Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate $\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$ I want to find the following limit:
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$$
This is what I do. I change the variable $t=-x$ and I have the following limit:
$$\lim_{t\rightarrow +\infty}{e^{\frac {1}{2+t}}\cdot{\frac{t^2-2t-1}{-t-2}}+t}=\lim_{t\rightarrow+\infty}{h(x)}$$
We have $e^{\frac{1}{2+t}}\rightarrow1$ for $t\rightarrow+\infty$
Therefore I think (this is the passage I'm less sure about)
$$h(x)\sim \frac{t^2-2t-1}{-t-2}+t=\frac{t^2-2t-1+t(-t-2)}{-t-2}=\frac{t^2-2t-1-t^2-2t}{-t-2}=\frac{-4t-1}{-t-2}\sim{\frac {-4t}{-t}}\rightarrow4$$
The solution should actually be $3$. Any hints on what I'm doing wrong?
| ALternatively:
$$\lim_{x\rightarrow -\infty}\left[{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x\right]=\\
\lim_{x\to-\infty} \left[e^{\frac{1}{2-x}} \left(x+4+\frac7{x-2}\right)-x\right]=\\
\lim_{x\to-\infty} \left[e^{\frac{1}{2-x}}\left(x+4\right)-x\right] +\underbrace{\lim_{x\to-\infty} e^{\frac{1}{2-x}} \cdot \frac7{x-2}}_{=0}=\\
\lim_{x\to-\infty} \left[e^{\frac{1}{2-x}}\left(x+4\right)-(x+4)+4\right]=\\
\lim_{x\to-\infty} \left[\frac{e^{\frac1{2-x}}-1}{\frac1{2-x}}\cdot \frac{x+4}{2-x}+4\right]=\\
1\cdot (-1)+4=3,$$
where it was used:
$$\lim_{x\to 0} \frac{e^x-1}{x}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How to integrate $ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ $ I want to know how to integrate this function I have tried many things many substitutions but none works.
I even try to expand the numerator by $\sin3x$ and $\cos3x$ properties and tried to convert higher powers to multiple angles
| $$ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ =\int \frac{3\sin x -4\sin^3x+4\cos^3x-3\cos x}{ \sin^3 x + \cos^3 x } dx=\int \frac{(\cos x -\sin x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)(1- \sin x\cos x) } dx=\int \frac{(\cos^2 x -\sin^2 x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)^2(1- \sin x\cos x) } dx=\int \frac{\cos 2x (1+ 2\sin 2x)}{ (1+ \sin 2x)(1- 0.5\sin 2x) } dx=0.5\int \frac{1+ 2\sin 2x}{ (1+ \sin 2x)(1- 0.5\sin 2x) } d(\sin 2x)$$ Hope you can finish from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$xy+yz+zx=32$
I'm stuck. What is the next step to this exercise?
My idea is to show the equation using one variable and after computing the derivative reach global extremes.
| Use the the method of Lagrange multipliers. The problem is
$$
\left\{
\begin{array}{rl}
\min & x\cdot y\cdot z \\
\mbox{such that} & x+y+z=10\\
& x^2 + y^2 + z^2 = 36
\end{array}
\right.
$$
By Lagrange multiplier theorem the critical points of $f (x, y, z)=x\cdot y\cdot z$ satisfying the constraints $g(x,y,z)= x+y+z-10=0$ and $h(x,y,z)=x^2+y^2+z^2-36=0$ is solution of
$$
\left\{
\begin{array}{rl}
\frac{\partial}{\partial x}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial y}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial z}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial \lambda}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial \mu}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\end{array}
\right.
$$
Here,
$
\mathcal{L}(x,y,z,\lambda,\mu)= f(x,y,z)-\lambda g(x,y,z)-\mu h(x,y,z)
$
In this case,
$$
\left\{
\begin{array}{rlcl}
y\cdot z-\lambda -2x\mu=&0 &\hspace{1cm} & (1)\\
x\cdot z-\lambda -2y\mu=&0 &\hspace{1cm} & (2)\\
x\cdot y-\lambda -2z\mu=&0 &\hspace{1cm} & (3)\\
x+y+z=&10 &\hspace{1cm} & (4)\\
x^2+y^2+z^2=&36 &\hspace{1cm} & (5)\\
\end{array}
\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Proof-Explanation : $(a+1)(ab^2+1) \leq b^3$ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that-
$(xz)^k\leq b^3 \implies (a+1)(ab^2+1) \leq b^3$ , I could not figure it out.
Here, $(xz)^k=(a+1)(ab^2+1)$ (see page $8$),$a, b \geq 2, $ and both are integers, $k\geq 3 $ (see theorem 11 on page $8$, please see the attached-paper for detail ).
If we simplify by Rearrangement inequality, we have that
$$(a+1)(ab^2+1)=a^2b^2+ab^2+a+1 $$
how this could be less than or equal to $b^3$?
| On page 8 there is the inequality
$$ b\geq (k^{k}a^{k-2})^{1/2} \label{12}\tag{12} $$
and due to the comment on page 9 we may assume that $k\geq 50$ and $a\geq 2^{49}$. Since $b\geq 1$, we see by \eqref{12} that
$$b^{3}\geq b^{2}(k^{k}a^{k-2})^{1/2} \label{*}\tag{*}$$
Now we have, since $b\geq 1$, that
\begin{align}
(xz)^{k}&=&(a+1)(ab^2+1)\\
&=&a^2b^2+a+ab^2+1\\
&\leq&b^2(a^2+a+a+1)\\
&=&b^2(a+1)^2.
\end{align}
Now we note that
$$(a+1)^2=a^2+2a+1\leq a^2+2a^2+a^2=4a^2,$$
whence we see that
$$(a+1)^{2}\leq 4a^{2}\leq k^{k/2}a^{\frac{k-2}{2}}$$
since $k\geq 50$. Therefore we have by \eqref{*} that
$$(xy)^{k}\leq b^{2}(a+1)^{2}\leq b^{2}(k^{k}a^{k-2})^{1/2}\leq b^{3},$$
as desired
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the proof that every real number in the unit interval is the limit of a sequence of dyadic numbers
Given $x \in (0,1)$, show there exists a sequence $(x_n) \subset \{0,1\}$ such that $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$.
After running into difficult in trying to solve this problem, I found this MSE post, which, to my chagrin, gave me more difficulties. I'm trying to follow John Ma's reasoning. Here's how I understand it:
If $x < \frac{1}{2}$, the choose $x_1 =0$ and therefore $|x-\frac{x_1}{2}| = |x| = x < \frac{1}{2}$, and in fact $x-\frac{x_1}{2} \ge 0$, so in this case we can find $x_1$ for which $0 \le x - \frac{x_1}{2} < \frac{1}{2}$. If, however, $x \ge \frac{1}{2}$, then choose $x_2 =1$, and $|x-\frac{x_1}{2}| < \frac{1}{2}$ holds if and only if $- \frac{1}{2} < x - \frac{1}{2} < \frac{1}{2}$ if and only if $0 < x < 1$. Since $0 < x < 1$ is in fact true, so must $|x-\frac{x_1}{2}| < \frac{1}{2}$. Also, we have $x - \frac{x_1}{2} \ge 0$ since we assumed $x \ge \frac{1}{2}$. Hence, in either case we found $x_1 \in \{0,1\}$ for which $0 \le x-\frac{x_1}{2} < \frac{1}{2}$ holds. Now assume that $x_1,...x_n \in \{0,1\}$ have been chosen such that $0 \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}$. From this we want to show that it is possible to choose $x_{n+1} \in \{0,1\}$. If $x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$, then choose $x_{n+1} = 0$ and we obtain $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$. If, however, $x - \sum_{k=1}^n \frac{x_k}{2^k} \ge \frac{1}{2^{n+1}}$, choose $x_{n+1} = 1$. Then $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} = (x - \sum_{k=1}^n \frac{x_k}{2^k}) - \frac{1}{2^{n+1}}$
But here's the problem: why is $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k}$ nonnegative? If that can't be shown, then I cannot appeal to Squeeze lemma to show $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$. I could really use some help. Thanks in advanced!
| (1).Let $x_1=0$ if $0<x<1/2$ and $x_1=1$ if $1/2\leq x<1.$ Let $y_1=x_1/2.$ We define the sequences $(x_n)_{n\in \Bbb N}$ and $(y_n)_{n\in \Bbb N}$ as follows:
(2). Suppose $y_n\leq x.$ Let $x_{n+1}=0$ if $x<y_n+2^{-(n+1)}.$ Let $x_{n+1}=1$ if $x\ge y_n+2^{-(n+1)}.$ And let $y_{n+1}=y_n+x_n2^{-(n+1)}.$ In both cases we have $y_n\leq x\implies y_{n+1}\leq x.$
And we have $y_1\leq x.$ So by induction we have $y_n\leq x$ for all $n$.
(3). We have $y_n\geq x- 2^{-n}\implies y_{n+1}\geq x-2^{-(n+1)}.$ Proof: Suppose $y_n\geq x-2^{-n}.$ Then
$\quad$(i).If $y_n\leq x-2^{-(n+1)}$ then $x_{n+1}=1$ so $\quad y_{n+1}=y_n+2^{-(n+1}\geq (x-2^{-n})+2^{-(n+1)}=x-2^{-(n+1)}.$
$\quad$(ii). If $y_n>x-2^{-(n+1)}$ then $x_{n+1}=0$ so $y_{n+1}=y_n>x-2^{-(n+1)}.$
In both (i) and (ii) we have $y_{n+1}\geq x-2^{-(n+1)}.$
And we have $y_1\geq x-2^{-1}.$ So $y_n\geq x-2^{-n}$ for all $n$ by induction.
(4). Since $x-2^{-n}\leq y_n\leq x$ for all $n,$ we have $x=\lim_{n\to \infty}y_n=\sum_{n=1}^{\infty}x_n2^{-n}.$
Remark. In (2) we use recursion and induction together, to define $x_{n+1}$ and $y_{n+1}$ recursively from $y_n$, with the inductive hypothesis that $y_n\leq x.$ If you wanted, you could instead say " If $y_n>x$ then let $x_{n+1}=1=y_{n+1}$ " ( or some other arbitrary values) for a purely recursive def'n, and then prove separately that $y_n\leq x$ for all $n$ by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Riemann sum of $x\cdot \ln(x)$ I did not find any information regarding this Riemann sum anywhere:
Riemann sum of $f(x)=\begin{cases} 0& x=0 \\ x\cdot \ln(x)& \text{otherwise}\end{cases}$ in the interval $[0, 1]$.
I don't want the answer given to me, I'm only looking for a hint that could guide me in the right direction. I have already proved that the function is continuous in this interval, i.e can be integrated in it also.
| The sum can be written as
$$S_n = \frac{1}{n^2}\sum_{k=1}^nk \log k -\frac{\log n}{n^2}\sum_{k=1}^nk = \frac{1}{n^2}\sum_{k=1}^nk \log k -\frac{\log n}{n^2}\frac{n(n+1)}{2} \\ = \frac{1}{n^2}\sum_{k=1}^nk (\log k- H_k) -\frac{\log n}{n^2}\frac{n(n+1)}{2} + \frac{1}{n^2}\sum_{k=1}^nk H_k $$
where $H_n = \sum_{k=1}^n \frac{1}{k}$
Using summation by parts, we have
$$\frac{1}{n^2}\sum_{k=1}^nk H_k = \frac{1}{n^2}H_n \sum_{k=1}^n k + \frac{1}{n^2}\sum_{k=1}^{n-1}\left(\sum_{j=1}^kj \right)(H_k - H_{k+1}) \\ = \frac{H_n}{n^2}\frac{n(n+1)}{2} - \frac{1}{4}\left(1 - \frac{1}{n}\right) $$
Hence,
$$S_n = -\frac{1}{n^2}\sum_{k=1}^nk (H_k - \log k) + \frac{H_n -\log n}{n^2}\frac{n(n+1)}{2} - \frac{1}{4}\left(1 - \frac{1}{n}\right)$$
Using the fact that $H_n - \log n \to \gamma$ as $n \to \infty$, we can show, with some additional work, that the contribution of the first and second terms cancels in the limit as $n \to \infty$ and we have
$$\lim_{n \to \infty} S_n = - \frac{1}{4} = \int_0^1 x \log x \, dx$$
Additional Work
Note that for any $\epsilon > 0$ there exists $N$ such that $\gamma - \epsilon < H_k - \log k < \gamma + \epsilon$ when $k > N$.
We have
$$\sum_{k=1}^nk (H_k - \log k) = \frac{1}{n^2}\sum_{k=1}^Nk (H_k - \log k)+ \frac{1}{n^2}\sum_{k=N+1}^nk (H_k - \log k) $$
Since you don't want the full answer -- see if you can finish from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving $\frac{1}{1-x} \circ \frac{1}{1-x} = 1 - \frac{1}{x}$ from a series point of view It's an elementary exercise in grade school algebra that
$$ \frac{1}{1-\frac{1}{1-x}} = 1 - \frac{1}{x} $$
However from the series point of view it's not at all obvious. There are two different series expressions for $\frac{1}{1-x}$ which are
$$ \sum_{k=0}^{\infty} x^k = 1 + x+ x^2 + ... \ |x| < 1 $$
$$ - \sum_{k=1}^{\infty} \frac{1}{x^k} = -\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... \ |x| > 1 $$
and attempting to compose yields troubles: (there are 4 cases to analyze here)
$$ 1+ (1+x+x^2 ... ) + (1 + x + x^2 + ...)^2 + ... $$
This leads to coefficient blow up, and even with using zeta function values to renormalize infinities it leads to an expression that seems meaningless (or I should say, is very "difficult" to interpret).
$$ 1 + (-\frac{1}{x} - \frac{1}{x^2} ... ) + (-\frac{1}{x} - \frac{1}{x^2} ...)^2 ... $$
actually simplifies to the correct expression $1- \frac{1}{x}$ (so series can confirm the identity for: $|x|<1, 1 > |x-1|$)
$$ - \frac{1}{1 + x + x^2 ... } - \frac{1}{(1 + x + x^2 ... )^2 } ... $$
Is again intractable without referencing the geometric series formula.
$$ - \frac{1}{ - \frac{1}{x} - \frac{1}{x^2} ... } - \frac{1}{(-\frac{1}{x} - \frac{1}{x^2} ... )^2} ... $$
Is even more horrific (I nicknamed this expression Harmonic Hell).
My worry here is only 1 of these 4 compositions could be simplified into the correct target expression, how to correctly manipulate the other 3 to yield the target expression, after all there are specific domain, range combinations that I am losing when I only consider any one of these pairs of series (yet the expression $1 - \frac{1}{x}$ is true globally).
Motivation
This is part of a toy problem: Action of 3x3 invertible matrices on $\mathbb{C}$? where I began to wonder if it was possible to find an action of $3\times 3$ matrices on the complex plane.
My program of research was the following:
*
*Interpret mobius transformations as literally pairs of laurent series which accept a quadruple of parameters $a,b,c,d$ corresponding to elements of a $2\times 2$ rotation matrix.
*Prove the action property (that composing these series yields a new series of the same form, with parameters respecting $2\times 2$ matrix multiplication), [this is where i'm stuck hence this question]
*Look to now construct series that respect the action of $3\times 3$ matrix multiplication, perhaps inspired by the completion of (2).
| For your problem with power series for $\, f(x) := 1/(1-x), \,\, g(x) := 1-1/x, \,$ you would like to center each power series about the same number when the functions are composed. In this case the common center is $\, \omega, \,$ a primitive sixth root of unity because
$\, \omega = f(\omega) = g(\omega) \,$ is a fixed point of both $\,f\,$
and $\,g.\,$ Thus, let $\, y := x-\omega \,$ be the local variable.
Check that the two power series expansions in powers of $\, y\,$ are
$$ f(x) = \omega + \omega^2 y - y^2 - \omega y^3 -\omega^2 y^4
+ y^5 + O(y^6) \, =
\frac{\omega + \omega^2 y - y^2} {(1 + y^3)}, \tag1$$
$$ g(x) = \omega - \omega y + y^2 + \omega^2 y^3 - \omega y^4
+ y^5 + O(y^6) \, = \omega + \frac{y}{\omega(\omega+y)}. \tag2$$
The radius of convergence for both series is $\,1\,$ centered at $\, \omega \,$ and includes $\, 0<x<1. \,$
Check with composition the following equations:
$$ f(f(x)) = g(x), \,\, f(g(x)) = x, \,\,
g(f(x)) = x, \,\, g(g(x)) = f(x). \tag3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Continuity of $f(x) = \frac 1 x - \frac 1 {x+1} + \frac{1}{x+2} - \cdots$
Is function
$$f(x) = \frac 1 x - \frac 1 {x+1} + \frac{1}{x+2} - \frac 1{x+3} + \cdots$$
continuous on $(0, \infty)$ ?
I think I should extend the continuity of each term to $f$ but I cannot show that series converges uniformly.
| $$\left( \frac 1 x - \frac 1 {x+1} \right) + \left( \frac 1 {x+2} - \frac 1 {x+3} \right) + \cdots + \left( \frac 1 {x+2n} -\frac 1 {x+2n-1} \right)$$ $$=\frac 1 {x(x+1)}+\frac 1 {(x+2)(x+3)}+\cdots +\frac 1 {(x+2n) (x+2n+1)}$$ $$\leq \frac 1 {x(x+1)}+\frac 1 {2^{2}}+ \frac 1 {4^{2}}+\cdots+\frac 1 {(2n)^{2}} $$.
This gives uniform convergence of the partial sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sin^8\theta+\cos^8\theta=\frac{17}{32}$, find the value of $\theta$ using de Moivre's theorem. If
$$\sin^8\theta+\cos^8\theta=\frac{17}{32}$$
find the value of $\theta$ using de Moivre's theorem. I tried a lot exapanding using Binomial theorem and taking real part and equating it given value.
| HINT
According to the formula
\begin{align*}
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\end{align*}
We can rewrite the given expresion as
\begin{align*}
\sin^{8}(\theta) + \cos^{8}(\theta) & = [\sin^{4}(\theta) + \cos^{4}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = [(\sin^{2}(\theta)+\cos^{2}(\theta))^{2} - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = [1 - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = \left[1 - \frac{\sin^{2}(2\theta)}{2}\right]^{2} - \frac{\sin^{4}(2\theta)}{8} = 1 -\sin^{2}(2\theta) + \frac{\sin^{4}(2\theta)}{8} = \frac{17}{32}
\end{align*}
Therefore, if we make the substitution $y = \sin(2\theta)$, the given problem is equivalent to
\begin{align*}
& 4y^{4} - 32y^{2} + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2}) + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2} + 16) - 49 = 0 \Longleftrightarrow\\
& 4(y^{2}-4)^{2} = 49 \Longleftrightarrow y^{2} - 4 = \pm\frac{7}{2} \Longleftrightarrow y^{2} = 4\pm\frac{7}{2} \Longleftrightarrow y = \pm\frac{1}{\sqrt{2}} = \pm\sin\left(\frac{\pi}{4}\right)
\end{align*}
Can you proceed from here?
EDIT
According to the Newton's binomial theorem as well as the De Moivre's formula, we have
\begin{align*}
\cos(8\theta) + i\sin(8\theta) = [\cos(\theta) + i\sin(\theta)]^{8} = \sum_{k=0}^{8}\binom{8}{k}[\cos(\theta)]^{k}[i\sin(\theta)]^{8-k}
\end{align*}
As a consequence, you can obtain the value of the expression $\sin^{8}(\theta) + \cos^{8}(\theta)$ in terms of sines and cosines and solve the
corresponding equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use cosine rule to determine the length of a side of a triangle. Cannot replicate textbook solution Given a triangle of lengths $4$, $5$ and $x$, where the sides of length $4$ and $5$ meet at an angle of $60^{\circ}$, I must calculate the length of side $x$.
The cosine rule is: $$a^2 = b^2 + c^2 - 2bc\cos(\alpha)$$
So, applying to my problem:
$$\begin{align}
a^2&=b^2+c^2-2bc\cos(\alpha) \\
x^2&=4^2+5^2-2(4)(5)(\cos(60^{\circ})) \\
x^2&=16+25-40\cos(60^{\circ}) \\
x^2&=41-40\cos(60^{\circ}) \\
x^2&=41-40(0.9524) \\
x^2&=41-38.1 \\
x^2&=2.9 \\
x&=\sqrt{2.9}
\end{align}$$
The textbook solution says I should arrive at $x = \sqrt{21}$, not $x=\sqrt{2.9}$. The textbook does not provide working to the solution, only the final answer.
Where did I go wrong?
| Hint: $\cos(60^o)=0.5$ maybe you mixed radians ($60^o=1/3\pi$) with degrees on your calculator?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925421",
"timestamp": "2023-03-29T00:00:00",
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} |
Simplify $\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$ to $\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$ I'm trying to bring this expression:
$$\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$$
To this one:
$$\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$$
Where $\log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
| We have
$$\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log 2=\overbrace{\frac{5}{6}\log\left(\frac{5}{4}\right)+\color{red}{\frac{1}{6}\log\left(\frac{5}{4}\right)}} -\overbrace{\color{red}{\frac{1}{6}\log\left(\frac{5}{4}\right)}- \frac{1}{6}\log 2}=$$
$$=\log\left(\frac{5}{4}\right)-\frac16 \log\left(\frac{5}{2}\right)$$
indeed recall that $\log A+ \log B= \log AB$ and therefore
$$-\frac{1}{6}\log\left(\frac{5}{4}\right)- \frac{1}{6}\log 2=-\frac16\left[\log\left(\frac{5}{4}\right)+\log 2\right]=-\frac16\log\left(\frac{5}{4}\cdot 2\right)=-\frac16\log\left(\frac{5}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Solving $\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0$
I'm looking for a more elegant way to find the real solution
$$\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0.$$
Here is my way:
First write the equation as
$$\sin\left(\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+x+2)}\right)\cos\left(\frac{x}{2(x^2+1)}-\frac{1}{2(x^2+x+2)}\right)=0.$$
Then the real solution can be find by
$$\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+x+2)}=0$$
which gives $x=-1$.
| It must be $$\frac{x}{2(x^2+1)}+\frac{1}{(2x^2+x+1)}=k\pi$$
or
$$\frac{x}{2(x^2+1)}+\frac{1}{(2x^2+x+1)}=\frac{(2k+1)\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove inequality $(a^2+b^2+c^2)^3\ge6(a^3+b^3+c^3)^2$ when $a+b+c=0$? I know the identity $a^3+b^3+c^3-3abc = 1/2 (a+b+c) [(a-b)^2+(b-c)^2+(c-a)^2]$.
So when it comes to this problem where $a+b+c=0$, you get $(a^2+b^2+c^2)^3\ge54(abc)^2$ when $a+b+c=0$
(because $ a^3+b^3+c^3-3abc=0$).
But if you use average inequality, you can only get $(a^2+b^2+c^2)^3\ge27(abc)^2$.
So what kind of technique works on this problem?
| After substitution $c=-a-b$ we need to prove that
$$4(a^2+ab+b^2)^3\geq27a^2b^2(a+b)^2,$$ which is true because
$$4(a^2+ab+b^2)^3-27a^2b^2(a+b)^2=(a-b)^2(2a+b)^2(a+2b)^2\geq0.$$
Indeed.
let $a^2+b^2=2uab$.
Thus,
$$4(a^2+ab+b^2)^3-27a^2b^2(a+b)^2=4(2uab+ab)^3-27a^2b^2(2uab+2ab)=$$
$$=2a^3b^3(2(2u+1)^3-27(u+1))=2a^3b^3(16u^3+24u^2-15u-25)=$$
$$=2a^3b^3(16u^3-16u^2+40u^2-40u+25u-25)=2a^3b^3(u-1)(16u^2+40u+25)=$$
$$=2a^3b^3(u-1)(4u+5)^2=(a^2+b^2-2ab)(2(a^2+b^2)+5ab)^2=(a-b)^2(2a+b)^2(b+2a)^2.$$
There is a nice proof by AM-GM.
Indeed, since $a+b+c=0$, we can assume that $ab\leq0.$
Thus, by AM-GM we obtain:
$$(a^2+b^2+c^2)^3=(a^2+b^2+(a+b)^2)^3=(2(a+b)^2-ab-ab)^3\geq$$
$$\geq\left(3\sqrt[3]{2(a+b)^2\cdot(-ab)^2}\right)^3=54a^2b^2(a+b)^2=$$
$$=6(3ab(a+b))^2=6(a^3+b^3-(a+b)^3)^2=6(a^3+b^3+c^3)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding a geometric series for $\frac{1}{(1 + x)^2}$ I'm asked to find a geometric series for $f(x) = \frac{1}{(1 + x)^2}$, I integrate it first and I get $F(x) = \int{\frac{1}{(1 + x)^2}dx} = \frac{1}{-2(1 - (-x))}$
Since $\frac1{1-x} = \sum_{n=0}^\infty x^n$, $F(x) = \frac{-1}2\sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{2}x^n$
If I derive that, $f(x) = \sum_{n=0}^\infty\frac{(-1)^{n+1} n x^{n-1}}{2} = \sum_{n=1}^\infty\frac{(-1)^n (n + 1) x^n}{2}$
But my textbook says that $f(x) = \sum_{n=0}^\infty (-1)^n (n + 1) x^n$
Obviously those 2 things are pretty different (notice the index). What did I get wrong?
| Original question:
find a geometric series for $$f(x) = \frac{1}{1+x^2}$$
As you wrote, $$\frac{1}{1-y} = \displaystyle\sum_{n=0}^\infty y^n$$
Substituting $y=-x^2$, we have
$$f(x)= \frac{1}{1+x^2} = \displaystyle\sum_{n=0}^\infty (-1)^n x^{2n}$$
Edited question:
find a geometric series for $$g(x) = \frac{1}{\left(1+x\right)^2}$$
Recall that $$(1+x)^2 = 1+2x+x^2$$
As you wrote, $$\frac{1}{1-y} = \displaystyle\sum_{n=0}^\infty y^n$$
Substituting $y=-2x-x^2$, we have
$$g(x)= \frac{1}{\left(1+x\right)^2}=\frac{1}{1-\left(-2x-x^2\right)} = \displaystyle\sum_{n=0}^\infty (-1)^n \left(2x+x^2\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation: $\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$ How would I go about solving the equation below:
$\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$
After appyling sin to both sides I end up with:
$(2x^2-1)(\sqrt{1-x^2}^2 - x^2) + 2x \sqrt{1 -(2x^2-1)^2}\sqrt{1-x^2}= -1$
| Hint:
As $\arcsin(-y)=-\arcsin y$
$$\iff-2\arcsin x=\dfrac\pi2-\arcsin(1-2x^2)=\arccos(1-2x^2)$$
Now $0\le\arccos(1-2x^2)\le\pi\implies0\le-2\arcsin x\le\pi\iff0\ge\arcsin x\ge-\dfrac\pi2$
Now let $\arcsin x=y\implies x=\sin y,0\le-2y\le\pi$
$$\arccos(1-2x^2)=\arccos(\cos2y)=\begin{cases}2y &\mbox{if }0\le2y\le\pi\\
-2y& \mbox{if } 0\le-2y\le\pi \end{cases}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\sum_{n=0}^\infty \frac1{(4n^2 - 1)^2}$ How do I find the value of the following infinite series?
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} $$
My attempt at a solution:
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} = \sum_{n=0}^\infty \frac{1}{((2n-1)(2n+1))^2} = \sum_{n=0}^\infty \left(\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)\right)^2 = \frac{1}{4}\sum_{n=0}^\infty \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)^2 $$
I then tried to compute the partial sums of this series, but with no luck. Does anyone else know how to do it?
| Note that$$(\forall n\in\mathbb{N}):\frac1{(4n^2-1)^2}=\frac1{4(2n+1)}-\frac1{4 (2n-1)}+\frac1{4(2n+1)^2}+\frac1{4(2n-1)^2}.$$It is clear that the series$$\sum_{n=0}^\infty\frac1{4(2n+1)}-\frac1{4(2n-1)}$$is a telescoping series, whose sum is $\frac14$. On the other hand\begin{align}\sum_{n=0}^\infty\frac{1}{4 (2 n+1)^2}+\frac{1}{4(2 n-1)^2}&=\frac14+2\sum_{n=0}^\infty\frac{1}{4 (2 n+1)^2}\\&=\frac14+\frac12\sum_{n=0}^\infty\frac1{(2n+1)^2}.\end{align}But, since$$\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8,$$we have that the sum of your series is$$\frac14+\frac14+\frac{\pi^2}{16}=\frac12+\frac{\pi^2}{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked to prove (by induction) that:
$$
\sum^n_{i=1}i^2 = \frac{n(n+1)(2n+1)}{6}
$$
In the inductive step I add $(n+1)^2$ to either side and then try to work my way through, multiplying out the factors:
$$
\frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\
= \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\
= \frac{(n^2+n)(2n+1)+6(n+1)^2}{6} \\
= \frac{2n^3+n^2+2n^2+n+6(n^2+2n+1)}{6} \\
= \frac{2n^3+n^2+2n^2+n+6n^2+12n+6}{6} \\
= \frac{2n^3+9n^2+13n+6}{6}
$$
I think everything up to this point is pretty trivial, but it's around this point that I get a bit lost. I know I need to factor the terms in the 3-degree polynomial in the numerator somehow, but I'm having a hard time with the actual mechanics given the lack of obvious common factors & the fact that 13 is prime.
I'm pretty sure the final result should look something like:
$$
\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} \\
= \frac{(n+1)(n+2)(2n+3)}{6}
$$
Multiplying the expected result out I can see the two statements are equal and I'm confident that
$$
\frac{2n^3+9n^2+13n+6}{6} = \frac{(n+1)(n+2)(2n+3)}{6}
$$
but I'm just not quite sure how to factor the polynomial myself to arrive at the final result.
Any hints on how to proceed from here, or what I need to be reading up on to get my head around this?
| Step-by-step simplification
$$2n^3 + 9n^2+13n+6$$ $$= 2n(n^2 + 4n + 4) + (n^2+5n+6)$$$$=2n(n+2)^2+(n+2)(n+3)$$$$=(n+2)(n+3+2n(n+2))$$$$=(n+2)(n+3+2n^2+4n)$$$$=(n+2)(2n^2+5n+3)$$$$=(n+2)(2n(n+1) + 3(n+1))$$$$ = (n+2)(n+1)(2n+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Double summation with improper integral So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx .$$
Here is my approach:
First evaluate the integral:
$$ \frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx.$$
This can be done using integration by parts and we get:
$$ \frac{1}{b!} \frac{b}{a} \int_0^\infty \frac{x^{b-1}}{e^{ax}}\ dx.$$
We can do this $ b $ times until we get:
$$ \frac{1}{b!} \frac{(b)(b-a).....(b-b+1)}{a^b} \int_0^\infty \frac{x^{b-b}}{e^{ax}}\ dx.$$
and hence we end up with:
$$ \frac{1}{b!} \frac{b!}{a^b}\qquad\left(\frac{-1 \ e^{-ax}}{a}\Big|_0^\infty\right) = \frac{1}{a^{b+1}}.$$
Now we can apply the sum of GP to infinity formula and we get:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \frac{1}{a^{b+1}} = \sum_{a=2}^\infty \frac{\frac{1}{a^{2}}}{1-\frac{1}{a}}.$$
This is a telescoping series and we end up with $$ \frac{1}{a-1} = \frac{1}{2-1} = 1.$$
Do you guys have any other ways of solving this problem? Please do share it here.
| since $\frac{x^b}{e^{ax} b!}$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
\begin{align}
&\sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx \\
&=\int_{0}^\infty \sum_{a=2}^\infty e^{-ax} \sum_{b=1}^\infty \frac{x^{b}}{ \ b!} \ dx \\
&= \int_{0}^\infty \underbrace{\left(\sum_{a=2}^\infty (e^{-x})^a\right)}_{\text{geometric series}} \overbrace{\left(\sum_{b=0}^\infty \frac{x^{b}}{ \ b!}-1\right)}^{\text{series definition of $e^x$}} \ dx \\
&= \int_{0}^\infty \frac{1}{e^x(e^x-1)}(e^{x}-1)dx \\
&= \int_0^\infty e^{-x} dx \\&= 1.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 4,
"answer_id": 1
} |
Find the limit of $\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}$
Find the following limit: \begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}
I have tried to divide the numerator and denominator by $\sqrt{x}$, but it did not work.
I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work.
I have tried to multiply by the conjugates of the numerator only but it did not work.
So what shall I do?
| Using a substitution and a derivative:
Set $x=t^2$:
$$ \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} = \lim_{t \rightarrow 2} \frac{\sqrt{1 + 2t^2} -3}{t - 2} = f'(2) \mbox{ for } f(t) = \sqrt{1 + 2t^2}$$
So,
$$f'(t) = \frac{2t}{\sqrt{1 + 2t^2}} \Rightarrow \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} = f'(2) = \frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a recursive definition of this sequence I always have some difficulty with problems of this type, and I was wondering if there was a typical trick that makes it reasonable.
Let $W_n$ be the number of words of length $n$ formed with the letters $A$ and $B$ such that
*
*There are no words that contain sequences of $A$s with length exactly 2
*There are no words that contain sequences of $B$s with length exactly 2 or 3.
Examples of words: $AAAB, ABABABA, ABBBBAAABA$, and of non-words: $AAB, ABBBA$, etc.
These "recursion with constraints" type problems always cause me difficulty. I want to write $W_n = f(W_{n-1}, W_{n-2}, \ldots, W_{n-k})$ for some fixed $k$. I tried to break it into cases based on what the last letter is, so $W_n = A_n + B_n$ where $A_n$ is the number of words of length $n$ that end in $A$ and $B_n$ defined similarly. Then getting recursive definitions for $A_n$ in terms of some $B_{n-k_i}$ and $W_{n-k_j}$ and likewise recursive definitions for $B_n$ produces a whole mess of terms that don't cancel even when you get these $W_{n-11}$ terms showing up. Surely I'm missing the trick?
| We can write down an infinite recursion and then simplify.
Let $A_n$ be the number of such sequences ending in A, and $B_n$ be the number of such sequences ending in B. We count the empty string in both.
Then, for $n \ge 1$, we have
$$
A_n = B_{n-1} + B_{n-3} + B_{n-4} + B_{n-5} + \dotsb
$$
because we can add A or AAA or AAAA or AAAAA to the end of a string ending in B, but not both. So for $n \ge 2$, we have
\begin{align}
A_n - A_{n-1} &= (B_{n-1} + B_{n-3} + B_{n-4} + B_{n-5} + \dotsb) - (B_{n-2} + B_{n-4} + B_{n-5} + B_{n-6} + \dotsb) \\
&= B_{n-1} - B_{n-2} + B_{n-3}. \\
A_n &= A_{n-1} + B_{n-1} - B_{n-2} + B_{n-3}.
\end{align}
We can do the same thing for $B_n$, except that $B_n$ won't have an $A_{n-3}$ term in the recurrence. For $n\ge 2$, we get
\begin{align}
B_n - B_{n-1} &= (A_{n-1} + A_{n-4} + A_{n-5} + \dotsb) - (A_{n-2} + A_{n-5} + A_{n-6} + \dotsb) \\
&= A_{n-1} - A_{n-2} + A_{n-4}. \\
B_n &= A_{n-1} + B_{n-1} - A_{n-2} + B_{n-4}.
\end{align}
This is already enough to make the problem a finite linear recurrence, but we can still try to simplify it further.
For $n \ge 1$, we have $W_n = A_n + B_n$, so we have
$$
W_n = 2W_{n-1} - W_{n-2} +B_{n-3} + A_{n-4}
$$
by adding the two equations. Unfortunately, we still have a mismatched $A$ and $B$ to deal with here.
Expanding $B_{n-3}$ and $A_{n-4}$ out using their recurrences, we get
\begin{align}
W_{n} &= 2W_{n-1} - W_{n-2} + (W_{n-4} - A_{n-5} + A_{n-7}) + (W_{n-5} - B_{n-6} + B_{n-7}) \\
&= 2W_{n-1} - W_{n-2} + W_{n-4} + W_{n-5} - A_{n-5} - B_{n-6} + W_{n-7}
\end{align}
which again has a mismatched $A$ and $B$, but now in the reverse order. So if we subtract $W_{n-2}$ from both sides, we get
\begin{align}
W_n - W_{n-2} &= (2W_{n-1} - W_{n-2} + W_{n-4} + W_{n-5} - A_{n-5} - B_{n-6} + W_{n-7}) \\&\quad - (2W_{n-3} - W_{n-4} + B_{n-5} + A_{n-6}) \\
&= 2W_{n-1} - W_{n-2} - 2W_{n-3} + 2W_{n-4} - W_{n-6} + W_{n-7}
\end{align}
and the final recursion is $W_n = 2W_{n-1} - 2W_{n-3} + 2 W_{n-4} - W_{n-6} + W_{n-7}$. (This may not fly for small $n$ - in particular, $W_0 \ne A_0 + B_0$ - but for $n> 7$ all the manipulations above are valid.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$ Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$
I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$
A tangent on $f(x)=\sqrt{1-x}$ at $x={1\over 3}$ is $$y=-{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$
So we have, for all $x\in[0,1]$:
$$\sqrt{1-x} \leq -{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$ and we are done...
I wonder if there is elegant method avoiding calculus?
|
We may assume that $x+y+z=1$, so we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$
By the RMS-AM (root-mean square - arithmetic mean) inequality:
$$
\frac{\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}}{3} \le \sqrt{\frac{\left(\sqrt{1-x}\right)^2+\left(\sqrt{1-y}\right)^2+\left(\sqrt{1-z}\right)^2}{3}} = \sqrt{\frac{2}{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$ If $a.b,c \in \mathbb{R^+}$ and $ab+bc+ca=1$
Then Prove $$S=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$$
My try we have $$S=\sum \frac{a}{\sqrt{a^2+ab+bc+ca}}=\sum \frac{a}{\sqrt{a+b}\sqrt{a+c})}$$
any hint here?
| Let $a=\tan\left(\dfrac{\alpha}{2}\right), b=\tan\left(\dfrac{\beta}{2}\right), a=\tan\left(\dfrac{\gamma}{2}\right) \ \ (\alpha+\beta+\gamma=\pi)$
$$\Leftrightarrow \sin\left(\dfrac{\alpha}{2}\right)+\sin\left(\dfrac{\beta}{2}\right)+\sin\left(\dfrac{\gamma}{2}\right) \le 3\sin\left(\dfrac{\pi}{6}\right)\le \dfrac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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finding extreme points for Lagrangian with multiple inequality constraints I am trying to find maximum of
\begin{equation}
f(x, y) = x^2 - xy + y - 4x
\end{equation}
\begin{equation}\label{constraints}
\text{s.t. } 0 \leq x \leq 2 \text{ and } 0 \leq y \leq 1
\end{equation}
I got an advice to first check for the interior solution (it has none there is one saddle point at [1,-2] and afterwards check each constraint separately, but I am not sure if what I did is correct and how to proceed with finding extreme points at the boundaries
\begin{equation}
\begin{aligned}
0=x; 0=y \\
0=x; 1=y \\
2=x; 0=y \\
2=x; 1=y \\
\end{aligned}
\end{equation}
Now we can set up 4 Lagrangians to check for each boundary:
\subsubsection*{First Lagrangian:}
\begin{equation}
\mathcal{L} = x^2 - xy + y - 4x - \lambda_1(-x) - \lambda_2 (-y)
\end{equation}
with FOCs
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{L} }{\partial x} & = 2x - y -4 + \lambda_1 = 0\\
\frac{\partial \mathcal{L}}{ \partial y} & = -x + 1 + \lambda_2 = 0 \\
x&=0\\
y&=0
\end{aligned}
\end{equation}
hence here $x=y=0$ and $\lambda_1 = 4$ and $\lambda_2=-1$
\subsubsection*{Second Lagrangian:}
\begin{equation}
\mathcal{L} = x^2 - xy + y - 4x - \lambda_1(-x) - \lambda_2 (-y)
\end{equation}
with FOCs
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{L} }{\partial x} & = 2x - y -4 + \lambda_1 = 0\\
\frac{\partial \mathcal{L}}{ \partial y} & = -x + 1 + \lambda_2 = 0 \\
x&=0\\
y&=1
\end{aligned}
\end{equation}
hence here $x=y=0$ and $\lambda_1 = 3$ and $\lambda_2=-1$
\subsubsection*{Third Lagrangian:}
\begin{equation}
\mathcal{L} = x^2 - xy + y - 4x - \lambda_1(-x) - \lambda_2 (-y)
\end{equation}
with FOCs
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{L} }{\partial x} & = 2x - y -4 + \lambda_1 = 0\\
\frac{\partial \mathcal{L}}{ \partial y} & = -x + 1 + \lambda_2 = 0 \\
x&=0\\
y&=1
\end{aligned}
\end{equation}
hence here $x=y=0$ and $\lambda_1 = 3$ and $\lambda_2=-1$
\subsubsection{Fourth Lagrangian}
\begin{equation}
\mathcal{L} = x^2 - xy + y - 4x - \lambda_1(-x) - \lambda_2 (-y)
\end{equation}
with FOCs
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{L} }{\partial x} & = 2x - y -4 + \lambda_1 = 0\\
\frac{\partial \mathcal{L}}{ \partial y} & = -x + 1 + \lambda_2 = 0 \\
x&=2\\
y&=1
\end{aligned}
\end{equation}
| Since $f$ is a quadratic function having a saddle point it is not convex.
In order to find the maximum of $f$ on the boundary $\partial R$ of the given rectangular domain $R$ just pull back $f$ to the four edges of $R$, i.e., consider the four auxiliary functions
$$\eqalign{&\phi_0(y):=f(0,y)=y,\quad\phi_2(y)=f(2,y)=-y-4\qquad\qquad(0\leq y\leq 1),\cr &\psi_0(x):=f(x,0)=(x-2)^2-4,\quad \psi_1(x):=f(x,1)=(x-2.5)^2-5.25\qquad(0\leq x\leq2)\ .\cr}$$
Since all four of these functions are monotone on the relevant intervals we can conclude that
$$\max_{(x,y)\in R}f(x,y)=\max\bigl\{f(0,0),f(2,0),f(0,1),f(2,1)\bigr\}=1\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$ If $a,b,c \in \mathbb{R+, }$ Then Prove that $$\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$$
My try:
Consider $$P=\frac{a}{3\sqrt{3}}+\frac{b}{4\sqrt{4}}+\frac{c}{5\sqrt{5}}$$
BY Cauchy Scwartz Inequality we have
$$P \le \sqrt{3} \times \sqrt {\left(\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \right)}$$
any way proceed here?
| For $\lambda = a+b+c$, Cauchy-Schwarz inequality gives $$\lambda^2 \leqslant (3^3 + 4^3+ 5^3)(\frac{a^2}{3^3} + \frac{b^2}{4^3} + \frac{c^2}{5^3})$$ and you can check that $3^3+4^3+5^3=6^3$.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How check if the sequence $x_n=(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing or not? I want to check if the sequence $(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing. I tried computing $\frac{x_{n+1}}{x_n}$ to check if the ratio is less than 1 or greater than 1, but I am unable to simplify:
$$\frac{x_{n+1}}{x_n} = \frac{n^{\frac{1}{n}}\left(n+2\right)^{\frac{1}{n+1}}}{\left(n+1\right)^{\frac{2n+1}{n\left(n+1\right)}}}$$
I also tried $x_{n+1} - x_n = \left(\frac{n+2}{n+1}\right)^{\frac{1}{n+1}}-\left(\frac{n+1}{n}\right)^{\frac{1}{n}}$
Is there any other way to check the monotonic behaviour of the sequence $x_n$?
Also I would like to know if I can check $y_n=(1-\frac{1}{n})^{\frac{1}{n}}$ using similar arguments?
| The sequence decreases if and only if $\frac{\ln\left(1+\frac1n \right)}{n}$ decreases.
let $f(x) = \frac{\ln \left( 1+\frac1x \right)}{x}$ then
\begin{align}
f'(x) &= \frac{x \cdot \frac{1}{1+\frac1x}\cdot \left( -\frac1{x^2} \right) - \ln \left( 1 +\frac1x \right)}{x^2} \\
&= \frac{-\frac1{x+1}-\ln\left( 1+\frac1x\right)}{x^2}
\end{align}
which is negative for $x>0$.
If we consider $$g(x) = \frac{\ln \left( 1-\frac1x \right)}{x}$$
we have
\begin{align}
g'(x) &= \frac{x \cdot \frac{1}{1-\frac1x}\cdot \left( \frac1{x^2} \right) - \ln \left( 1 -\frac1x \right)}{x^2} \\
&= \frac{\frac1{x-1}-\ln\left( 1-\frac1x\right)}{x^2}
\end{align}
which is positive for $x>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Then value of $\alpha^2 +4\alpha$ in Infinite series
If $\displaystyle \alpha = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty.$
Then value of $\alpha^2 +4\alpha$ is
Try: Let $$S = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$
$$S+1 = 1+\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$
Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{6\cdot 3!}+\cdots \cdots$$
So $\displaystyle nx=\frac{5}{6}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{35}{27}$
So $$\frac{nx(nx-x)}{2}=\frac{5}{12}\cdot \frac{5-6x}{6}=\frac{35}{27}$$
So $\displaystyle x=-\frac{41}{18}$ and $\displaystyle n=-\frac{15}{41}$
I am getting $\displaystyle S+1=\bigg(1-\frac{41}{18}\bigg)^{-\frac{15}{41}}$
but answer of $\alpha^2+4\alpha = 23$
which is not possible from my answer. could some help me how can i solve it, thanks
| Using
\begin{align}
\prod_{k=1}^{n} (2 k +3) &= 2^{n} \, \prod_{k=1}^{n} \left(k + \frac{3}{2}\right) = \frac{2^{n} \, \left(n + \frac{3}{2}\right)!}{\left(\frac{3}{2}\right)!} \\
(a)_{n} &= \frac{\Gamma(n+a)}{\Gamma(a)} \\
\sum_{n=0}^{\infty} \frac{(a)_{n} \, x^n}{n!} &= (1-x)^{-a}
\end{align}
then the following is determined:
\begin{align}
\alpha &= \sum_{k=1}^{\infty} \frac{\prod_{s=1}^{k} (2s + 3) }{(k+1)! \, 3^{k}} \\
&= \frac{1}{\Gamma(5/2)} \, \sum_{k=1}^{\infty} \frac{\Gamma\left(k + \frac{5}{2}\right) }{(k+1)!} \, \left(\frac{2}{3}\right)^{k} \\
&= \frac{3}{2} \, \frac{\Gamma(3/2)}{\Gamma(5/2)} \, \sum_{k=2}^{\infty} \frac{(3/2)_{k}}{k!} \, \left(\frac{2}{3}\right)^{k} \\
&= \left(1 - \frac{2}{3}\right)^{-3/2} - 1 - 1 \\
&= 3^{3/2} - 2.
\end{align}
Now,
$$\alpha^{2} + 4 \alpha = \alpha (\alpha + 4) = (3^{3/2} -2)(3^{3/2}+2) = 3^{3} - 2^{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How is the sequence $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ going to converge to 1 Having some trouble understanding that if $x_{1} = 4$ and the sequence where n is defined as $x_{n+1} = \frac{(x_n)^{2} + 5}{ 6}$ how is it going to converge to 1.
I have solved using the L as limit and using the quadratic i get two possibilites that are 5, or 1.
| By induction, $1< x_n\le 4$ for all $n$. Indeed, this is true for $n=1$, and if $1< x_n\le 4$, then $1< x_n^2\le 16$ and so $x_{n+1}=\frac{x_n^2+5}{6}$ is $> \frac{1+5}6=1$ and $\le \frac{16+5}6<4$.
With these bounds in mind, we have
$$x_{n+1}-x_n=\frac{x_n^2-6x_n+5}6=\frac{(x_n-1)(x_n-5)}6<0$$
because the first factor in the numerator is positive and the second is negative.
So $\{x_n\}_n$ is bounded and strictly decreasing, hence convergent. You already found out that the only possible limits are $1$ and $5$. However, $5$ can be excluded according to the above observations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Can $pk+1$ divide $(p-k)^2$? Let $p>3$ be a prime, $0<k<p$. Then is it possible that $pk+1\mid(p-k)^2$?
For $k=1$, since $(p-1)^2\equiv(-2)^2\equiv4\pmod{p+1}$, and $p>3$, this is not possible.
And if $pk+1\mid(p-k)^2$ then $k^2-3pk+p^2-1\geq0$, so $k\leq\frac{3p-\sqrt{5p^2+4}}2$.
This arises from another question. I am running out of ideas, so any help is sincerely appreciated, thanks in advance.
| Claim: If $a,b$ are positive integers and $q(a,b) := \frac{(a-b)^2}{ab+1}$ is an integer, it is a square number. This may be proved by Vieta jumping as follows:
Let $k \geqslant 1$ be an integer, which is not a square. Assume that there are positive integers $A,B$ such that $q(A,B) = k$, and choose $A$ and $B$ so that $A \geqslant B$ and that $A+B$ is minimal.
By expanding the statement $q(A,B) = k$, we find that $A$ is a solution to the quadratic $X^2 - B(2+k)X + B^2 - k = 0$. Since its discriminant is positive, there is another solution $C$ which may be expressed using Vieta's formulas as
\begin{align*}
C = B(2+k) - A = \frac{B^2 - k}{A} < A.
\end{align*}
The first representation of $C$ shows that it's an integer, and the other one shows that it's non-zero, since $k$ isn't a square. Also, since
\begin{align*}
\frac{(C-B)^2 }{CB + 1} = k > 0,
\end{align*}
we see that $C$ has to be a positive integer. But then
\begin{align*}
C+ B < A+B,
\end{align*}
contradicting the minimality of $A+B$. $\blacksquare$
This means that if $q(p,t)$ is an integer, there is some integer $m$ with
\begin{align*}
q(p,t) = \frac{(p-t)^2}{pt + 1} = m^2,
\end{align*}
and by estimating $q(p,t)$ (using $1 \leqslant t \leqslant p-1$) it's easy to see that $m^2 \leqslant p-1$.
Rearranging this equation gives
\begin{align*}
p^2 - 2pt + t^2 = m^2 p t + m^2,
\end{align*}
so
\begin{align*}
p \mid p^2 - (2 + m^2) p t = m^2 - t^2 = (m-t)(m+t).
\end{align*}
Given the bounds on $m$ and $t$, the only possibilities here are that $m = t$ or that $m+t = p$. In the first case, our equation is
\begin{align*}
p^2 - 2pt + t^2 = t^3 p + t^2 &\Leftrightarrow p^2 - 2pt - t^3 p = 0 \\
&\Leftrightarrow p = 2t + t^3
\end{align*}
so $t \mid p$, meaning $t = 1$, so $p = 3$, which is impossible. In the second case, we have $m = p-t$, so our equation is
\begin{align*}
(p-t)^2 = p^2 - 2pt + t^2 = (p-t)^2 p t + (p-t)^2 &\Leftrightarrow 1 = pt + 1,
\end{align*}
forcing $t = 0$, which is impossible.
| {
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"url": "https://math.stackexchange.com/questions/2947956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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} |
What is the probability of 4th player to make it to the final? There are 8 players in a tennis tournament. At the beginning of the tournament, the players are matched at random. For each subsequent round, the losers are eliminated and the remaining players are matched at random.
What is the probability that the 4th make it to the finals?
The 4th one only reach the final when it is not allowed to play with first 3 players and the two second best players get eliminated in the second round, and it will only happen when in the first round the three best players play with each other so the number of favourable cases are
$$\frac{4C1}{7C1} \cdot \frac{3C2}{3C2+3C1 \cdot 3C1} \cdot \frac{1}{4C2}$$
| Let's begin by counting the number of possible ways the players could be paired in the first round. The top-ranked player could be paired with any of the seven lower-ranked players. That leaves six players. The top-ranked player among them could be paired with any of the five lower-ranked players left. That leaves four players. The top-ranked player among them could be paired with any of the three lower-ranked players left. The final two players must play each other. Therefore, there are
$$7 \cdot 5 \cdot 3 \cdot 1$$
possible ways to match the players in the first round.
To make it to the finals, the fourth-ranked player must be paired with one of the four lower-ranked players in the first round, then be paired with another of those players in the second round. For that to happen, two of the other three players ranked beneath the fourth-ranked player must play each other in the first round.
With that in mind, we count the favorable cases in the first round. The fourth-ranked player must be paired with one of the four lower-ranked players. Two of the other three lower-ranked player must play each other. That leaves four players, including the top-ranked player. The top-ranked player could be paired with any of three remaining lower-ranked players. The remaining two players must play each other. Hence, the number of favorable ways to match the players in the first round is
$$4 \cdot \binom{3}{2} \cdot 3 \cdot 1$$
Hence, the probability of a favorable outcome in the first round for the fourth-ranked player is
$$\frac{4 \cdot \binom{3}{2} \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1}$$
Assuming that this favorable scenario occurs, there are four players left, including the fourth-ranked player, two higher-ranked players, and one lower-ranked player. To reach the final, the fourth-ranked player must draw the lower-ranked player, which occurs with probability $1/3$.
Hence, the probability that the fourth-ranked player reaches the final is
$$\frac{4 \cdot \binom{3}{2} \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} \cdot \frac{1}{3} = \frac{4}{35}$$
Why is your answer incorrect?
If I understand what you are doing correctly, the first term represents the probability that the fourth-ranked player plays a lower-ranked player in the first round, the second term represents the probability that two of the other three lower-ranked players play each other in the first round, and the third term represents the probability that the fourth-ranked player plays the winner of that match in the second round.
Your first term is correct.
For the second term, There are $5 \cdot 3 \cdot 1$ ways to pair the players who have not already been assigned. There are $\binom{3}{2}$ ways to select which two of the three remaining lower-ranked players play each other and three ways to select which of the higher-ranked players the other lower-ranked player will draw. That gives
$$\frac{\binom{3}{2} \cdot 3}{5 \cdot 3 \cdot 1}$$
However, it looks like you wanted to add the case in which no two of the three remaining lower-ranked players play each other to the case in which that does occur. If no two of the three remaining lower-ranked players play each other, each must play one of the three highest-ranked which can occur in $3!$ ways. That gives
$$\binom{3}{2} \cdot 3 + 3! = 5 \cdot 3 \cdot 1$$
In the second round, four players are left, including the top player. The top player can be matched with any of the three remaining lower-ranked players. The two remaining players must play each other. Hence, there are
$$3 \cdot 1$$
ways to pair the players in the second round. If those players include the fourth-ranked player, two higher-ranked players, and one lower-ranked player (the favorable case for the fourth-ranked player), the probability that the fourth-ranked player draws the lower-ranked player is
$$\frac{1}{3 \cdot 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $A^2 +B^2+C^2=D^2$ using the following diagram (tetrahedron) Slicing a corner off a square gives a right-angled triangle, as shown in the diagram below.
The lengths of the sides of this triangle are related by Pythagoras’s theorem: $a^
2 + b^
2 = c^
2$
. Show
that this two-dimensional setup generalises to three dimensions in the following way. Slice a corner
off a cube, as shown in the diagram below. This gives a tetrahedron in which three of the faces are
right-angled triangles, while the fourth is not. Let’s call the areas of the three right-angled faces $A,
B, C$ and the area of the fourth face $D$.
$A
^2 + B^
2 + C^
2 = D^2$
.
Can anyone explain to me what formula/how to go about doing this question? thank you.
| Let the sliced corner be at the origin and the fourth face lie in the plane $x/a + y/b + z/c = 1$. The foot of the altitude drawn to the fourth face is at $\lambda (1/a, 1/b, 1/c), \,\lambda = 1/(1/a^2 + 1/b^2 + 1/c^2)$. Therefore
$$9 V^2 = A^2 a^2 = B^2 b^2 = C^2 c^2 = D^2 h^2, \\
A^2 + B^2 + C^2 =
D^2 h^2 \left( \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \right) =
D^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Computing the value of $\frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3}\ldots=$? I have tried converting this series into a telescopic sum whose terms could cancel out but haven't succeeded in that effort. How should I proceed further?
| The hint:
Use the telescopic sum and $$(n+2)^2+n=(n+4)(n+1).$$
Now, $$\sum_{n=1}^{+\infty}\frac{1}{(n+2)^2+n}=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{13}{36}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n} \ $$ and that this sum is either less than $2$ or equal to $2$ using C-S.
| Hint.
Define
$$
f_n(x) = \sum_{k=1}^{2n+1}x^{n+k-1}=x^n\left(\frac{x^{2n+1}-1}{x-1}\right)
$$
hence
$$
\int f_n(x) dx = \int x^n\left(\frac{x^{2n+1}-1}{x-1}\right)dx = \frac{x^{n+1}}{n+1}+\cdots+\frac{x^{3n+1}}{3n+1}
$$
and now apply the Cauchy-Schwartz inequality on $u(x) = x^n, v(x) = \frac{x^{2n+1}-1}{x-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Trigonometry problem $\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$
What is the value of:
$$\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$$
I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:
$\large{{\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ}=\\{\sin(90^\circ+10^\circ)+\cos(60^\circ+10^\circ)\over\cos(90^\circ-10^\circ)-\cos(30^\circ+10^\circ)}=\\{\sin90^\circ\cos10^\circ+\cos90^\circ\sin10^\circ+\cos60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ\over\cos90^\circ\cos10^\circ+\sin90^\circ\sin10^\circ-\cos30^\circ\cos10^\circ+\sin30^\circ\sin10^\circ}=\\{\cos10^\circ+{1\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over\sin10^\circ-{\sqrt3\over2}\cos10^\circ+{1\over2}\sin10^\circ}=\\{{3\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over{3\over2}\sin10^\circ-{\sqrt3\over2}\cos10^\circ}}$
Not sure what I should do further with this.
| It is:
$$\frac{\sin 100^\circ +\cos 70^\circ}{\cos 80^\circ-\cos 20^\circ}=\frac{\sin 80^\circ +\sin 20^\circ}{\cos 80^\circ-\cos 20^\circ}=\frac{2\sin 50^\circ\cos 30^\circ}{-2\sin 50^\circ\sin 30^\circ}=-\sqrt{3}.$$
| {
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"source": "stackexchange",
"question_score": "4",
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Prove that $3\cdot 5^{2n+1} +2^{3n+1}$ is divisible by $17$ for all $n ∈ \mathbb{N}$
Use mathematical induction to prove that $3\cdot 5^{2n+1} +2^{3n+1}$
is divisible by $17$ for all $n ∈ \mathbb{N}$.
I've tried to do it as follow.
If $n = 1$ then $392/17 = 23$.
Assume it is true when $n = p$. Therefore $3\cdot 5^{2p+1} +2^{3p+1} = 17k $ where $k ∈ \mathbb{N} $. Consider now $n=p+1$. Then
\begin{align}
&3\cdot 5^{2(p+1)+1} +2^{3(p+1)+1}=\\
&3\cdot 5^{2p+1+2} + 2^{3p+1+3}=\\
&3\cdot5^{2p+1}\cdot 5^{2} + 2^{3p+1}\cdot 2^{3}.
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
| Having shown that
$$3\cdot5^{2p+1}+2^{3p+1}=17k$$
we have (continuing from the last line)
$$3\cdot5^{2p+1}5^{2}+2^{3p+1}2^{3}=17(3\cdot5^{2p+1})+8(3\cdot5^{2p+1}+2^{3p+1})=17(3\cdot5^{2p+1}+8k)$$
so the next expression is also divisible by 17, as required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding value of infinite series limit
Finding value of $\displaystyle \lim_{n\rightarrow \infty}\frac{2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+........+\sqrt{2}}}}}_{\bf{n\; times}}}{4^{-n}}$
Try: I am trying to convert it into $\cos$ ine series sum
Let $$\displaystyle \sqrt{2+2\cos \theta } = 2\cos \frac{\theta}{2}$$ and
$$\displaystyle \sqrt{2+\sqrt{2+2\cos \theta}} = 2\cos \frac{\theta}{4}$$
$$\displaystyle \sqrt{2+\sqrt{2+\sqrt{2+2\cos \theta}}} = 2\cos \frac{\theta}{8}$$
could some help me how i write $$\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+........+\sqrt{2}}}}}_{\bf{n\; times}}$$ into cosine series form. thanks
| $$
\eqalign{
& \underbrace {\sqrt 2 }_1 = 2\cos \left( {\pi /4} \right) \cr
& \underbrace {\sqrt {2 + \sqrt 2 } }_2 = \sqrt {2 + 2\cos \left( {\pi /4} \right)} = \cr
& = \sqrt {2\left( {1 + \cos \left( {\pi /4} \right)} \right)} = \sqrt {2\left( {1 + \cos ^2 \left( {\pi /8} \right) - \sin ^2 \left( {\pi /8} \right)} \right)} = \cr
& = \sqrt {4\cos ^2 \left( {\pi /8} \right)} = 2\cos \left( {\pi /8} \right) \cr
& \quad \quad \vdots \cr
& \underbrace {\sqrt {2 + \sqrt { \cdots + \sqrt 2 } } }_n = 2\cos \left( {\pi /2^{n + 1} } \right) \cr}
$$
Then probably you are looking for the value of $\lim_ {n \to \infty}{(2-\sqrt{2+\sqrt{\cdots}})4^n}$.
| {
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"url": "https://math.stackexchange.com/questions/2957628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational?
The number $x$ defined below is rational or irrational?
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$
From: IMO 1973 - Longlist
My attempt (my real question is at the end):
the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-ac-bc)$ when $a+b+c=0$, leads to $$a^3+b^3+c^3=3abc \tag{1}$$
Now considering
$$a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2},c=-x$$
from (1) it is true that
$$x^3-3x-2\sqrt{5}=0 \tag{2}$$
That is the number $x$ is a root from (2).
Note: By trial and error I've found that answer is $x=\sqrt{5}$ (the other 2 roots are complex), that is irrational. But my question is more subtle.
Question: Can I conclude just inspecting (2), judging by the coefficient $2\sqrt{5}$, that $x$ is irrational, without actually solving the equation? In a math contest that might be helpful, if possible, as it would avoid extra steps.
| Yes you can.
Notice that we have :
$$x^3-3x -2\sqrt{5} = 0 \Leftrightarrow x(x^2-3) = 2\sqrt{5}$$
Hence if $x$ is rational, $x^2-3$ is rational so $x(x^2-3)$ is rational.
Hence because $2\sqrt{5}$ is irrational, that’s a contradiction and $x$ is not rational.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Difficulty with factorizing determinant I came across this determinant and its just proved difficult to express as a product of in linear factors
$$\begin{vmatrix} 1 & 1 & 1 \\a^2 & b^2 & c^2 \\(b+c)^2 & (c+a)^2 & (a+b)^2 \end{vmatrix}$$
I have tried the basic operation combining rows and column but I can't just get to the answer $2(a-b)(b-c)(a+b+c)(c-a)$.
I really need a guide or hints to proceed.
| Let us call $\Delta$ this determinant. Do the following operation: replace the last column by its difference with the second one ($C_3\leftarrow C_3-C_2$) to get
$$
\Delta=\begin{vmatrix} 1 & 1 &0\\a^2 & b^2 & c^2-b^2 \\(b+c)^2 & (c+a)^2 & (a+b)^2-(c+a)^2 \end{vmatrix}.
$$
Since $(a+b)^2-(c+a)^2=\left(b-c\right)(2a+b+c)$, we can use multilinearity of the determinant to get
$$
\Delta=(b-c)\begin{vmatrix} 1 & 1 &0\\a^2 & b^2 & -(b+c) \\(b+c)^2 & (c+a)^2 & 2a+b+c \end{vmatrix}.
$$
Then We do $C_2\leftarrow C_2-C_1$ to get
$$
\Delta=(b-c)(a-b)\begin{vmatrix} 1 & 0&0\\a^2 & -(a+b) & -(b+c) \\(a+c)^2 & a+b+2c & 2a+b+c \end{vmatrix}.
$$
Expanding with respect to the first line gives
$$
\Delta=(b-c)(a-b)\begin{vmatrix} -(a+b) & -(b+c) \\ a+b+2c & 2a+b+c \end{vmatrix}.
$$
Doing $L_2\leftarrow L_2+L_1$ gives
$$
\Delta=(b-c)(a-b)\begin{vmatrix} -(a+b) & -(b+c) \\ 2c & 2a \end{vmatrix}
=2(b-c)(b-a)\begin{vmatrix} a+b & b+c \\ c & a \end{vmatrix}.
$$
Finally, do $C_2\leftarrow C_2-C_1$ to get the wanted result.
| {
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"timestamp": "2023-03-29T00:00:00",
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what is the probability that drawn balls have the same number? In the urn, there is one ball with number $1$, $2$ with number $2$, and so on until $n$ balls with the number $n$.
From the urn, we draw two balls.
Calculate the probability that the two drawn balls have the same number.
| Overall there are $N=1+2+\cdots+n=\frac{n(n+1)}{2}$ balls.
You need to calculate:
$$\begin{align}P(2\cap 2)+P(3\cap 3)+\cdots +P(n\cap n)=\\
\frac{2}{N}\cdot \frac1{N-1}+\frac{3}{N}\cdot \frac2{N-1}+\cdots+\frac{n}{N}\cdot \frac{n-1}{N-1}=\\
\frac{1}{N(N-1)}\sum_{k=2}^n k(k-1)=\\
\frac{1}{N(N-1)}\sum_{k=1}^n k(k-1)=\\
\frac1{N(N-1)} \left[\sum_{k=1}^n k^2-\sum_{k=1}^nk\right]=\\
\frac1{N(N-1)} \left[N\cdot \frac{2n+1}{3}-N\right]=\\
\frac{2(n-1)}{3(N-1)}=\\
\frac{4(n-1)}{3(n(n+1)-2)}=\\
\frac{4}{3(n+2)}.\\
\end{align}$$
| {
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How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$ So, I was watching this video by blackpenredpen where he mentions that $$\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$$ so I wanted to try and prove it myself.
Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$
But$\sqrt {a-\sqrt {a+\underbrace{\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}_x}}=x$
$\therefore \sqrt {a-\sqrt {a+x}}=x$
$a-\sqrt {a+x}=x^2$
$x^2-a=-\sqrt {a+x}$
$x^4-2ax^2+a^2=a+x$
$x^4-2ax^2-x+a^2-a=0$
Note that this is of the form $y^4+py^2+qy+r=0$ so we can use Ferrari-Cardano.
We need to find a $z$ such that $(2z-p)y^2-qy+(z^2-r)$ has a discriminant of $0$. The discriminant of $(2z-p)y^2-qy+(z^2-r)$ is equal to $q^2 - 4(2z - p)(z^2 - r),$ which simplifies to $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$
Substituting values from $x^4-2ax^2-x+a^2-a=0$ into $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$ gives us
$8z^3-4\cdot(-2a)\cdot z^2-8\cdot(-a)\cdot z+\left(4\cdot (-2a) \cdot (-a) - (-1)^2 \right)=0 \implies 8z^3+8az^2+8az+(8a^2-1)=0$
Using Cardano's formula, or in my case Wolfram Alpha, we get that $$z_1 = \frac {\sqrt [3]{-16 a^3 - 144 a^2 + 3 \sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}}{6\sqrt[3]2} - \frac {192 a - 64 a^2}{48\cdot 2^{\frac 23} \sqrt [3]{-16 a^3 - 144 a^2 + 3\sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}} - \frac a3$$
I simply can not solve that quintic and continue as it is already too cluttered. Was there a mistake in my problem or is there any other way to do it? Also, I am sincerely sorry but I am not sure how to tag this question.
Edit $1:$
After Ross Millikan's answer, I snooped around in the comment section of the video and found someone who found that it is true using alternating root series. Was his proof correct as $\frac {\sqrt {4a-3}-1}2$ does not seem to have real values for $a \lt \frac 34$? Thank you!
| It is CORRECT!
$x^4-2ax^2-x+a^2-a=0$ can be factorised as $(x^2-x-1)*(x^2+x+1-a)$. You'll get the desired answer from the right factor $(x^2+x+1-a)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Your evil probability professor has an urn with 9 balls. Your evil probability professor has an urn with 9 balls: 2 red, 3 white and 4 blue. He draws two balls from the urn without replacement. Let X be the number of red balls drawn and Y the number of white balls.
a) Determine the joint probability mass function of X and Y.
b) Are X and Y independent random variables?
c) Compute the covariance between X and Y.
For point A:
$P(0,0)=\frac{4}{9} \cdot \frac{3}{8} = \frac{1}{6}$ That is correct for the solution.
$P(0,1)=\frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{4}{8} = \frac{1}{3}$ That is correct for the solution.
$P(1,0)=\frac{2}{9} \cdot \frac{4}{8} + \frac{4}{9} \cdot \frac{2}{8}= \frac{2}{9}$ That is correct for the solution.
$P(1,1)=\frac{2}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{2}{8}= \frac{1}{6}$ That is correct for the solution.
$P(2,0)=\frac{2}{9} \cdot \frac{1}{8} = \frac{1}{36}$ That is correct for the solution.
$P(0,2)=\frac{3}{9} \cdot \frac{2}{8} = \frac{1}{12}$ That is correct for the solution.
For point B to check the independancy I have just to check if for example $P(X=0,Y=0) = P(X=0) \cdot P(Y=0)$.
$\frac{1}{6} \neq (\frac{7}{9} \cdot \frac{6}{8}) \cdot (\frac{6}{9} \cdot \frac{5}{8})$.
So X and Y are not independent.
For point C I know that the covariance $Cov(X,Y)=E[X \cdot Y]-E[X]\cdot E[Y]$, but how can compute the expectations, do I have to figure put with distribution is? How can I do it?
Can someone help me?
Thanks in advance, Fabio!
| Another way to check dependence (i.e. part B only) is that if $X,Y$ are independent then for any values $x,y$ we have $P(Y=y) = P(Y=y | X=x)$. However, clearly if there are 2 red balls then there cannot be any white balls, i.e. $P(Y > 0) > 0$ but $P(Y > 0 | X=2) = 0$. This might be a more intuitive example demonstrating dependence.
| {
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Solve this question Question : $$\begin{align}a+b+c &= 0\\
a^3 +b^3 +c^3 &= 12\\
a^5 +b^5 +c^5 &= 40\end{align}$$
Then , $a^4 + b^4 + c^4 = ?$
My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think
$3abc = 12$ and got $abc = 4$
After that I am unable to link it with any other equation
| Hint: $$(a+b+c)^4=a^4+b^4+c^4+4a^3(b+c)+4b^3(a+c)+4c^3(a+b)+6(a^2b^2+b^2c^2+c^2a^2)+12abc(a+b+c)=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of an angle at the same point in a parametric curve
What is the value of $\cos\theta$ for the line defined by the parametric equation $(x, y, z)=\left(-t^2+4t+1,\:3\cos\left(\pi t\right),\:t^3-21t\right)$ at the point where it crosses itself?
So I let the two variables be $s$ and $t$. So when the line crosses itself, $t=5$ and $s=-1$. The derivative of the above equation is $\left(-2t+4,\:-3\pi \sin\left(\pi t\right),\:3t^2-21\right)$. So I plugged in the values for $s$ and $t$ and got that for $t$ the vector was $(-6, 0, 54)$ and for $s$ the vector was $(6, 0, -18)$.
So to calculate $\cos\theta$, I did the following:
$\frac{\left(-6,\:0,\:54\right)\cdot \left(6,\:0,\:-18\right)}{\sqrt{\left(-6\right)^2+\left(54\right)^2}\sqrt{6^2+\left(-18\right)^2}}$ which gave me a value of $\frac{-14}{\sqrt{205}}$, which is incorrect.
Any help?
| Your point is correct
*
*$-t^2+4t+1=-s^2+4s+1 \implies t^2-s^2=4(t-s) \implies t+s=4$
*$t^3-21t=s^3-21s \implies t^3-s^3=21(t-s) \implies t^2+st+s^2=21$
that is $t=5$ ans $s=-1$ and also the evaluation for the tangent vectors and the final step, indeed we obtain
$$\frac{\left(-6,\:0,\:54\right)\cdot \left(6,\:0,\:-18\right)}{\sqrt{\left(-6\right)^2+\left(54\right)^2}\sqrt{6^2+\left(-18\right)^2}}=\frac{-1008}{72\sqrt{205}}=\frac{-14}{\sqrt{205}}$$
| {
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Prove that $1/n+1/(n+1)+\dots+(1/2n)>2/3$
Prove that $\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$
I tried to use mathematical induction, but I'm not able to prove that:
$$
\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3}.
$$
My method was:
Assumption:
$$\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$$
$$\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}$$
$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n}$$
Now in order to prove the thesis I have to prove that
$$\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n} > \frac{2}{3}$$
But it's a contradiction. Did I make a mistake somewhere? How can I solve this problem? I'd appreciate your help.
| There is also a "non-inductive" way to show the inequality.
You may transform the sum into a Riemann-sum and then estimate this sum by an integral:
$$1/n+1/(n+1)+\dots+(1/2n) = \frac{1}{n}\left(\frac{1}{1+\frac{0}{n}} + \frac{1}{1+\frac{1}{n}} + \cdots + \frac{1}{1+\frac{n}{n}} \right) =\boxed{\sum_{k=0}^n \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n}}$$
Now, note that $f(x) = \frac{1}{1+x}$ is strictly decreasing $[0,1]$. So, you have
*
*$\frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} = \int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{1}{1+\frac{k}{n}} \; dx > \int_{\frac{k}{n}}^{\frac{k+1}{n}} \frac{1}{1+x}\; dx$
This gives
$$\boxed{\sum_{k=0}^n \frac{1}{1+\frac{k}{n}} \cdot \frac{1}{n} > \int_0^1 \frac{1}{1+x}\; dx = \ln 2 > \frac{2}{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Upper bound $f(n) \leq c\alpha^n$ with recurrence $f(n) = 2 \cdot f(n-1) + f(n-2)$ with $f(1) = 1$ and $f(2) = 2$ I have following recurrence: $f(n) = 2 \cdot f(n-1) + f(n-2)$ with $f(1) = 1$ and $f(2) = 2$.
How can I find an upper bound with the form $f(n) \leq c\alpha^n$?
I know that in this task the value of $c$ does not have to be determined.
| As it was suggested in the comments, there is a technique to solve this type of problems (here is an example, you will find a lot more on MSE). By noting $a_n=f(n)$, we have a homogeneous recurrence
$$a_{n}=2a_{n-1}+a_{n-2}$$
which can be solved using characteristic polynomial technique. In this case, the polynomial is
$$x^2-2x-1=0$$
It has $1-\sqrt{2}$ and $1+\sqrt{2}$ as solutions, thus the general term of the recurrence is
$$a_n=A(1-\sqrt{2})^n+B(1+\sqrt{2})^n \tag{1}$$
Given initial conditions $a_1=1, a_2=2$ we have
$$1=A(1-\sqrt{2})+B(1+\sqrt{2})$$
$$2=A(1-\sqrt{2})^2+B(1+\sqrt{2})^2$$
leading to $A=-\frac{\sqrt{2}}{4}$ and $B=\frac{\sqrt{2}}{4}$
or
$$a_n=\frac{\sqrt{2}}{4}\left(1+\sqrt{2}\right)^n-\frac{\sqrt{2}}{4}\left(1-\sqrt{2}\right)^n \tag{2}$$
As a result
$$f(n)<\frac{\sqrt{2}}{2}\left(1+\sqrt{2}\right)^n$$
because $|1-\sqrt{2}|<1$ and $\left(1-\sqrt{2}\right)^n \rightarrow 0$ when $n\rightarrow\infty$.
| {
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How can I deduce exact values of $\cos\left(\frac{\pi}{12}\right)$ and $\sin\left(\frac{\pi}{12}\right)$? I have been struggling on the problem for some time; I would like a gentle tip on answering the question:
Given the equation $z^4 = 1 + \sqrt{3}i $ where $ z \in \mathbb{C}$ , deduce the exact values of $\cos\left(\frac{\pi}{12}\right)$ and $\sin\left(\frac{\pi}{12}\right)$.
So far I am proceeding in the following manner:
I pose $Z' = z^2$
therefore I have $Z'^2 = 1 + \sqrt{3}i $
I calculate that $Z'$ = $\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$
Right now I have to solve for $Z' = z^2 \iff z^2= \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$
After solving this equation I get that $ z = \frac{1}{2}\sqrt{\sqrt{6}+1} + \frac{1}{2}\sqrt{1-\sqrt{6}}i$
If I am taking that $\cos\left(\frac{\pi}{12}\right) = \Re(z)$ and $\sin\left(\frac{\pi}{12}\right) = \Im(z)$, certainly my solution is wrong.
| You start with $z^4=2(\cos\frac\pi3+i\sin\frac\pi3)$, so your final $z$ should be equal to $\sqrt[4]2(\cos\frac\pi{12}+i\sin\frac\pi{12})$. So, to get your cosine and sine divide by $\sqrt[4]2$ in the end. But your $z$ is wrong (square it); I get $\frac12\sqrt{2\sqrt2+\sqrt6}+\frac i2\sqrt{2\sqrt2-\sqrt6}$
| {
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For every natural number $m$, $2^{2m}-1$ is divisible by $3$ Can anyone tell me if I missed any small details in this proof?
(basis) $2^{2(1)}-1 = 3*1$ which is divisible by 3.
(induction) Fix $m>1$ so $p<m$. Hence for every natural number $p,$ $2^{2p}-1=3p.$ then for every natural number $(p+1),$ $2^{2(p+1)}-1=3(p+1)$ then
$2^{2(p+1)}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2(3p+1)-1=3(p+1)$ by induction hypothesis
$2(3p+1)+2^2-3=3(p+1)$
$6p+3=3(p+1)$
$3(p+1)=3(p+1)$
Thus, the statement is true for the natural number n whenever it is true for any natural number less than n.
| An option:
$m \ge 1.$
$2^{2m}-1 = (2^2)^m -1=$
$4^m -1= (1+3)^m-1.$
In the binomial expansion of $(1+3)^m$ all terms except the first term have a factor $3$, i.e.
$(1+3)^m=1^m +3k$, where $k \in \mathbb{Z^+}$.
Hence
$2^{2m}-1 = (1^m +3k) -1=3k .$
| {
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Prove $(1+\sqrt{2})^k= \sqrt{n}+\sqrt{n-1}$
Prove that for every $k \in \mathbb{N}$ there exists $n\in \mathbb{N}$ such that $(1+\sqrt{2})^k= \sqrt{n}+\sqrt{n-1}$
I tried to prove it using induction, but I could not move to the next step after assuming the result hold for some $k=m$.
| This follows from $(1-\sqrt{2})^k\,(1+\sqrt{2})^k=(-1)^k$ for every integer $k$. So, if $a_k$ and $b_k$ are integers such that $(1+\sqrt{2})^k=a_k+b_k\sqrt{2}$, then $a_k-b_k\sqrt{2}=(1-\sqrt{2})^k$, making $$a_k^2-2\,b_k^2=(-1)^k\,.$$
Let now $k$ be a nonnegative integer, so $a_k$ and $b_k$ are both nonnegative. If $k$ is even, set $n:=a_k^2=2b_k^2+1$. If $k$ is odd, set $n:=a_k^2+1=2\,b_k^2$. This gives
$$(1+\sqrt{2})^k=\sqrt{n}+\sqrt{n-1}\,.$$
Similar things happen with other quadratic radicals as well. For an example, for every positive integer $k$, there exists a positive integer $n$ such that $$(2+\sqrt{3})^k=\sqrt{n}+\sqrt{n-1}\,,$$ and the same goes for $(2+\sqrt{5})^k$. For another example, for every positive integer $k$, there exists a positive integer $n$ such that $$(5+2\sqrt{6})^k=\sqrt{n}+\sqrt{n-1}\,.$$ For one last example, for every positive integer $k$, there exists a positive integer $n$ such that $$(8+3\sqrt{7})^k=\sqrt{n}+\sqrt{n-1}\,.$$
This question is related to Pell's Equation $x^2-dy^2=\pm1$, where $d\in\mathbb{Z}_{>0}$ is non-square.
It turns out that, for any positive integer $p$ and for any nonnegative integer $k$, $$\left(\sqrt{p+1}+\sqrt{p}\right)^k=\sqrt{p_k+1}+\sqrt{p_k}$$
for some nonnegative integer $p_k$. Clearly, $p_0=0$ and $p_1=p$. (Obviously, this implies also that $\left(\sqrt{p+1}+\sqrt{p}\right)^{-k}=\sqrt{p_k+1}-\sqrt{p_k}$ for all nonnegative integers $k$.)
The case $k$ is even follows by observing that
$$\left(\sqrt{p+1}+\sqrt{p}\right)^2=\left(2p+1+\sqrt{4p(p+1)}\right)^2\,,$$
and so
$$\left(\sqrt{p+1}+\sqrt{p}\right)^k=\left(2p+1+\sqrt{4p(p+1)}\right)^{\frac{k}{2}}\,.$$
It is not difficult to verify that, for any integer $m$,
$$\left(2p+1+\sqrt{4p(p+1)}\right)^m=a_m+b_m\sqrt{4p(p+1)}$$
for some integers $a_m$ and $b_m$. This implies
$$\left(2p+1-\sqrt{4p(p+1)}\right)^m=a_m-b_m\sqrt{4p(p+1)}\,.$$
Thus, $a_m^2-4p(p+1)\,b_m^2=1$. Consequently, we take
$$p_k=4p(p+1)\,b_{\frac{k}{2}}^2
\text{ which gives }p_k+1=a_{\frac{k}{2}}^2\,,$$
when $k$ is even.
Now, we solve the case where $k$ is odd. We note that
$$(\sqrt{p+1}+\sqrt{p})^k=\left(2p+1+\sqrt{4p(p+1)}\right)^{\frac{k-1}{2}}\,(\sqrt{p+1}+\sqrt{p})\,.$$
Hence,
$$(\sqrt{p+1}+\sqrt{p})^k=\left(a_{\frac{k-1}{2}}+b_{\frac{k-1}{2}}\,\sqrt{4p(p+1)}\right)\,(\sqrt{p+1}+\sqrt{p})\,.$$
Ergo,
we can take
$$p_k:=\left(a_{\frac{k-1}{2}}+2(p+1)\,b_{\frac{k-1}{2}}\right)^2\,p$$
so that
$$p_k+1=\left(a_{\frac{k-1}{2}}+2p\,b_{\frac{k-1}{2}}\right)^2\,(p+1)\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence.
Let $n\in \mathbb N$ and:
$$
x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n
$$
Show that $\{x_n\}$ is a bounded sequence.
This sequence appears a bit tricky because it involves harmonic series. Below are steps I take.
Lower bound:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \ge \sqrt{\sum_{k=1}^n\left(\frac{k}{k}\right)^2} = \sqrt{\sum_{k=1}^n1}=\sqrt{n}\implies\\
\implies x_n \ge \sqrt n - \sqrt n \ge 0
$$
Lower bound is simple.
Upper bound:
To get rid of radical lets use Cauchy-Schwarz (note the below is incorrect as shown in Zvi's answer):
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \le \sqrt{\left(\sum_{k=1}^n\left(\frac{k+1}{k}\right)\right)^2} =\sum_{k=1}^n\left(\frac{k+1}{k}\right) = n+\sum_{k=1}^n{1\over k} = n + H_n
$$
So this doesn't show $x_n$ is bounded above. I've tried another approach:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}
$$
Consider nominator:
$$
\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=n+\sum_{k=1}^n{2\over k}+\sum_{k=1}^n{1\over k^2}-n = \sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}
$$
For denominator:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n = \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n
$$
So $x_n$ is:
$$
x_n = \frac{\sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}}{ \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n }
$$
But i don't see how to proceed from this point. What else could i try? How to show $x_n$ is bounded above?
Please note the precalculus tag.
| Even if $\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n + 2H_n + H_n^{(2)}$ involves (generalized) harmonic numbers, it is pretty straightforward to notice$^{(*)}$ that $H_n=O(\log n)$ and $H_{n}^{(2)}=O(1)$ as $n\to +\infty$, hence
$$ \sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n\left(1+O\left(\frac{\log n}{n}\right)\right),\tag{1}$$
$$ \sqrt{\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2} = \sqrt{n}\left(1+O\left(\frac{\log n}{n}\right)\right)=\sqrt{n}+O\left(\frac{\log n}{\sqrt{n}}\right)\tag{2}$$
and the wanted limit is zero.
$(*)$ $H_n \leq 1+\int_{1}^{n}\frac{dx}{x}=1+\log(n)$ and $H_{n}^{(2)}\leq 1+\sum_{k=1}^{n}\frac{1}{k(k+1)}\leq 2$.
| {
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Can we proof that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?; given $0 < a < b < 1$ I have found in some test problem:
Given $0 < a < b < 1$, can we conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?
I divide the combination of a and b into a few cases. Then, see what happens.
case 1 : $a \rightarrow 0$ and $b \rightarrow 0$
When $a$ approaches to zero and $b$ approaches to $a$ (which is zero), the expression will be evaluated as $\sqrt{0 + 0}$ and $\sqrt{0} + \sqrt{0}$. Since $0 = 0$, this case cannot conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 2 : $a \rightarrow 1$ and $b \rightarrow 1$
When $b$ approaches to one and $a$ approches to $b$ (which is one), the expression will be evaluated as $\sqrt{1 + 1}$ and $\sqrt{1} + \sqrt{1}$. Since
$\sqrt{2} < 2$, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 3 : $a \rightarrow 0$ and $b \rightarrow 1$
When $a$ approaches to zero and $b$ approches to one, the expression will be evaluated as $\sqrt{0 + 1}$ and $\sqrt{0} + \sqrt{1}$. Since
$1 = 1$, then this case cannot concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 4 : $a \rightarrow b$
When $a$ approaches to $b$ the expression will be evaluated as $\sqrt{b + b}$ and $\sqrt{b} + \sqrt{b}$. Since $\sqrt{2b} < 2\sqrt{b} $, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
From above cases, am I still missing some point? If not, How should I write the conclusion from that pieces of thinking mathematically? Because I instinctively believe that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$ should be true (and not $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$).
| This proof works for all $a,b\geq 0$. Consider two vectors in the $2$-dimensional Euclidean plane $\mathbb{R}^2$ (with usual Euclidean norm $\Vert\bullet\Vert$): $u=(\sqrt{a},0)$ and $v=(0,\sqrt{b})$. We have $\Vert u\Vert+\Vert v\Vert\geq\Vert u+v\Vert$ by the triangle inequality. This gives
$$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}.$$
Since the equality occurs if, and only if, $u$ and $v$ are parallel, we conclude that the only equality cases are (1) $a=0$ and (2) $b=0$. Hence, if $a,b>0$,
$$\sqrt{a}+\sqrt{b}>\sqrt{a+b}.$$
| {
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"url": "https://math.stackexchange.com/questions/2982025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Woodbury's formula with block-diagonal matrix From Woodbury's formula, it is easy to show that
\begin{align*}
(a \mathbf{I}_n + b J_n)^{-1} = \frac{1}{a}\mathbf{I}_n - \frac{b}{a(a+bn)}J_n
\end{align*}
where $\mathbf{I}_n$ is the $n\times n$ identity matrix and $J_n$ is the $n\times n$ matrix of 1's. Is there a similarly nice form for
\begin{align*}
(a\mathbf{I}_n + b J_{n_1, \cdots, n_k} + c J_n)^{-1}
\end{align*}
where $J_{n_1, \cdots, n_k} = \text{BlockDiag}(J_{n_1}, \cdots, J_{n_k})$ so that $\sum_{i=1}^{k}n_i = n$?
| Woodbury's matrix identity is a generalization of the Sherman–Morrison formula. The Sherman–Morrison formula tells you how to compute the inverse of $A$ plus a rank-$1$ matrix if you already have the inverse of $A$ computed. The Woodbury matrix identity tells you how to compute the inverse of $A$ plus a rank-$r$ matrix if you already have the inverse of $A$ computed.
Here, you can check that $J_{n_1,\ldots,n_k}$ is a rank-$k$ matrix and $J_n$ is a rank-$1$ matrix. So we are trying to compute the inverse of $aI_n$ plus a rank-$(k+1)$ matrix.
Let $A = aI_n$, $U = \begin{bmatrix}\vec{1}_{n_1}& & & & \vec{1}_{n_1} \\ & \vec{1}_{n_2} & & & \vec{1}_{n_2} \\ & & \ddots & & \vdots \\ & & & \vec{1}_{n_k} & \vec{1}_{n_k}\end{bmatrix}$, and $C = \text{diag}(\underbrace{b,\ldots,b}_{k \ b's},c)$.
You can check that $aI_n+bJ_{n_1,\ldots,n_k}+cJ_n = A+UCU^T$. Thus,
\begin{align*}
(aI_n+bJ_{n_1,\ldots,n_k}+cJ_n)^{-1} &= (A+UCU^T)^{-1}
\\
&= A^{-1}-A^{-1}U(C^{-1}+U^TA^{-1}U)^{-1}U^TA^{-1}
\\
&= \dfrac{1}{a}I_n-\dfrac{1}{a}I_nU\left(C^{-1}+U^T\dfrac{1}{a}I_nU\right)^{-1}U^T\dfrac{1}{a}I_n
\\
&= \dfrac{1}{a}I_n-\dfrac{1}{a^2}U\left(C^{-1}+\dfrac{1}{a}U^TU\right)^{-1}U^T
\end{align*}
It is not hard to compute $U^TU = \begin{bmatrix}n_1 & & & & n_1\\ & n_2 & & & n_2 \\ & & \ddots & & \vdots \\ & & & n_k & n_k \\ n_1 & n_2 & \cdots & n_k & n\end{bmatrix}$, so $C^{-1}+\dfrac{1}{a}U^TU = \begin{bmatrix}\tfrac{n_1}{a}+\tfrac{1}{b} & & & & \tfrac{n_1}{a}\\ & \tfrac{n_2}{a}+\tfrac{1}{b} & & & \tfrac{n_2}{a} \\ & & \ddots & & \vdots \\ & & & \tfrac{n_k}{a}+\tfrac{1}{b} & \tfrac{n_k}{a} \\ \tfrac{n_1}{a} & \tfrac{n_2}{a} & \cdots & \tfrac{n_k}{a} & \tfrac{n}{a}+\tfrac{1}{c}\end{bmatrix}$.
Computing the inverse of $C^{-1}+\dfrac{1}{a}U^TU$ can be done using the Sherman-Morrison formula $$(D+vv^T)^{-1} = D^{-1} + \dfrac{D^{-1}vv^TD^{-1}}{1+v^TD^{-1}v}$$ where $D = \text{diag}(\tfrac{n_1}{a}+\tfrac{1}{b},\ldots,\tfrac{n_k}{a}+\tfrac{1}{b},\tfrac{n}{a}+\tfrac{1}{c})$ and $v = \begin{bmatrix}\tfrac{n_1}{a} & \tfrac{n_2}{a} & \cdots & \tfrac{n_k}{a} & 0\end{bmatrix}^T$.
After doing this tedious computation, computing $\dfrac{1}{a}I_n-\dfrac{1}{a^2}U\left(C^{-1}+\dfrac{1}{a}U^TU\right)^{-1}U^T$ shouldn't be too much harder.
Unfortunately, I'm too tired to do this right now, but hopefully this answer is helpful.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$
My Work
I was able to prove identity using half angle formula and $\cos3\theta $ expansion.
Since
$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$
$$\therefore \cos4\theta=\cos3\theta$$
I cannot prove the final part.
Please help me. Thanks in advance.
| Since you have obtained sufficient information regarding the route you take to solve the problem, I am giving a different solution. It is a sketch.
Let $\omega:=\exp\left(\dfrac{2\pi\text{i}}{7}\right)$. Show that $\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$. This implies
$$\begin{align}\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)&=\frac{\omega+\omega^{-1}}{2}+\frac{\omega^2+\omega^{-2}}{2}+\frac{\omega^{3}+\omega^{-3}}{2}\\&=\frac{(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1)-\omega^3}{2\omega^3}=-\frac12\,.\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area bounded by circle $x^2+y^2=4$, $y=x^2+x+1$ and the curve $ \left\lfloor \sin^2\frac{x}{4}+\cos\frac{x}{4} \right\rfloor $ and the $x$ axis is Prove that the area bounded by circle $x^2+y^2=4$,the parabola $y=x^2+x+1$ and the curve $ \left\lfloor \sin^2\frac{x}{4}+\cos\frac{x}{4} \right\rfloor $ and the $x$ axis is $\sqrt3+\frac{2\pi}{3}-\frac{1}{6}.$
In this question,$x^2+y^2=4$ is a circle centered at $(0,0)$ and radius is $2$ and parabola $y=x^2+x+1$ faces upwards whose vertex is $(\frac{-1}{2},\frac{3}{4})$.
$I=2\int_{0}^{1}\sqrt{4-y^2}dy-\int_{-1}^{0}1-(x^2+x+1)dx=\sqrt3+\frac{2\pi}{3}-\frac{1}{6}$
Thanks i now understood after help from RobertZ.
| Hint. Inside the circle we have that $x\in [-2,2]$ which implies that $\cos(x/4)\in [0,1]$. Therefore
$$\sin^2\frac{x}{4}+\cos\frac{x}{4}=1+\cos\frac{x}{4}(1-\cos\frac{x}{4})\in [1,1+\frac{1}{4}]$$
Hence $\left\lfloor \sin^2\frac{x}{4}+\cos\frac{x}{4} \right\rfloor=1$ for $x\in [-2,2]$, which means that inside the circle, that curve is simply an horizontal segment.
Now make a drawing and try to find the area as an integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252$ Show that $1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252$ without calculate the value of $1 + 3^{-2} + 5^{-2} + ...$
Can I do in this way?
$$\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx$$
$$\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+3)^{-1}+\frac12(7)^{-1} = \frac1{14}$$
$$\lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+1)^{-1}+\frac12(5)^{-1} = \frac1{10}$$
$$\therefore \frac1{14} \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \frac1{10}$$
$$\therefore 1.22253... \le 1 + 3^{-2} + 5^{-2} + ... \le 1.25111... $$
$$\therefore 1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252 $$
| Idea: Perhaps this will help:
$$ {1\over n^2}\leq {1\over n\cdot(n-1)} = {1\over n-1}- {1\over n}$$
and vice versa:$${1\over n}- {1\over n+1}={1\over n\cdot(n+1)} \leq {1\over n^2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding number of solutions when condition is given. Q - Find number of solutions when it is given that Re(z²) = 0 and |z| = a $\sqrt{2}$ , where z is a complex number and a>0.
First I assumed $z = x + iy$ and then squared it and equated the real part to $0$.
I don't know how to approach after that.
Please guide.
| Let $z=re^{i\theta}$. Then
*
*$\Re(z^2)=0$, implies $r^2\cos(2\theta)=0$.
*Also $|z|=a\sqrt{2}$ implies $r=a\sqrt{2} \neq 0$.
From these two, we get $\cos(2\theta)=0$. This implies $2\theta=\frac{(2n+1)\pi}{2}$ or $\theta=\frac{(2n+1)\pi}{4}$.
Thus $z=a\sqrt{2}e^{i\frac{(2n+1)\pi}{4}}$, where $n \in \Bbb{Z}$. Now you can count the distinct solutions out of these as:
\begin{align*}
z& =a\sqrt{2}e^{i\frac{\pi}{4}}=a\left(1+i\right)\\
z&=a\sqrt{2}e^{i\frac{3\pi}{4}}=a\left(-1+i\right)\\
z&=a\sqrt{2}e^{i\frac{5\pi}{4}}=a\left(-1-i\right)\\
z&=a\sqrt{2}e^{i\frac{7\pi}{4}}=a\left(1-i\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\cos^8x.\sec^6y,\frac12,\sin^8x.\csc^6y$ in AP if $\cos^4x.\sec^2y,\frac12,\sin^4x.\csc^2y$ in A.P If $\cos^4x.\sec^2y,\dfrac{1}{2},\sin^4x.\csc^2y$ are in A.P, then prove that $\cos^8x.\sec^6y,\dfrac{1}{2},\sin^8x.\csc^6y$ in AP.
My Attempt
$$
\cos^4x.\sec^2y+\sin^4.x\csc^2y=\frac{\cos^4x}{\cos^2y}+\frac{\sin^4x}{\sin^2y}=1\\
\implies\sin^2y.\cos^4x+\cos^2y.\sin^4x=\sin^2y.\cos^2y
$$
$$
\cos^8x.\sec^6y+\sin^8x.\csc^6y=\frac{\cos^8x}{\cos^6y}+\frac{\sin^8x}{\sin^6y}
$$
How do I know that the given terms are in A.P, G.P or H.P ?
| Substitute $\sin^2y=1-\cos^2y$ and $\sin^2x=1-\cos^2x$, and factor the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995054",
"timestamp": "2023-03-29T00:00:00",
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Differences between numbers of the form abcdef and acdef which are perfect square. $541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?
| Of course,
$$
\begin{aligned}
\overline{abcdef} - \overline{acdef}
&=
\overline{ab0000} - \overline{a0000}
\\
&=10000\cdot(\overline{ab} - \overline{a})
\\
&=100^2\cdot(9a+b)\ .
\end{aligned}
$$
So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:
for a in [1..9]:
for b in [0..9]:
ab_a = 9*a + b
if ab_a.is_square():
print ( "a=%s b=%s %s-%s = %s^2"
% (a, b, 10*a+b, a, sqrt(ab_a)) )
And the results are:
a=1 b=0 10-1 = 3^2
a=1 b=7 17-1 = 4^2
a=2 b=7 27-2 = 5^2
a=3 b=9 39-3 = 6^2
a=4 b=0 40-4 = 6^2
a=5 b=4 54-5 = 7^2
a=7 b=1 71-7 = 8^2
a=8 b=9 89-8 = 9^2
a=9 b=0 90-9 = 9^2
Here i was using sage.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the following inequality holds when $x>0$ $\require{cancel}$
Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$
I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the inequality yields
$$1+\frac{x}{2}-\frac{x^2}{8}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<1+\frac{x}{2}$$
which is equivalent to
$$\begin{cases}
\cancel{1+\frac{x}{2}-\frac{x^2}{8}}\cancel{<1+\frac{x}{2}-\frac{x^2}{8}}+\frac{x^3}{16 (\xi+1)^{5/2}}\quad(1)\\
\cancel{1+\frac{x}{2}}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<\cancel{1+\frac{x}{2}}\qquad\,\,\,\,\,\, (2)
\end{cases}$$
Inequality $(1)$ holds for the values given by the problem statement. Instead for $(2)$
$$\frac{x^3}{16 (\xi+1)^{5/2}}<\frac{x^2}{8}$$
But $$\displaystyle{\max_{0\leq\xi\leq x}\Bigg\{\frac{1}{16 (\xi+1)^{5/2}}\Bigg\}}=\displaystyle{\min_{0\leq\xi\leq x}{\Big\{16 (\xi+1)^{5/2}}\Big\}}$$
which occurs at $\xi=0$, and thus, for $(2)$, it is left to prove that
$$\frac{x^3}{16}<\frac{x^2}{8}$$
but this happens for $x<0\,\vee\,0<x<2$.
Instead, if I use $\xi=x$ then inequality $(2)$ becomes
$$\frac{x^3}{16 (x+1)^{5/2}}<\frac{x^2}{8}$$
which holds for $-1<x<0\,\vee\,x>0$, and thus coincides with the restriction given by the problem.
Would it be correct to take $\xi=x$ rather than $0$? Is this approach correct at all?
| Well the inequality can be proved using algebra by squaring it. A proof via Taylor's theorem is as follows. We have a number $c\in(0,x)$ such that $$\sqrt {1+x}=1+\frac{x}{2}-\frac{x^2}{8}(1+c)^{-3/2}$$ The desired inequality follows because $0<(1+c)^{-3/2}<1$.
Your approach is similar but unnecessarily uses third derivatives and complicates the situation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Does $a_{n}/a_{n-1}$ converge to the golden ratio for all Fibonacci-like sequences? Yesterday a friend challenged me to prove that
$$\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}=\varphi\; ,$$
where $\varphi$ is the golden ratio, for the Fibonacci series.
I started rewriting the limit as
$$\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}=\lim_{n\rightarrow\infty}\frac{a_{n-1}+a_{n-2}}{a_{n-1}}=\lim_{n\rightarrow\infty}1+\frac{a_{n-2}}{a_{n-1}}\; .$$
If the sequence $b_n=\frac{a_n}{a_{n-1}}$ is convergent,
$$\lim_{n\rightarrow\infty}\frac{a_{n-2}}{a_{n-1}}=\left(\lim_{n\rightarrow\infty}\frac{a_n}{a_{n-1}}\right)^{-1}\; .$$
Renaming the desired limit $x$, we obtain the quadratic equation
$$x=1+\frac{1}{x}$$
$$x^2-x-1=0$$
if $x\neq 0$. Therefore, if $b_n$ is convergent, it must be equal to $\frac{1+\sqrt{5}}{2}$ or $\frac{1-\sqrt{5}}{2}$.
Since $a_n>0$, $b_n>0, \forall n$, so the limit must be equal to $\varphi=\frac{1+\sqrt{5}}{2}$.
This proof made me think that I didn't make use of the initial values of the sequence, so it must hold true for any sequence where $a_{n}=a_{n-1}+a_{n-2}$. The first question is, is $a_{n}/a_{n-1}$ convergent for all Fibonacci-like sequences?
The second and most intriguing for me is, is there any Fibonacci-like sequence where the limit is $\frac{1-\sqrt{5}}{2}$? Since this solution is negative, $a_n$ should change its sing with each $n$, but I couldn't find any values for $a_0$ and $a_1$, which would lead me to this case. If the answer to this question is no, what mathematical sense does this negative solution have?
| One way to look at this problem is that if $a_{n+1}=a_n+a_{n-1}$, then we we have
$$\begin{pmatrix}0&1\\1&1\end{pmatrix}\begin{pmatrix}a_{n-1}\\a_n\end{pmatrix}=\begin{pmatrix}a_n\\a_{n+1}\end{pmatrix}.$$
The eigenvalues of the matrix
$$\begin{pmatrix}0&1\\1&1\end{pmatrix}$$
are $$\dfrac{1+\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}.$$
These have corresponding eigenvectors $v_1,v_2$ which span $\mathbb{R}^2$. This leads us to the conclusion that $$\begin{pmatrix}a_1\\a_2\end{pmatrix}=cv_1+bv_2,$$
and if $c\not=0,$ then $v_1$ will dominate the sequence, and we can show the ratios converge to $(1+\sqrt{5})/2.$ This leads us to the conclusion that if we want a Fibonacci like sequence to have ratios converging to $(1-\sqrt{5})/2$, then we must have
$$\begin{pmatrix}a_1\\a_2\end{pmatrix}=bv_2$$
for some non zero $b\in\mathbb{R}$. So to determine all such sequences we simply have to have an eigenvector $v_2$ corresponding to $(1-\sqrt{5})/2$. One such eigenvector is
$$\begin{pmatrix}1\\\dfrac{1-\sqrt{5}}{2}\end{pmatrix}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Finding an explicit isomorphism between $\mathbb R^4 / \ker \ T$ and $\mathbb R^2$ I'm wondering if I have a valid answer to this.
It is exactly (e) of the following:
I first state that the two vector spaces are isomorphic because they have equal dimension. I then define a linear transform $S: \mathbb R^4 / Ker \ T \to \mathbb R^2$ by
$$S \begin{pmatrix}
0 \\
a \\
b \\
0 \\
\end{pmatrix} \to
\begin{pmatrix}
a \\
b \\
\end{pmatrix}: a,b \in \mathbb R$$
Where $\begin{pmatrix}
0 \\
a \\
b \\
0 \\
\end{pmatrix}$ is an arbitrary element in the departure space.
If this mapping translates basis to basis, this is an isomorphism.
$$
S \begin{pmatrix}
0 \\
a \\
b \\
0 \\
\end{pmatrix} \to
\begin{pmatrix}
a \\
b \\
\end{pmatrix} \in \mathbb R^2 \implies S\left(a \begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix} +
b\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix}\right) = a \ \vec e_1 + b \vec \ e_2 $$
$$
\text{ as
$S\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix} =
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}$ and
$
S\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\
\end{pmatrix} =
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}$}$$
Is this a valid answer?
| Your answer is not valid, since you don't even mention $F^4/\ker T$ in it. An explicit isomorphism would be$$\begin{array}{rccc}\Psi\colon&F^4/\ker T&\longrightarrow&F^2\\&\begin{pmatrix}a\\b\\c\\d\end{pmatrix}+\ker T&\mapsto&T\begin{pmatrix}a\\b\\c\\d\end{pmatrix}.\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
bound on $\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert $? Question: How to get a good bound on $\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert $?
Context: I want to show that $\frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $\Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $\Bbb R^2 - K$ , the absolute value of the function is small.
Attempt:
$\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert
\leq \frac{|x+2y-2|}{x^2+y^2+1} \leq \frac{\sqrt{x^2+y^2}+2(\sqrt{x^2+y^2})-2}{x^2+y^2+1} \leq \frac{3\sqrt{x^2+y^2}-2}{x^2+y^2}
$
This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?
But I am not sure how to get rid of the $-2$ in the numerator.
| How about
$$\left| \dfrac {x+2y-2}{x^2+y^2+1} \right| \le \dfrac {|x|+|2y|+|2|}{|x|+|y|+1} \le \dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$
This should hold when $|x|, |y| > 1$.
You can take $K$ to be the unit square. You can check that $f\left(- \dfrac 14, - \dfrac 12 \right) < -2$, so indeed the function has an absolute minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $ab(\frac1a)b^2cb^{-3} = c$? I'm working through some textbook exercise and am unable to solve the following exercise:
Use the basic rules of algebra to simplify the following expression:
$$ab\frac{1}{a}b^2cb^{-3}$$
Within the chapter of the book in question the author has covered:
*
*Associative, commutative and distributive properties of an equation
*Expanding brackets
*Factoring
*Quadratic factoring
*Completing the square
The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.
How can one arrive at $c$ with all the steps in between?
| $$a \cdot b \cdot \frac{1}{a} \cdot b^2 \cdot c\cdot b^{-3}$$ $$ = a \cdot b\cdot \frac{1}{a} \cdot b^{2} \cdot c\cdot \frac{1}{b^{3}}$$ $$ = a \cdot \frac{1}{a} \cdot b\cdot b^{2} \cdot \frac{1}{b^{3}} \cdot c $$ $$ = \frac{a}{a} \cdot b^{3} \cdot \frac{1}{b^{3}} \cdot c$$ $$ = \frac{a}{a} \cdot \frac{b^{3}}{b^{3}} \cdot c$$$$ = 1 \cdot 1 \cdot c $$$$ = c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Find all real numbers $x,y,z\in [0,1]^3$ such that $(x^2+y^2)\sqrt{1-z^2}\ge z$.... Such that:
$$(x^2+y^2)\sqrt{1-z^2}\ge z$$
and
$$(z^2+y^2)\sqrt{1-x^2}\ge x$$
and
$$(x^2+z^2)\sqrt{1-y^2}\ge y$$
Since $x,y,z$ $\in ]0,1[^3$
then , there are some real numbers $a,b,c$ such that
$\cos a=x, \cos b=y , \cos c=z$
After some manipulations , we find that :
$$\frac{1}{1+\tan^2 a}+\frac{1}{1+\tan^2 b}\ge \frac{1}{\tan c}$$
.... same for other inequalities
I don't know what i must do now
| The hint:
$$(x^2+y^2)\sqrt{1-z^2}\geq z$$ it's
$$(x^2+y^2)^2(1-z^2)\geq z^2$$ or
$$(x^2+y^2)^2\geq((x^2+y^2)^2+1)z^2,$$ which gives
$$z^2\leq\frac{(x^2+y^2)^2}{(x^2+y^2)^2+1}\leq\frac{(x^2+y^2)^2}{2(x^2+y^2)}=\frac{x^2+y^2}{2}.$$
By the same way $$y^2\leq\frac{x^2+z^2}{2}$$ and $$x^2\leq\frac{y^2+z^2}{2},$$ which gives
$$x=y=z$$ and all inequalities the are equalities.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the maximal $n$ satisfying $a_n \geq \frac{1}{10}$
*Let $a_n$ be the $n$-th term of the following sequence $$\frac{1}{1},\frac{1}{4},\frac{3}{4},\frac{1}{9},\frac{3}{9},\frac{5}{9},\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{1}{25},...$$
From what could I start resolving this problem ? I have found the sequence pattern that the numerator always the odd number which always start again from number 1 if the denominator change pattern. the pattern of the denominator is the square of 1, 2, 3 and etc
I have also found the pattern for the series that
*
*$a_1 = 1$
*$a_3= 2$
*$a_6 = 3$
*$a_{10} = 4$
*$a_{15}= 5$
*$a_{21} = 6$
And etc.
But I stucked on this formula, I have no idea to find the maximum n, can anyone give me some suggestion and steps for solving this problem ?
| The terms of your sequence which are immediately followed by a smaller term are
$$\frac11,\ \frac34,\ \frac59,\ \frac7{16},\ \frac9{25}\ \dots$$
The $k^\text{th}$ one of these is given by the formula $\frac{2k-1}{k^2}$. The last term in your sequence which exceeds or equals $\frac1{10}$ will correspond to the greatest value of $k$ satisfying the inequality $\frac{2k-1}{k^2}\ge\frac1{10}$ or equivalently
$$f(k)=k^2-20k+10\le0$$
Solving the quadratic equation, the zeros of $f(k)$ are $10\pm\sqrt{90}$, so $f(k)\le0$ for $10-\sqrt{90}\le k\le10+\sqrt{90}$. Since the greatest integer below $10+\sqrt{90}$ is $19$, the last term of your sequence above $\frac1{10}$ is
$$\frac{2\cdot19-1}{19^2}=\frac{37}{361}$$
and the position of this term in your sequence is $1+2+3+\cdots+19=190$,
so the short answer to your question is
$$\boxed{a_{190}=\frac{37}{361}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\sum_{k=0}^{n} \frac{(-1)^kk}{4k^2-1}$ Calculate $\sum_{k=0}^{n} \frac{(-1)^k k}{4k^2-1}$
$\sum_{k=0}^{n} \frac{(-1)^k k}{4k^2-1}=\sum_{k=0}^{n} \frac{(-1)^k k}{(2k-1)(2k+1)}=\sum_{k=0}^{n} (-1)^k k\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$ after that I get this $\sum_{k=0}^{n} (-1)^k \frac{k}{2k-1}+(-1)^{n+1}\frac{n+1}{2n+1}+\sum_{k=0}^{n}\frac{1}{2} \frac{(-1)^k}{2k+1} $. But that does not help me so much I do not know how to continue, i try everything after I still get the same, can you help me? $\sum_{}^{}$ $\sum_{}^{1}$
| $$\sum_{k=0}^n\frac{(-1)^kk}{4k^2-1}=\frac{1}{4}\sum_{k=0}^n(-1)^k\left[\frac{1}{2k-1}+\frac{1}{2k+1}\right]$$
$$=\frac{1}{4}\left[\frac{1}{-1}+\frac{1}{1}-\frac{1}{1}-\frac{1}{3}+\frac{1}{3}+\frac{1}{5}-\cdots+\frac{(-1)^n}{2n-1}+\frac{(-1)^n}{2n+1}\right]$$
$$=\frac{1}{4}\left[\frac{1}{-1}+\frac{(-1)^n}{2n+1}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt:
we have $x+2y=100-5z$
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$
we need to collect coefficient of $100-5z$ in the above given by
$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$
Total number of solutions is
$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$
$$S(z)=540.5$$
what went wrong in my analysis?
| Note that the number of non-negative integer solutions of the following equation
$$ x+y=n $$
is $n+1$. Here $n$ is a non-negative integer. Clearly $5|(x+2y)$. Let
$$x+2y=5k\tag{1}$$
where $0\le k\le 20$. For (1), if $k$ is odd, then so is $x$, and if $k$ is even, then so is $x$.
Case 1: $k$ is odd. Let $k=2n-1$ and $x=2m-1$. Then $1\le n\le 10$ and (1) becomes
$$ m+y=5n-2 $$
whose number of non-negative integer solutions is $5n-1$.
Case 2: $k$ is even. Let $k=2n$ and $x=2m$. Then $0\le n\le 10$ and (1) becomes
$$ m+y=5n $$
whose number of non-negative integer solutions is $5n+1$.
Thus the number of non-negative integer solutions is
$$ \sum_{n=1}^{10}(5n-1)+\sum_{n=0}^{10}(5n+1)=551 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the equation system $I:x^2-y^2=a;II: 2xy= b$ has always a solution $(x,y)\in \mathbb {R}^2$ The statement must be false because if $b=0$ then $x$ or $y$ must be $0$. If $x$ is $0$ we get from $I$ that $-y^2=a\iff y^2 = -a$. Wouldn't that be a contradiction to the statement? Because if $a>0$ then there is no solution for the equation System, vice versa if $y=0$ and $a<0$.
For the general case I have assumed that $b\neq 0$ and therefore from $II$ we get $y=\frac{b}{2x}$, substituting with $I \Longrightarrow x^2-(\frac{b}{2x})^2=a\iff 4x^4 - b^2 = 4x^2a \iff x^4 -ax^2 - \frac{b^2}{4}=0 (*)$
Completing the square
$(*)\iff x^4 - 2\frac{ax^2}{2}-\frac{b^2}{4} \iff x^4 - 2\frac{ax^2}{2} +\frac{a^2}{4}-\frac{b^2}{4}-\frac{a^2}{4}\iff (x^2 - \frac{a}{2})^2+\frac{-b^2-a^2}{4}=0$
$\Rightarrow x^2-\frac{a}{2}=\sqrt{\frac{b^2+a^2}{4}} \Rightarrow x^2=\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2} \Rightarrow x = \pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}}$
$II\Longrightarrow y= \frac{b}{2(\pm \sqrt{\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2}})}$
Can somebody tell my whether this Solutions are Right or not because when I have tried to verify it I end up with that term:
$\frac{5a^2+a(5\sqrt\frac{b^2+a^2}{4})(4\sqrt{\frac{b^2+a^2}{4}}+\frac{a}{2})}{4}$ which should be equal to $a$
Thank you for your time.
| Geometric solution
If $\;a,b\neq 0\;$ the equations define two hyperbolas centered both in $0,$ where
*
*$I:x^2-y^2=a\;$ has asymptotes $\;y=\pm x,$ and lies in the left and right sectors ($a>0$) or in upper and lower sectors ($a<0$) delimited by the asymptotes;
*$II: 2xy= b\;$ has asymptotes $\;x=0,y=0\;$ and lies in the first and third quadrants ($b>0$) or in the second and fourth if $b<0$ (the quadrants are sectors delimited by asymptotes).
If $a=0$ or $b=0,$ the corresponding equation defines the two asymptotes of $I$ or of $II.$
Thus the two objects $I$ and $II$ necessarilly intersect in $2$ points if at least one of them is a hyperbola, and in one point in the degenerate case $a=0=b.$ The coordinates $(x,y)$ of these points are solutions of the system.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$ I need some help with this exercise.
Given that $$\tan\alpha=2$$
calculate the value of:
$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$$
I've tried everything but I always end up having something irreducible in the end. Maybe this is an easy question, but I'm new at trigonometry. If anyone can help me by providing a step-by-step solution to this, I would be really thankful! :)
| Notice $\boxed{\sin \alpha = 2\cos \alpha}$ and $$\cos ^2\alpha = {1\over 1+\tan^2\alpha} ={1\over 5}$$
so we have
$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}=\frac{8\cos^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{6\cos\alpha +2\cos\alpha}$$
$$= \frac{6\cos^{3}\alpha + 3\cos\alpha}{8\cos\alpha} = \frac{6\cos^{2}\alpha + 3}{8} = {21\over 40}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Seeking Methods to solve $ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$ I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Methods that can be used to solve it:
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$$
My method:
Let
$$ I(t) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(t\sin(x)\right)}{\sin(x)}\:dx$$
Thus,
\begin{align}
I'(t) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\left(t^2\sin^2(x) + 1\right)\sin(x)}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{t^2\sin^2(x) + 1}\:dx \\
&= \left[\frac{1}{\sqrt{t^2 + 1}} \arctan\left(\sqrt{t^2 + 1}\tan(x) \right)\right]_{0}^{\frac{\pi}{2}} = \sqrt{t^2 + 1}\frac{\pi}{2}
\end{align}
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t) + C$$
Now
$$I(0) = C = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(0\cdot\sin(x)\right)}{\sin(x)}\:dx = 0$$
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t)$$
And finally,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
| $$\begin{align}
\int_0^{\pi/2}\frac{\arctan \sin(x)}{\sin(x)}dx
&=\int_0^{\pi/2}\frac{1}{\sin(x)}\sum_{n=0}^\infty \frac{(-1)^n \sin^{2n+1}(x)}{2n+1}dx\\
&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^{\pi/2}\sin^{2n}(x)dx\\
&=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}\cdot \frac{(2n-1)!!}{(2n)!!}\\
&=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2^{2n-1}(2n+1)}\cdot \binom{2n-1}{n} \\
&=\frac{\pi}{2}+\frac{\pi}{2}\cdot (\sinh^{-1}(1)-1) \\
&=\frac{\pi}{2}\ln(1+\sqrt{2}) \\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 3
} |
Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $\dfrac{1}{4\sqrt{2}}$.
| First replace $1/x=h$ to find
$$L=\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)=\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$
Let $\sqrt{1+\sqrt{1+h^4}}=y+\sqrt2\implies1+\sqrt{1+h^4}=(\sqrt2+y)^2=2+2\sqrt2y+y^2$
$$1+h^4=(1+y(2\sqrt2+y))^2=1+2y(2\sqrt2+y)+y^2(2\sqrt2+y)^2$$
$$L=\lim_{y\to0}\dfrac{1+2y(2\sqrt2+y)+y^2(2\sqrt2+y)^2-1}y=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What's wrong with this Penrose pattern? I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in
*
*Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.
*Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman
The orthonormal basis is chosen as
$$
M=\sqrt{\frac{2}{5}} \begin{bmatrix}
\cos 0 & \cos \frac{2\pi}{5} & \cos \frac{4\pi}{5}& \cos \frac{6\pi}{5}& \cos \frac{8\pi}{5} \\
\sin 0 & \sin \frac{2\pi}{5} & \sin \frac{4\pi}{5}& \sin \frac{6\pi}{5}& \sin \frac{8\pi}{5} \\
\cos 0 & \cos \frac{4\pi}{5} & \cos \frac{8\pi}{5}& \cos \frac{12\pi}{5}& \cos \frac{16\pi}{5} \\
\sin 0 & \sin \frac{4\pi}{5} & \sin \frac{8\pi}{5}& \sin \frac{12\pi}{5}& \sin \frac{16\pi}{5} \\
\frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}}\\
\end{bmatrix}
$$
Each row presents a basis vector, i.e.
$$
M_i\cdot M_j=0, \;\;\textrm{for } i<j.$$
and
$$||M_i||=1, \;\;\textrm{for } 1\leq i \leq 5.
$$
$M$ consists of the parallel operator (representing the physical space)
$$
A=\begin{bmatrix}
M_1\\
M_2 \\
\end{bmatrix}=
\begin{bmatrix}
\cos 0 & \cos \frac{2\pi}{5} & \cos \frac{4\pi}{5}& \cos \frac{6\pi}{5}& \cos \frac{8\pi}{5} \\
\sin 0 & \sin \frac{2\pi}{5} & \sin \frac{4\pi}{5}& \sin \frac{6\pi}{5}& \sin \frac{8\pi}{5} \\
\end{bmatrix}
$$
and the perpendicular operator
$$
B=\begin{bmatrix}
M_3\\
M_4 \\
M_5 \\
\end{bmatrix}=\begin{bmatrix}
\cos 0 & \cos \frac{4\pi}{5} & \cos \frac{8\pi}{5}& \cos \frac{12\pi}{5}& \cos \frac{16\pi}{5} \\
\sin 0 & \sin \frac{4\pi}{5} & \sin \frac{8\pi}{5}& \sin \frac{12\pi}{5}& \sin \frac{16\pi}{5} \\
\frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}}\\
\end{bmatrix}
$$
The 5D lattice points are integer combinations of basis such as
$$
p=i \begin{bmatrix}
1\\
0\\
0\\
0\\
0\\
\end{bmatrix}
+ j\begin{bmatrix}
0\\
1\\
0\\
0\\
0\\
\end{bmatrix}
+\dots, \;\; i,j,\dots \in \mathbb{Z}
$$
A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, \;\; v\in hypercube
$$
so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.
The resultant 2d projection $Ap$ is
Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows
Is this a mistake or an alternative view of the same tiling?
Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)
where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?
| I wrote to Prof. J.H.H.Perk, this is a quote from his email:
You want to make a different choice (the translation on lattice $p$), so that no three grid lines (3) pass through the same point. This requires that at least three of the five gammas are nonzero. Generically then you get a regular Penrose tiling, but you can still get singular Penrose tilings like the one for all gammas zero. Those choices are everywhere dense, like rationals are dense within the reals, but only infinitesimal part of them.
Which means the 5D lattice points are
$$
p=\gamma_1 \begin{bmatrix}
1\\
0\\
0\\
0\\
0\\
\end{bmatrix}
+ \gamma_2 \begin{bmatrix}
0\\
1\\
0\\
0\\
0\\
\end{bmatrix}
+ \gamma_3 \begin{bmatrix}
0\\
0\\
1\\
0\\
0\\
\end{bmatrix}
+ \gamma_4 \begin{bmatrix}
0\\
0\\
0\\
1\\
0\\
\end{bmatrix}
+ \gamma_5 \begin{bmatrix}
0\\
0\\
0\\
0\\
1\\
\end{bmatrix}
$$
However, I have not figured out the correct $\gamma_1,\dots,\gamma_5 \in \mathbb{R}$.
If $\gamma_i=0$ (the lattice is not shifted), the 3D window is as follows,
which contains ten points. This is against the case of Penrose pattern whose window should contain five points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
} |
Request for crazy integrals I'm a sucker for exotic integrals like the one evaluated in this post. I don't really know why, but I just can't get enough of the amazing closed forms that some are able to come up with.
So, what are your favorite exotic integral identities, and how do you prove them?
| Here are some links to a few integrals: 1
(Big list, but not all of them got the right answer).
From AoPS: 2, 3 , 4. Some that are solvable with Feynman's trick: here.
As for my favourites (most of them appeared on Romanian Mathematical Magazine), some are: $$I_1=\int_0^\frac{\pi}{2} \frac{\arctan(\tan x\sec x)}{\tan x +\sec x}dx=\frac{\pi}{2}\ln 2 -\frac{\pi}{6}\ln(2+\sqrt 3)$$
$$I_2=\int_0^\infty \exp\left(-\frac{3x^2+15}{2x^2+18}\right)\cos\left(\frac{2x}{x^2+9}\right)\frac{dx}{x^2+1}=\frac{\pi}{e}$$
$$I_3=\int_0^1 \frac{\ln^2 (1+x) (\ln^2 (1+x) +6\ln^2(1-x))}{x}dx=\frac{21}{4}\zeta(5)$$
$$I_4=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
$$I_5=\int_0^\infty \frac{1-\cos x}{8-4x\sin x +x^2(1-\cos x)}dx=\frac{\pi}{4}$$
$$I_6=\int_0^\infty \frac{\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt{3}}-\frac{2}{3}G+\frac{\pi}{12}\ln(2+\sqrt{3})$$
$$I_7=\int_0^\infty \frac{\ln(1+x)}{x^4-x^2+1}dx=\frac{\pi}{6}\ln(2+\sqrt 3)+\frac23 G -\frac{\pi^2}{12 \sqrt 3}$$
$$I_8=\int_0^1 \frac{\ln(1-x^2)\ln(1+x^2)}{1+x^2}dx=\frac{\pi^3}{32}-3G\ln 2+\frac{\pi}{2}\ln^22.$$
$$I_{9}=\int_0^{\frac{\pi}{4}} \ln\left(2+\sqrt{1-\tan^2 x}\right)dx = \frac{\pi}{2}\ln\left(1+\sqrt{2}\right)+\frac{7\pi}{24}\ln2-\frac{\pi}{3}\ln\left(1+\sqrt{3}\right)-\frac{G}{6}$$
$$I_{10}=\int_{-\infty}^\infty \frac{\sin \left(x-\frac{1}{x}\right) }{x+\frac{1}{x}}dx=\frac{\pi}{e^2}$$
$$I_{11}=\int_{-\infty}^\infty \frac{\cos \left(x-\frac{1}{x}\right) }{\left(x+\frac{1}{x}\right)^2}dx=\frac{\pi}{2e^2}$$
$$I_{12}=\int_0^1 \frac{\ln(1-x)\ln(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2} G$$
$$I_{13}=\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x}dx=\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96}$$
$$I_{14}=\int_0^\frac{\pi}{4} \operatorname{arcsinh} (\sin x) dx=G-\frac58\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$
$$I_{15}=\int_0^\frac{\pi}{2} x \arcsin \left(\sin x-\cos x\right)dx=\frac{\pi^3}{96}+\frac{\pi}{8}\ln^2 2$$
$$I_{16}=\int_0^\infty \int_0^\infty \frac{\ln(1+x+y)}{xy\left((1+x+y)(1+1/x+1/y)-1\right)}dxdy=\frac72 \zeta(3)$$
Where $G$ is Catalan's constant and $\operatorname{Cl}_2 (x)$ is the Clausen function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 7,
"answer_id": 4
} |
Taylor series of $\ln\frac{1+x}{1-x}$
Let $f(x)=\ln\frac{1+x}{1-x}$ for $x$ in $(-1,1)$. Calculate the Taylor series of $f$ at $x_0=0$
I determined some derivatives:
$f'(x)=\frac{2}{1-x^2}$; $f''(x)=\frac{4x}{(1-x^2)^2}$; $f^{(3)}(x)=\frac{4(3x^2+1)}{(1-x^2)^3}$; $f^{(4)}(x)=\frac{48x(x^2+1)}{(1-x^2)^4}$; $f^{(5)}(x)=\frac{48(5x^2+10x^2+1)}{(1-x^2)^5}$
and their values at $x_0=0$:
$f(0)=0$; $f'(0)=2$; $f''(0)=0$; $f^{(3)}(0)=4=2^2$
$f^{(4)}(0)=0$; $f^{(5)}(0)=48=2^4.3$; $f^{(7)}(0)=1440=2^5.3^2.5$
I can just see that for $n$ even, $f^{(n)}(0)=0$, but how can I generalize the entire series?
| \begin{align}
\ln \frac{1+x}{1-x} &= \ln (1+x) - \ln (1-x) \\
&= \sum_{n=0}^\infty \frac{(-1)^n}{n+1}x^{n+1} - \sum_{n=0}^\infty \frac{(-1)^n}{n+1}(-x)^{n+1}\\
&= \sum_{n=0}^\infty \frac{(-1)^n}{n+1}x^{n+1} + \sum_{n=0}^\infty \frac{1}{n+1}x^{n+1}\\
&= \sum_{n=0}^\infty \frac{(-1)^n+1}{n+1}x^{n+1}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}+1}{n}x^n\\
&=\sum_{n=1}^\infty \frac{2}{2n-1}x^{2n-1}\\
\end{align}
Remark: Your observation that all the even terms vanishes is due to this is an odd function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3035412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find existence of limit Show that $\left(\frac{1}{|x|} +\frac{1}{|y |}\right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=r\cos \alpha$ and $y=r\sin \alpha$, $r>0$.But I got stuck as $\left(\frac{1}{|\cos \alpha|} +\frac{1}{ |r\sin \alpha|}\right) \left(\frac{1}{r}\right)$ lies between $\frac{1}{r}$ and infinity.
| We have:
$(x,y) \rightarrow 0.$
Let $\epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < \delta_n$ $(=\epsilon_n)$ implies
$|(x^2+y^2)^{1/2}| < \epsilon_n =1/n$, or
$n=1/\epsilon_n \lt \dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} \ge |x|$, and
$(x^2+y^2)^{1/2} \ge |y|.$
Then
$\dfrac{2}{(x^2+y^2)^{1/2}} \le \dfrac{1}{|x|}+\dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| \lt \delta_n$ implies
$2n =2/\epsilon_n \lt \dfrac{2}{(x^2+y^2)^{1/2}} \le $
$\dfrac{1}{|x|} +\dfrac{1}{|y|}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3038078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find all the solutions of $z^2-(1+3i)z-8-i=0$ I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.
I want to find all the roots of:
$$z^2-(1+3i)z-8-i=0$$
There are two ways I tried to approach this.
*
*Quadratic formula:
$$\begin{align} z_1,z_2 &=-\frac{p}{2}\pm\sqrt{\left( \frac{p}{2}\right)^2-q} \\ & \implies z_1,z_2=\frac{1+3i}{2}\pm\sqrt{\left( \frac{(-1-3i)}{2}\right)^2-(-8-i)}\\ &= \frac{1+3i}{2} \pm\sqrt{\frac{24+10i}{4} }=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{align}$$
*Completing the square:
$$\begin{aligned} z^2-(1+3i)z-8-i &=0 \\ & \iff \left(z-\left( \frac{1+3i}{2}\right) \right)^2-8-i=\left( \frac{1+3i}{2}\right)^2 \\ & \iff u^2=\frac{24+10i}{4} \\ & \iff u=\pm\frac{\sqrt{24+10i}}{2} \\ & \iff z_{1,2}=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{aligned} $$
Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:
$$z_1=-2+i \\ z_2=3+2i$$
I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?
| To find square roots $\;\pm(a+ ib)\;$ of $\;24+10i,\;$ solve the equation
$$(a+ib)^2=24+10i.$$
Real and imaginary parts of RHS and LHS are equal, and also absolute values:
$$\begin{aligned}a^2-b^2&=24\\ 2ab&=10\\a^2+b^2&=26\end{aligned}$$
We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $\pm(5+i).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
range of $3x^2-2xy$ subjected to $x^2+y^2=1$ If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$\displaystyle x^2+y^2\geq 2xy$
$\displaystyle -2xy\geq -(x^2+y^2)$
$\displaystyle 3x^2-2xy\geq 2x^2-y^2$
this will not help more
how do i solve it help me please
| The range that you're after is $\left[\frac{3-\sqrt{13}}2,\frac{3+\sqrt{13}}2\right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2\pm x\sqrt{1-x^2}.$$So, for each $x\in[-1,1]$, let$$f(x)=3x^2-2x\sqrt{1-x^2}\text{ and let }g(x)=3x^2+x\sqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $\left[\frac{3-\sqrt{13}}2,\frac{3+\sqrt{13}}2\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Evaluating $I(s)=\int_0^{\frac{\pi}{2}}\ln(1+s\tan(\theta))\cot(\theta)d\theta$ Recently I've encountered this interesting integral$$I(s)=\int_0^{\frac{\pi}{2}}\ln(1+s\tan(\theta))\cot(\theta)d\theta=\int_0^{\infty}\frac{\ln(1+sx)}{x}\frac{dx}{x^2+1}$$
I was wondering if this integral has a closed form in terms of polylogarithms or other special functions and I would be grateful if someone could provide the answer or a method for finding it.
| To evaluate your integral I will do this by using Feynman's trick of differentiating under the integral sign.
Let
$$I(s) = \int_0^{\frac{\pi}{2}} \ln (1 + s \tan \theta ) \cot \theta \, d\theta, \qquad s > 0.$$
Note that $I(0) = 0$. Differentiating with respect to $s$ we have
$$I'(s) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{1 + s \tan \theta}.$$
To find the resulting integral, a $t$-substitution of $t = \tan \frac{\theta}{2}$ can be used. Here $d\theta = 2/(1 + t^2) \, dt$ and $\tan \theta = 2t/(1 - t^2)$. Thus
\begin{align}
I'(s) &= 2 \int_0^1 \frac{1 - t^2}{(1 + t^2)(1 + 2st - t^2)} \, dt\\
&= 2 \int_0^1 \left [\frac{1}{1 + s^2} \cdot \frac{1 - st}{1 + t^2} + \frac{s}{1 + s^2} \cdot \frac{t - s}{t^2 - 2st - 1} \right ] \, dt\\
&= \frac{2}{1 + s^2} \int_0^1 \frac{dt}{1 + t^2} - \frac{s}{1 + s^2} \int_0^1 \frac{2t}{1 + t^2} \, dt + \frac{s}{1 + s^2} \int_0^1 \frac{2t - 2s}{t^2 - 2st - 1} \, dt\\
&= \frac{2}{1 + s^2} \big{[}\tan^{-1} t \big{]}_0^1 -\frac{s}{1 + s^2} \big{[} \ln (1 + t^2) \big{]}_0^1 + \frac{s}{1 + s^2} \big{[} \ln |t^2 - 2st - 1| \big{]}_0^1\\
&= \frac{\pi}{2(1 + s^2)} + \frac{s \ln s}{1 + s^2}.
\end{align}
Integrating up with respect to $s$:
\begin{align}
I(s) &= \frac{\pi}{2} \int \frac{ds}{1 + s^2} + \frac{1}{2} \int \frac{2s}{1 + s^2} \cdot \ln s \, ds\\
&= \frac{\pi}{2} \tan^{-1} s + \frac{1}{2} \ln (s) \ln (1 + s^2) - \frac{1}{2} \int \frac{\ln (1 + s^2)}{s} \, ds, \tag1
\end{align}
where in the second of the integrals integration by parts has been used. The final integral appearing in (1) can be found in terms of the dilogarithm $\operatorname{Li}_2 (z)$. To find this set: $-x = s^2$, then $ds = -dx/(2\sqrt{-x})$ and yields
$$\int \frac{\ln (1 + s^2)}{s} ds = \frac{1}{2} \int \frac{\ln (1 - x)}{x} \, dx = - \frac{1}{2} \operatorname{Li}_2 (x) + C = -\frac{1}{2} \operatorname{Li}_2 (-s^2) + C,$$
where the integral definition for the dilogarithm has been used. Thus (1) becomes
$$I(s) = \frac{\pi}{2} \tan^{-1} s + \frac{1}{2} \ln (s) \ln (1 + s^2) + \frac{1}{4} \operatorname{Li}_2 (-s^2) + C.$$
To find the constant $C$, as $I(0) = 0$, we see that $C = 0$. Thus
$$I(s) = \frac{\pi}{2} \tan^{-1} s + \frac{1}{2} \ln (s) \ln (1 + s^2) + \frac{1}{4} \operatorname{Li}_2 (-s^2), \qquad s > 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find invertible matrix $Q$ From the book "Bilinear forms and their matrices" by Prof. Joel Kamnitzer:
Consider: $$ A=\pmatrix{0 & 4 \\ 4 & 2} $$
After doing simultaneous row and column operations we reach:
$$ Q^tAQ= \pmatrix{-8 & 0 \\ 0 & 2} $$
And then he implies that $ Q= \pmatrix{1 & 0 \\ -2 & 1 }$ , which indeed verifies the result, but I do not know how he found that matrix. So I found the post "How to diagonalize a matrix" and tried to follow those steps. However the eigenvalues I found involve $\sqrt{17}$, and that number does not go away once I try to diagonlize the matrix. So I must be doing something wrong.
So, how can one find Q? A general solution is of course welcome too.
| Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left(
\begin{array}{rr}
0 & 4 \\
4 & 2 \\
\end{array}
\right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = \left(
\begin{array}{rr}
0 & 4 \\
4 & 2 \\
\end{array}
\right)
$$
==============================================
$$ E_{1} = \left(
\begin{array}{rr}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
$$
$$ P_{1} = \left(
\begin{array}{rr}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rr}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rr}
2 & 4 \\
4 & 0 \\
\end{array}
\right)
$$
==============================================
$$ E_{2} = \left(
\begin{array}{rr}
1 & - 2 \\
0 & 1 \\
\end{array}
\right)
$$
$$ P_{2} = \left(
\begin{array}{rr}
0 & 1 \\
1 & - 2 \\
\end{array}
\right)
, \; \; \; Q_{2} = \left(
\begin{array}{rr}
2 & 1 \\
1 & 0 \\
\end{array}
\right)
, \; \; \; D_{2} = \left(
\begin{array}{rr}
2 & 0 \\
0 & - 8 \\
\end{array}
\right)
$$
==============================================
$$ P^T H P = D $$
$$\left(
\begin{array}{rr}
0 & 1 \\
1 & - 2 \\
\end{array}
\right)
\left(
\begin{array}{rr}
0 & 4 \\
4 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rr}
0 & 1 \\
1 & - 2 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
2 & 0 \\
0 & - 8 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rr}
2 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rr}
2 & 0 \\
0 & - 8 \\
\end{array}
\right)
\left(
\begin{array}{rr}
2 & 1 \\
1 & 0 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
0 & 4 \\
4 & 2 \\
\end{array}
\right)
$$
..............
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
| Hint:
Try to write
$$28+16 \sqrt 3=(a+b\sqrt 3)^4$$
for suitable $a$ and $b$.
You obtain, reordering the terms
\begin{align}
(a+b\sqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)\sqrt 3.\\
\end{align}
Can you choose $a$ and $b$ so that
$$a^4+18a^2b^2+9b^4=28,\quad ab(a^2+3b^2)=4?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Last two digits of odd products Is there any proof where we can find the last two digits of odd number product.
$$1\cdot3\cdot5\cdot7\cdot\ldots\cdot99 = a_i$$ Find the last two digits of $a_i$. The answer would be $75$ as any multiples of $5$ will always end with $5$.
Are there any solid proof or trick to find the last two digits of the above product?
| $1,4 + 1=5, 2*8+1=9,....,4*24+1=97$ are are $1 \pmod 4$
$3,7,......, 4*24+3=99$ are all $-1\pmod 4$.
So $N = 1\cdot 3.... \cdot 97\cdot 99 \equiv 1^{25}(-1)^{25}\equiv -1\equiv 3\pmod 4$.
And obviously $N \equiv 0 \pmod {25}$.
So by Chinese Remainder theorem we can always solve $N\pmod{100 = 4*25}\equiv 75$.
...
Most statements of CRT give the formula for solving but I can never remember the proper variables so I always do it by hand each time.
In this case $3 + 4k = 25j;0\le k < 24; 0\le j < 4;$. So $ 3 + 4k= j + 4(6j)$. $j = 3; k=6j=18$
and $N \equiv 3+4k = 25j = 75 \pmod {100}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
| Hint $\ x^3-3x^2+5x-2\,\bmod\, \color{#c00}{x^2-2x+3}\, =\, \color{#0a0}1\ $ (and $= \color{#90f}2$ for the other). So we seek a polynomial with roots $\color{#0a0}1$ and $\color{#90f}2,\,$ e.g. $\ (x-\color{#0a0}1)(x-\color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$\begin{align}
&\ \ \ 1\ {-}3\ \ \ \, 5\,\ {-}2\\
&\color{#c00}{{-}1\,\ \ \ 2\ {-}3}\\
& \ \ \ \ \ \ {-}1\ \ \ \ 2\ \ {-}2\\
&\color{#c00}{\ \ \ \ \ \ \ \ \ 1\ {-}2\ \ \ \ \ 3}\\
&\qquad\qquad\quad\ \ \color{#0a0}1
\end{align}\qquad\qquad\quad$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the volume of intersection between cylinders
Find the volume of intersection of the cylinder
{$ x^2 +
y^2 \leq 1 $} , {$ x^2 + z^2 \leq 1$}, {$ y^2 + z^2 \leq 1$}.
i am having tough time finding the volume how do i solve this kind of questions ? .
my trial :
i will move to the cylinder coordinates of the xy cylinder let :
$x^2 + y^2 = r^2 $
$ z = z $
$0\leq\theta \leq 2\pi$
solving the inequalties i get :
$ 0 \leq r^2 \leq 1$
$ -\sqrt{1-\frac{r^2}{2}}\leq z \leq \sqrt{1-\frac{r^2}{2}}$
$0\leq\theta \leq 2\pi$
the integral is :
$ \int_{z=-\sqrt{1-\frac{r^2}{2}}}^{z=\sqrt{1-\frac{r^2}{2}}}\int_{r=0}^{r=1}\int_0^{2\pi} dz \ dr \ d{\theta}$ = $ \frac{4\pi}{\sqrt{2}}\frac{(2-r^2)^{\frac{3}{2}}}{-3} |_{r=0}^{r=1}$
| Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.
What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $\sqrt{1-x^2} - 1/\sqrt{2}$, over the interval $x \in [0,1/\sqrt{2}]$. This is simply $$\int_{x=0}^{1/\sqrt{2}} x \left(\sqrt{1 - x^2} - 1/\sqrt{2}\right) \, dx = \left[-\frac{(1-x^2)^{3/2}}{3} - \frac{x^2}{2\sqrt{2}}\right]_{x=0}^{1/\sqrt{2}} = \frac{8 - 5 \sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2\sqrt{2} + 2(8-5\sqrt{2}) = 8(2 - \sqrt{2}).$$ I have left the reasoning as an exercise for the reader.
An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $\sqrt{1-x^2} - x$, giving $$\frac{V}{48} = \int_{x=0}^{1/\sqrt{2}} x \left(\sqrt{1-x^2} - x\right) \, dx.$$ Again, the details are left for the reader.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
If $f(4xy)=2y[f(x+y)+f(x-y)]$ and $f(5)=3$, find $f(2015)$
Suppose the function $f:\Bbb R\to\Bbb R$ satisfies the following conditions:
$$\begin{align}
f(4xy)&=2y[f(x+y)+f(x-y)] \\[4pt]
f(5)&=3
\end{align}$$
Find the value of $f(2015)$.
I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.
| Firstly, it’s trivial that
$$f(0) = f(4\cdot 0\cdot 0) = 2\cdot 0 \cdot [f(0) + f(0)] = 0$$
Next, we see that
$$0 = f(0) = f(4\cdot 0 \cdot y) = 2y\cdot [f(y) + f(-y)]\text{,}$$
which implies that $f(-y) = -f(y)$ for all $y\neq 0$.
Afterwards, one notices that
$$f(4xy) = f(4yx)$$
Hence,
$$2y\cdot [f(x+y) + f(x-y)] = 2x\cdot [f(x+y) + f(y-x)]$$
$$\Leftrightarrow (x-y)\cdot f(x+y) = (x+y)\cdot f(x-y)$$
$$\Leftrightarrow f(x+y) = \frac{x+y}{x-y}\cdot f(x-y)$$
If we substitute now $x=1010$ and $y=1005$, we get that
$$f(2015) = \frac{2015}{5}\cdot 3 = 1209$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Second order taylor series expansion of $\cos(x+2y)e^{2x}$ around $(2, -1)$ I'm asked to find the second order taylor series expansion of $\cos(x+2y)e^{2x}$ around $(2, -1)$. It's from an old exam of a professor who doesn't give solutions to them, so I don't know if my results are correct. So I woud like that you tell me if my results are correct, and if not, where I made a mistake. Thank you.
As first partial derivatives, I get $$f_x=-\sin(x+2y)e^{2x}+2\cos(x+2y)e^{2x}$$ $$f_y=-2\sin(x+2y)e^{2x}$$ As second partial derivatives, I get
$$\begin{align}
f_{xx}&=-\cos(x+2y)e^{2x}-2\sin(x+2y)e^{2x}-2\sin(x+2y)e^{2x}+4\cos(x+2y)e^{2x}\\
&=3\cos(x+2y)e^{2x}-4\sin(x+2y)e^{2x}
\end{align}$$
$$f_{yy}=-4\cos(x+2y)e^{2x}$$ $$f_{xy}=f_{yx}=-2\cos(x+2y)e^{2x}-4\sin(x+2y)e^{2x}$$
So now, if I plug everything into the formula, I get
$$e^{4}+2e^{4}(x-2)+\frac{1}{2!}(3e^{4}(x-2)^2-4e^{4}(x-2)(y+1)-4e^{4}(y+1)^2)=e^{4}(1+2(x-2)+\frac{1}{2!}(3(x-2)^2-4(x-2)(y+1)-4(y+1)^2))$$
Are my results correct ? I just proceeded as usual, using the common two-variables taylor series expansion. Thanks for your help !
| It is correct. You can also obtain the expansion by using the Taylor series of $\sin, \cos,\exp$:
\begin{align}
\cos(x+2y)e^{2x} &= \cos((x-2) + 2(y+1))e^{4+2(x-2)}\\
&= e^{4}\big[\cos(x-2)\cos(2(y+1)) - \sin(x-2)\sin(2(y+1))\big]e^{2(x+2)}\\
&= e^{4}\left[\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(x-2)^{2n}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}2^{2n}(y+1)^{2n}\right)\right.\\
&- \left.\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(x-2)^{2n+1}\right)\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}2^{2n+1}(y+1)^{2n+1}\right)\right]\left(\sum_{n=0}^\infty \frac{2^n}{n!}(x-2)^{n}\right)\\
&= e^{4}\left[1+2(x-2) + \left(\frac{(-1)^1}{2!} - \frac{(-1)^1}{1!}\frac{2^1}{1!}\right)(x-2)^2 - \frac{(-1)^0}{0!}2^1\cdot \frac{2}{2!}(x-2)(y+1) + \frac{(-1)^1}{2!}2^2(y+2)^2 + \cdots\right]\\
&= e^4\left[1+2(x-2)+\frac32 e^{4}(x-2)^2-2e^{4}(x-2)(y+1)-2e^{4}(y+1)^2 + \cdots\right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$. I have managed to solve it in one way, but I became very interested in this failed attempt.
$$
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x
= \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x
- \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x
$$
We only have to show that those two on the right are equal. And numerical evaluations seem to suggest that they both are in fact $-\frac{\pi}{4}$ but I don't know how to break these down.
I am currently really interested in proving this
$$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$
Anyway, here's my trivial solution using $u = \frac1x$:
$$
\begin{align}
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x & = \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = \int_\infty^1 \frac{\frac{1}{u^2} \ln(\frac1u)}{(1+\frac{1}{u^2})^3} \frac{-1}{u^2} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = -\int_1^\infty \frac{\ln(u)}{u(u+\frac{1}{u})^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = - \int_1^\infty \frac{u^2 \ln(u)}{(1+u^2)^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = 0
\end{align}
$$
I'm sure there are many more interesting methods for cracking this integral, since it's so closely related to the popular $\int_0^\infty \frac{\ln(x)}{1+x^2} {\rm d}x = 0$. Please share them if you do come up with any.
| $$\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx \stackrel{(*)}{=}\frac{\pi(1-\alpha^2)}{16\cos\frac{\pi \alpha}{2}} $$
$(*)$: we use the substitution $\frac{1}{1+x^2}=u$, the Beta function and the reflection formula for the $\Gamma$ function.
This holds for any $\alpha$ such that $-3<\text{Re}(\alpha)<3$, and since the RHS is an even function, the origin is a stationary point, i.e.
$$\color{red}{0}=\frac{d}{d\alpha}\left.\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx\right|_{\alpha=0}\stackrel{\text{DCT}}{=}\int_{0}^{+\infty}\frac{x^{2}\log x}{(1+x^2)^3}\,dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.
Let $x$ and $y$ be real numbers such that
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$
Prove that
$$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$$
| Hint:
Like greedoid, WLOG let $x=\cot2A,y=\cot2B,0<2A,2B<\pi$ (Reference )
$$1=(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=\cot A\cot B$$
$\implies\tan A=\cot B$
$x=\cot2A=\dfrac{1-\tan^2A}{2\tan A}=\dfrac{1-\cot^2B}{2\cot B}=\cdots=-\cot2B=-y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Matrices and Probability question Question:
Let Matrix $A$ be a non-singular Matrix and satisfies with Matrix $B$ such that $A^2B=A$ . Given $S=\{-1,0,1\}$ and
$$ A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$
in which $a,b,c,d\in S $ . What is the probability that $det(A+B)=detA + detB $
I did manage to work out and get the equation $a^2+d^2=-2bc$ and calculate $\frac{8}{81}$? But i'm not sure if its the correct answer.
| Yes, you are correct!
Since $A$ is non-singular:
$$A^2B=A \iff AB=I \iff \\
B=A^{-1}=\frac{1}{ad-bc}
\begin{pmatrix}
d & -b\\
-c & a
\end{pmatrix};\\
A+B=\begin{pmatrix}a+\frac{d}{ad-bc}&b-\frac{b}{ad-bc}\\ c-\frac{c}{ad-bc}&d+\frac{a}{ad-bc}\end{pmatrix}.$$
So:
$$\det(A)+\det(B)=ad-bc+\frac{1}{(ad-bc)^2}(ad-bc)=\frac{(ad-bc)^2+1}{ad-bc};\\
\det(A+B)=ad+\frac{a^2+d^2}{ad-bc}+\frac{ad}{(ad-bc)^2}-bc+\frac{2bc}{ad-bc}-\frac{bc}{(ad-bc)^2};\\
\det(A)+\det(B)=\det(A+B) \Rightarrow \\
(ad-bc)^2+1=(ad-bc)^2+a^2+d^2+2bc+1 \Rightarrow \\
a^2+d^2=-2bc.$$
Note that $ad\ne bc$ and the RHS must be positive. Then the favorable outcomes are:
$$\begin{array}{r|r|r|r}
N&a&b&c&d\\
\hline
1&-1&-1&1&-1\\
2&1&-1&1&1\\
3&-1&1&-1&-1\\
4&1&1&-1&1
\end{array}$$
There are total $3^4$ possible ways for $a,b,c,d\in \{-1,0,1\}$.
Hence the required probability is:
$$P=\frac{n(\text{favorable})}{n(\text{total})}=\frac4{81}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that base is twice the height if base angles of a triangle are $22.5^\circ$ and $112.5^\circ$
The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Show that the base is twice the height.
My Attempt
$$
h=c.\sin22.5^\circ=c.\cos 67.5^\circ\\
=b\sin 67.5^\circ=b\cos 22.5^\circ
$$
$$
a=c\cos22.5^\circ- b\sin22.5\circ=\frac{h}{b}-\frac{h}{c}=h\cdot\frac{c-b}{bc}
$$
I have no clue of how to prove this.
| By the law of sines, $$\frac{a}{\sin{45^{\circ}}}=\frac{b}{\sin{22.5^{\circ}}}$$
By the double angle formula, this is equivalent to $$\frac{a}{2\sin{22.5^{\circ}}\cos{22.5^{\circ}}}=\frac{b}{\sin{22.5^{\circ}}}\implies\frac{a}{2\cos{22.5^{\circ}}}=b$$
From the smaller right triangle we see that $$\frac{h}{b}=\cos{22.5^{\circ}}\implies h=b\cos{22.5^{\circ}}$$
Combining the results gives $a=2h$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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how to show $ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})$ converge/diverge $$ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}\left(e^{\frac{1}{n^2}} - 1\right)\left(\sqrt{n^4 - 8}\right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too.
I know that $\sin(\frac{1}{n}) < \frac{1}{n}$. I have tried to show with Cauchy that it diverges.
| Because $\sin(x)$ is concave on $\left[0,\frac\pi2\right]$, for $x\in\left[0,\frac\pi2\right]$,
$$
\frac{\sin(x)-\sin(0)}{x-0}\ge\frac{\sin\left(\frac\pi2\right)-\sin(0)}{\frac\pi2-0}\tag1
$$
Therefore,
$$
\sin(x)\ge\frac{2x}\pi\tag2
$$
Since $e^x$ is convex,
$$
\begin{align}
\frac{e^x-e^0}{x-0}
&\ge\lim_{t\to0}\frac{e^t-e^0}{t-0}\\
&=e^0\\[6pt]
&=1\tag3
\end{align}
$$
Therefore,
$$
e^x\ge1+x\tag4
$$
and that
$$
\begin{align}
x-\sqrt{x^2-a}
&=\frac{a}{x+\sqrt{x^2-a}}\\
&\le\frac ax\tag5
\end{align}
$$
Therefore,
$$
\sqrt{x^2-a}\ge x-\frac ax\tag6
$$
Thus, applying $\color{#C00}{(2)}$, $\color{#090}{(4)}$, and $\color{#00F}{(6)}$, for $n\ge2$, we have
$$
\begin{align}
\frac{\color{#C00}{\sin\left(\frac1n\right)}}{\log(n)}\color{#090}{\left(e^{\frac1{n^2}}-1\right)}\color{#00F}{\sqrt{n^4-8}}
&\ge\frac{\color{#C00}{\frac2{\pi n}}}{\log(n)}\color{#090}{\left(\frac1{n^2}\right)}\color{#00F}{\left(n^2-\frac8{n^2}\right)}\\
&=\frac2\pi\frac1{n\log(n)}\left(1-\frac8{n^4}\right)\\
&\ge\frac1\pi\frac1{n\log(n)}\tag7
\end{align}
$$
The series $\sum\limits_{n=2}^\infty\frac1{n\log(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $\sum\limits_{n=2}^\infty\frac1{n\log(n)}$ using $(7)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $ Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$
What I have tried is firstly using the inequality:
$\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, for any $x, y$ and any $a,b > 0$.
Using this inequality we obtain $\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} \geq \frac{a + b + c}{2}$, and then we have:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq \frac{a + b + c}{2} + \frac{36}{a + b + c} = \frac{(a + b + c)^2 + 72)}{2(a + b + c)} $$.
Using $ab + bc + ca = 1$, we would then have to prove that:
$$a^2 + b^2 + c^2 + 74 - 40(a + b + c) \geq 0 $$ and then I tried replacing in this inequality $c = \frac{1 - ab}{a + b}$, but I didn't get anything nice.
I also tried rewriting the lhs:
$$\frac{a^2}{b + c} = \frac{a^2(ab + bc + ca)}{b + c} = a^3 + \frac{a^2bc}{b + c}$$
And this would result in: $a^3 + b^3 + c^3 + abc(\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}) + \frac{36}{a + b + c} \geq 20$, but I didn't know how to continue from here.
Do you have any suggestions for this inequality?
| Your first step gives a wrong inequality. Try $c\rightarrow0+$ and $a=b=1$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\frac{\sum\limits_{cyc}a^2(a+b)(a+c)}{9uv^2-w^3}+\frac{12}{u}\geq20.$$
Now, we see that it's a linear inequality of $w^3$ because $\sum\limits_{cyc}a^2(a+b)(a+c)$ is fourth degree and the condition does not depend on $w^3$.
Indeed, $$\sum_{cyc}a^2(a+b)(a+c)=\sum_{cyc}a^2(a(a+b+c)+bc)=$$
$$=(a^3+b^3+c^3)(a+b+c)+(a+b+c)abc=(27u^3-27uv^2+3w^3)3u+3uw^3.$$
Hence, we need to prove that
$$\frac{(27u^3-27uv^2+3w^3)3u+3uw^3}{9uv^2-w^3}+\frac{12}{u}\geq20,$$ which is a linear inequality of $w^3$ after full expanding.
Thus, it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
Thus, we need to prove that
$$\frac{a^2}{\frac{1}{a}}+\frac{\frac{1}{a^2}}{a}+\frac{36}{a+\frac{1}{a}}\geq20$$ or
$$(a-1)^4(a^4+4a^3+11a^2+4a+1)\geq0;$$
2. Two variables are equal.
Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a<1$.
Thus, we need to prove that:
$$\frac{2a^2}{a+\frac{1-a^2}{2a}}+\frac{\left(\frac{1-a^2}{2a}\right)^2}{2a}+\frac{36}{2a+\frac{1-a^2}{2a}}\geq20$$ or
$$(1-a)(1+a+3a^2-157a^3+415a^4-225a^5+381a^6-99a^7)\geq0,$$
which is smooth.
We can use also the following way.
$$\sum_{cyc}\frac{a^2}{b+c}=\sum_{cyc}\frac{a^2(ab+ac+bc)}{b+c}=a^3+b^3+c^3+\sum_{cyc}\frac{a^2bc}{b+c}=$$
$$=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc+\sum_{cyc}\frac{a^2bc}{b+c}\geq$$
$$\geq(a+b+c)^3-3(a+b+c).$$
Id est, it's enough to prove that
$$(a+b+c)^3-3(a+b+c)+\frac{36}{a+b+c}\geq20.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Least Squares solution for a symmetric singular matrix I want to solve this system by Least Squares method:$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\5\\-2\end{pmatrix} $$ This symmetric matrix is singular with one eigenvalue $\lambda1 = 0$, so $\ A^t\cdot A$ is also singular and for this reason I cannot use the normal equation: $\hat x = (A^t\cdot A)^{-1}\cdot A^t\cdot b $.
So I performed Gauss-Jordan to the extended matrix to come with $$\begin{pmatrix}1 & 2 & 3\\\ 0 & 1 & 2 \\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\3\\-1\end{pmatrix} $$
Finally I solved the $\ 2x2$ system: $$\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix} $$ taking into account that the best $\ \hat b\ $ is $\begin{pmatrix}1\\3\\0\end{pmatrix}$
The solution is then $\ \hat x = \begin{pmatrix}-5\\3\\0\end{pmatrix}$
Is this approach correct ?
EDIT
Based on the book 'Lianear Algebra and its applications' from David Lay, I also include the Least Squares method he proposes: $(A^tA)\hat x=A^t b $
$$A^t b =\begin{pmatrix}5\\9\\13\end{pmatrix}, A^tA = \begin{pmatrix}14 & 20 & 26 \\
20 & 29 & 38 \\
26 & 38 & 50\end{pmatrix}$$
The reduced echelon from the augmented is:
$$ \begin{pmatrix}14 & 20 & 26 & 5 \\
20 & 29 & 38 & 9 \\
26 & 38 & 50 & 13 \end{pmatrix} \sim \begin{pmatrix}1 & 0 & -1 & -\frac{35}{6} \\
0 & 1 & 2 & \frac{13}{3} \\
0 & 0 & 0 & 0
\end{pmatrix} \Rightarrow \hat x = \begin{pmatrix}-\frac{35}{6} \\ \frac{13}{3} \\ 0 \end{pmatrix}$$ for the independent variable case that $z=\alpha , \alpha=0 $
| The RLS solution is given by
$$ \hat x = A^+ \, b$$
where $A^+$ is the pseudo inverse of $A$.
As $A$ is not full rank, it is not possible effectively to calculate it with the simple formula you mentioned.
However, it is still possible to calculate it with numerical solutions, for example based on the SVD.
By doing so, we get:
$$ \hat x = \begin{pmatrix}-41/12\\-1/2\\29/12\end{pmatrix} $$
For a value $A \, \hat x$ equal to:
$$ A \, \hat x = \begin{pmatrix}17/6\\4/3\\-1/6\end{pmatrix} $$
Your method does not seem to work. I can only give my interpretation of what happens:
The issue is your choice of the "best" $\hat b$ value. The vector that you considered is not directly related to the real $b$ vector, but to something obtained after some manipulations on the rows of the linear system matrix.
Difficult in this situation to rely it to a selection of a "best" $\hat b$ vector
| {
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Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
| Write $$\frac{\sqrt{8}(1+\sqrt{3x})}{(1-\sqrt{3x})(1+\sqrt{3x})}$$ and this is
$$\frac{2\sqrt{2}(1+\sqrt{3x})}{1-3x}$$
| {
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"question_score": "3",
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} |
Manipulating this $\frac{x-y}{z-y}$ to $1+\frac{x-z}{z-y}$ There is probably a very easy explanation for this that is lost on me. Came across a formula that was manipulated into another form and it was presented as a given, so I am trying to figure out how that was done.
Original:
$\frac{x-y}{z-y}=1+\frac{x-z}{z-y}$
So I began to break it down:
$\frac{x-y}{z-y}=\frac{x}{z-y}-\frac{y}{z-y}$
$\frac{x-y}{z-y}=\frac{x}{z-y}-(\frac{z}{z-y}-\frac{y}{y})$ <-- maybe this is wrong, but it works and I don't know why.
Opening up the bracket:
$\frac{x-y}{z-y}=\frac{x}{z-y}-\frac{z}{z-y}+1$
$\frac{x-y}{z-y}=1+\frac{x-z}{z-y}$
As I typed through all this, I see that $\frac{y}{z-y}=\frac{z}{z-y}-1$ is just true, and I get it when I simplify. I don't get how someone could see that to begin and wish to expand a formula like that.
Thank you for your time and patience with my high school level math question.
| A super easy way to get what you want is to just add it in (then subtract it so that you don't change the expression's value).
$$\begin{align}
\frac{x-y}{z-y} + (1+\frac{x-z}{z-y}) - (1+\frac{x-z}{z-y}) &= 1+\frac{x-z}{z-y}+\frac{(x-y)-(z-y)-(x-z))}{z-y}\\
&=1+\frac{x-z}{z-y}\\
\end{align}$$
It's especially useful for doing messy trig proofs, e.g..
$$\begin{align}
\frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} &= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - (2\sin x - 1) + (2\sin x - 1)\\
&= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - \frac{(2\sin x - 1)(\sin x - 2)}{\sin x - 2} + (2\sin x - 1)\\
&= \frac{2\sin^2 x - 5\sin x + 2}{\sin x - 2} - \frac{2\sin^2 x -5\sin x + 2}{\sin x - 2} + (2\sin x - 1)\\
&= 0+ (2\sin x - 1)\\
&= 2\sin x - 1\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Some weird results in complex number computing The question I met is to show that if $z=\cos (\theta)+i\sin(\theta)$ with $i=\sqrt{-1}$,then $ Re(\frac{z-1}{z+1})=0$
In the normal way, we found that: $$\frac{z-1}{z+1}=\frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}\\
=\frac{\bigl(\cos(\theta)-1+i\sin(\theta)\bigl)\bigl(\cos(\theta)+1-i\sin(\theta)\bigl)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\
=\frac{\cos^2(\theta)+\cos(\theta)-\cos(\theta)-1+\sin^2(\theta)+2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\
=\frac{2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}$$
So $Re(\frac{z-1}{z+1})=0$
If we do it in another way:
$$
\frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}=\frac{-2\sin^2(\frac{\theta}{2})+2i\cos(\frac{\theta}{2})\sin(\frac{\theta}{2})}{2cos^2(\frac{\theta}{2})+2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\
=-\tan(\frac{\theta}{2})\frac{\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\
=-\tan(\frac{\theta}{2})\frac{\cos\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)+i\sin\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\
=-\tan(\frac{\theta}{2})\bigl(\cos(-\theta-\frac{\pi}{2})+i\sin(-\theta-\frac{\pi}{2})\bigl)
$$
So the real part of it will be $-\tan(\frac{\theta}{2})\cos(-\theta-\frac{\pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
| You should be more careful when applying transformations of sines into cosines.
It's much easier than that:
$$
\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}=
i\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)
$$
You could write this in trigonometric form
$$
=\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)
\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)
=\cos\left(\frac{\theta}{2}+\frac{\pi}{2}\right)+
i\sin\left(\frac{\theta}{2}+\frac{\pi}{2}\right)
$$
and now you can go and chase for your mistake.
Another way to do the same: the conjugate of your number $w=\frac{z-1}{z+1}$ is
$$
\bar{w}=\frac{\bar{z}-1}{\bar{z}+1}
$$
but, since $|z|=1$, we have $\bar{z}=z^{-1}$; therefore
$$
\bar{w}=\frac{z^{-1}-1}{z^{-1}+1}=\frac{1-z}{1+z}=-w
$$
From $\bar{w}=-w$ it follows that $\operatorname{Re}(w)=0$.
If you want to find the imaginary part in term of $\theta$, you can consider $z=u^2$, where $u=\cos(\theta/2)+i\sin(\theta/2)$; then
$$
w=\frac{z-1}{z+1}=\frac{u^2-1}{u^2+1}=\frac{u-u^{-1}}{u+u^{-1}}
=\frac{u-\bar{u}}{u+\bar{u}}=
\frac{2i\sin(\theta/2)}{2\cos(\theta/2)}=i\tan\frac{\theta}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $n \ge 20$, there are at least $0.6 \cdot \frac{2^n}{n}$ primes in $\left[2^{n-1},2^n - 1\right]$. From the prime number theorem, one can deduce the following inequality:
For $x \ge 355991$, if $\pi(x) = \left|\{p \le x:p \text{ is prime}\}\right|$
then we have
$$\frac{x}{\ln(x)}\left(1+\frac{1}{\ln(x)}\right) < \pi(x) <
\frac{x}{\ln(x)}\left(1+\frac{1}{\ln(x)}+\frac{2.51}{(\ln(x))^2}\right)$$
Form here I have to deduce that for $n \ge 20$ one has that there are at least $0.6 \cdot \frac{2^n}{n}$ primes in $\left[2^{n-1},2^n - 1\right]$.
My try
Observation: $\pi(2^n -1) = \pi(2^n)$ since $2^n$ is not prime.
So the number of primes in $\left[2^{n-1},2^n - 1\right]$ is given by $\pi\left(2^n\right)-\pi\left(2^{n-1}\right)$.
We have:
$$\pi\left(2^{n-1}\right) <\frac{2^{n-1}}{\ln\left(2^{n-1}\right)}\left(1+\frac{1}{\ln\left(2^{n-1}\right)}+\frac{2.51}{\left(\ln\left(2^{n-1}\right)\right)^2}\right) \implies \\
-\frac{2^{n-1}}{\ln\left(2^{n-1}\right)}\left(1+\frac{1}{\ln\left(2^{n-1}\right)}+\frac{2.51}{\left(\ln\left(2^{n-1}\right)\right)^2}\right) < -\pi\left(2^{n-1}\right) $$
$$\frac{2^n}{\ln\left(2^n\right)}\left(1+\frac{1}{\ln\left(2^n\right)}\right) < \pi\left(2^n\right)$$
adding up these two:
$$\frac{2^n}{\ln\left(2^n\right)}\left(1+\frac{1}{\ln\left(2^n\right)}\right) - \frac{2^{n-1}}{\ln\left(2^{n-1}\right)}\left(1+\frac{1}{\ln\left(2^{n-1}\right)}+\frac{2.51}{\left(\ln\left(2^{n-1}\right)\right)^2}\right) < \pi\left(2^n\right) - \pi\left(2^{n-1}\right)$$
At this point I could do various simplifications but I cannot find the right approximation. Any ideas?
Related:
How many all prime numbers p with length of bits of p = 1024 bits?
Other sources:
This slides provide a rough estimation.
Here there seems to be an accurate deduction referenced (see section 4)
In particular, they refer to corollary 3 in this paper which reads:
For $x \ge 20.5$, $$\frac{3 \cdot x}{5 \cdot \log(x)} < \pi(2x) - \pi(x)$$
I would still thank he person with the patience to put everything together and give a well-reasoned answer.
| You almost did it:
$$\frac{2^n}{\ln\left(2^n\right)}\left(1+\frac{1}{\ln\left(2^n\right)}\right) - \frac{2^{n-1}}{\ln\left(2^{n-1}\right)}\left(1+\frac{1}{\ln\left(2^{n-1}\right)}+\frac{2.51}{\left(\ln\left(2^{n-1}\right)\right)^2}\right)=\\
\frac{2^n}{n}\left(\color{red}{\frac{1}{\ln2}}+
\color{blue}{\frac{1}{n(\ln2)^2}}-\color{red}{\frac{n}{2(n-1)\ln{2}}}-
\color{blue}{\frac{n}{2(n-1)^2(\ln{2})^2}}-
\frac{n\cdot2.51}{2(n-1)^3(\ln{2})^3}\right)=\\
\frac{2^n}{n}\left(\color{red}{\frac{n-2}{(n-1)\ln{4}}}+
\color{blue}{\frac{n^2-4n+2}{2(n-1)^2 n(\ln{2})^2}}-
\frac{n\cdot2.51}{2(n-1)^3(\ln{2})^3}\right)=\\
\frac{2^n}{n}\cdot\frac{1}{\ln{4}}\left(\left(1-\frac{1}{n-1}\right)+
\frac{1}{n\ln{2}}\left(1-\frac{2n-1}{(n-1)^2}\right)-
\frac{n\cdot 2.51}{(n-1)^3(\ln{2})^2}\right)>...$$
Now, it is not too difficult (checking 1st derivative and monotonicity) to show that
$$1-\frac{2n-1}{(n-1)^2}>\frac{1}{2}, \forall n\geq6$$
and
$$-\frac{n^2\cdot 2.51}{(n-1)^3(\ln{2})^2} > -\frac{1}{2}, \forall n\geq14$$
thus
$$...>\frac{2^n}{n}\cdot\frac{1}{\ln{4}}\left(\left(1-\frac{1}{n-1}\right)+
\frac{1}{2n\ln{2}}-\frac{1}{2n}\right)=\\
\frac{2^n}{n}\cdot\frac{1}{\ln{4}}\left(\left(1-\frac{1}{n-1}\right)+
\frac{1}{2n}\left(\frac{1}{\ln{2}}-1\right)\right)>\\
\frac{2^n}{n}\cdot\frac{1}{\ln{4}}\left(1-\frac{1}{n-1}+
\frac{0.4426}{2n}\right) > ...$$
and again, checking 1st derivative and monotonicity we have
$$\frac{1}{\ln{4}}\left(1-\frac{1}{n-1}+
\frac{0.4426}{2n}\right)>0.6, \forall n\geq6$$
thus, finally
$$...> 0.6 \frac{2^n}{n}$$
It is also worth mentioning this paper, Ramanujan's proof of Bertrand’s postulate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that:
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx - \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.
| We will find a general reduction formula for the integral
$$I_n=\int\frac{dx}{(ax^2+b)^n}$$
Integration by parts with
$$dv=dx\Rightarrow v=x\\ u=\frac1{(ax^2+b)^n}\Rightarrow du=\frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bn\int\frac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=\frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=\frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=\int\frac{dx}{ax^2+b}$$
Let $$x=\sqrt{\frac{b}a}\tan u\Rightarrow dx=\sqrt{\frac{b}a}\sec^2u\, du$$
So $$I_1=\sqrt{\frac{b}a}\int\frac{\sec^2u}{b\tan^2u+b}du$$
$$I_1=\frac1{\sqrt{ab}}\int\frac{\sec^2u}{\sec^2u}du$$
$$I_1=\frac1{\sqrt{ab}}\int du$$
$$I_1=\frac{u}{\sqrt{ab}}$$
$$I_1=\frac1{\sqrt{ab}}\arctan\sqrt{\frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.
Edit: Whatever I'll just give you the answer.
Plugging in $a=1,\ b=1,\ n=3$ we have
$$I_{3}=\frac{x}{4(x^2+1)^{2}}+\frac{3}{4}I_{2}$$
Note that
$$I_{2}=\frac{x}{2(x^2+1)}+\frac{1}{2}I_{1}$$
$$I_{2}=\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x+C$$
So
$$I_{3}=\frac{x}{4(x^2+1)^{2}}+\frac{3x}{8(x^2+1)}+\frac{3}{8}\arctan x+C$$
| {
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Find $N = 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$ Please show how to solve this step by step, because I don't even have an idea to begin with.
$$N = 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$$
| You can see that the expression is found within itself. The trick is to manipulate that fact...
$$N+1 = 2 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$$
$$N+1 = 2+\frac{1}{N+1}$$
Can you solve the quadratic and continue from here?
P.S. You will get two values for $N$ by solving the quadratic. Only one of them is correct...
| {
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If $\tan 2\alpha \cdot \tan \alpha = 1$, then what is $\alpha$? Different methods give different answers.
If $\tan 2\alpha\cdot\tan \alpha = 1$, then what is $\alpha$?
I tried two methods but got two different answers.
Method 1:
$$\begin{align}
\tan 2\alpha\cdot\tan \alpha = 1
&\implies
\frac{2\tan \alpha}{1 - \tan ^2 \alpha}\;\tan \alpha = 1 \tag{1a}\\[6pt]
&\implies 2\tan ^2\alpha = 1 - \tan ^2\alpha \tag{1b}\\[6pt]
&\implies \tan ^2 \alpha = \frac{1}{3} \tag{1c}\\[6pt]
&\implies \tan \alpha = \pm\frac{1}{\sqrt 3} \tag{1d}\\[6pt]
&\implies \tan \alpha = \tan\left(\pm\frac{\pi}{6}\right) \tag{1e}\\[6pt]
&\implies \alpha = n\pi \pm \frac{\pi}{6} \;\text{where}\; n \in \mathbb{Z} \tag{1f}
\end{align}$$
Method 2:
$$\begin{align}
\tan 2\alpha \cdot \tan \alpha = 1
&\implies \tan 2\alpha = \frac{1}{\tan \alpha} \tag{2a}\\[6pt]
&\implies \tan 2\alpha = \cot \alpha \tag{2b}\\[6pt]
&\implies \tan 2\alpha = \tan\left(\frac{\pi}{2} - \alpha\right) \tag{2c}\\[6pt]
&\implies 2\alpha = n\pi + \frac{\pi}{2} - \alpha \text{?} \tag{2d}\\[6pt]
&\implies \alpha = \frac{1}{3}\left(n\pi + \frac{\pi}{2}\right)\;\text{where}\; n \in \mathbb{Z} \tag{2e}
\end{align}$$
Which one is correct? Is there any mistake in the above solutions?
| Your first method is correct.
In the second method, notice that the equation
$$\tan 2\alpha = \frac{1}{\tan\alpha}$$
is not valid when $\alpha = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$ or when $\alpha = \frac{\pi}{4} + \frac{n\pi}{2}, n \in \mathbb{Z}$ or when $\alpha = n\pi, n \in \mathbb{Z}$. Therefore, in your solution
$$\alpha = \frac{\pi}{6} + \frac{n\pi}{3}, n \in \mathbb{Z}$$
$n \neq 3k + 1$, $k \in \mathbb{Z}$ since that would imply
$$\alpha = \frac{\pi}{6} + k\pi + \frac{\pi}{3} = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$
which is not valid.
If we replace $n$ by $3k$, we obtain
$$\alpha = \frac{\pi}{6} + k\pi, k \in \mathbb{Z}$$
If we replace $n$ by $3k - 1$, we obtain
$$\alpha = \frac{\pi}{6} + k\pi - \frac{\pi}{3} = -\frac{\pi}{6} + k\pi, k \in \mathbb{Z}$$
which agrees with your first solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If $z=\cos\theta + i\sin \theta$ prove $\frac{z^2-1}{z^2+1}=i\tan\theta$
If $z=\cos\theta + i\sin \theta$ prove $$\frac{z^2-1}{z^2+1}=i\tan\theta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$
$$\frac{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)-1}{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)+1}$$
$$\frac{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta) -1}{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta)+1}$$
$$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}$$
I understand how I can do it with using $z=e^{i \theta}$, however I want to solve it using double angle identities.
| Your approach will work with double angle formulae, but this is quicker: since $z=\exp i\theta$, $\frac{z-1/z}{z+1/z}=\frac{2i\sin\theta}{2\cos\theta}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Eigenvalue of (some) $ 4 \times 4 $ symmetric matrices $$A=\pmatrix{
0 & 3 & 2 & 0 \\
3 & 0 & 0 & 2 \\
2 & 0 & 0 & 3 \\
0 & 2 & 3 & 0 \\
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
| We can look at the matrices $A=\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}$ and $B=\begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}$. Each of these can be split into two $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ reflection blocks, so they have eigenvalues $1$ and $-1$ each with multiplicity two. Specifically, the eigenvectors are $\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}0\\0\\1\\1\end{bmatrix}$ for $A$ and the eigenvalue $1$, $\begin{bmatrix}1\\-1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}0\\0\\1\\-1\end{bmatrix}$ for $A$ and the eigenvalue $-1$, $\begin{bmatrix}1\\0\\1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\\1\end{bmatrix}$ for $B$ and the eigenvalue $1$, and finally $\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}$ for $B$ and the eigenvalue $-1$. Linear combinations of each pair will also work, of course, such as $\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$ for the eigenvalue $1$ in both $A$ and $B$, or $\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix}$ for the eigenvalue $-1$ in both $A$ and $B$.
The matrix we care about is a linear combination $M=3A+2B$, so the two common eigenvectors $\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\-1\\-1\\1\end{bmatrix}$ we found are still eigenvectors, with eigenvalues $3+2=5$ and $-3-2=-5$ respectively. We need two more - and it turns out we can get more out of our eigenvector lists. $\begin{bmatrix}1\\1\\-1\\-1\end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $1$, and an eigenvector for $B$ with eigenvalue $-1$, so it's an eigenvector for $M$ with eigenvalue $3-2=1$. Similarly, $\begin{bmatrix}1\\-1\\1\\-1\end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $-1$ and for $B$ with eigenvalue $1$, so it's an eigenvector for $M$ with eigenvalue $-3+2=-1$.
We have found four linearly independent eigenvectors, with four different eigenvalues. As $M$ is a $4\times 4$ matrix, that's everything.
The strategy implied here isn't something that will work very reliably, but when it does and we can decompose a matrix into pieces with known eigenvectors, we can do a lot.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it possible that $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$ same as $a^{2} - 1 \equiv 0 \pmod{m}$? $a^{2} - 1 \equiv 0\pmod{m\cdot a}$ same as $a^{2} - 1 \equiv 0 \pmod{m}$?
Given that $a$ and $m$ are relatively prime, is it possible for the above equation to hold true? My rational behind this is the following:
Because $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$, this implies that $m\cdot a\mid \left(a^{2} - 1\right)$, which implies $a^{2} - 1 = l\cdot m\cdot a, l \in \mathbb Z$
Is it wrong to then say that: let $t = l\cdot a, t \in \mathbb Z$. Hence, we have $a^{2} - 1 = t\cdot m$, thus implying that $a^{2} - 1 \equiv 0 \pmod {m}$
Note that $a$ is a positive integer in this case
EDIT: What I meant to ask was does $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$ imply $a^{2} - 1 \equiv 0 \pmod{m}$?
| $a^2 -1 \pmod {ma} \implies ma|a^2 - 1 \implies m|a^2-1$ so that one direction always holds if it ever holds.
However $ma|a^2 - 1 \implies \exists k: a^2 - 1 = mak$ so $ a -\frac 1a =mk\in \mathbb Z$ which implies $a \equiv \pm 1$ so $a^2 - 1 \pmod {ma}$ only if $a = \pm 1$.
So, yes, IF $a^2 -1 \pmod {ma}$ then $a^2 -1 \pmod m$. BUT $a^2 - 1 \pmod{ma}$ only when $a = \pm 1$.
The other way:
$a^2 -1 \equiv \pmod m \implies a^2 - 1\equiv km \pmod {am}$ for some integer $k: 0\le k< m$ (and chinese remainder theorem says that such a $k$ is unique) but there is no reason to assume $k =0$ (and lots of reason to assume it isn't).
So so any $a^2 -1 \pmod {m}$ where $a\not \equiv \pm 1 \pmod m$ will be a counter example.
Fo instance $5^2 -1 \equiv 0 \pmod {24}$ and $\gcd(a,5) = 1$ but $5^2 -1 \not \equiv 0 \pmod {5*24 = 120}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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