Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Using quotient rule to differentiate Use the quotient rule to differentiate. I want to know if I'm doing this correctly:
$$
f(x)=\frac {2x}{x^4+6}
$$
First, I find $f$ prime of $x$ and $g$ prime of $x$:
$$
f'(x) = 2
$$$$
g'(x) = 4x^3
$$
After using the rule, I end up with:
$$
\frac {2x^4+12-8x^4}{(x^4+6)^2} = \frac {-6x^4+12}{(x^4+6)^2}
$$
Would this be the final answer if I'm correct? Or do I need to expand the denominator?
| As a check, you can use the chain rule (if you've learned it already) and the product rule. (I use it because for a long time I could never remember the order of the quotient rule!)
Let $f(x) = 2x(x^4 + 6)^{-1}$. To find the derivative, apply the product rule: $f'(x) = (2x)'(x^4 + 6)^{-1} + 2x\left[(x^4 + 6)^{-1}\right]'$. Keep going, applying the chain rule:
$$
\begin{aligned}
\ 2(x^4+6)^{-1}+(2x)(-1)(x^4+6)^{-2}(4x^3) &= \frac{2}{x^4+6} - \frac{8x^4}{(x^4+6)^2} \\
\ &= \frac{2(x^4+6)-8x^4}{(x^4+6)^2} \\
\ &= \frac{12-6x^4}{(x^4+6)^2} \\
\end{aligned}
$$
It might be a little longer for this problem. However, you can always use it to check your answer on a test (or if you ever forget the quotient rule).
| {
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"url": "https://math.stackexchange.com/questions/687634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using Lagrange multipliers to solve for minimum I am having troubles with one part of this homework problem. Hopefully somebody can help me out:
Find the minimum and maximum values of the function $f(x,y)=x^2+y^2$ subject to the given constraint $x^4+y^4=18$.
Using Lagrange multipliers, I can easily solve for the maximum:
$f_x(x,y)=2x$ and $f_y(x,y)=2y$.
If we call the second equation $g$, then:
$g_x=4x^3$ and $g_y=4y^3$.
Then we apply the Lagrange multiplier:
$2x=\lambda 4x^3$ and $2y = \lambda 4y^3$
By solving for $x$ and $y$ and plugging in to $g$, we get $\lambda=1/\sqrt{36}=\pm1/6$.
To find the maximum, I will use the positive $1/6$, and solve for $x^2=3$ and $y^2=3$, resulting in a maximum of $6$, which the system spits out as correct.
For a minimum, I originally thought $0$ because $x^2$ and $y^2$ must be positive numbers, but that is not correct. Then I noticed that since $x^2=1/(2\lambda)$, and $y$ also, then when $\lambda$ is negative that would be the minimum, resulting in $-6$, which is also not correct.
So long story short, how do I find the minimum value in this case?
Thanks!
| When you solve for $\lambda$ using the systems:
$\begin{cases} 2x = \lambda 4x^3 & (1) \\
2y = \lambda 4y^3 & (2) \\
x^4 + y^4 = 18 & (3)
\end{cases}$
You cancel out $x$ in (1) to get $x^2 = 1/(2\lambda)$, you missed a case that $x=0$.
If $x=0$, then $y^4 = 18$, yields $y = \pm {18}^{\frac{1}{4}}$.
The minimum value of $f(x,y)$ would be $\sqrt{18}$, occurs at $(0,\pm {18}^{\frac{1}{4}})$ or $(\pm {18}^{\frac{1}{4}},0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the maximum or minimum value of the quadratic function by completing the square. Find the maximum or minimum function of the quadratic function by completing the squares. State the value of $x$ at which the function is maximum or minimum.
$y=3x^2+7x+9$
I already posted similar question on this topic before, and got great solutions, but as soon as I try another sum my answer doesn't come out right.
Have a look at this: find max or min by completing the square
Maybe I have misunderstood something, can you please tell me the actual and perfect way for solving such problems?
I tried in this way:
$y=3(x^2+\frac{7}{3}x+\frac{9}{3})$
$y=3[(x+\frac{7}{9})^2+\frac{7}{9}+3]$
Is this the right way? If I proceed, $x$ = $-\frac{7}{9}$ and my book says it is not right! Help please :'(
| By using that $(x+a)^2=x^2+2ax+a^2$, you have that $$\begin{align*}y&=3\left(x^2+(7/3)x+3\right)=3\left(x^2+2(7/6)x+(7/6)^2-(7/6)^2+3\right)\\&=3\left(\left(x+(7/6)\right)^2+3-(7/6)^2\right)\end{align*}$$
(where $a=7/6$). So the minimum is for $x=-(7/6)$ since a square is always $\ge0$. For the named $x$ the term in the square is $0$ so it attends it's minimum value.
| {
"language": "en",
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Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}-\frac{3}{\sqrt{n+3}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}=\sum _{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\frac{4}{\sqrt{n+4}}\right)$$
$$\cdots$$
And is there some easy cancellation that I have missed on the right hand side?
Mathematica:
Clear[s, i, n, j]
s = 1/2;
i = 1;
j = 0;
Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s - 2/(n + 2)^s + 1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s + 1/(n + 0)^s,
{n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s +
1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
N[Accumulate[Sqrt[Range[4]]], 20]
| For the first one,
$$\begin{align*}
\sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)
&= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}-\sum_{n=1}^\infty\frac{1}{\sqrt{n+1}} \\
&= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}-\sum_{n=2}^\infty\frac{1}{\sqrt{n}} \\
&= \frac{1}{\sqrt{1}}\\
&= \sqrt{1}.
\end{align*}$$
For the second,
$$\begin{align*}
\sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)
&=\sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)
+\sum_{n=1}^\infty\left(\frac{2}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)\\
&= \sqrt{1} + 2\sum_{n=2}^\infty\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\\
&= \sqrt{1} + \frac{2}{\sqrt{2}} \\
&= \sqrt{1} + \sqrt{2}.
\end{align*}$$
And so on. It would probably be faster to set up an induction using the ideas from the second case than to prove the third and fourth cases individually.
| {
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"timestamp": "2023-03-29T00:00:00",
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Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false. What is the correct way to use the formula?
| Note how the last term in the factorization of $x^n-y^n$ has $x$ raised to the $0$th power. That means you have to stop when $x$ is raised to the power of $0$.
$$x^2-y^2=(x-y)(x^1y^0+x^0y^1)$$
$$\boxed{x^2-y^2=(x-y)(x+y)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $yy'+x=\sqrt{x^2+y^2}$? I tried this:
$yy'+x=\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\frac{1}{y}\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\sqrt{(\frac{x}{y})^2+1}$
Substitution: $v=\frac{y}{x}$
$v'x+v=-\frac{1}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2+ v\sqrt{(\frac{1}{v})^2+1}}{v}$
$v'x=-\frac{1-v^2+\sqrt{1+v^2}}{v}$
If I have to separate at this point, it's going to be pretty awkward to work with.
Is there a better way?
| Once you set $z=x^2+y^2$, then things get extremely simple. Indeed:
$$
z'=2x+2yy'=2\sqrt{x^2+y^2}=2z^{1/2},
$$
and hence (assuming that $z>0$)
$$
z^{-1/2}z'=2,
$$
or
$$
\big(z^{1/2}\big)'=1,
$$
or
$$
z^{1/2}=x+c
$$
for some constant $c$, and finally
$$
z=(x+c)^2,
$$
equivalently
$$
x^2+y^2=(x+c)^2,
$$
and hence
$$
y^2=2cx+c^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Indeterminate form limits question $$\lim_{x\to 0}\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x} $$
This is an indeterminate limit.
I want help in solving this problem. Thanks in advance
| If we replace in the expression $-1$ by $1$ we have in this case the indeterminate form $\frac 00$ and we calculate the limit using the Taylor series:
$$\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x}=\frac{\exp(x\log10) -\exp(x\log2) - \exp(x\log5) + 1 } {x\tan x}\sim_0\frac{1+x\log10+\frac{x^2\log^210}{2}-1-x\log2-\frac{x^2\log^22}{2}-1-x\log5-\frac{x^2\log^25}{2}+1}{x^2}\sim_0\frac{\log^210-\log^22-\log^25}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What are integers $x$ and $n$ such as $ x^4+x^3+x^2+x+1 = n^2$? What are the integer numbers $x$ such as $x^4+x^3+x^2+x+1$ is the square of an integer?
$0$ and $-1$ are obvious answers, but how many are there?
| Idea: Bound between known perfect squares.
For $ x \neq 0$, we have (expand and verify)
$$ ( x^2 + \frac{x}{2} ) ^ 2 < x^4 + x^3 + x^2 + x + 1 < (x^2 + \frac{x}{2} + 1)^2. $$
Hence, the only possibility for a perfect square is for $x$ to be odd, and the square is equal to $ ( x^2 + \frac{x}{2} + \frac{1}{2})^2$. This gives us (expand and simplify) $ \frac{1}{4} (-x^2 + 2x + 3) =0$, hence $x=-1, 3$.
Adding in $x=0$, we get 3 solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/697882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the product of the series? Can anybody please help me to evaluate the following product?
$$\prod_{n=2}^\infty\dfrac{n^3 - 1}{n^3 + 1} $$
| Notice $$\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$ and $(n+1)^2-(n+1)+1=n^2+n+1$. Therefore we have a telescoping product
$$\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}=\frac{1\cdot \color{blue}{7}}{\color{red}{3}\cdot 3}\cdot\frac{2\cdot \color{blue}{13}}{\color{red}{4}\cdot \color{blue}{7}}\cdot \frac{\color{red}{3}\cdot \color{blue}{21}}{\color{red}{5}\cdot \color{blue}{13}}\cdots=\frac{1\cdot 2}{3}=\frac{2}{3}$$
If nervous about doing cancelling like this in an infinite product (which is a valid point!), do it carefully for the partial products and end up with the same.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How find this sum $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}$ Find the sum
$$\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}$$
My idea: since
$$\dfrac{1}{n(n+1)(n+2)\cdots(n+2014)}=\dfrac{1}{2014}\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+2013)}-\dfrac{1}{(n+1)(n+2)\cdots(n+2014)}\right)?$$
| Hint
The telescoping sum is useful
$$\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}=\frac1{2014}\dfrac{(n+2014)-n}{n(n+1)(n+2)(n+3)\cdots(n+2014)}=\frac 1{2014}\left(\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2013)}-\dfrac{1}{(n+1)(n+2)(n+3)\cdots(n+2014)}\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve the differential equation: $(y^2-xy)dx+x^2dy=0$ $$(y^2-xy)dx+x^2dy=0$$
Need step by step answer.
Thank you in advance for your help
| With some manipulation, we have
$$(y^2 - xy)dx + x^2dy = 0$$
$$(xy - y^2)dx = x^2dy$$
$$\begin{align}\frac{dy}{dx} &= \frac{xy - y^2}{x^2}\\
&= \frac{y}{x} - \left(\frac{y}{x}\right)^2\end{align}$$
Now, let $v = \frac{y}{x} \implies y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$ (product rule). Then,
$$v + x\frac{dv}{dx} = v - v^2$$
$$x\frac{dv}{dx} = -v^2$$
$$\frac{1}{v^2}\frac{dv}{dx} = -\frac{1}{x}$$
Integrating both sides W.R.T $x$,
$$\int\frac{1}{v^2}\frac{dv}{dx}\cdot dx = \int\frac{-1}{x}dx$$
$$\int\frac{1}{v^2}dv = \int\frac{-1}{x}dx$$
$$-v^{-1} = -\ln{|x|} - C$$
where $C$ is the constant of integration. Then,
$$v^{-1} = \ln{|x|} + C$$
$$v = \frac{1}{\ln{|x|} + C}$$
But $v = \frac{y}{x}$, so
$$\frac{y}{x} = \frac{1}{\ln{|x|} + C}$$
$$y = \frac{x}{\ln{|x|} + C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \cos(3x) \sin(2x) \, dx$.
Evaluate the indefinite integral
\begin{align}
\int \cos(3x) \sin(2x) \, dx
\end{align}
Please see my work attempt as my answer below.
| Employing trigonometric identities, we have
\begin{align}
\int \cos3x \sin2x \, dx &= \int \cos(2x+x) \sin2x \, dx \\
&= \int [\cos2x\cos x - \sin2x\sin x ]\sin2x \, dx \\
&= \int \cos2x \cos x \sin2x - \sin^2 2x \sin x \, dx \\
&= \int (\cos^2 x - \sin^2 x) \cos x (2 \sin x \cos x) - (2 \sin x \cos x)^2 \sin x \, dx \\
&= \int 2 \cos^4 x \sin x - 2 \cos^2 x \sin^3 x - 4 \sin^3 x \cos^2 x \, dx \\
&= \int 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x \, dx \\
&= \int 2 \cos^4 x \sin x \, dx - \int 6 \sin^3 x \cos^2 x \, dx \\
\end{align}
For the first term, we invoke the substitution rule by letting $u = \cos x$. Then $du = -\sin x \, dx$. Thus,
\begin{align}
\int 2 \cos^4 x \sin x \, dx &= \int 2u^4 (-du) \\
&= -\frac{2}{5}u^5 + C \\
&= -\frac{2}{5} \cos^5 x + C
\end{align}
Similarly for the second term, let $v = \cos(x)$. Then $dv = -\sin(x) \, dx$. Thus,
\begin{align}
\int 6 \sin^3 x \cos^2 x \, dx &= \int 6 \sin^2 x \cos^2 x \sin x \, dx \\
&= \int 6 (1 - \cos^2 x) \cos^2 x \sin x \, dx \\
&= \int 6 (1-v^2)v^2 (-dv) \\
&= \int -6v^2 + 6v^4 dv \\
&= -2v^3 + \frac{6}{5}v^5 + C \\
&= -2 \cos^3 x + \frac{6}{5} \cos^5 x + C
\end{align}
Combining the two terms together, we get
\begin{align}
\int \cos 3x \sin 2x \, dx &= \left(-\frac{2}{5} \cos^5 x + C\right) - \left(-2 \cos^3 x + \frac{6}{5} \cos^5 x + C \right) \\
&= -\frac 85 \cos^5 x + 2 \cos^3 x + C
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integration of $1/(1+\sin x)$ I solved it using $t=\tan(\frac{x}{2})$ substitution and got $-2/(1+\tan(x/2))+C$, but in my math book solution is $\tan(x/2-\pi/4)+C$. Are those the same expressions and if they are, how do I transform from one to another, or are one(or both) solutions incorrect ?
| Here is the integral with the Weierstrass substitution:
$$
\begin{align}
\int\frac{\mathrm{d}x}{1+\sin(x)}
&=\int\frac1{1+\frac{2z}{1+z^2}}\frac{2\,\mathrm{d}z}{1+z^2}\\
&=\int\frac{2\,\mathrm{d}z}{1+2z+z^2}\\
&=\frac{-2}{1+z}+C\\
&=\frac{-2}{1+\tan(x/2)}+C
\end{align}
$$
so your answer is correct. Now consider
$$
\begin{align}
\tan(x/2-\pi/4)
&=\frac{\tan(x/2)-1}{1+\tan(x/2)}\\
&=\frac{-2}{1+\tan(x/2)}+1
\end{align}
$$
Here is another approach, without Weierstrass substitution:
$$
\begin{align}
\int\frac{\mathrm{d}x}{1+\sin(x)}
&=\int\frac{1-\sin(x)}{1-\sin^2(x)}\mathrm{d}x\\
&=\int\frac{\mathrm{d}x}{\cos^2(x)}+\int\frac{\mathrm{d}\cos(x)}{\cos^2(x)}\\[4pt]
&=\tan(x)-\sec(x)+C
\end{align}
$$
and again
$$
\begin{align}
\tan(x)-\sec(x)
&=\frac{\sin(x)-1}{\cos(x)}\\
&=\frac{\frac{2\tan(x/2)}{1+\tan^2(x/2)}-1}{\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}}\\
&=\frac{-1+2\tan(x/2)-\tan^2(x/2)}{1-\tan^2(x/2)}\\
&=\frac{\tan(x/2)-1}{1+\tan(x/2)}\\
&=\frac{-2}{1+\tan(x/2)}+1
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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How show that this sum $\sum^{n}_{i=0} \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}=0$ show that :
$$\sum^{n}_{i=0} \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}=0$$
This problem is from : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239364&sid=2fbf367cb9fab8240df03e632a41085a#p239364
Evaluate a finite sum with four factorials
I think this problem have integral methods,But I can't ,Thank you
| I will assume that $n+k \geq 1$. (That is, $n$ and $k$ are not both zero simultaneously.) Let $p_{n}(x)$ be the polynomial defined by
$$ p(x) = \displaystyle \sum_{i=0}^{n} \frac{(-1)^{i}}{n+k+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} x^{m+i} . $$
Our goal is to evaluate $p(1)$. As a first step, we rewrite $p(x)$ as
\begin{align*}
p(x)
&= \sum_{i=0}^{n} \frac{(-1)^{i}}{n+k+i} \frac{1}{i!(n-i)!} \frac{d^{n}}{dx^{n}} x^{m+n+i} \\
&= \frac{d^{n}}{dx^{n}} \left( \frac{1}{n!} \sum_{i=0}^{n} \binom{n}{i} \frac{(-1)^{i}}{n+k+i} x^{m+n+i} \right) . \tag{1}
\end{align*}
Keep manipulating the expression inside the bracket of $\text{(1)}$, we get
\begin{align*}
\frac{1}{n!} \sum_{i=0}^{n} \binom{n}{i} \frac{(-1)^{i}}{n+k+i} x^{m+n+i}
&= \frac{x^{m-k}}{n!} \sum_{i=0}^{n} \binom{n}{i} (-1)^{i} \frac{x^{n+k+i}}{n+k+i} \\
&= \frac{x^{m-k} }{n!} \sum_{i=0}^{n} \binom{n}{i} (-1)^{i} \int_{0}^{x} t^{n+k+i-1} \, dt \\
&= \frac{x^{m-k}}{n!} \int_{0}^{x} t^{n+k-1} (1 - t)^{n} \, dt. \tag{2}
\end{align*}
Now we evaluate the integral $\text{(2)}$. Note that if $m \geq 1$, integration by parts shows that for any $m, n \geq 0$ the following identity holds:
\begin{align*}
\int_{0}^{x} t^{m} (1 - t)^{n} \, dt
&= \frac{m!n!}{(m+n+1)!} \left( 1 - \sum_{j=n+1}^{m+n+1} \binom{m+n+1}{j} x^{m+n+1-j} (1-x)^{j} \right).
\tag{3}
\end{align*}
Since $n + k \geq 1$, we can apply $\text{(3)}$ to $\text{(2)}$ and we get
\begin{align*}
\frac{x^{m-k}}{n!} \int_{0}^{x} t^{n+k-1} (1 - t)^{n} \, dt
&= \frac{(n+k-1)!}{(2n+k)!} \left( x^{m-k} - \sum_{j=n+1}^{2n+k} \binom{2n+k}{j} (1-x)^{j} x^{2n+m-j} \right)
\end{align*}
Then $\text{(1)}$ now reduces to
\begin{align*}
p(x)
&= \frac{(n+k-1)!}{(2n+k)!} \frac{d^{n}}{dx^{n}} \left( x^{m-k} - \sum_{j=n+1}^{2n+k} \binom{2n+k}{j} (1-x)^{j} x^{2n+m-j} \right) \\
&= \left( \frac{(n+k-1)!}{(2n+k)!} \prod_{j=0}^{n-1} (m-k-j) \right) \, x^{m-k-n} - q(x),
\end{align*}
where $q(x)$ is the polynomial given by
$$ q(x) = \frac{(n+k-1)!}{(2n+k)!} \sum_{j=n+1}^{2n+k} \binom{2n+k}{j} \frac{d^{n}}{dx^{n}} (1-x)^{j} x^{2n+m-j}. $$
But since $(1-x) \mid \frac{d^{i}}{dx^{i}} (1-x)^{j}$ whenever $i \leq n < j$, it follows that $(1-x) \mid q(x)$ and hence $q(1) = 0$. Therefore we obtain
$$ p(1) = \frac{(n+k-1)!}{(2n+k)!} \prod_{j=0}^{n-1} (m-k-j) \underset{m \geq k}{=} \frac{(n+k-1)!(m-k)!}{(2n+k)!(m-k-n)!}. $$
In particular, if $n > m-k$ then $p(1) = 0$. Clearly, if $ 1 \leq k \leq m \leq n$ then $n > m - k$ is automatically satisfied and the desired conclusion follows as corollay.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
$a,b\in\mathbb{N}$ and $ab>2$.Suppose,lcm$(a,b)=L,\gcd(a,b)=G$ and $a+b\mid L+G$.Prove that $\dfrac{(a+b)}{4}(a+b)\ge (L+G)$ and $a,b$ are two odd... $a,b\in\mathbb{N}$ and $ab>2$.Suppose,$\text{lcm}(a,b)=L,\gcd(a,b)=G$ and $a+b\mid L+G$.Prove that $\dfrac{(a+b)}{4}\cdot(a+b)\ge (L+G)$. Also prove that equality occurs when $a,b$ are consecutive odd integers.
Hint: There is an exception when $a,b$ are both equal to 2.
I could not approach this problem at all. Please help.
| Without loss of generality, $a\leq b$.
If $\gcd(a,b)\geq2$, then $G+L\leq a+\frac{ab}2\leq a\frac b2+\frac{ab}2=\frac{4ab}4\leq\frac{(a+b)^2}4$
The inequality is sharp except for $a=b=2$ in which case $G+L=\frac{(a+b)^2}4$.
If $\gcd(a,b)=1$, we have $a+b\mid L+G=ab+1=(a+b)b-b^2+1$, so $${a+b\mid b^2-1=(b+1)(b-1)}\tag{1}$$
We want to prove $L+G=ab+1\leq\frac{(a+b)^2}4\Longleftrightarrow 4ab+4\leq(b+a)^2\Longleftrightarrow4\leq(b-a)^2$.
Surely $a<b$. If $b=a+1$, then by $(1)$ we have $2a+1\mid(a+2)a$, but $\gcd(2a+1,a)=1$, so $2a+1\mid a+2$ which is impossible ($a=1\Longrightarrow b=2$ and $ab=2$).
Therefore $b\geq a+2$, so the inequality holds and is sharp except for $b=a+2$, in that case $b$ must be odd, because $\gcd(a,b)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of the series? I need to find the sum of this series, where k is from 1 to infinity?
$$\sum_{k=1}^\infty\frac{k^3+6k^2+11k+5}{(k+3)!}$$
| HINT:
As the numerator is cubic, we need to express $\displaystyle k^3+6k^2+11k+5$
as $1\cdot(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3$ where $b_i$s are arbitrary constants
Observe that we have to start with $k+3$ as it is there in the factorial of the denominator.
Had the numerator been square polynomial like $A k^2+Bk+C$ with the denominator $(k+r)!$
we had to start with $A\cdot (k+r)(k+r-1)+a_1(k+r)+a_2$
Here we have
$$\frac{k^3+6k^2+11k+5}{(k+3)!}=\frac{(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3}{(k+3)!}$$
$$=\frac1{(k)!}+b_1\frac1{(k+1)!}+b_2\frac1{(k+2)!}+b_3\frac1{(k+3)!}$$
Multiplying either sides by $(k+3)!$ we get $$k^3+6k^2+11k+5=(k+3)(k+2)(k+1)+b_1(k+3)(k+2)+b_2(k+3)+b_3$$
Methods to determine $b_i$s:
Method $\#1:$
Set $k+3=0$ to find $b_3$
$k+2=0$ to find $b_2$ in terms of $b_3$ which is known
and so on
Method $\#2:$
Compare the coefficients of the different powers of $k$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| For a more informal answer...
$$\begin{align*}\sqrt{(x+a)(x+b)}-x&= \sqrt{x^2 + (a+b)x + ab} - x \\&= \sqrt{\left(\dfrac{a+b}{2} + x\right)^2-\frac{(a+b)^2}{4} +a b} - x \end{align*}$$
The expression $-\frac{1}{4} (-a-b)^2+a b$ is constant and is not important as $x \to \infty$ because the expression under the square root is dominated by the squared term.
Therefore
$$\begin{align*}\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)&= \\ \lim_{x\to\infty} \left(x + \dfrac{a+b}{2}\right) - x &= \dfrac{a+b}{2} \end{align*}$$
To add justification, consider $$\sqrt{ax^n + bx^m} = (ax^n + bx^m)^{1/2}, \quad n \geq m.$$
This is equal to $$(ax^n)^{1/2} + \frac{1}{2}\dfrac{bx^m}{\sqrt{ax^n}} - \dfrac{1}{8}\dfrac{b^2x^{2m}}{(\sqrt{ax^n})^3} ... $$
If we consider the case $n = 2, m = 1$ it is clear that the first two terms are non zero as $x \to \infty$ which is why we can't say that $\sqrt{x^2 - x} - x \approx 0 $ for large $x$. On the other hand, the only term left in the case where $n = 2$ and $m = 0$ as $x \to \infty$ is the first term because every other term has a power of $x$ in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
} |
Is there a general formula for the sum of a quadratic sequence? I tried Googling "formula for sum of quadratic sequence", which did not give me anything useful. I just want an explicit formula for figuring out a sum for a quadratic sequence. For example, how would you figure out the sum of $2+6+12+20+\dots+210$? Can someone please help? Thanks
For those of you who do not know, a quadratic sequence is a sequence where the differences of the differences between the terms are constant. Let's use $2+6+12+20+\dots$ as an example. The differences between the terms are $4$, $6$, $8$, etc. The difference between the differences of the terms is $2$. So the sequence will continue like $2+6+12+20+30+42+56+72+\dots$
| In this particular case,
$$\begin{align}
&2+6+12+20+\cdots+210\\\\
&=2(1+3+6+10+\cdots+105)\\\\
&=2\left[\binom 22 +\binom 32 +\binom 42+\binom 52+\cdots \binom {15}2\right]\\
&=2\sum_{r=2}^{15}\binom r2\\
&=2\binom {16}3\\
&=\color{red}{1120}\qquad\blacksquare
\end{align}$$
This is the sum of triangular numbers (where the difference of the difference is constant) and the result is a pyramidal number (all scaled by 2). The summation can be shown as
$$\begin{align}
&2\cdot (1)+\\
&2\cdot (1+2)+\\
&2\cdot (1+2+3)+\\
&2\cdot (1+2+3+4)+\\
&\quad \vdots\qquad\qquad\qquad\ddots\\
&2\cdot (1+2+3+\cdots+15)
\end{align}$$
If we tried to sum a series where the difference of the difference of the difference is constant, i.e. sum of pyramidal numbers, the result would be a pentatope number. And so on...
An example of the summation of pyramidal numbers, extending from the original question, would be
$$2+8+20+40+\cdots+910 =2 \sum_{r=1}^{15} \binom r3=2 \binom {16}4=1820\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/710484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Relationship between coefficients of a polynomials and its roots. The roots of the equation $x^4 -3x^2 + 5x - 2 = 0$ are $\alpha$, $\beta$, $\gamma$ and $\delta$. $\alpha^n + \beta^n + \gamma^n + \delta^n$ is denoted by $S(n)$. Find values of $S(2)$ and $S(4)$ and of $S(3)$ and $S(5)$.
Hence, find the value of
$\alpha^2 (\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3)$
This topic is a rather strange one, and I can't seem to locate the formulas I need, and I need them for my exam. If you do use any formula (except $\sum\alpha$, $\sum\alpha\beta$, $\sum\alpha\beta\gamma$ and $\alpha\beta\gamma\delta$), please state it. Thank you in advance.
| Denote
$A(1) = \alpha+\beta+\gamma+\delta = 0$;
$A(2) = \alpha\beta+...+\gamma\delta = -3$;
$A(3) = \alpha\beta\gamma +...+\beta\gamma\delta = -5$;
$A(4) = \alpha\beta\gamma\delta = -2$.
Applying Newton's identities, we obtain
$S(1) = A(1) = 0$;
$S(2) = A(1) S(1) - 2A(2) = 0+6=6$;
$S(3) = A(1) S(2) - A(2)S(1) + 3A(3) = 0+0-15=-15$;
$S(4) = A(1) S(3) - A(2)S(2) + A(3)S(1) - 4A(4) = 0+18+0+8=26$;
$S(5) = A(1) S(4) - A(2)S(3) + A(3)S(2) - A(4)S(1) + 5A(5) = 0-45-30+0+0=-75$
(one can obtain last identity, when consider equation $x^5-3x^3+5x^2-2x=0$ with $5$ roots $\alpha,\beta,\gamma,\delta,0$, where $A(5)=0$, and previous $A(n)$ are the same).
Easy to see, that
$$\alpha^2 (\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3)$$
$$=\alpha^2 (\alpha^3+\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \beta^3+\gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \gamma^3+\delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3 + \delta^3) - \alpha^5-\beta^5-\gamma^5-\delta^5$$
$$
= (\alpha^2+\beta^2 + \gamma^2 + \delta^2)(\alpha^3+\beta^3 + \gamma^3 + \delta^3)
- \alpha^5-\beta^5-\gamma^5-\delta^5 = S(2)S(3)-S(5).
$$
$$
= 6\cdot (-15) - (-75) = -90+75=-15.
$$
Note:
equation $x^4-3x^2+5x-2=0$ has $2$ real and $2$ complex roots:
$x_1\approx -2.344470$;
$x_2\approx 0.578277$;
$x_{3,4} \approx 0.883096 \pm 0.833866i$.
| {
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"source": "stackexchange",
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Proof Verification: Show sequence is bounded and find limit: $x_1 \gt 1$ and $x_{n + 1} = 2 - \frac{1}{x_n}$ Came across the following exercise in Bartle's Elements of Real Analysis and am a little unsure about my solution. Would be extremely grateful if someone could verify it for me.
Let $x_1 \in \Bbb R$ satisfy $x_1 \gt 1$ and let $x_{n + 1} = 2 -
\frac{1}{x_n}$ for $n \in \Bbb N$. Show that the sequence $(x_n)$ is
monotone and bounded. What is the limit?
My Attempt:
Part 1 - (Showing monotony and bound)
We will prove that $1 \lt x_{n +1} \lt x_n$ for each $n \in \Bbb N$. $x_2 = 2 - \frac 1 {x_n} \gt 2 - 1 = 1$ and $x_2 = 2 - \frac 1 {x_2} = \dfrac{2x_2 - 1}{x_2} = 1 + \dfrac{x_2 -1}{x_2} \lt 1 + x_2 - 1 = x_2$ whence we have that $1 \lt x_2 \lt x_1$. Now suppose $1 \lt x_{n +1} \lt x_n$ for an arbitrary natural number $n$.
$$x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } \gt 2 - 1 = 1$$
$$ x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } = \frac{2x_{n + 1} - 1}{x_{n + 1}} = 1 + \frac{x_{n + 1} - 1}{x_{n + 1}} \lt 1 + x_{n + 1} - 1 = x_{n + 1} $$
Therefore we have $ 1 \lt x_{n + 2} \lt x_{n + 1} $ and hence induction is complete.
So the sequence is monotone decreasing and is bounded below by $1$. By the Monotone Convergence Theorem $\lim (x_n) = x$ exists in $\Bbb R$.
Part 2 - (Finding Limit)
We know that $x_nx_{n + 1} = 2x_n - 1$ and that $(x_n)$ converges and so does $(x_{n + 1})$ to the same limit since it can be interpreted as a subsequence of $(x_n)$. Then the sequence $(x_nx_{n + 1})$ will also converge to the limit $\lim {(x_n)} \lim {(x_{n+ 1})} = x^2 $. The sequences $(y_n)$ and $(z_n)$ given by $y_n = 2$ and $ z_n = -1 $ for $n \in \Bbb N$ also converge to $2$ and $(-1)$ respectively so $\lim (2x_n - 1) = 2x - 1$. Whence we have that $$x^2 = 2x - 1 \implies x = 1$$
*
*Am a little unsure about two areas of the proof - one, the part which uses induction since I found both steps can be shown using the
same method and two, the computation of $\lim (x_n)$. So any criticism
is more than welcome.
*If there is an easier way to prove this I would very much like to know since this is just the first exercise and I do not want to follow
a wrong path.
| To take this out of unanswered list:
We will prove that $1 \lt x_{n +1} \lt x_n$ for each $n \in \Bbb N$. $x_2 = 2 - \frac 1 {x_n} \gt 2 - 1 = 1$ and $x_2 = 2 - \frac 1 {x_2} = \dfrac{2x_2 - 1}{x_2} = 1 + \dfrac{x_2 -1}{x_2} \lt 1 + x_2 - 1 = x_2$ whence we have that $1 \lt x_2 \lt x_1$. Now suppose $1 \lt x_{n +1} \lt x_n$ for an arbitrary natural number $n$.
$$x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } \gt 2 - 1 = 1$$
$$ x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } = \frac{2x_{n + 1} - 1}{x_{n + 1}} = 1 + \frac{x_{n + 1} - 1}{x_{n + 1}} \lt 1 + x_{n + 1} - 1 = x_{n + 1} $$
Therefore we have $ 1 \lt x_{n + 2} \lt x_{n + 1} $ and hence induction is complete.
So the sequence is monotone decreasing and is bounded below by $1$. By the Monotone Convergence Theorem $\lim (x_n) = x$ exists in $\Bbb R$.
We know that $x_nx_{n + 1} = 2x_n - 1$ and that $(x_n)$ converges and so does $(x_{n + 1})$ to the same limit since it can be interpreted as a subsequence of $(x_n)$. Then the sequence $(x_nx_{n + 1})$ will also converge to the limit $\lim {(x_n)} \lim {(x_{n+ 1})} = x^2 $. The sequences $(y_n)$ and $(z_n)$ given by $y_n = 2$ and $ z_n = -1 $ for $n \in \Bbb N$ also converge to $2$ and $(-1)$ respectively so $\lim (2x_n - 1) = 2x - 1$. Whence we have that $$x^2 = 2x - 1 \implies x = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What would be the sum of Trigonometric Functions I was trying to remember the functions provided on the site: http://www.purplemath.com/modules/idents.htm#restatement
From there I came to know about some of the function, basically the Sum of the trigonometric functions. But I was only able to find the Sin(x) +/- Sin(y) and Cos(x) +/- Cos(y) functions.
I wanted to know, is it possible to use +/- operators on Sin(x) and Cos(x) such as: Sin(x) - Cos(x) ?
| In general, we have (this method is known as the R-Alpha method):
$$a\sin\theta - b\cos\theta = \sqrt{a^2 + b^2}\sin\left(\theta - \tan^{-1}\frac{b}{a}\right)$$
For your case, we get
$$\sin x - \cos x = \sqrt2\sin\left(\theta - \frac{\pi}{4}\right)$$
Here's the proof: Let
$$a\sin\theta - b\cos\theta = R\sin(\theta - \alpha)$$
for some real number $R, \alpha$.
By using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we have
$$a{\color{red}\sin}{\color{red}\theta} - b{\color{blue}\cos}{\color{blue}\theta} = R{\color{red}\sin}{\color{red} \theta} \cos \alpha - R{\color{blue}\cos}{\color{blue} \theta} \sin \alpha$$
Equating the coefficients of ${\color{red}\sin}{\color{red}\theta}$ and ${\color{blue}\cos}{\color{blue}\theta}$, we have:
$$a = R\cos\alpha$$
$$b = R\sin\alpha$$
Divide both equations to get:
$$\tan\alpha = \frac{b}{a} \implies \alpha = \tan^{-1}\frac{b}{a}$$
Now, to find $R$, we take $a^2 + b^2$:
$$\begin{align}a^2 + b^2 &= (R\cos\alpha)^2 + (R\sin\alpha)^2 \\&= R^2(\cos^2\alpha + \sin^2\alpha) \\&= R^2\end{align}$$
so we have: $$R = \sqrt{a^2 + b^2}$$
To conclude:
$$a\sin\theta - b\cos\theta = \sqrt{a^2 + b^2}\sin\left(\theta - \tan^{-1}\frac{b}{a}\right)$$
In fact, you can use the same approach for $a\sin\theta + b\cos\theta$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Writing a system of ODEs in polar coordinates I have this system of equations:
$$\dot{x}=x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}} \\
\dot{y}=x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}}$$
How can I get this in polar coordinates ? I know that $r^2=x^2+y^2$, but how can I find $\dot{r}$ or $\dot{\theta}$ ?
| Setting
$$
x=r\cos\theta,\ y=r\sin\theta,
$$
we have
\begin{eqnarray}
\dot{r}\cos\theta-r\dot{\theta}\sin\theta&=&\dot{x}=r(\cos\theta-\sin\theta)-r^3\cos\theta+r\cos\theta\sin\theta,\\
\dot{r}\sin\theta+r\dot{\theta}\cos\theta&=&\dot{y}=r(\cos\theta+\sin\theta)-r^3\sin\theta-r\cos^2\theta.
\end{eqnarray}
It follows that
$$
\dot{r}=r-r^3,\quad \dot{\theta}=1-\cos\theta.
$$
| {
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"url": "https://math.stackexchange.com/questions/721297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$ Ok, so I know that since the numerator has a higher power that long division is needed. So after doing that, the main fraction is $\frac{-6x-36}{x^2 + 12x + 36}$. I think that's right. But my problem is that after you factor the denominator they're both equal to $(x+6)$. So how would you use partial fraction decomposition?
| \begin{array}{rcrrrrrrr}
& & x & -12 \\
& & -- & --- & --- & --- & \\
x^2+12x+36 & ) & x^3 \\
& & x^3 & 12x^2 & 36x \\
& & -- & --- & --- \\
& & & -12x^2 & -36x \\
& & & -12x^2 & -144x & -432 \\
& & & --- & --- & --- \\
& & & & 108x & 432 \\
\end{array}
And so
$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108x+432}{(x+6)^2}$$
where
\begin{align}
\dfrac{108x+432}{(x+6)^2} &= \dfrac{A}{x+6} + \dfrac{B}{(x+6)^2} \\
108x + 432 &= A(x+6) + B
\end{align}
$$ \text{Let $x=-6$ and you get $B = -216$.}$$
$$ \text{Let $B = -216$ and you get}$$
\begin{align}
108x + 432 &= A(x+6) - 216 \\
A(x+6) &= 108x + 648 \\
A &= 108
\end{align}
And so
$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108}{x+6} - \dfrac{216}{(x+6)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the 2nd order taylor polynomial of f(x,y)? I'm just computing the 2nd order taylor polynomial for $f(x,y) = tan(x + 3y + \frac{\pi}{4})$ centered at (3,-1) and wondering if I have done this correctly or if anyone has any suggestions on how I can improve my answer (I've never done this before with mulitvariables so just want to be sure I am on the right track):
I have that: $$f(3,-1) = tan(\frac{\pi}{4}) = 1$$
$$f_{x}(3,-1) = sec^2(\frac{\pi}{4}) = 2$$
$$f_{y}(3,-1) = 3sec^2(\frac{\pi}{4}) = 6$$
So the gradient vector for $f$ is $(2,6)$
The letting $u = x + 3y + \frac{\pi}{4}$, I get that $\frac{du}{dx}= 1 $ and $\frac{dy}{dx}= 3 $
So, $$f_{x}(u) = sec^2(u)$$
$$f_{xx}(u) = 2tan(u)sec^2(u)\frac{du}{dx}$$
$$f_{xy}(u) = 2tan(u)sec^2(u)\frac{du}{dy}$$ and similarly,
$$f_{y}(u) = 3sec^2(u)$$
$$f_{yx}(u) = 6tan(u)sec^2(u)\frac{du}{dx}$$
$$f_{yy}(u) = 6tan(u)sec^2(u)\frac{du}{dy}$$
Substituting back for $u$ and plugging in $(3,-1)$ I get:
$$f_{xx}(3,-1) = 2tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot1 = 4$$
$$f_{xy}(3,-1) = 2tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot3 = 12$$ and similarly,
$$f_{yx}(3,-1) = 6tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot1 = 12$$
$$f_{yy}(3,-1) = 6tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot3 = 36$$
So,
$$T_{2}((x,y),(3,-1)) = 1+(2,6) \cdot (x-3,y+1) + \frac{1}{2} \bigl( \begin{smallmatrix}
x-3 & y+1\\
\end{smallmatrix} \bigr) \bigl( \begin{smallmatrix}
4 & 12\\
12 & 36
\end{smallmatrix} \bigr) \bigl( \begin{smallmatrix}
x-3 \\
y+1
\end{smallmatrix} \bigr)$$
Performing the calculations I get:
$$= 1+2x-6+6y+6+\frac{1}{2}[(x-3)(4x-12+12y+12)+(y+1)(12x-36+36y+36)]$$
$$ = 2x^2 + 18y^2 + 12xy + 2x + 6y + 1$$
Is this correct and have I done the steps correctly? Many thanks in advance!
| For what it's worth after two years, yes, you did everything correctly.
To back that up, the Mathematica expression for Taylor Series
f[x_,y_]=Tan[x+3y+Pi/4];
Normal[Series[f[(x-3)t+3,(y+1)t-1],{t,0,2}]] /. t-> 1
yields indeed
$$1+2 x+2 x^2+6 y+12 x y+18 y^2.$$
At least that's one unanswered question less.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Optimization with a constrained function Okay so I understand how to find points of extrema when for example,
We have $3x^2 + 2y^2 + 6z^2$ subject to the constaint $x+y+z=1$. I followed the method of the Lagrange multiplier and resulted in the point $(\frac13, \frac12, \frac16)$. I know this point is correct.
However, I don't know how to prove that it's a minimum value. Any ideas?
| An elementary proof would be as follows: notice that by "completing the square",
$$3x^2 - 2x + 2y^2 - 2y + 6z^2 - 2z \\
= 3\left(\left(x - \frac{2}{6}\right)^2 - \left(\frac{2}{6}\right)^2\right) + 2\left(\left(y - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2\right) + 6\left(\left(z - \frac{2}{12}\right)^2 - \left(\frac{2}{12}\right)^2\right)\\
= 3\left(x - \frac{2}{6}\right)^2 + 2\left(y - \frac{1}{2}\right)^2 + 6\left(z - \frac{2}{12}\right)^2 - \frac{1}{3} - \frac{1}{2} - \frac{1}{6}\\
\ge -1$$
(In case you're confused: $-\frac{1}{3} - \frac{1}{2} - \frac{1}{6} = -1$)
But $x + y + z = 1$, so we have
$$3x^2 + 2y^2 + 6z^2 -2 \ge -1$$
$$3x^2 + 2y^2 + 6z^2 \ge 1$$
It is quite evident (from the first inequality) that equality is achieved when $x = \frac{1}{3}, y = \frac{1}{2}, z = \frac{1}{6}$.
A more advanced method: using the Cauchy-Schwarz inequality, we have
$$(x + y + z)^2 \le \left(3x^2 + 2y^2 + 6z^2\right)\left(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\right)$$
But $x + y + z = 1$, so we can substitute this and simplify a bit to get:
$$3x^2 + 2y^2 + 6z^2 \ge 1$$
Equality holds when $3x = 2y = 6z$. From this, it can be trivially deduced that indeed, $(x,y,z) = \left(\frac{1}{3}, \frac{1}{2}, \frac{1}{6}\right)$ minimizes the expression with the given constrain.
While not exactly nor directly relevant to Lagrange Multipliers, this does provides a simpler alternative for certain optimization problems.
| {
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Vector triangle lines intersection proof
In this triangle, the distance between $M$ and $A$ is twice the distance between $M$ and $B$. And the measure of AN is the third part of the measure of $CN$. Write $X$ in function of $A$, $AB$ and $AC$.
I don't know how to write vector notation, but A is a point, and the rest are vectors. So I started trying to write $X$ like this:
$X = A + AM + MX$
And I know that $AM = 2MB$ so $AM = \frac{2}{3}AB$, then:
$X = A + \frac{2}{3}AB + MX$
I just need to get rid of the $MX$ but when I try to write it in function of other things, I Always get some that can't be written in function of $A$, $AB$ and $AC$.
The answer is $X = A + \frac{3}{5}AB + \frac{1}{10}AC$
Thanks!
| Note that:
$$X= A + \frac{2}{3} AB + MX \quad (1)$$
$$X= A + \frac{1}{4} AC + NX \quad (2)$$
$$MX = \lambda MC = \lambda(MA +AC)= \lambda (- \frac{2}{3} AB +AC) \quad (3)$$
and
$$NX = \mu NB = \mu(NA +AB)= \mu (- \frac{1}{4} AC + AB). \quad (4)$$
Substituting $(3)$ in $(1)$ and $(4)$ in $(2)$ we get:
$$X= A + ( \frac{2}{3}- \frac{2 \lambda}{3})AB + \lambda AC \quad (5)$$
and
$$X= A + ( \frac{1}{4} - \frac{\mu}{4})AC + \mu AB \quad (6).$$
From $(5)$ and $(6)$ you get your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to get the characteristic equation from a recurrence relation of this form? I've been getting the characteristic equation from relations of the form
$$U_n=3U_{n-1}-U_{n-3}$$
Thanks to this question I made before: How to get the characteristic equation?
Now, I recently have been asked to get such equation from this:
$$f(n) = \begin{cases}5 & \text{ if n = 0} \\ f(n-1) + 3 + 2n + 2^n \end{cases}$$
I tried to apply the same idea. I took:
$$F_n=F_{n-1} + 3 + 2n + 2^n$$
Changed the subscripts to exponents...
$$F^n=F^{n-1} + 3 + 2n + 2^n$$
Replaced $F$ by the desired variable...
$$x^n=x^{n-1} + 3 + 2n + 2^n$$
Divided by the smallest exponent (I guess it is $n-1$):
$$x=1+ 3^{-x+1} + 2n^{-x+1} + 2^{n-x+1}$$
And this doesn't look right. As you can see, the terms $3$, $2n$ and $2^n$ messed up the whole thing.
What did I do wrong, and how can I get the characteristic equation from this?
| Use the machinery of generating functions. Define $F(z) = \sum_{n \ge 0} f(n) z^n$, and write your recurrence as:
$$
f(n + 1) = f(n) + 5 + 2 n + 2^{n + 1}
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0} z^n &= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2} \\
\sum_{n \ge 0} 2^n z^n
&= \frac{1}{1 - 2 z}
\end{align}
Pulling all together:
$$
\frac{F(z) - 5}{z}
= F(z) + \frac{5}{1 - z} + \frac{2 z}{(1 - z)^2} + \frac{2}{1 - 2 z}
$$
Solving for $F(z)$, as partial fractions:
$$
F(z) = \frac{5 - 13 z + 8 z^2 - 2 z^3}{1 - 5 z + 9 z^2 - 7 z^3 + 2 z^4}
= \frac{2}{1 - 2 z} + \frac{1}{(1 - z)^2} + \frac{2}{(1 - z)^3}
$$
Using the (generalized) binomial theorem:
\begin{align}
f(n) &= 2 \cdot 2^n + \binom{-2}{n} (-1)^n + 2 \cdot \binom{-3}{n} (-1)^n \\
&= 2^{n + 2} + \binom{n + 2 - 1}{1} + 2 \cdot \binom{n + 3 - 1}{2} \\
&= 2^{n + 1} + n + 1 + \frac{(n + 2) (n + 1)}{2} \\
&= 2^{n + 1} + \frac{n^2 + 5 n + 4}{2}
\end{align}
Maxima's help with the algebra is gratefully acknowledged.
| {
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How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$
| To start, this polynomial is called a biquadratic equation.
First, factor out the polynomial.
$$x^4 - 2x^3 - 6x^3 + 12x^2 + 12x^2 - 24x - 8x + 16 = 0$$
$$x^3(x - 2) - 6x^2(x - 2) + 12x(x - 2) - 8(x - 2) = 0$$
$$(x^3 - 6x^2 + 12x - 8)(x - 2) = 0$$
Then after, continue the factorization of the polynomial.
$$(a + b)^2 = a^2 + 2ab + b^2$$
$$(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$$
$$(a + b)^4 = a^4 + 4a^3b + 6a^2 b^2 + 4ab^3 + b^4$$
$$x^4 + 4 \times x^2 \times (-2) + 6 \times x^2 \times (-2)^2 + 4 \times x \times (-2)^3 + (-2)^4 = 0$$
So $\mathbf{(x - 2)^4 = 0}$.
| {
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In $\triangle ABC$, I is the incenter. Area of $\triangle IBC = 28$, area of $\triangle ICA= 30$ and area of $\triangle IAB = 26$. Find $AC^2 − AB^2$ In $\triangle ABC$, I is the incenter. Area of $\triangle IBC = 28$, area of $\triangle ICA = 30$ and area of $\triangle IAB = 26$. Find $AC^2 − AB^2$.
Here is a sketch that I drew:
From the given areas ,
$r*AC=60$
$r*AB=52$
$r*BC=56$
So that I've three equations with 4 unknowns .
Also,
$A=rs$
$2A=168=r(AB+BC+AC)$
But this can't be used as it can be derived from the other three equations.
So, what should I do? Any hints are apreciated.
(This is not class-homework , I'm solving sample questions for a competitive exam )
| we note $AC=b,BC=a,AB=c$, we will find $b^2-c^2$. We have
$$ar=56,br=60,cr=52.$$
then we have the proportion of $a$ and $b,c$:
$$b=\frac{60}{56}a,c=\frac{52}{56}a$$
Then we have the equality:
$$a=\frac{56}{112}(b+c)=\frac{1}{2}(b+c)$$
$$a=\frac{56}{8}(b-c)=7(b-c)$$
Now we should recall the formula of triangle's area:
$S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}$
Then we have
$$84=\sqrt{\frac{a+b+c}{2}\cdot\frac{b+c-a}{2}\cdot\frac{a+c-b}{2}\cdot\frac{a+b-c}{2}}\\=\frac{1}{4}\sqrt{[(b+c)^2-a^2][a^2-(b-c)^2]}\\=\frac{1}{4}\sqrt{[(b+c)^2-\frac{1}{4}(b+c)^2][49(b-c)^2-(b-c)^2]}\\=\frac{1}{4}\sqrt{\frac{3}{4}(b+c)^2\cdot48(b-c)^2}\\\frac{3}{2}|b^2-c^2|\\=\frac{3}{2}(b^2-c^2)$$(because $b>c$).
then we have what we want.
| {
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"timestamp": "2023-03-29T00:00:00",
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
| $GM\le AM\le RMS\implies GM\times AM\le RMS^2\implies a+b+c\le a^2+b^2+c^2$.
| {
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Find domains of convergence of the series I tried to solve this question but the final solution which I obtain is not the as same as in the text book
Find domains of convergence of the series
$$\sum_{n=1}^ \infty \frac{1.3.5...(2n-1)}{n!}(\frac{1-z}{z})^n$$
The book answer is $|z-\frac{4}{5}|\lt \frac{2}{3}$
But my answer is $$|\frac{1-z}{z}|\lt \frac{1}{2}$$
Is this true ??
How can I obtain the solution as in the book .
| Your solution is correct :
$$\left|\frac{1-z}z\right|<\frac12\iff|2-2z|<|z|$$
Now put $\;z=x+iy\;,\;\;x,y\in\Bbb R\implies 2-2z=(2-2x)-2yi\;$ , so::
$$|2-2z|^2<|z|^2\iff(2-2x)^2+4y^2<x^2+y^2\iff$$
$$4-8x+4x^2+4y^2<x^2+y^2\iff3x^2-8x+3y^2+4<0\iff$$
$$3\left(x-\frac43\right)^2-\frac{16}3+3y^2+4<0\iff 3\left(x-\frac43\right)^2+3y^2<\frac43\iff$$
$$\left(x-\frac43\right)^2+y^2<\left(\frac23\right)^2\iff\left|z-\frac43\right|<\frac23$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/740918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the range of a function How to find the range of the following trigonometric function $\sin^2x-5\sin x-6$. Can some one help me out. Thank you
| $$\sin^2x-5\sin x-6=\frac{(2\sin x)^2-2\cdot2\sin x\cdot5+5^2-24-25}4=\frac{(2\sin x-5)^2-49}4$$
Now, for real $\displaystyle x,-1\le\sin x\le1\iff -2\le\sin x\le2\iff-7\le2\sin x-5\le-3$
$\displaystyle2\sin x-5\le-3\implies(2\sin x-5)^2\ge9$
$\displaystyle2\sin x-5\ge-7\implies(2\sin x-5)^2\le49$
Can you take it home from here?
| {
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Solve Partial differential equation(geometric optics) Solve $x^2((u_x)^2+(u_y)^2)=1$ , $u(x,0)=0$
Use the characteristic equation
The solution is
$u(x,y)=-\ln\dfrac{\sqrt{x^2+y^2}+y}{x}$
I drove
$\dfrac{dx}{dt}=2x^2p$
$\dfrac{dy}{dt}=2x^2q$
$Z=2t$
$\dfrac{dp}{dt}=\dfrac{2}{x}$
$dq=0$
From above equations, I got
$q=\dfrac{1}{s}$
$p=\sqrt{-\dfrac{1}{s^2}+\dfrac{1}{x^2}}$
Then I lost. Please help
| $x^2((u_x)^2+(u_y)^2)=1$
$(u_x)^2+(u_y)^2=\dfrac{1}{x^2}$
$(u_y)^2=\dfrac{1}{x^2}-(u_x)^2$
$u_y=\pm\sqrt{\dfrac{1}{x^2}-(u_x)^2}$
$u_{xy}=\pm\dfrac{-\dfrac{2}{x^3}-2u_xu_{xx}}{2\sqrt{\dfrac{1}{x^2}-(u_x)^2}}$
$u_{xy}=\mp\dfrac{x^3u_xu_{xx}+1}{x^3\sqrt{\dfrac{1}{x^2}-(u_x)^2}}$
Let $v=u_x$ ,
Then $v_y=\mp\dfrac{x^3vv_x+1}{x^3\sqrt{\dfrac{1}{x^2}-v^2}}$ with $v(x,0)=0$
$x^3vv_x\pm x^3\sqrt{\dfrac{1}{x^2}-v^2}~v_y=-1$ with $v(x,0)=0$
$-x^3v_x\mp\dfrac{x^3}{v}\sqrt{\dfrac{1}{x^2}-v^2}~v_y=\dfrac{1}{v}$ with $v(x,0)=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dv}{dt}=\dfrac{1}{v}$ , letting $v(0)=0$ , we have $v^2=2t$
$\dfrac{dx}{dt}=-x^3$ , letting $x(0)=x_0$ , we have $x^2=\dfrac{x_0^2}{2x_0^2t+1}=\dfrac{x_0^2}{x_0^2v^2+1}$
$\dfrac{x^2}{1-x^2v^2}=x_0^2$
$\dfrac{dy}{dt}=\mp\dfrac{x^3}{v}\sqrt{\dfrac{1}{x^2}-v^2}=\mp\dfrac{\sqrt{x_0}}{\sqrt{2t}(2x_0^2t+1)^\frac{3}{2}}$ , letting $y(0)=f(x_0^2)$ , according to http://integrals.wolfram.com/index.jsp?expr=c%5E%281%2F2%29%2F%28%282x%29%5E%281%2F2%29%282c%5E2x%2B1%29%5E%283%2F2%29%29&random=false, we have $y=f(x_0^2)\mp\sqrt{\dfrac{2x_0t}{2x_0^2t+1}}=f\left(\dfrac{x^2}{1-x^2v^2}\right)\mp\sqrt xv\sqrt[4]{1-x^2v^2}$
$v(x,0)=0$ , i.e. $y(x,v=0)=0$ :
$f(x^2)=0$
$f(x)=0$
$\therefore y=\mp\sqrt xv\sqrt[4]{1-x^2v^2}$
$y^4=x^2v^4(1-x^2v^2)$
$x^4v^6-x^2v^4+y^4=0$
| {
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Find the taylor series by $f$ at $x=a$ I don't know how to expand taylor series, I was wondering if anyone could guide me in the right direction, I've tried a few pages on the internet, but still cannot understand how to expand.
The example I've been given is;
Find the Taylor series generated by $f=\frac{2}{x^2}$ at $x=a$ where $a=1$
| As stated by Michael, the notation $f^n(a)$ means the $n^{th}$ derivative of $f$ evaluated at $a$. A typical place to start is to list some derivatives of $f$ and evaluate them at $x=a$. For example, $f^0(a)$, the "zeroth" derivative, is just the function $f$, $f^1(a)$ is the first derivative of $f$, and so forth:
\begin{align*}
f^0(a)&=\frac{2}{a^2} \\
f^1(a)&=-\frac{4}{a^3} \\
f^2(a)&=\frac{12}{a^4} \\
f^3(a)&=-\frac{48}{a^5} \\
f^4(a)&=\frac{240}{a^6} \\
\ldots &= \ldots
\end{align*}
We need enough of these to spot a pattern. We assemble this all together by the formula with the sum index starting at $n=0$:
\begin{align*}
&\sum_{n = 0}^{\infty} \frac{f^n (a)}{n!} (x-a)^n \\=& \frac{\frac{2}{a^2}}{0!}(x-a)^0+\frac{-\frac{4}{a^3}}{1!}(x-a)^1+\frac{\frac{12}{a^4}}{2!}(x-a)^2 +\frac{-\frac{48}{a^5}}{3!}(x-a)^3+\frac{\frac{240}{a^6}}{4!}(x-a)^4 +\ldots \\
=&\frac{2}{a^2}-\frac{4(x-a)}{a^3}+\frac{12(x-a)^2}{2a^4}-\frac{48(x-a)^3}{6a^5}+\frac{240(x-a)^4}{24a^6}-\ldots \\
=&\frac{2}{a^2}-\frac{4(x-a)}{a^3}+\frac{6(x-a)^2}{a^4}-\frac{8(x-a)^3}{a^5}+\frac{10(x-a)^4}{a^6} -\ldots+\ldots.
\end{align*}
The pattern should now be clear to where most people could write the next term after visual inspection of the prior terms. The first 5 terms were more than sufficient to see it.
You can then substitute $1$ for $a$, or any other nonzero number $a$ you wish to get the Taylor series representation at that point.
| {
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Prove that $\sum_{k=0}^n k^2{n \choose k} = {(n+n^2)2^{n-2}}$ Prove that: $$\sum_{k=0}^n k^2{n \choose k} = {(n+n^2)2^{n-2}}$$ i know that: $$\sum_{k=0}^n {n \choose k} = {2^n}$$ how to get the (n + n^2)?
| Alternative proof using no calculus, no combinatorics, just pure algebraic manipulation starting from the special case of the binomial theorem: $$\sum_{k=0}^n \binom{n}{k} = 2^n.$$ We also note the identity $$\begin{align*} \binom{k}{r} \binom{n}{k} &= \frac{k!}{r! \, (k-r)!} \cdot \frac{n!}{k! \, (n-k)!} = \frac{n!}{r! \, (n-r)!} \cdot \frac{(n-r)!}{(k-r)! \, (n-k)!} \\ &= \binom{n}{r} \binom{n-r}{k-r}. \end{align*}$$
Then $$S_1 = \sum_{k=0}^n k \binom{n}{k} = \sum_{k=1}^n \binom{k}{1}\binom{n}{k} = \binom{n}{1} \sum_{k=1}^n \binom{n-1}{k-1} = n 2^{n-1}.$$
Similarly, $$S_2 = \sum_{k=0}^n k(k-1) \binom{n}{k} = 2 \sum_{k=2}^n \binom{k}{2} \binom{n}{k} = 2 \binom{n}{2} \sum_{k=2}^n \binom{n-2}{k-2} = n(n-1) 2^{n-2}.$$
Consequently, $$\sum_{k=0}^n k^2 \binom{n}{k} = S_1 + S_2 = n(n+1)2^{n-2}.$$ In this manner, we can recursively evaluate sums of the form $$\sum_{k=0}^n k^m \binom{n}{k},$$ using the general identity $$\sum_{k=r}^n \binom{k}{r}\binom{n}{k} = \binom{n}{r} 2^{n-r}.$$
| {
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Finding argument of a complex number How do you evaluate the following
$$\text{Arg}\{\sin\frac{8\pi}{5} + i(1 + \cos\frac{8\pi}{5})\}$$
| $\displaystyle1+\cos\frac{8\pi}5=1-\cos\frac{3\pi}5>0$
and $\displaystyle\sin\frac{8\pi}5=\sin\left(\pi+\frac{3\pi}5\right)=-\sin\frac{3\pi}5<0$
So using the definition of atan2, $$\text{Arg}\left[\sin\frac{8\pi}{5} + i(1 + \cos\frac{8\pi}{5})\right]=\pi+\arctan\left(\frac{1-\cos\frac{3\pi}5}{-\sin\frac{3\pi}5}\right)$$
Now, $\displaystyle\frac{1-\cos2A}{-\sin2 A}=\frac{2\sin^2A}{-2\sin A\cos A}=-\tan A=\tan(-A)$
Here $\displaystyle2A=\frac{3\pi}5$
$$\implies \text{Arg}\{\sin\frac{8\pi}{5} + i(1 + \cos\frac{8\pi}{5})\}=\pi+\left(-\frac{3\pi}{10}\right)=\frac{7\pi}{10}$$
| {
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Solving a trigonometric equation Can someone help me to solve this problem?
Find all number pairs $x,y$ that satisfy the equation:
$$\tan^4(x) + \tan^4(y) + 2\cot^2(x)\cot^2(y) = 3 + \sin^2(x+y)$$
| By AM-GM inequality:
$$(\tan x)^4 + (\tan y)^4 + (\cot x)^2\cdot (\cot y)^2 + (\cot x)^2\cdot (\cot y)^2 \geq 4 \geq 3 + (\sin(x+y))^2
$$
So the equation occurs when: $\sin(x+y) = 1, -1$, and $\tan x = \tan y, -\tan y$, and $\tan x = \cot x, -\cot x$. So $\tan x = 1, -1 = \tan y$. You can look at cases.
| {
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Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| Another approach:
Rewrite
$$
\begin{align}
4\sin\theta +3\cos\theta &= 5\\
4\sin\theta&=5-3\cos\theta.\tag1
\end{align}
$$
Square $(1)$, yields
$$
16\sin^2\theta=25-30\cos\theta+9\cos^2\theta.\tag2
$$
Use identity $\sin^2\theta=1-\cos^2\theta$ and substitute to $(2)$.
$$
\begin{align}
16(1-\cos^2\theta)&=25-30\cos\theta+9\cos^2\theta\\
25\cos^2\theta-30\cos\theta+9&=0\\
(5\cos\theta-3)^2&=0\\
\cos\theta&=\frac35.
\end{align}
$$
Consequently, $\sin\theta=\dfrac45$. Thus,
$$
4\cos\theta-3\sin\theta=4\left(\frac35\right)-3\left(\frac45\right)=\Large\color{blue}0.
$$
| {
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Supremum of an expression involving $\sin$ Fix $n,m$ two different positive integers. If $n,m$ are both sufficiently big, does
$$
\sup_{x \in \mathbb{R}} |\sin(nx)-\sin(mx)| = 2 \,\ ?
$$
I've tried to come up with some sequence $(x_k)_{k \geq 1}$ that would show this, but I did not succeed.
| Using the difference of sines formula, we get
$$
\sin(nx)-\sin(mx)=2\sin\left(\frac{n-m}2x\right)\cos\left(\frac{n+m}2x\right)\tag{1}
$$
For $m^2\ne n^2$, we would like to arrange
$$
\frac{n-m}{2\pi}x\equiv\frac12\pmod{1}\tag{2}
$$
and
$$
\frac{n+m}{2\pi}x\equiv0\pmod{1}\tag{3}
$$
for then, $|\sin(nx)-\sin(mx)|=2$.
To satisfy $(2)$ and $(3)$, we need some $x\in\mathbb{R}$ and $j,k\in\mathbb{Z}$ so that
$$
(n-m)x=(2j+1)\pi\tag{4}
$$
and
$$
(n+m)x=2k\pi\tag{5}
$$
Thus,
$$
\frac{n+m}{n-m}=\frac{2k}{2j+1}\tag{6}
$$
Therefore, $(2)$ and $(3)$ have a solution if $n+m$ has more factors of $2$ than $n-m$.
Suppose $n+m$ has no more factors of $2$ than $n-m$. Then
$$
\frac{n+m}{n-m}=\frac{2j+1}{k}\tag{7}
$$
Furthermore, suppose
$$
\left|\,\frac{n-m}{2\pi}x-\frac{2p+1}2\,\right|\le\epsilon\tag{8}
$$
Then
$$
\begin{align}
\left|\,\frac{n+m}{2\pi}x-\frac{2j+1}{k}\frac{2p+1}2\,\right|
&=\left|\,\frac{n+m}{2\pi}x-\frac{n+m}{n-m}\frac{2p+1}2\,\right|\\
&\le\frac{n+m}{n-m}\epsilon\tag{9}
\end{align}
$$
and therefore, the triangle inequality and $(9)$ imply
$$
\begin{align}
\left|\,\frac{n+m}{2\pi}x-q\,\right|
&\ge\left|\,q-\frac{2j+1}{k}\frac{2p+1}2\,\right|-\left|\,\frac{n+m}{2\pi}x-\frac{2j+1}{k}\frac{2p+1}2\,\right|\\
&\ge\frac1{2k}-\frac{n+m}{n-m}\epsilon\tag{10}
\end{align}
$$
Therefore,
$$
\left|\,\frac{n-m}{2\pi}x-\frac{2p+1}2\,\right|\le\frac{n-m}{4kn}
\implies\left|\,\frac{n+m}{2\pi}x-q\,\right|\ge\frac{n-m}{4kn}\tag{11}
$$
Applying $(11)$ to $(1)$, we get
$$
|\sin(nx)-\sin(mx)|\le2\cos\left(\frac{n-m}{4kn}\pi\right)\tag{12}
$$
Summary
If $n+m$ has more factors of $2$ than $n-m$, there is an $x$ so that $|\sin(nx)-\sin(mx)|=2$. Otherwise, $|\sin(nx)-\sin(mx)|$ is bounded away from $2$.
Example 1:
If $n=6$ and $m=2$, then $8$ has one more factor of $2$ than $4$. Plugging into $(4)$, $(5)$, and $(6)$ yields $j=0$, $k=1$, and $x=\frac\pi4$:
$$
\left|\,\sin\left(6\cdot\frac\pi4\right)-\sin\left(2\cdot\frac\pi4\right)\,\right|=2
$$
Example 2:
If $n=2$ and $m=1$, then $3$ has no more factors of $2$ than $1$. Plugging into $(7)$ yields $j=1$ and $k=1$. Then $(12)$ says that
$$
\begin{align}
|\sin(2x)-\sin(x)|
&\le2\cos\left(\frac\pi8\right)\\
&\doteq1.84775906502257
\end{align}
$$
Mathematica says that the maximum of $|\sin(2x)-\sin(x)|$ is approximately $1.760172593046$.
| {
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Is a Mersenne-Prime always of the form $3n + 1$? Examples are:
$M_3 = 7 = 3\times 2 + 1$
$M_5 = 31 = 3\times 10 + 1$
$M_7 = 127 = 3\times 42 + 1$
$M_{13} = 8191 = 3\times2730 + 1$
| $2^{2n+1} - 1 \equiv 1 \pmod 3$ is true for all odd numbers $2n + 1$.
$2^{2n+1} \equiv 2 \pmod 3$
$2^{2n} \equiv 1 \pmod 3$
$4^n \equiv 1^n \equiv 1 \pmod 3$
Alternatively, you could show the cyclic pattern of the powers of two $\pmod 3$: $1, 2, 1, 2, 1, 2, \dots$, so it's clear that $2^k \equiv 1$ for $k$ even and $2$ for $k$ odd.
| {
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Spectral Norm of $2\times 2$ symmetric matrix Consider a $2\times 2$ symmetric matrix, in this case, is there some closed formula for its spectral norm ?
By spectral norm I mean the induced 2-norm, there is a definition here.
Thanks.
| It's the square root of the largest (in magnitude) eigenvalue of $A^2$. Since $A^2$ is $2 \times 2$, the eigenvalues are given by the roots of a quadratic, which you can work out. You might not like the "max", in which case you can avoid it by replacing it with
$$
max(x, y) = \frac{x+y+ |y-x|}{2}
$$
in your formula.
If
$$
A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}
$$
then $A^2$ is
$$
A^2 = \begin{bmatrix} a^2 + b^2 & b(a+c) \\ b(a+c) & c^2 + b^2 \end{bmatrix}
$$
and its characteristic polynomial is
$$
p(x) = x^2 - (a^2 + 2b^2 + c^2) x + (ac-b^2)^2.
$$
The roots are then
$$
x = \frac{- (a^2 + 2b^2 + c^2) \pm \sqrt{ (a^2 + 2b^2 + c^2)^2 - 4(ac-b^2)^2}}{2}
$$
The greater root (in absolute value) is the one with the $-$ sign. (No need for the max trick in this case, because the coefficient of $x$ in the quadratic is so obviously negative). So that leads to your answer: the 2-norm is
$$
\sqrt{\frac{ (a^2 + 2b^2 + c^2) + \sqrt{ (a^2 + 2b^2 + c^2)^2 - 4(ac-b^2)^2}}{2}}
$$
| {
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Line integral - should I parametrize the square? I have the following $1-\text{form}$ defined: $$\omega = \displaystyle\frac{2xy}{(1-x^2)^2+y^2}\mathrm{dx}+\displaystyle\frac{1-x^2}{(1-x^2)^2+y^2}\mathrm{dy}$$ I'd like to find $\displaystyle\oint_{\gamma}\omega$ where $\gamma$ is the square defined by the points $(2,-1),(2,1),(0,1),(0,-1)$ counter clockwise.
Since the square is a closed path, if $\omega$ is exact then $\displaystyle\oint_\gamma \omega =0$, right?
This is something I have proved using $$\int\omega=y\int\frac{2x}{(1-x^2)^2}+y^2 dx \\ \text{change variables: }u= (1-x^2)\implies du=-2x \\ =-y\int \frac{1}{(u/y)^2+1}du \\ \text{2nd change of variables: } z=u/y \implies dz=du/y \\ =-\int \frac{1}{z^2+1}dz = - \arctan(z)\\=-\arctan\left(\frac{1-x^2}{y}\right) + C(y)$$
And it seems to be enough if I set $C(y)=0$ since
$$\frac{\partial}{\partial y} \left(-\arctan\frac{1-x^2}{y}\right)=-\frac{1}{\frac{(1-x^2)^2}{y^2}+1}\frac{\partial}{\partial y}\left(\frac{1-x^2}{y}\right) \\ =-\frac{y^2}{(1-x^2)^2+y^2}\left(-\frac{1-x^2}{y^2}\right) = \frac{1-x^2}{(1-x^2)^2+y^2}$$
Is this enough to show that $\displaystyle\oint_\gamma \omega =0$ ?
| Have you thought of Green's theorem? It's quite simple to solve if you use Green's theorem.
| {
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$p$ divides the sum of the quadratic residues $\bmod p$ Could you help me at the following exercise?
Show that, if $p>3$ is a prime,then $p$ divides the sum of the quadratic residues $\bmod p$.
| Let $S$ be the sum of the quadratic residues. As $k$ runs from $1$ to $p-1$, the number $k^2$ runs (twice) over the quadratic residues modulo $p$. Thus
$$2S\equiv 1^2+2^2+\cdot +(p-1)^2.$$
By using the formula for the sum of the first $n$ consecutive squares, we find that
$$2S\equiv \frac{(p-1)(p)(2p-1)}{6}\pmod{p},$$
and therefore
$$12S\equiv (p-1)(p)(2p-1)\pmod{p}.$$
Since $p$ divides $(p-1)(p)(2p-1)$ but $p\gt 3$, it follows that $p$ divides $S$.
Another way: Let $g$ be a primitive root of $p$. Then the quadratic residues are congruent, modulo $p$, to $g^2,g^4,\dots, g^{p-1}$. Thus if $S$ is the sum of the quadratic residues, we have
$$S\equiv g^2+g^4+\cdots +g^{p-1}\pmod{p}.$$
Multiply both sides by $g^2-1$. We get
$$(g^2-1)S\equiv g^2(g^{p-1}-1)\pmod{p}.$$
By Fermat's Theorem, the right-hand side is congruent to $0$ modulo $p$. And if $p\gt 3$, then $g^2\not\equiv 1\pmod{p}$, and therefore $S\equiv 0\pmod{p}$.
| {
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Maximize $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$ Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$.
(Source : Singapore Math Olympiad 2012, Senior section, Round 1, question 29).
I tried using the fact that $2x +13, 8z + 12\ge 0$ to deduce that $y \le 11$, but I couldn't continue from there on. The answer should be an integer, since only integer answers were allowed.
The competition was designed for 15/16 year olds. A simple yet elegant solution would be nice.
For reference of the original problem,
| If $y$ is allowed to be negative, then there is no maximum.
Take $ x =k - \frac{13}{2}$, $ y = -k - \frac{5}{3}$, $ z= 3 + \frac{13}{2} + \frac{5}{3}$
The important part of the value looks like $ \sqrt{2k} + \sqrt[3]{-3k} + C$, which tends to infinity as $k\rightarrow \infty$.
Otherwise, if $y$ is positive, see Math110's solution.
| {
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If $(1 + 2i)$ and $(3 - 2i)$ are two roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$, then $a$ =? Consider the polynomial $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$ where $a, b, c, d$ are real
numbers. If $(1 + 2i)$ and $(3 - 2i)$ are two roots of this polynomial then what is the
value of a?
Well, I know only the 4 roots, which are obvious from what are given (the conjugates serve as the other two), but what next?
| Let the roots be $a_1$, $a_2$, $a_3$, $a_4$, $a_5$.
Now any polynomial, $(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)=0$ implies
$$(x^5)+(x^4)(a_1+a_2+a_3+a_4+a_5)+(x^3)(a_1a_2+a_2a_3+\cdots)+(x^2)(a_1a2a_3+a_2a_3a_4+\cdots)+(x^1)(a_1a_2a_3a_4+a_2a_3a_4a_5+\cdots)+(x^0)(a_1a_2a_3a_4a_5)=0$$
So, $a_1a_2a_3a_4a_5=4$, thus $(1+2i)(1-2i)(3-2i)(3+2i)a_5=4$ and so $a_5=\frac{4}{65}$.
Now $a_1+a_2+a_3+a_4+a_5=a$, meaning $(1+2i)+(1-2i)+(3+2i)+(3-2i)+\frac{4}{65}=a$, therefore $a=\frac{524}{65}$.
| {
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Why is $ \int \frac{\sin x (b-a\cos x)}{(b^2+a^2-2ab \cos x)^{3/2}}\,dx = \frac{a-b\cos x}{b^2 \sqrt{a^2-2ab\cos x + b^2}}$? Why is $$ \int \frac{\sin x (b-a\cos x)}{(b^2+a^2-2ab \cos x)^{3/2}}\,dx = \frac{a-b\cos x}{b^2 \sqrt{a^2-2ab\cos x + b^2}}\text{ ?}$$
Constant of integration omitted.
| The integral can be written as
$$
\int \frac{\sin x (b-a\cos x)}{(a^2+b^2-2ab \cos x)^{\Large\frac{3}{2}}}\,dx=\int \frac{a\cos x-b}{\sqrt{(a^2+b^2-2ab \cos x)^3}}\,d(\cos x).
$$
Let $y=\cos x$, then the integral becomes
$$
\int \frac{ay-b}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy=\int \frac{ay}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy-\int \frac{b}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy.
$$
The right integral in RHS can easily be solved using substitution $u=a^2+b^2-2ab y$ and $du=-2ab\ dy$.
For the left integral in RHS, let $y=\dfrac{(a^2+b^2)}{2ab}\sin^2\theta$ and $dy=\dfrac{(a^2+b^2)}{ab}\sin\theta\cos\theta\ d\theta$, then
\begin{align}
\int \frac{ay}{\sqrt{(a^2+b^2-2ab y)^3}}\,dy&=\frac{\sqrt{a^2+b^2}}{2ab^2}\int \frac{\sin^3\theta}{\cos^2\theta}\,d\theta\\
&=\frac{\sqrt{a^2+b^2}}{2ab^2}\int \frac{\sin\theta(1-\cos^2\theta)}{\cos^2\theta}\,d\theta\\
&=\frac{\sqrt{a^2+b^2}}{2ab^2}\left[\int \frac{\sin\theta}{\cos^2\theta}\,d\theta-\int\sin\theta\ d\theta\right]\\
&=-\frac{\sqrt{a^2+b^2}}{2ab^2}\left[\int \frac{1}{\cos^2\theta}d(\cos\theta)+\int\sin\theta\ d\theta\right].
\end{align}
From here, the last integral should be easy to evaluate. I leave it to you Kappa for the rest.
| {
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Elementary problems Find $n$ such that $n$ is a positive integer satisfying the following equation.
$$2(2^2)+3(2^3)+4(2^4)+\ldots+n(2^n)=2^{n+10}$$
Can anybody help ? I can't believe this is an elementary school problem...
| Let $\displaystyle S=2\cdot2^2+3\cdot2^3+\cdots+(n-1)2^{n-1}+n\cdot2^n$
$$\implies2S=2\cdot2^3+3\cdot2^4+\cdots+(n-1)2^{n}+n\cdot2^{n+1}$$
$$S-2S=2\cdot2^2+(3-2)2^3+(4-3)2^4+\{n-(n-1)\}2^n-n\cdot2^{n+1}$$
$$\implies -S=2\cdot2^2-n\cdot2^{n+1}+\underbrace{(2^3+2^4+\cdots+2^n)}_{\text{Geomteric Series}}$$
$$\implies-S=2\cdot2^2-n\cdot2^{n+1}+2^3\cdot\frac{(2^{n-2}-1)}{(2-1)}$$
$$\implies S=(n-1)2^{n+1}$$ which needs to be $\displaystyle2^{n+10}\implies n-1=2^9$
Reference : Arithmetico-geometric sequence
| {
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Prove that $\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}$ The question is:
Prove $$\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}.$$
I had no idea how to do it & proceeded by making RHS = $\tan60^\circ$ but this doesn't make any help. Please help me solving this.
| Consider $\tan x+\tan(x+60^\circ)+\tan(x+120^\circ)$, then
\begin{align}
\tan x+\tan(x+60^\circ)+\tan(x+120^\circ)&=\tan x+\frac{\tan x+\tan60^\circ}{1-\tan x\tan60^\circ}+\frac{\tan x+\tan120^\circ}{1-\tan x\tan120^\circ}\\
&=\tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}+\frac{\tan x-\sqrt{3}}{1+\sqrt{3}\tan x}\\
&=\tan x+\frac{\tan x+\sqrt{3}\tan^2 x+\sqrt{3}+3\tan x}{1-3\tan^2 x}\\
&+\frac{\tan x-\sqrt{3}\tan^2 x-\sqrt{3}+3\tan x}{1-3\tan^2 x}\\
&=\frac{\tan x(1-3\tan^2 x)}{1-3\tan^2 x}+\frac{8\tan x}{1-3\tan^2 x}\\
&=\frac{9\tan x-3\tan^33 x}{1-3\tan^2 x}\\
&=3\left(\frac{3\tan x-\tan^33 x}{1-3\tan^2 x}\right)\\
&=3\tan3x
\end{align}
See triple-angle formula of trigonometric identities. Plugging in $x=10^\circ$, we get
\begin{align}
\tan 10^\circ+\tan(10^\circ+60^\circ)+\tan(10^\circ+120^\circ)&=3\tan(3\cdot10^\circ)\\
\tan 10^\circ+\tan70^\circ+\tan130^\circ&=3\tan(30^\circ)\\
\tan 10^\circ+\tan70^\circ+\tan(180^\circ-50^\circ)&=3\cdot\frac{1}{\sqrt{3}}\\
\tan 10^\circ+\tan70^\circ-\tan50^\circ&=\sqrt{3}&\qquad\blacksquare
\end{align}
It took me ages to get the prove this identity. ᕙ(^▽^)ᕗ
| {
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How prove this inequality $\sum_{k=1}^{n}\frac{2k-1}{k\binom{n}{k}}\ge \frac{n}{2^{n-1}}$
let $1\le k\le n,k,n\in N^{+}$, show that
$$\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\ge \dfrac{n}{2^{n-1}}$$
I know this $$\sum_{k=1}^{n}(2k-1)=n^2$$
and $$\sum_{k=1}^{n}k\binom{n}{k}=n\cdot 2^{n-1}$$
I want Use Cauchy-Schwarz inequality .
$$\left(\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\right)\left(\sum_{k=1}^{n}k\binom{n}{k}\right)\ge (\sum_{k=1}^{n}\sqrt{2k-1})^2$$
then
$$\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\ge\dfrac{(\sum_{k=1}^{n}\sqrt{2k-1})^2}{n\cdot 2^{n-1}}$$
Now we must prove
$$\sum_{k=1}^{n}\sqrt{2k-1}\ge n?$$
maybe can use integral inequality to prove it.
I can't prove this.Thank you
| First prove that: for $x≥0$ and $y≥0$, $(x+y)^{\frac{1}{2}}≤x^{\frac{1}{2}}+y^{\frac{1}{2}}$. Define $f(x)=(x+1)^{\frac{1}{2}}-x^{\frac{1}{2}}-1$. Note that $f'(x)<0$. Then for $x >0$, we have $f(x)<0$. Therefore $(x+1)^{\frac{1}{2}}<x^{\frac{1}{2}}+1$. Replace $x$ by $x/y$ we have, $(x+y)^{\frac{1}{2}}<x^{\frac{1}{2}}+y^{\frac{1}{2}}$. As the result holds for $x=y=0$, we have $(x+y)^{\frac{1}{2}}≤x^{\frac{1}{2}}+y^{\frac{1}{2}}$, for $x≥0$ and $y≥0$. For induction,
$$(x_{1}+...+x_{n})^{\frac{1}{2}}≤x^{\frac{1}{2}}_{1}+...+x^{\frac{1}{2}}_{n}.$$
So, for $x_{k}={2k-1}$ we have, $\sum_{k=1}^{n}\sqrt{2k-1}≥\sqrt{\sum_{k=1}^{n}{2k-1}}=\sqrt{n^{2}}=n.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to solve $x=tx'^2+x'^3$ How about the differential equation $x=tx'^2+x'^3$. Can somebody solve it?
my solution:suppose $\dot x$=p,then $x=tp^2+p^3$.
Then, p=$p^2$+2tp$\dot p$+3$p^2\dot p$, then we can use these function to solve p, but I just find the final result is so complex. I don't know how to continue? Or, does somebody have other kinds of methods? Thank you
| A parametric solution isn't that horrible.
If $x(t)$ is not constant, then $p(t) = \dot{x}(t)$ isn't identically zero and one can rewrite your equation as
$$\frac{dp}{dt} = \frac{1-p}{2t + 3p}\tag{*1} = -\frac{\tilde{p}}{2\tilde{t}+3\tilde{p}}
\quad\text{ where }\quad
\begin{cases}\tilde{p} &= p - 1\\\tilde{t} &= t+\frac32\end{cases}
$$
Introduce an auxiliary parameter $s$ such that
$$\frac{dt}{ds} = \frac{d\tilde{t}}{ds} = 2\tilde{t}(s) + 3\tilde{p}(s)$$
Equation $(*1)$ can be recasted into a matrix form:
$$\frac{d}{ds} \begin{bmatrix}\tilde{p}(s)\\\tilde{t}(s)\end{bmatrix} =
\begin{bmatrix}-1 & 0\\3 & 2\end{bmatrix} \begin{bmatrix}\tilde{p}(s)\\\tilde{t}(s)\end{bmatrix}
$$
The matrix
$\begin{bmatrix}-1 & 0\\3 & 2\end{bmatrix}$ has eigenvalues $-1$ and $2$ with corresponding eigenvectors $\begin{bmatrix}1\\-1\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$.
From this, we find
$$
\begin{bmatrix}p(s)\\t(s)\end{bmatrix} =
\begin{bmatrix}1\\{\small -\frac32}\end{bmatrix}
+ Ae^{-s} \begin{bmatrix}1\\-1\end{bmatrix}
+ \frac{B}{2}e^{2s} \begin{bmatrix}0\\1\end{bmatrix}
\tag{*2}
$$
for suitable chosen constant $A$ and $B$. If $p(t)$ is not the constant function $1$, then $A \ne 0$ and we can absorb it into $e^{-s}$ and set it to $\text{sign}(A) = \pm 1$. After suitable rescaling of $B$ and let $u = \mp e^{-s}$, we can transform $(*2)$ as
$$
\begin{cases}
p(u) &= 1 - u\\
t(u) &= -\frac32+ u + \frac{B}{2u^2}
\end{cases}
$$
As a function of $u$, we now have
$$
\frac{d x(u)}{du} = \frac{d x(t)}{dt} \frac{dt(u)}{du} = (1 - u)\left(1 - \frac{B}{u^3}\right)
= 1 - u + \frac{B}{u^2} - \frac{B}{u^3}
$$
and hence
$$
\begin{cases}
x(u) &= u - \frac{u^2}{2} - \frac{B}{u} + \frac{B}{2u^2} + C\\
t(u) &= -\frac32 + u + \frac{B}{2u^2}
\end{cases}\tag{*3}
$$
where $C$ is yet another integration constant.
Substitute $(*3)$ into the expression
$x(t) - tx'(t)^2 - x'(t)^3$, we find it equal to the constant $C - \frac{B-1}{2}$.
This fixes $C$ to $\frac{B-1}{2}$ and the original ODE has parametric solutions of the form:
$$
\begin{cases}
x(u) &= u - \frac{u^2}{2} - \frac{B}{u} + \frac{B}{2u^2} + \frac{B-1}{2}
= \frac{(B-u^2)(1-u)^2}{2u^2}\\
t(u) &= -\frac32 + u + \frac{B}{2u^2}
\end{cases}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}$ $$
I_1:=\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}, \qquad \alpha \geq -1.
$$
I am trying to prove this integral $I_1$. We can write
$$
\int_0^{\pi/2} \ln(\alpha(1/\alpha+\sin^2 x))dx=\int_0^{\pi/2} \left(\ln \alpha+\ln (\frac{1}{\alpha}+\sin^2 x)\right)dx=\frac{\pi}{2} \ln \alpha+I_2
$$
where
$$
I_2=\int_0^{\pi/2}\ln (\frac{1}{\alpha}+\sin^2 x) \,dx
$$
however I am not sure what that will do for us.... I also tried differentiating wrt $\alpha$ but didn't get placed. How can we prove $I_1$ result? Thanks
| Let $ \displaystyle I(a) = \int_{0}^{\pi /2} \ln(1+ a \sin^{2}x) \, dx$.
Then differentiating under the integral sign, $$I'(a) = \int_{0}^{\pi /2} \frac{\sin^{2} x}{1+a \sin^{2} x} \, dx = \int_{0}^{\pi /2} \frac{1}{a+ \csc^{2} x} \, dx .$$
Now let $u = \cot x$.
Then
$$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{1}{a+1+u^{2}} \frac{1}{1+u^{2}} \, du \\ &= \frac{1}{a} \int_{0}^{\infty} \left(\frac{1}{1+u^{2}} - \frac{1}{1+a+u^{2}} \right) \, du \\ &= \frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{1+a} \int_{0}^{\infty} \frac{1}{1+\frac{u^{2}}{1+a}} \, du \right) \\ &=\frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{\sqrt{1+a}} \int_{0}^{\infty} \frac{1}{1+v^{2}} \, dv \right) \\ &= \frac{\pi}{2a} \left(1 - \frac{1}{\sqrt{1+a}} \right). \end{align}$$
Then integrating back,
$$ \begin{align} I(a) &= \frac{\pi}{2} \int \frac{1}{a} \left(1 - \frac{1}{\sqrt{1+a}} \right) \, da \\ &= \frac{\pi}{2} \int \frac{1}{u^{2}-1} \left(1 - \frac{1}{u} \right) 2u \, du \\ &= \pi \int \frac{1}{1+u} \, du \\ &= \pi \ln \left(1+ \sqrt{1+a} \right) + C. \end{align}$$
And since $I(0) = 0$, $C = -\pi \ln 2$.
Therefore,
$$I(a) = \pi \ln \left(\frac{1 +\sqrt{1+a}}{2} \right) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 3
} |
TaylorSeries of complete elliptic integral of the first kind I want compute $K(k)$ as a Taylor Series; $k\in\mathbb{R}$ and $\vert k \vert < 1$.
Can someone help me?
$$
K(k):= \int^{\frac{\pi}{2}}_0 \dfrac{dt}{\sqrt{1-k^2 sin^2t}}
$$
Results so far:
$$
K(k):= \int^{\frac{\pi}{2}}_0 \dfrac{dt}{\sqrt{1-k^2 sin^2t}} = \int^{\frac{\pi}{2}}_0 (1-k^2 sin^2t)^{-\frac{1}{2}}dt
$$
With using binomial Series we get
$$
\int^{\frac{\pi}{2}}_{0} \sum^\infty_{\Phi=0} {-\frac{1}{2} \choose \Phi}(-k^{2\Phi}){\sin^{2\Phi}{t}} \ dt = \sum^\infty_{\Phi=0} {-\frac{1}{2} \choose \Phi}(-k^{2\Phi})\int^{\frac{\pi}{2}}_{0}\sin^{2\Phi}t \ dt
$$
For $\Phi$ even:
$$
\int^{\frac{\pi}{2}}_{0}\sin^{2\Phi}t \ dt = \frac{\pi}{2}\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{n-1}{n} = S
$$
thus we get:
$$
1. \sum^\infty_{\Phi=0} {-\frac{1}{2} \choose \Phi}(-k^{2\Phi})\cdot S
$$
now i need some help to compute 1. as taylor series, can someone help?
Thanks! Landau.
| The generating function for central binomial coefficient is given by mathworld site.
\begin{align*}
\int_{0}^{\frac \pi 2} \frac{1}{\sqrt{1 - 4 \left( \frac {k^2} 4 \sin^2 (t) \right )}} dt &= \int_0^{\frac \pi 2 } \sum_{n=0}^{\infty} \binom{2n}{n} \left(\frac {k^2} 4 \sin^2 (t) \right )^n dt \\
&= \sum_{n=0}^{\infty} \binom{2n}{n} \frac{k^{2n}}{4^n} \cdot \frac 1 2 \cdot \beta (n+1/2, 1/2) \\
&= \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2 2^{2n} }\cdot \frac 1 2 \left( \frac{\Gamma(n+1/2)\Gamma(1/2)}{\Gamma(n+1)} \right ) k^{2n}\\
&= \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2 2^{2n} }\cdot \frac 1 2 \left( \frac{(2n)! \sqrt{\pi} \sqrt{\pi}}{(n!)^2} \right ) k^{2n} \\
&= \frac{ \pi }{2 } \sum_{n=1}^\infty \left( \frac{(2n)!}{(n!)^{2} 2^{2n}} \right )^2 k^{2n}\\
&= \frac{\pi}{2} \sum_{n=0}^\infty P_{2n}(0)k^{2n}
\end{align*}
Where $P_{n}(0)$ is Legendre polynomial. Seems that wolf gives the sum of the right side as EllipticK[k^2]. Also it is given here on wiki.
| {
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"url": "https://math.stackexchange.com/questions/777592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Working out the value of $a^4+b^4$ If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$
My approach:
$$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$
$$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$
$$=(5)^4-6(24)-4ab(a^2-b^2)$$
$$=(5)^4-6(24)-8(a+b)(a-b)$$
$$=(5)^4-6(24)-8(5)(a-b)$$
I am a little stuck now and any help will be appreciated.
| \begin{align}
a^4+b^4 &= (a+b)^4-4a^3b-6a^2b^2-4ab^3
\\&= (a+b)^{4} - 4 ab (a^{2} + b^{2}) - 6 (ab)^{2}
\\&=
(a+b)^{4} - 4 ab ((a+b)^{2} - 2 ab) - 6 (ab)^{2}
\\&= (a+b)^{4} - 4 ab (a+b)^{2} + 2 (ab)^ {2}.
\end{align}
As noted in a comment, you made a sign error (it happens) in your calculations, otherwise you would have got here yourself.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Logarithm base transformation I am trying to solve a problem which, I think, revolves around base transformation of logarithms. It goes like this:
$\log_5\,{\log_6\,{\frac{6x-1}{x+1}}} < \log_\frac{1}{5}\,{\log_\frac{1}{6}\,{\frac{x+1}{6x-1}}}$
I tried transforming "first" logarithms to base 5, yielding
$\log_6\,{\frac{6x-1}{x+1}}< \frac{1}{\log_\frac{1}{6}\,{\frac{x+1}{6x-1}}}$ (if I am right, of course..) Further transformation to base 6 leaves me helpless with :
$\log_6\,{\frac{6x-1}{x+1}}<- \frac{1}{\log_6\,{\frac{x+1}{6x-1}}}$
Thanks.
| You need to use the following two properties of the logarithm: $\log_b{y} = -\log_{1/b}{y}$ and $\log_b{y} = -\log_b{1/y}$.
Using these properties the right hand side becomes:
$\log_{1/5} \log_{1/6} \frac{x+1}{6x-1} = \log_{1/5} \log_{6} \frac{6x-1}{x+1} = -\log_5 \log_6 \frac{6x-1}{x+1}$.
So your inequality now becomes $\log_5 \log_6 \frac{6x-1}{x+1} < 0$ which is quite trivial to solve.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the solution of the boundary value problem correct? I have to solve the following boundary value problem:
$$ u_{xx}+u_{yy}=0, 0<x<a, 0<y<b \tag1$$
$$u(0,y)=u(a,y)=0, 0<y<b $$
$$u(x,0)-u_y(x,0)=0, u(x,b)=f(x), 0<x<a$$
With the method of separation of variables I have found the following:
The solution is of the form $ u(x,y)=X(x)Y(y)$
$$(1) \Rightarrow X''Y+XY''=0 \Rightarrow \frac{X''}{X}+\frac{Y''}{Y}=0 \Rightarrow \frac{X''}{X}=- \frac{Y''}{Y}=- \lambda$$
So we get the following problems:
$$\left.\begin{matrix}
X''+ \lambda X=0, 0<x<a\\
X(0)=X(a)=0
\end{matrix}\right\}\tag{*}$$
$$\left.\begin{matrix}
Y''- \lambda Y=0, 0<y<b\\
Y(0)-Y'(0)=0
\end{matrix}\right\}\tag{**}$$
$$u(x,b)=X(x)Y(b)=f(x)$$
$(*) \Rightarrow $ The eigenvalues are $\lambda_n =\left(\frac{n \pi}{a}\right)^2$ and the eigenfunction are $ X_n(x)= \sin\left(\frac{n \pi x}{a}\right)$.
$$(**) \Rightarrow Y(y)=A_n \sinh\left(\frac{n \pi y}{a}\right)+B_n \cosh\left(\frac{n \pi y}{a}\right)$$
$$Y'(y)=\frac{n \pi}{a} A_n \cosh\left(\frac{n \pi y}{a}\right)+\frac{n \pi}{a}B_n \sinh\left(\frac{n \pi y}{a}\right)$$
$$Y(0)-Y'(0)=0 \Rightarrow B_n=\frac{n \pi }{a} A_n$$
So $Y_n(y)=A_n \sinh\left(\frac{n \pi y}{a}\right)+\frac{n \pi }{a} A_n \cosh\left(\frac{n \pi y}{a}\right)$
$$u(x,y)=\sum_{n=1}^{\infty}{X_n(x)Y_n(y)}$$
$$ u(x,y)=\sum_{n=1}^{\infty}{A_n\left[ \sinh\left(\frac{n \pi y}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi y }{a}\right)\right] \sin\left(\frac{n \pi x}{a}\right)}$$
$$u(x,b)=f(x) \Rightarrow $$
$$\sum_{n=1}^{\infty}{A_n\left[ \sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)\right] \sin\left(\frac{n \pi x}{a}\right)}=f(x) \Rightarrow $$
$$A_n\left[ \sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)\right] =\frac{2}{a} \int_0^a{f(x) \sin\left(\frac{n \pi x}{a}\right)}dx $$
$$ \Rightarrow A_n=\frac{2 \int_0^a{f(x) \sin\left(\frac{n \pi x}{a}\right)}dx}{a \sinh\left(\frac{n \pi b}{a}\right)+ n \pi \cosh\left(\frac{n \pi b}{a}\right)}$$
Is my solution correct??
| Yes, this is the standard process and you carried it out correctly. I like to write the complete answer as
$$u(x,y)=\sum_{n=1}^{\infty}{b_n\frac{\sinh\left(\frac{n \pi y}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi y }{a}\right)}{\sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)} \sin\left(\frac{n \pi x}{a}\right)} \tag{1}$$
where $b_n$ are the Fourier (sine) coefficients of $f$. Then it's easy to check that all required properties hold: $x=0$ gives $0$, $x=a$ also gives $0$, at $y=0$ we get
$$u(x,y)=\sum_{n=1}^{\infty}{b_n\frac{ \frac{n \pi }{a}}{\sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)} \sin\left(\frac{n \pi x}{a}\right)}$$ for both the function and for its $y$-derivative. And when $y=a$, the sum (1) turns into the sine series of $f$, which is what the coefficients $b_n$ are for.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Integral question I would like to ask for help with $$\int \cos^3 (x/9) \,dx$$
I got $$9\sin(x/9) - 3\sin^3(x/9) + c$$ but the book gives another answer which I'm not sure how they got to.
| First use the substitution
$=\int \cos ^3\left(u\right)9du$
$=\int \:9\cos ^3\left(u\right)du$
Now take out the Constant
$=9\int \cos ^3\left(u\right)du$
Now use the reduction rule:
$\int \cos ^n\left(x\right)dx=\frac{\sin \left(x\right)\cos ^{n-1}\left(x\right)}{n}+\frac{n-1}{n}\int \cos ^{n-2}\left(x\right)dx$
$\int \cos ^3\left(u\right)du=\frac{\sin \left(u\right)\cos ^2\left(u\right)}{3}+\frac{2\int \cos \left(u\right)du}{3}$
$=9\left(\frac{\sin \left(u\right)\cos ^2\left(u\right)}{3}+\frac{2\int \cos \left(u\right)du}{3}\right)$
Substitute Back u
$=9\left(\frac{\sin \left(\frac{x}{9}\right)\cos ^2\left(\frac{x}{9}\right)}{3}+\frac{2\sin \left(\frac{x}{9}\right)}{3}\right)$
$=\frac{3\sin \left(\frac{x}{9}\right)\left(\cos \left(\frac{2x}{9}\right)+5\right)}{2}$
Now add the Constant
$=\frac{3\sin \left(\frac{x}{9}\right)\left(\cos \left(\frac{2x}{9}\right)+5\right)}{2}+C$
Hope this Helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
| Here is a high school level solution using L'Hôpital's rule,
$ \lim{\sqrt{n^2+n} -n} = \infty - \infty $
We can write the limit relation in following manner to be able to use L'Hôpital's rule.
$ \lim{~n(\sqrt{1+\frac{1}{n}} -1)} = \infty \times 0 $
rewriting the expression as:
$\lim{\frac{(\sqrt{1+\frac{1}{n}} -1)}{\frac{1}{n}}} $
now can take the derivatives of numerator and denominator as following and solve the problem:
$ \lim{\frac{\frac{-1}{2n^\frac{3}{2}\sqrt{n+1}}}{\frac{-1}{n^2}}} = \lim~\frac{1}{2\sqrt{1+\frac{1}{n}}}$
where $\lim~\frac{1}{2\sqrt{1+\frac{1}{n}}} = \frac{1}{2}$
| {
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"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 6
} |
Point on $z = \frac{1}{xy}$ closest to origin Where $x>0$ and $y>0$.
I want to work with the square of the distance formula from the origin, so I went with $f(x,y) = x^2 + y^2 + \frac{1}{(xy)^2}$.
Then I found the first partial derivatives:
$f_x = 2x - \frac{2}{x^3y^2}$ and $f_y = 2y - \frac{2}{y^3x^2}$.
Set them equal to $0$ to find the critical points, and multiply both sides by respective $x$ and $y$ to get a common term to set equal to each other:
$2xy = \frac{2}{x^3y}$ and $2xy = \frac{2}{y^3x}$ so $\frac{2}{x^3y} = \frac{2}{y^3x}$. Simplfying, we get $y^2 = x^2$, or $y = x$ since we are given that $x>0$ and $y>0$.
I'm not sure what to do from here.
| Alternative solution: The Inequality of arithmetic and geometric means
(with $n=3$) states that for non-negative real numbers $a, b, c$
$$
\frac{a + b + c}3 \ge (abc)^{1/3}
$$
with equality if and only if $a=b=c$. It follows that if $(x, y, z)$ is a point
on the surface $z = 1/(xy)$ then
$$
\frac{x^2 + y^2 + z^2}3 \ge (x^2y^2z^2)^{1/3} = 1
$$
and therefore $x^2 + y^2 + z^2 >= 3$ with equality if and only if $x=y=z$ ($= 1$).
| {
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Value of $ \cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $? I am a little weak in trigonometry. I have two questions:
*
*Find the value of $\cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $
*Find the value of $\sin 28^{\circ}+ \cos 17^{\circ} + \cos 28^{\circ} + \sin 17^{\circ} $
I am asking these questions because:
1. I am weak and unable to solve these.
2. I want to know the difference in these questions.
| Observe that $$\cos3(\underbrace{60^\circ-x})=\cos(180^\circ-3x)=-\cos3x$$
and $$\cos3(\underbrace{60^\circ+x})=\cos(180^\circ+3x)=-\cos3x$$
and $$\cos3(\underbrace{180^\circ-x})=\cos(360^\circ+180^\circ-3x)=\cos(180^\circ-3x)=-\cos3x$$
As $\displaystyle\cos3x=4\cos^3x-3\cos x,$
and as $\displaystyle-\cos3x=\cos(180^\circ-3x)=\cos(180^\circ+3x)=\cos3(180^\circ-x)$
the roots of $\displaystyle4\cos^3y-3\cos y=-\cos3x\iff4\cos^3y-3\cos y+\cos3x=0$
will be $\displaystyle\cos(60^\circ-x),\cos(60^\circ+x),\cos(180^\circ-x)$ (no two are in general same, right?)
Using Vieta's formula $\displaystyle\cos(60^\circ-x)+\cos(60^\circ+x)+\cos(180^\circ-x)=\frac04$
Here $\displaystyle x=8^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\cos(\frac{B-C}2)\ge \sqrt{\frac{2r}{R}}$? For any triangle $ABC$, prove that:
$$\cos(\frac{B-C}2)\ge \sqrt{\frac{2r}{R}}$$
I have tried many approaches but none seems to work. I noted that $\cos(\frac{B-C}2)=\frac{AM}{2R}$, where $M$ is point of intersection of the circumcircle and bisector of $\angle A$, but it wasn't much helpful. Although, Euler's inequality seems to be helpful, it isn't. And expanding out $\cos(\frac{B-C}2)$ gets very complicated.
Does anyone have some elegant methods?
| It is known that $$\frac{r}{4R}= \sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}.$$
and consequently$$\frac{2r}{R}= 8\sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}.$$
$$\cos \frac{B-C}{2}\geq\sqrt{\frac{2r}{R}}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq\frac{2r}{R}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq 8\sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq 4\sin \frac{A}{2}(\cos \frac{B-C}{2}-\cos \frac{B+C}{2})\Leftrightarrow$$
$$\cos^2 \frac{B-C}{2}\geq 4\sin \frac{A}{2}(\cos \frac{B-C}{2}-\sin \frac{A}{2})\Leftrightarrow(\cos \frac{B-C}{2}-\sin \frac{A}{2})^2\geq 0.$$
| {
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How to find the Square Root of a Polynomial $4x^4 + 4x^3 - 11x^2 -6x + 9$
How do you find the square root of this polynomial? I really don't understand. Please provide an easy-to-understand explanation. Thanks.
| If this is the square of some polynomial $ax^2+bx+c$, we find
$$\begin{align}4x^4 + 4x^3 - 11x^2 -6x + 9&= (ax^2+bx+c)^2\\&=a^2x^4+2abx^3+(b^2+2ac)x^2+2bcx+c^2\end{align}$$
By equating coefficients, we find $a^2=4$, hence (as we may play with the sign) wlog. $a=2$. Next, $4=2ab$ gives us $b=1$. Then $-6=2bc$ gives us $c=-3$. We check that all wiorks out, i.e. that $-11=b^2+2ac$ and $c^2=9$.
Alternatively, you can make use of the fact that the derivative of $g(x)^2$ is $2g(x)g'(x)$. Hence if $f(x)=g(x)^2$, then $g$ is a common divisor of $f$ and $f'$. Use the euclidean algorithm to find $\gcd(f,f')$:
$$\begin{align}\gcd(f,f')&= \gcd(4x^4 + 4x^3 - 11x^2 -6x + 9,16x^3+12x^2-22x-6)\\&=\gcd(4x^4 + 4x^3 - 11x^2 -6x + 9,8x^3+6x^2-11x-3)\\
&=\gcd(8x^3+6x^2-11x-3,2x^2+x-3)\\
&=\gcd(2x^2+x-3,0)\\
&=2x^2+x-3\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $\arctan(\frac{\tan \theta}{2}) \approx \frac{1}{2 - \theta} - \frac{1}{2 + \theta}$ for small $\theta$? In answering this question, I needed to show
that
$\arctan \left( \frac{\tan \theta}{2} \right) \approx \frac{\theta}{2}$
when $\theta$ was small.
So, naturally,
I computed the first few terms of the Taylor series for $\arctan \left( \frac{\tan \theta}{2} \right)$.
Sure enough,
the first three terms are $0 + \frac12 \theta + 0 \theta^2$.
But what I didn't expect was the striking pattern in the next few terms:
$$
\arctan \left( \frac{\tan \theta}{2} \right)
= \frac12 \theta
+ \frac{1}{8} \theta^3
+ \frac{1}{32} \theta^5
+ \frac{11}{1920} \theta^7 + \cdots
$$
It clearly seems to be the case that
\begin{align*}
\arctan \left( \frac{\tan \theta}{2} \right)
&\approx \sum_{n \ge 0} \left(\frac{\theta}{2}\right)^{2n+1} \\
&= \frac{\theta / 2}{1 - \theta^2 / 4} \\
&= \frac{1}{2 - \theta} - \frac{1}{2 + \theta}. \\
\end{align*}
Here is a plot:
Honestly, I have no idea how to explain this, and in particular I'm having trouble assigning any meaning to $\frac{1}{2 - x}$ at all when $x$ is
an angle.
So, my question: is there any intuitive geometric reason why this approximation works? If not, is there any other concise reason?
Obviously one can arrive at this result via algebraic manipulation (like I did above) but I am interested in a more intuitive explanation.
| The approximation you noticed is the Padé approximant of order $(m,n) = (1,2)$ of $f(\theta) = \tan^{-1} \bigl( \frac{1}{2} \tan \theta \bigr)$.
| {
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} |
Evaluate the given definite integral How to evaluate the integral $$\int_0^1 (1-x^{2/3})^{3/2} \, dx.$$ Does some substitution work?
| You can also use $x=\sin^3\theta$ and $dx=3\sin^2\theta\cos\theta\ d\theta$, then
$$
\begin{align}
\int_0^1\left(1-x^{\large\frac23}\right)^{\large\frac32}\ dx&=\int_0^{\large\frac\pi2}\left(1-\sin^2\theta\right)^{\large\frac32}\cdot 3\sin^2\theta\ \cos\theta\ d\theta\\
&=3\int_0^{\large\frac\pi2}\cos^4\theta(1-\cos^2\theta)\ d\theta\\
&=3\int_0^{\large\frac\pi2}(\cos^4\theta-\cos^6\theta)\ d\theta.
\end{align}
$$
The last integral can be solved by using integration by reduction formula, we will obtain
$$
\begin{align}
\int_0^1\left(1-x^{\large\frac23}\right)^{\large\frac32}\ dx&=3\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta-3\int_0^{\large\frac\pi2}\cos^6\theta\ d\theta\\
&=3\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta-3\left(\left.\frac16\cos^5\theta\sin\theta\right|_0^{\large\frac\pi2}+\frac56\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta\right)\\
&=\frac12\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta.
\end{align}
$$
The rest I leave it to you.
| {
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} |
How to find out the equation whose roots are trigonometric? Find out the equation whose roots are $\sin^2 (2\pi/7), \sin^2 (4\pi/7), \sin^2 (8\pi/7)$. I wanted to solve this by firstly finding the equation ,the roots of which are $\sin 2\pi/7,\sin 4\pi/7, \sin 8\pi/7$ .Then i would find the equation wanted in the equation by using $Y = x^2$ or, $X = Y^1/2$; $X$ is the root of the 2nd equation i.e. $\sin 2\pi/7\ldots$ & $Y$ is the root of 1st equation. Now i would put the value of $X$ in the 2nd equation i.e. $Y^1/2$ & wld get the equation required. But i don't know how to find an equation whose roots are trigonometric. Plz help.
| Like
factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
or
Prove that $\frac{1}{4-\sec^{2}(2\pi/7)} + \frac{1}{4-\sec^{2}(4\pi/7)} + \frac{1}{4-\sec^{2}(6\pi/7)} = 1$
$\displaystyle z^3+z^2-3z-1=0\ \ \ \ (1),$ has the roots $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$
Now using $\displaystyle\cos2A=2-\sin^2A\iff\sin^2A=\frac{1-\cos2A}2,$
$\displaystyle\sin^2\frac{8\pi}7=\dfrac{1-\cos\frac{16\pi}7}2$
and again$\displaystyle\cos\frac{16\pi}7=\cos\left(2\pi+\frac{2\pi}7\right)=\cos\frac{2\pi}7$ etc.
Can you take it home form here?
See also: Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$
| {
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Enumerating sums of integers So I found a question which seems to be really easy to answer but after thinking a lot about it, well, I've have to say, I came up with nothing.
So here's the question:
Assume the sum of $a+b+c+d+e+f$ that each of the numbers can be $0,3,4,5$, how many different summations can exist?
The only point I came up with is that the integer $0$ doesn't change the summation therefore different permutations of a $0$ count as one as I said the reason above.
And sorry if the Tags are irrelevant and my Math language is crappy I'm a newbie!
Please feel free to edit my tags I really do not know which one to add.
Any help would be highly appreciated in advance.
| Denote the set of all possible sums by $S$. Clearly $S\subset\{0,1,2,3,\cdots,30\}$.
Then take care of each element in $\{0,1,\cdots,30\}$.
$30=5+5+5+5+5+5$
$29=5+5+5+5+5+4$
$28=5+5+5+5+5+3$
$27=5+5+5+5+4+3$
$26=5+5+5+5+3+3$
Replacing one 5 by 0, we can obtain $25,24,23,22,21$. Repeat this process for 4 times until we get
$10=0+0+0+0+5+5$
$9=0+0+0+0+5+4$
$8=0+0+0+0+5+3$
$7=0+0+0+0+4+3$
$6=0+0+0+0+3+3$.
Then we see $S=\{0,3,4,5\}\cup\{6,7,\cdots,30\}$.
| {
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"source": "stackexchange",
"question_score": "3",
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Curves triangular numbers. Sometimes you have to deal with this equation: $X^2+aX+Y^2+bY=Z^2+cZ$
$a,b,c$ - integer coefficients.
I wrote below - to start a particular solution of Diophantine equations.
To do this, use the solutions of Pell's equation: $p^2-2k(k-1)s^2=1$
I turned solutions such.
$X=(k-1)((a+b)k+a-c)s^2-(ck-b)ps$
$Y=k((a+b)k+c-a-2b)s^2-(ck+a-c)ps$
$Z=(2ck^2+(a-b-2c)k+c-a)s^2-((a+b)k-b)ps$
And more:
$X=(c-a-b)p^2+(k(3c-2a-2b)+2a+b-2c)ps+(k-1)((2c-a-b)k+a-c)s^2$
$Y=(c-a-b)p^2+(k(3c-2a-2b)+a-c)ps+k((2c-a-b)k+a-c)s^2$
$Z=(c-a-b)p^2+(k(4c-3a-3b)+2a+b-2c)ps+(2k-1)((2c-a-b)k+a-c)s^2$
Probably will have to write a more general formula. But I think that's not a problem.
Here is another question. When trying to solve the more general equation:
$nX^2+aX+qY^2+bY=jZ^2+cZ$
Always turns in decisions that need to share the decision itself by type: $w=n+q-j$
That is, the formula is written as a fraction. Certainly at Pell's equation infinitely many solutions. And of course no matter how large the number it was not always a chance that the fraction reduced. But still not clear how all of these decisions to allocate whole? Although there may be another idea to solve this equation and all that there is no need to do?
The coefficients should have this appearance. Because they will appear as unknown for other equations.
| The interesting thing is that the equation: $aX^2+bX+cY^2+qY=jZ^2+dZ$
If we use the equation Pell: $p^2-rs^2=1$
Coefficient is defined as follows: $r=(a+c)jk^2-2cjkt+(j-a)ct^2$
$k,t$ - integers asked us. Then the solutions are of the form:
$X=(qt-dk)ps+((b+q)jk^2-(dc+qj)kt-(b-d)ct^2)s^2$
$Y=((d-b)t-dk)ps+((b+q)jk^2+(ad-(b+2q)j)kt+(j-a)qt^2)s^2$
$Z=(qt-(b+q)k)ps+((a+c)dk^2+((b-2d)c-aq)kt-(b-d)ct^2)s^2$
And another formula.
$X=\frac{1}{a+c-j}[(d-b-q)p^2+(((a+c+j)d-2(b+q)j)k+((j+c-a)q+2(b-d)c)t)ps+$
$+(j((a+c)d-(b+q)j)k^2+(2bcj+(j+c-a)jq-(a+c+j)cd)kt-$
$-c((b-d)c+(j-a)q)t^2)s^2]$
...........................................................................................................................................
$Y=\frac{1}{a+c-j}[(d-b-q)p^2+((a+c+j)d-2(b+q)j)k+(2(j-a)q+$$(j+c-a)(b-d))t)ps+$
$+(j((a+c)d-(b+q)j)k^2+(((a+c-j)a-2cj)d+(j+c-a)jb+2j(j-a)q)kt+$
$+(a-j)((b-d)c+(j-a)q)t^2)s^2]$
...........................................................................................................................................
$Z=\frac{1}{a+c-j}[(d-b-q)p^2+$$((2(a+c)d-(a+c+j)(b+q))k+((j+c-a)q+2(b-d)c)t)ps+$
$+((a+c)((a+c)d-(b+q)j)k^2+((a+c+j)cb-2(a+c)cd+(2cj-(a+c-j)a)q)kt-$
$-c((b-d)c+(j-a)q)t^2)s^2]$
In the second formula should be chosen so Pell's equation and its solution so that the fraction decreased and turned to an integer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving cubic equation? I am trying to figure out the following cubic root thing.
$ax^3+bx^2+cx+d=0$
I set up
$x=y-\frac{3}{ba}$
Then I plug in for x
$a(y-\frac{3}{ba})^3+b(y-\frac{3}{3a})^2+c(y-\frac{3}{ba})=0$
The issue I am having trouble with is the simplification
I try to multiply it all but I gets messy.
Maybe the binomial theorem can be used
this is supposed to go down into the depressed cubic which is.
$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^3}{27a^2}-\frac{bc}{2a})$
| I think you may have the substitution wrong. Try
$$x=y-\frac{b}{3a}$$
then
$$a\left(y-\frac{b}{3a}\right)^3 +b\left(y-\frac{b}{3a}\right)^2 +c\left(y-\frac{b}{3a}\right)+d=0$$ gives
$$ay^3 -by^2 +\frac{b^2}{3a}y -\frac{b^3}{27a^2}+by^2-\frac{2b^2}{3a}y +\frac{b^3}{9a^2}+cy-\frac{bc}{3a}+d=0$$
or
$$ay^3 +\left(c-\frac{b^2}{3a} \right) y +\frac{2b^3}{27a^2}-\frac{bc}{3a}+d=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the limit of $\lim_{x\to\infty}(\sqrt{\frac{x^3}{x+2}}-x)$. How do I evaluate this limit:
$$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)$$
I tried to evaluate this using rationalizing the denominator, numerator and L'Hospital rule for nearly an hour with no success.
| HINT :
Rewrite:
$$
\frac{x^3}{x+2}=x^2-2x+4-\frac{8}{x+2}
$$
and
$$
\sqrt{\frac{x^3}{x+2}}-x=\left(\sqrt{\frac{x^3}{x+2}}-\sqrt{x^2}\right)\cdot\frac{\sqrt{\frac{x^3}{x+2}}+\sqrt{x^2}}{\sqrt{\frac{x^3}{x+2}}+\sqrt{x^2}}.$$
| {
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"source": "stackexchange",
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Expected value of number of draws We have $5$ number in a bag: $(1,3,5,7,9)$. We draw one from the bag and then put it back. We do this until the sum of the numbers can be divided by $3$. Whats the expected value of the number of draws?
My idea was to solve it with Markov-chains:
States: $0,1,2$. So numbers$\mod 3$.
The matrix will be:
$\begin{pmatrix}
2/5 & 2/5 & 1/5 \\
1/5 & 2/5 & 2/5 \\
2/5 & 1/5 & 2/5
\end{pmatrix}$
Then we have an equation system:
$k_1=1+2/5k_1+2/5k_2$, $\ k_2=1+1/5k_1+2/5k_2$ and $k_3=0$.
Even if I solve this, I'm not sure how to continue. Thanks for help.
| We may also consider an absorbing markov chain:
$$
A=\left(\begin{array}{rrrr}
0 & \frac{1}{5} & \frac{2}{5} & \frac{2}{5} \\
0 & \frac{2}{5} & \frac{1}{5} & \frac{2}{5} \\
0 & \frac{2}{5} & \frac{2}{5} & \frac{1}{5} \\
0 & 0 & 0 & 1
\end{array}\right)
$$
Expected time to absorbing state is then given by
$$(I-Q)^{-1}\cdot c$$
where
$$
I=\left(\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right),
Q=\left(\begin{array}{rrr}
0 & \frac{1}{5} & \frac{2}{5} \\
0 & \frac{2}{5} & \frac{1}{5} \\
0 & \frac{2}{5} & \frac{2}{5}
\end{array}\right), c= \left(\begin{array}{r}
1 \\
1 \\
1
\end{array}\right)$$
which is 3.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Integral of $\sqrt{a^2+b^2t^2}$ I'm trying to calculate mass of some line and this is the integral needed to be solved.
Wolfram shows me some way with the fuction sec and reduction methods and I don't know how to use these. is there any other way that I miss?
$$\int\sqrt{a^2+b^2t^2}dt$$
| Let $t=\dfrac ab\tan\theta\;\Rightarrow\;dt=\dfrac ab\sec^2\theta\ d\theta$, the integral turns out to be
$$
\begin{align}
\int\sqrt{a^2+b^2t^2}\ dt&=\int\sqrt{a^2+a^2\tan^2\theta}\cdot\dfrac ab\sec^2\theta\ d\theta\\
&=\dfrac {a^2}b\int\sec^3\theta\ d\theta\\
&=\dfrac {a^2}b\int\frac{1}{\cos^3\theta}\cdot\frac{\cos\theta}{\cos\theta}\ d\theta\\
&=\dfrac {a^2}b\int\frac{\cos\theta}{\cos^4\theta}\ d\theta\\
&=\dfrac {a^2}b\int\frac{\cos\theta}{(1-\sin^2\theta)^2}\ d\theta.
\end{align}
$$
Now, let $u=\sin\theta\;\Rightarrow\;du=\cos\theta\ d\theta$, then
$$
\dfrac {a^2}b\int\frac{\cos\theta}{(1-\sin^2\theta)^2}\ d\theta=\dfrac {a^2}b\int\frac{1}{(1-u^2)^2}\ du.
$$
The last integral admits a decomposition by partial fractions:
$$
\dfrac {a^2}b\int\frac{1}{(1-u^2)^2}\ du=\dfrac {a^2}{4b}\int\left(\frac{1}{1-u}+\frac{1}{(1-u)^2}+\frac{1}{1+u}+\frac{1}{(1+u)^2}\right)\ du
$$
Integrating term-by-term, we will obtain
$$
\begin{align}
\int\sqrt{a^2+b^2t^2}\ dt&=\dfrac {a^2}{4b}\left(\ln\left|\frac{1+\sin \theta}{1-\sin \theta}\right|+2\sec \theta\tan \theta\right)+C\\
&=\dfrac {a^2}{4b}\left(\ln\left|\frac{\sqrt{a^2+b^2t^2}+bt}{\sqrt{a^2+b^2t^2}-bt}\right|+\frac{2bt\sqrt{a^2+b^2t^2}}{a^2}\right)+C\\
&=\dfrac {a^2}{4b}\left(\ln\left|\frac{\sqrt{a^2+b^2t^2}+bt}{\sqrt{a^2+b^2t^2}+bt}\cdot\frac{\sqrt{a^2+b^2t^2}+bt}{\sqrt{a^2+b^2t^2}-bt}\right|+\frac{2bt\sqrt{a^2+b^2t^2}}{a^2}\right)+C\\
&=\dfrac {a^2}{2b}\left(\ln\left|\frac{\sqrt{a^2+b^2t^2}+bt}{a}\right|+\frac{bt\sqrt{a^2+b^2t^2}}{a^2}\right)+C.\\
\end{align}
$$
| {
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total least squares derivation with matrices Taken from a computer vision book: "to minimize the sum of the perpendicular distances between points and lines, we need to minimize $$ \sum_i (ax_i + by_i +c)^2$$ subject to $a^2 +b^2 =1$. Now using a Lagrangian multiplier $\lambda$, we have a solution if $$ \left( \begin{array}{ccc}
\overline{x^2} & \overline{xy} & \overline{x} \\
\overline{xy} & \overline{y} & \overline{y} \\
\overline{x} & \overline{y} & 1 \end{array} \right)\left[ \begin{array}{c}
a \\
b \\
c
\end{array} \right] = \lambda \left[ \begin{array}{cc}
2a\\
2b \\
0
\end{array} \right]$$
How is the book getting these matrices?
Also, the notion is that $\overline{u} = \frac{\sum u_i}{k}$. (Yeah, I don't know what $k$ stands for. I can only assume this is an average.)
It goes onto say that $c = -a\overline{x} - b\overline{y}$, and that we can substitute this back to get the eigenvalue problem $$\left[ \begin{array}{cc}
\overline{x^2} -\overline{x}~\overline{x} & \overline{xy} - \overline{x}\overline{y}\\
\overline{xy} - \overline{x}\overline{y} & \overline{y^2} - \overline{y} ~\overline{y} \\
\end{array} \right] \left[\begin{array}{cc}
a\\
b
\end{array} \right] = \mu \left[ \begin{array}{cc}
a\\
b
\end{array} \right].$$
I don't see what they substituted into, and how the answer is derived.
| We want to minimize
$$
S=\sum_i(ax_i+by_i+c)^2\tag{1}
$$
Varying $(1)$ yields
$$
\frac12\delta S=
\begin{bmatrix}\delta a&\delta b&\delta c\end{bmatrix}
\left(\sum_i\begin{bmatrix}x_i\\y_i\\1\end{bmatrix}\begin{bmatrix}x_i&y_i&1\end{bmatrix}\right)
\begin{bmatrix}a\\b\\c\end{bmatrix}\tag{2}
$$
Since, $a^2+b^2=1$, we restrict the variations to those so that
$$
0=\begin{bmatrix}\delta a&\delta b&\delta c\end{bmatrix}\begin{bmatrix}a\\b\\0\end{bmatrix}\tag{3}
$$
We want to find $a,b,c$ that cancel $(2)$ under the constraints in $(3)$. Orthogonality implies that we need
$$
\left(\sum_i\begin{bmatrix}x_i\\y_i\\1\end{bmatrix}\begin{bmatrix}x_i&y_i&1\end{bmatrix}\right)
\begin{bmatrix}a\\b\\c\end{bmatrix}
=\lambda\begin{bmatrix}a\\b\\0\end{bmatrix}\tag{4}
$$
Equation $(4)$ is the first equation in the question.
The bottom row of $(4)$ says
$$
\left(\sum_i\begin{bmatrix}x_i&y_i&1\end{bmatrix}\right)
\begin{bmatrix}a\\b\\c\end{bmatrix}=0\tag{5}
$$
The top two rows of $(4)$ say
$$
\left(\sum_i\begin{bmatrix}x_i\\y_i\end{bmatrix}\begin{bmatrix}x_i&y_i&1\end{bmatrix}\right)
\begin{bmatrix}a\\b\\c\end{bmatrix}
=\lambda\begin{bmatrix}a\\b\end{bmatrix}\tag{6}
$$
we can rewrite $(6)$ as
$$
\left(\sum_i\begin{bmatrix}x_i\\y_i\end{bmatrix}\begin{bmatrix}x_i&y_i\end{bmatrix}\right)
\begin{bmatrix}a\\b\end{bmatrix}
+\left(\sum_i\begin{bmatrix}x_i\\y_i\end{bmatrix}\right)c
=\lambda\begin{bmatrix}a\\b\end{bmatrix}\tag{7}
$$
and $(5)$ as
$$
\left(\sum_i\begin{bmatrix}x_i&y_i\end{bmatrix}\right)
\begin{bmatrix}a\\b\end{bmatrix}+\left(\sum_i1\right)c=0\tag{8}
$$
If we set $n=\sum\limits_i1$ and substitute $c$ computed from $(8)$ into $(7)$, we get
$$
\left(\sum_i\begin{bmatrix}x_i\\y_i\end{bmatrix}\begin{bmatrix}x_i&y_i\end{bmatrix}-\frac1n\sum_i\begin{bmatrix}x_i\\y_i\end{bmatrix}\sum_i\begin{bmatrix}x_i&y_i\end{bmatrix}\right)
\begin{bmatrix}a\\b\end{bmatrix}
=\lambda\begin{bmatrix}a\\b\end{bmatrix}\tag{9}
$$
Equation $(9)$ is the last equation in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/801836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a^4+b^4\in\mathbb Q$ and $a^3+b^3\in\mathbb Q$ and $a^2+b^2\in\mathbb Q$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$.
If $\begin{cases}a^4+b^4\in\mathbb Q\\ a^3+b^3\in\mathbb Q\\ a^2+b^2\in\mathbb Q\end{cases}$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$. It is given that $a,b\in\mathbb R$.
The proof of the latter would simply follow from the former, and vice versa. So I think a better question would be:
Prove one of these statements: $a+b\in\mathbb Q$ or $ab\in\mathbb Q$.
The problem is from the selection to IMO.
I've tried a whole lot of things, including the identities: $$a^4+b^4=(a+b)(a^3+b^3)-ab(a^2+b^2)\\ a^3+b^3=(a+b)(a^2-ab+b^2)\\ a^2+b^2=(a+b)^2-2ab\\ (a+b)^3=a^3+b^3+3ab(a+b)\\ \text{etc...}$$
Even if one could solve the problem using these identities, doing it would most likely be quite tedious imho... Any observations would be greatly appreciated. Thanks.
| (Note, added later: This answer was posted before the OP added the stipulation that $a,b\in\mathbb{R}$.)
This is more plodding than ronno's slick hint, but perhaps it shows more of the thought process:
$$(a^2+b^2)^2=a^4+2(ab)^2+b^4\implies (ab)^2\in\mathbb{Q}$$
$$(a^2+b^2)^3=a^6+3(ab)^2(a^2+b^2)+b^6\implies a^6+b^6\in\mathbb{Q}$$
$$(a^3+b^3)^2=a^6+2(ab)^3+b^6\implies (ab)^3\in\mathbb{Q}$$
$$(ab)^2\in\mathbb{Q}\land (ab)^3\in\mathbb{Q}\implies ab\in\mathbb{Q}$$
It would be nice to prove $a+b\in\mathbb{Q}$ as well, but you can't: $a=1+\sqrt{-3}, b=-1+\sqrt{-3}$ is a counterexample. What you can show is
$$a^3+b^3=(a+b)(a^2-ab+b^2)\implies a+b\in\mathbb{Q}\lor a^2-ab+b^2=0$$
which is to say, $a+b$ is rational unless $a^3=-b^3$ with $a\not=-b$.
| {
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"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$
Let $a,b,c>0$ such that
$$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$
Show that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$
My try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$
so
$$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$
we only prove
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$
and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such
$$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$
$$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$
$$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$
and let $$p=a+b+c,q=ab+bc+ac,r=abc$$
so
$$2p^2+21\ge 7q$$
and we only prove
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$
maybe this is not true.
But this not usefull to solve this problem .
Thank you
| By AM-GM, $1=\dfrac{3}{a^2+2}+\dfrac{3}{b^2+2}+\dfrac{3}{c^2+2} \geq \dfrac{9}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}$
By AM-GM, $7 = \dfrac{3a^2}{a^2+2}+\dfrac{3b^2}{b^2+2}+\dfrac{3c^2}{c^2+2} \geq \dfrac{9\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}$
As everything is positive, we can safely divide the inequalities with each other.
$\dfrac{7}{1} \geq \dfrac{\dfrac{9\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}}{\dfrac{9}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}}$
$\Rightarrow 7 \geq (abc)^{\dfrac{2}{3}}$
$\Rightarrow \dfrac{1}{\sqrt[3]{abc}} \geq \dfrac{1}{\sqrt{7}}$
By AM-GM, $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq \dfrac{3}{\sqrt[3]{abc}}\geq \dfrac{3}{\sqrt{7}}$
$\therefore \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq \dfrac{3}{\sqrt{7}}$
| {
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"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
Partial fraction decomposition,how? I need to decompose this fraction:
$${x^2+1\over (x-1)^3\cdot(x+3)}$$
I tried to write it up like this: $${A\over (x-1)}+{B\over (x-1)^2}+{C\over (x-1)^3}+{D\over (x+3)}$$
But now i get $$A\cdot(x-1)^5\cdot (x+3) +B\cdot (x-1)^4\cdot (x+3) +C\cdot(x-1)^3\cdot (x+3) +D\cdot (x-1)^6 =x^2+1$$ which is not correct i think.
What should i do?
| I think your expression should be:
$$
A\cdot(x-1)^2\cdot (x+3) +B\cdot (x-1)^1\cdot (x+3) +C\cdot (x+3) +D\cdot (x-1)^3 =x^2+1
$$
because the LCM of $(x-1)$, $(x-1)^2$, $(x-1)^3$ and $(x+3)$ is $(x+3)(x-1)^3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Showing that a sequence of Picard iterates converges I have a sequence of functions:
$$y_{n}(x) = 1 + \int \limits_0^x 1 + t^2 + y_{n-1}^2(t)\,\mathrm dt$$
With $y_0 = 1$.
I'm trying to show that this converges in a box $-1 \le x \le 1$ and $-10 \le y \le 10$. However when I try to show this converges in this box, the bound on the integral gives $|y-1| \le xM$ where $M$ is the maximum of $1 + t^2 + y^2$ in the box, however this maximum is $102$ so the iterates only converge in the box when $|y-1| \le 102$ but this isn't true?
Am I missing something obvious here?
Thanks
| A theorem for finding an interval of convergence for a Picard iteration scheme is usually stated as:
Given the IVP
\begin{equation*}
y^{\prime} = f(x,y), \quad y(x_{0}) = y_{0},
\end{equation*}
suppose $f$ and $df/dy$ are continuous in the rectangle $R$ given by
\begin{equation*}
x_{0} \le x \le x_{0}+a, \quad \left| y-y_{0}\right| \le b
\end{equation*}
compute the values
\begin{equation*}
M = \max_{(x,y)\in R}\left|f(x,y)\right|, \quad \alpha = \min\left(a, \frac{b}{M}\right).
\end{equation*}
Then the Picard iterates converge to a unique solution on the interval $[x_{0},x_{0}+\alpha]$.
(Note, the interval given by this theorem may be smaller that the actual interval of convergence).
In your case we have that,
\begin{alignat*}{2}
x_{0} &= 0 \\
a &= 1, \\
b &= 10,\\
M &= \max_{R}\left| 1 + x^{2} + y^{2}\right| \\
&= 1 + a^{2} + b^{2} \\
&= 102 &&\Rightarrow \\
\alpha &= \min\left(1, \frac{10}{102}\right)\\
&\le 1 \\
&= a &&\Rightarrow
\end{alignat*}
which falls short of the desired result.
Alternatively, we can try looking at the iterates to see if we can learn more about their radius of convergence.
For $y_{n}(x) = 1 + \int_{0}^{x}1+t^{2}+y^{2}_{n-1}(t)dt$, assume the differential equation has the form
\begin{alignat*}{2}
y(x) = 1 + \int_{0}^{x}f(t,y)dt &&\Rightarrow \\
f(t,y) = 1 + t^{2}+y^{2}(t)
\end{alignat*}
and
\begin{equation*}
y(0) = 1 = y_{0}
\end{equation*}
Now,
\begin{alignat*}{2}
y_{1}(x) &= 1 + \int_{0}^{x}f(t,y_{0})dt \\
&= 1 + \int_{0}^{x}1+t^{2}+1 dt \\
&= 1 + 2x + \frac{1}{2}x^{2},
\end{alignat*}
\begin{alignat*}{2}
y_{2}(x) &= 1 + \int_{0}^{x}f(t,y_{1})dt \\
&= 1 + \int_{0}^{x}1+t^{2}+(1 + 2t + \frac{1}{2}t^{2})^{2} dt \\
&= 1 + 2x + 2x^{2}+2x^{3}+\frac{1}{2}x^{4}+\frac{1}{20}x^{5}
\end{alignat*}
and for $y_{3}$ I get something like:
\begin{equation*}
y_{3}(x) = 1 + 2x + 2x^{2} + \frac{7}{3}x^{3} + 3x^{4} + \frac{13}{5}x^{5}+\frac{202}{120}x^{6} + \frac{52}{70}x^{7}+\frac{22}{80}x^{8}+\cdots
\end{equation*}
So, it seems (loosely) for some constant $M$ large enough
\begin{alignat*}{2}
M\sum_{k=0}^{\infty}x^{k} &\ge \lim_{k\to\infty}y_{k}(x) &&\ge 1+\sum_{k=1}^{\infty}\frac{x^{k}}{k} \Rightarrow\\
M\frac{1}{1-x} &\ge \lim_{k\to\infty}y_{k}(x) &&\ge 1-\log(1-x)
\end{alignat*}
both of which are valid for $\left|x\right| < 1$, but not for the closed interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/807062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Dilogarithm in closed form Is there a closed form expression for
\begin{align}
e^{\Large\frac{i\pi}3} \text{Li}_{2}\left( \frac{e^{\Large\frac{i\pi}3} }{2}\right) + e^{-\Large\frac{i\pi}3} \text{Li}_{2}\left( \frac{e^{-\Large\frac{i\pi}3} }{2}\right)
\end{align}
in terms of known constants such as: $\pi$, $\ln 2$, $\ln 3$, $e$, $\gamma$, etc.
The expression to be calculated can be placed into series form and is equivalent to calculating the series
\begin{align}
\sum_{n=1}^{\infty} \frac{ \cos\left( \frac{(n+1)\pi}{3} \right) }{2^{n-1} \ n^{2}}
\end{align}
in closed form.
| Letting $z_{\pm}=\frac{e^{\pm\frac{i\pi}{3}}}{2}$, we find,
$$\frac{z_{\pm}}{z_{\pm}-1}=\mp\frac{i}{\sqrt{3}},$$
and
$$\ln{\left(1-z_{\pm}\right)}=\frac12\ln{\left(\frac34\right)}\mp i\frac{\pi}{6},\\
\implies \ln^2{\left(1-z_{\pm}\right)}=\frac14\ln^2{\left(\frac43\right)}-\frac{\pi^2}{36}\pm i \frac{\pi}{6}\ln{\left(\frac43\right)}.$$
Then using Landen's dilogarithm identity,
$$\begin{align}
\operatorname{Li}_{2}{\left(z_{\pm}\right)}
&=-\operatorname{Li}_{2}{\left(\frac{z_{\pm}}{z_{\pm}-1}\right)}-\frac12\ln^2{\left(1-z_{\pm}\right)}\\
&=-\operatorname{Li}_{2}{\left(\mp\frac{i}{\sqrt{3}}\right)}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)}\pm i \frac{\pi}{12}\ln{\left(\frac34\right)}.
\end{align}$$
The real and imaginary components of $\operatorname{Li}_{2}{\left(z_{+}\right)}$ are then,
$$\Re{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}=-\Re{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)}\\
=-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)},$$
and,
$$\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}=-\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}+\frac{\pi}{12}\ln{\left(\frac34\right)}.$$
The term $\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}$ may be further simplified with the aid of the dilogarithmic identity,
$$\operatorname{Li}_{2}{\left(z\right)}-\operatorname{Li}_{2}{\left(-z\right)}+\operatorname{Li}_{2}{\left(\frac{1-z}{1+z}\right)}-\operatorname{Li}_{2}{\left(-\frac{1-z}{1+z}\right)}=\frac{\pi^2}{4}+\ln{\left(z\right)}\ln{\left(\frac{1+z}{1-z}\right)}.$$
Setting $z=-\frac{i}{\sqrt{3}}$,
$$\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}-\operatorname{Li}_{2}{\left(\frac{i}{\sqrt{3}}\right)}+\operatorname{Li}_{2}{\left(e^{\frac{i\pi}{3}}\right)}-\operatorname{Li}_{2}{\left(-e^{\frac{i\pi}{3}}\right)} = \frac{\pi^2}{12}+i\frac{\pi}{6}\ln{(3)}.$$
$$\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}-\operatorname{Li}_{2}{\left(\frac{i}{\sqrt{3}}\right)} = i\frac{\pi}{6}\ln{(3)}+i\frac{5\pi^2}{9\sqrt{3}}-i\frac{5}{6\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$
$$\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]} = \frac{\pi}{12}\ln{(3)}+\frac{5\pi^2}{18\sqrt{3}}-\frac{5}{12\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$
$$\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]} = -\frac{\pi}{6}\ln{(2)}-\frac{5\pi^2}{18\sqrt{3}}+\frac{5}{12\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$
The quantity of interest is then:
$$\begin{align}
A
&=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2z_{-}\operatorname{Li}_{2}{\left(z_{-}\right)}\\
&=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2\,\overline{z_{+}}\operatorname{Li}_{2}{\left(\overline{z_{+}}\right)}\\
&=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2\,\overline{z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}}\\
&=4\Re{\left[z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}\\
&=\Re{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}-\sqrt{3}\,\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}\\
&=-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{7\pi^2}{24}-\frac18\ln^2{\left(\frac34\right)}+\frac{\sqrt{3}\,\pi}{6}\ln{(2)}-\frac{5}{12}\psi^{(1)}{\left(\frac13\right)}.\\
\end{align}$$
I don't think the sum of dilogarithm trigamma terms in the final line can be simplified to a combination of the basic constants suggested.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Indefinite Integration of $(\arctan(x))^2$ $$
\mbox{Hello. I was wondering how to integrate this:}\quad
\int\arctan^{2}\left(x\right)\,{\rm d}x
$$
Do I first do a u-substitution first ?.
| This integral is done by using integration by parts. We have:
\begin{eqnarray}
&&\int (\arctan(x))^2 dx = \int x^{'} (\arctan(x))^2 dx =\\
&& x \arctan(x)^2 - \int x \cdot 2 \arctan(x) \cdot \frac{1}{1+x^2} dx =\\
&& x \arctan(x)^2 - \log(1+x^2) \arctan(x) + \int \frac{\log(1+x^2)}{1+x^2} dx =\\
&& x \arctan(x)^2 - \log(1+x^2) \arctan(x) + \frac{1}{2 \imath}\int \left(\log(x+\imath)+\log(x-\imath)\right) \left(\frac{1}{x-\imath}-\frac{1}{x+\imath}\right) dx =\\
&& x \arctan(x)^2 - \log(1+x^2) \arctan(x) + \frac{1}{2 \imath}\left( \frac{1}{2} \log(x-\imath)^2 - \frac{1}{2} \log(x+\imath)^2\right) + \\
&&\frac{1}{2 \imath}\int \left( \frac{\log(x+\imath)}{x-\imath} - \frac{\log(x-\imath)}{x+\imath}\right)dx =\\
&&x \arctan(x)^2 - \log(1+x^2) \arctan(x) + Im\left( \frac{1}{2} \log(x-\imath)^2 +\log(2 \imath) \log(x-\imath) - Li_2(\frac{x-\imath}{-2 \imath})\right)
\end{eqnarray}
| {
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Which is the normal vector?? Apply the divergence theorem over the region $1 \leq x^2+y^2+z^2 \leq 4$ for the vector field $\overrightarrow{F}=-\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p^3}$, where $p=(x^2+y^2+z^2)^\frac{1}{2}$.
$$$$
The divergence theorem is the following:
$$\iiint_D \nabla{\overrightarrow{F}} \cdot dV=\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma$$
How can I calculate the integral: $$\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma$$ ??
Which is the normal vector $\hat{n}$??
EDIT:
In my notes it is:
$f=x^2+y^2+z^2$
$\displaystyle{\hat{n}=\pm \frac{\bigtriangledown f}{|\bigtriangledown f|}=\pm \frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}}$
$"+": p=2$
$"-" \text{ for } p=1$
I haven't understood the signs.. Could you explain me why it's $\pm$? And also why is it $"+"$ when $p=2$ and $"-"$ when $p=1$?
| First you need two surfaces to bind your region, the two surfaces are $x^2+y^2+z^2 = 4$ and $x^2+y^2+z^2=1$.First calculate the surface with the outer surface as
$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \vec F(x, y, \sqrt{4-x^2-y^2})\cdot \frac{\nabla(x^2+y^2+z^2)}{|\nabla(x^2+y^2+z^2)\cdot \hat k|}dydx - \int_{-2}^2\int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}} \vec F(x, y, -\sqrt{4-x^2-y^2})\cdot \frac{\nabla(x^2+y^2+z^2)}{|\nabla(x^2+y^2+z^2)\cdot (-\hat k)|}dydx$$
Note that the first integral is for upper surface of sphere and second integral is for lower surface. Calculate the gradient first in those two integrals, and replace $z = \sqrt{4-x^2-y^2}$ on upper surface of sphere and $z = -\sqrt{4-x^2-y^2}$ on lower surface.
For inner surface, the expression is similar, just note the direction of normal surface. It's inward. And it's better to use spherical coordinate as Elya suggested on other answer.
| {
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} |
Simplifying $\sqrt[4]{161-72 \sqrt{5}}$ $$\sqrt[4]{161-72 \sqrt{5}}$$
I tried to solve this as follows:
the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:
$$a^4+30 a^2 b^2+25 b^4=161$$
$$4 a^3 b+20 a b^3=-72$$
In an attempt to solve this, I first tried to factor stuff and rewrite it as:
$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$
$$4 a b \left(a^2+5 b^2\right)=-72$$
Then letting $p = a^2 + 5b^2$ and $q = ab$ you get
$$4 p q=-72$$
$$p^2+10 q^2=161$$
However, solving this yields messy roots. Am I going on the right path?
| A (more complicated) approach that works on any nested radical, would be to use the Zippel Denesting Theorem.
$\sqrt[4]{161-72\sqrt{5}}$ is a fourth power exponent in $\mathbb{Q}(\sqrt{5})$ so setting the radical equal to its primitive root of unity and finding its roots gives us the simplification.
So we have: $\sqrt[4]{161-72\sqrt{5}}=x\iff x^4+72\sqrt{5}-161=0\iff (\sqrt{5}-2-x)(x+\sqrt{5}-2)(4\sqrt{5}-9-x^2)=0$ with the first one giving the correct denesting of $\sqrt{5}-2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Let $x_{n+1} = x_n + 1/(x_1 + x_2 +\ldots + x_n)$ with $x_1 = 1$. Show that $x_n\sim\sqrt{2\log(n)}$. As the title states we have a sequence defined by
$$x_{n+1} = x_n + \dfrac{1}{x_1 + x_2 + \cdots + x_n}$$
with $x_1 = 1$.
The first few terms are: $1, 2, \frac{7}{3}, \frac{121}{48} \cdots$
Any ideas would be appreciated.
| Repeatedly integrating by parts yields
$$
\int\sqrt{2\log(x)}\,\mathrm{d}x=x\sqrt{2\log(x)}-x\sum_{k=0}^{n-1}\frac{(2k-1)!!}{\sqrt{2\log(x)}^{\,2k+1}}-\int\frac{(2n-1)!!\,\mathrm{d}x}{\sqrt{2\log(x)}^{\,2n+1}}\tag{1}
$$
and
$$
\int\frac{\mathrm{d}x}{\sqrt{2\log(x)}}=\frac{x}{\sqrt{2\log(x)}}+x\sum_{k=1}^{n-1}\frac{(2k-1)!!}{\sqrt{2\log(x)}^{\,2k+1}}+\int\frac{(2n-1)!!\,\mathrm{d}x}{\sqrt{2\log(x)}^{\,2n+1}}\tag{2}
$$
Therefore, using the Euler-Maclaurin Sum Formula and $(1)$ and $(2)$, for $n\gt1$, we get
$$
\left(\sum_{j=1}^n\sqrt{2\log(j)}\,\mathcal{I}(j)\right)^{-1}=\frac1{n\sqrt{2\log(n)}}\,\mathcal{I}(n)\tag{3}
$$
where $\mathcal{I}(n)=1+O\left(\frac1{\log(n+1)}\right)$.
Furthermore,
$$
\int\frac{\mathrm{d}x}{x\sqrt{2\log(x)}^{\,k}}=\frac1{2-k}\sqrt{2\log(x)}^{\,2-k}+C\tag{4}
$$
Thus, applying the Euler-Maclaurin Sum Formula to $(3)$ and $(4)$ gives
$$
\sum_{k=2}^n\left(\sum_{j=1}^k\sqrt{2\log(j)}\,\mathcal{I}(j)\right)^{-1}
=\sqrt{2\log(n)}\,\mathcal{I}(n)\tag{5}
$$
Using the definition of $x_n$, we get
$$
\begin{align}
x_{n+1}
&=2+\sum_{k=2}^nx_{k+1}-x_k\\
&=2+\sum_{k=2}^n\left(\sum_{j=1}^kx_j\right)^{-1}\tag{6}
\end{align}
$$
Equation $(5)$ and $(6)$ show that
$$
x_{n+1}=\sqrt{2\log(n)}+O\left(\frac1{\sqrt{\log(n)}}\right)\tag{7}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How do I evaluate $\lim_{A\to\infty}\int_{-A}^A \left(\frac{\sin x}{x}\right)^2e^{itx}\,dx$? How do I compute
$$
\lim_{A\to\infty}\int_{-A}^A \left(\frac{\sin x}{x}\right)^2e^{itx}\,dx
$$
for $t\in\mathbb{R}$ and $i$ the imaginary unit?
| Let $ \displaystyle I(t) = \int_{-\infty}^{\infty} \frac{\sin^{2}x}{x^{2}} e^{itx} \ dx$.
Then
$$\begin{align} I(t) &= \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \cos tx \ dx \\ &= 2 \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \cos tx\ dx \\ &= \int_{0}^{\infty} \frac{\left(1- \cos 2x \right) \cos tx}{x^{2}} \\ &= \int_{0}^{\infty}\left(\frac{\cos tx}{x^{2}} - \frac{\cos(t+2)x}{2x^{2}} - \frac{\cos(t-2)x}{2x^{2}} \right) \ dx \ . \end{align} $$
Integrating by parts,
$$ \begin{align} \int \frac{\cos ax}{x^{2}} \ dx &= - \frac{\cos ax}{x} - a \int \frac{\sin ax}{x} \ dx \ \\ &= - \frac{\cos ax}{x} - a \ \text{Si}(ax) \end{align}$$
where $\text{Si}(x)$ is the sine integral.
http://mathworld.wolfram.com/SineIntegral.html
Therefore,
$$I(t) = - \frac{\cos tx}{x} - t \ \text{Si}(tx) + \frac{\cos (t+2)x}{2x} + \frac{(t+2) \ \text{Si}\big((2+t)x\big)}{2} + \frac{\cos (t-2)x}{2x}$$
$$ + \frac{(t-2) \ \text{Si}\big((2-t)x\big)}{2} \Bigg|^{\infty}_{0}$$
$$ = -t \ \text{sgn} (t) \frac{\pi}{2} + \frac{t+2}{2} \text{sgn}(t+2) \frac{\pi}{2} + \frac{t-2}{2} \text{sgn}(t-2) \frac{\pi}{2}$$
where I used the fact that
$$ \lim_{x \to \infty} \text{Si} (ax) = \int_{0}^{\infty} \frac{\sin ax}{x} \ dx = \text{sgn}(a) \frac{\pi}{2} \ .$$
Then you can check to see what happens for different values of $t$.
If $t > 2$,
$$I(t) = - t \frac{\pi}{2} + \frac{t+2}{2} \frac{\pi}{2} + \frac{t-2}{2} \frac{\pi}{2} = 0$$
If $t=2$,
$$ I(2) = -2 \frac{\pi}{2} +2 \frac{\pi}{2} + 0 = 0$$
If $0 < t < 2$,
$$ I(t) = -t \frac{\pi}{2} + \frac{t+2}{2} \frac{\pi}{2} - \frac{t-2}{2} \frac{\pi}{2} = \frac{\pi}{2} (2-t)$$
If $t=0$,
$$ I(0) = 0 + \frac{\pi}{2} + \frac{\pi}{2} = \pi$$
And $I(t)=I(-t)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Direct proportion - sum of numbers and square root of sum of the squared numbers Is it true for all the cases that if $x + y > a+b$, then $\sqrt{x^2 + y^2} > \sqrt{a^2 + b^2}$? In other words - is there a direct proportion between sum of numbers and square root of the sum of those numbers squared?
| No, take $x=y=3,a=5,b=0$, then $x+y=6, a+b=5, \sqrt{x^2+y^2}=\sqrt {18} =3\sqrt 2\approx 4.242, \sqrt{a^2+b^2}=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$?
How to integrate $$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx$$ ?
I have:
$$\int \frac{\cos x}{\sqrt{\sin2x}} \,dx = \int \frac{\cos x}{\sqrt{2\sin x\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\cos x}{\sqrt{\sin x}\sqrt{\cos x}} \,dx = \frac{1}{\sqrt2}\int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx = \frac{1}{\sqrt2}\int \sqrt{\cot x} \,dx \\
t = \sqrt{\cot x} \implies x = \cot^{-1} t^2 \implies \,dx = -\frac{2t\,dt}{1 + t^4}$$
so I have:
$$-\sqrt2 \int \frac{t^2 \,dt}{1 + t^4}$$
I tried partial integration on that but it just gets more complicated. I also tried the substitution $t = \tan \frac{x}{2}$ on this one: $\frac{1}{\sqrt2}\int \sqrt{\frac{\cos x}{\sin x}} \,dx$
$$= \frac{1}{\sqrt2}\int \sqrt{\frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1+t^2}}} \frac{2\,dt}{1+t^2} = \frac{1}{\sqrt2}\int \sqrt{\frac{1 - t^2}{2t}} \frac{2\,dt}{1+t^2} = \int \sqrt{\frac{1 - t^2}{t}} \frac{\,dt}{1+t^2}$$
... which doesn't look very promising.
Any hints are appreciated!
| I have a half answer (exactly as you did).
Put $\sin 2x=2\sin x\cos x$. Then you integral is:
$$\int\dfrac{\cos x}{\sqrt{2\sin x }\sqrt{\cos x}}dx=\int\dfrac{\sqrt{\cos x}}{\sqrt{2\sin x }}dx=\dfrac{\sqrt{2}}{2}\int\sqrt{\cot x}\,dx.$$
For the second half of the answer, see this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
Find the value of $a$ such that at least one root of the equation $f(x)=x^2 - (a-3)x + a =0$ is greater than $2$. As the title says.
To solve this problem I took two cases and solved them separately:
1. when the $x$ coordinate of the vertex is greater than $2$ and $f(2)>0$;
2. when $f(2)<0$.
However I wasn't able to find the answer as the two conditions have no common solution (obviously they won't have because in one case $f(2)>0$ and in the other one $f(2)<0$.
The inequalities which I got after solving for the two cases separately were:
1. $a<10$ and $a<7$
2. $10 < a$
And help would be appreciated. :)
| The roots of that equation are
$$
\frac{(a-3) \pm \sqrt{a^2-6a+9-4a}}{2}
$$
so specifically we need
$$
\frac{(a-3) + \sqrt{a^2-6a+9-4a}}{2} > 2 \implies \sqrt{a^2-10a+9} > 7 - a
$$
Here's where I made a mistake at first. If $a < 7$ then we have
$$
a^2 - 10a + 9 > 49 - 14a + a^2 \implies a > 10
$$
so that this route gets us no solutions, now if $a > 7$ we just need to have $\sqrt{a^2 - 10a + 9}$ to be real which mean $a^2 - 10a + 9 > 0$. Notice that the roots to this second polynomial are
$$
\frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2} = 9,1
$$
Now since we have the restriction that $a > 7$ we need to see whether the polynomial is positive or negative in $[7,9]$ and likewise from $[9,\infty)$. Plugging in $a=8$ we see that
$$
8^2-10(8)+9 = -16 + 9 < 0
$$
so $a \not\in [7,9]$. Now checking $a=10$ we have
$$
10^2 -10(10) + 9 = 9 > 0
$$
so that indeed $a > 9$.
Now we need to do something similar for the other root
$$
\sqrt{a^2-10a+9} < a - 7
$$
Now if $a < 7$ this doesn't make sense since a square root can only be positive. Otherwise if $a > 7$ then we have
$$
a^2 - 10a + 9 < a^2 - 14a + 49 \implies 4a < 40 \implies a < 10
$$
as long as $a^2 - 10a + 9 \ge 0 \implies a \ge 9$
So our other root gives us a solution if $a \in [9,10)$. Now combining these we get that $a \in [9, \infty)$.
Sorry for the issues before
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/820441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Math Olympiad Algebra Question If,
$$ax + by = 7$$ $$ax^2 + by^2 = 49$$ $$ax^3 + by^3 = 133$$ $$ax^4 + by^4 = 406$$
Then find the value of$-$
$$2014(x+y-xy) - 100(a+b)$$
I came across this question in a Math Olympiad Competition and I am not sure how to solve it. Can anyone help? Thanks.
| From $ax+by=7$, we have $ax=7-by, by=7-ax$.
Noting
$$ ax^2+by^2=x\cdot ax+y\cdot by=x(7-by)+y(7-ax)=7(x+y)-(a+b)xy, $$
from $ax^2+by^2=49$, we obtain
$$ \tag{$*$} 7(x+y)-(a+b)xy=49. $$
Similarly,
$$ ax^2=49-by^2,by^2=49-ax^2, ax^3=133-by^3,by^3=133-ax^3, $$
from which, we have
$$ ax^3+by^3=x\cdot ax^2+y\cdot by^2=x(49-by^2)+y(49-ax^2)=49(x+y)-(ax+by)xy $$
and hence $49(x+y)-7xy=133$
or
$$ \tag{$**$} 7(x+y)-xy=19. $$
Finally,
$$ ax^4+by^4=x\cdot ax^3+y\cdot by^3=x(133-by^3)+y(133-ax^3)=133(x+y)-(ax^2+by^2)xy=133(x+y)-49xy $$
and hence $133(x+y)-49xy=406$ or
$$\tag{$***$}\quad\quad\quad 19(x+y)-7xy=58. $$
From $(*), (**), (***)$, it is easy to see
$$x+y=\frac{5}{2},xy=-\frac{3}{2},a+b=21 $$
and hence
$$ 2014(x+y-xy)-100(a+b)=5956. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Integration by substitution - $\int (x+1)\sqrt{5x-6} \text{dx}$ $\int (x+1)\sqrt{5x-6} \text{dx}$
My working so far:
$u=5x-6$
$du=5dx$
$\int\dfrac{x+1}{5}\sqrt u \ \text{du}$
Substituting x
$\int \dfrac{(u+6)/5)+(5/5)}{5}\cdot\sqrt u \ \text{du}$
$\int (u+11)\sqrt u \ \text{du}$
I've got the same powers right when I integrate that, but I have my coefficients wrong. The answer is:
$\dfrac{2}{75}(5x-6)^{5/2}+\dfrac{22}{75}(5x-6)^{3/2}+c$
Where am I going wrong?
| Let the integral in question be
\begin{align}
I = \int (x+1) \sqrt{5x-6} \ dx
\end{align}
then making the substitution $u = 5x-6$ the integral becomes
\begin{align}
I &= \int \left( \frac{u+11}{5} \right) \ \sqrt{u} \ \frac{du}{5} \\
&= \frac{1}{25} \int \left[ u^{3/2} + 11 u^{1/2} \right] du \\
&= \frac{1}{25} \left[ \frac{2}{5} \ u^{5/2} + \frac{22}{3} u^{3/2} \right]
\end{align}
and upon back substitution the result becomes
\begin{align}
I = \frac{2}{125} (5x-6)^{5/2} + \frac{22}{75} (5x-6)^{3/2} + c.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question. Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 + \sin(\theta)^2$
If this is incorrect thinking can someone please put me on the right path?
Thanks
| Set $\theta=\pi/4$; then $\sin\theta=\cos\theta=1/\sqrt{2}$; therefore
$$
\frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta}=
\frac{1}{1/2}+\frac{1}{1/2}=2+2=4\ne1.
$$
We might ask when equality holds:
$$
\frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta}=1
$$
is equivalent to
$$
\frac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}=1
$$
which becomes
$$
\sin^2\theta\cos^2\theta=1.
$$
We can multiply both sides by $4$ getting
$$
4\sin^2\theta\cos^2\theta=4.
$$
or
$$
\sin^2(2\theta)=4
$$
which has no solution. So, not only $\theta=\pi/4$ is a counterexample, but all values of $\theta$ are (excluding integer multiples of $\pi/2$ that make the left hand side undefined).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Can some one explain to me what is going on here - power of complex number So here is the question and the work to solve it, but I have no idea how one knows to do the first step or what the first step is...
$$ \begin{align}
(6-i\sqrt{12})^{12} &= \left[\sqrt{48}\left(\cos\left(\frac{\pi}{6}\right) - i\sin\left(\frac{\pi}{6}\right)\right)\right]^{12}\\
&= (\sqrt{48})^{12} \left[\cos\left(\frac{12\pi}{6}\right) - i\sin\left(\frac{12\pi}{6}\right)\right]\\
&=48^6
\end{align}
$$
| Some of the other posts have given general theory let be just point out how to analyse the question. We have $6-i\sqrt{12}$ we want to write it as $\sin$ and $\cos$ so we can use DeMovire. We need first to find its length,
$$\sqrt{6^2+\sqrt{12}^2} =\sqrt{48}=4\sqrt{3}$$
the normalized number is
$$\frac{6-i\sqrt{12}}{4\sqrt{3}}=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$
Thus we need $\theta$ such that
$$\cos \theta =\frac{\sqrt{3}}{2}$$
and
$$\sin \theta =-\frac{1}{2}$$ from our knowledge of trig and the unit circle we see that
$\theta=\frac{\pi}{6}$. The rest you know...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/830906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is there a simpler/better proof of this simple trigonometric property? The sine function has the following nice property : for any
$x,y$, we have $\sin(x)+\sin(y)=\sin(x+y)$ iff at least one of
$x,y,x+y$ is $0$ modulo $2\pi$.
I sketch below my current proof of it, which I find somewhat
unsatisfying. Does anyone know a better proof ?
Let $\xi=\sin(x)$ and $\eta=\sin(y)$. Then
$\sin(x+y)=\xi\cos(y)+\eta\cos(x)$, so
$$
\begin{array}{lcl}
\sin^2(x+y)&=&\xi^2(1-\eta^2)+\eta^2(1-\xi^2)+2\xi\eta\cos(x)\cos(y) \\
&=& \xi^2+\eta^2-2\eta^2\xi^2+2\xi\eta\cos(x)\cos(y)
\end{array} \tag{1}$$
Whence
$$
\big(\sin^2(x+y)-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2=
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \tag{2}
$$
If $\sin(x)+\sin(y)=\sin(x+y)$, it follows then that $A(\xi,\eta)=0$
where
$$
\begin{array}{lcl}
A(\xi,\eta)&=&\big((\xi+\eta)^2-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta-2\eta^2\xi^2\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta\big(1-\eta\xi\big)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-\eta\xi\big)^2-(1-\xi^2)(1-\eta^2) \Bigg) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-2\eta\xi+\eta^2\xi^2\big)-\big(1-\xi^2-\eta^2+\eta^2\xi^2\big) \Bigg) \\
&=& 4(\xi\eta)^2 \big(\xi-\eta\big)^2 \\
\end{array} \tag{3}$$
So one of $\xi,\eta,\xi-\eta$ must be zero. A little more case analysis then shows
that in the end one of $x,y,x+y$ must be
zero modulo $2\pi$ as wished.
| Here is a different derivation of Daniel Fischer's beautiful formula,
We have the well known,
$$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$$
And
$$\sin (x+y)=2\sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x+y}{2}\right)$$
Now
$$\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)=
2\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$
so
$$\sin x+\sin y-\sin (x+y)=4\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/834321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Simple solving Skanavi book exercise: $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}$ Simple way to solve this exercise
$$
x = \sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}
$$
| You are given two nested radicals. To denest each of them, we can try to find two numbers $u,v\in\mathbb {Q} $ such that
\begin{equation*}
9+\sqrt{80}=\left( u+\sqrt{v}\right) ^{3},\qquad 9-\sqrt{80}=\left( u-\sqrt{v
}\right) ^{3}.\tag{1}
\end{equation*}
The product
\begin{equation*}
\left( u+\sqrt{v}\right) \left( u-\sqrt{v}\right) =u^{2}-v=\sqrt[3]{9+\sqrt{
80}}\sqrt[3]{9-\sqrt{80}}=1
\end{equation*}
implies that
\begin{equation*}
v=u^{2}-1,
\end{equation*}
and
\begin{equation*}
9+\sqrt{80}=\left( u+\sqrt{u^{2}-1}\right) ^{3}=4u^{3}-3u+\sqrt{\left(
u^{2}-1\right) \left( 4u^{2}-1\right) ^{2}}.\tag{2}
\end{equation*}
A solution is
\begin{cases}
4u^{3}-3u=9 \\
\left( u^{2}-1\right) \left( 4u^{2}-1\right) ^{2}=80\tag{3}
\end{cases}
Since
\begin{eqnarray*}
4u^{3}-3u-9 &=&4\left( u-\frac{3}{2}\right) \left( u^{2}+\frac{3}{2}u+\frac{3
}{2}\right),\\
( u^{2}-1) ( 4u^{2}-1) ^{2}-80
&=&16u^{6}-24u^{4}+9u^{2}-81 \\
&=&16\left( u+\frac{3}{2}\right) \left( u-\frac{3}{2}\right) \left( u^{2}-
\frac{3}{2}u+\frac{3}{2}\right) \left( u^{2}+\frac{3}{2}u+\frac{3}{2}\right)
,
\end{eqnarray*}
the single solution of $(3)$ is
\begin{equation*}
u=\frac{3}{2}.\tag{4}
\end{equation*}
Consequently, $v=u^{2}-1=\frac{5}{4}$,
\begin{equation*}
9+\sqrt{80}=\left( \frac{3}{2}+\frac{\sqrt{5}}{2}\right) ^{3},\qquad 9-\sqrt{
80}=\left( \frac{3}{2}-\frac{\sqrt{5}}{2}\right) ^{3}\tag{5}
\end{equation*}
and
\begin{eqnarray*}
x &=&\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=\frac{3}{2}+\frac{\sqrt{5}}{2}+\frac{3}{2}-\frac{\sqrt{5}}{2}=3.\tag{6}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$ If $A + B + C = \pi$,
then prove $\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 4 \cdot \sin(\frac{B - C}{2}) \cdot \sin(\frac{C - A}{2}) \cdot \sin(\frac{A - B}{2})$
My try:
$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = \sin(A - C) + \sin(B - A) + \sin(C -B)$
$\sin(A - C) + \sin(B - A) + \sin(C -B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \cos(A - \frac{C + B}{2}) + \sin(C - B)$
$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B) = 2 \cdot \sin(\frac{B -C}{2}) \cdot \sin(\frac{A}{2}) + \sin(C - B)$
I am confused what steps should follow?
| Use the product-to-sum identities on the RHS a couple of times:
\begin{align}
4\sin\frac{B-C}{2}\sin\frac{C-A}{2}\sin\frac{A-B}{2}
&= 4\cdot\frac{1}{2}\left(\cos\frac{B+A-2C}{2} - \cos\frac{B-A}{2}\right)\sin\frac{A-B}{2} \\
&= 2\left(\cos\frac{B+A-2C}{2}\sin\frac{A-B}{2} - \cos\frac{B-A}{2}\sin\frac{A-B}{2}\right) \\
&= \sin(A-C) - \sin(B-C) - \sin 0 + \sin(B-A) \\
&= \sin(A-C) + \sin(C-B) + \sin(B-A).
\end{align}
Then, as you observed in the original post, since $A+B+C=\pi$, this is equal to
$$\sin(B + 2C) + \sin(C + 2A) + \sin(A + 2B).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After simplification, we get $$\displaystyle \lim_{x\rightarrow 1}\frac{(1-x^{n+1})\cdot(1-x^{n+2})\cdot(1-x^{n+3})\cdots(1-x^{2n})}{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)}$$
Now I did not understand How can I solve after that, Help me
Thanks
| We can use the identity: $$1 - x^m = (1-x)(1 + x + \cdots+ x^{m-1})$$
in every single factor, there. All of those $1-x$ will cancel, and we'll be left with $$\lim_{x \to 1} \frac{(1+x+\cdots+ x^n)(1+x+\cdots+ x^{n+1})\cdots (1+x+\cdots +x^{2n - 1})}{(1+x)(1+x+x^2)\cdots(1+x+x^2+\cdots +x^{n-1})} = \frac{(n+1)(n+2)\cdots 2n}{1\cdot 2 \cdot 3 \cdots n}$$
To put this nicely, see that $$\frac{(n+1)(n+2)\cdots 2n}{1\cdot 2 \cdot 3 \cdots n} = \frac{1\cdot 2 \cdot 3 \cdots n \cdot (n+1)\cdot (n+2)\cdots 2n}{ 1\cdot 2 \cdot 3 \cdots n \cdot 1\cdot 2 \cdot 3 \cdots n} = \frac{(2n)!}{(n!)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$ I wish to compute the following limit
$$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$$
Wolfram Alpha gives the answer as $1/2$. I've tried L'Hopital's rule, but I can't seem to get it to work.
| Try
$$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})= \lim_{n \rightarrow \infty} \sqrt{n+1}\frac{n-1}{\sqrt{n^3 + n} + \sqrt{n^3 + 1}}=\lim_{n \rightarrow \infty} \sqrt{1+1/n}\frac{1-1/n}{\sqrt{1 + 1/n^2} + \sqrt{1 + 1/n^3}}=1/2$$
| {
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"url": "https://math.stackexchange.com/questions/840048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$ Evaluation of $\displaystyle \int \frac{\sqrt[3]{x+\sqrt[4]{x}}}{\sqrt{x}}dx$
$\bf{My\; Try::}$ Let $x=t^4\;,$ Then $dx = 4t^3dt$
So Integral is $\displaystyle \int\frac{\sqrt[3]{t^4+t}}{t^2} \cdot 4t^3dt$
So Integral is $\displaystyle 4\int t^{\frac{7}{3}}\cdot (1+t^{-3})^{\frac{1}{3}}$
Now How can i solve after that
Help me
Thanks
| Let $\mathcal{I}$ be the integral. You can actually evaluate it using the substitution $x = t^4$.
$$\mathcal{I} =
\int\frac{\sqrt[3]{x + \sqrt[4]{x}}}{\sqrt{x}}dx
= \int\frac{\sqrt[3]{(t^3 + 1)t}}{t^2}4t^3dt
= 4\int\sqrt[3]{1+t^3}t^{4/3}dt
$$
For $|x| < 1$, we can expand the integrand at RHS using following expansion
$$\frac{1}{(1-t)^\gamma} = \sum_{k=0}^\infty \frac{(\gamma)_k}{k!} t^k$$
where $(\gamma)_k = \gamma(\gamma+1)\cdots(\gamma+k-1)$ is the rising Pochhammer symbol. This gives us
$$
\mathcal{I}
= 4\int\left(\sum_{k=0}^\infty \frac{(-1)^k (-\frac13)_k}{k!}t^{3k}\right)t^{4/3}dt
= 4 \sum_{k=0}^\infty \frac{(-1)^k (-\frac13)_k}{k!}\frac{t^{3k+7/3}}{3k+7/3}
$$
Using another identity
$$\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}$$
We can transform above expression to
$$\mathcal{I}
= \frac{12}{7} t^{\frac73}
\sum_{k=0}^\infty \frac{(-\frac13)_k}{k!}\frac{(\frac79)_k}{(\frac{16}{9})_k}(-t^3)^k
$$
The expansion in the right is that for a
hypergeometric function ${}_2F_1$. As a result, up to an integration constant, we have
$$\mathcal{I}
= \frac{12}{7} t^{\frac73} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -t^3 \right)
= \frac{12}{7} x^{\frac{7}{12}} {}_2F_1\left( -\frac13, \frac79 ; \frac{16}{9}; -x^{\frac34} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/842271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x.$$
In order to find the local extrema of the function, we set $f'(x) = 0$. Therefore,
\begin{align}
4 - \frac{9\pi^2}{x^2} + \cos x &= 0 \\
4x^2 - 9\pi^2 + x^2 \cos x &= 0 \\
x^2 (4 + \cos x) &= 9\pi^2.
\end{align}
However, I'm not sure what to do from here or if, indeed, I'm doing it right at all. Any help would be appreciated.
| Let's try breaking the equation into two parts; I'll explain below.
So we have $\displaystyle f'(x)= \underbrace{4-\frac{9\pi^2}{x^2}}_{0}+\underbrace{\cos x}_{0}=0$. Note the underbraces; I set the terms equal to $0$ (so we can have $0+0=0$):
$$4-\frac{9\pi^2}{x^2}=0 \text{ and } \cos x =0$$
From $\displaystyle4-\frac{9\pi^2}{x^2}=0$, we can algebraically solve it and get $\displaystyle x=\pm\frac{3\pi}{2}$. Also, $\displaystyle \cos \left( \pm\frac{3\pi}{2} \right)=0$.
Putting it altogether, at $\displaystyle x=\pm\frac{3\pi}{2}$,
$$\left( 4-\frac{9\pi^2}{x^2} \right)+(\cos x)=0+0=0$$
which verifies.
(This isn't the best way to solve it, but it's a lot like trying to solve $x+\sin x = 0$. You can't solve it analytically, but we can easily guess $x=0$ is the solution because $0+\sin 0 = 0+0=0$.)
Back to your problem. If we have $\displaystyle x=\pm\frac{3\pi}{2}$, well we can drop the negative value and just have $\displaystyle x=\frac{3\pi}{2}$. Why? Well the domain stated in your problem, $0 < x < \infty$, allows only positive real numbers.
Now you can plug in $\displaystyle \frac{3\pi}{2}$ into your $f(x)$. You will get $f\left( \frac{3\pi}{2} \right)=12\pi - 1$ as your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/843107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$A$ is not similar to a diagonal matrix over the reals Let $A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix} $ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes.
We know that $A$ is similar to a diagonal matrix over the reals(complexes) if there exist $D$ diagonal matriz and $P$ invertible matrix both $n \times n$ with real entries (complex entries) such that $A = PDP^{-1}$
I find that the inverse of $A$ is $A^{-1} = \frac{1}{2}\begin{bmatrix} -7 & 1 & 4 \\ -8 & 2 & 4 \\ -10 & 0 & 6 \end{bmatrix}$ and i diagonalize the matrix $A$ and got $A = PBP^{-1}$ with $ P = \begin{bmatrix} 1 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 0 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 2 & 1 & 1 \end{bmatrix} $ , $ P^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -1 + 3i & 1-\frac{i}{2} & \frac{1}{2}-\frac{3i}{2} \\ -1 - 3i & 1+\frac{i}{2} & \frac{1}{2}+\frac{3i}{2} \end{bmatrix} $ and finally i got $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix}$ then A is diagonalizable but my question is how can i conclude that $A$ is not similar to a diagonalize matrix over the reals? , please some help for this.
| Suppose for contradiction that $A$ were diagonalizable over the reals.
There would be some real $D = \begin{bmatrix} \beta_1 & 0 & 0 \\ 0 & \beta_2 & 0 \\ 0 & 0 & \beta_3\end{bmatrix}$ similar to $A$.
Furthermore, the $\beta_i$ are real roots of $2-x +2 x^2-x^3$.
Thus, $\beta_1=\beta_2=\beta_3=2$
Hence $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$
As a result, what can you say about $\operatorname{Trace(A)}$ and $\operatorname{Trace(D)}$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/844724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
E(X), var(X) and std dev (X) For some reason, this question didn't show properly in my previous post.
X is given by its Distribution function:
$$
F(x) =\begin{cases}
0 &x \leq 0\\
\frac{x^2}{4} &0 < x \leq 2\\
1 &x > 2
\end{cases}
$$
I'm trying to find $E(X)$, $Var(X)$ and $Std\;dev (X)$
Thanks.
| These calculations are under the assumption that $X$ is a continuous variable and that $F(X)$ is the CDF (and not the PDF, at which point $E[X]$ is not defined due to a diverging integral). First we calculate the PDF
$$f(x) = \frac{dF(x)}{dx} = \frac{x}{2}, \quad 0<x<2$$
and $f(x) = 0$ otherwise.
The definition of the expectational value is
\begin{aligned}
E[X] & = \int_{0}^{\infty}xf(x)dx = \\
& = \int_{0}^{2}x\frac{x}{2}dx = \\
& = \frac{4}{3}.
\end{aligned}
The variance is defined as
\begin{aligned}
\text{Var}[X] & = E[X^2]-E[X]^2 = \\
& = \int_{0}^{2}x^2\frac{x^2}{2}dx - \left(\frac{4}{3}\right)^2\\
& = \frac{16}{5} - \frac{16}{9} = \\
& = \frac{64}{45}.
\end{aligned}
The variance is the standard deviation squared, which means that
\begin{aligned}
\sigma[X] = \sqrt{\text{Var}[X]} = \sqrt{\frac{64}{45}} = \frac{8}{3\sqrt{5}}.
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/844882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to solve $\frac{dy}{dx} = \frac{x^2-y^2}{x^2(y^2+1)}$ I tried to solve this using the solution of a first order differential equation but I don't think this can be reduced to that form.
How to approach this problem and find $y$?
Please help.
| Hint:
$\dfrac{dy}{dx}=\dfrac{x^2-y^2}{x^2(y^2+1)}$
$\dfrac{dx}{dy}=\dfrac{x^2(y^2+1)}{x^2-y^2}$
With reference to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=180:
Let $u=\dfrac{x}{y}$ ,
Then $x=yu$
$\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$
$\therefore y\dfrac{du}{dy}+u=\dfrac{y^2u^2(y^2+1)}{y^2u^2-y^2}$
$y\dfrac{du}{dy}=\dfrac{y^2(y^2+1)u^2}{y^2(u^2-1)}-u$
$y\dfrac{du}{dy}=\dfrac{y^2((y^2+1)u^2-u(u^2-1))}{y^2(u^2-1)}$
$\dfrac{du}{dy}=\dfrac{y((y^2+1)u^2-u(u^2-1))}{y^2(u^2-1)}$
Let $v=y^2$ ,
Then $\dfrac{du}{dy}=\dfrac{du}{dv}\dfrac{dv}{dy}=2y\dfrac{du}{dv}$
$\therefore2y\dfrac{du}{dv}=\dfrac{y((y^2+1)u^2-u(u^2-1))}{y^2(u^2-1)}$
$2\dfrac{du}{dv}=\dfrac{(v+1)u^2-u(u^2-1)}{v(u^2-1)}$
$(u^2v+u^2-u(u^2-1))\dfrac{dv}{du}=2(u^2-1)v$
$\left(v-u+1+\dfrac{1}{u}\right)\dfrac{dv}{du}=2\left(1-\dfrac{1}{u^2}\right)v$
This belongs to an Abel equation of the second kind.
Let $w=v-u+1+\dfrac{1}{u}$ ,
Then $v=w+u-1-\dfrac{1}{u}$
$\dfrac{dv}{du}=\dfrac{dw}{du}+1+\dfrac{1}{u^2}$
$\therefore w\left(\dfrac{dw}{du}+1+\dfrac{1}{u^2}\right)=2\left(1-\dfrac{1}{u^2}\right)\left(w+u-1-\dfrac{1}{u}\right)$
$w\dfrac{dw}{du}+\left(1+\dfrac{1}{u^2}\right)w=2\left(1-\dfrac{1}{u^2}\right)w+2u-2-\dfrac{4}{u}+\dfrac{2}{u^2}+\dfrac{2}{u^3}$
$w\dfrac{dw}{du}=\left(1-\dfrac{3}{u^2}\right)w+2u-2-\dfrac{4}{u}+\dfrac{2}{u^2}+\dfrac{2}{u^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/845035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$? Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$
My effort:
Since $a+b+c =1$, we can write $$ (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1 +(ab+bc+ca) - abc = (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?
| You can rewrite $(1-a)(1-b)(1-c)$ as $abc(\frac1a +\frac1b + \frac1c -1)$, meaning that your inequality is equivalent to proving that $$\frac 1a + \frac 1b + \frac 1c \geq 9,$$
an inequality that is simpler to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/845607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I need help with a double sum I need help with the following exercise:
determining the value of the next sum:
$$\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } $$
First time I face a double sum.
I have not the slightest idea how to begin to work it
| Let
$$S = \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } $$
Using the fact that
$$\sum_{m = 1}^\infty\sum_{n = 1}^\infty f(m,n) = \sum_{m = 1}^\infty\sum_{n = 1}^\infty f(n,m) $$
we can rewrite $S$ as
$$S = \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } $$
Summing up, we have
$$\begin{align}2S &= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left({{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } + { \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n{ 3 }^{ m }+m{ 3 }^{ n }) } }\right) \\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{3^nm^2n + 3^mn^2m}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn(m3^n + n3^m)}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn}{3^{m + n}}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left(\frac{m}{3^m}\cdot\frac{n}{3^n}\right)\end{align}$$
But $\sum_{m = 1}^\infty\sum_{n = 1}^\infty x_my_n = \left(\sum_{m = 1}^\infty x_m \right)\left(\sum_{n = 1}^\infty y_n \right)$. Hence,
$$\begin{align}2S &= \left(\sum_{m = 1}^\infty \frac{m}{3^m}\right)\left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)\\
&= \left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)^2\end{align}$$
To find this sum, we apply a classical trick: notice that for $|x| < 1$,
$$\frac{1}{1-x} = 1 + x + x^2 + \dots$$
Differentiating and multiplying both sides by $x$,
$$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + \dots = \sum_{n = 1}^\infty nx^n$$
Substituting $x = \frac{1}{3}$,
$$\frac{3}{4} = \sum_{n = 1}^\infty \frac{n}{3^n}$$
So we deduce that
$$2S = \left(\frac{3}{4}\right)^2$$
and finally,
$$\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } = S = \frac{9}{32}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
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