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Can I simplify $n\cdot r\cdot \sin(90^\circ-\frac{180^\circ}{n})\sqrt{r^2-r^2\sin^2(90^\circ-\frac{180^\circ}{n})}$ further? I need to write a simplified formula for this:
$$A_i = n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$
I am not very confident that I know enough trigonometry identities to simplify this completely. Other than converting $\sin(90^\circ-x)$ to $\cos(x)$, I am not sure there isn’t anything I’m missing, with the radical sign in there and everything.
$n$ and $r$ are variables and natural numbers. $A_i$ is a value based on $n$ and $r$.
This is in degrees, if that wasn’t clear.
| $$n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$
$$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\cos^2\left(\frac{180^\circ}{n}\right)}$$
$$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\left(1-\cos^2\left(\frac{180^\circ}{n}\right)\right)}$$
$$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\sin^2\left(\frac{180^\circ}{n}\right)}$$
$$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)r\sin\left(\frac{180^\circ}{n}\right)$$
$$ =n\cdot r^2\cdot \cos\left(\frac{180^\circ}{n}\right)\sin\left(\frac{180^\circ}{n}\right)$$
$$= \frac{1}{2}nr^2\sin\left(\frac{360}{n}\right)$$
Is this what you are looking for?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $n$ such that there are $11$ non-negative integral solutions to $12x+13y =n$
What should be the value of $n$, so that $12x+13y = n$ has 11 non-negative integer solutions?
As it is a Diophantine equation, so we check whether the solution exists? it exists if $gcd(12,13)|n$ that is $1|n$ and hence integer solutions to $12x+13y = n$ exist.
To find the solutions explicitly, we write $1 $ as the linear combinations of $12$ and $13$ that is $1 = 12(1) + 13(-1)$, next we write $n = 12(n) + 13(-n)$ so that the integer solutions to $12x+13y = n$ is $x = n + t(12)$ and $y = -n + t(13)$. To check for non-negative integral solutions, $x \geq 0$ and $y \geq 0$ so that $t \geq -\frac{n^2}{12}$ and $t \geq \frac{n^2}{13}$ so $t \geq \frac{n^2}{13}$. But now I am struggling how to relate this to the 11 number of non-negative integral solutions?.Any help?
| $12x + 13y = n$
$12(-n)+13(n)=n$ implies $x(t)=-n+13t$ and $y(t)=n-12t$
$x(t) \ge 0 \implies t \ge \dfrac{n}{13}$
$y(t) \ge 0 \implies t \le \dfrac{n}{12}$
So $\dfrac{n}{13} \le t \le \dfrac{n}{12}$
The existence of $11$ solutions implies
$$\text{$\dfrac{n}{13} \le t_0$ and $t_0 + 10 \le \dfrac{n}{12}$} \tag{1}$$
That is,
$$\dfrac{n}{13} \le t_0 \le \dfrac{n-120}{12}\tag{2}$$
for some integer $t_0$.
We find $12n \le 156t_0 \le 13n - 1560$
Or $$0 \le 1560t_0-12n \le n - 1560 \tag{3}$$
So $n \ge 1560$.
For $n = 1560$, condition $(3)$ becomes $0 \le 1560t_0-12n \le 0$.
So $t_0 = 120$.
It follows that $12x+13y = 1560$ has solutions,
$\{(0, 120), (13, 108), (26, 96), (39, 84), (52, 72), (65, 60), (78, 48), (91, 36), (104, 24), (117, 12), (130, 0)\}$
That is to say, $n = 1560$.
| {
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If $\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$, then what's the value of $4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$? I was trying to solve this problem:
If
$$\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$$
then what is the value of
$$S=4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$$
I tried to solve this problem using a change of variables
$\cos^3y=a\cos x$ and $\sin^3y=b \sin x$.
So:
$$\frac{S}{4}=a+b$$
$$\frac{\cos^6y}{a^2}+\frac{\sin^6y}{b^2}=1$$
and
$$\frac{\cos^2y}{a} + \frac{\sin^2y}{b}=-1$$
I tried to manipulate these equations, but it gives no result
| http://comentario.fariasbrito.com.br/vest/index.php?vid=68 There you go (it's the seventh question).
It's a pretty interesting problem, but I think when you're searching problems from IME or ITA, you should first check the sites of courses that are about them if the problems aren't from before 2000 (like the one I sent and Poliedro).
@edit: A person told me that I should outline the solution here, so here I go:
$$S = \frac{(3\sin (x+y) + \sin(x-3y))}{\sin x \cos x}$$
(as found by D.R.)
Then defining $a = 3\sin (x+y) + \sin(x-3y)$, $b = \sin x \cos x$ and seeing that $\frac{\cos x}{\cos y} + \frac{\sin x}{\sin y} = -1 \Rightarrow \sin y \cdot \cos x + \sin x \cdot \cos y = -\sin y \cdot \cos y \Rightarrow 2\sin(x+y) +\sin 2y = 0 = c$ we get:
$$S = \frac ab$$
$$a-4b-2c = a-4b = 3\sin (x+y) + \sin(x-3y) - 2\sin 2x - 4\sin (x+y) - 2\sin2y = -\sin (x+y) + \sin(x-3y) - 2\sin 2x - 2\sin2y = 2\sin (-2y)\cdot \cos (x-y) - 2\cdot 2\sin (x+y) \cdot \cos(x-y)$$
Since $2\sin(x+y) = -\sin 2y$ we get:
$$a - 4b = -2\sin 2y \cdot \cos (x-y) + 2\sin 2y \cdot \cos (x-y) = 0 \Rightarrow a = 4b$$
Thus: $$S = \frac {4b}{b} \Rightarrow S = 4$$
In the link there's another solution that is bigger, but I think it should be more acessible to people that don't want the trick of using $a, b$ and $c$.
| {
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Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\int \frac{\sqrt{\sin^2(u)}}{1-\cos(u)}( -\sin(u))du=\int \frac{-\sin(u)^2}{1-\cos(u)}du=\int \frac{-(1 - \cos^2(u)) }{1-\cos(u)}du$$ , someone could help because i block here because i don't get the same resultat with this website
| Substitute
$$t=\sqrt{\frac{1+x}{1-x}}$$ then we get
$$x=\frac{t^2-1}{t^2+1}$$
$$dx=\frac{4t}{(t^2+1)^2}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
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Using the epsilon- N definition of the limit verify that: $ \lim_{n\to\infty} \frac{n}{\sqrt{9n^2+4n+3}} = \frac{1}{3}$
Here is what I have so far:
$\mathrm{x}_\mathrm{n}=\frac{n}{\sqrt{9n^2+4n+3}},$ $a=\frac{1}{3}$
$ \forall{\epsilon}$ $\exists{\mathrm{N}_\mathrm{\epsilon}}$ $|$ $\forall{n}\in{\mathbb{N}}$ $n>\mathrm{N}_\mathrm{\epsilon} \Rightarrow|\mathrm{x}_\mathrm{n}-a|<\epsilon$
$|\frac{n}{\sqrt{9n^2+4n+3}}-\frac{1}{3}|\Rightarrow$
Then I applied the got a common denominator and applied the algebraic formula $a-b=\frac{a^2-b^2}{a+b}$ and ended up with the following but I cannot see how I should simplify it any further before I analyze to verify the inequality is less than epsilon. Here is where I am stuck in the simplification:
$$|\frac{4n+3}{9n\sqrt{9n^2+4n+3}+3(9n^2+4n+3)}|$$
| I can only guess that the task is about showing that $$ \lim_{n \to \infty} \frac{n}{\sqrt{9n^2+4n+3}} = \frac{1}{3} $$
Here's one way:
It holds that $\frac{n}{\sqrt{9n^2+4n+3}} = \frac{n}{\sqrt{n^2\left(9+\frac{4}{n}+\frac{3}{n^2}\right)}} = \frac{n}{\sqrt{n^2}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}} $
Since $n \ge 1$ it holds that
$ \frac{n}{\sqrt{n^2}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}} =\frac{1}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}$
which converges $\to \frac{1}{3}$ for $n \to \infty$
Edit:
This is another approach using $\varepsilon$ kriteria:
We want to show that for every $\varepsilon > 0 $ there is an $n_0 \in \mathbb{N}$ such that $$ \vert \frac{n}{\sqrt{9n^2+4n+3}} -\frac{1}{3} \vert < \varepsilon $$ for all $n \ge n_0$.
It holds that $ \vert \frac{1}{2n} \vert < \frac{\varepsilon}{2} $
now begin with
$$ \left\vert \frac{n}{\sqrt{9n^2+4n+3}} -\frac{1}{3} \right\vert = \left\vert \frac{3n -\sqrt{9n^2+4n+3}}{3\sqrt{9n^2+4n+3}} \right\vert = \left\vert \frac{9n^2-(9n^2+4n+3)}{3\sqrt{9n^2+4n+3}(3n+\sqrt{9n^2+4n+3})} \right\vert = \left\vert \frac{-4n-3}{9n\sqrt{9n^2+4n+3}+3(9n^2+4n+3)} \right\vert < \left\vert \frac{-4n-3}{9n^2} \right\vert = \vert -1 \vert \left\vert\frac{4n+3}{9n^2} \right\vert = \left\vert\frac{4n+3}{9n^2} \right\vert $$
Now making use of the triangle inequality leads to
$$\left\vert\frac{4n+3}{9n^2} \right\vert \le \left\vert\frac{4n}{9n^2} \right\vert + \left\vert\frac{3}{9n^2} \right\vert \le \left\vert\frac{4}{9n} \right\vert + \left\vert\frac{3}{9n} \right\vert < \left\vert\frac{1}{2n} \right\vert + \left\vert\frac{1}{2n} \right\vert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$
I'm sure there are better or at least more elegant solutions but i think it suffices the task.
| {
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Derivation of the pdf of RV $X (\left|X \right| + 1)$ How to derive the probability density function $p_Y(y)$ in case, when $Y = X (\left|X \right| + 1)$, where $X \sim \mathrm{Uniform}(-1, 1)$.
I know that if $X < 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(1- \sqrt{1 - 4y} \right)$ and also if $X \geq 0$ and $(\left|X \right| + 1) X < y$ then $X < \dfrac{1}{2} \left(\sqrt{1 + 4y} - 1 \right)$. But I do not know how to find CDF in this case. How correctly to combine these conditions into one?
Thank you very much!
| Your results so far are correct and you are almost there.
Note that the density of $X$ is symmetric with respect to zero as in $\Pr\{ X < 0 \} = \Pr\{ X > 0 \} = \frac12$. Meanwhile, $X < 0$ actually means $-1 < X < 0$ by definition, and similarly $X > 0$ implicitly means $0 < X < 1$.
Also note that $Y$ has the same sign as $X$. That is, $Y < 0$ whenever $X < 0$ and when $X$ is positive then so is $Y$.
Last thing to note is that the density of $Y$ is also symmetric with respect to zero. This is due to $|X|+1$ having the range of $[1,2]$ and the product with the symmetric $X$ just produces a copy on the negative side. Thus we already have $\Pr\{ Y < 0 \} = \Pr\{ Y > 0 \} = \frac12$ before doing any calculation.
On the negative $X$ side, we have
\begin{align}
\Pr\bigl\{ Y < y ~\big| ~X < 0 \bigr\} &= \Pr\Bigl\{ -1 < X < 0 ~~\& ~~X < \frac{1}{2} \left(1- \sqrt{1 - 4y} \right)\Bigr \} \\
& = \frac{1}{2} \left(1- \sqrt{1 - 4y} \right) - (-1)
\end{align}
where you already knew that $r_m \equiv \frac12 \left(1- \sqrt{1 - 4y} \right)$ is smaller than zero, hence we do the $r_m -(-1)$ to take the length from the left end point (negative one) to $r_m$ as the probability for the uniform distribution.
Putting $\Pr\{ X < 0 \}$ back, unconditionally we have on the negative $Y$ part (since $Y$ follows the sign of $X$)
$$F_Y(y) = \frac12 \cdot \frac12 \left( 3 - \sqrt{1 - 4y} \right) \qquad \text{for}~~ y < 0$$
On the positive side, the conditional CDF (ignoring the negative side) is
\begin{align}
\Pr\bigl\{ Y < y ~\big| ~X > 0 \bigr\} &= \Pr\Bigl\{ 0 < X < 1 ~~\& ~~X < \frac12 \left(-1 + \sqrt{1 + 4y} \right) \Bigr \} \\
& = \frac12 \left(-1 + \sqrt{1 + 4y} \right) - 0
\end{align}
where we take the length between the left end point (now zero) and $r_p \equiv \frac12 \left(-1 + \sqrt{1 + 4y} \right)$ which you already knew is between $0$ and $1$.
Now, unconditionally we are no longer ignoring the negative side and have to put things on top of it. Namely, we already have accumulated the $1/2$ probability mass from $y<0$.
\begin{align}
F_Y(y) &= \Pr\bigl\{ Y < \frac12 \bigr\} + \Pr\bigl\{ X > 0 \bigr\} \cdot \Pr\bigl\{ Y < y ~\big| ~X > 0 \bigr\} \\
&= \frac12 + \frac12 \cdot \frac12 \left(-1 + \sqrt{1 + 4y} \right) \\
& = \frac14 \left( 1 + \sqrt{1 + 4y} \right) \qquad \text{for}~~ y > 0
\end{align}
As mentioned before, the density of $Y$ is symmetric (even) while the cumulative function is rotationally symmetric (odd) with respect to $(0,\frac12)$. This can be emphasized by using the sign function $Sgn(\cdot)$, which is odd, and the absolute value that is even.
\begin{align}
f_Y(y) &\equiv \frac{d F_Y(y) }{ dy } = \frac1{2 \sqrt{ 1 + 4 |y| } } & &\text{for} ~~ -2 < y < 2 \\
F_Y(y) &= \frac12 - \frac{ Sgn(y) }4 \Bigl( 1 - \sqrt{1 + 4 |y|} \Bigr) & &\text{for} ~~ -2 < y < 2
\end{align}
The density written in this form is obviously an even function. As for the CDF, note that the sign fucntion (odd) multiplying the big parenthesis (even) is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$
Using Integration by parts
$$ I =\frac{2}{b}\bigg[\sin (x)\cdot \ln|a-b\cos x|\bigg]\bigg|^{\pi}_{0}-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln|a-b\cos x|dx$$
$$I=-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln(a-b\cos x)dx$$
Could some help me to solve it, Thanks in advance
| $b<a$ or the integral will not converge.
If you don't want to use complex analysis, Try the substitution:
$t = \tan \frac {x}{2}\\
x = 2\arctan t\\
dx = \frac {2}{1+t^2}\ dt\\
\sin x = \frac {2t}{1+t^2}\\
\cos x = \frac {1-t^2}{1+t^2}$
Due to the discontinutity of the $\tan$ function, this will work better if we integrate from $-\pi$ to $\pi.$ And since this is a still a full period for the integrand that is okay.
The limits of integration will be $(-\infty, infty)
$\int_{-\infty}^{\infty} \frac {8t^2}{(t^2+1)^2(a(1+t^2) + b(1-t^2))}\ dt\\
\int_{-\infty}^{\infty} \frac {8t^2}{(t^2+1)^2( (a+b) + (b-a)t^2)}\ dt$
Which looks like it is asking for a partial fraction decomposition.
$\frac {At + B}{(t^2 + 1)} + \frac {Ct + D}{(t^2 + 1)^2} + \frac {Et+F}{(a+b)t^2 + (b-a)}$
I get:
$\frac {2(b-a)}{a^2} \frac {1}{(t^2 + 1)} + \frac {4}{a} \frac {1}{(t^2 + 1)^2} + \frac {2(b^2-a^2)}{a^2} \frac {1}{(a+b)t^2 + (b-a)}$
$\frac {2(b-a)}{a^2} \arctan {t} + \frac {2}{a} (\frac {t}{1+t^2} + \arctan t) + \frac {2\sqrt {b^2-a^2}}{a^2}\arctan \sqrt {\frac {b-a}{a+b}} t$
$(\frac {(b-a)}{a^2} + \frac {1}{a}+ \frac {\sqrt {b^2-a^2}}{a^2}) 2\pi$
And double check my algebra, because I could well have dropped something along the way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorization of $(1+x+x^2+x^3)^2 - x^3$
Factorize : $(1+x+x^2+x^3)^2 - x^3$
I've tried to expand it but the equation will be even more complicated, anyone can give me some hints to solve it without expanding it (or it is necessary to expand it)?
| A strange but efficient way that uses the geometric series formula 3 times:
$$(1+x+x^2+x^3)=\frac{1-x^4}{1-x}$$
Your polynomial is then:
$$p(x)=(1+x+x^2+x^3)^2-x^3=\frac{(1-x^4)^2-x^3(1-x)^2}{(1-x)^2}$$
$$=\frac{1-2x^4+x^8-x^3+2x^4-x^5}{(1-x)^2}=\frac{1-x^3-x^5+x^8}{(1-x)^2}=$$
$$=\frac{(1-x^3)(1-x^5)}{(1-x)(1-x)}=(1+x+x^2)(1+x+x^2+x^3+x^4)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integral of the form $\int_a^b \frac{\ln(c+dx)}{P(x)}dx$ I found here a "great theorem" which states that: $$\int_a^b \frac{\ln(c+dx)}{P(x)}dx =\frac{\ln((ad+c)(bd+c))}{2}\int_a^b\frac{dx}{P(x)}$$
I don't know how to prove this, but I am pretty sure that we should work by symmetry with a substitution of the form $\frac{mx+n}{sx+p}$, then add the result with the initial integral.
An easier case which shows this idea is the well-known integral $\int_0^1 \frac{\ln(1+x)}{1+x^2}dx$ which can be dealt with the substitution $\frac{1-x}{1+x}$ which produces $\int_0^1 \frac{\ln 2 -\ln(1+x)}{1+x^2}dx$ and adding this with the initial integral simplifies the logarithm.
In our case, after finding the magic substitution we will have: $$\int_a^b \frac{\ln(c+dx)}{P(x)}dx=\int_a^b \frac{\ln((ad+c)(bd+c)) - \ln(c+dx)}{P(x)}dx$$
Unfortunately I dont know what $P(x)$ is, but it's hard to believe that it can be any polynomial of the form $x^2+sx+p$. I would appreciate some help to prove this "great theorem".
| \begin{align}J=\int_a^b \frac{\ln(c+dx)}{P(x)}dx\end{align}
Formally,
1)"Clean up" the logarithm. Perform the change of variable $u=c+dx$,
\begin{align}J=\frac{1}{d}\int_{c+da}^{c+db} \frac{\ln u}{P\left(\frac{u-c}{d}\right)}du\end{align}
2) Change of the bounds of the integral to new ones, m,M such that $m\times M=1$.
Perform the change of variable $v=\frac{1}{\sqrt{(c+db)(c+da)}}u$
\begin{align}J&=\frac{1}{d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln\left( v\sqrt{(c+db)(c+da)}\right)}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv\\
&=\frac{1}{d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln v}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv+\\
&\frac{\ln\left((c+db)(c+da)\right)}{2d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{1}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv
\end{align}
In the latter integral perform the change of variable $z=\dfrac{v\sqrt{(c+db)(c+da)}-c}{d}$,
\begin{align}J&=\frac{1}{d}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln v}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv+\frac{\ln\left((c+db)(c+da)\right)}{2}\int_a^b\frac{1}{P(z)}\,dz\end{align}
If for all $v$ real, $v^2 P\left(\frac{\frac{1}{v}\sqrt{(c+db)(c+da)}-c}{d}\right)=P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)$ then ,
\begin{align}\int_{\sqrt{\frac{c+da}{c+db}}}^{\sqrt{\frac{c+db}{c+da}}} \frac{\ln v}{P\left(\frac{v\sqrt{(c+db)(c+da)}-c}{d}\right)}dv=0\end{align}
(Perform the change of variable $w=\dfrac{1}{v}$)
Thus,
\begin{align}J=\frac{\ln\left((c+db)(c+da)\right)}{2}\int_a^b\frac{1}{P(z)}\,dz\end{align}
PS:
Don't expect to use this formula with P a polynomial of degree>2 or even equal to 1.
| {
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Prove $\frac{2n+\sin(n)}{n+2}$ converges to 2. Here's my attempt.
Let $\epsilon >0.$ Then $N \geq \frac{5}{\epsilon}$, with $n \geq N \implies$
$| \frac{2n+\sin(n)}{n+2} -2|=| \frac{2n+\sin(n)-2n-4}{n+2}|=| \frac{\sin(n)-4}{n+2}|= \frac{|\sin(n)-4|}{n+2} \leq \frac{5}{n+2} \leq \frac{5}{n} \leq \epsilon$
| Alternatively,
$$
\frac{2n-1}{n+2}
\le
\frac{2n+\sin(n)}{n+2}
\le
\frac{2n+1}{n+2}
$$
and both bounds go to $2$ as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Showing that the inequality $\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}$ stands.
If $\displaystyle u_{n}=\int_{0}^{1} t^n \sqrt{t+1} dt$, where $n\geq 1$, prove that
$$\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}.$$
I have already proven the second inequality, but couldn't the first one. I managed to show that $\dfrac{1}{n+1} \leq u_n$, and I wrote a recursive formula by using the $\displaystyle \int_{0}^{1} \dfrac{t^n}{\sqrt{t+1}} dt$ integral, it got worse. Seems to be some easier way.
| Since $t\to\sqrt{t+1}$ is concave, it follows that, for $t\in [0,1]$,
$$1+(\sqrt{2}-1)t\leq \sqrt{t+1}\leq \sqrt{2}.$$
Hence
$$\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}\leq \int_{0}^{1} t^n \sqrt{t+1} dt\leq \frac{\sqrt{2}}{n+1}.$$
It remains to show that
$$
\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2}\leq
\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}$$
that is
$$\frac{\sqrt{2}-1}{(n+1)(n+2)}\leq \frac{1/2}{n^2}$$
which holds for all $n\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1} {3+x^2+z^2}\leq \frac {3}{5} . $ Let $x, y, z>0$ s.t. $x+y+z=3$.
Show that $$\frac {1}{3+x^2+y^2} + \frac {1}{3+y^2+z^2} +\frac{1 } {3+x^2+z^2}\leq \frac {3}{5}\ . $$
My idea: $$3 + x^2 + y^2 \geq 1 + 2x+ 2y=7-2z $$
I notice that $f (t)=\frac {1}{7-2t} $ is a convex function but it's uselessness.
Also I have some troubles with the next inequality $$\frac {a}{a^2+bc} +\frac {b}{b^2+ac}+\frac {c}{c^2+ab}\geq \frac {3}{2} \frac{a+b+c}{a^2+b^2+c^2} $$
My idea is to multiply and to apply Muirhead's inequality. But there are too much terms.
| The first inequality.
We need to prove that
$$\sum_{cyc}\frac{1}{3+x^2+y^2}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{3+x^2+y^2}-\frac{1}{3}\right)\leq\frac{3}{5}-1$$ or
$$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+3}\geq\frac{6}{5}.$$
Now, by C-S twice we obtain:
$$\sum_{cyc}\frac{x^2+y^2}{x^2+y^2+3}\geq\frac{\left(\sum\limits_{cyc}\sqrt{x^2+y^2}\right)^2}{\sum\limits_{cyc}(x^2+y^2+3)}=$$
$$=\frac{\sum\limits_{cyc}\left(2x^2+2\sqrt{(x^2+y^2)(x^2+z^2)}\right)}{2(x^2+y^2+z^2)+9}\geq$$
$$\geq\frac{\sum\limits_{cyc}\left(2x^2+2(x^2+yz)\right)}{2(x^2+y^2+z^2)+(x+y+z)^2}=\frac{\sum\limits_{cyc}(4x^2+2xy)}{\sum\limits_{cyc}(3x^2+xy)}\geq\frac{6}{5}$$ because the last inequality it's $$\sum_{cyc}(x-y)^2\geq0.$$
| {
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"url": "https://math.stackexchange.com/questions/2954303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expanding $4\sin^3(x)$ using the Complex Exponential How would I go about proving:
$$
4 \sin^3(x)=3\sin(x)-\sin(3x)
$$
Using the complex exponential, i.e.
$$
e^{ix}=\cos(x)+i\sin(x)
$$
| $$(\cos x+i\ \sin x)^n=\cos nx+i\ \sin nx....(1)$$
Since we want $\sin^3x$ we will expand $(\cos x+i\ \sin x)^3$
$(\cos x+i\ \sin x)^3=\cos^3x-3\cos x\sin^2x+i\ (3\cos^2x\sin c-\sin^3x)$
From $(1)$ we have
$$\cos3x+i\ \sin3x=\cos^3x-3\cos x\sin^2x+i(3\cos^2x\sin x-\sin^3 x)$$
Now equate the imaginary parts and we get
$$\sin3x=3\cos^2x\sin x-\sin^3x$$
$$\sin3x=3(1-\sin^2x)\sin x-\sin^3x$$
$$\sin3x=3\sin x-3\sin^2x\sin x-\sin^3 x$$
$$\sin3x=3\sin x-4\sin^3x$$
Therefore, $$4\sin^3x=3\sin x-\sin3x$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed-form expression for infinite series related to a Gaussian Consider the following infinite series, where $x$ is indeterminate and $r$ is held constant:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^2} + \frac{x^3}{r^3} + ...$
It is relatively easy to see that the above, for $\frac{x}{r} < 1$, converges to
$\displaystyle \frac{1}{1-\frac{x}{r}}$
Now suppose we modify the above to this:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^{2^2}} + \frac{x^3}{r^{3^2}} + ...$
which we can rewrite as
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^4} + \frac{x^3}{r^9} + ...$
Does there then exist a well-known, closed-form expression for this series?
If not for general $r$, then perhaps for certain special values of $r$? For example, if we set $r=e$ above, then we get
$\displaystyle 1 + \frac{x}{e} + \frac{x^2}{e^4} + \frac{x^3}{e^9} + ...$
which we can rewrite as
$\displaystyle 1 + xe^{-1^2} + x^2e^{-2^2} + x^3e^{-3^2} + ...$
So that we can see that for x=1, this becomes a series of evenly spaced points on a Gaussian function.
Does there exist a closed-form expression for any of these?
| Interestingly, if we change the definition slightly, we get something related to the Jacobi theta function.
If we start with this series:
$\displaystyle 1 + \frac{x}{r} + \frac{x^2}{r^4} + \frac{x^3}{r^9} + ...$
We can make the following substitutions:
$q = \frac{1}{r}$
$x = \exp(2\pi i z)$
to obtain
$\displaystyle 1 + q\exp(2\pi i z) + q^4\exp(4\pi i z) + q^9\exp(6\pi i z) + ...$
$ = \sum_0^\infty q^{n^2} \exp(2\pi i n z)$
If we simply change the bottom bound from $0$ to $\infty$, we get
$\theta_3(z;q) = \sum_{-\infty}^\infty q^{n^2} \exp(2\pi i n z)$
So it is easy to write the Jacobi theta function in terms of the function I described; it is probably possible to write it the other way as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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minimum value of $\bigg||z_{1}|-|z_{2}|\bigg|$
If $z_{1}\;,z_{2}$ are two complex number $(|z_{1}|\neq |z_{2}|)$ satisfying
$\bigg||z_{1}|-4\bigg|+\bigg||z_{2}|-4\bigg|=|z_{1}|+|z_{2}|$
$=\bigg||z_{1}|-3\bigg|+\bigg||z_{2}|-3\bigg|.$Then minimum of $\bigg||z_{1}|-|z_{2}|\bigg|$
Try: Let $|z_{1}|=a$ and $|z_{2}|=b$ and $a,b\geq 0$ and $a\neq b$
So $$|a-4|+|b-4|=|a|+|b| = |a-3|+|b-3|.$$
Let $A(0,0)$ and $B(3,3)$ and $C(4,4)$ and $P(a,b)$
Then $PA=PB=PC.$ and we have to find $|a-b|$
so i did not understand how can i conclude it,
could someone please explain me
Thanks
| The minimum for $|a-b|$ is at $a=b$
$$
|a-c|-a = |b-c|-b \Rightarrow a+b=c
$$
so the intersections for
$$
a+b = 4 \cap a = b \Rightarrow a = 2\\
a+b = 3 \cap a = b \Rightarrow a = \frac 32
$$
so the minimum is at $a = b = \frac 32$ or $a = b = 2$ which is $0$
Attached in red $a+b=4$ in blue $a+b=3$ and in lightblue $a = b$
| {
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"timestamp": "2023-03-29T00:00:00",
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Bound on $\sum\limits_{n=0}^{x}{\sin{\sqrt{n}}}$ Using Desmos and Mathematica, I was able to find a function $g(x)$ that seemingly estimated the function
$$f(x)=\sum_{n=0}^{x}{\sin{\sqrt{n}}}$$
I found that
$${g(x)=2\sqrt{x}*\sin{\left({{\sqrt{4x+{\pi}^2}-\pi}\over{2}}\right)}}\approx f(x)$$
Furthermore, I found that the difference between these two functions, or $|{f(x)-g(x)}|$, never seemed to exceed some constant $C\approx 0.464568$
Is there some way to prove the conjecture below?
$g(x)<f(x)<(g(x)+C)$ for all $x>14$
Or perhaps a weaker version considering I have no good definition for $C$:
$g(x)<f(x)<(g(x)+{1\over2})$ for all $x>14$.
Further questions:
Is there a closed form for $C$?
Is there a closed form for $f(x)$?
| (Too long for a comment)
We have
\begin{align*}
f(x)
= \sum_{k=1}^{n} \int_{0}^{\sqrt{k}} \cos x \, dx
= \int_{0}^{\infty} \left( \sum_{k=1}^{n} \mathbf{1}_{\{ x < \sqrt{k}\}} \right) \cos x \, dx.
\end{align*}
Notice that $\sum_{k=1}^{n} \mathbf{1}_{\{ x < \sqrt{k}\}} = (n - \lfloor x^2 \rfloor)_+$, where $a_+ = \max\{a, 0\}$ denotes the positive part of $a$. So
\begin{align*}
f(x)
&= \int_{0}^{\sqrt{n}} (n - \lfloor x^2 \rfloor) \cos x \, dx \\
&= \int_{0}^{\sqrt{n}} \left(n - x^2 + \frac{1}{2} \right) \cos x \, dx + \int_{0}^{\sqrt{n}} \left(x^2 - \lfloor x^2 \rfloor - \frac{1}{2} \right) \cos x \, dx
\end{align*}
The first integral is easy to evaluate:
$$ \int_{0}^{\sqrt{n}} \left(n - x^2 + \frac{1}{2} \right) \cos x \, dx
= -2\sqrt{n}\cos\sqrt{n} + \frac{5}{2}\sin\sqrt{n}. $$
To evaluate the second integral, we substitute $x^2 \mapsto x$ to write
\begin{align*}
\int_{0}^{\sqrt{n}} \left(x^2 - \lfloor x^2 \rfloor - \frac{1}{2} \right) \cos x \, dx
&= \sum_{k=0}^{n-1} \int_{0}^{1} \left(x - \frac{1}{2}\right) \frac{\cos\sqrt{x+k}}{2\sqrt{x+k}} \, dx \\
&= \sum_{k=0}^{n-1} \int_{0}^{\frac{1}{2}} x \left( \frac{\cos\sqrt{k+\frac{1}{2}+x}}{2\sqrt{k+\frac{1}{2}+x}} - \frac{\cos\sqrt{k+\frac{1}{2}-x}}{2\sqrt{k+\frac{1}{2}-x}} \right) \, dx
\end{align*}
Although I cannot prove at this point, numerical evidence suggests that this integral converge as $n\to\infty$. Assuming this, it follows that
Conjecture. $c = \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left(x^2 - \lfloor x^2 \rfloor - \frac{1}{2} \right) \cos x \, dx $ converges. Consequently,
$$ f(n) = -2\sqrt{n}\cos\sqrt{n} + \frac{5}{2}\sin\sqrt{n} + c + o(1) $$
Indeed, the following figure is the plot of the difference $f(n) - \left( -2\sqrt{n}\cos\sqrt{n} + \frac{5}{2}\sin\sqrt{n} \right)$ for $n \leq 10^4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}=2-\text{something}.$$
First, find the "something" and then use that form of the equation to prove the assertion.
I can't seem to figure out the form that this equation can be written as. Also, once I find the form how would I do the proof.
I understand it involves using a Basic Step and an Induction Step
| This question is circling in rounds.
Because, as everybody knows, the sum tends to $2$ so that in
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}+ r_n=2,$$ the expression of $r_n$ must be exact!
If you try with an inequality such as
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}+ r_n<2,$$
induction will require $r_n\ge\dfrac{n+1}{2^{n+1}}$ to absorb the next term, but if $r_n$ is not tight sooner or later the sum with the remainder will exceed $2$.
Robert Z. found a nice workaround.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that f(x) is a quadratic function and that f(x) is only positive when x lies between -1 and 3, find f(x) if f(-2) = -10. What i did:
$f(x)=ax^2+bx+c$
$f(-2)=4a-2b+c=-10$
$f(0) =c > 0$
$f(1) =a+b+c > 0$
$f(2) =4a+2b+c > 0$
I thought using $b^2-4ac = 0$ for $f(-2)$ but its wrong since I am getting c = -6.
ANS: $-2x^2+4x+6$
| So $-1$ and $3$ are zeroes of this function so we can write it in factor form:
$$ f(x)=a(x+1)(x-3)$$ Now use $f(-2)=10$ to get $a$.
| {
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Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}$$ The linked proof relies upon this trigamma identity. Now by rewriting the integral as:
$$I=\int_0^1 \frac{\sqrt{x}\ln x}{x^2-2\cos\left(\frac{\pi}{3}\right)x+1}dx$$
And using that: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$I=\frac{1}{\sin \left(\frac{\pi}{3}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{3} (n+1) \right)\int_0^1 x^{n+1/2} \ln x dx$$
$$\text{Since} \ \int_0^1 x^p \ln x dx= -\frac{1}{(p+1)^2}$$
$$I=-\frac{2}{\sqrt 3} \sum_{n=0}^\infty \frac{\sin\left((n+1)\frac{\pi}{3}\right)}{(n+1+1/2)^2}=-\frac{8}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} $$
And well by using the previous link we can deduce that the series equals to $\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$, where $G$ is Catalan's constant.
I thought this might be a coefficient of some Fourier series, or taking the imaginary part of $\left(\sum_{n=1}^\infty \frac{e^{i\frac{n\pi}{3}}}{(2n+1)^2}\right)$, but I was not that lucky afterwards.
Is there a way to show the result without relying on that trigamma identity? Another approach to the integral would of course be enough.
| Integrate by parts
\begin{align}
\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx
=& \int_0^1 \ln x \>d\left( \tan^{-1} \frac{\sqrt x} {1-x}
-\frac1{\sqrt3 } \tanh^{-1} \frac{\sqrt {3x}} {1+x} \right)\\
=&\frac1{\sqrt3 }I_1 - I_2\tag1
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\tanh^{-1} \frac{\sqrt {3x}} {1+x}}x dx\\
I_2 & = \int_0^1 \frac{\tan^{-1} \frac{\sqrt {x}} {1-x}}x dx
= \underset{\sqrt x\to x}{\int_0^1 \frac{\tan^{-1} \sqrt x}x dx }
+ \underset{\sqrt {x^3}\to x}{\int_0^1 \frac{\tan^{-1} \sqrt {x^3}}x }dx \\
&= \left(2+ \frac23 \right) \int_0^1 \frac{\tan^{-1} x}x dx = \frac83G\tag2
\end{align}
Evaluate $I_1$ with $J(a) = \int_0^1 \frac{\tanh^{-1} \frac{2a\sqrt {x}} {1+x}}x dx$
$$J’(a) =\int_0^1 \frac{2(\frac1{\sqrt x}+\sqrt x)dx}{(x+1)^2-(2a\sqrt x)^2}
=\frac{2 \tan^{-1}\frac{\sqrt{(1-a^2)x}}{1-x}\bigg|_0^1} {\sqrt{1-a^2}}=\frac\pi{\sqrt{1-a^2}}
$$
Then
$$I_1 = J(\frac{\sqrt3}2)=\int_0^{ \frac{\sqrt3}2}J’(a)da
=\int_0^{ \frac{\sqrt3}2} \frac\pi{\sqrt{1-a^2}} da =\frac{\pi^2}3\tag3
$$
Plug (2) and (3) into (1) to obtain
$$\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx = \frac{\pi^2}{3\sqrt3}-\frac83G$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "17",
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For which real $a$ does$ \sum_{n=1}^{\infty} \left\{ e-\left(1+\frac{1}{n}\right)^{n+a} \right\}$ converge or diverge? For which real $a$ does
$ \sum_{n=1}^{\infty} \left\{ e-\left(1+\frac{1}{n}\right)^{n+a} \right\}$
converge or diverge?
This is a generalization of
$\sum_{n=1}^{\infty} \left\{ e-(1+\frac{1}{n})^n \right\}$. is this converge or diverge
where it is shown that
the sum diverges for
$a=0$.
My conjecture is that
the sum converges for
$a = \frac12$
and diverges for
$a \ne \frac12$.
| I think you are correct.
Note that
$$\begin{align}\log \left(1 + \frac{1}{n} \right)^{n +a} &= \frac{n+a}{n+1/2}(n + 1/2)\log\left(1 + \frac{1}{n} \right)\\&=\frac{n+a}{n+1/2}(n + 1/2)\left(\frac{1}{n} - \frac{1}{2n^2} +\mathcal{O}\left(\frac{1}{n^3} \right) \right) \\ &= \frac{n+a}{n+1/2}\left(1 + \mathcal{O}\left(\frac{1}{n^2} \right) \right) \\ &= \left(1 + \frac{a - 1/2}{n + 1/2} \right)\left(1 + \mathcal{O}\left(\frac{1}{n^2} \right) \right)\end{align}$$
We have convergence if $a = 1/2$ and divergence otherwise, since
$$e- \left(1 + \frac{1}{n} \right)^{n +a}= e - \exp \left(1 + \frac{a - 1/2}{n + 1/2} + \mathcal{O}\left(\frac{1}{n^2}\right)\right) \\ = e\frac{ 1/2-a}{n + 1/2} + \mathcal{O}\left(\frac{1}{n^2} \right)$$
| {
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - ....) = 1 + \log {G(x)}$$
Solving for $G(x)$ we have,
$$G(x) = e^{-\frac {1}{2}x + \frac {1}{3}x^2 - ...}$$
The difficulty starts here, for in order to get the desired Taylor expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ I have to plug in the entire expansion $-\frac {1}{2}x + \frac {1}{3}x^2 - ...$ for the variable in the Taylor expansion for $e$. Is there any alternative way? My main question is : what is the rth term of the desired expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ in its closed form?
| This is a slight rewriting of a problem solved in Math Overflow 77389. That answer can be put in the form
$$ (1+x)^{1/x} = e\,\sum_{n=0}^\infty a_n x^n \text{ with } a_n=\sum_{k=0}^n \frac{S_1(n+k, k)}{(n+k)!} \sum_{m=0}^{n-k}\frac{(-1)^m}{m!} $$
$S_1(n,k)$ is the Stirling number of the first kind, StirlingS1[n,k], in Mathematica.
| {
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Sum of polynomials over a ring Please refer to question 1. in the below link (pg 469)
Ques link
I think the question is incorrect as $f(x)$ contains the term $-5$ which doesn't belong to the ring $Z8[x]$ or I should go ahead by changing $f(x)$ to $f(x)= 4x+3$ and similarl changes for $g(x)$ and them add both using modulus funtion ?
| $-5$ is an element of $\mathbb{Z}_8$.
As you mentioned we have $-5=3=11=19=27=-13=-21=...$ in $\mathbb{Z}_8$.
$f(x)=4x-5=4x+3$
$g(x)=2x^2-4x+2=2x^2+4x+2=2(x^2+2x+1)=2(x+1)^2$
We get:
$(f+g)(x)=4x+3+2x^2+4x+2=2x^2+8x+5=2x^2+5$
and
$(f\cdot g)(x)=(4x+3)(2(x+1)^2)=8x(x+1)^2+6(x+1)^2\stackrel{8=0}{=}6(x+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
| Notice that with
$$f(x):=\frac{1+x}{1-x}$$ we have
$$f(-x)=\frac1{f(x)}$$ so that
$$\arctan\sqrt{f(-x)}=\text{arccot}\sqrt{f(x)}=\frac\pi2-\arctan\sqrt{f(x)}.$$
This shows that to an additive constant, the initial function is odd and its derivative must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R=0$.
For $p \equiv 3 \mod 4$, how can we find $R$? Does $R$ depend on the value of $p$?
| We can also use the symmetric polynomials.
Let $A=(1^2+1)(2^2+1) \cdots ((p-1)^2+1)(p^2+1)$. Then, $A \equiv (1^2+1)(2^2+1) \cdots ((p-1)^2+1) \mod p$, since $p^2+1\equiv 1 \mod p$.
Consider the symmetric polynomial $\displaystyle\prod_{i=1}^{\frac{p-1}2}(1+x_i)=\sum_{j=0}^{\frac{p-1}2}e_j(x)$, where $e_j$ is the $j$-th elementary symmetric polynomial.
Then $\displaystyle A\equiv(\sum_{j=0}^{\frac{p-1}2}e_j(y))^2$, where $y_i$ belongs to the set $QR:=\{1^2,\cdots,(\frac{p-1}2)^2\}$.
Now notice that $QR$ consists of roots of $x^{(p-1)/2}-1$ (over $\mathbb F_p$). So $e_j(y)$ is $(-1)^j$ times the coefficient of $x^{\frac{p-1}2-j}$ in $x^{(p-1)/2}-1$. Thus $e_0(y)=1,\,e_{(p-1)/2}(y)=(-1)^{\frac{p+1}2}$, and $e_j(y)=0$ otherwise.
Therefore $A\equiv(1+(-1)^{(p+1)/2})^2\pmod p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Sum to $n$ terms the series $\frac{1}{3\cdot9\cdot11}+\frac{1}{5\cdot11\cdot13}+\frac{1}{7\cdot13\cdot15}+\cdots$.
Q:Sum to n terms the series :
$$\frac{1}{3\cdot9\cdot11}+\frac{1}{5\cdot11\cdot13}+\frac{1}{7\cdot13\cdot15}+\cdots$$
This was asked under the heading using method of difference and ans given was
$$S_n=\frac{1}{140}-\frac{1}{48}\left(\frac{1}{2n+3}+\frac{1}{2n+5}+\frac{1}{2n+7}-\frac{3}{2n+9} \right)$$
My Approach:First i get $$U_n=\frac{1}{(2n+1)(2n+7)(2n+9)}$$
In order to make $U_n$ is the reciprocal of the product of factors in A.P i rewrite it
$$U_n=\frac{(2n+3)(2n+5)}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}=\frac{(2n+7)(2n+9)-48}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}=\frac{1}{(2n+1)(2n+3)(2n+5)}-\frac{48}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}$$
Then i tried to make $U_n=V_n-V_{n-1}$ in order to get $S_n=V_n-V_0$.But i really don't know how can i figure out this.Any hints or solution will be appreciated.Thanks in advance.
| Let
$$S_n=\sum_{k=1}^n\frac{1}{(2k+1)(2k+7)(2k+9)}.$$
By the partial fraction decomposition:
$$\frac{1}{(2k+1)(2k+7)(2k+9)}=\frac{1/48}{2k+1}
-\frac{1/12}{2k+7}+\frac{1/16}{2k+9}$$
Then, after letting $O_n=\sum_{k=1}^n\frac{1}{2k+1}$, we have that
\begin{align}
S_n&=\frac{O_n}{48}-\frac{O_{n+3}-O_3}{12}+\frac{O_{n+4}-O_4}{16}\\
&=\frac{4O_3-3O_4}{48}+\frac{O_n-4O_{n+3}+3O_{n+4}}{48}\\
&=\frac{1}{140}-\frac{1}{48}\left(O_{n+3}-O_n-3(O_{n+4}-O_{n+3})\right)\\
&=\frac{1}{140}-\frac{1}{48}\left(\frac{1}{2n+3}+\frac{1}{2n+5}+\frac{1}{2n+7}-\frac{3}{2n+9} \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $a$, $b$, $c$, $d$ such that $(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$ How would I find $a$, $b$, $c$, and $d$ in :
$$(ax+b)^2(x+c) = 4x^3 + dx^2 + 55x - 100$$
I have already worked out that $a=2$, but I may be wrong. I am not sure how to find the values of the other letters.
Thanks in advance!
| Just do it:
$(ax+b)^2(x+c)$ expands to $(a^2x^2 + 2abx + b^2)(x+c) = a^2x^3 + 2abx^2 + b^2 x + a^2cx^2 + 2abcx + b^2c= a^2x^3 + (2ab + a^2c)x^2 + (b^2 +2abc)x + b^2 c$
So
$a^2x^3 + (2ab + a^2c)x^2 + (b^2 +2abc)x + b^2 c= 4x^3 + dx^2 + 55x - 100$
So you get three sets of equations:
$a^2 = 4$
$2ab + a^2c = d$
$b^2 +2abc=55$
and $b^2c = -100$.
So solve them:
So $a^2 = 4 \implies a = \pm 2$. Now if $\pm2, b_1, c_1, d_1$ is a solutions set then $\mp2, -b_1,c_1, d_1$ is also a solution set so, wolog, we'll assume $a = 2$. And
$2ab + a^2c = d\implies d =4(b+c)$
$b^2 +2abc=55\implies b^2 + 4cb - 55 = 0 \implies b = \frac {- 4c \pm \sqrt{16c^2 + 220}}{2}= -2c \pm \sqrt{c^2 - 55}$
And finally $b^2c = -100$ means $c(-2c \pm \sqrt{c^2 - 55})^2 = -100$
$c(4c^2 \pm 4\sqrt{c^2 - 55} + c^2 - 55) = -100$
Which I don't envy you solving.
But $c < 0$ (because $cb^2 =-100 < 0$) and$c^2 > 55 $ so $|c|(4c^2 + c^2 - 55) > 100$ so that "$\pm$" must be "$-$"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Primes of the form $p=2a^2-1$ satisfying $p^2=2b^2-1$. Are there any prime numbers $p$ for which there exist integers $a$ and $b$ such that
$$p=2a^2-1\qquad\text{ and }\qquad p^2=2b^2-1,$$
other than $p=7$?
The fact that $p^2=2b^2-1$ implies that
$$(p+b\sqrt{2})(p-\sqrt{2}b)=-1,$$
and hence that $p+b\sqrt{2}$ is a unit in $\Bbb{Z}[\sqrt{2}]^{\times}=\langle-1,1+\sqrt{2}\rangle$, from which it quickly follows that
$$p+b\sqrt{2}=(1+\sqrt{2})^{2n+1},$$
for some integer $n\geq0$. This line of thought shows that $p$ must be of the form
\begin{eqnarray*}
p&=&\frac{(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}}{2},\\
p&=&\begin{pmatrix}1\\0\end{pmatrix}
\begin{pmatrix}0&1\\-1&6\end{pmatrix}^n\begin{pmatrix}1\\7\end{pmatrix},\\
p&=&a_n\quad\text{ where }\quad a_0=1, a_1=7\quad\text{ and }\quad a_{n+1}=6a_n-a_{n-1},
\end{eqnarray*}
for some integer $n\geq0$.
Checking up to $n=150$ yields no solutions, but unfortunately I'm not computer-savvy enough to check quickly. I am also at a loss how to prove that there are no other such primes. Any ideas are welcome.
| Clearly $p\ne 2$ and $p>b>a$. Write $$ p(p-1)= 2(b-a)(a+b)$$
Case 1: $$p\mid a+b\implies a+b = kp \implies k<2$$
So $a+b=p$ and $p-1 = 2(b-a)$. From here we get $p=4a-1$ so $$4a-1 = 2a^2-1\implies a=2$$
so $p= 7$.
Case 2: $$p\mid b-a\implies p \leq b-a <p $$
Which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$ Q:Solve the equation $2x^3+x^2-7x-6=0$ given that the difference of two roots is $3$.My book solve it leting the roots of the equation be $\alpha,\alpha+3,\beta$ then find the equation whose roots are $\alpha-3,\alpha,\beta-3$.And i know how to find the equation whose roots are diminished by $3$ and they get it $2x^3+19x^2+53x+36=0.$
Hence $(x+1)$ is a common factor of $2x^3+x^2-7x-6=0$ and $2x^3+19x^2+53x+36=0.$And they showed all the roots are: $-1,2,\frac{-3}{2}$Now my Question is "Is there exist any easier way to solve it?" Because in this process i need a lot of work in order to find the new equation and find GCD/HCF of these two equation.Any hints or solution will be appreciated.
Thanks in advance.
| Just to give another approach, if the roots are $\alpha$, $\alpha+3$, and $\beta$, then, by Vieta, we have
$$\alpha+(\alpha+3)+\beta=-1/2$$
and
$$\alpha(\alpha+3)+\alpha\beta+(\alpha+3)\beta=-7/2$$
The first of these can be rewritten as $2\alpha+\beta=-1/2-3=-7/2$, which means the second implies
$$\alpha^2+3\alpha+2\alpha\beta+3\beta=2\alpha+\beta$$
This rewrites as
$$\alpha^2+\alpha+(2\alpha+2)\beta=0$$
or
$$(\alpha+1)(\alpha+2\beta)=0$$
Now either $\alpha=-1$, or else $\alpha=-2\beta$. But if $\alpha=-2\beta$, then $2\alpha+\beta=-7/2$ implies $\beta=7/2$, which, by the Rational Root Theorem, is not a possible root for $2x^3+x^2-7x-6$, since $7\not\mid6$. So we must have $\alpha=-1$, so that $\beta=-7/2-2\alpha=-7/2+2=-3/2$, and thus the roots are $-1$, $2$, and $-3/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find bases B3 and B2 for $\Bbb R^3$ and $\Bbb R^2$ given a linear transformation and its matrix A linear transformation T is defined by
T: $\Bbb R^3$ $\rightarrow$ $\Bbb R^2$ $\Rightarrow$ T$\begin{pmatrix}x\\y\\z\end{pmatrix}$ = $\begin{pmatrix}2x+y\\y+2z\end{pmatrix}$
Find bases $\mathscr B_3'$ and $\mathscr B_2'$ for $\Bbb R^3$ and $\Bbb R^2$ respectively such that $Mat_{\mathscr B_3' , \mathscr B_2'}$ (T) = $\begin{pmatrix} 1&0&-1\\0&1&2 \end{pmatrix}$
So far I have attempted to answer this by producing an arbitrary basis for $\Bbb R^3$, and then using the definition of the transformation matrix, express it as a linear combination of the basis vectors that will make up the basis for $\Bbb R^2$. If my arbitrary basis for $\Bbb R^3$ is {$\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$,$\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix}$,$\begin{pmatrix}a_3\\b_3\\c_3\end{pmatrix}$}
then I can simplify this to $(\alpha_2+2\alpha_3) \begin{pmatrix}2a_2+b_2\\b_2+2c_2\end{pmatrix}$ + $(\alpha_1-\alpha_3) \begin{pmatrix}2a_1+b_1\\b_1+2c_1\end{pmatrix}$
I'm not sure what else to try or how to proceed if this is the correct approach
| I do not think your approach is correct.
We can simplify the answer by taking the ordered bases
$$\begin{align}
\mathscr B_3' & = \left\{
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}0\\1\\0\end{pmatrix},
\begin{pmatrix}0\\0\\1\end{pmatrix}\right\},\\\\
\mathscr B_2' & = \left\{
\begin{pmatrix}\beta_{11}\\ \beta_{12} \end{pmatrix},
\begin{pmatrix}\beta_{21}\\ \beta_{22} \end{pmatrix}\right\}\end{align}$$
and solving for $\mathscr B_2'.$ The simplification is that with our choice for $\mathscr B_3',$ coordinate matrices in $\Bbb R^3$ have the same entries as vectors in $\Bbb R^3,$ so we do not have to worry about converting from one to the other. Then with $[ \cdot ]_{\mathscr B_2'}$ denoting the coordinate matrix with respect to ordered basis $\mathscr B_2',$ we have
$$\begin{align}
\left[ T \begin{pmatrix} x\\ y\\ z\end{pmatrix} \right]_{\mathscr B_2'} & = \mbox{Mat}_{\mathscr B_3', \mathscr B_2'}(T) \begin{pmatrix} x\\ y\\ z\end{pmatrix}\\\\
\left[ \begin{pmatrix} 2x + y\\ y + 2z\end{pmatrix} \right]_{\mathscr B_2'} & = \begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & 2\end{pmatrix} \begin{pmatrix} x\\ y\\ z\end{pmatrix}\\\\
& = \begin{pmatrix} x - z\\ y + 2z\end{pmatrix}\\\\
\begin{pmatrix} 2x + y\\ y + 2z\end{pmatrix} & =
(x - z) \begin{pmatrix} \beta_{11}\\ \beta_{12}\end{pmatrix} +
(y + 2z) \begin{pmatrix} \beta_{21}\\ \beta_{22}\end{pmatrix}\end{align}$$
To solve for the four $\beta$s, equate the coefficients of $x, y,$ and $z$ and solve to get
$$\mathscr B_2' = \left\{
\begin{pmatrix} 2\\ 0\end{pmatrix},
\begin{pmatrix} 1\\ 1\end{pmatrix}\right\}.$$
As amd stated in his answer to your previous question, other bases will satisfy the hypotheses of your question.
| {
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"url": "https://math.stackexchange.com/questions/2976723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence to $\sqrt{2}$ It is a very good way to approximate $\sqrt{2}$ using the following;
Let $D_{k}$ and $N_{k}$ be the denominator and the numerator of the $k$th term, respectively.
Let $D_1=2$ and $N_1=3$, and for $k\geq2$, $D_k=D_{k-1}+N_{k-1}$, and $N_{k}=D_{k-1}+D_{k}$.
To clarify, here is the sequence;
$\frac{3}{2},\frac{2+2+3}{2+3},\frac{2+3+2+3+2+2+3}{2+3+2+2+3},\cdots$ that is $\frac{3}{2},\frac{7}{5},\frac{17}{12},\cdots$
*
*Why the sequence converges to $\sqrt{2}$ even the initial numerator and denominator were other positive integers?
*How to find the $j$th term (i.e. $\frac{N_{j}}{D_{j}}$) without finding the preceding terms, say the $45$th term?
*Is there a similar way to approximate the square root of any other positive integer, say $\sqrt{3}$?
I do know many ways to approximate square roots, such as Newton's method. But I am asking about a similar way.
| You have a pair of linear recurrence relations. You can write it as $$\begin {pmatrix} N_k\\D_k \end {pmatrix}=\begin {pmatrix} 1&2\\1&1 \end {pmatrix}\begin {pmatrix} N_{k-1}\\D_{k-1} \end {pmatrix}$$ You find the eigenvalues and eigenvectors of the matrix. The eigenvector $\begin {pmatrix} \sqrt 2\\1 \end {pmatrix}$corresponding to the eigenvalue $\sqrt 2 +1$ will dominate, so you will converge to $\sqrt 2$ for any starting condition that does not have this eigenvector with a $0$ coefficient.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many Eisenstein integers modulo 3 are there? I'm trying to find a system of representatives for the Eisenstein integers modulo 3. What would the set S be and how can I determine the number of solutions in S of the equation x^2=0 mod 3?
| The ring of Eisenstein integers is isomorphic to $\Bbb{Z}[X]/(X^2+X+1)$, so the ring of Eisenstein integers mod 3 is isomorphic to
\begin{eqnarray*}
(\Bbb{Z}[X]/(X^2+X+1))/(3)&\cong&(\Bbb{Z}[X]/(3))/(X^2+X+1)\\
&\cong&\Bbb{F}_3[X]/(X^2+X+1)\\
&\cong&\Bbb{F}_3[X]/((X-1)^2).
\end{eqnarray*}
This is a ring of $9$ elements, so counting the solutions to $x^2=0$ is quickly done. In fact we have
$$(aX+b)^2=a^2X^2+2abX+b^2=2a(a+b)X+(b^2-a^2)=(a+b)(2aX+a-b),$$
which quickly shows that the solutions are precisely the elements of the form $a(X-1)$.
Alternatively, without any reference to ring theory:
Two Eisenstein integers $x=a\omega+b$ and $y=c\omega+d$ are congruent modulo $3$ if and only if $a\equiv c\pmod{3}$ and $b\equiv d\pmod{3}$. From this it follows that a system of representatives is, for example, the set
$$S=\{a\omega+b:\ 0\leq a,b<3\}.$$
For the second part of your question, we are looking for Eisenstein integers $x$ satisfying $x^2\equiv0\pmod{3}$. Write $x=a\omega+b$ with $a$ and $b$ integers. Then because $\omega^2=-\omega-1$ we find that
\begin{eqnarray*}
x^2=(a\omega+b)^2&=&a^2\omega^2+2ab\omega+b^2\\
&=&(2ab-a^2)\omega+(b^2-a^2)
\end{eqnarray*}
So to get $x^2\equiv0\pmod{3}$ we need that both
$$2ab-a^2\equiv0\pmod{3}\qquad\text{ and }\qquad b^2-a^2\equiv0\pmod{3}.$$
The latter implies that $b\equiv\pm a\pmod{3}$. If $b\equiv a\pmod{3}$ then the first congruence becomes $a^2\equiv0\pmod{3}$ and hence $a\equiv b\equiv0\pmod{3}$. If $b\equiv-a\pmod{3}$ then the first congruence becomes $-3a^2\equiv0\pmod{3}$, which holds for all $a$. So we get the solutions
$$0,\ \omega-1,\ 2\omega-2.$$
That is, all integer multiples of $\omega-1$.
| {
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If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$
I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refering to has marked the answer as $19\pi/24$.
Am I doing something wrong?
| First of all, your values for $A$ and $B$ satisfy the given formulae and $B$ is positive, so their proposed minimum for $B$ is definitely wrong.
One thing you did miss in your reasoning is that there are two families of solutions to $\sec\left(\theta\right) = \frac{2}{\sqrt{3}}$; there should be a $\pm$ with $\frac{\pi}{6}$ in your formula for $A+B$. However, it turns out that you get the minimum value for $B$ from the family of solutions you gave, and the minimum positive value of $B$ is indeed $\frac{7\pi}{24}$.
Explicitly, $$A-B = \frac{\pi}{4} + n\pi,$$ $$A+B = \pm \frac{\pi}{6} + 2m\pi,$$ so $$B = \frac{\pi}{2}\left(\pm\frac{1}{6} - \frac{1}{4} + \left(2m - n\right)\right),$$ and then we see that $$B = \frac{\pi}{2}\left(-\frac{1}{12} + k\right), \quad k \in \mathbb{Z}$$ or $$B = \frac{\pi}{2}\left(\frac{-5}{12} + k\right), \quad k \in \mathbb{Z},$$ and the smallest positive value of $B$ is then $\frac{7\pi}{24}$.
EDIT: I see that you have/somebody has now edited your question to include the $\pm$.
| {
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Transition Matrix from B to C
If $B=\{ 2+x,1+2x\}$ and $C=\{ 1+x, 1-x\}$ are 2 basis for $P_1$, and $v=-3x+4$ find $[v]_B$, $_BP_C$ and $[v]_C$.
my attempt:
Since $B$ is a basis for $V$, then any $v\in B$ can be written "uniquely" as a linear combination of the vectors of $B$.
then $v=-3x+4$ can be written uniquely in the form:
$-3x+4=c_1(2+x)+c_2(1+2x)$ where $c_1$ and $c_2$ are scalars.
$-3x+4=c_1(2+x)+c_2(1+2x)$
$=2c_1+c_2+(c_1+2c_2)x$
Then
$2c_1+c_2=4$
$c_1+2c_2=-3$
and by solving this system, we have
$c_1=\frac{11}{3}$ and $c_2=\frac{-10}{3}$
so, $[v]_B=$ $\left[ {\begin{array}{c}
\frac{11}{3} \\
\frac{-10}{3}
\end{array} } \right]
$
The transition matrix from $B$ to $C$ is
$_BP_C=$ $\left[ {\begin{array}{cc}
[2+x]_C & [1+2x]_C
\end{array} } \right]
$
we can find $[2+x]_c$ as follows
$2+x=\alpha_1(1+x)+\alpha_2(1-x)$
$=(\alpha_1+\alpha_2)+(\alpha_1-\alpha_2)x$
and hence
$\alpha_1+\alpha_2=2$
$\alpha_1-\alpha_2=1$
and by solving this system we get that
$\alpha_1=\frac{3}{2}$ and $\alpha_1=\frac{1}{2}$
Similarly,
$1+2x=\alpha_1(1+x)+\alpha_2(1-x)$
$=(\alpha_1+\alpha_2)+(\alpha_1-\alpha_2)x$
and hence
$\alpha_1+\alpha_2=1$
$\alpha_1-\alpha_2=2$
and by solving this system we get that
$\alpha_1=\frac{3}{2}$ and $\alpha_1=\frac{-1}{2}$
$_CP_B=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{-1}{2} & \frac{1}{2}
\end{array} } \right]
$
or,
$_CP_B=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{1}{2} & \frac{-1}{2}
\end{array} } \right]
$
what is the true?
and we know that
$[v]_C=_BP_C[v]_B$
$[v]_C=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{1}{2} & \frac{-1}{2}
\end{array} } \right]$
$\left[ {\begin{array}{c}
\frac{11}{3} \\
\frac{-10}{3}
\end{array} } \right]
$
$= \left[ {\begin{array}{c}
\frac{1}{2} \\
\frac{7}{2}
\end{array} } \right]
$
and if I use $_CP_B=\left[ {\begin{array}{cc}
\frac{3}{2} & \frac{3}{2} \\
\frac{-1}{2} & \frac{1}{2}
\end{array} } \right]
$
I got that
$= \left[ {\begin{array}{c}
\frac{1}{2} \\
\frac{7}{2}
\end{array} } \right]
$
what is the true please? Thanks.
| $b_1 = 2+x = a_{11} c_1 + a_{21} c_2\\
2+x = a_{11}(1+x) + a_{21}(1-x)\\
a_{11}+a_{21} = 2\\
a_{11}-a_{21} = 1\\
a_{11} = \frac 32, a_{21} = \frac 12$
and
$b_2 = 1+2x = a_{12} c_1 + a_{22} c_2\\
a_{12} + a_{22} = 1\\
a_{12} - a_{22} = 2\\
a_{12} = \frac 32, a_{22} = - \frac 12$
$_CP_B = \begin{bmatrix} \frac 32 &\frac 32\\\frac 12&-\frac 12\end {bmatrix}$
Here is a different way to think about it.
$B = \begin {bmatrix} 2&1\\1&2 \end{bmatrix}$
Will transform vectors represented in the basis $B$ into the standard basis.
$C = \begin {bmatrix} 1&1\\1&-1 \end{bmatrix}$
Will transform vectors represented in the basis $C$ into the standard basis.
$C^{-1} = \begin {bmatrix} \frac {1}{2}&\frac {1}{2}\\\frac {1}{2}& -\frac {1}{2} \end{bmatrix}$
Will reverse that and take vectors in the standard basis to the basis $C.$
$C^{-1}B$ will take a vector in $B$, translated to the standard basis and then transform that into the basis $C.$
$\begin {bmatrix} \frac {1}{2}&\frac {1}{2}\\\frac {1}{2}& -\frac {1}{2} \end{bmatrix}\begin {bmatrix} 2&1\\1&2 \end{bmatrix}= \begin {bmatrix} \frac 32 & \frac 32 \\ \frac 12 & -\frac 12 \end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all $a$ and $b$ that make this value rational: $ \left(\sqrt 2 + \sqrt a\right)/\left(\sqrt 3 + \sqrt b\right)$
$$ \frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b} $$ Find all integers $a$ and $b$ that make this value rational.
Looking at it, you will easily see that $a=3$, $b=2$ will make this value $1$.
Are there more possible $a$ and $b$?
| Note that $b=3$ does not give any solution since $$\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt 3}=\frac{\sqrt{6}+\sqrt{3a}}{6}$$and $\sqrt{6}+\sqrt{3a}$ cannot be rational number (See this question for why). Therefore, we can rationalize it.
Rationalize the value and expanding gives$$\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b}=\frac{(\sqrt 2 + \sqrt a)(\sqrt 3 - \sqrt b)}{3-b}=\frac{\sqrt 6 - \sqrt{2b}+\sqrt{3a}-\sqrt{ab}}{3-b}$$which is rational if and only if $\sqrt 6 - \sqrt{2b}+\sqrt{3a}-\sqrt{ab}$ is rational. Let $\sqrt 6 - \sqrt{2b}+\sqrt{3a}-\sqrt{ab}=P$. I'll omit rational numbers in the following calculations because they don't change $P$'s rationality.
$$\sqrt 6+\sqrt{3a}=P+\sqrt{ab}+\sqrt{2b}$$
Squaring both sides,
$$2\sqrt{18a}=P'+2P\sqrt{ab}+2P\sqrt{2b}+2b\sqrt{2a}$$
Where $P'$ is remaining rational part of the equation. It is just $(6-2b)\sqrt{2a}-2P\sqrt{2b}=P'+2P\sqrt{ab}$, and squaring both sides again,
$$-8P(6-2b)\sqrt{ab}=P''+4PP'\sqrt{ab}$$
and therefore, $\sqrt{ab}$ is rational or, as an exceptional case, $-8P(6-2b)=4PP'$ or $PP'=4P(b-3)$.
Let's consider the case where $PP'=4P(b-3)$ first.
From $b \neq 3$, $P=(\sqrt 2 + \sqrt a)(\sqrt 3 - \sqrt b)\neq 0$. Therefore, $P'=4(b-3)$.
Squaring both sides of $\sqrt 6+\sqrt{3a}=P+\sqrt{ab}+\sqrt{2b}$ and considering rational term gives $P'=P^2+ab+2b-6-3a$. Applying $P'=4(b-3)$ and processing gives $P^2=(a-2)(3-b)$. That is, either $P=0,a=2$ or $a>2, b<3$ or $a<2, b>3$.
$P=0,a=2$ gives $b=3$, which we ruled out.
For $a>2, b<3$, either $b=0$, $b=1$ or $b=2$. $b=0$ means $\sqrt6+\sqrt{3a}=P$, so no solution.
$b=1$ means $P=\sqrt{2(a-2)}$, so $(\sqrt 2 + \sqrt a)(\sqrt 3 - 1)=\sqrt{2(a-2)}$. This case gives the solution $a=6$, but $\frac{\sqrt 2 + \sqrt 6}{\sqrt 3 + \sqrt 1}$ is not rational.
$b=2$ means $P=\sqrt{a-2}$, so $(\sqrt 2 + \sqrt a)(\sqrt 3 - \sqrt{2})=\sqrt{a-2}$. This case gives the solution $a=3$.
For $a<2, b>3$, either $a=0$ or $a=1$. $a=0$ means $\sqrt6-\sqrt{2b}=P$, which means $b=3$, which we ruled out.
$a=1$ means $P=\sqrt{b-3}$, so $(\sqrt 2 + 1)(\sqrt 3 - \sqrt{b})=\sqrt{b-3}$. This case gives the solution $b=3$, which we ruled out.
Therefore, one solution $(a, b)=(3, 2)$ exists in this case.
Now, what is left is the case where $\sqrt{ab}$ is rational.
From $(6-2b)\sqrt{2a}-2P\sqrt{2b}=P'+2P\sqrt{ab}$, $(6-2b)\sqrt{2a}-2P\sqrt{2b}$ is rational. Once again, by extension of above question, either $(6-2b)\sqrt{2a}$ and $2P\sqrt{2b}$ are rational or $(6-2b)\sqrt{2a}=2P\sqrt{2b}$ or $a(3-b)^2=P^2b$.
Firstly, let's consider the case $a(3-b)^2=P^2b$.
Since $b=0$ does not give any solution, we can divide both sides by $b$ to get $P=(3-b)\sqrt{\frac{a}{b}}$. Now we get$$(\sqrt2+\sqrt{a})(\sqrt3-\sqrt b)=(3-b)\sqrt{\frac{a}{b}}$$and since $b \neq 3$, we can divide each side by $\sqrt3-\sqrt b$ $$\sqrt2+\sqrt{a}=(\sqrt3+\sqrt b)\sqrt{\frac{a}{b}}$$simplifying gives$$\sqrt2=\sqrt{\frac{3a}{b}}$$which is simply $2b=3a$.
$2b=3a$ and $ab$ must be square number. However, this is impossible because $ab=\frac{3a^2}{2}$ cannot be square number.
Now let's consider the case where $(6-2b)\sqrt{2a}$ and $2P\sqrt{2b}$ are rational. It easily follows that $\sqrt{2a}$ is rational, and $a=2k^2$ for some nonnegative integer $k$. Then $P=(k+1)(\sqrt{6}-\sqrt{2b})$ and $\sqrt{6}-\sqrt{2b}$ must be rational. It means $b=3$, which we ruled out.
Therefore, $(a, b)=(3, 2)$ is the only pair of integers which makes $\frac{\sqrt 2 + \sqrt a}{\sqrt 3 + \sqrt b}$ a rational number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I calculate the value of $\tan(1/2)$ for the De Moivre formula I have to solve $(2+i)^3$ using the trigonometric representation. I calculated the modulus but I don't know how to calculate $\varphi$ when it is equal to $\tan(1/2)$.
How do I calculate $\varphi$
Also, is there a fast way to solve these in analogous way if we know that the only thing that changes is modulus power and $\varphi$ product?
$(2+i)^2,\quad (2+i)^3,\quad (2+i)^4,\quad (2+i)^5,\quad (2+i)^6,\quad (2+i)^7$
| There is no real advantage in computing powers this way over the algebraic method. There would be an advantage if the argument is a “known angle”.
You can surely write $2+i=\sqrt{5}(\cos\varphi+i\sin\varphi)$, where
$$
\cos\varphi=\frac{2}{\sqrt{5}}\qquad\sin\varphi=\frac{1}{\sqrt{5}}
$$
Then, yes,
$$
(2+i)^3=5\sqrt{5}(\cos3\varphi+i\sin3\varphi)
$$
but now the problem is to compute $\cos3\varphi$ and $\sin3\varphi$, that's no easier than using the binomial theorem from the outset:
$$
(2+i)^3=2^3+3\cdot2^2i+3\cdot2i^2+i^3=8+12i-6-i=2+11i
$$
To wit
$$
\cos3\varphi=\cos^3\varphi-3\cos\varphi\sin^2\varphi
=\frac{8}{5\sqrt{5}}-3\cdot\frac{2}{\sqrt{5}}\frac{1}{5}=\frac{2}{5\sqrt{5}}
$$
and
$$
\sin3\varphi=3\cos^2\varphi\sin\varphi-\sin^3\varphi=
3\cdot\frac{4}{5}\frac{1}{\sqrt{5}}-\frac{1}{5\sqrt{5}}=\frac{11}{5\sqrt{5}}
$$
As you can see, there are exactly the same computations, with added denominators that cancel with the modulus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct? Is only one correct? Is there an exception that I am missing?
| If you already know that $x(1-x) \le 1/4$ for all $x$, then it also holds for $ax$. Divide the inequation you get by $a$ and you obtain the result you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Sine Fourier Series for $\min(x, 1 - x)$. I am currently taking a class on differential equations and only Analysis and Linear Algebra I and II were expected. In those subjects we never covered Fourier series.
In an exercise I am now expected to solve
$$
\sum_{k = 1}^{\infty} a_k \sin(\pi k x)
= \min(x, 1 - x)
$$
for the coefficients $a_k$ and for $x \in (0,1)$.
I'd be interested if anyone could share an approach, which someone with no previous knowledge of Fourier series could understand.
for reference: the given differential equation is $u_t - u_{xx} = 0$ for $t > 0$ and $x \in (0,1)$ with $u(0,t) = u(1,t) = 0$ and $u(x,0) = \min(x, 1 - x)$. for $x \in (0,1)$
EDIT 4 - Finally Correct
\begin{align*}
a_k
& = 2 [1 - (-1)^k] \int_{0}^{\frac{1}{2}} x \sin(k \pi x) dx
= 2 [1 - (-1)^k] \left[ \frac{\sin(k \pi x)}{(k \pi )^2} - \frac{x}{k \pi } \cos(k \pi x) \right]_{x = 0}^{\frac{1}{2}} \\
& = 2 [1 - (-1)^k] \left( \frac{\sin(\frac{1}{2} k \pi)}{(k \pi )^2} - \frac{1}{2 k \pi } \cos\left(\frac{1}{2} k \pi\right) - \left( \frac{\sin^2(0)}{(k \pi )^2} - \frac{0}{k \pi } \cos(0) \right) \right) \\
& = \frac{[1 - (-1)^k]}{k \pi} \left[ \frac{2 \sin(\frac{1}{2} k \pi)}{k \pi} - \cos\left(\frac{1}{2} k \pi\right) \right]
\end{align*}
Therefore we have $a_2k = 0$.
For $a_{2k + 1}$ we have
\begin{align*}
a_{2k + 1}
& = \frac{1 - (-1)^{2k + 1}}{(2k + 1) \pi} \left( \frac{2 \sin(\frac{1}{2} (2k + 1) \pi)}{(2k + 1) \pi} - \cos\left(\frac{1}{2} (2k + 1) \pi\right) \right) \\
& = \frac{2}{(2k + 1) \pi} \left[ \frac{2 \overbrace{\sin\left(\frac{\pi}{2} + k \pi\right)}^{= \cos(k \pi)}}{(2k + 1) \pi} - \underbrace{\cos\left(\frac{\pi}{2} + k \pi\right)}_{= - \sin(k\pi) = 0.} \right] \\
& = \frac{4 (-1)^k}{\left((2k + 1)\pi \right)^2}.
\end{align*}
| The sine functions $\sin(k \pi x)$ are orthogonal. Thus forming the same scalar product on both sides of the sine series expansion $\sum_n a_n\sin(nπx)=\min(x,1-x)$, that is, multiplying with $\sin(k \pi x)$ and integrating, filters out the term with the coefficient $a_k$.
$$
a_k\int_0^1\sin^2(k \pi x)\,dx=\int_0^1\min(x,1-x)\,\sin(kπx)\,dx\\
\iff a_k\int_0^1[1-\cos(2k \pi x)]\,dx=2\int_0^{1/2}[1-(-1)^k]\,x\sin(kπx)\,dx\\
$$
One can see that for even $k$ the integrand on the right is an odd function relative to the axis $x=\frac12$, thus $a_{2m}=0$ for $m\in \Bbb N$. The integral for odd $k$ is a standard application for partial integration. Use
$$
\int_0^1\cos(2πkx)\,dx=0 ~~\text{ and }~~\int x\sin(ax)\,dx =-\frac{x\cos(ax)}a+\frac{\sin(ax)}{a^2}
$$
to get
$$
a_{2m+1}=4\;\left[-\frac{x\cos((2m+1)πx)}{(2m+1)π}+\frac{\sin((2m+1)πx)}{((2m+1)π)^2}\right]_{x=0}^{x=\frac12}
=\frac{\sin(mπ+\fracπ2)}{((m+\frac12)π)^2}
=\frac{(-1)^m}{((m+\frac12)π)^2}
$$
#/usr/bin/gnuplot
min(a,b)=(a<b)?a:b
c(k,x)=4*sin(k*pi*x)/(k*pi)**2
plot [0:1] min(x,1-x) lt 8 lw 2, c(1,x)-c(3,x)+c(5,x)-c(7,x)+c(9,x)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984136",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$. Find the inverse Laplace of $ \ \frac{2}{(s-1)^3(s-2)^2}$.
Answer:
To do this we have to make partial fractions as follows:
$ \frac{2}{(s-1)^3(s-2)^2}=\frac{A}{S-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)}+\frac{J}{(s-1)^3}+\frac{K}{(s-2)^2}+\frac{L}{s-2}$
Am I right so far?
Does there is any easy way?
| Another way is using convolution
\begin{align}
\frac{2}{(s-1)^3(s-2)^2}
&= \frac{2}{(s-1)^3}\cdot\frac{1}{(s-2)^2} \\
&= {\cal L}\left(t^2e^t\right){\cal L}\left(te^{2t}\right) \\
&= \int_0^xt^2e^t(x-t)e^{2x-2t}\ dt \\
&= e^{2x}\int_0^xe^{-t}(xt^2-t^3)\ dt \\
&= \color{blue}{2e^{2x}(x-3)+e^x(x^2+4x+6)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & 8 \\
4 & 3 & -9 & 9 \\
2 & 3 & -5 & 7 \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & -13 & 9 & -17 \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_3\,\leftrightarrow\, R_4}}{R_2\,\leftrightarrow\, R_3}}{\overset{R_1\,\leftrightarrow\,R_2}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & -13 & 9 & -17
\end{array}\right]$$
$$\overset{R_4\to R_4-R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -3 & 1 & -5 \\
0 & 10 & 8 & -12
\end{array}\right]
\overset{\overset{\large{R_3\to R_3+R_2}}{R_4\to R_4-5R_2}}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 2 & -3 & 1 \\
0 & -1 & -2 & -4 \\
0 & 0 & 23 & -17
\end{array}\right]
\overset{\overset{\large{R_2\to R_2+2R_3}}{R_3\to-R_3}}{\large\longrightarrow}$$
$$\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 0 & -7 & -7 \\
0 & 1 & 2 & 4 \\
0 & 0 & 23 & -17 \\
\end{array}\right]
\overset{R_2\,\leftrightarrow\,R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7 & 12 \\
0 & 1 & 2 & 4 \\
0 & 0 & -7 & -7 \\
0 & 0 & 23 & -17 \\
\end{array}\right]$$
However, the answer in the book $(3, 2, 1)$ fits the system.
Was there an arithmetical mistake, or do I misunderstand something fundamentally?
| You do (in the third matrix): $$L3-L4=(0, -3, 1 \mid -5)-(0, -13, 9 \mid -19)=(0, 10, -8 \mid 12)$$ but you have $(0, 10, 8 \mid -12)$ instead.
| {
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"question_score": "13",
"answer_count": 3,
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Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$ This question is I am working on is the extension of the domain of zeta function.$\zeta(x)$=$1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+\frac{1}{5^x}+\frac{1}{6^x}+ \cdots$$\eta(x)$=$1-\frac{1}{2^x}+\frac{1}{3^x}-\frac{1}{4^x}+\frac{1}{5^x}-\frac{1}{6^x}+ \cdots$Note: The zeta $\zeta(x)$ function consists of positive terms only and the eta $\eta(x)$ function converges for all positive values for $x$ even when $x$ is less than $0$ where $0\lt x \lt 1$. Also in the equation $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$, the left hand side is only defined when $x\gt 1$ because that is the domain of the zeta series. The right hand side can be used to define the zeta function for $0\lt x \lt 1$ and the right hand side totally depends on $\eta(x)$ in the function because it is already defined for all positive values for $x$. Here is my attempt so far:$({1\over 1-2^{1-x}})$ $\eta (x)$$\qquad\qquad\qquad=({1\over 1-2^{1-x}})\cdot 1 -({1\over 1-2^{1-x}})(\frac{1}{2^x})+({1\over 1-2^{1-x}})(\frac{1}{3^x})-({1\over 1-2^{1-x}})(\frac{1}{4^x})+({1\over 1-2^{1-x}})(\frac{1}{5^x})-({1\over 1-2^{1-x}})(\frac{1}{6^x})+\cdots$$\qquad\qquad\qquad=(\frac{2^x}{2^x-2})-(\frac{1}{2^x-2})$$+$[$({2/3})^{x}\over{2^x-2}$]$-$($2^{-x}\over{2^x-2}$)$+$[$({2/5})^{x}\over{2^x-2}$]$-$($3^{-x}\over{2^x-2}$)$+\cdots$By adding each pair of terms from above we get(note: I attempted to input the info but its too difficult to do due to the complexity from combining the terms but I have the next step written down and after that I am stuck showing it). Need help showing that $\zeta (x)$= ${1\over 1-2^{1-x}}$ $\eta (x)$. Getting cancellations is hard
| As i said in comment:
$\zeta (x) - \eta (x) = \sum\limits_{n=1}^{\infty}\frac{1}{n^x} - \sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}=\sum\limits_{n=1}^{\infty}(\frac{1}{n^x} - \frac{(-1)^{n-1}}{n^x})=2\sum\limits_{n=1}^{\infty}\frac{1}{(2n)^x}=2^{1-x}\sum\limits_{n=1}^{\infty}\frac{1}{n^x}=2^{1-x}\zeta (x) $
Hence: $\zeta (x)(1 - 2^{1-x}) - \eta(x)=0\\
\zeta(x)=\frac{\eta(x)}{1-2^{1-x}}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Calculate the length of the closed curve $x^{2/3} + y^{2/3} = 4$ I realise that this function forms a closed curve, and the range of both $x$ and $y$ are: $-8 \leq x, y \leq 8$.
I began by differentiating the function implicitly, arriving at a expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$:
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} &: \frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \\
&\implies x^{-\frac{1}{3}} + y^{-\frac{1}{3}}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \\
&\implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y^{-\frac{1}{3}}}{y^{-\frac{1}{3}}} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}
\end{align}
Once that was done, I applied the formula for the arc length of a curve:
\begin{align}
&\int_{a}^{b}{\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}}\\
&=\int_{a}^{b}{\sqrt{1 + \left(\frac{y}{x}\right)^\frac{2}{3}}}\\
&=\int_{a}^{b}{\sqrt{1 + \frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}}}\\
&\vdots\\
&=2\int_{a}^{b}{\frac{1}{\sqrt[\leftroot{4}\uproot{1}3]{x}}} \\
&=3\left[x^\frac{2}{3}\right]_{a}^{b}
\end{align}
Now, I'd like to know what the bounds are.
As mentioned earlier, the function has a range (and domain) of $[-8, 8]$, but inserting them gives a definite integral with a result of zero, which is clearly incorrect. Using the bound $[0, 8]$ gives a result of $12$ which is more believable. However, I plotted the graph in Desmos, and I think the answer is definitely off by a large margin.
I believe (this was by sheer visual inspection) the length is approximately—but not equal to—the circumference of a circle with radius $8$, so I should be getting an answer closer to $16\pi \approx 50.$ I notice that $12 \times 4 = 48$; is the answer as simple as that?
Is there a more rigorous method of choosing the bounds of the integral?
| With parametric equations you have $$x=8cos^3 \theta, y=8\sin ^3 \theta $$
Then you integrate $$L=4 \int _{0}^{\pi/2} \sqrt {(\frac {dx}{d\theta})^2 +(\frac {dx}{d\theta})^2} d\theta $$
The derivatives are straight forward and you can finish it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does the integral $\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}$ diverge Does the integral
$$J:=\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}dx $$
diverge ?
If we integrate by parts we find
$$J=\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)} -\\
\int_{0}^{\infty}\sin{(x^2-x)}
\left( \frac{3x^2}{(1+x^2)(2x-1)}-\\
\frac{2x^4}{(1+x^2)^2(2x-1)}-\frac{2x^3}{(1+x^2)(2x-1)^2}\right)dx$$
If we integrate by parts again, we find that the latter integral reduces to
a sum of absolutely convergent integrals plus
the boundary term
$$ -\lim_{a\rightarrow +\infty}\,
\cos{(a^2-a)}\left(\frac{3a^2}{(1+a^2)(2a-1)^2}
-\\
\frac{2a^4}{(1+a^2)^2(2a-1)^2}-\frac{2a^3}{(1+a^2)(2a-1)^3}
\right)=0.$$
So the question simplifies to the existence of the limit
$\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)}$ which I believe does not exist.
Correct ?
| Also you can do that: an improper integral $\lim_{r\to\infty}\int_0^r f(x)\, dx$ converges if and only if for each $\epsilon>0$ there is some $M>0$ such that $\left|\int_a^b f(x)\, dx\right|<\epsilon$ for all pairs $a,b\ge M$.
Now note that $\cos(x)\ge\sqrt 2/2$ when $x\in(-\pi/4+2k\pi,\pi/4+2k\pi)$ for any $k\in\Bbb Z$. Now observe that
$$x^2-x=\alpha\,\text{ and }\, x,\alpha>0\implies x=\frac{1+\sqrt{1+4\alpha}}2$$
Now for $\alpha=2\pi n\pm\pi/4$ we set $a_{n\pm}:=\frac12(1+\sqrt{1+\pi(4n\pm 1)})$ and in your case you have that
$$\begin{align}I_n&:=\left|\int_{a_{n-}}^{a_{n+}}f(x)\, dx\right|\ge\frac{\sqrt2}2\int_{a_{n-}}^{a_{n+}}\frac{x^3}{1+x^2}\, dx\\
&\ge\frac{\sqrt 2}2\cdot(a_{n+}-a_{n-})\min_{x\in[a_{n-},a_{n+}]}\frac{x}2,&\text{because }\frac{x^3}{1+x^2}\ge \frac{x}2\text{ when }x\ge 1\\
&=\frac{\sqrt 2}2(a_{n+}-a_{n-})\cdot\frac{a_{n+}+a_{n-}}{a_{n+}+a_{n-}}\cdot\frac{a_{n-}}2\\
&\ge\frac{\sqrt 2}2\cdot\frac{a^2_{n+}-a^2_{n-}}{2a_{n+}}\cdot\frac{a_{n-}}2,&\text{ because } 2a_{n+}\ge a_{n-}+a_{n+}\\
&\ge\frac{\pi\sqrt 2}{16}\cdot\sqrt{\frac{1+\pi(4n-1)}{1+\pi(4n+1)}}\end{align}$$
Hence $\lim_{n\to\infty}I_n\ge\pi\sqrt 2/16$, so we can conclude that the integral doesn't converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate squared trigonometric function I'm trying to integrate $\int_a^b \left( \frac{1}{1+x^2} \right)^2 dx$
I know that $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$, but how can I integrate with the squared part?
I've tried substitution with no success.
| Hint. Note that
$$\left( \frac{1}{1+x^2} \right)^2 = \frac{1-x^2+x^2}{(1+x^2)^2}= \frac{1}{1+x^2} -\frac{x^2}{(1+x^2)^2}=\frac{1}{1+x^2} +\frac{x}{2}\left(\frac{1}{1+x^2}\right)'.$$
then integrate by parts the last term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Additional methods for integral reduction formula So I have successfully found a reduction formula for
$$I_{m,n}=\int\frac{dx}{\sin^m(ax)\cos^n(ax)}$$
Went as follows:
$$\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\int\csc^m(ax)\sec^n(ax)dx=\int\csc^m(ax)\sec^{n-2}(ax)\sec^2(ax)dx\\
\begin{vmatrix}u=\csc^m(ax)\sec^{n-2}(ax)\\du=-am\csc^m(ax)\cot(ax)\sec^{n-2}(ax)+a(n-2)\csc^m(ax)\sec^{n-2}(ax)\tan(ax)dx\end{vmatrix}\\
\begin{vmatrix}dv=\sec^2(ax)\quad v=\frac{1}{a}\tan(ax)\end{vmatrix}\\
\int udv=uv-\int vdu\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-\int(n-2)\csc^m(ax)\sec^{n-2}(ax)\tan^2(ax)\\
-m\csc^m(ax)\sec^{n-2}(ax)dx\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-\int(n-2)\csc^m(ax)\sec^n(ax)\\
-(n-2)\csc^m(ax)\sec^{n-2}(ax)-m\csc^m(ax)\sec^{n-2}(ax)dx\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-(n-2)\int\csc^m(ax)\sec^n(ax)dx\\
+(m+n-2)\int\csc^m(ax)\sec^{n-2}(ax)dx\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-(n-2)I_{m,n}+(m+n-2)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{\csc^m(ax)\sec^{n-2}(ax)\tan(ax)}{a(n-1)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{\csc^{m-1}(ax)\sec^{n-1}(ax)}{a(n-1)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{1}{a(n-1)\sin^{m-1}(ax)\cos^{n-1}(ax)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}$$
This is definitely not a check my proof question, but I was wondering if there are more elegant or creative ways of finding this reduction formula, or if IBP, u-sub, and identities are the only ways of going with formulas like these.
| The idea of this trick is similar to IBP but this ruduces a lot in calculations (because differentiation is "easier" than integration).
Take $f_{m,n}(x)= \csc^m(ax)\sec^n(ax)$. Take the derivative with respect to $x$:
\begin{align*}
f'_{m-1,n-1}&=a(n-1)\underbrace{(1-\cos^2(ax))f_{m,n}}_{f_{m-2,n}}-a(m-1)f_{m,n-2}\\
f'_{m-1,n-1}&=a(n-1)f_{m,n}-a(m+n-2)f_{m,n-2}.\\
\end{align*}
Integrate on both sides, we arrive the reduction formula
\begin{align*}
f_{m-1, n-1}=a(n-1) \color{blue}{I_{m,n}}-a(m+n-2) I_{m, n-2},
\end{align*}
equivalent to the last line of your question.
Similarly, we can find
$$f_{m+1, n-1}=a(n-1) \color{blue}{I_{m,n}}-a(m+1) I_{m+2, n-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), on the right hand side of the page. Am I incorrect with my use of the modulus operator after I integrated each side then?
$$\begin{align*}x\frac{dy}{dx}&=y-\frac{1}{y}\quad\quad(-1<y<1)\\
\int\frac{1}{y-\frac{1}{y}}\ dy&=\int\frac{1}{x}\ dx\\
\int\frac{y}{y^2-1}\ dy&=\int\frac{1}{x}\ dx\\
\frac{1}{2}\ln{|y^2-1|}&=\ln{|x|}+c\\
\sqrt{|y^2-1|}&=A|x|\ \text{, where $A=e^c$}\\
|y^2-1|&=A^2x^2\\
\implies y^2-1&=A^2x^2&1-y^2&=A^2x^2\\
y^2&=A^2x^2+1&y^2&=1-A^2x^2\\
\\
\therefore y&=\pm\sqrt{Kx^2+1}&y&=\pm\sqrt{1-Kx^2}\ \text{, where $K=A^2$}\\
\end{align*}$$
Thanks :)
| Your solutions are $\pm\sqrt{1\pm Kx^2}$, where $K$ is positive. But this does not differ from $\pm\sqrt{1+Kx^2}$ where $K$ is unconstrained.
From
$$\frac12\log|y^2-1|=\log|x|+C$$ you draw
$$\log|y^2-1|=\log(e^Cx^2)$$ and
$$|y^2-1|=e^Cx.$$
Then
$$y^2=1\pm e^Cx^2$$ which you can very well rewrite as
$$y^2=1+Dx^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Finding a cubic polynomial with Cayley-Hamilton Theorem I have two matrices:
$A = \begin{bmatrix}
1 &2 \\
-1 &4
\end{bmatrix}
$
and
$B = \begin{bmatrix}
0 &-1 \\
2 &3
\end{bmatrix}
$
I need to find a monic cubic polynomial $g$ such that $g(A)=g(B)=0$, the zero matrix.
I understand through the Cayley-Hamilton Theorem that both $A$ and $B$ satisfy their own characteristic equations. That is,
$f_A(A)=0\,$ for $\,f_A(t)=t^2-5t+6$
and
$f_B(B)=0\,$ for $\,f_B(t)=t^2-3t+2$
I can verify this computationally or simply by citing Cayley-Hamilton, so that's fine. However, I'm not sure how to combine these and find the cubic polynomial that's been requested so that both matrices evaluate the polynomial to $0$.
Am I missing something deeper about the Cayley-Hamilton Theorem, or am I just forgetting some basic algebra tricks?
| the least common multiple of $(t-3)(t-2)$ and $(t-1)(t-2)$ is
$$ (t-1)(t-2)(t-3) $$
As with natural numbers, the LCM of two polynomials is their product divided by their gcd. If you did not notice the common root, the euclidean algorithm can find the gcd
$$ \left( x^{2} - 5 x + 6 \right) $$
$$ \left( x^{2} - 3 x + 2 \right) $$
$$ \left( x^{2} - 5 x + 6 \right) = \left( x^{2} - 3 x + 2 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( - 2 x + 4 \right) $$
$$ \left( x^{2} - 3 x + 2 \right) = \left( - 2 x + 4 \right) \cdot \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x + 1 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x + 3 }{ 2 } \right) }{ \left( \frac{ - x + 1 }{ 2 } \right) } $$
$$ \left( x - 3 \right) \left( \frac{ 1}{2 } \right) - \left( x - 1 \right) \left( \frac{ 1}{2 } \right) = \left( -1 \right) $$
$$ \left( x^{2} - 5 x + 6 \right) = \left( x - 3 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 0 \right) $$
$$ \left( x^{2} - 3 x + 2 \right) = \left( x - 1 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x - 2 \right) } $$
$$ \left( x^{2} - 5 x + 6 \right) \left( \frac{ 1}{2 } \right) - \left( x^{2} - 3 x + 2 \right) \left( \frac{ 1}{2 } \right) = \left( - x + 2 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post is someone said that this integral cannot be solved without using special functions.
Another alternative solutions will be appreciated.
| No need multiple-integrals, no need Feynman's trick
\begin{align}J&=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{\text{IBP}}=-\int_0^1 \frac{(1-2x)\ln x}{1+x(1-x)}dx\\
&=-\int_0^{\frac{1}{2}} \frac{(1-2x)\ln x}{1+x(1-x)}dx-\int_{\frac{1}{2}}^1 \frac{(1-2x)\ln x}{1+x(1-x)}dx\\
&\overset{u=x(1-x)}=-\int_0^{\frac{1}{4}}\frac{\ln\left(\frac{1-\sqrt{1-4u}}{2}\right)}{1+u}du+\int_0^{\frac{1}{4}}\frac{\ln\left(\frac{1+\sqrt{1-4u}}{2}\right)}{1+u}du\\
&=-\int_0^{\frac{1}{4}}\frac{\ln\left(\frac{1-\sqrt{1-4u}}{1+\sqrt{1-4u}}\right)}{1+u}du\overset{z=\sqrt{1-4u}}=2\int_0^1 \frac{z\ln\left(\frac{1-z}{1+z}\right)}{z^2-5}dz\\
&=\int_0^1 \frac{\ln\left(\frac{1-z}{1+z}\right)}{z-\sqrt{5}}dz+\int_0^1 \frac{\ln\left(\frac{1-z}{1+z}\right)}{z+\sqrt{5}}dz\\
&\overset{x=\frac{1-z}{1+z}}=-2\int_0^1 \frac{\ln x}{(1+x)\left((\sqrt{5}+1)x+\sqrt{5}-1\right)}+2\int_0^1 \frac{\ln x}{(1+x)\left((\sqrt{5}-1)x+\sqrt{5}+1\right)}\\
&=-\frac{\sqrt{5}+1}{\sqrt{5}-1}\int_0^1 \frac{\ln x}{\frac{\sqrt{5}+1}{\sqrt{5}-1}x+1}dx-\frac{\sqrt{5}-1}{\sqrt{5}+1}\int_0^1 \frac{\ln x}{\frac{\sqrt{5}-1}{\sqrt{5}+1}x+1}dx+2\int_0^1 \frac{\ln x}{1+x}dx
\end{align}
Let $\rho=\frac{\sqrt{5}+1}{\sqrt{5}-1}$,
\begin{align}J&=\underbrace{-\rho\int_0^1 \frac{\ln x}{\rho x+1}dx}_{u=\rho x}-\underbrace{\frac{1}{\rho}\int_0^1 \frac{\ln x}{\frac{x}{\rho}+1}dx}_{u=\frac{1}{\rho}}+2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-\int_0^\rho \frac{\ln\left(\frac{u}{\rho}\right)}{1+u}du-\int_0^{\frac{1}{\rho}} \frac{\ln\left(\rho u\right)}{1+u}du+2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-\int_0^\rho \frac{\ln u}{1+u}du-\int_0^{\frac{1}{\rho}} \frac{\ln u}{1+u}du+\ln^2\rho+2\int_0^1 \frac{\ln x}{1+x}dx\\
&=\int_\rho^1 \frac{\ln u}{1+u}du-\underbrace{\int_1^{\frac{1}{\rho}} \frac{\ln u}{1+u}du}_{z=\frac{1}{u}}+\ln^2\rho\\
&=\int_\rho^1 \frac{\ln u}{1+u}du+\int_\rho^1\frac{\ln z}{z(1+z)}+\ln^2\rho\\
&=\int_\rho^1\frac{\ln u}{u}du+\ln^2\rho=\left[\frac{\ln^2 u}{2}\right]_\rho^1+\ln^2\rho=\frac{1}{2}\ln^2\rho\\
\end{align}
Moreover,
Let $\varphi=\dfrac{\sqrt{5}+1}{2}$, then,
\begin{align}\rho=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}+1\right)^2}{4}=\varphi^2\end{align}
Therefore,
\begin{align}J=\frac{1}{2}\ln^2\rho=\boxed{2\ln^2\varphi}\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3002309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the recursive equation $y^{(n+2)} +(n+1)y^{(n+1)} +\tfrac{n(n+1)}{2} y^{(n)}=0$ I encounter the problem when I try to get the Taylor series of $\arctan x$ at $x=1$. It seems that the method for expanding it at $x=0$ does not work anymore (which is obtained by observing that $\arctan' x=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^n x^{2n}$, and then integration over the convergence domain).
So I try to compute the derivatives at $x=1$: Note that if we set $y(x)=\arctan x$, then
$$y'=\frac{1}{1+x^2},\quad y''=\frac{-2x}{(1+x^2)^2}\implies
(1+x^2)y''=-2x y'.$$
By Leibniz rule, taking derivative of order $n$ at both side, we have
$$
y^{(n+2)}(1)+(n+1)y^{(n+1)}(1)+\frac{n(n+1)}{2} y^{(n)}(1)=0.
$$
It is easy to show
$$
y(1)=\pi/4,\quad y'(1)=1/2,\quad y''(1)=-1/2,\quad y'''(1)=1/2,\quad y''''(1)=0.
$$
I don't know how to get a general formula from the above recursive equation, any ideas?
In fact, I am also searching for a general theory about recursive equations, any reference there?
| If you simply want to find the Taylor series about $x=1$ of $\text{arctan}(x)$, then you can use $$\frac{\text{d}}{\text{d}x}\,\text{arctan}(x)=\frac{1}{1+x^2}=\frac{1}{1+(z+1)^2}=\frac{1}{2\text{i}}\left(\frac{1}{1-\text{i}+z}-\frac{1}{1+\text{i}+z}\right)\,,$$
where $z:=x-1$. This gives
$$\frac{\text{d}}{\text{d}x}\,\text{arctan}(x)=\frac{1}{2\text{i}}\,\left(\frac{\left(1+\frac{z}{1-\text{i}}\right)^{-1}}{1-\text{i}}-\frac{\left(1+\frac{z}{1+\text{i}}\right)^{-1}}{1+\text{i}}\right)\,.$$
Therefore, if $|z|<\sqrt{2}$, then
$$\frac{\text{d}}{\text{d}x}\,\text{arctan}(x)=\frac{1}{2\text{i}}\,\left(\frac{\sum\limits_{k=0}^\infty\,(-1)^k\,\left(\frac{z}{1-\text{i}}\right)^{k}}{1-\text{i}}-\frac{\sum\limits_{k=0}^\infty\,(-1)^k\,\left(\frac{z}{1+\text{i}}\right)^{k}}{1+\text{i}}\right)\,.$$
Simplifying this, we have
$$\frac{\text{d}}{\text{d}x}\,\text{arctan}(x)=\sum_{k=0}^\infty\,(-1)^k\,\left(\frac{\frac1{(1-\text{i})^{k+1}}-\frac{1}{(1+\text{i})^{k+1}}}{2\text{i}}\right)\,(x-1)^k$$
for $x\in\mathbb{C}$ such that $|x-1|<\sqrt{2}$.
Using $1\pm\text{i}=\sqrt{2}\,\exp\left(\pm\dfrac{\text{i}\pi}{4}\right)$, we conclude that
$$\frac{\text{d}}{\text{d}x}\,\text{arctan}(x)=\sum_{k=0}^\infty\,\frac{(-1)^k}{2^{\frac{k+1}{2}}}\,\sin\left(\frac{(k+1)\pi}{4}\right)\,(x-1)^k$$
for all complex numbers $x$ with $|x-1|<\sqrt{2}$.
Integrating the series above, we have
$$\text{arctan}(x)=\frac{\pi}{4}+\sum_{k=1}^{\infty}\,\frac{(-1)^{k-1}}{2^{\frac{k}{2}}\,k}\,\sin\left(\frac{k\pi}{4}\right)\,(x-1)^k\,.$$
for every $x\in\mathbb{C}$ such that $|x-1|<\sqrt{2}$. In this way, it follows that
$$y^{(k)}(1)=\begin{cases}\frac{\pi}{4}&\text{if }k=0\,,\\
\frac{(-1)^{k-1}\,(k-1)!}{2^{\frac{k}{2}}}\,\sin\left(\frac{k\pi}{4}\right)&\text{if }k=1,2,3,\ldots\,,
\end{cases}$$
where $y(x):=\text{arctan}(x)$.
However, if you really want to solve the recursion without the expansion above, then let $a_k:=\dfrac{y^{(k)}(1)}{(k-1)!}$ for $k=1,2,3,\ldots$. From
$$y^{(k+2)}(1)+(k+1)\,y^{(k+1)}(1)+\frac{k(k+1)}{2}\,y^{(k)}(1)=0\text{ for }k=1,2,3,\ldots\,,$$
we divide both sides by $(k+1)!$ to get
$$a_{k+2}+a_{k+1}+\frac{1}{2}\,a_k=0\text{ for }k=1,2,3,\ldots\,.$$
The characteristic polynomial of the recursion above is $\lambda^2+\lambda+\dfrac{1}{2}$, whose roots are $\dfrac{-1\pm\text{i}}{2}$. Therefore, for $k=1,2,3,\ldots$,
$$a_k=p\,\left(\frac{-1+\text{i}}{2}\right)^k+q\,\left(\frac{-1-\text{i}}{2}\right)^k$$
for some fixed $p,q\in\mathbb{C}$. Since $a_1=\dfrac{1}{2}$ and $a_2=-\dfrac{1}{2}$, we get $$p=+\frac{\text{i}}{2}\text{ and }q=-\frac{\text{i}}{2}\,,$$
whence
$$y^{(k)}(1)=\frac{\left(\frac{-1+\text{i}}{2}\right)^k-\,\left(\frac{-1-\text{i}}{2}\right)^k}{2\text{i}}\,(k-1)!=\frac{(-1)^{k-1}\,(k-1)!}{2^{\frac{k}{2}}}\,\sin\left(\frac{k\pi}{4}\right)$$
for $k=1,2,3,\ldots$.
| {
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Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
| To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 \choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7\choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 \choose 4}+{7\choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
| {
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"timestamp": "2023-03-29T00:00:00",
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Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$
How do you find the Laurent series for $f(z) = \dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?
I know that $f(z) = \dfrac{z-12}{z^2 + z - 6} = \dfrac{-2}{z-2} + \dfrac{3}{z+3}$
It is easy for me to extract a series for $\dfrac{3}{z+3}$, but have no idea how to do it for $\dfrac{-2}{z-2}$.
Please help? Thank you!
| Use the fact that\begin{align}\frac{-2}{z-2}&=\frac{-2}{-1+(z-1)}\\&=\frac2{1-(z-1)}\\&=-2\sum_{n=-\infty}^{-1}(z-1)^n\end{align}and that\begin{align}\frac3{z+3}&=\frac3{4+(z-1)}\\&=\frac34\times\frac1{1+\frac{z-1}4}\\&=-\frac34\sum_{n=-\infty}^{-1}\frac{(-1)^n}{4^n}(z-1)^n.\end{align}
| {
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Solve for $z$, which satisfy $\arg(z-3-2i) = \frac{\pi}{6}$ and $\arg(z-3-4i) = \frac{2\pi}{3}$. solve for $z$, which satisfy $\displaystyle \arg(z-3-2i) = \frac{\pi
}{6}$ and $\displaystyle\arg(z-3-4i) = \frac{2\pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $\tan\theta = \frac{\text{imaginary part}}{\text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$\tan30° = \frac{y-2}{x-3}$
$\frac{1}{\sqrt3} = \frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+\frac{5\sqrt3}{2}-2\sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
| $$\arg(z-3-2i)=\frac{\pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^\circ$ with positive $x$ axis.
And $$\arg(z-3-4i)=\frac{2\pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^\circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y=0$$ we get
$$12y+2\lambda+2\mu y=0 \tag{2}$$
For $$L_z=0$$ we get
$$8z+\lambda=0 \tag{3}$$
From $(1)$ and $(2)$ we get
$$x=\frac{4 \lambda}{1-2\mu}$$
$$y=\frac{8 \lambda}{6-\mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 \frac{\lambda^2}{(1-2\mu)^2}+4 \frac{\lambda^2}{(6-\mu)^2}=1 \tag{4}$$
$$\frac{4 \lambda}{1-2\mu}+\frac{16 \lambda}{6-\mu}+\frac{\lambda}{8}=4 \tag{5}$$
But its tedious to solve above equations for $\lambda$ and $\mu$
Any other approach?
| Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $\frac{11}2y^2+4z^2$. But then your first equation becomes $\lambda+4\mu x=0$, which is much simpler.
| {
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Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \cos(\frac{\pi}{3})$
$(\frac{2}{2}) \cdot \frac{3}{5} + \frac{1}{2} \cdot (\frac{5}{5})$
$\frac{6}{10} + \frac{5}{10}$
$\frac{11}{10}$
| You cannot separate out the $\cos$ function as you have done in step two.
You can remember this identity.
$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$
Here $\arccos(\frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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An inequality to finalize a proof Related to this Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ and my second answer I have to prove this :
Let $a,b,c$ be real positive numbers then we have : $$\sum_{cyc}\left(\frac{a^3}{13a^2+5b^2}\right)\left(\frac{b^3}{13b^2+5c^2}\right)\geq \frac{ab+bc+ca}{18^2}$$
My try :
With Chebychev's inequality we get :
$$\sum_{cyc}\left(\frac{a^3}{13a^2+5b^2}\right)\left(\frac{b^3}{13b^2+5c^2}\right)\geq \left(\frac{ab+bc+ca}{3}\right)\left(\sum_{cyc}\left(\frac{a^2}{13a^2+5b^2}\right)\left(\frac{b^2}{13b^2+5c^2}\right)\right) $$
Remains to prove :
$$\sum_{cyc}\left(\frac{a^2}{13a^2+5b^2}\right)\left(\frac{b^2}{13b^2+5c^2}\right)\geq \frac{3}{18^2}$$
Or with the right substitution :
$$\sum_{cyc}\left(\frac{x}{13x+5y}\right)\left(\frac{y}{13y+5z}\right)\geq \frac{3}{18^2}$$
And after this I don't know what to do...
Can someone help me ?
Thanks
| It's wrong. Try $a=b=1$ and $c=2$.
We have:
$$LS=\frac{1}{18\cdot33}+\frac{8}{33\cdot57}+\frac{8}{57\cdot18}=\frac{155}{11286}$$ and
$$RS=\frac{5}{324}.$$
| {
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Find the value of $a$
Find $a$ for which $f(x) = \left(\frac{\sqrt{a+4}}{1-a} -1\right)x^5-3x+\ln5 \;$ decreases for all $x$ with $a\neq 1$ and $a\geq -4$.
My try:
For $f(x)$ to decrease
$$5x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) -3 <0\implies x^4\left(\frac{\sqrt{a+4}}{1-a} -1\right) < 3/5 $$
How can I proceed further?
| This holds true for every $x$ only if $$\frac{\sqrt{a+4}}{1-a} -1\le 0$$or $${\sqrt{a+4}\over 1-a}\le 1$$Surely $a>1$ is one region of answer. For $a<1$ we have $$\sqrt {a+4}\le 1-a\to a+4\le a^2-2a+1\to a^2-3a-3\ge 0\to\\a\ge {3+\sqrt {21}\over 2}\\a\le {3-\sqrt {21}\over 2}$$therefore the region of the answer is $$[-4, {3-\sqrt {21}\over 2}]\cup (1,\infty)$$
| {
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Solution to infinite product $\prod_{p-primes}^{\infty} \frac{p}{p-1}$ I want to find the $\prod_{p-primes}^{\infty} \frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$$
of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $\sum_0^\infty \frac{1}{2^n}$$\sum_0^\infty \frac{1}{3^n}$$\sum_0^\infty \frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $\sum_1^\infty log(a_n)$ seems to converge, but hey I could be wrong.
| An alternative approach: $\frac{p}{p-1}\geq 1+\frac{1}{p}$ gives
$$ \prod_{p}\frac{p}{p-1}\geq \prod_{p}\left(1+\frac{1}{p}\right) =\!\!\!\!\!\!\!\! \sum_{\substack{n\geq 1\\n\text{ squarefree}}}\!\!\!\!\!\!\!\frac{1}{n} $$
but since the set of squarefree numbers has a positive density in $\mathbb{N}$ ($\frac{6}{\pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that
$$\sum_{\substack{1\leq n\leq N\\n\text{ squarefree}}}\!\!\!\!\frac{1}{n}=O(1)+\frac{6}{\pi^2}\sum_{n=1}^{N}\frac{1}{n}=\frac{6}{\pi^2}\log(N)+O(1) $$
so the original product is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is my solution incorrect for solving these quadratic equations? $$\frac2x -\frac5{\sqrt{x}}=1 \qquad \qquad 10)\ \frac3n -\frac7{\sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = \sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$\displaystyle \frac{2}{y^{2}} - \frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$\displaystyle y = -\frac{5}{2} \pm \frac{\sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $\displaystyle y = \frac{58}{4} \pm \frac{10\sqrt{33}}{4}$
But this isn't the correct solution.
Also for $\#10$ I do the same thing:
Let $y = \sqrt n$ then $y^2 = n$
So I have $\displaystyle \frac{3}{y^2} - \frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $\displaystyle y = -\frac{3}{2}, \frac{1}{3}$
Then $n = \frac{9}{4}, \frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1)\ (x-7)^2 -13(x-7) +36=0 \qquad \qquad 4)\ 3(w/6)^2 -8(w/6) +4=0 \\
2)\ (1-3x)^2 -13(1-3x) +36=0 \qquad \qquad 5)\ 3(w^2-2)^2 -8(w^2-2) +4=0 \\
3)\ x^4 -13x^2 +36=0 \qquad \qquad 6)\ \frac{3}{p^2} -\frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
| You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $\sqrt{x}$, which is your substitution, can only be positive.
| {
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For pair of st. lines , length of line joining feet of perpendiculars from $(f,g)$ to them is$\sqrt {4.\frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$ Consider a pair of straight lines through the origin,
$$ax^2+2hxy+by^2=0$$
This can be written as,
$$y=m_{1,2}x$$
where $m_{1,2}=-\frac {a}{h±\sqrt {h^2-ab}}$. Now, suppose a point $(f,g)$ whence perpendiculars are drawn to both the lines . Then the eqations of the perpendiculars are,
$$y=-\frac {1}{m_{1,2}}x + C_{1,2}$$
where $C_{1,2}=g-\frac {h±\sqrt {h^2-ab}}{a}f$. Finally , we can find the coordinates of the feet of perpendiculars using the formula of intersection coordinates:
$$(x,y)=(\frac {c_1-c_2}{m_2-m_1},\frac {c_1m_2-c_2m_1}{m_2-m_1})$$
Hence using the distance formula, the distance between the feet of perpendiculars is,
$$s=\sqrt {(\frac {C_2m_2}{1+m_2^2}-\frac {C_1m_1}{1+m_1^2})^2+(\frac {C_2m_2}{m_2+\frac {1}{m_2}}-\frac {C_1m_1}{m_1+\frac {1}{m_1}})^2}$$
I am unable to proceed further to get the desired result i.e. $\sqrt {4.\frac {(h^2-ab)(f^2+g^2)}{(a-b)^2+4h^2}}$. Initially I felt the need of simplification, but that too didn't work , when even my simplified expression $s=\sqrt {\frac {(1+m_1^2)(f-gm_2)^2+(1+m_2^2)(f-gm_1)^2-2(f-gm_1)(f-gm_2)(1+m_1m_2)}{(1+m_1^2)(1+m_2^2)}}$ needed a CAS to be rendered in the desired form. Any help is welcome.
| Write the lines like $(Ax+By)(A'x+B'y)=AA'x^2+2\frac{AB'+A'B}{2}xy+BB'y^2=0$
Then the system of one line and its perpendicular through $(f,g)$ is $\langle Ax+By,Ay-Bx-Ag+Bf\rangle$ and has solution $(x,y)=(\frac{B}{A^2+B^2}(Bf-Ag),\frac{A}{A^2+B^2}(Ag-Bf))$ and similarly for the primed, which gives the distance $$\sqrt{\frac{(AB'-A'B)^2(f^2+g^2)}{(A^2+B^2)(A'^2+B'^2)}}.$$ Now $h^2-ab=\frac{(AB'-A'B)^2}{4}$ and $(a-b)^2+4h^2=(A^2+B^2)(A'^2+B'^2)$ and were done.
| {
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Solve differential equation $\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$ I want to solve the following first- order nonlinear ordinary differential equation:
$\frac{dx}{dt}=a-b x-cx(1-x)=cx^2-x(b+c)+a$
where a,b and c are constants. I rewrote the equation:
$\leftrightarrow 1=\frac{1}{cx^2-x(b+c)+a}\frac{dx}{dt}\\
\leftrightarrow \int 1dt=\int \frac{1}{cx^2-x(b+c)+a} dx\\
\leftrightarrow t+k= \int \frac{1}{cx^2-x(b+c)+a} dx\\
\leftrightarrow t+k= \int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx
$
for some arbitrary number k. How do I solve the last integral? Wolfram-Alpha tells me that it is
$\frac{2tan^{-1}(\frac{-c-b+2cx}{\sqrt{-c^2-b^2-2cb+4ca}})}{\sqrt{-c^2-b^2-2cb+4ca}}$
But I don't know how to calculate that on my own.
| After completing the square the integral has the form:
$\int \frac{1}{c(x-\frac{b+c}{2c})^2+a-\frac{(b+c)^2}{4c}} dx=\frac{1}{c}\int \frac{1}{(x-\frac{b+c}{2c})^2+\frac{a}{c}-\frac{(b+c)^2}{4}} $
By defining $y:=x-\frac{b+c}{2c}$ and $p^2:=\frac{a}{c}-\frac{(b+c)^2}{4}$ we get:
$\frac{1}{c}\int \frac{1}{(x-\frac{b+c}{2c})^2+\frac{a}{c}-\frac{(b+c)^2}{4}}=\frac{1}{c}\int\frac{1}{y^2+p^2}=\frac{1}{c}\frac{arctan(\frac{y}{p})}{p}+k_1=\frac{1}{c}\frac{arctan(\frac{x-\frac{b+c}{2c}}{\sqrt{\frac{a}{c}-\frac{(b+c)^2}{4}}})}{\sqrt{\frac{a}{c}-\frac{(b+c)^2}{4}}}+k_1$
for some arbitrary number $k_1$.
| {
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limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6}\sqrt{(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\require{cancel} \cancel{x^3} \sqrt{(9-\frac {5}{x^2})}}{\require{cancel} \cancel{x^3}(1-\frac {2}{x}+\frac{1}{x^3})}} =
\frac {3}{1} = +3$$
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
I'll be glad to get your help!
Thank you.
|
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
Because $\sqrt{a^2}=a$ is only true if $a \ge 0$; for $a \le 0$, you have $\sqrt{a^2}=-a$.
You can summarize this as follows (and remember by heart!), for all $a$ you have:
$$\boxed{\sqrt{a^2}=|a|}$$
Apply this to $a=x^3$.
| {
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Primes dividing $x^2+xy+y^2+1$
Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y \in \Bbb Z$ ?
I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y \in \Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|(x^{2} + y^{2} + 1) $. :
the sets $\{-1-x^2 \mid x \in \Bbb F_p\}$ and $\{y^2 \mid y \in \Bbb F_p\}$ have both cardinality $(p+1)/2$, so they must have a non-empty intersection, i.e. an element of the form $z=-1-x^2=y^2$.
But the proof does not work to show that all but finitely many (or at least infinitely many) primes divide $x^2+xy+y^2+1$ for some integers $x,y$. So what can I do?
| If $p\ne2$ then $p\mid(x^2+xy+y^2+1)$ iff $p\mid(4x^2+4xy+4y^2+4)$
that is iff $p\mid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
are $y$ and $z$ with $p\mid(z^2+3y^2+4)$.
| {
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Find $\lim_{n\rightarrow\infty} \left(\frac 1e\left( 1+\frac1n+\frac c{n^2}\right)^n\right)^n$
Find
$$\lim_{n\rightarrow\infty} \left(\frac 1e\left( 1+\frac1n+\frac c{n^2}\right)^n\right)^n.$$
According to Wolfram the limit is $e^{c-1/2}$. I have simplified the expression to
$$ \exp \lim \left[ n^2 \ln \left(1 + \frac{1}{n} + \frac{c}{n^2} \right) - n\right]. $$
I want to approximate $ \ln \left(1 + \frac{1}{n} + \frac{c}{n^2}\right) $ with
a Taylor polynomial with $ n=3 $, $ x_0 =0 $ and $ f(x) = \ln (1+x) $.
I have tried
$$ f(1/n) = \frac{1}{n} - \frac{\left(\frac{1}{n}\right)^2}{2} +
\frac{\left(\frac{1}{n}\right)^3}{3} - b_n$$
with $ |b_n| < \frac{1}{4n^4} $.
Then
$$ e^{\lim\left[ n - \frac{1}{2} + \frac{c}{3n} + b_n n^2 - n \right] }.$$
So $ b_n n^2 $ will go to 0 and the $ n $'s will cancel out.
I'm not sure how to get $ -\frac{1}{2} + c $ in the exponent because of the $ 3n $ in the
denominator.
| Put $x=\dfrac 1n+\dfrac c{n^2}$, then
\begin{align*}
n^2f(x)-n&=\color{blue}{n^2}\left(\dfrac 1n+\color{blue}{\dfrac c{n^2}}-\frac12\left(\dfrac 1n+\dfrac c{n^2}\right)^2+o(\frac1{n^2})\right)-n\\
&=\color{blue}c-\frac12+o(1).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$ I am in the middle of proving that
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$
And I have reduced the series to
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt=\int_{-1}^0\frac{\log(2u+i\sqrt3+1)}u\mathrm du+\int_{-1}^0\frac{\log(2u-i\sqrt3+1)}u\mathrm du-2\log2\int_{-1}^0\frac{\mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.
| Firstly observe that, for $x$ real,
\begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\\
&=x^2-x+1
\end{align}
\begin{align}J&=\int_0^1 \frac{\ln(x^2-x+1)}{x(x-1)}\,dx\\
&=-\int_0^1 \frac{\ln(x^2-x+1)}{1-x}\,dx-\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=1-x$,
\begin{align}J&=-2\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx\\
&=-2\int_0^1 \frac{\ln\left(\frac{x^3+1}{x+1}\right)}{x}\,dx\\
&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{\ln\left(x^3+1\right)}{x}\,dx\\
&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{x^2\ln\left(x^3+1\right)}{x^3}\,dx\\
\end{align}
In the latter integral perform the change of variable $y=x^3$,
\begin{align}J&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-\frac{2}{3}\int_0^1
\frac{\ln\left(x+1\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\ln\left(1-x^2\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{x\ln\left(1-x^2\right)}{x^2}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=-\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=-\frac{2}{3}\Big[\ln x\ln(1-x)\Big]_0^1 -\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=-\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=\frac{2}{3}\zeta(2)\\
&=\frac{2}{3}\times \frac{\pi^2}{6}\\
&=\boxed{\frac{\pi^2}{9}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Proving a three variables inequality Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+c\ge3\sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)\ge63$$The last thing I tought about was that I have both $a+b+c\ge9$ and $a+b+c\ge3\sqrt[3]{abc}$ so if I somehow related them I would have $\sqrt[3]{abc} \ge 3 \rightarrow abc\ge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
| I think you mean $$a,b,c>0$$ in this case we have
$$\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ so $$a+b+c\geq 9$$ and we get also $$\frac{bc+ac+ab}{3}\geq \sqrt[3]{(abc)^2}$$ so $$abc\geq 27$$ and $$ab+ac+bc\geq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1\geq 27+27+10=64$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$$\displaystyle =\displaystyle \lim\limits _{x\rightarrow \infty }\dfrac{\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }\right)}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }} =$
$\displaystyle =\lim\limits _{x\rightarrow \infty }\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}$
What is the next step should be? Please help!
| $$
\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}-\sqrt{2x^4}=
\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4}}}=
\dfrac{x^{2}\sqrt{x^{4} +1} -x^{4}}{\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4} \ }}\cdot\frac{x^{2}\sqrt{x^{4} +1} +x^{4}}{x^{2}\sqrt{x^{4} +1} +x^{4}}\\=\frac{x^4}{\big(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} +\sqrt{2x^{4}}\big)\big(x^{2}\sqrt{x^{4} +1}+x^4\big)}\\=\frac{1}{\big(\sqrt{1 +\sqrt{1+x^{-4}}} +\sqrt{2}\big)\big(\sqrt{x^{4}+1}+x^2\big)}\,\to\,0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Minimizing distance from a point to a parabola Problem: The point on the curve $x^2 + 2y = 0$ that is nearest the point $\left(0, -\frac{1}{2}\right)$ occurs at what value of y?
Using the distance formula, I get my primary equation: $L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
However, when using the secondary equation $x^2 + 2y = 0$ to write $L^2$ as a function of a single variable and then minimizing $L^2$, thus minimizing $L$, I am getting conflicting answers.
Let $D = L^2 = (x-0)^2 + \left(y-\left(-\frac{1}{2}\right)\right)^2$
Approach 1: substituting $y = -\frac{x^2}{2}$ into our primary equation,
$$D = (x-0)^2 + \left(-\frac{x^2}{2}-\left(-\frac{1}{2}\right)\right)^2$$
$$\frac{dD}{dx} = 2x + 2\left(\frac{x^2}{2}-\frac{1}{2}\right)(x)$$
$$\frac{dD}{dx} = x(x^2+1)$$
Therefore, $D$ has a critical point at x = 0. Using the first derivative test, $\frac{dD}{dx} < 0$ for all $x<0$ and $\frac{dD}{dx}>0$ for all $x>0$, therefore a minimum exists at $x = 0$. At $x=0$, $y = 0$.
Approach 2: substituting $x^2 = -2y$ into our primary equation,
$$D = -2y + \left(y-\left(-\frac{1}{2}\right)\right)^2$$
$$\frac{dD}{dy} = -2 + 2\left(y+\frac{1}{2}\right)$$
$$\frac{dD}{dy} = 2y-1$$
Therefore, $D$ has a critical point at $y = \frac{1}{2}$. Using the first derivative test, $\frac{dD}{dy} < 0$ for all $y<\frac{1}{2}$ and $\frac{dD}{dy}>0$ for all $y>\frac{1}{2}$, therefore a minimum exists at $y = \frac{1}{2}$.
Why do the results differ? What faulty assumption(s) have I made? Thanks in advance.
| Note that $x\in (-\infty,+\infty), \ y\in (-\infty, 0]$. Your approach $2$ has implicit constraint on $y\le 0$ and you must check the border too. Hence $y=0$ is an optimal point.
Also note for $y=\frac 12$, $x^2=-1$, which gives complex roots as in the approach $1$: $x^2+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$
Solve this equation:
$$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\log_{4}x}}{\sqrt{3}} = \sqrt{x}$$
$$\Leftrightarrow 3^{\log_{4}x}.3+ 3^{\log_{4}x} = \sqrt{3x}$$
$$\Leftrightarrow 4.3^{\log_{4}x}= \sqrt{3x}$$
At this step, I can't continue. Please help me!
| You may also continue as follows:
$$\begin{eqnarray*}
4 \cdot 3^{2 \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 9^{\log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 4^{\log_4{9} \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
(\sqrt{x})^{\log_4{9} -1} &= & \frac{\sqrt{3}}{4} \Leftrightarrow\\
x & = & \left( \frac{\sqrt{3}}{4}\right)^{\frac{2}{\log_4{9}-1}} \approx 0.0571725
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Derivative of piecewise function with $\sin\frac{1}{x}$ term I was going through my calculus book, and I am not sure I understand this part
$f(x) = \begin{cases} \frac{x^2}{4}+x^4\sin(\frac{1}{x}) &\text{if $x\neq0$ } \\ 0 &\text{if $x=0$ } \end{cases}$
$ f'(x) = \begin{cases} \frac{x}{2}-x^2\cos(\frac{1}{x})+4x^3\sin(\frac{1}{x}) &\text{if $x\neq0$ } \\ 0 &\text{if $x=0$ } \end{cases}$
$ f''(x) = \begin{cases} \frac{1}{2}+12x^2\sin(\frac{1}{x})-\sin(\frac{1}{x})-6x\cos(\frac{1}{x}) &\text{if $x\neq0$ } \\ \frac{1}{2} &\text{if $x=0$ } \end{cases}$
So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $\frac{1}{2}$, or rather why does the $\sin\frac{1}{x}$ term go to 0?
| hint
$$f''(0)=\lim_{x\to 0,x\ne 0}\frac{f'(x)-f'(0)}{x-0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039786",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$ As part of a recent question I posted, I decided to try and generalise for a power of $2$ to any $r \in \mathbb{R}$. As part of the method I took, I had to solve the following integral:
\begin{equation}
I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx
\end{equation}
I believe what I've done is correct, but I'm concerned that I may missed something (in particular whether it holds for all $r \neq 0$). So, here I have two questions (1) Is what I've done correct? and (2) What other methods can be employed that doesn't rely on complex analysis?
Here is the method I took:
First make the substitution $u = x^{\frac{1}{r}}$ to arrive at
\begin{equation}
I = \frac{1}{n} \int_{0}^{\infty} \frac{1}{1 + u} \cdot u^{1 -\frac{1}{r}}\:du
\end{equation}
We now substitute $t = \frac{1}{1 + u}$ to arrive at:
\begin{align}
I &= \frac{1}{r} \int_{1}^{0} t \cdot \left(\frac{1 - t}{t}\right)^{\frac{1}{r} -1}\frac{1}{t^2}\:dt = \frac{1}{r}\int_{0}^{1}t^{-\frac{1}{r}}\left(1 - t\right)^{ \frac{1}{r} - 1}\:dt \\
&= \frac{1}{r}B\left(1 - \frac{1}{n}, 1 + \frac{1}{r} - 1\right) = \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right) \\
&= \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right)
\end{align}
Wheer $B(a,b)$ is the Beta function. Using the relationship between the Beta and Gamma function we arrive at:
\begin{equation}
I = \frac{1}{r} \frac{\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)}{\Gamma\left(\frac{r - 1}{r} + \frac{1}{r}\right)} = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)
\end{equation}
And so, we arrive at:
\begin{equation}
I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)
\end{equation}
for $r > 1$
As per KemonoChen's comment and others, we can employ Euler's Reflection Formula to position this result for $\frac{1}{r} \not \in \mathbb{Z}$ Here, as $r \in \mathbb{R}, r > 1 \rightarrow \frac{1}{r} \not \in \mathbb{Z}$ and so our formula holds.
\begin{equation}
I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right) = \frac{\pi}{r\sin\left(\frac{\pi}{r} \right)}
\end{equation}
Thank you also to Winther, Jjagmath, and MrTaurho's for their comments and corrections/clarifications.
| NOT A FULL SOLUTION:
I've been working with special cases of the integral.
Here we will consider $r = 2m$ where $m \in \mathbb{N}$. In doing so, we observe that the roots of the denominator are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
\begin{align}
x^{2m} + 1 = 0 \rightarrow x^{2m} = e^{\pi i}
\end{align}
By De Moivre's formula, we observe that:
\begin{align}
x = \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) \mbox{ for } j = 0\dots 2m - 1
\end{align}
Which we can express as the set
\begin{align}
S &= \Bigg\{ \exp\left({\frac{\pi + 2\pi \cdot 0}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi \cdot 1}{2m} i} \right),\dots,\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 2)}{2m} i} \right)\\
&\qquad\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 1)}{2m} i} \right)\Bigg\} \\
\end{align}
Which can be expressed as the set of 2-tuples
\begin{align}
S &= \left\{ \left( \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi(2m - 1 - j )}{2m} i} \right)\right)\: \bigg|\: j = 0 \dots m - 1\right\}\\
& = \left\{ (z_j, c\left(z_j\right)\:|\: j = 0 \dots m - 1 \right\}
\end{align}
From here, we can factor $x^{2m} + 1$ into the form
\begin{align}
x^{2m} + 1 &= \prod_{r \in S} \left(x + r_j\right)\left(x + c(r_j)\right) \\
&= \prod_{r \in S} \left(x^2 + \left(r_j + c(r_j)\right)x + r_j c(r_j)\right) \\
&= \prod_{r \in S} \left(x^2 + 2\Re\left(r_j\right)x + \left|r_j \right|^2\right)
\end{align}
For our case here $\left|r_j \right|^2 = 1$ and $\Re\left(r_j\right) = \cos\left({\frac{\pi + 2\pi j}{2m} } \right)$. Hence,
\begin{align}
\frac{1}{x^{2m} + 1} = \prod_{j = 0}^{m - 1}\frac{1}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1}
\end{align}
From here, to evaluate the integral we must employ Partial Fraction Decomposition:
\begin{align}
\frac{1}{x^{2m} + 1} = \prod_{j = 0}^{m - 1}\frac{1}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1} = \sum_{j = 0}^{m - 1}\frac{\alpha_j + \beta_jx}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1}
\end{align}
And solve for $\alpha_j$ and $\beta_j$. Putting the coefficents to the side we can find general expressions for the integral:
\begin{align}
\frac{1}{x^{2m} + 1} = \prod_{j = 0}^{m - 1}\frac{1}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1}
\end{align}
From here, to evaluate the integral we must employ Partial Fraction Decomposition:
\begin{align}
\int_{0}^{\infty}\frac{1}{x^{2m} + 1}\:dx &= \int_{0}^{\infty}\sum_{j = 0}^{m - 1}\frac{\alpha_j + \beta_jx}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1}\:dx \\
&= \sum_{j = 0}^{m - 1}\left[ \int_{0}^{\infty}\frac{\alpha_j}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1}\:dx + \int_{0}^{\infty}\frac{\beta_jx}{ x^2 + 2\cos\left({\frac{\pi + 2\pi j}{2m} } \right)x + 1}\:dx\right] \\
&=\sum_{j = 0}^{m - 1}\left[ I_1 + I_2\right]
\end{align}
Evaluating each individually:
\begin{align}
\int_{0}^{\infty} \frac{\alpha_j}{ x^2 + 2\cos\left(\frac{\pi + 2\pi j}{2m} \right)x + 1}\:dx &= \left[ \csc\left(\frac{\pi + 2\pi j}{2m} \right)\arctan\left(\frac{(x - 1)\tan\left(\frac{\pi + 2\pi j}{4m} \right)}{x + 1} \right)\right]_{0}^{\infty} \\
&= \csc\left(\frac{\pi + 2\pi j}{2m} \right)\left( \frac{\pi + 2\pi j}{2m}\right)
\end{align}
Now if we consider the second integral on it's own we find that as a general expression the integral doesn't converge. This doesn't take away from this method, but to proceed we must solve for the unknown coefficients to proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Proof by mathematical induction that, for all non-negative integers $n$, $7^{2n+1} + 5^{n+3}$ is divisible by $44$. I have been trying to solve this problem, but I have not been able to figure it out using any simple techniques. Would someone give me the ropes please?
Prove for $n=1$
$7^3 + 5^4 = 968 = 44(22)$
Assume $F(k)$ is true and try $F(k+1)+-F(k)$
$F(k) = 7^{2k+1} + 5^{k+3}$
$F(k+1) = 7^{2k+3} + 5^{k+4}$
$49*7^{2k+1} + 5*5^{k+3} +- [7^{2k+1} + 5^{k+3}]$
Neither addition nor subtraction gives us a chance to take $44$ in front of the bracket, the best approach it has come to my mind is to subtract $F(k)$ which gives us:
$4[12*7^{2k+1} + 5^{5+3}]$
And now we need to prove that the inside is divisible by $11$. I have not been able to do so.
| Base case : n=1✓
Hypothesis : $44$ divides $7^{2n+1} +5^{n+3}$.
Step $n+1$:
$7^{2(n+1)+1}+ 5^{(n+1)+1}=$
$7^{2n+1}7^2+ 5^{n+1}5=$
$7^{2n+1}(44+5)+5^{n+1}5=$
$(44)7^{2n+1} +5(7^{2n+1}+5^{n+1}).$
First term divisible by $44$, so is the second term by hypothesis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Find the geometric locus $z \in \mathbb C$ so that $\frac{z+2}{z(z+1)}\in \mathbb R$
Find the geometric locus of the set of $z \in \mathbb C$ so that
$$\frac{z+2}{z(z+1)}\in \mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)
My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).
Hints and solutions are welcomed.
| Using
$$
z = \sqrt{x^2+y^2}e^{\arctan\frac yx}
$$
we have
$$
\frac{z+2}{z(z+1)} = \frac{\rho_1 e^{i\phi_1}}{\rho_2 e^{i\phi_2}\rho_3 e^{i\phi_3}}
$$
and we seek for
$$
\arctan\frac{y}{x+2}-\arctan\frac yx -\arctan\frac{y}{x+1} = 0
$$
then
$$
\tan\left(\arctan\frac{y}{x+2} -\arctan\frac{y}{x+1}\right) = \frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b}
$$
so
$$
\tan\left(\arctan\frac{y}{x+2} -\arctan\frac{y}{x+1}\right) = \frac{y}{y^2+x^2+3x+2}
$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Global extrema of $f(x,y)= e^{-4x^2-9y^2}(2x+3y)$ on the ellipse $4x^2+9y^2 \leq 72$ I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 \leq 72$
So for the inside, I have $\nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-36xy-54y^2)e^{-4x^2-9y^2})= (0, 0)$. I get two equations $$-16x^2-24xy+2=0$$ $$-54y^2-36xy+3=0$$ Adding $\frac{-2}{3}$ of the second to the first gives $-16x^3+36y^2=0$,$\to$ $9y^2=4x^2$ $\to$ $ y = \pm \frac{2x}{3}$. Putting that in our original function, I get $e^{-8x^2}(4x)$ and $e^{-8x^2}(0)=0$ So deriving the first (because the second gives $0$) and setting it equal to $0$, I get $x=\pm \frac{1}{16}$ implies $y=\pm \frac{1}{24}$
For the border, I use Lagrange : $$\nabla f = \lambda \nabla g$$ and get : $$-2 e^{-4 x^2 - 9 y^2} (8 x^2 + 12 x y - 1) = \lambda 8x $$ and $$e^{-4 x^2 - 9 y^2} (-36 x y - 54 y^2 + 3)=\lambda 18y$$
So $$\frac{-2(8x^2+12xy-1)}{8x}= \frac{(-36xy-54y^2+3)}{18y}$$
After rearranging, I get $-24x^2+12xy=-54y^2+1$ which seems a bit ugly. According to WolframAlpha, the solution for that is $y=\frac{2x}{3}$ which is one of the value (the other being $-\frac{2x}{3}) $ for the inside of the region that I got before, which I find a bit strange. So my questions are :
1) Is what I did until now even correct ?
2) If not, what did I do wrong and how should I proceed instead ?
3) If yes, how should I proceed now ?
4) Also for the inside values I got (i.e. $x=\pm \frac{1}{16}$, $y =\pm \frac{1}{24} $) : I just need to plug them into the function and at the end find the greatest and smallest values from all candidates, is that correct (I'm quite sure, just want to get a confirmation) ?
Thanks for your help !
| From the lagrangian
$$
L(x,y,\mu,\epsilon) = f(x,y) + \mu(g(x,y)-72+\epsilon^2)
$$
here $\epsilon$ is a slack variable to transform the inequality constraint into an equality.
The stationary conditions are
$$
\left\{
\begin{array}{rcl}
8 \lambda x-8 (2 x+3 y) x+2& = & 0 \\
18 \lambda y-18 (2 x+3 y) y+3& =& 0 \\
\epsilon ^2+4 x^2+9 y^2-72 & = & 0 \\
2 \epsilon \lambda & = & 0 \\
\end{array}
\right.
$$
here $\lambda = \mu e^{4x^2+9y^2}$. After solving we obtain
$$
\begin{array}{ccccc}
x & y & \lambda & \epsilon & f(x,y) \\
-3 & -2 & -\frac{143}{12} & 0 & -\frac{12}{e^{72}} \\
-\frac{1}{4} & -\frac{1}{6} & 0 & \sqrt{\frac{143}{2}} & -\frac{1}{\sqrt{e}} \\
\frac{1}{4} & \frac{1}{6} & 0 & \sqrt{\frac{143}{2}} & \frac{1}{\sqrt{e}} \\
3 & 2 & \frac{143}{12} & 0 & \frac{12}{e^{72}} \\
\end{array}
$$
NOTE
Solutions with $\epsilon = 0$ are solutions at the boundary and solutions with $\epsilon \ne 0$ are interior at the feasible region.
Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.
| {
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Find a diagonal form of the quadratic form $f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j$ Find a diagonal form of the quadratic form
$$f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j.$$
It turned out to be such a problem:
How to change quadratic form $f(x) = \sum_{i = 1}^nx_i^2 + \sum_{i < j}x_ix_j$ into diagonal form?
We can solve it with congruent transformation, although it is a little complex.
Besides, I tried to transform it into a diagonal form by orthogonal transformation, since its quadratic matrix has the same diagonal elements and the same non-diagonal elements(I thought that I can find a simple way to find its eigenvalues and eigenvectors, but failed. Maybe you can try it).
Is there any other way?
| This is too long for the comment box.
The essence of your exercise is given in the following example.
Suppose we have $f(x, y, z) = x^2 + y^2 + z^2 + xy + xz + yz.$ We complete the square by getting rid of one of the variables, say we pick the first one ($x$ in my notation). So you get,
$$\begin{align*}
f(x, y, z) &= (x^2 + xy + xz) + y^2 + z^2 + yz \\
&= \left( x + \dfrac{y}{2} + \dfrac{z}{2} \right)^2 - \frac{y^2}{4} - \frac{z^2}{4} -\frac{yz}{2} + y^2 + z^2 + yz\\
&=u^2 + \dfrac{3y^2}{4} + \dfrac{yz}{2} + \dfrac{3z^2}{4}
\end{align*},$$
where $u = x + \dfrac{y}{2} + \dfrac{z}{2}.$ Now, we can get rid of the second variable, $y$ in my case, by, again, completing the square
$$\begin{align*}
f(x, y, z) &= u^2 + \left( \dfrac{\sqrt{3}y}{2} + \frac{z}{2\sqrt{3}} \right)^2-\dfrac{z^2}{12}+\dfrac{3z^2}{4} = u^2 + v^2+\dfrac{5}{6}z^2
\end{align*},$$
which is a required representation of $f.$
The case with $x_1, \ldots, x_p$ is just guessing a pattern and using induction (I am guessing).
| {
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Find the value of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$
Find the value of this :$$1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\frac{1}{25}....$$
Try: We can write the above series as
$${S} = \int^{1}_{0}\bigg[1-x^6+x^8-x^{14}+x^{16}-x^{22}+\cdots\bigg]dx$$
$$S = \int^{1}_{0}(1-x^6)\bigg[1+x^{8}+x^{16}+\cdots \cdots \bigg]dx$$
So $$S = \int^{1}_{0}\frac{1-x^6}{1-x^{8}}dx = \int^{1}_{0}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now i am struck in that integration.
Did not understand how to solve it
could some help me to solve it. Thanks in advance
| The integral is tedious. We have, by Sophie Germain, that $x^4+1=(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x+1)$. Now, partial fraction and arctan gives the indefinite to be $\frac{1}{4}\left(2\tan^{-1}x+\sqrt{2}\left(\tan^{-1}\left(\sqrt{2}x+1\right)-\tan^{-1}\left(1-\sqrt{2}x\right)\right)\right)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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An inequality for convex function from $\mathbb R$ to $\mathbb R$ If $f(x): \mathbb R \to \mathbb R$ is a convex function, prove that
$$f(x_1) + f(x_2) + f(x_3) + 3 f(\frac{x_1 + x_2 + x_3}{3}) \geq 2 f(\frac{x_1 + x_2}{2}) + 2 f(\frac{x_1 + x_3}{2}) + 2 f(\frac{x_2 + x_3}{2})$$
What I have tried:
If two of $x_1$, $x_2$, $x_3$ are equal, without loss of generality, assuming $x_2 = x_3$, then the inequality degenerate to
\begin{align}
f(x_1) + 3 f(\frac{x_1 + 2 x_2}{3}) \geq 4 f(\frac{x_1 + x_2}{2})
\end{align}
Since $f$ is convex, we have
$
\frac{1}{4} f(x_1) + \frac{3}{4} f(\frac{x_1 + 2 x_2}{3})
\geq f(\frac{1}{4} x_1 + \frac{3}{4} \frac{x_1 + 2 x_2}{3})
\geq f(\frac{x_1 + x_2}{2})
$, thus the inequality holds.
Then we consider the case where $x_1$, $x_2$, $x_3$ are distinct. Without loss of generality, we assume $0 = x_1 < x_2 = \lambda \leq \frac{1}{2} < x_3 = 1$. Then the inequality degenerates to
\begin{align}
\frac{1}{2} f(0) + \frac{1}{2} f(\lambda) + \frac{3}{2} f(\frac{\lambda + 1}{3}) + \frac{1}{2} f(1) \geq f(\frac{\lambda}{2}) + f(\frac{1}{2}) + f(\frac{\lambda + 1}{2})
\end{align}
If $x_2 = \frac{1}{2}$, it further degenerates to
$\frac{1}{2} f(0) + \frac{1}{2} f(0.5) + \frac{1}{2} f(0.5) + \frac{1}{2} f(1) \geq f(0.25) + f(0.75),$
which is obviously true. Thus we can only consider $0 < \lambda < \frac{1}{2}$ (then $0 < \frac{\lambda}{2} < \lambda < \frac{\lambda + 1}{3} < \frac{1}{2} < \frac{\lambda + 1}{2} < 1$).
| A proof of Popoviciu's inequality follows from Jensen's Inequality for convex functions. Wlog, assume $x \le y \le z$ there are two possibilities. Either $\displaystyle y < \frac{x+y+z}{3}$ or the reverse.
In the first case, $\displaystyle \frac{x+y+z}{3} < \frac{x+z}{2}< z$ and $\displaystyle \frac{x+y+z}{3} < \frac{y+z}{2} < z$, thus, there are positive real numbers $\lambda_1,\lambda_2 >0$ such that,
$\displaystyle \frac{x+z}{2} = \lambda_1 z + (1-\lambda_1)\left(\frac{x+y+z}{3}\right)$ and $\displaystyle \frac{y+z}{2} = \lambda_2 z + (1-\lambda_2)\left(\frac{x+y+z}{3}\right)$
It is easy to verify that $\lambda_1 + \lambda_2 = \dfrac{1}{2}$.
Now, using Jensen's inequality one has:
$\displaystyle f\left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}$
$\displaystyle f\left(\frac{x+z}{2}\right) = \lambda_1 f(z) + (1-\lambda_1)f\left(\frac{x+y+z}{3}\right)$
$\displaystyle f\left(\frac{y+z}{2}\right) = \lambda_2 f(z) + (1-\lambda_2)f\left(\frac{x+y+z}{3}\right)$
Adding the three inequalities gives the Popoviciu's inequality. The second case where $\displaystyle y > \frac{x+y+z}{3}$ we simply interchange the roles of $x$ and $z$ and the rest is similar.
| {
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"url": "https://math.stackexchange.com/questions/3046440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify the determinant of a $4 \times 4$ matrix. I have to find the determinant of the following 4x4 matrix:
$\quad A=\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$det\begin{bmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{bmatrix}=\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\xrightarrow{3R_3-R_1}\begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&8&0\\0&0&0&-4\end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $\quad 0 \quad 0 \quad 2 \quad 0 \quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
| It should be: $$\begin{vmatrix}3&0&1&0\\0&2&0&0\\1&0&3&0\\0&0&0&-4\end{vmatrix}\rightarrow \begin{vmatrix}3&0&1&0\\0&2&0&0\\0&0&\color{red}{3-{1\over 3}}&0\\0&0&0&-4\end{vmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can $3p^4-3p^2+1$ be square number? I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?
| Partial solution if $p$ is prime.
Write $$3p^4-3p^2+1=n^2\implies 3p^2(p^2-1) = (n-1)(n+1)$$
If $p\ne 2$ (which is not a solution) then $p^2\mid n-1$ or $p^2\mid n+1 $
First case: If $p^2\mid n-1$ then $n+1\mid 3p^2-3$ so $ n-1= p^2k$ and $n+1\leq 3p^2-3$.
If $k\geq 3$ then $$3p^2-3\geq n+1 >n-1 \geq 3p^2$$ which is impossible. So $k\leq 2$
$\bullet$ If $k=2$ then $n= 2p^2+1$ so $$2p^2+2\mid 3p^2-3 \implies 2p^2+2\mid 2(3p^2-3)-3(2p^2+2) = -12$$
So $p^2+1\mid 6 \implies p^2+1\in \{1,2,3,6\}$ which is impssible.
$\bullet$ If $k=1$ then $n= p^2+1$ so $$p^2+2\mid 3p^2-3 \implies p^2+2\mid (3p^2-3)-3(p^2+2) =-9 $$
So $p^2+2\mid 9 \implies p^2+2\in \{1,3,9\}$ which is impossible again.
Second case: If $p^2\mid n+1$ then $n-1\mid 3p^2-3$ so $ n+1= p^2k$ and $n-1\leq 3p^2-3$.
Again, if $k\geq 3$ then $$3p^2-3\geq n-1 = n+1-2\geq 3p^2-2$$ which is impossible. So $k\leq 2$
$\bullet$ If $k=2$ then $n= 2p^2-1$ so $2p^2-2\mid 3p^2-3$ which is impossible.
$\bullet$ If $k=1$ then $n= p^2-1$ so $$p^2-2\mid 3p^2-3 \implies p^2-2\mid (3p^2-3)-3(p^2-2) =3 $$
So $p^2-2\mid 3 \implies p^2-2\in \{-1,1,3\}$ which is impossible again.
So the answer is negative if $p$ is prime.
| {
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How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^2+1})}^{3/2}-{\frac{1}{2}}{\sqrt {x^2+1}}+C$$for some constant $C$
| You made a small mistake on the last line; integrating $\frac{1}{2}u^{1/2}-\frac{1}{2}u^{-1/2}$ should give $\frac{1}{3}u^{3/2}-u^{1/2}+C$, not $\frac{2}{3}u^{3/2}-\frac{1}{2}u^{1/2}+C$. When you want to double-check an antiderivative, it's worth differentiating it to see if you get the expected integrand.
| {
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Solving $\log_5 (2x+1)=\log_3 (3x-3)$. I am trying to resolve the equation $$\log_5 (2x+1) = \log_3 (3x-3)$$ and then of sketch the functions $y=\log_5 (2x+1)$ and $y=\log_3 (3x-3)$ get the solution $x=2$. There is an method that do not use the graphic method? Thanks for your suggestions. Any ideas will be appreciated.
| If you want a purely algebraic solution, although rather tedious, you can solve the equation as follows:
$$\log_5(2x+1) = \frac{\log_3(2x+1)}{\log_3 5} \tag{1}$$
$$\frac{\log_3(2x+1)}{\log_3 5} = \log_3(3x-3) \tag{2}$$
$$\left(3^{\log_3(2x+1)}\right)^{\frac{1}{\log_3 5}} = 3x-3 \tag{3}$$
$$(2x+1)^{\frac{1}{\log_3 5}} = 3x-3 \tag{4}$$
$$(2x+1)^{\log_5 3} = 3x-3 \tag{5}$$
$$\log_{2x+1} (3x-3) = \log_5 3 \tag{6}$$
$$\boxed{x = 2} \tag{7}$$
$(1)$: Change of bases: $\log_a b = \frac{\log_c b}{\log_c a}$
$(2)$: Rewriting the equation
$(3)$: Rewriting as an exponential: $c = \log_a b \iff a^c = b$
$(4)$: Simplifying
$(5)$: Inversion of base and argument: $\frac{1}{\log_a b} = \log_b a$
$(6)$: Rewriting as a logarithm: $a^b = c \iff \log_a b = c$
$(7)$: Solution to $2x+1 = 5$ and $3x-3 = 3$
But of course, you have to check for any extra solutions. Since $\log_a b^c = c\log_a b$ and $\log_{a^c} b = \frac{1}{c}\log_a b$, you have $$\log_{a^ c} b^c = \log_a b$$
$$\implies \log_5 3 = \log_{5^n} 3^n$$
$$\implies\log_{2x+1} (3x-3) = \log_{5^n} 3^n$$
$$\implies 2x+1 = 5^n; \quad 3x-3 = 3^n$$
$$3x-3 = 3^n \implies x-1 = 3^{n-1} \implies x = 3^{n-1}+1$$
Plugging this in the first equation, you have
$$2\left(3^{n-1}+1\right) = 5^n \implies 2\left(3^{n-1}\right)+3 = 5^n$$
It’s fairly obvious the only solution is $n = 1$ (since the RHS grows at a much faster rate), for which we already solved the equation. Hence, it’s the only solution.
| {
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With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.
| $|z| =|z+1/z-1/z| \le$
$ |z+1/z| +|1/z| =1+|1/z|;$
$r:=|z|$;
$r-1/r \le 1;$
$r^2-r -1\le 0;$
$(r-1/2)^2 -1/4-1\le 0;$
$r-1/2 \le (1/2)√5;$
$r \le (1/2)(1+√5)$.
Is the maximum attained?
| {
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How to get the value of $a + b + c$? $(0 \leq a < b < c) \in Z$,
$a + b + c + ab + ac + bc + abc = 1622$
$a + b + c = ?$
I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$
Now $bc + b + c = 1622$.
But I found that was useless and got stuck.
| Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$
$$(1+a)(1+b)(1+c)=1623=1\cdot3\cdot541$$
$$\implies a=0,\ b=2,\ c=540.$$
Note that $0\leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria
| {
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How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
$= \lim\limits_{t \to 0} \dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}$
$= \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}$
Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?
Or any other alternative way that uses only the fact that $\lim\limits_{x \to 0} \dfrac{ \sin x}{x} = 1$?
Thanks :)
| $\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}=2\lim \limits_{x \to \frac{π}{3}} \dfrac{\sin x - \sin \cfrac \pi 3}{\cos \frac{3x}{2}}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6) \sin (\cfrac x2 - \cfrac \pi 6)}{\sin (\frac \pi 2-\frac {3x}{2})}$
$=4\lim \limits_{x \to \frac{π}{3}} \dfrac{\cos (\cfrac x2+\cfrac \pi 6) (\cfrac x2 - \cfrac \pi 6)}{\frac \pi 2-\frac {3x}{2}}$
$=-4/3\lim \limits_{x \to \frac{π}{3}} {\cos (\cfrac x2+\cfrac \pi 6) }$
$=-4/3\cos \pi/3=-2/3$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\frac12(a+b)\cos\frac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.
We first expand
$$\sin a + \sin b = 2 \sin\frac12(a+b) \cos\frac12(a-b)$$
Then we add $\sin(a+b)$, which is $\sin a \cos b + \cos a \sin b$.
We now have:
$$2 \sin\frac12(a+b) \cos\frac12(a-b) + \frac12 \left(\sin(a-b) + \sin (a+b)\right) + \frac12 \left( \sin(b-a) + \sin (a+b)\right)$$
From there, we can't see how to obtain the right-hand side of the equation which is
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$
| \begin{align}
\sin a + \sin b + \sin (a+b)&=2 \sin\left(\frac{a+b}2\right) \cos\left(\frac{a-b}2\right)+2 \sin\left(\frac{a+b}2\right) \cos\left(\frac{a+b}2\right)\\
&=2 \sin\left(\frac{a+b}2\right)\left(\cos\left(\frac{a-b}2\right)+\cos\left(\frac{a+b}2\right)\right)
\end{align}
Now use the identity $$\cos A +\cos B=2\cos \left(\frac{A+B}2\right)\cos \left(\frac{A-B}2\right)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding sum multinomial
I did put x=$w, w^2 ,i ,-i$ but nothing of type is fetting
formed. How come 1/2 is remaining constant.
That means because of some substitution,
$2a_o= a_1+ a_2$ is happening.
Also tried putting x=ix.
| We consider
\begin{align*}
f(x)&:=(x^{2016}+x^{2008}+2)^{2010}=\sum_{j=0}^na_jx^j\\
g(x)&:=\frac{1}{3}\left(f(x)+f(xe^{2\pi i/3})+f(xe^{4\pi i/3}\right)=\sum_{{j=0\ }\atop{\ \ j\equiv 0(3)}}^n a_jx^j
\end{align*}
We obtain
\begin{align*}
&\color{blue}{a_0-\frac{1}{2}a_1-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4-\frac{1}{2}a_5+\cdots}\\
&\qquad=\sum_{{j=0\ }\atop{\ \ j\equiv 0(3)}}^n a_j-\frac{1}{2}\sum_{{j=0\ }\atop{\ \ j\not\equiv 0(3)}}^n a_j\\
&\qquad=g(1)-\frac{1}{2}\left(f(1)-g(1)\right)\\
&\qquad=\frac{3}{2}g(1)-\frac{1}{2}f(1)\\
&\qquad=\frac{1}{2}\left(f(e^{2\pi i/3})+f(e^{4\pi i/3})\right)\\
&\qquad=\frac{1}{2}\left(e^{2\pi i/3\cdot2016}+e^{2\pi i/3\cdot2008}+2\right)^{2010}\\
&\qquad\qquad+\frac{1}{2}\left(e^{4\pi i/3\cdot2016}+e^{4\pi i/3\cdot2008}+2\right)^{2010}\\
&\qquad=\frac{1}{2}\left(1+\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)+2\right)^{2010}\\
&\qquad\qquad+\frac{1}{2}\left(1+\left(-\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)+2\right)^{2010}\\
&\qquad=\frac{1}{2^{2011}}\left(5+i\sqrt{3}\right)^{2010}+\frac{1}{2^{2011}}\left(5-i\sqrt{3}\right)^{2010}\\
&\qquad\,\,\color{blue}{<0}
\end{align*}
where the last line was calculated with the help of Wolfram Alpha. We conclude option (2) is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Remainder of the polynomial: What is wrong with this approach?
Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided
by $x^2+1$.
$$
P(x)=(x^2+1)\cdot Q(x)+R(x)
$$
When $x^2=-1$,
$$
P(x)=-5+x+2x+1+1 = 3x-3
$$
Exactly, why the method above, does not work for the following question?
Find the remainder when $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x$ is divided
by $x^3-x$.
$$
P(x)=(x^3-x)\cdot Q(x)+R(x)=x(x+1)(x-1)\cdot Q(x) + R(x)
$$
$P(1)=5$, but $P(-1)=-5$, and $P(0)=0$
However, if I factor $P(x)$ by $x$,
$$
P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)
$$
and consider $P(1)$ and $P(-1)$ without the factored $x$, then I get $5x$ in both cases. (The correct answer.)
Question: I can solve the question by slightly different methods where there is no confusion or ambiguity involved. I just like to know the above method does not work with this problem. (And by “exactly why” I mean to ask, how am I supposed to foresee that the above method should not be applied directly.)
| The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving used Real Based Methods: $\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt$ In working on integrals for the past couple of months, I've come across different cases of the following integral:
\begin{equation}
I\left(x,a,k,n,m\right) = \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt
\end{equation}
Where $x,a\in \mathbb{R}^{+}$.
Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.
Interested in special cases too!
The method I took:
First I wanted to bring the 'a' out the front:
\begin{equation}
I(x,a,k,n,m) = \int_0^x \frac{t^k}{\left(a\left[\left(a^{-\frac{1}{n}}t\right)^n + 1\right]\right)^m}\:dt = \frac{1}{a^m} \int_0^x \frac{t^k}{\left(\left(a^{-\frac{1}{n}}t\right)^n + 1\right)^m}\:dt
\end{equation}
Here let $u = a^{-\frac{1}{n}}t$ Thus,
\begin{equation}
I(x,a,k,n,m) = \frac{1}{a^m} \int_0^{a^{-\frac{1}{n}}x} \frac{\left(a^{\frac{1}{n}}u\right)^k}{\left(u^n + 1\right)^m}a^{\frac{1}{n}}\:du = a^{\frac{k + 1}{n} - m}\int_0^{a^{-\frac{1}{n}}x} \frac{u^k}{\left(u^n + 1\right)^m}\:du = a^{\frac{k + 1}{n} - m}I(a^{-\frac{1}{n}}x,1,k,n,m)
\end{equation}
From here I will use $I$ in place of $I(x,a,k,n,m)$ for ease of typing. The next step is to make the substitution $w = u^n$ to yield:
\begin{equation}
I = a^{\frac{k + 1}{n} - m}\int_0^{ax^n} \frac{w^\frac{k}{n}}{\left(w + 1\right)^m}\frac{\:dw}{nw^{\frac{n - 1}{n}}} = \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_0^{ax^n} \frac{w^{\frac{k + 1}{n} - 1}}{\left(w + 1\right)^m}\:dw
\end{equation}
Here make the substitution $z = \frac{1}{1 + w}$ to yield:
\begin{align}
I &= \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_1^{\frac{1}{1 + ax^n}} z^m \left(\frac{1 - z}{z}\right)^{\frac{k + 1}{n} - 1}\left(-\frac{1}{z^2}\right) \:dz = \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_{\frac{1}{1 + ax^n}}^1 z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz \\
&= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[\int_0^1 z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz - \int_0^{\frac{1}{1 + ax^n}} z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz \ \right] \\
&= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left( \frac{1}{1 + ax^n}; m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right]
\end{align}
Where $B(a,b)$ is the Beta Function and $B(x; a,b)$ is the Incomplete Beta Function.
And so, we arrive at:
\begin{equation}
\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(\frac{1}{1 + ax^n}; m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right]
\end{equation}
Here we observe that for convergence:
\begin{equation}
m - \frac{k + 1}{n} \gt 0,\quad \frac{k + 1}{n} \gt 0,\quad n \neq 0
\end{equation}
Note: when $x \rightarrow \infty$ we have:
\begin{equation}
\int_0^\infty \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)
\end{equation}
Update: Today I realised that we can use this result for another integral:
\begin{equation}
\int_0^\infty \frac{\ln(t)}{\left(t^n + 1\right)^m}\:dt
\end{equation}
This is achieved through a simple use of Feynman's Trick. Here we consider the case when $x \rightarrow \infty$ and $a = 1$. We see that
\begin{align}
\frac{d}{dk}\left[ \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt \right]&= \frac{d}{dk}\left[\frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right] \\
\int_0^\infty \frac{t^k \ln(t)}{\left(t^n + 1\right)^m}\:dt &= \frac{1}{n^2}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\left[\psi^{(0)}\left(\frac{k + 1}{n}\right) - \psi^{(0)}\left(m - \frac{k + 1}{n}\right) \right]
\end{align}
Thus,
\begin{equation}
\lim_{k \rightarrow 0} \int_0^\infty \frac{t^k \ln(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k \rightarrow 0}\frac{1}{n^2}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\left[\psi^{(0)}\left(\frac{k + 1}{n}\right) - \psi^{(0)}\left(m - \frac{k + 1}{n}\right) \right]
\end{equation}
And finally:
\begin{equation}
\int_0^\infty \frac{ \ln(t)}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n^2}B\left(m - \frac{1}{n}, \frac{1}{n} \right)\left[\psi^{(0)}\left(\frac{1}{n}\right) - \psi^{(0)}\left(m - \frac{1}{n}\right) \right]
\end{equation}
Note: In the case where $m = 1$ we arrive:
\begin{align}
\int_0^\infty \frac{ \ln(t)}{\left(t^n + 1\right)^1}\:dt &= \frac{1}{n^2}B\left(1 - \frac{1}{n}, \frac{1}{n} \right)\left[\psi^{(0)}\left(\frac{1}{n}\right) - \psi^{(0)}\left(1 - \frac{1}{n}\right) \right] \\
&= \frac{1}{n^2} \Gamma\left(\frac{1}{n} \right)\Gamma\left(1 - \frac{1}{n} \right) \cdot -\pi\cot\left(\frac{\pi}{n}\right) \\
&= \frac{1}{n^2} \frac{\pi}{\sin\left(\frac{\pi}{n}\right)}\cdot -\pi\cot\left(\frac{\pi}{n}\right)
\end{align}
Thus:
\begin{equation}
\int_0^\infty \frac{ \ln(t)}{t^n + 1}\:dt = -\frac{\pi^2}{n^2} \operatorname{cosec}\left(\frac{\pi}{n} \right)\cot\left(\frac{\pi}{n}\right)
\end{equation}
| NOT A SOLUTION:
I've found some special cases on this site that I will list (this will evolve as I find more special (but generalised) cases:
*
*Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
*Evaluate the integral $ \int _0^{+\infty} \frac{x^m}{(a+bx^n)^p}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Showing $\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$ Since $\log(nm) = \log(n) + \log(m)$, and $\sum_{k=1}^n \frac{1}{k} \approx \log n$ for large $n$, we would expect that $$\sum_{k=1}^{nm} \frac{1}{k} \approx \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{m} \frac{1}{k}$$
when $n,m$ are large.
I'm wondering if this approximation can be demonstrated through discrete means. That is to say, manipulations of rational fractions and/or elementary number-theoretical considerations, without using the $\log n$ approximation.
| Let $$S = \sum_{k=1}^n \frac{1}k + \sum_{k=1}^m \frac{1}k.$$
Note that we can write $\sum_{k=1}^{nm} \frac{1}k$ as
$$\begin{align*}
1 + \cdots + \frac{1}n \\
\frac1{n+1} + \cdots + \frac{1}{2n} \\
\cdots \\
\frac{1}{n(m-1)} + \cdots + \frac{1}{nm}
\end{align*} $$
Label the rows of the above table from $1$ upto $m$. Note that all the numbers in the $j$th row are $\ge \frac{1}{jn}$ so the sum of the numbers in the $j$th row is at least $n \cdot \frac{1}{jn} = \frac{1}j$. Summing this from $j=2$ to $m$ gives us
$$ \sum_{k=1}^{nm} \frac{1}k \ge \left(\sum_{k=1}^n \frac{1}k + \sum_{j=1}^m \frac{1}j \right) - 1 = S-1.$$
Similarly, each entry in the $j$th row is at most $\frac{1}{(j-1)n}$ so summing from $j=2$ to $m$ gives
$$ \sum_{k=1}^{nm} \frac{1}k \le \left(\sum_{k=1}^n \frac{1}k + \sum_{j=1}^m \frac{1}j \right) - \frac{1}m = S-\frac{1}m.$$
Therefore,
$$ \frac{1}m \le \sum_{k=1}^n \frac{1}k + \sum_{k=1}^m \frac{1}k - \sum_{k=1}^{nm} \frac{1}k \le 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?
HERE IS ONE
Let $n\in\Bbb{N}$ be fixed such that $a_n=\left(1-\frac{1}{\sqrt{n}}\right)^n.$ Then,
\begin{align} a_n&=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&=\exp\left[ -n\left(\frac{1}{\sqrt{n}}+\frac{1}{2n}+\sum^{\infty}_{k=3}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right)\right]\\&=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&\equiv\exp \left(-\sqrt{n}\right)\exp \left(-\frac{1}{2}\right)\end{align}
Choose $b_n=\exp \left(-\sqrt{n}\right)$, so that
\begin{align} \dfrac{a_n}{b_n}\to\exp \left(-\frac{1}{2}\right).\end{align}
Since $b_n \to 0$, there exists $N$ such that for all $n\geq N,$
\begin{align} \exp \left(-\sqrt{n}\right)<\dfrac{1}{n^2}.\end{align}
Hence, \begin{align}\sum^{\infty}_{n=N}b_n= \sum^{\infty}_{n=N}\exp \left(-\sqrt{n}\right)\leq \sum^{\infty}_{n=N}\dfrac{1}{n^2}<\infty,\end{align}
and so, $\sum^{\infty}_{n=1}b_n<\infty\implies \sum^{\infty}_{n=1}a_n<\infty$ by Limit comparison test.
| Noting that $\ln(\tfrac{1}{1-x}) = x + x^2/2 + O(x^3)$,
\begin{align*}
\frac{\left(\,1-\frac{1}{\sqrt{n}}\,\right)^n}{e^{-\sqrt{n}}}
%
&= \frac{\exp \left(\, -n\ln \frac{1}{1-\frac{1}{\sqrt{n}}}\,\right)}{e^{-\sqrt{n}}}\\
%
&= \frac{\exp \left(\, -n\left(\tfrac{1}{\sqrt{n}} + \tfrac{1}{2n} + O\left(\tfrac{1}{n^{3/2}}\right) \right)\,\right)}{e^{-\sqrt{n}}}\\
%
&= e^{-1/ 2} \cdot \underbrace{e^{-O\left(\tfrac{1}{n^{1/2}}\right)}}_{\to\,1}
\to e^{-1/ 2}
\end{align*}
So $\sum_{n=1}^\infty \left(\,1-\frac{1}{\sqrt{n}}\,\right)^n$ and $\sum_{n=1}^{\infty} e^{-\sqrt{n}}$ converge or diverge together by the limit comparison test. Given that
$$\int_1^\infty e^{-\sqrt{t}}\;dt
= \left[\, -2e^{-\sqrt{x}}(\sqrt{x}+1)\,\right]_1^\infty
= \frac{4}{e} < \infty$$
we conclude that both of the latter series then converge by the integral test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ = $\frac{15*\sqrt[4]{5}*\sqrt[4]{25}}{\sqrt[4]{5}}$ =
(cancel out $\sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15\sqrt[4]{25}$$
Where did I go wrong and how can I arrive at $15\sqrt{5}$?
| It turns out that $\sqrt[4]{25}=\sqrt{5}$. This is because $25=5^2$, so that $\sqrt[4]{25}=\sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$ So, you are correct, as is the book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan(u)}}{\sec^2(u)}du.$$
Wolfram gave an answer of $ \dfrac{\pi}{8\sqrt{2}},$ but how would one get to that answer?
| We could do it with contour integration.
take the contour from 0 to R along the real axis.
$\int_0^R \frac {x^2}{(x^4+1)^2} \ dx$
The quater circle.
$\int_0^{\frac \pi 2} \frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) \ dt$
$\lim_\limits{R\to \infty} \frac {R^2e^2it}{(R^4e^{4it}+1)^2}(iRe^{it}) = 0$
And down the imaginary axis.
$\int_R^0 \frac {(e^{\frac {\pi}{2} i} x)^2}{((e^{\frac {\pi}{2} i} x)^4+1)^2} (e^{\frac {\pi}{2} i})\ dx\\
\int_R^0 \frac {-x^2}{x^4+1)^2} (i)\ dx\\
$
$(1+i)\int_0^\infty \frac {x^2}{(x^4+1)^2} \ dx = (2\pi i) \text{ Res}_{\left(x=e^{\frac\pi4i}\right)}\frac {x^2}{(x^4+1)^2}$
The pole is of order 2.
$\frac {d}{dx}\frac {x^2}{(x^3 + x^2e^{\frac \pi4 i} + xe^{\frac {2\pi}{4} i}+e^{\frac {3\pi}{4} i})^2} = \frac {2x(x^3 + x^2e^{\frac \pi4 i} + xe^{\frac {2\pi}{4} i}+e^{\frac {3\pi}{4} i}) - 2x^2(3x^2 + 2xe^{\frac \pi4 i} + e^{\frac {2\pi}{4} i})}{(x^3 + x^2e^{\frac \pi4 i} + xe^{\frac {2\pi}{4} i}+e^{\frac {3\pi}{4} i})^3}$
Evaluated at $e^{\frac {\pi}{4} i}$
$\frac {4}{(4e^{\frac {3\pi}4 i})^3} = \frac {1}{16e^{\frac {\pi}4 i}}$
$\int_0^\infty \frac {x^2}{(x^4+1)^2} \ dx = \frac {2\pi i}{16\sqrt 2 i} = \frac {\pi}{8\sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Changing variable - Integration goes wrong. I was trying to do the integration
$$I=\frac{\pi}{2}\int_0^\pi \frac{dx}{a^2\cos^2x + b^2\sin^2x}$$
If I divide throughout by $\cos^2x$ and use substitution ($t=\tan x$),
I obtain$$I=\frac{\pi}{2}\int_0^0\frac{dt}{a^2+(bt)^2} $$
This would be evaluated to be $0$.
But however its actual answer is $$\frac{\pi^2}{2ab}$$
which can be obtained by using properties of definite integrals to change the limit to $0$ to $\frac{\pi}{2}$ and then splitting it from $0$ to $\frac{\pi}{4}$ and $\frac{\pi}{4}$ to $\frac{\pi}{2}$ ,dividing throughout by $\cos^2x$ and $\sin^2x$ respectively and then substituting ($t=\tan x$) and ($t=\cot x$) respectively.
Can anyone please tell me what is wrong in the first approach.
| The reason why your substitution doesn't work is due to the singularity at $\frac {\pi}2$. Indeed, $\tan\left(\frac {\pi}2\right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$\begin{align*}\int\limits_0^{\pi}\frac {\mathrm dx}{a^2\cos^2x+b^2\sin^2x} & =\int\limits_0^{\pi/2}\frac {\mathrm dx}{a^2\cos^2x+b^2\sin^2x}+\int\limits_{\pi/2}^{\pi}\frac {\mathrm dx}{a^2\cos^2x+b^2\sin^2x}\\ & =\int\limits_0^{\pi/2}\frac {\mathrm dx}{a^2\cos^2x+b^2\sin^2x}+\int\limits_0^{\pi/2}\frac {\mathrm dx}{a^2\sin^2x+b^2\cos^2x}\\ & =\int\limits_0^{\infty}\frac {\mathrm dx}{a^2+b^2x^2}+\int\limits_0^{\infty}\frac {\mathrm dx}{b^2+a^2x^2}\\ & =\frac 1{ab}\left[\arctan\left(\frac {bx}a\right)+\arctan\left(\frac {ax}b\right)\right]\,\Biggr\rvert_0^{\infty}\end{align*}$$So now, we get that$$\frac {\pi}2\int\limits_0^{\infty}\frac {\mathrm dx}{a^2\cos^2x+b^2\sin^2x}\color{blue}{=\frac {\pi^2}{2ab}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac{25}{29}=\frac{20}{29}\\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
| You can first deduce
$$
\tan\frac{B}{2}=\tan\left(\frac{\pi}{2}-\frac{A+C}{2}\right)=\cot\frac{A+C}{2}=
\frac{1-\tan\frac{A}{2}\tan\frac{C}{2}}{\tan\frac{A}{2}+\tan\frac{C}{2}}=\frac{2/3}{37/30}
=\frac{20}{37}
$$
Therefore
$$
\sin A=\frac{2(5/6)}{1+25/36}=\frac{60}{61}
$$
Similarly,
$$
\sin B=\frac{1480}{1769}\qquad \sin C=\frac{20}{29}
$$
By the sine law,
$$
\frac{a+c}{2}=\frac{b}{2\sin B}(\sin A+\sin C)=
b\frac{1769}{2960}\left(\frac{60}{61}+\frac{20}{29}\right)=b
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Logarithmic integral $ \int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x $ At various places e.g.
Calculate $\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$
and
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
logarithmic integrals are connected to Euler-sums. In view of the last link I'm wondering about the following integral
$$
\int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x \, .
$$
I see I can throw it into Wolfram Alpha and get some disgusting anti-derivative with Li's up to ${\rm Li}_3$. Anyway is there some manually more tractable way to solve this?
I have tried two things of which both don't seem to lead anywhere so far.
For the first one:
I expressed $\frac{x}{x^2+1}$ by it's Mellin transform $\frac{\pi/2}{\cos\left(\frac{\pi s}{2}\right)}$ and interchanged the integral order
$$
\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} {\rm d}s \, \frac{\pi/2}{\cos\left(\frac{\pi s}{2}\right)} \left( -\frac{{\rm d}}{{\rm d}s} \right)\int_0^1 {\rm d}x \, x^{-s} \log(x+1)
$$
where the constant $c>-1$ is right of the first pole of the cosine at $s=-1$ and the contour can be closed in a circle on the left hand side of the plane to use the residue theorem. The $x$-integral is equal to
$$
\int_0^1 {\rm d}x \, x^{-s} \log(x+1) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1-s)} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{1-s} \left( \frac{1}{n} - \frac{1}{n+1-s} \right)\\
= {\frac {\Psi \left( 1-s/2 \right) - \Psi \left( 3/2-s/2 \right) }{2(1-s)}} + {\frac {\log \left( 2 \right) }{1-s}}
$$
where $\Psi$ is the Digamma function, related to the harmonic numbers $H_n$. Deriving with respect to $s$ and picking up the residue $(-1)^k$ of the Mellin transform at $s=-2k-1$ ($k=0,1,2,3,...$) one obtains
$$
\sum_{k=0}^\infty (-1)^{k+1} \Bigg\{ {\frac {\Psi \left( 3/2+k \right) - \Psi \left( 2+k \right) }{ 8\left( 1+k \right) ^{2}}} - {\frac {\Psi' \left( 3/2+k \right) - \Psi' \left( 2+k \right) }{8(1+k)}} + {\frac {\log \left( 2 \right) }{ 4\left( 1+k \right) ^{2}}} \Bigg\}
$$
where $\Psi'$ is the derivative of the Digamma function related to $H_{n,2}$. The terms with integral argument presumably can be evaluated in closed form, but I'm wondering if the half-integer argument terms can be also evaluated just by algebraic manipulations?
Second:
I tried to find closed form for the integral by partial integration
\begin{align}
I(a) &=\int_0^1 \frac{\log(x) \log(x+1)}{x+a} \, {\rm d}a \\
&=-\frac{\log(2)}{a+1} - \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a)} \, {\rm d}x + \int_0^1 \frac{x\left(\log(x)-1\right) \log(x+1)}{(x+a)^2} \, {\rm d}x \\
&=-\frac{\log(2)}{a+1} - \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a)} \, {\rm d}x - \int_0^1 \left( \frac{\log(x+1)}{x+a} - \frac{a\log(x+1)}{(x+a)^2} \right) + I(a) + a I'(a)
\end{align}
and thus
$$
I(a) = \int_\infty^a \frac{{\rm d}a'}{a'} \Bigg\{ \frac{\log(2)}{a'+1} + \int_0^1 \frac{x\left(\log(x)-1\right)}{(x+1)(x+a')} \, {\rm d}x + \int_0^1 \left( \frac{\log(x+1)}{x+a'} - \frac{a'\log(x+1)}{(x+a')^2} \right) {\rm d}x \Bigg\}
$$
of which many terms are easy to integrate, but there is one combination which seems very difficult, namely something like
$$
\int \frac{{\rm Li}_2(a')}{a'+1} \, {\rm d}a' \, .
$$
$a=\pm i$ at the end. Any insights?
| Integrate as follows
\begin{align}
& \hspace{5mm}\int_0^1 \frac{x \ln x\ln(1+x) }{1+x^2}dx\\
&= \frac12 \int_0^1 \frac{x \ln x\ln(1-x^2) }{1+x^2}\overset{x^2\to x}{dx}
-\frac12\int_0^1 \frac{x \ln x\ln\frac{1-x}{1+x}}{1+x^2}\overset{\to x}{dx}\\
&= \frac18 \int_0^1 \frac{\ln x\ln(1-x) }{1+x}dx
-\frac14\int_0^1 \frac{\ln x\ln\frac{1-x}{1+x}}{1+x}{dx}\\
&\overset{ibp} = -\frac18\int_0^1 \frac{\ln x\ln(1+x) }{1-x}dx
+\frac18\int_0^1 \frac{\ln (1-x)\ln(1+x)-\ln^2(1+x) }{x}dx\tag1
\end{align}
Evaluate
\begin{align}
& \int_0^1 \frac{\ln (1-x)\ln(1+x)-\ln^2(1+x) }{x}dx\\
=& \frac12\int_0^1 \frac{\ln^2(1-x^2)}xdx
- \frac12\int_0^1 \frac{\ln^2(1-x)}xdx
- \frac32\int_0^1 \frac{\ln^2(1+x)}xdx\\
=& \frac12\cdot \zeta(3)-\frac12\cdot 2\zeta(3)-\frac32\cdot \frac14\zeta(3)=-\frac78\zeta(3)
\end{align}
and
\begin{align}
& \int_0^1 \frac{\ln x\ln(1+x) }{1-x}dx\\
\overset{ibp}=& \>\ln(1+x)\left(\int_0^x \frac{\ln t}{1-t}dt\right)\bigg|_0^1-\int_0^1\frac1{1+x} \left(\int_0^x \frac{\ln t}{1-t}\overset{t= xy}{dt}\right)dx\\
=& -\ln(2)\zeta(2)-\int_0^1 \int_0^1 \frac{x\ln x}{(1+x)(1-xy)}dy dx
- \int_0^1 \int_0^1 \frac{x\ln y}{(1+x)(1-xy)}\overset{x\leftrightarrows y}{dy dx}\\
=& -\ln(2)\zeta(2)+ \int_0^1 \int_0^1 \left(\frac{\ln x}{(1+x)(1+y)}- \frac{\ln x}{1-xy} \right) dy dx\\
=& -\ln(2)\zeta(2)- \frac12\ln2 \zeta(2) + \int_0^1 \frac{\ln x\ln(1-x)}{x} \overset{ibp}{dx}\\
=& -\frac32\ln2\zeta(2)+\frac12\int_0^1 \frac{\ln^2x}{1-x}dx
= -\frac32\ln2\zeta(2)+\zeta(3)
\end{align}
Substitute the two integral results evaluated above into (1) to arrive at
$$\int_0^1 \frac{x \ln x\ln(1+x) }{1+x^2}dx= \frac3{16}\ln2\zeta(2)-\frac{15}{64}\zeta(3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$
Use this identity to (4) too
and simplify (3),
we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$
And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$
But got stuck...
| Hint: $ac=x, bc=y, ad=u, bd=v$, then the equations are
$x+y+u+v=1$
$x^2+y^2+u^2+v^2=9$
$x^3+y^3+u^3+v^3=7$
$x^4+y^4+u^4+v^4=25$
Use Newton-Girard to compute the elementary polynomials.
Then you have the polynomial $P(z)= (z-x)(z-y)(z-u)(z-v)$ with variable $z$.
Solve the quartic equation $P(z)=0$, and there you have the values $x,y,u,v$ in some order.
Note that not in any order: $xv=yu$ must be true, see the definition of these variables.
Of course, once you have $x,y,u,v$, it is easy to compute $a,b,c,d$.
P.S. By this way we can get:
$$\{x,y,u,v\}=\{-1,2,\sqrt2,-\sqrt2\},$$ which gives $abcd=-2.$
Up to symmetry, the solution is $(a,b,c,d)= (t, -\sqrt{2}t, -\frac{1}{t}, -\frac{\sqrt{2}}{t})$ for any $t\neq 0$.
(By up to symmetry, I mean you can switch $a$ and $b$, you can switch $c$ and $d$, and you can switch the pair $(a,b)$ with $(c,d)$, so there are $8$ symmetries.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 3
} |
Non zero solution of $3x\cos(x) + (-3 + x^2)\sin(x)=0$ How can I find exact non-zero solution of $3x\cos(x) + (-3 + x^2)\sin(x)=0$. Simple analysis and the below plot show that the equation has an infinite number of non-zero solutions.
| There's probably no closed-form solution for the real roots you're looking for, although you have a chance with the complex solutions.
Let's notice that $3x \cos x$ and $\left( x^2 - 3 \right) \sin x$ have the same trigonometric argument. That means that the Harmonic Addition Theorem can be used: we can rewrite this as one trigonometric function.
HAT states that
\begin{align}a \cos x + b \sin x &= A \cos \left(x + B \right),
\end{align}
where $a^2 + b^2 = A^2$ and $-\frac{b}a = \tan B$.
So, substituting $3x \cos x$ for $a$ and $x^2 - 3$ for $b$, we find that $A^2 = x^4 + 3x^2 + 9$ and $\frac{3 - x^2}{3x} = \tan B$.
This of course has infinitely many solutions, although the factor of $A$ only contributes some complex roots. Ignoring questions of signs and branch choice, let's declare that $A = \sqrt{x^4 + 3x^2 + 9}$ and $B = \tan^{-1} \frac{3 - x^2}{3x}$, or in other words
$$3x \cos x + (x^2 - 3) \sin x = \sqrt{x^4 + 3x^2 + 9} \cos \left( x + \tan^{-1} \frac{3 - x^2}{3x} \right).$$
The problem can therefore be rewritten as finding the zeroes of $\sqrt{x^4 + 3x^2 + 9}$ and $\cos \left( x + \tan^{-1} \frac{3 - x^2}{3x} \right)$, although I doubt you'll find any good closed form for the latter.
This WolframAlpha plot looks like it agrees with your image.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find maximum value of $\frac xy$
If $x^2-30x+y^2-40y+576=0$, find the maximum value of $\dfrac xy$.
First I completed the squares and got $(x-15)^2+(y-20)^2=7^2$, which is the equation of a circle.
I think I need to use some properties but I don't know what to do next.
| Let $\frac{x}{y}=k$.
Thus, $x=ky$ and the equation
$$k^2y^2-30ky+y^2-40y+576=0$$ or
$$(k^2+1)y^2-2(15k+20)y+576=0$$ has real roots, which says that the discriminant must be non-negative.
Id est, $$(15k+20)^2-576(k^2+1)\geq0$$ or
$$351k^2-600k+176\leq0$$ or
$$\frac{44}{117}\leq k\leq\frac{4}{3}.$$
The value $\frac{4}{3}$ occurs for $$y=\frac{15k+20}{k^2+1}=14.4$$ and $$x=14.4\cdot\frac{4}{3}=19.2,$$ which says that $\frac{4}{3}$ is a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
| Use that $$\tan(3x)=\frac{\tan(x)+\tan(2x)}{1-\tan(x)\tan(2x)}$$ and then
$$\tan(x+x)=\frac{2\tan(x)}{1-\tan^2(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove all elements $x_n \geq \sqrt{2}.$ Given $x_1 = 2,$ and $$x_{n+1} = \frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg),$$ show that for all $n \in \mathbb{N},$ $x_n \geq \sqrt{2}.$
I tried the following. Suppose, for contradiction, that $x_{n+1} < \sqrt{2}.$ Then, $$\frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg) < \sqrt{2},$$ and $$x_n + \frac{2}{x_n} < 2\sqrt{2}.$$ Given that $x_n \geq \sqrt{2},$ it follows that $$\sqrt{2} + \frac{2}{x_n} \leq x_n + \frac{2}{x_n} < 2\sqrt{2},$$ and $$\sqrt{2} + \frac{2}{x_n} < 2\sqrt{2}.$$ So, $$\frac{2}{x_n} < \sqrt{2},$$ and $$\sqrt{2} < x_n.$$ No contradiction. I've tried proving it directly, by squaring both sides and by leaving the expression as is, and I consistently find that the $x_n$ and the $x_n^{-1}$ are together problematic.
I also tried something that allowed me to arrive at the following: $$x_{n+1}x_n \geq 2.$$ But, I don't think that's helpful.
Help?:)
| This is immediate from AM-GM: $\frac{1}{2}\bigg(x + \frac{2}{x} \bigg)$ is the arithmetic mean of $x$ and $2/x$, which is always greater than or equal to the geometric mean $\sqrt{x\cdot\frac{2}{x}}=\sqrt{2}$ (for $x>0$).
Alternatively, if you don't know AM-GM, you can reach the same conclusion easily with a bit of calculus. Letting $f(x)=x + \frac{2}{x}$, we have $f'(x)=1-\frac{2}{x^2}$ which is negative for $0<x<\sqrt{2}$ and positive for $x>\sqrt{2}$. It follows that $f(x)$ is minimized (on $(0,\infty)$) at $x=\sqrt{2}$, so $\frac{1}{2}f(x)\geq\frac{1}{2}f(\sqrt{2})=\sqrt{2}$ for all $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
We throw $5$ dice: What is the probability to have $4$ different numbers? We throw $5$ dice: What is the probability to have $4$ different numbers?
I know it is $$\frac{6\cdot 5\cdot 4\cdot 3}{6^5}.$$
I wanted to use an other argument, but it look to not work : I take $\binom{6}{4}$ numbers. Then I have $$\frac{6\cdot 5\cdot 4\cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways.
*
*If I want all dice different, this argument works: I take $\binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $\frac{5\cdot 5\cdot 4\cdot 3\cdot 2}{5!}5!=6\cdot 5\cdot 4\cdot 3\cdot 2$ possibilities, that is the correct answer.
So why doesn't it work with the previous situation ?
| Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.
Suppose the dice are five different colors so that we can distinguish between them. There are $\binom{5}{2}$ ways for two of the five dice to display the same number and $\binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display. The number of possible ways those three dice can display the numbers is $5 \cdot 4 \cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list). Hence, the number of favorable outcomes is
$$\binom{5}{2} \cdot 6 \cdot 5 \cdot 4 \cdot 3$$
from which we obtain the probability
$$\frac{\binom{5}{2} \cdot 6 \cdot 5 \cdot 4 \cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Convergence of $\int_0^{\frac{\pi}{2}} \frac{\exp({-1/x)}}{\sqrt{ \sin x}}\, \mathrm{d}x$ Let $$f(x)=\frac{\exp(-1/x)}{\sqrt{ \sin x}}\, \mathrm{d}x.$$
$f(x)$ seems to not have any problems at $x=\frac {\pi}{2}$, but at $x=0$.
So I should understand behavior of $f(x)$ at $x=0$ by evaluating $\lim_{x\to0+}f(x)$.But I am stuck here as I can't apply Taylor expansions to $e^{-\frac{1}{x}}$ and using l'Hospital's rule gives me nothing.
| Using the continuity of the square root, we can conclude
\begin{equation*}
I
:=\lim_{x \searrow 0} \frac{e^{-\frac{1}{x}}}{\sqrt{\sin(x)}}
= \sqrt{ \lim_{x \searrow 0} \frac{e^{-\frac{2}{x}}}{\sin(x)}}
= \sqrt{ \lim_{x \searrow 0} \frac{1}{e^{\frac{2}{x}} \sin(x)}}
\end{equation*}
Now, we can apply L'Hôpital to the following limit
\begin{equation*}
L
:= \lim_{x \searrow 0} e^{\frac{2}{x}} \sin(x)
= \lim_{x \searrow 0} \frac{\sin(x)}{e^{-\frac{2}{x}}}
= \lim_{x \searrow 0} \frac{ \frac{d}{d x}\sin(x)}{\frac{d}{d x} e^{-\frac{2}{x}}}
= \lim_{x \searrow 0} \frac{\cos(x)}{\frac{2}{x^2} e^{-\frac{2}{x}}}
= \lim_{x \searrow 0} \frac{x^2}{2} \cos(x) \cdot e^{\frac{2}{x}}.
\end{equation*}
And now, because the limits of the factors exist and one is finite we have by the product rule
\begin{equation*}
L = \frac{1}{2} \underbrace{\lim_{x \searrow 0} \cos(x)}_{= 1} \cdot \left(\lim_{x \searrow 0} x^2 \cdot e^{\frac{2}{x}}\right)
= \frac{1}{2} \underbrace{\left(\lim_{x \searrow 0} x^2 \cdot e^{\frac{2}{x}}\right)}_{:= \widetilde{L}}
\end{equation*}
Since $x \mapsto x^2$ is monotone and continuous, we have
\begin{equation*}
\widetilde{L}
= \big( \underbrace{\lim_{x \searrow 0} x \cdot e^{\frac{1}{x}}}_{=: \widehat{L}} \big)^2
\end{equation*}
Now, by L'Hôpital, we have
\begin{equation*}
\widehat{L}
= \lim_{x \searrow 0} \frac{e^{\frac{1}{x}}}{\frac{1}{x}}
= \lim_{x \searrow 0} \frac{\frac{d}{dx} e^{\frac{1}{x}}}{\frac{d}{dx} \frac{1}{x}}
= \lim_{x \searrow 0} \frac{-\frac{1}{x^2} e^{\frac{1}{x}}}{-\frac{1}{x^2}}
= \lim_{x \searrow 0} e^{\frac{1}{x}}
= \infty.
\end{equation*}
Therefore, we have $L = \infty$ and so $I = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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