9,\,y=4$. A geometrical investigation can give us better insights as depicted below.
$\hspace{2cm}$
| We have
$$\begin{align*}
\sqrt{x} + y &= 7 \\
x + \sqrt{y} &= 11
\end{align*}$$
Under the constraints
$$0\le x \le 11,\quad 0\le y \le 7$$
A contour plot shows a single point of intersection
We can isolate the square root of $x$ in first equation, square both sides (this causes us to have extraneous roots that we have to eliminate) and then do the same in second equation, substitute and arrive at
$$y=(11−(49−14y+y^2))^2$$
Solving this, we find four roots for $y$, including the only valid root at $y = 4$ and solving for $x$, we arrive at $x = 9$.
To verify this, we can use a Groebner Basis and eliminate either variable and if we first eliminate $x$
$$y^4-28 y^3+272 y^2-1065 y+1444 = (y-4) \left(y^3-24 y^2+176 y-361\right) = 0$$
We could have also chosen to eliminate $y$
$$x^4-44 x^3+712 x^2-5017 x+12996 = (x-9) \left(x^3-35 x^2+397 x-1444\right) = 0$$
Using either of these, we have four real roots, but only one meets the constraints and original equations $(x, y) = (9, 4)$.
We can verify this result using Wolfram Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
Given ellipse of axes $a$ and $b$, find axes of tangential and concentric ellipse at angle $t$ Let’s say I have an ellipse with horizontal axis $a$ and vertical axis $b$, centered at $(0,0)$.
I want to compute $a’$ and $b’$ of a smaller ellipse centered at $(0,0)$, with the axes rotated by some angle $t$, tangent to the bigger ellipse and $\frac{a’}{b’}=\frac{a}{b}$.
| Let the oblique ellipse be
$$
\frac{x^2}{a^2}+
\frac{y^2}{b^2}-1=k
\left(
\frac{x\sin \theta}{a}-
\frac{y\cos \theta}{b}
\right)^2$$
which touches the standard ellipse at $(a\cos \theta,b\sin \theta)$.
This can be easily verified by plugging the point $(x',y')=(a\cos \theta,b\sin \theta)$ in the both ellipses and also have the same tangent line by referring to the useful fact in the very last of the post.
By finding the eigenvalues $(\lambda_1 <\lambda_2)$ of the matrix $\mathbb{M}$, namely
\begin{align}
\mathbb{M} &=
\begin{pmatrix}
A & \frac{B}{2} \\ \frac{B}{2} & C
\end{pmatrix} \\
&=
\begin{pmatrix}
\dfrac{1-k\sin^2 \theta}{a^2} & \dfrac{k\sin \theta \cos \theta}{ab} \\
\dfrac{k\sin \theta \cos \theta}{ab} & \dfrac{1-k\cos^2 \theta}{b^2}
\end{pmatrix}
\end{align}
Using Mathematica, we can get
$$\frac{\lambda_2}{\lambda_1}=\frac{a^2}{b^2} \implies
k=\frac{(a^4-b^4)\cos 2\theta}{(a^2\cos^2 \theta+b^2\sin^2 \theta)^2}$$
which gives smaller ellipse if $k<0 \implies \cos 2\theta<0$ providing $a>b$.
Due to similarity, the new semi-axes are scaled by the same factor which can be determined by the determinant of the matrix $\mathbb{M}$.
Now,
\begin{align}
\lambda_1 \lambda_2 &= \det \mathbb{M} \\
\frac{1}{a'^2 b'^2} &= AC-\frac{B^2}{4} \\
&= \frac{1-k}{a^2 b^2} \\
\frac{ab}{a'b'} &= \sqrt{1-k} \\
&=
\frac{a^2\sin^2 \theta+b^2\cos^2 \theta}
{a^2\cos^2 \theta+b^2\sin^2 \theta} \\
\frac{a'}{a} &= \frac{b'}{b} \\
&=
\sqrt{
\frac{a^2\cos^2 \theta+b^2\sin^2 \theta}
{a^2\sin^2 \theta+b^2\cos^2 \theta}}
\end{align}
Using the result of another answer of mine, the angle of rotation is given by
\begin{align*}
t &= \tan^{-1}
\left(
\frac{C-A}{B} \pm \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \:
\right) \\
t_1 &= \tan^{-1}
\left(
-\frac{2ab}{a^2+b^2} \cot 2\theta
\right) \tag{semi-axis $a$} \\
t_2 &= \tan^{-1}
\left(
\frac{a^2+b^2}{2ab} \tan 2\theta
\right) \tag{semi-axis $b$}
\end{align*}
In principle, $\theta$ can be solved in terms of $t$ explicitly. The work is tedious and verified experimentally by using Mathematica.
\begin{align}
C & \equiv 2ab\cos t \\
S & \equiv (a^2+b^2)\sin t+\sqrt{(a^2+b^2)^2\sin^2 t+C^2} \\
\sqrt{1-k} &= \frac{a^2 S^2+b^2 C^2}{a^2 C^2+b^2 S^2}
\end{align}
Finally by rotational transformation, the oblique ellipse can be expressed as
$$
\frac{(x\cos t+y\sin t)^2}{a^2}+
\frac{(y\cos t-x\sin t)^2}{b^2}=
\frac{a^2 C^2+b^2 S^2}{a^2 S^2+b^2 C^2}$$
Further application
Illustration of spiral galaxy
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x',y')$ is given by
$$ax'x+h(y'x+x'y)+by'y+g(x+x')+f(y+y')+c=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question on Cardano's Method of Solving Cubic Polynomial Equations I'm having trouble with part of a question on Cardano's method for solving cubic polynomial equations. This is a multi-part question, and I have been able to answer most of it. But I am having trouble with the last part. I think I'll just post here the part of the question that I'm having trouble with.
We have the depressed cubic equation :
\begin{equation}
f(t) = t^{3} + pt + q = 0
\end{equation}
We also have what I believe is the negative of the discriminant :
\begin{equation}
D = 27 q^{2} + 4p^{3}
\end{equation}
We assume $p$ and $q$ are both real and $D < 0$. We also have the following polynomial in two variables ($u$ and $v$) that results from a variable transformation $t = u+v$ :
\begin{equation}
u^{3} + v^{3} + (3uv + p)(u+v) + q = 0
\end{equation}
You also have the quadratic polynomial equation :
\begin{equation}
x^{2} + qx - \frac{p^{3}}{27} = 0
\end{equation}
The solutions to the 2-variable polynomial equation satisfy the following constraints :
\begin{equation}
u^{3} + v^{3} = -q
\end{equation}
\begin{equation}
uv = -\frac{p}{3}
\end{equation}
The first section of this part of the larger question asks to prove that the solutions of the quadratic equation are non-real complex conjugates. Here the solutions to the quadratic are equal to $u^{3}$ and $v^{3}$ (this relationship between the quadratic polynomial and the polynomial in two variables was proven in an earlier part of the question). I was able to do this part. The second part of this sub-question is what I'm having trouble with.
The question says, let :
\begin{equation}
u = r\cos(\theta) + ir\sin(\theta)
\end{equation}
\begin{equation}
v = r\cos(\theta) - ir\sin(\theta)
\end{equation}
The question then asks the reader to prove that the depressed cubic equation has three real roots :
\begin{equation}
2r\cos(\theta) \text{ , } 2r\cos\left( \theta + \frac{2\pi}{3} \right) \text{ , } 2r\cos\left( \theta + \frac{4\pi}{3} \right)
\end{equation}
In an earlier part of the question they had the reader prove that given :
\begin{equation}
\omega = \frac{-1 + i\sqrt{3}}{2}
\end{equation}
s.t. :
\begin{equation}
\omega^{2} = \frac{-1 - i\sqrt{3}}{2}
\end{equation}
and :
\begin{equation}
\omega^{3} = 1
\end{equation}
that if $(u,v)$ is a root of the polynomial in two variables then so are :
$(u\omega,v\omega^{2})$ and $(u\omega^{2},v\omega)$. I think that the part of the question I'm having trouble with is similar. I suspect that :
\begin{equation}
2r \cos\left( \theta + \frac{2\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{1}
\end{equation}
and :
\begin{equation}
2r \cos\left( \theta + \frac{4\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{2}
\end{equation}
I have derived that :
\begin{equation}
\omega = \cos(\phi) + i\sin(\phi)
\end{equation}
where $\phi = \frac{2\pi}{3}$. Also :
\begin{equation}
\omega^{2} = \cos(2\phi) + i \sin(2\phi)
\end{equation}
So that the goal of the question may be to prove equations $(1)$ and $(2)$. I have tried to do this but haven't been able to.
Am I approaching this question in the correct way ? If I am approaching it the right way can someone show me how to use trigonometric identities to prove equations #1 and #2 ?
| Let $w(\alpha) = \cos \alpha + i\sin \alpha$. Then
$$w(\alpha) w(\beta) = (\cos\alpha + i \sin \alpha)(\cos \beta + i\sin \beta) \\ =\cos\alpha \cos \beta - \sin \alpha \sin \beta +i(\cos\alpha \sin \beta + \sin \alpha \cos \beta) = \cos(\alpha + \beta) + i \sin(\alpha + \beta) \\= w(\alpha + \beta) .$$
An easier way to see this is to write $w(\alpha) = e^{i\alpha}$. Then
$$w(\alpha) w(\beta) = e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)} = w(\alpha + \beta) .$$
We have
$$u\omega = rw(\theta)w(\phi) = rw(\theta+\phi) ,$$
$$u\omega^2 = rw(\theta)w(2\phi) = rw(\theta+2\phi) .$$
Moreover, since $v = \overline u$ and $\omega^2 = \overline \omega$, we get
$$v\omega^2 = \overline u \cdot \overline \omega = \overline{u\omega} ,$$
thus
$$u\omega + v\omega^2 = 2\Re (u\omega) = 2r\cos(\theta + \phi) = 2r\cos(\theta + 2\pi/3) .$$
Similarly
$$v\omega = \overline u \cdot \overline {\omega^2} = \overline{u\omega^2},$$
thus
$$u\omega^2 + v\omega = 2\Re (u\omega^2) = 2r\cos(\theta + 2\phi) = 2r\cos(\theta + 4\pi/3) .$$
Edited:
In my opinion it is an odd aproach to apply Cardano's formula and then translate the result into a trigonometric form. A direct approach is via angle trisection. By Moivre's formula we have
$$\cos\phi + i\sin\phi = (\cos(\phi/3) + i\sin(\phi/3))^3$$
which gives
$$\cos \phi = \cos^3(\phi/3) -3\cos(\phi/3)\sin^2(\phi/3)\\ = \cos^3(\phi/3) -3\cos(\phi/3)(1- \cos^2(\phi/3)) = 4 \cos^3(\phi/3) - 3 \cos(\phi/3) .$$
Writing $\theta = \phi/3$ and $x = 2\cos \theta$ gives us the cubic angle trisection equation
$$x^3 - 3x = 2\cos \phi \tag{1}.$$
By construction it has the obvious solution $x_0 = 2\cos \theta$. But since $\cos \phi = \cos (\phi + 2\pi) = \cos (\phi + 4 \pi)$, it also has the solutions $x_1 = 2 \cos((\phi + 2\pi)/3) = 2\cos (\theta + 2\pi/3)$, $x_2 = 2 \cos((\phi + 4\pi)/3) = 2\cos (\theta + 4\pi/3)$.
Under the assumption that $p, q$ are real and $D = 27q^2 + 4 p^3<0$ it is possible to reduce the general equation
$$t^3 + pt + q = 0 \tag{2}$$
to the angle trisection equation (1). Since $D < 0$, we must have $p < 0$. Note that therefore $D < 0$ is equivalent to $27q^2/(-4p^3) < 1$.
Let us write $t = cx$. Then
$$x^3 + (p/c^2)x = -q/c^3 .$$
With $c = \sqrt{-p/3} > 0$ we get
$$x^3 -3x = 2(-q/2c^3) .$$
But
$$(-q/2c^3)^2 = q^2 /4(-p/3)^3 = 27q^2/(-4p^3) < 1$$
which means that
$$-q/2c^3 \in (-1,1) .$$
Therefore $\phi = \arccos(-q/2c^3)$ is a well-defined number in $(0,2\pi)$ and we get the cubic equation (1) with solutions $x_k$ as above. Therefore the solutions of (2) are
$$t_k = 2\sqrt{-p/3}\cos(\phi/3 + 2k\pi/3) , k = 0,1,2 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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EGMO 2014/P3 : Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ for any positive integers $a,b$ satisfying $a+b=n$.
My progress: Really beautiful but hard problem!
For $k=1$,we can take $n=2^{p-1}$, where p is an odd prime. Let's say for some $a+b=n$ and write $a=2^ke$ and $b=2^kf$ with $e, f$ odd and $0\le k<p-1$. If $d(n)|d(a^2+b^2)$, then$$p|d \left ( 2^{2k+1}\cdot \dfrac{e^2+f^2}{2} \right )=2^{2k+2}\cdot d\left ( \dfrac{e^2+f^2}{2} \right )$$.
So $p|d\left (\dfrac{e^2+f^2}{2}\right) $ .Now, for $p$ to divide $d\left (\dfrac{e^2+f^2}{2}\right) $, we should have $\left (\dfrac{e^2+f^2}{2}\right)=l^{p-1}\cdot x, l $ is a prime and $gcd(l,x)=1$. But note that both 2 and 3 does not divide $\left (\dfrac{e^2+f^2}{2}\right)$. But Max$(a^2+b^2)=4^{p-1}<5^{p-1}$ .
So we are done for $k=1$ .
I thought that this would be almost same for $k>1$ , but I am not able to prove.
I have conjectured that for any $k$ we can take $n = 2^{p-1}j$ such that $j$ has only $k-1$ primes.
But no progress!Please, if possible post hints rather than solution.
Thanks in advance.
| $\boxed{\text{Complete solution}}$
(The merit of the following solution is that it gives an explicit construction for $n$ with given $k$ satisfying the conditions.)
Let $p_m$ denote the $m^{th}$ prime with $p_1=2,p_2=3,\ldots$ and so on. Take, for $k>1$, $$n=2^{p-1}p_2p_3\cdots p_k$$ for some suitable prime $p$ and work on it. Then $d(n)=2^{k-1}p$ and $\omega(n)=k$. The key observation is that $$d(n)\mid d(a^2+b^2)\implies p\mid d(a^2+b^2)\implies q^{p-1}\mid a^2+b^2$$ for some prime $q$. Now proceed considering different cases of $q$.
Case 1 ($q>4$)
Since $q^{p-1}\mid(a^2+b^2)$ then we have $$q^{p-1}\leq(a^2+b^2)\leq (a+b)^2=n^2=4^{p-1}p_2^2p_3^2\cdots p_k^2$$ Since $q>4$ we can choose sufficiently large prime $p$ such that $$q^{p-1}>4^{p-1}p_2^2p_3^2\cdots p_k^2$$ which is a contradiction! Hence for sufficiently large prime $p$, $n$ satisfies the condition.
Case 2 ($q=3$)
Since $-1$ is not a quadratic residue modulo $3$, $3^{p-1}\mid a^2+b^2$ implies $3^{p-1}\mid a^2,3^{p-1}\mid b^2$. This implies $3^{\frac{p-1}{2}}\mid a$ and $3^{\frac{p-1}{2}}\mid b$ which gives $$3^{\frac{p-1}{2}}\mid (a+b)=n$$ Take $p>3$ then we get $v_3(n)\geq 2$ but by construction $v_3(n)=1$. So for $p>3$, $n$, as constructed, satisfies the conditions.
Case 3 ($q=2$)
then we get $2^{p-1}\mid a^2+b^2$ and also by construction $2^{p-1}\mid n^2=(a+b)^2=a^2+b^2+2ab$. This implies $2^{p-2}\mid ab$. Then write $a=2^r\alpha$ and $b=2^s\beta$ where $r,s$ are both odd. Then $p-1=v_2(n)=v_2(a+b)=\mathrm{min}(r,s)$. Therefore $r\geq p-1$ and $s\geq p-1$. This implies $v_2(ab)=r+s\geq 2(p-1)$. Or $v_2(2ab)\geq 2p-1$. On the other hand $v_2(n^2)=2p-2$. So $v_2(a^2+b^2)=v_2(n^2-2ab)=\mathrm{min}(v_2(2ab),v_2(n^2))=2p-2$. Now try to prove why this will lead you to a contradiction!
Remark:
For establishing that there are infinitely many $n$ for a given $k$ we can consider numbers of the form $$2^{p-1}p_{m+2}p_{m+3}\cdots p_{m+k}$$ for $m\geq0$ and suitable primes $p$. The proof will be analoguous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $a_n=100a_{n-1}+134$, find least value of n for which $a_n$ is divisible by $99$
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $n$ for which $a_{n}$ is divisible by $99 ?$
We have to find. $a_n \equiv 0\pmod{99}$
$$ a_n=100a_{n-1}+134 \\ \implies a_n-a_{n-1}\equiv 35 \pmod{99}$$
Now how do I proceed from here?
I Concluded some results, verified for smaller values, which proved to be wrong. How does doing this get me the $n$? Or am I even correct to proceed like this?
The (unofficial) solution isn't very good either:
$a_{3}=253534 \\
a_{4}=25353534 \\
\therefore a_{n}=2 \underbrace{535353 \ldots 53}_{(n-1) \text { Times } 53}4$
Now, $a_n \rightarrow$ divisible by $99 \Rightarrow$ by $\ 9 \ \& \ 11$ both.
Sum of digits $=6+8(n-1)$
To be divisible by 9
$\mathrm{n}=7,16,25,34,43,52,61,70,79,88, \ldots$
$a_{7}=2\underbrace{535353535353 }_{6 \text { Times } 53} 4$
But $a_{7} \rightarrow$ Not divisible by 11 .
$a_{16}=2\underbrace{5353535353 \ldots \ldots 53 }_{15 \text { Times } 53}4$
Similarly, $a_{16} \rightarrow$ Not divisible by 11 .
Now, $n=88$
$a_{88}=2 \underbrace{5353 \ldots \ldots 53}_{87 \text { Times } 53}4$
Divisibility by $11 \rightarrow|(2+3+3 \ldots \ldots)-(5+5+\ldots+ 4)|$
$$
\begin{array}{l}
=|263-439| \\
=176
\end{array}
$$
$\therefore$ Least $n=88$
Hints are more appreciated than the solution.
| Your method can work quite well. For all $1 \lt i \le n$, note $a_i - a_{i-1}\equiv 35 \pmod{99}$ means each $a_i$ is congruent to $35$ more than the previous one of $a_{i-1}$. Thus, starting from $a_1$ and repeating this $n - 1$ times, you get
$$a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$$
Since $99 = 9(11)$, you can split \eqref{eq1A} into
$$35n - 11 \equiv 0 \pmod{9} \implies 8n - 2 \equiv 0 \pmod{9} \implies 4n \equiv 1 \pmod{9} \tag{2}\label{eq2A}$$
$$35n - 11 \equiv 0 \pmod{11} \implies 2n \equiv 0 \pmod{11} \implies n \equiv 0 \pmod{11} \tag{3}\label{eq3A}$$
Note \eqref{eq3A} means $n = 11k, \; k \in \mathbb{Z}$. You can thus use this to determine $k$ from \eqref{eq2A}.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{2} (x-1) x + y$ is a bijection. (on p.45 Munkres Topology 2nd Edition) I am reading "Topology 2nd Edition" by James R. Munkres.
On p.45, Munkres leaves it to the readers to show that $g$ is bijection:
Show that $g(x, y) = \frac{1}{2} (x-1) x + y$ is a bijection from $\{(x, y) \in \mathbb{Z}_{+} \times \mathbb{Z}_{+} \mid y \leq x\}$ to $\mathbb{Z}_{+}$.
I proved the above fact, but I am not sure my proof is right or not.
And if my proof is right, please give me a better proof.
My proof:
$g$ is injective:
Let $(x, y), (x^{'}, y^{'}) \in \mathbb{Z}_{+} \times \mathbb{Z}_{+}$ and $y \leq x$ and $y^{'} \leq x^{'}$.
Let $(x, y) \neq (x^{'}, y^{'})$.
If $x \neq x^{'}$, then $x < x^{'}$ or $x > x^{'}$. Without loss of generality, we can assume $x < x^{'}$.
Then, $x+1 \leq x^{'}$, because if $x+1 > x^{'}$, then $0 < x^{'} - x < 1$ and $x^{'} - x \in \mathbb{Z}_{+}$.
But there is no element $x \in \mathbb{Z}_{+}$ such that $0 < x < 1$.
$$\frac{1}{2}(x-1)x+y \leq \frac{1}{2}(x-1)x+x =\frac{1}{2}x(x+1)\leq \frac{1}{2} (x^{'}-1)x^{'}<\frac{1}{2} (x^{'}-1)x^{'}+y^{'}.$$
If $x = x^{'}$ and $y \neq y^{'}$, then $y < y^{'}$ or $y > y^{'}$. Without loss of generality, we can assume $y < y^{'}$.
$\frac{1}{2}(x-1)x+y = \frac{1}{2}(x^{'}-1)x^{'}+y < \frac{1}{2}(x^{'}-1)x^{'}+y^{'}$.
So, $g$ is injective.
$g$ is surjective:
We prove by induction.
$1 = \frac{1}{2} (1 - 1) 1 + 1$ and $1 \leq 1$.
Assume that $n = \frac{1}{2} (x - 1) x + y$ and $y \leq x$.
If $y < x$, then $n+1 = \frac{1}{2} (x - 1) x + (y+1)$ and $y+1 \leq x$.
If $y = x$, then $n+1 = \frac{1}{2} (x - 1) x + y+1 = \frac{1}{2} (x - 1) x + x+1 = \frac{1}{2} x (x + 1) + 1$ and $1 < x+1$.
So, $g$ is surjective.
| Your proofs are fine.
Here is a quicker proof of injectivity. Suppose $T_{x-1}+y=T_{x'-1}+y'$ for $y\le x$ and $y'\le x'$ where $T_x=x(x+1)/2$ is the $x$th triangular number. Without loss of generality let $x<x'$ and thus $$T_{x-1}+y\ge T_x+1\implies y\ge(T_x-T_{x-1})+1=x+1$$ which is a contradiction. Hence $x=x'$ from which it follows that $y=y'$.
A quicker proof of surjectivity is that any positive integer can be written as a triangular number $T_{x-1}$ plus the remainder which is at most $x$, since $T_{x-1}+x=T_x$.
| {
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Prove that $\ddot{x}+(2x^2+\dot{x}^2-1)\dot{x}+x=0$ has a periodic solution. Prove that $\ddot{x}+(2x^2+\dot{x}^2-1)\dot{x}+x=0$ has a periodic solution.
Here's my partial solution: Let $\dot{x}=y$. Then $\ddot{x}=\dot{y}=-(2x^2+y^2-1)y-x=-2x^2y-y^3+y-x$.
So we have $\dot{x}=y$ and $\dot{y}=-2x^2y-y^3+y-x$.
Convert to polar coordinates:
$r\dot{r}=x\dot{x}+y\dot{y}=xy+y(-2x^2y-y^3+y-x)=xy-2x^2y^2-y^4+y^2-xy$.
$r\dot{r}=-2r^4cos^2\theta sin^2\theta-r^4sin^4\theta +r^2sin^2\theta$ which reduces to
$\dot{r}=-2r^3cos^2\theta sin^2\theta-r^3sin^4\theta +rsin^2\theta$.
Now I don't know how to proceed to deduce that we have a periodic solution. Any ideas?
| Hint.
$$
\cases{
\dot x_1 = x_2\\
\dot x_2 = -x_1-(2x_1^2+x_2^2-1)x_2
}
$$
or
$$
\cases{
\dot x_1 x_1= x_1x_2\\
\dot x_2 x_2= -x_1x_2-(2x_1^2+x_2^2-1)x_2^2
}
$$
adding the equations
$$
\frac 12(x_1^2+x_2^2)' = -(2x_1^2+x_2^2-1)x_2^2
$$
| {
"language": "en",
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If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$.
So far I've got a minimum of $\sqrt {3}$. Can anyone confirm this? However, I've been having trouble actually proofing that this is the lower bound. Typically, I've solved problems where I need to prove an inequality as true, but this problem is a bit different asking for the minimum of an inequality instead, and I'm not sure how to show that $\sqrt {3}$ is the lower bound of it. Any ideas?
| For $a=b=c=\frac{1}{\sqrt3}$ we obtain a value $\sqrt3$.
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{ab}{c}\geq\sqrt{3(a^2+b^2+c^2)}$$ or
$$\sum_{cyc}a^2b^2\geq\sqrt{3a^2b^2c^2(a^2+b^2+c^2)}$$ or
$$\sum_{cyc}(a^4b^4-a^4b^2c^2)\geq0$$ or
$$\sum_{cyc}c^4(a^2-b^2)^2\geq0$$ and we are done!
| {
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Calculation of $\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}dx=$
$$\int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}\,dx=\text{?}$$
My work :
$$\int\frac{e^{-\sin^2(x)} \tan^3(x)}{\cos(x)} \, dx = \left\{\left(\int e^{-\sin^2(x)}\tan^3(x)\,dx\right)\frac{1}{\cos(x)}\right\}-\int\frac{\sin(x)}{\cos^2(x)}\left(\int e^{-\sin^2(x)}\tan^3(x) \, dx\right)$$
where
$$\int e^{-\sin^2(x)} \tan^3(x)\,dx=\frac{e^{-\sin^2(x)}}{2\cos^2(x)}+c$$
Proof
\begin{align*}
\int e^{-\sin^2(x)}\tan^3(x) \,dx &=\int \frac{e^{-\sin^2(x)}(-\sin^2(x))(-2\sin(x)\cos(x)}{2\cos^4(x)}\,dx\\
&=\int \frac{e^{-\sin^2(x)}(-\sin^2(x))(\cos^2(x))^{'}}{2\cos^4(x)}dx\\
&=\int \frac{e^{\cos^2(x)-1}(\cos^2(x)-1)(\cos^2(x))^{'}}{2\cos^4(x)} \, dx\\
&=\frac{1}{2}\int e^{t-1}\frac{t-1}{t^2} \, dt & (t=\cos^2(x))\\
&=\frac{1}{2}\int e^{t-1}(\frac{1}{t}-\frac{1}{t^2}) \, dt\\
&=\frac{1}{2} \left(\int \frac{e^{t-1}}{t}dt+\frac{e^{t-1}}{t}-\int \frac{e^{t-1}}{t} \, dt\right)\\
&=\frac{1}{2}\frac{e^{t-1}}{t}\\
&=\frac{1}{2}\frac{e^{\cos^2(x)-1}}{\cos^2(x)}+c
\end{align*}
So:
$$\int \frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}dx=\frac{1}{2} \frac{e^{\cos^2(x)-1}}{\cos^3(x)}-\frac{1}{2} \int \frac{e^{\cos^2(x)-1}\sin(x)}{\cos^4(x)} \, dx=\frac{1}{2} \frac{e^{\cos^2(x)-1}\sin(x)}{\cos^4(x)}-\frac{1}{2}J$$
where:
$$J=\int\frac{e^{\cos^2(x)-1}\sin(x)}{\cos^4(x)} \, dx$$
But I have not found a way to calculate $J$ after so many attempts
Any help please and thanks in advance
| You may try this.
Let $t=\sin^2(x)$, then $$ \int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}\,dx=\int2 \dfrac{e^{-t}t^2 }{(1-t)^2}dt,$$
Let $1-t=-u$, then $$ \int\frac{e^{-\sin^2(x)}\tan^3(x)}{\cos(x)}= \int \dfrac{2e^{1-u}(1+u)^2}{u^2}du.$$
NB: Think about $ \Gamma$ -function.
| {
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To prove $a+b+c \ge ab+bc+ca$ when $abc=1$ The actual question is not in the heading, it's this-
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$$ {a \over \sqrt{7+b+c}} + {b \over \sqrt{7+c+a}} + {c \over \sqrt{7+a+b}} \geq 1$$
To complete a proof of this inequality after a deft application of Hölder's Inequality, it's enough to prove that
$$\tag{1} (a+b+c)^3\geq7(a+b+c)+2(ab+bc+ca)$$
and it turns to
$$ \tag{2} (a+b+c)^3\geq7(a+b+c)+{2\over 3}(a+b+c)^2 $$ as $ a+b+c\geq3 $.
According to my book, this solution is provided.
But I am not able to understand how ($1$) $\Rightarrow$ ($2$). Can someone explain me?
For this to happen, $a+b+c\geq ab+bc+ca$.
Thanks!
| We need to prove
$$\tag{1} (a+b+c)^3\geq7(a+b+c)+2(ab+bc+ca).$$
But from know inequality $3(ab+bc+ca) \leqslant (a+b+c)^2.$ Therefore the proof is completed if
$$\tag{2} (a+b+c)^3\geq7(a+b+c)+\frac23(a+b+c)^2.$$
Let $x = a+b+c \geqslant 3\sqrt[3]{abc} = 3,$ ienquality $(2)$ become
$$x^3 \geqslant 7x+\frac{2x^2}{3},$$
or
$$\frac{x(3x+7)(x-3)}{3} \geqslant 0.$$
Which is true.
Note. Must be $(2) \Rightarrow (1) $
| {
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Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2 for which there exists $a,b,c$ such that $\left (2^m\right )^2=4^m=a^2+b^2+c^2$. Since $4^m$ is divisible by 4, $a^2+b^2+c^2$ has to be divisible by 4 too. This is only possible if $a^2\equiv b^2\equiv c^2\equiv 0\pmod 4$, so we can write $a=2a',b=2b',c=2c'$ with $a',b',c'\in \mathbb{N}$. But then we get $\left (2^{m-1}\right )^2=4^{m-1}=a'^2+b'^2+c'^2$, so $m=1$, otherwise $2^m$ wouldn't be the smallest power of two with this property. It is easy to check that $n=2$ doesn't work, so for $n=2^m$ the statement doesn't hold.
*I suspect (but can't prove) that for all other values the statement holds. It would be enough to prove that for all odd primes $p$ there exists $a,b,c$ such that $p^2=a^2+b^2+c^2$, since for all other values of $n$ there exist some $p,m$ such that $n=pm$. Then we get $n^2=(pm)^2=(ma)^2+(mb)^2+(mc)^2$.
| Some Pythagorean triples:
$3^2+4^2=5^2$
$5^2+12^2=13^2$
So: $3^2+4^2+12^2=13^2$
Generalize that:
$(3t)^2+(4t)^2+(12t)^2=(13t)^2$
$n=13t$ , $t> 0 $
| {
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How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$ I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$ doesn't work.
|
For the sake of a different method:
$$
\begin{cases}
P(x)-P(x-1)=x^2 \\
P'(x)-P'(x-1)=2x \\
P''(x)-P''(x-1)=2
\end{cases}
$$
or writing derivatives explicitly:
$$
\begin{cases}
a\big(x^3-(x-1)^3\big)+b\big(x^2-(x-1)^2\big)+c\big(x-(x-1)\big)=x^2~~~~~~~(1) \\
3a\big(x^2-(x-1)^2\big)-2b\big(x-(x-1)\big)=2x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2) \\
6a\big(x-(x-1)\big)=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (3)
\end{cases}
$$
Beginning from the end, it's very easy to solve for $a,b,c$:
*
*First, find $\boxed{a=\frac{1}{3}}$ from the $(3)$.
*Second, plug $a$ into $(2)$ and find $\boxed{b=\frac{1}{2}}$.
*Third, plug $a$ and $b$ into $(1)$ and find $\boxed{c=\frac{1}{6}}$
Each step yields a trivial linear equation.
| {
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Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$ Prove:
$$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$
First obvious sub is $t = 1 + x^{3}$:
$$
\frac{1}{3}\int_{1}^{9}{\left(\,{t - 1}\,\right)}^{-2/3}\, t^{-1/2}\, \mathrm{d}t
$$
From here I tried many things like $\frac{1}{t}$, $t-1$, and more. The trickiest part is the bounds! Reversing it from the answer the integral should be like
$$
\frac{1}{6}\int_{0}^{1}
x^{-2/3}\left(\,{1 - x}\,\right)^{-5/6}\,\mathrm{d}t
$$
I'm not sure where the $1/2$ comes from and the $0$ to $1$ bounds. Any idea or tip please ?.
| An elementary solution: Consider the substitution
$$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$
$t$ is monotonic decreasing on $0<x<2$, and $$\tag{1}\frac{{dx}}{{\sqrt {1 + {x^3}} }} = -\frac{{dt}}{{3\sqrt {1 + {t^3}} }}$$
this can be verified by explictly computing $(dt/dx)^2$ and compare it to $9(1+t^3)/(1+x^3)$. When $x=2, t=-1$, so
$$\int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{1}{3}\int_{ - 1}^\infty {\frac{1}{{\sqrt {1 + {t^3}} }}dt} $$
I believe you now have no difficulty to solve last integral via Beta function.
A conceptual solution: Consider the elliptic curve $E:y^2=x^3+1$, $P=(2,3),Q=(0,1)$ on $E$, $\omega = dx/y$ is the invariant differential on $E$. For the multiplication-by-$3$ isogeny $\phi:E\to E$, we have $3P=(-1,0), 3Q=O$. So $3\int_0^2 \omega \cong \int_{-1}^\infty \omega$ up to an element of $H_1(E,\mathbb{Z})$.
$t$ given above is the $x$-component of $\phi$ and $(1)$ is equivalent to $\phi^\ast \omega = 3\omega$.
The $P$ above is $6$-torsion, if we consider $4$ or $5$-torsion instead, we obtain
results like
$$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732 $$
$$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \times 14$
*$4 \times 14$
I think we have
$$\det \left[\begin{array}{ll}6 & 2 \\ 2 & 6\end{array}\right] = 4 \times 8$$
$$\det \left[\begin{array}{lll}6 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 6\end{array}\right] = 4^{2} \times10$$
and for any $n$ we have $\det(A_n)= 4^{n-1} \times (6+2×(n-1))$ so "1" is true.
| As, dimension of nullspace of $(A-4I)$ is $4$. So, Geometric multiplicity of eigenvalue $4$ is $4$, as matrix $A$ is symmetric, so, $A$ must be diagonalizable, and hence, Algebraic multiplicity and Geometric multiplicity of eigenvalue $4$ is same, so, Algebraic multiplicity of eigenvalue $4$ is $4$.
And, as each row sum is $14$, so, $14$ is another eigenvalue of algebraic multiplicity $1$.
Det($A$)=multiplication of eigenvalues= $4^4×14$
| {
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On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
$$ \Rightarrow 2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq (b+c-2a)^3 $$
But couldn't take it any further, but it looks as if Hölder's Inequality may help. And also I don't think that it is a "warm-up" problem.
Any help will be appreciated.
| If $b+c<2a$ then $$LHS \geq 0 \ge RHS.$$
If $b+c \geq 2a.$ We write inequality as
$$4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3,$$
or
$$4(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \geq (b+c-2a)^3.$$
Because $a+b+c = 3a+(b+c-2a) \geq b+c-2a,$ so we will show that
$$4(a^2+b^2+c^2-ab-bc-ca)\geq (b+c-2a)^2,$$
equivalent to
$$3(b-c)^2 \geq 0.$$
Which is true. Equality holds when $ a=b=c$ or $ a=0,\,b=c.$
Note. We have
$$a^3+b^3+c^3-3abc-{2\left(\frac{b+c}{2}-a\right)^3}$$
$$=(b-c)^2+\frac{3\,a}{2} \Big[(a-b)^2+(b-c)^2+(c-a)^2\Big] \geq 0.$$
| {
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Let $A, B$ be skew-symmetric matrices such that $AB = -BA$. Show that $AB = 0$
Let $A, B$ be skew-symmetric matrices such that $AB = -BA$. Show that $AB = 0$.
I know that $AB$ is skew-symmetric,because $$(AB)^t=B^tA^t=BA=-AB$$
but I don't know how show that $AB=0$.
| Let
$$ A = \begin{pmatrix}
0 & 0 & 0 & -1 \\
0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}
\quad\text{and}\quad
B = \begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}.$$
Then
$$ AB = -BA = \begin{pmatrix}
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix} $$
but this is obviously non-zero.
Remark. This example comes from the matrix representation of the quaternion, see the Wikipedia article for instance.
Addendum. In general, let $F$ be a field whose characteristic is not equal to 2, and let $A$ and $B$ be $n\times n$ skew-symmetric matrices over $F$ such that $AB=-BA$. Then we claim the following:
*
*If $n \leq 3$, then either $A = \mathbf{0}$ or $B = \mathbf{0}$.
*If $n \geq 4$, it is possible to have $AB \neq \mathbf{0}$.
When $n \geq 4$, the assertion is obvious by the first part of this answer. Also, the cases of $n=1, 2$ are easy to tackle by a direct computation. So we move on to the case of $n = 3$.
Suppose that $n = 3$. Then $A$ and $B$ takes the form
$$ A = \begin{pmatrix} 0 & a_1 & a_2 \\ -a_1 & 0 & a_3 \\ -a_2 & -a_3 & 0 \end{pmatrix} \quad\text{and}\quad B = \begin{pmatrix} 0 & b_1 & b_2 \\ -b_1 & 0 & b_3 \\ -b_2 & -b_3 & 0 \end{pmatrix}. $$
Then a direct computation tells that $AB = -BA$ is equivalent to
$$ a_i b_j + a_j b_i = 0, \qquad \forall i, j \in \{1, 2, 3 \}. \tag{*} $$
We claim that this implies either $A=\mathbf{0}$ or $B=\mathbf{0}$, hence the conclusion still holds. To this end, we assume $A \neq \mathbf{0}$ without losing the generality. Then there exists $i_0 \in \{1,2,3\}$ such that $a_{i_0} \neq 0$. Then by $\text{(*)}$ with $i = j = i_0$, we have $b_{i_0} = 0$. Now plugging $i = i_0$ to $\text{(*)}$ shows that $b_j = 0$ for any $j$, and therefore $B=\mathbf{0}$ as desired.
| {
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Find the number of points of non differentiability in $(x^2-3x+2)(|x^3-6x^2+11x-6|)$ The expression is factorized as $$(x-2)(x-1)(|(x-1)(x-2)(x-3)|)$$
I expected the points to be $1,2,3$, put looking at the graph, it’s only $x=3$
I kinda figured out the reason for this, because the expression would end up simplifying to
$\pm (x-2)^2(x-1) ^2|x-3|$ but I am not finding this very convincing. What can be a concrete reason for this?
| $$F(x)=(x-1)(x-2)|(x-1)(x-2)(x-3)|.$$ Only $x=3$ is the point of non-differentiability. At $x=1,2$ is it differentiable. At these points, we have $(x-a)|x-a|$ which is differentiable at $x=a$.
Note that $g(x)=(x-a)|x-a|= (x-a)(a-x), ~if~ x<a$ and $g(x)=(x-a)^2, ~if ~x\ge a$, hence both the left and right derivatives are equal and zero at $x=a$.
| {
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
| WLOG let $a \geq b \geq c$ , then $\sqrt[3]a \geq \sqrt[3]b \geq \sqrt[3]c$. Let $x = \sqrt[3]a, y = \sqrt[3]b , z =\sqrt[3]c$ , then $x \geq y\geq z > 0$ and the inequality is equivalent to :
$$
x^{3n+3} + y^{3n+3} + z^{3n+3} \geq x^{3n+1}yz + y^{3n+1}xz + z^{3n+1}xy
$$
by Muirhead's inequality, since the sequence $(3n+3,0,0)$ majorizes $(3n+1,1,1)$, we know that
$$
\sum_{\mathrm{sym}} x^{3n+3}y^0z^0 \geq \sum_{\mathrm{sym}} x^{3n+1}y^1z^1
$$
which upon division by $2$ gives the result we require. Note that $n$ natural is required for the inequality to hold.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find the complex integral: $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$. Problem with integral formula.... Question I am trying to find the complex integral: $\displaystyle\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$.
My Attempt (and eventual question): $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz=\frac{1}{2}\int_\infty^\infty\frac{z^6}{(z^4+1)^2}dz$. Now, the singularities of $z^4+1$ are of the form $z_k=e^{\frac{i(\pi+2\pi k)}{4}}$, where $k=0,1,2,3$. Drawing the contour in the upper half plane, we see tht the only two singularities in our contour are $z_0=e^{\frac{i\pi}{4}}$ and $z_1=e^{\frac{i3\pi}{4}}$.
Let $f(z)=\frac{g(z)}{h(z)}$ where $g(z)=z^6$ and $h(z)=(z^4+1)^2$. Then, $h'(z)=8z^3(z^4+1)$. So, the value of the integral is $2\pi i\frac{1}{2}\sum_{k=0}^1\frac{z_k^6}{8z_k^3(z_k^4+1)}=\frac{\pi i}{8}\Big(\frac{e^{i\frac{3\pi}{2}}}{e^{i\frac{3\pi}{4}}(e^{i\pi}+1)}+\frac{e^{i\frac{\pi}{2}}}{e^{i\frac{\pi}{4}}(e^{i\pi}+1)}\Big)$. But, $e^{i\pi}+1=0$, so I must have done something wrong....? Any insight would be great! Thank you.
| What do you mean by "complex integral"? This is a real integral. Okay, you can evaluate it using residues.
Consider $C$, the closed curve consisting of the line segment from $-R$ to $R$ union with semicircle of radius $R$ in the upper half plane
$$\oint_C \frac{z^6}{(z^4+1)^2}\, dz =
\int_0^\pi \frac{R^6 e^{6i\theta} R i e^{i\theta} \, d\theta}{(1+ R^4 e^{4i\theta})^2} + \int_{-R}^R \frac{x^6dx}{(1+x^4)^2}$$
$$\left| \int_0^\pi \frac{R^6 e^{6i\theta} R i e^{i\theta} \, d\theta}{(1+ R^4 e^{4i\theta})^2}\right| < \int_0^{\pi} \frac{R^7 \, d\theta}{(R^4-1)^2}<\frac{\pi}{R-2} $$
The integral along the semicircle goes to zero as $R\to\infty$ and we are left with
$$\int_{-\infty}^\infty \frac{x^6dx}{(1+x^4)^2} = 2\pi i \left[ \text{Res}_{z=e^\frac{i \pi}{4}} \frac{z^6}{(z^4+1)^2} + \text{Res}_{z=e^\frac{i 3\pi}{4}} \frac{z^6}{(z^4+1)^2} \right] = \frac{3\pi}{4\sqrt{2}} $$
$$\int_0^\infty \frac{x^6}{(1+x^4)^2}= \frac{3\pi}{8\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $a>1$, show that $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$, $x \geq 1$ I'm self-learning the analysis I "by Herbert Amann" and got stuck in this problem. It's in Chapter IV Taylor's theorem.
For $a>1$ and $x\geq 1$ show that $$\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$$
This is what I've tried:
Let $f(t)=\frac{1}{1+tx}$ which is convex and,
$f(a) \geq f(1)+f'(1)(a-1)$
so $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{x}{(1+x)^2}(a-1) \leq \frac{a-1}{4}$
| Let
$$f : x \mapsto \frac{1}{1+x} - \frac{1}{1+ax}$$
$f$ is differentiable on $[1, +\infty)$ and
$$f'(x)= -\frac{1}{(1+x)^2}+\frac{a}{(1+ax)^2}$$
so $$f'(x)\geq 0 \Leftrightarrow \frac{a}{(1+ax)^2} \geq \frac{1}{(1+x)^2}$$
$$\Leftrightarrow \sqrt{a}(1+x) \geq 1+ax \Leftrightarrow \frac{1-\sqrt{a}}{\sqrt{a}-a} \geq x$$
So the maximum of $f$ is
$$f\left( \frac{1-\sqrt{a}}{\sqrt{a}-a}\right) = \frac{1}{1+\frac{1-\sqrt{a}}{\sqrt{a}-a}} - \frac{1}{1+a\frac{1-\sqrt{a}}{\sqrt{a}-a}} = \frac{\sqrt{a}-a}{1-a} - \frac{1-\sqrt{a}}{1-a}$$ $$= \frac{a+1-2\sqrt{a}}{a-1}= \frac{(\sqrt{a}-1)(\sqrt{a}-1)}{(\sqrt{a}+1)(\sqrt{a}-1)}=\frac{\sqrt{a}-1}{\sqrt{a}+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)^2$, which expands to $(x^2+y^2+z^2) + 2\cdot (xy+yz+xz)$, but I couldn't go anywhere from there (other than substituting xy, yz, xz, but that's not enough information.
| HINT. You can take the equations pairwise. For example, you have $xy=1$ and $yz=2$. So for example,
$$
\dfrac{1}{2}= \dfrac{xy}{yz}=\dfrac{x}{z}
$$
But then $z=2x$. But you know $xz=3$. Can you substitute and find $x$ or $z$? Can you repeat this process for $y$? Knowing their values it should be routine to find $x^2+y^2+z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a closed form for $\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}?$ I found
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx=I\tag1.$$
Mathematica failed to find $I$, so I am not sure if there is closed form for it. I am just giving it a try here.
First idea came to my mind is to use the Fourier series of $-\ln(\cos x)=\ln(2)+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$ and we have
$$I=8\ln(2)\underbrace{\int_0^{\pi/2}x^2\cot x\ dx}_{\frac32\ln(2)\zeta(2)-\frac78\zeta(3)}+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x^2 \cot x\cos(2nx)\ dx.$$
I got stuck here. Any help would be much appreciated.
Proof of $(1)$
from here we have
$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$
replace $x$ by $\sqrt{x}$ we get
$$\sum_{n=1}^\infty\frac{2^{2n}x^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$
multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=2\int_0^1\frac{\arcsin^2(\sqrt{x})\ln(1-x)}{x}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx$$
| $$S=-8 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^4}{90}-\frac{1}{3} \log ^4(2)+\frac{4}{3} \pi ^2 \log ^2(2)$$
Proof $1$. This. Proof $2$. This. Proof $3$. This. Bonus: $$\small \int_0^{\frac{\pi }{2}} x^3 \cot (x) \log (\cos (x)) \, dx=\frac{3}{2} \pi \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{16} \pi \zeta (3) \log (2)-\frac{\pi ^5}{120}+\frac{1}{16} \pi \log ^4(2)-\frac{1}{8} \pi ^3 \log ^2(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding Equilibrium Points of Dynamical System I am wondering how to find the equilibrium points from these $2$ equations
\begin{align}
\frac{\mathrm{d}x}{\mathrm{d}t} & =
0.4x\left(1 - \frac{x}{1000} + \frac{y}{1000}\right)
\\[2mm]
\frac{\mathrm{d}y}{\mathrm{d}t} & =
0.1y\left(1 + \frac{x}{1000} - \frac{y}{500}\right)
\end{align}
*
*From the $2$nd equation, the $y$ value I got is $y = 0$ and
$x = -1,000$
*and from the $1$st equation I got $x = 0$ and
$y = -1,000$
So I am wondering is it use the $1$st equation to get the
$\left(x,y\right)$ value or use the $2$nd equation to get the $\left(x,y\right)$ value ?
| We want to simultaneously find the zeros of
$$0.4x\left(1-\frac{x}{1000}+\frac{y}{1000}\right) = 0\\ 0.1y\left(1+\frac{x}{1000}-\frac{y}{500}\right) = 0$$
We can see that each equation is of the form $a b = 0$, so each of those being zero, will satisfy the equation and we just need for them to do that simultaneously for both equations.
We can set $x = 0$ in the first equation and substitute that in the second equation
$$0.1y\left(1+\frac{0}{1000}-\frac{y}{500}\right) = 0$$
From this we get two $y$ values of
$$y = 0, 500$$
We can set $y = 0$ in the second equation and substitute that in the first equation
$$0.1x\left(1-\frac{x}{1000}+\frac{0}{1000}\right) = 0$$
From this we get two $x$ values of
$$x = 0, 1000$$
So that takes care of the $x$ and $y$ terms by themselves. Now, ignoring that term, we have
$$0.4\left(1-\frac{x}{1000}+\frac{y}{1000}\right) = 0\\ 0.1\left(1+\frac{x}{1000}-\frac{y}{500}\right) = 0$$
Solving the first for $y$
$$y = x - 1000$$
Note: We could have solved for $x$ or also used the 2nd equation - try it.
Substituting that into the second
$$\frac{1}{10} \left(-\frac{1}{500} (x-1000)+\frac{x}{1000}+1\right)=0$$
Solving this gives $x = 3000$, which means $y = 2000$.
So, our solution set is
$$(x, y) = (0,0), (0,500), (1000, 0), (3000, 2000)$$
We can use a contour plot to see these points of intersection
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
| By binomial approximation
$$(x^2+1)^{\frac{3}{2}}=(x^2)^{\frac{3}{2}}\left(1+\frac1{x^2}\right)^{\frac{3}{2}} = x^3+\frac32 x +O\left(\frac1{x}\right)\implies \frac{2(x^2+1)^{\frac{3}{2}}}{3} = \frac{2x^{3}}{3}+x+O\left(\frac1{x}\right)$$
therefore
$$x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=O\left(\frac1{x}\right)\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
If $ \bigtriangleup ABC$: $\angle CAB = \frac{\pi}{2}$, with height $AD$ and median $AK$. Prove $\angle BAD = \angle BCA = \angle KAC.$
If $\triangle ABC$ is a triangle and $\angle CAB = \frac{\pi}{2}$, with height $AD$ and median $AK$; suppose that $D$ is between $B$ and $K$.
*
*Prove that $\angle BAD = \angle BCA = \angle KAC$.
*Then, prove that $\angle BCA= \frac\pi 8$ if $|AD|=|DK|$.
*Conclude that
$$\sin\frac \pi 8=\frac{\sqrt{2-\sqrt 2}}{2};\quad \cos\frac \pi 8=\frac{\sqrt{2+\sqrt 2}}{2};\quad \operatorname{tg}\frac \pi 8= \sqrt 2 - 1$$
I already achieved the draw, but I don't really know how to start.
I know that $\bigtriangleup DBA \sim \bigtriangleup DAC$
|
Consider the circumcircle of $\triangle ABC$. Since $\angle A=\frac{\pi}{2}$, it subtends the diameter, thus $K$ is the circumcenter and $$KA=KB=KC\tag{1}$$
*
*Since $\triangle KCA$ is isosceles, $\angle KCA=\angle KAC$.
In $\triangle ABD$ $\ \ \angle D=\frac{\pi}{2}$, thus $\angle BAD=\frac{\pi}{2}-\angle ABD$, but $\frac{\pi}{2}-\angle ABC=\angle ACB$, thus $\angle BAD=\angle ACB=\angle KAC$, QED.
*In $\triangle ADK$ $\ \ \angle D=\frac{\pi}{2}$, thus $|AD|=|DK|$ $\Rightarrow$ $\angle A=\angle K=\frac{\pi-\angle D}{2}=\frac{\pi}{4}$.
Since $\frac{\pi}{4}=\angle AKD=\angle KAC+\angle KCA$ and $\angle KAC=\angle KCA$, thus $\angle ACK=\frac{\pi}{8}$, QED.
*In $\triangle ADC$ $\ \ \angle D=\frac{\pi}{2}$ and $AK=KC=AD\sqrt{2}$ thus $$\tan \frac{\pi}{8}=\frac{AD}{DK+KC}=\frac{AD}{AD+AD\sqrt{2}}=
\frac{1}{1+\sqrt{2}}=\sqrt{2}-1,$$
the other functions of $\frac{\pi}{8}$ are done by using
$$\frac{1}{\cos^2\theta}=1+\operatorname{tg}^2\theta,\quad
\frac{1}{\sin^2\theta}=1+\operatorname{ctg}^2\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all 3 number solutions for $x[(x-2)^2+1]=6$
Find all 3 number solutions for $x[(x-2)^2+1]=6$
I used trial and error method to find integer solutions for $x$, and found that 1 possible solution is $x=3$.
However, there are 2 other non-integer solutions and I do not know how to find them.
I appreciate any help, thanks.
| Looking for integer solutions, the equation $x[(x-2)^2+1]=6$ is equivalent to
$$\begin{cases}x=2,\\(x-2)^2+1=3, \end{cases}\qquad\text{or}\qquad\begin{cases}x=3,\\(x-2)^2+1=2. \end{cases}$$
The second equation in the first system implies that $(x-2)^2\equiv -1\mod 3$. Unfortunately, the only squares mos. $3$ are $0$ and $1$, so this first system has no solution.
The second equation in the second system means $(x-2)^2=1$, i.e. $x-2=\pm 1\iff x=3\;\text{ or }\;x=1 $. Only $x=3$ is compatible with the first equation.
Therefore there is a single integer solution. For the other solutions, we can expand the l.h.s. to obtain the cubic equation, divisible by $x-3$:
$$x^3-4x^2+5x-6=0\iff (x-3)(x^2-x+2)=0$$
The quadratic equation $x^2-x+2=0$ has complex conjugate roots:
$$x=\frac{1\pm i\sqrt 7}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where I made a mistake during factoring $x^6+x^5+x^4+x^3+x^2+x+1$? In order to factor the expression, due to symmetry of coefficients if $r_1,r_2,r_3$ are zeros of $x^6+x^5+x^4+x^3+x^2+x+1$ then $\frac{1}{r_1}, \frac{1}{r_2} , \frac{1}{r_3}$ are also zeros. So we can rewrite:
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2-(r_1+\frac{1}{r_1})x+1)+(x^2-(r_2+\frac{1}{r_2})x+1)(x^2-(r_3+\frac{1}{r_3})x+1)$$
And it can be written as:$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)(x^2+cx+1)$$
$$=x^6+(a+b+c)x^5+(3+ab+ac+bc)x^4+(abc+a+b+c)x^3+(3+ab+ac+bc)x^2+(a+b+c)x+1$$
In orther to find$a,b,c$ I should solve the system :
${ \begin{cases}{a+b+c=1} \\ {ab+ac+bc=-2} \\ {abc=0}\end{cases} }$
from the equation $abc=0$ We have $a=0$ (Since our derivation was symmetric in $a,b,c$ it doesn't matter to put $a=0$ or $b=0$ or $c=0$).
${ \begin{cases}{b+c=1} \\ {bc=-2}\end{cases} }$
and from that we have $b=2$ , $c=-1$ , $a=0$
So I get :
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+1)(x^2+2x+1)(x^2-x+1)=(x^2+1)(x+1)^2(x^2-x+1)$$
But my answer is wrong because if we multiply the equation by $x-1$ We should obtain $x^7-1$. But:
$$(x-1)(x^2+1)(x+1)^2(x^2-x+1)=(x-1)(x+1)(x^2+1)(x+1)(x^2-x+1)=(x^4-1)(x^3+1) \ne x^7-1$$
I checked my answer several times but I couldn't find my mistake.
| To factor over $\mathbb{R}$, note that:
$$
\begin{align}
x^6+\cdots+1
&=\frac{1}{x-1}\left(x^7-1\right)
\end{align}
$$
where $x^7-1$ has its roots being the complex seventh roots of unity. One such root is of course $1$, while the other six pair up with their complex conjugate:
$$
\begin{align}
&x^6+\cdots+1\\
&=\frac{1}{x-1}(x-1)\left(x-e^{2\pi i/7}\right)\left(x-e^{-2\pi i/7}\right)\left(x-e^{4\pi i/7}\right)\left(x-e^{-4\pi i/7}\right)\left(x-e^{6\pi i/7}\right)\left(x-e^{-6\pi i/7}\right)\\
&=\left(x^2-2\cos(2\pi/7)x+1\right)\left(x^2-2\cos(4\pi/7)x+1\right)\left(x^2-2\cos(6\pi/7)x+1\right)
\end{align}
$$
This is the full factorization over $\mathbb{R}$.
Any factorization over $\mathbb{Q}$ would have to be composed of polynomials where complex conjugate roots are paired. So a factorization over $\mathbb{Q}$ is either trivial (the original polynomial is irreducible) or with all other factorings under consideration, at least one of those quadratic polynomials would be present, and one of $\cos(2\pi/7)$, $\cos(4\pi/7)$, $\cos(6\pi/7)$ is rational. But the only way that the cosine of a rational multiple of $\pi$ can be a rational number is if said cosine is $0$, $\pm\frac12$, or $\pm1$. (This last fact not trivial, just something I've memorized. A proof is accessible though.) And of course those three cosine values are not in $\{0,\pm\frac12,\pm1\}$. Therefore the original polynomial must be irreducible.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple number theory in one unknown Given $n$ is an integer and $0 \leq n \leq 2000$, how many $n$ are there such that $\lfloor \sqrt {n} \rfloor$ divides $n$ and $\lfloor \sqrt {n+1} \rfloor$ divides $n+1$?
I'm new in number theory and I'm getting trouble assuming $\lfloor \sqrt {n} \rfloor$ as $k$ such as $k$ is also an integer. I really think this is the best method since $\lfloor \sqrt {n} \rfloor$ is the smallest integer in this problem but I still don't get the sense of my following steps. I love to see some original solutions to this so that I can choose the best method that suits to me.
I know it's so simple to you guys.I really appreciate any help and thank you in advance :)
PS. I already know how to. If $m²$ is a perfect square which has a factor of $m$ then the number before it $m² - 1$ has a factor of $m-1.$ This two consecutive numbers both have a factor which are consecutive as well. Problem solve.
| For a positive integer $ n \geq 1 $, let $ d_{n} = gcd(\left \lfloor{\sqrt{n}}\right \rfloor, \left \lfloor{\sqrt{n+1}}\right \rfloor) $. Then since from the hypothesis we know that $ \left \lfloor{\sqrt{n}}\right \rfloor \mid n $ and $ \left \lfloor{\sqrt{n+1}}\right \rfloor \mid n+1 $ we get the following:
$$ d_{n} \mid \left \lfloor{\sqrt{n}}\right \rfloor \mid n $$ $$ d_{n} \mid \left \lfloor{\sqrt{n+1}}\right \rfloor \mid n+1 $$ hence $ d_{n} \mid n $ and $ d_{n} \mid n+1 $ so $ d_{n} \mid (n+1-n) $ thus $ d_{n} \mid 1 $. Therefore either $ \left \lfloor{\sqrt{n}}\right \rfloor = \left \lfloor{\sqrt{n+1}}\right \rfloor = 1 $ or $ \left \lfloor{\sqrt{n}}\right \rfloor \geq 2 $ and $ d_{n} = 1 $.
If $ \left \lfloor{\sqrt{n}}\right \rfloor = \left \lfloor{\sqrt{n+1}}\right \rfloor = 1 $ then clearly $ n \in \{1,2\} $. Now, if $ \left \lfloor{\sqrt{n}}\right \rfloor \geq 2 $ and $ d_{n} = 1 $, what $ d_{n} = 1$ tells us is that $ \left \lfloor{\sqrt{n}}\right \rfloor $ and $ \left \lfloor{\sqrt{n+1}}\right \rfloor $ are relatively prime since by definition $ d_{n} = gcd(\left \lfloor{\sqrt{n}}\right \rfloor, \left \lfloor{\sqrt{n+1}}\right \rfloor) $. But clearly $$ \left \lfloor{\sqrt{n}}\right \rfloor + 1 \geq \left \lfloor{\sqrt{n+1}}\right \rfloor \geq \left \lfloor{\sqrt{n}}\right \rfloor $$ where the RHS inequality is obvious whereas the LHS one follows from the fact that $ \sqrt{n}+1 \geq \sqrt{n+1} $(which can be easily verified by squaring). Therefore we must have that $$ \left \lfloor{\sqrt{n+1}}\right \rfloor = \left \lfloor{\sqrt{n}}\right \rfloor + 1$$.
Let $ k = \left \lfloor{\sqrt{n}}\right \rfloor $. Then $$ k \leq \sqrt{n} < k+1 $$ So $$ k^{2} \leq n < (k+1)^{2} $$ Hence $$ n+1 \leq (k+1)^{2} $$
On the other hand, we have that $ \left \lfloor{\sqrt{n+1}}\right \rfloor = \left \lfloor{\sqrt{n}}\right \rfloor + 1$ hence $ \left \lfloor{\sqrt{n+1}}\right \rfloor = k+1$ so $ k+1 \leq \sqrt{n+1} $ which yields $ (k+1)^{2} \leq n+1 $. Therefore, from this and the above inequality, we must have that $$ n+1 = (k+1)^{2} $$ meaning that $$ n = (k+1)^{2}-1 $$ Let's verify that actually all positive integers of the form $ n=m^{2}-1 $ for $ m \geq 2 $ verify the hypothesis. So let $ n = m^{2}-1 $ for some positive integer $ m \geq 2 $. For $ m \geq 2 $ we have $$ m^{2} > n=m^{2}-1 > (m-1)^{2}$$ So $$ m > \sqrt{n} > m-1 $$ Then we must have that $\left \lfloor{\sqrt{n}}\right \rfloor = m-1 $ which clearly divides $ n = m^{2}-1=(m-1)(m+1) $. Furthermore $ n+1 = m^{2} $ hence $ \sqrt{n+1} = m $ so $ \left \lfloor{\sqrt{n+1}}\right \rfloor = m $ as well which clearly divides $ n+1 = m^{2} $.
Therefore all positive integers $ n $ of the form $ m^{2}-1$ for some positive integer $ m \geq 2 $ satisfy the hypothesis. We also saw that $ n \in \{1,2 \} $ also satisfies the hypothesis. Furthermore, $ 0 $ is also a solution as per the definition of divisibility for natural numbers i.e. we say $ k \mid n $ where $ k,n \in \mathbb{N} $ if there exists $ m \in \mathbb{N} $ such taht $ k \cdot m = n $. In the case of $ n = 0 $, all natural numbers $ k $ divide $ n $ including $ 0 $ itself since for $ n = 0 $ there exists(infinitely many, actually all natural numbers) $ m \in \mathbb{N} $ such that $ 0 = 0 \cdot m $.
To conclude, we have that $ n \in \{0,1,2\} \cup \{ m^{2}-1| m \in \mathbb{N}, m \geq 2 \} $ which answers the question of finding all natural numbers $ n $ that satisfy the hypothesis. If we are to restrict to $ n \leq 2000 $, we must find the largest positive integer $ m $ such that $ m^{2}-1 \leq 2000 $ which can be easily found to be $ 44 $. Therefore the natural numbers less than or equal to $ 2000 $ that satisfy the hypothesis are $$ n \in \{0,1,2,2^{2}-1, 3^{2}-1, 4^{2}-1, \dots , 44^{2}-1 \} = \{0,1,2,3,8,15, \dots 1935 \} $$ hence there are $ 46 $ solutions to the problem.
| {
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Recurrence and modular arithmetic incorrect solution
Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$. (Source: 2017 AIME I)
This is my solution:
We wish to find the least $n$ such that $a_n\equiv 0\pmod{99},$ with the recurrence relation $a_n \equiv a_{n-1} + n \pmod{99}.$ Also, for every $n > 10,$ $a_n = \sum_{k=10}^n k = \frac{10 + n}{2} \cdot (n-9),$ so we wish to find the least $n$ such that $$(10 + n)(n - 9)2^{-1} \equiv 50(10+n)(n - 9) \equiv 0 \pmod{99}.$$ $$50(n^2+n-90) \equiv 50(n^2+n+9) \equiv 50n(n+1)+450 \equiv 0 \\ \Longleftrightarrow 50n(n+1) \equiv 45 \Longleftrightarrow n(n+1) \equiv 45\cdot 50^{-1} \equiv 90 \Longleftrightarrow n(n+1) \equiv 90.$$ Then $n\equiv 9 \pmod{99}$, so the least $n>10$ is $108$.
It seems that $n=108$ actually does work in the sense that $99 \mid a_{108}$, but the actual answer is
45
How should I edit my solution to give the minimum value? I suspect that somewhere along the second line my solution became a little suspect, I'm not sure why it gives the wrong answer.
| You did well through the conclusion $n(n+1)\equiv90\bmod99$, but your solution of that was lacking.
I would solve that congruence using the Chinese remainder theorem;
$99$ is the product of $11$ and $9$, which are relatively prime.
$n(n+1)\equiv0\bmod9$ and $n(n+1)\equiv2\bmod11$.
$n\equiv0 $ or $8\bmod9$ and $n\equiv1$ or $9\bmod11$.
Putting these together, we get $n\equiv45, 9, 89$, or $53\bmod99$.
Do you get it now?
| {
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Solving $\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$ for real $x$
Solve the equation in the Real number system:
$$\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$$
I have attempted using $(A-B)^3 = A^3 - B^3 - 3.A.B.(A-B)$ with $A = \sqrt[3]{x+1}$ , $B = \sqrt[3]{x-1}$ and $(A-B) = \sqrt[3]{x^2-1}$, however I end up getting $-3.(\sqrt[3]{x^2-1})^2 = x^2-3$ which is not equivalent to the original one since 0 is solution only of $-3.(\sqrt[3]{x^2-1})^2 = x^2-3$. Could someone explain to me why does it not work here?
By the way, the answer according to the book is $\left\lbrace \frac{\sqrt 5}{2},-\frac{\sqrt 5}{2} \right\rbrace$ but I could not get there.
| We can use $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
$a^2+b^2+c^2-ab-ac-bc=0$ for $a=b=c$ only.
Since $$\sqrt[3]{x+1}=- \sqrt[3]{x-1} =-\sqrt[3]{x^2-1}$$ gives $x=0$ and $0$ is not a root of the equation, we need to remove the number $0$ and we obtain:
$$x+1-(x-1)-(x^2-1)-3\sqrt[3]{x+1}\cdot\sqrt[3]{x-1}\cdot\sqrt[3]{x^2-1}=0$$ or
$$3-x^2=3\sqrt[3]{(x^2-1)^2}$$ or $$x^2(x^4+18x^2-27)=0,$$ which gives $$\left\{\sqrt{6\sqrt3-9},-\sqrt{6\sqrt3-9}\right\}$$
| {
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Find the argument of $z = {\left( {2 + i} \right)^{3i}}$ $z = {\left( {2 + i} \right)^{3i}}$
My approach is as follow
$z = {\left( {2 + i} \right)^{3i}} = {\left( {{{\left( {2 + i} \right)}^3}} \right)^i} = {\left( {8 + {i^3} + 12i - 6} \right)^i} = {\left( {2 + 11i} \right)^i}$
$\ln z = i\ln \left( {2 + 11i} \right)$
$2 + 11i = r\cos \theta + ir\sin \theta \Rightarrow r = 5\sqrt 5 ;\theta = {\tan ^{ - 1}}\frac{{11}}{2}$
$ \Rightarrow \ln z = i\ln \left( {2 + 11i} \right) = i\ln \left( {r{e^{i\theta }}} \right) = i\left( {\ln \left( r \right) + \ln \left( {{e^{i\theta }}} \right)} \right) = i\left( {\ln 5\sqrt 5 + i\theta } \right) = - \theta + i\left( {\ln 5\sqrt 5 } \right)$
$ \Rightarrow z = {e^{\left( { - \theta + i\left( {\ln 5\sqrt 5 } \right)} \right)}} = {e^{ - \theta }}{e^{i\left( {\ln 5\sqrt 5 } \right)}} \Rightarrow r{e^{i\phi }}$
The modulus is $r = {e^{ - {{\tan }^{ - 1}}\frac{{11}}{2}}}$, Argument $= \phi = \ln 5\sqrt 5 $
How do we convert this argument into angle
| Another possible approach:
Taking principal logarithm both side of $z=(2+i)^{3i}$,
$Log z=3i[\ln \sqrt{2^2+1^2}+i ~tan^{-1}(1/2)]=3i[\ln \sqrt 5+i ~tan^{-1}(1/2)]$.
Thus $Log z=-3 tan^{-}(1/2)+i~3 \ln \sqrt 5$ and so $z=e^{-3tan^{-1}(1/2)} \cdot e^{i3 \ln \sqrt 5}$.
This gives us $z=e^{-3tan^{-1}(1/2)} (cos(3 \ln \sqrt 5)+i ~sin (3 \ln \sqrt 5))$.
Thus, $\text{Arg}(z)=tan^{-1}(tan(3 \ln \sqrt{5}))=3 \ln \sqrt 5$, because $\forall x \in (-\pi/2,~\pi/2), \ tan^{-1}(tan ~x)=x.$
| {
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Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$ Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$
Using AM- HM inequality,
$\left(\sum_{k=1001}^{3001} k\right)\left(\sum_{k=1001}^{3001} \frac{1}{k} \right) \geq(2001)^{2}$
But $\sum_{k=1001}^{3001} k=(2001)^{2}$
Hence, $\sum_{k=1001}^{3001} \frac{1}{k}>1$
How to prove the next part?(Is there any inequality I am missing)
| Since $$\frac{1}{(1002+i)(3000-i)}<\frac{1}{1002\cdot3000}$$ for any $i\in[1,998]$ and $$\frac{1}{1001\cdot3001}+\frac{1}{2001^2}<\frac{2}{1002\cdot3000},$$we obtain:
$$\sum_{k=1001}^{3001}\frac{1}{k}=\sum_{k=1}^{1001}\left(\frac{1}{k+1000}+\frac{1}{3000-k+2}\right)=$$
$$=4002\sum_{k=1}^{1001}\frac{1}{(k+1000)(3000-k+2)}<4002\sum_{k=1}^{1001}\frac{1}{1002\cdot3000}=\frac{4002\cdot1001}{1002\cdot3000}<\frac{4}{3}.$$
| {
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If $a$, $b$, $c$ are the roots of $x^3-6x^2+3x+1=0$, find all possible values of $a^2b+b^2c+c^2a$
Let $a$, $b$, $c$ be the roots of
$$x^3 - 6x^2 + 3x + 1 = 0$$ Find all possible values of $a^2 b + b^2 c + c^2 a$. Express all the possible values, in commas.
I've already tried to bash out all the roots, Vieta's Formula, and try to manipulate the equation to become easier to work with. However, Vieta's didn't get me anywhere, and I couldn't find a way to make the equation simpler or anything. Any hints to start this problem?
| Hint : Let $A=a^2 b + b^2 c + c^2 a$ and $B=a^2 c + b^2 a + c^2 b$. Now calculate $A+B$ and $AB$.
Now consider the quadratic whose roots are $A$ and $B$.
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Experienced mathematicians simplifying messy algebra $$\frac{pa}{n}\left(p\frac{a-1}{N-1}+q\frac{b+1}{N+1}\right)+p\left(1-\frac{a}{N}\right)\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+\frac{qb}{N}\left(p\frac{a+1}{N+1}+q\frac{b-1}{N-1}\right)+q\left(1-\frac{b}{N}\right)\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)=\frac{(p+q)(pa+qb)}{N}$$
The ways people with different experience in math simplify algebra ranges massively. It is something assumed trivial, so any techniques or tricks in doing so are only found by finding them yourself. Any googling will land you at an elementary school homepage.
As a first year, my attempt at the above simplification was to set denominators equal, which made for two pages of tedious workings. I would be very interested in hearing how people with more experience in math than myself would approach this simplification, or other daunting simplifications in general.
To prevent this question being too broad, I also asked my professor what he would do, he suggested that I "factorise groups of terms which had obvious common factors". The meaning of these words independently is obvious, but could somebody explain the better approach he is alluding to, as I can only see approaches equivalent to matching denominators?
| $$\frac{pa}{N}\left(p\frac{a-1}{N-1}+q\frac{b+1}{N+1}\right)+p\left(1-\frac{a}{N}\right)\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+\frac{qb}{N}\left(p\frac{a+1}{N+1}+q\frac{b-1}{N-1}\right)+q\left(1-\frac{b}{N}\right)\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)= \\
\frac{pa}{N}\left(\frac{pa}{N-1} -\frac{p}{N-1}+\frac{qb}{N+1} + \frac{q}{N+1}\right)+\left(p-\frac{pa}{N}\right)\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+\frac{qb}{N}\left(\frac{pa}{N+1}+\frac{p}{N+1}+\frac{qb}{N-1} - \frac{q}{N-1}\right)+\left(q-\frac{qb}{N}\right)\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)$$
So what I did so far: I just broke the terms with $a+1,a-1,b+1,b-1$ in two parts,one with the letter and one with the $1$. Note that this way i can simplify the minus terms in the second and fourth parentheses with those in the first and third respectively. Which gives:
$$ \frac{pa}{N}\left(-\frac{p}{N-1}+ \frac{q}{N+1}\right)+p\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+\frac{qb}{N}\left(+\frac{p}{N+1}- \frac{q}{N-1}\right)+q\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)$$
Now, I would like to have $\frac{1}{N}$ outside so:
$$ \frac{1}{N}\left[pa\left(-\frac{p}{N-1}+ \frac{q}{N+1}\right)+pN\left(\frac{pa}{N-1}+\frac{qb}{N+1}\right)+qb\left(+\frac{p}{N+1}- \frac{q}{N-1}\right)+qN\left(\frac{pa}{N+1}+\frac{qb}{N-1}\right)\right]$$
Now if you observe carefully you may see that the $N-1,N+1$ will go away but to make it more clear:
$$ \frac{1}{N}\left[pa\left(-\frac{p}{N-1}\right)+ pa\left(\frac{q}{N+1}\right)+pN\left(\frac{pa}{N-1}\right) +pN\left(\frac{qb}{N+1}\right)+qb\left(+\frac{p}{N+1}\right)- qb\left(\frac{q}{N-1}\right)+qN\left(\frac{pa}{N+1}\right)+qN\left(\frac{qb}{N-1}\right)\right]$$
Now pair the first term with the third and we get $p^2a$, the second with the seventh and we get $qpa$, the fourth with the fifth and we get $qpb$ and
the sixth with the eighth and we get $q^2b$. Putting it all together:
$$\frac{1}{N}(p^2a + qpa + qpb + q^2b) = \\
\frac{1}{N}[pa(p + q) + qb(p + q)] = \\
\frac{(pa+qb)(p + q)}{N}$$
| {
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Determine dlog in quotient rings of polynomial rings Question:
Determine $\operatorname{dlog}_x (x^2 + 1)$ in $\Bbb Z_5[x]/\langle\,x^3 + x + 1\,\rangle$
So I know the elements of $F = \Bbb Z_5[x]/\langle\,x^3 + x + 1\,\rangle $ are of the form $ax^2 + bx + c \bmod x^3 + x + 1$ ($a, b, c \in \Bbb Z_5$). I know how to calculate inverse of elements in F and all, but I dont know how to solve the discrete log problem. Any solution, partial or complete, would be great. Thanks...
| *
*$x^1 = x$.
*$x^2= x^2$.
*$x^3 \equiv -x-1 \equiv 4x+4$. Here we use that $x^3 + x +1 \equiv 0$ in this field (assuming it is a field, haven't checked), and coefficients are mod 5.
*$x^4 \equiv 4x^2 + 4x$.
*$x^5 \equiv 4x^3 + 4x^2 \equiv 4(-x-1)+ 4x^2 \equiv 4x^2 + x +1$.
*$x^6 \equiv 4x^3 + x^2 + x \equiv 4(-x-1) + x^2 +x \equiv x^2 + 2x + 1$.
*$x^7 \equiv x^3 + 2x^2 + x \equiv (-x-1) + 2x^2 + x \equiv 2x^2 + 4$.
*$x^8 \equiv 2x^3 + 4x \equiv 2(-x-1) + 4x = 2x+3$.
*$x^9 \equiv 2x^2 + 3x$.
*$x^{10} \equiv 2x^3 + 3x \equiv 2(-x-1) + 3x \equiv x + 3$.
*$x^{11} \equiv x^2 + 3x$.
*$x^{12} \equiv x^3 + 3x^2 \equiv -x-1 + 3x^2 = 3x^2 + 4x + 4$.
*$x^{13} \equiv 3x^3 + 4x^2 + 4x \equiv 3(-x-1) + 4x^2 + 4x \equiv 4x^2 + x + 2$.
*$x^{14} \equiv 4x^3 + x^2 + 2x \equiv 4(-x-1) + x^2 + 2x \equiv x^2 + 3x +1$.
etc. Continue until we get $x^n \equiv x^2+1$, and the answer is $n$.
Or use a computer algebra package, or write your own program...
| {
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For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$ For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfrac{\sin^2\alpha}{\cos^2\alpha}+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}{\cos^2\alpha}.$
This doesn't seem to help much.
| You're almost there!
$$\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\overbrace{\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}}{\cos^2\alpha}$$
$$=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\cos^4(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$
$$=\dfrac{\sin^2\alpha+ \overbrace{\sin^2\alpha\cdot\cos^2\alpha+\cos^4(\alpha)}}{\cos^2\alpha}$$
$$=\dfrac{\sin^2\alpha+\cos^2(\alpha)(\cos^2(\alpha)+\sin^2(\alpha))}{\cos^2\alpha}$$
$$=\dfrac{\overbrace{\sin^2\alpha+\cos^2(\alpha)}}{\cos^2\alpha}$$
$$=\dfrac{1}{\cos^2\alpha}$$
$$=\boxed{\sec^2(\alpha)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For $\alpha\in(0^\circ;90^\circ)$ simplify $E=\frac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\frac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}$ For $\alpha\in(0^\circ;90^\circ)$ simplify $E=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}.$
My try: $E=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\dfrac{\sin^2\alpha-\cos^2\alpha}{\cos^2\alpha}}=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha\cdot\cos^2\alpha+\cos^3\alpha}{\sin^2\alpha-\cos^2\alpha}.$ Is there a better approach? Thank you in advance!
| $$\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\sin\alpha+\cos\alpha}{\tan^2\alpha-1}=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\cos^2\alpha(\sin\alpha+\cos\alpha)}{\sin^2\alpha-\cos^2\alpha}=$$
$$=\dfrac{\sin^2\alpha}{\sin\alpha-\cos\alpha}-\dfrac{\cos^2\alpha}{\sin\alpha-\cos\alpha}=\sin\alpha+\cos\alpha.$$
| {
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Finding an expression and proof for $\sqrt{1-4x}$ This is similar to $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{2n\choose n}x^n$$
However I want to find an expression the same way for $$\sqrt{1-4x}$$ rather than $$\frac{1}{\sqrt{1-4x}}$$
Here's my thoughts so far:
$$(1-4x)^\frac{1}{2}=\sum_{n\geq0}{\frac{1}{2}\choose n}(-4)^nx^n$$
but this is where I get stuck.
| Notice that $$\frac{d}{dx}\sqrt{1-4x}=\frac{-2}{\sqrt{1-4x}}$$
So now, $$\frac{-1}{2}\frac{-2}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n$$ and integrating with respect to $x$ yields $$\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}=-\frac{1}{2}\sqrt{1-4x} +C $$ so we get that $$\sqrt{1-4x}+C=-2\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1} $$ Letting $x=\frac{1}{4}$, we find that
$$C=-2\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^{n+1}(n+1)}=-\frac{1}{2}\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^n(n+1)}$$
Putting it all together, $$\begin{align}\sqrt{1-4x}&=\frac{1}{2}\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^n(n+1)} - 2\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}\\&=\frac{1}{2}\sum_{n\geq 0} \binom{2n}{n}\frac{1-(4x)^{n+1}}{4^n(n+1)}\end{align}$$
| {
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Find the smallest 3 digit number which satisfies the following conditions
The smallest 3 digit number n such that if the 3 digits are $a,b$ and $c$, then
$n = a+b+c+ab+bc+ac+abc$.
I tried $n = 100a + 10b + c$ (say)
Plugged that into the given equation but couldn't simplify it further. I have no idea how should I proceed here.
Any help would be appreciated.
| To find a lower bound for the answer we calculate necessary conditions for $a = 1$ first. Then we want to solve $$100 + 10b + c = 1 + b + c + b + bc + c+ bc = 1+ 2b + 2c+ 2bc,$$
i.e. $99 + 8b =c + 2bc = (1+2b)c $ and thus $$c = \frac{99+8b}{1+2b}.$$
From this we obtain $1+2b \mid 99 + 8b$, so $1+2b \mid 99+8b - 4\cdot (1+2b) = 95 = 5 \cdot 19$. Since $b \in \{0, \dots, 9\}$ this can only happen if $1+2b \in \{1,5, 19\}$, i.e. $b \in \{0, 2,9\}$. Now we see that $b \in \{0,2\}$ yields $c \in \{99, 23\}$, so those two possibilities do not work. For $b = 9$ we get $$c = \frac{99 + 72}{1 + 18} = \frac{171}{19} = 9$$ and hence $199$ is the smallest possible solution with $a>0$. As pointed out in the comments by lulu, this actually is a solution.
| {
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Implicit derivative of $x^2+y^2=(2x^2+2y^2-x)^2$ Hi there.
I have the following function and want to calculate $y'$:
$x^2+y^2=(2x^2+2y^2-x)^2$
I've used implicit differentiation to solve it and my answer is:
$y'=\frac{(-4x^3+3x^2-4xy^2+y^2)}{y(8x^2-4x+8y^2-1)}$
However, my calculation gets huge ( I use the chainrule) and I'm just curious if there's a smarter/more simple way to calculate this?
Thank you:)
| $$x^2+y^2=(2x^2+2y^2-x)^2$$
Differentiate with respect to the variable $x$:
$$2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)$$
$$x+yy'=(2x^2+2y^2-x)(4x+4yy'-1)$$
$$(x+yy')(1-4(2x^2+2y^2-x))=-(2x^2+2y^2-x)$$
$$(x+yy')=-\dfrac {(2x^2+2y^2-x)}{(1-4(2x^2+2y^2-x))}$$
Finally we get;
$$y'=\dfrac 2y\dfrac {(4x^3+4y^2x-3x^2-y^2)}{(1-4(2x^2+2y^2-x))}$$
| {
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"url": "https://math.stackexchange.com/questions/3812115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.
In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.
Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.
$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$
is what I have got so far, and from here I can’t continue.
Am I on the right direction? Any suggestions or hints will be much appreciated.
| $$z=-(x+y)$$
$$\implies 6 [x^3 + y^3 - (x+y)^3]^2 - [x^2 + y^2 +(x+y)^2]^3$$
$$=-8 x^6 - 24 x^5 y + 6 x^4 y^2 + 52 x^3 y^3 + 6 x^2 y^4 - 24 x y^5 - 8 y^6$$
$$=-2\,(x-y)^2 \, (2 x^2 + 5 x y + 2 y^2)^2\leq 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}\leq\\
\frac{1}{4\sqrt[4]{a^3bcd}}+\frac{1}{4\sqrt[4]{ab^3cd}}+\frac{1}{4\sqrt[4]{abc^3d}}+\frac{1}{4\sqrt[4]{abcd^3}} = \\
\frac{1}{4\sqrt{a}}+\frac{1}{4\sqrt{b}}+\frac{1}{4\sqrt{c}}+\frac{1}{4\sqrt{d}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4\sqrt{abcd}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}
\end{equation}$$
This is as far as I got. Could you please help me finish off my thought pattern and finish the question the way I was attempting to solve it?
| We can not finish it because the inequality, which you'll get after your step is wrong.
Indeed, the degree of the left side is $1$ and the degree of your expression is $\frac{3}{2},$
which says that after homogenization we'll get a wrong inequality for $d\rightarrow0^+$.
Indeed, it's enough to prove that:
$$\frac{a+b+c+d}{4}\geq\sum_{cyc}\frac{\sqrt{abc}}{4}$$ or
$$(a+b+c+d)\sqrt[8]{abcd}\geq\sqrt{abc}+\sqrt{abd}+\sqrt{acd}+\sqrt{bcd}.$$
The last inequality is homogeneous already and we can forget about the condition $abcd=1$.
Now, let $d\rightarrow0^+$ and $a=b=c=1$.
Thus, the left side close to $0$ and the right side close to $1$, which says that this inequality is wrong.
The solution by Tangent Line method.
By AM-GM
$$\sum_{cyc}\frac{1}{a^3+b+c+d}\leq\sum_{cyc}\frac{1}{a^3+3\sqrt[3]{bcd}}=\sum_{cyc}\frac{1}{a^3+\frac{3}{\sqrt[3]a}}.$$
Now, let $a=x^3$, $b=y^3$, $c=z^3$ and $d=t^3$.
Thus, $xyzt=1$ and we need to prove that:
$$\sum_{cyc}\left(\frac{x^3}{4}-\frac{x}{x^{10}+3}\right)\geq0.$$
Now, since by AM-GM
$$\frac{1}{3}x^{10}-\frac{10}{3}x+3\geq0$$ and $$\frac{2}{3}x^{10}-2x^4+\frac{4}{3}x\geq0,$$ after summing we obtain: $$x^{10}+3\geq2x^4+2x$$ and it's enough to prove that
$$\sum_{cyc}\left(\frac{x^3}{4}-\frac{x}{2x^4+2x}\right)\geq0$$ or
$$\sum_{cyc}\left(x^3-\frac{2}{x^3+1}\right)\geq0$$ or
$$\sum_{cyc}\left(a-\frac{2}{a+1}-\frac{3}{2}\ln{a}\right)\geq0,$$ which is true because
$$\left(a-\frac{2}{a+1}-\frac{3}{2}\ln{a}\right)'=\frac{(a-1)(2a^2+3a+3)}{2a(a+1)^2}.$$
The coefficient $\frac{3}{2}$ we can get by the following way.
Let $f(x)=x-\frac{2}{x+1}+\lambda\ln{x}$.
We see that $f(1)=0$.
We'll choose $\lambda$ such that also $f'(1)=0$.
Easy to see that it gives $\lambda=-\frac{3}{2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\lim\limits_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$
Prove that $$\lim_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$$
My attempt:
$$x^2 + 2y^2 = x^2+y^2 + y^2 \implies \lim_{x^2 + y^2 \to +\infty}x^2 +2y^2 = +\infty$$
Then, from Cauchy-Schwarz:
$$x^2 + 2y^2 \geq 2\sqrt2xy \geq 2xy $$
Thus,
$$x^2+2y^2 -2xy \geq 0$$
I think I am on the correct path, but I don't know how to proceed.
| Because $$x^2-2xy+2y^2\geq\frac{1}{3}(x^2+y^2)\rightarrow+\infty.$$
$$x^2-2xy+2y^2\geq\frac{1}{3}(x^2+y^2)$$ it's
$$3x^2-6xy+6y^2\geq x^2+y^2$$ or
$$2x^2-6xy+5y^2\geq0,$$ which is true because $$3^2-2\cdot5<0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that if $n$ divides $a^n-b^n$ then $n$ divides $\frac{a^n-b^n}{a-b}$
Let $a,b,n \in \mathbb Z^+$. Show that if $n$ divides $a^n-b^n$ then $n$ divides $\frac{a^n-b^n}{a-b}$.
This is from Apostol’s Introduction to Analytic Number Theory, Chapter $5$, exercise $13$.
It is trivial when $\gcd(n,a-b) = 1.$
It is also easy when $n$ divides $a-b$ since $\frac{a^n-b^n}{a-b}=\sum_{k=0}^{n-1} a^kb^{n-1-k} \equiv na^{n-1}\pmod n \equiv 0\pmod n$.
But after that, I am stuck. Any hint or help is appreciated, thank you.
| Let $d=\gcd(n,a-b)$. The explicit expression
$$\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1},$$
shows that, because $a\equiv b\pmod{d}$, we also have
$$a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\equiv na^{n-1}\equiv0\pmod{d}.$$
That is to say $d$ divides $\tfrac{a^n-b^n}{a-b}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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On the integer solutions of some exponential equations I am trying to find the non-negative integer solutions of the following exponential equations:
$2^a=3^b+1\quad\color{blue}{(1)}$
$3^a=2^b+1\quad\color{blue}{(2)}$
I have found a way to obtain the non-negative integer solutions of these two exponential equations, but I think there exists a most refined and shorter way.
Could you help me to get it?
I am going to solve the equation $\;(1)\;\;2^a=3^b+1\;$.
If $\;b=0\;$ then $\;a=1\;.$
If $\;b\ge 1\;$ then $\;a\;$ is even.
I prove it by contradiction. If $\;a\;$ were odd, it would exist $\;\alpha\in\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$ such that $\;a=2\alpha+1.$ Consequently, $\;2\cdot 4^\alpha=3^b+1\;,$ hence $\;2\left(3+1\right)^\alpha=3^b+1\;.$ So there exists $\beta\in\mathbb{N}_{0}\;$ such that $\;2\left(3\beta+1\right)=3^b+1\;,$ that is $\;1=3\left(3^{b-1}-2\beta\right),$ but the last equality contradicts the fact that $\;1\;$ is not a multiple of $\;3\;.$ Therefore $\;a\;$ is even otherwise it would lead to a contradiction.
Since $\;a\;$ is even, there exists $\;\alpha\in\mathbb{N}\;$ such that $\;a=2\alpha\;,\;$ so the equation $\;(1)\;$ turns into the following one:
$4^\alpha-1=3^b\;,$
$\left(2^\alpha+1\right)\left(2^\alpha-1\right)=3^b\;.$
Hence,
$\begin{cases}
2^\alpha+1 &= 3^h\\
2^\alpha-1 &= 3^{b-h}
\end{cases}\quad$
where $\;h\in\mathbb{N}\;$ and $\;h\le b\;.$
By subtracting the previous equalities, we get that
$2=3^{b-h}\left(3^{2h-b}-1\right)\;.$
Hence $\;h=b\;$ and $\;3^b-1=2\;$.
So $\;b=1\;,\;\alpha=1\;$ and $\;a=2\;.$
Consequently the non-negative integer solutions of the equation $(1)$ are $\;a=1\;,\;b=0\;$ and $\;a=2\;,\;b=1\;.$
Now I am going to solve the equation $\;(2)\;\;3^a=2^b+1\;$.
Since $\;a\in\mathbb{N}\;$ there exist $\;n\in\mathbb{N}_{0}\;$ and $\;a_1\in\mathbb{N}\;$ ($a_1$ odd) such that $\;a=2^n a_1\;.$
So the equation $(2)$ turns into the following one:
$\left(3^{2^n}\right)^{a_1}-1=2^b\;,$
$\left(3^{2^n}-1\right)\cdot\left[1+3^{2^n}+\left(3^{2^n}\right)^2+\ldots+\left(3^{2^n}\right)^{a_1-1}\right]=2^b.$
Since inside the square brackets there is a sum of a number odd $\;a_1\;$ of addends which are all odd, the sum inside the square brackets is odd, so the sum has to be equal to $\;1\;$ otherwise it could not be a factor of $\;2^b\;,\;$ consequently $\;a_1=1\;$ and $\;a=2^n\;.$
$3^{2^n}-1=2^b\;.$
If $\;n=0\;$, then $\;a=1\;$ and $\;b=1\;.$
If $\;n\ge1\;$ the last equality turns into the following one by decomposing the left-hand side into factors:
$\left(3^{2^{n-1}}+1\right)\left(3^{2^{n-2}}+1\right)\cdots\left(3^{2^1}+1\right)\left(3^{2^0}+1\right)\left(3^{2^0}-1\right)=2^b\;,$
Since $\;\left(3^{2^1}+1\right)=2\cdot 5\;$ and $\;5\;$ is not a factor of $\;2^b\;$, it follows that $\;n-1<1\;,\;$ hence $\;n=1\;,\;a=2\;$ and $\;b=3\;.$
Consequently the non-negative integer solutions of the equation $(2)$ are $\;a=1\;,\;b=1\;$ and $\;a=2\;,\;b=3\;.$
Could you help me to get a most refined and shorter solution to the two exponential equations?
| Here's a condensed proof:
If $a$ and $b$ are integers such that $2^a=3^b+1$ then reducing mod $3$ shows that $a$ is even if $b>0$, say $a=2c$. Then
$$3^b=2^a-1=2^{2c}-1=(2^c-1)(2^c+1),$$
and so the two factors are both powers of $3$. They differ by $2$ so they are $1$ and $3$, so $b=c=1$, and $a=2$. If $b=0$ then clearly $a=1$.
Similarly, if $a$ and $b$ are integer such that $3^a=2^b+1$ then reducing mod $4$ shows that $a$ is even if $b>1$, say $a=2c$. Then
$$2^b=3^{2c}-1=(3^c-1)(3^c+1),$$
and so the two factors are both powers of $2$. They differ by $2$ so they are $2$ and $4$, so $c=1$ and $a=2$ and $b=3$. It remains to check for solutions with $b\leq0$. A quick check yields only $(a,b)=(1,1)$.
| {
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"url": "https://math.stackexchange.com/questions/3827429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The polynomial $ x^7 + x^2 +1$ is divisible by (A) $ x^5 - x^4 + x^2 -x +1 \quad$ (B) $ x^5 + x^4 +1 \quad$ (C) $ x^5 + x^4 + x^2 +x +1\quad$
(D) $ x^5 - x^4 + x^2 +x +1$
My effort: Looking at the polynomial, I know that it will have only one real root, which is negative. All other 6 roots should be imaginary. And that's how it is in the four options as well. Without actually dividing each polynomials in the options, which looks a tedious way, is there a method for finding it?
| Polynomial $x^7+x^2+1$ has higher term coefficient $1$ and also constant term coefficient $1$.
Thus if we divide it by a polynomial of degree $5$ of the form $(x^5+\cdots+1)$ as $A,B,C,D$ proposition are, it will be $x^2+ax+1$, only the central coefficient is unknown, the other two are forced to $1$.
If you multiply this polynomial by $B$ or $C$ $\begin{cases}(x^2+ax+1)(x^5+x^4+1)\\(x^2+ax+1)(x^5+x^4+x^3+x^2+1)\end{cases}$
you can notice that there will be an unremovable (and not desired) term in $x^4$, thus these possibilities should be excluded.
So you have to decide between $A$ and $D$ $\begin{cases}(x^2+ax+1)(x^5-x^4+x^2-x+1)\\(x^2+ax+1)(x^5-x^4+x^2+x+1)\end{cases}$
In both case we have $(a-1)x^6$ therefore $a=1$ to get rid of it.
The term in $x^1$ is also easy to get $A:(a-1)x=0$ and $D:(a+1)x=2x$, and since there should be no term in $x$, then $D$ should be excluded.
You just need to verify that $A$ works via a final expansion, which is indeed the case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$ I have to evaluate the following limit
$$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$$
I tried to divide tan in $\frac{sin}{cos}$ or with Hopital but I can't understand where I'm making mistakes.
The final result is:
$\frac{b}{1-ab}$ if $ab \ne 1$
$- \infty$ if $ab=1$ and $a>0$
$+ \infty$ if $ab=1$ and $a<0$
| If you perform the trigonometric expansion, you should end with
$$y=x \tan \left(a x+\tan ^{-1}\left(\frac{b}{x}\right)\right)=x\frac{b \cos (a x)+x \sin (a x)}{x \cos (a x)-b \sin (a x)}$$ and now, there is a problem because of the denominator.
Using Taylor expansions, we have
$$y=x \frac {b+ \left(a-\frac{a^2 b}{2}\right)x^2+\frac{1}{24} a^3 (a b-4)x^4+O\left(x^5\right) } { (1-a b)x+\frac{1}{6} \left(a^3 b-3 a^2\right)x^3+O\left(x^5\right) }$$
Using the long division
$$y=\frac{b}{1-a b}+\frac{a (a b (a b-3)+3)}{3 (1-a b)^2}x^2+O\left(x^4\right)$$
So, if $ab=1$ we have a poblem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the equation $z^4=(2+3i)^4$ without expanding the right side. So I tried :
$z={ \left| r \right| }e^{i \theta}$
$r^4e^{4 \theta i}=(\sqrt{13} e^{(2k\pi+\tan ^{-1}(\frac{3}{2}))i})^4$
$r=\sqrt{13}$
$4\theta=4 \times {(2k\pi+\tan ^{-1}(\frac{3}{2}))}$
$\theta=2k\pi+\tan ^{-1}(\frac{3}{2})$
But I calculated the value of $\theta$ wrongly. How can I fix it?
| Hint:
Use the fact
$$x^2-a^2=(x-a)(x+a)$$
and
$$x^2+a^2=(x-ai)(x+ai)$$
so
$$z^4-(2+3i)^4=0$$
$$\left ( x^2-(2+3i)^2 \right )\left ( x^2+(2+3i)^2 \right )=0$$
$$\left ( x-(2+3i) \right )\left ( x+(2+3i) \right )\left ( x-(2+3i)i \right )\left ( x+(2+3i)i \right )=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
} |
Finding range of a equation with given condition. We are given three real number $$a,b,c$$. With the condition $$(a^2+1)(b^2+1)(c^2+1)=9$$
We have to find the number of integers in the range of $$(ab+bc+ca)$$
Ans given is 7.
| $(a^2+1)(b^2+1)(c^2+1)=((a+b)^2+(ab-1)^2)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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