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How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}} = \frac{2^{n+1}-1}{2^{n+1}}$.
What should the approach be here?
|
If $T(n) = \frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}$, then
$$T(n+1) - T(n) = \frac{1}{2^{n+1}}$$
hence if $T(n) = \frac{2^n-1}{2^n}$, then
$$T(n+1) = T(n) + \frac{1}{2^{n+1}} = \frac{2^n-1}{2^n}+\frac{1}{2^{n+1}} = \frac{2(2^n-1)}{2^{n+1}}+\frac{1}{2^{n+1}} = \frac{2^{n+1}-2+1}{2^{n+1}},$$
giving the desired formula.
Since $T(1) = \frac{1}{2}=\frac{2-1}{2}$, the result is established by induction.
|
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"url": "https://math.stackexchange.com/questions/141126",
"timestamp": "2023-03-29T00:00:00",
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|
How to approach an integral over $g(\cos(t))$ from $0$ to $2\pi$, where $g(x)$ is nasty? For notational convenience, let $f(t) = a^2 + 2 a b \cdot \cos(t) + b^2$, where $a,b$ are both positive real constants and $t$ will be the variable of integration, which is supposed to be carried out from $t=0$ to $t=2 \pi$. I want to find an expression (or approximations) for
\begin{align}
\int\limits_0^{2\pi} \frac{\exp(-c(24m^2 - 24m \sqrt{f(t)} + 7 f(t) - 4 a b \cdot \cos(t)))}{(m^2\cdot f(t))^{1/4}} dt
\end{align}
where $c,m$ are again positive real constants.
My first attempt was to use Mathematica, but without any result. Are there maybe any tricks how to approximate the above integral? Any comments or suggestions are most welcome! Thanks in Advance.
|
Because of symmetry, we will take the integral from $t=0$ to $t=\pi$ and double it. Let $f(t)=u$.
It follows that
$$du=-2abSIN(t)dt$$
Converting the limits ($t=0$ & $t=\pi$) from t to u using the equation $u=a^2+2abCOS(t)+b^2$ we get $$lower limit=a^2+2ab+b^2$$
$$upper limit=a^2-2ab+b^2$$
Solving for cos(t), we get $$cos(t)=\frac{u-a^2-b^2}{2ab}$$
$$t=ARCCOS\left(\frac{u-a^2-b^2}{2ab}\right)$$
Substituting these into the integral, we get
$$\int\limits_0^{\pi} \frac{e^{-c\left[24m^2 - 24m \sqrt{f(t)} + 7 f(t) - 4 a b \cdot \cos(t)\right]}}{[m^2 f(t)]^{1/4}} dt=\int_{a^2+2ab+b^2}^{a^2-2ab+b^2} \frac{e^{-c\left[24m^2-24m\sqrt{u}+7u-4ab\left(\frac{u-a^2-b^2}{2ab}\right)\right]}}{-2abSIN(t)(m^2u)^{\frac{1}{4}}}du$$
Because we are integrating from $0$ to $\pi$ we can use $$sin(t)=\sqrt{1-cos^2(t)}$$
We get
$$\int_{a^2+2ab+b^2}^{a^2-2ab+b^2} \frac{e^{-c\left[24m^2-24m\sqrt{u}+7u-4ab\left(\frac{u-a^2-b^2}{2ab}\right)\right]}}{-2ab\sqrt{1-\left(\frac{u-a^2-b^2}{2ab}\right)^2}m^{\frac 12}u^{\frac{1}{4}}}du$$
$$=\int_{a^2-2ab+b^2}^{a^2+2ab+b^2} \frac{e^{-c\left[24m^2-24m\sqrt{u}+5u+2(a^2+b^2)\right]}}{2ab\sqrt{1-\left(\frac{u-a^2-b^2}{2ab}\right)^2} m^{\frac 12}u^{\frac{1}{4}}}du$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\int{\frac{\sqrt{x^2+1}}{x+2}dx}$ I have got such integral $$\int{\frac{\sqrt{x^2+1}}{x+2}dx}$$ and with Maple I got something like this:
$$\int\frac{1}{2} + \frac{1+3u^2+4u^3}{-2u^2+2u^4-8u^3}du$$ And I want to know how to achive this changes.
I tried to use WolframAlpha, but there is scarier solution. This integral was for Gaussian quadrature method, so it's analytic solution is horrible.
|
*
*We can use the Euler substitution $t=\sqrt{x^{2}+1}-x$ to obtain a rational fraction in terms of $t$
$$\begin{eqnarray*}
I =\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }\mathrm{d}t.
\end{eqnarray*}$$
*Since the integrand is a rational fraction, we can expand it into partial fractions and integrate each fraction.
$$\begin{equation*}
\frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }=1-\frac{1}{t^{2}}+
\frac{4}{t}+\frac{20}{t^{2}-4t-1}.
\end{equation*}$$
Added. Detailed evaluation. From $t=\sqrt{x^{2}+1}-x$, we get $x=\dfrac{1-t^{2}}{2t}$ and $\dfrac{dx}{dt}=-\dfrac{t^{2}+1}{2t^{2}}$. So we have
$$\begin{eqnarray*}
I &=&\int \frac{\sqrt{x^{2}+1}}{x+2}\mathrm{d}x=\int \frac{t+\frac{1-t^{2}}{2t}}{\frac{1-t^{2}}{2t}+2}\left( -\frac{t^{2}+1}{2t^{2}}\right) \mathrm{d}t
\\
&=&\frac{1}{2}\int \frac{1+2t^{2}+t^{4}}{t^{2}\left( -1+t^{2}-4t\right) }
\mathrm{d}t.
\end{eqnarray*}$$
Expanding into partial fractions as above, we obtain
$$
\begin{eqnarray*}
2I &=&\int 1-\frac{1}{t^{2}}+\frac{4}{t}+\frac{20}{t^{2}-4t-1}\mathrm{d}t \\
&=&\int 1\mathrm{d}t-\int \frac{1}{t^{2}}\mathrm{d}t+4\int \frac{1}{t}
\mathrm{d}t+20\int \frac{1}{t^{2}-4t-1}\mathrm{d}t \\
&=&t+\frac{1}{t}+4\ln \left\vert t\right\vert -2\sqrt{5}\ln \frac{\sqrt{5}t-2
\sqrt{5}+5}{5-\sqrt{5}t+2\sqrt{5}}+C \\
&=&\sqrt{x^{2}+1}-x+\frac{1}{\sqrt{x^{2}+1}-x}+4\ln \left( \sqrt{x^{2}+1}
-x\right) \\
&&-2\sqrt{5}\ln \frac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C.
\end{eqnarray*}$$
Therefore the given integral is
$$\begin{eqnarray*}
I &=&\frac{1}{2}\left( \sqrt{x^{2}+1}-x\right) +\frac{1}{2}\frac{1}{\sqrt{
x^{2}+1}-x}+2\ln \left( \sqrt{x^{2}+1}-x\right) \\
&&-\sqrt{5}\ln \frac{\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) -2\sqrt{5}+5}{5-\sqrt{5}\left( \sqrt{x^{2}+1}-x\right) +2\sqrt{5}}+C.
\end{eqnarray*}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate square root of $i \Leftrightarrow z^2=i$ Let $z = r(\cos\theta+i\sin\theta)$. In my notes there was this example to calculate the square roots of $i$. What was done was:
$z = r(\cos\theta+i\sin\theta)\\z^2 = r^2(\cos(2\theta)+i\sin(2\theta))\\=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$
Im not entirely understand what they did above, what does the 2 lines below actually tell us? $\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$
I thought to calculate the square roots of iyou let $z=x+iy$ and work out $x,y$ from $(x+iy)^2=i$?
Also, how did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?
|
How did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?
This whole problem is easiest to do via Euler's identity: $$e^{i\theta} = \cos(\theta) + i\sin(\theta).$$ Now, any complex number $z$ can be
expressed as $z=re^{i\theta}$ where $r$ is some real
number and $0 \leq \theta < 2\pi$. It follow that
$z^2 = r^2e^{i2\theta} = r^2(\cos(2\theta)+i\sin(2\theta))$, and so if we want to solve the equation $z^2 = i$, then we have
$$z^2 = r^2e^{i2\theta} = r^2(\cos(2\theta)+i\sin(2\theta)) = i.$$
Now, $r^2\cos(2\theta)+ir^2\sin(2\theta) = 0 + i$ means that
$r^2\cos(2\theta) = 0$ and $r^2\sin(2\theta) = 1$. The latter equation
shows us that $r \neq 0$ and hence it must be that
$\cos(2\theta) = 0$, giving $2\theta = \pi/2$ and so $\sin(2\theta) = 1$. Substituting $\sin(2\theta) = 1$ into
the latter equation
reduces it to $r^2 = 1$, and so $r = \pm 1$.
Also, since $2\theta = \pi/2$, we get $\theta = \pi/4$.
Hence, the two square roots of $i$ are
$$i^{1/2} = \pm\left(\cos(\pi/4) + i\sin(\pi(4)\right) = \pm \frac{1+i}{\sqrt{2}}$$
|
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|
How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?
|
If $n\equiv 0\pmod 3$ Ok. If $n\equiv 1\pmod 3$, we have
\begin{equation}
n^{2} + 8 \equiv 1^{2} + 2\equiv 0\pmod 3.
\end{equation}
If $n \equiv 2\pmod 3$ we have
\begin{equation}
n^{2} + 8 \equiv 2^{2} + 2\equiv 6 \equiv 0\pmod 3.
\end{equation}
|
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|
Solving $|x-2| + |x-5|=3$
Possible Duplicate:
How could we solve $x$, in $|x+1|-|1-x|=2$?
How should I solve:
$|x-2| + |x-5|=3$
Please suggest a way that I could use in other problems of this genre too
Any help to solve this problem would be greatly appreciated.
Thank you,
|
Well, there are a couple of ways.
Method 1. By cases.
One is to consider cases: note that
$$\begin{align*}
|x-2|&=\left\{\begin{array}{ll}
x-2 & \text{if }x\geq 2\\
2-x &\text{if }x\lt 2
\end{array}\right.\\
|x-5|&=\left\{\begin{array}{ll}
x-5 &\text{if }x\geq 5\\
5-x &\text{if }x\lt 5
\end{array}\right.
\end{align*}$$
So, you consider what happens if $x\geq 5$, if $2\leq x\lt 5$, and if $x\lt 2$. In the first case, you have
$$x-2+x-5 = 3$$
which is the same as $2x-7 = 3$, or $2x=10$, or $x=5$. Since this satisfies $x\geq 5$, that is one solution.
If $2\leq x \lt 5$, then you get $$x-2 + 5-x = 3$$
which is always true. So all numbers between $2$ and $5$ work (check and see this is true).
And if $x\lt 2$, you get
$$2-x + 5-x = 3$$
which is the same as $7-2x = 3$, or $2x=4$; that is, $x=2$. But $x=2$ does not satisfy $x\lt 2$, so there are no solutions here.
So the solution is that $x$ satisfies the equation if and only if $2\leq x\leq 5$.
Method 2. The absolute value is a measure of distance. $|x-2|$ is how far $x$ is from $2$, and $|x-5|$ is how far $x$ is from $5$. you are trying to find all numbers whose distance from $2$ plus their distance from $5$ equal $3$. Note that no number greater than $5$ can work, because then their distance to $2$ is already greater than $3$. No number smaller than $2$ can work because their distance to $5$ is already greater than $3$. And any number between $2$ and $5$, inclusively, will work, because if $2\leq x \leq 5$, then adding the distance from $x$ to $2$ and from $x$ to $5$ will necessarily add up to $3$. So the answer is that $x$ satisfies the equation if and only if $2\leq x \leq 5$.
|
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|
Where is $f(x)={(x^2+2x-48)}/{x^2}$ increasing? decreasing? The question is to find where the graph is increasing or decreasing.
The original function is $f(x)={(x^2+2x-48)}/{x^2}$
I know I need to find the prime of this function and I think it is this after using the quotient rule:
$2(-x^2-x+49)/x^3$
Finally, in order to draw a graph I need to find the points on the x axis that are either undefined and/or the slope equals 0.
Immediately I know that 0 is an undefined point because it makes the prime graph undefined.
It's when I take the numerator and set it equal to 0 that trips me up.
$2(-x^2-x+49)=0$
$-x^2-x+49=0$
$(x)(x)=0$
I'm guessing I did something wrong finding this prime because the answers are all including the number 48. I'm stumped because I've been at this question for about 1 hour.
Thanks!
|
You computation of derivative is incorrect. It is easier to separate out each term and compute the derivative.
$$f(x) = \dfrac{x^2 + 2x - 48}{x^2} = 1 + \dfrac2{x} - \dfrac{48}{x^2}$$
Hence, $$f'(x) = 0 - \dfrac2{x^2} + \dfrac{2 \times 48}{x^3} = \dfrac{96-2x}{x^3}$$
Now setting, $f'(x) = 0$ gives us $96-2x = 0 \implies x = 48$.
EDIT
As FrankScience rightly points out in his comment, you do not really need to differentiate to figure out the behavior of this function. We have that
\begin{align}
f(x) & = \dfrac{x^2 + 2x - 48}{x^2}\\
& = -48 \left( \left(\dfrac1x \right)^2 - \dfrac1{24} \dfrac1x\right) + 1\\
& = -48 \left( \left(\dfrac1x \right)^2 - 2 \dfrac1x\dfrac1{48} + \left(\dfrac1{48} \right)^2 - \left(\dfrac1{48} \right)^2\right) + 1\\
& = -48 \left( \dfrac1x - \dfrac1{48}\right)^2 + \dfrac1{48} + 1\\
& = \dfrac{49}{48} -48 \left( \dfrac1x - \dfrac1{48}\right)^2
\end{align}
Hence, note that the functions hits the maximum at $x=48$, since for all $x \neq 48$, you always subtract a positive quantity from $\dfrac{49}{48}$.
Analyze what happens when $x \to \infty$, $x \to - \infty$, $x \to 0^{\pm}$, $x \to 48^{\pm}$ to draw conclusions.
|
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|
Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) = a^2 + b^2$$
$$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$
Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought?
|
I find it easier to go from polar to rectangular.
$r=2a\cos\theta+2b\sin\theta$; $r^2=2ar\cos\theta+2br\sin\theta$; $x^2+y^2=2ax+2by$; $x^2-2ax+y^2-2by=0$; $(x-a)^2+(y-b)^2=a^2+b^2$; voila, circle of radius $\sqrt{a^2+b^2}$ centered at $(a,b)$.
|
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|
Prove that the product of four consecutive positive integers plus one is a perfect square I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.
If $n = 1 + m$, where $m$ is the product of four consecutive positive
integers, prove that $n$ is a perfect square.
Now since $m = p(p+1)(p+2)(p+3)$;
$p = 0, n = 1$ - Perfect Square
$p = 1, n = 25$ - Perfect Square
$p = 2, n = 121$ - Perfect Square
Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?
|
Product of 4 consecutive numbers can be shown as
$$\begin{align}(x+1)(x+2)(x+3)(x+4)
&=(x+1)(x+4)(x+2)(x+3) \\
&=(x^2+5x+4) (x^2+5x+6) \\
&=\underset{A-B}{(x^2+5x+5 -1)} \underset{A+B}{(x^2+5x+5 +1)} \\
&=\underset{A^2-B^2}{(x^2+5x+5)^2 - 1}\end{align}$$
Where $(x^2+5x+5)^2$ is a perfect square.
Proved.
|
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|
The Fibonacci sum $\sum_{n=0}^\infty \frac{1}{F_{2^n}}$ generalized The evaluation,
$$\sum_{n=0}^\infty \frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}=\left(\frac{1-\sqrt{5}}{2}\right)^3+\left(\frac{1+\sqrt{5}}{2}\right)^2$$
was recently asked in a post by Chris here.
I like generalizations, and it turns out this is not a unique feature of the Fibonacci numbers. If we use the Pell numbers $P_m = 1,2,5,12,29,70,\dots$ then the sum is also an algebraic number of deg 2. In general, it seems for any positive rational b, then,
$$\sum_{n=0}^\infty \frac{1}{\frac{1}{\sqrt{b^2+4}}\left( \left(\frac{b+\sqrt{b^2+4}}{2}\right)^{2^n}-\left(\frac{b-\sqrt{b^2+4}}{2}\right)^{2^n}\right)}=1+\frac{2}{b}+\frac{b-\sqrt{b^2+4}}{2}$$
where Fibonacci numbers are just the case b = 1, the Pell numbers b = 2, and so on. (For negative rational b, then one just uses the positive case of $\pm\sqrt{b^2+4}$.)
Anyone knows how to prove/disprove the conjectured evaluation?
|
Your conjecture is indeed right. Before proving your conjecture, let us obtain an intermediate result first. Let us prove the following claim first.
CLAIM:
If we have a sequence given by the recurrence,
$$a_{n+2} = ba_{n+1} + a_n,$$ with $a_0 =0 $ and $a_1 = 1$, we then have
$$\boxed{\color{blue}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
Proof:
Let us write out a few terms of this sequence, we get
$$a_0 = 0, a_1 = 1, a_2 = b, a_3 = b^2 + 1, a_4 = b^3 + 2b, \cdots$$
The proof is by induction on $N$. For $N=1$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} = 1 + \dfrac1b$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_1}{a_2} = 1 + \dfrac2b - \dfrac1{b} = 1 + \dfrac1b$$
For $N=2$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} + \dfrac1{a_4} = 1 + \dfrac1b + \dfrac1{b^3 + 2b}$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_3}{a_4} = 1 + \dfrac2b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1{b^3+2b}$$ Hence, it holds for $N=1$ and $N=2$. Now lets go ahead with induction now. Assume the result is true for $N=m$ i.e. we have
$$\sum_{k=0}^{m} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}}$$
Now $$\sum_{k=0}^{m+1} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}}$$ Hence, we want to show that
$$ - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}} = -\dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}}$$
i.e.
$$\dfrac1{a_{2^{m+1}}} + \dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}} = \dfrac{a_{2^m-1}}{a_{2^m}}$$
i.e.
$$a_{2^m}(1+a_{2^{m+1}-1}) = a_{2^m-1} a_{2^{m+1}} \,\,\,\, (\star)$$
which can be verified using the recurrence. In fact $(\dagger)$, a slightly more general version of $(\star)$, which is easier to check is true.
$$a_{2k}(1+a_{4k-1}) = a_{2k-1} a_{4k} \,\,\,\, (\dagger)$$ i.e.
$$a_{2k-1} a_{4k} - a_{2k} a_{4k-1} = a_{2k} \,\,\,\, (\dagger)$$
Hence, we get that
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
Now letting $N \to \infty$, we see that your conjecture is indeed right. This is so since
from the recurrence we get that
$$\dfrac{a_{n+2}}{a_{n+1}} = b + \dfrac{a_n}{a_{n+1}}$$
If we have $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{a_{n+1}} = L$, then we get that
$$\dfrac1L = b + L$$ and since $L>0$, we have $L = \dfrac{\sqrt{b^2+4}-b}2$. Hence,
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = \lim_{N \to \infty} \displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \lim_{N \to \infty} \dfrac{a_{2^N-1}}{a_{2^N}} = 1 + \dfrac2b - L = 1 + \dfrac2b + \dfrac{b}2 -\dfrac{\sqrt{b^2+4}}2}}$$
EDIT
After some googling, I found out that a similar result is true for a more general class of recurrences of the form $$a_{n+1} = P a_n + Q a_{n-1}$$ See this article for more details.
Also, try googling Millin series for more details.
|
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|
$1 +1$ is $0$ ?
Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?
So:
$$
\begin{align}
1+1 &= 1 + \sqrt{1} \\
&= 1 + \sqrt{1 \times 1} \\
&= 1 + \sqrt{-1 \times -1} \\
&= 1 + \sqrt{-1} \times \sqrt{-1} \\
&= 1 + i \times i \\
&= 1 + (-1) \\
&= 1 - 1\\
&= 0
\end{align}
$$
I can't see anything wrong there, and I can't see anything wrong in $1+1=2$ too. Clearly, $1+1$ is $2$, but I really want to know where is the incorrect part in the above.
|
We have that $\sqrt{-1} \times \sqrt{-1} = (\sqrt{-1})^{2} = -1$ but $\sqrt{-1 \times -1} = \sqrt{1} = 1$. So $\sqrt{-1 \times -1} \neq \sqrt{-1} \times \sqrt{-1}$ which is the error.
|
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|
How to derive ellipse matrix for general ellipse in homogenous coordinates So lets say we have an ellipse with axes a and b and the rotation angle $\phi$ and center at $(0,0)$.
Now I apply the rotation to $x^2/a^2+y^2/b^2=1$ getting
$$x' = x\cos(\phi) + y\sin(\phi)$$
$$y' = y\cos(\phi) + x\sin(\phi)$$
$$x^2(b^2cos(\phi)^2 + a^2sin(\phi)^2) + y^2(b^2\sin(\phi)^2 + a^2\cos(\phi)^2) + (b^2-a^2)\sin(\phi)\cos(\phi)xy - a^2b^2$$
This is some kind of quadratic form but I need to derive the quadratic form which I'll be able to convert to ellipse matrix.
What are the next steps to do this?
Thanks
|
In general, an ellipse in a general position $[h,k]$ (what I needed) is implicitly given as
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
where $a$ and $b$ are semiaxes. Rotating the $x$ and $y$ coords yields (you're missing a $-$ sign in the first line)
$$x' = x\cos(\phi) - y\sin(\phi)$$
$$y' = x\sin(\phi) + y\cos(\phi)$$
Plugging this into ellipse equation, you get
$$\frac{([x\cos(\phi) - y\sin(\phi)]-h)^2}{a^2} + \frac{([x\sin(\phi) + y\cos(\phi)]-k)^2}{b^2} = 1$$
Now, you will need this in a $Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0$ form, as the ellipse matrix constitutes of the coefficients $A,B,C,D,E,$ and $F$. So you rewrite the previous, getting
$$-(a^2 b^2) + b^2 h^2 + a^2 k^2 - 2 b^2 h x \cos(\phi) - 2 a^2 k y \cos(\phi) + b^2 x^2 \cos(\phi)^2 + a^2 y^2 \cos(\phi)^2 - 2 a^2 k x \sin(\phi) + 2 b^2 h y \sin(\phi) + 2 a^2 x y \cos(\phi) \sin(\phi) - 2 b^2 x y \cos(\phi) \sin(\phi) + a^2 x^2 \sin(\phi)^2 + b^2 y^2 \sin(\phi)^2 = 1$$
The coefficients therefore are
$$\begin{array}
AA &=& b^2 \cos^2(\phi) + a^2 \sin^2(\phi) \\
B &=& a^2 \cos(\phi) \sin(\phi) - b^2 \cos(\phi) \sin(\phi)\\
C &=& a^2 \cos^2(\phi) + b^2 \sin^2(\phi)\\
D &=& -b^2 h \cos(\phi) - a^2 k \sin(\phi)\\
E &=& -a^2 k \cos(\phi) + b^2 h \sin(\phi)\\
F &=& -a^2 b^2 + b^2 h^2 + a^2 k^2\end{array}$$
and the final ellipse matrix
$$M = \begin{bmatrix}A & B & D \\
B & C & E \\
D & E & F
\end{bmatrix}$$
By multiplying $\begin{bmatrix}x&y&1\end{bmatrix}M\begin{bmatrix}x&y&1\end{bmatrix}^T$, you get the original $Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = 0$ equation, hence the matrix form.
|
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|
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?
|
One possible proof is to use the fact that for triangular $n\times n$ matrix $A$ it holds
$$\det(A) = \prod_{i=1}^n a_{i,i}$$
An example
$$\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
0 & \frac{3}{4} & \frac{1}{4} \\
0 & \frac{1}{4} & \frac{3}{4} \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
0 & \frac{3}{4} & \frac{1}{4} \\
0 & 0 & \frac{2}{3} \\
\end{pmatrix}
$$
Determinant of the last form is clearly positive. Try to generalize this result.
|
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|
Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$ I don't know how to find an explicit form for this sum, anyone can help me?
$$\sum_{k=-\infty}^{\infty}
{1 \over \left\vert\,x - k\,x_{\atop{ \small 0}}\,\right\vert}
$$
Here are the calculations I made, but don't bring me anywhere:
(original image)
$$\begin{align}\sum_{k=-\infty}^\infty\frac{1}{|x-kx_0|}&=\frac{1}{x}+\sum_{k=-\infty}^{-1}\frac{1}{|x-kx_0|}+\sum_{k=1}^\infty\frac{1}{|x-kx_0|}\\\\ & =\frac{1}{x}+\sum_{k'=1}^{\infty}\frac{1}{|x+k'x_0|}+\sum_{k=1}^\infty\frac{1}{|x-kx_0|}\end{align}$$
If $x\neq 0$,
$$=\frac{1}{x}+\sum_{x+k'x_0>0}\frac{1}{x+k'x_0}+\sum_{x+k'x_0<0}\frac{-1}{x+k'x_0}+\sum_{x-kx_0>0}\frac{1}{x-kx_0}+\sum_{x-kx_0<0}\frac{1}{kx_0-x}$$
$$=\frac{1}{x}+\sum_{k>-\frac{x}{x_0}}\frac{1}{x+kx_0}-\sum_{k<-\frac{x}{x_0}}\frac{1}{x+kx_0}+\sum_{k<\frac{x}{x_0}}\frac{1}{x-kx_0}+\sum_{k>\frac{x}{x_0}}\frac{1}{kx_0-x}$$
$$=\frac{1}{x}+\frac{1}{x_0}\left[\sum_{k>-\frac{x}{x_0}}\frac{1}{k+\frac{x}{x_0}}-\sum_{k<-\frac{x}{x_0}}\frac{1}{k+\frac{x}{x_0}}+\sum_{k<\frac{x}{x_0}}\frac{1}{-k+\frac{x}{x_0}}+\sum_{k>\frac{x}{x_0}}\frac{1}{k-\frac{x}{x_0}}\right]$$
and my prof's version (I'm not sure he could be so quick on the absolute value)
(original image)
$$\sum_{n=-\infty}^{+\infty}\frac{1}{|x-nx_0|}=\frac{1}{x}+\sum_{n=-\infty}^{-1}\frac{1}{|x-nx_0|}+\sum_{n=1}^\infty\frac{1}{|x-nx_0|}$$
$$=\frac{1}{x}+\sum_{m=+\infty}^{+1}\underbrace{\frac{1}{x-nx_0}}_{\substack{\text{change variable }m=-n,\\\\ \Large \frac{1}{x+mx_0}}}+\sum_{n=1}^\infty\frac{1}{nx_0-x}$$
$$=\sum_{n=1}^{+\infty}\underbrace{\frac{1}{nx_0-x}-\frac{1}{x+nx_0}}_{\Large\frac{x+nx_0-nx_0+x}{n^2x_0^2-x^2}}$$
$$\frac{1}{x}+2x\sum_{n=1}^{+\infty}\frac{1}{n^2x_0^2-x^2}$$
Thanks!
|
As noted by some users, the series below is the one for the case
$$\sum_{k=-\infty}^{+\infty}\frac{1}{z-k}$$
i.e, there are no absolute values.
I scanned too fast but the last thing you have is this
$$\pi \cot(\pi z)=\frac 1 z+2z \sum_{n=1}^\infty \frac{1}{z^2-n^2} $$
One option is to use
$$\frac{{\sin \pi z}}{{\pi z}} = \prod\limits_{k = 1}^\infty {\left( {1 - \frac{{{z^2}}}{{{k^2}}}} \right)} $$
Take logarithms and differentiate:
$$\eqalign{
& \log \sin \pi z - \log \pi z = \log \prod\limits_{k = 1}^\infty {\left( {1 - \frac{{{z^2}}}{{{k^2}}}} \right)} \cr
& \log \sin \pi z - \log \pi z = \sum\limits_{k = 1}^\infty {\log \left( {1 - \frac{{{z^2}}}{{{k^2}}}} \right)} \cr
& \pi \frac{{\cos \pi z}}{{\sin \pi z}} - \frac{1}{z} = \sum\limits_{k = 1}^\infty {\frac{{ - \frac{{2z}}{{{k^2}}}}}{{1 - \frac{{{z^2}}}{{{k^2}}}}}} \cr
& \pi \cot \pi z = \frac{1}{z} - 2z\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2} - {z^2}}}} \cr} $$
$$\pi \cot \pi z = \frac{1}{z} + 2z\sum\limits_{k = 1}^\infty {\frac{1}{{{z^2} - {k^2}}}} $$
I know virtually nothing about complex analysis, but what you start with is the decomposition of the cotangent into partial fractions in complex analysis, pretty much like one decomposes a polynomial with its roots, one does it with the singularities here.
|
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|
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
|
Expand $\frac{1}{1+x^{4}}$ into partial fractions. For this purpose you need
to factorize the polynomial in the denominator. You can proceed by writing it as a product of four linear terms
\begin{equation*}
x^{4}+1=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})
\end{equation*}
where $x_{1},x_{2},x_{3},x_{4}$ are its complex roots. Since
\begin{equation*}
x^{4}+1=0\Leftrightarrow x^{4}=-1=\cos \pi +i\sin \pi
\end{equation*}
the four roots are
\begin{eqnarray*}
x_{1} &=&\cos \left( \frac{\pi }{4}\right) +i\sin \left( \frac{\pi }{4}
\right) =\frac{\sqrt{2}}{2}(1+i) \\
x_{2} &=&\cos \left( \frac{3\pi }{4}\right) +i\sin \left( \frac{3\pi }{4}
\right) =\frac{\sqrt{2}}{2}(-1+i) \\
x_{3} &=&\overline{x}_{2} \\
x_{4} &=&\overline{x}_{1}.
\end{eqnarray*}
Rewrite $x^{4}+1$ as a product of quadratic terms, by grouping the factors $
(x-x_{1}),(x-x_{4})=(x-\overline{x}_{1})$ and $(x-x_{2}),(x-x_{3})=(x-
\overline{x}_{2})$
\begin{eqnarray*}
x^{4}+1 &=&\left[ (x-x_{1})(x-\overline{x}_{1})\right] \left[ (x-x_{2})(x-
\overline{x}_{2})\right] \\
&=&\left( x^{2}-\sqrt{2}+1\right) \left( x^{2}+\sqrt{2}x+1\right)
\end{eqnarray*}
Find the constants $A,B,C,D$ such that
\begin{equation*}
\frac{1}{\left( x^{2}-x\sqrt{2}+1\right) \left( x^{2}+x\sqrt{2}+1\right) }=
\frac{A+Bx}{x^{2}-x\sqrt{2}+1}+\frac{C+Dx}{x^{2}+\sqrt{2}x+1}.
\end{equation*}
To evaluate
\begin{equation*}
\int \frac{1}{x^{2}\mp \sqrt{2}x+1}dx
\end{equation*}
complete the square in the denominator (see this answer of mine)
\begin{equation*}
x^{2}\mp \sqrt{2}x+1=\left( x\mp \frac{\sqrt{2}}{2}\right) ^{2}+\left( \frac{\sqrt{
2}}{2}\right) ^{2}
\end{equation*}
and make the substitutions
\begin{equation*}
x\mp \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}t.
\end{equation*}
To compute the remaining integrals rewrite them as
\begin{eqnarray*}
\int \frac{x}{x^{2}\mp \sqrt{2}x+1}dx &=&\frac{1}{2}\int \frac{2x\mp \sqrt{2}}{
x^{2}\mp \sqrt{2}x+1}dx\pm\frac{\sqrt{2}}{2}\int \frac{1}{x^{2}\mp \sqrt{2}x+1}dx.
\end{eqnarray*}
|
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|
What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$
Here is my attempt:
Letting
$$f(z)=\frac{1}{az^2+bz+c}$$
The poles of $f(z)$ are located at
$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$
Then
$$
b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)=
\lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}=
\lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}
$$
Using L'Hopital's rule. Continuing, we have the limit is
$$
\lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}=
\frac{\pi\cot (\pi z_0)}{2az_0+b}
$$
For $z_0 \ne 0$
Similarly, we find
$$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$
Then
$$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\
-\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=
-\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)=
\frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}=
\frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}}
$$
Then we have
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$
Is this correct? I feel like I made a mistake somewhere. Could someone correct me? Is there an easier way to evaluate this sum?
|
Take $a=1,\ b=3, \ c=2$, then $z_0=-2, \ z_1=-1$, and so you have to compute $\cot(-\pi)$ and $\cot(-2\pi)$ which make no sense. However
$$
\sum_{n=0}^\infty\frac{1}{n^2+3n+2}=\sum_{n=0}^\infty(\frac{1}{n+1}-\frac{1}{n+2})
=\lim_{m\to \infty}(1-\frac{1}{m+2})=1.
$$
|
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|
Whats the sum of the length of all the sides of a triangle? You are given triangles with integer sides and one angle fixed at 120 degrees. If the length of the longest side is 28 and product of the remaining to sides is 240, what is the sum of all sides of the triangle?
I have tried to solve it using the formula given in the following link about integer triangle with 120 link
Integer triangles with a 120° angle can be generated by
|
Since the longest side must be opposite the largest angle in the triangle, the side which is 28 units is opposite the angle of degree measure of 120. Let us name the remaining sides $a$ and $b$. Using the cosine rule we get:
$$28^2=a^2+b^2-2ab\cos(120)$$
$$784=a^2+b^2-2ab\left(\frac{-1}{2}\right)$$
$$784=a^2+b^2+ab$$
Since $ab=240$
$$a^2+b^2=544$$
We thus have the following system of equations to solve:
$$\begin{align*}a^2+b^2&=544 \\ ab&=240\end{align*}$$
Solving for $b$ in the latter equation
$$b=\frac{240}{a}$$
Plugging into the first one gives:
$$a^2+\left(\frac{240}{a}\right)^2=544$$
Multiplying both sides by $a^2$ and letting $u=a^2$
$$u^2-544u+240^2=0$$
Using the quadratic equation
$$\begin{align*}u&=\frac{544\pm\sqrt{(-544)^2-4\cdot240^2}}{2} \\
u&=\frac{544\pm256}{2} \\
u&=144 \text{ or }400\end{align*}$$
But since $u=a^2$,
$$a^2=144 \text{ or } a^2=400$$
$$a=12 \text{ or } a=20$$
Using $b=\dfrac{240}{a}$ we thus get $a=12$ and $b=20$ or $a=20$ and $b=12$. In either case the sum of the sides is 60.
Edit: lhf pointed out a much quicker alternative:
When we came to the following stage
$$784=a^2+b^2+ab$$
Rather than solve for $a$ and $b$ etc. simply add $ab$ to both sides to get
$$784+ab=a^2+b^2+2ab$$
Since $a^2+b^2+2ab$ is a perfect square and $ab=240$
$$1024=(a+b)^2$$
$$a+b=32$$
Therefore the sum of all the sides is $a+b+28=32+28=60$
|
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|
Error Term in Passing from Summation to Integral I encountered the following in a paper and do not understand how the error term is being bounded. In what follows, $n$ and $k$ are large integer constants.
$$ \sum_{i=0}^{n-1} \ln\left(1 - \frac{i}{k}\right) = \int_0^n
\ln\left(1 - \frac{x}{k}\right) dx \pm e(n,k) $$ where the error term
$e(n,k)$ can be bounded by $$ e(n,k) \leq \int_0^n \ln\left(1 -
\frac{x}{k}\right) - \ln\left(1 - \frac{x+1}{k}\right) dx $$
|
Note that $\ln \left( 1- \dfrac{x}k \right)$ is a decreasing function of $x$ for all $x < k$. Hence, $$\int_a^{a+1} \ln \left( 1 -\dfrac{x}k \right) dx \leq \ln \left( 1 - \dfrac{a}k\right) \leq \int_{a-1}^{a} \ln \left( 1 -\dfrac{x}k \right) dx $$
Hence, taking $a=0$ to $a=n-1$ and adding them up, we get that
$$\int_0^{n} \ln \left( 1 -\dfrac{x}k \right) dx \leq \sum_{a=0}^{n-1}\ln \left( 1 - \dfrac{a}k\right) \leq \int_{-1}^{n-1} \ln \left( 1 -\dfrac{x}k \right) dx$$
Hence, we get that
$$0 \leq \underbrace{\sum_{a=0}^{n-1}\ln \left( 1 - \dfrac{a}k\right) - \int_0^{n} \ln \left( 1 -\dfrac{x}k \right) dx}_{e(n,k)} \leq \int_{-1}^{n-1} \ln \left( 1 -\dfrac{x}k \right) dx - \int_0^{n} \ln \left( 1 -\dfrac{x}k \right) dx$$
Hence, we get that $$\vert e(n,k) \vert \leq \int_0^{n} \left( \ln \left( 1 - \dfrac{x-1}k\right) - \ln \left( 1 - \dfrac{x}k\right)\right) dx$$
which is a slightly better bound than what you want. If you want the bound you have mentioned, note that $$\left( \ln \left( 1 - \dfrac{x-1}k\right) - \ln \left( 1 - \dfrac{x}k\right)\right)$$ is an increasing function of $x$ for $x < k+1$. Hence,
$$\int_0^{n} \left( \ln \left( 1 - \dfrac{x-1}k\right) - \ln \left( 1 - \dfrac{x}k\right)\right) dx \leq \int_1^{n+1} \left( \ln \left( 1 - \dfrac{x-1}k\right) - \ln \left( 1 - \dfrac{x}k\right)\right) dx\\ = \int_0^{n} \left( \ln \left( 1 - \dfrac{x}k\right) - \ln \left( 1 - \dfrac{x+1}k\right)\right) dx$$
Hence, $$\vert e(n,k) \vert \leq \int_0^{n} \left( \ln \left( 1 - \dfrac{x-1}k\right) - \ln \left( 1 - \dfrac{x}k\right)\right) dx \leq \int_0^{n} \left( \ln \left( 1 - \dfrac{x}k\right) - \ln \left( 1 - \dfrac{x+1}k\right)\right) dx$$
|
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|
Polynomials identity, factoring $x^{2^n}-1$ There is a proof that I can't solve.
Show that for any integer $k$, the following identity holds:
$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{k-1}})=1+x+x^2+x^3+\cdots+x^{2^k-1}$$
Thanks for your help.
|
We work with the left-hand side.
$$(1+x)(1+x^2)=(1+x)(1)+(1+x)(x^2)=1+x+x^2+x^3.$$
Thus
$$\begin{align}
(1+x)(1+x^2)(1+x^4)&=(1+x+x^2+x^3)(1+x^4)\\
&= (1+x+x^2+x^3)(1)+(1+x+x^2+x^3)(x^4)\\
&=1+x+x^2+x^3+x^4+x^5+x^6+x^7\end{align}.$$
Continue. We will be multiplying $1+x+\cdots+x^7$ by $1+x^8$. Multiplying $1+x+\cdots+x^7$ by $x^8$ gets us $x^8+x^9+\cdots+x^{15}$, so the full product is $1+x+\cdots+x^{15}$
The pattern is obvious. If we wish, we can use a formal induction argument.
|
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|
linear algebra: inverse of a matrix The inverse of the matrix
$A=\left( \matrix{1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} }\right)$
is
$A^{-1}=\left( \matrix{ 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 } \right)$.
Then, perhaps the matrix
$B=\left( \matrix{1 & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n+1} \\ ... & ... & & ...\\ \frac{1}{n} & \frac{1}{n+1} & ... & \frac{1}{2n-1} }\right)$
is invertible and $B^{-1}$ has integer entries. How can I prove it?
|
Here you will find your answer and many other things about Hilbert matrices : http://www.jstor.org/stable/2975779
|
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|
Squares of the form $x^2+y^2+xy$ How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?
Thanks !
|
I have shown here that:
All coprime triples $(a,b,c)$ so that $a^2 + ab + b^2 = c^2$ can be
enumerated, without duplication, by taking two positive integers
$m \ge n$, where $3$ does not divide $n$, and either $mn$ is odd and
$\gcd(m,n) = 1$, or $8$ divides $mn$ and $\gcd(m,n) = 2$, and by setting
$$
\begin{align}
a&=mn\tag{1a}\\[9pt]
b&=\frac{(3m+n)(m-n)}{4}\tag{1b}\\[9pt]
c&=\frac{3m^2+n^2}{4}\tag{1c}
\end{align}
$$
|
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|
Baby Rudin chapter 3 problem 12. Suppose $a_n > 0$ and $\sum a_n$ converges. Put $r_n = \sum_{m=n}^{\infty} a_m$.
1) Show that $\frac{a_m}{r_m} + ... + \frac{a_n}{r_n} > 1 - \frac{r_n}{r_m}$, if $m < n$, and deduce that $\sum \frac{a_n}{r_n}$ diverges.
2) Show that $\frac{a_n}{\sqrt[]r_n} < 2(\sqrt[]{r_n} - \sqrt[]{r_{n+1}})$ and deduce that $\sum \frac{a_n}{\sqrt[]r_n}$ converges.
I am stumped on this problem, do not know how to start. Any help would be great.
|
For problem 1, note that
$$\frac{a_i}{r_i}=\frac{r_i-r_{i+1}}{r_i}$$
which gives us
$$\frac{a_m}{r_m} + ... + \frac{a_n}{r_n}=\frac{r_m-r_{m+1}}{r_m}+\cdots+\frac{r_n-r_{n+1}}{r_n}>\frac{r_m-r_{m+1}+\cdots+r_n-r_{n+1}}{r_m}$$
and nice things happen when you cancel terms in the numerator.
For problem 2, note that
$$\frac{a_n}{\sqrt{r_n}}=\frac{r_n-r_{n+1}}{\sqrt{r_n}}=\frac{r_n}{\sqrt{r_n}}-\frac{\sqrt{r_{n+1}}}{\sqrt{r_{n}}}\frac{r_{n+1}}{\sqrt{r_{n+1}}}=\sqrt{r_n}-\frac{\sqrt{r_{n+1}}}{\sqrt{r_{n}}}\sqrt{r_{n+1}}$$
so we need to show that
$$-\frac{\sqrt{r_{n+1}}}{\sqrt{r_{n}}}\sqrt{r_{n+1}}<\sqrt{r_n}-2\sqrt{r_{n+1}}$$
which is equivalent to
$$r_{n+1}>2\sqrt{r_nr_{n+1}}-r_n.$$
Using the AM-GM inequality, we have that $2\sqrt{r_nr_{n+1}}< r_{n}+r_{n+1}$. Thus
$$r_{n+1}=r_{n+1}+r_n-r_n>2\sqrt{r_nr_{n+1}}-r_n$$
and the result follows.
|
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|
Verifying some trigonometric identities: $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$ Prove the following:
46. $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
I got as far as
Right Side: $\tan\theta\sin\theta$ to $\dfrac{\sin\theta}{\cos\theta}\dfrac{\sin\theta}{1}$ and then; $\dfrac{\sin^2\theta}{\cos\theta}$
Left Side:
$$\begin{align*}
\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}
&= \dfrac{\frac{1}{\sin^2\theta}-{\frac{\cos^2\theta}{\sin^2\theta}}}{\frac{\cos\theta}{\sin\theta}-{\frac{1}{\sin^2\theta}}}\\
&= \dfrac{\frac{1-\cos^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\
&= \dfrac{\frac{\sin^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\
&= \frac{1}{\frac{\cos\theta}{\sin^2\theta}}\\
&= \frac{\sin^2\theta}{\cos\theta}
\end{align*}$$
Thanks a lot!
|
There is no "cross cancelling". You are subtracting the fractions, not multiplying them.
$$\begin{align*}
\frac{\csc\theta}{\cot\theta} - \frac{\cot\theta}{\csc\theta} & = \frac{\csc^2\theta - \cot^2\theta}{\cot\theta\csc\theta}\\
&= \frac{\quad\frac{1}{\sin^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}\quad}{\frac{\cos\theta}{\sin\theta}\frac{1}{\sin\theta}}\\
&= \frac{\quad\frac{1 - \cos^2\theta}{\sin^2\theta}\quad}{\frac{\cos\theta}{\sin^2\theta}}.
\end{align*}$$
Can you take it from there?
|
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|
Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$ I need to evaluate this integral: $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$.
I've tried $t=\log(x+1)$, $t=x+1$, but to no avail. I've noticed that:
$\int_0^1 \frac{\log(x+1)}{1+x^2} dx = \int_0^1\log(x+1) \arctan'(x)dx =\left. \log(x+1)\arctan(x) \right|_{x=0}^{x=1} - \int_0^1\frac{\arctan(x)}{x+1}dx$
But can't get further than this.
Any help is appreciated, thank you.
|
$\displaystyle A=\int_0^1\dfrac{\Big(\log(1+x)\Big)^2}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$
$\displaystyle A=\int_0^1\dfrac{\left(\log\left(\dfrac{2}{1+x}\right)\right)^2}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2-\log(1+x)\Big)^2}{1+x^2}dx$
$\displaystyle A=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx-2\int_0^1\dfrac{\log 2\log(1+x)}{1+x^2}dx+A$
Therefore,
$\displaystyle \int_0^1\dfrac{2\log 2\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx$
Finally,
$\displaystyle \int_0^1\dfrac{\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\log 2}{2(1+x^2)}dx=\dfrac{\log 2}{2}\Big[\arctan x\Big]_0^1=\dfrac{\pi\log 2}{8}$
|
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|
Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how to do this type of questions..
|
$\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$
$=\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2-(a^3+a^2) & b-a & 1-1 \\ c^3+c^2-(a^3+a^2) & c-a &1-1\end{pmatrix} $ (applying $R_2'=R_2-R_1\ and\ R_3'=R_3-R_1$)
$=(b-a)(c-a)
\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^2+a^2+ab+b+a & 1 & 0 \\ c^2+a^2+ca+c+a & 1 & 0\end{pmatrix}$
$=(b-a)(c-a)\cdot1\cdot
\det\begin{pmatrix} b^2+a^2+ab+b+a & 1 & \\ c^2+a^2+ca+c+a & 1\end{pmatrix}$
$=(b-a)(c-a)(b-c)(a+b+c+1)$
|
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|
Integers of the form $a^2+b^2+c^3+d^3$ It's easy$^*$ to prove that if $n=3^{6m}(3k \pm 1)$ where $(m,k) \in \mathbb{N} \times \mathbb{Z}$, then $n=a^2+b^2+c^3+d^3$ with $(a,b,c,d) \in \mathbb{Z}^4$.
But how to prove that this is true if $n=3k$?
Thanks,
W
$^*$ Because $3k+1=0^2+(3k+8)^2+(k+1)^3+(-k-4)^3$, $3k+2=1^2+(3k+8)^2+(k+1)^3+(-k-4)^3$ and
$3^6(a^2+b^2+c^3+d^3)=(27a)^2+(27b)^2+(9c)^3+(9d)^3$.
|
You do not need the final $d^3.$ Every integer is the sum of two squares and a cube, as long as we do not restrict the $\pm$ sign on the cube.
TYPESET FOR LEGIBILITY:
Solution by Andrew Adler:
$$ 2x+1 = (x^3 - 3 x^2 + x)^2 +(x^2 - x - 1)^2 -(x^2 - 2x)^3 $$
$$ 4x+2 = (2x^3 - 2 x^2 - x)^2 +(2x^3 -4x^2 - x + 1)^2 -(2x^2 - 2x-1)^3 $$
$$ 8x+4 = (x^3 + x +2 )^2 +(x^2 - 2x - 1)^2 -(x^2 + 1)^3 $$
$$ 16x+8 = (2x^3 - 8 x^2 +4 x +2)^2 +(2x^3 -4x^2 - 2 )^2 -(2x^2 - 4x)^3 $$
$$ 16x = (x^3 +7 x - 2)^2 +(x^2 +2 x + 11)^2 -(x^2 +5)^3 $$
You can check these with your own computer algebra system. Please let me know if I mistyped anything.
Alright, our conjecture (Kaplansky and I) is that, for any odd prime $q,$ $x^2 + y^2 + z^q$ is universal. However, this is false as soon as the exponent on $z$ is odd but composite. The example we put in the article is
$$ x^2 + y^2 + z^9 \neq 216 p^3, $$
where $p \equiv 1 \pmod 4$ is a (positive) prime.
This defeated a well-known conjecture of Vaughan. We told him about it in time for him to include it in the second edition of his HARDY-LITTLEWOOD BOOK, where it is now mentioned on pages 127 ("There are some exceptions to this,") and exercise 5 on page 146.
|
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|
Evaluating $\int_\gamma \frac{\cos(z)}{z^3+2z^2} \ dz$ Let $\gamma(t)=e^{it},t \in [0,2\pi]$. We take a look at:
$$\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz$$
We let $$f(z)=\frac{\cos(z)}{z^2+2z}$$ and have
$$\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz=\int_\gamma \frac{f(z)}{z}dz$$
The problem is that $\lim_{z\rightarrow 0} f(z)=\infty$. Now, I can't use the Cauchy integral formula to evaluate the integral. What can be done?
|
$$\frac{\cos z}{z^2(z+2)}=\frac{1}{z^2}\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...\right)\left(\frac{1}{2}\frac{1}{1+\frac{z}{2}}\right)=$$
$$\frac{1}{2z^2}\left(1-\frac{z}{2}+\frac{z^2}{4}-...\right)\left(1-\frac{z^2}{4}+...\right)=\frac{1}{2z^2}-\frac{1}{4z}+...\Longrightarrow Res_{z=0}\left(\frac{\cos z}{z^2}\right)=-\frac{1}{4}$$
Thus, the integral equals $\,\displaystyle{-\frac{1}{4}2\pi i=-\frac{\pi i}{2}}$
|
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|
Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
|
Remember how to complete the square:
$$Ax^2+Bx=A\left(x+\frac{B}{2A}\right)^2-\frac{B^2}{4A^2}$$
So now
$$ax^2+bx+c=0 ---- \text{complete square}$$
$$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}=-c$$
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$
$$x_{1,2}+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
|
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What is a good technique for solving polynomials? Say for example:
$6x^{3}-17x^{2}-4x+3=0$
I sort of look at it and don't know where to start, other than just guessing what the first one would be and trying to do from there. Is there a good technique for approaching such polynomials?
|
$f(x)= 6x^3 -17x^2 -4x +3$
Always observe the polynomials behavior for different values of x
$f(-1)=-16$
$f(0)=3$
$f(1)=-12$
Here we see that there are two roots of $f(x)=0$ between $-1$ and $1$
Now the constant part of the term $6x^3$ is $6=(1)(2)(3)$
Check for $x=\frac{-1}{3}, \frac{1}{3},\frac{-1}{2}, \frac{1}{2}$
$f(\frac{1}{3})=0$
$f(x)$ is not an even function so discard $x=\frac{-1}{3}$
$f(\frac{-1}{2})=0$
$f(x)$ is not an even function so discard $x=\frac{1}{2}$
we see that two of the roots are $x=\frac{-1}{2}$ and $x=\frac{1}{3}$
Now $f(x)= (x +\frac{1}{2})(6x^2 -20x^2 +6)$
i.e. $f(x)= (x +\frac{1}{2})(x -\frac{1}{3})(6x -18)$
i.e. $f(x)= (6)(x +\frac{1}{2})(x -\frac{1}{3})(x -3)$
i.e. $f(x)= (1)(2)(3)(x +\frac{1}{2})(x -\frac{1}{3})(x -3)$
the solution set for $f(x)=0$ is $(\frac{-1}{2}, \frac{1}{3}, 3)$
|
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Solving $5^n > 4,000,000$ without a calculator
If $n$ is an integer and $5^n > 4,000,000.$ What is the least possible value of $n$? (answer: $10$)
How could I find the value of $n$ without using a calculator ?
|
By logarithm rules:
$$5^{n}>4\cdot10^{6}\iff n>\log_{5}2^{2}2^{6}5^{6}=\log_{5}2^{8}+\log_{5}5^{6}=\log_{5}2^{8}+6=\log_{5}256+6$$
Since these are relatively small numbers I assume it is ok to write
: $5^{3}=125$ thus clearly $3<\log_{5}256<4$ hence the minimal $n$
that satisfies this inequality is $4+6=10$
|
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|
Prove $\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$ How would I prove the following?
$$\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$$
I do not know how to do do the problem I do know $\sin(3x)$ can be $\sin(2x+x)$ and such yet I am not sure how to commence.
|
Using $\sin 2\theta=2\sin\theta\cos\theta$ and $\sin^2\theta=(1-\cos2\theta)/2$ we get
$$\cos^2x ~\sin^3x=(\cos x\sin x)^2\sin x=\left(\frac{\sin 2x}{2}\right)^2\sin x=\frac{1}{4}\frac{1-\cos 4x}{2}\sin x \tag{$\circ$}$$
With the sum rule we have $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha.$ Therefore we have the identity
$$\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}=\sin\beta\cos\alpha.$$
Apply with $\alpha=4x$ and $\beta=x$ to $(\circ)$.
|
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|
Determine the nature of a critical point (Lagrange multipliers)
Let $F \colon \mathbb R^2 \to \mathbb R$ be the function
$$
F(x,y):=xye^x + ye^y - e^x+1
$$
and denote with $C$ the set of zeroes of $F$, i.e. $C:=\{(x,y) \in \mathbb R^2 : F(x,y)=0\}$. Let also $f \colon \mathbb R^2 \to \mathbb R$ be a function which is $C^2$ in a neighbourhood of $(0,0)$ and such that
$$
\nabla f(0,0) = (-2,2), \qquad \qquad H_f(0,0) = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}.
$$
Question: is $(0,0)$ a minimum for $f$ on $C$?
Well, I show you what I have done so far. First of all, some routine calculations yield
$$
\nabla F(0,0) = (-1,1) \qquad \qquad H_F(0,0) = \begin{pmatrix} -1 & 1 \\ 1 & 2 \end{pmatrix}
$$
Indeed, the fact that
$$
\nabla f(0,0) = 2\nabla F(0,0)
$$
does agree with the theory of Lagrange multipliers: the gradients are parellel, so I think that $(0,0)$ is an extremum for $f$ on $C$. The problem is how to classify it without any information on $f$: we have only its hessian matrix, which is - I suppose - the key to solve this. Both $H_f(0,0)$ and $H_F(0,0)$ are indefinite.
How can we establish the nature of the critical point $(0,0)$?
Thanks in advance for your help.
|
In order to find the Hessian, we need to know the change in $f$ to second order in the deviation from $(0,0)$.
Expanding $F(x,y)$ around this point, we have up to second order
$$ F(x,y) = \nabla F(0,0)\begin{pmatrix} x \\y\end{pmatrix}
+ \tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_F \begin{pmatrix} x \\y\end{pmatrix}.$$
Similarly, expanding $f(x,y)$ we obtain
$$ f(x,y) = \nabla f(0,0)\begin{pmatrix} x \\y\end{pmatrix}
+ \tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_f \begin{pmatrix} x \\y\end{pmatrix}.$$
In the last step, we need to implement the constraint $F(x,y)=0$. To first order, we have $\nabla F(0,0) \begin{pmatrix} x \\y\end{pmatrix}=0$ and because $\nabla f \parallel \nabla F$, we have that $f(0,0)$ is an extremum. In particular, we have $y=x$ to first order.
To second order, we have
$$\tfrac12 \begin{pmatrix} x & y\end{pmatrix} H_F \begin{pmatrix} x \\y\end{pmatrix} = \tfrac12 \begin{pmatrix} x & x\end{pmatrix} H_F \begin{pmatrix} x \\x\end{pmatrix} = \frac{3}{2} x^2. $$
In order that $F(x,y)=0$ to second order, we need to compensate this via the first term, so we have $y= x - \frac{3}{2} x^2$.
Plugging this into the expansion of $f(x,y)$, we obtain a term from the gradiant and a term from the Hessian (Tunococ just considered the term from the Hessian). In total, we have
$$f(x,y) = -3 x^2 + 2 x^2 =-x^2.$$
So it is a local maximum.
|
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|
$1867k =\ldots 1992$, $k\in\mathbb{Z}^+$. Find the minimum value of k. The number $1867$ is multiplied by a positive integer $k$. The last four digits of the product are $1992$. Determine the minimum value of $k$.
$1867k =\ldots 1992$
|
Hint $\ $ Let $\rm\, k = \ldots\!dcba\,$ have undetermined digits. Multiplying and comparing digits yields
$$\rm\begin{eqnarray} 1867\,(\ldots\!dcba)\ &= &\rm\ (7+6\cdot10+8\cdot 10^2 + 10^3)\,(a + b\cdot 10 + c\cdot 10^2+ d\cdot 10^3 + (\cdots)\, 10^4)\\
& =&\rm\ 7a + (6a\!+\!7b)\, 10 + (8a\!+\!6b\!+\!7c)\, 10^2 + (a\!+\!8b\!+\!6c\!+\!7d)\,10^3+ (\cdots)\, 10^4\\
& =&\rm\, \ldots\!1992\end{eqnarray}$$
$10^0$ digit: $\rm\ mod\ 10\!:\ 2\equiv 7a\:\Rightarrow\: a\equiv 6,\:$ so $\rm\:7a = 42,\:$ with carry $\color{#C00}4$
$10^1$ digit: $\rm\ mod\ 10\!:\ 9\equiv 6\cdot6 + 7b+\color{#C00}4\equiv 7b\:\Rightarrow\:b\equiv 7,\,$ so sum $= 89,$ with carry $= \color{#0A0}8$
$10^2$ digit: $\rm\ mod\ 10\!:\ 9 \equiv 8\cdot6\!+\!6\cdot 7\!+\!7c\!+\!\color{#0A0}8 \equiv 7c\!+\!8\,\Rightarrow\,7c\equiv 1\,\Rightarrow\,c\equiv 3,\,$ sum $119,\,$ carry $\color{brown}{11}$
$10^3$ digit: $\rm\ mod\ 10\!:\ 1 \equiv 6 + 8\cdot 7 + 6\cdot 3+ 7d + \color{brown}{11}\equiv 7d +1\,\Rightarrow\, 7d\equiv 0\,\Rightarrow\,d\equiv 0$
Therefore $\rm\:a + b\cdot 10 + c\cdot 10^2+ d\cdot 10^3 = 376.\:$ Indeed $\rm\:1867\cdot 376 = 701992.$
Remark $\ $ Equivalently, we may solve it iteratively, examining it mod $10, 10^2, 10^3, 10^4.\:$ This works because the above system of modular equations has nice triangular form: with carries $\rm\:c_i$
$$\rm \left[\begin{array}{cccc} 7 & 6 & 8 & 1 \\ 0 & 7 & 6 & 8 \\ 0 & 0 & 7 & 6 \\ 0 & 0 & 0 & 7 \end{array}\right]\left[\begin{array}{c} d\\ \rm c\\ \rm b\\ \rm a\end{array}\right] + \left[\begin{array}{c}\rm c_3 \\\rm c_2\\ \rm c_1\\ 0\end{array}\right] \equiv \left[\begin{array}{c} 1\\9\\9\\2\end{array}\right]$$
The point of writing it out in this slight longer form is to explicitly exhibit this special structure - so to make clear the reason it works. This is perhaps better known in the analogous power series case, e.g. for computing inverses, Newton iteration, etc.
|
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|
Integral: $\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t$ How to prove the following:
$$\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t=\frac{\pi}{2}\left(\ln(2)+\frac{\pi}{2} i\right)$$
|
Let's evaluate the real and imaginary parts of the integral separately. Using $\Re(\ln(t+1)) = \ln(|t+1|)$ and $\Im(\ln(t+1)) = \pi [t < -1]$, where $[t<1]$ is the Iverson bracket:
$$ \begin{eqnarray}
\Re \int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t &=& \int_{-\infty}^\infty \frac{\ln(|t+1|)}{t^2+1} \mathrm{d} t = \frac{1}{2} \int_{-\infty}^\infty \frac{\ln((t+1)^2 )}{t^2+1} \mathrm{d} t\\
\Im \int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t &=& \int_{-\infty}^{-1} \frac{\pi}{t^2+1} \mathrm{d} t
\end{eqnarray}
$$
The integral for the imaginary part is easy, since $\int \frac{\mathrm{d} t}{t^2+1} = \arctan(t) + C$:
$$
\int_{-\infty}^{-1} \frac{1}{t^2+1} \mathrm{d} t = \left.\arctan(t)\right|_{-\infty}^{-1} = \arctan(-1) - \lim_{t\to -\infty} \arctan(t) = -\frac{\pi}{4} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{4}
$$
Now to the real part:
$$
\int_{-\infty}^\infty \frac{\frac{1}{2} \log((t+1)^2)}{t^2+1} \mathrm{d} t =
\int_{-\infty}^\infty \frac{\frac{1}{2} \log(t^2)}{t^2-2t +2} \mathrm{d} t =
\int_0^\infty \log(t^2) \frac{1}{2} \left( \frac{1}{(t-1)^2+1}+\frac{1}{(t+1)^2 + 1} \right) \mathrm{d} t = \int_0^\infty \log(t^2) \frac{t^2+2}{t^4+4} \mathrm{d} t \stackrel{u=t^2}{=} \int_0^\infty \frac{\log(u)}{2} \frac{u+2}{u^2+4} \frac{\mathrm{d} u}{\sqrt{u}} = \frac{1}{2} \left.\frac{\mathrm{d} }{\mathrm{d} s} \int_0^\infty u^{s-1} \frac{u+2}{u^2 + 4} \mathrm{d} u \right|_{s=\frac{1}{2}}
$$
The latter integral is a sum of Mellin-Barnes transforms of $\frac{1}{u^2+4}$:
$$
\int_0^\infty u^{s-1} \frac{u+2}{u^2 + 4} \mathrm{d} u = 2 \int_0^\infty u^{s-1} \frac{1}{u^2 + 4} \mathrm{d} u + \int_0^\infty u^{s} \frac{1}{u^2 + 4} \mathrm{d} u =
\frac{2^{s-2} \pi}{\sin\left(\frac{\pi s}{2}\right)} + \frac{2^{s-2} \pi}{\cos\left(\frac{\pi s}{2}\right)}
$$
Differentiating, and setting $s=\frac{1}{2}$ yields the result:
$$
\int_{-\infty}^\infty \frac{\frac{1}{2} \log((t+1)^2)}{t^2+1} \mathrm{d} t = \pi \log(2)
$$
Recombining the real and imaginary parts:
$$
\int_{-\infty}^\infty \frac{\ln(t+1)}{t^2+1} \mathrm{d} t = \pi \log(2) + i \frac{\pi^2}{4}
$$
|
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|
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
|
Hint $\rm\,\ n\,$ odd $\rm\:\:\! \Rightarrow\ 2\:|\:\color{#90f}{n^2 + 5}$
and, furthermore $\ \:\!\rm 8\:|\:\color{#0a0}{n^2}-\color{#c00}1,\ $ by $\ \rm mod\ 8\!:\ \color{#0a0}{odd^2} \equiv \{\pm 1,\pm 3\}^2\equiv \color{#c00}1$
multiplying $\rm \Rightarrow 2\cdot 8\:\!\:\!|\:\!(n^2\!-1)\:\!(\color{#90f}{n^2\!+\!5}).\,\ \small QED$
|
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|
I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
|
$S_{2n}+4S_{n}=n(2n+1)^2$
$S_{2n}=S_{n}+S_{(n+1,2n)}$ -------(A)
$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(2n)^2$
$S_{(n+1,2n)}=(n+1)^2+(n+2)^2+(n+3)^2+\cdots+(n+n)^2$
$S_{(n+1,2n)}=n(n)^2+(1^2+2^2+3^2+\cdots+n^2)+(2n)(1+2+3+\cdots+n)$
$S_{(n+1,2n)}=n^3+S_{n}+(2n)\frac{n(n+1)}{2}$
$S_{(n+1,2n)}=n^3+S_{n}+n^2(n+1)$
$S_{(n+1,2n)}=n^3+S_{n}+n^3+n^2$
$S_{(n+1,2n)}=S_{n}+2n^3+n^2$ -------(B)
we, have from (A) and (B),
$S_{2n}=S_{n}+S_{n}+2n^3+n^2$
$S_{2n}=2S_{n}+2n^3+n^2$
we, now have,
$6S_{n}+2n^3+n^2=n(2n+1)^2$
$6S_{n}+2n^3+n^2=n(4n^2+4n+1)$
$6S_{n}+2n^3+n^2=4n^3+4n^2+n$
$6S_{n}=2n^3+3n^2+n$
$6S_{n}=n(n+1)(2n+1)$
$S_{n}=\frac{n(n+1)(2n+1)}{6}$
|
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|
Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.10 Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\displaystyle \frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}} . $$
Thanks.
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We have over $(a,b,c)$:
$$\displaystyle LHS := \sum_{cyc} \frac{a}{\sqrt{a+2b}} = \frac{\sum_{cyc}\sqrt{a^2(b+2c)(c+2a)}}{\sqrt{(a+2b)(b+2c)(c+2a)}}$$
Using CS:
$$\displaystyle LHS \leq \sqrt{\frac{\left(a^2(b+2c)+b^2(c+2a)+c^2(a+2b)\right)\left(3(a+b+c)\right)}{(a+2b)(b+2c)(c+2a)}} \\
= \sqrt{3}\sqrt{\frac{a^2(b+2c)+b^2(c+2a)+c^2(a+2b)}{(a+2b)(b+2c)(c+2a)}} \\
= \sqrt{\frac{3}{2}}\sqrt{1-\frac{9abc}{2(a+2b)(b+2c)(c+2a)}} < \sqrt{\frac{3}{2}}$$
|
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|
Show $ \int_0^\infty\left(1-x\sin\frac 1 x\right)dx = \frac\pi 4 $ How to show that
$$
\int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=\frac{\pi}{4}
$$
?
|
Another way is to use Laplace transform to evaluate this improper integral. In fact, let $f(x)=x-\sin x$ and then $F(s)=\frac{1}{s^2}-\frac{1}{s^2+1}$. Thus
\begin{eqnarray*}
\mathcal{L}\big\{\frac{f(x)}{x}\big\}&=&\int_s^\infty F(s)ds=-\frac{\pi}{2}+\frac{1}{s}+\arctan s, \\
\mathcal{L}\big\{\frac{f(x)}{x^2}\big\}&=&\int_s^\infty(-\frac{\pi}{2}+\frac{1}{u}+\arctan u)du\\
&=&-1+s(\frac{\pi}{2}-\arctan s)+\ln\frac{\sqrt{s^2+1}}{s}, \\
\mathcal{L}\big\{\frac{f(x)}{x^3}\big\}&=&\int_s^\infty\left(-1+u(\frac{\pi}{2}-\arctan u)-\ln\frac{\sqrt{u^2+1}}{u}\right)du\\
&=&\frac{s}{2}-\frac{\pi s^2}{4}-\frac{1}{2}\arctan s+\frac{1}{2}s^2\arctan s+s\ln s-\frac{1}{2}s\ln(s^2+1).
\end{eqnarray*}
From this, one can obtain
\begin{eqnarray*}
&&\int_0^\infty (1-x\sin\frac{1}{x})dx=\int_0^\infty\frac{x-\sin x}{x^3}dx=\lim_{s\to 0^+}\mathcal{L}\big\{\frac{f(x)}{x^3}\big\}\\
&=&\lim_{s\to 0^+}(\frac{\pi}{4}+\frac{s}{2}-\frac{\pi s^2}{4}-\frac{1}{2}\arctan s+\frac{1}{2}s^2\arctan s+s\ln s-\frac{1}{2}s\ln(s^2+1))\\
&=&\frac{\pi}{4}.
\end{eqnarray*}
|
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|
Compute this limit of series: $\lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 $ Compute this limit of series:
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 $
I used the definition of the definite integral
$\displaystyle \int_{a}^{b}f(x)dx=\lim_{n\rightarrow \infty} S_{n}$
where $\displaystyle f(x)=x^2$, $\displaystyle [a,b]=[0,1]$; since the function is continuous in $\displaystyle [0, 1]$ then it is certainly integrated.
$\displaystyle \int_{0}^{1}x^2dx=\lim_{n\rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{n}\ \left(\frac{k}{n}\right)^2=\lim_{n\rightarrow \infty} \frac{1}{n^3}\sum_{k=1}^{n-1}k^2$
We have:
$\displaystyle\int_{0}^{1}x^2dx=\frac{1}{3}$
then
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 =\frac{1}{3}$
Any suggestions, please? This limit can be solved in other ways?
Thanks.
|
By induction, you can prove that
$\displaystyle \sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$
Thus
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{n^3}\sum_{k = 1}^{n - 1} k^2 = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2(n - 1) + 1)}{6} = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2n - 1)}{6} = \lim_{n \rightarrow \infty}\frac{2n^3 - 3n^2 + n}{6n^3}$
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{3} - \frac{1}{2n} + \frac{1}{6n^2} = \frac{1}{3}$
|
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|
Find the sum function of $\sum_{n=0}^{\infty}\frac{n(n-2)}{n+1}x^{n-1}$ series summation:
$$\sum_{n=0}^{\infty}\frac{n(n-2)}{n+1}x^{n-1}$$
where $-1 <x <1$
is there a convinient function that sums the above series?
(unsure but this may be an expanded taylor series?)
|
Write that as
$$\sum\limits_{n = 0}^\infty {\frac{{{n^2}}}{{n + 1}}{x^{n - 1}}} - \sum\limits_{n = 0}^\infty {\frac{{2n}}{{n + 1}}{x^{n - 1}}} $$
Now think about primitives and derivatives.
$$\eqalign{
& \sum\limits_{n = 0}^\infty {\frac{2}{{n + 1}}n{x^{n - 1}}} = f'\left( x \right) = \frac{d}{{dx}}\left[ {\sum\limits_{n = 0}^\infty {\frac{2}{{n + 1}}{x^n}} } \right] \cr
& \sum\limits_{n = 0}^\infty {\frac{n}{{n + 1}}n{x^{n - 1}}} = g'\left( x \right) = \frac{d}{{dx}}\left[ {\sum\limits_{n = 0}^\infty {\frac{n}{{n + 1}}{x^n}} } \right] \cr} $$
and $$\sum\limits_{n = 0}^\infty {\frac{n}{{n + 1}}{x^n}} = \sum\limits_{n = 0}^\infty {\frac{{n + 1 - 1}}{{n + 1}}{x^n}} = \sum\limits_{n = 0}^\infty {{x^n}} - \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n + 1}}} $$
Now use
$$\sum\limits_{n = 0}^\infty {\frac{{{x^{n + 1}}}}{{n + 1}}} = - \log \left( {1 - x} \right)$$
$$\sum\limits_{n = 0}^\infty {{x^n}} = \frac{1}{{1 - x}}$$
|
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|
Prove $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$ Please help me for proving this inequality $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$
|
A more complicated approach is to prove the result by induction. Let
$$S(n)=\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}.$$
We want to show that $S(n)\lt 1-\dfrac{1}{n}$ for every integer $n \ge 2$.
The result is clearly true when $n=2$. We show that for any $k\ge 2$, if the result is true when $n=k$, it is true when $n=k+1$.
We have
$$S(k+1)=S(k)+\frac{1}{(k+1)^2}.$$
By the induction assumption, $S(k)\lt 1-\dfrac{1}{k}$. It follows that
$$S(k+1)\lt 1-\frac{1}{k}+\frac{1}{(k+1)^2}.$$
But
$$\frac{1}{k}-\frac{1}{(k+1)^2}=\frac{k^2+k+1}{k(k+1)^2}\gt \frac{k^2+k}{k(k+1)^2}=\frac{1}{k+1}.$$
It follows that $S(k+1)\lt 1-\dfrac{1}{k+1}$.
Remark: If we try to prove the weaker result $S(n)\lt 1$ by induction, we get into trouble. For from the induction assumption $S(k)\lt 1$, we cannot conclude in any direct way that $S(k+1)\lt 1$. So with the above problem, we have the seemingly paradoxical fact that in an induction proof, a strong inequality can be easier to prove than a weaker inequality.
|
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|
Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$ Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})]$, in the region $|x|<1$.
|
\begin{align}
& \lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})] \\
&= \lim_{n\to\infty}\frac{(1-x)(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{(1-x^2)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{(1-x^4)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})}{1-x} \\
&= \cdots \cdots \cdots \\
&= \lim_{n\to\infty}\frac{(1-x^{2^n})(1+x^{2^n})}{1-x} \\
&= \lim_{n\to\infty}\frac{1-x^{2^{n+1}}}{1-x} \\
&= \frac{1}{1-x}\lim_{n\to\infty}{(1-x^{2^{n+1}})} \\
&= \frac{1}{1-x}\cdot 1 \\
&= \frac{1}{1-x} .
\end{align}
|
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|
Two unrelated equations, $w^2 = -\frac{15}{4} - 2i$ and $z^2 - (3-4i)z + (2-4i) = 0$? I am to solve the equation $z^2 - (3-4i)z + (2-4i) = 0$, and also have to help me that $w^2 = -\frac{15}{4} - 2i$.
I can ofcourse find $w$, but I fail to see how this helps me in solving $z^2 - (3-4i)z + (2-4i) = 0$. What am I missing? Is there a connection I fail to see?
|
$z=\frac{3-4i±\sqrt{(3-4i)^2-4\cdot 1\cdot (2-4i)}}{2}$
Now, $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i=1^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
So, $z=\frac{3-4i±(1-4i)}{2}=1$ or $2-4i$
$4w^2=-15-8i=(1)^2+(4i)^2-2\cdot 1\cdot 4i=(1-4i)^2$
$2w=±(1-4i)$
Alternatively, let $-\frac{15}{4}-2i=(a-ib)^2$
So, $-\frac{15}{4}-2i=a^2-b^2-i2ab$
$a^2-b^2=-\frac{15}{4}$ and $2ab=2$
$(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2=(a^2-b^2)^2+(2ab)^2=\frac{289}{16}$
$a^2+b^2=\frac{17}{4}$
$a^2=\frac{1}{4},b^2=4$
So,$a=±\frac{1}{2},b=±2$ as $a,b$ are of same sign.
The square root of $(3-4i)^2-4\cdot 1\cdot (2-4i)=-15-16i$ can be calculated in the same way.
|
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|
Inequality $(1+\frac1k)^k \leq 3$ How can I elegantly show that: $(1 + \frac{1}{k})^k \leq 3$ For instance I could use the fact that this is an increasing function and then take $\lim_{ k\to \infty}$ and say that it equals $e$ and therefore is always less than $3$
*
*Is this sufficient?
*What is a better wording than "increasing function"
|
Using binomial theorem, $(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$
Now, consider ${k\choose r}\frac{1}{k^r}=\frac{k!}{r!(k-r)!}\frac{1}{k^r}=\frac{(1)(1-\frac{1}{k})(1-\frac{2}{k})...(1-\frac{r-1}{k})}{r!}\lt \frac{1}{r!}$
Thus, $$(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$$ $$ \lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}$$ $$\lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}+\frac{1}{(k+1)!}+\cdots =e$$
Hence $$(1+\frac{1}{k})^k\lt e\lt 3$$
|
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|
Inequality under condition $a+b+c=0$ I don't know how to prove that the following inequality holds (under condition $a+b+c=0$):
$$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$
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This one made me struggle so much that I was close to go crazy. Therefore let me post my solution to have a relief from this burden..
First we have $$2a^2=\frac43a^2+\frac23a^2=\frac43a^2+\frac23(b+c)^2\leq \frac43(a^2+b^2+c^2);$$ where the last inequality follows from the arithmetic-quadratic mean.
Analogously $$\begin{split}2b^2&\leq \frac43 (a^2+b^2+c^2),\\ 2c^2&\leq \frac43(a^2+b^2+c^2).\end{split}$$ It follows that $$\sum_\text{cyc}\frac{(2a+1)^2}{2a^2+1}\geq3\sum_\text{cyc}\frac{(2a+1)^2}{4(a^2+b^2+c^2)+3}=3\left(1+\frac{4(a+b+c)}{4(a^2+b^2+c^2)+3}\right)=3.$$
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|
Relations between the roots of a cubic polynomial How do I solve the last two of these problems?
The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\dfrac{1}{\gamma+1} $. $\quad \quad \quad (2)$
For the cases $n=1$ and $n=2$, find the value of
$$\dfrac{1}{(\alpha+1)^n}+\dfrac{1}{(\beta+1)^n}+\dfrac{1}{(\gamma+1)^n}. \tag{2}$$
Deduce the value of $\dfrac{1}{(\alpha+1)^3}+\dfrac{1}{(\beta+1)^3}+\dfrac{1}{(\gamma+1)^3}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \,(2)$
Hence show that $\dfrac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}+\dfrac{(\gamma+1)(\alpha+1)}{(\beta+1)^2}+\dfrac{(\alpha+1)(\beta+1)}{(\gamma+1)^2}=\dfrac{73}{36} \quad \quad \quad \quad \quad \quad \quad (3)$
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Let $a=\frac1{(1+\alpha)}$ etc,
so, $a,b,c$ are the roots of $6t^3-7t^2+3t-1=0$
$\implies a+b+c=\frac 7 6, ab+bc+ca=\frac 3 6=\frac 12$ and $abc=\frac1 6$
So, $$a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$$
$$=(a+b+c)((a+b+c)^2-3(ab+bc+ca))+3abc$$
$$=(\frac 7 6)((\frac 7 6)^2-3(\frac 12))+3\frac1 6=\frac{73}{216}$$
Now, $$\frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=\frac{a^2}{bc}=\frac{a^3}{abc}=6a^3$$
So, $$\sum \frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=6(a^3+b^3+c^3)=6\left(\frac{73}{216}\right)=\frac{73}{36}$$
For the generalization of the sums of Powers of Roots, one may look here for the statement, here for the proof.
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|
Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $$
$$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $$
$$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $$
$$ 4 \cos^2 x = 2 \sqrt 3 \cos x $$
$$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $$
$$ 4 \cos x = 2 \sqrt 3 $$
$$ \cos x = \frac{2 \sqrt 3}{4} $$
$$ \cos x = \frac{\sqrt 3}{2} $$
The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:
$$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $$
$$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $$
Since I earlier on exponentiated both sides I have to check my solutions:
$$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $$
$$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $$
Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.
$$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $$
Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...
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There's a nice trick:
$$\sin x + \sqrt 3 \cos x = 1 \\\\ = 2 \left(\frac{1}{2}\sin x + \frac{\sqrt 3}{2} \cos x\right) \\\\ = 2\left(\cos\left(\frac{\pi}{3} + 2k\pi\right)\sin x + \sin\left(\frac{\pi}{3} + 2k\pi\right)\cos x \right)\\\\= 2\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1.$$
When is $$\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1/2$$
true?
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|
How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
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$$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\lim_{x\to 1} \frac{\sqrt{3+x}-2}{x-1}\cdot \lim_{x\to 1} \frac{x-1}{\sqrt[3]{7+x}-2}$$ $$=\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}$$
Use $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$$ which gives
$$\lim_{x+3\to 4} \frac{{(x+3)^{0.5}}-4^{0.5}}{(x+3)-4}\cdot \frac{1}{\lim_{(x+7)\to 8}\frac{({x+7})^{1/3}-8^{1/3}}{(x+7)-8}}=(0.5)4^{-0.5}\cdot\frac{1}{(1/3)8^{-2/3}}=3$$
see
|
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|
substitution in a non linear differential equation and to get a nicer form well I had this equation at the begining
$$
i \frac{\partial u}{\partial{z}} + \frac{1}{2 k_0} \frac{\partial^2 u}{\partial x^2} +\frac{1}{2}k_0 n_1 F(z) x^2 u-\frac{i[g(z) -\alpha(z)]}{2}u + k_0 n_2|u|^2 u = 0,
$$
If I substitute $X=x/w_0$, $Z=z/L_D$, $G=(g(z)-\alpha(z))L_D$, $U=u\sqrt{k_o n_2 L_D}$, $L_D=k_0w_0^2$ $w_0=(k_0^2n_1)^{-1/4}$
I am not getting this equation
$$
i\frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial X^2} + F(Z) \frac{X^2}{2} U - \frac{i}{2} G(Z) U + |U|^2 U = 0
$$
but getting 2nd term multiplied with $w_0$, could any one tell me am I right or wrong? I must say I have calculated it 3 times. please help. well replace $f(z)$ by $F(Z)$ and $g(z)$ by $G(Z)$
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Lets see. Taking the change of variables $X = \frac{x}{w_0}$, $Z = \frac{z}{L_D}$ and $U = \frac{u}{C}$, we have
\begin{align}
\frac{\partial}{\partial x} &= \frac{\partial X}{\partial x}\frac{\partial}{\partial X} = \frac{1}{w_0}\frac{\partial}{\partial X},\\
\frac{\partial^2}{\partial x^2} &= \frac{1}{w_0^2}\frac{\partial}{\partial X},\\
\frac{\partial}{\partial z} &= \frac{\partial Z}{\partial z}\frac{\partial}{\partial Z} = \frac{1}{L_D}\frac{\partial}{\partial Z}.
\end{align}
Substituting into the equation,
\begin{multline}
\frac{i}{C L_D} \frac{\partial U}{\partial Z} + \frac{1}{2 C k_0 w_0^2} \frac{\partial^2 U}{\partial x^2} + \frac{w_0^2 k_0 n_1}{2 C} f(L_D Z) X^2 U \\ - \frac{i \big[g(L_D Z) - \alpha(L_D Z)\big]}{2 C} U + \frac{k_0 n_2}{|C|^2 C} |U|^2 U = 0.
\end{multline}
Multiplying by $C L_D$ the hole equation,
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{L_D}{2 k_0 w_0^2} \frac{\partial^2 U}{\partial x^2} + \frac{w_0^2 k_0 n_1 L_D}{2} f(L_D Z) X^2 U \\ - \frac{i L_D\big[g(L_D Z) - \alpha(L_D Z)\big]}{2} U + \frac{k_0 n_2 L_D}{|C|^2} |U|^2 U = 0.
\end{multline}
Taking $L_D = k_0 w_0^2$,
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{w_0^4 k_0^2 n_1}{2} f(k_0 w_0^2 Z) X^2 U \\ - \frac{i k_0 w_0^2\big[g(k_0 w_0^2 Z) - \alpha(k_0 w_0^2 Z)\big]}{2} U + \frac{k_0^2 w_0^2 n_2}{|C|^2} |U|^2 U = 0.
\end{multline}
Let $w_0 = (k_0^2 n_1)^{-1/4}$, then $L_D = \frac{1}{\sqrt{n_1}}$ and
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{1}{2} f(n_1^{-1/2} Z) X^2 U \\ - \frac{i \big[g(n_1^{-1/2} Z) - \alpha(n_1^{-1/2} Z)\big]}{2 \sqrt{n_1}} U + \frac{k_0 n_2}{\sqrt{n_1} |C|^2} |U|^2 U = 0.
\end{multline}
Taking $C = \frac{\sqrt{k_0 n_2}}{n_1^{1/4}} = \sqrt{k_0 n_2 L_D}$,
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{1}{2} f(n_1^{-1/2} Z) X^2 U \\ - \frac{i \big[g(n_1^{-1/2} Z) - \alpha(n_1^{-1/2} Z)\big]}{2 \sqrt{n_1}} U + |U|^2 U = 0.
\end{multline}
Denoting
\begin{align}
F(Z) &= f\left(\tfrac{Z}{\sqrt{n_1}}\right),\\
G(Z) &= \frac{1}{\sqrt{n_1}}\left[g\left(\tfrac{Z}{\sqrt{n_1}}\right) - \alpha\left(\tfrac{Z}{\sqrt{n_1}}\right)\right]
\end{align}
we have
$$
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{1}{2} F(Z) X^2 U - \frac{i}{2} G(Z) U + |U|^2 U = 0.
$$
What we have done here is written the original equation in adimensional form. Judging by the form of the equation, $L_D$ has [time] dimensions, $w_0$ has [space] dimesions, and $u$ has [space/time²] dimensions.
Please doublecheck the math.
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|
Check my workings: Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition. Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.
Precalculations
My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that
$$0<|x+2|<\delta\ \ \text{implies}\ |3x^2+4x-2-2|<\epsilon$$
$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|$
$=|3(x+2)^2-4(x+2)|$
$\leq3|x+2|^2+4|x+2|$ by triangle inequality
$<3\delta^2+4\delta$
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
Proof
For all $\epsilon>0$, choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
$$\begin{align*}0<|x+2|<\delta\ \ \to\ \ &|3x^2+4x-2-2|<3\delta^2+4\delta\\&<3\left(\sqrt{\frac{\epsilon}{6}}\right)^2+4\delta\\&=\frac{\epsilon}{2}+4\delta\\&<\frac{\epsilon}{2}+4\frac{\epsilon}{8}\\&=\frac{\epsilon}{2}+\frac{\epsilon}{2}\\&=\epsilon\end{align*}$$
Therefore proven? Hehe. Not sure this will work or not.
My doubts lies in the steps.
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
And hey, I am looking out for other possible ways to do this question too.
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You made a mistake here:
$$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|=|3(x+2)^2-4(x+2)|$$
It should be $\,8\,$ instead $\,4\,$ in the RHS. All the rest you did is fine, fixing this little mistake.
I show you now how'd I do it:
$$|3x^2+4x-2-2|=|3(x+2)^2-8(x+2)|=$$
$$|x+2|\,|3x-2|\stackrel{\text{for}\,|x+2|<0.5\Longrightarrow |3x-2|<10}<10|x+2|$$
Thus, we're fine if
$$10|x+2|<\epsilon\Longrightarrow |x+2|<\frac{\epsilon}{10}$$
Thus we can choose
$$\delta =\min\left(\frac{\epsilon}{10}\,,\,\frac{1}{24}\right)$$
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|
How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
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Consider the substitution $x=4\sin y$. Then $dx=4\cos y\ dy$, and
$$\frac 1{(16-x^2)^2} = \frac 1{(16\cos^2 y)^2} = \frac 1{(4\cos y)^4} $$
So,
$$\int\frac 1{(16-x^2)^2}dx = \int\frac{4\cos y}{(4\cos y)^4} dy$$
So, finally you need $\displaystyle\int\frac1{\cos^3}$.
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|
Finding the derivative of $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ Find the derivative
How do I tackle this? My answer is totally different from the correction model, but I have tried for half an hour to show my answer in lateX but I don't know how to, it's too complicated, so, can someone please give a step to step of the solution? the correction model's solution is $ \dfrac {2-3x^2}{\sqrt{x} (x^2+2)^2} $
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Use the quotient rule:
\begin{align}
h'(x) & = \frac{(x^2+2)\frac{d}{dx}(2\sqrt{x}) - 2\sqrt{x}\frac{d}{dx}(x^2+2)}{(x^2+2)^2} \\[10pt]
& = \frac{(x^2+2)\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot 2x}{(x^2+2)^2}.
\end{align}
Now clear out fractions by multiplying the top and bottom both by $\sqrt{x}$:
$$
\frac{(x^2+2)-2x\cdot2x}{\sqrt{x}(x^2+2)^2}.
$$
Finally, do the routine simplifications of the numerator and you get exactly what you say is "the correction model's solution".
|
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|
what is the limit of the sequence $x_0=a$, $x_1=b$ define $$x_{n+1}=\left(1-\frac{1}{2n}\right)x_n+ \frac{1}{2n} x_{n-1}$$
we need to find the limit of $\{x_n\}_n$ as $n\rightarrow \infty$.
Let $l$ be this limit , I tried taking the limit on both sides getting $l=l$ as $n\rightarrow \infty$.
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As $x_{n+1}=(1-\frac{1}{2n})x_{n}+\frac{1}{2n}x_{n-1}$, we can get that:
$$\begin{align*}
x_{n+1}-x_{n}&=-\frac{1}{2n}(x_{n}-x_{n-1})\\
x_{n}-x_{n-1}&=-\frac{1}{2(n-1)}(x_{n-1}-x_{n-2})\\
&\vdots\\
x_{2}-x_{1}&=-\frac{1}{2}(x_{1}-x_{0})\\
\end{align*}$$
then,by computation $$x_{n+1}-x_{n}=\frac{(-\frac{1}{2})^{n}}{n!}(b-a)$$
$$\begin{align*}
x_{n}&=\sum_{k=0}^{n-1}(x_{k+1}-x_{k})+x_{0}\\
&=a+(b-a)\sum_{k=0}^{n-1}\frac{(-\frac{1}{2})^{k}}{k!}\\
\end{align*}$$
$$\begin{align*}
\lim_{n\rightarrow\infty}x_{n}
&=a+(b-a)\sum_{k=0}^{\infty}\frac{(-\frac{1}{2})^{k}}{k!}\\
&=a+(b-a)e^{-\frac{1}{2}}\\
&=(1-e^{-\frac{1}{2}})a+e^{-\frac{1}{2}}b
\end{align*}$$
|
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|
Analytic Geometry question I can't solve An isosceles triangle $ABC$ has 2 given vertices, $A(3,2)$ and $C (7,14$). The slope of AB is $\dfrac{1}{2}$. What are the coordinates of B?
I could figure out that line AB = $\dfrac{1}{2}x + \dfrac{1}{2} $
I found that the length of AC = is $\sqrt{160}$
But I haven't got a clue as to finding the coordinates of B.. can someone give me a hint?
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Put $B(x,y)$. I solve your problem with assume the triangle $ABC$ isosceles at $A$.
Because $AB =AC$, then $AB =\sqrt{160}$ or
$$x^2+y^2-6x-4y-147 = 0.$$
The coordinates of the point $B$ are solutions of the system
$$x^2+y^2-6x-4y-147 = 0, \quad y = \dfrac12x + \dfrac12.$$
We get $B(3 - 8\sqrt{2}, 2(1-2\sqrt{2})$ or $B(3 + 8\sqrt{2}, 2(1+2\sqrt{2})$.
|
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|
Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
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Yet another way is to start with:
$$
\sum_{0 \le k \le n} z^k = \frac{1 - z^{n + 1}}{1 - z}
$$
differentiate thrice:
$$
\frac{\mathrm{d}}{\mathrm{d} z}
\left( z \frac{\mathrm{d}}{\mathrm{d} z}
\left(
z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1 - z^{n + 1}}{1 - z}
\right)
\right)
= \frac{1 + 4 z
+ z^2
- z^n (1 - 3 n^2 - n^3)
- z^{n + 1} (4 - 6 n^2 - 3 n^3)
- z^{n + 2} (1 - 3 n + 3 n^2 + 3 n^3)
+ z^{n + 3} n^3}
{(1 - z)^4}
$$
Taking the limit as $z \to 1$ of this mess (l'Hôpital to the rescue) again gives:
$$
\sum_{0 \le k \le n} k^3 = \frac{n^2 (n + 1)^2}{4}
$$
The above courtesy of my CAS, maxima. Any transcription errors are mine only.
|
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|
Proof of an inequality about $\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$ I've encountered an inequality pertaining to the following expression:
$\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$, where $z$ is a complex number.
After writing $z$ as $x + iy$ we have the inequality when $y \gt 1$ and $|x| \le \frac{1}{2} $:
$|\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}|
\le C + C\sum_{n=1}^{\infty}\frac{y}{y^2+n^2}$
The "proof" of the inequality is given as follows:
$\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2} = \frac{1}{x+iy} +\sum_{n=1}^{\infty}\frac{2(x+iy)}{x^2 - y^2 - n^2 + 2ixy}$
But I fail to see how the inequality follows.
|
As
$$1\leq y\leq|x+iy|\leq{y\over2}+y={3\over2} y$$
we have
$${1\over |x+iy|}\leq 1$$
and
$$|n^2+y^2-x^2-2ixy|\geq |n^2+y^2-x^2|\geq\Bigl(1-{1\over8}\Bigr)(n^2+y^2)\qquad(n\geq1)\ .$$
It follows that
$$\left|{1\over z}+\sum_{n=1}^\infty{2z\over z^2-n^2}\right|\leq 1+\sum_{n=1}^\infty {3y \over{7\over8}(n^2+y^2)}=1+{24\over7}\sum_{n=1}^\infty {y \over n^2+y^2}\ .$$
Therefore the stated inequality is true with $C=4$, say.
|
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|
How to find constants a and b in this function? $\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax+b}-2}x = \frac 5{12}$
How do you solve for constants $a$ and $b$?
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You can use $(x^3-y^3)=(x-y)(x^2+xy+y^2)$ with $x=\sqrt[3]{ax+b},y=2$ and multiply the term $\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)$ into nominator and denominator of the limit: $$\displaystyle \lim_{x\to0} \frac{\big(\sqrt[3]{ax+b}-2\big)\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)} = \lim_{x\to0} \frac {ax+b-8}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}\\= \lim_{x\to0} \frac {x(a+\frac{b-8}{x})}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}=\lim_{x\to0} \frac {(a+\frac{b-8}{x})}{\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}$$ which should be definite and equals to $5/12$. So the term $(a+\frac{b-8}{x})$ should be certain and then $b$ must be $8$ at $x=0$. Now put $b=8$ in your fraction: $$\lim_{x\to0} \frac {a}{\big(\sqrt[3]{(ax+8)^2}+2\sqrt[3]{ax+8}+4\big)}$$ which is $5/12$. I think it is easy to find $a$.
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|
Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
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Assuming $f$ is sufficiently smooth, repeated application of the fundamental theorem of the calculus gives (finite Taylor expansion)
$$f(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \int_0^x (x-t)^2 \frac{(f'''(t)-f'''(0))}{2!}\, dt$$
Using the fact that $\sin' = \cos, \cos' = -\sin$, we can then expand $f(x) = \sin x$ as
$$ \sin x = x -\frac{x^3}{6} + \int_0^x (x-t)^2 \frac{(-\cos t +1)}{2!}\, dt$$
Let $\epsilon>0$, and choose $\delta>0$ such that if $|t|< \delta$, then $|1-\cos t| < \epsilon$. Then, replacing $1-\cos t$ by $\epsilon$ and integrating, we have the estimate
$$ \left|\sin x - x + \frac{x^3}{6} \right| \leq \epsilon \frac{|x|^3}{6}$$
If $0 < |x| < \delta$, then dividing through by $|x^3|$ gives:
$$ \left| \frac{\sin x - x}{x^3} +\frac{1}{6} \right| \leq \frac{\epsilon}{6} < \epsilon$$
The desired limit follows.
|
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|
If I add a constant $c$ to each fraction's numerator and denominator in a sequence of fractions, how is the sequence affected? Given a sorted ascending sequence of fractions, if I add a constant $c$ to each fraction's numerator and denominator, how is the sequence affected?
For example, if I have a sequence in ascending order:
$$\frac12, \frac38, \frac57, \frac ab,\ldots$$
if I add a constant $c=1$
$$\frac{1+c}{2+c}, \frac{3+c}{8+c}, \frac{5+c}{7+c}, \frac{a+c}{b+c},\ldots$$
will my sequence still be sorted in ascending order?
Thank you!
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You need to check that $\frac{a}{b} < \frac{x}{y} \implies \frac{a+c}{b+c} < \frac{x+c}{y+c}$. The latter statement is equivalent to $(a+c)(y+c) < (x+c)(b+c)$ and this to $ay + ac + yc < xb + xc + bc$ as $ay < xb$ is given, you only need to see that $a-b < x-y$. This need not be true.
Consider $\frac{a}{b} = \frac{1}{2}$, $\frac{x}{y} = \frac{3}{5}$, $c=2$. Then $\frac{1+2}{2+2} = \frac{3}{4}$ and $\frac{3+2}{5+2} = \frac{5}{7}$. Finally $3 \cdot 7 > 5 \cdot 4$.
Note that I assumed everything positive. With negative values it is even more obvious. Consider $\frac{1}{2}$ with $\frac{2}{101}$ and $c = -1$.
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|
Finding $\int\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x$. In Arnold's Mathematical Methods of Classical Mechanics he uses,
$$\int_0^1\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x = \left\{\frac{\pi}{2-b} : 0\leqslant b < 2\right\}$$ but he doesn't explain how to get it.
Via Mathematica (wolfram alpha works also), $$\int\frac{1}{\sqrt{x^b - x^2}}\,\text{d}x = -\frac{2x^{b/2}\sqrt{1-x^{2-b}}\arcsin{\big(x^{(2-b)/2} }\big)}{(2-b)\sqrt{x^b-x^2}}.$$
(Besides using a CAS) I am not sure how to solve the general case or the specific improper integral.
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Write it in the following way:
$$\int_0^1\frac{dx}{\sqrt{x^b-x^2}}=\int_0^1 dx\,\frac{x^{-b/2}}{\sqrt{1-x^{2-b}}}.$$
Now make a trigonometric substitution $x^{1-\tfrac{b}{2}}=\sin\theta$ so that $\left(1-\frac{b}{2}\right)x^{-b/2}dx=\cos\theta\,d\theta$ and the integral becomes (ignore the bounds for a moment)
$$=\int d\theta\,\frac{\cos\theta}{\left(1-\frac{b}{2}\right) \cos \theta}=\left.\frac{1}{1-\frac{b}{2}}\sin^{-1}\left(x^{1-b/2}\right)\right|_0^1=\frac{\pi}{2-b}.$$
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|
How to find the roots of $x³-2$? I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\sqrt[3] 2$ and $\sqrt[3] {2}e^{\frac{2\pi}{3}i}$ but I don't why.
The first one is easy to find, but the another two roots?
I need help
Thank you
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If $\omega^3 = 1$ and $x^3 = 2$ then $(\omega x)^3 = \omega^3 x^3 = 2$.
Possible values of $\omega$ are $e^{\frac{1}{3}2 i \pi}$, $e^{\frac{2}{3}2 i \pi}$ and $e^{\frac{3}{3}2 i \pi}$. This is because $1 = e^{2 i \pi} = (e^{\frac{1}{k} 2 i \pi})^k$.
So the solutions of $x^3 - 2 = 0$ are $e^{\frac{1}{3}2 i \pi} \sqrt[3]{2}$, $e^{\frac{2}{3}2 i \pi} \sqrt[3]{2}$ and $\sqrt[3]{2}$.
|
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|
A geometry problem proposed at national olympiad. Let $ABC$ be a triangle with $\angle ABC=90^{\circ}$. We have: $$1) BH \perp AC;
$$ $$2)AD \text{ the bisector of } \angle{A} \text{ and } AD\cap BH=\{Q\},D\in BC;$$
$$3) CE \text{ the bisector of } \angle C \text{ and } CE \cap BH =\{P\},E \in AB;
$$
$$4) CE \cap AD ={I};$$
$$5) NE=NP;$$
$$6) QM=MD;$$
Prove that: $$NM \parallel AC .$$
This problem was proposed this year to National Olympiad from Romanian.
The solution can be check here: http://onm2012.isjcta.ro/doc/9_barem.pdf .
What I cannot understand is the the following relation:
$$ \frac{QA}{QD}=\frac{c^2}{a^2}\cdot \frac{b+c}{c}.$$
Thanks :)
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See the figure below:
Let $K$ be a point such that $K \in BQ$ and $\angle QKD$ is a right angle.
Using similarity and angle bisector theorem we get:
$$\frac{AH}{HC}=\frac{c^2}{a^2} \quad(1)$$
and
$$\frac{DB}{DC}=\frac{c}{b}. \quad(2)$$
From equation $(2)$ we conclude that
$$\frac{DB}{BC}=\frac{c}{b+c}. \quad(3)$$
Note that $\triangle BDK \sim \triangle BCH$, therefore
$$\frac{DB}{BC}=\frac{KD}{HC}. \quad(4) $$
From $(3)$ and $(4)$ we get:
$$\frac{KD}{HC}=\frac{c}{b+c}. \quad(5)$$
Dividing $(1)$ by $(5)$ we get:
$$\frac{AH}{KD}=\frac{c^2}{a^2} \cdot \frac{b+c}{c}. \quad(6)$$
But as $\triangle AHQ \sim \triangle DKQ$, we can conclude that
$$\frac{QA}{QD}=\frac{AH}{KD}= \frac{c^2}{a^2} \cdot \frac{b+c}{c}$$
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|
If the order of $a$ modulo $p$ is 6, is it true that the order of $(1-a)$ modulo $p$ is also 6? Given a prime $p$ and an integer $a$. If the order of $a$ modulo $p$ is 6, is it true that the order of $(1-a)$ modulo $p$ is also 6? If so prove it, if not give a counterexample.
It seems right. So I aimed to show that
$(1-a)^6 \equiv 1 \pmod p$, and also
$(1-a)^i \ne 1 \pmod p$ for all $i = 1, 2, 3, 4, 5$
Am I on the right track? Or is there any other way of thinking?
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Note that $a$ has order $6$ iff $a$ is a solution of the congruence $x^6-1\equiv 0\pmod{p}$, but does not have order $\lt 6$.
We can factor $x^6-1$ as
$$x^6-1=(x^3-1)(x^3+1)=(x^2-1)(x^2+x+1)(x^2-x+1).$$
If $a^3-1\equiv 0\pmod p$ or $a+1\equiv 0\pmod{p}$, then $a$ has order $\lt 6$. So any element of order $6$ must be a solution of the congruence $x^2-x+1\equiv 0\pmod{p}$.
Conversely, let $p$ be a prime $\ge 7$. Suppose that $a^2-a+1\equiv 0\pmod p$. It is clear that $a\not\equiv 1\pmod{p}$. And we cannot have $a\equiv -1\pmod{p}$ unless $(-1)^2-(-1)+1\equiv 0\pmod{p}$, which forces $p=3$. And we cannot have $a^2+a+1\equiv 0\pmod{p}$, for that would imply that $(a^2+a+1)-(a^2-a+1)\equiv 0\pmod{p}$. This is impossible, since $2a\equiv 0\pmod{p}$ implies that $a\equiv 0\pmod{p}$, and thus that $a^2-a+1\not\equiv 0\pmod{p}$.
So for $p \ge 7$, the elements of order $6$ are precisely the solutions of the congruence $x^2-x+1\equiv 0\pmod p$. Suppose that $a$ is such a solution. Then $(1-a)^2-(1-a)+1=(1-2a+a^2)-(1-a)+1=a^2-a+1\equiv 0\pmod{p}$, and therefore $1-a$ is also a solution of $x^2-x+1\equiv 0\pmod{p}$. (The congruence has no more than $2$ solutions.)
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Structure of $x^2 + xy + y^2 = z^2$ integer quadratic form The pythagorean triples $x^2 + y^2 = z^2$ can be solved in integers using rational parameterization of solutions to $x^2 + y^2 = 1$.
It goes through $(1,0)$, then consider the line $y = -k (x - 1)$ so that $x^2 + k^2(x-1)^2 = 1$
We get $(1+ k^2 )x^2 - 2k^2 x + k^2 = 1$ or $x = \frac{k^2-1}{k^2+1}$ and $y=\frac{-2k}{k^2+1}$ and $z=1$
Then set $k= m/n$, for $(x,y,z) = (k^2-1,2k,k^2+1)=(m^2-n^2,2mn,m^2+n^2)$
What happens for 60-degree angle triangles $x^2+xy+y^2 = z^2$ ? We could look for rational solutions to
$x^2 + xy + y^2 = 1 $ and $(1,0)$ works again. Intersect with the line $y= k (x-1)$ ...
get integer solutions: $(m^2-n^2, -m^2+2mn, m^2 - m n + n^2)$ A similar derivation was obtained earlier on math.StackExchange
In the case of pythagorean triple we can build new solutions $(m^2-n^2,2mn,m^2+n^2)$ from old using the maps
$$ (m,n) \mapsto (2m-n,m) \text{ or } (2m+n,m) \text{ or } (m+2n,n)$$
Can I find something similar to the Pythagorean triple tree for this quadratic form, $x^2 + xy + y^2 = z^2$?
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Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$
more:
$X=(a+4)p^2-2ps$
$Y=3p^2-4ps+s^2$
$Z=(2a+5)p^2-(a+4)ps+s^2$
more:
$X=8s^2-4ps$
$Y=p^2-(4-2a)ps+a(a-4)s^2$
$Z=-p^2+4ps+(a^2-8)s^2$
For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.
$X=3s^2+2ps$
$Y=p^2+2ps$
$Z=p^2+3ps+3s^2$
more:
$X=3s^2+2ps-p^2$
$Y=p^2+2ps-3s^2$
$Z=p^2+3s^2$
In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.
$X=s^2-bp^2$
$Y=ap^2+2ps$
$Z=bp^2+aps+s^2$
$p,s$ - integers asked us.
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Integer Solutions to $x^2+y^2=5z^2$ I'm looking for a formula to generate all solutions $x$, $y$, $z$ for $x^2 + y^2 = 5z^2$.
Any advice?
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Ok, so I am assuming rational solutions. This method can yield a parametrization of all integer solutions without too much work.
Note that $(1,2)$ lie on the circle $x^2 + y^2 = 5$. Let $x_0 = 1, y_0 = 2$. Now suppose $x^2 + y^2 = 5$. Let $m = x - x_0, n = y - y_0$, then we have $$m^2 + 2mx_0 + x_0^2 + n^2 + 2ny_0 + y_0^2 = 5$$
And thus $m^2 + 2mx_0 + n^2 + 2ny_0 = 0$. Let $\lambda = \frac{m}{n}$. Then we have: $$n^2\lambda^2 + 2n\lambda x_0 + n^2 + 2ny_0 = 0$$
$$n\lambda^2 + 2\lambda x_0 + n + 2y_0 = 0$$
$$n = \frac{-2y_0 - 2\lambda x_0}{1 + \lambda^2}$$
Plugging in $x_0 = 1, y_0 = 2$:
$$n = \frac{-4 - 2 \lambda}{1 + \lambda^2}$$
Thus it follows $$(x,y) = \left (1 + \frac{-4\lambda - 2\lambda^2}{1 + \lambda^2}, 2 + \frac{-4 - 2 \lambda}{1 + \lambda^2} \right )$$
where $\lambda$ is an arbitrary number in $\mathbb{Q}$.
Now for $x^2 + y^2 = 5z^2$, we simply need: $$(x,y,z) = \left (z + \frac{-4z\lambda - 2z\lambda^2}{1 + \lambda^2}, 2z + \frac{-4z - 2z \lambda}{1 + \lambda^2},z \right )$$
If you want solutions in $\mathbb{Z}$, it takes only a little more work to finish.
EDIT: So either I made a massive reading failure or the author changed the title. So here's how to finish.
A slightly neater form to work with is $$(x,y,z) = \left (1 + \lambda^2 -4\lambda - 2\lambda^2, 2 + 2\lambda^2 -4 - 2\lambda, 1 + \lambda^2 \right )$$
Letting $\lambda = \frac{m}{n}$, we see that:
$$(x,y,z) = \left (m^2 - n^2 -4mn, 2n^2 -2m^2 - 2mn, m^2 + n^2 \right )$$
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Proof that a sum $x$ is found in$ [1, n (n + 1) / 2]$ squence In the subset sum existance problem, we have a sequence of integers. We are given an integer $x$ that we should look for a possible subsequence that sum to $x$. For example, if we have the sequence $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $x = 39$ then we have a subsequence $\{ 10, 9, 8, 7, 5 \}$ that sums up to $39$.
I want to prove that any sequence $[1, n]$ have a subsequence sum for any $x \in [1, n(n + 1) / 2]$. In the previous example, $n = 10$ so that any $x$ between $1$ and $55$ inclusive can be found. I have made a program and tested that this is true. Now I want to determine a proof.
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Here is an alternative approach: consider the case for $n = 5$.
Then $n(n+1)/2 = 5(6)/2 = 15$.
Observe that we can write the numbers $1, \ldots, 15$ as follows:
$1, 2, 3, 4, 5,$
$1+5, 2+5, 3+5, 4+5,$
$1+4+5, 2+4+5, 3+4+5,$
$1+3+4+5, 2+3+4+5,$
$1+2+3+4+5$.
The number of elements summed for entries in the above rows, respectively, are: $1, 2, 3, 4, 5$.
By construction, they are distinct and increasing by one, beginning with $1$ and ending with $15$.
The above can be similarly written out for the general $n$ instead of $15$.
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Computing the indefinite integral $\int x^n \sin x\,dx$ $\newcommand{\term}[3]{
\sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!}
}$
I am trying to prove that for $n \in\mathbb N$,
$$
\int x^n \sin x \, dx
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{(n-1)}{k}{n-2k-1}
$$
I started with differentiation, and this is what I got:
If we define $f(x)$ as
$$
f(x)
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{n-1}{k}{n-2k-1}
$$
then we have
$$
\begin{align*}
f’(x)
&= \cos x \term{n}{k+1}{n-2k-1}
- \sin x \term{n}{k+1}{n-2k} \\
&\qquad
+ \sin x \term{(n-1)}{k}{n-2k-2}
+ \cos x \term{(n-1)}{k}{n-2k-1} \\[8pt]
&= \cos x \left[
\term{n}{k+1}{n-2k-1}
+ \term{(n-1)}{k}{n-2k-1}
\right] \\
&\qquad
+ \sin x \left[
\term{(n-1)}{k}{n-2k-2}
- \term{n}{k+1}{n-2k}
\right]
\end{align*}
$$
I don't know how to go on, because of the different limits of the sum with $\lfloor{n/2}\rfloor$ and $\lfloor{(n-1)/2}\rfloor$.
|
I finally got my proof with differentiation finished, too.
If
$$f(x) = \sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k}{n!\over(n-2k)!}\cos x+\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^kx^{n-2k-1}{n!\over(n-2k-1)!}\sin x$$
with $n\in \Bbb N$.
then
$$\begin{align}f'(x) &= \left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}\right)\cos x\\&{}\quad+\left(\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k}x^{n-2k-2}{n!\over(n-2k-2)!}-\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k}{n!\over(n-2k)!}\right)\sin x\end{align}$$
$$\begin{align}&= \left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}\right)\cos x\\&{}\quad+\left(\sum_{k=2}^{\lfloor{(n-1)/2}\rfloor+2}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}-\sum_{k=1}^{\lfloor{n/2}\rfloor+1}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}\right)\sin x\end{align}$$
$$\begin{align}&=\left(\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}\right)\cos x\\ &{}\quad+\left(\sum_{k=2}^{\lfloor{(n-1)/2}\rfloor+2}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}-\sum_{k=2}^{\lfloor{n/2}\rfloor+1}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}\right)\sin x\\ &{}\quad+ x^n\sin x\end{align}$$
Now we have to show that (1) $$ \sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!} = 0$$ and (2) $$\sum_{k=2}^{\lfloor{(n-1)/2}\rfloor+2}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!}-\sum_{k=2}^{\lfloor{n/2}\rfloor+1}(-1)^{k}x^{n-2k+2}{n!\over(n-2k+2)!} = 0$$
For even numbers:
$$ \lfloor{n/2}\rfloor = n/2 $$
$$ \lfloor{(n-1)/2}\rfloor = (n-2)/2 $$
For odd numbers:
$$ \lfloor{n/2}\rfloor = (n-1)/2 $$
$$ \lfloor{(n-1)/2}\rfloor = (n-1)/2 $$
Also $n!$ is only defined for $$n\ge 0$$
(1) The odd case is trivial because of $$\lfloor{(n-1)/2}\rfloor = \lfloor{n/2}\rfloor$$
Even: We need $$ n -2k -1 \ge 0 $$ so
$$ k \le \lfloor{(n-1)/2}\rfloor = (n-2)/2 $$
We get:
$$ \sum_{k=0}^{(n-2)/2}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!}-\sum_{k=0}^{(n-2)/2}(-1)^{k+1}x^{n-2k-1}{n!\over(n-2k-1)!} = 0$$
With the same argumentation you can show (2).
It remains:
$$= 0\cos x+0\sin x+ x^n\sin x= x^n\sin x$$
|
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find $f(x)$ when $3f(x-6)-2f(x-9)=x^2-54$ I can easily show that with the assumption $f$ is a polynomial $f(x)=x^2$. But without that assumption how can I prove that $f(x)=x^2$???. I have tried many change of variables $x=u+k$ but to no result. I am lost here
|
Let $f(x)$ be polynomial of $x$.
Let $n\ge 3,$ be the smallest power of $x$ whose coefficient$(a)$ may be $\ne 0$.
The coefficient of $x^n$ in $3f(x−6)−2f(x−9)$ is $3a-2a=a$
But the coefficient of $x^n$ where $n\ge 3$ in $x^2−54$ is $0\implies a=0$
So, $f(x)$ can not contain any higher powers $(\ge 3)$ of $x.$
So, $f(x)=bx^2+cx+d$
$3f(x−6)−2f(x−9)=x^2−54$
$b(x-h)^2+c(x-h)+d=x^2(b)+x(c-2bh)+(d-ch+bh^2)$
The coefficient of $x^2$ in $3f(x−6)−2f(x−9)$ is $3b-2b=b$
But the coefficient of $x^2$ in $x^2−54$ is $1\implies b=1$
The coefficient of $x$ in $f(x−h)$ is $c-2bh=c-2h$
So, the coefficient of $x$ in $3f(x−6)−2f(x−9)$ is $3\{c-2(-6)\}-2\{c-2(-9)\}=c$
But the coefficient of $x$ in $x^2−54$ is $0\implies c=0$
The constant term in $f(x−h)$ is $d-ch+bh^2=d+h^2$
The constant term in $3f(x−6)−2f(x−9$ is $3\{d+(-6)^2\}-2\{d+(-9)^2\}=d-54$
The constant term in $x^2−54$ is $-54\implies d-54=-54\implies d=0$
So, $f(x)=x^2+0x+0=x^2$
|
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|
How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq \frac{21}{4}$$
|
For the second and the third question,a generalized version is presented here:
Let $a,b,c>0$ and $a+b+c=3$,then
$$f(a,b,c)=(\sum\frac{ab}{c})+\lambda abc\geq3+\lambda$$
for $1\leq \lambda\leq9/4$,where the constant $9/4$ is optimal.
It is easy to see that 9/4 is optimal(a simple comparation between $f(1,1,1)$ and $f(2,1/2,1/2)$).
Proof:
Without loss of generality,we assume that $a\geq b\geq c$.Then $1\leq a<3$.
$$f(a,b,c)=a(\frac{b}{c}+\frac{c}{b})+\frac{bc}{a}+\lambda abc=\frac{a(b+c)^2}{bc}+(\lambda a+\frac{1}{a})bc-2a$$
Recall that $a+b+c=3$,then
$$f(a,b,c)=\frac{a(3-a)^2}{bc}+(\lambda a+\frac{1}{a})bc-2a$$
Given $p,q>0$,it is obvious that function $u(x)=px+q/x$ is monotone decreasing on interval $(0,\sqrt{q/p}]$.We take $(\lambda a+\frac{1}{a},a(3-a)^2)$ as $(p,q)$,then $f(a,b,c)$ is a monotone decreasing function(for $bc$) on the interval $(0,\sqrt{q/p}]$ with $a$ fixed.
AM-GM inequality suggests that $$bc\leq\frac{(b+c)^2}{4}=\frac{(3-a)^2}{4},$$ and it is natural to test whether $(b+c)^2/4\leq\sqrt{q/p}$,i.e.,
$$\frac{(3-a)^2}{4}\leq\frac{a(3-a)}{\sqrt{\lambda a^2+1}}$$
It suffices to show that $$\frac{(3-a)^2}{16}\leq\frac{a^2}{\lambda a^2+1},$$i.e.,
$$\lambda\frac{(3-a)^2}{16}+\frac{1}{\lambda a^2+1}\leq 1$$.
Because $1\leq a<3$,$LHS\leq\lambda/4+1/(\lambda+1)\leq 1$ for every $1\leq \lambda\leq 3$
Hence $f(a,b,c)$ achieve its minimum(for fixed $a$) when $bc=(3-a)^2/4$.
$$\min_{a fixed}f(a,b,c)=(\lambda a+\frac{1}{a})\frac{(3-a)^2}{4}+2a$$.
We just need to show that
$$(\lambda a+\frac{1}{a})\frac{(3-a)^2}{4}+2a\geq 3+\lambda$$
for every $1\leq \lambda\leq 9/4$,which is equivalent to show that
$$\lambda(a-1)^2(a^2-4a+\frac{9}{\lambda})\geq 0$$
for every $1\leq \lambda\leq 9/4$.
It suffices to prove that $(a^2-4a+\frac{9}{\lambda})\geq 0$ for for every $1\leq \lambda\leq 9/4$,and it is quite obvious.
Q.E.D.
|
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|
The Length of a Bisector How can I prove that :
$$b_{A}=\frac{1}{b+c} \cdot \sqrt{bc\left[(b+c)^2-a^2\right]} ?$$ where $b_{A}$ is the length of the bisector from $A$ .
Thanks :)
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Let $P$ be the point where the bisector of $\angle A$ meets $BC$. Let $x=BP$ and $y=PC$. Then $x+y=a$. Moreover, by a standard theorem on angle bisectors, we have
$\dfrac{x}{c}=\dfrac{y}{b}$, or equivalently $bx=cy$.
We have two linear equations in two unknowns. Solve for $x$ and $y$. We obtain
$$x=\frac{ac}{b+c}\qquad\text{and}\qquad y=\frac{ab}{b+c}.$$
Now let $\theta=\angle APB$ and $\phi=\angle APC$. Then $\cos\phi=-\cos\theta$. By the Cosine Law on $\triangle BPA$, we have
$$c^2=b_A^2+x^2 -2b_A \,x\cos\theta.\tag{$1$}$$
Similarly
$$b^2=b_A^2 +y^2+2b_A \,y\cos\theta.\tag{$2$}$$
Multiply both sides of $(1)$ by $y$, and both sides of $(2)$ by $x$, and add. We get
$$yc^2+xb^2=(x+y)b_A^2+xy(x+y).$$
Substituting our values for $x$ and $y$, and noting that $x+y=a$, we obtain
$$\frac{abc^2}{b+c}+\frac{acb^2}{b+c}=ab_A^2+\frac{a^3bc}{(b+c)^2}.$$
This simplifies to
$$bc=b_A^2+\frac{a^2bc}{(b+c)^2},$$
or equivalently
$$b_A^2=\frac{1}{(b+c)^2}\left(bc\left[(b+c)^2-a^2\right]\right),$$
which is what we wanted to show.
|
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|
The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left(t + \frac{\pi}{3}\right)$
The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left( t + \frac{\pi}{3}\right)$
(a) At what time(s) in the interval $[0,2\pi]$ do the particles meet?
(b) What is the farthest apart that the particles every get?
(c) When in the interval $[0,2\pi]$ is the distance between the particles changing the fastest?
I am studying for the AP Calculus BC exam and this is a problem out of the Calculus Problem book. I know the answers: (a) $\frac{\pi}{3}, \, \frac{4\pi}{3}$ (b) 1 (c) $\frac{\pi}{3}, \, \frac{4\pi}{3}$
I am have trouble with (a)
I rewrote
$$\sin\left(t + \frac{\pi}{3}\right) \Rightarrow \frac{\sqrt{3}}{2}\sin t + \frac{1}{2}\cos t$$
I get this which I can't seem to solve for the answer given:
$$\left(2 - \sqrt{3}\right)\sin t = \cos t $$
For part (b), I would maximized the distance formula. $d(x)=\sqrt{x_1^2 + x_2^2}$
Set the 1st derivative to zero and use the 2nd derivative test to find the max.
for part (c) I would find the maximum of $d^{\prime}(x)$
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(a)
For coincidence, $\sin(t+\frac{\pi}3)=\sin t$
So, $t+\frac{\pi}3=n\pi+(-1)^nt$ where $n$ is any intgere.
If $n=2m$(even), $t+\frac{\pi}3=2m\pi+t\implies 2m\pi=\frac{\pi}3$ which is impossible.
If $n=2m+1$(odd), $t+\frac{\pi}3=(2m+1)\pi-t\implies t=(6m+2)\frac{\pi}6=\frac{(3m+1)\pi}3$
Putting $m=0,t=\frac{\pi}3$
Putting $m=1,t=\frac{4\pi}3$
(b) We need to maximize $\mid\sin(t+\frac{\pi}3)-\sin t\mid$
Now, $\sin(t+\frac{\pi}3)-\sin t=\sin t(\cos \frac{\pi}3-1 )+\cos t \sin\frac{\pi}3=\sin(\frac{\pi}3-t) $
So, the distance will be maximum if $\sin(\frac{\pi}3-t)$ is minimum/maximum.
But $-1\le \sin(\frac{\pi}3-t)\le 1$
So, the distance will be maximum $(=1)$
if $\sin(\frac{\pi}3-t)=\pm1\implies \cos(\frac{\pi}3-t)=0\implies \frac{\pi}3-t=(2r+1)\frac{\pi}2,t=\frac{\pi}3-(2r+1)\frac{\pi}2$ where $r$ is any integer.
(c)The change of distance=$$\mid\frac{d\{\sin t-\sin(t+\frac{\pi}3)\}}{dt}\mid=\mid\cos t-\cos(t+\frac{\pi}3)\mid$$
Now, $\cos t-\cos(t+\frac{\pi}3)=\cos t(1-\cos\frac{\pi}3)-\sin t\sin \frac{\pi}3=\cos(t+\frac{\pi}3) $
For the distance between the particles changing the fastest,
$\cos(t+\frac{\pi}3)=\pm 1$ as $-1\le \cos(\frac{\pi}3-t)\le 1$
$\implies \sin(t+\frac{\pi}3)=0$
$\implies t+\frac{\pi}3=s\pi$ where $s$ is any integer.
So, $t=s\pi-\frac{\pi}3$
Putting $s=0,t=-\frac{\pi}3$ which is not $[0,2\pi]$
Putting $s=1,t=\frac{2\pi}3$
Putting $s=2,t=\frac{5\pi}3$
Putting $s=3,t=\frac{8\pi}3>2\pi$ so is not acceptable.
|
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|
Find $x$ such that $\arctan(3/2)+\dots=\arctan x$ Find $x$ such that
$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$
|
\begin{align}
& {} \quad \tan(a+b+c+d) \\
& = \frac{\tan a+\tan b+\tan c+\tan d\ \overbrace{ - \tan a\tan b\tan c - \cdots}^\text{4 terms}}{1-\ \underbrace{\tan a \tan b- \cdots}_\text{6 terms} +\tan a\tan b\tan c\tan d}
\end{align}
Therefore
\begin{align}
& {} \quad \tan\Big(\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3)\Big) \\
& = \frac{\frac32 + \frac54 + \frac{-5}2 +\frac{-8}3 - (\text{sum of four terms, each a product of three numbers})}{1 - (\text{sum of six terms, each a product of two numbers}) + (\text{a product of four numbers}))}
\end{align}
You get a number. Then you say $\tan x=\text{that number}$, and find the tangent.
|
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|
Combinatorics: how many ways are there to form 100 with 1, 2 and 5
Possible Duplicate:
Making Change for a Dollar (and other number partitioning problems)
dollar notes in denominations of 1, 2 and 5. How many ways are there to form exactly $100 using just multiples of these notes?
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$x+2y+5z=100$
$0\le x\le100\implies 0\le 100-2y-5z\le100\implies 0\le 2y+5z\le 100$
$\implies 0\le y\le \lfloor \frac{100-5z}2\rfloor$
As $0\le 5z\le 100, 0\le z\le 20 $
If $z$ is odd $=2a+1, 0\le y\le \lfloor \frac{100-5(2a+1)}2\rfloor=47-5a, 47-5a+1=48-5a$ values of $z$
So, $0\le 2a+1\le 20\implies 0\le a\le9 $
If $z$ is even,$=2b, 0\le y\le \lfloor \frac{100-5(2b)}2\rfloor=50-5b, 50-5b+1=51-5b$ values of $z$
So, $0\le 2b\le 20\implies 0\le b\le10 $
So, the number of combinations =$\sum_{ 0\le a\le9}(48-5a)+\sum_{0\le b\le10}(51-5b)=48\cdot10+51\cdot 11-5\sum_{ 0\le a\le9}a-5\sum_{0\le b\le10}b=541$
|
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|
Infinite series and logarithm Is it true that:
$$\log_e 2 = \frac12 + \frac {1}{1\cdot2\cdot3} + \frac {1}{3\cdot4\cdot5}+ \frac{1}{5\cdot6\cdot7}+ \ldots$$
It was one of my homeworks . Thanks!
|
$$a_n = \dfrac1{(2n-1)2n(2n+1)} = \dfrac12 \left( \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1}\right)$$
and $$a_0 = \dfrac12$$
Hence,
\begin{align}
\sum_{k=0}^n a_k & = \dfrac12 + \dfrac12 \left(\dfrac11 - \dfrac12 - \dfrac12 + \dfrac13 + \dfrac13 - \dfrac14 - \dfrac14 + \dfrac15 + \cdots + \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1}\right)\\
& = \dfrac12 \left(1 + 1 - \dfrac12 - \dfrac12 + \dfrac13 + \dfrac13+ \cdots + \dfrac1{2n-1} + \dfrac1{2n-1} - \dfrac1{2n} - \dfrac1{2n} + \dfrac1{2n+1} \right)\\
& = 1 - \dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots + \dfrac12\dfrac{1}{2n+1}
\end{align}
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|
How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$? I was tackling through an olympiad practice book when I saw one of these problems:
If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ - \ 1 \over\sqrt{x}}$?
The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that I have the following:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$
|
When you know the answer, it is often easier...
Indeed, begin with $(x-1)/\sqrt{x} = 2\sqrt{2}$
It is equivalent to $x-1 = 2\sqrt{2x}$
Equivalent to $(x-1)^2 = 8x$
Equivalent to $x^2-2x+1 = 8x$
Equivalent to $x^2-10x+1 = 0$
Equivalent to $(x-a)(x-b) = 0$ where $a = 5+2\sqrt{6}$ and $b = 5-2\sqrt{6}$
Equivalent to {$x=a$ or $x=b$}
Of course, since these are equivalences, this suffices to prove the fact. However, if you want to write it in a more straightforward way, it suffices to go back from the end to the beginning:
Let $x = 5+2\sqrt{6}$. Then $(x-(5+2\sqrt{6})(x-(5-2\sqrt{6}))=0$. But this product can be expanded as $x^2-10x+1$. So, we have $x^2+1 = 10x$. At this point, you can generate various identities. For instance, you could conclude $x^2+2x+1 = 12x$. Therefore $(x+1)^2=12x$ and $x+1 = 2\sqrt{3x}$ and so $(x+1)/\sqrt{x} = 2\sqrt{3}$...
|
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|
Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea.
Thanks:)
|
Here is another proof using the Reverse Rearrangment Inequality.
Observe that with $ a \leq b \leq c$, we have $ bc \leq ca \leq ab $ and $a^2 \leq b^2 \leq c^2$. Hence, applying the RRI on these similarly ordered sequences, we conclude that
$ ( a^2 + bc ) ( b^2 + ca) ( c^2 + ab) \leq ( a^2 + ab)( b^2 + bc)(c^2 + ca) = abc ( a + b)(b+c)(c+a) $.
|
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|
Generalized permutation and combination: number of solutions How many solutions are there to the equation: $$x_1 + x_2 + x_3 + x_4 + x_5= 21\;,$$
where $x_i$ is a non-negative integer such that
$$0 \le x_1 \le 3;\; 1 \le x_2 < 4;\text{ and }x_3 \ge 15\;?$$
|
This should be the coefficient of $a^{21}$ in
$(1+a+a^2+a^3)(a+a^2+a^3)(a^{15}+a^{16}+a^{17}+a^{18}+...)(1+a+a^2+a^3+a^4+...)^2$
where $a$ is some real $0<a<1$
= coefficient of $a^5$ in
$(1+a+a^2+a^3)(1+a+a^2)(1+a+a^2+a^3+a^4+...)^3$
= coefficient of $a^5$ in
$(1-a^4)(1-a^3)(1-a)^{-6}$
= coefficient of $a^5$ in
$(1-a^4 - a^3 + a^7)(1-a)^{-6}$
= coefficient of $a^5$ in
$(1-a^4 - a^3 + a^7)(1+6a+\frac{6.7}{2!}a^2 +\frac{6.7.8}{3!}a^3 +\frac{6.7.8.9}{4!}a^4 + ... )$
$= \frac{6.7.8.9.10}{5!} - 6 - \frac{6.7}{2!}$
|
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|
If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
|
$\gcd(a+b, a^2-ab+b^2) = \gcd(a+b, (a+b)^2-3ab) = \gcd(a+b, 3ab)$ by the Euclidean algorithm. If the gcd was $d \ne 1 ,3$, then $d \mid a$ or $d \mid b$ in $3ab$ but then from $a+b$, $d$ would divide the other. Thus, the result follows.
|
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|
How to find the Laplace transform of $\frac{1-\cos(t)}{t^2}$? $$ f(t)=\frac{1-\cos(t)}{t^2} $$ $$ F(S)= ? $$
|
Let
$$F(s) = \int_{0}^{\infty} f(t) \, e^{-st} \, dt = \int_{0}^{\infty} \frac{1-\cos t}{t^2} e^{-st} \, dt. $$
The function $f(t)$ satisfies the bound $ f(t) = O(1 \wedge t^{-2})$, thus it is absolutely integrable and we can apply Leibniz's integral to obtain
$$ F''(s) = \int_{0}^{\infty} (1-\cos t) \, e^{-st} \, dt = \frac{1}{s} - \frac{s}{s^2 + 1}. $$
Integrating and using the condition $F'(\infty) = 0$, we have
$$ F'(s) = \log s - \log \sqrt{s^2 + 1}. $$
Thus we have
$$F(s) = \int \left\{ \log s - \log \sqrt{s^2 + 1} \right\} \, ds. $$
The first term is easily integrated to yield $s \log s - s$. For the second term, note that
\begin{align*}
-\int \log \sqrt{s^2 + 1} \, ds
&= - s \log \sqrt{s^2 + 1} + \int \frac{s^2}{s^2 + 1} \, ds \\
&= - s \log \sqrt{s^2 + 1} + s - \arctan s + C.
\end{align*}
Combining, we obtain
$$ F(s) = s \log s - s \log \sqrt{s^2 + 1} - \arctan s + C. $$
But since $F(\infty) = 0$, we must have $C = \frac{\pi}{2}$ and therefore
\begin{align*}
F(s)
&= s \log s - s \log \sqrt{s^2 + 1} - \arctan s + \frac{\pi}{2} \\
&= s \log \bigg( \frac{s}{\sqrt{s^2 + 1}} \bigg) + \arctan \left(\frac{1}{s}\right).
\end{align*}
|
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|
Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
|
Cancelation:
$$
\begin{array}{cccccccccccccccc}
x & (x^3 & + & x^2 y & + & xy^2 & + & y^3) \\
& & -y & (x^3 & + & x^2 y & + & xy^2 & + & y^3) \\[25pt]
= & x^4 & + & x^3 y & + & x^2y^2 & + & xy^2 \\
& & - & x^3y & - & x^2y^2 & - & xy^2 & - & y^3 \\[25pt]
= & x^4 & & & & & & & - & y^4
\end{array}
$$
|
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}
|
How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$? How do I integrate this guy? I've been stuck on this for hours..
$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$
|
Let $x = \sin^2y$.
(It seems we've started like @Sasha here and like @sos440.)
Then
$$\begin{eqnarray*}
I &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x(1-x)} \\
&=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
+ \underbrace{\frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{1-x}}_{x\to 1-x}
\qquad (\textrm{partial fractions}) \\
&=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
\qquad (\textrm{integral linked above}) \\
&=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)}{x} \left(-\sum_{k=1}^\infty \frac{x^k}{k}\right)
\qquad (\textrm{Taylor expansion for }\log(1-x) ) \\
&=& -\frac{1}{8} \sum_{k=0}^\infty \frac{1}{k+1} \underbrace{\int_0^1 dx\, x^k \log x}_{-1/(k+1)^2}
\qquad (\textrm{standard integral involving log}) \\
&=& \frac{1}{8} \sum_{k=0}^\infty \frac{1}{(k+1)^3} \\
&=& \frac{\zeta(3)}{8}.
\end{eqnarray*}$$
Addendum:
Note that $\csc(y)\sec(y) = \cot y + \tan y$.
Then
$$\begin{eqnarray*}
I &=& \frac{1}{2} \int_{0}^{\pi/2} dy\,
\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y} \\
&=& \frac{1}{2} \int_{0}^{\pi/2} dy\,
\frac{\ln(\sin y)\ln(\cos y)}{\tan y}
+ \underbrace{\frac{1}{2} \int_{0}^{\pi/2} dy\,
\frac{\ln(\sin y)\ln(\cos y)}{\cot y}}_{y\to \pi/2-y} \\
&=& \int_{0}^{\pi/2} dy\,
\frac{\ln(\sin y)\ln(\cos y)}{\tan y}.
\end{eqnarray*}$$
This is exactly the integral linked above.
|
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|
Manipulation of a binomial coefficient In obtaining a formula for the Catalan numbers I have got the expression $-\frac{1}{2}\binom{1/2}{n}(-4)^n$. All my efforts to show that this simplifies to $\frac{1}{n}\binom{2n-2}{n-1}$ have not succeeded. Is there some mistake in the original expression, and if not, how do I simplify it?
Thanks for your time.
|
First we show that
$$\binom{1/2}{n}={1\over n!}(1/2-1)(1/2-2)...(1/2-(n-1))=$$$$={1\over n!}(-1/2)(-3/2)...(-(2n-3)/2)=$$
$$={1\over n!}{(-1)^{n-1}\over2^n}1\cdot3\cdots(2n-3)\frac{2\cdot4\cdot6...(2n-4)}{2^{n-2}(n-2)!}=$$
$$={1\over n!}{(-1)^{n-1}\over2^{2n-2}}\frac{(2n-3)!}{(n-2)!}\frac{(2n-2)(2n-1)2n}{(n-1)n\cdot2^2\cdot(2n-1)}=$$
$$={1\over n!}{(-1)^{n-1}\over2^{2n}}\frac{(2n)!}{n!(2n-1)}=$$
$$={(-1)^{n-1}\over4^n(2n-1)}\frac{(2n)!}{n!n!}={(-1)^{n-1}\over4^n(2n-1)}\binom{2n}{n}$$ then we get
$$\frac{-1}{2}\binom{1/2}{n}(-4)^n=\frac{-1}{2}{(-1)^{n-1}\over4^n(2n-1)}\binom{2n}{n}(-1)^n4^n=$$
$$={(-1)^{2n}\over2(2n-1)}\binom{2n}{n}={1\over2(2n-1)}{2n(2n-1)\over n(n-1)}\binom{2n-2}{n-2}=$$
$$={1\over n-1}\binom{2n-2}{n-2}={1\over n-1}\binom{2n-2}{n}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of a spiral in $\mathbb{C}$ Does the series
$$\sum_{k=0}^{\infty}\frac{i^k}{k!}$$converge, and if so, what is the value of it?
|
Alternately (and equivalently to several of the other answers), if you didn't know the Euler formula but did know the power series for sin and cos, you could reason as follows:
$i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=i^0=1$. Therefore, the numerators of the power series are periodic of period 4; what's more, they split off naturally into a real part (the even members) and an imaginary part (the odd members):
$$\begin{align*}
\sum_{k=0}^{\infty}\frac{i^k}{k!} &= 1+\frac{i}{1!}+\frac{i^2}{2!}+\frac{i^3}{3!}+\frac{i^4}{4!}+\frac{i^5}{5!}+\cdots \\
&= 1+\frac{i}{1!}+\frac{-1}{2!}+\frac{-i}{3!}+\frac{1}{4!}+\frac{i}{5!}+\cdots \\
&=\left(1+\frac{-1}{2!}+\frac{1}{4!}+\cdots\right)+\left(\frac{i}{1!}+\frac{-i}{3!}+\frac{i}{5!}+\cdots\right) \\
&=\left(1-\frac{1}{2!}+\frac{1}{4!}+\cdots\right)+i\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}+\cdots\right) \\
&=\cos(1)+i\cdot\sin(1)\\
\end{align*}$$
|
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|
Finding Laurent Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ at $z=2$. I need to find the Laurent Series for $z=2$, $z=i$, and $1<|z|<2$ for the following function:
$$ f(z)= \frac{z^2 -2z +5}{(z-2)(z^2+1)}.$$
I was thinking if I could get something like $$\frac{1}{1 - (z-2)}$$ I could use geometric series.
Like in this example for $z=-1$:
$$f(z)=\frac{1}{1-z} =\frac{1}{2-(z+1)} =\frac{1}{2\left(1-\frac{z+1}{2}\right)}=\frac{1}{2}\sum \left(\frac{z+1}{2}\right)^n.$$
Unfortunately, I'm not getting anywhere with that approach. What am I missing?
edit: can I rewrite it like this:
$$f(z)=\frac{1}{z-2}-\frac{i}{z+i}+\frac{i}{z-i}$$
and then use Taylor?
|
*
*$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)}=\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}=\frac{1}{z-2}-\frac{2}{z^2+1}=\frac{1}{z-2}-\frac{2}{(z-2)^2+4z-3}=\frac{1}{z-2}-\frac{2}{(z-2)^2+4(z-2)+5}=\frac{1}{z-2}-\frac{2}{5}[1+\{\frac{4}5(z-2)+\frac{1}5(z-2)^2\}]^{-1}.$
Hence etc.
*
*$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)}=\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}=\frac{1}{z-2}-\frac{2}{z^2+1}=\frac{1}{z-2}-\frac{2}{(z-i)(z+i)}=\frac{i}{z-i}+(\frac{1}{z-2}-\frac{i}{z+i})=\frac{i}{z-i}+\frac{1}{z-i-2+i}-\frac{i}{z-i+2i}=\frac{i}{z-i}+\frac{1}{i-2}(1+\frac{z-i}{i-2})^{-1}+\frac{1}{2}(1+\frac{z-i}{2})^{-1}.$
Hence etc.
|
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|
$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} =
-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$
|
Hint: Use partial fractions,
$$\sum_{n=2}^{\infty} \frac {2}{(n^3-n)3^n} = \sum_{n=2}^\infty \frac {1}{3^n} \left( \frac {1}{n-1} - \frac {2}{n} + \frac {1}{n+1}\right)$$
Now, shift the indexing up/down 1 as necessary. Remember to check the power of 3.
|
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"question_score": "6",
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|
Proof of tangent half identity
Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$
I was unable to find any proofs of the above formula online. Thanks!
|
There are several ways to proceed, apart from what Ayman gave
Approach 1: If you know the tangent t-formulas: For $t = \tan \frac {\theta} {2}$,
$$ \sin \theta = \frac {2t}{1+t^2}, \cos \theta = \frac {1-t^2}{1+t^2}$$
Substitute these into the equation and simplify.
Approach 2: (my preference)If you know that $\tan \frac {\theta}{2} = \frac {\sin \theta}{1+\cos \theta} = \frac { 1- \cos \theta}{\sin \theta}$, then by the equivalence class of fractions, we can sum across the numerator and denominator to get
$$\tan \frac {\theta}{2} = \frac {\sin \theta + 1 - \cos \theta}{ 1 + \cos \theta + \sin \theta}$$
Of course, you could also take the difference across the numerator and denominator to get
$$\tan \frac {\theta}{2} = \frac {\sin \theta - 1 + \cos \theta}{ 1 + \cos \theta - \sin \theta}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Help figuring out all the alternative solutions to the integrals of sine and cosine I always worry a lot when doing integrals with trigonometric functions because there's always many ways to write the final answer. I am trying to figure out the general pattern for the various different solutions.
The integral
$$ \int \sin^3\left(x\right)\cos^2\left(x\right)\, \mathrm{d}x $$
My answer (which never seems to be the same with wolfram alpha's):
$$\frac{\cos^5\left(x\right)}{5} + \frac{-\cos^3\left(x\right)}{3} + C$$
wolfram alpha lists many more:
$$\cos^3\left(x\right)\left(\frac{1}{10}\cos\left(2x\right)-\frac{7}{30}\right) + C$$
$$\frac{1}{240}\left(-30\cos\left(x\right) - 5\cos\left(3x\right)+3\cos\left(5x\right)\right)+C$$
$$\frac{\cos^5\left(x\right)}{80} -\frac{\cos^3\left(x\right)}{48}-\frac{\cos\left(x\right)}{8}-\frac{1}{8}\sin^2\left(x\right)\cos^3\left(x\right) + \frac{1}{16}\sin^4\left(x\right)\cos\left(x\right)+\frac{1}{16}\sin^2\left(x\right)\cos\left(x\right) + C$$
Could anyone help me understand what integration procedure would result in those answers?
|
Putting $\cos x=z,dz=-\sin xdx$
$$ \int \sin^3\left(x\right)\cos^2\left(x\right)\, \mathrm{d}x=\int (1-z^2)z^2(-dz)=\int z^4 dz-\int z^2 dz=\frac{z^5}5-\frac{z^3}3+c=\frac{\cos^5x}5-\frac{\cos^3x}3+C $$
(i)
$$\frac{\cos^5x}5-\frac{\cos^3x}3=\frac{\cos^3x}{15}(3\cos^2x-5)=\frac{\cos^3x}{30}(6\cos^2x-10)=\frac{\cos^3x}{30}\{3(1+\cos2x)-10\}$$ as $\cos2x=2\cos^2x-1$
$$\implies \frac{\cos^5x}5-\frac{\cos^3x}3=\frac{\cos^3x(3\cos2x-7)}{30}$$
(ii)
Using the Euler's Identity $e^{iy}=\cos y+i\sin y\implies e^{-iy}=\cos (-y)+i\sin(-y)=\cos y-i\sin y\implies 2\cos y=e^{iy}+e^{-iy} $,
$(2\cos x)^5=(e^{ix}+e^{-ix})^5=(e^{5ix}+e^{-i5x})+\binom5 1(e^{3ix}+e^{-i3x})+\binom5 2(e^{ix}+e^{-ix})=2\cos5x+10\cos 3x+20\cos x$
Similarly, $(2\cos x)^3=(e^{ix}+e^{-ix})^3=e^{3ix}+e^{-3ix}+3(e^{ix}+e^{-ix})=2(\cos 3x+3\cos x)$
Put the values of $\cos5x,\cos 3x$ is the integration result to get $$\int \sin^3\left(x\right)\cos^2\left(x\right)\, \mathrm{d}x=\frac{\cos5x}{80}-\frac{\cos3x}{48}-\frac{\cos x}8+ c$$ which is the last alternative form of wolfram alpha
(iii) As $\sin3x=3\sin x-4\sin^3x,\cos2x=2\cos^2x-1,$ $$\sin^3x\cos^2x=\frac{(3\sin x-\sin3x)(1+\cos2x)}8=\frac{3\sin x-\sin3x+3\sin x\cos2x-\sin3x\cos2x}8$$
Applying $2\sin A\cos B=\sin(A+B)+\sin(A-B),$
$$\sin^3x\cos^2x=\frac{6\sin x-2\sin3x+3(\sin 3x-\sin x)-(\sin5x+\sin x)}{16}=\frac{\sin3x}{16}-\frac{\sin5x}{16}+\frac{\sin x}8$$
Now we can use $$\int\sin mx dx=-\frac{\cos mx}m+C$$
|
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|
Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
|
This is wrong
If you are familiar with majorization, observe that the function is schur concave. Thus, its maximum occurs at a point where all variables are equal, and since that point exists in the constraint set (i.e. $a,b,c\geq 0$ and $a^2+b^2+c^2=3$), $a=b=c=1$ is the maxima. Thus, the inequality comes.
|
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|
Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$
Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$
I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$
|
The identity $k^2-1=(k+1)(k-1)$ shows that
$$
\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{k^2-1}{k^2}=\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2,
$$
and the value of limit should follow.
|
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|
Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as:
$$\frac{x^4}{x(y^2+z^2)}+\frac{y^4}{y(z^2+x^2)}+\frac{z^4}{z(x^2+y^2)}$$ and then I try to apply Cauchy-Buniakowsky but still nothing.
I need a proof/idea without derivatives.
thanks for your help.
|
By Cauchy-Schwarz you want to show that $\sum x(y^2+z^2) \leq 6$. But this is equivalent to
$$\begin{eqnarray}\sum x(3-x^2) \leq 6
\\ \Leftrightarrow 3\sum x \leq 6 + \sum x^3\end{eqnarray}$$
This then follows from $x^3 + 1 + 1 \ge 3x$ (AM-GM) and add up.
|
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|
Missing and parasite roots in the trigonometric equation. I have this equation:
$$\boxed{\cos(2x) - \cos(8x) + \cos(6x) = 1}$$
RIGHT
And its right solution from the textbook is:
$$
\begin{align}
\cos(2x)+\cos(6x)&=1+\cos(8x)\\\\
2\cos(4x)*\cos(2x)&=2\cos^2(4x)\\\\
\cos(4x)*(\cos(2x)-\cos(4x))&=0\\\\
\cos(4x)*2\sin(3x)*\sin(x)=&0
\end{align}
$$
*
*$\cos(4x) = 0 \implies 4x = \frac{\pi}{2} + \pi k \implies \boxed{x=\frac{\pi}{8} + \frac{\pi k}{4}}$
*$\sin(3x) = 0 \implies 3x = \pi k \implies \boxed{x=\frac{\pi k}{3}}$
*$\sin(x) = 0 \implies \boxed{x=\pi k}$
WRONG
And this is the way how I have been tried to solve it (with different approach) and here I need a help to determine where I'm wrong:
$$
\begin{align}
\cos(2x) - \cos(8x) + \cos(6x) &= 1\\
\cos(2x) + \cos(6x) &= 1 + \cos(8x)
\end{align}
$$
Then using this formula:
$$\bbox[Beige]{\boxed{\cos\alpha + \cos\beta = 2\cos(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})}} \tag{1}$$
$$2\cos(4x)\cos(2x) = 1 + \cos(8x)$$
And then similarly:
$$\bbox[Beige]{\boxed{\cos2\alpha=2\cos^2\alpha-1}} \tag{2}$$
$$2\cos(4x)\cos(2x) = 1 + 2\cos^2(4x) - 1$$
After that I subtracted $1$ from $1$ in the right part and divided both right and left parts by $2$:
$$\cos(4x)\cos(2x) = \cos^2(4x)$$
Then once again I divided both parts. This time by $\cos(4x)$ and with the aim not to lose a root I checked whether $\cos(4x)$ can be equal to zero.
$$\cos(4x) = 0 \implies 4x = \frac{\pi}{2} + \pi k \implies \boxed{x=\frac{\pi}{8} + \frac{\pi k}{4}}$$
so here I found the first root. And using formula (2):
$$\cos(2x) = \cos(4x) \implies \cos(2x) = 2\cos^2(2x) - 1$$
then I simply introduced new variable and solved quadratic equation:
$$u = \cos(2x)$$
$$
2u^2 - u -1 = 0\\
u_1=1, \, u_2=-\frac{1}{2}
$$
So finally:
*
*$\cos(2x) = 1 \implies 2x = \pi k \implies \boxed{x=\frac{\pi k}{2}}$ (wrong)
*$\cos(2x) = -\frac{1}{2} \implies 2x = \pm \frac{2 \pi}{3} + 2\pi k \implies \boxed{x=\frac{\pi}{3} + \pi k}$ (wrong)
But none of the two last roots fits to the original equation. And I lost another 2 roots. Where is my mistake? Any suggestions will be greatly appreciated. Thanks for attention.
|
Just minor slippage. For example, from $\cos 2x=1$ you concluded that $2x=\pi k$. That is false, we need $k$ to be even: if for example $2x=\pi$, then $\cos 2x=-1$.
I don't understand your concern with $\frac{\pi}{3}$. It works.
|
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|
Is this function differentiable at $(0,0)$? Let $f: \mathbb R^2 \to \mathbb R$ be the function
$$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$
where $(x,y) \neq (0,0)$ and $f(0,0)=0$.
Is $f$ differentiable at $(0,0)$ and if so, how can I prove it?
|
We will prove that $f$ is differentiable at $(0,0)$ with derivative $df_{(0,0)}=0$.
Using $|\sin z|\leq |z|$, we have
$$
\frac{|f(x,y)|}{\sqrt{x^2+y^2}}\leq \frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}+\frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}.
$$
Now using Cauchy-Schwarz, we find
$$
\frac{|x|^3|x+y|}{(x^2+y^2)^{3/2}}\leq \frac{|x|^3\sqrt{2}\sqrt{x^2+y^2}}{(x^2+y^2)^{3/2}}=\frac{|x|^3\sqrt{2}}{x^2+y^2}\leq \frac{\sqrt{2}|x| (x^2+y^2)}{x^2+y^2}=\sqrt{2}|x|\leq \sqrt{2}\sqrt{x^2+y^2}
$$
So this term converges to $0$ as $\sqrt{x^2+y^2}$ tends to $0$.
Next observe that
$$
\frac{|y|}{(x^2+y^2)^{1/2}}\leq \frac{\sqrt{x^2+y^2}}{(x^2+y^2)^{1/2}}=1
$$
and
$$
\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}=\frac{y^2}{x^2+y^2}\cdot|y\ln(x^2+y^2)|\leq \sqrt{x^2+y^2}\ln(x^2+y^2).
$$
Finally,
$$
\frac{|y|}{(x^2+y^2)^{1/2}}\frac{|y^3\ln(x^2+y^2)|}{x^2+y^2}\leq \sqrt{x^2+y^2}\ln(x^2+y^2)
$$
so this term also converges to $0$ as $\sqrt{x^2+y^2}$ tends to $0$, using $\lim_{u\rightarrow 0^+}\sqrt{u}\ln u=0$.
So
$$
\lim_{(x,y)\rightarrow(0,0)} \frac{|f(x,y)-f(0,0)-0|}{\sqrt{x^2+y^2}}=0
$$
which proves exactly our claim.
|
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|
The number of ways to put numbers in squares If one put the numbers from $1$ to $7$ in the squares in stead of letters in which the sum of cells is in descending order like shown in the figure .Every number should be used once or twice or never .Also the order of numbers in every oblique row its sum appear always in descending order.
In how many way one can do that ?
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If the inequalities are strict, $Cxy$ can only be $431$ or $521$ Starting with $521$, we must have $b=6, A=7$ and $AbCD$ fails. So we must have $Cxy=431$. Then $cxz=432$ or $531$ If it is $531$ we must have $abcd=7651$, but then $A=7$ and again $AbCD$ fails. If $b=6, A=7$ and again $AbCD$ fails, so $b=5$. Then $a=7, d=3$ and we can have either $A=7,D=1$ or $A=6, D=2$, giving two solutions. $$\begin {array}{c c c} A & 7 & 6 \\ a & 7& 7\\b & 5 & 5 \\c&4&4\\C&4&4\\d&3&3\\D&1&2\\x&3&3\\y&1&1\\z&2&2\end {array}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Inclusion Exclusion principle question What is the number of surjective (onto) functions
from the set [3] to the set [3].
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We first count the complement. There are $2^5$ functions that "miss" $1$, also $2^5$ that miss $2$, and $2^5$ that miss $3$. Add. We get $3\cdot 2^5$.
But we have double-counted the functions that miss both $1$ and $2$, also the functions that miss $2$ and $3$, also the functions that miss $3$ and $1$. There is $1$ (or if you prefer, $1^5$) of each kind, so we subtract $3\cdot 1^5$.
Thus the total number of onto functions is $3^5-3\cdot 2^5+3\cdot 1^5$.
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|
Expression as a product of disjoint cycles Let $\alpha = (9312)(496)(37215) \in S_n, n \ge 9$. Express $\alpha$ as a product of disjoint cycles.
I know this is probably a really easy question, but my professor didn't elaborate on how to exactly do this and neither does my assigned text. If anyone could elaborate on the algorithm of going about this I would really appreciate it.
Thanks you
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Forgive me if the answer is a little sloppy, this is my first answer on stack exchange.
We begin by writing in the following format,
(37215)=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
5&1&7&4&3&6&2&8&9
\end{pmatrix}
Now, if we take the cycle $(496)$ composed of this, we get
(496)o (37215)=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
5&1&7&9&3&4&2&8&6
\end{pmatrix}
and then one more time
(9312)o ((496)o(37215))=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
5&2&7&3&1&4&9&8&6
\end{pmatrix}
Now, we can conveniently put this in one giant matrix(and breach some mathematical etiquette while we're at it) and write
\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\
5&1&7&4&3&6&2&8&9\\
\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\
5&1&7&9&3&4&2&8&6\\
\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow &\downarrow \\
5&2&7&3&1&4&9&8&6
\end{pmatrix}
This is not necessarily the quickest or most elegant approach, but serves as a good substitute while familiarity with cycle notation grows. In cycle notation, the answer is (15)(37964), because, (from the top row of the bottom matrix to the bottom row)1 goes to 5, which goes back to 1, and 3 goes to 7 which goes to 9 and so on.
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|
Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
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Introduce the notation $[a_0]=\sqrt{a_0}$, and $[a_0,a_1]=\sqrt{a_0+\sqrt{a_1}}$, and so on, including infinite lists:
$$[a_0,a_1,a_2,...]=\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}=\sqrt{a_0+[a_1,a_2,\ldots]}.$$
Generally $[a_0,a_1,\ldots]^2 = a_0 + [a_1,a_2,\ldots]$, so for constant-term lists we have a closed-form solution:
$$
[x,x,\ldots]^2=x+[x,x,\ldots] \implies [x,x,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{1+4x}.
$$
If $b_i \le a_i$ for each $i$, then clearly $[b_0,b_1,\ldots]\le[a_0,a_1,\ldots]$. What happens when a multiplicative factor is introduced? You have
$$k[a_0,a_1,a_2,...]=\sqrt{k^2 a_0 + k^2[a_1,a_2,...]}=[k^2a_0,k^4a_1,k^8a_2,\ldots]$$
In your case,
$[1,2,4,\ldots]=\sqrt{1+[2,4,8,\ldots]}=\sqrt{1+\sqrt{2}[1,1,1/2,1/16,\ldots]}$. Using the bounds
$$
\sqrt{2}=[1,1]\le[1,1,1/2,1/16,\ldots]\le[1,1,1,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{5},
$$
you have
$$
1.732 \approx \sqrt{3} \le [1,2,4,\ldots] \le \sqrt{1+\frac{1+\sqrt{5}}{\sqrt{2}}}\approx 1.813.
$$
Tighter bounds can be provided, of course, but this suffices to show that the limit exists.
|
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|
$k=2^n + 1$ is prime $\rightarrow n=2^m$ I am struggling with this proof. I want to prove the contrapositive, $n=2^ab \rightarrow 2^n + 1$ is composite.
My professor gave me a hint,
$n=2^ab$, $b=2r+1 \ge 3$ $\rightarrow$ $2^{2^a}+1 | 2^n+1$. I truly don't know what I am doing and need lots of explanation.
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Recall that if $b$ is odd, then
$$(x+1) \vert x^{b}+1$$
This can be seen immediately from the remainder theorem, since $(-1)^b + 1 = -1 + 1 = 0$. Equivalently, for odd $b$, we have
$$x^b+1 = (x+1)(1-x+x^2-x^3 \pm \cdots - x^{b-2} + x^{b-1})$$
Hence, if we write $n = 2^a \cdot b$, such that $2^a \Vert n$ i.e. $b$ is odd, we have $x^n+1 = x^{2^a \cdot b} + 1 = \left(x^{2^a}\right)^b + 1$ is divisible by $x^{2^a}+1$. Hence, for $x^n+1$ to be a prime, we need $n = 2^a$ i.e. $n$ is a power of $2$.
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating the following limit: $\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}} $ I am trying to calculate this limit:
$$
\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}
$$
I've tried using conjugate of both denominator and numerator but I can't get the right result.
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$$\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$$
$$=\frac{(1+\sqrt{x+1})\{x^2+1-(x+1)\}}{(\sqrt{x^2+1}+\sqrt{x+1})(1-(x+1))}$$
$$=\frac{(1+\sqrt{x+1})\{x(x-1)\}}{(\sqrt{x^2+1}+\sqrt{x+1})(-x)}$$
$$=\frac{(1+\sqrt{x+1})(1-x)}{(\sqrt{x^2+1}+\sqrt{x+1})}\text { if } x\ne0$$
As $x\to0,x\ne0$
So, $$\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}=\lim_{x\to0}\frac{(1+\sqrt{x+1})(1-x)}{(\sqrt{x^2+1}+\sqrt{x+1})}=\frac{(1+1)}{(1+1)}=1$$
Alternatively, as $\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$ is of the form $\frac00,$
Applying L'Hospital's Rule we get,
$$\lim_{x\to0}\frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}$$
$$=\lim_{x\to0}\frac{\frac x{\sqrt{x^2+1}}-\frac1{2\sqrt{x+1}}}{-\frac1{2\sqrt{x+1}}}$$
$$=\lim_{x\to0}\left(1-\frac{2x\sqrt{x+1}}{\sqrt{x^2+1}}\right)\text{ as }x+1\ne0$$
$$=1$$
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|
Are there rigorous mathematical definitions for these waves? My friend linked this .gif to me tonight, and asked me if I knew of any equations that might model these bottom two waves (the blue and green waves). Unfortunately, I am not far enough in my education to recognize if any such model exists. Are these waves modeled after some equation, or is this just some piece of eye candy?
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So, for the fist one obviously, it's sine function.
$$
h(t) = r \sin \omega t
$$
As for the second one,
$$
h(t) = \frac a2 \cdot \left \{ \begin{array}{lcc}
\tan \left (-\frac \pi 4 + \omega t \right )& \text{if} & 0 \le t \le \frac T4 \\
1 & \text{if} & \frac T4 \le t \le \frac T2 \\
\tan \left ( \frac {3\pi}4 + \omega t \right ) & \text{if} & \frac T2 \le t \le \frac {3T}4 \\
-1 & \text{if} & \frac {3T}4 \le t \le T
\end{array}\right .
$$
For the last one,
$$
h(t) = a \cdot \left \{ \begin{array}{lcc}
\frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t\right) & \text{if} & 0 \le t \le \frac T6 \\
\frac 12 + \frac 12 \left [\frac 12 + \frac {\sqrt 3}2 \tan \left( -\frac \pi 6 + \omega t - \frac \pi 3\right) \right ] & \text{if} & \frac T6 \le t \le \frac T3 \\
1 - \frac 12 \left [\frac 12 + \frac {\sqrt 3}2 \tan \left( -\frac \pi 6 + \omega t - \frac {2\pi} 3\right) \right ] & \text{if} & \frac T3 \le t \le \frac T2 \\
\frac {\sqrt 3}2 \tan \left( \pi - \omega t + \frac \pi 6 \right) & \text{if} & \frac T2 \le t \le \frac {2T}3 \\
-\frac 12 - \frac 12 \left [ \frac 12 + \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t - \frac {4 \pi}3 \right )\right ] & \text{if} & \frac {2T}3 \le t \le \frac {5T}6 \\
-1 + \frac 12 \left [ \frac 12 + \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t - \frac {5 \pi}3\right )\right ] & \text{if} & \frac {5T}6 \le t \le T
\end{array}\right .
$$
PS: For last two cases $a$ is a square or hexagon side. $T = \frac {2\pi}\omega$
Initial time angles are chosen according to this
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|
Evaluate $\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$ How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?
My attempt:
$$\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2} = \frac{1}{2}\int_{0}^{2\pi} \frac{d\theta}{(2+\cos\theta)^2}$$
To find the singularity, I solve:
$ (2+\cos\theta)^2 = 0 $ and therefore, $\cos\theta = -2$.
Substituting: $\cos z = \frac{e^{iz} + e^{-iz}}{2} = \frac{z + \frac{1}{z}}{2}$,
I find that $z = -2 + \sqrt{3} $ is the singular point that lies in the unit circle $|z| = 1$.
From this point, I have little idea how to go about solving this problem. I know I have to find the residue and then just sum them but to get the expression that would cancel out the pole is where I am currently stuck.
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Trigonometric substitution:
$$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1}dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}\Longrightarrow$$
$$\int\limits_0^\pi\frac{d\theta}{(2+\cos\theta)^2}=\int\limits_0^\infty\frac{2\,dx}{1+x^2}\frac{1}{\left(2+\frac{1-x^2}{1+x^2}\right)^2}=2\int\limits_0^\infty\frac{x^2+1}{(x^2+3)^2}dx=$$
$$2\int\limits_0^\infty\left(\frac{1}{3+x^2}-\frac{2}{(3+x^2)^2}\right)=\left.\frac{2}{\sqrt 3}\arctan\frac{x}{\sqrt 3}\right|_0^\infty-\left.4\left(\frac{x}{6(3+x^2)}+\frac{1}{6\sqrt 3}\arctan\frac{x}{\sqrt 3}\right)\right|_0^\infty=$$
$$=\frac{4}{3\sqrt 3}\frac{\pi}{2}=\frac{2\pi}{3\sqrt 3}$$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.