Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Simple definite integral I do not figure out how to solve
$$
L= \frac{1}{2}\int_1^4 \sqrt{(16t^2+t+4)}\;dt
$$
The key is probably in simplifying the polynomial equation, but I don't find a way that is really simplifying the integral. Can someone help?
Thank you.
| Not that easy. I will ignore the 1/2 factor and the extrema, by now.
$$16t^2 + t + 4 = \left(4t + \frac{1}{8}\right)^2 + \frac{255}{64}$$
Substitute then $z = 4t + \frac{1}{8}$ and $dz = 4 dt$ you have now:
$$\frac{1}{4}\int\sqrt{z^2 + \frac{255}{64}} dz$$
Now use the substitution $z = \frac{1}{8}\sqrt{255}\tan(s)$ and $dz = \frac{1}{8}\sqrt{255}\sec^2(s) dx$ this will lead to (make the math):
$$\frac{\sqrt{255}}{{32}}\int \frac{1}{8}\sqrt{255}\sec^3(s) ds = \frac{255}{256}\int \sec^3(s) dx$$
Now use the secant reduction formula:
$$\int \sec^m(x) dx = \frac{\sin(x)\sec^{m-1}(x)}{m-1} + \frac{m-2}{m-1}\int \sec^{m-2}(x) dx$$
and in your case $m = 3$. Remember that the integral of $\sec(x)$ is $\log(\tan(x) + \sec(x))$ and then you get:
$$\frac{255}{512}\tan(s)\sec(s) + \frac{255}{512}\log(\tan(s) + \sec(s))$$
Substitute back now. You had:
$$s = \arctan\left(\frac{8u}{255}\right)$$
and
$$u = \frac{1}{8} + 4y$$
obtaining:
$$\frac{1}{2}\left[\frac{1}{64}\sqrt{16t^2 + 4t + 4}(32t + 1) + \frac{255}{512}\log\left(\frac{8\sqrt{16t^2 + t + 4} + 32t + 1}{\sqrt{255}}\right)\right]_1^{4} = \text{do the math to obtain a number}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$n^{th}$ determinant Find determinant $D_n$ of matrix $$
\begin{bmatrix}
1 & 1 & \cdots & 1 & -n \\
1 & 1 & \cdots & -n & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & -n & \cdots & 1 & 1 \\
-n & 1 & \cdots & 1 & 1
\end{bmatrix}
$$
After multiplying first row by $-1$ and adding to $n-1$ rows:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & 0 & \cdots & -n-1 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & -n-1 & \cdots & 0 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Second and $n-1$ row are changing place:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Multiplying first row by $n$ and adding it to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & n+1 & \cdots & n+1 & 1-n^2
\end{vmatrix}
$$
Adding second to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & n+1 & (1+n)(2-n)
\end{vmatrix}
$$
Adding $n-1$ to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & 0 & (1+n)(3-n)
\end{vmatrix}
$$
Product on main diagonal gives $$D_n=(n+1)(3-n)(-n-1)^{n-2}$$
This is not correct. What is wrong with this upper triangular transformation?
| An easier way to do that:
*
*Step $1$: Add the $2$nd, $3$rd, $\ldots$, $n$th row to the first row to get
$$\begin{bmatrix}
-1 & -1 & \cdots & -1 & -1\\
1 & 1 & \cdots & -n & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & -n & \cdots & 1 & 1 \\
-n & 1 & \cdots & 1 & 1 \\
\end{bmatrix}$$
*Step $2$: Add the first row to the $2$nd, $\ldots$, $n$th row respectively to get:
$$\begin{bmatrix}
-1 & -1 & \cdots & -1 & -1\\
0 & 0 & \cdots & -n - 1 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & -n - 1 & \cdots & 0 & 0 \\
-n - 1 & 0 & \cdots & 0 & 0 \\
\end{bmatrix}$$
*Step $3$: Expand the determinant along the $n$th column to get
$$\det(D_n) = (-1)^{1 + n}(-1)\det(B) = (-1)^{n + 2}\det(B),$$
where $B$ is the anti-diagonal matrix
$$\begin{bmatrix}
0 & 0 & \cdots & -n - 1 \\
\vdots & \vdots & \ddots & \vdots \\
0 & -n - 1 & \cdots & 0 \\
-n - 1 & 0 & \cdots & 0 \\
\end{bmatrix}$$
For any $m \times m$ anti-diagnoal matrix $D$
$$D = \begin{bmatrix}
0 & \cdots & d \\
\vdots & \ddots & \vdots \\
d & \cdots & 0 \\
\end{bmatrix},$$
we have $$\det(D) = \begin{cases}
(-1)^{m/2}d^m & d \text{ is even},\\
(-1)^{(m - 1)/2}d^m & d \text{ is odd}.
\end{cases}$$
Therefore,
\begin{align}
\det(D_n) = & (-1)^{n + 2}\begin{cases}
(-1)^{(n - 1)/2}(-n - 1)^{n - 1} & n - 1 \text{ is even}, \\
(-1)^{(n - 1 - 1)/2}(-n - 1)^{n - 1} & n - 1 \text{ is odd}.
\end{cases} \\
= & \begin{cases}
(-1)^{\frac{5n + 1}{2}}(n + 1)^{n - 1} & n \text{ is odd}, \\
(-1)^{\frac{5n}{2}}(n + 1)^{n - 1} & n \text{ is even}.
\end{cases} \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this inequality involving trig functions: $\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$? Source - Larsen 1.3.12
Problem:
Prove using $y=\sin(x)$ graph, for $A,B,C$ as angles of triangle, that:
$$\frac{\sin B + \sin C}{2}\leq \sin\left(\frac{B+C}{2}\right)$$
Attempt:
(without graph)
Using sum-to-product formula, $\sin B+\sin C=2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$
So,
$$\frac{\sin B + \sin C}{2}=\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})$$
$$=\sin(\frac{180-A}{2})\cos(\frac{B-C}{2})$$
$$=\cos(\frac{A}{2})\cos(\frac{B-C}{2})$$
$$=\frac{\cos(\frac{A+B-C}{2})+\cos(\frac{A-B+C}{2})}{2}$$
$$=\frac{\cos(\frac{180-2C}{2})+\cos(\frac{180-2B}{2})}{2}$$
$$=\frac{\sin C+\sin B}{2}$$
And I think I arrived where I had begun.
Question:
How to solve this inequality? (without graph)
| Another proof, based on properties of the sine function is that it is true because, on $[0,\pi]$, sine is a concave function, since $\;(\sin)''(x)=-\sin x\le 0$ on this interval.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number Theory: Show that $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$ Show that for all $n\in\mathbb{N}$, $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$.
I think I've proved this problem, but I was unsure if my proof was correct:
Proof
Let $n=1$. Then, $10^3=1000\equiv1\pmod{3^3}$ since $3^2=9\mid999$ and $3\mid111\implies 3^3\mid999$; but, $3^4\not\mid999$.
Now, assume $10^{3^n}\equiv1\pmod{3^{n+2}},3^{n+3}\not\mid10^{3^n}-1$.
We want to prove that this is true for $n+1$:
$10^{3^{n+1}}\equiv1\pmod{3^{n+3}}$
$10^{3^{n+1}}-1\equiv0\pmod{3^{n+3}}$
$(10^{3^n})^3-1\equiv0\pmod{3^{n+3}}$
$(10^{3^n}-1)((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod{3^{n+3}}$.
By our assumption, $3^{n+2}\mid(10^{3^n}-1)$ but $3^{n+3}\not\mid10^{3^n}-1$, so $\gcd(3^{n+3},10^{3^n}-1)=3^{n+2}$.
So,
$((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod3$.
But $((10^{3^n})^2+10^{3^n}+1)$ is of the form $100\dots00100\dots001$, so the sum of its digits is $3$, so it can be divided by $3\implies ((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod3$. (Just added this:) However, it cannot be divided by $3^2=9$ since the sum of these digits is not divisible by $9$.
Thus, $10^{3^{n+1}}\equiv1\pmod{3^{n+3}}$ and $3^{n+4}\not\mid 10^{3^{n+1}}-1$.
So, by induction, $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$ for all $n\in\mathbb{N}$.
| You can simply apply one of the Lifting The Exponent (LTE) Lemmas.
Let $\upsilon_p(m)$ denote the exponent of the highest power of $p$ that divides $m$.
If $a,b\in\Bbb Z,n\in\Bbb Z^+$, $p$ is prime, $a\equiv b\not\equiv 0\pmod{p}$, then $$\upsilon_p\left(a^n-b^n\right)=\upsilon_p(a-b)+\upsilon_p(n)$$
In this case, $10\equiv 1\not\equiv 0\pmod{3}$, so
$$\upsilon_3\left(10^{3^n}-1\right)=\upsilon_3(10-1)+\upsilon_3\left(3^n\right)=2+n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$ $$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$
I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=\tan\theta$, but I ended up with $\sin\theta\cos^3\theta$ as my integrand. Can someone explain how to compute this?
| Notice that:
$$\begin{eqnarray*} \int_{-\infty}^{+\infty}\frac{x^2}{x^4+1}\,dx &=& 2\int_{0}^{+\infty}\frac{x^2}{x^4+1}\,dx\\ &=& 2\int_{0}^{1}\frac{x^2}{1+x^4}\,dx+2\int_{1}^{+\infty}\frac{x^2}{1+x^4}\,dx\\&=&2\int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx\\&=&2\left(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\ldots\right)\\&=&2\left(1+\frac{2}{4^2-1}-\frac{2}{8^2-1}+\frac{2}{12^2-1}-\ldots\right)\end{eqnarray*}$$
and, from the logarithmic derivative of the Weierstrass product for the sine and cosine function:
$$ \sum_{k\geq 0}\frac{1}{(2k+1)^2-x^2}=\frac{\pi}{4x}\,\tan\left(\frac{\pi x}{2}\right),$$
$$ \sum_{k\geq 1}\frac{1}{k^2-x^2}=\frac{1-\pi x\cot(\pi x)}{2x^2},$$
so by taking limits as $x\to\frac{1}{2}$ we get:
$$ \int_{-\infty}^{+\infty}\frac{x^2}{x^4+1}\,dx = \color{red}{\frac{\pi}{\sqrt{2}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the distribution law Throwing a cube twice,
let $X$ be the sum of the two throws
Find the distribution law
My attempt
$$
\begin{array}{c|lcr}
&2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/12 & 1/12 & 1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12 \\
\end{array}
$$
But shouldn't the sum be $1$? Currently the sum is $11/12$
| The probabilities are not equal. Have a look at the following table.
$$
\left.\begin{array}{c|c|c|c|c} & 1 & 2 & 3 & 4 & 5& 6\\\hline 1 & 2 & 3 & 4 & 5& 6& 7 \\\hline 2 & 3 & 4 & 5 & 6 & 7& 8\\\hline 3 & 4 & 5 & 6 & 7 & 8& 9\\\hline 4 & 5 & 6 & 7 & 8 & 9& 10\\\hline 5 & 6 & 7 & 8 & 9 & 10& 11\\\hline 6 & 7 & 8 & 9 & 10 & 11& 12\end{array}\right.
$$
It is easy to tell various probabilities of the outcomes of this experiment using the table above. For example, the probability that we get $7$ is $6/36=1/6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring $x^3-3x-1$ in terms of $\alpha$ unknown
I never got a satisfactory answer here: Factoring $x^3-3x-1\in \Bbb Q[x]$ in terms of a unknown root
But all context is contained below
I want to factor $x^3-3x-1\in \Bbb Q[x]$ in terms of an unknown root $\alpha$, i.e. over $\Bbb Q(\alpha)$.
So since $\alpha$ is a root of $x^3-3x-1$ we can divide it out, obtaining:
$(x-\alpha)(x^2+\alpha x+(\alpha^2 -3))$ and obtaining $\alpha^3-3\alpha-1=0$.
Now my next thought would be to apply the quadratic formula to the quadratic factor, but this doesn't yield anything useful, atleast not seemingly, even with me manipulating what I believe is the key, that $\alpha^3-3\alpha-1=0$
We can do other manipulations:
$$x^2+\alpha x+(\alpha^2-3))=x^2+\alpha x + \frac1\alpha=0\implies \alpha x^2 +\alpha^2 x +1 = 0$$
$$\implies \frac{-\alpha^2\pm \sqrt{\alpha^4-4\alpha}}{2\alpha}=-\frac{\alpha}{2}\pm \frac{\sqrt{3\alpha^2-3\alpha}}{2\alpha}$$
Now I am lost again. I am meant to get the other two roots as:
$$\alpha^2-\alpha-2,\quad 2-\alpha^2$$
| You can find the roots using Vieta's formulas. The roots $r_1, r_2$ satisfy $x^2+\alpha x + (\alpha^2-3)=0$, so $r_1+r_2=-\alpha$ and $r_1r_2=\alpha^2-3$. So, let's use undetermined coefficients:
$$r_1=a\alpha^2+b\alpha+c$$
$$r_2=-a\alpha^2+(b-1)\alpha-c$$
where $r_2$ is chosen to satisfy the first equation. The second one gives
$$r_1r_2=-a^2\alpha^4-a\alpha^3+(b^2-b-2ac)\alpha^2-c\alpha-c^2$$
$$=-a^2(3\alpha^2+\alpha)-a(3\alpha+1)+(b^2-b-2ac)\alpha^2-c\alpha-c^2$$
$$\alpha^2-3=\alpha^2(-3a^2+b^2-b-2ac)+\alpha(-a^2-3a-c)+(-a-c^2)$$
Hence we get the system $\{1=-3a^2+b^2-b-2ac, 0=-a^2-3a-c, -3=-a-c^2\}$. This has two solutions $\{a=-1,b=0,c=2\}$ and $\{a=-1, b=1, c=2\}$, the first of which corresponds to the desired roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$
How to solve the differential equation
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$$
Hints please.
| HINT:
$$y'(x)+\frac{1}{x}\tan(y(x))=\frac{1}{x^2}\tan(y(x))\sin(y(x))\Longleftrightarrow$$
$$y'(x)+\frac{\tan(y(x))}{x}=\frac{\tan(y(x))\sin(y(x))}{x^2}\Longleftrightarrow$$
Let $y(x)=\sin^{-1}(v(x))$, which gives $\frac{\text{d}y(x)}{\text{d}x}=\frac{\frac{\text{d}v(x)}{\text{d}x}}{\sqrt{1-v(x)^2}}$:
$$\frac{\frac{\text{d}v(x)}{\text{d}x}}{\sqrt{-v(x)^2+1}}+\frac{v(x)}{x\sqrt{-v(x)^2+1}}=\frac{v(x)^2}{x^2\sqrt{-v(x)^2+1}}\Longleftrightarrow$$
$$\frac{x\frac{\text{d}v(x)}{\text{d}x}+v(x)}{x\sqrt{-v(x)^2+1}}=\frac{v(x)^2}{x^2\sqrt{-v(x)^2+1}}\Longleftrightarrow$$
$$-\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)^2}-\frac{1}{xv(x)}=-\frac{1}{x^2}\Longleftrightarrow$$
Let $u=(x)=\frac{1}{v(x)}$, which gives $\frac{\text{d}u(x)}{\text{d}x}=-\frac{\frac{\text{d}v(x)}{\text{d}x}}{v(x)^2}$:
$$\frac{\text{d}u(x)}{\text{d}x}-\frac{u(x)}{x}=-\frac{1}{x^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation with complex numbers and fractions I want to solve this, but I do not know how to continue after solving the square in the denominator. I tried multiplying by denominators, but I got awkward results.
I appreciate your help.
| To simplify this I did the following. First expand out the square in the denominator.
$(1+i)^2=1+2i+i^2=2i\Rightarrow$
$z_2=\frac{-\frac{1}{\sqrt{3}}+2i}{5i+2i} + \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}-2i}$
Next multiply the second fractions numerator and denominator by the denominators conjugate.
$(\sqrt{3}-2i)(\sqrt{3}+2i)=3+4=7 \Rightarrow$
$z_2=\frac{-\frac{1}{\sqrt{3}}+2i}{7i} + \frac{\frac{\sqrt{3}+2i}{\sqrt{3}}}{7}=\frac{-\frac{1}{\sqrt{3}}+2i}{7i} + \frac{\sqrt{3}+2i}{7\sqrt{3}}$
Next we combine the fractions by multiplying them by the each others denominators.
$z_2=\frac{-7+14\sqrt{3}i+7\sqrt{3}i-14}{49\sqrt{3}i}=\frac{21\sqrt{3}i-21}{49\sqrt{3}i}=\frac{3\sqrt{3}i-3}{7\sqrt{3}i}$
Finally we simplify to get the answer.
$z_2=\frac{3}{7}-\frac{3}{7\sqrt{3}i}$
| {
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How do you factor $x^2-x-1$? I know you can't have all integers, but how do you factor this anyway?
Wolfram|Alpha gives me $-\frac{1}{4} (1+\sqrt{5}-2 x) (-1+\sqrt{5}+2 x)$.
Cymath gives me $(x-\frac{1+\sqrt{5}}{2})(x-\frac{1-\sqrt{5}}{2})$.
The closest I can get is $(x+1)(x-1)-x$.
So how do I get a nice answer like the ones listed above?
| Apply quadratic formula for the roots of $x^2-x-1=0$ as follows
$$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}=\frac{1\pm \sqrt 5}{2}$$
hence, one should have the following factors $$x^2-x-1=1\cdot \left(x-\frac{1+ \sqrt 5}{2}\right)\left(x-\frac{1- \sqrt 5}{2}\right)$$
or $$\frac{1}{4}(2x-1-\sqrt 5)(2x-1+\sqrt 5)$$
$$=-\frac{1}{4}(1+\sqrt 5-2x)(-1+\sqrt 5+2x)$$
So both the answers are correct
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve $ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$. Is my solution correct? Find the roots of the following equation, if any:
$$
\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
=
27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.
$$
My approach:
The following constraints should hold jointly for x:
*
*$1-x^2\geq0\iff x\in[-1,1]$
*$1+x\geq0 \iff x\geq-1$
*$1-x\geq0 \iff x\leq1$
*$\sqrt{1+x}\ne\sqrt{1-x}\iff x\ne0$
Consequently, $x\in[-1,0)\cup(0,1]$ and for such x's I solve and I reach at the following equation
$$
\left(\frac{x}{1+\sqrt{1-x^2}}\right)^3=27\implies
\frac{x}{1+\sqrt{1-x^2}}=3,
$$
which gives
$$
x = 3 + 3\sqrt{1-x^2}\implies
3\sqrt{1-x^2} = x-3\implies
9-9x^2=x^2-6x+9\implies
10x^2-6x=0\implies
x(10x-6)=0,
$$
and thus $x=0$ or $x=\frac{3}{5}$. The first one is rejected, but $x=\frac{3}{5}\in[-1,0)\cup(0,1]$.
However, if we plug $x=\frac{3}{5}$ into the original equation, we find out that this is impossible. So we should reject $x=\frac{3}{5}$. The original equation is impossible.
Is my approach correct? Thank you very much in advance.
| $
\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
$
is of the form $\frac{1-a}{1+a}$, which is $\leq1$.
$
27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}
$
is of the form $27\frac{a+b}{a-b}$, which is $\geq27$.
So there is no solution.
| {
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What is the difference between the two real numbers that satisfy this equation?
What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form
I tried guessing solutions but seeing how there are no common zeroes to both the left- and right-hand sides I don't know what to do.
| $$(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$$
$$\frac{(x+2)}{(x-2)}=\frac{x^2-1}{x^2-9}$$
$$1+\frac{4}{x-2}=1+\frac{8}{x^2-9}$$
$$4x^2-36=8x-16$$
$$x^2-2x-5=0$$
Now difference of roots is $$\frac{\sqrt{D}}{a}=\sqrt{24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ where $x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$ Consider the following equality:
$$x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$$
Solve for $x$.
The only thing I noticed is the first part like $(1-2)$,$(3-4)$ gives us $-1$ but then I just don't see what the trick behind this problem is.
| Hint: have you ever tried to look at a formula like this: $$ {a^3 - b^3 \over a-b} = (?)(?)$$ ? Just expand and observe the two occuring factors and look like the astronomers over long distances ...
[update]
As $ {a^3-b^3 \over a - b }= a^2+ab+b^2$ and thus $$ (a-b)(a^2+ab+b^2) = a^3-b^3 $$the problem-formula is equivalent $(a^3-b^3) + (b^3-c^3) + ... +(x^3-y^3) + (y^3-z^3) = a^3-z^3 $ and comes out to be $$x = 1^3 - 50^3 = -124999 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Limit problem with fifth root $$\lim_{x\to 0}\frac{x^2}{\sqrt[5]{1+5x}-1-x}$$
How to solve this limit without L'Hopital rule and Taylor series?
| Let's use the simple substitution $1 + 5x = t^{5}$ so that $x = (t^{5} - 1)/5$. Then as $x \to 0$ we have $t \to 1$. Therefore
\begin{align}
L &= \lim_{x \to 0}\frac{x^{2}}{\sqrt[5]{1 + 5x} - 1 - x}\notag\\
&= \lim_{t \to 1}\dfrac{\left(\dfrac{t^{5} - 1}{5}\right)^{2}}{t - 1 - \dfrac{t^{5} - 1}{5}}\notag\\
&= \frac{1}{5}\lim_{t \to 1}\frac{(t^{5} - 1)^{2}}{5(t - 1) - (t^{5} - 1)}\notag\\
&= \frac{1}{5}\lim_{t \to 1}\frac{(t - 1)^{2}}{5(t - 1) - (t^{5} - 1)}\cdot\left(\frac{t^{5} - 1}{t - 1}\right)^{2}\notag\\
&= \frac{1}{5}\lim_{t \to 1}\frac{t - 1}{5 - (t^{4} + t^{3} + t^{2} + t + 1)}\cdot\left(5\right)^{2}\notag\\
&= -5\lim_{t \to 1}\frac{t - 1}{(t^{4} + t^{3} + t^{2} + t) - 4}\notag\\
&= -5\lim_{t \to 1}\frac{t - 1}{(t^{4} - 1) + (t^{3} - 1) + (t^{2} - 1) + (t - 1)}\notag\\
&= -5\lim_{t \to 1}\dfrac{1}{\dfrac{t^{4} - 1}{t - 1} + \dfrac{t^{3} - 1}{t - 1} + \dfrac{t^{2} - 1}{t - 1} + 1}\notag\\
&= -5\cdot\frac{1}{4 + 3 + 2 + 1}\notag\\
&= -\frac{5}{10} = -\frac{1}{2}\notag
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Convergence of $\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}$ for $|z|= \frac{\sqrt{3}}{3}$ I have shown that the series $$\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}$$
converges for $|z|<\frac{\sqrt{3}}{3}$. Can anyone help me to study the convergence for $|z|= \frac{\sqrt{3}}{3}$?
| We will use the series
$$
\arctan(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}
$$
Using Dirichlet's Test, we get that, except for $z=\pm i\frac{\sqrt3}3$, the series converges. Using Abel's Theorem, we get that for $z=\frac{\sqrt3}3e^{i\theta}$,
$$
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^n3^n}{2n+1}\left(\frac{\sqrt3}3e^{i\theta}\right)^{2n+1}
&=\frac{\sqrt3}3\sum_{n=1}^\infty(-1)^n\frac{e^{i\theta(2n+1)}}{2n+1}\\
&=\frac{\sqrt3}3\left(\arctan\left(e^{i\theta}\right)-1\right)\\
&=\frac{\sqrt3}3\left(\frac i2\log\left(\frac{i+e^{i\theta}}{i-e^{i\theta}}\right)-1\right)
\end{align}
$$
which only blows up when $e^{i\theta}=\pm i$.
For $z=\pm i\frac{\sqrt3}3$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^n3^n}{2n+1}\left(\pm i\frac{\sqrt3}3\right)^{2n+1}
&=\pm i\frac{\sqrt3}3\sum_{n=1}^\infty\frac1{2n+1}
\end{align}
$$
which diverges by comparison to the Harmonic Series.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Trigonometric equation $\sin x+1=\cos x$ $$\sin x+1=\cos x,\quad x\in[-\pi,\pi]$$
How do you solve by squaring both sides? the solution is $x\in\{-\pi/2,0\}$ so the solutions $\pi$ and $-\pi$ are inadmissible, I do not understand how by subbing $-\pi$ back into both sides of the equations makes them unequal, and the same for positive $\pi$. Which equation are you subbing $\pi$ into to check, the original?
| $\sin x + 1 = \cos x$
$\sin \pm \pi + 1 = 1 \ne \cos \pm \pi = -1$ so $\pm \pi$ are not solutions.
$\sin -\pi/2 + 1 = -1 + 1 = 0 = \cos -\pi/2 $ so $-\pi/2$ is solution.
$\sin 0 + 1 = 0 + 1 = \cos 0 $ so $0$ is solution.
But how to solve?
$\sin x + 1 = \cos x$
$\sin x - \cos x = -1$
$(\sin x -\cos x)^2 = (-1)^2$
$\sin^2 x + \cos^2 - 2\cos x \sin x = 1$
$1 - 2\cos x \sin x = 1$
$\cos x \sin x = 0$
So $\cos x = 0 $ or $\sin x = 0$.
So $x =\pm \pi/2$ or $x = \{0, \pi\}$.
But note.
$\sin x + 1 = \cos x \le 1$ so $\sin x = \cos x - 1 \le 0$ and $\cos x = \sin x + 1 \ge -1 + 1 = 0$.
$x = \pi/2 \implies \sin x = 1 > 0$ so $-\pi/2$ is not solution.
$x = \pi \implies \cos x = -1 < 0$ so $\pi$ is not solution
Solutions are $-\pi/2$ or $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What's the integral of this fraction?
Find $\displaystyle \int \dfrac{\cos^2(x)\sin(x)+2\sin(x)}{\cos^2(x)+1}dx$.
I tried splitting it up, but I don't think that helps. What is the best way to solve this?
| Keep in mind that whenever you make any substitution, you will be dividing by the derivative of the substitution, so look to see if something cancels. Remember, the whole point of making substitutions is to simplify things. In this case, $u=\cos x$ seems like a good choice for a substitution. As a first step just multiply by one, e.g. by $$\frac{\frac{du}{dx}}{\frac{du}{dx}}$$
First we have
$$u=\cos x$$
$$\frac{du}{dx}=\color{green}{-\sin x}$$
Next the integral
$$\begin{array}{lll}
\displaystyle\int\frac{\cos^2x\sin x+2\sin x}{\cos^2x+1}dx&=&\displaystyle\int\frac{\cos^2x\sin x+2\sin x}{\cos^2x+1}\cdot\frac{\frac{dx}{1}}{1}\\
&=&\displaystyle\int\frac{\cos^2x\sin x+2\sin x}{\cos^2x+1}\cdot\frac{\frac{dx}{1}}{1}\cdot\frac{\frac{du}{dx}}{\frac{du}{dx}}\\
&=&\displaystyle\int\frac{\cos^2x\sin x+2\sin x}{\cos^2x+1}\cdot\frac{\frac{dx}{1}}{1}\cdot\frac{\frac{du}{dx}}{\color{green}{-\sin x}}\\
&=&\displaystyle-\int\frac{\cos^2x+2}{\cos^2x+1}\cdot\frac{\frac{dx}{1}}{1}\cdot\frac{\frac{du}{dx}}{1}\\
&=&\displaystyle-\int\frac{\cos^2x+2}{\cos^2x+1}du\\
&=&\displaystyle-\int\frac{u^2+2}{u^2+1}du\\
\end{array}$$
The reason why I do things this way is so that I can see what's "going on" instead of just following some preset algorithm. This way, i can see the "why" instead of just the "what".
Finishing off we have
$$\begin{array}{lll}
-\displaystyle\int\frac{u^2+2}{u^2+1}du &=&-\displaystyle\int\frac{(u^2+1)+1}{u^2+1}du\\
&=&-\displaystyle\int 1+\frac{1}{u^2+1}du\\
&=&\displaystyle -(u+\tan^{-1}u)+C\\
&=&\displaystyle -(\cos x+\tan^{-1}(\cos x))+C\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Solve the equation: $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$ How do I solve this equation:
$$ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$$
| \begin{align}
\sin x - 3\sin 2x + \sin 3x &= \cos x - 3\cos 2x + \cos3x\\
2\sin 2x \cos x - 3\sin 2x &= 2\cos 2x \cos x - 3\cos 2x \\
\sin 2x (2 \cos x - 3) &= \cos 2x (2\cos x - 3) \\
\end{align}
Now if $2 \cos x - 3=0$ then $\cos x=\frac 32$ which is not possible, hence as $2 \cos x - 3 \neq0$ then we divide both sides by $2 \cos x - 3$ to obtain $\sin 2x = \cos 2x$ which is solvable.
NOTE I used sum-to-product relations from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find the value of $\lim_{x \to 2} \frac {xf^2(x)-9}{(x-2)}$ Suppose that $$\lim_{x\to 2}\frac{x f(x)-3}{x-2}=5.$$
Then, what is the value of $$\lim_{x\to 2}\frac{x f^2(x)-9}{x-2}?$$
| You can see that in $\lim_{x\to 2}\frac{x f(x)-3}{x-2}=5$, the denominator goes to zero. Since we have a limit, the numerator should go to zero too and hence
$$\begin{align}
\lim_{x\to 2}x f(x)-3 &= 0 \\
\lim_{x\to 2}x f(x) &= 3 \\
\frac{\lim_{x\to 2}x f(x)}{\lim_{x\to 2}x} &= \frac{3}{\lim_{x\to 2}x} \\
\lim_{x\to 2}\frac{x f(x)}{x} &= \frac{3}{2}
\end{align}$$
Then the limit of $f(x)$ at $x=2$ will be
$$\lim_{x\to 2}f(x)=\frac{3}{2}$$
and then we can conclude that
$$\begin{align}
\lim_{x\to 2}f(x) \cdot \lim_{x\to 2}f(x) &= \frac{3}{2} \cdot \frac{3}{2} \\
\lim_{x \to 2}f(x) \cdot f(x) &= \frac{9}{4} \\
\lim_{x \to 2}f^2(x) &= \frac{9}{4} \\
\lim_{x \to 2}x \cdot \lim_{x \to 2}f^2(x) &= (\lim_{x \to 2}x) \cdot \frac{9}{4} \\
\lim_{x \to 2} xf^2(x) &= \frac{9}{2} \\
\lim_{x\to 2}xf^2(x)-9 &= -\frac{9}{2}
\end{align}$$
So we can conclude for the second limit that
$$\lim_{x\to 2^{\mp}}\frac{x f^2(x)-9}{x-2}= \pm \infty$$
as the numerator goes to $-\frac{9}{2}$ and the denominator goes to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there any error in my solution? Let $f(x)$ is continuous function for all real values of $x$ and satisfies $$\int^x_0 f(t)dt =\int^1_{x} t^2f(t)dt +\frac{x^{16}}{8}+\frac{x^6}{3}+a$$
Then the value of a is equal to:
(a) $\frac{-167}{840}$
(b) $\frac{-167}{840}$
(c) $\frac{-167}{840}$
(d) $\frac{-167}{840}$
Is there any error in my following solution:
For $x =1$, $$\int^1_0 f(1)dt =0+\frac{1}{8}+\frac{1}{3}+a = \frac{11}{24}+a$$
Differentiating both sides
$$f(x) = -x^2f(x) +2x^{15}+2x^5$$
$$\implies 2 \int^1_0 \frac{x^{15}+x^5}{1+x^2}dx =\frac{11}{24}+a$$
$$\implies a =-\frac{167}{840}$$
| Not quite sure what you are doing, although differentiating to find $f$ is the right way to go:
$$f(x) = -x^2 f(x) + 2 x^{15} + 2 x^5 \implies f(x) = 2 x^5 \frac{1+x^{10}}{1+x^2} = 2 x^5 (1-x^2+x^4-x^6+x^8)$$
Now plug in $x=0$ in the integral equation to find that
$$a = -2 \int_0^1 dx \,(x^7-x^9+x^{11}-x^{13}+x^{15}) = - \left (\frac14-\frac15+\frac16-\frac17+\frac18\ \right ) = -\frac{167}{840}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Is $x^6 + 1$ divisible by $x^2 + 1$? I used polynomial long division and found that the remainder was $x + 1$. So I concluded that $x^6 + 1$ is not divisble by $x^2 + 1$. But this is not true since $x^6 + 1$ can be factored as a sum of cubes of $x^2$ and $1$.
What can we conclude if we use long division on two polynomials and get a non-zero remainder?
| $$x^6+1=(x^2+1)(x^4-x^2+1)$$
as à consequence of $a^3+b^3=(a+b)(a^2-ab+b^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Why is this an eigenvector of this matrix? The matrix is
$$
\begin{bmatrix}
1 & 1 & -3 \\
2 & 0 & 6 \\
1 & -1 & 5
\end{bmatrix}
$$
I understand where the eigenvector $[-3, 0, 1]$ comes from because the eigenvalue is $\lambda = 2$ and when you substitute it into the matrix and then row reduce you get
$$
\begin{bmatrix}
1 & 0 & 3 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
But the answer also puts $[1, 1, 0]$ as an eigenvector. Any idea where this comes from?
| $$\det\Bigg(\begin{bmatrix}
1-\lambda & 1 & -3 \\
2 & -\lambda & 6 \\
1 & -1 & 5-\lambda
\end{bmatrix}\Bigg)=(1-\lambda)(-\lambda(5-\lambda)+6)-(2(5-\lambda)-6)-3(-2+\lambda)$$
$$\det(A-\lambda I) = (\lambda-2)^3=0$$
$$A-2 I =\begin{bmatrix}
-1 & 1 & -3 \\
2 & -2 & 6 \\
1 & -1 & 3
\end{bmatrix}$$
From here you can see that the bottom two rows are just multiples of the first row. So
$$\begin{bmatrix}
-1 & 1 & -3 \\
2 & -2 & 6 \\
1 & -1 & 3
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=\begin{bmatrix}
-1 & 1 & -3 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=0$$
This means that $$x=y-3z$$ So we get the eigenvector is
$$\begin{bmatrix}
y-3z \\
y \\
z
\end{bmatrix}=\begin{bmatrix}
y \\
y \\
0
\end{bmatrix}+\begin{bmatrix}
-3z \\
0 \\
z
\end{bmatrix}= y\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+z\begin{bmatrix}
-3 \\
0 \\
1
\end{bmatrix}$$
Which are your two linearly independant eigenvectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a=0$,then the number of integer roots in the equation is?
Let a,b,c,d be natural numbers.Now consider an equation
$x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a=0$,then the number of integer
roots in the equation is ?
Me: Plugged in $0,1,-1$.No use.At 0 the sign is negative and at infinity the sign is positive. At least one real root.Not sure if that will be integer though :/.Help please!
| Suppose that the equation has an integer root $x$. By @JimmyK4542's comment we have $x>0$; also, $x\mid a$ so $x\le a$. Clearly
$$x-(a+b+c)<0$$
and so
$$d=\frac{x^4-(a+b+c)x^3-(a+b+c)x^2-(a+b)x-a}{x^3}<0\ .$$
This is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ S_{n}=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+...........+\frac{x^{2^{n}}}{(x+1)(x^2+1)...(x^{2^{n}}+1)}$
If $\displaystyle S_{n}=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^{2^{2}}}{(x+1)(x^2+1)(x^{2^2}+1)}+...........+\frac{x^{2^{n}}}{(x+1)(x^2+1)...(x^{2^{n}}+1)}$
Then $\displaystyle \lim_{n\rightarrow \infty}S_{n} = \;,$ Where $x>1$
$\bf{My\; Try::}$ First we will calculate $\bf{r^{th}}$ term of the sequence.
So $$\displaystyle \bf{T_{r}} = \frac{x^{2^{r}}}{(x+1)(x^2+1)............(x^{2^{n}}+1)} = \frac{x^{2^{r}}(x-1)}{x^{2^{r+1}}-1}$$
So We get $$\displaystyle \bf{T_{r}} = \frac{x^{2^{r}}(x-1)}{(x^{2^r}-1)(x^{2^{r}}+1)}$$
Now I did not Understand How can I convert into Telescopic Sum.
Help me
Thanks
| $$\begin{align} S_{\infty}&=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^{2^{2}}}{(x+1)(x^2+1)(x^{2^2}+1)}+\ldots+\frac{x^{2^{n}}}{(x+1)(x^2+1)\cdots(x^{2^{n}}+1)}\\\
&=\frac{x+1-1}{x+1}+\frac{x^2+1-1}{(x+1)(x^2+1)}+\frac{x^{2^{2}}+1-1}{(x+1)(x^2+1)(x^{2^2}+1)}+\ldots+\frac{x^{2^{n}}+1-1}{(x+1)(x^2+1)\cdots}\\\
&=1-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{(x+1)(x^2+1)}+\frac{1}{(x+1)(x^2+1)}-\frac{1}{(x+1)(x^2+1)(x^{2^2}+1)}+\ldots\\\
&=1
\end{align}
$$
The last term becomes vanishingly small as we increase $n$, that's why we can ignore it and say that the limit of $S_n$ as $n\to\infty$ is unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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what is the value of $abc $ where $a > 1, b>1,c>1$ and $ (abc(abc-a-b-c)+ab+bc+ca-1) \equiv 0 \pmod{(abc)} $? Given that,
$a > 1, b>1,c>1,$
$a,b,c$ are distinct.
and
$ (abc(abc-a-b-c)+ab+bc+ca-1) \equiv 0 \pmod{(abc)} $
Find the value of $abc$.
| The only solutions are $(a,b,c)=(2,3,5)$ and its permutations, with $abc=30$.
We only need to check triples $(a,b,c)$ with $abc\le ab+bc+ca-1$, because otherwise $abc$ cannot divide $ab+bc+ca-1$, and the congruence condition does not hold. This is a finite problem, easy to solve. For example, without loss of generality, we may assume that $a$ is the smallest number of $a,b,c$. For $a=2$ we see that we need $b=3$, and then we obtain $6c\le 5(c+1)$, which gives $c\le 5$, so that $c=5$ (because $c=4$ is impossible, and $a,b,c>1$ are distinct).
| {
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"timestamp": "2023-03-29T00:00:00",
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Describe the nonzero integer solutions to the equation $a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 =0$ Can someone describe all the integer solutions to the above equation such that $abcdefg\neq 0$ ?
| For the equation.
$$x_1^3+x_2^3+x_3^3+x_4^3+x_5^3+x_6^3=x_7^3$$
You can write a fairly simple formula.
$$x_1=t^2-3(k+s)(p+t)-3p^2+2u$$
$$x_3=t^2-3(p+s)(k+t)-3k^2+2u$$
$$x_5=t^2-3(p+k)(s+t)-3s^2+2u$$
$$x_2=2t^2+3((k+s)(p-t)-2pt)+3p^2+u$$
$$x_4=2t^2+3((p+s)(k-t)-2kt)+3k^2+u$$
$$x_6=2t^2+3((p+k)(s-t)-2st)+3s^2+u$$
$$x_7=3(t^2-2(p+k+s)t+u)$$
where,
$$u=3(p^2+k^2+s^2)$$
Cube certainly look nice, but I prefer to solve such equations. Look cumbersome, but the solution much simpler. The sum of the cubes and the amount of combinations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588122",
"timestamp": "2023-03-29T00:00:00",
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How to prove $\left(1+\frac{1}{\sin a}\right)\left(1+\frac{1}{\cos a}\right)\ge 3+2\sqrt{2}$? Prove the inequality:
$$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$
| Here is another proof:
Assume $x^2+y^2=1$ then $(x+y)^2=1+2xy$ and
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{xy+x+y+1}{xy}=\frac{(x+y+1)^2}{(x+y)^2-1}=\frac{x+y+1}{x+y-1}$$
Now rearranging the inequality becomes $x+y\leq \sqrt{2}$. And it can be seen simply that this is the maximum value of $x+y$ given $x^2+y^2=1$, for example $z=x+y$ is a plane and its intersection with the cylinder $x^2+y^2=1$has a maximum at $x=y$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
| Sheer brute force is adequate:
$$(2a + b + c)(a + 2b + c)(a + b + 2c)$$
$$= 2(a^3 + b^3 + c^3) + 7(a^2b + ab^2 + a^2c+ac^2 + b^2c+bc^2) + 16(abc)$$
$$\geq 2(3abc) + 7(6abc) + 16(abc) = 64abc$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute $\lim\limits_{x \to 0} \left(\frac{e^{x^2} -1}{x^2}\right)^\frac{1}{x^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used? Thanks
$$\lim\limits_{x \to 0} \left(\frac{e^{x^2} -1}{x^2}\right)^\frac{1}{x^2}$$
Note: In a previous version of this question the limit was written as $\left(\frac{(e^{x})^2 -1}{x^2}\right)^\frac{1}{x^2}$.
| Let
$$y = \left(\frac{e^{2x} -1}{x^2}\right)^\frac{1}{x^2}.$$
Then
$$\ln y = \frac{1}{x^2} \ln\left(\frac{e^{2x} -1}{x^2}\right).$$
Notice that
$$\lim_ \limits{x \to 0^+} \ln\left(\frac{e^{2x} -1}{x^2}\right) = \lim_ \limits{x \to 0^+} \ln\left(\frac{1+2x+4x^2+\cdots-1}{x^2}\right)=\infty,$$
but
$$\lim_ \limits{x \to 0^-} \ln\left(\frac{e^{2x} -1}{x^2}\right) = \lim_ \limits{x \to 0^-} \ln\left(\frac{1+2x+4x^2+\cdots-1}{x^2}\right)$$
is undefined since you have $-\infty$ inside the $\log$.
Therefore the limit does not exist
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{(1+b)(1+c)}+\frac{b}{(1+a)(1+c)}+\frac{c}{(1+a)(1+b)} \geq \frac{3}{4}.$
Let $a,b,c$ be positive numbers that satisfy $abc = 1$, prove that $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$
Attempt
I tried doing $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} = \dfrac{1}{bc(1+b)(1+c)}+\dfrac{1}{ac(1+a)(1+c)}+\dfrac{c}{bc(1+a)(1+b)}$$ then using the fact that $a^2+b^2+c^2 \geq ab+bc+ca$ but that only seemed to get me $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{1}{bc+bc^2+c+1}+\dfrac{1}{ac+a^2c+c+1}+\dfrac{1}{ab+ab^2+b+1}.$$
| Ok, let's sum it:
$$\frac{(a+1)a+b(b+1)+c(c+1)}{(a+1)(b+1)(c+1)}$$, called it $(1)$
Now use Cauchy inequality:
$$(1) >= \frac{3((a+1)ab(b+1)c(c+1))^{\frac{1}{3}}}{(a+1)(b+1)(c+1)}$$, now
Derived all, what we can derive :)
And use Cauchy for $a+1,b+1,c+1$.
So continue it!
Hint: show that:
$$((a+1)(b+1)(c+1))^{\frac{2}{3}} >= 4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fourier series of $\frac{1}{5+4 \cos x}$ using contour integration The function
$$f(x)=\frac{1}{5+4 \cos x}$$
is periodic with the main period being $T=2\pi$. The graph is easily obtained, but here is a graph from Desmos as it looks better:
The function is even, so all the coefficients $b_{n}$ vanish: $$b_{n}=\frac{2}{t}\int_{-T/2}^{T/2} f(x)\sin\left ( \frac{2n\pi x}{T} \right )\,dx=0.$$
Using the residue theorem I found that
$$a_0 = \frac{2}{T} \int_{-T/2}^{T/2} f(x)\,dx =
\frac{1}{\pi i}\oint \frac{dz}{5z+2z^2+2}=\frac{2}{3}$$
and also
$$a_{n}=\frac{1}{2\pi i }\oint \frac{z^{2n}+1}{z^{n}(5z+2+2z^2)}\, dz = \operatorname*{Res}\limits_{z=-1/2} f(z)+ \operatorname*{Res}\limits_{z=0} f(z).$$
Furthermore
$$\operatorname*{Res}\limits_{z=-1/2} f(z) = \frac{1+\left ( \frac{-1}{2} \right )^{n}}{3\left ( \frac{-1}{2} \right )^{n}}.$$
The residue at $z=0$ isn't so easily obtained since there is a pole of order $n$, and when doing Laurent expansion it gets too messy and it seems that the coefficient $A_{-1}$ may not be easily derived from there. How do I effectively find that residue?
| I suggest that you write
$$
\frac{1}{2+5z+2z^2}=\frac{1}{(2+z)(1+2z)}=\frac{2}{3}\frac{1}{1+2z}-\frac{1}{6}\frac{1}{1+z/2}.
$$
Next, since $n$ is positive, the term $z^{2n}$ can be neglected (it will never contribute to the $z^{-1}$ coefficient in the Laurent expansion). Thus
$$
\begin{aligned}
\text{Res}_{z=0}f(z)&=\text{Res}_{z=0}\frac{1}{z^n}\Bigl(\frac{2}{3}\frac{1}{1+2z}-\frac{1}{6}\frac{1}{1+z/2}\Bigr)
\end{aligned}
$$
Now you only need to turn the geometric series around (using
$$
\frac{1}{1+a}=1-a+a^2-a^3+\cdots)
$$
and find the necessary coefficient in front of $z^{-1}$. I leave it for you to do the details.
Spoiler with result below:
I get, as a result, $$\text{Res}_{z=0}f(z)=(-1)^n\frac{1-4^n}{3\cdot 2^n}.$$
| {
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Evalute $\int\frac{3x-2}{x^2-4x+5} \, dx$ Evalute $$\int\frac{3x-2}{x^2-4x+5} \, dx$$
So, I tried using $x^2-4x+5 = t$, which unfortunately didn't help much.
I also tried making the denominator $(x-2)^2 +1$ and then using $x-2 = t$, which made everything messy. Any suggestions? It looks so simple, and I bet my first approach is right.
| The denominator is an irreducible quadratic. First thing you should do is write $$3x-2 = \tfrac{3}{2}(2x - \tfrac{4}{3}) = \tfrac{3}{2}(2x - 4) + 4,$$ so that the integrand becomes $$\frac{3}{2}\int \frac{2x-4}{x^2 - 4x + 5} \, dx + 4 \int \frac{dx}{x^2-4x+5}.$$ The first integral is trivial. The second readily yields to a substitution of the form you tried; i.e., $$x^2 - 4x + 5 = (x-2)^2 + 1,$$ and choose $u = x-2$.
| {
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Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
| Let me elaborate on partial fraction, and give you a way that doesn't involve solving linear systems.
We have
$$\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x+3}.$$
Let's multiply both sides by (x+1)(x-1)(x+3); we get
$$7x^2+1 = A(x-1)(x+3)+B(x+1)(x+3)+C(x-1)(x+1).$$
Since this equality must hold for all $x$, let us plug in some nice values of $x$:
*
*$x=-1$ yields $\ \ 7+1 = A(-1-1)(-1+3)$ and so $A=-2$
*$x=+1$ yields $\ \ 7+1 = B(+1+1)(+1+3)$ and so $B=+1$
*$x=-3$ yields $63+1 = C(+3-1)(+3+1)$ and so $C=+8$
With this we finally get
$$\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=-\frac{2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}.$$
This methods works every time you have (one or more) linear factors. You can solve for as many constants as you have linear factors.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of Multiple angle trigonometri ratios If $\tan^2{A}=1+2\tan^2B$ then prove that $\cos2B=1+2\cos2A$
I could not relate from $\tan$ to $\cos$
| $$\frac{\sin^2 A}{\cos^2A}=1+2\frac{\sin^2 B}{\cos^2B}$$
using the identity $$\tan A = \frac{\sin A}{\cos A}$$
now using $$\cos^2A+\sin^2A = 1$$
$$\frac{1-\cos^2 A}{\cos^2 A} = 1+ 2\frac{1-\cos^2 B}{\cos^2 B}$$
$$\frac{1}{\cos^2 A}-1 = 1+ \frac{2}{\cos^2 B} -2$$
$$2\cos^2 A=\cos^2 B$$
and using the identity $$\cos^2 A = \frac{1+ \cos 2A}{2}$$
$$1+\cos 2A = \frac{1+\cos 2B}{2}$$
and
$$1+2\cos 2A = \cos 2B$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$-\sin x -\cos x = -\sqrt{2}\sin (x+{\pi\over4})$
$-\sin x -\cos x = -\sqrt{2}\sin (x+{\pi\over4})$
How does the cosine disappear and how did sin x turn into $\sin (x+{\pi\over4})$?
| $$\sin x\cos a + \cos x \ sin a = \sin(x+a)$$
$$-\sin x -\cos x = - (\sin x +\cos x ) = -(\sin x\cos a /cos a + \cos x \ sin a / \ sin a)$$
choosing $\cos a = \ sin a$ , which means $a = \pi/4$, we have
$$-(\sin x\cos a /cos a + \cos x \ sin a / \ sin a) = -\sin(x+\pi/4)\times 2/\sqrt{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand? I know that
$x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1 = (x^4-x^3+x+1)(x^2+2x-1)$
but I would not know how to do that factoring without a software.
Some idea? Thank you!
| Note that $(x - 1/x)^{2k}$ gives $x^{2k} + x^{-2k}$ and other symmetric terms, and $(x - 1/x)^{2 k + 1} = x^{2 k + 1} - x^{- 2 k - 1}$ and other terms, it looks like due to the symmetry of the coefficients by dividing by $x^3$ you can reduce the degree to a cubic in $x - 1/x$, and go from there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that $\frac1{a^3(b+c)}+\frac1{b^3(a+c)}+\frac1{c^3(a+b)} \geq \frac32$
Let $a,b,c$ be positive real numbers with $abc = 1$. Prove that $$\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)} \geq \dfrac{3}{2}.$$
Attempt
We can rewrite it as $\dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}$. Then, using AM-GM we have $$\dfrac{bc}{2a^2\sqrt{bc}}+\dfrac{ac}{2b^2\sqrt{ac}} \dfrac{ab}{2c^2\sqrt{ab}} \geq \dfrac{bc}{a^2(b+c)}+\dfrac{ac}{b^2(a+c)}+\dfrac{ab}{c^2(a+b)}.$$
I get stuck here.
| we write $$\frac{abc}{a^3(b+c)}+\frac{abc}{b^3(a+c)}+\frac{abc}{c^3(a+b)}=\frac{bc}{a^2(b+c)}+\frac{ac}{b^2(a+c)}+\frac{ab}{c^2(a+b)}=\frac{(bc)^2}{a^2bc(b+c)}+\frac{(ac)^2}{ab^2c(a+c)}+\frac{ab)^2}{ab^2c(a+b)}$$ we use $abc=1$ again and we get
$$\frac{(bc)^2}{a(b+c)}+\frac{(ac)^2}{b(a+c)}+\frac{(ab)^2}{c(a+b)}\geq \frac{(bc+ac+ab)^2}{2(ab+bc+ca)}\geq bc+ac+ab \geq 3$$ by AM-GM
| {
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How to prove this inequality given the equation Given $ a>0 $ and $ a^5-a^3+a=3$, how can I prove the inequality: $ a^6 \ge 5 $ ?
I have tried factorizing the equation, solving for $ x, x^2, x^3 $ and then adding equations made together. Last but not least I tried, to solve for $x$ and then raise it to the power of $ x^6 $, but none of these seem to work
| I actually found a simpler solution to my problem:
$$a^5 -a^3 + a = 3 $$
Multiply by $a$:
$$a^6 - a^4 + a^2 = 3a $$
$$a^6 = a^4 - a^2 + 3a \ge 5 $$
Multiply by $a$:
$$ a^5 - a^3 + 3a^2 \ge 5a$$
Changing the first equation: $$a^5 - a^3 = 3-a$$
Then: $$3-a + 3a^2 \ge 5a$$
$$3a^2 + 3 - 6a \ge 0$$
Divide by $3$:
$$a^2 + 1 -2a \ge 0$$
$$(a-1)^2 \ge 0$$
| {
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Find general term of a recurrent sequence If $a_1=6$ and $a_{n+1}= \frac{1}{18}(2a_n-1+\sqrt{8a_n+1}), n\geq1$ expressed $a_n$ in the function of $n$.
My attempt: We calculated the first 5 terms of the series, but I was unable to determine the desired formula and then apply mathematical induction.
| We note that the terms are (perhaps surprisingly) all rational. As such, we posit that the square root will always evaluate to a rational number, i.e. that $8a_n+1=b_n^2$, where $b_n$ is rational. Let's work through the algebra of this substitution:
$$\frac{b_{n+1}^2-1}{8}=\frac{2(\frac{b_{n}^2-1}{8})-1+b_n}{18}$$
$$b_{n+1}^2-1=\frac{2(b_n^2-1)-8+8b_n}{18}$$
$$b_{n+1}^2=\frac{2b_n^2+8b_n+8}{18}=\frac{b_n^2+4b_n+4}{9}=(\frac{b_n+2}{3})^2$$
So, $b_{n+1}=\frac{b_n+2}{3}\implies b_{n+1}-1=(b_n-1)/3$. Some backtracking tells us that $b_1-1=7-1=6$, whence we see that $b_n-1=\frac{6}{3^{n-1}}$. We then see:
$$b_n=1+\frac{18}{3^n}$$
Some algebra using $a_n=\frac{b_n^2-1}{8}$ ultimately gets us to $a_n=\frac{3^{2-n}+3^{2(2-n)}}{2}$, and we are done here.
| {
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Cosine rule problem Question:
In triangle ABC, BC = a, AC= b, AB = c, and BD is perpendicular to AC.
If $\angle ABC = 60^{\circ}$, prove that $c = \dfrac{1}{2}a \pm \sqrt{b^2-\dfrac{3}{4}a^2}$
My approach: $b^2 = a^2 + c^2 - 2ac\cos B$
$b^2 = a^2 + c^2 - 2ac \cos{60}^\circ$
$b^2 = a^2 + c^2 - ac$
$b^2 = (a-c)^2+ac$
After this I don't know how to isolate the $c$ term.
| Your work is correct. Picking up your work from $b^2=a^2+c^2-ac$, subtract $b^2$ from both sides and rewrite to get
$$c^2-ac+a^2-b^2=0$$
which is quadratic in $c$. Notice this equation fits the standard quadratic form with $A=1$, $B=-a$ and $C=a^2-b^2$.
Using the quadratic formula, we get
$$c=\frac{a\pm\sqrt{a^2-4(a^2-b^2)}}{2}$$
$$c=\frac{a}{2}\pm\frac{\sqrt{4b^2-3a^2}}{2}$$
$$c=\frac{a}{2}\pm\sqrt{b^2-\frac{3}{4}a^2}\quad.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the infinite product $\sqrt\frac12\cdot\sqrt{\frac12+\sqrt\frac12}\cdot\sqrt{\frac12+\sqrt{\frac12+\sqrt\frac12}}\cdots$ Find $\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
Let $x=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
$\log x=\frac{1}{2}\log(\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}})+....$
I do not know how to solve it further.
| Begin by investigating the sequence
$$x_0:=0,\quad x_{k+1}:=\sqrt{{1\over2}+x_k}\quad(k\geq0)\ .$$
It has a certain limit $\xi$. Knowing $\xi$ draw conclusions about the limit
$$\lim_{n\to\infty}\>\prod_{k=1}^n x_k\ .$$
| {
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A right angled trapezium is circumscribed about circle.What is the radius of the circle,if the lengths of the bases are $a$ and $b.$ A right angled trapezium is circumscribed about circle.What is the radius of the circle,if the lengths of the bases(i.e. parallel sides ) are $a$ and $b.$
By using the property that the length of tangents from the external point to the circle are equal.We conclude that in the right angled trapezium $ABCD$,$AB+CD=BC+AD$ where $AD=a$ and $BC=b$ are parallel sides.
So $AB+CD=a+b$ but i do not know how to find the radius of the circle.
| Let $r$ be the radius of circle, with the center $O$, circumscribed in the trapezium $ABCD$, having $\angle ABC=\angle BCD=90^\circ$, $AB=a$ & $CD=b$ & drop a perpendicular $AE$ from vertex $A$ to the side $CD$, drop perpendiculars $OM$ & $ON$ from $O$ to the sides $AB$ & $CD$ respectively. Then apply Pythagorean theorem
in right $\triangle OMA$, $$\color{red}{OA^2}=AM^2+OM^2=(a-r)^2+r^2$$
in right $\triangle OND$, $$\color{blue}{OD^2}=DN^2+ON^2=(b-r)^2+r^2$$
in right $\triangle AOD$, $$AD^2=\color{red}{OA^2}+\color{blue}{OD^2}=(a-r)^2+r^2+(b-r)^2+r^2$$
$$AD^2=4r^2+a^2+b^2-2r(a+b)\tag 1$$
in right $\triangle AED$, $$AD^2=AE^2+DE^2=(2r)^2+(b-a)^2$$
$$AD^2=4r^2+a^2+b^2-2ab\tag 2$$
equating the values of $AD^2$ from (1) & (2),
$$4r^2+a^2+b^2-2r(a+b)=4r^2+a^2+b^2-2ab$$
$$r(a+b)=ab$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{r=\frac{ab}{a+b}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate $\int \frac{x-3}{\sqrt{1-x^2}}$
$$\int \frac{x-3}{\sqrt{1-x^2}} \mathrm dx$$
I know that $\int \frac{1}{1-x^2}\mathrm dx=\arcsin(\frac{x}{1})$ but how can I continue from here?
| Two hints:
*
*$$\frac{x-3}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}} - \frac{3}{\sqrt{1-x^2}}$$
*You had it a bit wrong: $$\int\frac{1}{\sqrt{1-x^2}} = \arcsin(x)$$
| {
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Show that $\varphi(\langle n, k\rangle) = \frac{1}{2}(n+k+1)(n+k)+n $ is injective How to show that the following function is an injective function?
$ \varphi : \mathbb{N}\times \mathbb{N} \rightarrow \mathbb{N} \\
\varphi(\langle n, k\rangle) = \frac{1}{2}(n+k+1)(n+k)+n$
I'm starting with $ \frac{1}{2}(a+b+1)(a+b)+a = \frac{1}{2}(c+d+1)(c+d)+c$, but how am I supposed to show from this equality that $\langle a, b\rangle = \langle c, d\rangle$, where $\langle a, b\rangle \in \mathbb{N}\times \mathbb{N}$ ?
| Note that
$$\begin{align}
\varphi(0, n+k) &= \frac 1 2 (n+k)(n+k+1) \\
&\le \varphi(n,k) \\
&= \frac 1 2 (n+k)(n+k+1) + n \\
&= n + \sum_{i \le (n+k)} i \\
&< \sum_{i \le (n+k+1)} i \\
&= \varphi(0, n+k+1).
\end{align}$$
Suppose $\varphi(a,b) = \varphi(c,d)$. If $a+b < c+d$, then
$$\begin{align}
\varphi(a,b) &< \varphi(0,a+b+1) \\
&\le \varphi(0,c+d) \\
&\le \varphi(c,d), \\
\end{align}$$
so $a+b \ge c+d$. Similarly, $a+b \le c+d$, so $a+b = c+d$. By definition of $\varphi$, we have $a=c$. It follows that
$$\begin{align}
(a+b)(a+b+1) &= (c+d)(c+d+1) \\
&= c^2 + 2cd + d^2 + c + d \tag{i}\\
&= (c+b)(c+b+1) \\
&= c^2 + 2cb + b^2 + c + b.\tag{ii}
\end{align}$$
Subtracting (ii) from (i),
$$\begin{align}
0 &= (c^2 + 2cd + d^2 + c + d) - (c^2 + 2cb + b^2 + c + b) \\
&= 2c(d-b) + (d^2-b^2) + (d - b) \\
&= 2c(d-b) + (b+d)(d-b) + (d - b) \\
&= (2c+b+d+1)(d-b),\\
\end{align}$$
so $(2c+b+d+1) = 0$ or $(d-b) = 0$. But $(2c+b+d+1) > 0$ as $b,c,d\in \Bbb N$, so we must have $d-b = 0$, which is to say, $b = d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$ Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$
$a.)\ 1 \\
\color{green}{b.)\ 13} \\
c.)\ 15 \\
d.)\ 16 $
$ 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)\\
=3(1-\sin (2x))^{2}+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=3(1-2\sin (2x)+\sin^{2} (2x))+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=9+3\sin^{2} 2x+4(\sin^{6} x+\cos^{6} x)\\
=9+12\sin^{2}x\times \cos^{2}x+4(\sin^{6} x+\cos^{6} x)\\$
I am stucked.
I look for a short and simple way.
I have studied maths upto $12$th grade.
| The answer should be the same for all $x$. So plug in $x = 0$,
$$3(-1)^4+6(1)^2+4(1)^6 = 13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding roots of a Complex Polynomial
Question
Given that x = 3 is a solution to $$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$ find the other two solutions.
What I have attempted;
If $x = 3$ is a root then $(x-3)$ is a factor
so $$ (x-3)(x^2 + Ax + B) = 0 $$
$$ x^3 + (A-3)x^2 + (B-3A)x - 3B = 0 $$
Equation Coefficients
$$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$
$$ x^2 :−3+A=−7−3i $$
$$ ⇒ A=−4−3i $$
$$\text{Constant: } -3B = -6(2+3i) $$
$$ ⇒ B=4+6i $$
$$ (x-3)(x^2 -(4+3i)x + 4+6i) = 0 $$
$$ x = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x = {4+3i\pm\sqrt{(4+3i)^2-4(4+6i)} \over 2} $$
$$ x = {4+3i\pm\sqrt{-9} \over 2} $$
$$ x = {4+3i±3i\over 2} $$
$$ x_1 = 3 $$
$$ x_2 = 2+3i $$
$$ x_3 = 2 $$
Hopefully this is correct , and if so I was wondering if there is an alternative method in solving this.
I was thinking because we are given x = 3 is a root and (x-3) is a factor we can do polynomial division with complex numbers? Is it possible to do polynomial division with complex numbers? If so can someone show how to get the same answer by using that method..
| If $a$ and $b$ are the other roots, then you can use Vieta's formulas for the sum and product of roots:
$$
3+a+b=7+3i,
\qquad
3ab=6(2+3i)
$$
and from those get a quadratic equation for $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$
Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$.
This question reminds me of rearrangement, but I can't really find two sequences that fit. Maybe there is a way using the triangle inequality, but I am unsure.
| Also, we can use SOS:
$$\sum_{cyc}(a^5-a^3bc)=\sum_{cyc}a^3(a^2-bc)=$$
$$=\frac{1}{2}\sum_{cyc}a^3((a-b)(a+c)-(c-a)(a+b))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)(a^3(a+c)-b^3(b+c))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)(a^2+b^2)+c(a^2+ab+b^2))\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
mysterious sum of two sequences Let
$$S_1 = \sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots$$
$$S_2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots$$
So
\begin{align}
S_1 - S_2 = {} & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\
& {} - \left[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \cdots \right] \\
= {} & 0 + 1 + 0 + \frac{1}{2} + 0 + \frac{1}{3} + 0 + \frac{1}{4} + \cdots \\
= {} & S_1
\end{align}
That makes $S_2$ zero, right?
The reason I am asking is because I was under the impression that $S_2 = \ln 2$.
What did I miss?
Is it that $S_1$ is divergent and $S_2$ is convergent and it's not allowed?
| You're saying $\infty - S_2 = \infty$; therefore $S_2=0$.
That is not valid. If $f$ approaches $\infty$ then the limits of $f$ and $f-5$ are both the same, but that doesn't mean $5=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve $\sin(3x)=\cos(2x)$
Question: Solve $\sin(3x)=\cos(2x)$ for $0≤x≤2\pi$.
My knowledge on the subject; I know the general identities, compound angle formulas and double angle formulas so I can only apply those.
With that in mind
\begin{align}
\cos(2x)=&~ \sin(3x)\\
\cos(2x)=&~ \sin(2x+x) \\
\cos(2x)=&~ \sin(2x)\cos(x) + \cos(2x)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos(x)\cos(x) + \big(1-2\sin^2(x)\big)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x)\big(1-\sin^2(x)\big)+\sin(x)-2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x)\\
\end{align}
edit
\begin{gather}
2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x) = 1-2\sin^2(x) \\
2\sin^3(x) - 3\sin(x) + 1 = 0
\end{gather}
This is a cubic right?
So $u = \sin(x)$,
\begin{gather} 2u^3 - 3u + 1 = 0 \\
(2u^2 + 2u - 1)(u-1) = 0
\end{gather}
Am I on the right track?
This is where I am stuck what should I do now?
| $$\cos2x=\sin3x=\cos\left(\dfrac\pi2-3x\right)$$
$$\iff2x=2m\pi\pm\left(\dfrac\pi2-3x\right)$$ where $m$ is any integer
Alternatively, $$\sin3x=\cos2x=\sin\left(\dfrac\pi2-2x\right)$$
$$3x=n\pi+(-1)^n\left(\dfrac\pi2-2x\right)$$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin\left(x-3\right)}}$ without L'Hôspital I have to calculate this limit $$\lim _{x\to 3}\left(3x-8\right)^{\frac{x}{\sin (x-3)}}$$
without L'Hôpital's rule.
These are my steps
$$\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\:=\:e^{\ln\left(\left(3x-8\right)^{^{\frac{x}{\sin\left(x-3\right)}}}\right)}=e\:^{\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}}$$
now i will caculate only $\frac{x\left(3x-8\right)}{\sin\left(x-3\right)}$
$$ t = x-3 \Rightarrow x = t+3$$
$\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\:=\lim \:_{t\to \:0}\:\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}$
And here im stack.
Help someone ?
Thanks.
| Note that
\begin{align}
\lim _{x\to 3}\left(\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}\right)\ &= \lim_{t\to 0}\frac{\left(t+3\right)\ln\left(3(t+3)-8\right)}{\sin\left(t\right)}\\
&= \lim_{t\to 0}\frac{\left(t+3\right)\ln\left(3t+9-8\right)}{\sin\left(t\right)}\\
&= \lim_{t\to 0}\frac{\left(3\right)\ln\left(1 + 3t\right)}{\sin\left(t\right)} \cdot \frac{t}{t} \cdot \frac{3t}{3t}\\
&= \lim_{t\to 0} 3\frac{3t}{t}\\
&= 9
\end{align}
thus
$$\lim_{x\to 3}\left(3x-8\right)^{\frac{x}{\sin (x-3)}} = \lim_{x \to 3} e^{\frac{x\ln\left(3x-8\right)}{\sin\left(x-3\right)}} = e^9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve in positive integers $x^2+3y^2=z^2$ and $x^2+y^2=5z^2$ Solve the following equations in postive integers:
*
*$x^2+3y^2=z^2$
*$x^2+y^2=5z^2$
I have solved them using this as a reference, but I am interested in other solutions, more "elegant" ones.
The equations are to be solved separate!!!
| In general, given one solution to,
$$a_1y_1^2+a_2y_2^2+\dots+ a_ny_n^2 = 0$$
then an infinite more can be found without scaling. For example,
$n=3$
$$ax_1^2+bx_2^2+cx_3^2 = (ay_1^2+by_2^2+cy_3^2)(az_1^2+bz_2^2+cz_3^2)^2$$
where,
$$x_1, x_2, x_3 = uy_1-vz_1,\; uy_2-vz_2,\; uy_3-vz_3\\
\text{and}\\
u,v = az_1^2+bz_2^2+cz_3^2,\; 2(ay_1z_1+by_2z_2+cy_3z_3)$$
If you have an initial solution $ay_1^2+by_2^2+cy_3^2 = 0$, then the identity gives you an infinite more with three free variables $z_1, z_2, z_3$.
$n=4$
$$ax_1^2+bx_2^2+cx_3^2+dx_4^2 = (ay_1^2+by_2^2+cy_3^2+dy_4^2)(az_1^2+bz_2^2+cz_3^2+dz_4^2)^2$$
where,
$$x_1, x_2, x_3, x_4 = uy_1-vz_1,\; uy_2-vz_2,\; uy_3-vz_3,\; uy_4-vz_4\\
\text{and}\\
u,v = az_1^2+bz_2^2+cz_3^2+dz_4^2,\; 2(ay_1z_1+by_2z_2+cy_3z_3+dy_4z_4)$$
Likewise, you now have four free variables $z_1, z_2, z_3, z_4$.
And so on for any $n$. I trust the pattern is easy to see?
Example:
For $x^2+y^2 = 5z^2$, we have $a,b,c = 1,1,-5$, and initial solution $y_1, y_2, y_3 = 1,2,1$. Using the formula, we get,
$$x = z_1^2 + z_2^2 - 5 z_3^2 - 2 z_1 (z_1 + 2 z_2 - 5 z_3)\\
y = 2 ( z_1^2 + z_2^2 - 5 z_3^2) - 2 z_2(z_1 + 2 z_2 - 5 z_3)\\
z = z_1^2 + z_2^2 - 5 z_3^2 - 2 z_3(z_1 + 2 z_2 - 5 z_3)$$
for three free variables $z_i$. Let, $z_2 = u + 2 v + 2 z_1,\; z_3 = v + z_1$, then $z_1$ cancels out and we recover W. Jagy's version in the other answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$ Finding $$\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$$
I think if $$\int_{1}^{\infty}x^2 \frac{2^{x-1}}{3^x}dx$$
exists that this sequence is convergent, but I doubt that this integral is equal to the number to which the sequence converges to. I do not know a way to solve this sequence, thought about doing geometric progression, but can't because of $k^2$
| The sequence can be written as $=\frac{1}{2}\sum_{1}^{\infty} k^2.a^k$ where $a=\frac{2}{3}$
Let $$S = \sum_{1}^{\infty} k^2.a^k\tag{1}$$
Multiply (1) by a,
$$aS = a^2+4a^3+9a^4+\cdot+\infty$$
Subtract now
$$(1-a)S = a+3a^2+5a^3+\cdot+\infty = 2a-a + 4a^2-a^2 + 6a^3-a^3 +\cdot+\infty$$
$$(1-a)S = 2a(1+2a+3a^2+\cdot+\infty) - a(1+a+a^2+\cdot+\infty)$$
$$(1-a)S = 2aS'-\frac{a}{1-a}$$
$$S' = 1+2a+3a^2+4a^3+\cdot+\infty$$
$$aS' = a+2a^2+3a^3+\cdot+\infty$$
$$(1-a)S' = 1+a+a^2+a^3+\cdot+\infty$$
$$S' = \frac{1}{(1-a)^2}$$
$$(1-a)S = \frac{2a}{(1-a)^2} - \frac{a}{1-a}$$
$$S = \frac{2a}{(1-a)^3} - \frac{a}{(1-a)^2}$$
Substituting the value of $a = \frac{2}{3}$
The sequence equals $\frac{1}{2}\sum_{1}^{\infty} k^2.a^k= \boxed{15}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove: $ 1\times3 +2\times4 + \cdots + n(n+2) = \frac{1}{6} \times n(n+1)(2n+7)$ using Induction I'm told to prove this by Mathematical Induction:
$ 1\times3 +2\times4 + \cdots + n(n+2) = \frac{1}{6} \times n(n+1)(2n+7)$
This is what I have so far:
BC:
Try $n=1$:
$ 1\times3 +2\times4 + \cdots + n(n+2) = \dfrac{1}{6} \times n(n+1)(2n+7)$
$ 3= \dfrac{1}{6} \times 2(9) = 3$
So Base case is true:
IH:
Let $n = k$.
$ 1\times3 +2\times4 + \cdots + k(k+2) = \frac{1}{6} \times k(k+1)(2k+7)$
IS: show that
$n \implies n+1$
I'm told to only work from one side, so I've attempted the left side (I was told I can't plug this into both sides).
We Have:
$ 1\times3 +2\times4 + \cdots + (k+1)(k+3) = ...$
I'm not sure where to go from here, any help would be greatly appreciated!!
| $$
\begin{align}
1 \times 3+2 \times 4+ \cdots +k(k+2)+(k+1)(k+3)
&=\frac{1}{6} \times k(k+1)(2k+7)+(k+1)(k+3)
\\[5pt]&=(k+1)\left(\frac{2k^2+7k}{6}+k+3\right)
\\[5pt]&=(k+1)\left(\frac{2k^2+7k+6k+18}{6}\right)
\\[5pt]&=(k+1)\left(\frac{2k^2+13k+18}{6}\right)
\\[5pt]&=\frac{(k+1)(k+2)(2(k+1)+7)}{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I calculate the sum $\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}$? I have to calculate the partial sum for an equation. How can I calculate the sum for
$$\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}\ \text{?}$$
Thanks.
| You should start with partial fraction decomposition by finding the roots of the denominator\begin{align}\frac{1}{16n^2-8n-3}&=\frac{1}{\left(4n-3\right)\left(4n+1\right)}=\frac14\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\end{align} In this way, the initial series can be written as telescoping series and therefore you can calculate the partial sum up to $N \in \mathbb N$ and then let $N\to+\infty$ as follows:
\begin{align}\sum_{n=1}^{N}&\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)=\\[0.2cm]&=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\left(\frac19-\frac1{15}\right)\dots+\left(\frac{1}{4N-3}-\frac1{4N+1}\right)\\[0.2cm]&=1+\left(\frac15-\frac15\right)+\left(\frac19-\frac19\right)+\dots+\left(\frac{1}{4N-3}-\frac1{4N-3}\right)-\frac1{N+1}\\[0.2cm]&=1-\frac{1}{4N+1}
\end{align}
So, returning to initial series
\begin{align}\sum_{n=1}^{+\infty}\frac14\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)&=\lim_{N\to+\infty}\frac14\sum_{n=1}^{N}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\\[0.2cm]&=\lim_{N\to+\infty}\frac{1}{4}\left(1-\frac1{4N+1}\right)=\frac14(1-0)=\frac14\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $F(x)=\int_0^xf(t)\,{\rm d}t$ where $f(x)= 1/(x-5)^2 + x^3$ Given the function $f(x)= 1/(x-5)^2 + x^3$, find $F(x)=\int_0^xf(t)\,{\rm d}t$. I'm not sure how to go about this problem since my function is in terms of $x$ and not $t$.
| Well, $f(t)= \frac{1}{(t-5)^2} + t^3.$
Thus, $F(x)=\displaystyle \int_\limits 0^x \left(\frac{1}{(t-5)^2} + t^3\right)\,{\rm d}t=[-(t-5)^{-1}+\frac{t^4}{4}]_0^x=-\frac{1}{x-5}+\frac{x^4}{4}-[-\frac{1}{0-5}+\frac{0^4}{4}]=-\frac{1}{x-5}+\frac{x^4}{4}+\frac{1}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of $1-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....$ $1-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....$
How can we find sum of above series upto infinite terms?
I don't know how to start and just need some hint.
| Note that
$$\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots$$
$$\frac{d}{dx}\left(\frac{1}{1 - x}\right) = 1 + 2x + 3x^2 + 4x^3 + \dots$$
$$x\cdot\frac{d}{dx}\left(\frac{1}{1 - x}\right) = x + 2x^2 + 3x^3 + 4x^4 + \dots$$
$$\frac{d}{dx}\left(x\cdot\frac{d}{dx}\left(\frac{1}{1 - x}\right)\right) = 1 + 2^2x + 3^2x^2 + 4^2x^3 + \dots$$
Now determine LHS and substitute in $x = -\frac{1}{5}$ (I used Wolfram):
$$-\frac{x + 1}{(x - 1)^3} = 1 + 2^2x + 3^2x^2 + 4^2x^3 + \dots$$
$$\frac{25}{54} = 1 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Finding a MacLaurin expansion of a function I am being asked to Find the MacLaurin expansion of the following function:
$f(x) = \frac{2x-8}{x^2-8x+12}$
I was not given a point about which to expand so I assume to use x=0. I know I can begin taking derivatives but this seems inefficient, but I do not see any way to reduce it to relate it to a more common series. Is there something I am missing, or is taking derivatives the only option?
| $\begin{array}\\
f(x)
&= \frac{2x-8}{x^2-8x+12}\\
&= \frac{2x-8}{(x-6)(x-2)}\\
&=2(x-4)\frac14\left( \frac1{x-6}-\frac1{x-2}\right)\\
&=\frac12(x-4)\left( \frac1{x-6}-\frac1{x-2}\right)\\
&=\frac12\left( \frac{x-4}{x-6}-\frac{x-4}{x-2}\right)\\
&=\frac12\left( \frac{x-6+2}{x-6}-\frac{x-2-2}{x-2}\right)\\
&=\frac12\left( 1+\frac{2}{x-6}-(1-\frac{2}{x-2})\right)\\
&=\frac12\left( \frac{2}{x-6}+\frac{2}{x-2}\right)\\
&= \frac{1}{x-6}+\frac{1}{x-2}\\
\end{array}
$
Now,
get the MacLaurin expansions
of
$\frac1{x-a}$
for
$a=2$ and $a=6$
and add them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to find pythagorean triples and n-tuples A pythagorean triple is any set of three positive integers $(a,b,c)$ where $a^2 + b^2 = c^2$
I'm wondering, is there a formula to find all pythagorean triples, and can it be generalized to $n$-tuples? (i.e $a_1^2+a_2^2+a_3^2+a_4^2+....+a_n^2=x^2$)
| If (a,b,c) is a pythagorian triple, so is (ka,kb,ck) so we only need concern ourselves with coprime triplets (if two have a common factor the third will as well).
Note: if $n = 2k+1 = m^2$ is an odd square then $k^2 + m^2 = k^2 + 2k + 1 = (k+1)^2$. So that is a method of generating infinite triplets: ie. for $i \in \mathbb Z$ $(2i^2 + 2i ,2i+1, 2i^2 + 2i+1 )$
Such solutions are all of coprime pythogarian triplets.
Wolog assume $a \le b$.
Then $a^2 = c^2 - b^2 = (c-b)(c+b)$. $\gcd((c - b),(c+b)) | 2c$ and $\gcd((c - b),(c+b)) | 2b$ so $gcd(c-b, c+ b) =\{1,2\}$. So $a^2 = (c-b)(c+b)$ means either both $(c + b)$ and $(c -b)$ are perfect squares or 2 times perfect squares. But $\gcd(a, c \pm b) =\{1,2\}$ for the same reasons. So $c -b$ = 1 or 2.
If $c - b = 2$ then $a^2 = c^2 - (c - 2)^2 = 4c - 4$ as $(a/2)^2 = c - 2$ which is pretty much impossible if $0 < a < c$.
So $c = b+1$.
Not sure how to extend to n-tuples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you go about solving this recurrence? How do you make an estimation for the substitution method, when the recursion tree did not help so much?
I have a recurrence
$$T(n) = 5\cdot T(n/3) + n (\log n)^2$$
And upon doing the recurrence tree, I got a rather difficult running time
$$ \sum_{j=1}^{\log(n)/\log(3)} \frac{5^{j-1} n\log_2^2[n/(3^{j-1})]}{(3^{j-1})^2} $$
How do you go about solving this recurrence?
| There is another closely related recurrence that admits an exact
solution. Suppose we have $T(0)=0$ and $T(1)=T(2)=1$ and for
$n\ge 3$
$$T(n) = 5 T(\lfloor n/3 \rfloor) + n \lfloor \log_3 n \rfloor^2.$$
It seems reasonable to use integer values here as the running time of
an algorithm is a function of discrete quantities.
Furthermore let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k.$$
Then we can unroll the recurrence to obtain the following exact
formula for $n\ge 3$
$$T(n) = 5^{\lfloor \log_3 n \rfloor}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor -1}
5^j \times (\lfloor \log_3 n \rfloor - j)^2 \times
\sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^{k-j}.$$
Now to get an upper bound consider a string of value two digits to
obtain
$$T(n) \le 5^{\lfloor \log_3 n \rfloor}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor -1}
5^j \times (\lfloor \log_3 n \rfloor - j)^2 \times
\sum_{k=j}^{\lfloor \log_3 n \rfloor} 2 \times 3^{k-j}.$$
This simplifies to
$$\frac{1457}{32} 5^{\lfloor \log_3 n \rfloor}
-\frac{9}{2} \lfloor \log_3 n \rfloor^2 3^{\lfloor \log_3 n \rfloor}
-\frac{45}{2} \lfloor \log_3 n \rfloor 3^{\lfloor \log_3 n \rfloor}
- 45\times 3^{\lfloor \log_3 n \rfloor}
\\ + \frac{1}{4} \lfloor \log_3 n \rfloor^2
+ \frac{5}{8} \lfloor \log_3 n \rfloor
+ \frac{15}{32}.$$
This bound is actually attained and cannot be improved upon, just like
the lower bound, which occurs with a one digit followed by zeroes to
give
$$T(n) \ge 5^{\lfloor \log_3 n \rfloor}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor -1}
5^j \times (\lfloor \log_3 n \rfloor - j)^2 \times
3^{\lfloor \log_3 n \rfloor-j}.$$
This simplifies to
$$16\times 5^{\lfloor \log_3 n \rfloor}
- \frac{3}{2} \lfloor \log_3 n \rfloor^2 3^{\lfloor \log_3 n \rfloor}
- \frac{15}{2} \lfloor \log_3 n \rfloor 3^{\lfloor \log_3 n \rfloor}
- 15 \times 3^{\lfloor \log_3 n \rfloor}.$$
Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$5^{\lfloor \log_3 n \rfloor}
\in \Theta\left(3^{\log_3 5 \times \log_3 n}\right)
= \Theta\left(n^{\log_3 5}\right).$$
This is in agreement with what the Master theorem would produce.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Factorize $2a^3 - b^3 - c^3$ I need to factorize the expression $2a^3 - b^3 - c^3$. I see that one zero is achieved when $a=b=c$, but I can't find the factor(s).
| Writing
$$
2a^3-b^3-c^3=a^3-b^3+a^3-c^3=(a-b)(a^2+ab+b^2)+(a-c)(a^2+ac+c^2)
$$
We can see that we have simple factorizations if $a=b$ or $a=c$ or $b=c$ and also for $b+a=c$. But in general we cannot find a factorization for all real values of $a,b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Complex number ( prove ) Let
$$
{x-yi\over{x+yi}}=a+bi\;\;.
$$
Prove $a^2+b^2=1$
I don't know how to start prove it, can anyone help me?
| ${x-yi\over{x+yi}}= {x-yi\over{x+yi}}{x-yi\over{x-yi}} ={x^2 - y^2 - 2xyi \over{x^2 + y^2}}={x^2 - y^2\over x^2 + y^2} + {-2xy\over x^2 + y^2}i$
So $a= {x^2 - y^2\over x^2 + y^2}$ and $b = {-2xy\over x^2 + y^2}$
So $a^2 = ({1\over x^2 + y^2})^2(x^4 - 2x^2y^2 + y^4)$ and $b^2 = ({1\over x^2 + y^2})^2(4x^2y^2)$
and $a^2 + b^2 = ({1\over x^2 + y^2})^2(x^4 + 2x^2y^2 + y^4)=({1\over x^2 + y^2})^2(x^2 + y^2)^2 = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)x^3-x^2-a}{(e^{1-\alpha x}-1)(x-1)}$ is $\frac{al(k\alpha-\beta)}{\alpha}$ If $\alpha,\beta$ are two distinct real roots of the equation $ax^3+x-1-a=0,(a\ne-1,0)$,none of which is equal to unity,then $\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)x^3-x^2-a}{(e^{1-\alpha x}-1)(x-1)}$ is $\frac{al(k\alpha-\beta)}{\alpha}$.Find $kl$.
Since $\alpha,\beta$ are the roots of the equation $ax^3+x-1-a=0$.So $a\alpha^3+\alpha-1-a=0$ and $a\beta^3+\beta-1-a=0$
$\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)x^3-x^2-a}{(e^{1-\alpha x}-1)(x-1)}$
$\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)\frac{1}{\alpha^3}-\frac{1}{\alpha^2}-a}{(e^{1-\alpha x}-1)(x-1)}$
I am stuck here.The numerator is tending to zero and the denominator is tending to zero,i applied L Hospital rule,but could not solve it.
| First of all, since we have that
$$ax^3+x-1-a=(x-1)(ax^2+ax+a+1)$$
and that $\alpha\not=1,\beta\not=1$, we can have that
$$a\alpha^2+a\alpha+a+1=a\beta^2+a\beta+a+1=0,$$
i.e.
$$(1+a)\left(\frac{1}{\alpha}\right)^2+a\cdot\frac{1}{\alpha}+a=(1+a)\left(\frac{1}{\beta}\right)^2+a\cdot\frac{1}{\beta}+a=0.$$
Since the numerator can be factored as
$$(1+a)x^3-x^2-a=(x-1)((1+a)x^2+ax+a)$$
we have
$$\lim_{x\to 1/\alpha}\frac{(x-1)((1+a)x^2+ax+a)}{(e^{1-\alpha x}-1)(x-1)}=\lim_{x\to 1/\alpha}\frac{(1+a)x^2+ax+a}{e^{1-\alpha x}-1}$$
Using l'Hôpital's rule once,
$$\begin{align}\lim_{x\to 1/\alpha}\frac{(1+a)x^2+ax+a}{e^{1-\alpha x}-1}&=\lim_{x\to 1/\alpha}\frac{2(1+a)x+a}{e^{1-\alpha x}(-\alpha)}\\&=\frac{2(1+a)(1/\alpha)+a}{e^{1-\alpha(1/\alpha)}(-\alpha)}\\&=\frac{-2-2a-a\alpha}{\alpha^2}\end{align}$$
So, noting that $\alpha+\beta=-a/a=-1\Rightarrow \beta=-\alpha-1$, we have
$$\frac{-2-2a-a\alpha}{\alpha^2}=\frac{al(k\alpha-\beta)}{\alpha}=\frac{al(k\alpha^2+\alpha^2+\alpha)}{\alpha^2}$$
$$\Rightarrow -2-2a-a\alpha=al(k\alpha^2+\alpha^2+\alpha)$$
Using $\alpha^2=-\alpha-1-\frac 1a$ and simplifying it
$$\alpha(2a+a^2-a^2l-al-alk)=a^2+a-lka^2-lka$$
$$\Rightarrow\alpha(2+a-al-l-lk)=a+1-lka-lk$$
Now using that if $p\alpha=q$, then $p^2(a\alpha^2+a\alpha+a+1)=0\Rightarrow aq^2+apq+p^2(a+1)=0$,
$$a(a+1-lka-lk)^2+a(a+1-lka-lk)(2+a-al-l-lk)+(a+1)(2+a-al-l-lk)^2=0$$
Expanding the LHS,
$$(k^2l^2+kl^2-3kl+l^2-3l+3)a^3+(3k^2l^2+4kl^2-10kl+3l^2-10l+10)a^2+(3k^2l^2+5kl^2-11kl+3l^2-11l+11)a+k^2l^2+2kl^2-4kl+l^2-4l+4=0$$
$$\Rightarrow k^2l^2+kl^2-3kl+l^2-3l+3=3k^2l^2+4kl^2-10kl+3l^2-10l+10=0$$
Now, noting that
$$3(k^2l^2+kl^2-3kl+l^2-3l+3)-(3k^2l^2+4kl^2-10kl+3l^2-10l+10)=0\Rightarrow (l-1)(kl-1)=0\Rightarrow l=1\quad\text{or}\quad kl=1$$
In either case, we have $k=l=1$, and this is sufficient.
Hence, the answer is $\color{red}{kl=1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can all of them be irrational ??
Assume that $\{x,y,x^2,y^2,xy\}$ are all irrational.
Can it be true that all of $\{x-y,x+y,x^2-y^2,x^2+y^2\}$ are irrational?
Details: $|x|\ne|y|$ and $x,y\in\mathbb R$.
In the question above I took $x$ to be $\sqrt{2} + 1$ and $y$ to be $\sqrt{2} - 1$ and thus both are irrational and their squares which are $x^2$ and $y^2$ are also irrational, so I get $x - y = (\sqrt{2} + 1) - ( \sqrt{2} - 1) = 2$ and $x^2 + y^2 = (\sqrt{2} + 1)^2 + ( \sqrt{2} - 1)^2 = 6$ to be rational.
But in the answer above it says none could be rational !! So, do I digress somewhere in my reasoning ?
| We need to show that, if any two of
$\{x-y,\; x+y,\; x^2-y^2,\; x^2 + y^2\}$
are rational then that at least one of
$\{x,\; y,\; x^2,\; y^2,\; xy \}$
is rational. Which means that the answer to the question is no.
There are $6$ ways to choose two objects out of $4$.
Case 1, 2, and 3: Any two of $\{x-y,\; x+y,\; x^2 - y^2\}$ are rational.
Since $(x-y)(x+y) = x^2 - y^2$, then if any two are rational, then all three are rational. It follows then, that $\dfrac{(x-y)+(x+y)}2 = x$ is rational in all three cases.
Case 4: $x-y$ and $x^2 + y^2$ are rational
Then $\dfrac{(x^2 + y^2) - (x-y)^2}2 = xy$ is rational.
Case 5: $x+y$ and $x^2 + y^2$ are rational
Then $\dfrac{(x+y)^2 - (x^2 + y^2)}2 = xy$ is rational.
Case 6: $x^2-y^2$ and $x^2 + y^2$ are rational
Then $\dfrac{(x^2-y^2) + (x^2 + y^2)}2 = x^2$ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The only positive integers that divide successive numbers of the form $n^2+3$ are $1$ and $13$ I stuck with this problem, I don't know how to start with.
Prove that the only positive integers that can divide successive numbers of the form $n^2+3$ are 1 or 13.
| We apply the euclidean algorithm on the polynomials:
$n^2+3,(n+1)^2+3$
We get:
$(n+1)^2+3=(n^2+3)+(2n+1)$
$2(n^2+3)=n(2n+1)+(-n+6)$
$2n+1=-2(-n+6)+13$
From here:
If $d$ divides $(n+1)^2+3$ and $(n^2+3)$ then $d$ divides $2n+1$.
If $d$ divides $n^2+3$ and $(2n+1)$ then $d$ divides $(-n+6)$.
If $d$ divides $2n+1$ and $-n+6$ then $d$ divides $13$.
Putting it all together we have that if $d$ divides $n^2+3$ and $(n+1)^2+3$ then $d$ divides $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many people are on the group? In a group of n people, two are randomly selected with probability 11/21 of being same gender. How many people are on the group if the number of males is twice as large as the number of females?
I've been trying to solve this problem by introducing a r.v X which represents the distribution of 2 males and of 2 females from M+F persons.
$$\frac{{M \choose 2}}{{M+F \choose 2}}+\frac{{F \choose 2}}{{M+F \choose 2}}=\frac{11}{21}$$
But how do I find n out of this?
| Let $x$ be the number of females. Then $2x$ is the number of males. The total population is $3x$. The probability of two males $$\Pr(MM) = \frac{2x}{3x} \frac{2x -1}{3x -1}=\frac{4x^2-2x}{9x^2-3x}.$$
Similarly, $$\Pr(FF)=\frac{x}{3x}\frac{x-1}{3x-1}=\frac{x^2-x}{9x^2-3x}.$$
The total probability should be
$$\Pr(MM \cup FF) = \frac{4x^2-2x+x^2-x}{9x^2-3x}=\frac{5x^2-3x}{9x^2-3x}=\frac{11}{21}.$$
Solving for $x$ yields $x=5$, which means there are $5$ females, $2\cdot 5=10$ males, and $15$ people in total.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the equation of diagonal If $ax^2+2hxy+by^2=0$ be the two sides of a parallelogram and $px+qy=1$ is one diagonal then prove that the other diagonal is $y(bp-hq)=x(aq-hp)$.
My solution is here;
$ax^2+2hxy+by^2=0$
Multiplying by a and adding h^2y^2,
$(ax+hy)^2=y^2(h^2-ab)
ax+hy=+/- y√(h^2-ab)$
so,
$ax+y√(h^2-ab) + hy=0,
ax-y√(h^2-ab) + hy=0$
are the two straight lines represented by the given equation and also O(0,0) is the point of intersection of these lines.
Now, How do I move further?
| You have got the two lines $L_1$ and $L_2$:
$$x = \dfrac{-h\pm\sqrt{h^2-ab}}{a}y.$$
Next step is to find the points of intersection $(x_1,y_1)$ and $(x_2,y_2)$ of these lines with the known diagonal $D_1$ with equation $\;px+qy=1$.
Finding $(x_1,y_1):\quad$ Substitute the equation for $D_1$ into the equation for $L_1$ to give:
$$ax_1 = \left(-h+\sqrt{h^2-ab}\right)\left(1-px_1\right)/q.$$
Simplify to:
$$x_1 = \dfrac{-h+\sqrt{h^2-ab}}{aq-ph+p\sqrt{h^2-ab}},\qquad\text{and then } \qquad y_1 = \dfrac{1-px_1}{q}.$$
Finding $(x_2,y_2):\quad$ Substitute the equation for $D_1$ into the equation for $L_2$ to give:
$$ax_2 = \left(-h-\sqrt{h^2-ab}\right)\left(1-px_2\right)/q.$$
Simplify to:
$$x_2 = \dfrac{-h-\sqrt{h^2-ab}}{aq-ph-p\sqrt{h^2-ab}},\qquad\text{and then } \qquad y_2 = \dfrac{1-px_2}{q}.$$
$$\\$$
Now, the unknown diagonal passes through the origin and also the point $(x_1+x_2,\;y_1+y_2)$. So it has equation:
\begin{align}
y &= \dfrac{y_1+y_2}{x_1+x_2} x \\
& \\
&= \left(\dfrac{\dfrac{1-px_1}{q} + \dfrac{1-px_2}{q}}{x_1+x_2}\right) x \\
& \\
&= \left(\dfrac{2}{q(x_1+x_2)} - \dfrac{p}{q}\right) x \\
& \\
&= \dfrac{aq-hp}{bp-hq} x \qquad\text{after substituting for $x_1,\; x_2$ and some tedious simplification.}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Substitution (integration) $$\int\frac{dx}{(x^{2}-36)^{3/2}}$$
My attempt:
the factor in the denominator implies
$$x^{2}-36=x^{2}-6^{2}$$
substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$
$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$
using trig identities:
$$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$
now using integration by parts:
$$\frac{1}{6}\int \sin^{-1}\theta$$
$$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$
using $uv-\int{vdu}$
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$
now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$
it is apparent that
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$
$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$
$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$
I have the following questions:
1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?
2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?
| Be careful: with the substitution you have
$$
(x^2-36)^{3/2}=(36\tan^2\theta)^{3/2}=216\tan^3\theta
$$
(at least in an interval where $\tan\theta$ is positive) so your integral becomes
$$
\int\frac{6\tan\theta\sec\theta}{216\tan^3\theta}\,d\theta=
\frac{1}{36}\int\frac{\cos\theta}{\sin^2\theta}\,d\theta=
-\frac{1}{36}\frac{1}{\sin\theta}+C
$$
If instead you set $x=6\cosh u$, you get $dx=6\sinh u$ and the identity $\cosh^2u-1=\sinh^2u$ brings the integral in the form
$$
\frac{1}{36}\int\frac{1}{\sinh^2u}\,du=
-\frac{1}{36}\frac{\cosh u}{\sinh u}+C=
-\frac{1}{36}\frac{x}{\sqrt{x^2-36}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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I want to solve $\int \frac{2}{x^2(x^2+1)^2}dx$ I want to solve this primitive $$I=\int \frac{2}{x^2(x^2+1)^2}dx.$$
I substitute $u=x^2$ then,
$$I=\int \frac{2}{x^2(x^2+1)^2}dx=\int \frac{du}{u^{3/2}(u+1)^2}=\cdots$$
How do I use partial fraction decomposition? Is there another idea to compute this primitive?
| HINT:
$$\int\frac{2}{x^2\left(x^2+1\right)^2}\space\text{d}x=$$
$$2\int\left[\frac{1}{x^2}-\frac{1}{x^2+1}-\frac{1}{(x^2+1)^2}\right]\space\text{d}x=$$
$$2\left[\int\frac{1}{x^2}\space\text{d}x-\int\frac{1}{x^2+1}\space\text{d}x-\int\frac{1}{(x^2+1)^2}\space\text{d}x\right]=$$
$$2\left[-\frac{1}{x}-\arctan(x)-\int\frac{1}{(x^2+1)^2}\space\text{d}x\right]=$$
Substitute $x=\tan(u)$ and $\text{d}x=\sec^2(u)\space\text{d}u$.
Then $(x^2+1)^2=(\tan^2(u)+1)^2=\sec^4(u)$ and $u=\arctan(x)$:
$$2\left[-\frac{1}{x}-\arctan(x)-\int\cos^2(u)\space\text{d}u\right]=$$
$$2\left[-\frac{1}{x}-\arctan(x)-\int\left[\frac{1}{2}+\frac{\cos(2u)}{2}\right]\space\text{d}u\right]=$$
$$2\left[-\frac{1}{x}-\arctan(x)-\int\frac{1}{2}\space\text{d}u+\int\frac{\cos(2u)}{2}\space\text{d}u\right]=$$
$$2\left[-\frac{1}{x}-\arctan(x)-\frac{1}{2}\int1\space\text{d}u+\frac{1}{2}\int\cos(2u)\space\text{d}u\right]=$$
Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.
$$2\left[-\frac{1}{x}-\arctan(x)-\frac{1}{2}\int1\space\text{d}u+\frac{1}{4}\int\cos(s)\space\text{d}s\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is this function bounded above? Consider nonconstant functions $f(x), g(x) \neq x$. Suppose there exist positive constants $k_1$ and $k_2$ such that
$k_{1} x \leq f(x) \leq k_{2} x$ and $\frac{1}{2}k_{1} x \leq g(x) \leq k_{2} x$. Does it follow that $\ \frac{f(x) + x}{\mid f(x) - g(x)\mid} < \infty$ ?
My attempt: Since $f(x) + x \leq (k_{2} + 1)x$ and $\mid f(x) - g(x)\mid \geq \frac{1}{2}k_{1}x $,
the upper bound evaluates to $\frac{2k_{2} + 2}{k_{1}} < \infty$, as required ?
| Let
\begin{align}
f(x) &= k x & \text{where $k_1 < k < k_2, k\neq -1$,} \\
g(x) &= f(x) - \frac12 k_1 x\arctan(x - 1). &
\end{align}
Then
$$
\frac{f(x) +x}{|f(x)-g(x)|}
= \frac{(k + 1)x}{\frac12 k_1 x \arctan(x - 1)}
= \frac{2(k+1)}{k_1 \arctan(x - 1)}.
$$
But $\lim_{x\to 1} \arctan(x - 1) = 0$, and $1/\arctan(x - 1)$
is unbounded in any neighborhood of $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding bivariate probability mass function (by counting?) Suppose that we role $d$ dice. Let $X, Y$ be random variables, where
$X = \#$ rolled by the die with the highest value.
$Y = \#$ rolled by the die with the second highest value.
By convention, we allow for the possibility that $X=Y$ in case more than one die has the same value (e.g. in a scenario where two dice roll a $6$ and $(d-2)$ dice roll values lower than $6$).
We want to find the probability mass function $P(X=x, Y=y)$. Clearly, $y \le x$, so this eliminates some cases.
\begin{array}{|c|c|c|c|c|c|c|}
\hline
& x = 1 & x = 2 & x = 3 & x= 4 & x = 5 & x=6 \\ \hline
y = 1 & & &\\ \hline
y = 2 & 0 & &\\ \hline
y = 3 & 0 & 0 &\\ \hline
y = 4 & 0 & 0 & 0\\ \hline
y = 5 & 0 & 0 & 0 & 0\\ \hline
y = 6 & 0 & 0 & 0 & 0 & 0\\ \hline
\end{array}
Basically we want to complete this table. There may be many elegant ways to do this, but I only thought of counting outcomes. For example, there is only one outcome such that $P(X=1, Y=1)$, namely the case in which all dice have the value $1$. This probability is $\displaystyle \frac{1}{6^d}$ (unless I am doing something terribly wrong). Therefore:
\begin{array}{|c|c|c|c|c|c|c|}
\hline
& x = 1 & x = 2 & x = 3 & x= 4 & x = 5 & x=6 \\ \hline
y = 1 & \displaystyle \frac{1}{6^d} & &\\ \hline
y = 2 & 0 & &\\ \hline
y = 3 & 0 & 0 &\\ \hline
y = 4 & 0 & 0 & 0\\ \hline
y = 5 & 0 & 0 & 0 & 0\\ \hline
y = 6 & 0 & 0 & 0 & 0 & 0\\ \hline
\end{array}
Of course, this was the easy part. Now, to compute $P(X=2, Y=1)$ we consider the outcomes in which one die has value $2$ and all the others have value $1$. Now this part is the one that I am not sure about (I am not very good --a.k.a. terrible -- with counting arguments). My idea is that since we have $d$ dice there are $d$ for us to get one die has value $2$ and all the others have value $1$. So eventually $P(X=2, Y=1) = \displaystyle \frac{d}{6^d}$. This is the same probability $P(X=3, Y=1)$, $P(X=4, Y=1)$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
& x = 1 & x = 2 & x = 3 & x= 4 & x = 5 & x=6 \\ \hline
y = 1 & \displaystyle \frac{1}{6^d} & \displaystyle \frac{d}{6^d} & \displaystyle \frac{d}{6^d}& \displaystyle \frac{d}{6^d} & \displaystyle \frac{d}{6^d} & \displaystyle \frac{d}{6^d}\\ \hline
y = 2 & 0 & &\\ \hline
y = 3 & 0 & 0 &\\ \hline
y = 4 & 0 & 0 & 0\\ \hline
y = 5 & 0 & 0 & 0 & 0\\ \hline
y = 6 & 0 & 0 & 0 & 0 & 0\\ \hline
\end{array}
Finally, to consider a harder case, something like $P(X=4, Y=3)$. This is the event in which one die has value $4$, one has value $3$ and the rest can have any value among $\{1, 2, 3\}$. There are $d$ ways to get a $4$ in one of the dice, $(d-1)$ ways to get a $3$ (since one die must be a $4$) and $3^{(d-2)}$ possibilities for the remaining dice. So the probability is eventually $\displaystyle \frac{d(d-1)3^{d-2}}{6^d}$.
Are these counting arguments correct (or, for that matter, my proposed approach) or am I missing something in this process? All comments are greatly appreciated.
| You've neglected the possibility of ties and are over counting events where multiple dice equal $y$.
You wish to calculate the probability that two dice are $x$ and $y$ and none of the remaining die are higher than $y$.
There are two cases to consider. When $x=y$ and when $x>y$
*
*When $x=y$ you want the probability that all dice are at most $x$, minus the probability that one dice equals $x$ and all the rest are less.
*When $x>y$ you want the probability that one die equals $x$ and all the rest are at most $y$ minus the probability that one die equals $x$ and all the rest are less than $y$.
$$\begin{align}\mathsf P(X=x,Y=y) & =\begin{cases} { \mathsf P(\bigcap\limits_{k\in\{1..d\}} Z_k\le x) \\- \prod\limits_{i=1}^d \mathsf P(Z_i=x,\bigcap\limits_{k\in\{1..d\}\setminus\{i\}} Z_k< x) }& : x\in\{1..6\}, y=x\\[2ex]\hdashline{ \prod\limits_{i=1}^d~\mathsf P(Z_i=x, \bigcap\limits_{k\in\{1..d\}\setminus\{i\}} Z_k\le y) \\ - \prod\limits_{i=1}^d~\mathsf P(Z_i=x, \bigcap\limits_{k\in\{1..d\}\setminus\{i\}} Z_k< y) } & : x\in\{2..6\}, y\in \{1..x-1\}\\[2ex]\hdashline 0 & :\textsf{elsewhere}\end{cases}\\[4ex]& =\begin{cases} {(\tfrac x 6)}^d - \tfrac d 6~{(\tfrac {x-1}6)}^{d-1} & : x\in\{1..6\}, y=x\\[2ex]\hdashline \tfrac d6~{(\tfrac y 6)}^{d-1} - \tfrac d6~{(\tfrac {y-1} 6)}^{d-1} & :x\in\{2..6\}, y\in \{1..x-1\}\\[2ex]\hdashline 0 & :\textsf{elsewhere}\end{cases}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difficult Integral $\int_0^{1/\sqrt{2}}\frac{\arcsin({x^2})}{\sqrt{1+x^2}(1+2x^2)}dx=$ I have a difficult integral to compute.I know the result, but need to know the method of calculation.
How prove this result? $$\int_0^{1/\sqrt{2}}\frac{\arcsin({x^2})}{\sqrt{1+x^2}(1+2x^2)}dx=\frac{\pi^2}{144}$$
| $\displaystyle J=\int_0^{\tfrac{1}{\sqrt{2}}}\dfrac{\arcsin (x^2)}{\sqrt{1+x^2}
(1+2x^2)}dx$
Perform change of variable $y=x^2$, one obtains:
$\displaystyle J=\int_0^{\tfrac{1}{2}}\dfrac{\arcsin x}{2\sqrt{x(1+x)}(1+2x)}dx$
$\displaystyle J=\left[\arcsin(x)\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)\right]_0^{\frac{1}{2}}-\int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx$
$\displaystyle J=\dfrac{\pi^2}{36}-\int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx$
Define on $[0,1]$, $\displaystyle F(a)=\int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(a\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx$
For all $a$ in $[0,1]$, $\displaystyle F'(a)=\int_0^{\tfrac{1}{2}}\dfrac{x}{\sqrt{x(1-x)}(a^2x+x+1)}dx$
For all $a$ in $[0,1]$, $F'(a)=2\left[\dfrac{\sqrt{2+a^2}\arctan\left(\dfrac{\sqrt{x}}{\sqrt{1-x}}\right)-\arctan\left(\dfrac{\sqrt{2+a^2}\sqrt{x}}{\sqrt{1-x}}\right)}{(1+a^2)\sqrt{2+a^2}}\right]_0^{\tfrac{1}{2}}$
For all $a$ in $[0,1]$, $F'(a)=\dfrac{\pi}{2}\dfrac{1}{1+a^2}-\dfrac{2\arctan\left(\sqrt{2+a^2}\right)}{(1+a^2)\sqrt{2+a^2}}$
Therefore,
$\displaystyle \int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx=\int_0^1 F'(a)da=\dfrac{\pi}{2}\int_0^1\dfrac{1}{1+a^2}da-2\int_0^1\dfrac{\arctan\left(\sqrt{2+a^2}\right)}{(1+a^2)\sqrt{2+a^2}}da$
Therefore,
$\displaystyle J=\dfrac{\pi^2}{36}-\dfrac{\pi^2}{8}+2\int_0^1\dfrac{\arctan\left(\sqrt{2+a^2}\right)}{(1+a^2)\sqrt{2+a^2}}da$
The last integral is the Ahmed's integral (see How to evaluate Ahmed's integral? )
Finally,
$J=\dfrac{\pi^2}{36}-\dfrac{\pi^2}{8}+2\times \dfrac{5\pi^2}{96}=\dfrac{\pi^2}{144}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$ How do I prove the following inequality
$$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$
without the help of induction?
Thanks for any help!!
| $f(x)=\frac{1}{n+x}$ is a convex function on $\mathbb{R}^+$, hence:
$$ \frac{f(1)+f(2)+\ldots+f(n)}{n}\geq f\left(\frac{1+2+\ldots+n}{n}\right) \tag{1} $$
implies:
$$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\geq \frac{n}{n+\frac{1+2+\ldots+n}{n}}=\frac{2n}{3n-1}.\tag{2}$$
On the other hand, the Hermite-Hadamard inequality gives:
$$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\leq \int_{n+\frac{1}{2}}^{2n+\frac{1}{2}}\frac{dx}{x} = \log\left(\frac{4n+1}{2n+1}\right)\leq \log 2.\tag{3}$$
An alternative approach is to prove first that the sequence $\{a_n\}_{n\geq 1}$ given by $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$ is increasing, then to compute its limit at $n\to +\infty$
and check it is less than $\frac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Representing a linear operator on $V$ with an element of $V \otimes V^*$ I got interested by the first sentence of this wikipedia subsection. It claims that any linear operator $f:V\to V$ can be represented by an element of $V\otimes V^*$ in a very concrete way: the element $w\otimes h$ maps a vector $v$ to $h(v)w$.
Clearly such an object is a linear operator, but given a basis I wanted to "see" the relationship between the matrix representing $f$ and the coefficients of $h$ and $w$. So I came up with an arbitrary matrix:
$$
M=\begin{pmatrix}
2 & 3 \\
5 & 7 \\
\end{pmatrix}
$$
and tried to determine $w$ and $h$ but found I couldn't.
I think I understand my mistake: this particular linear operator in this basis is represented by an element of $V\otimes V^*$ that isn't a pure tensor.
My question: Given that, is there an algorithmic way to find the right element of $V\otimes V^*$?
EDIT: Marko's answer gives the general solution but in the interim I found a solution for this particular matrix. I thought it might help someone in the future to see the grisly working-out, so here it is...
I noticed that the cases where the answer is a pure tensor are those where the matrix's columns are linearly dependent. So I wrote it like this:
$$
\begin{pmatrix}
2 & 3 \\
5 & 7 \\
\end{pmatrix}
=
\begin{pmatrix}
a & \lambda a \\
b & \lambda b \\
\end{pmatrix}
+
\begin{pmatrix}
c & \gamma c \\
d & \gamma d \\
\end{pmatrix}
$$
This was easy to solve by inspection:
$$
\begin{pmatrix}
2 & 3 \\
5 & 7 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}
+
\begin{pmatrix}
1 & 1 \\
3 & 3 \\
\end{pmatrix}
$$
I could then express each of these as a pure tensor:
$$
\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}
=\begin{pmatrix}
1 & 2 \\
\end{pmatrix}
\otimes
\begin{pmatrix}
1 \\
2 \\
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
1 & 1 \\
3 & 3 \\
\end{pmatrix}
=\begin{pmatrix}
1 & 1 \\
\end{pmatrix}
\otimes
\begin{pmatrix}
1 \\
3 \\
\end{pmatrix}
$$
So the answer I was looking for was
$$
\begin{pmatrix}
2 & 3 \\
5 & 7 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 2 \\
\end{pmatrix}
\otimes
\begin{pmatrix}
1 \\
2 \\
\end{pmatrix}
+
\begin{pmatrix}
1 & 1 \\
\end{pmatrix}
\otimes
\begin{pmatrix}
1 \\
3 \\
\end{pmatrix}
$$
| We see this in quantum mechanics. We can expand linear operator in terms of bra and ket.
$$ A = \sum |x><x|A|y><y| $$
Then $\sum |x><x|=1 $ is the identity matrix and $<x|A|y> $ are the matrix elements.
Then again. Physics is know for suggestive notations which may not work 100% of the time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Showing that this sequence is eventually decreasing I'm trying to show that this sequence $$a_n = \frac{3^n-7}{4^n+5}$$ is decreasing for all $n$ greater than some $N\in \Bbb N$.
All I can see to do is something like
$$a_{n+1} = \frac{3^{n+1}-7}{4^{n+1}+5} = \frac{3\cdot3^n-7}{4\cdot 4^n+5}\le \frac{3\cdot3^n-7}{4^n+5}$$
But that last expression is not less than $a_n$ for large $n$. Is there some better way to do this?
| Calculate $a_{n+1}-a_n$, then
$$\frac{3^{n+1}-7}{4^{n+1}+5}-\frac{3^n-7}{4^n+5}=\frac{-3^n 4^n +21\cdot 4^n + 10\cdot 3^n}{(4^{n+1}+5)(4^n+5)}. $$
If we show $-3^n 4^n +21\cdot 4^n + 10\cdot 3^n < 0$ for large $n$, then the proof is over. Since $3^n < 4^n$ for all $n\in\mathbb{N}$,
$$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n < -3^n4^n + 31\cdot 4^n=(31-3^n)4^n.$$
If $n\ge 4$, then $31-3^n < 0$, and so $(31-3^n)4^n< 0$. Therefore
$$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n <(31-3^n)4^n< 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $f(n) =\frac{(n+1)^n}{n^{n+1}}$ is Monotonic
Prove that for each $n \in N$,
$$f(n) =\frac{(n+1)^n}{n^{n+1}}$$
is monotonic.
First, I can tell that the function is decreasing. If I take $\frac{1}{n}$, the function looks like $\frac{1}{n}(1+\frac{1}{n})^n$. Can this help?
| Consider $$\frac{f(n)}{f(n+1)} = \frac{(n+1)^n}{n^{n+1}}\cdot \frac{(n+1)^{n+2}}{(n+2)^{n+1}}$$
$$ = \frac{(n+1)^n}{n^n \cdot n}\cdot \frac{(n+1)^n \cdot (n+1)^2}{(n+2)^n \cdot (n+1)}$$
$$ = \bigg(\frac{(n+1)^2}{n(n+2)}\bigg)^n \cdot\frac{(n+1)}{n}$$
Now for $n>0$ we have $$\frac{(n+1)^2}{n(n+2)}>1$$ And clearly $$\frac{n+1}{n} >1$$
So $$\frac{f(n)}{f(n+1)} >1 \Rightarrow f(n)>f(n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why is $x^2+1$ divisible by $10$ if $x$ has a $3$ or $7$ in the one's place? So I have the simple polynomial $x^2+1$. If I plug in ANY number that has a $3$ or a $7$ in the ones place $x^2+1$ is divisible by $10$. Why? Is there a reason for this?
So numbers like $3,7,13,17,23,27,\ldots$ when plugged into $x^2+1$ is divisible by 10. Why? Is there a reason for this?
| \begin{align*}
(10k + 3)^2 + 1 &= 100k^2 + 60 k + 9 + 1 = 10(10k^2 + 6k + 1)
\end{align*}
and similarly for $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results.
| Let $\theta =\arctan(11/2)$. Then, $\cos(\theta)=\frac{2}{5\sqrt 5}$.
Now, let $\alpha$ be given by $\alpha=3\arccos(2/\sqrt 5)$. Using the triple angle formula
$$\cos(3x)=\cos^3(x)-3\sin^2(x)\cos(x)$$
we find that
$$\begin{align}
\cos(\alpha)&=\cos(3\arccos(2/\sqrt 5))\\\\
&=\cos^3(\arccos(2/\sqrt 5))-3\sin^2(\arccos(2/\sqrt 5))\cos(\arccos(2/\sqrt 5))\\\\
&=\frac{8}{5\sqrt 5}-3\left(\frac15\right)\left(\frac{2}{\sqrt 5}\right)\\\\
&=\frac{2}{5\sqrt 5}
\end{align}$$
Therefore, $\alpha = \theta$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Showing Trigonometric Identity Prove that:
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Attempt:
\begin{align*}
L.H.S & = \cos^2\theta\sin^4\theta\\
& = \cos^2\theta\sin^2\theta\sin^2\theta\\
& = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\
& = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta)
\end{align*}
Now, what should I do?
| HINT: use that $$\cos(6\theta)=32\, \left( \cos \left( \theta \right) \right) ^{6}-48\, \left( \cos
\left( \theta \right) \right) ^{4}+18\, \left( \cos \left( \theta
\right) \right) ^{2}-1
$$
$$\cos(2\theta)=2\, \left( \cos \left( \theta \right) \right) ^{2}-1$$
$$\cos(4\theta)=8\, \left( \cos \left( \theta \right) \right) ^{4}-8\, \left( \cos
\left( \theta \right) \right) ^{2}+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Proof of limit of $x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)$ correct? I want to show that $$\lim_{x\to\infty} x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= 1 $$
But I'm not sure if it's correct:
Because $\arcsin:[-1;1]\to \mathbb{R}$ is continuous and differentiable on $(-1,1)$, the mean value theorem says that there is a $c$ in $\left(\frac{1}{x+1} , \frac{1}{x}\right)$ for $f(t)=\arcsin(t)$ such that
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= f'(c_x) {\frac{x}{1+x}} $$
then
$f'(c_x)=\frac{1}{(\sqrt{1-c_x})}(\frac{x}{x+1}))\leqslant \frac{1}{(\sqrt{1-c_x})}\times\frac{x}{x} = \frac{1}{(\sqrt{1-c_x})} $
with Sandwichlemma for $x\to\infty, c_x\to0$
so the limit of $(x^2(\arcsin{\frac{1}{x}} -\arcsin{\frac{x}{1+x}} )= 1$
| If one wishes to use the MVT, then for $f(x)=\arcsin(x)$, there exists $\xi\in (y,z)$ such that
$$\arcsin(z)-\arcsin(y)=\frac{1}{\sqrt{1-\xi^2}}(z-y)$$
Letting $z=\frac1x$ and $y=\frac1{x+1}$, we have
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)=\frac{1}{\sqrt{1-\xi^2}}\frac{x}{x+1}$$
Inasmuch as $\frac{1}{x+1}<\xi<\frac1x$, we have $\lim_{x\to \infty}\xi =0$ and therefore,
$$\lim_{x\to \infty}\left(x^2\arcsin\left(\frac{1}{x}\right)-x^2\arcsin\left(\frac{1}{x+1}\right)\right)=1$$
If one would like to use the squeeze theorem in a more direct manner, then we have
$$\frac{x+1}{\sqrt{x(x+2)}}<\frac{1}{\sqrt{1-\xi^2}}<\frac{x}{\sqrt{x^2-1}}$$
which shows that
$$\frac{x}{\sqrt{x(x+2)}}<\left(x^2\arcsin\left(\frac{1}{x}\right)-x^2\arcsin\left(\frac{1}{x+1}\right)\right)<\frac{x^2}{(x+1)\sqrt{x^2-1}}$$
whereupon application of the squeeze theorem yields the expected result.
| {
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"url": "https://math.stackexchange.com/questions/1650854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the shortest distance between two Parabolas Recently, a problem asked me to find the minimum distance between the parabolas $y=x^2$ and $y=-x^2-16x-65$.
I proceeded with the problem as thus.
Let $P(a,a^2), Q(b, -b^2-16b-65), a-b=x$.
$\therefore PQ^2=x^2+(2a^2+2ax+16a+x^2+16x+65)^2$.
$PQ^2=x^2+(2(a+\frac{x+8}{2})^2+\frac{(x+8)^2+2}{2})^2 \ge (x^2+(\frac{(x+8)^2+2}{2})^2)(1+\frac{1}{4}) \times \frac{4}{5}$
Applying Cauchy gives us that
$PQ^2 \ge (\frac{1}{4}x^2+3x+\frac{33}{2})^2 \times \frac{4}{5} \ge (\frac{15}{2})^2 \times \frac{4}{5}=75$
This implies that the answer is $\sqrt{75}$.
However, it took me a long time to find the values for Cauchy, and the calculations proved tedious.
What are other approaches to this problem?
EDIT: $(\frac{15}{2})^2 \times \frac{4}{5} \neq 75$, it`s $45$ actually!
| Suppose the closest points are $(a,a^2)$ and $(b,-b^2-16b-65)$.
As @brevan-ellefsen noted, the slopes of these parabolas are equal at these points, so $2a=-2b-16$, or $b=-a-8$.
The closest points are therefore $(a,a^2)$ and $(-a-8,-(-a-8)^2-16(-a-8)-65)$ for some value of $a$. The squared distance between these points is $$s(a)=(2a+8)^2+(a^2+(a+8)^2-16(a+8)+65)^2=65 + 32 a + 8 a^2 + 4 a^4.$$ This is smallest when $s'(a)=16(a^3+a+2)=(a+1)(a^2-a+2)=0$, or when $a=-1$. When $a=-1$, $s(a)=65 -32 + 8 + 4=45$, so the minimum distance between the parabolas is $\sqrt{45}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Number of terms in the expansion of $\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$
Number of terms in the expansion of $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$$
$\bf{My\; Try::}$ We can write $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n=\frac{1}{x^{2n}}\left(1+x+x^2\right)^n$$
Now we have to calculate number of terms in $$(1+x+x^2)^n$$ and divided each term by $x^{2n}$
So $$(1+x+x^2)^n = \binom{n}{0}+\binom{n}{1}(x+x^2)+\binom{n}{2}(x+x^2)^2+......+\binom{n}{n}(x+x^2)^n$$
and divided each term by $x^{2n}$
So we get number of terms $\displaystyle = 1+2+3+.....+n+1 = \frac{(n+1)(n+2)}{2}$
Is my solution is Right, If not then how can i calculate it
actually i don,t have solution of that question.
Thanks
| The number of terms in the expansion of $$\left(1+\frac1{x}+\frac1{x^2}\right)^n=\dfrac{(1+x+x^2)^n}{x^{2n}}$$
will be same as in $(1+x+x^2)^n$
As $a^3-b^3=(a-b)(a^2+ab+b^2),$
$$(1+x+x^2)^n=\left(\dfrac{1-x^3}{1-x}\right)^n =(1-x^3)^n(1-x)^{-n}$$
The highest & the lowest power of $x$ in $(1+x+x^2)^n$ are $0,2n$ respectively.
Clearly, all the terms present in $1\cdot(1-x)^{-n}$ (using Binomial Series expansion) as in $(1-x^3)^n(1-x)^{-n}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Easiest way to solve the equation $x^\frac 43 = \frac {16}{81}$ What would be the easiest way to solve this?
$$x^\frac 43 = \frac {16}{81}$$
I saw this in class and have no clue how did they get $$x = \frac 8{27}$$
| Given real number $x$ such that $x^\frac{4}{3} = \frac{16}{81}$:
$\sqrt[3]{x}^4 = (\frac{2}{3})^4$.
$\sqrt[3]{x} = \frac{2}{3}$ or $\sqrt[3]{x} = -\frac{2}{3}$.
$x = (\frac{2}{3})^3$ or $x = (-\frac{2}{3})^3 = -(\frac{2}{3})^3$.
To actively counter the common error made by the other respondents, note all the following for real $x$:
It is false that $(x^4)^\frac{1}{4} = x$. It is true that $(x^4)^\frac{1}{4} = |x|$.
It is false that $(x^\frac{4}{3})^\frac{3}{4} = x$. It is true that $(x^\frac{4}{3})^\frac{3}{4} = |x|$.
That is the reason why those respondents do not get the two solutions that I have shown above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
proving convergence and calculating sum of a series So I have this series:
$$\displaystyle\sum_{n=1}^{\infty}\frac{3n^2+15n+9}{n^4+6n^3+9n^2}$$
I noticed:
$$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}(\frac{3n^2+15n+9}{n^2+6n+9})$$
So can I say that it's convergent already or should I use a critera on the second fraction in the sum?
Also for sum I could use partial fractions or am I mistaken?
Any help would be appreciated.
| Consider
\begin{align}
a_n=\frac{3n^2+15n+9}{n^4+6n^3+9n^2} &\le \frac{3n^2+15n^2+9n^2}{n^2(n+3)^2} \\
&= \frac{27}{(n+3)^2}\\
&\le \frac{27}{n^2} = b_n
\end{align}
It is well know that
$$\sum_{n=1}^\infty \frac{1}{n^2}$$
converges. Since $0\le a_n\le b_n$ and $\sum b_n$ converges, we may conclude that $\sum a_n$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1656716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Proving that $\lceil x-\lceil y \rceil \rceil =\lfloor \lceil x \rceil - y \rfloor$ Having trouble proving this equality for all real numbers. I understand that the left side simplifies to $\lceil x \rceil$ - $\lceil y \rceil$, however I cannot for the life of me get the right side simplified to equal the left side.
$\lceil x-\lceil y \rceil \rceil =\lfloor \lceil x \rceil - y \rfloor$
| For any real numbers, $x, y$ there are integers $n_x$ and $n_y$ such that
$n_x \le x < n_x + 1$ and $n_y \le y < n_y + 1$.
Case 1: Neither $x$ nor $y$ are integers.
$\lceil x-\lceil y \rceil \rceil = \lceil x-(n_y + 1) \rceil= \lceil x\rceil-(n_y + 1)= n_x + 1 - n_y - 1 = n_x - n_y$
While $\lfloor \lceil x \rceil - y \rfloor = \lfloor n_x + 1 - n_y - (y - n_y) \rfloor = n_x - n_y + \lfloor 1 - (y - n_y) \rfloor = n_x - n_y + 0 = n_x - n_y$
Case 2: $x = n_x$ is an integer. $y$ is not.
$\lceil x-\lceil y \rceil \rceil = x-(n_y + 1)= n_x - n_y - 1 $
While $\lfloor \lceil x \rceil - y \rfloor = \lfloor n_x - n_y - (y - n_y) \rfloor = n_x - n_y + \lfloor - (y - n_y) \rfloor = n_x - n_y -1$
Case 3: $x$ is not and integer. $y = n_y$ is.
$\lceil x-\lceil y \rceil \rceil = \lceil x-n_y \rceil= \lceil x\rceil-n_y= n_x + 1 - n_y $
While $\lfloor \lceil x \rceil - y \rfloor = \lfloor n_x + 1 - n_y \rfloor = n_x +1 - n_y$
Case 4: $x = n_x$ and $y = n_y$ are both integers.
$\lceil x-\lceil y \rceil \rceil =\lceil x_n-y_n \rceil = x_n - y_n= \lfloor x_n - y_n \rfloor = \lfloor \lceil x \rceil - y \rfloor$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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obtaining Bernoulli numbers from determinant I am reading a paper entitled Bernoulli Numbers Via Determinants by Hongwei Chen and I'm confused about a particular step. The author sets up a system of equations via the following: first let $B_n$ represent the $n$-th Bernoulli number. Then
$$x=(e^x-1)\sum_{n=0}^\infty B_n\frac{x^n}{n!}$$
Letting $b_n=B_n/n!$ and expanding we get
$$x=\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)\left(b_0+b_1x+b_2x^2+...\right)$$
Immediately we can see that $b_0=1$. Using Cauchy Products, we then obtain an infinite sequence of equations which are coefficients of the powers of $x$. But we don't need infinite, we can look at the system for coefficients up to $x^n$. Therefore
\begin{cases}
b_1=-\frac{1}{2!} \\[2ex]
\frac{b_1}{2!}+b_2=-\frac{1}{3!} \\[2ex]
\frac{b_1}{3!}+\frac{b_2}{2!}+b_3=-\frac{1}{4!} \\[2ex]
\vdots \\[2ex]
\frac{b_1}{n!}+\frac{b_2}{(n-1)!}+...+b_n=-\frac{1}{(n+1)!} \\[2ex]
\end{cases}
Then the author goes on to state that applying Cramers rule produces
$$b_n=
\begin{vmatrix}
1 & 0 & 0 & \cdots & -\frac{1}{2!} \\
\frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\
\frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\
\end{vmatrix}
$$
Then
$$B_n=n!
\begin{vmatrix}
1 & 0 & 0 & \cdots & -\frac{1}{2!} \\
\frac{1}{2!} & 1 & 0 & \cdots & -\frac{1}{3!} \\
\frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & -\frac{1}{4!} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n!} & \frac{1}{(n-1)!} & \frac{1}{(n-2)!} & \cdots & -\frac{1}{(n+1)!} \\
\end{vmatrix}
$$
And finally,
$$B_n=(-1)^n n!
\begin{vmatrix}
\frac{1}{2!} & 1 & 0 & \cdots & 0 \\
\frac{1}{3!} & \frac{1}{2!} & 1 & \cdots & 0 \\
\frac{1}{4!} & \frac{1}{3!} & \frac{1}{2!} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{(n+1)!} &\frac{1}{n!} & \frac{1}{(n-1)!} & \cdots & \frac{1}{2!} \\
\end{vmatrix}
$$
Can someone please explain the last step? I just don't see how that last step comes from the previous. I figure it has something to do with properties of determinants, and i do know that $|cA|=c^n|A|$, but I don't see where that is coming from here.
| Here is the answer:
If $c$ is a constant and $A$ is an $n \times n$ matrix, then
$\det c A = c^n \det A$. Note that here $c$ multiplies all columns
not just the first column. Now, if $c$ multiplies only one column
then $\det [A_1, A_2, \cdots, c A_i, \cdots A_n]= c \det A$.
Now for the problem. The constant $c=-1$ only multiplies the first
column after being rolled from the last place ($n-1$ swaps).
Then we have the total sign is $(-1)^{n-1} (-1) = (-1)^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\int_0^\infty \frac{x^{k-1} + x^{-k-1}}{x^a + x^{-a}}dx = \frac{\pi}{a \cos(\frac{\pi k}{2a})}$. I need help in proving this identity
$$\int_0^\infty \frac{x^{k-1} + x^{-k-1}}{x^a + x^{-a}}dx = \frac{\pi}{a \cos(\frac{\pi k}{2a})}$$
for $0<k<a.$ It might be done using residues, but I don't know which contour to choose.
| \begin{align*}
I &= \int^\infty_0 \frac{x^{k - 1} + x^{-kn - 1}}{x^a + x^{-a}} \, dx\\
&= \int^\infty_0 \frac{x^{k - 1} + x^{-k - 1}}{x^{-a} (x^{2a} + 1)} \, dx\\
&= \int^\infty_0 \frac{x^{k + a -1} + x^{a - k - 1}}{x^{2a} + 1} \, dx
\end{align*}
Let $t = x^{2a}, x = t^{\frac{1}{2a}}, dx = \frac{1}{2a} t^{\frac{1 - 2a}{2a}} \, dt$ while the limits of integration remain unchanged. Thus
\begin{align*}
I &= \int^\infty_0 \frac{t^{\frac{k + a - 1}{2a}} + t^{\frac{a - k - 1}{2a}}}{1 + t} \cdot \frac{1}{2a} t^{\frac{1 - 2a}{2a}} \, dt\\
&= \frac{1}{2a} \int^\infty_0 \int^\infty_0 \frac{t^{\frac{k - a}{2a}} + t^{\frac{-a-k}{2a}}}{1 + t} \, dt\\
&= \frac{1}{2a} \int^\infty_0 \frac{t^{\frac{k + a}{2a} - 1}}{(1 + t)^{\frac{k + a}{2a} + \frac{a - k}{2a}}} \, dt + \frac{1}{2a} \int^\infty_0 \frac{t^{\frac{a - k}{2a} - 1}}{(1 + t)^{\frac{a - k}{2a} + \frac{k + a}{2a}}} \, dt\\
&= \frac{1}{2a} \mbox{B} \left (\frac{k + a}{2a}, \frac{a - k}{2a} \right ) + \frac{1}{2a} \mbox{B} \left (\frac{k - a}{2a}, \frac{a + k}{2a} \right )
\end{align*}
Here $\displaystyle{\mbox{B} (m,n)}$ is the beta function. Since $\mbox{B}(m,n) = \mbox{B}(n,m)$ we have
\begin{align*}
I &= \frac{1}{a} \mbox{B} \left (\frac{k + a}{2a}, \frac{a - k}{2a} \right )\\
&= \frac{1}{a} \cdot \frac{\Gamma \left (\frac{k + a}{2a} \right ) \Gamma \left (\frac{a - k}{2a} \right )}{\Gamma \left (\frac{k + a}{2a} + \frac{a - k}{2a} \right )}, \quad \mbox{since} \,\, \mbox{B}(m,n) = \frac{\Gamma (m) \Gamma (n)}{\Gamma (m + n)}\\
&= \frac{1}{a} \Gamma \left (\frac{1}{2} + \frac{k}{2a} \right ) \Gamma \left (\frac{1}{2} - \frac{k}{2a} \right ), \quad \mbox{since} \,\, \Gamma (1) = 1\\
&= \frac{\pi}{a \cos \left (\frac{k\pi}{2} \right )}
\end{align*}
as required to show. Note in the last line the following reflection formula for the gamma function has been used
$$\Gamma \left (\frac{1}{2} + x \right ) \Gamma \left (\frac{1}{2} - x \right ) = \frac{\pi}{\cos (x \pi)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proof of $n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$ by mathematical induction
Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$
My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.
(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6
We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$
I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?
| First, show that this is true for $n=1$:
$1(1^2+5)=6$
Second, assume that this is true for $n$:
$n(n^2+5)=6k$
Third, prove that this is true for $n+1$:
$(n+1)((n+1)^2+5)=$
$\color\red{n(n^2+5)}+3n^2+3n+6=$
$\color\red{6k}+3n^2+3n+6=$
$6\left(k+\frac{n(n+1)}{2}+1\right)$
Since either $n$ or $n+1$ is even, $\frac{n(n+1)}{2}$ is integer.
Please note that the assumption is used only in the part marked red.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $I=\iint\limits_D(x^2 - y^2)d\sigma, D:0\le y\le \sin x, 0 \le x \le \pi$ For this double integral $$I=\iint\limits_D(x^2 - y^2)d\sigma, D:0\le y\le \sin x, 0 \le x \le \pi$$
I am calculating it like this:
$$I = \int_0^\pi dx \int_0^{\sin x} (x^2 - y^2) dy = \int_0^\pi (x^2y - \frac13y^3)\bigg|_0^{\sin x} dx=\int_0^\pi (x^2\sin x-\frac13\sin^3 x)dx$$
$$=-\int_0^\pi x^2d\cos x -\frac13 \int_0^\pi (\sin x -\sin x\cos ^2x) dx$$
This becomes rather complex, am I doing it in right approach or have I got anywhere wrong? Is there a simpler way to solve this?
| You have:
$$=-\int_0^\pi x^2d\cos x -\frac13 \int_0^\pi (\sin x -\sin x\cos ^2x) dx$$
The first integral is (integrating by parts):
$$-\int_0^\pi x^2d\cos x=-x^2\cos x\big|_0^\pi+\int\limits_{0}^{\pi}{2x\cos xdx}=\pi^2+2x\sin x\big|_0^\pi-2\int\limits_{0}^{\pi}\sin xdx=\pi^2-4$$
The other two integrals are also clear:
$$-\frac13 \int_0^\pi \sin x dx=\frac{1}{3}\cos x\big|_0^\pi=-\frac{2}{3}$$
$$\frac{1}{3}\int\limits_{0}^{\pi}{\sin x\cos ^2xdx}=-\frac{1}{3}\int\limits_{0}^{\pi}{\cos^2xd\cos x}=-\frac{1}{3}\frac{\cos^3x}{3}\big|_0^\pi=\frac{2}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation $x^3+2x+1=2^n$ in positive integers Determine all pairs of positive integers $(x,n)$ which satisfy the condition
$$x^3+2x+1=2^n.$$
My work so far:
No solution exists for $n=1$. For $n=2$ we get $x=1$.
We show that no solutions exist for $n\ge3$. Suppose that $n \ge 3$. Obviously, $x$ is odd. Then $x^2+2\equiv 3 \pmod 8$. As with the original equation $x(x^2+2)\equiv -1 \pmod 8$, then $x\equiv 5\pmod 8$.
I can not get a contradiction.
| Suppose $n \ge 3$.
Note that $x \equiv 5 \pmod 8$.
Since $x^3+2x+1 \equiv 3x+1 \equiv 1 \pmod 3$ by Fermat's little theorem, $2^{n}$ divided by $3$ has a remainder of $1$. This implies that $n$ is a even number. Let $n=2k$.
By adding $2$ to each side, the equation can be factorized so that $$(x+1)(x^2-x+3)=2^{2k}+2.$$
Since $x^2-x+3 \equiv 23 \equiv 7 \pmod 8$, there exists a prime number $p$ where $p \equiv 5, 7\pmod8$ such that $p$ divides $x^2-x+3$. (If there is no such $p$, then $x^2-x+3 \equiv 3^{k} \pmod 8$. But $3^{k} \not \equiv 7 \pmod 8$). Since $x^2-x+3$ is a factor of $2^{2k}+2$, $p$ divides $2^{2k}+2$.
That means $(2^{k})^2 \equiv -2 \pmod p$. But $-2$ is a non-quadratic residue of $p$ since $p \not\equiv 1,3\pmod 8$. A contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an inequality of three sequences Let two sequences ($u_n$) and ($v_n$):
\begin{align}
u_n &= \sin(1/n^2) + \sin(2/n^2) + \ldots + \sin(n/n^2)\\[0.2cm]
v_n &= 1/n^2 + 2/n^2 + .... + n/n^2
\end{align}
In the previous parts of my exercise, I proved that $v_n$ is decreasing, convergent and its limit towards $+\infty$ is $\frac12$. I also proved that the following functions are all increasing and strictly positive in the interval $(0,+\infty)$
\begin{align}
f(x) &= x - \sin (x)\\
g(x) &= -1 + \tfrac12x^2 + \cos (x)\\
h(x) &= -x + \tfrac16x^3 + \sin (x)
\end{align}
Then, using these informations, I must prove the following inequality:
$$v_n - \frac16 \cdot\frac1{n^2} \leq u_n \leq v_n$$
before inferring that $u_n$ is convergent and $u_n$'s limit.
The problem is that I really don't know how to manage this part of my exercise : how to prove this inequality of sequences?
Thanks for your answers.
| You have done the difficult part which is to calculate the limit of the $v_n$ and to show that $h(x),f(x)\ge0$. Just as an observation: $h'(x)=g(x)$ and $h''(x)=g'(x)=f(x)$. Now, for any $1\le k\le n$ $$0\le h(k/n^2)=-\frac{k}{n^2}+\frac16\cdot\frac{k^3}{n^6}+\sin{\left(k/n^2\right)} \implies \frac{k}{n^2}-\frac{k^3}{6n^6}\le\sin{\left(k/n^2\right)}$$ and summing up over $k$ \begin{align}\sum_{k=1}^n\left(\frac{k}{n^2}-\frac{k^3}{6n^6}\right)\le \sum_{k=1}^n\sin{\left(k/n^2\right)}&\implies v_n-\frac{1}{6n^6}\cdot\frac{n^2(n+1)^2}{4}\le u_n \\[0.2cm]&\implies v_n-\frac16\cdot\frac{1}{n^2}\le u_n\end{align} Similarly $$0\le f\left(k/n^2\right)=k/n^2-\sin{\left(k/n^2\right)}\implies \sum_{k=1}^{n}\sin(k/n^2)\le \sum_{k=1}^nk/n^2 \implies u_n\le v_n$$ Putting these together and using the Squeeze Theorem you have that \begin{align}\lim_{n\to +\infty }\left(v_n-\frac16\cdot\frac1{n^2}\right)\le \lim_{n\to+\infty}u_n\le \lim_{n\to+\infty}v_n&\implies \frac12-0\le \lim_{n\to+\infty}u_n\le \frac12 \\[0.2cm]&\implies\lim_{n\to+\infty}u_n=\frac12\end{align}
Although the result already obtains with the lower bound $$v_n-\frac{1}{6n^6}\frac{n^2(n+1)^2}{4}$$ it is only a straightforward calculation to show that $$v_n-\frac{1}{6n^2}\le v_n-\frac{1}{6n^6}\frac{n^2(n+1)^2}{4}$$ for any $n\ge 1$ in order to bring it in the form that the exercise requires.
| {
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"timestamp": "2023-03-29T00:00:00",
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Newton-Raphson 2 variable method: Finding complex solutions Use the several variable version of the Newton-Raphson algorithm to find all intersection points (real and complex) of the 2 circles
$$(x-7)^2+(y-2)^2=100, \text{ and }$$ $$(x-11)^2+(y-5)^2=75$$
So I was able to find the real solution easily by plotting both circles and determining where they both intersect and using that estimate to use the Newton-Raphson algorithm.
However, finding the complex solutions have been difficult. I know I can just guess a point with complex values and use the Newton Raphson method. But all my guesses don't work out.
After asking my Professor for a method to efficiently find a guessing point. He suggested to use the modulus and note that
$$\omega=|f(x,y)|+|g(x,y)|=0$$ is true if and only if both $f(x,y)=0$ and $g(x,y)=0$, where
$$f(x,y)=(x-7)^2+(y-2)^2-100, \text{ and }$$ $$g(x,y)=(x-11)^2+(y-5)^2-75$$
So suppose that $(a+b*I,c+d*I)$ is a solution to both $f$ and $g$, but then I have two equations with four unknowns.
So can I have some at advice how to get those complex solutions?
| You are correct; set $x=a+i b$, $y=c+id$, replace in the equations and isolate the real and imaginary parts (setting them equal to zero).
Doing so, you get $$a^2-14 a-b^2+c^2-4 c-d^2-47=0 \tag 1$$ $$2 a b-14 b+2 c d-4 d=0\tag 2$$ $$a^2-22 a-b^2+c^2-10 c-d^2+71=0\tag 3$$ $$2 a b-22 b+2 c d-10 d=0\tag 4$$ You could solve these four equations using Newton-Raphson method.
What I found amazing is that the problem has analytical solutions.
$(1)-(3)$ leads to $$8 a+6 c-118=0 \tag 5$$ Similarly $(2)-(4)$ leads to $$8 b+6 d=0\tag 6$$ So, using $c=\frac{59-4 a}{3} $ and $d=-\frac{4 b}{3}$ and replacing we are let with $$a^2-22 a-b^2+94=0\tag 7$$ $$a b-11 b=0\tag 8$$ The last equation shows that either $a=11$ or $b=0$.
If $a=11$, $(7)$ gives $b^2+27=0$ that is to say $b=\pm 3 i \sqrt{3}$
If $b=0$, $(7)$ gives $a^2-22 a+94$ that is to say $a=11 \pm 3 \sqrt{3}$ For each pair, compute the corresponding $c$ and $d$ and the four solutions are $$\left\{a= 11,b= -3 i \sqrt{3},c= 5,d= 4 i \sqrt{3}\right\}$$ $$\left\{a=
11,b= 3 i \sqrt{3},c= 5,d= -4 i \sqrt{3}\right\}$$ $$\left\{a= 11+3
\sqrt{3},b= 0,c= 5-4 \sqrt{3},d= 0\right\}$$ $$\left\{a= 11-3 \sqrt{3},b=
0,c= 5+4 \sqrt{3},d= 0\right\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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By what rule or property this expression is equal to six? $$2^{-1} \equiv 6\mod{11}$$
Sorry for very strange question. I want to understand on which algorithm there is a computation of this expression. Similarly interested in why this expression is equal to two?
$$6^{-1} \equiv 2\mod11$$
| The notation $2^{-1} \pmod{11}$ means the multiplicative inverse of $2$ modulo $11$, that is, a number $x$ satisfying the equation $2x \equiv 1 \pmod{11}$.
Since $2 \cdot 6 \equiv 12 \equiv 1 \pmod{11}$, $2^{-1} \equiv 6 \pmod{11}$. The same calculation shows that $6^{-1} \equiv 2 \pmod{11}$.
The multiplicative inverse of $6$ exists modulo $11$ since $\gcd(6, 11) = 1$.
We can find the multiplicative inverse of $6$ modulo $11$ using the extended Euclidean algorithm.
\begin{align*}
11 & = 1 \cdot 6 + 5\\
6 & = 1 \cdot 5 + 1\\
5 & = 5 \cdot 1
\end{align*}
We solve for $1$ as a linear combination of $6$ and $11$.
\begin{align*}
1 & = 6 - 5\\
& = 6 - (11 - 6)\\
& = 2 \cdot 6 - 11
\end{align*}
Hence, $2 \cdot 6 \equiv 1 \pmod{11} \iff 6^{-1} \equiv 2 \pmod{11}$.
Let's see what happens when we use the extended Euclidean algorithm to solve for $2^{-1}$ modulo $11$.
The multiplicative inverse of $2$ exists modulo $11$ since $\gcd(2, 11) = 1$.
\begin{align*}
11 & = 5 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Solving for $1$ as a linear combination of $2$ and $11$ yields
$$1 = 11 - 5 \cdot 2$$
Thus, $$-5 \cdot 2 \equiv 1 \pmod{11} \iff 2^{-1} \equiv -5 \pmod{11}$$
What does this mean? It means that for each $t \in \mathbb{Z}$,
$$2(-5 + 11t) \equiv 1 \pmod{11}$$
In particular, if $t = 1$, we obtain
$$2 \cdot 6 \equiv 1 \pmod{11} \iff 2^{-1} \equiv 6 \pmod{11}$$
Why is $2(-5 + 11t) \equiv 1 \pmod{11}$ for $t \in \mathbb{Z}$? Observe that
$$2(-5 + 11t) \equiv -10 + 22t \equiv 1 - 11 + 22t \equiv 1 + 11(2t - 1) \equiv 1 \pmod{11}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compare five ways of solving cubic equation by iterations (nested expressions) Say we have a depressed cubic equation in the general form:
$$x^3-bx-c=0$$
There are basically five ways of solving it by iterations. Let's consider them in no particular order (the names are my own).
1) The continued fraction / nested radical method:
$$x=\sqrt{b+\frac{c}{x}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}}}$$
Take some $x_0$ and solve the recurrence:
$$x_n=\sqrt{b+\frac{c}{x_{n-1}}}$$
2) The continued fraction / nested square method:
$$x=-\cfrac{c}{b-x^2}=-\cfrac{c}{b-\cfrac{c^2}{(b-x^2)^2}}=-\cfrac{c}{b-\cfrac{c^2}{\left( b-\cfrac{c^2}{(b-x^2)^2} \right)^2}}$$
3) The branching continued fraction method:
$$x=\cfrac{b}{x}+\cfrac{c}{x^2}=\cfrac{b}{\cfrac{b}{x}+\cfrac{c}{x^2}}+\cfrac{c}{\left( \cfrac{b}{x}+\cfrac{c}{x^2} \right)^2}$$
It can also be expressed in several different ways, like this one:
$$x=\cfrac{c+bx}{x^2}=\cfrac{c+b\cfrac{c+bx}{x^2}}{\cfrac{(c+bx)^2}{x^4}}$$
But anyway, for some $x_0$:
$$x_n=\cfrac{b}{x_{n-1}}+\cfrac{c}{x_{n-1}^2}$$
4) Nested cubic roots method:
$$x=\sqrt[3]{c+bx}=\sqrt[3]{c+b\sqrt[3]{c+bx}}=\sqrt[3]{c+b\sqrt[3]{c+b\sqrt[3]{c+bx}}}$$
5) And nested cubes method:
$$x=\frac{1}{b} \left(-c+x^3 \right)=\frac{1}{b} \left(-c+\frac{1}{b^3} \left(-c+x^3 \right)^3 \right)$$
They all work in different ways, so I wonder, what are general rules of using one method or the other? How to get all three roots by using them?
For example we have the equation with three real roots $(-3.65617,0.42297,3.23319)$:
$$x^3-12x+5=0$$
Let's take $x_0=1$. Then methods 1, 3 and 4 will converge to the third root, while methods 2 and 5 will converge to the second root.
Changing $x_0$ to other values will not lead to the first root, but some methods will not converge (mainly, method 3).
| The nested cubic roots method, by affixing a complex cube root of unity $\zeta = e^{2\pi\,i/3}$, can give you all three roots when the cubic has only one real root,
$$x_k= \zeta^k\sqrt[3]{c+b\,\zeta^k\sqrt[3]{c+b\,\zeta^k\sqrt[3]{c+b\,\zeta^k\sqrt[3]{c+bx\dots}}}}$$
for $k=0,1,2$.
When the cubic is a casus irreducibilis, the method can still give two of the three real roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$ conditionally convergent $$1 + {1 \over 3} - {1 \over 2} + {1 \over 5} + {1 \over 7} - {1 \over 4} + {1 \over 9} + {1 \over 11} - {1 \over 6} + +-...$$
I want to show first that $S_{3n}$, $S_{3n+1}$ and $S_{3n+2}$ converges to the same limit, I show
$$S_{3n} = (1 + {1 \over 3} - {1 \over 2}) + ({1 \over 5} + {1 \over 7} - {1 \over 4}) + ({1 \over 9} + {1 \over 11} - {1 \over 6}) + ... + ({1 \over 4n-3} + {1 \over 4n-1} - {1 \over 2n})$$
but how to proceed from here?
| The series is
$$\sum_{n=1}^\infty\left(\frac1{4n-3}+\frac1{4n-1}-\frac1{2n}\right)=\sum_{n=1}^\infty\frac{8n-3}{2n(4n-1)(4n-3)}$$
and this thing is a converging positive series, since
$$\frac{8n-3}{2n(4n-1)(4n-3)}\le\frac4{(4n-1)(4n-3)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving Ordinary Differential Equation $y(x^4-y^2)dx+x(x^4+y^2)dy=0$ The Ordinary Differential Equation $y(x^4-y^2)dx+x(x^4+y^2)dy=0$ can be solve using Integrating Factor by Inspection. Here is the solution using the inspection method:
By expanding the equation, we have,$$x^4ydx-y^3dx+x^5dy+xy^2dy=0$$
$$x^4(ydx+xdy)+y^2(xdy-ydx)=0$$
Multiplying $\frac{1}{x^4}$,$$(ydx+xdy)+\frac{y^2}{x^2}\frac{xdy-ydx}{x^2}=0$$
$$d(xy)+\left(\frac{y}{x}\right)^2d\left(\frac{y}{x}\right)=0$$
$$xy+\frac{y^3}{3x^3}=C ,\ where\ C\ is\ a\ constant\ no.$$
$$3x^4y+y^3=Cx^3$$
I wanna know if it is also possible to use other method to solve this equation, because I tried it using the "General Method for finding the Integrating Factor" but my answer is not complete, and if it is, what are those possible method? If possible with your solution attached.
| Looking at the equation, I first change $y=x^2 \,z$ which makes $y'=2x \,z+x^2\,z'$. Replacing and simplifying leads to $$x \left(z^2+1\right) z'+z \left(z^2+3\right)=0$$ which is separable $$\frac{x'} x=-\frac{z^2+1}{z \left(z^2+3\right)}=-\frac{2 z}{3 \left(z^2+3\right)}-\frac{1}{3 z}$$ which integrates very simply and leads to $$x=\frac{C}{\sqrt[3]{z(z^2+3)}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of solutions of $8\sin(x)=\frac{\sqrt{3}}{\cos(x)}+\frac{1}{\sin(x)}$ Number of solution in $[0,2\pi]$ satisfying the equation
$$8\sin(x)=\frac{\sqrt{3}}{\cos(x)}+\frac{1}{\sin(x)}$$
Options are $5$ or $6$ or $7$ or $8$.
Doing some manipulations I reached $4\sin^2(x)\cos(x)=\sin(x+\pi/6)$ but I think this direction will not lead towards result. How to proceed?
| Recall that $\sin x = \sqrt{1 - \cos^2 x}$
\begin{align}
8\sin^2x \cos x &= \sqrt{3} \sin x + \cos x\\
8(1 - \cos^2 x) &= \sqrt{3} \sqrt{1 - \cos^2 x} + \cos x
\end{align}
Let $a = \cos x$
\begin{align}
8(1 - a^2)a &= \sqrt{3}\sqrt{1 - a^2} + a\\
8(1 - a^2)a - a &= \sqrt{3}{\sqrt{1 - a^2}}
\end{align}
Then square both sides(this will not introduce extraneous solutions since $\sqrt{1 - \cos^2 x}$ is defined for all real $x$, since $-1 \leq \cos x \leq 1$).
\begin{align}
64a^2(1 - a^2)^2 - 16a^2(1 - a^2) + a^2 &= 3 - 3a^2\\
16a^2(1 - a^2)(4(1 - a^2) - 1) + 4a^2 - 3 &= 0\\
\end{align}
Let $a^2 = y$
\begin{align}
16y(1 - y)(4 - 4y - 1) + (4y - 3) &= 0\\
16y(1 - y)(3 - 4y) + (4y - 3) &= 0 \\
-16y(1 - y)(4y - 3) + (4y - 3) &= 0 \\
(1 - 16y(1 - y))(4y - 3) &= 0\\
(16y^2 - 16y + 1)(4y - 3) &= 0\\
\end{align}
The discriminant of the quadratic is positive, so we have 2 real solutions for the left factor, and one for the right factor. That yields 3 solutions for $y$. Recall that if $a^2 = y$ then $a = \pm \sqrt{y}$. This doubles the amount of solutions. Therefore there are 6 for $a$ solutions in the range $0 \leq x \leq 2\pi$. We can then use our substitution $a = \cos x$ to show that each of these solutions appears once in the given range. Henceforth there are 6 solutions to the equation in the desired range.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int\frac{dx}{(A^2+x^2)^{3/2}}$ I have the following to integrate:
$$\int\frac{dx}{(A^2+x^2)^{3/2}}$$
Where, $A$ is a constant.
Not really sure what technique to use. Any hint would be helpful.
| let $x=A\tan\theta\implies dx=A\sec^2\theta\ d\theta$
$$\int \frac{1}{(A^2+x^2)^{3/2}}dx$$
$$=\int \frac{1}{(A^2+A^2\tan^2\theta)^{3/2}}(A\sec^2\theta\ d\theta)$$
$$=\int \frac{A\sec^2\theta\ d\theta}{A^{3}\sec^3\theta}$$
$$=\frac{1}{A^2}\int \cos \theta\ d\theta$$
$$=\frac{1}{A^2}\sin\theta+C$$
$$=\frac{1}{A^2}\sin\left(\tan^{-1}\frac{x}{A}\right)+C$$
$$=\frac{1}{A^2}\sin\left(\sin^{-1}\frac{x/A}{\sqrt{1+(x/A)^2}}\right)+C$$
$$=\color{red}{\frac{x}{A^2\sqrt{A^2+x^2}}+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673558",
"timestamp": "2023-03-29T00:00:00",
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$\cos x=12/13$, where $0 \lt x \lt 90$ degrees. What is the value of $\sin(2x)?$ apologies if this is an overly simplistic question to answer:
I have the value of $\cos x = \frac{12}{13}$, and I need to find the value of $\sin(2x)$, where $x$ is between $0$ and $90$ degrees (first quadrant).
I have $\sin(2x)=2\sin x(\frac{12}{13})$ but I am stumped trying to find the value of $\sin x$.
Any help would be greatly appreciated
| $\cos x =\frac{12}{13}$ and $\cos^2 x+\sin^2 x=1$ so
$$1-\sin^2x=\left(\frac{12}{13}\right)^2$$
Now, $\color{blue}{\sin(2x)}=2\sin x \cos x=2\sin x\cdot\frac{12}{13}=2\cdot \sqrt{1-\cos^2x}\cdot \frac{12}{13}=2\cdot\sqrt{1-(12/13)^2}\cdot \frac{12}{13}=\color{blue}{\frac{120}{169}}$
Note that I used:
$\sin(2x)=2\sin x \cos x$
and
$\sin x=\sqrt{1-\cos^2 x}$
$$\color{blue}{\sin x}=\sqrt{1-(12/13)^2}=\color{blue}{\frac{5}{13}}$$
Hope it helps
| {
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If a prime $p>2$ is expressible as $p=a^2+b^2$, then $4\mid (p-1)$ Show that if $p\in\mathbb{P}, p>2$ is a sum of squares, i.e $p=a^2+b^2$, then $p\equiv 1\pmod{4}$.
Have established that the remainder can't be either of $0,2$ for obvious reasons. So we must consider $1$ or $3$. Attempting assume by contradiction that $4\mid (p-3)$:
$$4k+3= a^2+b^2\Longleftrightarrow 4k + 2ab +3 = (a+b)^2$$
I can't find a contradiction, yet when I think about a prime divided by 4 giving the remainder of 3, for example 7, there is no way to express it as a sum of squares.
Another attempt at the same approach is $4k+3 = a^2+b^2$ so we have
$$4k-a^2=b^2-3$$
I believe I should somehow conclude one of the summands is not a square of an integer. How to proceed?
| Instead of cancelling out the remainders, I proved that if this is to happen then the remainder is 1.
Proof:
First of all we note that in all cases $p$ must be odd. So $a^2 + b^2$ must be odd. So $a$ and $b$ can not be both odd and even since then $a^2 + b^2$ would produce an even number. So one of $a$ and $b$ is odd and the other is even. Now let us take that $a$ is odd and $b$ that is even.
Now, $$a^2 + b^2 = (2m+1)^2 + (2n)^2 = 4m^2 + 4m + 1 + 4n^2 = 4(m^2 + m + n^2) + 1 = 4k + 1$$
But we know that when we divide $4k + 1$, we get a remainder of 1 so $p \equiv 1 \pmod 4$. Hence, proved.
Hope you found that useful.
| {
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If $x=\omega-\omega^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is?
If $x=\omega-\omega^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is? ($\omega$ represents the cube roots of unity not equal to $1$).
Directly substituting the given value will work. But there is no fun in that. The calculations will also get lengthy.
$$\omega-\omega^2-2=2\omega-1$$
I tried simplifying the given equation.
$$x^4+3x^3+2x^2-11x-6=x^3(x+3)+2\left(x+\frac{1}{2}\right)(x-6)$$
I got stuck here.
| Another approach might be to note that $ \ \omega^2 \ = \ \overline{\omega} \ $ , thus $ \ \omega \ - \ \omega^2 \ = \ 2 \ Im(\omega) \ = \ i \ \sqrt{3} \ $ . So $ \ x \ + \ 2 \ = \ i \ \sqrt{3} \ $ .
The polynomial can be decomposed as
$$x^4 \ + \ 3x^3 \ + \ 2x^2 \ - \ 11x \ - \ 6 \ $$
$$ = \ (x \ + \ 2)^4 \ - \ 5(x \ + \ 2)^3 \ + \ 8(x \ + \ 2)^2 \ - \ 15(x \ + \ 2) \ + \ 16 \ \ , $$
so its value for our choice of $ \ x \ $ is
$$ ( i \ \sqrt{3})^4 \ - \ 5( i \ \sqrt{3})^3 \ + \ 8( i \ \sqrt{3})^2 \ - \ 15( i \ \sqrt{3}) \ + \ 16 $$
$$ = 3^2 \ - \ [ (-i) \ 15 \sqrt{3}] \ + \ 8( - 3) \ - \ ( i \ 15 \sqrt{3}) \ + \ 16 $$
$$ = 9 \ -24 \ + \ 16 \ = \ 1 \ \ . $$
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.