Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Color integers of finite set $M\subseteq \mathbb{N}$ so that if different $a,b$ are the same color then $a+b$ is different color. $|M|_{\max}=?$
The integers $1, 2, \ldots, n$ are each colored with one of three colors, following the rule that if $a$, $b$ are distinct integers that both have the same color, $a+b$ must not be that color. What is the maximum possible value of $n$ for which this can be done?
Here is an example for $n=23$ (I can't find bigger $n$).
$$A=\{9,10,11,12,13,14,15,16,17,18\}$$
$$B=\{3,5,6,7,19,20,21\}$$
$$C=\{1,2,4,8,22\}$$
But is this maximum value?
| An unsatisfactory method, but:
By exhaustive search, there are only three solutions for $n=23$ :
$$\begin{align}
A &= \{1,2,4,8,11,22\},\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,16,17,18,20\}\end{align}$$
$$\begin{align}
A &= \{1,2,4,8,11,16,22\}\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,17,18,20\}\end{align}$$
$$\begin{align}
A &= \{1,2,4,8,11,17,22\}\\
B &= \{3,5,6,7,19,21,23\}\\
C &= \{9,10,12,13,14,15,16,18,20\}\end{align}$$
(Remarkably, they all share the same set $B$!)
None of these can be extended to a solution for $n=24$ (in all three cases, appending $24$ fails because $2+22=3+21=9+15=24$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Congruent number problem- one elementary statement We call a positive integer $n$ a congruent number if there is a rational right triangle with area $n$.
Equivalently, $n$ is a congruent number if there exists positive integers $x$ and $y$ such that
$$x^2\pm n=y^2.$$
Somewhere, I came across a statement:
If we take $a$ and $b$ such that from the set $\{a,b,a+b,a-b\}$ $3$ of the elements are squares, then the $4$th element is a congruent number.
I tried to prove it, but I can't find a way to do it. I think that the second definition is the one more suitable here.
For example, I tried to suppose that $a=x^2$, $b=y^2$ and $a+b=z^2$. Then $a-b=x^2-y^2$. I think this could be just an elementary exercise to find a relation from definition?
| I'll prove that if three of the four integers $\{\,a,b,a+b,a-b\,\}$ are squares, then the fourth is a congruent number. By the way, this result was first mentioned by Leonardo of Pisa, better known as Fibonacci.
First, let's get a correct definition of congruent: a positive integer $n$ is said to be congruent if there is a rational number $x$ such that both of the numbers $x^2\pm n$ are squares of rational numbers.
Next, a couple of lemmas we'll prove later:
Lemma 1. The positive integer $n$ is congruent if and only if $nk^2$ is congruent for all positive integer values of $k$.
Lemma 2. If $n$ is the area of a right triangle with integer sides then $n$ is congruent.
Now for the proof of the main result.
For any positive integers $a>b$, the right triangle with legs $2ab$ and $a^2-b^2$ has integer sides (the hypotenuse is $a^2+b^2$) and area $ab(a+b)(a-b)$, so $ab(a+b)(a-b)$ is congruent by Lemma 2. Since $ab(a+b)(a-b)$ is congruent, and three of the numbers $a$, $b$, $a+b$, $a-b$ are squares, Lemma 1 implies that the fourth of these numbers is congruent.
Now let's go back and prove the lemmas.
Proof of Lemma 1. If $n$ is congruent, then the numbers $x^2\pm n$ are both squares, so $k^2(x^2\pm n)=(kx)^2\pm nk^2$ are both squares, so $nk^2$ is congruent. And if $nk^2$ is congruent for all $k$, then it's congruent for $k=1$, so $n$ is congruent.
Proof of Lemma 2. If $a,b,c$ are positive integers with $a^2+b^2=c^2$ (so $a,b$ are legs of a right triangle with hypotenuse $c$), then both numbers $c^2\pm2ab$ are squares ($c^2+2ab=a^2+2ab+b^2=(a+b)^2$, etc.), so $2ab$ is congruent, but $2ab$ is four times the area of the triangle, so the area is congruent.
| {
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How to evaluate $\int_{0}^{+\infty} \cos(x^2)\cos(x)dx$ How do I evaluate $\int_{0}^{+\infty} \cos(x^2)\cos(x)dx$?
I don't know what to do. Should I use a contour integration?
| By the cosine addition formulas and symmetry we have
$$ \int_{0}^{+\infty}\cos(x)\cos(x^2)\,dx = \int_{0}^{+\infty}\cos\left(x^2-\tfrac{1}{4}\right)\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\cos\left(x-\tfrac{1}{4}\right)}{\sqrt{x}}\,dx $$
where the last integral is clearly converging by Dirichlet's test. Since the Laplace transform of $\cos\left(x-\frac{1}{4}\right)$ is $\frac{1}{1+s^2}\left[\sin\tfrac{1}{4}+s\cos\tfrac{1}{4}\right]$ and the inverse Laplace transform of $\frac{1}{\sqrt{x}}$ is $\frac{1}{\sqrt{\pi s}}$, the original integral equals
$$\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sin\tfrac{1}{4}+s\cos\tfrac{1}{4}}{(1+s^2)\sqrt{s}}\,ds $$
which by the substitution $s\mapsto u^2$ only depends on the elementary integrals $\int_{0}^{+\infty}\frac{du}{1+u^4}=\frac{\pi}{2\sqrt{2}}$ and $\int_{0}^{+\infty}\frac{u^2\,du}{1+u^4}=\frac{\pi}{2\sqrt{2}} $. Summarizing:
$$ \int_{0}^{+\infty}\cos(x)\cos(x^2)\,dx = \color{blue}{\frac{\sqrt{\pi}}{2}\,\sin\left(\frac{\pi+1}{4}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Cauchy Product starting from $1$ The definition of the Cauchy product from Wikipedia is defined as
$$\left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right)
=\sum_{k=0}^\infty\sum_{\ell=0}^ka_\ell b_{k-\ell}$$
Question: Does this apply if the summation runs from $1$ instead of $0$ for both $i,j$?
In particular, can it be said that
$$(\ln 2)^ 2=\left(\sum_{r=1}^\infty \frac {(-1)^{r+1}}r\right)^2
=\sum_{k=1}^\infty\sum_{\ell=1}^k \frac {(-1)^{\ell+1}}\ell\cdot \frac {(-1)^{k-\ell+1}}{k-\ell}\\
=\sum_{k=1}^\infty\sum_{\ell=1}^k \frac {(-1)^k}{\ell (k-\ell)} \text{?}$$
If not how can the Cauchy product be used to express $(\ln 2)^2$ as a double summation?
(NB - A related question is found here).
| What you need is $\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty a_{i,j} = \sum_{k=1}^\infty \sum_{\ell=1}^k a_{\ell,\,k+1-\ell}.$
$$
\begin{array}{c|ccccccccc}
& 1 & 2 & 3 & 4 & 5 & \longrightarrow j \\
\hline
1 & a_{11} & a_{12} & a_{13} & a_{14} & \cdots \\
2 & a_{21} & a_{22} & a_{23} \\
3 & a_{31} & a_{32} \\
4 & a_{41} \\
5 & \vdots \\
\downarrow \\
i
\end{array}
$$
Here are the terms in which $k=4:$
$$
\begin{array}{c|ccccccccc}
& 1 & 2 & 3 & 4 & 5 & \longrightarrow i \\
\hline
1 & & & & a_{14} \\
2 & & & a_{23} \\
3 & & a_{32} \\
4 & a_{41} \\
5 \\
\downarrow \\
j
\end{array}
$$
When $k=4,$ you have
$$
a_{14} + a_{23} + a_{32} + a_{41}.
$$
In each case, the sum of the two indices is $5,$ not $4\text{:}$ $\quad 1+4,\quad 2+3,\quad 3+2,\quad 4+1.$
So the sum $a_{14} + a_{23} + a_{32} + a_{41}$ is $\displaystyle \sum_{\ell=1}^4 a_{\ell,\,5-\ell}.$
Let us contrast that with the situation where you start with $0$ rather than with $1.$
$$
\begin{array}{c|ccccccccccc}
& 0 & 1 & 2 & 3 & 4 & 5 & \longrightarrow i \\
\hline
0 & & & & & a_{04} \\
1 & & & & a_{13} \\
2 & & & a_{22} \\
3 & & a_{31} \\
4 & a_{40} \\
5 \\
\downarrow \\
j
\end{array}
$$
Here, in each case, the sum of the two indices is $4\text{:} \quad 0+4,\quad 1+3, \quad 2+2, \quad 3+1, \quad 4+0.$
So you have $\displaystyle \sum_{\ell=0}^4 a_{\ell,\,4-\ell}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove only one solution exists for $f(\zeta)$ and $g(\zeta)$ using a convex function argument I want to prove that the following only have one solution, for $\zeta\in[0,1]$, at $\zeta =1$.
$$f(\zeta)=\frac{1}{1+\zeta}$$
$$g(\zeta) =
\frac{
(1-\zeta)(2-\zeta)\zeta
-
(2-\zeta)^2\log\left(2 - \zeta \right)
}
{\zeta(\zeta-4)(\zeta-1)^2}$$
These are plotted below. Note that $f(0)=1$, $\lim_{\zeta\rightarrow 0}g(\zeta)=\infty$ and $f(1)=1/2=g(1)$.
Both functions are convex and monotonically decreasing over this region, therefore if they intersect at $\zeta=1$ this is the only place they can? Can I shown they are both convex over $\zeta\in[0,1]$ by showing their second derivatives wrt $\zeta$ are +ve over this region? Or is this thinking too simplistic?
Note, the limit as $g(\zeta)$ as $\zeta\rightarrow 1$ is proven below:
\begin{align}
\lim_{\zeta\rightarrow 1} g(\zeta)
%%$
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{4\log(2 - \zeta) - 7\zeta - 2\zeta \log(2 - \zeta) + 3\zeta^2 + 4}
{4\zeta^3 - 18\zeta^2 + 18\zeta - 4}
\right)\\
%%%
&=
%%%
\lim_{\zeta\rightarrow 1}
\left(
\frac
{6\zeta - 2\log(2 - \zeta) - 9}
{12\zeta^2 - 36\zeta + 18}
\right)\\
%%%
&=
%%%
\frac
{6 - 9}
{12 - 36 + 18}\\
%%%
&=
%%%
\frac
{1}
{2}.
\end{align}
EDIT: So the second derivative of $g(\zeta)$ is pretty horrific. It does have a simplified form, but it is also pretty terrible.
http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))
$$g'' =
\frac
{(x - 1) x (x (x (x (x (2 x - 7) - 13) + 124) - 224) + 64)
- 2 (x (x (x (x (3 (x - 12) x + 184) - 472) + 588) - 304) + 64) \log(2 - x))}
{(x - 4)^3 (x - 1)^4 x^3}$$
It is clear from the numerics that it is greater than $0$ in the region $\zeta\in[0,1]$.
http://www.wolframalpha.com/input/?i=(d%5E2%2Fdx%5E2+(+(1-x)(2-x)x+-+(2-x)%5E2%5Clog(2+-x)+)%2F(x(x-4)(x-1)%5E2))+%3D+0
| we want to solve g(x)=f(x) for x in [0,1]
by doing the calculations we end up in this equation
$$ \ log(2-x)= \frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
$$h(x)=log(2-x)$$ and
$$ k(x)=\frac{4x^3-10x^2+6x}{(x+1)(2-x)^2}$$
h(x) is convex and k(x) is concave in [0,1] (1)
$$ k'(x)=\frac{-2(x^3+8x^2-17x+6)}{(x-2)^3(x+1)^2} $$
so k(1)=h(1)=0 and k'(1)=h'(1)=-1 (2)
from (1) and (2) we have that
$$ h(x)>=k(x)$$ for every x in [0,1] and k(x)=h(x) only for x=1
| {
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Find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$ I want to find the absolute minimum and maximum of $f(x,y)=x^2y+2xy+12y^2$ on the ellipse $x^2+2x+16y^2\leq{8}$
To calculate the critical points of $f$ I use the partial derivatives $f_x=2xy+2y$ and $f_y=x^2+2x+24y$ and the critical points will be those which $2xy+2y=0$ and $x^2+2x+24y=0$.
When I try to solve this two-equation system I don't get a point like $(a,b)$, I get the points $(0,y),(0,0),(-2,0)$ This, I guess means that the points $(0,0),(-2,0)$ and all the points of the segment from $(0,-y)$ to $(0,y)$ for $|y|\leq{\sqrt{\frac{1}{2}}}$ are critical points. Since $f'(0,0)=0$ and $f'(-2,0)=0, (0,0)$ and $(-2,0)$ are both the absolute minimums of the function. And the absolute maximum would be $(0,\frac{1}{\sqrt{2}})$ and $(0,\frac{-1}{\sqrt{2}})$ (because $f(0,y)=12y^2)$
Is this correct?
Do I still need to calculate the critical points on the boundary?
Thank you for your time.
| If $x=-1$ and $y=-\frac{3}{4}$ then we get a value $\frac{15}{2}$.
We'll prove that it's a maximal value.
Indeed, by the given $$(x+1)^2+16y^2\leq9.$$
Thus, $$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\leq$$
$$\leq|y|(x+1)^2+12y^2-y\leq|y|(9-16y^2)+12y^2-y.$$
Hence, it remains to prove that $$|y|(9-16y^2)+12y^2-y\leq\frac{15}{2}.$$
Consider two cases.
*
*$y\geq0.$
Thus, $0\leq y\leq\frac{3}{4}$ and we need to prove that
$$y(9-16y^2)+12y^2-y\leq\frac{15}{2}$$ or
$$(4y+3)(8y^2-12y+5)\geq0,$$
which is obvious even for all $y\geq0$.
*$y\leq0$.
Thus, $-\frac{3}{4}\leq y\leq0$ and we need to prove that
$$-y(9-16y^2)+12y^2-y\leq\frac{15}{2}$$ or
$$(4y+3)(8y^2-5)\leq0,$$
which is true for $-\frac{3}{4}\leq y\leq0$.
Now, about the minim.
Let $y\geq0$.
Thus, $$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\geq12y^2-y=12\left(y-\frac{1}{24}\right)^2-\frac{1}{48}\geq-\frac{1}{48}.$$
Let $y\leq0$.
Thus, $-\frac{3}{4}\leq y\leq0$ and
$$x^2y+2xy+12y^2=y(x+1)^2+12y^2-y\geq y(9-16y^2)+12y^2-y=$$
$$=-16y^3+12y^2+8y\geq\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}},$$
where the equality occurs for $y=\frac{1}{4}-\frac{1}{4}\sqrt{\frac{11}{3}}.$
Since $$-\frac{1}{48}>\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}},$$ we see that
$$\min\limits_{x^2+2x+16y^2\leq8}(x^2y+2xy+12y^2)=\frac{5}{2}-\frac{11}{6}\sqrt{\frac{11}{3}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine $ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \ $ Determine $ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \ $
Answer:
$ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5} \\ = \ \large \lim_{x \to \infty} \frac{40}{5-\frac{1}{x}}- \ \large \lim_{x \to \infty} \frac{2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \\ =8 - \frac{0 \cdot \lim_{x \to \infty} \sin (x}{5-0} \\ =8-0 \\ =8$
Am I right?
| Yes, you are right:
$$ \lim_{x \to +\infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}}=\frac{40-\lim\limits_{x\rightarrow+\infty}\frac{32\sin{x}}{2^{0.1x}}}{5-\lim\limits_{x\rightarrow+\infty}\frac{1}{x}}=8$$
| {
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If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$
I just learnt to prove If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$ a few days ago, and it was just basic application of CS. But I find this one really difficult. I can not apply CS here directly on the numbers $a+\frac{1}{a}$ and $b+\frac{1}{b}$ because CS is for squares. So some manipulation is needed. Anything from hint to full answer will be appreciated.
| By Holder and AM-GM we obtain:
$$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3=\frac{1}{4}\cdot(1+1)^2\left(\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3\right)\geq$$
$$\geq\frac{1}{4}\left(a+\frac{1}{a}+b+\frac{1}{b}\right)^3=\frac{1}{4}\left(1+\frac{1}{ab}\right)^3\geq\frac{1}{4}\left(1+\frac{1}{\frac{(a+b)^2}{4}}\right)^3=\frac{125}{4}.$$
Also we can use Jensen here.
Indeed, let $f(x)=\left(x+\frac{1}{x}\right)^3$.
Thus, $$f''(x)=\frac{6(x^6+x^2+2)}{x^5}>0$$ for all $x>0$.
Id est, $$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3\geq2\left(\frac{a+b}{2}+\frac{2}{a+b}\right)^3=\frac{125}{4}.$$
Also we can use AM-GM only.
$$\left(a+\frac{1}{a}\right)^3+\left(b+\frac{1}{b}\right)^3=5+\left(\frac{1}{ab}-1\right)^3-3ab\geq5+(4-1)^3-\frac{3}{4}=\frac{125}{4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve.
Thank you people, but I need the thinking process, not the answer.
| It's $$\left(\frac{x^2+2}{x}\right)^2-6\left(\frac{x^2+2}{x}\right)+8=0,$$
which gives
$$\frac{x^2+2}{x}=2$$ and $x\in\{1+i,1-i\}$ or
$$\frac{x^2+2}{x}=4,$$ which gives $x\in\{2+\sqrt2,2-\sqrt2\}$.
| {
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"url": "https://math.stackexchange.com/questions/2494962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to prove that this diophantine equation has only two solutions? Consider an equation of the form.
$$z^3=y^2+4$$
where z ,y $\in \Bbb N$
Hence, the solution of form (z,y) are (2,2) and (5,11).
So How to prove that this diophantine equation only has these two solutions.
| Suppose $z^3=y^2+4 = (2-yi)(2+yi)$, if the two factors $2-yi$ and $2+yi$ are coprime, then $(u+vi)^3 = 2+yi$, and the rest is easy. However, they need not be coprime, a more comprehensive analysis is necessary.
Let us first assume $z$ is even, letting $z=2z_1,y=2y_1$ gives $2z_1^3 = y_1^2+1$. Modulo $4$ shows that $z_1$ must be odd. Write $2z_1^3 = (y_1+i)(y_1-i)$. The two factors on the right cannot be coprime, since LHS is divisible by $2$ but not divisible by $4$. Let $a$ be their GCD, then $a|2$. Obviously $a$ cannot be $2$, so $a = 1+i$, a Gaussian prime factor of $2$. Therefore
$$z_1^3 = \frac{y_1+i}{1+i} \frac{y_1-i}{1-i} = \frac{(y_1-1)+(y_1+1)i}{2}\frac{(y_1-1)-(y_1+1)i}{2}$$ Now two factors on the right are coprime, let $(u+vi)^3 = \frac{(y_1-1)+(y_1+1)i}{2}$, equating real and imaginary part and cancel $y_1$ gives $$-1 = (u+v)(u^2-4uv+v^2)$$ Hence $$\begin{cases}u+v = -1 \\ u^2-4uv+v^2 = 1 \end{cases} \quad \text{or} \quad \begin{cases}u+v = 1 \\ u^2-4uv+v^2 = -1 \end{cases}$$ Only the first case yields integral solution: $(u,v)=(-1,0)$ or $(0,-1)$. Hence plugging back gives $y_1 = \pm 1, z_1 = 1$. This gives the solution $y=\pm 2,z=2$.
If $z$ is odd. Assume $a|(2-yi), a|(2+yi)$, then $a|4$. So $a = 1,1+i,2$ or $4$. But only $a=1$ is possible, because $z^3 = (2-yi)(2+yi)$ and $z$ is odd. Hence $(2+yi) = (u+vi)^3$, which gives $(u,v) = (-1,\pm 1)$ or $(2,\pm 1)$. The first case is ruled out since $z$ is odd. The second case gives a new solution $y=\pm 11, z=5$.
| {
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Prove that $\sqrt{(1+1/n)}$ is irrational for all $n\in \mathbb{N}$ I have trouble solving this probably simple problem:
Prove that $\sqrt{1+1/n}$ is irrational for all $n\in \mathbb{N}$.
| Here's another way.
$\sqrt{1 + \frac 1n} = \sqrt{\frac {n+1}{n}} = \sqrt{\frac {n(n+1)}{n^2}}= \frac 1n \sqrt{n(n+1)}$ which rational if and only if $\sqrt{n(n+1)}$ is rational.
Square roots of integers are rational only when that are integer square roots of perfect squares. So $\sqrt{n(n+1)}$ is rational only if $n(n+1)$ is a perfect square.
But if $n > 0$ (which it must be if $n$ is natural $\frac 1n$ is defined) then:
$n^2 = n*n < n*(n+1) < (n+1)(n+1) = (n+1)^2$.
If $n*(n+1) = k^2$ then $n < k < n+1$ and $k$ can not be an integer so $n(n+1)$ is not a perfect square.
[The exceptions are $n = 0$ or $n = -1$ then $n(n+1) = 0^2$. There reason the argument hold is if $n= 0$ then $n*n = n*(n+1) < (n+1)(n+1)$ so $n \le k < n+1$ so $k = n = 0$. or if $n = -1 < 0$ and $n+1 = 0$ $n*n > n*(n+1) = (n+1)(n+1)$ and so $k = n+1 = 0$.]
======
Ooooh..... this is cute:
Let $(1 +\frac 1n ) = \frac {a^2}{b^2}; a, b\in \mathbb Z; \gcd(a,b) = 1;b \ne 0$
Then $b^2(1 + \frac1n) = a^2$
$(b^2 - a^2) = \frac {b^2}{n}$.
$(b-a)(b+a) = \frac {b^2}{n}\in \mathbb Z$.
Let $p|\frac {b^2}{n}$ then $p|b$. But as $a,b$ are relatively prime. $p\not |b \pm a$ so $p\not \mid \frac {b^2}{n}$ after all and $\frac {b^2}{n} = 1$
$(b-a)(b+a) = 1$ so $(b-a) = (b+a) = \pm 1$ and $a = 0;(b = \sqrt{n})$ and $1 + \frac 1n = 0$. Which is not possible.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof: $2n! < 2^{2n} (n!)^{2}$ Could someone help with finishing this inequality induction proof.
Hypothesis: $2n! < 2^{2n}(n!)^2$
Step: $2(n+1)! < 2^{2(n+1)}((n+1)!)^2$
(2n+2)! < 2ˆ(2n + 1) * ((n + 1)!)ˆ2
| Note that $4n^2+6n+2 < 4n^2+8n+4$ so $\color{blue}{2(n+1)(2n+1)< 2^2(n+1)^2}$.
\begin{eqnarray*}
(2(n+1))!=2(n+1)(2n+1) \underbrace{\color{red}{(2n)!} < \color{blue}{2(n+1)}}_{ \text{by the } {\color{red}{ \text{inductive hypothesis}}}}\color{blue}{(2n+1)} \color{red}{2^{2n}}\underbrace { \color{red}{(n!)^2} < \color{blue}{2^2(n+1)^2}}_{ \text{using the above}} 2^{2n} (n!)^2=2^{2(n+1)} ((n+1)!)^2.
\end{eqnarray*}
| {
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How to solve a trigonometric equation involving double and triple angles? The problem is:
$\textrm{From the following equation}$
$$10\cos{2\omega}-13\cos{3\omega}+2\sin{\frac{5\omega}{2}}\sin{\frac{\omega}{2}}=0$$
$\textrm{find the value of:}$
$$\sec{\omega}+\sec{3\omega}$$
$\textrm{Consider:}$
$$\omega\neq (2K+1)\frac{\pi}{6}\;k\in \mathbb{Z}$$
I figured out some familiar expression in the third term of the equation and by applying prosthaphaeresis formula in the third term of the equation I got to this,
$10\cos{2\omega}-13\cos{3\omega}-(\cos3\omega-cos2\omega)=0$
$11\cos{2\omega}-14\cos{3\omega}=0$,
However there are second and double angles in the latter equation, therefore I decided to transform into their power equivalents shown below,
$11(\cos^{2}\omega-\sin^{2}\omega)-14(4\cos^{3}\omega-3cos\omega)=0$,
By using pitagorean identity in the earlier equation then I reached to this expression:
$11(\cos^{2}\omega-(1-\cos^2\omega))-14(4\cos^{3}\omega-3cos\omega)=0$,
$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3cos\omega)=0$,
$22\cos^{2}\omega-22-56\cos^{3}\omega+42\cos\omega=0$
to which is transformed into a cubic equation as follows (note I multiplied by $-1$):
$56\cos^{3}\omega-22\cos^{2}\omega-42\cos\omega+22=0$
I am not sure if my procedure is correct, moreover to solve a cubic equation entitles a problem since I don't know how to find the angle from there. How can I get to the answer?
| Consider the equation
$$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3\cos\omega)=0$$
and just multiply it by $2$:
$$11(4\cos^{2}\omega-2)-28(4\cos^{3}\omega-3\cos\omega)=0$$
From this equation and the assumption that $\omega \neq (2k+1)\frac{\pi}{6}$, we conclude that
$$\frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega}=\frac{28}{11}$$
But it is easy to check
$ \frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega} = \sec \omega + \sec 3\omega $.
| {
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Is it actually possible to solve $\frac{xy}{\ln(x^p)(x+y)^2}=c$ for $x$? Is it possible to solve the following for $x$
$$\frac{xy}{\ln(x^p)(x+y)^2}=c$$
where $p\in[0,1]$ and all other variables are positive real numbers ?
Without the $\ln(x^p)$ in the denominator, the solution simply is
$$x=-\frac{-y+2cy+y\sqrt{-4c+1}}{2c},\:x=-\frac{-y+2cy-y\sqrt{-4c+1}}{2c}$$
But does $\ln(x^p)$ make a solution impossible?
Is it possible to solve it $p$ assumed values like $1,0.5$ etc.?
| Probably, the best you can do (without numerical method) is to express $y$ in terms of $x$:
$$
\frac{(x+y)^2}{xy} = \frac{1}{c\ln(x^p)}
$$
$$
\frac{x}{y} + \frac{y}{x} + 2 = \frac{1}{c\ln(x^p)}
$$
Let $z=\frac{y}{x}$, then
$$
z+\frac{1}{z}=\frac{1}{c\ln(x^p)}-2
$$
$$
z=\frac{\frac{1}{c\ln(x^p)}-2\pm\sqrt{\frac{1}{c^2\ln^2 x^p}-\frac{4}{c\ln(x^p)}}}{2}
$$
$$
y=\frac{x(1-2c\ln(x^p)\pm\sqrt{1-4c\ln(x^p)}\;)}{2c\ln(x^p)}
$$
or
$$
y=\frac{x(1-2cp\ln(x)\pm\sqrt{1-4cp\ln(x)}\;)}{2cp\ln(x)}
$$
BTW, this means that $x\le e^{\frac{1}{4cp}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it okay to "ignore" small numbers in limits where $x$ approaches infinity? I got a limit:
$$\lim_{x\to\infty}\frac {(2x+3)^3(3x-2)^2} {(x^5 + 5)}$$
As far as $x$ approaches infinity, can I just forget about 'small' numbers (like $3$, $-2$ and $5$ in this example)? I mean is it legal to make a transition to:
$$\lim_{x\to\infty}\frac {(2x)^3(3x)^2} {x^5}$$
Or if it is not always okay — in what cases such transitions are okay?
| We have $$\lim_{x\to\infty}\frac{2x+3}{2x} = 1,\ \lim_{x\to\infty}\frac{3x-2}{3x} = 1,\ \lim_{x\to\infty}\frac{x^5+5}{x^5} = 1$$
and so $$\lim_{x\to\infty}\frac{(2x+3)^3(3x-2)^2}{x^5+5}= \lim_{x\to\infty}\left(\frac{\frac{(2x+3)^3}{(2x)^3}\cdot\frac{(3x-2)^2}{(3x)^2}}{\frac{x^5+5}{x^5}}\cdot\frac{(2x)^3(3x)^2}{x^5}\right) = \lim_{x\to\infty}\frac{(2x)^3(3x)^2}{x^5}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\theta$ in terms of $\cos\theta$.
Then show that $\cos3\theta=\cos2\theta$ can be written as $4z^3-2z^2-3z+1=0$ where $z=\cos\theta$.
Lastly, by solving the equation above for $z$, we are to find out the value of $\cos\frac{2\pi}{5}$.
I attempted the question with just subbing in $z=\cos\theta$ into the equation and I got: $$4\cos^3\theta-3\cos\theta=\cos2\theta$$ and don't really know where to go next.
| The trick here is to find $2$ representations of the complex exponential $$e^{ix} = \cos(x)+i\sin(x).$$
With this in mind, we have
$$e^{2ix} = \cos(2x)+i\sin(2x)$$ and
$$e^{3ix} = \cos(3x)+i\sin(3x) $$ along with
$$(e^{ix})^2 = (\cos(x)+i\sin(x))^2 = (\cos(x)^2 - \sin(x)^2) + i(2\sin(x)\cos(x))$$ and
$$(e^{ix})^3 = (\cos(x)+i\sin(x))^3 = (\cos(x)^3 -3\cos(x)\sin(x)^2) + i(3\cos(x)^2\sin(x) - sin(x)^3).$$
Equating the real part of $e^{2ix}$ yields
$$\cos(2x) = \cos(x)^2 - \sin(x)^2.$$
Using the Pythagorean identity, we see that $\sin(x)^2 = 1-\cos(x)^2$ and so we have
$$\cos(2x) = \cos(x)^2 - \sin(x)^2 = \cos(x)^2 - (1-\cos(x)^2) = 2\cos(x)^2-1.$$
Similarly, we may equate the real part of $e^{3ix}$ to get
$$\cos(3x) = \cos(x)^3 -3\cos(x)\sin(x)^2 = \cos(x)^3 - 3\cos(x)(1-\cos(x)^2) = 4\cos(x)^3 - 3\cos(x).$$
So, if we set $$\cos(2z) = \cos(3z)$$
then we have
$$2\cos(z)^2-1 = 4\cos(z)^3 - 3\cos(z)$$ which implies
$$4\cos(z)^3 - 2\cos(z)^2 -3 \cos(z) +1 =0, \text{ as desired}.$$
| {
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Integrate the given expression to find $f(1)$ If $$f(x)=\int \frac {x^2}{(x^2+2)(1+\sqrt{x^2+1})}dx$$ and $f(0)=0$ , then find $f(1)$
I multiplied and divided by $\sqrt{x^2+1}-1$ but later got stuck at $\int \frac{\sqrt{x^2+1}}{x^2+2}.dx$ How to solve this integral. Could someone help me with this?
| $$\int_{0}^{1}\frac{x^2\,dx}{(x^2+2)(1+\sqrt{1+x^2})}\stackrel{x\mapsto\sinh z}{=}\int_{0}^{\log(1+\sqrt{2})}\frac{\sinh^2(z)\cosh(z)\,dz}{(\sinh^2(z)+2)(1+\cosh z)} $$
equals, by the substitution $z=\log t$,
$$ \int_{1}^{1+\sqrt{2}}\frac{(1-t)^2 (1+t^2)}{t(1+6t^2+t^4)}\,dt $$
which can be computed by partial fraction decomposition. The roots of $t(1+6t^2+t^4)$ are given by $0$ and $\pm i(\sqrt{2}\pm 1)$ and $\int_{1}^{1+\sqrt{2}}\frac{dt}{t-\zeta}=\log\left(\frac{1+\sqrt{2}-\zeta}{1-\zeta}\right)$, hence by computing five residues and simplifying
$$ \int_{1}^{1+\sqrt{2}}\frac{(1-t)^2 (1+t^2)}{t(1+6t^2+t^4)}\,dt = \color{blue}{\log(1+\sqrt{2})-\tfrac{1}{\sqrt{2}}\left[\text{arctanh}\tfrac{1}{2}+\arctan\tfrac{1}{\sqrt{2}}\right]}.$$
| {
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Prove $ \frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40} $ Prove:
$$
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40}
$$
i have tried to write $1/40$ as $(1/40^{1/2007})^{2007}$ and prove $(1/40^{1/2007})^{2007}$ to be greater than $2007/2008$ but i quickly found out this is not true. Is there another way to manipulate this product? I think telescoping is possible but I just don't know how to do it; maybe split the fractions and work it out? Any help will be appreciated thank you.
| $$\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}=\sqrt{\frac{3\cdot (3 \cdot 5)\cdot(5 \cdot 7)\cdot... \cdot (2005\cdot 2007)\cdot2007}{4\cdot4^2 \cdot 6^2 \cdot ... \cdot2006^2\cdot 2008^2}}<\sqrt{\frac{3\cdot2007}{4\cdot2008^2}}<\frac{1}{40}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many solutions are there to $x+y+z=14$ where $x,y,z$ are all non-negative integers, $x \leq 5, y \leq 6, z \leq 7$? Through using brute force, I have got 15 triplets of solutions. Have I reached the right answer, and how can I not use brute force?
| For the given conditions, the maximum values of x, y and z are:
$$\begin{align}
x={} & 5 \\
y= {} & 6\\
z = {} & 7 \\
\sup(x+y+z) = {} & 18
\end{align}$$
Hence we must subtract 4 from $x$, $y$ and $z$ collectively such that $x+y+z=14$.
Of the 5 partitions of 4, all except 1+1+1+1 are possible allocations.
*
*4+0+0 - ${3\choose1}=3$ ways of allocating.
*3+1+0 - $3!=6$ ways of allocating.
*2+2+0- ${3\choose1}=3$ ways of allocating.
*1+1+2- ${3\choose1}=3$ ways of allocating.
Thus, as you calculated, $3+6+3+3=15$ integer solutions.
| {
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Prove that for $n\in N^*$ and $a,x \geq 0$ : $(a+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}a^nx$
Prove that for $n\in N^*$ and $a,x \geq 0$ : $(a+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}a^nx$
My proof :
$$(a+x)^{n+1} = \bigg(n\frac{a}{n} + x\bigg)^{n+1} = \bigg( \frac{a}{n} + \frac{a}{n} + \dots +\frac{a}{n} +x \bigg)^{n+1} $$
$$\geq$$
$$\bigg[ (n+1)\bigg(\frac{a}{n}\cdot\frac{a}{n}\cdot \dots \cdot \frac{a}{n}\cdot x\bigg)^{\frac{1}{n+1}} \bigg]^{n+1}$$
$$= $$
$$\bigg\{ (n+1) \bigg[ \bigg(\frac{a}{n}\bigg)^nx\bigg]^{\frac{1}{n+1}} \bigg\}^{n+1}$$
$$=$$
$$\frac{(n+1)^{n+1}}{n^n}a^nx$$
Wanted to ask if my approach seems correct and smooth enough, or if there's a better/easier way to prove it.
| use the AM-GM inequality, then
$$\frac{\frac{a}{n}+\frac{a}{n}+...+\frac{a}{n}+x}{n+1}\geq \sqrt[n+1]{\left(\frac{a}{n}\right)^n\cdot x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find value of $b$ and $c$ Suppose $x^{12} - 1$ is divisible by $x^4 + bx^2 + c$ . Find all possible values for $b$ and $c$ . I've tried to use $x^4 + bx^2 + c = 0$ for finding remainder and equaling it with zero but didn't help for finding $b$ and $c$.
| $$x^{12}-1=(x^6-1)(x^6+1)=(x^2-1)(x^4+x^2+1)(x^2+1)(x^4-x^2+1)=(x^4-1)(x^4+x^2+1)(x^4-x^2+1)$$
The situation is a bit trickier since those $3$ don't need to be the only possible solutions. Using the substitution suggested in comments by MathematicianByMistake $t=x^2$ we get $t^6-1=(t^3-1)(t^3+1)=(t-1)(t+1)(t^2+t+1)(t^2-t+1)$ needs to be divisible by $t^2+bt+c$, since $t^2-t+1$ and $t^2+t+1$ are irreducible you're left with $(t-1)(t+1)$ which can be combined only in one way to get $t^2-1$ hence those are the only $3$ choices.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a_n=(1-\frac{1}{n})^n$ is monotonically increasing sequence I try to solve it Bernoulli inequality but it too complicated, am I missing something easier?
My try-
$$\frac{a_n}{a_{n+1}}=\frac{(1-\frac{1}{n})^n}{(1-\frac{1}{n+1})^{n+1}}\\=(\frac{1}{1-\frac{1}{n+1}})(\frac{\frac{n-1}{n}}{\frac{n}{n+1}})^n\\=(\frac{1}{1-\frac{1}{n+1}})(1-\frac{1}{n^2})^n<1\\\iff (1-\frac{1}{n^2})^n<1-\frac{1}{n+1}$$
here is where that I want to use Bernoulli...
| You brought the wrong thing to the right hand side. Your inequality is equivalent to
$$1 + \frac{1}{n} < \biggl(1 + \frac{1}{n^2-1}\biggr)^n\,.$$
Now applying Bernoulli's inequality to the right hand side we get
$$\biggl(1 + \frac{1}{n^2-1}\biggr)^n \geqslant 1 + \frac{n}{n^2-1} > 1 + \frac{1}{n}\,.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int \sin^3(x)\sqrt{\cos(x)} dx$ I tried to solve $\int \sin^3(x)\sqrt{\cos(x)}\,dx$ by setting it equal to
$$\int \cos^{1/2}(x)\left(1-\cos^2x\right)\sin(x)\,dx $$
and then making $u=\cos(x)$ and $du=-\sin(x)$. I ended up with
$$\frac{-2\cos^{3/2}(x)}{3} + \frac{\cos^2(x)}{2} + C$$
but the book's answer is
$$\frac{2\cos^{7/2}(x)}{7} + \frac{2\cos^{3/2}(x)}{3} + C$$
Could you give a hint as to what I'm doing wrong? Here's my full work.
| $$\begin{align*}
\int \sin^3 x \sqrt{\cos x} \, dx
&= \int (1 - \cos^2 x) \cos^{1/2} x \sin x \, dx \\
&= \int (\cos^{1/2} x - \cos^{5/2} x) \sin x \, dx \\
&= \int -(u^{1/2} - u^{5/2}) \, du \\
&= \frac{2}{7} u^{7/2} - \frac{2}{3} u^{3/2} + C \\
&= \frac{2}{7} \cos^{7/2} x - \frac{2}{3} \cos^{3/2} x + C. \end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prime factorization of integers Find the prime factorization of the following integers:
$e) 2^{30} -1$
Click here for the solutions
I used the formulas:
$a^2-1^2=(a-1)(a+1)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$
And I became:
$2^{30} -1$
$=(2^{15}-1)(2^{15}+1)$
$=(2^5-1)(2^{10}+2^5+1)(2^5+1)(2^{10}-2^5+1)$
$=31*1057*32*993$
$=2^5*3*7*31*151*331$
What did I do wrong?
| the trick in c) to transform $2^{15}$ in $(2^5)^3$ and use $a^3-b^3$ formula
| {
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"timestamp": "2023-03-29T00:00:00",
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Area inside a rectangle
SEAMO 2016 paper E
I do not know how to start with this question and I have tried finding heights and bases of the triangle in terms of the sides of rectangle but i could not find the ideal pair
| Let $AC\cap EG=\{H\}$ and $AC\cap GF=\{I\}$.
Thus, $$\frac{AH}{HC}=\frac{AE}{GC}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3},$$
Which says that $$AH=\frac{2}{5}AC.$$
Also,
$$\frac{CI}{IA}=\frac{CG}{AF}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4},$$
Which says that $$CI=\frac{3}{7}AC.$$
Thus, $$HI=AC\left(1-\frac{3}{7}-\frac{2}{5}\right)=\frac{6}{35}AC,$$
which says
$$\frac{S_{\Delta IHG}}{S_{\Delta ADC}}=\frac{6}{35}\cdot\frac{1}{2}=\frac{3}{35}$$ and
$$S_{\Delta IHG}=\frac{3}{70}S_{ABCD}=1.5.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximising the area of a triangle with its vertices on a parabola.
Let $A\equiv (4,-4)$ and $B\equiv(9,6)$ be tw points on the parabola $y^2 = 4x$. Let $C$ be a point on the parabola between $A$ and $B$ such that the area of the triangle $\triangle ABC$ is maximal. What are the coordinates of $C$?
The parametric coordinates of C are $(t^2 , 2t)$. Using Shoelace formula, I get the area of the triangle to be $5t^2 -5t$. But this has no global maximum? In this case, what shluld I do?
| The shoelace formula for the points $(4, -4), (9,6), (t^2, 2t)$ gives
$$
A = \frac12\left|4\cdot 6 + 9\cdot 2t + t^2\cdot (-4) - (-4)\cdot 9 - 6\cdot t^2 -
2t\cdot 4 \right|\\
= \frac12|24 + 18t - 4t^2 + 36 - 6t^2 - 8t|\\
= |30 + 5t - 5t^2|
$$
Since we "[l]et $C$ be a moving point on the parabola between $A$ and $B$", that means that $t$ is between $-2$ (which makes $C = A$) and $3$ (which makes $C = B$). On this interval, $30 + 5t - 5t^2$ is positive, so we can drop the absolute value signs, and find a maximum at $t = 0.5$, which gives $C = (\frac14, 1)$ and area of $31.25$.
| {
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Find the domain and range of $y=-x^2+4x-3$ Find the domain and range of $y=-x^2+4x-3$
My Attempt:
$$y=f(x)=-x^2+4x-3$$
The given function is a polynomial of degree $2$ in $x$. $f(x)$ is defined for all $x\in R$, so the domain of $f=R$.
Again,
$$y=-x^2+4x-3$$
$$y=-(x^2-4x+3)$$
$$y=-(x-1)(x-3)$$
$$y=(x+1)(x-3)$$.
| Write $f(x)=-x^2+4x-3=1-(x-2)^2$, then
$f(x)=1-(x-2)^2\leq -1$
and the range is $(-\infty,1]$ and the domain is all the set $\mathbb{R}=(-\infty,\infty)$ because $f$ is a polynomial function. It's minimum will be always at the middle point of the two roots: $x_1=-1$ and $x_2=3$ (because $f(x)$ is a polynomial function of degree 2) which is $x=2$, so $f(2)=1$ is the minimum value of $f(x)$ at $x=2$.
If you isolate $x$ from $y=1-(x-2)^2$ you find a two partial inverses (reciprocal binary relation, the inverse relation of $f$ is not a function) for $f(x)$:
$$ x = 2\pm\sqrt{1-y}$$
and then the range of $f$ is the domain of the functions:
$$f^{-1}(y)=2\pm\sqrt{1-y}.$$
which is $(-\infty,1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Values of $a,b,c$ such that $\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$ Find the values of $a,b,c$ such that $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$
Here's what I have got so far
Using L'Hospital's rule,
$$\lim_{x \to 0}\frac{a+b-\cos x+x\sin x-c\cos x}{5x^4}=1$$
So,$a+b-1=0$
Again,
$$\lim_{x \to 0}\frac{\sin x+\sin x+x\cos x+c\sin x}{20x^3}=1$$
But the solution is given as $a=120,b=60,c=180$
There is no way that $a+b=1$
| Expand with Taylor:
$$
x(a+b-\cos x)-c\sin x=
x\left(a+b-1+\frac{x^2}{2!}-\frac{x^4}{4!}\right)
-c\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)+o(x^5)
$$
Thus you want
$$
\begin{cases}
a+b-1-c=0 \\[4px]
1/2+c/6=0 \\[4px]
-1/24-c/120=1
\end{cases}
$$
The second equation becomes $c=-3$, the third becomes $-5-c=120$.
No solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration by parts of $x^5\ln(x)$ I have to integrate the function: $x^5\ln(x)$
My Attempt
$$\int(x^5\ln(x))dx$$
$u=\ln(x)$ and $du=\frac{1}{x}dx$
$dv=x^5dx$ and
$v=\frac{x^6}{6}$
Using this I can then integrate the function using the method
$$uv-\int v du$$
Which then substituting I get:
$$\frac{x^6\ln(x)}{6}-\int\frac{x^6}{6}\frac{1}{x}dx$$
$$\frac{x^6\ln(x)}{6}-\int\frac{x^5}{6}dx$$
$$\frac{x^6\ln(x)}{6}-\frac{x^6}{36}+C$$
My question is if I used a different u instead of the first u which I used my first attempt how will getting the same answer look like? I tried setting $u=x^5$ and it just sort of repeated itself when integrating, I know the answer is the same but how does one get there?
What I mean
$u=x^5$ and $du=5x^4$
$dv=\ln(x)dx$ and $v=x\ln(x)-x$
Then using the method previously stated I get:
$$x^6\ln(x)-x^6-\int(x\ln(x)-x)(5x^4)dx$$
| how about
$u = x^3$
$du=3x^2 dx$
$\int(x^5\ln(x))dx = \int \frac{ux^2 \ln(u^\frac{1}{3})}{3x^2} du $
properties of logs to eliminate cube root
=$\frac{1}{3} \int \frac{u \ln(u)}{3} du $
=$ \frac{1}{9}\int u \ln(u) du $
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \int \frac{1}{2}u^2 / udu )$
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \int \frac{1}{2}u du) $
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \frac{1}{4}u^2) $
=$ \frac{1}{18}x^6\ln(x^3) - \frac{1}{36}x^6 $
=$ \frac{1}{6}x^6\ln(x) - \frac{1}{36}x^6 $
$ \frac{1}{6}x^6\ln(x) - \frac{1}{36}x^6 +c $
| {
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Finding the set of values of the real number $a$ for which $\sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a$ converges
Find the set of values of the real number $a$ for which $\displaystyle\sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a$ converges.
I had been stuck on this problem for a while but now I have managed to deduce following :
Using Maclaurin series for $\sin$ function,
$$\sin \frac 1n=\frac 1n - (\frac 1n)^3 \frac {1}{3!} + (\frac 1n)^5 \frac {1}{5!} - (\frac 1n)^7 \frac {1}{7!}+\dots$$
which implies
$$\frac 1n - \sin \frac 1n = (\frac 1n)^3 \frac {1}{3!} - (\frac 1n)^5 \frac {1}{5!} + (\frac 1n)^7 \frac {1}{7!}-\dots$$
and finally
$$0 \le (\frac 1n - \sin \frac 1n)^a \le ((\frac 1n)^3 \frac {1}{3!})^a$$
where the first inequality follows from the result : $\sin x \le x \; \forall x\gt 0$.
Hence
$$ 0 \le \sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a \le \sum_{n=1}^{\infty} ((\frac 1n)^3 \frac {1}{3!})^a=\sum_{n=1}^{\infty} (\frac {1}{3!})^a \frac 1{n^{3a}}=(\frac {1}{3!})^a\sum_{n=1}^{\infty}\frac 1{n^{3a}}.$$
Last inequality converges for $3a \gt 1.$
Thus the given series converges for $a \in (\frac 13, \infty).$
Are there any loopholes in this solution? What are other ways to think about this problem?
| For the divergence:
\begin{align*}
\sum_{n}\left(\frac{1}{n}-\sin\frac{1}{n}\right)^{a}&\geq\sum_{n}\left(\frac{1}{3!}\frac{1}{n^{3}}-\frac{1}{5!}\frac{1}{n^{5}}\right)^{a}\\
&=\frac{1}{(3!)^{a}}\sum_{n}\frac{1}{n^{3a}}\left(1-\frac{3!}{5!}\frac{1}{n^{2}}\right)^{a}\\
&\geq\frac{1}{(3!)^{a}}\frac{1}{2^{a}}\sum_{n\geq N}\frac{1}{n^{3a}}
\end{align*}
for large $N$, so it diverges for all $a\in(0,1/3]$. For $a\leq 0$ is clear, because the tail does not converge to zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ , where $ \ l=constant \ $
Answer
The given series is convergent clearly .
$ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \\ = \frac{1}{1^2+l^2}+\frac{1}{2^2+l^2}+......+\frac{1}{n^2+l^2}+........ \\ = \frac{1}{l} [ d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{n}{l})+.....] \\ = \frac{1}{l} \lim_{n \to \infty} \sum_{k=1}^{n}[d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{k}{l})] \\ = \frac{1}{l} \ d (\lim_{n \to \infty} \sum_{k=1}^{n}[ \tan^{-1}(\frac{1}{l})+ \tan^{-1}(\frac{2}{l})+........+ \tan^{-1}(\frac{k}{l})]) $
but now i can't proceed to find the sum of the series.
Help me out.
| Hint: Use the fact that
\begin{align}
\sum^\infty_{n=-\infty} \frac{1}{x^2+n^2} = \frac{\pi\coth(\pi x) }{x}.
\end{align}
| {
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Using the factorization of a quadratic equation If $a,b$ are the roots of the equation $x^2 -px +q=0$ and $a_1, b_1$ are the roots of the equation $x^2-qx +p=0$ then find the equation of quadratic whose roots are $\frac{1}{a_1b} + \frac{1}{ab_1}$ and $\frac{1}{aa_1} + \frac{1}{bb_1}$.
I know that the equation $Qx^2 + bx +c =0$ whose roots are $e,d$ can be expressed as $Q(x-e)(x-d)=0$. What I'm confused about in this question stated above is that when I try to write the new asked quadratic equation what do I take the $Q$ term to be?
| $$\left(\frac{1}{a_1b} + \frac{1}{ab_1}\right)+\left(\frac{1}{aa_1} + \frac{1}{bb_1}\right)=\frac{(a_1+b_1)(a+b)}{a_1b_1ab}=\frac{pq}{pq}=1.$$
$$\left(\frac{1}{a_1b} + \frac{1}{ab_1}\right)\left(\frac{1}{aa_1} + \frac{1}{bb_1}\right)=\frac{(ab_1+a_1b)(aa_1+bb_1)}{p^2q^2}=$$
$$=\frac{a^2q+b_1^2p+a_1^2p+b^2q}{p^2q^2}=\frac{(q^2-2p)q+(p^2-2q)p}{p^2q^2},$$
which by the Viete's theorem gives the answer:
$$x^2-x+\frac{p^3+q^3}{p^2q^2}-\frac{4}{pq}=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$ proof by induction
Prove with mathematical induction that for every $n \in \Bbb N$ it holds:
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$$
*
*Basis (P(1))
$\frac21=2^1$
*Inductive step - If it holds for P(n), then it also holds for P(n+1)
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1} \cdot \frac{2(n+1)} {2(n+1)-1} = 2^n \cdot \frac{2n+2} {2n+1}$$
But this is not true, right? Because there are more numbers between $\frac{2n}{2n-1}$ and $\frac{2n+2} {2n+1}$ or something like that? In numerator, a number will be for one bigger than the previous number, and that's why what I have written in inductive step isn't true because there's a missing $2n+1$ in the numerator.
Then I tried to write it like this.
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1} \cdot \frac{2n+1}{?}\cdot \frac{2n+2} {2n+1}$$
But then I don't know what to put in the denominator because in denominator, each number is for two bigger than the previous one, so the number should be $2n+1$ but that doesn't make any sense to me anymore.
I don't know what to do with this problem. How can this be proven?
| In the induction step, you assume
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$$
and wish to show that
$$\frac{(n+1)+1}1 \cdot \frac{(n+1)+2}3 \cdots \frac{2(n+1)}{2(n+1)-1}=2^{n+1}.$$
You have
\begin{align}\frac{(n+1)+1}1 \cdot \frac{(n+1)+2}3 \cdots \frac{2(n+1)}{2(n+1)-1}&=\frac{n+2}1 \cdot \frac{n+3}3 \cdots \frac{2n+2}{2n+1} \\&=\frac{1}{n+1} \cdot \frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1} \cdot \frac{2n+1}{2n+1} \cdot \frac{2n+2}{1} \\
&= \frac{1}{n+1} \cdot 2^n \cdot \underbrace{\frac{2n+1}{2n+1}}_{=1} \cdot (2n+2) \textrm{ (by the induction hypothesis)}\\
&= \frac{2n+2}{n+1} \cdot 2^n \\
&= 2 \cdot 2^n \\
&= 2^{n+1} \end{align}
| {
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Prove that $\displaystyle 1 <\cos A+\cos B+\cos C$ Prove that $\displaystyle 1 < \cos A+\cos B+\cos C \leq \frac{3}{2}$ where $A+B+C = \pi$
Attempt: $\cos A+\cos B+\cos C= 1+4\cos A\cdot \cos B\cdot \cos C$
Now $\displaystyle \cos A\cdot \cos B\cdot \cos C=\frac{1}{2}\cos A\left[\cos (B-C)-\cos A\right]\leq \frac{1}{2}\cos A(1-\cos A)\leq \frac{1}{2}\left[\frac{\cos A+1-\cos A}{2}\right]^2 = \frac{1}{8}$
so $\displaystyle \cos A+\cos B+\cos C\leq 1+\frac{1}{2} = \frac{3}{2}$
could some help me how to solve for minimum , thanks
| This question is as old as the OP's age ( probably )...Let's do the left inequality. $\cos A +\cos B+\cos C > 1\iff \cos A+\cos B -\left(1-\cos C\right) > 0\iff2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)-2\sin^2\left(\frac{C}{2}\right)>0\iff 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)-2\sin^2\left(\frac{C}{2}\right)>0\iff2\sin\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)-\sin\left(\frac{C}{2}\right)\right)>0\iff2\sin\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)-\cos\left(\frac{A+B}{2}\right)\right)>0\iff 4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)>0$. This last inequality is clearly true since each factor is $ > 0$. The right inequality is already done.
| {
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Computing: $\lim\limits_{x\to 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$ $$\lim_{x\to 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$$
If the limit is a finite non zero number, find n.
I tried using L'Hôpital's rule but failed. I differentiated it but not understanding what exactly should be done. Please help.
| Given that
$$ \cos'x|_{x=0} =\lim_{x\to 0}\frac{\cos x-1}{x} =\sin 0=0~~~and ~~~\lim_{x\to 0}\frac{e^x-1}{x} =(e^x)'|_{x=0}= 1$$
or by taylor series we have, $$\frac{\cos x-1}{x} \sim \frac{x}{2}~~~and ~~~\lim_{x\to 0}\frac{e^x-1}{x} \sim 1$$
and
$$ \frac{(\cos x-1)(\cos x-e^x)}{x^n} = \frac{(\cos x-1)(\color{red}{\cos x-1+1-e^x)}}{x^n} \\= \frac{1}{x^{n-2}}\left[\left(\frac{\cos x-1}{x}\right)^2 -\left(\frac{\cos x-1}{x}\right) \left(\frac{e^x-1}{x}\right) \right]\\\sim \frac{1}{x^{n-2}}\left[\frac{x^2}{4}-\frac{x}{2}\right] = \left[\frac{x^4}{4x^{n}}-\frac{x^3}{2x^n}\right] $$
we have
$$\lim_{x\to 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}=\begin{cases}0
\ &n= 0,1,2\\-\frac{1}{2}
\ &n=3\\diverges& n=4,5....\end{cases}$$
| {
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Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.
First substitute $x=\frac {a^2}u$ so$$\begin{align*}I & =-\int\limits_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}\\ & =\int\limits_0^{\infty}du\,\frac {2\log^2a}{a^2+x^2}-\int\limits_0^{\infty}du\,\frac {\log^2 u}{a^2+x^2}\end{align*}$$Hence$$\begin{align*}I & =\int\limits_0^{\infty}du\,\frac {\log^2a}{a^2+x^2}\\ & =\frac {\log^2a}{a^2}\int\limits_0^{\pi/2}dt\,\frac {a\sec^2t}{1+\tan^2t}\\ & =\frac {\pi\log^2a}{2a}\end{align*}$$However, when $a=e$ Wolfram Alpha evaluates the integral numerically as$$I\approx 2.00369$$however the input that I arrived at evaluates numerically$$\frac {\pi}{2e}\approx0.5778$$Where did I go wrong? And how would you go about solving this integral?
| $I =-\int_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}$
You are mixing $x's$ and $u's$
$I =\int_0^{\infty}\frac {\ln^2 x}{x^2+a^2}\ dx\\
x = \frac {a^2}{u}, dx = -\frac {a^2}{u^2} du\\
\int_{\infty}^0\frac {(\ln \frac {a^2}{u})^2}{\frac {a^4}{u^2}+a^2} (-\frac {a^2}{u^2})\ du\\
\int_0^{\infty}\frac {(2\ln a - \ln u)^2}{a^2+u^2} \ du$
Which is not
$\int_0^{\infty}\frac {2\ln^2 a - \ln^2 u}{a^2+u^2} \ du$
| {
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The ± sign in square root As I was doing a SAT question when I came across this question:
$\sqrt {x-a} = x-4$
If $a=2$,what is the solution set of the equation?
Options
*
*{$3,6$}
*{$2$}
*{$3$}
*{$6$} Correct Answer
I evaluated the equation and got $0=(x-3)(x-6)$If you put those number in the equation, you should get:
For 3:
$\sqrt {3-2} = 3-4$
Since $\sqrt {1} = ±1$
$±1 = -1$
For 6:
$\sqrt {6-2} = 6-4$
Since $\sqrt {4} = ±2$
$±2 = 2$
For the answer, they(SAT) evaluated $\sqrt {1}$ as $\sqrt {1} = -1$ and $\sqrt {4}$ as $\sqrt {4} = 2$ Why is it that $\sqrt {1}$ is equal to $-1$ and not $1$ and why $\sqrt {4}$ is equal to $2$ and not $-2$ Why isn't the solution set {$3,6$} a correct answer?
| By definition $\sqrt{}$ is always the non-negative root. Every positive number has exactly two square roots equal in magnitude, on positive and one negative.
$\sqrt{25} =5$ and $\sqrt {25} \ne -5$. But both $5$ and $-5$ are solutions to $x^2 = 25$.
To solve an equation $x^2 = k$ there will be two answers. One is $\sqrt k$ and $\sqrt k > 0$ and the other is $-\sqrt{k}$ and $-\sqrt {k} < 0$
So if you try to solve an equation by "squaring both sides", you will be changing the equation to allow for two different square roots that were not part of the original problem. This is called superfluous solutions.
So to solve
$\sqrt {x-2} = x- 4$
Is not just to solve $x-2 = (x-4)^2$ but is to ALSO solve $x - 4 \ge 0$.
So you did $\sqrt{x-2}^2 = (x-4)^2$. But that adds in the negative solution as well.
$x^2 - 8x + 16 = x-2$ and $x^2 - 9x + 18 = (x-6)(x-3)$ so both of those solve $x-2 = (x-4)^2$ but only one of them solve $\sqrt{x-2} = x-4$. (Because we must have $x-4 \ge 0$.)
$\sqrt{6-2} \overset?= 6-4$
$\sqrt{4} \overset?= 2$
$2 \overset \checkmark = 2$ check. $6$ is an answer and $6-4 > 0$.
$\sqrt{3-2} \overset?= 3-4$
$\sqrt{1} \overset?= -1$
$1 \ne -1$. No! $\sqrt{1} = 1$. $\sqrt{1} \ne -1$. And $3-2 < 0$.
| {
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Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:
Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.
This problem is seems to be related to Vieta's Theorem, but so far I have not used it. This is my working out:
\begin{align}
\frac {\alpha}{\beta} + \frac {\beta}{\alpha} & = \frac {\alpha^2}{\alpha \beta} + \frac {\beta^2}{\alpha \beta} \\
& = \frac {\alpha^2+\beta^2}{\alpha \beta}
\end{align}
Vieta's Theorem:
Let $p(x)=ax^2+bx+c$ be a quadratic polynomial with zeros $\alpha$,$~\beta$. Then
$$\frac {-b}{a}=\alpha + \beta \\
\frac{c}{a} = \alpha \cdot \beta $$
Well, it is clear that the question wants us to find $\alpha^2 + \beta^2$ and $\alpha \beta$.
\begin{align}
\alpha^2+\beta^2 & = \alpha^2+2\alpha \beta + \beta^2 -2\alpha\beta \\
& = (\alpha+\beta)^2-2(\alpha\beta)
\end{align}
We have:
\begin{align}
\alpha^2 + 3\alpha + 1 & = \beta^2 + 3\beta + 1 \\
0 & =\alpha^2 - \beta^2+3\alpha-3\beta \\
& = (\alpha+\beta)(\alpha-\beta)+3(\alpha-\beta) \\
& = (\alpha-\beta)(\alpha+\beta+3) \\
\therefore~ \alpha + \beta & = -3\tag{reject $\alpha=\beta$}
\end{align}
This is where I am stuck because I am unable to find $\alpha\beta$ by algebraic manipulation. I have thought about trying to find $\alpha$ and $\beta$ through the quadratic formula, but it seems quite tedious, so it is a last resort. Is there a method to finish this question off?
| Since $\alpha $ and $\beta $ are roots of $x^2+3x+1$ we know that $\alpha +\beta = -3$ and $\alpha \beta = 1$
Indeed $$ x^2+\color{red}{3}x+\color{blue}{1} =(x-\alpha)(x-\beta) = x^2 -\color{red}{(\alpha +\beta)}x+\color{blue}{\alpha \beta}$$
Hence,
$$\frac {\alpha}{\beta} + \frac {\beta}{\alpha} = \frac {\alpha^2+\beta^2}{\alpha \beta} = \frac {(\alpha +\beta)^2 -2\alpha \beta}{\alpha \beta} = \frac{(-3)^2-2}{1}=\color{blue}{7}$$
| {
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Find all prime numbers $p$ such that $16p+1$ is a perfect cube What I have attempted:
Suppose $16p+1=k^3$ where $k \in Z$ then $16p=k^3-1=(k-1)(k^2+k+1)$ so we can say that $k=17$ and thus $p=17^3+17+1=4931$ which is prime.
How would I find the remaining numbers?
| Since $k^3\equiv1\pmod{16}\implies k\equiv1\pmod{16}$, if $16p+1$ is a perfect cube, we must have
$$
\begin{align}
16p+1
&=(16k+1)^3\\
&=4096k^3+3\cdot256k^2+3\cdot16k+1
\end{align}
$$
Thus, we get $p=256k^3+48k^2+3k=(256k^2+48k+3)k$, which can only be prime if $k=1$, that is $p=307$ and thus
$$
17^3=16\cdot307+1
$$
is the only case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Simultaneous recurrence relations: $a_n = a_{n-1} (1 - b_n), b_n = b_{n-1} (1-a_n)$ I want to solve the following simultaneous recurrence relations:
\begin{align*}
a_n &= a_{n-1} (1-b_n) \\
b_n &= b_{n-1} (1-a_n).
\end{align*}
Initial conditions are $a_1 = b_1 = 1$ and $a_2 + b_2 = 1$.
The symmetric case, $a_2 = b_2 = 1/2$, has a simple solution: $a_n = b_n = 1/n$.
I have combined the equations through $a_n + b_n$, $a_n b_n$, and $a_n/b_n$ without getting closer to a solution.
If my algebra hasn't failed me, the two equations imply the following:
$$a_n a_{n-2} + a_n a_{n-1} a_{n-2} = a_{n-1}^2 + a_n a_{n-1}^2.$$
If it's of any help, it seems that $a_n$ approaches $2 - 1/a_2$ when $n$ tends to infinity for $a_2 \geq 1/2$.
| Since$$b_n=1-\frac{a_n}{a_{n-1}}$$
we get
$$1-\frac{a_n}{a_{n-1}}=\left(1-\frac{a_{n-1}}{a_{n-2}}\right)(1-a_n)$$
Multiplying the both sides by $a_{n-1}a_{n-2}$ gives
$$a_{n-1}a_{n-2}-a_na_{n-2}=a_{n-1}a_{n-2}-a_na_{n-1}a_{n-2}-a_{n-1}^2+a_na_{n-1}^2,$$
i.e.
$$a_{n-1}^2(a_n-1)=a_na_{n-2}(a_{n-1}-1)$$
Dividing the both sides by $a_{n}a_{n-1}$ gives
$$\frac{a_{n-1}(a_n-1)}{a_{n}}=\frac{a_{n-2}(a_{n-1}-1)}{a_{n-1}}$$
So, if we define $$c_n=\frac{a_{n-1}(a_n-1)}{a_n}$$
then we have
$$c_n=c_{n-1}=\cdots =c_2=\frac{a_1(a_2-1)}{a_2}=-\frac{b_2}{a_2}:=a$$
So,
$$\frac{a_{n-1}(a_n-1)}{a_n}=a\implies \frac{1}{a_n}+\frac{a}{a_{n-1}}=1\implies d_n-d_{n-1}=\frac{1}{(-a)^n}$$
where
$$d_n=\frac{1}{(-a)^n}\cdot\frac{1}{a_n}$$
So, for $n\ge 2$, we have
$$d_n=d_1+\sum_{k=2}^{n}\frac{1}{(-a)^k}=-\frac{1}{a}+\frac{a}{a+1}\left(\frac{1}{a^2}-\frac{1}{(-a)^{n+1}}\right)$$
which holds for $n=1$.
So, we get
$$a_n=\frac{a+1}{1-(-a)^n},\qquad b_n=\frac{a+1}{1-(-a)^n}(-a)^{n-1},$$
i.e.
$$\color{red}{a_n=\frac{1}{\displaystyle\sum_{k=0}^{n-1}\left(\dfrac{b_2}{a_2}\right)^k},\qquad b_n=\frac{\left(\dfrac{b_2}{a_2}\right)^{n-1}}{\displaystyle\sum_{k=0}^{n-1}\left(\dfrac{b_2}{a_2}\right)^k}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$'s General Solution I was looking for the general solution of the differential equation $$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$$
My work
We need to get the auxillary equation of the given differential equation above...
The auxillary equation would be:
$$r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$$
We need to get the roots of the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ using the rational zero theorem...
The rational zero theorem states that if $P(x)$ is a polynomial with integer coefficients and if is a zero of $P(x) ( P(\frac{p}{q} ) = 0 )$,
then $p$ is a factor of the constant term of $P(x)$ and $q$ is a factor of the leading coefficient of $P(x)$.
With that in mind....
The factors of $p$ are:
$$-1,12,1,-12$$
$$-2,6,2,-6$$
$$-3,4,3,-4$$
The factors of $q$ are:
$$1,1,-1,-1$$
Then the values of $\frac{p}{q}$ would be:
$$-1,1,2,-3,4,-4,12,-12,6,-6,-2,3$$
It was found out that only $r= -2$ and $r= 3$ could make the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ true, instead of $5$
Now....if I only got $2$ real and distinct roots out of possible $5$, then the solution of the differential equation $(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$
would be:
$$ y = c_1e^{-2x} + c_2e^{3x} + (unknown \space term) + (unknown \space term) + (unknown \space term) $$
I'm stuck....How do you get the solution of the given differential equation above?
UPDATE
After some feedback from intrepid answerers....I was able to factor out the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ into $(r+2)(r+2)(r-3)(r^2 +1)=0$, so I think the solution to the differential equation above would be:
$$ y = c_1e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + e^{x}(c_4\cos (-x) + c_5\sin (-x)) $$
or
$$ y = c_1e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + e^{x}(c_4\cos x - c_5\sin x) $$
Is my solution now correct?
| $(r+2)^2(r-3)(r^2+1)$
We have one double root and one pair of complex roots.
With complex roots:
$Ae^{it} + Be^{-it} = C_1 \cos t+ C_2\sin t$
are solutions
With roots of multiplicity.
$C_3^{-2t}+ C_4te^{-2t}$ will be solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this system of linear equations over $\mathbb{F}_{11}$? Given
$$\begin{pmatrix}0&1&2\\ 4&3&a\\ 1&2&4\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1\\ b\\ 4\end{pmatrix}$$
I need to find for what values $a,b \in \mathbb{F}_{11}$ there are solutions and what the general solution is (without determinan).
I created the following matrix
$$\begin{pmatrix}0&1&2&|1\\ 4&3&a&|b\\ 1&2&4&|4\end{pmatrix}$$
and after Gaussian elimination I got
$$\begin{pmatrix}4&3&a&|b\\ 0&5&10&|5\\ 0&0&a+5&|b\end{pmatrix}$$
How to find witch values of $a,b$ this system is trace?
| Writing $Ax=b$ for your system of linear equations, the determinant of $A$ is given by
$$
\det(A)=2(2a-27).
$$
Over the field $\mathbb{F}_{11}$ the determinant is non-zero if and only if $a\neq 8$ (we could write $a\neq -3=8$ as well). Then the unique solution is given by
$$
x=A^{-1}b=\frac{1}{a+3}\begin{pmatrix} 9(a + 7b + 7) \cr 3(a + 9b + 5) \cr b + 10 \end{pmatrix}.
$$
In case that $a=8$, we get a $1$-parameter solution, because the rank of $A$ is then equal to $2$. More precisely, we obtain $b=1$ and $x=(x_1,x_2,x_3)$ is a the general solution with $x_1=8(x_3 + 8)$ and
$x_2=5(x_3 + 5)$. The parameter $x_3$ is free, and can be any element of the finite field.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve a system of linear equations modulo n? For example,
$4x - 10y \equiv 8\pmod {20}$
$7x + 2y \equiv 5\pmod {20}$
It resembles linear diophantine equations and the Chinese Remainder Theorem, but I don't know how to actually solve it..
| \begin{eqnarray}
4x - 10y \equiv 8\pmod {20} \\
7x + 2y \equiv 5\pmod {20}
\end{eqnarray}
Method 1 - Take advantage of a unit coefficient (7)
\begin{align}
7x + 2y &\equiv 5\pmod {20} \\
21x + 6y &\equiv 15\pmod {20} \\
x + 6y &\equiv 15\pmod {20} \\
x &\equiv 14y + 15 \pmod{20} \\
\hline
4(14y + 15) - 10y &\equiv 8\pmod {20} \\
46y + 60 &\equiv 8 \pmod{20} \\
6y &\equiv 8 \pmod{20} \\
3y &\equiv 4 \pmod{10} \\
21y &\equiv 28 \pmod{10} \\
y &\equiv 8 \pmod{10} \\
y &= 8 + 10 t \\
\hline
x &\equiv 14(8 + 10 t) + 15 \pmod{20} \\
x &\equiv 112 + 140t + 15 \pmod{20} \\
x &\equiv 7 \pmod{20} \\
y &\equiv 8 \pmod{10} \\
\end{align}
Method 2 - Convert to prime-power moduli
\begin{eqnarray}
2y &\equiv 0\pmod 4 \\
3x + 2y &\equiv 1\pmod 4 \\
\hline
y &\equiv 0 \pmod 2 \\
\hline
3x &\equiv 1\pmod 4 \\
9x &\equiv 3\pmod 4 \\
x &\equiv 3 \pmod 4
\end{eqnarray}
\begin{eqnarray}
4x &\equiv 3\pmod 5 \\
2x + 2y &\equiv 0\pmod 5 \\
\hline
-4x &\equiv -3 \pmod 5 \\
x &\equiv 2 \pmod 5 \\
\hline
4 + 2y &\equiv 0\pmod 5 \\
2 + y &\equiv 0 \pmod 5 \\
y &\equiv 3 \pmod 5
\end{eqnarray}
$\left\{
\begin{array}{c}
x \equiv 3 \pmod 4 \\
x \equiv 2 \pmod 5 \\
\end{array}
\right\}
\implies
\left\{
\begin{array}{c}
5x \equiv 15 \pmod{20} \\
-4x \equiv -8 \pmod{20} \\
\end{array}
\right\}
\implies
x \equiv 7 \pmod{20}$
$\left\{
\begin{array}{c}
y \equiv 0 \pmod 2\\
y \equiv 3 \pmod 5 \\
\end{array}
\right\}
\implies
\left\{
\begin{array}{c}
-5y \equiv 0 \pmod{10} \\
6y \equiv 18 \pmod{10} \\
\end{array}
\right\}
\implies
y \equiv 8 \pmod{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The order of a modulo p.
If $p$is prime and $\operatorname{ord}_p(a)=4$, then $1+a+a^2+a^3≡ 0\bmod p$, where $\operatorname{ord}_p(a)$ is the order of $a$ modulo $p$.
I think it is true statement
$\operatorname{ord}_p(a)=4$, then $a^4≡1\bmod p$ so $a^4-1≡0\bmod p$
since $a^4-1=1+a+a^2+a^3$, then $a^4-1=1+a+a^2+a^3\equiv0\bmod p$
is that correct please?
| You can also find this directly since $4$ is conveniently small.
Since $p$ is prime, there are two roots of $1$: $\{1,-1\}$. Since $(a^2)^2\equiv 1$ but $a^2\not\equiv 1 \bmod p$ we must have $a^2\equiv \color{red}{-1} \bmod p$. Then $a^3\equiv a^2\cdot a \equiv \color{blue}{-a} \bmod p$.
Thus $1+a+a^2+a^3\equiv 1+a+ \color{red}{-1}+ \color{blue}{-a} \equiv 0\bmod p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
prove that $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ are positive real numbers such that $abc=1,$
prove that $$a^2+b^2+c^2 \geq a+b+c$$
i was thinking of using AM GM inequlaity.but donot have idea on which pairs to apply it.
any hint....
| By QM-AM for the first inequality and AM-GM for the second,
$$\frac{a^2+b^2+c^2}{3}\geq\left( \frac{a+b+c}{3} \right)^2\geq\frac{a+b+c}{3}\left( \sqrt[3]{abc} \right)$$
so since $abc=1, a^2+b^2+c^2\geq a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify expression $\frac{2\cos(x)+1}{4\cos(x/2+π/6)}$ How to simplify the following expression: $$\frac{2\cos(x)+1}{4\cos\left(\frac x2+\fracπ6\right)}$$
I got to: $ \dfrac{2\cos(x)+1}{4\cos\left(\dfrac x2\right)\cdot \dfrac{\sqrt3}2-\sin(x) \cdot \frac 12}$
| The denominator can be simplified to $$4(\cos(x/2)\cos(\pi/6)-\sin(x/2)\sin(\pi/6))=2(\sqrt3\cos(x/2)-\sin(x/2))$$
The numerator is $$2\cos(x)+1=2(2\cos^2(x/2)-1)+1=4\cos^2(x/2)-1$$
So you have $$\require{cancel}\frac{4\cos^2(x/2)-1}{2\sqrt3\cos(x/2)-2\sin(x/2)}=\frac{4\cos^2(x/2)-\cos^2(x/2)-\sin^2(x/2)}{2\sqrt3\cos(x/2)-2\sin(x/2)}\\=\frac{3\cos^2(x/2)-\sin^2(x/2)}{2\sqrt3\cos(x/2)-2\sin(x/2)}\\=\frac{(\sqrt3\cos(x/2)-\sin(x/2))(\sqrt3\cos(x/2)+\sin(x/2))}{2(\sqrt3\cos(x/2)-\sin(x/2))}\\=\frac{\cancel{(\sqrt3\cos(x/2)-\sin(x/2)}(\sqrt3\cos(x/2)+\sin(x/2))}{2\cancel{(\sqrt3\cos(x/2)-\sin(x/2))}}\\=\frac12(\sqrt3\cos(x/2)+\sin(x/2))$$
Edit:
As pointed out in the comments, this can be simplified further
$$=\left(\frac{\sqrt3}2\cos(x/2)+\frac12\sin(x/2)\right)=\cos(\pi/6)\cos(x/2)+\sin(x/2)\sin(\pi/6)\\=\cos\left(\frac x2-\frac\pi6\right)$$
Or equivalently, $\sin\left(\frac\pi3+\frac x2\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Finding $x$ values of $\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $
Solve $$\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $$ for $0^\circ<x<120^\circ$
I simplify it to
$$2 \cos 6x + 2 = 3 + \cos 3x $$
There is $\cos 6x $ and $\cos 3x$ how do I merge them together? To solve the equation from $0^\circ$ to $120^\circ$?
| Here's what I got...(EDIT: $3 \cos 3x$ is incorrect in the third line; the correct quantity is $\cos 3x$.).
$$\cos 6x + 1 = \frac {3}{2} + \frac {1}{2} \cos 3x$$
Multiplying by $2$ we get
$$2 \cos 6x + 2 = 3 + \cos 3x$$
Swapping constants and bringing over $\cos 3x$ we then get
$$2 \cos 6x - \cos 3x = 1$$
Using the double angle identity $\cos 6x = 2 \cos^2 3x - 1$ we get
$$4 \cos^2 3x - 2 - \cos 3x = 1$$
Adding 2 to each side gives us
$$4 \cos^2 3x - \cos 3x = 3$$
Now we can factor out $\cos 3x$ on the left to get
$$\cos 3x (4 \cos 3x - 1) = 3$$
We now have two equations. Since $\cos 3x = 3$ has no solution, we concentrate on $4 \cos 3x - 1 = 3$; this yields
$$4 \cos 3x = 4$$ or $$\cos 3x = 1$$
Taking the inverse cosine of both sides we get
$$3x = 0 + 360\times k$$ (k in degrees)
and then dividing by 3 we get
$$x = 0 + 120 \times k$$
For the condition $x \in (0, 120)$ we arrive at the answer, $$\bbox [yellow] {x=120}$$.
CHECK: $\cos 6(120) + 1 \rightarrow 1 + 1 \rightarrow 2$ and $\frac {3}{2} + \frac {1}{2} \cos 360 \rightarrow \frac {3}{2} + \frac {1}{2} \rightarrow \frac {4}{2} \rightarrow 2 \checkmark$
Alternative answer (as suggested by Lab Bhatterchajee):
Instead of factoring out as above, substitute $u = \cos 3x$ and subtract $3$ from both sides gives us
$$4u^2 - u + 3 = 0$$
Using the quadratic equation $x = \frac {-b \pm \sqrt {b^2-4ac}}{2a}$ we have
$$u = \frac {-1 \pm \sqrt {1^2-4(4)(-3)}}{(2)(4)}$$, which leaves us with
$$u = \frac {1 \pm 7}{8}$$
Bringing back $u = \cos 3x$ we have two solutions:
$$\cos 3x = -\frac {3}{4}, 1$$
Since we are looking for values from $0$ to $120$, we discard the negative value and continue on from $\cos 3x = 1$ from before - yielding the same answer as above, avoiding an extraneous solution.
| {
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"url": "https://math.stackexchange.com/questions/2560925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ I saw this problem somewhere recently and I was having some difficulty getting started on it.
The problem is twofold. The first is to evaluate:
$$\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$$
and once this is done, to explain what this has to do with the construction of a pentagon (maybe some other polygon?) using a compass and straight edge.
In terms of evaluating the series, I tried writing each $n$ as $m \cdot 2^k$ and evaluating the summation there since $2^k$ will alternate between + and - mod 5. However, this leads to a divergent series and I think this is not a valid thing to do since the original series is not absolutely convergent so we can't rearrange terms like that.
| Not an answer but too long for a comment.
Considering
$$f(i)=\sum_{k=0}^p \frac 1 {5k+i}=\frac{1}{5} \left(\psi ^{(0)}\left(\frac{i}{5}+p+1\right)-\psi
^{(0)}\left(\frac{i}{5}\right)\right)$$ then $5\left(f(1)-f(2)-f(3)+f(4)\right)$ write$$\psi ^{(0)}\left(p+\frac{6}{5}\right)-\psi ^{(0)}\left(p+\frac{7}{5}\right)-\psi
^{(0)}\left(p+\frac{8}{5}\right)+\psi ^{(0)}\left(p+\frac{9}{5}\right)-\psi
^{(0)}\left(\frac{4}{5}\right)+\psi ^{(0)}\left(\frac{3}{5}\right)+\psi
^{(0)}\left(\frac{2}{5}\right)-\psi ^{(0)}\left(\frac{1}{5}\right)$$ which, using generalized harmonic numbers, reduces to
$$5\left(f(1)-f(2)-f(3)+f(4)\right)=H_{p+\frac{1}{5}}-H_{p+\frac{2}{5}}-H_{p+\frac{3}{5}}+H_{p+\frac{4}{5}}+2 \sqrt{5}
\coth ^{-1}\left(\sqrt{5}\right)$$ Now, using the asymptotics of generalized harmonic numbers, we end with
$$\sum_{k=0}^p\left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)=\frac{2 \coth ^{-1}\left(\sqrt{5}\right)}{\sqrt{5}}-\frac{2}{125 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit.
It also allows to compute partial sums. Using $p=10$, the exact value is $\frac{2826908374432441763845}{6569974349001513017568}\approx 0.430277$ while the above development gives $\approx 0.430249$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
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Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?
My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And because
$$\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} \to \frac{1}{2}$$
there exists $n_0$ such that $\forall n > n_0: \left\vert \dfrac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} - \dfrac{1}{2}\right\vert < 1$
Hence, for $n \geq n_0$:
$$ \frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} < \frac{3}{2}$$
Thus: $$\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} < \frac{3}{2}\frac{1}{n^2}$$
and for $n_1$ sufficiently large, the sequence of terms is positive (as both the numerator and the denumerator converge to $+ \infty$, so for $n$ large enough the numerator and denumerator are positive, and hence the quotient is positive).
Let $N := \max\{n_0,n_1\}$
Then, because $\sum\dfrac{1}{n^2}$ converges, we conclude using the comparison test that the series
$$\sum_{n > N}^{+ \infty} \frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}}$$
converges, and hence, the given series converges.
Is this correct? (I tried to explain every single step). Do you have any comments? Is there an easier approach?
| Perhaps,
\begin{align*}
\left|\dfrac{n^{3}-5n^{2}+\pi}{2n^{5}-\sqrt{3}}\right|&\leq\dfrac{n^{3}+5n^{3}+\pi n^{3}}{2n^{5}-\sqrt{3}n^{5}}\\
&=\dfrac{6+\pi}{2-\sqrt{3}}\dfrac{1}{n^{2}}.
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2563715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
I think that the only solutions are $n = 0,1,2,4$, but I have no idea on how to prove it.
I tried to write $3^n-1$ as $1+3+3^2+...+3^{n-1}$ and manipulate the sum but found my self at the equally hard problem of finding the power of two dividing $3^k+1$
| The result is clear for $n = 0$. For $ n = 1, 2, 3 \ldots$ let the highest power of $2$ that divides $3^n - 1$ be $2^{p(n)}$. If $n$ is odd, say $n = 2m + 1$, then $3^n - 1 = (3 - 1)(3^{2m} + 3^{2m - 1} + \cdots + 1)$. The sum has an odd number of terms, so $p(n) = 1$. If $n$ is even, say $n = 2m$, then $3^n - 1 = (3^m - 1)(3^m + 1)$. By induction, $3^m \equiv 1 \mod 8$ if $m$ is even and $3^m \equiv 3 \mod 8$ if $m$ is odd. Hence $p(2m) = p(m) + 1$ if $m$ is even, $p(m) + 2$ if m is odd. By applying this repeatedly we get that if $n = 2^ab$, where $a > 0$ and $b$ is odd, then $p(n) = a + 2$. It follows easily that for $n > 0$ we have $p(n) \ge n$ iff $n = 1, 2, 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find the indefinite integral $\int\frac{dx}{(x^2+2x+5)^2}$ I need help with the indefinite integral
\begin{align}
& \int\frac{dx}{(x^2+2x+5)^2} \\[10pt]
= {} & \int\frac{dx}{((x+1)^2+4)^2} = \int\frac{du}{(u^2+4)^2} & & x+1=u,\quad du=dx \\[10pt]
= {} & \frac{1}{16} \int\frac{du}{(\frac{u^2}{4}+1)^2} \\[10pt]
= {} & \frac 1 8 \int \frac{ds}{(s^2+1)^2} = \text{?} & & \frac u 2 = s, \quad 2\,ds=du
\end{align}
or maybe there is an easier way ?
Any ideas ? thanks !
| There is a reduction process. Apply integration by parts.
$$\int \frac{1}{1+x^2}\,dx=\frac{x}{1+x^2}-\int x \,d\frac{1}{1+x^2}$$
$$=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}\,dx-2\int \frac{1}{(1+x^2)^2}\,dx$$
And rearranging,
$$2\int \frac{1}{(1+x^2)^2}\,dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}\,dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this.
The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question
| Hints:
The coefficient of the $n^\text{th}$ term can be written as: $$\frac{1\times 3 \times 5 … (1+4(n-2))}{2\times 4\times 6 …(2+4(n-2))}$$ $$=\frac{1\times 2 \times 3 … (1+4(n-2))\times (2+4(n-2))}{[2\times 4 \times … (2+4(n-2))]^2}$$ $$=\frac{(4n-6)!}{[(2n-3)!(2^{2n-3})]^2}$$ $$=\frac{(4n-6)!}{2^{4n-6}(2n-3)!^2}$$
The power of $x$ in the $n^{\text{th}}$ term with the number in the denominator can be written as: $$\frac{x^{2n-2}}{4n-4}$$
Thus, the $n^{\text{th}}$ term is: $$\frac{(4n-6)!}{2^{4n-6}(2n-3)!^2}\frac{x^{2n-2}}{4n-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solve $f(x+f(2y))=f(x)+f(y)+y$
Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for each $x$ and $y$ in $\mathbb{R}^+$, $$f(x+f(2y))=f(x)+f(y)+y$$
Note:
$f(x)=x+b$ is a solution for all $b\in\mathbb{R}^+$ but I can not prove it.
| Another solution is
$f(x) =-x/2$.
Here,
at the end of my ramblings,
is how I found it.
Inspired by the comment
that a solution is
$x+b$,
let
$f(x) = x+g(x)$.
Then
$x+2y+g(2y)+g(x+2y+g(2y))
=x+g(x)+y+g(y)+y
$
or
$$g(2y)+g(x+2y+g(2y))
=g(x)+g(y).
$$
Letting $y=0$,
this is
$g(0)+g(x+g(0))
=g(x)+g(0)
$
or
$g(x) = g(x+g(0))$.
If $g(0) = 0$,
this tells nothing.
If $g(0) \ne 0$,
$g$ is periodic
with period $g(0)$.
In particular,
$g(0)
= g(g(0))
= g(g(g(0)))
=...
$.
Setting $x=0$,
this is
$g(2y)+g(2y+g(2y))
=g(0)+g(y)
$.
This doesn't seem to help much.
If $x=2y$,
this is
$g(2y)+g(4y+g(2y))
=g(2y)+g(y)
$
or
$g(y)
=g(4y+g(2y))
$,
so
$g(y+g(0))
=g(4y+g(2y))
$.
If
$y+g(0)
=4y+g(2y)$,
then
$g(2y)
=g(0)-3y
$
or
$g(y)
=g(0)-3y/2
$.
This suggests trying
$f(x) = ax+b$.
Then
$a(x+2ay+b)+b
=a(x+y)+2b+y
$
or
$2a^2y+ab
=(a+1)y+b
$
or
$0
=(2a^2-a-1)y+b(a-1)
$.
Therefore
$0=(2a^2-a-1$
or
$a
=\dfrac{1\pm\sqrt{1+8}}{4}
=\dfrac{1\pm 3}{4}
=1, -\frac12
$.
If $a=1$
then any $b$ works.
If
$a=-\frac12$,
then
$b=0$,
so
$-x/2$
is a solution.
To check,
$f(x+f(2y))
=f(x-y)
=(y-x)/2
$
and
$f(x)+f(y)+y
=-x/2-y/2+y
=(y-x)/2
$.
That's all I have.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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} |
How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$? How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$?
I tried to find $(a_n-a_{n-1})$ but it didn't help.
| Let's show that for $\forall a_0>0$, the sequence is converging ...
Proposition 1. $3< a_n < 4$, for $\forall n \geq 2$.
If $a_0 > 0 \Rightarrow \frac{1}{a_0}>0 \Rightarrow \color{red}{a_1}=3+\frac{1}{a_0}\color{red}{>3}$. So $a_1 >0 \Rightarrow \frac{1}{a_1}>0 \Rightarrow \color{red}{a_2}=3+\frac{1}{a_1}\color{red}{>3}$ and, by induction
$$\color{red}{a_n > 3}, \forall n >0 \tag{1}$$
Also, from $(1)$ $a_n > 3 \Rightarrow 0< \frac{1}{a_n} < \frac{1}{3} < 1 \Rightarrow \color{red}{a_{n+1}}=3+\frac{1}{a_n}\color{red}{<4}$. But $(1)$ is true for $n > 0$, so (by induction again)
$$\color{red}{a_n < 4}, \forall n>1 \tag{2}$$
Altogether $$3< a_n < 4,\forall n \geq 2$$
Now, let's look at the function
$$f(x)=3+\frac{1}{x}$$
which is an important one, because $a_{n+1}=f(a_n)$.
Proposition 2. if $3\leq x \leq 4 \Rightarrow 3 \leq f(x) \leq 4$
$3\leq x \leq 4 \Rightarrow \frac{1}{3} \geq \frac{1}{x} \geq \frac{1}{4} \Rightarrow 4 > 3+ \frac{1}{3} \geq 3+\frac{1}{x} \geq 3+\frac{1}{3} > 3$ which is $3 < f(x) < 4$
Proposition 3 Function $f: [3,4] \rightarrow [3,4]$ is a contraction mapping.
From MVT
$$|f(x)-f(y)|= |f'(\varepsilon)| |x-y|=\left|\frac{1}{\varepsilon^2}\right| |x-y|$$
where $\varepsilon \in (x,y) \subset [3,4]$, thus
$$|f(x)-f(y)| \leq \frac{1}{9} |x-y|$$
Altogether, we can ignore the first 2 elements (Proposition 1) of the sequence, when the initial element is greater than $0$, they don't affect the convergence of the sequence. And according to the Banach fixed-point theorem, the sequence has a limit. More reading on this subject here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
$$...=...$$
$$2=2$$
$$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$
$$\sqrt2=\sqrt2\text{ hence proved.}$$
Cheers in advance :)
| $$$$
\begin{align}
\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= x \tag{A}\\
\bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)\bigg(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\bigg)
&= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
(2+\sqrt 3)-(2-\sqrt 3) &= \bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x \\
\bigg(\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3}\bigg)x &= 2\sqrt 3 \\
\sqrt{2+\sqrt3} + \sqrt{2-\sqrt3} &= \dfrac{2\sqrt 3}{x} \tag{B}\\
2\sqrt{2+\sqrt 3} &= \dfrac{2\sqrt 3}{x} + x \tag{B+A}\\
\sqrt{2+\sqrt 3} &= \dfrac{\sqrt 3}{x} + \dfrac x2 \tag{C}\\
2\sqrt{2-\sqrt 3} &= \dfrac{2\sqrt 3}{x} - x \tag{B-A}\\
\sqrt{2-\sqrt 3} &= \dfrac{\sqrt 3}{x} - \dfrac x2 \tag{D}\\
1 &= \dfrac{3}{x^2} - \dfrac{x^2}{4} \tag{CD} \\
4x^2 &= 12 - x^4 \\
x^4 + 4x^2 - 12 &= 0 \\
(x^2 - 2)(x^2 + 6) &= 0 \\
x &= \sqrt 2 \\
\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3} &= \sqrt 2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then have: $$\forall \epsilon > 0 \exists N\in\mathbb{N}_0 \forall n \geq N: \left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon$$
and after doing a bit of math we have $$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon \Leftrightarrow \left|\frac{1}{n}+2-2\right| < \epsilon \Leftrightarrow \frac{1}{n} < \epsilon \Leftrightarrow \frac{1}{\epsilon} < n $$ which means that we need a $N\in\mathbb{N}$ with $N > \frac{1}{\epsilon}$ and we have $\forall n\geq N$:$$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| = \dots=\frac{1}{n}\leq\frac{1}{N}<\epsilon\,.$$
And hence our sequence $a_n \longrightarrow a_\infty$ with $a_\infty = 2$$._{\,\,\square}$
Did I do this right?
| $$\frac{(n+1)^2 -n^2}{n}-2=\frac{n^2+2n+1-n^2}{n}-2=\frac{2n+1}{n}-2=\frac{1}{n}
\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Let $w = \dfrac{1 + i\sqrt{7}}{2}$, then $\dfrac{1 - i\sqrt{7}}{2} = \bar{w}$, its complex conjugate. We are required to find the value of $w^4 + \bar{w}^4$.
Observer that
$$w \bar{w} = 2 \quad {\rm and} \quad w + \bar{w} = 1.$$
Now
$$(w + \bar{w})^2 = w^2 + 2w\bar{w} + \bar{w}^2 \quad \Rightarrow \quad w^2 + \bar{w}^2 = 1 - 2 w \bar{w} = 1 - 2(2) = -3.$$
Squaring the last expression we have
\begin{align*}
(w^2 + \bar{w}^2)^2 &= (-3)^2\\
w^4 + 2w^2 \bar{w}^2 + \bar{w}^4 &= 9\\
\Rightarrow w^4 + \bar{w}^4 &= 9 - 2(w \bar{w})^2\\
&= 9 - 2(2)^2\\
&= 9 - 8\\
&= 1,
\end{align*}
as required to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Argue that $f$ is not uniformly continuous Let $f(x)=x^3$. We want to show that $f$ is not uniformly continuous.
Can someone please explain from $|f(x_n)-f(y_n)=\big|(n+\frac{1}{n})^3 - n^3)\big| = 3n+\displaystyle\frac{3}{n}$? How does $\big|(n+\frac{1}{n})^3-n^3\big| $ yield $3n+\displaystyle\frac{3}{n}$? ${}{}{}{}$
| The first part is just the Binomial Theorem:
$$
\left(n + \frac{1}{n}\right)^3 = n^3 + 3 n^2 \cdot \frac{1}{n} + 3 n \frac{1}{n^2} + \frac{1}{n^3} = n^3 + 3n + \frac{3}{n} + \frac{1}{n^3}
$$
So
$$
\left(n + \frac{1}{n}\right)^3 - n^3 = 3n + \frac{3}{n} + \frac{1}{n^3}
$$
The author seems to skip the $\frac{1}{n^3}$ term, but no matter. The expression is still $>3$, which is the point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
| \begin{align}\frac1{1-\cos x}-\frac2{x^2}&=\frac1{2\sin^2\left(\frac x2\right)}-\frac2{x^2}\\&=\frac{\frac{x^2}4}{\sin^2\left(\frac x2\right)}\times\frac2{x^2}-\frac2{x^2}\\&=\left(\left(\frac{\frac x2}{\sin\left(\frac x2\right)}\right)^2-1\right)\frac2{x^2}.\end{align}But $\frac x{\sin x}=1+\frac{x^2}6+o(x^3)$ and therefore$$\left(\frac{\frac x2}{\sin\left(\frac x2\right)}\right)^2=1+\frac{x^2}{12}+o(x^3).$$Therefore, your limit is $\frac2{12}=\frac16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equation with integers $x$, $y$ If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?
| We have: $$x + y = 14\sqrt{xy -48} \implies 196(xy-48)= (x+y)^2 \implies x^2+y^2-194xy+9408 =0$$
Now, rearranging the terms we get: $$ y^2-194xy = -x^2-9408 \implies 9409x^2-194xy +y^2=9408x^2-9408$$ $$ \implies (y-97x)^2 = 9408(x^2-1) \implies y = 97x \pm 56\sqrt{3(x^2-1)}$$
There are many solutions to this: $(1,97), (2,26)$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof for a prime number formula involving the prime counting function How would I go about proving this? I came across this when I was searching for interesting prime generating function. Or alternatively, could you kindly direct me to a source containing a complete proof for the following formula?
$$p_n=1+\sum^{2^n}_{m=1}\left\lfloor\left\lfloor\frac{n}{1+\pi(m)}\right\rfloor^{\frac{1}{n}}\right\rfloor.$$
Here $\pi(m)$ is the prime counting function.
| Lemma: For any $x \geqslant 0$ and $n \in \mathbb{N}_+$, $\bigl[[x]^{\frac{1}{n}}\bigr] = [x^{\frac{1}{n}}]$.
Proof: It is easy to see that $\bigl[[x]^{\frac{1}{n}}\bigr] \leqslant [x^{\frac{1}{n}}]$. Now, suppose $a = [x^{\frac{1}{n}}] \in \mathbb{N}$, then$$
x = (x^{\frac{1}{n}})^n \geqslant a^n \Longrightarrow [x] \geqslant a^n \Longrightarrow [x]^{\frac{1}{n}} \geqslant a \Longrightarrow \bigl[[x]^{\frac{1}{n}}\bigr] \geqslant a = [x^{\frac{1}{n}}].
$$
Thus, $\bigl[[x]^{\frac{1}{n}}\bigr] = [x^{\frac{1}{n}}]$.
Now back to the question. By the lemma,$$
\sum_{m = 1}^{2^n} \left[ \left[ \frac{n}{1 + π(m)} \right]^{\frac{1}{n}} \right] = \sum_{m = 1}^{2^n} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right].
$$
Note that $n < 2^n$ for any $n \geqslant 1$, thus for any $1 \leqslant m \leqslant 2^n$,$$
0 < \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \leqslant n^{\frac{1}{n}} < 2 \Longrightarrow \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] = 0 \text{ or } 1.
$$
By Bertrand's postulate, $p_n \leqslant 2^n$ for all $n \geqslant 1$, thus $π(m) \leqslant n - 1$ for $1 \leqslant m \leqslant p_n - 1$ and $π(m) \geqslant n$ for $p_n \leqslant m \leqslant 2^n$, which implies\begin{align*}
\sum_{m = 1}^{2^n} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] &= \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \leqslant n - 1}} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] + \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \geqslant n}} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right]\\
&= \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \leqslant n - 1}} 1 + \sum_{\substack{1 \leqslant m \leqslant 2^n\\π(m) \geqslant n}} 0 = \sum_{m = 1}^{p_n - 1} 1 + \sum_{m = p_n}^{2^n} 0 = p_n - 1.
\end{align*}
Therefore,$$
1 + \sum_{m = 1}^{2^n} \left[ \left[ \frac{n}{1 + π(m)} \right]^{\frac{1}{n}} \right] = 1 + \sum_{m = 1}^{2^n} \left[ \left( \frac{n}{1 + π(m)} \right)^{\frac{1}{n}} \right] = p_n.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that these perpendicular distances are in G.P. Let $BC$ be the chord of contact of the tangents from a point $A$ to the circle $x^2+y^2=1$. $P$ is any point on the arc $BC$. Let $PX, PY$ and $PZ$ be the lengths of the perpendiculars from P on the line $AB, BC$ and $CA $ respectively then prove that $PX, PY$, and $PZ$ are $G. P.$
My approach : I considered point $A$ to be $(\alpha, \beta)$. Using this I got the equation of common chord as $$\alpha x+\beta y-1=0$$. Now I considered point P to be $(\gamma, \delta)$. For this circle the pair of tangents can be given as $$(x^2+y^2-1)({\alpha}^2+{\beta}^2-1)=(\alpha x +\beta y -1)^2$$ . Simplifying this we get $$(x-\alpha) ^2 + (y-\beta)^2=(y\alpha-x\beta)^2$$. Now I can get the perpendicular distances to the pair of tangents and the common chord from their equations but I am stuck in finding the perpendicular distance in the case of pair of tangents because I am not able to factorise it into pair of two straight lines nor do I know any formula to find the perpendicular distances without factorising the expression. Any other method to solve this problem is also appreciated.
| I suspect there's an elegant synthetic solution to this, but after several hours of scribbling to no avail I resorted to trigonometry. My answer is very similar to mathlove's although I did derive it independently, and I think it's sufficiently different to be worth submitting.
WLOG, we can put $A$ on the +X axis. Let the angle that $AB$ makes with the -X axis be $\alpha$. Then
A is $(\sec\alpha, 0)$
B is $(\cos\alpha, \sin\alpha)$
C is $(\cos\alpha, -\sin\alpha)$
The equations of the tangents can be written in normal form:
AB is $x\cos\alpha + y\sin\alpha - 1 = 0$
AC is $x\cos\alpha - y\sin\alpha - 1 = 0$
Let P be $(\cos\theta, \sin\theta)$
Thus
$$\begin{align}\\
PY & = |\cos\theta - \cos\alpha|\\
PX & = |\cos\theta\cos\alpha + \sin\theta\sin\alpha - 1|\\
PZ & = |\cos\theta\cos\alpha - \sin\theta\sin\alpha - 1|\\
\end{align}$$
If $PX, PY, PZ$ are in geometric progression, $PX.PZ = PY^2$
$$\begin{align}\\
PX.PZ & = |(\cos\theta\cos\alpha - 1 + \sin\theta\sin\alpha)(\cos\theta\cos\alpha - 1 - \sin\theta\sin\alpha)|\\
& = |(\cos\theta\cos\alpha - 1)^2 - (\sin\theta\sin\alpha)^2\\
& = |(\cos\theta\cos\alpha - 1)^2 - (1 - \cos^2\theta)(1 - \cos^2\alpha)\\
& = |\cos^2\theta\cos^2\alpha - 2\cos\theta\cos\alpha + 1 - (1 - \cos^2\theta - \cos^2\alpha + \cos^2\theta\cos^2\alpha)|\\
& = |- 2\cos\theta\cos\alpha + \cos^2\theta + \cos^2\alpha|\\
& = (\cos\theta - \cos\alpha)^2\\
& = PY^2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5{x+3}+\frac 5{y+3}+\frac 5{z+3}-3\right]_{min}\Rightarrow \left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\left[ a+b+c\right]_{min}\Rightarrow a=b=c$
and from here, we get $$\frac 1{x+3}=\frac1{y+3}=\frac 1{z+3} \Rightarrow x=y=z$$
Finally, It must be $x=y=z=\frac 13$
and $$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}=\frac 32$$
Am I correct?
Note: I'm sorry for wrong mathematical symbols.
| You have to insert $(x,y,z)=(1/3,1/3,1/3)$ into the expression $\dfrac{5}{x+3}+\dfrac{5}{y+3}+\dfrac{5}{z+3}-3$ to get the value $\dfrac{3}{2}$ because you are looking for the minimum of $\dfrac{5}{x+3}+\dfrac{5}{y+3}+\dfrac{5}{z+3}-3$ but not $\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation using Rational Root Theorem and got $y = \frac 1 2$
$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$
Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $
$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $
But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$
Why don't I get the second solution correct?
| It's $$\cos2x(\cos2x+\cos4x)=\frac{1}{2}.$$
Now, let $\cos2x=t$.
Thus, $$t(t+2t^2-1)=\frac{1}{2}$$ or
$$4t^3+2t^2-2t-1=0$$ or
$$2t^2(2t+1)-(2t+1)=0$$ or
$$(2t^2-1)(2t+1)=0,$$ which gives
$$\cos4x=0$$ or $$\cos2x=-\frac{1}{2}$$ and we can write the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On $\sum_{n=0}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}+x$ and solvable Galois groups? In this comment, I hastily assumed that,
$$\sum_{\color{blue}{n=0}}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}$$
In fact, this is valid only for $b=1$ and, as Robert Israel pointed out, for general $b$ is missing a rational term. So for odd $b$ we have,
$$\begin{aligned}
\sum_{n=0}^\infty \frac{1}{n^2+n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{1^2-4c}\big)}{\sqrt{1^2-4c}}\\
\sum_{n=0}^\infty \frac{1}{n^2+3n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{3^2-4c}\big)}{\sqrt{3^2-4c}}-\frac1{c-2}\\
\sum_{n=0}^\infty \frac{1}{n^2+5n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{5^2-4c}\big)}{\sqrt{5^2-4c}}-\frac{2c-10}{(c-4)(c-6)}\\
\sum_{n=0}^\infty \frac{1}{n^2+7n+c} &= \frac{\pi \tan\big(\frac{\pi}2\sqrt{7^2-4c}\big)}{\sqrt{7^2-4c}}-\frac{3c^2-56c+252}{(c-6)(c-10)(c-12)}\end{aligned}$$
But I'm having trouble finding $b=9$.
Questions:
*
*What is the rational term for $b=9$?
*For $b=2m+1$ and $m>0$, is it true that the rational term has form $\displaystyle \frac{P_1(c)}{P_2(c)}$ where $P_1(c)$ is a polynomial of degree $m-1$, while $P_2(c)$ has degree $m$?
*This may be like throwing a curveball, but do the equations $P_1(c)=0$ and $P_2(c)=0$ have solvable Galois groups?
| It looks like the correct thing to look at is
$$\sum_{n=-\infty}^\infty\frac1{n^2+bn+c}.$$
This is surely expressible in terms of the digamma function.
Your rational fraction is surely an artefact of insisting on taking a one-sided sum. As an example,
$$\sum_{n=0}^\infty\frac1{n^2+9n+c}
=\sum_{m=-\infty}^{-9}\frac1{(-9-m)^2+9(-9-m)+c}
=\sum_{m=-\infty}^{-9}\frac1{m^2+9m+c}$$
and so
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+9n+c}
&=\frac12\sum_{n=-\infty}^\infty\frac1{n^2+9n+c}-\frac12
\sum_{n=-8}^{-1}\frac1{n^2+9n+c}\\
&=\frac12\sum_{n=-\infty}^\infty\frac1{n^2+9n+c}-
\left(\frac1{c-8}+\frac1{c-14}+\frac1{c-18}+\frac1{c-20}\right)
\end{align}
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
For $\alpha>1$ find $\lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1}$
For $\alpha>1$ find (w/o lhopital)$$
\lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1}
$$
Progress:
Used this:$$
x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})
$$ to get to this:
$$
\alpha-1=(\sqrt[2^n]{\alpha})^{2^n}-1^{2^n}=(\sqrt[2^n]{\alpha}-1)((\sqrt[2^n]{\alpha})^{2^n-1}+(\sqrt[2^n]{\alpha})^{2^n-2}+\dots+\sqrt[2^n]{\alpha}+1)
$$
and so$$
\sqrt[2^n]{\alpha}-1=\frac{\alpha-1}{((\sqrt[2^n]{\alpha})^{2^n-1}+(\sqrt[2^n]{\alpha})^{2^n-2}+\dots+\sqrt[2^n]{\alpha}+1)}
$$
| $\sqrt[2^n]{\alpha}=(1+(\alpha -1))^{2^{-n}}=1+(\alpha -1)2^{-n}+o(2^{-n})$
$\sqrt[n]{\sqrt[2^n]{\alpha}-1}=((\alpha -1)2^{-n}+o(2^{-n}))^{\frac{1}{n}}=(\alpha -1)^{\frac{1}{n}}(2^{-n})^{\frac{1}{n}}+o(1)=1\cdot 2^{-1}=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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For which values of $m$ and $n$ is the probability $P_{(A,A)} = 0.5$? Given are the classes $A$ and $B$.
Suppose there are $n$ instances from class $A$ and $m$ instances from class $B$.
Let $X$ be all instances. Thus, $X=\{~\{a\}_n, \{b\}_m~\}$.
I denote a pair $p$ as a combination of two instances $(i, j$) from $X$, for which $i \neq j$. Thus, pair $(i, j)$ is the same pair as $(j, i)$.
Let $P$ be all possible pairs $p$. Thus $|P| = \frac{|X|(|X|-1)}{2}$.
Let $P_{(A,B)}$ be the probability of a pair $p = (i, j)$ with $i$ of class $A$ and $j$ of class $B$ in $P$. (In other words, the number of pairs from class $A$ and $B$ over $|P|$.)
Question 1: For which values of $m$ and $n$ is the probability of $P_{(A,A)} =0.5$? A solution is $n=3, m=1$, but what is the next possible solution?
Question 2: What is a formula to compute all possible combinations?
Example 1:
Suppose you have three pingpong-balls. Two are orange (class $A$), one is white (class $B$).
Then $X = \{o_1, o_2, w_1\}$.
And $P = \{(o_1, o_2), (o_1, w_1), (o_2, w_1)\}$.
$P_{(A, A)} = \frac{|\{(o_1, o_2)\}|}{|P|} = \frac{1}{3}$.
Example 2:
Suppose you have four pingpong-balls. Three are orange (class $A$), one is white (class $B$).
Then $X = \{o_1, o_2, o_3, w_1\}$.
And $P = \{(o_1, o_2), (o_1, o_3), (o_1, w_1), (o_2, o_3), (o_2, w_1), (o_3, w_1)\}$.
$P_{(A, A)} = \frac{|\{(o_1, o_2), (o_1, o_3), (o_2, o_3)\}|}{|P|} = \frac{3}{6} = 0.5$.
| The question, in effect, asks for all possible positive integer pairs $(m, n)$ with$$
\frac{n(n - 1)}{(m + n)(m + n - 1)} = \frac{1}{2}. \tag{1}
$$
Denote $l = n - m$. After some manipulation, (1) becomes$$
(2l - 1)^2 - 8m^2 = 1. \tag{2}
$$
This is a Pell's equation. Because the fundamental solution of $x^2 - 8y^2 = 1$ is $(x_1, y_1) = (3, 1)$, by the theory of Pell's equation, solutions to $x^2 - 8y^2 = 1$ have the following recurrence relation:\begin{align*}
x_{k + 2} &= 6x_{k + 1} - x_k,\\
y_{k + 2} &= 6y_{k + 1} - y_k,
\end{align*}
and the initial conditions are $x_0 = 1$, $x_1 = 3$ and $y_0 = 0$, $y_1 = 1$. Thus the solutions to $x^2 - 8y^2 = 1$ are\begin{align*}
x_k &= \frac{1}{2} ((3 + 2\sqrt{2})^k + (3 - 2\sqrt{2})^k),\\
y_k &= \frac{1}{4\sqrt{2}} ((3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k).
\end{align*}
Note that for every $k$, $x_k^2 = 8y_k^2 + 1$ is an odd integer, so $x_k$ is an odd integer. Therefore, all solutions to (1) are\begin{align*}
&\mathrel{\phantom{=}} (m, n) = \left(y_k, \frac{1}{2} (x_k + 1) + y_k\right)\\
&= \left(\frac{1}{4\sqrt{2}} ((3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k), \frac{1 + \sqrt{2}}{4\sqrt{2}} (3 + 2\sqrt{2})^k - \frac{1 - \sqrt{2}}{4\sqrt{2}} (3 - 2\sqrt{2})^k + \frac{1}{2}\right)\\
&= \left(\frac{1}{4\sqrt{2}} ((1 + \sqrt{2})^{2k} - (1 - \sqrt{2})^{2k}), \frac{1}{4\sqrt{2}} ((1 + \sqrt{2})^{2k + 1} - (1 - \sqrt{2})^{2k + 1}) + \frac{1}{2}\right). \quad k \in \mathbb{N_+}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of x² from 0 to b - using archimedes sum of squares - Apostol's I'm reading Apostol's Calculus book and in the first chapter is presented the way archimedes found the sum of the square and how it can be used to calculate the integral of $x²$. But I'm not able to follow some steps of the proof.
from the book:
We subdivided the base in $n$ parts each with length $\ \frac{b}{n} $, a typical point corresponds to $\frac{kb}{n}$ where $k$ takes values from $k = 1, 2, 3, ..., n$
We can construct rectangles from for each $k th$ point:
$Base = \frac{b}{n}$
$Height = (\frac{kb}{n})^2$
$Area = Base * Height = \frac{b}{n} . (\frac{kb}{n})^2$
$Area = \frac{b^3}{n^3}.k^2 $
If we sum all the rectangles, we get a bit more than the area under the curve $x^2$
$S_{big} = \frac{b^3}{n^3}.(1² + 2² + 3² + ... + (n-1)² + n²)$
If we can construct smaller rectangles, using $n-1$ points, we we get a bit less than the area under the curve $x^2$.
$S_{small} = \frac{b^3}{n^3}.(1² + 2² + 3² + ... + (n-1)²)$
So the real area under the curve $x^2$ is between the two areas:
$S_{small} < A < S_{big}$
After a bit of algebra we get that:
$S_{big} = \frac{b^3}{n^3} . (\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6})$
$S_{small} = \frac{b^3}{n^3} . (\frac{n^3}{3} - \frac{n²}{2} + \frac{n}{6})$
To prove that $A$ is $\frac{b^3}{3}$ he uses this inequalities:
$1² + 2² + 3² + ... + (n-1)² < \frac{n³}{3} < 1² + 2² + 3² + ... + n²$
But I don't understand where the $\frac{n³}{3}$ came from. And can't follow the proof
EDIT:
After taking the average of the two expression:
$\frac{(\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6}) + (\frac{n^3}{3} - \frac{n²}{2} + \frac{n}{6})}{2} = \frac{n^3}{3} + \frac{n}{6}$
So I understan where $\frac{n³}{3}$ came from, but why is $\frac{n}{6}$ is thrown away?
| The central idea is that you have to find a unique number (independent of $ n$) which lies between every $S_{\text{small}} $ and $S_{\text{big}} $ both of which depend on $n$. First thing to note here is that there can't be two distinct numbers say $A, A'$ lying between these $S$'s. Without any loss of generality let $A<A' $ and then if $$S_{\text{small}}\leq A<A'\leq S_{\text{big}} $$ then we have $$S_{\text{big}} - S_{\text{small}} \geq A'-A>0$$ ie $$\frac{b^{3}}{n}\geq A'-A$$ Clearly this is a contradiction because we can choseo the positive integer $n$ as large as we want and we can definitely choose $n>b^3/(A'-A)$.
Next we need to show that $b^3/3$ lies between between $S_{\text{small}} $ and $S_{\text{big}} $. This is easily done by noting that $$1-\frac{3}{2n}+\frac{1}{2n^2}<1<1+\frac{3}{2n}+\frac{1}{2n^2}$$ and thus $A=b^3/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Showing something is even Show that if $x$ and $y$ are integers and $x^3+6=y^2$, then $x$ is odd.
I tried to do it by contradiction, with something like :
"Suppose for contradiction $x$ is even, then $x=2k$, and $x^3+6=y^2=8k^3+6=2(4k^3+3)$ which means $y=\sqrt{2(4k^3+3)}=\sqrt{2}\sqrt{4k^3+3}$ but I don't know how to conclude from there that $y$ is not gonna be an integer or some kind of contradiction...
Doing it the other way, with something like y is odd or y is even, I came to conclusion like $x$ is even... ($y$ is even, then $y^2$ is even, then $x^3+6$ is even, then $x^3$ is even, then $x$ is even, I clearly don't know what is wrong with that either...?)
| suppose that $x$ is even, $x=2z$, $x^3+6=8z^3+6=y^2$ if $y$ is odd impossible since$y^2$ is odd and $8z^3+6$ is even, if $y$ is even, then $y=2a$ you have $8z^3+6=4a^2$, this implies that $4$ divides $6$ contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Find the total number of 3 digit numbers in form of $abc$ such that HCF(a,b,c)=1. Find the total number of 3 digit numbers in form of $abc$ such that HCF(a,b,c)=1.
My approach:
Total number of cases=$9^3$
Considering that a,b,c are 0 to 9 and and for the HCF to be 1 there should be no common factor of 2 or 3 .
So total number of cases is equal to
$9^3-3^3-2^3=694 $
Now i might have made a total mess here but i what should be the shortest possible method and obviously the correct answer?
| For each leading digit from $1$ to $9$ there are $100$ possible integers. If the second and third digits both share a divisor greater than $1$ with the leading digit then we exclude that case.
So if $n$ digits share a divisor with the leading digit, we exclude $n^2$ cases.
$$
\begin{array}{c|crrr}
\text{Leading digit} & \text{Excluded digits} & \text{Size} & \text{Size}^2 & 100-\text{Size}^2 \\
\hline
1 & \text{None} & 0 & 0 & 100 \\
2 & 0,2,4,6,8 & 5 & 25 & 75 \\
3 & 0,3,6,9 & 4 & 16 & 84 \\
4 & 0,2,4,6,8 & 5 & 25 & 75 \\
5 & 0,5 & 2 & 4 & 96 \\
6 & \text{See below} & & & 63 \\
7 & 0,7 & 2 & 4 & 96 \\
8 & 0,2,4,6,8 & 5 & 25 & 75 \\
9 & 0,3,6,9 & 4 & 16 & 84 \\
Total & & & & 748 \\
\end{array}
$$
The leading digit of $6$ is a special case because both $2$ and $3$ are divisors. We can use Inclusion-Exclusion.
Exclude $\{0,2,4,6,8\}$ for $25$ cases. Exclude $\{0,3,6,9\}$ for $16$ cases. But $\{0,6\}$ has been double-counted so add back $4$ cases. $100-25-16+4=63$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral by change of variable $\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $ What is $$\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $$
Putting $\sin(x) = t$, then $\cos(x) = \sqrt{1-t^2}$ and $\cos(x)\,dx = dt$
At $x = 0$, $\sin(x) = 0$, $\therefore\ t = 0$
At $x = \pi$, $\sin(x) = 0$, $\therefore\ t = 0$
Integral becomes
$$\int_0^0 t^4\cdot\sqrt[5]{1-t^2}\,dt = 0$$
Integral should not evaluate to zero
What is going on here, i cannot find any mistake....
| $$\int_{0}^{\pi} \sin^4 x\cos^6 x dx =\int_{0}^{\pi} (\frac{1-\cos 2x}{2})^2 (\frac{1+\cos 2x}{2})^3 dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1-\cos 2x)^2 (1+\cos 2x)^3 dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1-2\cos 2x+\cos^2 2x) (1+3\cos 2x+3\cos^2 2x+\cos^3 2x) dx =$$
$$=\frac{1}{32}\int_{0}^{\pi} (1+\cos 2x-2\cos^2 2x-2\cos^3 2x+\cos^4 2x+\cos^5 2x) dx $$
$$\int_{0}^{\pi} (1) dx =\pi$$
$$\int_{0}^{\pi} (\cos 2x) dx =0$$
$$\int_{0}^{\pi} (\cos^2 2x) dx =\frac{\pi}{2}$$
$$\int_{0}^{\pi} (\cos^3 2x) dx =0$$
$$\int_{0}^{\pi} (\cos^4 2x) dx =\frac{3\pi}{8}$$
$$\int_{0}^{\pi} (\cos^5 2x) dx =0$$
$$=\frac{1}{32}(\pi -\pi+\frac{3\pi}{8})=\frac{3\pi}{256}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$ Solve $(y^2+2x^2y)dx+(2x^3-xy)dy=0$
My attempt
$Mdx + Ndy$ form
$M = (y^2+2x^2y)$, $N =(2x^3-xy) , M_y=2y+2x^2, N_x=6x^2-y$
I am not getting in of standard forms here...
Another attempt:
$2x^2 d(xy)+y(ydx-xdy)=0$,
$2d(xy)=yd(y/x)$ Any clue pls???
| $$(y^2+2x^2y)dx+(2x^3-xy)dy=0....(i)\\
\implies y^2dx+2x^2ydx+2x^3dy-xydy=0\\
\implies y(ydx-xdy)+x^2(2ydx+2xdy)=0$$
Which is the form of :
$$x^ay^b(mydx+nxdy)+x^{a'}y^{b'}(m'ydx+n'xdy)=0$$
So, comparing both equations we get:
$a=0, b=1, m=1, n=-1, a'=2, b'=0, m'=2, n'=2$
Now,
$$\frac{a+h+1}{m}=\frac{b+k+1}{n}\\
\implies \frac{h+1}{1}=\frac{1+k+1}{-1}\\
\implies h+k+3=0.....(ii)$$
and $$\frac{a'+h+1}{m'}=\frac{b'+k+1}{n'}$$
$$\implies \frac{2+h+1}{2}=\frac{k+1}{2}\\
\implies h-k+2=0.....(iii)$$
Solving equation (ii) & (iii) we get, $h=-\frac{5}{2}, k=-\frac{1}{2}.$
So,I.F.= $x^{-\frac{5}{2}}y^{-\frac{1}{2}}$
Multiplying equation (i) with the I.F. we get,
$$(x^{-\frac{5}{2}}y^\frac{3}{2}dx+2x^{-\frac{1}{2}}y^\frac{1}{2})dx+(2x^\frac{1}{2}y^{-\frac{1}{2}}-x^{-\frac{3}{2}}y^\frac{1}{2})dy=0$$
Therefore the solution is:
$$\int{Mdx}+\int{(\text{terms of N not containing x})dy}=c\\
\implies \int{x^{-\frac{5}{2}}y^\frac{3}{2}dx}+2\int{x^{-\frac{1}{2}}y^\frac{1}{2}dx}=c\\
\implies -\frac{2}{3}x^{-\frac{3}{2}}y^\frac{3}{2}+4(xy)^\frac{1}{2}=c\\
\implies 4(xy)^\frac{1}{2}-\frac{2}{3}\left(\frac{y}{x}\right)^\frac{3}{2}=c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Finding The Basis Of An Intersection Of Two Subspaces
Let $$U=\operatorname{Span}\{v_1,v_2,v_3\}, V=\operatorname{Span}\{v_4,v_5,v_6\}$$
Where $$v_1=(1,28,2,39),v_2=(2,28,2,39),v_3=(-1,28,2,39)\\v_4=(0,8,0,11),v_5=(0,31,1,43),v_6=(0,-3,0,-4)$$
Find a basis for $U\cap V$
I have forgot the algorithm for finding intersection of subspaces, but in the exercise I was given an algorithm that I have yet met and would like to understand, how and why it works.
The first step is to find an homogeneous system s.t the subspace is the solution set (Null space).
To do so for $U$ we look at $$\left(\begin{array}{ccc|c}
1 & 2 & -1 &x\\
28 & 28& 28&y\\
2 & 2 & 2& z\\
39& 39&39& w \\
\end{array}\right)\sim \left(\begin{array}{ccc|c}
1 & 2 & -1 &x\\
1 & 1& 1& \frac{y}{28}\\
0 & 0 & 0& \frac{z}{2}-\frac{y}{28}\\
0& 0&0& \frac{w}{39}-\frac{y}{28} \\
\end{array}\right)$$
So the matrix that $U$ is her solution set is $$ \begin{pmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39} \end{pmatrix}$$
Doing the same with $V$ we get
$$\left(\begin{array}{ccc|c}
0 & 0 & 0 &x\\
8& 31& -3&y\\
0 & 1 & 0& z\\
11& 43&-4& w \\
\end{array}\right)\sim \left(\begin{array}{ccc|c}
8 & 31 & -3 &y\\
0 & 1 & 0& z\\
0 & 0& 1& 8w-11y-3z\\
0& 0&0& x \\
\end{array}\right)$$
So the matrix that $V$ is her solution set is $$ \begin{pmatrix} 1&0&0&0\end{pmatrix}$$
To find the intersection we solution for $$\left(\begin{array}{cccc|c}
0 & -\frac{1}{28} & \frac{1}{2} &0 &0\\
0 & -\frac{1}{28}& &\frac{1}{39}& 0\\
1 & 0 & 0& 0 &0\\
\end{array}\right) $$
Which is $$ \begin{pmatrix} 0&28&2&39\end{pmatrix}$$
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?
b. Why looking at the null space of the vectors (why we put the vectors in horizontal and not in vertical) that we found give us the basis of intersection?
C. Is there another way to find the basis of intersection?
| For the point b, remember that the Null Space of the matrix V = empty set, because they are independent. What you found is the Null Space of the Transpose(V). Same thing about U. But for the intersection you must to swap rows and columns.
| {
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Given matrix times vector, find the inverse of the matrix times the same vector. Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.
My thought are the following:
Step 1:
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$
$V=8A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$
$\frac{1}{8} V=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$
Step 2:
Can I assume $AV=\lambda V=8*\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$ here? Hence, $V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$? Is this a correct step?
Step 3:
If step 2 is correct, then
$\frac{1}{8}V = \frac{1}{8} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} V=\begin{bmatrix} \frac{1}{8} \\ \frac{1}{8} \\ \frac{1}{8} \\\frac{1}{8}\end{bmatrix}$
Does the above inference sounds right?
| If B=A^-1 and b=<8,8,8,8> then from AV=b you get V=Bb and then BV=(B^2)b. I gess that this is the final result, if there are not other information about A.
| {
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Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| Suppose there exists an $x$ such that $f(x) = x$
$f(x) = \frac {x}{3} + 3 = x\\
x = \frac 92$
It would be a reasonable guess that we get something very close to $\frac {9}{2}$
$f(4) = 4 +\frac 13\\
f(4+ \frac 13) = 4 + \frac 13 + \frac 19\\
f(4+ \frac 13 + \frac 19) = 4 + \frac 13 + \frac 19 + \frac 1{27}$
Now we are on to something.
$f(4+ \sum_\limits{i=1}^{n} (\frac {1}{3})^i) = \frac {4}{3} + \frac{1}3\sum_\limits{i=1}^{n} (\frac {1}{3})^i + 3 =
4 + \sum_\limits{i=1}^{n+1}(\frac{1}3)^i$
$4 +\sum_\limits{i=1}^{n} \frac {1}{3}^n = 4 + \frac {(1-(\frac 13)^n)}{2}\\
4 +\sum_\limits{i=1}^{2018} \frac {1}{3}^n = 4 + \frac {(1-(\frac 13)^{2018})}{2}\\
4\frac 12 - \frac 12(\frac 13)^{2018}$
| {
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Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 + 3 + 5 + \dots &= x\\
1 + 1 + 2 + 3 + \dots &= x\\
1 + 2 + 3 + 5 + 8 + \dots &= 2x \text{ (shifting and adding)}
\end{align}$$
We in fact get the same sequence. But the new sequence is one less than the original sequence. So the new sequence is $x-1$.
But $x-1=2x$ which implies that $x=-1$.
So $1+1+2+3+5+8+\dots=x$ which means...
$1+1+2+3+5+8+13+21+\dots=-1$
Is this right or wrong? Can someone please tell? Thanks...
| This kind or reasoning (let $\sum{a_n}=x$ and so on) is valid iff the series converges. Otherwise is just a mathematical formalism, nice, but meaningless (at least in these context. Probably using Analytic continuation you can give some motivation to your results using generating functions).
Obviously the fibonacci's succession doesn't converge to $0$, and so the series cannot converge.
| {
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Finding side length and circumradius in a triangle
in a triangle $ABC$ if $2\sin B= \sin A $ and $a,b,c$ be the length of side $BC,CA$ and $AB$ and length of internal angle bisector through $A$ is $\displaystyle \frac{\sqrt{2}}{3}$ unit and the equation $25\cos^2(A-B)+x^2-40\cos(A-B)-2x+17=0$ has at least one solution ,
then value of $a$ and $R$(circumradius)
Try: $$(x-1)^2+\bigg(5\cos(A-B)-4\bigg)^2=0$$
so $x=1$ and $\displaystyle \cos(A-B) = \frac{4}{5}$
$5\cos A\cos B+5\sin A\sin B=5$
$5\cos A\cos B+10\sin^2 B=5$
could some help me to solve it, thanks
| We have $$2\sin B=\sin A\tag1$$$$\implies \cos^2B=\frac{3+\cos^2A}{4}\tag2$$$$\implies b=\frac a2\tag3$$
You already have
$$\cos(A-B)=\frac 45,$$
i.e.
$$\cos A\cos B+\sin A\sin B=\frac 45\tag4$$
From $(1)(4)$, we have
$$\cos A\cos B=\frac 45-\frac{\sin^2A}{2}$$
Squaring the both sides and using $(2)$ give
$$\cos^2A\cdot\frac{3+\cos^2A}{4}=\left(\frac 45-\frac{1-\cos^2A}{2}\right)^2\implies \cos^2A=\frac 15$$
from which
$$\sin A=\frac{2}{\sqrt 5},\qquad \sin B=\frac{1}{\sqrt 5}$$
follow.
Since $A\gt B$, we see that $B$ is acute. Also, from $(4)$, we have
$$\cos B=+\frac{2}{\sqrt 5},\qquad \cos A=+\frac{1}{\sqrt 5}$$
from which
$$\sin\frac A2=+\sqrt{\frac{1-\cos A}{2}}=\sqrt{\frac{\sqrt 5-1}{2\sqrt 5}}$$
and
$$\sin C=\sin(\pi-A-B)=\sin(A+B)=\sin A\cos B+\cos A\sin B=1\implies C=\frac{\pi}{2}$$
follow.
By the law of cosines,
$$\frac{a}{\sin A}=\frac{c}{\sin C}\implies c=\frac{\sqrt 5}{2}a\tag5$$
Let $D$ be the intersection point of the angle bisector of $\angle A$ with the side $BC$.
Now let us consider the area of $\triangle{ABC}$.
Since $[\triangle{ABC}]=[\triangle{ABD}]+[\triangle{ACD}]$ with $|\overline{AD}|=\frac{\sqrt 2}{3}$, we have
$$\frac{1}{2}ab=\frac{1}{2}b|\overline{AD}|\sin\frac A2+\frac 12c|\overline{AD}|\sin\frac A2$$
$$\implies \frac{1}{2}\cdot a\cdot\frac{a}{2}=\frac{1}{2}\cdot\frac a2\cdot \frac{\sqrt 2}{3}\cdot \sqrt{\frac{\sqrt 5-1}{2\sqrt 5}}+\frac 12\cdot \frac{\sqrt 5}{2}a\cdot \frac{\sqrt 2}{3}\cdot\sqrt{\frac{\sqrt 5-1}{2\sqrt 5}}$$
from which
$$\color{red}{a=\frac{2}{3}\sqrt{1+\frac{1}{\sqrt 5}}}$$
follows.
By the law of sines, we have $R=\frac{a}{2\sin A}$, so
$$\color{red}{R=\frac{\sqrt{5+\sqrt 5}}{6}}$$
| {
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$\exp(z^2)=i$ solution check Find all complex solutions to the equation $\exp(z^2)=i$.
Attempt
We have $z^2 = x^2-y^2+2ixy$ where $z=x+iy$. Then
$$exp(x^2-y^2)=1 \text{ and } 2xy=\frac{\pi}{2}$$
So $x^2=y^2$ and $xy = \frac{\pi}{4}$ Thus $x = y$ and so $x=y= \pm \frac{\sqrt{\pi}}{2}$
I just don't feel very confident about this for some reason.
| Your answer is fine except at the step where you say $2xy=\frac{\pi}{2}.$ In fact, you can have $2xy=\frac{\pi}{2}+2\pi k$ for any integer $k$, or $xy=\frac{\pi}{4}+\pi k.$
If $k\geq 0$ then you get $x=\pm\sqrt{\frac{\pi}{4}+\pi k}=\pm\frac{\sqrt{\pi}}{2}\sqrt{4k+1}$ and $y=x.$
If $k<0$ then you get $x=\pm\sqrt{-\pi k -\frac{\pi}{4}}=\pm\frac{\sqrt{\pi}}{2}\sqrt{-4k-1}$ and $y=-x.$
One neat thing you can do to simplify:
If $j$ is non-negative integer, then we can write:
$$\begin{align}x&=\pm\frac{\sqrt{\pi}}{2}\sqrt{2j+1}\\y=&(-1)^jx\end{align}$$
and this gets all solutions.
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Count the integer solutions of $x_1+x_2-x_3+x_4-x_5=3$ I am asking for help with solving this exercise:
Find the count of possible integer solutions for equation:
$$x_1+x_2-x_3+x_4-x_5=3$$
There are restrictions for possible values of $x$:
$$
\begin{aligned}
&0 < x_1 \le 6\\
-&8 \le x_2 < -2\\
&x_3 \le 1\\
&3 < x_4\\
&2 \le x_5 \le 8
\end{aligned}
$$
Note: We should be able to find the count of solutions by using permutations and Inclusion-Exclusion principle.
| By translations $x\to x+c$ and symmetries $x\to-x$ you can reformulate your problem as
Find all $x_1,x_2,x_3,x_4,x_5$ integers that satisfy
$$
1+x_3+x_4 = x_1+x_2+x_5$$
with bounds
$$0\le x_1\le 5\qquad 0\le x_2\le 5\qquad 0\le x_5\le 6$$
$$0\le x_3\qquad 0\le x_4$$
Now, call $p(n)$ the number of triples $(x_1,x_2,x_5)$ such that $x_1+x_2+x_5=n$ and such that
$$0\le x_1\le 5\qquad 0\le x_2\le 5\qquad 0\le x_5\le 6$$
hold. In the same way, let $q(n)$ the number of couples $(x_3,x_4)$ such that $x_3+x_4=n$ and such that
$$0\le x_3\qquad 0\le x_4$$
hold. Your solution is
$$
p(1)q(0) + p(2)q(1) + p(3)q(2) + \dots + p(16)q(15).
$$
We know that $q(n)=n+1$, but $p(n)$ is much harder to compute, and I think must be done by a computer.
Let's do it with permutations/inclusion-exclusion.
First of all, a Toy Problem
Find all $x_1,x_2,x_3,x_4,x_5$ positive integers that satisfy
$$c+x_3+x_4 = x_1+x_2+x_5$$
where $c$ is an integer (may be negative) and with bound
$$c + x_3+x_4\le n$$
where $n$ is a positive integer.
Like before, let $p(m)$ the number of triples $(x_1,x_2,x_5)$ such that $x_1+x_2+x_5=m$ and let $q(m)$ the number of couples $(x_3,x_4)$ such that $x_3+x_4=m$. We know that
$$
p(m) = \frac{(m+2)(m+1)}2 = \binom{m+2}2 \qquad q(m) = m+1 = \binom{m+1}1
$$
The solution is
$$TP(c,n) = q(-c)p(0) + q(1-c)p(1) + \dots +q(n-c)p(n) $$
$$ = \binom {1-c}1\binom 22 + \binom {2-c}1\binom 32 + \dots + \binom {n-c+1}1\binom {n+2}2 $$
If we return to our first problem, we can operate an inclusion-exclusion method.
*
*First of all, we notice that $x_1+x_2+x_5\le 16$, so we can consider the Toy Problem with $c=1$ and $n=16$, so we get $TP(1,16)$.
*We have to respect the bounds, though, so we have to subtract the wrong solutions. How many solution are there with $x_5>6$? If we substitute $x_5 = 7 +z_5$ we get again the Toy Problem with
$$-6+x_3 + x_4 = x_1+x_2+z_5$$
so $c=-6$ and $n=16 -7=9$, so $TP(-6,9)$. If we do the same for $x_1,x_2$ we get $TP(-5,10)$ twice.
*Now we subtracted too much. For example we subtracted twice the (wrong) solutions with $x_1>5,x_2>5$ and we have to add them again. With the same substitutions as before, we get $TP(-11,4)$. Doing the same for the other couples, we get $TP(-12,3)$ twice.
The final answer is thus
$$TP(1,16) - TP(-6,9) - 2\cdot TP(-5,10) + TP(-11,4) + 2\cdot TP(-12,3) $$
| {
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Prove Stirling numbers of the second kind, $S(n,n)=1$ Stirling numbers of the second kind,
$$
S(m,n)={m \brace n}=\frac{1}{n!}\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^m
$$
When $n=m$, I can easily find $S(n,n)=1$ if I set values for $n$. Ex:
$$
S(2,2)=\frac{2^2-2.1^2+0}{2!}=\frac{2}{2}=1
$$
How do i prove it for a general case ?.ie,
$$
S(n,n)=\frac{1}{n!}\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^n=1
$$
So we need to prove $\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^n=n!$
My Attempt:
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n=\binom{n}{0}n^n-\binom{n}{1}(n-1)^n+\binom{n}{2}(n-2)^n-\binom{n}{3}(n-3)^n+ .....\\=n^n-n.(n-1)^n+\frac{n(n-1)}{2}.(n-2)^n-\frac{n(n-1)(n-2)}{6}.(n-3)^n+ ......
$$
| Use the old coefficient trick
\begin{eqnarray*}
[y^n]: e^{yk} = \frac{k^n}{n!}.
\end{eqnarray*}
Now the sum can be written as (reverse the index)
\begin{eqnarray*}
\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} \frac{k^n}{n!} &=& [y^n]: \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} (e^{y})^{k} \\
&=& [y^n] : (e^y -1)^n = 1.
\end{eqnarray*}
| {
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How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?
| Let $f(x)=\frac{x}{1+x}.$
Then by Jensen inequality
$$
f(a)+f(b)=\frac{a}{1+a}+\frac{b}{1+b} \geq 2 f( \frac{a+b}{2})=\frac{a+b}{1+\frac{a+b}{2}} \geq \frac{a+b}{1+a+b}.
$$
| {
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What are the last 2 digits of $31^{41}$? By using a slick trick, I found that the last two digits were $31.$ However, I want to verify this using Fermat's little theorem or some alternate. How would I apply Fermat's Little Theorem?
| You changed the problem on me after I answered. So:
Two find the last two digits, just work in $\mathbb Z\bmod100.$
\begin{align}
33^1 & \equiv 31 \\
31^2 & \equiv 31\times 31 \equiv 61 \\
31^3 & \equiv 31\times61 \equiv 91 \\
31^4 & \equiv 31\times91 \equiv 21 \\
31^5 & \equiv 31\times21 \equiv 51 \\
31^6 & \equiv 31\times51 \equiv 81 \\
31^7 & \equiv 31\times81 \equiv 11 \\
31^8 & \equiv 31\times11 \equiv 41 \\
31^9 & \equiv 31\times41 \equiv 71 \\
31^{10} & \equiv 31\times71 \equiv 1
\end{align}
Therefore
$$
31^{41} = 31^{10}\times31^{10}\times31^{10}\times31^{10}\times31^1 \equiv 1\times1\times1\times1\times31 \equiv 31.
$$
Two find the last two digits, just work in $\require{cancel}\xcancel{\mathbb Z\bmod100.}$
$$
\xcancel{
\begin{align}
32^1 & \equiv 32 \\
32^2 & \equiv 32\times 32 \equiv 24 \\
32^3 & \equiv 32\times24 \equiv 68 \\
32^4 & \equiv 32\times68 \equiv 76 \\
32^5 & \equiv 32\times76 \equiv 32
\end{align}}
$$
After five steps you're back to $32.$
Every time you go four steps beyond where you had $32,$ you have $32$ again. So $41 = 1 + (1\times4)$ so you go through that cycle $10$ times, getting back to $32.$
| {
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Find $ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+\dots+\frac{n+1}{(2n)^2} $ I want to find the following limit:
$$ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2} $$
I guess it converges to 0 and I tried to prove it using the squeeze theorem but it doesn't seem to work.
Any ideas?
| Your limit can be rewritten as a Riemann sum + $a_n$ where $a_n\rightarrow 0$.
This is a Riemann sum:
\begin{align}
\lim_{n\rightarrow \infty} \frac{1}{(n+1)^2}+ \frac{2}{(n+2)^2}+\dots+\frac{n}{(2n)^2}&=\lim_{n\rightarrow \infty} \frac{1}{n}\left ( \frac{1/n}{(1+1/n)^2}+ \frac{2/n}{(1+2/n)^2}+\dots+\frac{n/n}{(1+n/n)^2}\right)\\
&=\int_{0}^1 \frac{x}{(1+x)^2} \text{d}x \\
&=\int_{0}^1 \frac{x+1}{(1+x)^2} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x\\
&=\int_{0}^1 \frac{1}{1+x} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x
\end{align}
Now,
\begin{align}a_n&=\frac{1}{n^2} + \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2} \\
&\leq \frac{1}{n^2} + \frac{1}{n^2}+ \frac{1}{n^2}+\dots+\frac{1}{n^2}\\
&=(n+1)\frac{1}{n^2}\rightarrow 0
\end{align}
| {
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Definite Trigonometric Integration
Calculate the integral : $$\ \int_{0}^{\pi}\frac{x}{a-\sin(x)}dx , \quad a>1 $$
I have tried trig substitutions, and replacing $\ x$ with $\ \pi-y $, along with other commonly used "tricks", but none seem to work.
| As mentioned in the comments, the substitution $x \to \pi - x$ yields :
$$\int_0^{\pi}\frac{xdx}{a-\sin x}=\frac{\pi}{2}\int_0^{\pi}\frac{dx}{a-\sin x}$$
I'll go over the calculation of the integral on the RHS.
We have :
$$ I ={\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x =-{\displaystyle\int}\dfrac{1}{\sin\left(x\right)-a}\,\mathrm{d}x $$
Solving the integral :
$$I= {\displaystyle\int}\dfrac{1}{\sin\left(x\right)-a}\,\mathrm{d}x $$
Let's prepare our integral for the Weierstrass Substitution :
$${\displaystyle\int}\dfrac{1}{\frac{2\tan\left(\frac{x}{2}\right)}{\tan^2\left(\frac{x}{2}\right)+1}-a}\,\mathrm{d}x$$
Substitute :
$$u=\tan\left(\dfrac{x}{2}\right)\to \mathrm{d}x=\dfrac{2}{\sec^2\left(\frac{x}{2}\right)}\,\mathrm{d}u =\dfrac{2}{u^2+1}\,\mathrm{d}u$$
The integral then becomes :
$$I=-\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{au^2-2u+a}\,\mathrm{d}u = -{\displaystyle\int}\dfrac{1}{\left(\sqrt{a}u-\frac{1}{\sqrt{a}}\right)^2+a-\frac{1}{a}}\,\mathrm{d}u$$
Substitute :
$$v=\dfrac{au-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}} \to \mathrm{d}u=\dfrac{\sqrt{a-\frac{1}{a}}}{\sqrt{a}}\,\mathrm{d}v$$
The integral then becomes :
$$I=-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}}}{\displaystyle\int}\dfrac{1}{v^2+1}\,\mathrm{d}v = -\dfrac{\arctan\left(v\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}}$$
Substituting back, we get :
$$I=-\dfrac{\arctan\left(\frac{au-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}} = \dots =-\dfrac{2\arctan\left(\frac{a\tan\left(\frac{x}{2}\right)-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}} $$
The integral is now solved, as by :
$$-{\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x =\dfrac{2\arctan\left(\frac{a\tan\left(\frac{x}{2}\right)-1}{\sqrt{a}\sqrt{a-\frac{1}{a}}}\right)}{\sqrt{a}\sqrt{a-\frac{1}{a}}}$$
Which by simplifying and rewriting, leads to :
$$\boxed{{\displaystyle\int}\dfrac{1}{a-\sin\left(x\right)}\,\mathrm{d}x= \dfrac{2\arctan\left(\frac{a\tan\left(\frac{x}{2}\right)-1}{\sqrt{a^2-1}}\right)}{\sqrt{a^2-1}}+C, \quad a>1}$$
Can you now, using this result and the result from the substitution, calculate the given indefinite integral :
$$\ \int_{0}^{\pi}\frac{x}{a-\sin(x)}dx , \quad a>1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this substitution to solve an integral correct? To solve the integral:
$\int\frac {x}{\sqrt{4-x^2}}\ dx$
I used the substitution:
$u = \sqrt{4-x^2}$
hence:
$\frac{du}{dx} = -\frac {x}{\sqrt{4-x^2}}$
$dx = -\frac {\sqrt{4-x^2}}{x}\ du$
So the integral becomes:
$\int\frac {x}{u}(-\frac {\sqrt{4-x^2}}{x})\ du$
$= \int\frac {x}{u}(-\frac {u}{x})\ du$
$= \int -1\ du$
$= -u+c$
$= -\sqrt{ 4-x^2}+c$
However the solution I have seen is:
Use the substitution $x = 2\sin(t)$, so $dx = 2\cos(t)\ dt$
So the integral becomes:
$\int \frac {2\sin(t)(2\cos(t))}{\sqrt{\cos(t)^2}}\ dt$
After some rearranging the integral becomes:
$\int 2\sin(t)\ dt$
Which gives:
$-2\cos(t)+c$
Then using $\sin(t) = \frac x2$:
$\cos(t)^2 = 1 - \sin(t)^2$
$\cos(t)^2 = 1 - \frac {x^2}{4} = \frac{4-x^2}{4}$
$\cos(t) = \frac{\sqrt{4-x^2}}{2}$
Which then gives the solution as:
$= -\sqrt{ 4-x^2}+c$
I feel like the method I used was more efficient as it didn't require the use of trig functions. I'm fairly confident my method is correct so why would the solution use trig functions? Was I just lucky that I decided $u = \sqrt{ 4-x^2}$ was a good substitution?
| Let $4-x^2=u$
Then $xdx=-\frac {du} 2$......
| {
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Infinite series summation
$$\sum^{\infty}_{n=0}\frac{1}{(4n+5)(4n+6)\cdots \cdots (4n+11)}$$
Try: $$\sum^{\infty}_{n=0}\frac{(4n+4)!}{(4n+11)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\frac{(4n+4)!\cdot 6!}{(4n+4+6+1)!} = \frac{1}{6!}\sum^{\infty}_{n=0}\int^{1}_{0}x^{4n+4}(1-x)^6dx$$
$$ \frac{1}{6!}\int^{1}_{0}\left(\sum^{\infty}_{n=0}x^{4n}\right)x^4(1-x)^6dx = \frac{1}{6!}\int^{1}_{0}\frac{x^4(1-x)^6}{1-x^4}dx$$
could some help me to solve it thanks
| Hint:
Another way:
$$\prod_{r=5}^{11}\dfrac{4n+11-(4n+5)}{4n+r}=\prod_{r=6}^{11}\dfrac1{4n+r}-\prod_{r=5}^{10}\dfrac1{4n+r}$$
See Telescoping series
| {
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Convergence of $\sum\limits_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $
I've got in my assignment to show if the following series converges or diverges.
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$
Attempt:
\begin{align*}
\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) &=\sum_{n=1}^\infty \left( 1 + \frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...-1-\frac{1}{n}\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)\\
&=\sum_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)
\end{align*}
At this point I'm lost. I tried using D'Alambert as follows:
$$\lim \frac{a_{n+1}}{a_n}=\lim\frac{\sqrt[n+1]{e}-1-\frac{1}{n+1}}{\sqrt[n]{e}-1-\frac{1}{n}}$$
which I tried to simplify with basic limit laws (hopefully correctly):
$$\lim\frac{\sqrt[n+1]{e}-1}{\sqrt[n]{e}-1} = 1$$
I don't know where to go from here. Thank you for all your help in advance.
| Thank you for the comment. By using the comparison test I easily proved the sum converges.
$$\sum_{n=1}^\infty (\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...) \le \sum_{n=1}^\infty \frac{e}{n^2}$$
| {
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If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$
Book's Answer
I have mentioned my problem in the image.Any and all help will be appreciated
| Note that
$$a_{n}=a_{n-1}^{2}-2\implies a_n^2=a_{n-1}^{4}-4a_{n-1}-4\implies a_n^2-4=a_{n-1}^{2}(a_{n-1}^{2}-4) $$
thus
$$a_{n-1}^2-4=a_{n-2}^{2}(a_{n-2}^{2}-4) \implies a_n^2-4=a_{n-1}^{2}a_{n-2}^{2}(a_{n-2}^{2}-4) $$
and so on, thus it can be easily proved by induction that
$$a_n^2-4=a_{n-1}^{2}a_{n-2}^{2}...a_2^2a_1^2(a_1^{2}-4) $$
thus
$$\sqrt{a_1^{2}-4}= \frac{\sqrt{a_n^2-4}}{a_1a_2...a_{n-2}a_{n-1}}\sim\frac{a_n}{a_1a_2...a_{n-2}a_{n-1}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is this approximation correct? I came across an interesting approximation today in an indian olympiad paper (page 6). It essentially boils down to:
$$V(x)=\frac{k}{\sqrt{x^2+4/3}}+\frac{k}{\sqrt{(x+1)^2+1/3}}+\frac{k}{\sqrt{(x-1)^2+1/3}}$$
$$\approx k\sqrt{\frac 34}(3+\frac {9}{16}x^2)$$
Even before I got down to attempt to solve this step, I realized a problem. expression 1 decreases with $x$, since $x$ is in the denominator of all three terms in it. But expression 2 increases with $x$, since $x$ is in the numerator. But both are supposed to be approximately equal. How do we justify this?
| Consider $$y(a,b)=\frac{1}{\sqrt{(x+a)^2+b}}$$ Using Taylor series around $x=0$, you get
$$y(a,b)=\frac{1}{\sqrt{a^2+b}}-\frac{a x}{\left(a^2+b\right)^{3/2}}+\frac{x^2 \left(2
a^2-b\right)}{2 \left(a^2+b\right)^{5/2}}+\frac{x^3 \left(3 a b-2 a^3\right)}{2
\left(a^2+b\right)^{7/2}}+O\left(x^4\right)$$ Using this for the complete expression with the proper values of $a,b$
$$V(x)=\frac{ 3\sqrt 3}2k\left(1+\frac{3 x^2}{16}+O\left(x^4\right) \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$ If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$
Try: $a+b+c=0,ab+bc+ca=1,abc=-1$
Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$
Could some help me to explain short way to calculate product,Thanks
| $a,b$ are the solutions of $x^3+x+1=0$, so we have
$$a^3+a=-1=b^3+b$$
and so,
$$(a^3-b^3)+(a-b)=0\implies (a-b)(a^2+ab+b^2+1)=0$$
Suppose here that $a=b$. Then, since $a+b+c=0\implies c=-2a$, we have
$$abc=-1\implies -2a^3=-1\implies a^3=\frac{1}{2}\implies a=-1-a^3=-\frac 32$$
However, $a=-\frac 32$ is not a solution of $x^3+x+1=0$.
So, we have $a\not=b$, so
$$a^2+ab+b^2+1=0\implies (a-b)^2=-3ab-1$$
Similarly, $$(b-c)^2=-3bc-1\qquad\text{and}\qquad (c-a)^2=-3ca-1$$
It follows that
$$\begin{align}&(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2\\\\&=(-3ab-1)(-3bc-1)+(-3bc-1)(-3ca-1)+(-3ca-1)(-3ab-1)\\\\&=9abc(a+b+c)+6(ab+bc+ca)+3\\\\&=9\cdot (-1)\cdot 0+6\cdot 1+3\\\\&=9\end{align}$$
and $$\begin{align}(a-b)^2(b-c)^2(c-a)^2&=(-3ab-1)(-3bc-1)(-3ca-1)\\\\&=-27(abc)^2-9abc(a+b+c)-3(ab+bc+ca)-1\\\\&=-27\cdot (-1)^2-9\cdot (-1)\cdot 0-3\cdot 1-1\\\\&=-31\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate limit with L'Hopital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.
I have done the following:
It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.
So, we can use L'Hopital's rule:
\begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ & =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ & =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}
We calculate the two limits separately
*
*$\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :
\begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1 \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq 2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}
*
*How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ?
| To use L'Hospital rule, one needs to check if $\lim_{x\rightarrow 0}\dfrac{f'(x)}{g'(x)}$ exists. So in this case the topologist's sine curve $\sin(1/x)$ does not have limit whenever $x\rightarrow 0$, so L'Hospital rule fails to apply.
Now use the trick like
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{x}{\sin x}=1
\end{align*}
and that
\begin{align*}
\lim_{x\rightarrow 0}x\cos(1/x)=0
\end{align*}
to conclude that the limit value is zero.
For the proof of $x\cos(1/x)\rightarrow 0$ whenever $x\rightarrow 0$:
\begin{align*}
|x\cos(1/x)|\leq|x|,
\end{align*}
and now use Squeeze Theorem to conclude that $\lim_{x\rightarrow 0}x\cos(1/x)=0$ because of that $\lim_{x\rightarrow 0}|x|=0$.
To claim that the topologist's sine curve does not have limit as $x\rightarrow 0$, simply let $a_{n}=\dfrac{1}{2n\pi}$ and $b_{n}=\dfrac{1}{2n\pi+\pi/2}$, so $a_{n},b_{n}\rightarrow 0$ but $\sin(1/a_{n})=0\rightarrow 0$ and $\sin(1/b_{n})=1\rightarrow 1$, the function has two distinct limit points whenever $x\rightarrow 0$, so the limit does not exist.
| {
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"url": "https://math.stackexchange.com/questions/2614160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Probability of someone picking the same toy For someone to pick the same toy:
Jon picks car1 and car2
Given 3 people
Hans
Thomas
Jon
And 6 toys:
Car1, Car2, Doll1, Doll2, Ball1, Ball2
Each person is given 2 toys, calculate the probability of someone getting the same toy. (there are two instances of each toy).
The way I tried to solve it is this way:
*
*Only Jon gets the same toys
*Only Thomas gets the same toys
*Only Hans gets the same toys
For 1-3, the probability is as follows:
Hans picks up any toy (5 toys are left)
Jon has to pick any toy that is not what Hans picked (4 options out of 5)
Thomas has to pick the toy which Hans picked (1 options out of 4)
Hans has to pick the same kind of toy he has (1 out of 3)
So the probability for scenario 1 is:
(4/5)*(3/4)*(1/3) = (1/5)
So for 1-3 the probability is 1/5
Now I have the probability of only 2 getting the same toy, which is not possible as the 3rd person by default also gets the same toy, so the probability is 0.
And the last option is that all 3 get the same instance of a toy.
(4/5)*(2/4)*(1/3)*(1/2) = (1/15)
Overall, the probability of at least one getting two instances of the same toy is: 1/15 + 3*(1/5)
But it seems I am wrong?
Any help will be appreciated.
| Someone picking a toy:
One in left hand one in right.
Left hand first, will chose between 6 options.
Then the right hand will have 5 options.
$5\cdot 6=30$ options
Do we care about what is in which hand? No... So:
In any case all the ways for the first "player" are $\frac{6\cdot 5}{2}=15$.
Second person has to chose between 4 toys anyway.. So, same logic:
All the ways: $\frac{4\cdot 3}{2}=6$ ways for the second "player".
Last one 1 way (only two toys left).
Does it matter if the "players" take first one toy and then the second?
No, same results.
Now:
For the wanted result,
A) the first has 3 ways between 15 options (to chose between one of the 3 pairs)
$P_A=\frac{3}{15}=\frac{1}{5}$
and the second has :
B1) if first got a pair : 2 ways from the four options... But we don't care (already true statement because of the first and the 2 ways was to chose between every pair left there)
Edit:
$P_B=P_A\cdot\frac{2}{6}=\frac{1}{5}\cdot\frac{1}{3}=\frac{1}{15}$
(We will not ask for this -first already made true our request-)
B2) first didn't got a pair : 1 way (first destroyed two of the pairs and only one left for the second)
$P_B=(1-P_A)\cdot\frac{1}{6}=\frac{4}{5}\cdot\frac{1}{6}=\frac{4}{30}=\frac{2}{15}$
Last player:
C1) If first or second (or both) did it, we don't care.
C2) If first and second failed The only option is that first and second destroyed the same two pairs... But then the third will pick the last pair anyway:
$P_C=P_{A->fail}\cdot P_{B->same\ pair\ as\ A}$
$P_C=(1-P_A)\cdot\frac{1}{6}=\frac{4}{5}\cdot\frac{1}{6}=\frac{4}{30}=\frac{2}{15}$
Answer:
first has 3\frac{1}{5} and if he didn't make it second has \frac{2}{15} to keep pair in his hands or \frac{4}{30} to leave pair to the third:
Edit:
$P=\frac{1}{5}+\frac{2}{15}+\frac{4}{30}=\frac{6}{30}+\frac{4}{30}+\frac{4}{30}=\frac{14}{30}=\frac{7}{15}$
About @kludg's comment:
About the same or different probabilities:
A) The probability for each player to be the only one with a pair is the same for all players and equal to ($\frac{2}{15}$) but:
*
*The probability of every player to catch a pair and be the one who will finishing the game is $\frac{1}{5}$ for the first player and then remains $\frac{1}{3}$ after the first player (inside the event left from first failure $=\frac{4}{5}$) where second and third have $\frac{1}{6}$ and \frac{1}{5} each one inside the event they remain ($\frac{4}{5}$ for the B and $\frac{4}{5}\cdot{5}{6}=\frac{2}{3}$ for the player C). Also, in the second step (Player B) the total possibilities inside the 80% of A's failure we have $\frac{1}{3}$ chances to find a (one) pair.
So, by 20% ($\frac{1}{5}$) the game finishing by the first player, then, if not the player B has $\frac{1}{6}$ chances to be the one with the pair... having a total chance $\frac{2}{15}$ and after him (in a case of both previous failure that comes by possibility $\frac{2}{3}$) by $\frac{1}{5}$. [Notice that $\frac{1}{5}\cdot\frac{5}{6}\cdot\frac{4}{5}=\frac{2}{15}$ again]
For somebody need another way of explanation:
--Player A: $P_A=\frac{1}{5}=\frac{3}{15}$ because we can consider he is taking first two toys... and $\frac{1}{5}$ is the paired toy he looks for.
--Player B: Player A have already failed and B is taking one of four toys that left and have to find the pair that is one of 3 that remain. But even the first toy he will take has to be deferent than both of Player A toys. So, he has two ways to take both toys non matching with A ($P_B=2\cdot\frac{2}{4}\cdot\frac{1}{3}=\frac{2}{12}=\frac{1}{6}$ being in an event of $1-P_{A}=\frac{4}{5}$). So, second player has $\frac{4}{30}=\frac{2}{15}$ total chances but also $\frac{1}{6}$ chances after A's failure.
--Player C: If both A,B have failed we are in a case with possibility $P_{fail\{A\&B\}}=\frac{4}{5}\cdot\frac{5}{6}=\frac{2}{3}$. But then he already have no choices... he will take the toys that left for him. The possibility of A and B to took the same (not-paired) pairs. What about these possibilities? failure of A ($\frac{4}{5}$) and the second has to take exactly the same toys that A got: $\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{6}$ again.
Again the total chance for the third player is $\frac{4}{5}\cdot\frac{1}{6}=\frac{2}{15}$ to be the one.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate length of one side given all other sides in a triangle.
All the lengths in red had been given in the problem statement. The task is to find $BC$. The green ones were calculated from the information in red.
Here's two ways I tried to solve it (and I would like feedback):
*
*
The green line is drawn parallel to $QC$. Suppose, the other unlabelled end of the green line be point $R$. So the line can be called $PR$.
In $\triangle BPR$,
$PC=QC=9, \space\space PC=3\times AP\\PB=6, \space\space PB=3 \times AQ\\\therefore \triangle BPR \sim \triangle APQ\\BD=4\times3, \space\space \boxed{BD=12}$
In quadrilateral $PQCD$, opposite sides are parallel to each other and it is therefore a parallelogram. So $DC=4$.
$BC=BD+DC\\BC=12+4\\\boxed{BC=16}$
*
In $\triangle ABC$, $AB=2+6=2(1+3)=\boxed{2\times 4}\\AC=3+9=3(1+3)=\boxed{3\times 4}\\BC=?$
In $\triangle APQ$, $AP=\boxed{2}\\AC=\boxed{3}\\PQ=4$
$\because \triangle ABC\sim \triangle APQ,\\BC=PQ\times4=4\times4=16,\\\boxed{BC=16}$
| Your question doesn't mention $PQ$ is parallel to $BC$ but your answer does.
So assuming that line $PQ$ is parallel to $BC$ then by similar triangles
( Considering the triangles $APQ$ and $ABC$)
$$\frac {AP}{AB}= \frac {AQ}{AC}=\frac {PQ}{BC}$$
Hence we get $$\frac {3}{12}=\frac {4}{BC}$$
$$\Rightarrow BC=16$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=e^{-2}$ without using de l‘Hôspital
$$\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=\frac{1}{e^2}.$$
But how do I prove this without using de l'Hôspital twice?
| Without change of variables, note that
$$ \left ( \frac{2n-1}{2n+3} \right ) ^n =\left ( \frac{2n+3-4}{2n+3} \right ) ^n=\left ( 1+\frac{-4}{2n+3} \right ) ^n=\sqrt{\left ( 1+\frac{-4}{2n+3} \right ) ^{2n+3}\left ( 1+\frac{-4}{2n+3} \right ) ^{-3}}\to \sqrt{e^{-4}\cdot1}=\frac1{e^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a combinatorial proof for $\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + ... + \binom{n-k}{0}$
Let $n$ and $k$ be integers with $n \geq k \geq 0$. Find a combinatorial proof for
$$\binom{n+1}{k} = \binom{n}{k} + \binom{n-1}{k-1} + \cdots + \binom{n-k}{0} .$$
My approach:
I was thinking to use the binomial formula as in $$2^n = \sum{\binom{n}{k}1^k1^{n-k}} .$$
I also tried to use Pascal's Identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$.
| It follows from Pascal's identity that\begin{align}\binom{n+1}k&=\binom nk+\binom n{k-1}\\&=\binom nk+\binom{n-1}{k-1}+\binom{n-1}{k-2}\\&=\binom nk+\binom{n-1}{k-1}+\binom{n-2}{k-2}+\binom{n-2}{k-3}\\&=\cdots\end{align}Can you take it from here?
| {
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"url": "https://math.stackexchange.com/questions/2626089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$
Find all real values of $ x>1$ which satisfy$$ \frac{x^2}{x-1} + \sqrt{x-1} +\frac{\sqrt{x-1}}{x^2} = \frac{x-1}{x^2} + \frac{1}{\sqrt{x-1}} + \frac{x^2}{\sqrt{x-1}}$$
Well I first observe that each factor has its own inverse involve in the equation and rewriting
gives $$\left( \frac{x^2}{x-1}-\frac{x-1}{x^2} \right) + \left(\sqrt{x-1} -\frac{1}{\sqrt{x-1}} \right)+\left(\frac{\sqrt{x-1}}{x^2} -\frac{x^2}{\sqrt{x-1}}\right) =0 $$
However after expanding all the terms it does not look wise to do so. Does anyone has an idea
| Let $$A=\frac{x^2}{x-1},\qquad B=\sqrt{x-1},\qquad C=\frac{\sqrt{x-1}}{x^2}$$
Then, we have
$$A+B+C=\frac 1A+\frac 1B+\frac 1C\qquad\text{and}\qquad ABC=1$$
from which we have
$$A+B+C=AB+BC+CA\qquad\text{and}\qquad ABC=1$$
Let $s:=A+B+C=AB+BC+CA$.
It follows from these that $A,B,C$ are the solutions of
$$t^3-st^2+st-1=0,$$
i.e.
$$(t-1)(t^2+t+1-st)=0$$
So, we see that either $A,B$ or $C$ has to be equal to $1$.
$A=\frac{x^2}{x-1}=1\implies x^2-x+1=0$ has no real solutions.
$B=\sqrt{x-1}=1\implies x=2$ which is sufficient.
$C=\frac{\sqrt{x-1}}{x^2}=1\implies x^4-x+1=0$.
Let $f(x):=x^4-x+1$. Then, $f'(x)=4x^3-1=0$. So, $f'(x)\gt 0$ for $x\gt 1$.
Since $f(1)=1\gt 0$, we see that $f(x)=0$ has no real solutions.
Therefore, $\color{red}{x=2}$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to remove absolute value from inequality So I know that $|x| \le 2 \iff x\le 2 \text{ and } x \ge -2$. But, I was wondering would the same rules apply for $|x+2| + |y| \le 5$. Would it be correct to say that this inequality is equivalent to $x + y \le 3 \text{ and }x + y \ge -7$?
| For a problem involving two variables such as $$ |x+2| + |y| \le 5$$ you need to look at 4 regions in the $xy-plane$
1) $y\le 0$ and $x+2 \le 0$. In this region your inequality is $-(x+2)-y \le 5$
2) $y\le 0$and $x+2 \ge 0$. In this region your inequality is $(x+2)-y \le 5$
3) $y\ge 0$ and $x+2 \ge 0$. In this region your inequality is $(x+2)+y \le 5$
4) $y\ge 0$ and $x+2 \le 0$. In this region your inequality is $-(x+2)+y \le 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding minimum value of logarithmic series Find the min value of $$\left\vert \log_{x_1} \left ( x_2 - \frac {1}{4}\right) +\log_{x_2} \left ( x_3 - \frac {1}{4}\right) +\log_{x_3} \left ( x_4 - \frac {1}{4}\right) +\cdot \cdot \cdot \cdot \cdot +\log_{x_{2016}} \left ( x_{2017} - \frac {1}{4}\right) +\log_{x_{2017}} \left ( x_{1} - \frac {1}{4}\right) \right\vert$$
Where $x_1,x_2,x_3,......, x_{2017}\in \left(\frac {1}{4}, 1\right)$
I am not getting any idea over where and how to start approaching this problem.
| Note that
$$\log_{x_i} \left ( x_j - \frac {1}{4}\right)=\frac {\log \left ( x_j - \frac {1}{4}\right)}{\log x_i}$$
thus by Rearrangement inequality
$$\left\vert \log_{x_1} \left ( x_2 - \frac {1}{4}\right) +\log_{x_2} \left ( x_3 - \frac {1}{4}\right) +\log_{x_3} \left ( x_4 - \frac {1}{4}\right) +\cdot \cdot \cdot \cdot \cdot +\log_{x_{2016}} \left ( x_{2017} - \frac {1}{4}\right) +\log_{x_{2017}} \left ( x_{1} - \frac {1}{4}\right) \right\vert=$$
$$=\left\vert\frac {\log \left ( x_2 - \frac {1}{4}\right)}{\log x_1}+\frac {\log \left ( x_3 - \frac {1}{4}\right)}{\log x_2}+...+\frac {\log \left ( x_1 - \frac {1}{4}\right)}{\log x_{2017}}\right\vert \ge \sum_{i=1}^{2017} \frac {\log \left ( x_i - \frac {1}{4}\right)}{\log x_i}$$
and equality holds when $\forall i,j\quad x_i=x_j$.
Thus we get the minimum when $\frac {\log \left ( x_i - \frac {1}{4}\right)}{\log x_i}$ is minimum that is 2 for $x_i=\frac12$.
Thus the minimum for the given sum is $\sum_{i=1}^{2017}2= 2 \cdot 2017=4034$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding value of $k$ in trigonometric equation.
$$\prod^{50}_{r=1}\tan\bigg[\frac{\pi}{3}\bigg(1+\frac{3^r}{3^{50}-1}\bigg)\bigg]=k\prod^{50}_{r=1}\cot\bigg[\frac{\pi}{3}\bigg(1-\frac{3^r}{3^{50}-1}\bigg)\bigg]$$
Try:$$\tan (60^\circ+x)=\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}$$
and $$\cot(60^\circ-x)=\frac{\sqrt{3}-\tan x}{\sqrt{3}+\tan x}$$
I did not understand how to solve it other then formule above mention. Help me
| $$X= \prod^{50}_{r=1}\tan \left(\frac{\pi}{3}\left(1+\frac{3^r}{3^{50}-1}\right)\right)$$ and $$Y = \prod^{50}_{r=1}\cot \left(\frac{\pi}{3}\left(1-\frac{3^r}{3^{50}-1}\right)\right)$$
Assuming $$a_r=\frac{\pi}{3}\cdot \frac{3^r}{3^{50}-1}$$.Then
$$\tan(\frac{\pi}3+a_r)=\frac{\sqrt 3+\tan a_r}{1-\sqrt 3\tan a_r}$$
and $$\cot(\frac{\pi}3-a_r)=\frac{1+\sqrt 3\tan a_r}{\sqrt 3-\tan a_r}$$
So $$\frac{\tan(\frac{\pi}3+a_r)}{\cot(\frac{\pi}3-a_r)}=\frac{3-\tan^2a_r}{1-3\tan^2a_r}=\frac{\tan 3a_r}{\tan a_r}=\frac{\tan a_{r+1}}{\tan a_r}$$
So $$XY=\prod_{r=1}^{50}\frac{\tan a_{r+1}}{\tan a_r}=\frac{\tan a_{51}}{\tan a_1}$$
Now $a_{51}-a_1=\pi$ and so $\displaystyle\frac XY=1$ and so $X=Y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question related to beta and gamma function I'm trying to derive the following integral.
$$\int_0^\infty \frac{x^8(1-x^6)}{(1+x)^{24}} \, dx.$$
What transformations can I use?
| Here is a way to evaluate the integral without invoking the Beta function.
Writing the integral as
$$\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx = \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_1^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx.$$
In the second of these integral, if a substitution of $x \mapsto 1/x$ is enforced we have
\begin{align*}
\int_0^\infty \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx &= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_0^1 \frac{\frac{1}{x^8} \left (1 - \frac{1}{x^6} \right )}{\left (1 + \frac{1}{x} \right )^{24}} \frac{dx}{x^2}\\
&= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx + \int_0^1 \frac{x^8 (x^6 - 1)}{(1 + x)^{24}} \, dx\\
&= \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx - \int_0^1 \frac{x^8 (1 - x^6)}{(1 + x)^{24}} \, dx\\
&= 0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fractions in Questions and Answers
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