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minimal value of $x^2+2y^2+5z^2$ with constraint. $x,y,z>0$, and $xy+yz+zx=1$.
I need to find the minimum value of
$x^2+2y^2+5z^2$
In general what can we say about the minimal value of $\frac{ax^2+by^2+cz^2}{xy+xz+yz}$, over all positive numbers with $a,b,c>0 $
| Using AM-GM, we have:
$$4x^2+9y^2 \ge 12xy$$
$$3y^2+12z^2 \ge 12yz$$
$$18z^2+2x^2\ge 12zx$$
Combining and using the constraint,
$$6x^2+12y^2+30z^2 \ge 12(xy+yz+zx) \implies x^2+2y^2+5z^2 \ge 2$$
Equality is when $x:y:z = 3:2:1$ and satisfying the constraint, i.e. when $x = \frac3{\sqrt{11}}, y = \frac2{\sqrt{11}}, z = \frac1{\sqrt{11}}$.
In general for $ax^2+by^2+cz^2$, you can construct similar inequalities using unknown coefficients to start with, which you solve to keep the equality condition and match LHS / RHS.
| {
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"timestamp": "2023-03-29T00:00:00",
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Logarithmic inequality Solve the inequality:
$$
\log_8(x^2-4x+3) < 1
$$
$$
\log_8(x^2-4x+3) < \log_8(8)
$$
$$
\log_8(x^2-4x+3) - \log_8(8) <0
$$
$$
\log_8 [(x^2-4x+3)/8] < 0
$$
Thats what I did for the question so far... and I'm confused as to what to do next. Can someone verify this for me?
| Firstly, $\log(n)$ is defined only if $n \in (0,\infty)$
$$
\begin{align}
\Rightarrow\quad x^2 - 4x + 3 &\in (0,\infty)\\
\Rightarrow\quad x^2 - 4x + 3 &> 0\\
\Rightarrow (x-3)(x-1) &> 0\\
\end{align}
$$
$$
\boxed{\therefore\quad x \in (-\infty, 1)\cup(3,\infty)
\quad\text{or}\quad x\in \mathbb R \sim [1,3]}\tag{1}
$$
Now, $\log_b(n) = y \iff n = b^y$,
$$
\begin{align}
\log_8(x^2-4x+3) &< 1\\
\Rightarrow x^2 - 4x + 3 &< 8^1\\
\Rightarrow x^2 - 4x - 5 &< 0\\
\Rightarrow (x+1)(x-5) &< 0
\end{align}
$$
$$
\boxed{\therefore x\in\left(-1,5\right)} \tag{2}
$$
Remember that both $(1)$ and $(2)$ have to be satisfied at the same time.
Now, $(1)\cap(2) \implies x\in(-1,1)\cup(3,5)$
| {
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"url": "https://math.stackexchange.com/questions/933218",
"timestamp": "2023-03-29T00:00:00",
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How would you integrate $\sqrt{1+\frac{1}{x^2}}$ I need to integrate $\sqrt{1+\frac{1}{x^2}}$
I've tried to let $u=1/x^2$ but end up with, $\int \frac{\sqrt{1+u^2}}{u^{3/2}}du$ . I attempted to then substitute $u=\tan\theta$ and lead me to $-\frac{1}{2}\int \frac{1}{(\sin\theta\cos\theta)^{3/2}}d\theta$.
Not sure how to progress from here. Is there any alternative way of doing this?
| $$\int\sqrt{1+\frac{1}{x^2}}\,dx=\int \frac{\sqrt{1+x^2}}{x}\,dx=\int \frac{x\sqrt{1+x^2}}{x^2}\,dx$$ Let $1+x^2=u^2 $ so that$$\int \frac{x\sqrt{1+x^2}}{x^2}\,dx=\int \frac{u^2}{u^2-1}\,du$$ hence we have that $$\int \frac{u^2}{u^2-1}\,du=u+\frac{1}{2}\ln |\frac{u-1}{u+1}|+c$$
| {
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Solving for $x$ given $y = 2x - 9$ and $y = 5$. When does the 5 come into play? I am doing some problems. It states "Isolate the left side for each of the following equations. Then solve for x, assuming the value of y is 5 in a all cases." I noticed that I can get two different answers, depending on when I decide to plug in the value of y. For example, this is what I get:
$$ y = 2x - 9 $$
$$ 9 + y = 2x -9 +9 $$
$$ 9 + y = 2x \frac{1}{2} $$
$$ \frac{1}{2} * 9 + y = x $$
$$ x = \frac{1}{2} * 9 + 5 $$
$$ x = \frac{9}{2} + 5 $$
$$ x = \frac{19}{2} $$
But, if 5 is plugged into y in the beginning, I've seen it done like this:
$$ 5 = 2x - 9 $$
$$ 5 + 9 = (2x - 9) + 9 $$
$$ 14 = 2x -9 + 9 $$
$$ 14 * \frac{1}{2} = 2x * \frac{1}{2} $$
$$ x = 7 $$
Which one is correct?
| The first is wrong because you forgot to multiply $y$ by $\tfrac12$. When you got into
$$
9 + y = 2x
$$
By the law of equations, you should do as follows:
$$
(9 + y) = 2x \\
\frac12(9+y) = \frac12 2x \\
\frac12(9+y) = x
$$
The you plug the value of $y$
$$
\frac12(9 + 5) = x \\
\frac12 14 = x \\
x = 7
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Am I differentiating this wrong? Differentiation is the opposite of Integration
$$\begin{align}\int \cos^2x dx\end{align}$$
$$\begin{align}-\frac{\cos^3x}{3\sin x}\end{align}$$
Now if we differentiate $-\frac{\cos^3x}{3\sin x}$ we should get $\cos^2x$. But I have always thought the differentiation goes like (and works out fine).
$$\begin{align}\frac{d}{dx}(-\frac{\cos^3x}{3\sin x})= \frac{-3 \cos^2x (\frac{d}{dx}\cos x)}{3\sin x}=\cos^2x\end{align}$$
And Tarda we get the answer as expected ($\cos^2x$). I normally never thought about it, but today this came to my mind. How come this isn't wrong. And it should be wrong I think. Why?
$$\begin{align}\frac{d}{dx}(-\frac{\cos^3x}{3\sin x})\end{align}$$
This is in the form $u/v$ , so we we need to use the chain rule/quotient rule, right? But here we havent used Quotient Rule which should be:
$$\begin{align}\frac{d}{dx}\frac{-\cos^3x}{3\sin x}=\frac{-9\sin x\cos^2x \sin x+3\cos^4x}{9\sin^2x}\end{align}$$
But Stilll even without using Quotient rule, I end u with the correct answer. Why is that ? we totally forget about the $3\sin x$ in denominator and differentiate only the top which is $-\cos^3x$ and still gets it right.
| You are wrong in $\int \cos^2 x dx = -\frac{\cos^3 x}{3\sin x}$. You seem to be confused with power rule ($\int x^n dx = \frac{x^{n+1}}{n+1} + c $, for $n \neq -1$), which is only valid for powers of $x$, not $f(x)$. Also, there's an extra $\sin x$ in denominator...
The correct way to evaluate $\int \cos^2 x dx$ is:
$\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \int \frac12 (1 + \cos 2x) dx = \frac 12(x + \frac{\sin 2x}{2}) + c$.
Now, if you differentiate this, you do get $\frac{1 + \cos 2x}{2}$ back, which equals $\cos ^2 x$.
Proof of $\cos^2 x = \frac{1+\cos 2x}{2}$:
Now, $1 + \cos 2x = \cos 0 + \cos 2x = 2\cos(\frac{0+2x}{2})\cos(\frac{0-2x}{2}) = 2\cos(x)\cos(-x) = 2\cos^2 x$, because $\cos (-x) = \cos x$.
Also, we know that $\cos2x=2\cos^2x-1$, then $\cos2x+1=\cos^2x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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computing integral without softwares: $\int \frac{2x+3}{x^2+\sqrt{1-x^2}}dx$ I was wondering if this integral can be solve without wolfram and others:
$$\int \frac{2x+3}{x^2+\sqrt{1-x^2}}dx$$
Thanks.
| This is a plan of attack, at least:
1) Separate into two terms, and to find $\displaystyle\int\frac{2x}{x^2+\sqrt{1-x^2}} dx$, let $u=x^2$ to get $\displaystyle\int\frac{1}{u+\sqrt{1-u}} du$, and then let $t=\sqrt{1-u}$ to obtain $$ 2\int\frac{t}{t^2-t-1} dt.$$
Now use partial fractions.
2) For the second term, let $x=\sin\theta$ to get $\displaystyle3\int\frac{1}{x^2+\sqrt{1-x^2}} dx=3\int\frac{\cos\theta}{\sin^{2}\theta+\cos\theta} d\theta$.
Next let $\theta=\tan\frac{t}{2}, \sin\theta=\frac{2t}{1+t^2}, \cos\theta=\frac{1-t^2}{1+t^2}, d\theta=\frac{2}{1+t^2} dt$ to get
$$\int\frac{2(1-t^2)}{4t^2+(1+t^2)(1-t^2)} dt=2\int\frac{t^2-1}{t^4-4t^2-1} dt.$$
Now use partial fractions.
| {
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Finding the volume of the area between two curves when rotated about the y-axis I keep going down a rabbit hole when answering this question:
Find the volume of the area lying in the first quadrant and bounded by the $y$-axis, the curve $y = x^3$ and the line $y = 3x + 2$, when it performs one revolution about the $y$-axis.
We need to find the ordinates first:
$$\begin{align}
x^3 & = 3x + 2 \\
0 & = x^3 - 3x + 2 \\
0 & = (x+1)^2(x-2)
\end{align}$$
This gives us the co-ordinates where both meet: $(-1, -1)$ and $(2, 8)$. But since we're only interested in the first quadrant, we'll use $y = 8$ and $y = 0$ as the ordinates to bound our area.
Let $A_1$ be the area between the $y$-axis and $y=x^3$, and let $A_2$ be the area between the $y$-axis and $y = 3x + 2$. The total volume, $V_T$, will be found by $V_1 - V_2$; where $V_1$ is the volumes found by revolving $A_1$, and $V_2$ is the volume found by revolving $A_2$.
$$\begin{align}
A_1 & = \pi(x^2) =\pi\left(\sqrt[3]{y}\right)^2=\pi\left(y^{\frac{2}{3}}\right) \\
A_2 & = \pi(x^2) = \pi\left[\left(\frac{y-2}{3}\right)^2\right] = \pi\left[\frac{y^2}{9} - \frac{4y}{9}+\frac{4}{9}\right]
\end{align}$$
Now using the volume of revolution equation:
$$\begin{align}
V_1 & = \pi \int _{8} ^{0} (y^{\frac{2}{3}})dy = \pi\left[\frac{3y^{\frac{5}{3}}}{5}\right]_0^8\\
& = \frac{3}{5}(32)\pi = \frac{96}{5}\pi\\
\\
V_2 & = \pi \int _{8} ^{0} \left(\frac{y^2}{9} - \frac{4y}{9}+\frac{4}{9}\right)dy = \pi\left[\frac{y^3}{27} - \frac{2y^2}{9}+\frac{4y}{9}\right]_0^8\\
& = \left( \frac{512}{27} - \frac{128}{9} + \frac{32}{9} \right)\pi = \frac{227}{27}\pi\\
\\
V_T & = V_1 - V_2 = \frac{1457}{135}\pi
\end{align}$$
But the expected answer is $\frac{56}{5}\pi$. Where am I going wrong?
| When calculating $V_2$, you need to integrate from $2$ to $8$ because for $0 \le y < 2$, we have $x=(y-2)/3<0$, which is left of the $y$-axis (plot the graph to see this more clearly).
| {
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Integrating $\int_{-\pi}^{\pi} \frac{ d\theta}{w - sin \theta}$ I know that the integral
$$\int_{-\pi}^{\pi} \frac{ d\theta}{w - sin \theta} = \frac{2\pi}{\sqrt{w^2-1}}$$
where w, is an arbitrary constant and at some point you must do the substitution $$u = tan( \frac{\theta}{2} )$$ Does any one know how to do the intermediate steps?
| This is a partial answer, assuming that $w>1$:
Let $u=\tan\left(\frac{\theta}{2}\right)$, so $\theta=2\arctan u$, $d\theta=\frac{2}{1+u^2}du$, $\sin\theta=\frac{2u}{1+u^2}$.
Then $\displaystyle\int\frac{d\theta}{w-\sin\theta}=\int\frac{1}{w-\frac{2u}{1+u^2}}\cdot\frac{2}{1+u^2}du=2\int\frac{1}{wu^2-2u+w}du$
$=\displaystyle\frac{2}{w}\int\frac{1}{u^2-\frac{2}{w}u+1}du=\frac{2}{w}\int\frac{1}{(u-\frac{1}{w})^2+(1-\frac{1}{w^2})}du$
$=\displaystyle\frac{2}{w}\left(\frac{1}{\frac{\sqrt{w^2-1}}{w}}\arctan\frac{u-\frac{1}{w}}{\frac{\sqrt{w^2-1}}{w}}\right)+C=\frac{2}{\sqrt{w^2-1}}\left(\arctan\frac{wu-1}{\sqrt{w^2-1}}\right)+C$.
Now since $u\rightarrow\pm\infty$ as $\theta\rightarrow\pm\pi$, we get
$\displaystyle\int_{-\pi}^{\pi}\frac{d\theta}{w-\sin\theta}=\frac{2}{\sqrt{w^2-1}}\left(\frac{\pi}{2}-(-\frac{\pi}{2})\right)=\frac{2\pi}{\sqrt{w^2-1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Under what conditions will one solution of $ax^2+bx+c = 0$ be the reciprocal of the other? Under what conditions will one solution of
$ax^2+bx+c = 0$ be the reciprocal of the other?
| Here's one way using the quadratic formula:
The roots of $ax^2+bx+c=0$ can be written as $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Let $x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ and let $x_2$ be the reciprocal of this, so $$x_2 = \frac{2a}{-b+\sqrt{b^2-4ac}}.$$
However, since $x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}$, we can set these equal:
$$\frac{2a}{-b+\sqrt{b^2-4ac}} = \frac{-b-\sqrt{b^2-4ac}}{2a}.$$
Multiplying through, we get
$$(2a)^2 = (-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac}) \\
4a^2 = b^2-b^2+4ac \\
4a^2 = 4ac \\
a = c.
$$
Here's another way:
If $x$ solves $ax^2+bx+c=0$, then it also solves $x^2+\frac{b}{a}x+\frac{c}{a} = 0$, which we obtain by just dividing through by $a$.
Let the roots of this polynomial be $x_1$ and $\frac{1}{x_1}$. Writing the polynomial in terms of factors, we get
$$(x-x_1)(x-\frac{1}{x_1}) = x^2+\frac{b}{a}x+\frac{c}{a}.$$
Expanding on the left hand side, this becomes
$$x^2-\left(x_1+\frac{1}{x_1}\right)x+1 - x^2+\frac{b}{a}+\frac{c}{a}.$$
Therefore, we must have $\frac{c}{a} = 1$, implying $c = a$. We do not need to care about $b$.
| {
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Tough trigonometric identity: $\cot 13^\circ\cot 23^\circ \tan 31^\circ\tan35^\circ\tan41^\circ = \tan 75^\circ$ Prove that $$\cot 13^o\cot 23^o \tan 31^o\tan35^o\tan41^o = \tan 75^o$$
I managed to rearrange it to the form $$\tan 31^o\tan 35^o\cot 49^o = \cot 15^o\tan 23^o\cot 77^o$$
and in this form we have the interesting property that the sum of arguments on both sides is equal, i.e. 31+35+49=15+23+77. I couldn't get past this stage, so I would appreciate any help.
EDIT: I found that $\tan x\tan(60^o-x)\tan (60^o+x)=\tan 3x$. Perhaps someone can use this to solve the problem. I myself haven't been able to. Thanks.
| First convert all ratios of the left hand side in tangent
Observe that $\tan(5\cdot77^\circ)=\cdots=\tan25^\circ$ etc.
So, let $\tan5x=\tan25^\circ\implies5x=180^\circ n+25^\circ\iff x=36^\circ n+5^\circ$ where $n$ is any integer
Like Sum of tangent functions where arguments are in specific arithmetic series or this
$$\tan5x=\frac{\binom51\tan x-\binom53\tan^3x+\binom55\tan^5x}{\binom50-\binom52\tan^2x+\binom54\tan^4x}$$
If $\tan5x=\tan25^\circ,$
$$\tan^5x-\cdots-\binom50\tan25^\circ=0$$
$$\prod_{r=0}^4\tan\left(36^\circ\cdot r+5^\circ\right)=\frac{\tan25^\circ}1$$
$r=0\implies\tan\left(36^\circ\cdot0+5^\circ\right)=\tan5^\circ$
$r=1\implies\tan\left(36^\circ\cdot1+5^\circ\right)=\tan41^\circ$
$r=2\implies\tan\left(36^\circ\cdot2+5^\circ\right)=\tan77^\circ=\cot13^\circ$
$r=3\implies\tan\left(36^\circ\cdot3+5^\circ\right)=\tan113^\circ=-\tan67^\circ=-\cot23^\circ$
$r=4\implies\tan\left(36^\circ\cdot4+5^\circ\right)=\tan149^\circ=-\tan31^\circ$
So, we need to show $$\tan35^\circ\frac{\tan25^\circ}{\tan5^\circ}=\tan75^\circ$$
which is readily available from your formula putting $x=25^\circ$ mentioned here (How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$)
| {
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Find the sum to n terms of the series Find the sum to n terms of the series
$$\frac {\sin x}{\cos x+\cos2x} + \frac {\sin2x}{\cos x+\cos4x} +
\frac {\sin3x}{\cos x + \cos6x} +\dotsb $$
How can I solve this?
Here is what I did for the first term:
$$\frac {\sin\bigl(\frac{3x}{2} - \frac x2\bigr)}
{2\cos\bigl(\frac{x}{2}\bigr)\cos\bigl(\frac {3x}{2}\bigr)}$$
After opening $N^r$ by $\sin(A-B) = \sin A\cos B - \cos A\sin B$ we get
$$ \frac 12\left(\tan\frac {3x}{2} - \tan\frac {x}{2}\right)$$
But this doesn't work out for the remaining terms. Am I missing something here or is this method completely wrong? Any hints? Help please!
edit : Please give a hint. I want to do it on my own :)
| So I figured out the answer :
$ S = \frac {sinx}{cosx+cos2x} + \frac {sin2x}{cosx+cos4x}+ \frac {sin3x}{cosx+cos6x} + \quad ...$
Multiply both sides by $2sin\frac x2$
Take the first term :
$$\frac {2sin\frac x2sinx}{cosx+cos2x}$$
$$\frac {cos\frac{x}2-cos\frac {3x}2} {2cos\frac {3x}2cos\frac x2}$$
$$\frac 12(sec \frac {3x}2- sec \frac{x}2)$$
Now by same method we extend this to $n\ terms$ and add them
Finally we get,
$$2sin\frac x2S = \frac 12(sec (2n+1)\frac x2- sec \frac x2)$$
$$\therefore \quad S = \frac 14cosec\frac x2(sec(2n+1)\frac x2 - sec\frac x2)$$
This is the answer.
| {
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Expressing $1 + \cos(x) + \cos(2x) +... + \cos(nx)$ as a sum of two terms Question in title, my progress:
let $z = \cos(x) + i\sin(x)$
then $1 + \cos(x) + \cos(2x) +\dots + \cos(nx) = Re(1 + z + z^2 +\dots + z^n) = Re\left (\dfrac{1-z^{n+1}}{1-z} \right)$
by geometric series;
multiplying $\dfrac{1-z^{n+1}}{1-z}$ by $\overline{1-z}$ we get $1 + \cos(x) + \cos(2x) +\dots + \cos(nx) = Re \left ( \dfrac{(1-z^{n+1})(\overline{1-z})}{|1-z|^2} \right )$
but I am not sure how to proceed from here.
edit: this is for a complex analysis course, so i'd appreciate a hint using complex analysis without using the exponential function
| Use $\sin(a) \cos(b) = \frac{1}{2} \sin(b+a) - \frac{1}{2} \sin(b-a)$ and multiply your sum with $\sin\left(x/2\right)$.
$$
\sum_{m=0}^{n} \sin\left(\frac{x}{2}\right) \cos(m x) = \frac{1}{2} \sum_{m=0}^n \left\{\sin \left(\left(m+1-\frac{1}{2}\right)x \right) - \sin\left( \left(m-\frac{1}{2}\right)x \right) \right\}
$$
The sum telescopes, i.e. $\sum_{m=0}^n \left(f(m+1)-f(m)\right) = f(n+1)-f(0)$, hence
$$
\sum_{m=0}^{n} \sin\left(\frac{x}{2}\right) \cos(m x) = \sin \left(\left(n+\frac{1}{2}\right)x \right) - \sin\left(\frac{x}{2} \right)
$$
now divide by $\sin\left(\frac{x}{2}\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
*How do I proceed?
| Using $e^{i x} = \cos(x) + i \sin(x)$ then for $z_{1} = e^{2 \pi i/3}$ it is seen that
\begin{align}
z_{1}^{10} + \frac{1}{z_{1}^{10}} &= e^{20 \pi i/3} + e^{- 20 \pi i/3} = e^{6 \pi i + 2 \pi i/3} + e^{-6 \pi i - 2\pi i/3} \\
&= e^{2\pi i/3} + e^{-2\pi i/3} = 2 \cos(2\pi /3) = -1
\end{align}
| {
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How to solve this differential equation: $x^2dy-y^2dx+xy^2(x-y)dy=0$
$$x^2dy-y^2dx+xy^2(x-y)dy=0$$
What I tried:
$$\frac{x^2}{y^2} \frac{dy}{dx}+x(x-y)\frac{dy}{dx}=1\\$$
Let $h=-1/x, \; k=-1/y,\; dh=1/x^2 \, dx, \; dk=1/y^2 \,dy$
$$\frac{dk}{dh}+\frac{(k-h)}{k^2} \frac{dk}{dh}=1\\
\frac{dk}{dh}+\frac hk=1+\frac1k\\
he^{\int-1/k^2\; dk}=\int\left(1+\frac1k\right)e^{\int-1/k^2\; dk}dk\\
he^{-y}=\int\frac{1-y}{y^2}e^{-y}dy=\int\left(\frac 1{y^2}-\frac1y\right)e^{-y}dy$$
Which probably is unsolvable?I tried using IBP on RHS.Dont use Ricatti Eqn(Not in my course)
Answer is:
$$\large\ln\left|\frac{x-y}{xy}\right|=\frac{y^2}2+\mathcal C$$
| $\left(\dfrac{x^2}{y^2} +x(x-y)\right)=\dfrac{dx}{dy}$
$x=vy\implies \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$
$\therefore \left(\dfrac{x^2}{y^2} +x(x-y)\right)=\dfrac{dx}{dy} \\\implies v+y\dfrac{dv}{dy}=v^2+v^2y^2-vy^2 \\ \implies y\dfrac{dv}{dy}=v^2+v^2y^2-vy^2-v\\ \implies y\dfrac{dv}{dy}=(1+y^2)v(v-1) \\\implies \displaystyle \int\dfrac{dv}{v(v-1)}=\displaystyle \int\left(y+\dfrac{1}{y}\right)dy \\ \implies \displaystyle \int\dfrac{v-(v-1)}{v(v-1)} dv=\displaystyle \int\left(y+\dfrac{1}{y}\right)dy\\\implies \displaystyle\int\dfrac{dv}{v-1}-\displaystyle\int\dfrac{dv}{v}=\displaystyle\int y\ dy + \displaystyle\int\dfrac{dy}{y} \\\implies \ln \left(1-\dfrac{1}{v}\right)=\dfrac{y^2}{2}+\ln y+ c \\\implies \ln \left(\dfrac{x-y}{xy}\right)= \dfrac{y^2}{2}+c$
| {
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Find the last two digit of $3^{3^{100}}$. Find the last two digit of $3^{3^{100}}$.
I know how to calculate if I have $3^{100}$. That I will use euler's theorem. which gives me $3^{40}\equiv 1 \pmod{100}$. And so on... but if I have $3^{3^{100}}$ what should I do?
I Tried:
Infact, i need $3^{3^{100}}\equiv x\pmod{ 100}$
For this, I need $3^{100}\equiv y\pmod{\phi{(100)}}$
So i got $y=1$ by using eulers thorem for the abouve cogruence.
That is $(3,40)=1\implies 3^{16}\equiv 1\pmod{40}$
So, i got $(3^{16})^63^4\equiv 1\pmod{40}$
Using this in first congruence i got $3^1\equiv 3\pmod{100}$.
So, the answer is $03$. Is it correct?
I did't use chinese remainder theorem and all. Is there any mistake in my arguments?
| If we can find $3^{3^{100}}$ in mods 4 and 25 we're done by the Chinese Remainder Theorem.
Mod 4: Note that $3\equiv -1\pmod 4$, hence $3^{3^{100}}\equiv (-1)^{3^{100}}\equiv (-1)^{2k+1}\equiv -1\equiv 3\pmod 4$.
Mod 25: Since $\varphi(25)=20$ we have $3^{3^{100}}\equiv 3^{3^{100}\pmod {20}}\pmod {25}$. Similarly, $3^{100}\equiv 3^{100\pmod 8}\pmod {20}$, and $100\equiv 4\pmod 8$. Hence, working backwards we get:
$3^{100}\equiv 3^{100\pmod 8}\equiv 3^4\equiv 1\pmod {20}$,
$3^{3^{100}}\equiv 3^{3^{100}\pmod {20}}\equiv 3^1\equiv 3\pmod {25}$.
Thus, threading the congruences together we have:
$3^{3^{100}}=4a+3=25b+3$, or $4a=25b$. So since $\gcd(4,25)=1$ we need $a=25k$. Therefore $3^{3^{100}}=4(25k)+3=100k+3$, so the answer is $\boxed{03}$.
| {
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$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty}\frac{\frac{\sqrt{6x^2 - 2}}{x}}{\frac{x+2}{x}}
=
\lim\limits_{x\to-\infty}\frac{{\sqrt{\frac{6x^2 - 2}{x^2}}}}{\frac{x+2}{x}}=\sqrt{6}
$$
My answer is $\sqrt{6}$, is my working wrong or there are actually another approach? Thank you.
| If $x\lt 0$, then
$$\frac{\sqrt{6x^2-2}}{x}\not=\sqrt{\frac{6x^2-2}{x^2}}$$
because the left-hand-side is negative and the right-hand-side is positive.
Since $-x\gt 0\Rightarrow -x=\sqrt{(-x)^2}=\sqrt{x^2}$, we have
$$\frac{\sqrt{6x^2-2}}{x}=\frac{\sqrt{6x^2-2}}{-(-x)}=\frac{\sqrt{6x^2-2}}{-\sqrt{(-x)^2}}=-\sqrt{\frac{6x^2-2}{(-x)^2}}=-\sqrt{\frac{6x^2-2}{x^2}}.$$
| {
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Are there any positive integers $a, b, c, d$ such that both $(a, b, c)$ and $(b, c, d)$ are Pythagorean triples? Pythagorean triple is a triple of integers $(a, b, c)$ such that $a^2+b^2=c^2$. Is there any Pythagorean triple such that, not only $a^2+b^2$, but also $b^2+c^2$ is a square number? If not, how to prove it?
I tried to prove non-existence the following way: If true, it would mean that there is a pair of integers such that both sum and difference of their squares is a square number. Let's call these integers $a$ and $b$ and $a<b$. Then, there are integers $c$ and $d$ such that:
\begin{align}
&a^2+b^2=c^2 \\
&a^2-b^2=d^2
\end{align}
Multiplying those equations gives:
\begin{equation}
a^4=(cd)^2+b^4
\end{equation}
This is similar to Fermat's Last Theorem for $n=4$, but using it only shows that $cd$ can't be square number, not that there are no integer solutions.
| For the $(a,b,c)$ and $(b,c,d)$ Pythagorean triples to work, we note:
*
*$a,b,c$ is primitive (otherwise reduce by the gcd)
*likewise $b,c,d$ is primitive
*Using the traditional solution for a triple:
$2mn$ ; $m^2-n^2$ ; $m^2+n^2$, then $c$ is odd
*if $c$ is odd in the $(b,c,d)$ triple, then $b$ must be even = $2mn$
*Then we have ($2mn$, $m^2+n^2$, $d$ ) for the second triple
*Then $(2mn)^2$ + $(m^2+n^2)^2$ = $m^4 + 6m^2n^2 + n^4 = d^2$ which is a $\square$
But H.C. Pocklington of St John's College in 1913 proved that
6 for the coefficient of $x^2y^2$ in the general equation $x^4 + dx^2y^2+y^4 = z^2$ is impossible by prime moduli, in this case 6 != 7 mod 8
See H. C. Pocklington. "Some diophantine impossibilities" Proc. Camb. Phil. Soc, 17: pf 110 – 118, 1914.
| {
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Diophantine equation $(x^2-1)^2-4y^2=0$ We have the diophantine equation $(x^2-1)^2-4y^2=0$, where $x$ and $y$ are positive integers.
Is the only solution $x=1$, $y=0$ or can there be infinitely many solutions?
| Note that $$\begin{align}(x^2-1)^2-4y^2=0&\iff (x^2-1-2y)(x^2-1+2y)=0\\&\iff x^2-1-2y=0\ \ \text{or}\ \ x^2-1+2y=0.\end{align}$$
So, considering $x^2-1-2y=0\iff x^2-1=2y$ gives us, for example,
$$(x,y)=(2a+1,2a^2+2a)$$
for positive integers $a$.
Hence, we know that there are infinitely many solutions.
| {
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Proof by Induction that $3^n ≥ 1+2^n$ Use the PMI to prove the following for all natural numbers:
$3^n ≥ 1+2^n$.
I have already verified the base case but am having trouble doing so with the inductive case. Thanks!
| Suppose the result holds for the case $n$ : $3^n \geq 1 + 2^n$. Now, to complete the induction proof, we must show $ 3^{n+1} \geq 1 + 2^{n+1} $.
$$ 3^{n+1} = 3^n \cdot 3 \geq(1 + 2^n)3 = 3 + 2^n \cdot 3 = 1 + 2 + 2^n \cdot 3 = 1 + 2 + (1+2)2^n = 1 + 2 + 2^{n+1} + 2 > 1 + 2^{n+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I solve this recurrence relation? Given a recursive relation
$$a_n = \begin{cases}
(1 - 2b_n)a_{n-1} + b_n, & n > 1 \\
\frac{1}{2}, & n =1
\end{cases}
$$, how can I expression $a_n$ in term of $b_i, i \in \{1, 2, \dots n\}$?
| Suppose that $a_n=\frac{1}{2}$ for some $n$. Then according to the recurrence:
\begin{align*}
a_{n+1}&=(1-2b_{n+1})a_n+b_{n+1} \\
&= \frac{1}{2}-b_{n+1}+b_{n+1} \\
&= \frac{1}{2}
\end{align*}
Since $a_1=\frac{1}{2}$, by induction this shows that $a_n=\frac{1}{2}$ for all $n$.
| {
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How to solve $10^{x^2+x}+\log{x} = 10^{x+1}$? In one of my recent exam, I was ask to solve this:
$$
10^{x^2+x}+\log{x} = 10^{x+1}
$$
My attempt to solve it was:
$$
10^{x^2+x}+\log{x} = 10^{x+1} \\
\log{x}=10^{x+1}-10^{x^2+x} \\
\log{x} = 10^{x+1}(1-10^x) \\
\log(\log{x})=(x+1)+\log(1-10^x) \\
$$
At this point I got stuck because I don't know how to solve an equation with double logs.
| In your manipulation of the equation you factored the powers of $10$ incorrectly. Correction:
$$10^{x^2+x} + \log(x) = 10^{x+1}\\\log(x) = 10^{x+1}(1-10^{x^2 - 1})$$
We know that $x > 0$ for $\log(x)$ to be defined.
Case ($0 < x < 1$): $\log(x) < 0 < 10^{x+1}(1-10^{x^2-1})$
Case ($x = 1$): $\log(1) = 0 = 10^2(1-10^0)$
Case ($x > 1$): $\log(x) > 0 > 10^{x+1}(1-10^{x^2-1})$
Thus the only solution is $\boxed{x = 1}$
| {
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Minimum volume cone. What would be the radius and the altitude of a right circular cone that circumscribes a sphere with a radius 8 cm if the volume of the cone is to be minimized?
Here is my rough sketch;
My idea is to write some characteristics of the cone as a function of the radius of the circle, minimize it with differential calculus and connect those characteristics to the base radius and height of the cone, however I'm stuck at step 1.
| Let $y=AC$ and $r=BD$, so $h=y+8$.
By similar triangles, $\displaystyle\frac{r}{y+8}=\frac{8}{\sqrt{y^2-64}}$ $\;\;$so $\displaystyle r^2=\frac{64(y+8)^2}{y^2-64}=\frac{64(y+8)}{y-8}$ and
$V=\displaystyle\frac{1}{3}\pi r^2h=\frac{64\pi}{3}\cdot\frac{(y+8)^2}{y-8}$.
Then $\displaystyle V^{\prime}(y)=\frac{64\pi}{3}\cdot\frac{(y+8)(y-24)}{(y-8)^2}=0$ if $y=24$,
sor $r=\sqrt{\frac{64(32)}{16}}=\sqrt{64(2)}=8\sqrt{2}$ $\;\;$ and $\;\;h=24+8=32$.
(By the first-derivative test, $y=24$ gives a minimum.)
Alternate Solution:
Let $\theta=\angle ABD$, so $r=8\cot\frac{\theta}{2}=8(\csc\theta+\cot\theta)$ and $h=r\tan\theta=8(\csc\theta+\cot\theta)\tan\theta$.
Then $V=\frac{1}{3}\pi r^{2}h=\frac{512\pi}{3}(\csc\theta+\cot\theta)^3\tan\theta$,
so $V^{\prime}(\theta)=\frac{512\pi}{3}\left[(\csc\theta+\cot\theta)^{3}\sec^{2}\theta+3(\csc\theta+\cot\theta)^{2}(-\csc\theta\cot\theta-\csc^{2}\theta)\tan\theta\right]$
$\hspace{.6 in}=\frac{512\pi}{3}(\csc\theta+\cot\theta)^2\left[(\csc\theta+\cot\theta)\sec^{2}\theta-3(\csc\theta\cot\theta+\csc^{2}\theta)\tan\theta\right]=0$
$\implies(\csc\theta+\cot\theta)\sec^{2}\theta=3(\csc\theta\cot\theta+\csc^{2}\theta)\tan\theta$
$\implies\displaystyle\frac{1+\cos\theta}{\sin\theta\cos^{2}\theta}=\frac{3(1+\cos\theta)}{\cos\theta\sin\theta}\implies\cos\theta=\frac{1}{3}\implies\sin\theta=\frac{\sqrt{8}}{3}$,
so $\csc\theta+\cot\theta=\frac{1+\cos\theta}{\sin\theta}=\sqrt{2}\;\;$ and $\;\;r=8\sqrt{2},\;\;$ $h=r\tan\theta=(8\sqrt{2})\cdot\sqrt{8}=32$.
| {
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Proving by induction $5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$ is divisible by $4$ I want to prove the following twice. Once by induction then again by any other method.
$$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$$ is a multiple of 4 for all nonnegative integers n.
Let n=0 , since it is the first nonnegative integer
$$5^{3(0)}+2*5^{2(0)}-5^{0}-2 = 0 $$
Factoring gives us $(5-1)(5+1)(5+2)$
| $$5^{3n} + 2 \cdot5^{2n} - 5^{n} - 2=5^n\cdot5^{2n}+2 \cdot5^{2n} - 5^{n} - 2=$$
$$5^{2n}( 5^{n} +2)-( 5^{n}+2)=( 5^{n} + 2)(5^{2n} - 1)=$$
$$=( 5^{n} + 2)(5^{n} + 1)(5^{n} -1)$$
| {
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Find the largest value of x given the equation... Find the largest value of $x$ for which $x^2 + y^2 + z^2 = x + y + z$.
What I did was subtract the RHS, to get $$x^2 - x + y^2 - y + z^2 - z = 0$$
$$x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} + z^2 - z + \frac{1}{4} = \frac{3}{4}$$
$$(x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 + (z-\frac{1}{2})^2 = \frac{3}{4}$$
Hence the given equation can be rewritten as the formula for a sphere centered at $(x,y,z) = (\frac{1}{2},\frac{1}{2},\frac{1}{2})$ with a radius of $\frac{\sqrt3}{2}$.
Now for $x$ to be at its largest value, that would mean $y = z = \frac{1}{2}$ (right?).
And if that is true, then $\boxed{x = \frac{\sqrt{3}+1}{2}}$.
Please let me know if I am correct, and if not, please help me understand how to achieve the correct answer. Thanks!
Edit: Thanks everyone for your responses!
| Yes, you are correct. The point where $(y - a)^2$ and $(z - b)^2$ are zero is where we would find the maximum $x$. Anything greater than zero would make the $x$ part of the equation smaller.
Nice thinking!
| {
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Prove that $\sum\limits_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$ This sum $\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} $showed up as I was computing the expected value of a random variable.
My calculator tells me that $\,\,\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$.
How can I show that?
I know how to find the value of the sum $\,\displaystyle \sum_{n=1}^\infty \frac{n^2}{2^n},\,$ but I can't deal with $\displaystyle \sum_{n=1}^\infty \frac{n^3}{2^n}$
| As hinted by @AlexandreHalm,
$$\sum_{k=0}^\infty x^k=\frac1{1-x}.$$
Then deriving on $x$ and multiplying by $x$,
$$\sum_{k=0}^\infty kx^k=\frac x{(1-x)^2},$$
$$\sum_{k=0}^\infty k^2x^k=\frac{2x^2}{(1-x)^3}+\frac x{(1-x)^2},$$
$$\sum_{k=0}^\infty k^3x^k=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac x{(1-x)^2}.$$
Evaluating at $x=\frac12$,
$$\sum_{k=0}^\infty\frac{k^3}{2^k}-\sum_{k=0}^\infty\frac{k^2}{2^k}=26-6=20.$$
| {
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Evaluate a linear system of three equations Solve for $x, y\ \text{and}\ z\ $:
$x-3z=10\\
-x+y+2z=7\\
2x+2y-5z=-8$
My working:
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
-1 & 1 & 2 & 7 \\
2 & 2 & -5 & -8
\end{array}\right) =
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 3 & 0 & -11
\end{array}\right) \begin{array}{l}
\\
R_1 + R_2 \\
R_3 + R_2 - R_1
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 0 & 3 & -62
\end{array}\right) \begin{array}{l}
\\
\\
R_3 - 3R_2
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -52 \\
0 & 1 & 0 & -7 \\
0 & 0 & 1 & -24
\end{array}\right) \begin{array}{l}
R_1 + R_3 \\
R_2 + R_3/3 \\
R_3/3
\end{array}$$
Yet plugging these solutions into the original equation does not work:
$-52+0-3 \times (-24)=20 \ne 10$ and
$-(-52)-7+2\times(-24)=-3\ne7$
What did I do wrong?
| The last line should read
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 0 & 3 & -62
\end{array}\right) \begin{array}{l}
\\
\\
R_3 - 3R_2
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -52 \\
0 & 1 & 0 & -11/3 \\
0 & 0 & 1 & -62/3
\end{array}\right) \begin{array}{l}
R_1 + R_3 \\
R_2 + R_3/3 \\
R_3/3
\end{array}$$
| {
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Does the series $\sum_{n=1}^\infty \frac{n+1}{n^3+10n}$ converge? Using the ratio test, we evaluate:
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty}\left| \frac{(n+1) + 1}{(n+1)^3 + 10(n+1)} \cdot \frac{n^3+10n}{n+1} \right| = \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{(n+1)^4 + 10(n+1)^2} \right|$$
$$ < \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{n^4 + 10n^2} \right| = 1$$
Hence $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ and the series converges.
Is this an appropriate solution?
| $$
\sum_{n=1}^\infty \frac{n+1}{n^3+10n} \leq \sum_{n=1}^\infty \frac{n+n}{n^3} = 2 \sum_{n=1}^\infty \frac{1}{n^2} < \infty.
$$
| {
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Differentiate the following function: $y = \frac{2x^2 + 6\sqrt{x} }{9x}$ My answer is $- \dfrac{1}{9 \sqrt{x} }$, however, the program I am using states that I am wrong. Where have I went wrong?
| $y = \frac{2x^{2}}{9x} + \frac{6x^\frac{1}{2}}{9x}=\frac{2x}{9} + \frac{2}{3}x^{-\frac{1}{2}}$.
Then $\frac{dy}{dx}=\frac{2}{9} -\frac{2}{3}⋅-\frac{1}{2}x^{-\frac{3}{2}}$
| {
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Probability of a random triangle containing the center of a polygon Consider a regular polygon of $2n+1$ sides. Let three random vertices be chosen at random to get a triangle. The probability that the chosen triangle contains the center of the polygon is 5/14.
What is n?
I'm confused as to how to proceed on the cases that we will get. They do not form any definite pattern.
| The probability seems to work out for a regular $9$-gon (so $n=4$).
Consider a circle with $9$ equally spaced points marked. Let's label the points $0, 1, 2, \dots, 8$ going clockwise around the circle. These points form our regular $9$-gon. When we choose $3$ of these vertices and form a triangle, the triangle contains the center if and only if the triangle is acute. This in turn occurs if and only if as you go from one chosen vertex to the next you never skip more than $3$ vertices.
By symmetry, we can freely choose vertex $0$. And then we choose two other vertices $a$ and $b$ with $a<b$. We need the spacing between these consecutive vertices to be at most $4$. This means that $a$ must be $1$, $2$, $3$, $4$.
For $a=1$ we must have $b=5$ to contain the center.
For $a=2$ we must have $b=5$ or $6$ to contain the center.
For $a=3$ we must have $b=5$ or $6$ or $7$.
For $a=4$ we must have $b=5$ or $6$ or $7$ or $8$.
This gives us a total of $10$ good triangles out of ${8\choose 2}=28$ triangles that use vertex $0$.
So the probability of containing the center is $\frac{10}{28}=\frac{5}{14}$.
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"url": "https://math.stackexchange.com/questions/975432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$ I need some starting ideas, hints for proving that
$$2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$$
I already checked that with Mathematica that numerically says that
$$\int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \approx 2.577632915067858 $$
| This can be done with 4 observations:
$$
\lim_{x\rightarrow 0^+} \frac{(1+x)^{1+x}}{x^x} = 1
$$
$\frac{(1+x)^{1+x}}{x^x}$ has value $4$ at $x=1$.
$\frac{(1+x)^{1+x}}{x^x}$ has value $\frac{\sqrt{27}}{2}$ at $x=\frac{1}{2}$.
At $x=1$, $\frac{d^2}{dx^2} \frac{(1+x)^{1+x}}{x^x} = 4 \ln 2$.
And $\frac{d^2}{dx^2} \frac{(1+x)^{1+x}}{x^x} < 0 $ on $(0,1)$, that is, the function is concave downward in the relevant interval.
Consider the trapezoid with corners at the $(0,0), (0,1), (1,4), (1,0)$ which has area $2.5$: Because $f(x) = \frac{(1+x)^{1+x}}{x^x}$ is concave downward, the line between $(0,1)$ and $(1,4)$ always lies on or above $y=f(x)$. This shows that the integral
$$
I > 2.5
$$
Now consider the trapezoid with corners at the $(0,0), (0,4-4\ln 2), (1,4), (1,0)$
which has area $4 - 2 \ln 2$. Because $f(x) = \frac{(1+x)^{1+x}}{x^x}$ is concave downward, the line between $(0,4-2\ln 2)$ and $(1,4)$ is always on or above $y=f(x)$. This shows that the integral
$$
I < 4 - 2 \ln 2
$$
Since $4 - 2 \ln 2 < 2.62$ this shows that the integral is between 2.50 and 2.62.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Proof of: if $x^2+y^2=2xy$ then $x=y$ I am trying to prove $x^2+y^2=2xy$ then $x=y$
What I have done is suppose $x^2+y^2=2xy$
then $x^2+y^2+(-2xy)=0\iff
x^2+(-xy)+(-xy)+y^2=0
\iff (x+(-y))\cdot x+(x+(-y))\cdot-y=0
\iff (x+(-y))^2=0$
i then square root both sides but i'm not sure if that's mathematically correct?
which gives me: $(x+(-y))=0$
so $x=y$
| The claim as written is incorrect:
$$
y=\sqrt{2},x=2+\sqrt{2}\implies x^2-y^2=4+4\sqrt{2}+2-2=2\times\sqrt{2}(2+\sqrt{2})=2xy.
$$
Perhaps the problem was meant to be $x^2+y^2=2xy$ implies $x=y$. In this case
$$
0=x^2+y^2-2xy=(x-y)^2\implies x=y.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $2^{10}+5^{12}$ is composite
Prove that $2^{10}+5^{12}$ is composite
I need to solve this using only high school mathematics.
Any ideas?
| $2^{10}+5^{12}=(2^5)^2+(5^6)^2$ $=(2^5+5^6)^2 - 2\cdot2^5\cdot5^6$ $=(2^5+5^6)^2 - (2\cdot5)^6$ $=(2^5+5^6)^2 - (10^3)^2$ $=(2^5+5^6 - 10^3)(2^5+5^6 + 10^3)$ $=(15657 - 1000)(15657 + 1000)$ $=14657\times 16657$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Closed-form of $\int_0^1 B_n(x)\psi(x+1)\,dx$ Is there a closed-form of the following integral?
$$I_n = \int_0^1 B_n(x)\psi(x+1)\,dx,$$
where $B_n(x)$ are the Bernoulli polynomials and $\psi(x)$ is the digamma function.
The motivation of the problem was this question. In this answer achille hui showed that for all $n>0$
$$\int_0^1 B_n\left(\left\{ \frac1x \right\}\right) \frac{dx}{x}
= -\int_0^1 B_n(u) \psi(1+u) du,$$
where $\{x\}$ is the fractional part of $x$.
By looking at specific $n$ values we could observe interesting patterns.
$$\begin{align}
I_0 & = 0\\
I_1 & = 1-\frac 12 \ln(2\pi) \\
I_2 & = -\frac 12 + 2 \ln A \\
I_3 & = \frac{1}{12} - \frac{3}{4\pi^2}\zeta(3) = \frac{1}{12} - \frac{1}{8}\frac{\zeta(3)}{\zeta(2)}\\
I_4 & = \frac{1}{45} - 4\zeta'(-3)\\
I_5 & = -\frac{13}{360} + \frac{15}{4\pi^4}\zeta(5) = -\frac{13}{360} + \frac{1}{24}\frac{\zeta(5)}{\zeta(4)}\\
I_6 & = -\frac{1}{252}-6\zeta'(-5)\\
I_7 & = \frac{47}{1260} - \frac{315}{8\pi^6}\zeta(7) = \frac{47}{1260} - \frac{1}{24}\frac{\zeta(7)}{\zeta(6)}\\
I_8 & = -\frac{8}{1575}-8\zeta'(-7)\\
I_9 & = -\frac{1703}{25200} -\frac{2835}{4\pi^8}\zeta(9) = -\frac{1703}{25200} -\frac{3}{40}\frac{\zeta(9)}{\zeta(8)}\\
I_{10} & = \frac{2461}{83160} - 10\zeta'(-9)\\
\dots
\end{align}$$
where $A$ is the Glaisher–Kinkelin constant, $\zeta(3)$ is the Apéry's constant and in general $\zeta$ is the Riemann zeta function and $\zeta'$ is its derivative.
| The complete answer for the question with $\,n\ge 1\,$ is the union of the answers of Start wearing purple and Olivier Oloa (the sign of the right side of the equation $(*)$ is wrong there), we get:
$$\int\limits_0^1 B_n(x)\psi(x+1)dx = (-1)^{n-1} \left(n \,\zeta’(1-n)- B_n H_{n-1}\right) + \sum\limits_{k=1}^n {\binom n k} \frac{B_{n-k}}{k}$$
A well-functioning trick is to invert $(1)$ of Start wearing purple’s answer by using:
$$b_{m-1}=\sum\limits_{v=0}^{m-1} {\binom m v} a_v \enspace \Leftrightarrow \enspace a_m=\frac{1}{m+1}\sum\limits_{v=0}^m {\binom {m+1} {v+1}} B_{m-v}b_v $$
Note 1:
$$\sum\limits_{k=1}^n {\binom n k} \frac{B_{n-k}}{k}=\int\limits_0^1\frac{B_n(x)-B_n(0)}{x}dx$$
Note 2:
$$n\zeta’(1-n)- B_n H_{n-1}=\frac{d}{dt}B_t(1)|_{t=n}-H_n B_n(1)$$
$\hspace{4cm}$ Informations about $\,B_t(x)\,$ can be found here .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$
*
*The below problem is taken from Joseph Edwards book Integral Calculus for beginners.
How does one show:
$$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\} $$
*
*Splitting $\sin{(5x)}$ as $\sin{(4x+x)}$ doesn't seem to be of much help since then we have a big term in the denominator after expansion.
| Note
\begin{align}
&\frac{\sin x}{\sin5x}=\frac1{4\left(\cos2x- \cos\frac{2\pi}{5}\right) \left(\cos2x- \cos \frac{4\pi}{5}\right)}\\=&\ \frac1{2\sqrt5}\bigg(\frac1{\cos2x- \cos\frac{2\pi}{5} }
-\frac1{\cos2x- \cos\frac{4\pi}{5} }
\bigg)\\
=&\ \frac1{4\sqrt5}\bigg(\frac1{\sin(\frac\pi5+x)\sin(\frac\pi5-x) }-\frac1{\sin(\frac{2\pi}5+x)\sin(\frac{2\pi}5-x) }
\bigg)\\
=&\ \frac15\sin\frac\pi5\left(\cot(\frac\pi5+x)+\cot(\frac\pi5-x) \right)-\frac15\sin\frac{2\pi}5\left(\cot(\frac{2\pi}5+x)+\cot(\frac{2\pi}5-x) \right)
\end{align}
Then, apply $(\ln \sin t)’= \cot t$ to integrate
$$\int \frac{\sin x}{\sin5x}= \frac15\sin\frac{2\pi}5\>\ln \frac{\sin(\frac{2\pi}5-x) }{\sin(\frac{2\pi}5+x) } - \frac15\sin\frac\pi5\>\ln \frac{\sin(\frac\pi5-x) }{\sin(\frac\pi5+x) }+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $5$ divides it for any number $x$ and $y$ because it only has to divide the digits number which is $5$.
To see if it divides by $9$, the sum of the digits must be divisible by $9$, and as for $11$, the alternating sum of digits must be divisible by $11$.
So, to test it by $9$, then, $2+7+3+x+4+9+y+5 = 30+x+y$ has be divisble by $9$. So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are $x= \{0,1,2,3,4,5,6\}$ and $y=\{0,1,2,3,4,5,6\}$.
But $11$ is what gets me. To make it divisible by $11$, then the alternating sums must be divisible by $11$. So then, $-2+7-3+x-4+9-y+5= 12+x-y$ must be divided by $11$. Then, the only numbers $x$ and $y$ that make it divisible by $11$ must be $x=\{0,1,2,\ldots,8\}$ and $y=\{1,2,\ldots,9\}$, if I did my math right.
Next, I assume we must then find numbers in $x$ and $y$ that satisfy the division by $9$ and $11$. So, if that is the case, then I know that since the number must be divisible by $9$ and $11$, then the numbers $x$ and $y$ must fit between $0$ to $6$ because that's $6$ is the highest number $x$ or $y$ can be in the sequence to divide the number. So, how do I check to see which numbers both satisfy division of $9$ and $11$?
| You are wrong considering $x,y\in\left[0,6\right]$.
$30+x+y$ is divisible by $9$. That's correct, but it may be $30+x+y=36$ or $30+x+y=45$. It cannot be $30+x+y=54$, since the greatest value of $\left(x+y\right)$ is $9+9=18$.
From these two options you have $x+y=6$ or $x+y=15$. Remember this.
From the division by $11$, you achieved that $12+x-y$ must be divisible by $11$. That's also correct.
Note that you cannot get $12+x-y=22$, since the maximum value $\left(x-y\right)$ can reach is $9-0=9$, so you have to use negative values of $\left(x-y\right)$ like $12+x-y=11$, i.e, $x-y=-1$. It means $x$ and $y$ are two consecutive numbers where $y=x+1$; it also means that one of them is odd and the other is even, because of this, there is no way $x+y=6$, it must be $x+y=15$.
Now, you have a linear system
$$\begin{cases}x+y=15\\x-y=-1\end{cases}$$
Solve this and you will have your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Floor function inequality $\lfloor a\sqrt{2}\rfloor\lfloor b\sqrt{7}\rfloor <\lfloor ab\sqrt{14}\rfloor$ Let $a,b$ be positive integers. Show that $\lfloor a\sqrt{2}\rfloor\lfloor b\sqrt{7}\rfloor <\lfloor ab\sqrt{14}\rfloor$.
[Source: Russian competition problem]
| Let $a\sqrt{2} = m + \alpha$ where $m$ is an integer and $0 < \alpha < 1$, and $b\sqrt{7} = n + \beta$ where $n$ is an integer and $0 < \beta < 1$. (The fractional parts are positive since $\sqrt{2}$ and $\sqrt{7}$ are irrational.) We are to prove that
$$ (m+ \alpha)(n + \beta) - mn \geq 1.$$
We have
$$\alpha = a\sqrt{2} - m = \frac{2a^2 - m^2}{a\sqrt{2} + m} = \frac{2a^2 - m^2}{2m + \alpha} \geq \frac{1}{2m + \alpha},$$
since $2a^2 - m^2$ is necessarily a positive integer. From this it is easy to see that
$$m \geq \frac{1}{2}\left(\frac{1}{\alpha} - \alpha\right).$$
We similarly prove $\beta \geq 1/(2n + \beta)$, hence
$$n \geq \frac{1}{2}\left(\frac{1}{\beta} - \beta\right).$$
Now we have
$$
\begin{align*}
(m + \alpha)(n + \beta) - mn &= \alpha n + \beta m + \alpha \beta \\
&\geq \frac{\alpha}{2}\left(\frac{1}{\beta} - \beta\right) + \frac{\beta}{2}\left(\frac{1}{\alpha} - \alpha\right) + \alpha \beta \\
&=\frac{1}{2}\left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right)\\
&= 1 + \frac{(\alpha - \beta)^2}{2\alpha\beta}\\
&\geq 1,
\end{align*}
$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/986337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding the square roots of a complex number.
Express $z=4\sqrt2(1+i)$ in modulus/argument form. Hence find the two
square roots of $z$ and mark their representations on an Argand
Diagram.
So far I've worked out the mod/arg form of the complex number which is just
$$
z = 8(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) \\
$$
Then used $\alpha = \frac{\theta +360k}{n}$ where $k$ is $0,1,2,...,n-1$ and got two results for the argument. The first being $\frac{-7\pi}{8}$ and the other being $\frac{\pi}{8}$.
My final answers were
$$
z_1 = 2\sqrt2(\cos\frac{-7\pi}{8} + i\sin\frac{-7\pi}{8}) \\
z_2 = 2\sqrt2(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8})
$$
For some reason, the answers from the book were
$$
z_1 = 2\sqrt2(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) \\
z_2 = -2\sqrt2(\cos\frac{\pi}{8} + i\sin\frac{\pi}{8})
$$
What was it that I did wrong in my calculations? Thanks in advance!
| Both are correct. In fact, they are the same.
$$ 2\sqrt 2 (\cos\frac{-7\pi}{8} + i\sin\frac{-7\pi}{8}) = -2 \sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) $$
Moreover, $-2\sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8})$ looks better.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Dividing a polynomial with $(x^2+1)^2$ I have been given that a polynomial $f(x)$ with real coefficients is divisible by $(x^2+1)$, and that when $f'(x)$ is divided by $(x^2+1)$, we get a remainder of $(x+1)$. I need to prove that $2f(x)+(x-1)(x^2+1)$ is divisible by $(x^2+1)^2$.
What I did was to express $f(x)$ as $f(x)=(x^2+1)g(x)$, for some polynomial $g$ (1),and $f'(x)=((x^2+1))q(x) + (x+1)$, for some polynomial $q$ (2).
From (1) I deduced another expression of $f'(x)$, (3): $$f'(x)=(x^2+1)g'(x) + 2x(g(x))$$
I then expressed the given expression as:
$$2f(x)+(x-1)(x^2+1) = 2(x^2+1)g(x) + (x-1)(x^2+1)$$ (4)
To find an expression for $g(x)$, I then equated (2) and (3) above, to get
$$ 2g(x) = (x^2+1)(q(x)-g'(x) + x+1)(1/x)$$
Letting $(q(x)-g'(x) + x+1)(1/x) = h(x)$, I simplified to:
$$2g(x)= (x^2+1)h(x)$$
then here I stopped, realizing that $h(x)$ is maybe not even a polynomial.
Any alternative method or advancement of this one will be very appreciated!
| You can use the rule for a linear factor that $a$ is a double-root, and $(x-a)^2$ a factor, of $p(x)=0$ if and only if $p(a)=p'(a)=0$.
Use $a=i$ and $a=-i$. $(x-i)^2(x+i)^2=(x^2+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
maximum, complex quadratic function, Is my solutions correct? I'm trying to compute $\max_{|z| \le 1} |(z+2)(z-1)|$.
Here's how I do it:
$\{z \in \mathbb{C} \ | \ |z| \le 1 \}$ is compact and $f(z) = (z+2)(z-1)$ is continuous, so it suffices to look for $\max_{|z| = 1} |(z+2)(z-1)|$.
Let $z = \cos \phi + i \sin \phi, \ \ 0 \le \phi \le 2 \pi$
Now, $|(z+2)(z-1)| = \sqrt{sin^2 \phi + (\cos \phi -1)^2} \sqrt{\sin^2 \phi + (\cos \phi +2)^2} = \sqrt{1+ 2 \cos \phi} \cdot 2 \sqrt{cos \phi +1} = 2 \sqrt{-2 cos^2 \phi - \cos \phi +1}$.
So we are left with a quadratic function with real coefficients and the leading coefficient is negative, so we can compute the vertex of the parabola $f(\cos \phi) = -2 cos^2 \phi - \cos \phi +1$ and $q =-\frac{ \Delta }{4a} = - \frac{9}{8}$ and $p = - \frac{b}{2a} = - \frac{1}{4}$.
So we have maximum value equal to $- \frac{9}{8}$ for $\cos \phi = - \frac{1}{4}$ (then $\sin \phi = \sqrt{1 - \frac{1}{16}} = - \frac{1}{4} \sqrt{15}$)
So it seems that $\max_{|z| \le 1} |(z+2)(z-1)| = \frac{9}{8}$ for $z = - \frac{1}{4} + i - \frac{1}{4} \sqrt{15}$.
Is it correct?
| setting $z=x+yi$ then we get
$|(x+2+yi)(x-1+yi)|=\sqrt{(x+2)^2+y^2}\sqrt{(x-1)^2+y^2}$ where $y^2=1-x^2$ thus you will have only one variable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Inequality of area of two triangles Let $ABC$ be a triangle with sides $a,b,c$ and $A_1B_1C_1$ be another triangle with sides $a+\frac{b}2$, $b+\frac{c}2$, $c+\frac{a}2$. Prove that:
$$\frac94[ABC]\le[A_1B_1C_1]$$
I tried using Heron's formula, but it is not getting me to the answer. Can anyone help? :)
| By Heron's formula,
$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$[A_1B_1C_1]=\frac{1}{4}\sqrt{\frac{3(a+b+c)}{2}(b+c-a+\frac{a+c-b}{2})(a+c-b+\frac{a+b-c}{2})(a+b-c+\frac{b+c-a}{2})}$
Set $b+c-a=x$, $a+c-b=y$, $a+b-c=z$,
Enough to show:
$\frac{27}{8}xyz\leq(x+\frac{y}{2})(y+\frac{z}{2})(z+\frac{x}{2}) $,
i.e
$27xyz\leq(x+x+y)(y+y+z)(z+z+x) $,
Use AM-GM for 3 terms of RHS.
For example, $x+x+y\geq 3\sqrt[3]{x^2y}$.
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $ Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $
Let $\epsilon > 0$ be arbitrary. I want to find $N$ such that $n \in \mathbb{N}$ guarantees $ \left | \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} - 0 \right | < \epsilon$.
$\Leftrightarrow \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon \Leftrightarrow n > \frac{4}{3} \cdot \frac{1}{\epsilon} \Rightarrow n \geq N$
Take $N = \left \lfloor \frac{4}{3} \cdot \frac{1}{\epsilon} \right \rfloor + 1$
In the above proof, I'm confused as to how $\frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} < \epsilon$
$\Leftarrow \frac{1}{n} \cdot \frac{4}{3} < \epsilon$. How did $\frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}$ become $\frac{4}{3} $?
| Since $n \in \mathbb N \implies n \geq 1$, we know that:
$$
\frac{1}{n} \leq 1 \iff 3 + \frac{1}{n} \leq 4
$$
as well as that:
$$
\frac{-1}{n} \geq -1 \iff 4 - \frac{1}{n} \geq 3
$$
Now to make a fraction smaller, we can either decrease the numerator or increase the denominator. Thus, it follows that:
$$
\epsilon > \frac{1}{n} \cdot \frac{4}{3}
\geq \frac{1}{n} \cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using differentiation under integral sign to calculate $\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$ I want to calculate the integral $$\int^{\pi/2}_0\frac{\log(1+\sin\phi)}{\sin\phi}d\phi$$ using differentiation with respect to parameter in the integral $$\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$$
I know that I have to solve from differentiate under the integral and I must use a suitable substitution for integrands involving trigonometric functions but I can't complete the solution. Could you help me?
| Let's consider the integral
\begin{align}I(\alpha)&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\sin\,\phi)}{\sin\,\phi}\;d\phi\quad\Rightarrow\quad\phi\mapsto \frac{\pi}{2}-\phi\\
&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\cos\,\phi)}{\cos\,\phi}\;d\phi, \qquad 0 < \alpha < \pi.\end{align}
Differentiating $I(\alpha)$ with respect to $\alpha$, we have
\begin{align}
{I}'(\alpha) &= \int_0^{\Large\frac{\pi}{2}} \frac{\partial}{\partial\alpha} \left(\frac{\ln(1 + \cos\alpha \cos \phi)}{\cos \phi}\right)\,d\phi \\
&=-\int_0^{\Large\frac{\pi}{2}}\frac{\sin \alpha}{1+\cos \alpha \cos \phi}\,d\phi \\
&=-\int_0^{\Large\frac{\pi}{2}}\frac{\sin \alpha}{\left(\cos^2 \frac{\phi}{2}+\sin^2 \frac{\phi}{2}\right)+\cos \alpha\,\left(\cos^2\,\frac{\phi}{2}-\sin^2 \frac{\phi}{2}\right)}\,d\phi \\
&=-\frac{\sin\alpha}{1-\cos\alpha} \int_0^{\Large\frac{\pi}{2}} \frac{1}{\cos^2\frac{\phi}{2}}\frac{1}{\left[\left(\frac{1+\cos \alpha}{1-\cos \alpha}\right) +\tan^2 \frac{\phi}{2} \right]}\,d\phi \\
&=-\frac{2\,\sin\alpha}{1-\cos\alpha} \int_0^{\Large\frac{\pi}{2}}\,\frac{\frac{1}{2}\,\sec^2\,\frac{\phi}{2}}{\left[\,\left(\dfrac{2\,\cos^2\,\frac{\alpha}{2}}{2\,\sin^2\,\frac{\alpha}{2}}\right) + \tan^2\,\frac{\phi}{2} \right]} \,d\phi \\
&=-\frac{2\left(2\,\sin\,\frac{\alpha}{2}\,\cos\,\frac{\alpha}{2}\right)}{2\,\sin^2\,\frac{\alpha}{2}}\,\int_0^{\Large\frac{\pi}{2}}\,\frac{1}{\left[\left(\dfrac{\cos \frac{\alpha}{2}}{\sin\,\frac{\alpha}{2}}\right)^2\,+\,\tan^2\,\frac{\phi}{2}\,\right]}\,d\left(\tan\,\frac{\phi}{2}\right)\\
&=-2\cot \frac{\alpha}{2}\,\int_0^{\Large\frac{\pi}{2}}\,\frac{1}{\left[\,\cot^2\,\frac{\alpha}{2} + \tan^2\,\frac{\phi}{2}\,\right]}\,d\left(\tan \frac{\phi}{2}\right)\,\\
&=-2\,\left.\tan^{-1} \left(\tan \frac{\alpha}{2} \tan \frac{\phi}{2} \right) \right|_0^{\Large\frac{\pi}{2}}\\
&=-\alpha
\end{align}
Therefore:
$$I(\alpha) = C - \frac{\alpha^2}{2}$$
However by definition, $I\left(\frac{\pi}{2}\right) = 0$, hence $C = \dfrac{\pi^2}{8}$ and
$$I(\alpha) = \frac{\pi^2}{8}-\frac{\alpha^2}{2}.$$
The integral we want to evaluate is
$$I(0) = \int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\sin\,\phi)}{\sin\,\phi}\;d\phi=\frac{\pi^2}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
Equivalence of summations Show that $$\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2}\equiv \frac{1}{n}\sum^{n}_{i=1}x_{i}^{2} - \bar{x}^{2}.$$ Note that $\bar{x} = \frac{1}{n}\sum^{n}_{i=1} x_{i}$. So I have started by:
\begin{align}
\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2} &= \frac{1}{n}\sum^{n}_{i=1} (x_{i}^{2} - 2x_{i}\bar{x} + \bar{x}^{2}) \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n}\sum_{i=1}^{n}x_{i}\bar{x} + \frac{1}{n}\sum^{n}_{i=1}\bar{x}^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n}\sum_{i=1}^{n}x_{i}\left(\frac{1}{n}\sum^{n}_{i=1}x_{i}\right) + \frac{1}{n}\sum_{i=1}^{n}\left(\frac{1}{n}\sum^{n}_{i=1}x_{i}\right)^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n^{2}}\left(\sum_{i=1}^{n}x_{i}\right)^{2} + \frac{1}{n^{3}}\left(\sum^{n}_{i=1}x_{i}\right)^{2}
\end{align}
Not really sure what to do next, any hints?
| When you go from the 3rd line of your calculation to the last line, you made a mistake. Thus the $\frac{1}{n^3}$ should really be $\frac{1}{n^2}$.
\begin{align}
\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2} &= \frac{1}{n}\sum^{n}_{i=1} (x_{i}^{2} - 2x_{i}\bar{x} + \bar{x}^{2}) \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n}\sum_{i=1}^{n}x_{i}\bar{x} + \frac{1}{n}\sum^{n}_{i=1}\bar{x}^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n}\sum_{i=1}^{n}x_{i}\left(\frac{1}{n}\sum^{n}_{i=1}x_{i}\right) + \frac{1}{n}\sum_{i=1}^{n}\left(\frac{1}{n}\sum^{n}_{i=1}x_{i}\right)^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n^{2}}\left(\sum_{i=1}^{n}x_{i}\right)^{2} + \frac{1}{n^{2}}\left(\sum^{n}_{i=1}x_{i}\right)^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2}- \frac{1}{n^{2}}\left(\sum_{i=1}^{n}x_{i}\right)^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \bar{x}^{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd. Prove: If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
I know that the $n$th row in pascal's triangle correspond to the coefficients of $(x+y)^n$:
$$\begin{align}(x+y)^n = {n \choose 0} x^n + {n \choose 1} x^{n-1} y + {n \choose 2}x^{n-2}y^2+{n \choose 3} x^{n-3} + \cdots + {n \choose n-1} xy^{n-1}+{n \choose n} y^n \end{align}.$$
I have to show that each coefficient $\begin{align} {n \choose 0},{n \choose 1},{n \choose 2},\ldots,{n \choose n-1},{n \choose n} \end{align}$ is of the form $2b+1$, but I'm not sure how to do this with $n =2^k-1$.
| As the answer above says, $\displaystyle{2^{n}-1\choose k-1}+{2^{n}-1\choose k} = {2^n \choose k}$. So if ${2^n \choose k}$ is even for $k \neq 0,2^n$, both terms are even or both terms are odd. First, let's prove that ${2^n \choose k}$ is even.
$$\displaystyle{2^n \choose k} = \frac{2^n!}{k!(2^n-k)!}$$
For $k = 1, {2^n \choose 1} = 2^n$ even. Now assume ${2^n \choose k}$ is even for some k, now:
$$\displaystyle{2^n \choose k} = \frac{2^n!}{k!(2^n-k)!} = \frac{2^n!(k+1)}{(k+1)!(2^n-k-1)!(2^n-k)} = {2^n \choose k+1}\frac{k+1}{2^n-k}$$
Now, if $k$ is either odd or even, $\displaystyle\frac{k+1}{2^n-k}$ is odd. As $\displaystyle{2^n \choose k}$ is even, it has to happen that $\displaystyle{2^n \choose k+1}$ must be even. Therefore by induction, $\displaystyle{2^n \choose k}$ is even for all $k \neq 0,2^n$.
Now, by induction as $\displaystyle{2^{n}-1\choose k-1}+{2^{n}-1\choose k} = {2^n \choose k}$, note that, as the first two terms ($k = 0,1$) are odd, then the third term must also be odd, and so on.
Therefore ${2^n-1 \choose k}$ is odd for all $k$ between $0$ and $2^n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Rewrite the expression in the form $A \sin(x+C)$ Rewrite the following expression in the form $A \sin(x+C)$
$$4 \sin x + 4\sqrt{3} \cos x$$
This is what I have so far, and I'm not even sure it's the right approach. I just dont understand this concept as a whole:
$$A \cos(c)\sin(x) + A\sin(c)\cos(x)$$
$$A\cos (c)=4$$
$$A\sin (c)=4\sqrt{3}$$
$$\sin^2(x) + \cos^2(x) = 1$$ ??
| rewrite your term in the form
$\sqrt{4^2+4^2\cdot 3}\left(\frac{4}{\sqrt{4^2+4^2\cdot3}}\sin(x)+\frac{4\sqrt{3}}{\sqrt{4^2+4^2\cdot3}}\cos(x)\right)$
now we have
$\cos(\phi)=\frac{4}{\sqrt{4^4+4^2\cdot 3}}$ and
$\sin(\phi)=\frac{4\sqrt{3}}{\sqrt{4^2+4^2\cdot3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
When do we specify restrictions in rational expressions or equations In the following equation $x \neq 0$ apply's according to wolframalpha (expanded form).
$$\frac{x^3 + x^2}{2x} = \frac{x^2 + x}{2}$$
Yet in the following equation it does not...
$$\frac{ab^2}{b} = 5 \iff ab = 5$$
Why don't we specify the restriction $b \neq 0$ here? And when do we specify restrictions in general?
| I think that
$$\dfrac{x^3 + x^2}{2x} = \dfrac{x^2 + x}{2} \Leftrightarrow
\begin{cases}
2x\cdot \dfrac{x^3 + x^2}{2x}=2x\cdot \dfrac{x^2 + x}{2}\\
2x\neq 0
\end{cases}$$
Also $$\dfrac{ab^2}{b} = 5 \iff
\begin{cases}
b\cdot\dfrac{ab^2}{b}=b\cdot 5\\
b\neq 0 \end{cases}\iff
\begin{cases}
ab^2=5b\\
b\neq 0 \end{cases}\iff\begin{cases}
b\cdot (ab-5)=0\\
b\neq 0\end{cases}\iff ab=5$$
I a general way: $$\dfrac{A}{B}=C\iff \begin{cases}
B\cdot \dfrac{A}{B}=B\cdot C\\
B\neq 0 \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$ - is there any other way to solve this? Here's a problem that I think I've managed to solve:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$$
Here's how I did it:
$(\sqrt{2^n+n^2}-\sqrt{2^n+1})\frac{\sqrt{2^n+n^2}+\sqrt{2^n+1}}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}=\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}$
And now you can see that (for sufficiently large $n$):
$0\le\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}\le\frac{n^2}{\sqrt{2^n}}$
Limit of $0$ clearly is $0$, and this is also true for $\frac{n^2}{\sqrt{2^n}}$, so that means that:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}=0$$
The thing is I got kind of lucky I've solved it and I'm interested if there are any other ways to solve this limit (not involving any derivatives nor integrals). Does anybody see any clever solution other than mine? And also - is my solution correct?
| If you know about Taylor expansions, you can do the following: as long as $\frac{a_n}{2^n}\xrightarrow[n\to\infty]{}0$,
$$
\sqrt{2^n + a_n} = \sqrt{2^n}\sqrt{1 + \frac{a_n}{2^n}} = \sqrt{2^n}\left(1+\frac{a_n}{2\cdot2^n}+o\!\left(\frac{a_n}{2^n}\right)\right) = 2^{\frac{n}{2}}+\frac{a_n}{2^{\frac{n}{2}+1}}+o\!\left(\frac{a_n}{2^n}\right)
$$
Applying this to your case with first $a_n=n^2$, then $a_n=1$, you get that the difference is
$$
2^{\frac{n}{2}}+\frac{n^2}{2^{\frac{n}{2}+1}}+o\!\left(\frac{n^2}{2^n}\right) - 2^{\frac{n}{2}}
$$
as $\frac{1}{2^n} = o\!\left(\frac{n^2}{2^n}\right)$. I.e., the difference is equivalent to $\frac{n^2}{2^{\frac{n}{2}+1}}$, which goes to 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
| Denote $\sqrt{14+4\sqrt{10}}=a, \sqrt{14-4\sqrt{10}}=b$, then
$$a^2+b^2=28, ab=6$$
So $(a-b)^2=16$ and $a-b=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Convergence of a summation How do I find out what this summation converges to? I dont even know how I'd start o_o
I assume the first part converges to infinity but dont know how the cos works in this case.
$$\sum^{\infty}_{k=1} 3^{-k}\cos{k}$$
| What you have is:
$$
\sum_{k=1}^\infty \frac{\cos k}{3^k}
$$
As Daniel mentioned, Euler's formula will be useful here:
You will notice that $\cos x = \frac{1}{2}(e^{ix}+e^{-ix})$. Thus:
\begin{align}
\sum_{k=1}^\infty \frac{\cos k}{3^k} &= \frac{1}{2}\sum_{k=1}^\infty \frac{e^{ik}+e^{-ik}}{3^k} \\
&= \frac{1}{2}\left(\sum_{k=1}^\infty \frac{e^{ik}}{3^k} + \sum_{k=1}^\infty \frac{e^{-ik}}{3^k}\right) \\
&= \frac{1}{2}\left(\sum_{k=0}^\infty \frac{e^{ik}}{3^k} - 1 + \sum_{k=0}^\infty \frac{e^{-ik}}{3^k} - 1\right) \\
&= \frac{1}{2}\left(\sum_{k=0}^\infty \left(\frac{e^i}{3}\right)^k + \sum_{k=0}^\infty \left(\frac{e^{-i}}{3}\right)^k - 2\right) \\
&= \frac{1}{2}\left(\frac{1}{1-\frac{e^i}{3}} +\frac{1}{1-\frac{e^{-i}}{3}}\right) - 1 \\
&= \frac{1}{2}\left(\frac{3}{3-e^i} +\frac{3}{3-e^{-i}}\right) - 1 \\
\end{align}
Notice that in the geometric series $\left|\frac{e^i}{3}\right| < 1$ and $\left|\frac{e^{-i}}{3}\right| < 1$.
Now, Euler's formula will show that $e^i = \cos(1) + i\sin(1)$ and $e^{-i} = \cos(-1) + i\sin(-1) = \cos(1)-i\sin(1)$. Thus:
\begin{align}
\sum_{k=1}^\infty \frac{\cos k}{3^k} &= \frac{1}{2}\left(\frac{3}{3-e^i} +\frac{3}{3-e^{-i}}\right) - 1 \\
&= \frac{3}{2}\left(\frac{1}{3-\cos(1)-i\sin(1)} +\frac{1}{3-\cos(1)+i\sin(1)}\right) - 1 \\
&= \frac{3}{2}\left(\frac{3-\cos(1)+i\sin(1) + 3-\cos(1)-i\sin(1)}{(3-\cos(1)+i\sin(1))(3-\cos(1)-i\sin(1))}\right) - 1 \\
&= \frac{3}{2}\left(\frac{6-2\cos(1)}{10-6\cos(1)}\right) - 1 \\
&= \frac{9-3\cos(1)}{10-6\cos(1)} - 1 \\
&= \frac{9-3\cos(1) - (10 - 6\cos(1)}{10-6\cos(1))} \\
&= \frac{3\cos(1)-1}{10-6\cos(1)} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Please the box method for factoring trinomial of the form $ax^2+bx=c$ I was given this method for factoring trinomials of the form $ax^2 + bx + c$
This is the method:
Find numbers $p$ and $q$ such as $a\cdot c=p\cdot q$ and $b=p+q$
With $p$ and $q$ $(GCF(a,q)x + GCF(c,p))(GCF(a,p)x+GCF(c,q))= ax^2+bx+c$
I have no idea how to prove this method or why it works.
Please help me.
| The greatest common factor is also called the greatest common divisor. I will call it the gcd for greatest common divisor.
There is an implicit assumption here that $\gcd(a, b, c) = 1$. For instance,
\begin{align*}
8x^2 + 24x + 18 & = 8x^2 + 12x + 12x + 18\\
& = 4x(2x + 3) + 6(2x + 3)\\
& = (4x + 6)(2x + 3)\\
& = 2(2x + 3)^2
\end{align*}
so $p = q = 12$, $\gcd(8, 12) = 4$, $\gcd(18, 12) = 6$, $ac = 8 \cdot 18 = 144 = 12 \cdot 12 = pq$ and $b = 24 = 12 + 12 = p + q$. However, $a = 8 \neq \gcd(8, 12)\gcd(8, 12) = \gcd(a, p)\gcd(a, q)$.
There also seems to be an implicit assumption that $c \neq 0$. Consider the factorization
$$6x^2 + 5x = x(6x + 5) = (x + 0)(6x + 5)$$
of $6x^2 + 5x = 6x^2 + 5x + 0$. Observe that $ac = 6 \cdot 0 = 0$ and that $b = 5$. If $pq = ac = 0$ and $p + q = b = 5$, then $p = 5$ and $q = 0$ or $p = 0$ and $q = 5$. Since $\gcd(0, 0)$ is undefined, the formula does not make sense if $c = 0$.
Assume $a, b, c, s, t, u, v$ are integers such that
$$ax^2 + bx + c = (sx + t)(ux + v)$$
with $a, c\neq 0$ and $\gcd(a, b, c) = 1$.
We will show that $a = su$, $b = sv + tu$, and $c = tv$. Since the equation $$ax^2 + bx + c = (sx + t)(ux + v)$$
is an algebraic identity, it must hold for each real number $x$. In particular, it holds when $x = 0, 1, -1$. Since it holds when $t = 0$,
$$c = tv$$
Since the identity holds when $t = 1$,
$$a + b + c = (s + t)(u + v) = su + sv + tu + tv$$
Since $c = tv$, we obtain
$$a + b = su + sv + tu$$
Since the identity holds when $t = -1$,
$$a - b + c = (-s + t)(-u + v) = su - sv - tu + tv$$
Since $c = tv$, we obtain
$$a - b = su - sv - tu$$
Adding the equations for $a + b = su + sv + tu$ and $a - b = su - sv - tu$ yields
$$2a = 2su$$
Hence,
$$a = su$$
Substituting $su$ for $a$ in the equation $a + b = su + sv + tu$ shows that
$$b = sv + tu$$
Take $p = sv$ and $q = tu$. Then
$$pq = svtu = (su)(tv) = ac$$
and
$$p + q = sv + tu = b$$
Note that if $\gcd(s, t) = d > 1$. Then $s = ds_0$, $t = dt_0$ for some integers $s_0, t_0$, so
\begin{align*}
a & = su = ds_0u\\
b & = sv + tu = ds_0v + dt_0u = d(s_0v + t_0u)\\
c & = tv = dt_0v
\end{align*}
which implies that $\gcd(a, b, c) \geq d > 1$, contrary to our assumption that $\gcd(a, b, c) = 1$. Hence, $\gcd(s, t) = 1$. A similar argument shows that $\gcd(u, v) = 1$.
Therefore,
\begin{align*}
\gcd(a, p) & = \gcd(su, sv) = s\\
\gcd(a, q) & = \gcd(su, tu) = u\\
\gcd(c, p) & = \gcd(tv, sv) = v\\
\gcd(c, q) & = \gcd(tv, tu) = t\\
\end{align*}
Thus,
\begin{align*}
ax^2 + bx + c & = (sx + t)(ux + v)\\
& = [\gcd(a, p)x + \gcd(c, q)][\gcd(a, q)x + \gcd(c, p)]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| rewrite it in the form $(\sqrt{x}-\sqrt{y})\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}=y-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? While Solving some integral problem, I encountered the following infinite series:
$$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$
I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.
| We have, $\displaystyle B_{2n}(x) = (-1)^{n-1}\frac{2(2n)!}{(2\pi)^{2n}} \sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}$
Cubing and integrating both sides,
\begin{align*}&\int_0^1 \left((-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}(x)\right)^3\,dx \\&= \int_0^1 \left(\sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}\right)^3\,dx \\&= \sum\limits_{k_1,k_2,k_3 = 1}^{\infty} \int_0^1 \frac{\cos \left(2\pi k_1x \right)\cos \left(2\pi k_2x \right)\cos \left(2\pi k_3x \right)}{k_1^{2n}k_2^{2n}k_3^{2n}}\,dx\\&= \frac{1}{4}\sum\limits_{k_1,k_2,k_3 = 1}^{\infty} \dfrac{\displaystyle \sum\limits_{\{\pm\}} \int_0^1 \cos \left(2\pi (k_1\pm k_2 \pm k_3)x \right)\,dx}{k_1^{2n}k_2^{2n}k_3^{2n}}\\&= \frac{3}{4}\sum\limits_{k_1,k_2= 1}^{\infty} \frac{1}{k_1^{2n}k_2^{2n}(k_1+k_2)^{2n}}\end{align*}
Since, $\displaystyle \int_0^1 \cos \left(2\pi (k_1 + k_2 - k_3)x \right)\,dx = \begin{cases}1 & \text{ when } k_3 = k_1 + k_2\\ 0 &\text{ otherwise }\end{cases}$
The case $n = 1$,
$$\sum\limits_{k_1,k_2= 1}^{\infty} \frac{1}{k_1^{2}k_2^{2}(k_1+k_2)^{2}} = \frac{4\pi^6}{3}\int_0^1 \left(B_2(x)\right)^3\,dx = \frac{1}{3}\frac{\pi^6}{945} = \frac{1}{3}\zeta(6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 1
} |
Integration question does this substitution work? $$\int_{0}^{3} \frac{1}{\sqrt{3-x}}dx$$
I'm a little lost here do i let $u = 3-x$, so that $du/dx = 1$, $du=dx$
$$\int_0^3 \frac{1}{\sqrt{u}}du = u^{-\frac{1}{2}} du$$
$$ \left[\frac{1}{\frac{1}{2}} u^{\frac{1}{2}}\right]_0^3 = \left[2(3-x)^{\frac{1}{2}}\right]_0^3$$
$$ = 2\sqrt{0}-2\sqrt{3} = -2\sqrt{3}$$
hi sorry i meant 3-x
| If you substitute $u = 3-x$, $du = -dx$. The bounds change as well so $x =0$ means $u=3$ and $x=3$ means $u=0$
$$\int_0^3 \frac{dx}{\sqrt{3-x}} = \int_3^0 \frac{-du}{\sqrt{u}} = \int_0^3 \frac{du}{\sqrt{u}} = \left[ 2\sqrt{u} \right]_0^3 = 2(\sqrt{3} - 0) = 2\sqrt{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Trigonometric Limits compute the following limit:
$$\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\ \mbox{as}\ x\to0.$$
How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
but without used ´Hospital
| Consider first $$\cos (y+x)-\cos y=\cos y\cdot(\cos x-1)-\sin x\sin y=-2\cos y\cdot\sin^2\frac x2-\sin x \sin y$$
And $$\sin(a+x)-\sin a=\sin a\cdot(\cos x-1)+\cos a\sin x=-2\sin a\cdot\sin^2 \frac x2+\cos a\sin x$$
Then (expressing the numerator as a double-difference ie a difference of differences) $$\cos (a+2x)-2\cos (a+x)+\cos a = (\cos (a+2x)-\cos(a+x))-(\cos (a+x)-\cos a)$$
We apply the first identity to the two components of this to obtain $$-2\cos(a+x)\cdot \sin^2 \frac x2-\sin x \cdot\sin(a+x)+2\cos a\sin^2\frac x2+\sin x\sin a =$$using the second identity too $$=2\sin^2\frac x2\left(2\cos a\sin^2\frac x2+\sin x\sin a \right)-\sin x\left(-2\sin a\sin^2 \frac x2+\cos a\sin x\right)$$
Now the basic limit here is $$\lim _{x\to 0} \frac {\sin x}{x}=1$$ and on dividing by $x^2$ and taking the limit, the terms in $\sin^4 \frac x2$ and $\sin x\sin^2\frac x2$ go to zero, leaving the final term which goes to $-\cos a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+v)}=\frac{1}{vv!}$$
I am struggling to find a solution for this but no luck yet. How can I analyze it to get to second part?
| \begin{eqnarray}
\frac{1}{n(n+1)...(n+v)}&=&\frac{1}{v}\frac{(n+v)-n}{n(n+1)...(n+v)}\\
&=&\frac{1}{v}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
\end{eqnarray}
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)...(n+v)}
=\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]$$
$$\frac{1}{v}\sum_{n=1}^{k}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
=\frac{1}{v}[\frac{1}{v!}-\frac{1}{(k+1)...(k+v)}]$$
Let $k\rightarrow\infty$, then
$\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
=\frac{1}{v}\frac{1}{v!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine whether this set of vectors is linearly independent For $n \geq 1$ determine if the set $S$ is linearly independent.
The set $S$ is these vectors:
$$
v_1 \equiv(\,1,2,\ldots,n\,)\,,\quad
v_2 \equiv(\,1,2^2,3^2,\ldots,n^2\,)\,,\ldots\,
v_n \equiv(\,1,2^n,3^n,\ldots,n^n\,)
$$
I know for vectors to be linearly independent we usually multiply by a scalar to each vector and see if it equals the zero vector
(i.e. $av_1 + bv_2 + cv_3=0$ ), but I've attempted it that way but am having trouble solving it when its a series involving $n$'s.
All help would be appreciated
| If $n=3$ it is
$$\left|\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 2^2 & 2^3 \\ 3 & 3^2 & 3^3 \end{array}\right|=\left|\begin{array}{rrr} 1 & 1 & 1 \\ 0 & 2 & 2^2 \\ 0 & 2\cdot 3 & 2\cdot 3^2 \end{array}\right|=\left|\begin{array}{rr} 2 & 2^2 \\ 2\cdot 3 & 2\cdot 3^2 \end{array}\right|=2^2\cdot 3 \left|\begin{array}{rr} 1 & 2 \\ 1 & 3 \end{array}\right|\ne 0.$$
If $n=4$ it is
$$\left|\begin{array}{rrrr}1 & 1 & 1 & 1 \\ 2 & 2^2 & 2^3 & 2^4 \\ 3 & 3^2 & 3^3 & 3^4 \\4 & 4^2 & 4^3 & 4^4 \end{array}\right|=\left|\begin{array}{rrrr}1 & 1 & 1 & 1 \\ 0 & 2 & 2^2 & 2^3 \\ 0 & 2\cdot 3 & 2\cdot 3^2 & 2\cdot 3^3 \\ 0& 3\cdot 4 & 3\cdot 4^2 & 3\cdot 4^3 \end{array}\right|=\left|\begin{array}{rrr} 2 & 2^2 & 2^3 \\ 2\cdot 3 & 2\cdot 3^2 & 2\cdot 3^3 \\ 3\cdot 4 & 3\cdot 4^2 & 3\cdot 4^3 \end{array}\right| \\=2^2\cdot 3^2 \cdot 4\left|\begin{array}{rrr} 1 & 2 & 2^2 \\ 1 & \cdot 3 & 3^2 \\ 1 & 4 & 4^2 \end{array}\right|\ne 0.$$
You can get the solution for arbitrary $n$ using induction.
| {
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"url": "https://math.stackexchange.com/questions/1017698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the exact closed from expression of $1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2$ I know the above expression equals to $\frac{n(2n−1)(2n+1)}{3}$, but how exactly can i come up with something from scratch?
| Most basic way in my opinion:
$$f(n)=1^2 + 3^2 + 5^2 + · · · (2n-1)^2+ (2n + 1)^ 2$$
$$f(n-1)=1^2 + 3^2 + 5^2 + · · · + (2(n -1)+ 1)^ 2=1^2 + 3^2 + 5^2 + · · · + (2n -1)^ 2$$
$$f(n)-f(n-1)=(2n + 1)^ 2$$
$$f(n)=an^3+bn^2+cn$$
$$f(n-1)=a(n-1)^3+b(n-1)^2+c(n-1)$$
$$an^3+bn^2+cn-a(n-1)^3-b(n-1)^2-c(n-1)=4n^2+4n+1$$
If you put $(n-1)^3=n^3-3n^2+3n-1$,
$(n-1)^2=n^2-2n+1$
Then you can find $a,b,c$ values easily
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Confusion about the expansion of $\;x(2x - 3)(3x + 4)$ For this algebraic product:
$\;x(2x - 3)(3x + 4)$
What I expected to do is:
$$\begin{align} x(2x - 3)(3x + 4) & = 2x^2 -3x(3x + 4) \\ & = 2x^2 -9x^2 -12x \\ & = -7x^2 -12x\end{align}$$
However, what the book I'm reading does is:
$$\begin{align} x(2x - 3)(3x + 4) & = (2x^2 - 3x)(3x + 4) \\ &= 6x^3 +8x^2 -9x^2 -12x \\ & = 6x^3 -x^2 -12x\end{align}$$
So, in this part: $x(2x - 3)$, why the product become between parenthesis??
| First off, to address a misunderstanding expressed in your comment:
$$a - b(c + d) \neq (a - b)(c + d)$$ Rather $$(a-b)(c+d) = a(c+d) - b(c+d) = ac + ad -bc -bd$$
Now, we address the posted problem. If proceeding left to right, $x$ needs to multiply both $2x$ and $-3$, to give $(2x^2 - 3x)$, all of which needs to then multiply $(3x + 4)$. So think of first multiplying the two factors $x$ and $2x- 3$ to get $$x(2x - 3) = 2x^2 - 3x$$ So now, we have multiply this partial product with the factor $3x + 4$ to get $$\begin{align}(2x^2 - 3x)(3x + 4) & = 2x^2(3x + 4) - 3x(3x + 4) \\ \\ & = 6x^3 + 8x^2 - 9x^2 -12 x \\ \\ &= 6x^3 - x^2 - 12 x\end{align}$$
Another way to approach this is to expand the two right-most factors first, then multiply that expansion by $x$:
$$\begin{align} x(2x - 3)(3x + 4) & = x(2x\cdot 3x+ 2x\cdot 4 - 3\cdot 3x - 3\cdot 4) \\ \\ &= x(6x^2 -x -12) \\ \\ & = 6x^3 - x^2 -12x\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculation of $ \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ Calculation of $\displaystyle \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ and $\displaystyle \cos\left(\frac{2\pi}{7}\right)\times \cos\left(\frac{4\pi}{7}\right) \times \cos\left(\frac{6\pi}{7}\right)$ is
$\bf{My\; Try::}$ Let $\displaystyle \frac{2\pi}{7}=\phi\;,$ Then $3\phi = 2\pi-4\phi\Rightarrow \cos \left(3\phi\right) = \cos \left(2\pi-4\phi\right)=\cos \left(4\phi\right)$
So $4\cos^3 \phi -3\cos \phi = 2\left[2\cos^2 \phi - 1\right]^2-1=2\left[4\cos^4 \phi+1-4\cos^2 \phi\right]-1 = 8\cos^4 \phi-8\cos^2 \phi+1$
So $8\cos^4\phi-4\cos^3 \phi-8\cos^2 \phi+3\cos \phi+1=0$
Now I did not understand how can i solve after that
Help me
Thanks
| To complete Tim Raczkowski's solution we will the address the second part of the problem. Using the identical notation as above, we have
$$\cos(2\pi/7)\cdot \cos(4\pi/7)\cdot \cos(6\pi/7)=$$$$(\omega+1/\omega)(\omega^{2}+1/\omega^{2})(\omega^{3}+1/\omega^{3})/8=\omega(\omega^2+1)(\omega^4+1)(\omega^6+1)/8=$$$$(\omega^{13}+\omega^{11}+\omega^9+2\omega^7+\omega^5+\omega^3+\omega)/8=(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+2)/8=1/8$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Prove a limit for $g(x)$ with the definition of limit only We have the function $g(x) = x^3+1$
Prove, with the definition of limit only, that the limit $L=9$ is indeed the limit of the function when $x=2$.
I started with: $\lvert g(x) - L \rvert$ : $\lvert x^3+1-9 \rvert = \lvert x^3-8 \rvert = \lvert (x-2)(x^2+2x+4) \rvert < \epsilon $
From the definition of the limit we know: $ \rvert x-2 \lvert < \delta$, so therefore: $\lvert (x-2)(x^2+2x+4) \rvert < \delta (x^2+2x+4) $
How do I proceed from here? do I need to find a condition for $ x^2+2x+4 $ ?
Thanks guys
| HINT: $x^2+2x+4<19$ for $1<x<3$, hence we can take $\delta=\varepsilon/19$ for $\varepsilon<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$
Show
$\forall \ x \in \mathbb{R}:\quad x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$
My attemps:
Case $x=-1$
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 + x} - \dfrac 1 4$$
let $g(x)=\dfrac{1 + x^{7}}{1 + x}-\dfrac{1}{4} \quad \forall x\in \mathbb{R} \backslash \{-1\}$
then
$
\begin{align*}
g'(x)&=\dfrac {(1+x^7)'(1+x)-(1+x)'(1+x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac {(7 x^6)(1+x)-(1+x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac {7 x^6+7 x^7-1-x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac{7 x^6+6 x^7-1}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}
\end{align*}
$
To determine the sign of the numerator $(7 x^6+6 x^7-1)$
once time let :$ h(x)=7 x^6+6 x^7-1$
then
$$h'(x)=42x^5+42x^6=42(1+x)x^5$$
$$h'(x)=0 \Longleftrightarrow x=-1 \text{or} x=0$$
thus $h$ admits a minimum on the point $x=0$
and a maximum on the point $x=-1$
or $h(-1)=0$ and $h(0)=-1$
$\lim_{x\to -\infty}h(x)=-\infty$ and $\lim_{x\to +\infty}h(x)=+\infty$
as $h(-1)=0$ and $h(0)=-1$ by Intermediate value theorem there is $u \in(-1, 0)$ such that h(u) = 0.
i'm stuck here
*
*am i on my way ?
*is there any other ways to solve it
| You can write your expression as
$$x(x-1)(x^4+x^2+1)+\frac{3}{4}\ .$$
Now consider $x(x-1)(x^4+x^2+1)$: the third factor is always positive, hence the whole assume negative values only for $x\in(0,1)$. In this interval $x(x-1)$ is negative and $x^4+x^2+1<3$. The minimum of $x(x-1)$ in $(0,1)$ is $-\frac{1}{4}$ for $x=\frac{1}{2}$ hence you have
$$x(x-1)(x^4+x^2+1)>-\frac{3}{4}$$
For $x\not\in(0,1)$ $x(x-1)(x^4+x^2+1)\geq 0$ so it's ok.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\frac{d^n}{dx^n} (\sin^4 x + \cos^4 x) = 4^{n-1}\cos (4x + \frac{n\pi}{2})$
*
*Question
Prove that $\frac{d^n}{dx^n} (\sin^4 x + \cos^4 x) = 4^{n-1}\cos (4x + \frac{n\pi}{2})$
*
*My attempt
*
*First calculate $\frac{d}{dx} (\sin^4 x + \cos^4 x)$, that is,
$$\frac{d}{dx} (\sin^4 x + \cos^4 x) =4\sin^3 x \cos x - 4\cos^3 x \sin x $$
$$= 4\sin x \cos x(\sin^2 x - \cos^2 x)$$
$$\tag {n=1} = -2\sin 2x \cos 2x = - \sin 4x$$
*Now, using the value of $n=1$, I calculate the derivatives for a few more values of $n$:
$$\tag {n=2} -4\cos 4x$$
$$\tag {n=3} 4^2\sin 4x$$
$$\tag {n=4} 4^3\cos 4x$$
$$\tag {n=2} -4^4\sin 4x$$
*
*From this I observe the consistency of the $4^{n-1}$ factor.
*Now I will expand $\cos (4x + \frac{n\pi}{2})$, which results in $$\cos 4x \cos \frac{n\pi}{2} - \sin 4x \sin \frac{n\pi}{2}$$
*From what I know of of $\sin$ and $\cos$ functions, this fits the pattern I observe in calculating the derivatives of $\sin^4 x + \cos^4 x$
*
*My concerns
I feel uncomfortable putting so much weight on observing a pattern, instead I feel I should be able to put it down in terms of mathematics.
One more thing, I don't feel comfortable with
From what I know of of $\sin$ and $\cos$ functions, this fits the pattern I observe in calculating the derivatives of $\sin^4 x + \cos^4 x$.
, there must be a better way to relate $$\frac{d^{n-1}}{dx^{n-1}} (-\sin 4x) $$ with $$\cos (4x + \frac{n\pi}{2})$$
I would appreciate any hints, suggestions, and alternative approaches.
Keep in mind that this should be approached with the tools of elementary introductory derivatives and trigonometry.
| Hint: Note that if
$$\cos (4x + \frac{n\pi}{2})=\cos 4x \cos \frac{n\pi}{2} - \sin 4x \sin \frac{n\pi}{2}$$
If $n$ is odd, then $n=4k-3$ or $n=4k-1$, $\forall k\in\mathbb{N}$, then $\cos\frac{n\pi}{2}=0$: $$\sin{\frac{(4k-3)\pi}{2}}=1$$ and
$$\sin{\frac{(4k-1)\pi}{2}}=-1$$
Now If $n$ is even, then $n=4k$ or $n=4k-2$, $\forall k\in\mathbb{N}$ then $\sin\frac{n\pi}{2}=0$:
$$\cos{\frac{(4k)\pi}{2}}=\cos{(2k\pi)}=1$$
and
$$\cos{\frac{(4k-2)\pi}{2}}=\cos{[(2k-1)\pi]}=-1$$
| {
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"url": "https://math.stackexchange.com/questions/1023520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equation $((x+y i)-\frac{1}{x+y i})/{(2 i)} = 2$
Solve the equation $$\frac{\left((x+y i)-\frac{1}{x+y i}\right)}{2 i} = 2$$
So far, I got $(0, 2-\sqrt{3}i)$ and $(0, 2+ \sqrt{3}i)$ as solutions for $x$ and $y$. Do I require $2$ more solutions?
| $$\frac{(x+iy)-\dfrac1{(x+iy)}}{2i}=2$$
$$\iff(x+iy)-\dfrac1{(x+iy)}=2(2i)$$
$$\iff(x+iy)^2-1=4i(x+iy)$$
$$\iff x^2-y^2-1+(2xy)i=(4x)i-4y$$
Equating the real parts, $x^2-y^2-1=-4y$
Equating the imaginary parts, $2xy=4x\iff2x(y-2)=0$
If $x=0, -y^2-1=-4y\iff y^2-4y+1=0,\implies y=?$
If $y-2=0\iff y=2, x^2-2^2-1=-4(2)\implies x^2=?$
But if $x$ is real, $x^2\ge0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Indefinite integral evaluation of fraction Here's the indefinite integral to evaluate:
$F(x) = \int \left(\frac{3x + 4}{x + 3}\right)dx$
Here are my steps (verbose):
$$\begin{align}
u & = x + 3 \\
(x + 3)dx & = du \\
\left(\frac{d}{dx}(x + 3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + \frac{d}{dx}(3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + 0\right)dx & = du \\
\left( \frac{dx}{dx} + 0 \right)dx & = du \\
(1 + 0)dx & = du \\
(1)dx & = du \\
dx & = \left( \frac{1}{1}\right)du \\
dx & = du \\
F(\ ) & = \int \left( \frac{(3x + 4)}{u} \right)du \\
u & = x + 3 \\
x + 3 & = u \\
x & = u - 3 \\
F(u)& = \int \left( \frac{3(u - 3) + 4}{u} \right)du \\
& = \int \left( \frac{3u - 9 + 4}{u} \right)du \\
& = \int \left( \frac{3u - 5}{u} \right)du \\
& = \int \left( \frac{3u}{u} - \frac{5}{u} \right)du \\
& = \int \left( 3 - \frac{5}{u} \right)du \\
& = \int (3)du - \int \left( \frac{5}{u} \right)du \\
& = \int (3)du - 5*\int \left( \frac{1}{u} \right)du \\
& = \int (3)du - 5*ln \lvert u \rvert \\
& = \int (3*u^0)du - 5*ln \lvert u \rvert \\
& = 3*\int (u^0)du - 5*ln \lvert u \rvert \\
& = 3* \frac{1}{0 + 1}*u^{0 + 1} - 5*ln \lvert u∣ \\
& = 3* \frac{1}{1}*u^1 - 5*ln \lvert u \rvert \\
& = 3u - 5*ln \lvert u \rvert \\
u & = x + 3 \\
F(x) & = 3(x + 3) - 5*ln\lvert x + 3\rvert \\
& = 3x + 9 - 5*ln\lvert x + 3\rvert \\
& = 3x - 5*ln\lvert x + 3\rvert + 9 \\
& = \bf [3x - 5*ln\lvert x + 3\rvert + 9 + C] \\
\end{align}$$
This result agrees with both symbolab and wolfram.
Here's the solution approach provided:
Since the degree of the numerator is not less than the degree of the denominator, we must divide. A clever way to do this is shown below.
$$\begin{align}
\int \left( \frac{3x+4}{x+3} \right)dx &= \int \left( \frac{3x+9−5}{x+3} \right)dx \\
&= \int \left( \frac{3(x+3)}{x+3} − \frac{5}{x+3} \right)dx \\
&= \int \left( 3 − \frac{5}{x+3} \right)dx \\
\end{align}$$
Now the integration is straightforward:
$\int \left(3 −\frac{5}{x+3} \right)dx = 3x−5*ln\lvert x + 3\rvert +C$
#
Although the partial fraction solution appears to be mathematically sound, nonetheless it loses real information: "+9"
Furthermore, I realize that if I were to take the derivative of either solution, I would end up with the same results as both constants "+9" and "C" would resolve to 0.
That stated, my concern is that using a method that is inherently prone to losing information is dangerous as it could result in actual computational error especially if instead of an indefinite integral, one was expected to evaluate a definite integral as in determining the actual area under the functions curve, in which case the missing "+9" would result in an erroneous outcome.
Would my argument above justify contesting the method provided?
| Nice work!
About your misgivings: Evaluating the definite integral $F(x) \Big|_b^a$ will yield $$ 3x - 5\ln\lvert x + 3\rvert + 9 + C\Big|_b^a = \Big(3a - 5\ln|a+3| + 9 +C\Big) - \Big( 3b - 5\ln |b + 3| + 9 + C \Big)$$
$$= 3a - 5\ln|a+3| +9 +C - 3b + 5\ln |b + 3| - 9 - C]$$
You can see that upon simplification, we encounter $9 - 9$, $C-C$.
Constants of integration all cancel in the subtraction $F(a) - F(b)$, and as a result, they play no roll in definite integration.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.
Edit :
The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.
| Here is what I have so far, so hopefully someone can take it further..
The above integral has a form of
$$
-\frac{1}{8}\int \frac{df}{dx}\frac{dg}{dx}dx
$$
Where
$$
f(x) = \ln\left(1-2x^2\right)\\
g(x) = \ln\left(\frac{2}{3}\sqrt{\left(2-x^2\right)\left(1+x^2\right)} +1\right)
$$
But as always double check the manipulation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
} |
GCD of polynomials by using Euclid's algorithm
Let $g = x^2 +6x -7$ and $f = x^4 - 1$. Find the GCD of $f$ and $g$.
So I started by evaluating $f/g$ and the result is $q = x^2-6x+43, r = -300x+300$.
I tried to follow the algorithm one step further but I can't see how the result ends up to be $x-1$.
| As you say, euclidean division yields
\begin{align*}
(x^4-1)
&= (x^2 - 6x + 43)(x^2 + 6x - 7) + (-300x+300)\\
&= q(x)\cdot(x^2 + 6x - 7) + r(x)
\end{align*}
Then $\gcd(x^4-1,\,x^2+6x-7)=\gcd(x^2+6x-7, -300x+300)$. You can apply again euclidean division to conclude
$$\gcd(x^4-1,\,x^2+6x-7) = -300x+300$$
But you already knew this. What you may have missed is that the greatest common divisor is unique up to units. In particular
$$-300x+300 = -300\cdot(x-1)$$
Now, since 300 is a unit, we can say:
$$\gcd(x^4-1,\,x^2+6x-7) = x-1$$
The point is that the constant term is not important. You can divide by $-300$ and you still have a greatest common divisor.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{a+b+c}{2}$ If $a,b,c$ are positive real numbers, not all equal, prove that $$\dfrac{bc}{b+c}+\dfrac{ca}{c+a}+\dfrac{ab}{a+b}<\dfrac{a+b+c}{2}$$
Progress:
$\frac{bc}{b+c}-\frac{b+c}{2}=$
| Since $(b+c)^2 \geq 4bc$, we have
$$\dfrac{bc}{b+c} \leq \dfrac{b+c}4$$
Use the same inequality for the rest and obtain what you want.
| {
"language": "en",
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"source": "stackexchange",
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Range of values of $t$ for which $ 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ Calculation of Range of values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ where $\displaystyle t \in \left[-\frac{-\pi}{2}\;,\frac{\pi}{2}\right]$
$\bf{My\; Try::}$ Given $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\Rightarrow 6\sin t\cdot x^2-4\sin t \cdot x-2\sin t = 1-2x+5x^2$
$\displaystyle \Rightarrow (6\sin t-5)x^2+2(1-2\sin t)x-(2\sin t+1) = 0$ , Now for calculation of value of $2\sin t\;,$
equation must have real roots. So $\bf{D\geq 0}$.
So $4(1-2\sin t)^2+4(2\sin t+1)\cdot (6\sin t-5)\geq 0$
$\displaystyle (1-2\sin t)^2+(2\sin t+1)\cdot(6\sin t-5)\geq 0$
So we get $16\sin^2 t-8\sin t-4\geq 0\Rightarrow 4\sin^2 t-2\sin t-1\geq 0$
Now HGow can I solve after that, Help me
Thanks
| Hint: If we assume that $x$ can be any real number, then we know that $2\sin t$ can take values between -2 and 2. So the problem is then to find what possible values of the right hand side are, assuming it can't take all possible values. Given those, work out the arcsin of the values and the possible range of $t$ from that. You can factorize $3x^2-2x-1=(3x+1)(x-1)$ but $1-2x+5x^2$ doesn't have any real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse
(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\quad ;u_n > 0 \\
\Leftrightarrow \ln \left( {u_n } \right) = \ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right)^{\frac{1}{n}} \\
\Leftrightarrow \ln \left( {u_n } \right) = \frac{1}{n}\ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right) \\
\Leftrightarrow \ln \left( {u_n } \right) = \frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n} \\
\Leftrightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\ln \left( {u_n } \right)} \right) = \mathop {\lim }\limits_{n \to + \infty } \left[ {\frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n}} \right] = 0 \\
\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1 \\
\end{array}
(2)
$$
\begin{array}{l}
t_n = \frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }} \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{1 \times 3 \times \cdots (2n - 1)(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{{1 \times 3 \times \cdots (2n - 1)}} \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{1} \\
= \mathop {\lim }\limits_{n \to + \infty } \left( {2\left( {\frac{n}{{n + 1}}} \right)^{n + 1} + \frac{{n^n }}{{\left( {n + 1} \right)^{n + 1} }}} \right) = 2e^{ - 1} ;\quad \left( {\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{x}{n}} \right)^n = e^x } \right) \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{\frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }}}} = 2e^{ - 1} \\
\end{array}
$$
(3) $$\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$$
$$\left( 4 \right)\mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = !?
$$
| You can rewrite the definition of $v_n$ as
$$
v_1=2\\[2ex]
v_{n+1}=\frac{(2n+1)(2n+2)}{n}v_n
$$
so
$$
\frac{v_{n+1}}{v_n}=\frac{(2n+1)(2n+2)}{n}
$$
Another way to see this is noting that
$$
v_n=\frac{(2n)!}{(n-1)!}
$$
For $t_n$ you have
\begin{align}
t_n
&=\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{n^n}\\
&=\frac{1}{n^n}
\frac{(1\cdot3\cdot\ldots\cdot(2n-1))(2\cdot 4\cdot\ldots\cdot 2n)}
{2\cdot 4\cdot\ldots\cdot 2n}\\
&=\frac{1}{n^n}\frac{(2n)!}{2^n\cdot n!}\\
\end{align}
so
$$
\frac{t_{n+1}}{t_n}=
\frac{1}{(n+1)^{n+1}}\frac{(2n+2)!}{2^{n+1}\cdot (n+1)!}
n^n\frac{2^n\cdot n!}{(2n)!}
=\frac{1}{2}\frac{n^n}{(n+1)^{n+1}}\frac{(2n+2)(2n+1)}{n+1}
$$
Thus we need to look at
$$
\lim_{n\to\infty}\frac{n^n}{(n+1)^{n+1}}(2n+1)
$$
and it's easier doing the limit of the inverse:
$$
\lim_{n\to\infty}\frac{(n+1)^{n+1}}{n^n}\frac{1}{2n+1}=
\lim_{n\to\infty}\frac{(n+1)^n}{n^n}\frac{n+1}{2n+1}=
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\frac{n+1}{2n+1}=\frac{e}{2}
$$
so your given limit is $2/e$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $ \forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0 $ How to proof simply that
$$
\forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0
$$
| $x^2+y^2+(x-1)(y-1)>0\iff x^2+y^2+xy-x-y+1>0\iff2x^2+2y^2+2xy-2x-2y+2>0\iff(x-1)^2+(y-1)^2+(x+y)^2>0$ which is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037503",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Eisenstein integers and applications to Diophantine equations Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?
I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.
| It is known that the Eisenstein integers $\mathbb{Z}[\omega]$ is an unique factorization domain and it has six units
$$\pm 1, \pm \omega, \pm \omega^2$$
Over $\mathbb{Z}[\omega]$, the numbers $7, 13, 19$ factorize into its prime factors as
$$\begin{cases}
7 &= (3 + \omega)(3 + \omega^2)\\
13 &= (4 + \omega)(4 + \omega^2)\\
19 &= (5 + 2\omega)(5 + 2\omega^2)
\end{cases}$$
This mean if we want to factorize $1729 = 7 \times 13 \times 19$ over $\mathbb{Z}[\omega]$ as
$$1729 = ( x + y\omega )(x + y\omega^2) = x^2 - xy + y^2
\quad x, y \in \mathbb{Z}
$$
the corresponding factor $x + y\omega$ must have the form
$$x + y\omega = u A B C\quad\text{ with }\quad
\begin{cases}
A &= 3 + \omega &\text{or}& 3 + \omega^2\\
B &= 4 + \omega &\text{or}& 4 + \omega^2\\
C &= 5 + 2\omega &\text{or}& 5 + 2\omega^2
\end{cases}
$$
and $u$ is one of above six units.
There are 8 possible choices of $A,B,C$. For each choice of $A,B,C$,
multiply by one of the six units allow one to obtain an pair of $x,y$
that satisfies $x \ge y \ge 0$:
*
*$ABC = (3+\omega)(4+\omega)(5+2\omega) = 43+40\omega$.
*$ABC = (3+\omega)(4+\omega)(5+2\omega^2) = 45+8\omega$.
*$ABC = (3+\omega)(4+\omega^2)(5+2\omega) = 48+23\omega$.
*$ABC = (3+\omega)(4+\omega^2)(5+2\omega^2) = 32-15\omega \implies -\omega^2 ABC = (47+32\omega)$
*$ABC = (3+\omega^2)(4+\omega)(5+2\omega) = 47+15\omega$.
*$ABC = (3+\omega^2)(4+\omega)(5+2\omega^2) = 25-23\omega \implies -\omega^2 ABC = 48+25\omega$
*$ABC = (3+\omega^2)(4+\omega^2)(5+2\omega) = 37-8\omega \implies -\omega^2 ABC = 45+37\omega$
*$ABC = (3+\omega^2)(4+\omega^2)(5+2\omega^2) = 3-40\omega \implies -\omega^2 ABC =
43+3\omega$
As a result, there are $8$ pairs of $(a,b)$ that solves the original problem:
$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the determinants of the given matrices Consider scalars $a,b,c,d,e,f$ such that
$\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 1 & e\\
c & 1 & f\\ \end{array} \right) = 7$ and $\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 2 & e\\
c & 3 & f\\ \end{array} \right) = 11$
What is $\det\left( \begin{array}{ccc}
a & 3 & d\\
b & 3 & e\\
c & 3 & f\\ \end{array} \right)$ and $\det\left( \begin{array}{ccc}
a & 3 & d\\
b & 4 & e\\
c & 5 & f\\ \end{array} \right)$?
For the first one, I was thinking that the determinant would be $3\cdot 7 = 21$, and I think the determinant of the second one would still just be $11$ (since it's just a multiple of the second determinant matrix). Does this seem right? How can I solve these kinds of problems?
| hint: $\left(\begin{array}{l} 3 \cr 4 \cr 5\end{array} \right) = 2 \left( \begin{array}{l} 1 \cr 1 \cr 1 \end{array} \right) + \left( \begin{array}{l} 1 \cr 2 \cr 3\end{array}\right).$
determinant is a linear function of every column.
| {
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"question_score": "3",
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Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$ It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed up. And I found that the rational solutions of $w^2+x^2+y^2=1$ is
$$w = \frac{2s}{1+s^2+t^2}$$
$$x = \frac{2t}{1+s^2+t^2}$$
$$y = \frac{1-s^2 - t^2}{1+s^2+t^2}$$
where $s$ and $t$ are both rational numbers. From this I found out that
$$(2mnq^2)^2 + (2pqn^2)^2 + (n^2q^2 - m^2q^2-n^2p^2)^2 = (n^2q^2 + m^2q^2+n^2p^2)^2.$$
But it helps little because I cannot prove that the solutions have to be in such form.
Can any one help?
| In General it is possible in various ways to write the solution to this equation:
$$x^2+y^2+z^2=q^2$$
I like such kind.
$$x=2a^2s^2-2abs^2\pm{2apbs}$$
$$y=2a^2s^2+2abs^2\pm2apbs$$
$$z=p^2b^2-a^2s^2+s^2b^2\pm2apbs$$
$$q=p^2b^2+3a^2s^2+s^2b^2\pm2apbs$$
If you want you can write infinitely many formulas are not the problem. You can even choose a particular form. It is of interest how to solve equations such as this:
$$ax^2+bxy+cxz+dy^2+jyz+rz^2=wq^2$$
$a,b,c,d,j,r,w - $ any specified coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding $\lim_{x\rightarrow 0}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}$ We know that $$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\tan({x}/{2})$$ so we can change the above function to another form as follow $$\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=\frac{(x/2)}{\tan(x/2)}$$ If we use the Taylor series of $\tan(x/2)$ we will get
$$\frac{(x/2)}{\tan(x/2)}=\frac{(x/2)}{\frac{x}{2}+\frac{x^3}{24}+\frac{x^5}{240}+....}=\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}$$ Now we will find the limit for the two same functions
$$\lim_{x\rightarrow 0^{+}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^{-}}\frac{1}{1+\frac{x^2}{12}+\frac{x^4}{120}+...}=1$$
$$\lim_{x\rightarrow 0^+}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=1$$
$$\lim_{x\rightarrow 0^-}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=-1$$
My question is "Which solution is right?"
| You are comparing two different functions because
$$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\sqrt{\tan^2\left(\frac{x}{2}\right)} = \Bigg|\tan\left(\frac{x}{2}\right)\Bigg| \not = \tan\left(\frac{x}{2}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
| $$\left(\frac{1}{2}\right)^{\frac{1}{3}}=\frac{\sqrt[3]1}{\sqrt[3]2}=\frac1{\sqrt[3]2}$$
$$\left(\frac{1}{3}\right)^{\frac{1}{2}}=\frac{\sqrt1}{\sqrt3}=\frac1{\sqrt3}$$
Now it is obvious that
$$\sqrt[3]2<\sqrt3$$
Thus
$$\frac1{\sqrt[3]2}>\frac1{\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 3
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Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$.
Is there an easy way to obtain this closed-form?
Also, if there are any good references for binomial coefficient identities like these I'd appreciate it. I searched some but did not find any similar to this.
| One way is by considering generating functions.
Denoting $\displaystyle a_n = \sum_{k=0}^{n} 2^k{n-k\choose k}$
(where, we use the notation $\displaystyle \binom{n}{m} = 0$ for $m > n$)
$\displaystyle \begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty x^n\sum_{k=0}^n2^k\binom{n-k}{k} =\sum_{k=0}^\infty\sum_{n=k}^\infty2^kx^n\binom{n-k}{k}=\sum_{k=0}^\infty\sum_{n=0}^\infty2^kx^{n+k}\binom{n}{k}\\
&=\sum_{k=0}^\infty2^kx^k\frac{x^k}{(1-x)^{k+1}}=\frac{1}{1-x}\sum_{k=0}^\infty2^k\left(\frac{x^2}{1-x}\right)^k\\
&=\frac{1}{1-x}\frac{1}{1-\frac{2x^2}{1-x}}\\
&=\frac{1}{1-x-2x^2}\end{align}$
Using partial fractions to decompose:
$\displaystyle \frac{1}{1-x-2x^2} = \frac{1}{3}\left(\frac{2}{1-2x}+\frac{1}{1+x}\right) = \sum_{n=0}^{\infty} \frac{2^{n+1}+(-1)^n}{3}x^n$
Thus, $\displaystyle a_n = \frac{2^{n+1}+(-1)^n}{3}$
| {
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"url": "https://math.stackexchange.com/questions/1042028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Number theory: $x^y + 1 = y^x$ Today a friend told me the equality:
$2^3 + 1 = 3^2$, and I wondered if there exist more solutions to the general problem $$x^y + 1 = y^x$$ where $x$ and $y$ are integers. Some research led me to the result for $x^y = y^x$, which has no integer solutions except $x = 2$ and $y = 4$ (assumed $x\neq y$). Is this a related result, or do more integer solutions exist?
| Consider function $f:(0,\infty)\rightarrow \mathbb{R}, f(x)=(1+\frac{n}{x})^x$ which is strictly increasing and $f(x)<e^n$.
I. $x<y, y=x+n, n>0$, integer
$$\begin{align} x^y+1=y^x
\implies & x ^{x+n}+1=(x+n)^x \\
\implies & x^n+\frac{1}{x^x}=(1+\frac{n}{x})^x<e^n \\
\implies & x<e \\
\implies & x=1, x=2\end{align}$$
For $x=1 \implies y=2$.
For $x=2 \implies 2^y+1= y^2 \implies y=3$
II. $x>y, x=y+k, k>0$, integer, using the idea of the situation I obtain the equation has no solution other than $(1,0)$.
The same method can solve equations:
*
*$x^y=y^x$
*$x^y+y=y^x+x$
*$x^y+x=y^x+y$
where $x>0, y>0$ integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Investigate convergence of the following series Investigate the corvergence of the following series?
$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13}+\frac{1}{15}-\frac{1}{8}+ \ldots$
$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}+ \ldots$
I need a hit about which cirterion to use in each of those cases and i do not know how to find a general term of those twos sequence?
| For the first one, you have consecutive triples
$$\frac1{4n-1}-\frac1{2n}+\frac1{4n+1}=\frac{1}{2n(4n-1)(4n+1)}\sim\frac1{32n^3}$$
The sum is thus convergent.
For the record, the denominators of the terms in the sum are, up to sign, this sequence in OEIS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
I need help with a limit problem I need help with the following limit:
$\lim_{n \to \infty}\left [ (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right ) \right ],a\in \mathbb{N}$
I am completely lost.
| Let us consider $$A= (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right )$$ In a first time, let us focus on $(1+x)^a$ and use the generalized binomial theorem for generating a Taylor-like series when $x$ is small. So $$(1+x)^a=1+a x+\frac{1}{2} (a-1) a x^2+\frac{1}{6} (a-2) (a-1) a x^3+O\left(x^4\right)$$ Now, replace $x$ by $\frac{2}{n^2}$ $$\left ( 1+\frac{2}{n^2} \right )^a - 1 =\frac{2 a}{n^2}+\frac{2 (a-1) a}{n^4}+\frac{4 (a-2) (a-1) a}{3 n^6}+\cdots$$ Multiply by $(n^2+1)$, expand and simplify; we arrive to $$A=2 a+\frac{2 a^2}{n^2}+\frac{2 \left(2 a^3-3 a^2+a\right)}{3
n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ which shows the limit and how it is approached.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$ How can I prove this equality?
$$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$
| Let $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=y\implies x+y=y^2\implies y^2-y-x=0\implies y=\dfrac{1\pm\sqrt{1+4x}}2$
As $x>1,y>1\implies y=\dfrac{1+\sqrt{1+4x}}2$
$\sqrt{x-\sqrt{x-\sqrt{x-\cdots}}}=z\implies x-z=z^2\iff z^2+z-x=0\implies z=\dfrac{-1\pm\sqrt{1+4x}}2$
As $x>1, z<1\implies z=\dfrac{-1+\sqrt{1+4x}}2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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} |
contractive and find its limit The real sequence $\{P_n\}$ is defined as $p_1$=2, $p_{n+1}$=$\frac{2}{1+p_n}$, n $\in N$. prove that $\{p_n\}$ is contractive, deduce that it converges, and find its limit.
first we need to show that $\frac{2}{3}$ $\leq$ $p_n$$\leq$2
pf:
as $p_1$ = 2, $p_2$=$\frac{2}{3}$,.. so that $\frac{2}{3}$ $\leq$ 1 $\leq$2
now we have $\vert \frac{2}{1+p_{n+1}} - \frac{2}{1+p_n} \vert$ =...some crazy math here..=
since $p_1$=2 and $p_n\lt$1, for all$ $ $n\geq2$ by induction we have 2$\frac{\vert p_n - p_{n+1}\vert}{(1+p_n)(1+p_{n+1})}$ $\lt$ $\frac{2}{2}$* $\frac{2}{2}$ $\vert p_n - p_{n+1} \vert$ = $\vert p_n - p_{n+1} \vert$ therefore $p_n$ is a contractive sequence.
is that good or need to fix my $\lt$ $\frac{2}{2}$* $\frac{2}{2}$
| You are quite close. If we can show that $p_k\ge \frac{2}{3}$ for all $k$, we will have, by a calculation similar to yours, that
$$|p_{n+1}-p_n|=\left|\frac{2}{1+p_n}-\frac{2}{1+p_{n-1}}\right|=\frac{2|p_n-p_{n-1}|}{(1+p_n)(1+p_{n-1})}.$$
But since $p_k\ge \frac{2}{3}$ for all $k$, it follows that $(1+p_k)(1+p_{n-1})\ge \frac{25}{9}$. This implies that
$$|p_{n+1}-p_n|\le \frac{18}{25}|p_n-p_{n-1}|,$$
which proves contractivenes.
To show that $p_k\ge \frac{2}{3}$ for all $k$, we show the stronger result $\frac{2}{3}\le p_k\le 2$. This is done by induction.
Suppose that $p_k\le 2$. We show that $p_{k+1}\ge \frac{2}{3}$. This is obvious.
Suppose that $\frac{2}{3}\le p_k$. Then $p_{k+1}\le \frac{2}{1+\frac{2}{3}}=\frac{6}{5}\le 2$.
This completes the induction step.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$ My Work
$x = \tan\theta$
$dx = \sec^2\theta d\theta$
$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$
$\int \sec\theta \tan\theta ln(tan\theta)$
$u = \ln(\tan\theta)$
$du = \frac{\sec^2 \theta}{\tan \theta}$
$dv = \sec \theta \tan \theta d\theta$
$v = \sec \theta$
$\sec\theta\ln\tan\theta - \int \frac{\sec^3\theta}{\tan \theta}$
$\sec\theta\ln\tan\theta - \int \sec^2\theta \csc\theta d\theta$
I'm stuck after here. Parts doesn't look particularly appealing. I don't see an easy substitution. Brain is pretty tired at this point. Anyone know what to do?
| You can evaluate the integral without using secant substitution. Here is the way.
Since we know that
\begin{gather*}
\mathrm{d}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}}\mathrm{d} x,
\end{gather*}
we can calculate, by differentiation by parts, as follows,
\begin{align*}
&\quad \frac{x\ln(x)}{\sqrt{x^2-1}} \mathrm{d} x=\ln(x)\mathrm{d} \sqrt{x^2-1}=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{\sqrt{x^2-1}}{x}\mathrm{d} x\\
&= \mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x^2-1}{x\sqrt{x^2-1}}\mathrm{d} x\\
&=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x}{\sqrt{x^2-1}}\mathrm{d} x+\frac{1}{x\sqrt{x^2-1}}\mathrm{d} x\\
&=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)+\frac{1}{x^2\sqrt{1-\left(\frac{1}{x}\right)^2}}\mathrm{d} x\\
&=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)-\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^ 2}}\mathrm{d} \left(\frac{1}{x}\right)\\
&=\mathrm{d}\left(\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)\right),
\end{align*}
hence we have
\begin{gather*}
\int\frac{x\ln(x)}{\sqrt{x^2-1}}\mathrm{d} x=\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)+C.
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $SL(2, 3) $ a subgroup of $SL(2, p)$ for $ p>3$? As the title says, I was wondering whether $SL(2,3)$ is a subgroup of $SL(2,p)$ for $p>3$. I know that it is for $p=5$ (it can be found explicitly using the quaternionic representation), and I have some evidence that it is for the other $p$'s, but I'm not sure how I would go about proving it, or indeed if it's true!
If anybody knows one way or another, please let me know. If it's true, it would be helpful to give a hint about how I might think about proving it, but please don't give a proof as I'd like to figure it out myself.
| $SL_2(3)$ is the semidirect product of the quaternion group
$Q = \langle i,j\mid i^4, j^4, i^2=j^2, i^{-1}ji=j^{-1}\rangle$
with a cyclic group $Z_3 = \langle z\rangle$ of order $3$ via
$i^z = j$ and $j^z = ij$.
Hint: Find a subgroup of $SL_2(p)$ ($p$ any odd prime) isomorphic to
$Q$, and then extend it to full group $SL_2(3)$ by using this
little gem from an answer by Geoff Robinson: The units of the
Hurwitz quaternion
are the quaternions plus 16 elements that can be written as
$\frac{\pm 1\pm i\pm j\pm k}{2}$ (with $k = ij$).
As we are in odd characteristic, $\frac{1}{2}$ exists and we have our
candidate for $SL_2(3)$.
To simplify calculations these two easy observations might be useful:
An element of the form $\frac{\pm 1\pm i\pm j\pm k}{2}$ can be inverted
by flipping the signs of $i$, $j$ and $k$, but keeping $1$'s.
This can be best seen by observing that the mixed terms cancel out, and
only those contributing to the coefficient of $1$ add up.
The coefficient at $1$ of the square of an element of the form
$\frac{\pm 1\pm i\pm j\pm k}{2}$ is $-1$, so one expects elements of
order $3$ to contain the summand $-1$ and those of order $6$ the summand
$+1$.
Proof: Let $p$ be an odd prime.
The square of the matrix
$i = \left(\begin{array}{cc} 0 & 1\\-1& 0\end{array} \right)$ is
$\left(\begin{array}{cc} -1 & 0\\0& -1\end{array} \right)$.
Its order is $4$ and its determinant $1$.
For $j$ let's take an arbitrary
$\left(\begin{array}{cc} a & b\\c& d\end{array} \right) \in SL_2(p)$
with $ad-bc = 1$.
Using the equation $i^{-1}ji = j^{-1}$ we get
$\left(\begin{array}{cc} 0 & -1\\1& 0\end{array} \right)
\left(\begin{array}{cc} a & b\\c& d\end{array} \right)
\left(\begin{array}{cc} 0 & 1\\-1& 0\end{array} \right) =
\left(\begin{array}{cc} -c & -d\\a& b\end{array} \right)
\left(\begin{array}{cc} 0 & 1\\-1& 0\end{array} \right) =
\left(\begin{array}{cc} d & -c\\-b& a\end{array} \right) \stackrel{!}{=}
\left(\begin{array}{cc} d & -b\\-c& a\end{array} \right)$, hence $b=c$.
On the other hand $j^2 = i^2$ yields
$\left(\begin{array}{cc} a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{array}
\right) \stackrel{!}{=}
\left(\begin{array}{cc} -1 & 0\\0& -1\end{array} \right)$,
which holds in case $a = -d$.
So in case $a^2+b^2 = -1$ we get the candidate
$j = \left(\begin{array}{cc} a & b\\b& -a\end{array} \right)$
for $Q = \langle i, j\rangle$.
As there are $\frac{p+1}{2}$ squares in $\mathbb F_{p}$, by the
Cauchy-Davenport theorem
all $p$ elements of $\mathbb F_{p}$ are sums of two squares.
So pick solutions $a$ and $b$ of $a^2+b^2 = -1$, and we found $Q$.
[If $p = -1 \bmod 4$ there exist also solutions with $a=0$ or $b=0$
which we seemed to avoid when choosing $a = -d$ before.]
Defining $z = \frac{-1+i+j+k}{2} = -\frac{(1-j)\cdot(1-i)}{2}$ we get an
element of $SL_2(p)$ of order $3$, as
$det(z) = det(1-j)\cdot det(1-i)\cdot(-\frac{1}{2})^2
= \left|\begin{array}{cc} 1-a & -b\\-b& 1+a\end{array} \right| \cdot
\left|\begin{array}{cc} 1 & -1\\1& 1\end{array} \right| \cdot\frac{1}{4} =
(1-a^2-b^2)\cdot 2\cdot\frac{1}{4} = 1$ and
$z^2 = \frac{-1-i-j-k}{2} = z^{-1}$.
Now $i^z = z^{-1}\cdot i\cdot z =
\frac{-1-i-j-k}{2}\cdot i\cdot\frac{-1+i+j+k}{2} =
\frac{1-i-j+k}{2} \cdot\frac{-1+i+j+k}{2} =
j\cdot\frac{-1-i-j-k}{2}\cdot\frac{-1+i+j+k}{2} = j\cdot z^{-1}\cdot z = j$
and $j^z = z^{-1}\cdot j\cdot z =
\frac{-1-i-j-k}{2}\cdot j\cdot\frac{-1+i+j+k}{2} =
\frac{1+i-j-k}{2}\cdot\frac{-1+i+j+k}{2} =
k\cdot\frac{-1-i-j-k}{2}\cdot\frac{-1+i+j+k}{2} = k\cdot z^{-1}\cdot z = k$
finish the proof.
Final remark: For $p=3$ the proof gives us the semidirect product $Q\rtimes Z_3$ as subgroup of $SL_2(3)$, hence proving the statement in the first sentence of my answer, as both groups have the same order.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
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} |
How to find the coefficient of a power in a power series?? How can I find the coefficient of $x^{80}$ in the power series $$(1+x+x^{2}+x^{3}+x^{4}+\cdots)(x^{2}+x^{4}+x^{6}+x^{8}+\cdots)(1+x^{3}+x^{5})\,?$$
Is there a general method to this?
| We can use the fact that
\begin{align}
1+x+x^2+\dotsb&=\frac{1}{1-x}\\
x^2+x^4+x^6+\dotsb&=\frac{x^2}{1-x^2}
\end{align}
so the expression is
$$
\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}
$$
and we can try doing partial fraction decomposition:
$$
\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}=
x^4+x^3+3x^2+3x+5+\frac{6x^2+2x-5}{(1-x)^2(1+x)}
$$
so we want to decompose the fraction into
$$
\frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x}
$$
which gives $A=3/2$, $B=-25/4$, $C=-1/4$. Therefore your product can be written as
$$
x^4+x^3+3x^2+3x+5
+\frac{3}{2}\sum_{k\ge0}(k+1)x^k
-\frac{25}{4}\sum_{k\ge0}x^k
-\frac{1}{4}\sum_{k\ge0}(-1)^kx^k
$$
and the coefficient of $x^{80}$ can be read directly as
$$
\frac{3}{2}\cdot 81-\frac{25}{4}-\frac{1}{4}=115
$$
Note that this method gives all coefficients.
I've used the formula
$$
\frac{1}{(1-x)^2}=\sum_{k\ge0}(k+1)x^k
$$
that can be deduced by differentiating
$$
\frac{1}{1-x}=\sum_{k\ge0}x^k
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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fixed points for the following system I'm trying to find the fixed point for the system (see document attached) but it seems so hard and I don't know what Im doing wrong. Can somebody help me with this. I need to find the to look for the value where the bifurcation occur (for that process i need to evaluate the jacobian matrix for the system on the fixed points and the looks for the delta, which is really easy) but this is the only part where I'm stuck. Please see document attached
| We need to simultaneously solve:
$$\begin{align*} a(1 - u) - u v^2 = 0 \\ u v^2 -(a + k) v = 0 \end{align*}$$
From the first equation we get:
$$u = \dfrac{a}{a+v^2}$$
Substituting this into the second equation yields:
$$-(a+k) v + \dfrac{a v^2}{a + v^2} = -(a+k)(v)(a + v^2) + av^2 = 0$$
Expanding and simplifying, yields:
$$ -a^2 v-a k v-a v^3+a v^2-k v^3=0$$
The three roots (critical or fixed points) to this cubic are:
$$v = 0, \dfrac{a-\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}, \dfrac{a+\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}$$
Now, we need to find the corresponding $u$ values using either equation. We have:
$$u = \dfrac{a}{v^2 + a}$$
*
*At $v = 0$, we get $u = 1$.
*At $v = \dfrac{a-\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}, u = \dfrac{a+\sqrt{a \left(-4 a^2-8 a k+a-4 k^2\right)}}{2 a}$
*At $v = \dfrac{a+\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}, u = \dfrac{a-\sqrt{a \left(-4 a^2-8 a k+a-4 k^2\right)}}{2 a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Formula for a periodic sequence of 1s and -1s with period 5 I've been playing with periodic sequences of 1s and -1s lately. This is what I came up with:
\begin{eqnarray*}
-(-1)^n& = &1, -1, 1, -1,\ldots\quad\textrm{(Period 2)}\\
\left(-(-1)^n\right)^{\frac{n+2}{2}} & = &1, 1, 1, -1, 1, 1, 1, -1,\ldots\quad\textrm{(Period 4)}\\
\left(\left(-(-1)^n\right)^{\frac{n+2}{2}}\right)^{\frac{n+4}{4}}& = &1, 1, 1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, 1, -1, \ldots\quad\textrm{(Period 8)}
\end{eqnarray*}
One can easily find a similar formula for a periodic sequence with period $ 2^n, n\in\mathbb{N} $.
I also found a formula:
$$
(-1)^{2 \sin\left(\frac{(2n-1)\pi}{6}\right)},
$$
which gives a sequence $ -1, 1, -1, -1, 1, -1,\ldots $ with period 3.
My question is: Is there a formula for a periodic sequence of 1s and -1s with period 5? If there is, what is it?
I know about this formula for a periodic zero and one sequence with period $ N $:
$$
\sum\limits_{k = 1}^N\cos\left(-2\pi\frac{n(k-1)}{N}\right)/N = 0, 0, 0, \ldots, 1.
$$
However, it requires to sum up $ N $ expressions to count the $ n\textrm{th} $ term, which is why I don't like it. I would also like the formula not to contain functions like floor or modulus.
| We have
$$
\frac{4\cos\left(\frac{2\pi k}5\right)+4\cos\left(\frac{4\pi k}5\right)-3}5=
\left\{\begin{array}{}
-1&\text{if }k\not\equiv0\pmod5\\
+1&\text{if }k\equiv0\pmod5\\
\end{array}\right.\tag{1}
$$
Explanation
The roots of $z^5-1$ are $z=e^{2\pi ik/5}$ for $k\in\{0,1,2,3,4\}$. Vieta says that the coefficient of $z^4$ in $z^5-1$ the sum of the roots of $z^5-1$. That is, the sum of the roots is $0$. Taking the real part of the roots yields
$$
1+\cos\left(\frac{2\pi}5\right)+\cos\left(\frac{4\pi}5\right)+\cos\left(\frac{6\pi}5\right)+\cos\left(\frac{8\pi}5\right)=0\tag{2}
$$
When $k\not\equiv0\pmod5$, $k$ is invertible $\bmod5$. Therefore,
$$
\left(\cos\left(\frac{2\pi k}5\right),\cos\left(\frac{4\pi k}5\right),\cos\left(\frac{6\pi k}5\right),\cos\left(\frac{8\pi k}5\right)\right)\tag{3}
$$
is a permutation of
$$
\left(\cos\left(\frac{2\pi}5\right),\cos\left(\frac{4\pi}5\right),\cos\left(\frac{6\pi}5\right),\cos\left(\frac{8\pi}5\right)\right)\tag{4}
$$
Therefore,
$$
1+\cos\left(\frac{2\pi k}5\right)+\cos\left(\frac{4\pi k}5\right)+\cos\left(\frac{6\pi k}5\right)+\cos\left(\frac{8\pi k}5\right)=0\tag{5}
$$
when $k\not\equiv0\pmod5$. When $k\equiv0\pmod5$,
$$
1+\cos\left(\frac{2\pi k}5\right)+\cos\left(\frac{4\pi k}5\right)+\cos\left(\frac{6\pi k}5\right)+\cos\left(\frac{8\pi k}5\right)=5\tag{6}
$$
Since $\cos(x)$ is an even function with period $2\pi$, $(5)$ and $(6)$ become
$$
1+2\cos\left(\frac{2\pi k}5\right)+2\cos\left(\frac{4\pi k}5\right)
=\left\{\begin{array}{}
0&\text{if }k\not\equiv0\pmod5\\
5&\text{if }k\equiv0\pmod5\\
\end{array}\right.\tag{7}
$$
Equation $(1)$ is simply a scaled and translated version of $(7)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining the last two digits of $229^{10} +37^{10}$ Determine the last two digits of: $229^{10}+37^{10}.$
I do not want to use the Euler-totient function or the carmichael function please!
Thanks
| We need $229^{10}+37^{10}\pmod{100}$
As $100=4\cdot25$ where $(4,25)=1$
$229\equiv1\pmod4\implies229^{10}\equiv1;37\equiv1\pmod4\implies37^{10}\equiv1$
$\implies229^{10}+37^{10}\equiv1+1\pmod4\equiv2\ \ \ \ (1)$
Again, $229\equiv4\pmod{25}\implies229^{10}\equiv4^{10}$
But $4^{10}=(2^2)^{10}=2^{20}$
and $37\equiv12\pmod{25}\equiv2^23\implies37^{10}\equiv(2^23)^{10}\equiv2^{20}3^{10}$
$\implies229^{10}+37^{10}\equiv2^{20}(1+3^{10})\pmod{25}$
Again, $3^3=27\equiv2\pmod{25}\implies3^{10}=3(3^3)^3\equiv3(2)^3\pmod{25}\equiv-1$
$\implies229^{10}+37^{10}\equiv1-1\pmod{25}\equiv0\ \ \ \ (2)$
Now apply Chinese Remainder Theorem on $(1),(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the limit of function using Taylor series Good evening,
I'm somehow stuck on solving some easy exercises :
$$\lim_{x\to\infty} x^{3/2}\bigl(\sqrt{x+1}+\sqrt{x-1}-2\,\sqrt{x}\bigr)$$
| $$\sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x}+\sqrt{x+1}},\qquad \sqrt{x-1}-\sqrt{x}=-\frac{1}{\sqrt{x}+\sqrt{x-1}},$$
hence:
$$\sqrt{x+1}+\sqrt{x-1}-2\sqrt{x}=\frac{\sqrt{x-1}-\sqrt{x+1}}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x-1})}=\frac{-2}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x-1})(\sqrt{x-1}+\sqrt{x+1})}\approx\frac{-2}{(2\sqrt{x})^3}$$
and the limit is $\color{red}{-\frac{1}{4}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A faster way of calculating this determinant? I'm doing a problem involving Cramer's rule, and one of the determinants I have to work with is as follows:
\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3
\end{vmatrix}
So I started off by getting the matrix to a triangular matrix so I can just take the product of the diagonal cells by doing this:
\begin{equation}
R_1 \times-c + R_2
\end{equation}
\begin{equation}
R_3 \times -\frac{1}{c^3} + R_1
\end{equation}
\begin{equation}
R_2 \times(\frac{1-\frac{b^3}{c^3}}{b-c})+R_1
\end{equation}
I then got this matrix
\begin{vmatrix}
((1-\frac{a^3}{c^3})-(a-c)(\frac{1-\frac{b^3}{c^3}}{b-c}))&0&0\\
a-c&b-c&0\\
a^3&b^3&c^3\\
\end{vmatrix}
By summing the diagonal cells, I got this as a final answer:
\begin{equation}
(c-a)(a-b)(b-c)(a+b+c)
\end{equation}
However, it was a lengthy process, and I can't help but thinking this is not the type of calculations I can afford the time in a written exam, especially if I have to repeat this four times solving three linear equations with Cramer's rule.
I'd really appreciate it if you have any thoughts as to speed this process up.
Just as an added detail, the original question was (translated from Chinese):
With regards to the following set of equations
\begin{equation}
x + y + z = 1\\
ax + by + cz = d\\
a^3x + b^3y + c^3z = d^3
\end{equation}
*
*Under what conditions can Cramer's rule be used?
*Please solve the set of equations with Cramer's rule.
For the first question, the immediate thought that pops up is that
\begin{equation}
det(A) ≠ 0
\end{equation}
For the second question, it's simply
\begin{equation}
\frac{\begin{vmatrix}A_1\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}},
\frac{\begin{vmatrix}A_2\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}},
\frac{\begin{vmatrix}A_3\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}}
\end{equation}
which is the part I'm having trouble solving quickly.
Again, appreciate any hints or thoughts on this. Thanks.
| You can compute the determinant of a generic $3\times 3$ matrix using a neat trick, if we have:
$$\mathbf{A}=\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$
Then we have the sum of the diagonals (highlighted in green) minus the sum of the antidiagonals (highlighted in red) as follows:
$$\det(\mathbf{A}) = \underbrace{aei}_{\begin{pmatrix}{\color{green} a} & b & c \\ d & {\color{green} e} & f \\ g & h & {\color{green} i}\end{pmatrix}} + \underbrace{bfg}_{\begin{pmatrix}a & {\color{green}b} & c \\ d & e & {\color{green} f} \\ {\color{green} g} & h & i\end{pmatrix}} + \underbrace{cdh}_{\begin{pmatrix}a & b & {\color{green} c} \\ {\color{green} d} & e & f \\ g & {\color{green} h} & i\end{pmatrix}} - \underbrace{afh}_{\begin{pmatrix}{\color{red} a} & b & c \\ d & e & {\color{red} f} \\ g & \color{red} h & i\end{pmatrix}} - \underbrace{bdi}_{\begin{pmatrix}a & \color{red} b & c \\ \color{red} d & e & f \\ g & h & \color{red} i\end{pmatrix}} - \underbrace{ceg}_{\begin{pmatrix}a & b & \color{red}c \\ d & \color{red} e & f \\ \color{red} g & h & i\end{pmatrix}}$$
Thus if we have your matrix:
$$\mathbf{M}=\begin{pmatrix}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{pmatrix}$$
Then:
$$\det(\mathbf{M})=bc^{3}+ca^{3}+ab^{3}-cb^{3}-ac^{3}-ba^{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$ Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$. For which $x$ does it converge to $f(x)$?
This is what I have been able to do so far
$$f(x)=\arctan(x)=\int \frac{1}{1+x^2}\, dx=\int \frac{1}{1+x^2+1-1}\,dx$$
Any hints please???
| I bet the OP was asking to find the values of the derivatives in $x=1$.
This is clearly the same as finding the derivatives in $x=1$ for $f(x)=\frac{1}{x^2+1}$. Since:
$$\frac{1}{x^2+1}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right),\tag{1}$$
we have that:
$$\frac{d^n}{dx^n}\frac{1}{x^2+1}=\frac{n!(-1)^n}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)\tag{2}$$
so:
$$ f^{(n)}(1) = n!(-1)^n\cdot \Im\left(\frac{1}{(1-i)^{n+1}}\right)\tag{3}$$
and since $(1-i)=\sqrt{2}e^{-i\pi/4}$ we have:
$$ f^{(n)}(1) = n! (-1)^n \sin\left(\frac{\pi(n+1)}{4}\right)2^{-\frac{n+1}{2}}\tag{4}$$
and:
$$ \frac{1}{x^2+1}=\sum_{k=0}^{+\infty}\frac{f^{(k)}(1)}{k!}(x-1)^k = \sum_{k=0}^{+\infty} (-1)^k \sin\left(\frac{\pi(k+1)}{4}\right)2^{-\frac{k+1}{2}}(x-1)^k.\tag{5} $$
The radius of convergence is given, as usual, by the distance from the closest singularity, the simple pole in $x=i$, giving $\rho=\sqrt{2}$. Integrating termwise we get:
$$ \arctan x = \frac{\pi}{4}+\sum_{k=0}^{+\infty} \frac{(-1)^k}{k+1} \sin\left(\frac{\pi(k+1)}{4}\right)2^{-\frac{k+1}{2}}(x-1)^{k+1}.\tag{6} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
More rigorous method for this elementary problem? The problem is:
Find all real values of $x$ such that $$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x=2\sqrt{3}$$
One solution I received was as follows:
$5+2\sqrt{6}$ can be expressed as $(\sqrt{3}+\sqrt{2})^2$, and $5-2\sqrt{6}$ as $(\sqrt{3}-\sqrt{2})^2$. This means we want to solve $$(\sqrt{3}+\sqrt{2})^{2x}+(\sqrt{3}-\sqrt{2})^{2x}=2\sqrt{3}$$
for $x$. Note that $$(\sqrt{3}+\sqrt{2})+(\sqrt{3}-\sqrt{2})=2\sqrt{3}$$
so $x=\dfrac{1}{2}$ is a solution.
The problem with the solution is that it doesn't prove that $\dfrac{1}{2}$ is the only solution, not to mention its absolute inelegancy. Is there a more mathematically rigorous way of doing this?
I don't mind if higher algebra is used. I'm currently attempting to self-teach myself group theory, so including some of that in a solution would do me good.
| Answer is attributed to ah-huh-moment's answer.
Since $5+2\sqrt{6}=\dfrac{1}{5-2\sqrt{6}}$ letting $y=5-2\sqrt{6}$ gives
$$\frac{1}{y^x}+y^x=2\sqrt{3}$$
which becomes
$$y^{2x}-2\sqrt{3}y^x+1=0$$
so
$$y^x=\sqrt{3}\pm \sqrt{2}$$
Plugging $y=5-2\sqrt{6}$ back in and taking $\log$ of both sides give
$$x=\log_{5-2\sqrt{6}}(\sqrt{3}\pm\sqrt{2})=\pm \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| Equation 2 is the equation of an ellipse. Basically, any equation
\begin{eqnarray}
Ax^2 + Bxy + Cy^2 + Dx + Ey + C = 0
\end{eqnarray}
can be translated, by a change of variable, into:
\begin{eqnarray}
\left(\frac{\tilde{x}}{a}\right)^2 + \left(\frac{\tilde{y}}{b}\right)^2 = 1
\end{eqnarray}
Leading to:
\begin{eqnarray}
\tilde{x} & = & a\cos(\theta) \\
\tilde{y} & = & b\sin(\theta) \\
\tilde{y} & = & b\sqrt{1 - \frac{\tilde{x}^2}{a^2}}
\end{eqnarray}
Now, for the transformation from $(x, y)$ to $(\tilde{x}, \tilde{y})$, we need to find the rotation angle $\Theta$ and translation $(0,0) \mapsto (x_c, y_c)$ so that the ellipse is centered on $0$ and its minor and major axis are aligned with the axis of our base. Back to the generic equation of an ellipse:
\begin{eqnarray}
A & = & a^2 \sin(\Theta)^2 + b^2 \cos(\Theta)^2 \\
B & = & 2(b^2 - a^2)\sin(\Theta)\cos(\Theta) \\
C & = & a^2 \cos(\Theta)^2 + b^2 \sin(\Theta)^2 \\
D & = & -2Ax_c - By_c \\
E & = & -Bx_c - 2 Cy_c \\
F & = & Ax_c^2 + Bx_cy_c + Cy_c^2 - a^2b^2 \\
\end{eqnarray}
With $D$ and $E$, you can find $(x_c, y_c)$. Here, there are no term $D$ or $E$, because your ellipse is already centered on $0$. $A+C = a^2 + b^2$ and $F = a^2b^2$, so you deduce $a$ and $b$. Then $B= (b^2-a^2)sin(2\Theta)$, and you find $\theta$. So
\begin{eqnarray}
\tilde{x} & = & x\cos(\Theta) - y\sin(\theta) \\
\tilde{y} & = & x\sin(\Theta) + y\cos(\theta)
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| We can use $$(x+y+z)^2=2\sum xy+(x^2+y^2+z^2)\ \ \ \ (1)$$
$$\sum x^3-3xyz=(x+y+z)(\sum x^2-\sum yz)$$
$$\iff\sum x^3-3xyz=(x+y+z)\{(x+y+z)^2-3\sum xy\}\ \ \ \ (2)$$
to find $xyz=c$(say)$,\sum xy=b$(say)
Then $x,y,z$ are the roots of $t^3-(1)t^2+bt-c=0$
$\implies t^4=t^3-bt^2+ct$
$\implies\sum x^4=\sum x^3-b\sum x^2+c\sum x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ Show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea:
Let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+(a^2+1)(b^2+1)\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-\dfrac{4ab(a+b+a^2b+b^2a)^2}{4ab}\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-(a+b+a^2b+b^2a)^2\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a+b)^2+(ab-1)^2+(a+b)^2(1+ab)^2
\end{align*}
so I think this idea is not good, maybe this have good methods,because this result is nice
| Notice for $|t| < 1$, $\frac{1}{\sqrt{1 - 2xt + t^2}}$ is the generating function for the Legendre polynomials:
$$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{*1}$$
It is known that for $x \in [-1,1]$, $|P_n(x)| \le P_n(1) = 1$. This means as long as $t$ is fixed and $|t| < 1$, the absolute values of the $n^{th}$ term is bounded by a geometric series $|t|^n$. As a result, the convergence in $(*1)$ is absolute and uniform for $x$ over $[-1,1]$.
Recall the orthogonality relation for Legendre polynomials:
$$\int_{-1}^1 P_n(x) P_m(x) dx =
\begin{cases} \frac{2}{2n+1}, & n = m\\ 0, & n \ne m\end{cases}
$$
We find
$$\begin{align}
&\int_{-1}^1 \frac{dx}{\sqrt{1-2ax+a^2}\sqrt{1-2bx+b^2}}\\
=& \int_{-1}^1 \left(\sum_{n=0}^\infty P_n(x) a^n\right)\left(\sum_{m=0}^\infty P_m(x) b^m\right) dx\\
=& \sum_{n=0}^\infty \sum_{m=0}^\infty a^n b^m \int_{-1}^1 P_n(x) P_m(x) dx\\
=& \sum_{n=0}^\infty \frac{2}{2n+1} (ab)^n\\
=& \sum_{k=0}^\infty \frac{1}{(k+1)}\left( (\sqrt{ab})^k + (-\sqrt{ab})^k \right)\\
=& - \frac{ \log(1 - \sqrt{ab})}{\sqrt{ab}} + \frac{ \log(1 + \sqrt{ab})}{\sqrt{ab}}\\
=& \frac{1}{\sqrt{ab}}\log\left(\frac{1+\sqrt{ab}}{1-\sqrt{ab}}\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Spaces of polynomials
Let $p_{1}(x)=-2x+2$, $p_{2}(x)=x+2$, $p_{3}(x)=x^{2}+2x+3$, $p_{4}(x)=x^{2}-x+3$.
a) From the above four polynomials, determine a linearly independent subset that spans the polynomials.
b) What is the dimension of the (sub)space that this basis spans?
c) Does this basis span the space of the second degree polynomials? If not, describe the subspace it spans in terms of the coefficients of a polynomial that belongs to this subspace?
I said since the Wronskian is 0, the functions are linearly dependent. We see three non-zero rows there. This means three of the functions form a linearly independent set of functions. So, {$p_{1}(x),p_{2}(x),p_{3}(x)$} is subset that spans the space of second degree polynomials and is a basis for this space. The dimension is three since there are three functions that form the basis. But parts a and c yelling at me: "Wrong!". Can somebody help me?
| (a) Let $\vec p_i$ be the coefficient vector of $p_i(x)$. Then,
$$M=
\begin{bmatrix}
\vec p_1 & \vec p_2 & \vec p_3 & \vec p_4
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 1 & 1 \\
-2 & 1 & 2 & -1 \\
2 & 2 & 3 & 3
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}
.$$
Columns $1$, $2,$ and $3$ are pivot columns $\Rightarrow T =\{p_1, p_2, p_3 \}$ is a linearly independent subset that spans $\{p_1, p_2, p_3, p_4 \} \Rightarrow T$ forms a basis for $\operatorname{Span} \{p_1, p_2, p_3, p_4 \}$.
(b) $\dim (\operatorname{Span} T) = \operatorname{Rank} M = 3.$
(c) Let $p \in \mathbb{P}^2$. Then, $p$ is of the form $ax^2+bx+c$. The corresponding system to solve is
$$
\begin{bmatrix}
\vec p_1 & \vec p_2 & \vec p_3 & \vec p
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 1 & a \\
-2 & 1 & 2 & b \\
2 & 2 & 3 & c
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 0 & 0 & \frac{1}{6} a + \frac{1}{6}c - \frac{1}{3} b \\
0 & 1 & 0 & - \frac{5}{3} a + \frac{1}{3}c + \frac{1}{3} b \\
0 & 0 & 1 & a
\end{bmatrix}.
$$
We can see that the system is consistent, and $$p = \left( \frac{1}{6} a + \frac{1}{6}c - \frac{1}{3} b \right)p_1 +
\left( -\frac{5}{3} a + \frac{1}{3}c + \frac{1}{3} b \right) p_2 + a p_3. $$
Hence, $\operatorname{Span} T = \mathbb{P}^2$. But, we showed that $T$ was linearly independent $\Rightarrow T$ forms a basis for $\mathbb{P}^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do I get $0.098765432098765432...$ when I divide $8$ by $81$? I got this remarkable thing when I divided $16$ by $162$, or, in a simplified version, $8$ by $81$. It's $0.098765432098765432\cdots$, or more commonly known as $0.\overline{098765432}$, with all the one-digit numbers going backwards...except for $1$. Yeah, it's missing the $1$. One, how do I get this remarkable outcome and two, why is it missing the $1$?
| More generally, in base $b$ (any integer $>1$)
$$ \dfrac{b-1}{b^2} + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} =
\dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}$$
so that
$$ \eqalign{
\dfrac{b-1}{b^2}& + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} +
\dfrac{0}{b^{b}} + \dfrac{b-1}{b^{b+1}} + \dfrac{b-2}{b^{b+3}} + \ldots
= \sum_{j=0}^\infty b^{-j(b-1)} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}\cr
&= \dfrac{b^{b-1}}{b^{b-1}-1} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}= \dfrac{(b-2)}{(b-1)^2}}$$
What you have is the case $b=10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| Let $ x = \sin \theta$
$$
\int \frac{1}{\sin \theta} \cdot \sqrt \frac{1-\sin \theta}{1 + \sin \theta} \cdot \cos \theta \; d \theta \;\;= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \sqrt\frac{(1-\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)} \; d\theta
\\
= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \frac{1-\sin\theta}{\cos\theta}\; d\theta \;\; = \;\; \int (\csc\theta \; - 1) \; d\theta
$$
Then convert back in terms of x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
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What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 \ \text{mod} \ 7= & &6 \\
555 \ \text{mod} \ 7= & &2 \\
5555 \ \text{mod} \ 7= & &4 \\
55555 \ \text{mod} \ 7= & &3 \\
555555 \ \text{mod} \ 7= & &0 \\
5555555 \ \text{mod} \ 7= & &5 \\
55555555 \ \text{mod} \ 7= & &6 \\
555555555 \ \text{mod} \ 7= & &2 \\
5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$.
$$\begin{array}
& 1 \ \text{number =} &5 \\
7 \ \text{numbers =} &5 \\
13 \ \text{numbers =} &5 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
985 \ \text{numbers =} &5 \\
991 \ \text{numbers =} &5 \\
997 \ \text{numbers =} &5 \\
998 \ \text{numbers =} &6 \\
999 \ \text{numbers =} &2 \\
\color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!
| $$\frac{5}{9}\left(10^{1000}-1\right) \equiv \frac{-2}{2}\left(10^{4}-1\right) \equiv 1-3^4 \equiv -80 \equiv 4\pmod{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 5
} |
Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$ Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):
$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$$
Transforming it into $2 \sin \frac{3\alpha-180}4 \cos \frac{\beta - \gamma}4 + \sin \frac{3\alpha}2 =\frac{3}2$
and
$\cos \frac{\beta - \gamma}4 =\cos \frac{180- \alpha}4 + 2 \sin \frac{\beta}4 \sin \frac{\gamma}4$ is as far as I came.
| i think the answer is $\alpha = 100^\circ, \beta = 40^\circ$ and $\gamma = 40^\circ$
here is my attempt at the solution. i will use a change of variable
$$b = \beta+{1 \over 2}\gamma,\ c = \gamma + {1 \over 2}\beta$$ and the following inverse relations
$$\alpha = 180^\circ - {2 \over 3}(b+c), \ \beta = {4 \over 3}(b - {c \over 2})\ , \gamma = {4 \over 3}(c - {b \over 2}) $$ this change of variable transforms
$\sin({\alpha - \beta \over 2}) + \sin({\alpha - \gamma \over 2}) + \sin{3\alpha \over 2} = {3 \over 2}$ into a symmetric $2\pi$-periodic equation
$$f(b,c)=\cos b + \cos c - \cos(b+c) = {3 \over 2}.$$ observe that $f = 1$ on the boundary and $f(b,b) = 2\cos b - \cos 2b = -2\cos^2 b + 2 \cos b +1$ is maximum $3/2$
at $b = \pi/3 \text{ and } 5\pi/3.$
we will show that the global maximum of $f(b,c)$ on the square $[0,2\pi] \times [0,2\pi]$ is ${3 \over 2}$ at $b = c = \pi/3.$
at a local extremum, the partial derivatives
$$f_b = -\sin b + \sin(b+c) = 0,\ f_c = -\sin c + \sin(b+c) = 0$$
so we need to solve $$\sin b = \sin c = \sin(b+c).$$
the solutions are $$c = b, c = \pi - b, c = 3\pi - b $$
only the case $c = b$ is relevant. we have $\sin b = \sin 2b$ an the solution is
$\cos b = 1/2, \sin b = 0 $
now we can transform the $$b = c = 60^\circ$$ to the values for $\alpha, \beta$ and $\gamma$ claimed at the top of the post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.
| Since $x$ is the sum of three square roots, $x$ is positive. Thus there exists a positive real $y$ such that $x=y^2-2$.
The equation then becomes
$$y^2-2=\sqrt{y^2-1}+\sqrt{y^2}+\sqrt{y^2+1}$$
$$y^2-y-2=\sqrt{y^2-1}+\sqrt{y^2+1}$$
Squaring gives
$$y^4-2y^3-3y^2+4y+4=2y^2 + 2\sqrt{y^4-1}$$
$$y^4-2y^3-5y^2+4y+4=\sqrt{4y^4-4}$$
Squaring again
$$y^8-4 y^7-6 y^6+28 y^5+17 y^4-56 y^3-24 y^2+32 y+16=4y^4-4$$
$$y^8-4 y^7-6 y^6+28 y^5+13 y^4-56 y^3-24 y^2+32 y+20=0$$
This octic has six real roots, of which five are extraneous. The approximate result is $$y \approx 3.56021$$
Using $x=y^2-2$ this gives $$x \approx 10.67507$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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