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Bennett's Inequality to Bernstein's Inequality Bennett's Inequality is stated with a rather unintuitive function,
$$
h(u) = (1+u) \log(1+u) - u
$$
See here. I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,
$$
h(u) \ge \frac{ u^2 }{ 2 + \frac{2}{3} u}
$$
and plugging it back into Bennett's Inequality. However, I can't see where this expression comes from. Could someone point me in the right direction?
| Here is how the approximation
could be derived.
$\begin{align}
h(u)
&=(1+u)\log(1+u)−u\\
&=(1+u)\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+u\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^{n+1}}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=2}^{\infty} \dfrac{(-1)^{n}u^{n}}{n-1}−u\\
&=u+\sum_{n=2}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=2}^{\infty} \dfrac{(-1)^{n}u^{n}}{n-1}−u\\
&=\sum_{n=2}^{\infty} u^n\left(\dfrac{(-1)^{n-1}}{n}
+\dfrac{(-1)^{n}}{n-1}\right)\\
&=\sum_{n=2}^{\infty} (-1)^{n}u^n\left(\dfrac{-1}{n}
+\dfrac{1}{n-1}\right)\\
&=\sum_{n=2}^{\infty} (-1)^{n}u^n\left(\dfrac{1}{(n-1)n}\right)\\
&=\sum_{n=2}^{\infty} \dfrac{(-1)^{n}u^n}{(n-1)n}\\
&= \dfrac{u^2}{2}-\dfrac{u^3}{6}+\dfrac{u^4}{12}
-\dfrac{u^5}{20}\pm ...\\
\end{align}
$
If we just look at the first two terms,
$\dfrac{u^2}{2}-\dfrac{u^3}{6}
=\dfrac{u^2}{2}(1-\dfrac{u}{3})
$,
and since
$\dfrac1{1+z}
=1-z+z^2 \pm ...
$,
$1-\dfrac{u}{3}
\sim \dfrac1{1+u/3}
$.
Therefore
$h(u)
\sim \dfrac{u^2}{2}\dfrac1{1+u/3}
= \dfrac{u^2}{2+2u/3}
$.
This shows how the approximation
could be derived.
The next step is to see
how accurate the approximation is.
Let
$h^*(u) = \dfrac{u^2}{2}\dfrac1{1+u/3} = \dfrac{u^2}{2+2u/3} $.
Expanding the approximation,
$\begin{align}
h^*(u)
&\sim \dfrac{u^2}{2}\dfrac1{1+u/3}\\
&=\dfrac{u^2}{2}(1-\dfrac{u}{3}+\dfrac{u^2}{9}
-\dfrac{u^3}{27}\pm ...)\\
&=\dfrac{u^2}{2}-\dfrac{u^3}{6}+\dfrac{u^4}{18} -\dfrac{u^5}{54}\pm ... \\
\end{align}
$
Therefore
$\begin{align}
h(u)-h^*(u)
&=(\dfrac{u^4}{12} -\dfrac{u^5}{20}\pm ...)
-(\dfrac{u^4}{18} -\dfrac{u^5}{54}\pm ...)\\
&=u^4(\dfrac{1}{12}-\dfrac{1}{18}) -u^5(\dfrac{1}{20}-\dfrac{1}{54})\pm ...\\
&=\dfrac{u^4}{36} -\dfrac{17u^5}{540}\pm ... \\
\end{align}
$
It certainly looks like
$h(u) > h^*(u)$.
Once you have this conjecture,
you can try to prove it.
In this case,
a proof can be developed
by looking at the successive terms
in the difference
and showing that the terms are decreasing
and alternating in sign.
Note that if we let
$h^{**}(u)
= \dfrac{u^2}{2}-\dfrac{u^3}{6}
$,
$h(u)-h^{**}(u)
=\dfrac{u^4}{12} -\dfrac{u^5}{20}\pm ...
$,
and this appears to have an error
about three times as large.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to find the maximum of this equation $x(1 - (1 - \frac{1}{x})^K)$ For x in reals and K some positive integer, can you find the maximum of this function analytically?
$f(x) = x(1 - (1 - \frac{1}{x})^K)$
| When $K = 1$, $f(x) = 1$. So assume $K > 1$ (not necessarily an integer).
The derivative of your function is
$$ 1 - \left(1 - \frac{1}{x}\right)^K - \frac{K}{x}\left(1 - \frac{1}{x}\right)^{K-1} =
1 -\left(1 + \frac{K-1}{x}\right)\left(1 - \frac{1}{x}\right)^{K-1} . $$
The derivative approaches $1$. In particular, when $x$ is large enough, this derivative is at least $1/2$ (say). Therefore the function grows to infinity with $x$.
Edit: In fact, as André points out, the derivative approaches $0$, and what is even worse, it's not even clear whether it remains positive. In fact, approximately $(1-1/x)^{K-1} = 1-(K-1)/x+(K-1)(K-2)/2x^2$ and so the derivative is approximately
$$1 - \left(1 + \frac{K-1}{x}\right)\left(1 - \frac{K-1}{x} + \frac{(K-1)(K-2)}{2x^2}\right) = \frac{K(K-1)}{2x^2} + \frac{(K-1)^2(K-2)}{2x^3}.$$
So it does appear that the function is ever-increasing. At the limit, we get the value $K$, as mentioned in the comments.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2) (x-4)^2 (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.
| $$y= (x-2) (x-4)^2 (x-6) + 6$$
$let , t = x-4$
$$y= t^2(t-2)(t+2)+ 6$$
$$y= t^4-4t^2+6 $$
$$y=(t^2-2)^2+2$$
$$y(min)=2$$
attained when $t^2=2$
$(x-4)^2=2$ which gives, $ x=4-\sqrt2$ and $ x=4+\sqrt2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Are there any $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square? Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?
I tried to simplify
\begin{align*}
n^4+n^3+n^2+n+1 &= n^2(n^2+1)+n(n^2+1)+1\\
&= (n^2+n)(n^2+1)+1 \\
&= n(n+1)(n^2+1)+1
\end{align*}
Then I assumed that the above expression is a square; then
$$ n(n+1)(n^2+1)+1 = k^2$$
$$
\begin{align*}
n(n+1)(n^2+1) &= (k^2-1) \\
&= (k+1)(k-1)
\end{align*}
$$
Then trying to reason with prime factors, but cannot find a concrete proof yet.
|
Here is my answer!
H. Bensom, Germany
| {
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Can someone please explain the cube to sphere mapping formula to me? I am wondering if anyone could explain how the following formula works, it is supposed to take the input as a point on a cube then map that to points on a sphere, please go gentle on me, I'm in 9th grade O_O
$$\begin{bmatrix}x' \\ y' \\ z'\end{bmatrix}=\begin{bmatrix} x\sqrt{1-\frac{y^2}{2}-\frac{z^2}{2}+\frac{y^2z^2}{3}} \\ y\sqrt{1-\frac{z^2}{2}-\frac{x^2}{2}+\frac{z^2x^2}{3}} \\ z\sqrt{1-\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^2y^2}{3}} \end{bmatrix}$$
Thanks.
| Thanks for the explanation by joriki. The above formula has been explained in 1. But the formula has some limitations. The cube length can be 2 (-1 to 1 in all directions). From the above concept, mentioned by joriki or 1 we can generalize for the cube of any length (i.e 2a (-a to a in all directions)). Here 2a is the side length of the cube.
$a^6-(a^2-x^2 )(a^2-y^2)(a^2-z^2)=x^2 a^2 (a^2-\frac{z^2}{2}-\frac{y^2}{2}+\frac{y^2 z^2}{3a^2})+y^2 a^2 (a^2-\frac{z^2}{2}-\frac{x^2}{2}+\frac{x^2 z^2}{3a^2})+z^2 a^2 (a^2-\frac{x^2}{2}-\frac{y^2}{2}+\frac{y^2 x^2}{3a^2})$
So,
$x'=xa \sqrt{a^2-\frac{z^2}{2}-\frac{y^2}{2}+\frac{y^2 z^2}{3a^2}}$,
$y'=ya \sqrt{a^2-\frac{z^2}{2}-\frac{x^2}{2}+\frac{x^2 z^2}{3a^2}}$,
$z'=za \sqrt{a^2-\frac{x^2}{2}-\frac{y^2}{2}+\frac{y^2 x^2}{3a^2}}$.
From the above equation, we can easily separate the (x',y',z') as explained in 1. The above equation will readily find the mapping of the generalized length of the cube to a sphere. I hope this will be helpful.
Form the above equation we can also find the evenly distribution of points within the sphere by varying the grid size of the cube. By discretizing the cube into different grids and simultaneously mapping into the sphere will give the evenly distribution of points within the sphere.
Kindly ignore some typos.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to prove that a solution of the equation is consider the equation
$ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} - \cdots-4x-4=0$
for $n = 1$ :: solution is : $x = 4$
for $n = 2$, ($x^2 - 4 x - 4 = 0$) :: solution is : $x = 4.8$
for $n = 3$, ($x^3 - 4x^2 - 4 x - 4 =0 $) :: solution is : x = $4.96$
how to prove that as $n \to\infty$ (what ever it means) the solution is $x = 5$.
| $ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} - \cdots 4x-4=0 $
$ x^n$ - 4{ $x^{n-1} + x^{n-2} + x^{n-3} + \cdots+x+1$}=0
Using the property
$1+x+x^2......x^n-1 $= $\frac{1-x^n}{1-x}$
$ x^n$ - 4{$\frac{1-x^n}{1-x}$} =0
$x^{n+1} - 5x^n + 4 $= 0
$x - 5 + \frac{4}{x^n} $= 0
now if n tends to infinity x will tend to five.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework:
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
| Integrate by parts to reduce to table integral:
$$
\int_0^1 \sqrt{1+x^2} \mathrm{d} x = \left. x \sqrt{1+x^2} \right|_0^1 - \int_0^1 \frac{x^2 {\color\green{+1-1}}}{\sqrt{1+x^2}} \mathrm{d} x = \sqrt{2} - \int_0^1 \sqrt{1+x^2} \mathrm{d} x + \int_0^1 \frac{\mathrm{d} x}{\sqrt{1+x^2}}
$$
Now solving the equation for $\int_0^1 \sqrt{1-x^2} \mathrm{d} x$, and using table anti-derivative $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x)$:
$$
\int_0^1 \sqrt{1+x^2} \mathrm{d} x = \frac{1}{2} \left( \sqrt{2} + \operatorname{arcsinh}(1) \right) = \frac{1}{2} \left( \sqrt{2} + \log(1+\sqrt{2}) \right) \approx 1.1478
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding $y$ from $y'=\sqrt{5x+2y-3}$. I'd love your help with finding the function $y$ from the following differential equation:
$y'=\sqrt{5x+2y-3}$.
I tried to use $z=5x+2-3$, so $z'=5+2y'$ , and $y'=\frac{z'}{2}-2.5$
and from the equation $\frac{z'}{2}-2.5=y'=\sqrt z$, and then $z'=2 \sqrt z+5$, so $\frac{dz}{2 \sqrt z +5} = dx$, but using integration here is difficult and won't lead me to $y$.
I tries to use substitution in other ways like $z=\sqrt{5x+2y-3}$, or $z^2= 5x+2y-3$ but then again I got stuck in the middle..
Any suggestion?
Thanks a lot
| Integration here is simple
$$
\int\frac{dz}{2\sqrt{z}+5}=\{t=\sqrt{z}\}=\int\frac{2t dt}{2t+5}=\int\left(1-\frac{5}{2t+5}\right)dt=t-\frac{5}{2}\int\frac{d(2t+5)}{2t+5}=
$$
$$
t-\frac{5}{2}\ln(2t+5)+C=\sqrt{z}-\frac{5}{2}\ln(2\sqrt{z}+5)+C
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution? Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1 $$
non-numerically?
How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists an analytic solution?
| If you write your equation as, say, $f(x)=t$ where $f(x) = e^x - 2 x^4 - x^3$, you can use the Lagrange inversion formula to get power series solutions. A convenient starting point is $f(1)=e-3$. Taking $a=1$ and $b=f(a) = e-3$, the Lagrange inversion formula says that for $t$ near $b$, one solution of $f(x)=t$ is
$$ g(t) = a + \frac{t-b}{f'(a)} + \sum_{n=2}^\infty \lim_{w \to a} \frac{d^{n-1}}{dw^{n-1}}
\left(\frac{w-a}{f(w)-b}\right)^n \frac{(t-b)^n}{n!}$$
which in this case is
$$ g(t) = 1 + \frac{t-b}{e-11} + \frac{-e+30}{(e-11)^3} \frac{(t-b)^2}{2} + \frac{2 e^2 - 115 e +2106}{(e-11)^5} \frac{(t-b)^3}{6} + \frac{-6e^3+470 e^2 - 16237 e + 232608}{(e-11)^7} \frac{(t-b)^4}{24} + \ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Radius of convergence of $\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$ Radius of convergence of
$$\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$$
The answer was to let $$c_n = \frac{(n^{2}(-2))^n}{4^n(2n+1)!}$$
But how do I get the $-2$? Its probably from $(3-2x)^n$. But how do I remove the $3$ and $x$ parts?
Also later into the answer, $$(\lim_{n\to\infty} ... (\lim_{n\to\infty} (1+\frac{1}{n})^n)^2) = \frac{e^2}{8}$$
What gives $e$ btw?
| I've seen something similar around. Since you need the $x$ part to be of the form $(x-a)^n$, you have to extract the $2$ in the $x$:
$$(3-2x)^n=(-2)^n \cdot (x-3/2)^n$$
Now you have
$$c_n = \frac{(-2n^2)^n}{4^n(2n+1)!}$$
$$c_n = \frac{(-2)^nn^{2n}}{4^n(2n+1)!}$$
You now need to evaluate the limit:
$$\lim\limits_{n \to \infty} \left| \frac{(-2)^{n+1}{(n+1)}^{2(n+1)}}{(-2)^nn^{2n}}\frac{4^n(2n+1)!}{4^{n+1}(2n+3)!}\right|=$$
$$\lim\limits_{n \to \infty}\left| \frac{(-2){(n+1)}^{2n}{(n+1)}^2}{n^{2n}}\frac{1}{4(2n+2)(2n+3)}\right|=$$
$$\frac{1}{2}\lim\limits_{n \to \infty} \left|{\left(\frac{{n+1}}{n}\right)^n}^2\frac{n^2+2n+1}{4n^2+10n+6}\right|=$$
$$\frac{1}{2}\lim\limits_{n \to \infty}\left| {\left(1+\frac{{1}}{n}\right)^n}^2\frac{n^2+2n+1}{4n^2+10n+6}\right|=$$
This limits shouldn't be hard for you, I guess.
$$\displaystyle \frac{1}{2}e^2\frac{1}{4}=\frac{e^2}{8}$$
| {
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Is the permutation $\alpha : \mathbb Z_{11} \rightarrow \mathbb Z_{11}$ given by $\alpha(x)=4x^2-3x^7$ a cycle or not? The permutation $\alpha : \mathbb Z_{11} \rightarrow \mathbb Z_{11}$ given by $\alpha(x)=4x^2-3x^7$ is a cycle? If it´s not a cycle then write it as a product of disjoint cycles.
| Of course martini's comment is very true, as the easiest thing to do would just be computing $\alpha(x) \bmod 11$ for $x = 0, 1, \ldots, 10$, and seeing if we get a cycle. But let us try to be clever, and try to obtain the cycles without doing 'brute force' calculations. (Maybe it's too much, in which case you can just skip to the bottom...)
First, modulo $11$, by Euler's theorem we have that $x^{10} \equiv 1$ for $x \neq 0$, but also $x^5 = +1$ if $x$ is a quadratic residue, and $x^5 = -1$ if $x$ is a quadratic nonresidue in $\mathbb{Z}_{11}^{*}$. So:
$$\alpha(x) = 4x^2 - 3x^7 \equiv 4x^2 + 8x^7 = 4x^2(1 + 2x^5) \equiv 4x^2(1 + 2(\pm1))\equiv \begin{cases} x^2, & \text{if }x \in \square, \\ 7x^2, & \text{if } x \notin \square. \end{cases}$$
Since $7 \equiv -4 \bmod 11$ and $4 \in \square$ we have $7 \notin \square$, and since $7^2 \not\equiv 1 \bmod 11$ it follows that $7$ is a generator of $\mathbb{Z}_{11}^{*}$. And in terms of powers of $7$, the function $\alpha$ is simple:
$$\alpha(7^k) \equiv \begin{cases} 7^{2k} & \text{if } k \text{ even} \\ 7^{2k+1} & \text{if } k \text{ odd} \end{cases}$$
Also note that $7^{10 + k} \equiv 7^k$ by Euler, so we can reduce the exponents modulo $10$. So in terms of the generator $7$, we get the following cycles:
$$7^0 \to 7^0 \\ 7^1 \to 7^3 \to 7^7 \to 7^5 \to 7^{1} \\ 7^2 \to 7^4 \to 7^8 \to 7^6 \to 7^2 \\ 7^9 \to 7^9 \\ (0 \to 0)$$
To get actual numbers out of this we do need to do some calculations now. But notice that to obtain the result that we have two cycles of length $4$ and three cycles of length $1$, we arguably did not have to do any calculations. In any case, these cycles correspond to the following numbers.
$$1 \to 1 \\ 7 \to 2 \to 6 \to 10 \to 7 \\ 5 \to 3 \to 9 \to 4 \to 5 \\ 8 \to 8 \\ (0 \to 0)$$
So as a product of cycles, we get $(1)(7,2,6,10)(5,3,9,4)(8)(0)$.
| {
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Riemann sum of $\sin(x)$ I would like to calculate the Riemann sum of $\sin(x)$. Fun starts here:
$$R = \frac{\pi}{n} \sum_{j=1}^n \sin\left(\frac{\pi}{n}\cdot j\right)$$
What would be the simplest way to calculate the sum of $\sin\left(\frac{\pi}{n}\cdot j\right)$, so that one could proceed to evaluating the limit and thus getting the value of the Riemann sum, in other words - the integral?
There maybe a way using $\mathbb{C}$?
| Use
$$
2 \sin\left(\frac{\pi}{2 n} \right) \sin\left(\frac{\pi}{n} \cdot j \right) = \cos\left( \frac{\pi}{2n} (2j-1) \right) - \cos\left( \frac{\pi}{2n} (2j+1) \right)
$$
Thus the sum telescopes $\sum_{j=1}^n \left(g(j) - g(j+1) \right) = g(1) - g(n+1) $:
$$
R_n =\frac{\pi}{n} \sum_{j=1}^n \sin\left(\frac{\pi}{n} \cdot j \right) = \frac{\pi}{2 n \sin\left( \frac{\pi}{2n} \right)} \left( \cos\left( \frac{\pi}{2n} \right) - \cos\left( \frac{\pi}{2n} (2n+1)\right) \right) = \frac{\pi}{n} \cdot \frac{1}{\tan\left( \frac{\pi}{2n} \right)}
$$
The large $n$ limit is easy:
$$
\lim_{n \to \infty} R_n = 2 \lim_{n \to \infty} \frac{\pi}{2 n} \cdot \frac{1}{\tan\left( \frac{\pi}{2n} \right)} = 2 \lim_{x \to 0} \frac{x}{\tan(x)} = 2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial fraction with a constant as numerator I am trying to express this as partial fraction:
$$\frac{1}{(x+1)(x^2+2x+2)}$$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
| $$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}.
$$
Then you need to find $A$, $B$, and $C$.
The polynomial $x^2+2x+2$ factors as $(x+1+i)(x+1-i)$, where $i$ is a square root of $-1$. You could go on to write
$$
\frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2} = \frac{A}{x+1} + \frac{D}{x+1+i} + \frac{E}{x+1-i},
$$
and the numbers $D$ and $E$ might not be real.
Later edit in response to a comment:
$$
\frac{1}{(x+1)(x^2+2x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}.
$$
There are several methods for finding $A$, $B$, and $C$.
Either of two methods begins by multiplying both sides by the denominator, getting
$$
1 = A(x^2+2x+2) + (Bx+C)(x+1).
$$
In one method, you can make the second term vanish by setting $x=-1$:
$$
1 = A\Big((-1)^2+2(-1)+2\Big) + \Big(\text{the second term, which is now }0\Big).
$$
Solving this for $A$ gives $A=1$, and then you write
$$
1 = 1(x^2+2x+2) + (Bx+C)(x+1).
$$
Now you can let $x=1$ in order to make the term involving $B$ vanish:
$$
1 = 1(0^2+2\cdot0 + 2) + C(0+1).
$$
This gives $C=-1$. Then we have
$$
1 = (x^2+2x+2) + (Bx-1)(x+1).
$$
We can no longer make anything vanish without making $B$ disappear, so let $x=\text{some number that won't make the arithmetic too messy}$. If we let $x=1$, we get
$$
1 = (1^2+2\cdot1+2) + (B2-1)(1+1).
$$
This gives us $B=-1/2$.
Another method is this: We had
$$
1 = A(x^2+2x+2) + (Bx+C)(x+1).
$$
Now multiply this out and collect like terms, where "like" means they are coefficients of the same powers of $x$:
$$
0x^2+0x^1 = (A+B)x^2 + (2A+B+C)x + (2A+C).
$$
Then equate coefficients of common powers of $x$:
$$
\begin{align}
A+B & = 0 \\
2A+B+C & = 0 \\
2A+C & = 1
\end{align}
$$
Then solve that system for $A$, $B$, and $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Counting the number of squares on an $n\times n$ Board Yesterday I was asked by a friend how many squares are in a chess board of $8\times 8$.
I thought about 64 immediately, but of course this is not the solution, there are much more.
So the number of squares is: $$8\times 8 + 7\times 7 + 6\times 6 + 5\times 5 + 4\times 4 + 3\times 3 + 2\times 2 + 1\times 1=1^2 + 2^2 + 3^2 + 4^2...+ 8^2$$
I came across this formula: $$\frac{n(n + 1)(2n + 1)} 6$$
It produces the sum of squares in $n\times n$ board.
My question is, how did he reached this formula? Did he just guessed patterns until he reached a match, or there is a mathematical process behind?
If there is a mathematical process, can you please explain line by line?
Thanks very much.
Btw: Couldn't find matching tags for this question, says I can't create.
| The first step is to recognize that there are $8^2$ squares of size $1$ by $1$, $7^2$ squares of size $2$ by $2$ and so on. That justifies the total number being, as you say, $1^2+2^2+3^2+\ldots 8^2$. Sums of powers are calculated by Faulhaber's formula. There are several ways to derive them. One way is to know or suspect that $\sum_{k=1}^n k^p$ should be of degree $p+1$. So for squares, we want $\sum_{k=1}^n k^2=an^3+bn^2+cn+d$. Then if we evaluate it at $n+1$, we get $\sum_{k=1}^{n+1} k^2=a(n+1)^3+b(n+1)^2+c(n+1)+d$. Subtracting, we get $(n+1)^2=a((n+1)^3-n^3)+b((n+1)^2-n^2)+c((n+1)^1-n^1)$ and equating the coefficients gets the desired formula. You can prove the formula rigorously by induction
| {
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Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ...
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}+\sqrt{2}i)^{-2011}
=\cos{(-2011\theta)} + i \sin{(-2011\theta)} = e^{i(-2011)\theta}$$
Is it correct?
The given answer is
...
$$\arg{z} = \frac{3\pi}{4}, \qquad z=2e^{i \frac{3\pi}{4}}$$
$$LHS
= (\frac{1}{z})^{2011}=2^{-2011}e^{-2011(\frac{3\pi}{4})i}
= 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})}
= 2^{-2011}e^{-\frac {\pi i}{4}}
= 2^{-2011} \color{blue}{\frac{1 - i}{\sqrt{2}}}
= ...
$$
Why is the red $1508\pi i$ removed in the following step?
How do I get the blue $\frac{1 - i}{\sqrt{2}}$ from the prev step?
| $$
\frac{1}{-\sqrt{2}+i\sqrt{2}}\frac{-\sqrt{2}-i\sqrt{2}}{-\sqrt{2}-i\sqrt{2}}=\frac{-\sqrt{2}-i\sqrt{2}}{4}=\frac12\frac{-1-i}{\sqrt{2}}
$$
which has absolute value $\dfrac12$ and argument $\dfrac{5\pi}{4}$, so
$\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}$ has absolute value $1/2^{2011}$ and argument $\frac{5\pi}{4}\cdot2011\equiv-\frac{\pi}{4}\pmod{2\pi}$. Therefore,
$$
\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}=\frac{1}{2^{2011}}\frac{1-i}{\sqrt{2}}
$$
If, as the title suggests, the exponent is $-2011$, we simply need to take the reciprocal:
$$
\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{-2011}=2^{2011}\frac{1+i}{\sqrt{2}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem about solving infinity limit with square root (I)
$$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$
$$\lim_{x \to \infty } \, \left(x\sqrt{1+1/x}-x\sqrt{1-1/x}\right)=$$
$$\lim_{x \to \infty } \, \left(x\sqrt{1}-x\sqrt{1}\right)=\lim_{x \to \infty } \, \left(x-x\right)=0$$
(II)
$$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$
$$\lim_{x \to \infty } \, \left(\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)*\frac{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}\right)=$$
$$\lim_{x \to \infty } \, \frac{2x}{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}=$$
$$\lim_{x \to \infty } \, \frac{2x}{\left(x\sqrt{1+1/x}+x\sqrt{1-1/x}\right)}=$$
$$\lim_{x \to \infty } \, \frac{2x}{\left(x\sqrt{1}+x\sqrt{1}\right)}=\lim_{x \to \infty } \, \frac{2x}{2x}=1$$
I found these two ways to evaluate this limit. I know the answer is 1. The first one is surely wrong. The question is: why? What is wrong there?
| By the same erroneous logic you'd conclude $\rm\:2\: =\: x(1+1/x)-x(1-1/x)\to\: 0$
In your example $\rm\:\displaystyle \sqrt{1+\frac{1}x}-\sqrt{1-\frac{1}x}\ =\ \frac{1}x + \frac{1}{8\: x^3} +\: \cdots\:$ which makes the error clear.
If the dominant terms in a sum of series cancel, then you need to look at subsequent terms.
| {
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How to prove $|F(x)|\leq\frac{M(b-a)^2}{8}$ $f(x)$ is derivable in $[a,b]$, $|f^{'}(x)|\leq M$.
$\int_a^b f(x)dx=0$.
Let $F(x)=\int_a^x f(t)dt$.
Try to prove $|F(x)|\leq\frac{M(b-a)^2}{8}$
I want to use Taylor expansion at $f(\xi)=0$, but I can't continue.
| I can prove that $\left | F( \frac{a+b}{2})\right | \le \frac{M(b-a)^2}{8}$.
By Taylor expansion at $ x_0 = \frac{a+b}{2}$,we have
$$F(x) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(x-\frac{a+b}{2}) + \frac{F''(\xi)}{2}(x-\frac{a+b}{2})^2$$
Substituting $x=a$ and $x=b$ into the formula above,we get
$$F(a) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(a-\frac{a+b}{2}) + \frac{F''(\xi_1)}{2}(\frac{a-b}{2})^2$$
$$F(b) = F(\frac{a+b}{2}) + F'(\frac{a+b}{2})(b-\frac{a+b}{2}) + \frac{F''(\xi_2)}{2}(\frac{a-b}{2})^2$$
To the sum of the two formulas above,with $F(a)=F(b)=0$,we have
$$\left |F(\frac{a+b}{2}) \right | \le \left | \frac{F''(\xi_1)+F''(\xi_2)}{4} \right |(\frac{a-b}{2})^2 \le \frac{M(b-a)^2}{8}$$
But I cannot get better ideas to prove that for any $x \in [a,b]$,$\left |F(x) \right | \le \frac{M(b-a)^2}{8}$
| {
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"url": "https://math.stackexchange.com/questions/135135",
"timestamp": "2023-03-29T00:00:00",
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Solving an equation with nested ceilings How would I go about solving something like the following for $x$ in terms of $y$?
$$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$
I've never had to solve an equation with a ceiling before, never mind nested ceilings. Even Wolfram Alpha Pro is timing out to the point of not being able to go anywhere.
| $$1+\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 2$$
$$\left \lceil \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \right \rceil = 1$$
$$0 \lt \log_{x-1} \left ( \left \lceil \frac{y}{x} \right \rceil \right ) \le 1$$
assuming $x \gt 2$:
$$1 \lt \left \lceil \frac{y}{x} \right \rceil \le x-1$$
$$1 \lt \frac{y}{x} \le \lfloor x-1 \rfloor$$
$$x \lt y \le x \lfloor x-1 \rfloor.$$
Note if $1 \lt x \lt 2$ then you need to solve $x-1 \le \left \lceil \frac{y}{x} \right \rceil \lt 1$, which would require you to find an integer between $0$ and $1$. You may have problems with the logarithm base $x-1$ if $x=2$ or $x \le 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A highschool factoring problem $x+y+z=0$
$x^3+y^3+z^3=9$
$x^5+y^5+z^5=30$
$xy+yz+zx=?$
I solved this problem by setting $xy+yz+zx=k$ and using the cubic equation with roots $x,y,z$. But is there any other methods?
| Here is one way of making progress, which uses the cubic as part of the solution. There are other routes which involve knowing some standard factorisations.
First note that $z=-(x+y)$ from the first equation and substitute in the second, obtaining:
$$-3x^2y-3xy^2 = 9$$
Divide by 3 to get:
$$-xy(x+y) = xyz = 3$$
Now $x,y,z$ are the roots of the cubic equation $t^3+kt-3 = 0$,
and therefore satisfy $t^5+kt^3-3t^2=0$
Substitute $x,y,z$ successively into this equation and add to get
$$30+9k-3(x^2+y^2+z^2) = 0$$
And use $0=(x+y+z)^2=x^2+y^2+z^2+2k$ to finish.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding area of a triangle with integrals I am suppose to find the area of a triangle using integrals with vertices 0,0 1,2 and 3,1
This gives me
$y= 2x$
$y=\frac{1}{3}x$
$y= \frac{-1}{2}x+\frac{5}{2}$
for my slopes
I know that I can calculate the area of the first part by finding
$$\int_{0}^{1}2x-\frac{1}{3}x$$
and the second part by
$$\int_{1}^{3}\frac{-1}{2}x+\frac{5}{2}-\frac{1}{3}x$$
The anti derivative of the top is $x^2 - x^2/3$
and the other one is $-x^2/4 + 5x/2 - x^2/6$
I am not sure what I am doing wrong but I do not get the proper answer of 5/2
| Working with what you have:
$$\int_{0}^{1}2x-\dfrac{1}{3}x dx$$
$$\int_{0}^{1}\dfrac{5x}{3}dx$$
$$\dfrac{5}{3}\int_0^{1}x dx$$
$$=\dfrac{5}{6}$$
The second integral before being evaluated from $1$ to $3$ is:
$$\dfrac{5x}{2} - \dfrac{5x^{2}}{12}$$
When evaluated at $x = 3$, you get $\dfrac{15}{4}$. When evaluated at $x=1$, you get $\dfrac{25}{12}$.
Get a common denominator and subtract them now. $\dfrac{45}{12} - \dfrac{25}{12} = \dfrac{20}{12} = \dfrac{5}{3}$.
Now simply add the two evaluated integrals:
$$\dfrac{5}{6} + \dfrac{5}{3} = \dfrac{5}{2}$$
Edit: When you said, and the other one is $\dfrac{-x^{2}}{4} + \dfrac{5x}{2} - \dfrac{x^{2}}{6}$, you were correct. However, your first antiderivative was incorrect. When you see something in the original integrand that can be simplified (i.e. your terms with just an $x$ in it), as a general rule, DO IT.
| {
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"timestamp": "2023-03-29T00:00:00",
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Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$
Thanks a lot.
| Well known equality :
$$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\leqslant \frac{1}{\sqrt{2n}}$$
Proof of right side. Having
$$\frac{1}{2}<\frac{2}{3},\frac{3}{4}<\frac{4}{5},\frac{5}{6}<\frac{6}{7},\cdots,\frac{2n-3}{2n-2}<\frac{2n-2}{2n-1},\frac{2n-1}{2n}<1$$
and multiplying we obtain
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}<\frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7} \cdots \frac{2n-2}{2n-1}$$
Now multiply both sides on $ \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}$ gives
$$ \left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\right)^2<\frac{1}{2n}$$
and we obtain
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} <\frac{1}{\sqrt{2n}}$$
Analogical way works for left side.
| {
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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?
| We want to show that our sum is equal to $1-\frac{1}{2^n}$. The base step is easy. Let's do the induction step.
Suppose that the result holds when $n$ is the particular number $k$. We want to show that the result holds for the "next" $n$, namely $k+1$.
Look at the sum
$$\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^k}+\frac{1}{2^{k+1}}.$$
This is equal to
$$\left(\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^k}\right)+\frac{1}{2^{k+1}}.$$
By the induction hypothesis this is
$$\left(1-\frac{1}{2^k}\right)+\frac{1}{2^{k+1}},$$
which is
$$1-\left(\frac{1}{2^k}-\frac{1}{2^{k+1}}\right).$$
But $\dfrac{1}{2^k}-\dfrac{1}{2^{k+1}}=\dfrac{1}{2^{k+1}}$ (bring to the common denominator $2^{k+1}$), which completes the induction step.
Remarks: $1$. If we are not looking for a formal induction, note that our sum is equal to
$$\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\cdots+\left(\frac{1}{2^{n-1}}-\frac{1}{2^n}\right).$$
Now remove the parentheses, and watch our sum collapse. The only survivors are the $1$ at the beginning and the $-\dfrac{1}{2^n}$ at the end.
$2.$ Amoroso ($A$) lives $1$ km from his beloved, who, naturally, lives at $B$. First, $A$ walks $\frac{1}{2}$ km towards $B$. Now $A$ is $\frac{1}{2}$ km from $B$.
Next, $A$ walks $\frac{1}{4}$ km towards $B$. Now, $A$ is $\frac{1}{4}$ km from $B$. Next, $A$ walks $\frac{1}{8}$ km towards $B$. Now $A$ is $\frac{1}{8}$ from $B$.
This pattern continues, until $A$ walks $\frac{1}{2^n}$ towards $B$. Now $A$ is $\frac{1}{2^n}$ from $B$. So the sum $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}$ of his travels is $1-\frac{1}{2^n}$.
| {
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If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. Prove that if $x, y$ are rational numbers and
$$ x^5 +y^5 = 2x^2y^2$$
then $1-xy$ is a perfect square.
| Here is a proof I really like:
Using "polar coordinates", we have $x=r \cos\theta$ and $y=r\sin\theta$, where $r\geq 0$ and $\theta\in [0,2\pi)$. If $xy=0$, then $1-xy=1$ which is clearly a perfect square of a rational number. Therefore, we assume that $xy\neq 0$, which implies that $x^5 +y^5\neq 0$ thanks to $x^5 +y^5 = 2x^2y^2$. In particular, we have $r>0$ and $\cos^5\theta+\sin^5\theta\neq 0$. Therefore, from the equation $x^5 +y^5 = 2x^2y^2$, we get
$$r=\frac{2\cos^2\theta \sin^2\theta}{\cos^5 \theta+\sin^5\theta}.$$
Hence
$$1-xy=1-r^2\cos\theta\sin\theta=1-\left(\frac{2\cos^2\theta\sin^2\theta}{\cos^5\theta+\sin^5\theta}\right)^2\cos\theta\sin\theta=1-\frac{4\cos^5\theta\sin^5\theta}{(\cos^5\theta+\sin^5\theta)^2}=\frac{(\cos^5\theta-\sin^5\theta)^2}{(\cos^5\theta+\sin^5\theta)^2}=\left(\frac{\cos^5\theta-\sin^5\theta}{\cos^5\theta+\sin^5\theta}\right)^2
=\left(\frac{(r\cos\theta)^5-(r\sin\theta)^5}{(r\cos\theta)^5+(r\sin\theta)^5}\right)^2=\left(\frac{x^5-y^5}{x^5+y^5}\right)^2$$
which is a perfect square of a rational number, since $x$ and $y$ are rational by assumption.
| {
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List the $x$-axis intercepts for this trigonometric function
Sketch the graphs of each of the following for $x$ in [0,2$\pi$]. list the $x$-axis intercepts of each graph for this interval.
$$y=\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+1$$
I tried to solve the equation $y = 0$ by performing the following steps:
$$\begin{align*}
-1&=\sqrt{2} \cos\left(x-\frac{\pi}{4}\right) \\\\
-\frac{1}{\sqrt{2}}&=\cos\left(x-\frac{\pi}{4}\right) \\\\
-\frac{1}{\sqrt{2}}+\frac{\pi}{4}&=\cos(x) \end{align*}$$
I'm unable to proceed further. Can you please explain this to me in a step-by-step fashion? And I think that after you've got the answer you have to look the unit circle to get the exact answer. Please also show me how to look for the answer from unit cirle.
Thanks so much!
| As mentioned in the comments, you can't add $\frac{\pi}{4}$ to both sides, you can think of it as being "trapped" inside the cosine. Your work is correct up to your second step:
\begin{align*} -1&=\sqrt{2} \cos\left(x-\frac{\pi}{4}\right) \\\\ -\frac{1}{\sqrt{2}}&=\cos\left(x-\frac{\pi}{4}\right) \end{align*}
Now, this is when you look at the unit circle. You will notice that $\cos(x) = -\frac{1}{\sqrt{2}}$ when $x = \frac{3\pi}{4}$ and $x = \frac{5\pi}{4}$. This means that we have to solve the following equations:
$$ x-\frac{\pi}{4} = \frac{3\pi}{4} \quad \quad \mbox{and} \quad \quad x-\frac{\pi}{4} = \frac{5\pi}{4}. $$
Doing this gives you $x = \pi$ and $x = \frac{3\pi}{2}$.
| {
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Puzzle: Cumulative Sum Divisible by 10 If we sum the first $4$ positive integers, we get $4 + 3 + 2 + 1 = 10$, which I think is pretty cool.
I'm interested in seeing solutions to the following puzzle: If we take the cumulative sum of the first $N$ positive integers, how many times along the way will the sum be divisible by $10$?
In other words, define:
$$f(N) = \left|\left\{n \leq N \;:\; \sum_{i=1}^n i = \frac{n(n+1)}{2} \text{ is divisible by $10$}\right\} \right|$$
Is there a nice expression for $f(N)$?
| You need $20$ to divide $n(n+1)$. Note $n$ and $n+1$ are of opposite parity. Hence, $4$ divides either $n$ or $n+1$. $5$ divides either $n$ or $n+1$ since $5$ is a prime.
Hence, we have $4$ cases which gives us as discussed below.
$1$. $4|n$ and $5|n$. This case is trivial. Since $(4,5)=1$, we get that $20|n$ and hence $n \equiv 0\bmod 20$.
$2$. $4|n$ and $5|(n+1)$. This means that $n = 4k_4$ and $n+1 = 5k_5$. Hence, we get that $5k_5 - 4k_4 = 1$.
Solving such general congruence fall under the Chinese remainder theorem, as TMM points out in the comments, and is certainly something which you should look at. In this case, we solve it as follows. Clearly, $(k_4,k_5) = (1,1)$ is a solution. In general, if $ax+by$ has integer solutions and $(x_0,y_0)$ is one such integer solution, then all integer solutions are given by $$(x,y) = \displaystyle \left( x_0 + k \frac{\text{lcm}[\lvert a \rvert,\lvert b \rvert]}{a}, y_0 - k \frac{\text{lcm}[\lvert a \rvert,\lvert b \rvert]}{b} \right)$$ where $k \in \mathbb{Z}$. Hence, we get that $(k_4,k_5) = (1+5k,1+4k)$. This gives us that $n = 20k+4$ i.e. $n \equiv 4 \bmod 20$.
$3$. $4|(n+1)$ and $5|n$. This means that $n = 5k_5$ and $n+1 = 4k_4$. Hence, we get that $4k_4 - 5k_4 = 1$. Clearly, $(k_4,k_5) = (4,3)$ is a solution. Hence, we get that $(k_4,k_5) = (4+5k,3+4k)$. This gives us that $n=20k+15$ i.e. $n \equiv 15 \bmod 20$.
$4$. $4|(n+1)$ and $5|(n+1)$. This case is again trivial. Since $(4,5) = 1$, we get that $20|(n+1)$ and hence $n \equiv 19 \bmod 20$.
To summarize, the solutions are given by
\begin{cases}
n \equiv 0\bmod 20 &(\text{ if }4 \text{ divides }n \text{ and }5 \text{ divides }n)\\
n \equiv 4\bmod 20 & (\text{ if }4 \text{ divides }n \text{ and }5 \text{ divides }n+1)\\
n \equiv 15\bmod 20 & (\text{ if }4 \text{ divides }n+1 \text{ and }5 \text{ divides }n)\\
n \equiv 19\bmod 20 & (\text{ if }4 \text{ divides }n+1 \text{ and }5 \text{ divides }n+1)
\end{cases}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
} |
Need help with an integral need help with calculating this:
$$\int_{0}^{2\pi}\frac{1-x\cos\phi}{(1+x^2-2x\cos\phi)^{\frac{3}{2}}}d\phi$$
Thanks in advance!
| Let's assume $-1<x<1$, for simplicity.
Changing integration variable $\phi \to 2 \phi$, and using $\cos(2\phi) = 1- 2 \sin^2(\phi)$ we get:
$$\begin{eqnarray}
\int_0^{2\pi} \frac{1-x \cos(\phi)}{\left(1+x^2-2x \cos(\phi)\right)^{3/2}}\mathrm{d} \phi &=& 2 \int_0^{\pi} \frac{1-x \cos(2\phi)}{\left(1+x^2-2x \cos(2\phi)\right)^{3/2}}\mathrm{d} \phi \\
&=& 4 \int_0^{\pi/2} \frac{1-x + 2 x \sin^2(\phi)}{\left((1-x)^2 + 4 x \sin^2(\phi)\right)^{3/2}} \mathrm{d} \phi \\
&=& \frac{4}{(1-x)^2} \int_0^{\pi/2} \frac{1 + \frac{2 x}{1-x} \sin^2(\phi)}{\left(1 + \frac{4 x}{(1-x)^2} \sin^2(\phi)\right)^{3/2}} \mathrm{d} \phi
\end{eqnarray}
$$
Letting $m=-\frac{4 x}{(1-x)^2}$ and $a= \frac{2x}{1-x}$:
$$\begin{eqnarray}
\int_0^{\pi/2}\frac{1 + a \cdot \sin^2(\phi)}{\left(1-m \cdot \sin^2(\phi)\right)^{3/2}} \mathrm{d}\phi &\stackrel{t = \sin^2(\phi)}{=}& \int_0^1 \frac{1+a t}{(1-m t)^{3/2}} \frac{\mathrm{d}t}{2\sqrt{t(1-t)}}\\
\end{eqnarray}
$$
Using
$$
\frac{1+a t}{(1-m t)^{3/2}} \frac{1}{2\sqrt{t(1-t)}}= \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{a+m}{1-m} \frac{\sqrt{t(1-t)}}{\sqrt{1-m t}}\right) + \frac{a+m}{2m(1-m)} \frac{\sqrt{1-m t}}{\sqrt{t(1-t)}} - \frac{a}{2m} \frac{1}{\sqrt{t(1-t)}\sqrt{1-m t}}
$$
we arrive at
$$
\int_0^1 \frac{1+a t}{(1-m t)^{3/2}} \frac{\mathrm{d}t}{2\sqrt{t(1-t)}} = \frac{a+m}{m(1-m)} E(m) - \frac{a}{m} K(m)
$$
Substituting $m = -\frac{4x}{(1-x)^2}$ and $a=\frac{2x}{1-x}$ and combining terms:
$$
\int_0^{2\pi} \frac{1-x \cos(\phi)}{\left(1+x^2-2x \cos(\phi)\right)^{3/2}}\mathrm{d} \phi = \frac{2}{1+x} \operatorname{E}\left(\frac{-4x}{(1-x)^2}\right) + \frac{2}{1-x} \operatorname{K}\left(\frac{-4x}{(1-x)^2}\right)
$$
Here is a numerical check with Mathematica:
In[27]:= With[{x = 0.78}, {NIntegrate[(
1 - x Cos[a])/(1 + x^2 - 2 x Cos[a])^(3/2), {a, 0, 2 Pi}],
2 EllipticE[-4 x/(1 - x)^2]/(1 + x) +
2 EllipticK[-4 x/(1 - x)^2]/(1 - x)}]
Out[27]= {13.2161, 13.2161}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Linear algebra exam questions Pick out the true statements:
*
*There exist $n\times n$ matrices $A$ and $B$ with real entries such that $(I-(AB-BA)^n) = 0$.
*If $A$ is symmetric and positive definite matrix then $tr(A)^n\geq n^n \det(A)$.
:(
I am stucked, unable to solve this problem.
| For the first statement, if $n=2$, let $A=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2}\end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -1 & -1\end{pmatrix}$
Then
$$AB-BA= \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& \frac{1}{2}\end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2}& +\frac{1}{2}\end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 1& 0\end{pmatrix}$$
I think this can be generalized to $2m \times 2m$ matrices by using
$\begin{pmatrix} A & 0 & 0& ..&0 \\
0 & A & 0& ..&0 \\
. & .& . & ..& . \\
0 & 0 & 0& ..&A \\\end{pmatrix}$ and $\begin{pmatrix} B & 0 & 0& ..&0 \\
0 & B & 0& ..&0 \\
. & .& . & ..& . \\
0 & 0 & 0& ..&B \\\end{pmatrix}$
For the second statement $A=I$...
If the problem asks instead to prove that
$$tr(A)^n \geq n^n \det(A)$$
then let $\lambda_1,..,\lambda_n$ be the eigenvalues. Then they are real and positive (WHY?), thus by $AM-GM$ we have:
$$\frac{tr(A)}{n}=\frac{\lambda_1+...+\lambda_n}{n} \geq \sqrt[n]{\lambda_1 \cdot ... \lambda_n}=\sqrt[n]{\det(A)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the Cartesian equation corresponding to $r = \frac{5}{3-2\cos(\theta)}$
Find the Cartesian equation corresponding to $r =
\frac{5}{3-2\cos(\theta)}$
I got it into the form:
$(x^2 + y^2)(3-2x)^2 = 25$
and can see that maybe the equation of a circle will appear, but I can't seem to get much further without complicating things. I have a note saying that completing the square was used but I can't see how or where.
| Rearranging we get that $3r - 2r\cos(\theta) = 5$. Note that $r = \sqrt{x^2 + y^2}$ and $r \cos(\theta) = x$.
This gives us $3 \sqrt{x^2+y^2} -2x = 5$.
Hence,
\begin{align}
9(x^2+y^2) & = (2x+5)^2\\
9x^2+9y^2 & = 4x^2 + 20x + 25\\
5x^2 -20x + 9y^2 & = 25\\
5(x-2)^2 + 9y^2 & = 45\\
\frac{(x-2)^2}{3^2} + \frac{y^2}{\left(\sqrt{5} \right)^2} & = 1
\end{align}
The above is an ellipse with center $(2,0)$ with semi-major axis along $x$ of length $3$ and semi-minor axis along $y$ of length $\sqrt{5}$.
EDIT
In general, $$r(\theta) = \dfrac{a(1-e^2)}{1 \pm e \cos(\theta)}$$ represents an ellipse whose semi-major axis is $a$ and eccentricity $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ What is $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ equal to (in degrees)?
Here is how I tried to solve it:
I assumed $x = \tan 15 (x= 0.27)$, so I rewrote it as:
$x\cdot 2x \cdot 3x \cdot 4x \cdot 5x = 120x^5$
$0.27^5 = 0.001323$
$120 \cdot 0.001323 = 0.16$
But Google Calculator gives me $-1.19576279$
What is the right way to calculate this trig expression?
| Note that if $0\leq \alpha\leq \frac{\pi}{2}$ ($0^{\circ}$ to $90^{\circ}$), then
$$\cos\left(\frac{\pi}{2}-\alpha\right) = \cos\frac{\pi}{2}\cos\alpha + \sin\frac{\pi}{2}\sin\alpha = \sin\alpha.$$
In terms of degrees, $\cos(90-\alpha) = \sin(\alpha)$.
So if you look at $\tan(15^{\circ})$ and $\tan(75^{\circ}) = \tan(90^{\circ}-15^{\circ})$, we have:
$$\tan(15)\tan(75) = \frac{\sin(15)}{\cos(15)}\;\frac{\sin(75)}{\cos(75)} = \frac{\sin(15)}{\sin(75)}\;\frac{\sin(75)}{\sin(15)} = 1.$$
Similarly with $\tan 30 \tan 60$. So the entire product is just equal to $\tan(45)$. And $\tan(45^{\circ})$ is equal to...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What equation intersects only once with $f(x)=\sqrt{1-(x-2)^2}$ Being $f(x)=\sqrt{1-(x-2)^2}$ I have to know what linear equation only touches the circle once(only one intersection), and passes by $P(0,0)$.
So the linear equation must be $y=mx$ because $n=0$.
I have a system of 2 equations:
\begin{align}
y&=\sqrt{1-(x-2)^2}\\
y&=mx
\end{align}
So I equal both equations and I get \begin{align}
mx&=\sqrt{1-(x-2)^2}\\
m&=\frac{\sqrt{1-(x-2)^2}}{x}
\end{align}
$m$ can be put in the $y=mx$ equation, which equals to:
\begin{align}
y&=\left(\frac{\sqrt{1-(x-2)^2}}{x}\right)x\\
&=\sqrt{1-(x-2)^2}
\end{align}
But that equation has $\infty$ intersections, and I want only the equation who has $1$ interception.
What is the good way to know this? And how can it be calculated?
| You were doing fine up through $mx=\sqrt{1-(x-2)^2}$. Starting there, square both sides to get $m^2x^2=1-(x-2)^2$, and rewrite this as $(m^2+1)x^2-4x+3=0$. Think of this as a quadratic in $x$; solving yields
$$x=\frac{4\pm\sqrt{16-12(m^2+1)}}{2(m^2+1)}=\frac{2\pm\sqrt{4-3(m^2+1)}}{m^2+1}\;,$$
which has two solutions or none unless $4-3(m^2+1)=0$, i.e., $m^2=\frac13$, and $m=\pm\frac1{\sqrt3}$.
However, the equation $f(x)=\sqrt{1-(x-2)^2}$ describes only the upper half of the circle with centre $\langle 2,0\rangle$ and radius $1$, so the line through the origin must have positive slope in order to intersect the semicircle. Its equation is therefore $$y=\frac{x}{\sqrt3}\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/147874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int\frac{x^{1/2}}{1+x^2}\,dx.$
Compute $$\int\frac{x^{1/2}}{1+x^2}\,dx.$$
All I can think of is some integration by substitution. But ran into something scary. Anyone have any tricks?
| $$I = \int \dfrac{x^{1/2}}{1+x^2} dx$$ Let $x = t^2$. We get $$I = \int \dfrac{2t^2 dt}{1+t^4}$$ Now factorize $(1+t^4)$ as $(t^2 + \sqrt{2}t+1)(t^2 - \sqrt{2}t+1)$ and use partial fractions.
$$I = \dfrac{1}{\sqrt{2}} \int \left( \dfrac{t}{t^2 - \sqrt{2} t+1} - \dfrac{t}{t^2 + \sqrt{2} t+1}\right)$$
Now $$\int \dfrac{t}{(t-a)^2 + b^2} dt = \int \dfrac{t-a+a}{(t-a)^2 + b^2} dt = \int \dfrac{t-a}{(t-a)^2 + b^2} dt + \int \dfrac{a}{(t-a)^2 + b^2} dt \\= \frac12 \log((t-a)^2+b^2) + \frac{a}{b} \arctan \left( \dfrac{t-a}{b}\right)$$
In our case, $a= \pm \dfrac1{\sqrt{2}}$ and $b = \dfrac1{\sqrt{2}}$.
Hence, the integral is $$\frac1{\sqrt{2}} \left( \frac12 \log(t^2 - \sqrt{2}t + 1) + \arctan(\sqrt{2}t+1) - \frac12 \log(t^2 + \sqrt{2}t + 1) + \arctan(\sqrt{2}t-1) \right) + C$$
Now plug in $t = \sqrt{x}$ to get the integral in terms of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/148275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Is $k=3$ the only solution to $\sum_{r=1}^n r^k =\left( \sum_{r=1}^nr^1 \right)^2 $? $f(k) = \sum_{r=1}^{n} r^k$. Find an integer $x$ that solves the equation $f(x) = \bigl(f(1)\bigr)^2$.
Problem credit: http://cotpi.com/p/2/
I understand why $x = 3$ is a solution. $1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2 = (1 + 2 + \dots + n)^2$. But how can we prove that there is no other solution?
Will we have other solutions if $x$ is real? What if $x$ is complex?
| More generally, one can look for expressions of the form
$f(k) = \sum_{j=1}^{\lfloor(n-1)/2\rfloor} a_j f(j) f(n+1-j)$. It seems to work for all odd $k$. For example:
$$\eqalign{f \left( 3 \right) &= \left( f \left( 1 \right) \right) ^{2}\cr
f \left( 5 \right) &=4\,f \left( 1 \right) f \left( 3 \right) -3\,
\left( f \left( 2 \right) \right) ^{2}\cr
f \left( 7 \right) &=6\,f \left( 1 \right) f \left( 5 \right) -15\,f
\left( 2 \right) f \left( 4 \right) +10\, \left( f \left( 3 \right)
\right) ^{2}\cr
f \left( 9 \right) &=8\,f \left( 1 \right) f \left( 7 \right) -28\,f
\left( 2 \right) f \left( 6 \right) +56\,f \left( 3 \right) f \left(
5 \right) -35\, \left( f \left( 4 \right) \right) ^{2}\cr
f \left( 11 \right) &=10\,f \left( 1 \right) f \left( 9 \right) -45\,f
\left( 2 \right) f \left( 8 \right) +120\,f \left( 3 \right) f
\left( 7 \right) -210\,f \left( 4 \right) f \left( 6 \right) +126\,
\left( f \left( 5 \right) \right) ^{2}\cr
}$$
Hmm, looks like $$f(2k+1) = \frac{1}{2}\sum_{j=1}^{2k-1} (-1)^{j+1} {2k \choose j} f(j) f(2k-j) $$
Ought to be easy to prove...
EDIT: yes, it is. By the binomial theorem
$$ \sum_{j=1}^{2k-1} \sum_{q=1}^n \sum_{r=1}^n (-1)^{j+1} {2k \choose j} q^j r^{2k-j}
= \sum_{q=1}^n \sum_{r=1}^n \left(r^{2k} + q^{2k} - (r-q)^{2k}\right)$$
and note that $$\sum_{q=1}^n \sum_{r=1}^n (r-q)^{2k} = \sum_{s=1}^n 2 (n-s) s^{2k} =
2 n f(2k) - 2 f(2k+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Perrin numbers in terms of the generalized hypergeometric function? Given the roots of $x^3=x^2+1$, we have sequence A001609,
$M(n) = x_1^n+x_2^n+x_3^n = \,_3F_2\left(\frac{-n}{3}, \frac{1-n}{3}, \frac{2-n}{3};\; \frac{1-n}{2}, \frac{2-n}{2};\; -\frac{3^3}{2^2}\right) = 1, 1, 4, 5, 6, 10, 15, 21,\dots$
for $n = {1,2,3,\dots}$
Question: Given $y^3=y+1$, is there any similar generalized hypergeometric formula for the Perrin numbers?
$P(n) = y_1^n+y_2^n+y_3^n = 0,2,3,2,5,5,7,10,\dots$
The closest I found is the binomial sum,
$ \begin{aligned}P(n) &= n\sum_{k=1}^{n/2} \frac{\binom k{n-2k}}{k} = 0,2,3,2,5,5,7,10,\dots\end{aligned}$
where both start with $n = 1,2,3,\dots$ Anyone knows how to translate that into the generalized hypergeometric function?
| A solution for calculating P(n) when n is prime using the hypergeometric function can be found at Perrin088.org (Chapter 15)
This equation is derived from an incomplete beta function giving the nth term of the Perrin sequence when n is prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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?
| Using induction it is obvious that the statement is true for $n=1$.
Now suppose that it is true for $n=k$,
then we have $k(k^2+8)=3m$, where m is an integer.
Considering the case where $n=k+1$, we have;
$$(k+1)[(k+1)^2+8]=k(k^2+2k+9)+k^2+2k+9$$
$$=k(k^2+8+2k+1)+k^2+2k+9=k(k^2+8)+k(2k+1)+k^2+2k+9$$
$$=k(k^2+8)+3k^2+3k+9=3m+3k^2+3k+9=3(m+k^2+k+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 5
} |
Computing $ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$ I would like to compute the integral:
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx$$
where $$ a<b$$ and $$ p\in\mathbb{N}$$
$$ \int_a^b \frac{x^p}{\sqrt{(x-a)(b-x)}} \mathrm dx=\frac{2}{b-a} \int_a^b \frac{x^p \mathrm dx}{\sqrt{1-\left(\frac{2x-(a+b)}{b-a}\right)^2}} $$
The substitution $x\rightarrow \frac{2x-(a+b)}{b-a}$ gives:
$$ \frac{1}{2^p} \int_{-1}^{1} \frac{((b-a)x+a+b)^p}{\sqrt{1-x^2}} \mathrm dx$$
I tried to make an integration by parts:
$$ \frac{1}{2^p}\left(2^{p-1}\pi(a^p+b^p)-p(b-a) \int_{-1}^1 \arcsin(x)((b-a)x+a+b)^{p-1} \mathrm dx \right)$$
What about the integral $$ \int_{-1}^1 \arcsin(x)((b-a)x+a+b)^{p-1} \mathrm dx $$?
| I would probably find it easiest to reduce the given integral to Euler's integral for Gauss's hypergeometric function:
$$
\int_a^b \frac{ x^p \mathrm{d} x}{\sqrt{(b-x)(x-a)}} \stackrel{x = a+(b-a) u}{=} \int_0^1 \frac{(a+(b-a)u)^p }{\sqrt{u(1-u)}} \mathrm{d} u = \\ a^p \int_0^1 u^{-\frac{1}{2}} \left(1-u\right)^{-\frac{1}{2}} \left( 1- \left(1-\frac{b}{a}\right) u \right)^p \mathrm{d} u
$$
This is the case of the Euler integral representation of the Gauss's hypergeometric function with parameters $\alpha = \frac{1}{2}$, $\gamma=1$ and $\beta = -p$ and $z=1-\frac{b}{a}$, hence
$$
\int_a^b \frac{ x^p \mathrm{d} x}{\sqrt{(b-x)(x-a)}} = a^p \cdot \underbrace{\mathrm{B}\left(\frac{1}{2}, \frac{1}{2}\right)}_{\pi} \cdot {}_2 F_1\left(\frac{1}{2}, -p ; 1; 1 - \frac{b}{a}\right)
$$
The hypergeometric series terminates for $p\in \mathbb{N}$, and the result coincides with that of @Marvis, but the above result extends to all the complex values of $p$:
In[29]:= With[{a = 2.4, b = 3.6, p = 3.4},
a^p \[Pi] Hypergeometric2F1[1/2, -p, 1, 1 - b/a]]
Out[29]= 142.388
In[30]:= With[{a = 2.4, b = 3.6, p = 3.4},
NIntegrate[x^p/Sqrt[(b - x) (x - a)], {x, a, b}]]
Out[30]= 142.388
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Ratio and number theory The question goes as follows:
Let $K$ be a three digit number such that the ratio of the number to the sum
of its digit is least. What is the difference between the hundreds and
the tens digits of $K$?
Now I was able to do this question by trial and error, assuming hundredth digit place to be 1 and unit as well as tens digit to be 9
So the number is 199, but I am not able to do it logically, any way to do it?
| Let the number be $100a + 10 b +c$, where $1 \leq a \leq 9$ and $0 \leq b,c, \leq9$. Hence, we want to minimize
$$L=\dfrac{100a + 10b + c}{a+b+c} = 1 + \dfrac{99a+9b}{a+b+c}$$
This means that you should choose $c$ to be maximum as possible since $c$ appears only in the denominator and the term is positive. Hence, $c = 9$. Hence, we want to minimize $$L=\dfrac{100a + 10b + c}{a+b+c} = 1 + \dfrac{99a+9b}{a+b+9} = 1 + \dfrac{90a - 81 + 9a+ 9b + 81}{a+b+9}$$
$$L = 1 + \dfrac{90a-81}{a+b+9} + 9 = 10 + 9 \left(\dfrac{10a-9}{a+b+9} \right)$$
Now again you should choose $b$ to be maximum as possible since $b$ appears only in the denominator and the term is positive. Hence, set $b=9$. Hence, we want to minimize $$L = 10 +9 \left( \dfrac{10a-9}{a+18} \right) = 10 +9 \left( \dfrac{10a + 180 -189}{a+18} \right) = 10 + 90 - \dfrac{9 \times 189}{a+18}$$
Now you ned to choose $a$ to be minimum as possible since $a$ appears in the denominator and the term is negative. Hence, set $a = 1$. Hence, $$L = 100 - \dfrac{9 \times 189}{19} = \dfrac{1900 - 9 \times 189}{19} = \dfrac{199}{19}$$ The number is $199$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$ I've encountered the following problem: Evaluate $$\lim_{x\to 0^+}\cfrac{x^{x^x}}{x}.$$
This is readily a "$\frac{0}{0}$" form, so I used L'Hopital's rule, but it got seriously messy, and fast. Can anyone recommend an alternative approach?
| All we need is a nice enough series expansion for $x^x$ about $0$, which can be obtained by rewriting $x^x$ as $\exp \left( x \log (x)\right)$ and looking at the Taylor series of $\exp \left( y\right)$. Now if we look at $$f(x) = \log \left( \dfrac{x^{x^x}}{x} \right) = \log \left( x^{x^x}\right) - \log x = \log (x) \left(x^x - 1 \right) \\= \log (x) \left(-1 + \left(1 + x \log (x) + \dfrac{x^2\log^2(x)}{2!} + \dfrac{x^3\log^3(x)}{3!} + \dfrac{x^4\log^4(x)}{4!} + \cdots\right) \right)\\=x \log^2(x) \left( 1 + \dfrac{x\log(x)}{2!} + \dfrac{x^2\log^2(x)}{3!} + \dfrac{x^3\log^3(x)}{4!} + \cdots\right)$$
Hence, the limit of $f(x)$ as $x \to 0$ is $0$. Hence, $$\lim_{x \rightarrow 0} \dfrac{x^{x^x}}{x} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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$3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
| May be you know this already but it's not clear to me from your post that you have the progression from the original problem statement to the final two factors exactly correct.
$3x^3 = 24 \\ \Rightarrow 3x^3 - 24 = 0 \\ \Rightarrow 3\left(x^3 - 8\right) = 0 \\ \Rightarrow x^3 - 8 = 0 \\ \Rightarrow \left(x-2\right)\left(x^2+2x+4\right) = 0 \\
\Rightarrow x-2 = 0, \ \ x^2+2x+4 = 0$
The first factor results in $x = 2$. The second factor must be solved using the quadratic formula or by completing the square since it is not factorable (Quadratic Formula on Wikipedia).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 think your substitutions from the first line to the second aren't quite right. It looks like you used $x=a\cos\theta$ and $y=b\sin\theta$, but you probably wanted $$\begin{align}x&=r\cos\theta\\y&=r\sin\theta.\end{align}$$
Using those, $$\begin{align}
(x-a)^2 + (y - b)^2 &= a^2 + b^2 \\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=a^2+b^2 \\
r^2\cos^2\theta+a^2-2ar\cos\theta+r^2\sin^2\theta+b^2-2br\sin\theta&=a^2+b^2 \\
r^2(\sin^2\theta+\cos^2\theta)+(a^2+b^2)-2r(a\cos\theta+b\sin\theta)&=a^2+b^2 \\
r^2-2r(a\cos\theta+b\sin\theta)&=0 \\
r^2=2r(a\cos\theta+b\sin\theta) \\
r=0\;\;\text{ or }\;\;r=2a\cos\theta+2b\sin\theta
\end{align}$$
and since $r=0$ describes only the pole, which is also contained in the graph of $r=2a\cos\theta+2b\sin\theta$, $$r=2a\cos\theta+2b\sin\theta$$ is sufficient to describe the circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Find the square root of the polynomial My question is:
Find the square root of the polynomial-
$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
| Let $t=\dfrac xy +\dfrac yx$.
Then $t^2=\dfrac {x^2}{y^2} +\dfrac{y^2}{x^2}+2$.
Your "polynomial" becomes finally an actual polynomial:
$$t^2-2-2t+3=t^2-2t+1=(t-1)^2=\left(\frac xy+\frac yx -1\right)^2.$$
So, the two square roots are $$\pm \left(\frac xy+\frac yx -1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx$ So, from here $$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)} dx$$
I divided by cos(x) and I got
$$\int \frac{\tan(x)}{2\cos^2(x)+1} dx$$
But I'm stuck here.
I tried to substitute $t=\cos(x)$
$$\int \frac{-1}{t\cdot(2t^2+1)} dt$$
Any help would be greatly appreciated.
| $$
\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big)
$$
$$
= \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du)
$$
Then use partial fractions.
Later edit in response to comments:
$$
\int \frac{1}{3u^3 + (1-u^2)u}\,(-du) = \int\frac{-du}{u(2u^2 + 1)}
= \int \frac{A}{u} + \frac{Bu+C}{2u^2+1} \, du
$$
Two logarithms plus an arctangent.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Summation of a series. I encountered this problem in Physics before i knew about a thing called Taylor Polynomials My problem was that i had to sum this series :
$$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n}$$
basically $$1,-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5},-\frac{1}{6},\frac{1}{7}.....$$
So now i know that there is something called a taylor polynomial that says that
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}....$$
So the above summation boils down to $\ln 2$.
What if i never knew the exansion then how would I calculate it?
Earlier I tried solving it like so ,
divide it into two different sets i.e.
$$\text{1 and $\dfrac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13} \ldots$ and $-\dfrac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\ldots$}$$
I said Hey! the first set would contain stuff like, $$\frac{1}{3^n},\frac{1}{5^n},\ldots$$ each of them would probably be reduced to a sum like so $$\sum^\infty_{n=1}\frac1{a^n}=\frac1{a-1}$$ and further become $$\sum^\infty_{a=3}\frac1{a-1}$$ which would subtract all the numbers in the other even set giving 1 as the answer which is wrong .
Where did I go wrong and how could I proceed even without knowing Taylor polynomials
| $\left[ - \sum\limits_{k = 0}^\infty {\ln \left( {(1 - q)(x{q^k} - \alpha ) + 3q - 1} \right)} + \sum\limits_{k = 0}^\infty {\ln \left( {\frac{{(1 - q)}}{2}(x{q^k} - (\alpha + 1))} \right)} \right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
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?
| This can be solved by probability theory: Let your matrix be $\Sigma$ with diagonal elements all 1 and the off-diagonal elements all equal to $\rho$.
This is the variance-covariance matrix of a random vector $X$ where
all components have variance 1 and all pairs of distinct elements have correlation $\rho$, where $|\rho| \le 1$ is necessary for this having any hope of being a variance-covariance matrix (which implies the determinant is nonnegative)
Are there any other restrictions? Yes, we can calculate the variance of the sum of the elements of $X$, which clearly must be positive:
$$\mbox{var}\left(\sum_{i=1}^n X_i\right) = 1^T\Sigma 1 = \sum_i \sum_j \Sigma_{ij} =n + n(n-1) \rho >0$$
(where 1 stands for the vector with all ones)
First cancel one factor $n$, then solve the inequality, you get
$\rho > -\frac{1}{n-1}$, and your value 1/2 clearly satisfies this!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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!
| The sum is divergent.
Let's assume $x/x_0\notin\mathbb{Z}$ to avoid a trivial divergence.
Note that for $k\ne 0$
$$\begin{eqnarray*}
|x-k x_0| &=& |k| |x/k-x_0| \\
&\le& |k|(|x/k| + |x_0|) \\
&\le& |k|(|x|+|x_0|).
\end{eqnarray*}$$
We have used the triangle inequality and the fact that
$1/|k| \le 1$ for $|k|\ge 1$.
Thus,
$$\begin{eqnarray*}
\sum_{k=-\infty}^\infty \frac{1}{|x-k x_0|}
&\ge& \frac{1}{|x|} + \frac{2}{|x| + |x_0|} \sum_{k=1}^\infty \frac{1}{k}.
\end{eqnarray*}$$
The sum diverges since the harmonic series diverges.
I suspect the sum you are actually interested in is the one dealt with by @PeterTamaroff.
Addendum:
I have tracked down the sign error.
Assuming $x/x_0 < 1$, so that $|n x_0 -x| = n x_0 - x$,
the relevant term in the sum is
$$\frac{1}{n x_0 -x} + \frac{1}{n x_0 + x}
= \frac{2 x_0 n}{n^2 x_0^2 - x^2} \sim \frac{1}{n}.$$
Note the relative sign is plus, not minus.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
| $$\left. \dfrac{(x+2)^2}5 \right \vert_0^1 = \dfrac{(1+2)^2}5 - \dfrac{(0+2)^2}5 = \dfrac{3^2}5 - \dfrac{2^2}5 = \dfrac{9}5 - \dfrac45 = \dfrac{9-4}5 = \dfrac55 = 1$$
Note that we can integrate $\displaystyle \int_{x_1}^{x_2} (x+a) dx$ in seemingly two different ways.
The first method is to treat $x+a$ together as one object i.e. $$\displaystyle \int (x+a) dx = \dfrac{(x+a)^2}2 + c_1$$
The second method is to treat $x+a$ as two separate objects i.e. $$\displaystyle \int (x+a) dx = \int x dx + \int a dx = \dfrac{x^2}2 + ax + c_2$$
It might seem that both are different. However, note that the first answer can be re-written as $$\dfrac{(x+a)^2}2 + c_1 = \dfrac{x^2}2 + ax + \dfrac{a^2}2 + c_1.$$ Now this looks more closely like the second. The only difference in fact is that the constants are different. They are in fact related as $c_2 = c_1 + \dfrac{a^2}2$. While performing a definite integral, the constants cancel off and hence both ways should give us the same answer.
As an exercise, we will integrate what you have by treating $x$ and $2$ separately.
\begin{align}
\int_0^1 \dfrac{2(x+2)}5 dx & = \dfrac25 \int_0^1 (x+2)dx = \dfrac25 \int_0^1 xdx + \dfrac25 \int_0^1 2dx = \dfrac25 \cdot \left. \dfrac{x^2}2 \right \vert_{0}^1 + \dfrac25 \cdot 2 \cdot \left(1 - 0 \right)\\
& = \dfrac25 \cdot \left(\dfrac{1^2}2 - \dfrac{0^2}2 \right) + \dfrac45 = \dfrac25 \cdot \dfrac12 + \dfrac45 = \dfrac15 + \dfrac45 = 1
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/160542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Need to find $\lim_{x\to4}\bigl((1/\sqrt x)-(1/2)\bigr)/(x-4)$ This one is frustrating $$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$$
| $\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$.
We substition $\sqrt {x}=t$, hance we:
If $ x\longrightarrow 4\Rightarrow t\longrightarrow 2. $
From here for the given limits have:
$\lim\limits_{x\to 4} \frac{(1 / \sqrt{x}) - \frac12}{x-4}$=$\lim\limits_{t\to 2} \frac{\frac{1}{t} - \frac12}{t^2-4}$=$\lim\limits_{t\to 2} \frac{\frac{2-t}{2t}}{t^2-4}$=$\lim\limits_{t\to 2} \frac{2-t}{2t(t-2)(t+2)}$=$-\lim\limits_{t\to 2} \frac{t-2}{2t(t-2)(t+2)}$=$-\lim\limits_{t\to 2} \frac{1}{2t(t+2)}$=$-\frac{1}{16}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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?
| Here is my answer in terms of the digamma function (cf. Abramowitz and Stegun). Using the decomposed version of the sum given by Eric Naslund♦, we find:
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}=
\frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=0}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right)=
\frac{\psi (-z_0)-\psi(-z_1)}{\sqrt{b^{2}-4ac}}
$$
I note that this answer is in fact quite similar to the sum's closed form from $-\infty$ to $\infty$.
Here is an alternate solution:
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}=
\frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}
$$
If $c_n$ is the $n^{th}$ term of the second version, then
$$\frac{c_{n+1}}{c_n}=\frac{(n-z_0)(n-z_1)}{(n-z_0+1)(n-z_1+1)}$$
which shows that this sum can be written as a hypergeometric function
$$
\frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}=
\frac{1}{az_0z_1}\sum_{n=0}^\infty \frac{\Gamma(n-z_0)\Gamma(n-z_1)n!}{\Gamma(n-z_0+1)\Gamma(n-z_1+1)n!}=
\frac{1}{z_0z_1a}{_3}F_2(-z_0, -z_1, 1;1-z_0, 1-z_1;1)
$$
| {
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"url": "https://math.stackexchange.com/questions/161259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$ It's a question (not hw) I bumped into few years back. Couldn't make any real progress with. Maybe you can help?
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$$
Thanks.
| For your original question, it might be worth noting that
$$\frac{1}{\sqrt{k^2}+\sqrt{k(k+1)}+\sqrt{(k+1)^2}}=\frac{1}{2k+1+\sqrt{k(k+1)}}=\frac{2k+1-\sqrt{k(k+1)}}{4k^2+4k+1-k^2-k}=\frac{2k+1-\sqrt{k(k+1)}}{3k^2+3k+1}$$
Thus
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=\frac{5-\sqrt{6}}{19}+\frac{7-2\sqrt{3}}{37}+\frac{9-2\sqrt{5}}{61} $$
As a side not, a much nicer expression is:
$$\frac{1}{\sqrt{4}+2\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+2\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+2\sqrt{20}+\sqrt{25}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small.
| Maybe this is easier:
$$\eqalign{
& \int\limits_0^1 {\frac{{{x^4}\log x}}{{{x^2} - 1}}dx} = \int\limits_0^1 {\frac{{\left( {{x^4} - 1} \right)\log x}}{{{x^2} - 1}}dx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr
& = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr} $$
$$ = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} = \left[ {\left( {\frac{{{x^3}}}{3} + x} \right)\log x} \right]_0^1 - \int\limits_0^1 {\left( {\frac{{{x^2}}}{3} + 1} \right)dx} = - \frac{{10}}{9}$$
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \int_0^1 \frac{\log x}{x-1}dx-\frac 1 2 \int_0^1 \frac{\log x}{x+1}dx$$
Now those two evaluate in terms of $\pi$, since we get
$$-\int_0^1 \frac{\log x}{1-x}dx=-\int_0^1 \log x \sum_{k=0}^\infty x^kdx= $$
$$=-\sum_{k=0}^\infty\int_0^1 x^k\log x dx=\sum_{k=0}\frac{1}{(k+1)^2}=\frac{\pi^2}{6} $$
Since
$$\int_0^1 x^k\log x dx=-\frac{1}{(k+1)^2}$$
And for the other, we get the similar:
$$\int_0^1 \frac{\log x}{1+x}dx=\int_0^1 \log x \sum_{k=0}^\infty (-1)^kx^kdx= $$
$$=\sum_{k=0}^\infty\int_0^1 \log x (-1)^kx^kdx=\sum_{k=0}^\infty(-1)^k\int_0^1 x^k\log x dx=\sum_{k=0}\frac{(-1)^{k+1}}{(k+1)^2}=-\frac{\pi^2}{12} $$
So we have that
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \left( \frac{\pi^2}{6}+\frac{\pi^2}{12} \right)$$
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac{\pi^2}{8}$$
and finally
$$I=\frac{\pi^2}{8}-\frac{10}{9}\approx 0.1225 < 0.125$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Integral - using Euler Substitution I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot get this result. Below is my solution of the problem. I've checked it many times, must be something very obvious that I'm missing:
(original image)
Euler Substituion
$\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}$
Let $\sqrt{x^2+x+1}=t-x$.
$x^2+x+1=t^2-2xt+x^2$
$x(1+2t)=t^2-1\implies x=\dfrac{t^2-1}{1+2t}$
$dx=\left(\dfrac{t^2-1}{1+2t}\right)'dt=\dfrac{2t(1+2t)-(t^2-1)2}{(1+2t)^2}=\dfrac{2t+4t^2-2t^2+2}{(1+2t)^2}=\dfrac{2(t^2+t+1)}{(1+2t)^2}$
$\sqrt{x^2+x+1}=t-x=t-\dfrac{t^2-1}{1+2t}=\dfrac{t^2+t+1}{1+2t}$
$\implies\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}=2\int\frac{\frac{t^2+t+1}{(1+2t)^2}\;dt}{\frac{t^2-1}{1+2t}\cdot\frac{t^2+t+1}{1+2t}}=2\int \frac{1}{t^2-1}\,dt$
$\dfrac{1}{t^2-1}=\dfrac{1}{(t+1)(t-1)}=\dfrac{A}{t+1}+\dfrac{B}{t-1}\implies \begin{eqnarray}&&At-A+Bt+B=1\\&&A+B=0\implies A=-B\\ &&B-A=1\implies B=\frac{1}{2},A=-\frac{1}{2}\end{eqnarray}$
$\implies \displaystyle 2\int \frac{1}{2}\frac{1}{2t-1}\,dt-2\int\frac{1}{2}\frac{1}{t+1}\,dt=\int\frac{1}{t-1}\,dt-\int\frac{1}{t+1}\,dt=$
$=\ln|t-1|-\ln|t+1|=\ln\left|\dfrac{t-1}{t+1}\right|$
$t-x=\sqrt{x^2+x+1}\implies t=\sqrt{x^2+x+1}+x$
$\implies \ln\left|\dfrac{t-1}{t+1}\right|=\ln\left|\dfrac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|$
I'll appreciate any help.
Thanks in advance!
| What you have done is correct. All you need to do is to rewrite it a different form.
$$\begin{align}
\ln \left( \sqrt{x^2+x+1} + x - 1\right) & = \ln \left( \dfrac{\left(\sqrt{x^2+x+1} + x - 1 \right) \left(\sqrt{x^2+x+1} - x + 1 \right)}{\left(\sqrt{x^2+x+1} - x + 1 \right)}\right)\\
& = \ln \left(\left(x^2 + x + 1 - (x^2 - 2x + 1) \right) \right) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right)\\
&= \ln(x) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right)
\end{align}
$$
Hence, $$\begin{align}
\ln \left( \dfrac{\sqrt{x^2+x+1} + x - 1}{\sqrt{x^2+x+1} + x + 1}\right) & = \ln \left( \sqrt{x^2+x+1} + x - 1\right) - \ln \left(\sqrt{x^2+x+1} + x + 1 \right)\\ & = \ln(x) - \ln \left(\sqrt{x^2+x+1} - x + 1 \right) -\ln \left(\sqrt{x^2+x+1} + x + 1 \right)\\
& = \ln(x) - \left(\ln \left(\sqrt{x^2+x+1} - x + 1 \right) +\ln \left(\sqrt{x^2+x+1} + x + 1 \right) \right)\\
& = \ln(x) - \ln \left( \left(\sqrt{x^2+x+1}+1 \right)^2 - x^2\right)\\
& = \ln(x) - \ln \left( x^2 + x + 1 + 1 +2 \sqrt{x^2+x+1} - x^2\right)\\
& = \ln(x) - \ln \left( 2 \sqrt{x^2+x+1} + x + 2\right)
\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/162515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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finding the rational number which the continued fraction $[1;1,2,1,1,2,\ldots]$ represents I'd really love your help with finding the rational number which the continued fraction $[1;1,2,1,1,2,\ldots]$ represents.
With the recursion for continued fraction $( p_0=a_0, q_0=1, p_{-1}=1, q_{-1}=o), q_s=a_sq_{s-1}+q_{s-2},p_s=a_sp_{s-1}+p_{s-2}$. I found out the $p_k=1,2,5,7,12,32, q_k=1,1,3,4,7,18 $ (anything special about these series? perhaps I did a mistake?), and I know that $r=\lim_{c_k}=\lim\frac{p_k}{q_k}$, but I can't see anything special about $p_k, q_k$ or the realtion between them, Any help?
Thanks a lot!
| We have that
$$\Large x=1+\frac{1}{1+\frac{1}{2\,+\,\frac{1}{1\,+\,\frac{1}{1\,+\,\frac{1}{2\,+\,\cdots}}}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{x}}}$$
so that
$$\frac{1}{x-1}=1+\frac{1}{2+\frac{1}{x}}$$
hence
$$\frac{1}{\frac{1}{x-1}-1}=\frac{1}{\frac{1}{x-1}-\frac{x-1}{x-1}}=\frac{1}{\frac{2-x}{x-1}}=\frac{x-1}{2-x}=2+\frac{1}{x}$$
and thus
$$x^2-x=2x(2-x)+(2-x)$$
$$x^2-x=-2x^2+4x+2-x$$
$$3x^2-4x-2=0$$
The roots of this equation are
$$x=\frac{4\pm\sqrt{40}}{6}=\frac{4\pm 2\sqrt{10}}{6}$$
but we know it can't be $x=\frac{4-2\sqrt{10}}{6}$ since that number is negative, so we can conclude that $x=\frac{4+2\sqrt{10}}{6}$. Note that this number is not rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Summing over a cyclic subgroup of a multiplicative group mod n Let $x$ be a unit in $\mathbb Z/ n \mathbb Z$ of multiplicative order $m$.
I am trying to determine when it is that
$$
\sum_{i=0}^{m-1} x^i \equiv 0 \mod n .
$$
Is this kind of situation something that has been studied? If so, I would be very interested in any suggestions of books that discuss this kind of problem.
| (Below I treat $x$ as an integer whose reduction $\bmod n$ is a unit of order $m$. This allows me to treat $\frac{x^m - 1}{x - 1}$ as an integer and consider its reduction $\bmod n$ without dividing by a zero divisor.)
As usual, the Chinese Remainder Theorem is your friend. We can reduce to the case that $n$ is a prime power $p^k$. If $p \nmid x-1$, then the geometric series argument works and the sum is just $0 \bmod p^k$.
If $p | x-1$, the situation gets a little more complicated. We will need to introduce the following fundamental notion: for an integer $n$, the $p$-adic valuation $\nu_p(n)$ is the greatest $k$ such that $p^k | n$.
Theorem (lifting the exponent): Let $p$ be an odd prime and let $x, y$ be integers relatively prime to $p$ such that $p | x-y$. Then
$$\nu_p(x^n - y^n) = \nu_p(x - y) + \nu_p(n).$$
Proof. Induction. See these notes.
Applying the above it follows that for odd $p$, if $p | x-1$ then
$$\nu_p \left( \frac{x^m - 1}{x - 1} \right) = \nu_p(m).$$
Thus $\frac{x^m - 1}{x - 1} \equiv 0 \bmod p^k$ if and only if $p^k | m$. Hence:
If $n$ is odd then $\frac{x^m - 1}{x - 1} \equiv 0 \bmod n$ if and only if for every prime factor $p$ of $n$, either $p \nmid x-1$ or $p^k | m$.
Okay, so what if $p = 2$? In the above notes you will also find the following result.
Theorem (lifting the exponent at $2$): Let $x, y$ be odd integers such that $4 | x-y$. Then
$$\nu_2(x^n - y^n) = \nu_2(x - y) + \nu_2(n).$$
So if $4 | x-1$ then the conclusion is the same as above. Otherwise, write
$$x = 1 + 2y$$
where $y$ is odd. If $m$ is odd, then
$$x^m = 1 + 2my + ...$$
where the remaining terms are divisible by $4$, and we conclude that $\nu_2(x^m - 1) = 2$, hence $\nu_2 \left( \frac{x^m - 1}{x - 1} \right) = 0$, so the sum is not divisible by $2^k$. If $m = 2 \ell$ is even, then $4 | x^2 - 1$, hence
$$\nu_2(x^{2\ell} - 1) = \nu_2(x^2 - 1) + \nu_2(\ell) = \nu_2(x + 1) + \nu_2(m).$$
This gives
$$\nu_2 \left( \frac{x^m - 1}{x - 1} \right) = \nu_2(y + 1) + \nu_2(m).$$
Thus $\frac{x^m - 1}{x - 1} \equiv 0 \bmod 2^k$ if and only if $2^k | m(y+1)$, and this can occur. So if $n$ is not odd then the above argument still applies to all the odd prime factors of $n$ and we conclude that
if $n = 2^k o, k \ge 1$ where $o$ is odd, then $\frac{x^m - 1}{x - 1} \equiv 0 \bmod n$ if and only if $\frac{x^m - 1}{x - 1} \equiv 0 \bmod o$ and either $x \equiv 1 \bmod 4$ and $2^k | m$ or $x \equiv 3 \bmod 4$ and $2^k | m \left( \frac{x+1}{2} \right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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proving there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$ by induction $n$ is a given positive integer.
how to prove by using mathematical induction that there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$, where $b$ is odd and at least one of $a, c$ is even
| First check directly that the result holds for $n=1$.
We do the induction step. Let $x$ be an integer such that
$$ax^2+bx+c\equiv 0\pmod{2^{n-1}}.$$
We would like to show that there is an integer $y$ such that $ay^2+by+c\equiv 0\pmod{2^n}$.
If $2^n$ already divides $ax^2+bx+c$, we are finished. So we need only worry about the case where $ax^2+bx+c\equiv 2^{n-1}\pmod{2^n}$.
Try for a $y$ of the shape $x+t$. So we want
$$a(x+t)^2+b(x+t)+c\equiv 0\pmod{2^n}.$$
But $ax^2+bx+c\equiv 2^{n-1}\pmod{p}$. So we want the difference
$$[a(x+t)^2+b(x+t)+c]-[ax^2+bx+c]$$
to be congruent to $2^{n-1}$ modulo $2^n$. That is, we want to choose $t$ so that
$$a(2xt+t^2)+bt\equiv 2^{n-1}\pmod{2^n}.$$
Note that $t=2^{n-1}$ works, since $b$ is odd and $2t$ and $t^2$ are each divisible by $2^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/165711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Want to test the series for convergence/divergence $\sum_n \frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Want to test the series $$\frac{\sqrt{2} - 1}{3^3 - 1} + \frac{\sqrt{3} - 1}{4^3 - 1} + \frac{\sqrt{4} - 1}{5^3 - 1} + \cdots$$ for convergence/divergence.
My attempt is
$U_n =\frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Now I wish to find $V_n $ which is less than (or greater than $U_n $) such that its convergence/divergence is known so that I can find the behaviour of $U_n $
| Here, $\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq \frac{1}{n^2}$. For n=1, it is true, and then the denominator of given series increases more rapidly than numerator and this rate is greater than the rate in $1/n^2$, so the given series is bounded by $1/n^2$ and hence $\sum_{1}^{\infty}\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq (\sum_{1}^{\infty}\frac{1}{n^2}=\pi^2/6)$. Thus the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Write down the sum of sum of sum of digits of $4444^{4444}$ Let $A = 4444^{4444}$;
Then sum of digits of $A = B$;
Then sum of digits of $B = C$;
Then sum of digits of $C = D$;
Find $D$.
What should be the approach here?
| It's well-known that, because $10 \equiv 1 \bmod 9$, and therefore $10^k = 1 \bmod 9$ for all $k>0$, we have that the sum of digits of $n$, $S(n) \equiv n \bmod 9$.
So what is $4444^{4444} \bmod 9$? The above equivalence gives us that $4444 \equiv 7 \bmod 9$.
Now we need the order of $7$ modulo $9$, the smallest $s$ such that $7^s \equiv 1 \bmod 9$. This is easy to find by examination: $7^2 \equiv 4 \bmod 9$ and so $7^3 \equiv 28 \equiv 1 \bmod 9$.
So $4444^{4444} \equiv 7^{4444} \equiv 7 \cdot (7^3)^{1481} \equiv 7 \ \bmod 9$, and the same equivalence is true for $B,C$ and $D$.
Now roughly how big is $A$? Since $\log_{10}4444 \approx 3.64777$, we know that $\log_{10}A \approx 16210.7$, that is $A$ has $16211$ digits. This gives us that $B \le 16211\times 9 = 145899$. So if $B>100000$, the first two digits sum to no more than $5$, which means that $C \le 45$ (that maximum being when $B=99999$).
Finally we can use our knowledge that $C \equiv 7 \bmod 9$ and observe that $C$ must be one of the values $\{7,16,25,34,43\}$ and thus that $S(C) = \fbox{D = 7}$.
Additional thoughts:
We get exactly the same answer, $D=7$, for $A=55555^{55555}$ . Impressively, a slight variation on the same process also works to get $D=9$ for the case $A=999999999^{999999999}$ (nine nines).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Condensing logarithms
Simplify:
$2\log_{10}\sqrt{x}+3\log_{10}x^{\frac{1}{3}}$
I got to this: $2\log_{10}x^{\frac{1}{2}}+3\log_{10}x^{\frac{1}{3}}$.
Now, usually you bring the exponent the the front and that would yield:
$$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x=\log_{10}x+\log_{10}x=2\log_{10}x$$
And that's it?
| $$ \frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x \neq 2\log_{10}x+3\log_{10}x$$
$1/2 * 2 = 1$ so do $1/3 * 3.$
So the correct thing is
$$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x = \log_{10}x+\log_{10}x = 2\log_{10}x$$
Bring the exponent inside the $\log$ if you like.
The other way around: bring $1/2$ into $\log_{10} \sqrt{x}$ to become $\log_{10} \sqrt{x}^2 = ??.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{2}\right)$$
But, what I do not know is how to factor a trinominal (reverse of the FOIL method)
I know that $(a+b)(c+d)=(ac+ad+bc+bd)$. But coming up with the reverse does not make sense to me. If someone can only tell me how to factor a trinomial that would be great.
| Using the discriminant of $ax^3+bx^2+cx+d=8x^3+27$:
$$\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2=0-0+0-0-27(8^2)(27^2) < 0$$
Thus there is only one real root, and that is, as Alex Becker found, $x=-3/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Computing a Laurent series Let $$f(z) = \frac{1}{(2z-1)(z-3)} $$. Compute the Laurent series about the point z = 1 in the annular domain $$ \frac{1}{2} < |z-1| < 2$$
My attempt:
I broke f(z) up into the partial fraction decomposition:
$$ -\frac{2}{5(2z-1)} + \frac{1}{5(z-3)} = -\frac{2}{5}*\frac{1}{(1-\frac{(z+\frac{1}{2})}{2})} +\frac{1}{5}*\frac{1}{1-(z-2)} = $$
$$-\frac{2}{5}\sum_{n=0}^\infty(-1)^{n}\frac{(z+1)^{n}}{2^n}-\frac{1}{5}\sum_{n=0}^\infty(z-2)^n $$
And that was my answer. But I was told I was wrong, and I'm not sure where I went wrong in there. So if someone could point out where I went wrong, it would be greatly appreciated!
| $$\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\left(\frac{1}{z-3}-\frac{1}{z-\frac{1}{2}}\right)=-\frac{1}{10}\frac{1}{1-\frac{z-1}{2}}-\frac{1}{5(z-1)}\frac{1}{\left(1+\frac{1}{2(z-1)}\right)}$$
Now only check that
$$\left|\frac{z-1}{2}\right|<1\,\,,\,\,|2(z-1)|^{-1}<1$$
and you'll be able to use the developments
$$\frac{1}{1-z}=1+z+z^2+...=\sum_{n=0}^\infty z^n\,\,,\,\,\frac{1}{1+z}=1-z+z^2-...=\sum_{n=0}^\infty(-1)^{n}z^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent. I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.
I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor oscillates between -1, 0 and 1.
Here is my attempt to prove that it is divergent:
Suppose that the sequence is convergent. This should lead to a contradiction. If it is convergent, then, for every $\epsilon>0$, there is an $N>0$ such that if $n>N$, then $\left|\dfrac{n}{n+1}\sin\dfrac{n\pi}{2} -L\right|<\epsilon$.
In particular, for $\epsilon = \frac{1}{2}$:
$$-\frac{1}{2}<\frac{n}{n+1}\sin\frac{n\pi}{2} -L<\frac{1}{2} \text{ for every integer }n>N$$
Notice that the sequence of values of $\sin\dfrac{n\pi}{2}$ is $1, 0, -1, 0, 1, 0, -1, \cdots$ for $n=1, 2, 3, 4, 5, 6, 7, \cdots$. If $n = 1, 5, 9 \cdots$, $\sin\frac{n\pi}{2}$ is $1$, so, if we choose a particular $n$ from this list of values, the value of $a_n$ is $\dfrac{n}{n+1}$. In this case, $\sin\dfrac{(n+2)\pi}{2}$ will be $-1$, therefore $a_{n+2} = -\dfrac{n+2}{n+3}$. So, we have that:
$$-\dfrac{1}{2}<\dfrac{n}{n+1}-L<\dfrac{1}{2} \text{ and } -\dfrac{1}{2}<-\dfrac{n+2}{n+3}-L<\dfrac{1}{2}$$
Rearranging the terms:
$$\dfrac{1}{2}>-\dfrac{n}{n+1}+L>-\dfrac{1}{2} \text{ and } \dfrac{1}{2}>\dfrac{n+2}{n+3}+L>-\dfrac{1}{2}$$
$$\dfrac{n}{n+1}+\dfrac{1}{2}>L>\dfrac{n}{n+1}-\dfrac{1}{2} \text{ and } \dfrac{1}{2}-\dfrac{n+2}{n+3}>L>-\dfrac{n+2}{n+3}-\dfrac{1}{2}$$
The first inequality says that $L$ is positive, but the second inequality says that $L$ is negative; if this is correct so far, then I found a contradiction, proving that the sequence is divergent.
My question is: is this correct or is there some inconsistency?
| Claim for you to prove: if a sequence converges then any subsequence converges and to the very same limit as the whole sequence.
Well, what about choosing two very special subsequences here and proving they don't have the same limit? In fact, one of the "obvious" subsequences doesn't even have a limit...
| {
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"url": "https://math.stackexchange.com/questions/173445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of given expression What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$
I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible
| Here is an 'algebra solution':
$\frac {x^2 + x + 1 } {x^2 - x + 1 } = \frac{(x+1)^2-(x+1)+1}{(x+1)^2-3(x+1)+3} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{x^2-x+1} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{(x-\frac{1}{2})^2+\frac{3}{4}}$.
Since the last term is greater than zero when $x\neq -1$, we see that the minimum is
$\frac{1}{3}$.
Of course, this is cheating since I know the answer from J.D.'s solution and this suggests the way to expand the expression.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove $\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}$ How would I solve the following double angle identity.
$$
\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}
$$
So far my work has been.
$$
\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B+\sin A\sin B}
$$
But what would I do to continue.
| What i get is, how to solve the problem?? Is that correct then here u are:
$$\dfrac{\dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y + \sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y} +\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y} {\cos x\cos y}+ \dfrac{\sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\tan x+\tan y}{1+\tan x\tan y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 ?
| Taking square roots of both sides we solve $5^r>2000=25\cdot80$. The right side is approximated from below by $5^2\cdot5^2\cdot3$ so we want $5^{r-4}>3$ or $r=5$, so $n=2r=10$. Check $n=9$ is too small: $5^9<2^9\cdot 3^9<2^9\cdot 3^5\cdot 3^4=512\cdot 243\cdot 81<125,000\cdot 100<4,000,000$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 10
} |
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.
| $2\sin x+\sin3x-\sin5x$
$=2\sin x-2\cos4x\sin x$ applying $\sin2C-\sin2D=2\sin(C-D)\cos(C+D)$ formula
$=2\sin x(1-\cos4x)$
$=4\sin x \cdot \sin^22x$, applying $\cos2A=1-2\sin^2A$
$=4\sin x (2\sin x\cos x)^2$
$=16\sin^3x\cos^2x$
Alternatively, we know $e^{iy}=\cos y+i\sin y$ => $e^{-iy}=\cos y - i\sin y$
So, $\cos y=\frac{e^{iy}+e^{-iy}}{2}$ and $\sin y=\frac{e^{iy} - e^{-iy}}{2i}$
$\sin^3x\cos^2x$
$=(\frac{e^{ix}-e^{-ix}}{2i})^3(\frac{e^{ix} + e^{-ix}}{2})^2$
$=\frac{-1}{32i}(e^{ix}-e^{-ix})^2(e^{ix} + e^{-ix})^2(e^{ix}-e^{-ix})$
$=\frac{-1}{32i}((e^{ix}-e^{-ix})(e^{ix} + e^{-ix}))^2(e^{ix}-e^{-ix})$
$=\frac{-1}{32i}(e^{2ix}-e^{-2ix})^2(e^{ix}-e^{-ix})$
$=\frac{1}{32i}(2(e^{ix}-e^{-ix}) + (e^{3ix}-e^{-3ix}) - (e^{5ix}-e^{-5ix}))$
$=\frac{1}{32i}(2(2i\sin x) + (2i\sin3x) - (2i\sin5x))$
$=\frac{2\sin x+\sin3x-\sin5x}{16}$
Alternatively, $\sin^3x\cos^2x$
$=\frac{1}{8}(4\sin^3x\cdot2\cos^2x)$
$=\frac{1}{8}(3\sin x-\sin3x)(1+\cos2x)$ as $\sin3A=3\sin A-4\sin^3A$ and $\cos2A=2\cos^2A-1$
$=\frac{1}{8}(3\sin x-\sin3x+3\sin x\cos2x-\sin3x\cos2x)$
$=\frac{1}{16}(6\sin x-2\sin3x+6\sin x\cos2x-2\sin3x\cos2x)$
$=\frac{1}{16}(6\sin x-2\sin3x+3(\sin3x-\sin x)-(\sin5x+\sin x))$ (applying $2\sin A\cos B=\sin(A+B)+\sin(A-B))$
$=\frac{1}{16}(2\sin x+\sin3x-\sin5x)$
| {
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"url": "https://math.stackexchange.com/questions/179294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$ I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:
Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality:
$$ (1+x)^n \ge 1+nx$$
Proof:
Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.
Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$.
Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$
Proof of Inductive Step:
$$\begin{align*}
(1+x)^k &\ge 1+kx \\
(1+x)(1+x)^k &\ge (1+x)(1+kx)\\
(1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\
1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\
\Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square
\end{align*}$$
| What you have is perfectly acceptable. The calculations could be organized a little more neatly:
$$\begin{align*}
(1+x)^{k+1}&=(1+x)(1+x)^k\\
&\ge(1+x)(1+kx)\\
&=1+(k+1)x+kx^2\\
&\ge1+(k+1)x\;,
\end{align*}$$
since $kx^2\ge 0$. This completes the induction step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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$\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$. Find the positive integer solutions (x,y). $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$.
Find the positive integer solutions (x,y).
| You can make it $(x-12)(y-48)=12\cdot 48$ so $y-48$ can be $1,3,9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/182882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The probability of bricks arranged randomly. If someone have $4$ red bricks and $8$ blue bricks and arranges them randomly in a circle,
what is the probability that two red bricks are not side by side?
Is the sample space $\frac{12!}{4!\,8!} \,=\, 495$ ?
and the answer $P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}$ ?
| Suppose there $r$ red bricks, and $b$ blue bricks. Counting clock-wise, let
$$
(k_1, k_2, \ldots, k_r)
$$
be the tuples, representing the number of blue bricks between two red bricks. The number of favorable configurations is:
$$
N_F = \sum_{k_1=1}^{b-r+1} \sum_{k_2=1}^{b-r+1} \cdots \sum_{k_r=1}^{b-r+1} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{t-t^{b-r+2}}{1-t} \right)^r = [t^{b-r}] \left( \frac{1-t^{b-r+1}}{1-t} \right)^r = [t^{b-r}] \left(1-t\right)^{-r} = \binom{b-1}{r-1}
$$
and the number of total configurations:
$$
N_T = \sum_{k_1=0}^{b} \sum_{k_2=0}^{b} \cdots \sum_{k_r=0}^{b} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{1-t^{b+1}}{1-t} \right)^r= [t^b] \left(1-t\right)^{-r} = \binom{b+r-1}{b}
$$
Hence the probability:
$$
p =\frac{N_F}{N_T} = \frac{\binom{b-1}{r-1}}{\binom{b+r-1}{b}} = \frac{b! (b-1)!}{(b-r)! (b+r-1)!}
$$
With $b=8$ and $r=4$, we have $N_F = \binom{7}{3} = 35$ and $N_T = \binom{11}{8} = 165$, thus
$$
p = \frac{N_F}{N_T} = \frac{35}{165} = \frac{7}{33}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove Lagrange trigonometric identity I would to prove that
$$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+
\frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$
given that
$$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$
where $z\neq 1$.
I put $z=e^{i\theta}$. I already got in left hand side cos exp in real part, but there is a problem in the right hand side, I can't split imaginary part and real part. Please help me. Thanks in advance.
| $2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2r+1)\theta}{2} - \sin\frac{(2r-1)\theta}{2}$
Putting r=1,2,....,n,
$2\sin\frac{\theta}{2}\cos \theta=\sin\frac{3\theta}{2} - \sin\frac{\theta}{2}$
$2\sin\frac{\theta}{2}\cos 2\theta=\sin\frac{5\theta}{2} - \sin\frac{3\theta}{2}$
...
$2\sin\frac{\theta}{2}\cos n\theta=\sin\frac{(2n+1)\theta}{2} - \sin\frac{(2n-1)\theta}{2}$
Adding we get,
$\sum_{1≤r≤n}2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2n+1)\theta}{2} - \sin\frac{\theta}{2}$
Divide both sides by $2\sin\frac{\theta}{2}$, we shall get,
$\sum_{1≤r≤n}\cos r\theta=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} - \frac{1}{2}$ (Assuming $\sin\frac{\theta}{2}≠0$ or $\theta≠2s\pi$ where s is any integer.)
Or, $1+\sum_{1≤r≤n}\cos r\theta=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} + \frac{1}{2}$ (adding 1 to both sides )
Just observe that for $\sum_{r}\cos(A+2rB)$ where A,B are constants and r is an integer, we need to multiply with $2\sin B$ as
$2\cos(A+2rB)\sin B=sin(A+(2r+1)B) - sin(A+(2r-1)B)$
Putting different ranges of values of r & adding them, we shall get their sums in the compact form.
In the current problem, $A=0, 2B=\theta, 1≤r≤n$
Also as, $2\sin B\sin(A+2rB) = cos(A+(2r-1)B) - cos(A+(2r+1)B)$
This can be used for $\sum_{r}\sin(A+2rB)$.
Also using DonAntonio's approach, we know
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i} \implies e^{ix}-e^{-ix}=2i\sin x$$
So,
$$\begin{align}
\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1} &=\frac{e^{(n+1)i\theta/2}\left(e^{(n+1)i\theta/2}-e^{-(n+1)i\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)} \\[0.5em]
&=e^{ni\theta/2}\frac{2i\sin\frac{(n+1)\theta}{2}}{2i\sin\frac{\theta}{2}} \\[0.5em]
&=\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\left(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}\right)\end{align}$$
Its real part is
$$\begin{align}
\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{n\theta}{2}
&=\frac{2\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{2\sin\frac{\theta}{2}} \\
&=\frac{\sin\frac{(2n+1)\theta}{2}+\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\\
&=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}}+\frac{1}{2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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A little help with algebra (homework) I was brushing up on my algebra a little, and I've seen that I've become rusty with some of the concepts. Indications for solving these would perhaps be better suited rather than full answers, as the goal is to figure it out myself.
*
*Show that for any $a, b \in \mathbb{R}$ (without induction, just using algebra) that $a^3 - b^3 = (a - b)(a^2 + 2ab + b^2)$ and something similar but for $a^3 + b^3$.
*Factorize the following: $(3x + 1)^2 - (x+3)^2$ and other such expressions
And if someone could provide a concise and simple reference for the rules of expansions, factorizing and related topic, I would be most appreciative.
| The simplest way to verify an identity like $a^3-b^3=(a-b)(a^2+2ab+b^2)$ is to multiply out the righthand side and verify that you do indeed get the lefthand side. If you try it in this case, however, you’ll fail:
$$\begin{align*}
(a-b)(a^2+2ab+b^2)&=a(a^2+2ab+b^2)-b(a^2+2ab+b^2)\\
&=\left(a^3+2a^2b+ab^2\right)-\left(a^2b+2ab^2+b^3\right)\\
&=a^3+2a^2b+ab^2-a^2b-2ab^2-b^3\\
&=a^3-b^3+\left(2a^2b-a^2b\right)+\left(ab^2-2ab^2\right)\\
&=a^3-b^3+a^2b-ab^2\;,
\end{align*}$$
and $a^2b-ab^2=ab(a-b)$ certainly isn’t guaranteed to be zero. (It’s zero if and only if either $a=0,b=0$, or $a=b$.) Thus, in general $a^3-b^3\ne(a-b)(a^2+2ab+b^2)$. The correct identity is $$a^3-b^3=(a-b)(a^2+ab+b^2)\;,$$ as you can check by multiplying out the righthand side: this time everything will cancel out except $a^3-b^3$.
The trick to factorizing an expression like $(3x + 1)^2 - (x+3)^2$ is to recognize that it has the form $a^2-b^2$, where $a=3x+1$ and $b=x+3$, and to recall the basic factorization formula $$a^2-b^2=(a-b)(a+b)\;.$$
A few of the standard basic formulas can be found here, together with a link to a practice page. Here is the start of a set of three pages on the topic, with examples. Googling on factoring formulas, with or without quotes, will turn up many more such resources.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$. $1-\frac{1}{n} \lt x \le 3 + \frac{1}{n}$, $\forall n \in \mathbb{N}$. what is the range of $x$.
NOTE: $1-\frac{1}{n} \lt 3 + \frac{1}{n}$
| when $n=1$,
$1-\frac{1}{n}=0$, $3+\frac{1}{n}=4$
when $n=1$, we have, $0 \lt x_{1} \le 4$
when $n=2$, we have, $\frac{1}{2} \lt x_{2} \le 3+\frac{1}{2}$
when $n \to \infty, \frac{1}{n} \to 0$, and then we have, $1 \lt x_{\infty} \le 3$
we can now conclude that, $0 \lt x \le 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\tan^{-1}[(\cos x - \sin x)/(\cos x + \sin x)]$
Write the following functions in simplest form:
$$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right), \quad 0<x<\pi$$
Please help me to solve this problem. I have been trying to solve this from last 3 hours. I can solve simple inverse trigonometric functions
| Let's multiply the numerator and denominator by $\ \cos\bigl(\frac {\pi}4\bigr)=\sin\bigl(\frac {\pi}4\bigr)\ $ :
$$\tan^{-1}\left(\frac{\sin\bigl(\frac {\pi}4\bigr)\cos(x)-\cos\bigl(\frac {\pi}4\bigr)\sin(x)}{\cos\bigl(\frac {\pi}4\bigr)\cos(x)+\sin\bigl(\frac {\pi}4\bigr)\sin(x)}\right),\quad 0<x<\pi$$
$$=\tan^{-1}\left(\frac{\sin\left(\frac {\pi}4-x\right)}{\cos\left(\frac {\pi}4-x\right)}\right),\quad 0<x<\pi$$
$$=\begin{cases}
&\frac {\pi}4-x&\quad\text{if}\ \ 0 < x < \frac{3\pi}4\\
&\text{not defined}&\quad\text{if}\quad x = \frac{3\pi}4\\
&\frac {5\pi}4-x&\quad\text{if}\ \ \frac{3\pi}4< x < \pi\\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to approximate the value of $ \arctan(x)$ for $x> 1$ using Maclaurin's series? The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$
$$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right) $$
Does it mean that I cannot approximate $\arctan(2)$ using this series? Also I'm getting radius of convergence $|x| < 1$ using ratio test. How do I get $|x| \leqslant 1$?
| Use, for $x>0$
$$
\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2}
$$
For $x<0$ use the parity.
Added: Use
$$
\tan\left(\frac{\pi}{2} - \phi\right) = \frac{\sin\left(\frac{\pi}{2}-\phi\right)}{\cos\left(\frac{\pi}{2}-\phi\right) } = \frac{\cos(\phi)}{\sin(\phi)} = \frac{1}{\tan(\phi)}
$$
Thus, for $x = \tan(\phi)$:
$$
\tan\left(\frac{\pi}{2} - \phi\right) = \frac{1}{x}, \quad \frac{\pi}{2} - \phi = \arctan\left(\frac{1}{x}\right), \quad
\frac{\pi}{2} - \arctan(x) = \arctan\left(\frac{1}{x}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
| Hint $\rm\quad S_n =\, \sum c_k n^k\ \Rightarrow\ S_{2n} + 4\, S_n\, =\: \sum\ (2^k\!+\!4)\ c_k\ =\ 4\, n^3+4\,n^2 + n,\:$ therefore
$$\rm S_n\, =\ \frac{4}{2^{\color{#C00}3}\!+\!4} n^\color{#C00}3 +\frac{4}{2^\color{#0A0}2\!+\!4}n^{\color{#0A0}2} + \frac{1}{2^\color{brown}1\!+\!4} n^{\color{brown}1}\ =\ \frac{n^3}3+\frac{n^3}2 + \frac{n}6\ =\ \frac{n(n+1)(2n+1)}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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prove by induction : $\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}=n$ How to prove the following question by using induction?
$${\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\frac{(1-a^n)(1-a^{n-1})(1-a^{n-2})}{1-a^3}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}=n}$$
and I am interested in solutions without using Induction. Can anyone show me the steps?
Thanks for your help.
| I applied Andre Nicolas's method. Please see steps below.
$$H_a(n)=\frac{(1-a^n)}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\frac{(1-a^n)(1-a^{n-1})(1-a^{n-2})}{1-a^3}+...+\frac{(1-a^n)(1-a^{n-1})...(1-a)}{1-a^n}$$
$$H_a(n+1)=\frac{(1-a^{n+1})}{1-a}+\frac{(1-a^{n+1})(1-a^{n})}{1-a^2}+\frac{(1-a^{n+1})(1-a^{n})(1-a^{n-1})}{1-a^3}+...+\frac{(1-a^{n+1})(1-a^{n})...(1-a)}{1-a^{n+1}}$$
$$H_a(n+1)-H_a(n)=a^n+a^{n-1}(1-a^n)+a^{n-2}(1-a^n)(1-a^{n-1})+...+\frac{(1-a^{n+1})(1-a^{n})...(1-a)}{1-a^{n+1}}$$
$$H_a(n)-H_a(n-1)=a^{n-1}+a^{n-2}(1-a^{n-1})+a^{n-3}(1-a^{n-1})(1-a^{n-2})+...+\frac{(1-a^{n})(1-a^{n-1})...(1-a)}{1-a^{n}}$$
$$(1-a^n)(H_a(n)-H_a(n-1))+a^n=a^n+a^{n-1}(1-a^n)+a^{n-2}(1-a^n)(1-a^{n-1})+a^{n-3}(1-a^n)(1-a^{n-1})(1-a^{n-2})+...+(1-a^n)\frac{(1-a^{n})(1-a^{n-1})...(1-a)}{1-a^{n}}$$
$$(1-a^n)(H_a(n)-H_a(n-1))+a^n=H_a(n+1)-H_a(n) \tag1$$
If we select $H_a(n)=n+c$ (where c is constant), $H_a(n)=n+c$ satisfies Equation (1).
But we know that $H_a(1)=\frac{(1-a^1)}{1-a}=1$, so $c$ should be zero
Note: I could not find a method to solve Equation (1) and to find the solution directly $H_a(n)=n+c$. If someone knows how to prove that I will be happy to see it.
| {
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"url": "https://math.stackexchange.com/questions/188918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $ \int_0^\infty\left(1-x\sin\frac 1 x\right)dx = \frac\pi 4 $ How to show that
$$
\int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=\frac{\pi}{4}
$$
?
| Use
$$
\int \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x = x - \int \sin\left(\frac{1}{x}\right) \mathrm{d} \frac{x^2}{2} = x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{1}{2} \int \cos\left(\frac{1}{x}\right) \mathrm{d}x
$$
Integrating by parts again $\int \cos\left(\frac{1}{x}\right) \mathrm{d}x = x \cos\left(\frac{1}{x}\right) - \int \sin\left(\frac{1}{x}\right) \frac{\mathrm{d}x}{x} $:
$$
\int \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x = x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{x}{2} \cos\left(\frac{1}{x}\right) + \frac{1}{2} \int \sin\left(\frac{1}{x}\right) \frac{\mathrm{d}x}{x}
$$
Thus:
$$ \begin{eqnarray}
\int_0^\infty \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x &=& \left[x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{x}{2} \cos\left(\frac{1}{x}\right)\right]_{0}^{\infty} + \frac{1}{2} \int_0^\infty\sin\left(\frac{1}{x}\right) \frac{\mathrm{d}x}{x} =
\\
&=& 0 + \frac{1}{2} \int_0^\infty \frac{\sin{u}}{u} \mathrm{d} u = \frac{\pi}{4}
\end{eqnarray}
$$
where the last integral is the Dirichlet integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $|x|<1$.
| Here is one way. One can write the product as the telescoping product
$$
\frac{1-x^2}{1-x}\times \frac{1-x^4}{1-x^2}\times \frac{1-x^8}{1-x^4}\times \dotsb\frac{1-x^{2^{n+1}}}{1-x^{2^n}}\times \dotsb.
$$
so that the $n$th partial product (indexed from $0$) equals
$$
\frac{1-x^{2^{n+1}}}{1-x}=1+x+\dotsb+x^{2^{n+1}-1}
$$
which converges in the topology of formal power series to $(1-x)^{-1}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to prove Cauchy-Schwarz Inequality in $R^3$? I am having trouble proving this inequality in $R^3$. It makes sense in $R^2$ for the most part. Can anyone at least give me a starting point to try. I am lost on this thanks in advance.
| You know that, for any $x,y$, we have that
$$(x-y)^2\geq 0$$
Thus
$$y^2+x^2\geq 2xy$$
Cauchy-Schwarz states that
$$x_1y_1+x_2y_2+x_3y_3\leq \sqrt{x_1^2+x_2^2+x_3^3}\sqrt{y_1^2+y_2^2+y_3^3}$$
Now, for each $i=1,2,3$, set
$$x=\frac{x_i}{\sqrt{x_1^2+x_2^2+x_3^2}}$$
$$y=\frac{y_i}{\sqrt{y_1^2+y_2^2+y_3^2}}$$
We get
$$\frac{y_1^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_1^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{y_1}{\sqrt{y_1^2+y_2^2+y_3^2}}$$
$$\frac{y_2^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_2^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_2}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{y_2}{\sqrt{y_1^2+y_2^2+y_3^2}}$$
$$\frac{y_3^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_3^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_3}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{y_3}{\sqrt{y_1^2+y_2^2+y_3^2}}$$
Summing all these up, we get
$$\frac{y_1^2+y_2^2+y_3^2}{{y_1^2+y_2^2+y_3^2}}+\frac{x_1^2+x_2^2+x_3^2}{{x_1^2+x_2^2+x_3^2}}\geq 2\frac{x_1y_1+x_2y_2+x_3y_3}{\sqrt{y_1^2+y_2^2+y_3^2}\sqrt{x_1^2+x_2^2+x_3^2}}$$
$$\sqrt{y_1^2+y_2^2+y_3^2}\sqrt{x_1^2+x_2^2+x_3^2}\geq {x_1y_1+x_2y_2+x_3y_3}$$
This works for $\mathbb R^n$. We sum up through $i=1,\dots,n$ and set
$$y=\frac{y_i}{\sqrt{\sum y_i^2}}$$
$$x=\frac{x_i}{\sqrt{\sum x_i^2}}$$
Note this stems from the most fundamental inequality $x^2\geq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality. $a^7b^2+b^7c^2+c^7a^2 \leq 3 $
Let $a,b,c$ be positive real numbers such that $a^6+b^6+c^6=3$. Prove that
$$a^7b^2+b^7c^2+c^7a^2 \leq 3 .$$
| Frank Science has showed that the inequality is equivalent to $$ 3(\sum a^7b^2)^2 \leqslant (\sum a^6)^3\tag{0}\\ $$
Use AM-GM for $a^{12}b^{6}$, $a^{12}b^{6}$ and $a^{18}$ we have the following:
$ 3a^{14}b^4 \leqslant 2a^{12}b^6+a^{18}\\ $
Combine with 2 other similar inequalities we have
$$3(\sum a^{14}b^4) \leqslant \sum a^{18}+2(\sum a^{12}b^6)\tag{1}$$
Use AM-GM again this time we have the following:
$ 6a^7b^9c^2 \leqslant 2(abc)^6+a^{12}b^6+3a^6b^{12}\\ $
Combine with 2 other similar inequalities we have
$$6(\sum a^7b^9c^2) \leqslant 6(abc)^6+\sum a^{12}b^6+3(\sum a^6b^{12}) \tag{2}$$
Summing $(1)$ and $(2)$ we have $(0)$
| {
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Polynomial factors Why must $x^2 + x + 1$ be a factor of $x^5+x^4+x^3+x^2+x+1$?
I know that when we divide $x^5+x^4+x^3+x^2+x+1$ by $x^3+1$ we get $x^2 + x + 1$, but is there an argument/theorem or anything that could tell that $x^2+x+1$ must divide $x^2 + x + 1$?
| $(x^2+x+1)(x-1)=x^3-1$ and $(x^5+x^4+x^3+x^2+x+1)(x-1)=x^6-1$. Then note that $x^6-1=(x^3-1)(x^3+1)$. Thus in effect it is the third binmoial formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric identity and roots of a polynomial. Prove that
$$(\operatorname{cosec} A–\sin A) (\sec A–\cos A) = \frac {1}{\tan A + \cot A} $$
Also help me with this question please
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2–2x–15$ then find a quadratic polynomial whose series [roots?] are $2\alpha$ and $2\beta$
| Have:
$\csc A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Now:
$(\csc A-\sin a)(\sec A-\cos A)=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)=(\frac{1-\sin^2{A}}{\sin a})(\frac{1-\cos^2{a}}{\cos A})=\frac{\cos^2 A}{\sin A}\cdot\frac{\sin^2 A}{\cos A}=\cos A\sin A=\frac{\cos A\sin A}{1}=\frac{\cos A\sin A}{\cos^2 A+\sin^2 A}=\frac{\frac{\cos A\sin A}{\cos A\sin A}}{\frac{\cos^2 A+\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos^2 A}{\cos A\sin A}+\frac{\sin^2 A}{\cos A\sin A}}=\frac{1}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}=\frac{1}{\cot A+\tan A}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/199814",
"timestamp": "2023-03-29T00:00:00",
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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...
| Note that the equation $\sin x + \sqrt 3 \cos x = 1$ is not equivalent to the equation $(\sin x)^2=(1−\sqrt{3} \cos x)^2$.
| {
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How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
| Substition
$x=4\sin y$ $\Rightarrow$ $dx=4\cos y dy$
$\sin y=\frac{x}{4}$$\Rightarrow$ $y=\arcsin\frac{x}{4}$
$\int\frac{1}{(16-x^2)^2}dx=\int\frac{1}{(4^2-x^2)^2}dx$=$\int\frac{4\cos y dy}{(16-(4\sin y)^2)^2}$=$\int\frac{4\cos y dy}{(16-16\sin^2y)^2}$ = $\int\frac{4\cos y dy}{16^2(1-\sin^2 y)^2} $=$\frac{1}{64} \int\frac{\cos y dy}{(\cos^2 y)^2}$=$\frac{1}{64}\int\frac{dy}{\cos^3 y}$=$\frac{1}{64}\int{\sec^3 y dy}$=$\frac{1}{64}\cdot\frac{\sin y}{2\cos^2 y} +\frac{1}{2}\int\frac{dy}{\cos y}$=$\frac{1}{32}\cdot\frac{\sin y}{\cos^2 y} +\frac{1}{2}\ln|\tan(\frac{y}{2}+\frac{\pi}{4})|$=$\frac{1}{32}\cdot\frac{\sin y}{1-\sin^2 y} +\frac{1}{2}\ln|\tan(\frac{y}{2}+\frac{\pi}{4})|$=$\frac{1}{32}\cdot\frac{\frac{x}{4}}{1-(\frac{x}{4})^2 } +\frac{1}{2}\ln|\tan(\frac{\arcsin\frac{x}{4}}{2}+\frac{\pi}{4})|$=$\frac{x}{8(16-x^2)} +\frac{1}{2}\ln|\tan(\frac{\arcsin\frac{x}{4}}{2}+\frac{\pi}{4})|+C$
We have implement the formula:
1) $\int\frac{dx}{\cos^n x}$=$\frac{\sin x}{(n-1)\cos^{n-1} x} +\frac{n-2}{n-1}\int\frac{dx}{\cos^{n-2}x}$
and
2) $\int\frac{dx}{\cos x}$=$\ln|\tan(\frac{x}{2}+\frac{\pi}{4})|$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the span of two vectors is T-invariant Show that the span of \begin{bmatrix}1\\1\\0\\1\end{bmatrix} and \begin{bmatrix}1\\0\\2\\0\end{bmatrix} is a $T$-invariant subspace of the linear map given by
\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}
I tried to take some general vector in the span and multiply it by the matrix in the hope of getting something that was clearly a linear combination of my two original spanning vectors, but this did not work, that is, the vector was clearly not in the span.
So how am I meant to show $T$-invariance?
Note: Apologies for the formatting, my first time
| As the map given by $x\mapsto Tx$ is linear, it suffices to prove that the image of the spanning vectors is again in the span of the two vectors. This is done by the following computation:
\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\0\\1\end{bmatrix}&=\begin{bmatrix}4\cdot 1+(-2)\cdot 1+(-1)\cdot 0+(-1)\cdot 1\\3\cdot 1+(-1)\cdot 1+(-1)\cdot 0+(-1)\cdot 1\\(-2)\cdot 1+2\cdot 1+2\cdot 0+0\cdot 1\\1\cdot 1+(-1)\cdot 1+0\cdot 0+1\cdot 1\end{bmatrix}=\begin{bmatrix}1\\1\\0\\1\end{bmatrix}\\
&=1\cdot \begin{bmatrix}1\\1\\0\\1\end{bmatrix}+0\cdot\begin{bmatrix}1\\0\\2\\0\end{bmatrix}\end{align*}
\begin{align*}\begin{bmatrix}4&-2&-1&-1\\ 3&-1&-1&-1\\-2&2&2&0\\1&-1&0&1\end{bmatrix}\begin{bmatrix}1\\0\\2\\0\end{bmatrix}&=\begin{bmatrix}4\cdot 1+(-2)\cdot 0+(-1)\cdot 2+(-1)\cdot 0\\3\cdot 1+(-1)\cdot 0+(-1)\cdot 2+(-1)\cdot 0\\(-2)\cdot 1+2\cdot 0+2\cdot 2+0\cdot 0\\1\cdot 1+(-1)\cdot 0+0\cdot 2+1\cdot 0\end{bmatrix}=\begin{bmatrix}2\\1\\2\\1\end{bmatrix}\\
&=1\cdot \begin{bmatrix}1\\1\\0\\1\end{bmatrix}+1\cdot \begin{bmatrix}1\\0\\2\\0\end{bmatrix}
\end{align*}
Another way to see this (but unnecessary complicated is to see that the spanned set is the kernel of $(T-I)^2$ and as such is $T$-invariant. (This is suggested by the theory of Jordan normal form.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$ $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$$
Attempt
This is indeterminate of the form $\frac{0}{0}$. Applying L'Hopital's rule twice results in,
$$\lim_{x \to 0} \frac{e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x}{6x}.$$
Wolfram Alpha says the two sided limit does not exist.
Question
Can I deduce the result by splitting it up into,
$$\frac{1}{6}\lim_{x \to 0} \left(e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x \right)\left(\frac{1}{x}\right).$$
And saying that the limit does not exist because $\frac{1}{x}$ does not have a two sided limit?
| $\lim_{x\to 0}\frac{e^x-1}{x}=1\implies e^x\approx 1+x $when $x\to 0$,thus, $e^{\sin^2x}\approx 1+\sin^2x$ as $x\to 0$
Hence, $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}=\lim_{x \to 0} \frac{1+ \sin^2 x - \cos x}{x^3}$$
Now you can easily apply L'Hopital's rule to conclude that limit doesn't exist.
| {
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Prove that $\lim_{x \rightarrow 0} \sqrt{9-x} + 3 = 6$ Prove that $\lim_{x \rightarrow 0} \sqrt{9-x} + 3 = 6$
So far my difficulty is trying to find a $\delta$ that will allow for this function to be less than $\epsilon$. I keep getting that $|\sqrt{9-x}+3-6|=|\sqrt{9-x}-3| < \epsilon$, but from here I do not know how to deal with finding $\delta$. Thanks in advance and any help is appreciated.
| Firstly, observe that for any $ x \in (- \infty,9] $, we have
\begin{equation}
|\sqrt{9 - x} - 3||\sqrt{9 - x} + 3| = |(9 - x) - 9| = |x|.
\end{equation}
Fix an $ \epsilon > 0 $. Then choosing any $ x $ that satisfies $ |x| < \min(9,3 \epsilon) $, we obtain
\begin{align}
|\sqrt{9 - x} - 3| &= \frac{|x|}{\sqrt{9 - x} + 3} \\
&= |x| \cdot \frac{1}{\sqrt{9 - x} + 3} \\
&< 3 \epsilon \cdot \frac{1}{3} \\
&= \epsilon.
\end{align}
Hence, we can set $ \delta := \min(9,3 \epsilon) $.
| {
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general equation of a tangent line to a hyperbola Suppose that there is a hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$.
I would like to figure out an equation that describes tangent line to this hyperbola.
How would I be able to do this using calculus? My calculus trials are bring me some gibberish answers.
I did use the calculus below, but what I want is how to derive the general form equation of $y=mx \pm \sqrt{a^2m^2-b^2}$ where $m$ refers to the slope of a tangent line.
| (1a)
Let $y=mx+c$ be a tangent of $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
Let $(h,k)$ be the point of intersection.
So, $k=mh+c, b^2h^2-a^2k^2=a^2b^2\implies b^2h^2-a^2(mh+c)^2=a^2b^2$
or,$h^2(b^2-a^2m^2)- 2a^2mch -a^2b^2-a^2c^2=0 $
This is a quadratic equation in $h,$ for tangency, the roots need to be same, to make the two points of intersection coincident.
$\implies (- 2a^2mc)^2=4\cdot (b^2-a^2m^2)(-a^2b^2-a^2c^2) $
$a^2m^2c^2=a^2m^2b^2+a^2m^2c^2-b^4-b^2c^2$ cancelling out $a^2$ as $a\ne0$
$0=a^2m^2b^2-b^4-b^2c^2$
$0=a^2m^2-b^2-c^2$ cancelling out $b^2$ as $b\ne0$
$\implies c^2=a^2m^2-b^2\implies c=\pm\sqrt{a^2m^2-b^2}$
(1b)
Alternatively, we can take the central conic to be $Ax^2+By^2=1$
Applying the same method we get, $ABc^2=A+m^2B$
Here $A=\frac 1{a^2},B=-\frac 1{b^2}\implies c^2=a^2m^2-b^2$
(2a)
Using calculus as André Nicolas has already done,
we find the equation of the tangent to be $$\frac {xx_1}{a^2}-\frac{yy_1}{b^2}-1=0$$
Comparing with $mx-y+c=0$ we get, $$\frac{x_1}{ma^2}=\frac{y_1}{b^2}=\frac{-1} c$$
Now $(x_1,y_1)$ lies on the given curve, so $\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$, eliminating $x_1,y_1$, we shall get the desired result.
(2b) we know , the parametric equation of the given curve is $x=a\sec t,y=b\tan t$
So, the equation of the tangent becomes $\frac{x\sec t}a-\frac{y\tan t}b-1=0$
Comparing with $mx-y+c=0$ we get, $$\frac{\sec t}{ma}=\frac{\tan t}b=\frac {-1}c$$
So, $\sec t=-\frac{ma}c,\tan t=-\frac b c$
Now use eliminate $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/214977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
| First, an identity. For $x\ne0$,
$$
\begin{align}
\frac{2\sin(x)-\sin(2x)}{x^3}
&=\frac{2\sin(x)-2\sin(x)\cos(x)}{x^3}\tag{1a}\\
&=2\frac{\sin(x)}x\frac{1-\cos(x)}{x^2}\tag{1b}\\
&=\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\sin(2x)=2\sin(x)\cos(x)$
$\text{(1b)}$: algebra
$\text{(1c)}$: $1-\cos(x)=\frac{\sin^2(x)}{1+\cos(x)}$
In this answer, it is shown geometrically that $\lim\limits_{x\to0}\frac{\sin(x)}x=1$. Thus, given $\epsilon\gt0$, we can choose $\delta\gt0$ so that if $0\lt|x|\le\delta$, then
$$
\frac2{1+\cos(x)}\frac{\sin^3(x)}{x^3}=[1-\epsilon,1+\epsilon]_\#\tag2
$$
where $[a,b]_\#$ is a real number in the interval $[a,b]$.
For $0\lt|x|\le\delta$, apply $(1)$ and $(2)$ to $x/2^{k+1}$ and divide by $2^{2k+3}$:
$$
\frac{2^{k+1}\sin\left(x/2^{k+1}\right)-2^k\sin\left(x/2^k\right)}{x^3}=\frac{[1-\epsilon,1+\epsilon]_\#}{2^{2k+3}}\tag3
$$
$\lim\limits_{x\to0}\frac{\sin(x)}x=1$ implies $\lim\limits_{k\to\infty}2^k\sin\left(x/2^k\right)=x$. Thus, if we sum $(3)$ in $k$, we get that for $0\lt|x|\le\delta$,
$$
\frac{x-\sin(x)}{x^3}=\frac{[1-\epsilon,1+\epsilon]_\#}6\tag4
$$
Therefore, the Squeeze Theorem yields
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/217081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
How to verify this trigonometric identity? Please help,
$9\cos^2 B − 9\sin^2 B = 18\cos^2 B − 9$
Thanks
| \begin{align*}
\text{LHS} &= 9\cos^2 B − 9\sin^2 B \\
&= 9(\cos^2 B-\sin^2 B) \\
&= 9(\cos^2 B - (1-\cos^2 B)) \\
&= 9(2\cos^2 B- 1) \\
&= 18\cos B - 9 \\
&= \text{RHS}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{n\to\infty} (1+\frac{1}{2}+...+\frac{1}{n})\frac{1}{n}$
Find the following limit:
$$\lim_{n\to\infty} \left(1+\frac{1}{2}+...+\frac{1}{n}\right)\frac{1}{n}$$
My intuition says that this goes to zero, because $1/n$ goes much faster to zero than the harmonic series go to infinity, but how can I prove this?
| Here is a completely elementary
and self-contained proof
(subject to accepting that
$\frac{k}{2^k} \to 0$
as $k \to \infty$).
Let
$a_n
= \frac{1}{n}\sum_{k=1}^n \frac1{k}
$.
Then
$na_n
= \sum_{k=1}^n \frac1{k}
$.
Therefore
$\frac1{n+1}
=(n+1)a_{n+1}-na_n
=n(a_{n+1}-a_n)+a_{n+1}
$
or
$a_{n+1}-a_n
=\frac1{n}(\frac1{n+1}-a_{n+1})
$.
Since,
for $n > 1$,
$a_n
\ge \frac1{n}(1+\frac12)
= \frac{3/2}{n}
$,
$a_{n+1}-a_n
<-\frac{1}{2n(n+1)}
$
so
$a_n$
is a decreasing sequence.
We also have
$\begin{array}\\
(2n)a_{2n}-na_n
&=\sum_{k=1}^{2n} \frac1{k}-\sum_{k=1}^n \frac1{k}\\
&=\sum_{k=n+1}^{2n} \frac1{k}\\
\text{so}\\
(2n)a_{2n}-na_n
&< 1\\
\text{and}\\
(2n)a_{2n}-na_n
&> \frac12\\
\end{array}
$
Therefore
$a_{2n}
< \frac1{2n}(1+na_n)
= \frac1{2n}+\frac12 a_n
$
or
$\begin{array}\\
a_n
&>2a_{2n}-\frac1{n}\\
&>2(2a_{4n}-\frac1{2n})-\frac1{n}\\
&=4a_{4n}-\frac1{n}-\frac1{n}\\
&=4a_{4n}-\frac{2}{n}\\
&=4(2a_{8n}-\frac1{4n})-\frac{2}{n}\\
&=8a_{8n}-\frac1{n}-\frac{2}{n}\\
&=8a_{8n}-\frac{3}{n}\\
\end{array}
$
By induction
we can show that
$a_n
> 2^ka_{2^kn}-\frac{k}{n}
$
or
$a_{2^kn}
<2^{-k}(a_n+\frac{k}{n})
$.
Since both
$\frac{k}{2^k}
\to 0
$
and
$\frac{a_n}{2^k}
\to 0
$
as
$k \to \infty$,
$a_{2^kn}
\to 0
$
as
$k \to \infty$.
Since
$a_n$
is decreasing and positive,
$a_n \to 0$
as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/221114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Frobenius Method to solve $x(1 - x)y'' - 3xy' - y = 0$ So, Im trying to self-learn method of frobenius, and I would like to ask if someone can explain to me how can we solve the following DE about $ x = 0$ using this method.
$$ x(1 - x)y'' - 3xy' - y = 0 $$
| The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.
Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$. Then we get that $$y'(x) = \sum_{n=0}^{\infty} na_n x^{n-1}$$
$$3xy'(x) = \sum_{n=0}^{\infty} 3na_n x^{n}$$
$$y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}$$
$$xy''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-1} = \sum_{n=0}^{\infty} n(n+1)a_{n+1} x^{n}$$
$$x^2y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n}$$
The ODE is $$xy'' - x^2 y'' -3xy' - y = 0$$
Plugging in the appropriate series expansions, we get that
$$\sum_{n=0}^{\infty} \left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_n\right)x^n = 0$$
Hence, we get that
$$n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n \implies a_{n+1} = \dfrac{n+1}{n}a_n$$
First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.
Hence, the solution is given by
$$y(x) = a_1 \left(x+2x^2 + 3x^3 + \cdots\right)$$
This power series is valid only within $\vert x \vert <1$. In this region, we can simplify the power series to get
\begin{align}
y(x) & = a_1 x \left(1 + 2x + 3x^2 + \cdots \right)\\
& = a_1 x \dfrac{d}{dx} \left(x + x^2 + x^3 + \cdots \right)\\
& = a_1 x \dfrac{d}{dx} \left(\dfrac{x}{1-x}\right)\\
& = a_1 \dfrac{x}{(1-x)^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
primitive residue classes modulo 32 $\mathbb Z_{32}^*$ is the primitive residue classes modulo 32. How is it possible to show that $\mathbb Z_{32}^*$ is generated by 5 and -1, without showing it for every element of $\mathbb Z_{32}^*$=$\{1,3,5,7,9,11,...,31\}$
| Let $n > 1$ be an integer.
We consider the group $G = (\mathbb{Z}/2^n\mathbb{Z})^*$.
Clearly $|G| = 2^{n-1}$.
Let $e \ge 2$ be an integer.
Let $k$ be an odd integer.
$(1 + 2^e k)^2 = 1 + 2^{e+1}k + 2^{2e} k^2 = 1 + 2^{e+1}(k + 2^{e-1}k^2)$.
Note that $k + 2^{e-1}k^2$ is odd.
Hence, by induction on $m \ge 1$, $(1 + 2^e k)^{2^m} = 1 + 2^{e+m}k'$ for some odd integer $k'$.
Since $5 = 1 + 2^2$, $5^{2^m} = 1 + 2^{m + 2}k'$, where $k'$ is odd.
Hence the order of $5$ in $G$ is $2^{n-2}$.
Let $H$ be the subgroup of $G$ generated by $5$.
Suppose $-1 \in H$.
There exists integer $m$ such that $-1 \equiv 5^m$ (mod $2^n$).
Since $5 \equiv 1$ mod $4$, $-1 \equiv 1$ (mod $4$).
This is a contradiction.
Hence $G$ is generated by $-1$ and $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with Trigonometry homework - prove an identity I need to prove the following identity:
$\sin^2 2\alpha-\sin^2 \alpha = \sin 3\alpha \sin \alpha$
What I have tried, is to work on each side of the identity. I have started with the left side:
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\sin 2\alpha+\sin \alpha)\\
&=(2\sin\alpha \cos\alpha-\sin \alpha)(2\sin \alpha \cos \alpha+\sin \alpha)\\
&=\sin\alpha(2\cos\alpha-1)\sin \alpha(2\cos \alpha+1)\\
&=\sin^2 \alpha(2\cos^2 \alpha-1)\\
&=\sin^2\alpha \cos 2\alpha
\end{align}
Then, I moved on to the right side of the identity:
\begin{align}
\sin 3\alpha \sin \alpha&=\sin(2\alpha+\alpha) \sin \alpha\\
&=\sin \alpha(\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha)\\
&=\sin \alpha[2 \sin \alpha \cos^2 \alpha+\sin \alpha(\cos^2 \alpha-\sin ^2 \alpha)]\\
&=\sin^2 \alpha(3\cos^2\alpha-\sin^2\alpha)\\
&=\sin^2\alpha(4\cos^2\alpha+1)
\end{align}
I'm not sure how to continue. Any tips?
| There is a funny identity that goes as $$\sin(x+y) \sin(x-y) = \sin^2(x) - \sin^2(y)$$
Take $x = 2\alpha$ and $y = \alpha$ to get what you want.
Proof:
\begin{align}
\sin(x+y) \sin(x-y) & = \left( \sin(x) \cos(y) + \sin(y) \cos(x) \right) \left( \sin(x) \cos(y) - \sin(y) \cos(x) \right)\\
& = \sin^2(x) \cos^2(y) - \sin^2(y) \cos^2(x)\\
& = \sin^2(x) (1-\sin^2(y)) - \sin^2(y) (1-\sin^2(x))\\
& = \sin^2(x) - \sin^2(x) \sin^2(y) - \sin^2(y) + \sin^2(y) \sin^2(x)\\
& = \sin^2(x) - \sin^2(y)
\end{align}
EDIT:
As for the mistakes in your answer, I have highlighted the corrections in red.
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\sin 2\alpha+\sin \alpha)\\
&=(2\sin\alpha \cos\alpha-\sin \alpha)(2\sin \alpha \cos \alpha+\sin \alpha)\\
&=\sin\alpha(2\cos\alpha-1)\sin \alpha(2\cos \alpha+1)\\
&=\sin^2 \alpha( \color{red}{4\cos^2 \alpha-1})
\end{align}
\begin{align}
\sin 3\alpha \sin \alpha&=\sin(2\alpha+\alpha) \sin \alpha\\
&=\sin \alpha(\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha)\\
&=\sin \alpha[2 \sin \alpha \cos^2 \alpha+\sin \alpha(\cos^2 \alpha-\sin ^2 \alpha)]\\
&=\sin^2 \alpha(3\cos^2\alpha-\sin^2\alpha)\\
&=\sin^2\alpha(\color{red}{4\cos^2\alpha-1})
\end{align}
EDIT
For more funny identities: Funny identities
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Lost in second derivative simplification My book has the following problem:
Find the second derivative of y for $x^2 + xy + y^2 = 3$
I follow them this far:
$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ x + 2y }$
But then how do they get here?
$y'' = -\dfrac{ 18 }{ (x + 2y)^3}$
Updated
Assuming the $x^2 + xy + y^2$ is substituted in, then shouldn't it read
$y'' = -\dfrac{ 18 } { x + 2y } = -(\dfrac{ 3 }{ (x + 2y)})^3$
or something like that? I don't see how the bottom gets cubed but the top stays 18
| I must have a different method than your book since I never reached the same step as you, but here is my derivation if it helps:
$x^2 + xy + y^2 = 3$
Find $\dfrac{dy}{dx}$:
$2x + [(x)(y') + (y)(1)] + 2y(y') = 0$
$2x + xy' + y + 2yy' = 0$
$(2x + y) + y'(x + 2y) = 0$
Solve for $y'$:
$y' = -\dfrac{ 2x + y }{ x + 2y }$
Find $y''$:
$y'' = -\dfrac{ (x + 2y)(2 + y') - (2x + y)(1 + 2y') }{ (x + 2y)^2 }$
$y'' = -\dfrac{ (2x + xy' + 4y + 2yy') - (2x + 4xy' + y + 2yy') }{ (x + 2y)^2 }$
$y'' = -\dfrac{ 2x + xy' + 4y + 2yy' - 2x - 4xy' - y - 2yy' }{ (x + 2y)^2 }$
$y'' = -\dfrac{ -3xy' + 3y }{ (x + 2y)^2 }$
Substitute for $y'$:
$y'' = -\dfrac{ -3x(-\dfrac{ 2x + y }{ x + 2y }) + 3y }{ (x + 2y)^2 }$
$y'' = -\dfrac{ \dfrac{ 6x^2 + 3xy }{ x + 2y } + 3y }{ (x + 2y)^2 }$
$y'' = -\dfrac{ \dfrac{ 6x^2 + 3xy + 3y(x + 2y) }{ x + 2y } }{ (x + 2y)^2 }$
$y'' = -(\dfrac{ 6x^2 + 3xy + 3xy + 6y^2 }{ x + 2y })(\dfrac{ 1 }{ (x + 2y)^2 })$
$y'' = -\dfrac{ 6x^2 + 6xy + 6y^2 }{ (x + 2y)^3 }$
$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ (x + 2y)^3 }$
As other users noted, replace $x^2 + xy + y^2 = 3$ from the original equation:
$y'' = -\dfrac{ 6(3) }{ (x + 2y)^3 }$
$y'' = -\dfrac{ 18 }{ (x + 2y)^3 }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/227269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Binomial sum of a sequence I have the following sequence:
$$
\sum\limits_{k=1}^n k\binom{n-1}{k-1}
$$
What is the sum of this sequence.
| $\sum_{k=1}^n \binom{n-1}{k-1}\cdot k$
$=\sum_{k=1}^n \binom{n-1}{k-1}+\sum_{k=1}^n \binom{n-1}{k-1}\cdot(k-1)$
Now, $ \binom{n-1}{k-1}\cdot(k-1)=\frac{(n-1)!}{\{n-1-(k-1)\}!(k-1)!}\cdot(k-1)$
$=(n-1)\frac{(n-2)!}{\{n-2-(k-2)\}!(k-2)!}=(n-1)\binom{n-2}{k-2}$
$\sum_{k=1}^n \binom{n-1}{k-1}\cdot k$
$=\sum_{k=1}^n \binom{n-1}{k-1}+(n-1)\sum_{k=1}^n\binom{n-2}{k-2}$
$=\sum_{r=0}^{n-1} \binom{n-1}{r}+(n-1)\sum_{s=0}^{n-2}\binom{n-2}{s}$ as $\binom{n}{r}=0$ if $r<0$ or $r>n$
$=(1+1)^{n-1}+(n-1)(1+1)^{n-2}=2^{n-2}(2+n-1)=(n+1)2^{n-2} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/228255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
limit of adjacent series Suppose $U_n = \prod_{k=2}^n \cos\left(\frac{\pi}{2^k}\right)$ and $V_n = U_n\cdot\cos\left(\frac{\pi}{2^n}\right)$.
How can I prove that $U_n$ and $V_n$ are adjacent?
What is the limit of $U_n$ and $v_n$?
Note: $W_n = U_n\cdot\sin\left(\frac{\pi}{2^n}\right)$ and $W_n$ is geometrical.
| By the double angle formula, we have that
$$\sin x=2\sin \frac x 2 \cos \frac x 2 $$
Using the double angle formula once more
$$\sin x=2\cdot 2 \sin \frac x 4\cos \frac x 4 \cos \frac x 2 $$
You should realize that by induction,
$$\sin x=2^n \sin \frac x {2^n} \prod_{k=1}^n \cos \frac x {2^k} $$
This means that, for $x\neq 0$.
$$\frac{\sin x}x=\frac{2^n}x \sin \frac x {2^n} \prod_{k=0}^n \cos \frac x {2^k} $$
$$\frac{\sin x}{x}= \frac{\sin \dfrac x {2^n}}{\dfrac x {2^n}} \prod_{k=0}^n \cos \frac x {2^k} $$
Now let $x=\pi /2$. We get
$$\frac{{\sin \frac{\pi }{2}}}{{\frac{\pi }{2}}} = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 0}^n {\cos } \frac{\pi }{{{2^{k + 1}}}}$$
or
$$\frac{2}{\pi } = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 2}^{n + 1} {\cos } \frac{\pi }{{{2^k}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing the indefinite integral $\int x^n \sin x\,dx$ $\newcommand{\term}[3]{
\sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!}
}$
I am trying to prove that for $n \in\mathbb N$,
$$
\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$.
| Integrate by parts twice, and use two inductions (odd and even case). You may be able to unite these if you are skillful.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Number of consecutive zeros at the end of $11^{100} - 1$. How many consecutive zeros are there at the end of $11^{100} - 1$?
Attempt
Trial and error on Wolfram Alpha shows using modulus shows that there are 4 zeros (edit: 3 zeros, not 4). Otherwise, I have no idea even where to start.
| Find the highest power of 2 in its factorization and find the highest power of 5 in its factorization. If you take the minimum of these two numbers, you will get the highest power of 10 in the factorization.
Now, to find these highest powers, as you have already mentioned, we use modulus.
$11 \equiv 1 \mod 2$ so that $11^{100} - 1 \equiv 1^{100} - 1 \equiv 0 \mod 2$
Thus, we know that it contains at least one 2.
$11 \equiv -1 \mod 4$ so that $11^{100} - 1 \equiv (-1)^{100} - 1 \equiv 0 \mod 4$
Thus, it contains at least $2^2$.
$11 \equiv 3 \mod 8$ so that $11^2 \equiv 1 \mod 8$ so that $11^{100} - 1 \equiv 1^{50} - 1 \equiv 0 \mod 8$
Thus, we have $2^3$.
$11 \equiv -5 \mod 16$ so that $11^2 \equiv 9 \mod 16$ so that $11^4 \equiv 1 \mod 16$ so that $11^{100} - 1 \equiv 1^{25} - 1 \equiv 0 \mod 16$.
Thus, we have $2^4$. We definitely do not have $2^5$ as
$11^2 \equiv -7 \mod 32$ so that $11^4 \equiv 17 \mod 32$ so that $11^8 \equiv 1 \mod 32$ so that $11^{100} - 1 \equiv 11^4 \cdot 11^{96} - 1 \equiv 17 \cdot 1 - 1 \mod 32 \equiv 16 \mod 32$.
So, your answer is at most 4. Now, do the same for 5.
$11 \equiv 1 \mod 5$ so that $11^{100} - 1 \equiv 1^{100} - 1 \equiv 0 \mod 5$.
Thus, there is at least one 5. Keep going... the calculations will get more difficult as your powers get higher. Or, you could do the same thing directly with 10. But, the numbers would be bigger much faster and thus the calculations might be more difficult.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/231350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
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}$$
| For the first one:
Set $\left(\sqrt{\frac{bc}{a}},\sqrt{\frac{ca}{b}},\sqrt{\frac{ab}{c}}\right)\rightarrow (x,y,z)$
Thus $xy+yz+zx=3$ and we must prove $$\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\ge \frac{3}{2}$$
It's equivalent to $\sum x^2+3\ge 3x^2y^2z^2+\sum y^2z^2$
$$\Leftrightarrow (x+y+z-xyz)(x+y+z+3xyz)\ge 12$$
Set $p=x+y+z, \,\ q=xy+yz+zx=3, \,\ r=xyz$
We need to prove that $(p-r)(p+3r)\ge 12\Leftrightarrow (pq-3r)(pq+9r)\geq 4q^3$
$$\Leftrightarrow p^2q^2+6pqr\ge 4q^3+27r^2$$
Since $\sum x^3y+\sum xy^3\ge 2\sum x^2y^2\implies p^2q+3pr\ge 4q^2\implies p^2q^2+3pqr\ge 4q^3 \,\ (\star )$
Since $(x+y+z)(xy+yz+zx)\ge 9xyz\implies pq\ge 9r\implies 3pqr\ge 27r^2 \,\ (\star \star)$
$(\star) +(\star \star)\implies $ the desired result
Done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/235636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the area of largest rectangle that can be inscribed in an ellipse The actual problem reads:
Find the area of the largest rectangle that can be inscribed in the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to find the derivative I don't think I'm doing it right.
| Suppose that the upper righthand corner of the rectangle is at the point $\langle x,y\rangle$. Then you know that the area of the rectangle is, as you say, $4xy$, and you know that $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\;.\tag{1}$$
Thinking of the area as a function of $x$, we have $$\frac{dA}{dx}=4x\frac{dy}{dx}+4y\;.$$ Differentiating $(1)$ with respect to $x$, we have
$$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\;,$$ so $$\frac{dy}{dx}=-\frac{b^2x}{a^2y}\;,$$ and $$\frac{dA}{dx}=4y-\frac{4b^2x^2}{a^2y}\;.$$
Setting this to $0$ and simplifying, we have $y^2=\dfrac{b^2x^2}{a^2}$. From $(1)$ we know that $$y^2=b^2-\frac{b^2x^2}{a^2}\;.$$ Thus, $y^2=b^2-y^2$, $2y^2=b^2$, and $\dfrac{y^2}{b^2}=\dfrac12$. Clearly, then, $\dfrac{x^2}{a^2}=\dfrac12$ as well, and the area is maximized when
$$x=\frac{a}{\sqrt2}=\frac{a\sqrt2}2\quad\text{and}\quad y=\frac{b}{\sqrt2}=\frac{b\sqrt2}2\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/240192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 4
} |
Evaluating $\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$ Evaluate
$$\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$$
| $$\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} = e^{-2} - \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$
Hence, $$1 - e^2 \left(\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} \right) = e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$
Hence,
$$f(n) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)\right)$$
$$f(n) = e n \cdot n! \left( \sum_{k=0}^{\infty} \dfrac{(-2)^k}{(k+n+1)!}\right) = \dfrac{n}{n+1} \underbrace{e \cdot (n+1)! \left( \sum_{k=0}^{\infty} \dfrac{(-2)^k}{(k+n+1)!}\right)}_{g(n)}$$
$$g(n) = e \left(1 + \dfrac{(-2)}{(n+2)} + \dfrac{(-2)^2}{(n+2)(n+3)} + \cdots \right)$$
Hence, $\lim g(n) = e$ and hence $\lim f(n) = e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/242201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Inequality with logarithm Hi can you help me with the following
Show that for every $u \ge 0$
$$(1 + u)\log(1 + u) - u \ge \cfrac {u^2} {2 + 2u/3}$$
Obviously i tried to see something clear from Taylor series of $\log(1+x)$ but didn't see something clear.
Thank you!
| Reading the book Concentration Inequalities, I found the exact inequality in Ex. 2.8. I solved it using series expansion (of the $\log$ function) and a carefully chosen geometric series thanks to an inequality proved by induction. Both are tricky and less straightforward than the direct approach. I'm posting this for fun and for OP's desire to "see something clear from Taylor series [of the $\log$ function around $1$]".
For any $x>0$, note that $0<x/(1+x)<1$. The first step allows any "large" $x$ to "fit into the Taylor expansion" by considering its multiplicative reciprocal and inserting a minus sign in front of the $\log$ sign.
$$
\begin{aligned}
& \log(1+x) \\
=& -\log\left(1-\frac{x}{1+x}\right) \\
=& \frac{x}{1+x} + \frac12\left(\frac{x}{1+x}\right)^2 + \frac13\left(\frac{x}{1+x}\right)^3 + \cdots
\end{aligned}
$$
Next, retrieve the LHS by
*
*multiplying $1+x$ on both sides
*moving $x$ from RHS to LHS (with a change of sign)
*on RHS, drag the factor $x$ out of the parenthesis "()" so that it's more pleasing to our eyes.
$$
\begin{aligned}
(1+x)\log(1+x)-x =& x \left[ \frac12\left(\frac{x}{1+x}\right) + \frac13\left(\frac{x}{1+x}\right)^2 + \frac14\left(\frac{x}{1+x}\right)^3 + \cdots \right] \\
=& \frac{x^2}{2(1+x)} \left[ 1 + \frac23\left(\frac{x}{1+x}\right) + \frac24\left(\frac{x}{1+x}\right)^2 + \cdots \right] \\
=& \frac{x^2}{2(1+x)} \sum_{k=0}^\infty \frac{2}{k+2} \left(\frac{x}{1+x}\right)^k
\end{aligned}
$$
The following shows that LHS can be bounded below by a geometric series with common ratio $$\frac23\left(\frac{x}{1+x}\right),$$ so that RHS is shrunk to something simpler than infinite series and more similar to the question.
Lemma For any non-negative integer $k$, we have
$$\frac{2}{k+2} \ge \left( \frac23 \right)^k.$$
Justification by induction
*
*Base case $k=0$: both sides equal to one
*Induction hypothesis $k=n$: $$\frac{2}{n+2} \ge \left( \frac23 \right)^n.$$
*Inductive step $k=n+1$: $$\frac{2}{n+3} = \frac{2}{n+2} \frac{n+2}{n+3} \ge \left( \frac23 \right)^n \left( 1-\frac{1}{n+3} \right) \ge \left( \frac23 \right)^n \left( 1-\frac13 \right)=\left( \frac23 \right)^{n+1}.\square$$
$$
\require{cancel}
\begin{aligned}
(1+x)\log(1+x)-x
=& \frac{x^2}{2(1+x)} \sum_{k=0}^\infty \frac{2}{k+2} \left(\frac{x}{1+x}\right)^k \\
\ge& \frac{x^2}{2(1+x)} \sum_{k=0}^\infty \left( \frac23 \right)^k \left(\frac{x}{1+x}\right)^k \\
=& \frac{x^2}{2(1+x)} \frac{1}{1-\frac23\left(\frac{x}{1+x}\right)} \\
=& \frac{x^2}{2\cancel{(1+x)}} \frac{3\cancel{(1+x)}}{3(1+x)-2x} \\
=& \frac{x^2}{2} \frac{3}{x+3} \\
=& \frac{x^2}{2 \left( 1+x/3 \right)}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/242303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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