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Solution for recurrence $T(n+1) = T(n) + \lfloor \sqrt{n+1}\rfloor $ ould someone please give me an idea as to how the solve the following.
$$T(1) = 1$$
$$T(n+1) = T(n) + \lfloor\sqrt{n+1}\rfloor$$
I converted the recurrence to $T(n) = T(n-1) + \lfloor\sqrt{n}\rfloor$ and then tried to solve it using the method taught in the "linear recurrences solving from Mathematics for computer science course" at MIT and got the solution as $T(n) = n^\left(3/2\right) + 1$.
But the page I found this problem in asks for $T(n^2)$ and gives the answer as $\frac n6\left(4n^2 - 3n + 5\right)$, which I'm not able to derive.
Can someone please help me solve this recurrence?
Thanks
| When $n=1$, it is straightforward to see $T(1) = 1 = \frac{1}{6}(4 \times 1^2 - 3\times 1 + 5)$.
Assume $T(n^2) = \frac{n}{6}(4 n^2 - 3n + 5)$.
For $m = n^2, n^2+1, \ldots , (n+1)^2-2$, we have
$$T(m+1)-T(m) = \lfloor \sqrt{m+1} \rfloor = n$$
Summing up this relation over these $2n$ values of $m$ we have
$$T((n+1)^2 - 1) - T(n^2) = 2n^2$$
Therefore
$$T((n+1)^2) = T((n+1)^2 - 1) + (n+1) = T(n^2) + 2n^2 + n + 1$$
$$T((n+1)^2) = \frac{n}{6}(4 n^2 - 3n + 5) + 2n^2 + n + 1 = \frac{n+1}{6}(4 (n+1)^2 - 3(n+1) + 5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all $y,z\in\mathbb{R}^+$ it is true that $(y + z) (1 - 8 y z + z^2 + y^2 (1 + 9 z^2))\ge0$ Prove that for all $y,z\in\mathbb{R}^+$ it is true that
$$(y + z) (1 - 8 y z + z^2 + y^2 (1 + 9 z^2))\ge0$$
It is obvious that $y+z>0$. Then I tried to reduce $1 - 8 y z + z^2 + y^2 (1 + 9 z^2)$ to sum of squares. It seems to be hard, so I tried to expand it and I got
$$y + y^3 + z - 7 y^2 z - 7 y z^2 + 9 y^3 z^2 + z^3 + 9 y^2 z^3\ge0$$
but what next? Is it possible to reduce it to sum of squares?
| Hint: Use completing the square for $-8yz+z^2+y^2$ which gives
$$-8yz+z^2+y^2=y^2+z^2-2yz-6yz=(y-z)^2-6yz$$
Now, \begin{align}
1-8yz+z^2+y^2+9y^2z^2&= 1+(y-z)^2-6yz+9y^2z^2\\
&=(y-z)^2+(3yz)^2-2.3yz.1+1\\
&=(y-z)^2+(3yz-1)^2
\\
&\geq 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Write ODE in Polar Coordinates I want to write this ODE system in polar coordinates (r,$\theta$).
$$\dot x =x-y-x^3 $$
$$\dot y = x+y-y^3$$
| We use:
$$x = r \cos \theta, y = r \sin \theta, r^2 = x^2 + y^2, \theta = \tan^{-1}\left(\frac{y}{x}\right)$$
From:
$$x^2 + y^2 = r^2 \implies x x' + y y' = r r'$$
We have:
$$\begin{align} x x' + y y' &= x(x-y-x^3) + y(x+y-y^3) \\ &= r \cos \theta(r \cos \theta-r \sin \theta - (r \cos \theta)^3) + r \sin \theta(r \cos \theta + r \sin \theta -(r \sin \theta)^3) \\&= r r'\end{align}$$
We also have:
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) \implies \theta' = \dfrac{xy'-yx'}{r^2}$$
Make the appropriate substitutions, do some trig simplifications and cleanup to derive $r'$ and $\theta'$.
Update
You should end up with:
$$r' = -\frac{1}{4} r \left(r^2 \cos (4 \theta)+3 r^2-4\right)$$
$$\theta' = -\left(\left(r^2-1\right) \cos (2 \theta)+\sin (2 \theta)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Sum $\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$ A series is given as follows
$$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$
Can you give me hints to get started finding its value? Thanks.
| $$\begin{align}
\sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i}
&=\frac12\sum_{n=1}^{\infty}\frac1{2^nn!}\prod_{i=1}^n\frac{3i-1}{3}\\
&=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\frac{1-3i}{3}\\
&=\frac12\sum_{n=1}^{\infty}\frac1{(-2)^nn!}\prod_{i=1}^n\left(\frac13-i\right)\\
&=\left[\frac12\sum_{n=1}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)\right]_{x=-1/2}\\
\end{align}$$
Now note that for $n\geq1$, $\prod_{i=1}^n\left(\frac13-i\right)=f^{(n)}(0)$, where $f(x)=(x+1)^{-2/3}$. We can manipulate the expression a little further to see it as a Taylor series for $f$.
$$\begin{align}
\sum_{n=1}^{\infty}\frac12\prod_{i=1}^n\frac{3i-1}{6i}
&=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}\prod_{i=1}^n\left(\frac13-i\right)-\frac12\right]_{x=-1/2}\\
&=\left[\frac12\sum_{n=0}^{\infty}\frac{x^n}{n!}f^{(n)}(0)\right]_{x=-1/2}-\frac12\\
&=\frac12f(-1/2)-\frac12\\
&=\frac12(1/2)^{-2/3}-\frac12\\
&=\sqrt[3]{1/2}-\frac12\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Find the set of all $\alpha$ such that Matrix A is invertible and calculate the inverse for all $\alpha$ $A=\begin{pmatrix}
0 & 1 & -1 & 2\\
2 & -1 & 3 & 0 \\
\alpha & 0 & 1& 0 \\
3 & -1 &4 & 0
\end{pmatrix}$
I know that a Matrix is invertible only if $det(A)\not=0$
$det(A)=0\begin{vmatrix}
-1 & 3 & 0 \\
0 & 1 & 0 \\
-1 & 4 & 0 \\
\end{vmatrix}-1\begin{vmatrix}
2 & 3 & 0 \\
\alpha & 1 & 0 \\
3 & 4 & 0 \\
\end{vmatrix}-1\begin{vmatrix}
2 & -1 & 0 \\
\alpha & 0 & 0 \\
3 & -1 & 0 \\
\end{vmatrix}-2\begin{vmatrix}
2 & -1 & 3 \\
\alpha & 0 & 1 \\
3 & -1 & 4 \\
\end{vmatrix}$
$=-[2\begin{vmatrix}
1 & 0\\
4 & 0\\
\end{vmatrix}-3\begin{vmatrix}
\alpha & 0\\
3 & 0\\
\end{vmatrix}]-[2\begin{vmatrix}
0 & 0\\
-1 & 0\\
\end{vmatrix}+\begin{vmatrix}
\alpha & 0\\
3 & 0\\
\end{vmatrix}]-2[2\begin{vmatrix}
0 & 1\\
-1 & 4\\
\end{vmatrix}+\begin{vmatrix}
\alpha & 1\\
3 & 4\\
\end{vmatrix}+3\begin{vmatrix}
\alpha & 0\\
3 & -1\\
\end{vmatrix}]$
$=-2(2+4\alpha-3-3\alpha)=-4-8\alpha+6+6\alpha=-2\alpha+2$
$-2\alpha+2=0 \iff \alpha=1$
So the matrix is invertible for all $\alpha \in \Bbb R| \alpha\not=0$
But how do I calculate the inverse of A for ALL other $\alpha$ without spending the rest of my life on this question? Am I missing something here?
Thanks in advance?
| $A^{-1}=\dfrac{1}{det(A)}co(A)^T$
It is fairly easy (a bit long but...) to calculate the comatrix of $A$ (matrix of cofactors or adjugate matrix)...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If a prime number is reversed, and then appended to itself, why is the result always a composite number? $2 \Rightarrow 22$ which is a composite number.
$37 \Rightarrow 3773$ which is a composite number.
$523 \Rightarrow 523325$ which is a composite number.
$8123 \Rightarrow 81233218$ which is a composite number.
If you take any prime, reverse it, and append it to itself, the result always seems to be a composite number. I have checked this for $2 \leq n \leq 999999$.
Is there a reason for this, or is it simply a coincidence?
| As John's answer notes, every palindromic number with an even number of digits is divisible by $11$, because the alternating sum of its digits is zero (and thus a multiple of $11$).
For example, for $81233218$, we have: $$8-1+2-3+3-2+1-8 = 0,$$ and so $81233218$ is divisible by $11$.
The reason why this divisibility rule works is most easily seen using a bit of modular arithmetic.
Specifically, by definition, two numbers $a$ and $b$ are equivalent modulo $m$ (which we write as $a \equiv b \pmod m$) if and only if their difference $a-b$ is divisible by $m$. Thus, in particular, a number $n$ is divisible by $m$ if and only if $n \equiv 0 \pmod m$.
Now, the reason this definition is handy is because we can "do arithmetic under the modulus $m$": in particular, if $a \equiv a' \pmod m$ and $b \equiv b' \pmod m$, then $a+b \equiv a'+b' \pmod m$ and $ab \equiv a'b' \pmod m$. Thus, if we're only interested in the result of a calculation modulo some number $m$ (like, say, if we just want to know whether $n \equiv 0 \pmod{11}$ for some number $n$), and the calculation only involves operations that work equivalently "under the modulus" (such as addition and multiplication, and any combination thereof), then we can freely add or subtract multiples of $m$ from any intermediate values to make them more convenient (which often means "closer to zero").
In particular, from the definition it clearly follows that: $$10 \equiv -1 \pmod{11},$$ since $10 - (-1) = 10 + 1 = 11$. We also know that the numerical value of a base-$10$ number is mathematically obtained by multiplying its lowest digit with $1$, the second digit with $10$, the third digit with $10^2 = 100$, etc., and adding the results together. For example: $$81233218 = 10^7 \cdot 8 + 10^6 \cdot 1 + 10^5 \cdot 2 + 10^4 \cdot 3 + 10^3 \cdot 3 + 10^2 \cdot 2 + 10 \cdot 1 + 8.$$
But since $10 \equiv -1 \pmod{11}$, if we only want to calculate the value of the number modulo $11$, we can replace $10$ with $-1$ in this calculation! Thus: $$81233218 \equiv (-1)^7 \cdot 8 + (-1)^6 \cdot 1 + (-1)^5 \cdot 2 + (-1)^4 \cdot 3 + (-1)^3 \cdot 3 + (-1)^2 \cdot 2 + (-1) \cdot 1 + 8 \pmod{11}.$$
And, of course, we can easily see that the powers of $-1$ simply alternate between $-1$ and $1$, so this expression simplifies down to: $$81233218 \equiv -8 + 1 -2 + 3 - 3 + 2 - 1 + 8 = 0 \pmod{11}.$$
(Ps. The same trick also works in bases other than $10$, showing that any number palindromic in base $b$, with an even number of base-$b$ digits, is necessarily divisible by $b+1 = 11_b$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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"answer_id": 1
} |
Inequality $\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$
If $x>0$, $y>0$, $z>0$ and $xyz = 1$ then
$$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$$
I tried using $\displaystyle x = \frac{a}{b},y = \frac{b}{c}$ and $\displaystyle z = \frac{c}{a}$ substitution,
$\displaystyle \sum_{cyc} \frac{\sqrt{a^2+b^2}}{b} \le \sqrt{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \\ \displaystyle \iff 6+ 2\sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le \sum_{cyc}\frac{a^2+b^2}{b^2} + 4\sum_{cyc}\frac{b}{a}$
So it would suffice if we showed $\displaystyle \sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le 2\sum_{cyc}\frac{b}{a}$
Squaring again does not lead anywhere nice. Is the last inequality true ? If not is there a better way of showing the result ?
| WLOG replace the variables by their cubes, while still $xyz=1$:
$$\sum_\text{cyc}\sqrt{1+x^6}\le \sqrt2\sum_\text{cyc}x^3$$
We can homogenize and use the Cauchy-Schwarz inequality:
$$\biggl(\sum_\text{cyc}\sqrt{x^2}\sqrt{y^2z^2+x^4}\biggr)^{\!2}\le\Bigl(\sum_\text{cyc}x^2\Bigr)\Bigl(\sum_\text{cyc}(y^2z^2+x^4)\Bigr)\implies\\ \sum_\text{cyc}\sqrt{1+x^6}=\sum_\text{cyc}\sqrt{x^2(y^2z^2+x^4)}\le\sqrt{\Bigl(\sum_\text{cyc}x^2\Bigr)\Bigl(\sum_\text{cyc}(y^2z^2+x^4)\Bigr)}$$
So it suffices to prove
$$\Bigl(\sum_\text{cyc}x^2\Bigr)\Bigl(\sum_\text{cyc}(y^2z^2+x^4)\Bigr)\le 2\Bigl(\sum_\text{cyc}x^3\Bigr)^{\!2}$$
Whose expanded and eliminated version is
$$2\sum_\text{cyc}x^4y^2+2\sum_\text{cyc}x^2y^4+3x^2y^2z^2\le \sum_\text{cyc}x^6+4\sum_\text{cyc}x^3y^3$$
Which is after symmetrization equivalent to
$$0\le \Bigl(\sum_\text{sym}x^6-\sum_\text{sym}x^4y^2\Bigr)+3\Bigl(\sum_\text{sym}x^3y^3-\sum_\text{sym}x^4y^2\Bigr)+\Bigl(\sum_\text{sym}x^3y^3-\sum_\text{sym}x^2y^2z^2\Bigr)$$
The last difference is just AM-GM, so we can forget it. Now observe that
\begin{align*}\sum_\text{sym}x^6-\sum_\text{sym}x^4y^2&=\sum_\text{sym}(x-y)x^4(x+y)=\sum_\text{cyc}(x-y)(x^4-y^4)(x+y)\\&=\sum_\text{cyc}(x-y)^2(x^3+x^2y+xy^2+y^3)(x+y)\\
\sum_\text{sym}x^4y^2-\sum_\text{sym}x^3y^3&=\sum_\text{sym}(x-y)x^3y^2=\sum_\text{cyc}(x-y)(x^3y^2-x^2y^3)\\&=\sum_\text{cyc}(x-y)^2x^2y^2\end{align*}
So we are done if $0\le (x^3+x^2y+xy^2+y^3)(x+y)-3x^2y^2$ and one can easily check it's the same as
$$0\le (x^2-y^2)^2+x^2y^2+2x^3y+2xy^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I have a Hard time solving this system of nonlinear equations $x^2+y^2-z^2=20$
$x^4+y^4-z^4=560$
$x^3+y^3+z^3=3xyz$
I know the fact that if $x^3+y^3+z^3=3xyz$ then $x+y+z=0$ (coming from Euler's identity) and first equation can be written as
$(x+y-z)^2-2(xy-xz-yz)=20$ and since $x+y=-z$ then $(xy-xz-yz)=2z^2-10$.
And the second equation becomes
$(x^2+y^2-z^2)^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$.
Since we know $x^2+y^2-z^2=20$, therefore $20^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$ or
$(x^2y^2-2x^2z^2-2y^2z^2)=80$.
Now I am stuck! Help please!
| since$$x^3+y^3+z^3=3xyz\Longrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0$$
$$\Longrightarrow x+y+z=0 ,\text {or} ,x=y=z(\text{impossible})$$
let $z=-x-y$,then you have
$$x^2+y^2-z^2=20\Longrightarrow x^2+y^2-(x+y)^2=20\Longrightarrow xy=-10$$
since
$$x^4+y^4=(x^2+y^2)^2-2x^2y^2=(20+z^2)^2-200=z^4+40z^2+200$$
so
$$z^4+40z^2+200-z^4=560\Longrightarrow z^2=9\Longrightarrow z=\pm 3$$
then It is easy to find $x,y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Conditional distribution on the unit circle and a square Let (X,Y) be uniformly distributed over $B=\{(x,y) \in \mathbb{R}^2: x^2+y^2 \leq 1 \}$ resp. $Q=[-1,1]^2$.
Now I want to calculate the conditional distributions and of Y given X=x.
And then the same for $R=\sqrt(X^2+Y^2)$ and $\varphi = arctan(Y/X)$ (conditional distributions of $\varphi$ given R=r).
At first I calculated the densities:
For the unit circle I get:
f(x,y) = $\begin{cases} 1/\pi~~~ if~ x^2+y^2 \leq 1\\
0 ~~~~~~ else
\end{cases}$.
$f_Y(y) = \frac{2 \sqrt(1-y^2)}{\pi}$
$f_X(x)= \frac{2 \sqrt(1-x^2)}{\pi}$
For the square I get:
$f(x,y)= \begin{cases} 1/4~~~if (x,y) \in Q\\
0~~~~~~else
\end{cases}$.
$f_Y(y)=f_X(x)=1/2$.
How do I get now the conditional distributions?
And how do I get the conditional distributions of $\varphi$?
| For the disc, $B$,
$$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} = \dfrac{1/\pi}{2\sqrt{1-y^2}/\pi} = \dfrac{1}{2\sqrt{1-y^2}}\qquad\text{for $|x|\leq \sqrt{1-y^2},$ otherwise $0$.}$$
Symmetrically,
$$f_{Y|X}(y|x) = \dfrac{1}{2\sqrt{1-x^2}}\qquad\text{for $|y|\leq \sqrt{1-x^2},$ otherwise $0$.}$$
So both these conditional distributions are uniform, which is as we would expect.
For $R,\varphi,\;$ we can use the "change of variable" method. Since $R=\sqrt{X^2+Y^2}$ and $\varphi=\arctan{\left(Y/X\right)}$, we have
$$X = R\cos\varphi \\
Y = R\sin\varphi.$$
So our Jacobian is
$$J=\begin{vmatrix}
\dfrac{\partial{x}}{\partial{r}} & \dfrac{\partial{y}}{\partial{r}} \\
\dfrac{\partial{x}}{\partial{\varphi}} & \dfrac{\partial{y}}{\partial{\varphi}} \\
\end{vmatrix}
=\begin{vmatrix}
\cos\varphi & \sin\varphi \\
-r\sin\varphi & r\cos\varphi \\
\end{vmatrix}
= r(\cos^2 \varphi + \sin^2 \varphi ) = r.$$
$$f_{R,\varphi}(r,\varphi) = f_{X,Y}(x(r,\varphi),\;y(r,\varphi))\; |J| = \dfrac{r}{\pi}\qquad\text{for $0\leq r \leq 1$ and $0\leq \varphi \leq 2\pi$}$$
$$f_R(r) = \int_0^{2\pi}{\dfrac{r}{\pi}\;d\varphi} = \left[ \dfrac{r\varphi}{\pi} \right]_0^{2\pi} = 2r.$$
$$f_{\varphi}(\varphi) = \int_0^1{\dfrac{r}{\pi}\;dr} = \left[ \dfrac{r^2}{2\pi} \right]_0^1 = \dfrac{1}{2\pi}.$$
Note that $f_{R,\varphi}(r,\varphi) = f_R(r)f_{\varphi}(\varphi)$ so $R,\varphi$ are independent. Therefore, $f_{\varphi|R}(\varphi|r)=f_{\varphi}(\varphi)$ and $f_{R|\varphi}(r|\varphi) = f_{R}(r)$.
$$\\$$
For the square, $Q$, we use a similar method.
$$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} = \dfrac{1/4}{1/2} = \dfrac{1}{2}\qquad\text{for $|x|\leq 1,$ otherwise $0$.}$$
Symmetrically,
$$f_{Y|X}(y|x) = \dfrac{1}{2}\qquad\text{for $|y|\leq 1,$ otherwise $0$.}$$
This means $X,Y$ are independent.
For $R,\varphi$, we have the same Jacobian, $J$, as above. However, the ranges of the r.v.'s are different and more difficult. Firstly, the joint distribution:
$$f_{R,\varphi}(r,\varphi) = f_{X,Y}(x(r,\varphi),\;y(r,\varphi))\; |J| = \dfrac{r}{4}\qquad\text{for $0\leq \varphi \leq 2\pi$ and} \\
0\leq R \leq \begin{cases}
\dfrac{1}{\cos{\varphi}}, & \text{if $k\pi - \dfrac{\pi}{4}\leq \varphi\leq k\pi + \dfrac{\pi}{4},\quad k\in\mathbb{Z}$} \\
\dfrac{1}{\sin{\varphi}}, & \text{if $k\pi + \dfrac{\pi}{4}\leq \varphi\leq k\pi + \dfrac{3\pi}{4},\quad k\in\mathbb{Z}.$} \\
\end{cases}
$$
If $0\leq r\leq 1,\quad f_R(r) = \int_0^{2\pi}{\frac{r}{4}\;d\varphi} = \left[\dfrac{r\varphi}{4}\right]_0^{2\pi} = \dfrac{\pi r}{2}.$
If $1\leq r\leq \sqrt{2},\quad f_R(r) = 8\int_{\arccos{(1/r)}}^{\pi/4}{\frac{r}{4}\;d\varphi} = \left[2r\varphi\right]_{\arccos{(1/r)}}^{\pi/4} = \dfrac{\pi r}{2} - 2r\arccos{(1/r)}.$
If $0\leq r\leq 1$,
$$f_{\varphi|r}(\varphi,r) = \dfrac{f_{R,\varphi}(r,\varphi)}{f_R(r)} = \dfrac{r/4}{\pi r/2} = \dfrac{1}{2\pi}.$$
If $1\leq r\leq \sqrt{2}$,
$$f_{\varphi|r}(\varphi,r) = \dfrac{r/4}{\pi r/2 - 2r\arccos(1/r)} = \dfrac{1}{2\pi - 8\arccos(1/r)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$
Show that
$$\sqrt[3]{3\sqrt{21} + 8} - \sqrt[3]{3\sqrt{21} - 8} = 1$$
Playing around with the expression, I found a proof which I will post as an answer.
I'm asking this question because I would like to see if there are alternative solutions which are perhaps faster / more direct / elementary / elegant / methodical / insightful etc.
| Sometimes you can get lucky de-nesting a radical. This is one of those times.
Write $\sqrt[3]{3\sqrt{21}\pm 8}$ as $\frac12\;\sqrt[3]{24\sqrt{21}\pm 64}$, and consider expressing the radicand as a perfect cube:
$$24\sqrt{21}\pm 64 = (\;p + q \sqrt{21}\;)^3 =p^3 + 3 p^2 q \sqrt{21} + 63 p q^2 + 21 q^3 \sqrt{21}$$
so that
$$p\left(\; p^2 + 63 q^2 \;\right) = \pm 64 \qquad q \left(\;p^2 + 7 q^2\;\right)\cdot 3\sqrt{21} = 8 \cdot 3\sqrt{21}$$
Clearly, we can take $p = \pm 1$ and $q = 1$. Then,
$$\begin{align}
\sqrt[3]{3\sqrt{21}+8} - \sqrt[3]{3\sqrt{21}-8}
&= \frac{1}{2}\left(\; \sqrt[3]{(1 + \sqrt{21} )^3} - \sqrt[3]{(-1 + \sqrt{21} )^3} \;\right) \\[6pt]
&= \frac{1}{2}\left(\;1 + \sqrt{21} - (-1 + \sqrt{21})\;\right) \\[6pt]
&= 1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem :
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$
Solution :
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other
Thanks
| An easy inspection shows that $f$ is nonconstant. Let $f(x)=x^ng(x)$, with $n\geq0$ and $g(0)\ne0$. Then $f(x)f(1/x)=g(x)g(1/x)$ and $f(x)+f(1/x)=x^ng(x)+\frac{g(1/x)}{x^n}$, hence
$x^ng(x)g(1/x)=x^{2n}g(x)+g(1/x)$. Let $m$ be the degree of $g$, and consider the reciprocal polynomial of $g$, that is $x^mg(1/x)$, who also has degree $m$. We have
$$x^ng(x)x^mg(1/x)=x^{2n+m}g(x)+x^mg(1/x)\,.$$
Note that the LHS has degree $n+2m$, and the summands on the RHS have degrees $2n+2m$ and $m$ respectively. Now $2n+2m=m$ implies $n=0$, and $2n+2m>m$ implies that the degree of the RHS is equal to $2n+2m$, so $2n+2m=n+2m$ and again $n=0$. Thus $f(x)$ has the form $f(x)=a_0+a_1x+\cdots+a_mx^m$, with $m>0$ and $a_0a_m\ne0$, and the equality above becomes $f(x)h(x)=x^mf(x)+h(x)\,,$ where $h(x)=x^mf(1/x)=a_m+a_{m-1}x+\cdots+a_0x^m$. We can rewrite this equality as
$$\bigl(f(x)-1\bigr)h(x)=x^mf(x)\,.$$
Since $0$ is not a root of $h(x)$ then $x^m$ divides $f(x)-1$, and since both $x^m$ and $f(x)-1$ have degree $m$, it follows that $f(x)-1=cx^m$ for some $c\ne0$. Replacing again yields $c(x^m+c)=cx^m+1$, which implies $c=\pm1$. Finally $64=f(4)-1=4^mc$, which in turn forces $c=1$ and $m=3$. We conclude then that $f(x)=x^3+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of inequality $x^2+3\sin x-3x \geq 0$ I'm looking for a simple (or any) way to prove the following inequality:
$$x^2+3\sin x-3x \geq 0\quad \text{for all }x\in \mathbb R$$
Progress
I have looked at the function corresponding to the left hand side: it has a root at zero, but its derivatives do not seem to give me useful information.
| For $x\le 0$ we have $\sin x\ge x$ and hence $$x^2+ 3\sin x-3x\ge x^2\ge 0.$$ If $x\ge 4$ we have
$$ x^2+ 3\sin x-3x\ge x^2-3x-3=(x-4)(x+1)+1>0.$$
Hence we need only consider $0<x<4$.
Taylor says $$\begin{align}f(x)
&=f(a)+f'(a)(x-a)+\frac12f''(a)(x-a)^2+\frac16f'''(\eta)(x-a)^3\\\end{align}$$
with $\eta$ between $a$ and $x$. Here, with $f(x)=3\sin x-3x+x^2$ and $a=0$ (so that $f'(x)=3\cos x-3+2x$, $f''(x)=-3\sin x+2$, $f''(x)=-3\cos x$)
$$ \tag1f(x) =x^2\cdot\left(1-\frac x2\cos\eta\right)$$
and with $a=\pi$
$$ \tag2f(x) =\underbrace{(\pi-3)\pi +(2\pi-6)(x-\pi)}_{=:g(x)}+(x-\pi)^2\left(1-\frac{x-\pi}2 \cos\eta\right)$$
The right hand side in $(1)$ is positive for $0<x<2$. The last summand in $(2)$ is nonnegative for $\pi-2\le x\le \pi+2$, so especially for $2\le x<4$.
Hence to complete the proof ist suffices to show $g(x)>0$ for $x\ge 2$. This follows from $g(\pi/2)=0$ together with $\frac\pi2<2$ and $2\pi-6>0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How find this diophantine equation integer solution $a^3+b^3=(2ab+1)^2$
Find this following diophantine equation integer solution
$$a^3+b^3=(2ab+1)^2$$
I think this equation only have two following solution
$$(a,b)=(1,0),(0,1)$$
maybe this equation have no other solution? because can see wolframalpha.com
I think following idea is usefull
$$a^3+b^3=(a+b)(a^2-ab+b^2)=(2ab+1)^2$$
since
$(a+b,a^2-ab+b^2)=1$ so
$a+b,a^2-ab+b^2$ are square numbers?
| First look for the solutions of the equation $z^2=a^3+b^3$ which are given for instance in
http://www.cecm.sfu.ca/~nbruin/thesis.pdf, lemma 3.2.6.
Then for each family (give values to $s,t$), in order to find the possibles $a,b$ and check if $2ab+1=z.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to write $4.13\bar{9}$ in explicit form I want to write $4.13\bar{9}$ in an explicit form.
I know how to solve a problem similar to this, for a number like $2.\overline{17}$, where the bar starts at the first number after the comma:
I know that
$$2.\overline{17}=$$
$$2+\frac{17}{100}\sum_{n=1}^\infty(\frac{1}{100})^{n-1}=$$
$$2+\frac{17}{100}\frac{1}{1-\frac{1}{100}=}$$
$$2+\frac{17}{99}=\frac{215}{99}.$$
But I'm not 100% sure how to solve a problem where the bar starts at the third number after the comma.
Do I start the sum from $n=1$, like I did above? Do I start it for $n=3$ ?
I mean, would I write
$$4+\frac{13}{100}+\sum_{n=3}^\infty{\frac{9}{10^n}}$$or
$$4+\frac{13}{100}+\sum_{n=1}^\infty{\frac{9}{10^n}}$$
?
I can't seem to understand which one to use.
| $4.13\bar{9} = 4 + \frac{13}{100} + \frac{9}{1000} + \frac{9}{10000} + \frac{9}{100000} + \ldots = 4 + \frac{13}{100} + \sum_{n=3}^{\infty} \frac{9}{10^n} = 4 + \frac{13}{100} + \frac{\frac{9}{1000}}{(1 - \frac{1}{10})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx$
Evaluate the integral
$$\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx$$
My Attempt:
Let $f(x) = \frac{ax+b}{(x^3+3x+1)^2}.$ Now differentiate both side with respect to $x$, and we get
$$
\begin{align}
f'(x) &= \frac{(x^3+3x+1)^2\cdot a-2\cdot (x^3+3x+1)\cdot (3x^2+3)\cdot (ax+b)}{(x^3+3x+1)^4}\\
&= \frac{5x^3+3x-1}{(x^3+3x+1)^3}\cdot\frac{(x^3+3x+1)\cdot a-6\cdot (x^2+1)\cdot (ax+b)}{(x^3+3x+1)^3}\\
&= \frac{5x^3+3x-1}{(x^3+3x+1)^3} \frac{-5ax^3+6bx^2-3ax+(a-6b)}{(x^3+3x+1)^3}\\
&= \frac{5x^3+3x-1}{(x^3+3x+1)^3}
\end{align}
$$
for $a = -1$ and $b = 0$. Thus, by the Fundamental Theorem of Calculus,
$$
\begin{align}
\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx &= \int f'(x)\,dx\\
&= f(x)\\
&= -\frac{x}{(x^3+3x+1)^2}+\mathcal{C}
\end{align}
$$
How we can solve the above integral directly (maybe by using the substitution method)?
| Here is another solution... I also happen to think that it cannot be done with substitution.
$$\begin{aligned}
\int \frac{5x^3+3x-1}{\left ( x^3+3x+1 \right )^3}\,dx &=\int \frac{-x^3-3x-1+6x^3+6x}{\left ( x^3+3x+1 \right )^3}\,dx \\
&= \int \frac{-x^3-3x-1+2x\left ( 3x^2+3 \right )}{\left ( x^3+x+1 \right )^3}\,dx\\
&= \int \frac{-\left ( x^3+3x+1 \right )^2+2x\left ( x^3+3x+1 \right )\left ( 3x^2+3x \right )}{\left ( x^3+3x+1 \right )^4}\,dx\\
&= \int \frac{-(x)'\left ( x^3+3x+1 \right )^2+x\left[ \left ( x^3+3x+1 \right )^2 \right]'}{\left ( x^3+3x+1 \right )^4}\,dx\\
&= \int \left [ -\frac{x}{\left ( x^3+3x+1 \right )^2} \right ]' \,dx = -\frac{x}{\left ( x^3+3x+1 \right )^2}+c, \; \; c \in \mathbb{R}
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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Solution to $\sqrt{x^2-5}+3>|x-1|$ I tried many ways to solve this but I just can't figure it out...
$$\sqrt{x^2-5}+3>|x-1|$$
| You can eliminate the square roots or absolute values provided you know the signs of the expressions under the square roots or between the absolute value signs.
What you can use : if $a,b\ge 0$, $a>b\iff a^2 >b^2$ ; if $a,b\le 0$, $a>b\iff a^2 <b^2$ ; if $a> 0 >b$, well… $a>b$.
Here it's fairly simple: first the square root is defined if and only if $x\ge \sqrt 5$ or $x\le-\sqrt 5$.
In the first case, $x >1$ so the inequation amounts to $\sqrt{x^2-5} > x- 4$.
Two subcases:
*
*either $x<4$, and the LHS ($\ge 0$) is greater than the RHS ($<0$).
*or $x\ge 4$, and $\sqrt{x^2-5} > x- 4\iff x^2-5 > (x-4) ^2 \iff 8x >21$,, which is indeed true if $x\ge4$.
In the second case, the inequation amounts to $\sqrt{x^2-5} > -(x+2)$. Since $x<-\sqrt 5$, we know $-(x+2)>0$,
so that $\sqrt{x^2-5} > -(x+2)\iff 4x <-9 $, so that $ x < -\dfrac94\bigl(<-\sqrt 5\bigr)$.
As a conclusion we have the following solutions:
$$ x< -\dfrac94\quad \text{or}\quad x\ge \sqrt 5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{cos(x/n)-1}{x^4}dx$ Show that the following limit exists and compute it:
$$\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx$$
Attempt: By using the integration by parts, I get the following result:
$$\begin{align*}\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx&=\lim_{k\to\infty}\int_{1}^{k}\frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\\\\
&=\lim_{k\to\infty}\left\{\left(\frac{k}{n}\right)^4\sin\left(\frac{k}{n}\right)+4\left(\frac{k}{n}\right)^3\cos\left(\frac{k}{n}\right)\right.\\
&\quad\quad\quad\quad-12\left(\frac{k}{n}\right)^2\sin\left(\frac{k}{n}\right)-24\left(\frac{k}{n}\right)\cos\left(\frac{k}{n}\right)\\
&\quad\quad\quad\quad+24\sin\left(\frac{k}{n}\right)+\frac{n^3}{3k^3}\\
&\quad\quad\quad\quad\left.-\left[\frac{1}{n^4}\sin\left(\frac{1}{n}\right)+\frac{4}{n^3}\cos\left(\frac{1}{n}\right)\right.\right.\\
&\quad\quad\quad\quad\left.\left.-12\left(\frac{1}{n}\right)^2\sin\left(\frac{1}{n}\right)-\frac{24}{n}\cos\left(\frac{1}{n}\right)\right.\right.\\
&\quad\quad\quad\quad\left.\left.+24\sin\left(\frac{1}{n}\right)+\frac{n^3}{3}\right]\right\}\end{align*}$$
But I could not compute this limit. And how can we show the existence of the limit $\displaystyle\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx$? Thanks!
| In order to avoid the use of dominated convergence theorem, we can notice that:
$$I=\int_{1}^{+\infty}\frac{1-\cos(x/n)}{x^4}\,dx = 2\int_{1}^{+\infty}\left(\frac{\sin\frac{x}{2n}}{x^2}\right)^2\,dx=2\int_{0}^{1}y^2\sin^2\frac{1}{2ny}\,dy $$
satisfies $I\leq\frac{1}{2n^2}$, since $|\sin x\,|\leq x$, while:
$$ I = \frac{2}{n^3}\int_{0}^{n}\left(y\sin\frac{1}{2y}\right)^2\,dy\geq\frac{2}{n^3}\int_{1}^{n}\left(y\sin\frac{1}{2y}\right)^2\,dy\geq\frac{2}{n^3(n-1)}\left(\int_{1}^{n}y\sin\frac{1}{2y}\,dy\right)^2, $$
due to Cauchy-Schwarz inequality, and:
$$ \int_{1}^{n}y\sin\frac{1}{2y}\,dy \geq \int_{1}^{n}\left(\frac{1}{2}-\frac{1}{48x^2}\right)\,dx \geq \frac{n}{2}-\frac{25}{48} $$
due to the fact that $\sin z \geq z-\frac{z^3}{6}$ for $z\in (0,1]$. The two inequalities together give:
$$ \frac{(n-25/24)^2}{n(n-1)}\leq 2n^2 I \leq 1, $$
so the wanted limit is $-\frac{1}{2}$ by squeezing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if the sum of digits of a number is divisible by 3, so is the number itself. Here is the proof of the converse:
Iff a number $n$ is divisible by $3$, then the sum of its digits is also divisible by $3$.
Proof:
We know $n \mod 3 = 0$. By the basis representation theorem, $n$ can be re-written as $n_k10^k + n_{k-1}10^{k-1} + \cdots + 10n_1 + n_0 \equiv 0 \mod 3$. By the modular arithmetic addition rule, we can take $n_k10^k \mod 3 + n_{k-1}10^{k-1} \mod 3 + \cdots + 10n_1 \mod 3 + n_0 \mod 3$. But then we just get $n_k + n_{k-1} + \cdots + n_1 + n_0$, since $10 \mod 3 \equiv 1$.
Now suppose that the sum of the digits was divisible by 3. How can I prove that the representation in base $10$ is divisible by $3$?
| I hope this will suffice:
$$
n
= (d_{m-1}\cdots d_0)_{10}
= \sum_{k=0}^{m-1} d_k 10^k
= \sum_{k=0}^{m-1} d_k (10^k - 1) + d_k
= \sum_{k=0}^{m-1} d_k (10^k - 1) + S(n)
$$
with $S(n)$ being the digit sum in base $10$ and
$$
10^k - 1 = 9 \frac{10^{k} - 1}{10-1} = 9 \sum_{i=0}^{k-1} 10^i
= \sum_{i=0}^{k-1} 9 \, 10^i
= (\underbrace{9\cdots 9}_{k \times})_{10}
$$
is divisible by $3$, it depends on the divisibility of $S(n)$.
The same statement should hold for division by $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding ordered pairs $(a,b)$ which are positive integers for $\sqrt{8+\sqrt{32+\sqrt{768}}}=a\cos\frac {\pi}b$
Finding ordered pairs $(a,b)$ for $\sqrt{8+\sqrt{32+\sqrt{768}}}=a\cos\frac {\pi}b$
I did:
$768=2^8\cdot 3$
So:
$$\sqrt{32+\sqrt{768}}=\sqrt{32+2\sqrt{2^6\cdot3}}=\sqrt{8}+\sqrt{24}$$
Now:
$$\sqrt{8+\sqrt{32+\sqrt{768}}}=\sqrt{8+2\sqrt2+2\sqrt6}=?$$
| $$\sqrt{8+\sqrt{32+\sqrt{768}}}$$
$$ \sqrt{8+\sqrt{32+16\sqrt{3}}}$$
$$ \sqrt{8+\sqrt{32+32\cos(\frac{\pi}{6})}}$$
$$ \sqrt{8+4\sqrt{2+2\cos(\frac{\pi}{6})}} $$
$$ \sqrt{8+8\cos(\frac{\pi}{12})} $$
$$ 4\cos(\frac{\pi}{24}) $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A calculus proof for the general term of the Fibonacci sequence Let $a_0=1$,$a_1=1$ and $a_n=a_{n-1} + a_{n-2}$ for $n \geq 2$, I would like to prove:
$$a_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n + 1}- \left(\frac{1-\sqrt{5}}{2}\right)^{n + 1} \right)$$
for $n\in \mathbb{N^*}$.
I used induction to prove it (not sure if it is a correct way), is there a way to find it doing some calculus ? I couldn't come to that result.
| Well there is no specific proof which makes use of calculus (meaning differentiation / integration) however there is a proof which does involve analysis in an indirect way. Here it goes.
The basic fact from analysis will be some concepts related to power series. A power series of type $\sum a_{n}x^{n}$ has a radius of convergence $R \geq 0$ such that the series is convergent if $|x| < R$ and divergent if $|x| > R$. The power series defines a continuous and infinitely differentiable function $f(x)$ for $|x| < R$. Also if two power series $\sum a_{n}x^{n}$ and $\sum b_{n}x^{n}$ are such that they are equal for all values of $x$ in some interval $(-R, R)$, then $a_{n} = b_{n}$ for all non-negative integers $n$. This last fact will be the main idea used in this proof.
Let $a_{n}$ denote $n^{\text{th}}$ Fibonacci number and consider the power series $F(x) = \sum_{n = 0}^{\infty}a_{n}x^{n}$. Clearly we can see that if this series converges in certain interval $(-R, R)$ then we have the following relations for $|x| < R$: $$F(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + \cdots\tag{1}$$$$xF(x) = a_{0}x + a_{1}x^{2} + a_{2}x^{3} + \cdots \tag{2}$$$$x^{2}F(x) = a_{0}x^{2} + a_{1}x^{3} + a_{2}x^{4} + \cdots\tag{3}$$ Now these equations clearly give us $$F(x) - xF(x) - x^{2}F(x) = a_{0} + (a_{1} - a_{0})x$$ as rest of the terms cancel because of the recurrence relation $a_{n + 2} = a_{n + 1} + a_{n}$. Putting $a_{0} = 1, a_{1} = 1$ we get $$F(x) = \frac{1}{1 - x - x^{2}}$$ If $\alpha = (1 + \sqrt{5})/2, \beta = (1 - \sqrt{5})/2$ then $\alpha, \beta$ are roots of $t^{2} - t - 1 = 0$ and hence we can write $$1 - x - x^{2} = (1 - \alpha x)(1 - \beta x)$$ Thus we have $$F(x) = \frac{1}{(1 - \alpha x)(1 - \beta x)} = \frac{A}{1 - \alpha x} + \frac{B}{1 - \beta x}$$ where $A, B$ need to be found out. Clearly we have $$1 = A(1 - \beta x) + B(1 - \alpha x)$$ and putting $x = 1/\alpha$ we get $A = \dfrac{\alpha}{\alpha - \beta}$. Similarly $B = -\dfrac{\beta}{\alpha - \beta}$. Thus we have $$F(x) = \frac{1}{\alpha - \beta}\left(\frac{\alpha}{1 - \alpha x} - \frac{\beta}{1 - \beta x}\right)\tag{4}$$ Next we use another result from power series namely that $$\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \cdots$$ for $|x| < 1$. Using this result we can see that $$\sum_{n = 0}^{\infty}a_{n}x^{n} = F(x) = \frac{1}{\alpha - \beta}\left(\sum_{n = 0}^{\infty}(\alpha^{n + 1} - \beta^{n + 1})x^{n}\right)$$ and hence the coefficients in both power series must be equal and we have $$a_{n} = \frac{\alpha^{n + 1} - \beta^{n + 1}}{\alpha - \beta} = \frac{1}{\sqrt{5}}\left\{\left(\frac{1 + \sqrt{5}}{2}\right)^{n + 1} - \left(\frac{1 - \sqrt{5}}{2}\right)^{n + 1}\right\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A die is rolled 3 times. What is the probability that a five is rolled at least twice? The probability of not getting a five is $(\frac56)^3$, and I figure the probability of getting at least one 5 is $1-(\frac56)^3$, but I don't know how to figure out if it is rolled at least twice. Thoughts? Thanks in advance!
| Since you're performing multiple experiments each with exactly two outcomes (the dice is a five vs. the dice is not a five), this is a classic example of a Bernoulli trial.
The standard formula to calculate the probability is then to use (from the Binomial distribution):
$$
P\{\text{that A occurs exactly k times in n trials}\} = \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) p^k (1 - p)^{n-k}
$$
where event $A$ is the trial and $p$ is the probability for the success of a single trial ($\frac{1}{6}$).
In your case, you want to know the probability of it happening at least twice, so you need to sum the chances for it happening exactly twice and exactly three times:
$$
P\{\text{that A occurs exactly 2 times in 3 trials}\} + P\{\text{that A occurs exactly 3 times in 3 trials}\}
$$
Therefore,
$$
\begin{align}
P&= \left( {\begin{array}{*{20}c} 3 \\ 2 \\ \end{array}} \right) p^2 (1 - p)^{(3-2)} + \left( {\begin{array}{*{20}c} 3 \\ 3 \\ \end{array}} \right) p^3 (1 - p)^{(3-3)} \\
&= \left( {\begin{array}{*{20}c} 3 \\ 2 \\ \end{array}} \right) \cdot \frac{1}{6}^2 \cdot \frac{5}{6}^{1} + \left( {\begin{array}{*{20}c} 3 \\ 3 \\ \end{array}} \right) \cdot \frac{1}{6}^3 \cdot \frac{5}{6}^{0}\\
&= 3 \cdot \frac{1}{36} \cdot \frac{5}{6} + 1 \cdot \frac{1}{216} \cdot 1\\
&= \frac{15}{216} + \frac{1}{216}\\
&= \frac{16}{216}\\
&= \frac{2}{27} = 0,074... = 7,41 \% \\
\end{align}
$$
As you can see, this method of calculation can unfortunately become very tedious for larger examples, but it has the advantage of being a consistent approach for any problems surrounding Bernoulli trials. The intuitive approaches mentioned in some of the other answers can be quicker for small cases where all the possible outcomes can be manually considered.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
Reducing $\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}$ for $n>2$
Simplify
$$\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}$$ for $n>2$.
The answer is $(a+b+c)^n$, but I can't seen to get it.
Can someone help me?
Thanks
| $$\frac{(b-c)\times a^n - (a-c)\times b^n + (a-b)\times c^n}{(a-b)(a-c)(b-c)}=$$
$$\frac{a^n-b^n+c^n}{1}=$$
$$\boxed{\frac{a^n}{(a - b) (a - c)} - \frac{b^n}{(a - b) (b - c)} + \frac{c^n}{(a - c) (b - c)}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Convergence of $\sum\limits_{n=1}^\infty \frac {n+1}{2^n}$. Test Convergence of $$\sum\limits_{n=1}^\infty \dfrac {n+1}{2^n}$$
Attempt: $$\sum\limits_{n=1}^\infty \dfrac {n+1}{2^n} = \sum\limits_{n=1}^\infty \dfrac {n }{2^n} + \sum\limits_{n=1}^\infty \dfrac {1}{2^n}$$
The second summation is definitely convergent. So, we need to just investigate if the first summation is convergent or not.
Let $$X = \sum\limits_{n=1}^\infty \dfrac {n }{2^n}$$
Is there a way to test convergence of this summation without the integral test?
Thank you very much for your help.
| Let:
$$ S_N = \sum_{n=1}^{N}\frac{n+1}{2^n}. $$
We have:
$$\begin{eqnarray*} \frac{S_N}{2}&=&S_N-\frac{S_N}{2}=\sum_{n=1}^{N}\frac{n+1}{2^n}-\sum_{n=1}^{N}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{N}\frac{n+1}{2^n}-\sum_{n=2}^{N+1}\frac{n}{2^{n}}\\&=&1+\sum_{n=2}^{N}\frac{1}{2^n}-\frac{N+1}{2^{N+1}}=\frac{3}{2}-\frac{N+3}{2^{N+1}}.\end{eqnarray*}$$
Since $\frac{N+2}{2^{N+1}}\to 0$ as $N\to +\infty$, we have $S_N\to\color{red}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Why is $ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d} \geq 0.5 $ with $a+b+c+d = 1$? For positive real numbers $a,b,c,d>0$ it seems to be true that:
if $$a+b+c+d = 1$$
then
$$ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d} \geq 0.5 $$
I can't think of a way to prove this statement.
Any help will be appreciated!
| Another approach using $4ab\le(a+b)^2$:
\begin{align*}\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}&=\left(a-\frac{ab}{a+b}\right)+\left(b-\frac{bc}{b+c}\right)+\left(c-\frac{cd}{c+d}\right)+\left(d-\frac{da}{d+a}\right)\\
&=1-\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{cd}{c+d}+\frac{da}{d+a}\right)\\
&\ge1-\frac14\Bigl((a+b)+(b+c)+(c+d)+(d+a)\Bigr)=\frac12\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Abelian torsion group of rational points of an elliptic curve I want to find the abelian group of rational points $E(\mathbb{Q})_{\text{torsion}}$ of the elliptic curve $y^2=x^3+8$.
$$E(\mathbb{Q})_{\text{torsion}}=\{P \in E(\mathbb{Q}) | P \text{ of finite order }\}$$
From Lutz-Nagell theorem we have the following:
Let $E|_{\mathbb{Q}} , y^2=x^3+ax+b, a, b \in \mathbb{Z}$ and $P=(x, y) \in E(\mathbb{Q})$.
We suppose that $P$ is of finite order.
Then $x, y \in \mathbb{Z}$ and $y=0$ (that corresponds to the points of order $2$ ) or $y^2 \mid D(f)=4a^3+27b^2$.
We have also the following:
If $P(x, y) \in E(\mathbb{Q})$ and $x \notin \mathbb{Z}$ or $y \notin \mathbb{Z}$, then $P$ has infinite order.
$$$$
So, to find the points $P$ of finite order of the curve $y^2=x^3+8$ we do the following:
For $y=0 \Rightarrow x^3+8=0 \Rightarrow x=-2$, $$P=(-2, 0) \in E(\mathbb{Q})$$ of order $2$
Let $f(x)=x^3+8$, then $D(f)=27 \cdot 8^2=2^6 \cdot 3^3$.
$$y^2 \mid 2^6 \cdot 3^3$$
$$2^i, i=0, 2, 4, 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3^j, j=0, 2$$
$$y^2 \mid D(f) \Rightarrow y^2=1, 3^2, 2^2, 2^2 \cdot 3^2, 2^4, 2^4 \cdot 3^2, 2^6, 2^6 \cdot 3^2$$
$$ \Rightarrow y=\pm 1 , \pm 3, \pm 2, \pm 2 \cdot 3, \pm 2^2 , \pm 2^2 \cdot 3 , \pm 2^3, \pm 2^3 \cdot 3 \\ \Rightarrow y=\pm 1 , \pm 3, \pm 2, \pm 6, \pm 4 , \pm 12 , \pm 8, \pm 24 $$
For $y=\pm 1 \Rightarrow x^3+7=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 3 \Rightarrow x^3-1=0 \Rightarrow x=1$, so there are two points with integer coordinates, $P_1=(1, 3), P_2=(1, -3)$.
Does this mean from Lutz-Nagell theorem that these two points are of finite order??
For $y=\pm 2 \Rightarrow x^3+4=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 6 \Rightarrow x^3-28=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 4 \Rightarrow x^3-8=0 \Rightarrow x=2$, so there are two points with integer coordinates, $P_3=(2, 4), P_4=(2, -4)$.
Which is the order of these points??
For $y=\pm 12 \Rightarrow x^3-136=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 8 \Rightarrow x^3-56=0 \Rightarrow x \notin \mathbb{Z}$.
For $y=\pm 24 \Rightarrow x^3-568=0 \Rightarrow x \notin \mathbb{Z}$.
| For this curve the torsion points are $P = (-2, 0)$ of order $2$ and the point at infinity. The reason why your points $P_1, ..., P_4$ are not torsion points are due to the fact that the Nagell-Lutz Theorem only gives you the possible candidates for the torsion points. In other words, Nagell-Lutz is not a if and only if-theorem.
When checking the list of possible torsion points, then if a point $P$ has order greater than $12$ then that point has infinite order. So you need to check if $nP = \mathcal{O}$, for $2 \leq n \leq 12$. If not, then $P$ has infinite order. This is due to the celebrated theorem of Barry Mazur, see wikipedia.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is $f:\mathbb{R}^2\to\mathbb{R}$ differentiable on $(0,0)$? Let $f:\mathbb{R}^2\to\mathbb{R}$ such that $f(x,y) = (x^2+y^2)\sin(\frac{1}{x^2+y^2})$ for $(x,y)\ne (0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$. Is $f$ differentiable on $(0,0)$?
So let's first evaluate the partial derivatives:
$\frac{\partial f}{\partial x} = 2x\sin(\frac{1}{x^2+y^2}) + (x^2+y^2)\cos(\frac{1}{x^2+y^2})\frac{-2x}{(x^2+y^2)^2} = 2x\left( \sin(\frac{1}{x^2+y^2}) - \frac{\cos(\frac{1}{x^2+y^2})}{x^2+y^2} \right)$
By symmetry;
$\frac{\partial f}{\partial y} = 2y\sin(\frac{1}{x^2+y^2}) + (x^2+y^2)\cos(\frac{1}{x^2+y^2})\frac{-2x}{(x^2+y^2)^2} = 2y\left( \sin(\frac{1}{x^2+y^2}) - \frac{\cos(\frac{1}{x^2+y^2})}{x^2+y^2} \right)$
| By continuity $f(0,0)=0$ (because $0$ is the limit of $f(x,y)$ when $(x,y)\rightarrow (0,0)$ Now let us look at $$lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}$$ This limit is also $0$ and therefore $f$ is differentiable at the origin
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
| $p_0 = (0, 0), p_i = (i/n, y_n), p_n = (1,0)$
$A = \frac{1}{2} \cdot \sum_{i=1}^n{(y_i + y_{i-1})} \cdot \frac{1}{n} = 1$
$L = \sum{|p_{i+1} - p_i|}$
$\frac{\partial A}{\partial y_i} = \frac{\partial }{\partial y_i} \frac{1}{2n} \cdot (y_i + y_{i-1} + y_{i+1} + y_i) = \frac{1}{n}$
$\frac{\partial L}{\partial y_i} = \frac{\partial }{\partial y_i} \bigg[\sqrt{\frac{1}{n^2} + (y_i - y_{i-1})^2} + \sqrt{\frac{1}{n^2} + (y_{i+1} - y_i)^2}\bigg] = $
$ = \frac{y_i - y_{i-1}}{\sqrt{\frac{1}{n^2} + (y_i - y_{i-1})^2}} + \frac{y_i - y_{i+1}}{\sqrt{\frac{1}{n^2} + (y_{i+1} - y_i)^2}} = \lambda / n$
Lagrange multiplier
Does anyone know how make this a differential equation? $n \rightarrow \infty, 1/n \rightarrow dx$
$\frac{\delta_i}{\sqrt{(dx)^2 + (\delta_i)^2}} - \frac{\Delta_i}{\sqrt{(dx)^2 + (\Delta_i)^2}} = \lambda \cdot dx$
$\frac{\delta_i^2}{{(dx)^2 + (\delta_i)^2}} = \lambda^2 \cdot (dx)^2 + \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}} + 2 \frac{\Delta_i}{\sqrt{(dx)^2 + (\Delta_i)^2}} \lambda \cdot dx $
$\bigg[\frac{\delta_i^2}{{(dx)^2 + (\delta_i)^2}} - \lambda^2 \cdot (dx)^2 - \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}}\bigg]^2 = 4 \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}} \lambda^2 \cdot (dx)^2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 12,
"answer_id": 8
} |
Showing that $\alpha \beta$ is the root of a polynomial Assuming that $\alpha, \beta$ are distinct roots of $P(x) = x^4+bx^3-1 = 0$, where $b \in \mathbb R$, show that $\alpha \beta$ is the root of $Q(x) = x^6+x^4+b^2x^3 -x^2 -1$.
I have already noticed that $P(0) < 0$ and $\deg P = 4$, so (from the intermediate value theorem) there are at least 2 real roots. It seems that the third and fourth root are complex, but I'm not sure, if this could help.
| $P(\alpha)=0$ and $P(\beta)=0$ therefore $\alpha^4=1-b\alpha^3$ and $\beta^4=1-b\beta^3$; multiplying by $\alpha^2$ (respectively$\beta^2$) we get $\alpha^6=-b\alpha^3+\alpha^2-b\alpha+b$ (resp $\beta^6=-b\beta^3+\beta^2-b\beta+b$) substitute the sixth and the fourth powers of $\alpha$ and $\beta$ in the expression of $$Q(\alpha\beta)=\alpha^6\beta^6+\alpha^4\beta^4+b^2\alpha^3\beta^3-\alpha^2\beta^2-1$$ simplify and you should get zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate $\lim_{x\to\infty} ((x^5+x^4)^{1/6}-(x^5-x^4)^{1/6})$ I've been struggling with the following:
$$\lim_{x\to\infty} ((x^6+x^5)^{1/6}-(x^6-x^5)^{1/6})$$
Tried factoring out $x^{5/6}$ and then using L'hopital- which got me nowhere, tried multiplying by the conjugate, but it got messy- so either I'm scared of the algebra or there's a better way.
| One more way (without L'Hospital):
After getting
$$
\lim_{x \to 0} \frac{(1+x)^{\frac{1}{3}} - (1-x)^{\frac{1}{3}}}{x((1+x)^{\frac{1}{6}} + (1-x)^{\frac{1}{6}})}
$$
multiply both numerator and denominator by $(1+x)^{\frac{2}{3}} +(1-x)^{\frac{2}{3}} + ((1+x)(1-x))^{\frac{1}{3}}$ to get $a^3 -b^3$ in the numerator:
$$
\lim_{x \to 0} \frac{2}{((1+x)^{\frac{1}{6}} + (1-x)^{\frac{1}{6}})((1+x)^{\frac{2}{3}} +(1-x)^{\frac{2}{3}} + ((1+x)(1-x))^{\frac{1}{3}})}
$$
Now take the limit to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Complicated integration How can this be integrated? :
$$\int_{b}^{a} x \left ( \frac{D}{a-b} \right ) \left ( \frac{a-x}{a-b} \right )^{D-1}dx$$
The solution is :
$$\frac{a+(D)(b)}{D+1}$$
| \begin{align}
S &= \int_{b}^{a} x \left( \frac{D}{a-b} \right) \left( \frac{a-x}{a-b} \right)^{D-1} \,\mathrm{d}x \\\\
&= \frac{D}{\left(a-b\right)^D} \int_b^a x(a-x)^{D-1}\,\mathrm{d}x\\\\
&= \frac{D}{\left(a-b\right)^D} \int_b^a x(a-x)^{D-1}\,\mathrm{d}x
\end{align}
Let $y=a-x$ so that $x=a-y$, and
\begin{align}
S&= \frac{D}{\left(a-b\right)^D} \int_b^a x(a-x)^{D-1}\,\mathrm{d}x \\\\
&= \frac{D}{\left(a-b\right)^D} \int_{a-b}^0 (y-a)y^{D-1}\,\mathrm{d}y\\\\
&=\frac{D}{\left(a-b\right)^D} \left(\int_{a-b}^0 y^D\,\mathrm{d}y - a\int_{a-b}^0 y^{D-1}\,\mathrm{d}y\right)\\\\
&=\frac{D}{\left(a-b\right)^D} \left(a\int_0^{a-b} y^{D-1}\,\mathrm{d}y - \int_0^{a-b} y^D \,\mathrm{d}y\right)\\\\
&=\frac{D}{\left(a-b\right)^D} \left(\frac{a}{D}(a-b)^D-\frac{1}{D+1} (a-b)^{D+1}\right)\\\\
&=D\left(\frac{a}{D}-\frac{a}{D+1}+\frac{b}{D+1}\right)\\\\
&=D\left(\frac{aD+a}{D^2+D}-\frac{aD}{D^2+D}+\frac{bD}{D^2+D}\right)\\\\
&=\boxed{\displaystyle\frac{a+Db}{D+1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The degree of $\sqrt{2} + \sqrt[3]{5}$ over $\mathbb Q$ I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I'm trying to construct a polynomial with root $\sqrt{2} + \sqrt[3]{5}$ and coefficients in $\mathbb Q$ of degree $6$, and then show that it is irreducible over $\mathbb Q$.
I managed to find that it is a root of the polynomial $x^6 - 6x^4 - 10x^3 +12x^2 - 60x +17$. This is where I run in to some problems. I don't know how to show that this is irreducible over $\mathbb Q$ (or $\mathbb Z$). The only criteria we have learned is Eisenstein's criteria, which clearly does not apply here. How else can I show that the degree of this number over $\mathbb Q$ is $6$?
| It's enough to show that $x^3-5$ is irreducible over $\mathbb{Q}(\sqrt{2})$. If it weren't, then it would have a root in $\mathbb{Q}(\sqrt{2})$. Since it has a unique real root, we must have $\sqrt[3]{5}\in \mathbb{Q}(\sqrt{2})$. So $(a+b\sqrt{2})^3=5$, and therefore $(a-b\sqrt{2})^3=5$, hence $a+b\sqrt{2}=a-b\sqrt{2}=\sqrt[3]{5}$, and so $\sqrt[3]{5}\in \mathbb{Q}$, contradiction.
$\bf{Added:}$ I am uncomfortable with my answer, since it assumes $\mathbb{Q}(\sqrt{2}+ \sqrt[3]{5}) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5})$, which is the crux of the problem. If that is proved, then the degree has to be divisible by $2$, and $3$, so it is necessarily $6$. So let's just show that $\alpha = \sqrt{2}+ \sqrt[3]{5}$ generates both roots. Now, we have the equality $(\alpha - \sqrt{2})^3= 5$ from which we can obtain $\sqrt{2}$ as a rational fraction in $\alpha$. In detail:
$$\alpha^3 - 3 \alpha^2 \sqrt{2} +3 \alpha \cdot 2 - 2 \sqrt{2} = 5$$
or
$$\frac{\alpha^3 +6 \alpha - 5 }{3 \alpha^2 + 2} = \sqrt{2}$$
By the way, the equation that we get for $\alpha$
$$(x^3 + 6 x -5)^2 - 2(3x^2 + 2)^2=0$$
is the equation in the posting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Pythagorean Triple divisible by $5$ Show that, if x, y and z are integers such that $x^2+y^2 = z^2$,then at least one of $x,y,z$ is divisible by $5$. I was able to show that at least one of $x$ or $y$ is divisible by $2$. Can someone please show me how to prove it for $5$?
| The only squares in $Z_5$ are $0$, $1$, and $4$.
Assume that none of $x$, $y$ and $z$ is divisible by 5. Then, we also have $x^2$, $y^2$ and $z^2$ non-divisible by 5. This means that the possible values of $x^2$ + $y^2$ - $z^2$ ($mod$ $5$) are:
$$(1 + 1 - 1 = 1);(1 + 1 - 4 = -2 = 3);(1 + 4 - 1 = 4);(1 + 4 - 4 = 1);(4 + 1 - 1 = 4);(4 + 1 - 4 = 1);(4 + 4 - 1 = 7 = 2); (4 + 4 - 4 = 4)$$
$0$ is none of the above values. A contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$ I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$
where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS:
$$
\begin {align}
a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\
&=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\
&=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\
&=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc]
\end{align}$$
where $$s=\frac{a+b+c}{2}$$
I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
But I'm stuck there. Any help please?
| $t=a^2(s-a)+b^2(s-b)+c^2(s-c)-abc=a^2(s-a)+b^2(s-a)+c^2(s-a)+a(b^2+c^2-bc)-(b^3+c^3)\\b^3+c^3=(b+c)(b^2+c^2-bc)\\t=(s-a)(a^2+b^2+c^2)-(b^2+c^2-bc)(b+c-a),b+c-a=2(s-a)\\t=(s-a)(a^2+b^2+c^2-2(b^2+c^2-bc))=(s-a)(a^2-(b^2+c^2-2bc))=(s-a)(a^2-(b-c)^2)=(s-a)(a+c-b)(a+b-c)=4(s-a)(s-b)(s-c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $A^{100}+A^5$ Let $A= \begin{pmatrix} 1&-1 \\ 2&-2 \\ \end{pmatrix}$. We are to find $A^{100} +A^5$
My attempt : I found characteristic equation , $x^2+x=0$, so eigenvalues are $0$ and $-1$ hence its diagnol matrix is $$D= \begin{pmatrix} 0&0 \\ 0&-1 \\ \end{pmatrix}$$ So $D^{100}+D^5$ is null matrix.
But I am getting something else if I use method given here To Find $A^{50}$ by achille hui.
Please help in this matter. Thanks
| If its characteristic equation is $x^2 + x = 0$, then that means that $x^2 = -x$, ie, $A^2 = -A$.
Thus, $A^{100} = (A^2)^{50} = (-A)^{50} = A^{50} = (A^2)^{25} = (-A)^{25} = -A^{25}$
At this point you've got an odd power, so simply factor out $-A$, so
$$\cdots = -A\cdot A^{24} = -A\cdot (A^2)^{12} = -A\cdot (-A)^{12} = -A\cdot A^{12} = -A\cdot (A^2)^6 = -A\cdot(-A)^6 = -A\cdot A^6$$
$$\cdots = -A\cdot (A^2)^3 = -A\cdot (-A)^3 = (-A)^4 = A^4 = (A^2)^2 = (-A)^2 = A^2 = -A$$
Thus, this shows that $A^{100} = -A$.
Doing the same thing with $A^5$ gives $A^5 = A\cdot (A^2)^2 = A\cdot (-A)^2 = A\cdot A^2 = A\cdot (-A) = -A^2 = -(-A) = A$.
Thus, indeed $A^{100} + A^5 = 0$.
This was the dumb way, but it's a method that's easily generalizable.
In this case, it's easier if you consider the powers of $A$:
$A^2 = -A$
$A^3 = A\cdot A^2 = A\cdot -A = -A^2 = A$
$A^4 = A\cdot A^3 = A\cdot A = A^2 = -A$
at which point you'll notice that $A^k = A$ if $k$ is odd, and otherwise $A^k = -A$ if $k$ is even. Thus, $A^{100} = -A$ and $A^5 = A$, so $A^{100} + A^5 = -A + A = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $x,y,z>0$ such that $x^2+y^2+z^2=1$. Prove that $x^2yz+y^2zx+z^2xy\le \frac13$ Let $x,y,z>0$ such that $x^2+y^2+z^2=1$. Prove that $x^2yz+y^2zx+z^2xy\le \dfrac13$
My attempt:
I tried using AM-GM and also weighted AM-GM but both seems to be unyielding. So, please help. Thank you.
| Apply AM-GM and Cauchy-Schwarz inequalities:
$LHS =xyz(x+y+z) \leq \dfrac{(x+y+z)^4}{27}\leq \dfrac{(3(x^2+y^2+z^2))^2}{27}=\dfrac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof by Induction: $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ I'm having some troubles trying to prove the Exercise 13, page 41 of Apostol's Calculus I, which is the one used to explain some features of integration in the next pages.
It says:
Prove that $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ if $n\geq 1$. Then use this to prove that
$$2\sqrt{m}-2 < \sum_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1 \qquad \text{if } m \geq 2$$
(At this moment I'm stuck with the first part of the problem).
Proof (by Induction):
$$P(n): 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})\qquad \text{for }n\geq 1$$
Base Case: $P(1)$
$$\begin{align*}
2(\sqrt{1+1} - \sqrt{1}) &< \frac{1}{\sqrt{1}} < 2(\sqrt{1}-\sqrt{1-1})&\\ 2(\sqrt{2} - \sqrt{1}) &<\hspace{5pt} 1 \hspace{6pt} < 2
\end{align*}$$
which is true.
Induction Hypothesis: Assume $P(k)$ is true for a positive integer $k \geq 1$:
$$\underbrace{2(\sqrt{k+1} - \sqrt{k})}_{a} < \frac{1}{\sqrt{k}} < 2(\sqrt{k}-\sqrt{k-1})\qquad (1)$$
Induction Step: Prove $P(k+1)$:
$$2(\sqrt{k+2} - \sqrt{k+1}) < \frac{1}{\sqrt{k+1}} < \underbrace{2(\sqrt{k+1} - \sqrt{k})}_{a}$$
This was my strategy:
Since we assumed $P(k)$ is true, by multiplying the set of inequalities $(1)$ by $\frac{\sqrt{k}}{\sqrt{k+1}}$ we get
$$2\left(\sqrt{k} - \frac{k}{\sqrt{k+1}}\right) < \frac{1}{\sqrt{k+1}} < \underbrace{2\left(\frac{k}{\sqrt{k+1}}-\sqrt{\frac{k-1}{k+1}}\right)}_{b}$$
Then, if we can show that $b < a$, by transitivity we have that $P(k+1)$ is true.
And there is where I'm stuck. I'm not sure what I'm doing wrong, but by setting the inequality $b<a$, it seems that the assumption I made doesn't work:
$$\begin{align*}\frac{k}{\sqrt{k+1}} - \sqrt{\frac{k-1}{k+1}} <&\ \sqrt{k+1} - \sqrt{k}\\
\frac{k}{\sqrt{k+1}} - \sqrt{1- \frac{2}{k+1}} <&\ \sqrt{k+1} -
\sqrt{k}\end{align*}$$
The RHS approaches $0$ while the LHS approaches $k$, which reverses the inequality, so I think it's not the correct approach.
Could someone shade me some light on it?
| Since $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}},$$ we have $$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}.$$ Since $$\frac{1}{2\sqrt{n}} = \frac{1}{\sqrt{n} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$$ we also have $$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the limit of $a_n = \frac{2n^3+n^2}{(n+2)^3}$ is $2$. if $a_n = \dfrac{2n^3+n^2}{(n+2)^3}$, prove that the limit of $a_n$ (as $n$ tends to infinite) is $2$ using the definition of a limit.
My attempt was $\left |\dfrac{(2(n^3))+(n^2)}{(n+2)^3} - 2\right |=\left |\dfrac{(2(n^3))+(n^2)-2((n+2)^3)}{(n+2)^3}\right |$ and tried to simplify it as much as I could (though I couldn't get very far) then set it to less than Epsilon.
I'm not sure if that was the right approach though. Would anyone be able to help me out on this? Any help is greatly appreciated. Thank you!
| $$
\left|\frac{2n^3 +n^2}{(n+2)^3} - 2\right| \le \left|\left( \frac{n}{n+2} \right)^3 \left(2+\frac1{n}-2 \right)\right| + 2\left|\left( \frac{n}{n+2} \right)^3 -1 \right| \\
= \frac1{n+2}\left( \left| \left(\frac{n}{n+2} \right)^2 \right|+4\left|\left(\frac{n}{n+2} \right)^2 +\left(\frac{n}{n+2} \right) + 1 \right|\right)
\lt \frac{13}{n+2} \rightarrow 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I feel like I have to find a way to cancel out the $n$'s but I'm not sure how to do so with the $\sin^2(11n)$ The limit is from $n$ to $\infty$, I tried dividing everything by $n^3$ but I don't know how to deal with that
$$ \frac{-4n^3+\sin^2(11n)}{n^3+17} $$
| You can try splitting up the fraction and then multiplying through one of the terms by $\frac{n^{-3}}{n^{-3}}$. $$\begin{align} \frac{-4n^3+\sin^2(11n)}{n^3+17} =\frac{\sin^2(11n)}{n^3+17} -\frac{4n^3}{n^3+17}\cdot\frac{n^{-3}}{n^{-3}} \\ =\frac{\sin^2(11n)}{n^3+17} -\frac{4}{1+\frac{17}{n^3}} \end{align}$$ You should be able to see that $\frac{\sin^2(11n)}{n^3+17}$ converges to zero as $n \to \infty$ because the numerator $\sin^2(11n)$ is bounded between $0$ and $1$, hence the denominator will grow exponentially while the numerator is bounded. Then the term $-\frac{4}{1+\frac{17}{n^3}}$ is nice because $\frac{17}{n^3}$ converges to zero as $n \to \infty$, so $-\frac{4}{1+\frac{17}{n^3}} \to -\frac{4}{1+0} = -4$. In conclusion, $$\lim_{n \to \infty} \frac{-4n^3+\sin^2(11n)}{n^3+17} = \lim_{n \to \infty}\frac{\sin^2(11n)}{n^3+17} -\frac{4}{1+\frac{17}{n^3}} = -4 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the smallest value for $n$ such that $ \sum_{k=1}^{n}30\left(\frac{1}{3}\right)^{k-1} > 44.8 $ I had a test today where I was asked to find the smallest value for $n$ such that:
$$ \sum_{k=1}^{n}30\left(\frac{1}{3}\right)^{k-1} > 44.8 $$ This question really frustrated me but this is how I went about:
This is a geometric sequence where $a = 30$ and $ r = \frac{1}{3} $ where $a$ is the first term and $r$ is the common ratio.
$$S_{n} = \frac{a(1-r^{n})}{1-r}$$
$$ \frac{30\left(1-\left(\frac{1}{3}\right)^{n}\right)}{1-\frac{1}{3}} > 44.8 $$
$$ 30\left(1-\left(\frac{1}{3}\right)^{n}\right) > 67.2 $$(Multiplied the right by $\frac{3}{2}$)
$$ -\left(\frac{1}{3}\right)^{n} > 1.24 $$
$$ \left(\frac{1}{3}\right)^{n} < -1.24 $$
But one cannot proceed as$ \left(\frac{1}{3}\right)^{n}$ cannot be negative. Also, I was told to use logs on both sides but one cannot have the log of a negative number. Where did I go wrong?
| $\frac{30\left(1-\left(\frac{1}{3}\right)^{n}\right)}{1-\frac{1}{3}} > 44.8$
...this can be written as: $45(1-\frac{1}{3^n}) > 44.8$. And then:
$$
45(1-\frac{1}{3^n}) > 44.8 \\
-\frac{1}{3^n} > - \frac{1}{225} \\
225 < 3^n \qquad \text{for } n \geq 5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Sine Law Homework Question I am given a triangle with two triangles inside of it, and am asked to solve for angles $x$ and $y$.
I have illustrated the triangle here:
My process:
$$\frac{25}{\sin10.5} = \frac{66}{\sin B}$$
We then solve for angle $B$ (the top one) and we get $28.8$ degrees.
$$\begin{align}
28.8 + 10.5 & = 39.3 \\
180 - 39.3 & = 140.7
\end{align}$$
Therefore, the angle on the left, $y$, is equal to $140.7$. Then, by the property of isosceles triangles, the one on the right is also $140.7$. $180 - 140.7 = 39.3$, giving us angle $X$.
$$\begin{align}
\angle X & = 39.3^\circ \\
\angle Y & = 140.7^\circ
\end{align}$$
The textbook answers shows that this is wrong. Am I wrong, or is the textbook?
| When you solved the Law of Sines, you should have obtained two solutions for $m\angle ABC$. The one you chose does not produce a triangle, so you have to use the other solution.
Consider the diagram below.
By the Law of Sines,
$$\frac{25}{\sin(10.5^\circ)} = \frac{66}{\sin\theta}$$
Solving for $\sin(\theta)$ yields
$$\sin\theta = \frac{66\sin(10.5^\circ)}{25}$$
One solution of the equation is
$$\theta = \arcsin\left[\frac{66\sin(10.5^\circ)}{25}\right] = 28.8^\circ$$
where the answer has been rounded to one decimal place. Since $\sin\alpha = 180^\circ - \sin\alpha$, the other solution is
$$\theta = 180^\circ - \arcsin\left[\frac{66\sin(10.5^\circ)}{25}\right] = 151.2^\circ$$
where the answer has been rounded to one decimal place.
If $\theta = 28.8^\circ$, then since the sum of the measures of $\triangle ABC$ is $180^\circ$,
$$y = 180^\circ - 10.5^\circ - 28.8^\circ = 140.7^\circ$$
We were given that $\overline{AB} \cong \overline{DB}$, so $\triangle ABD$ is isosceles. By the Isosceles Triangle Theorem, $\angle BAD \cong \angle BDA$. Thus,
$$m\angle BDA = m\angle BAD = y = 140.7^\circ$$
which is impossible since the Angle Sum Theorem for Triangles implies that a triangle cannot contain two obtuse angles. Hence, we must discard the solution $\theta = 28.8^\circ$.
If $\theta = 151.2^\circ$, then
$$y = 180^\circ - 10.5^\circ - 151.2^\circ = 18.3^\circ$$
Thus, by the Isosceles Triangle Theorem,
$$m\angle ADB = m\angle BAD = y = 18.3^\circ$$
Since the sum of the measures of $\triangle ABD$ is $180^\circ$,
$$m\angle ABD = 180^\circ - m\angle BAD - m\angle ADB = 180^\circ - 2 \cdot 18.3^\circ = 143.4^\circ$$
Observe that $\angle BDC$ is an exterior angle of $\triangle ABD$. By the Exterior Angle Theorem,
$$x = m\angle BDC = m\angle BAD + m\angle ABD = 18.3^\circ + 143.4^\circ = 161.7^\circ$$
As confirmation, observe that
$$\varphi = m\angle CBD = m\angle ABC - m\angle ABD = 151.2^\circ - 143.4^\circ = 7.8^\circ$$
and that
$$m\angle CBD + m\angle BCD + m\angle BDC = 7.8^\circ + 10.5^\circ + 161.7^\circ = 180^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $3*5*17*257-3^{16}$ using factorization formulas? look at this:
$2*4*10*82*6562-3^{16}$
It's easy to calculate it with elementary arithmetic.
but how to calculate it using factorization formulas?
| Hints: $$3 \cdot 5 \cdot 17 \cdot 257 = (2-1)(2+1)(2^2+1)(2^4+1)(2^8+1) = (2^2-1)(2^2+1)(2^4+1)(2^8+1) = \dots$$
$$2 \cdot 4 \cdot 10 \cdot 82 \cdot 6562 = (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can anyone explain how to do this? Find the limit of
$\lim\limits_{(x,\ y)\to(0,\ 0)}f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2} $, using squeeze theorem.
So far i have done is :
- |$\frac{1}{2}x$ +2|<= $\frac{x^3+4x^2+2y^2}{2x^2+y^2} $ <= |$\frac{1}{2}x$ +2|
But the limit of the LHS = -2 and RHS = 2.
The answer is 2
| $$f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2}=\frac{x^3}{2x^2+y^2}+\frac{2(2x^2+y^2)}{2x^2+y^2}$$
Then $$f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2}=\frac{x^3}{2x^2+y^2}+\frac{2(2x^2+y^2)}{2x^2+y^2}\le \frac{x^3}{2x^2}+2 \le \frac{x}{2}+2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int\sqrt{1+9t^4}\:dt$ I need to find
\begin{align}
\int\sqrt{1+9t^4}\:dt.
\end{align}
What I have so far:
\begin{align}
\int\sqrt{1+9t^4}\:dt & =\int\sqrt{1+\left(3t^2\right)^2}\:dt,\tag{1}
\end{align}
now let $3t^2=\tan\left(\theta\right)\implies \displaystyle t=\sqrt{\frac{\tan\left(\theta\right)}{3}},\:\:dt=\frac{1}{2}\left(\frac{\tan\left(\theta\right)}{3}\right)^{-1/2}\left(\frac{\sec^{2}\left(\theta\right)}{3}\right)$, which gives me
\begin{align}
\int\left[1+\tan^2\left(\theta\right)\right]^{1/2}\frac{\sqrt{3}\sec^2\left(\theta\right)}{6\sqrt{\tan\left(\theta\right)}}\:d\theta\tag{2}&=\frac{\sqrt{3}}{6}\int\frac{\sec^3\left(\theta\right)}{\sqrt{\tan\left(\theta\right)}}\:d\theta\\
& = \frac{\sqrt{3}}{6}\int\frac{\sec^2\left(\theta\right)\sec\left(\theta\right)\:d\theta}{\sqrt{\tan\left(\theta\right)}}\tag{3},
\end{align}
now let $u=\tan\left(\theta\right),\:\:du=\sec^{2}\left(\theta\right)\:d\theta$, which gives us
\begin{align}
\frac{\sqrt{3}}{6}\int\frac{\sqrt{1+u^{\color{red}{2}}}\:du}{\sqrt{u}}\tag{4},
\end{align}
and where do I go from here? Or perhaps I'm just going in circles and haven't made any progress with this result.
| Let's write $\sqrt{1+9t^4}$ as
$$
\sqrt{1+9t^4} = \frac{3t^2\sqrt{1+9t^4}}{3t^2}=3t^2\sqrt{1/(9t^4)+1}
$$
Now we can use the fractional power binomial expansion.
\begin{align}
(1/(9t^4)+1)^{1/2}&= 1 + \frac{1}{18t^4}-\frac{1}{5184t^8} +\cdots\\
3t^2(1/(9t^4)+1)^{1/2}&= 3t^2 + \frac{1}{6t^2}-\frac{1}{1728t^6} +\cdots
\end{align}
Now let's integrate
\begin{align}
\int\sqrt{1+9t^4}dt &= t^3 - \frac{1}{6t} + \frac{1}{1080t^5} -\frac{1}{34992t^9} + \cdots\\
&= 3\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!t^{3-4n}}{(4n-3)(2n-1)36^n(n!)^2}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$ I cannot seem to figure out how to solve the following problem:
$$\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$$
I have tried by using integration by parts.
I set $u = \sin(t)$, $v = \tan^{-1}(x)$ and $dv = \dfrac{1}{1+t^2}dt$. Thus,
$$\int_0^x \dfrac{\sin(t)}{1+t^2}dt = \left[\sin(t)\cdot \tan^{-1}(t)\right]_0^x - \int_0^x \cos(t)\cdot \tan^{-1}(t)dt$$
Is this correct or not? I would like to get some hints, I am currently stuck.
Best regards
| $\int_0^x\dfrac{\sin(t)}{1+t^2}dt$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n+1}}{(2n+1)!(t^2+1)}dt$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^n(t^2+1-1)^n}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(t^2+1)^k}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(t^2+1)$
$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\dfrac{\sinh1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\left[\dfrac{\sinh1\ln(t^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^k}{2(2n+1)!k!k(n-k)!}\right]_0^x$
$=\dfrac{\sinh1\ln(x^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!((x^2+1)^k-1)}{2(2n+1)!k!k(n-k)!}$
| {
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"timestamp": "2023-03-29T00:00:00",
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To find the maximum and minimum value of x such that it satisfies a polynomial Find the maximum and minimum value of $x$, where:
*
*$x+y+z=4$
*$x^2+y^2+z^2 =6$
I thought I could use these values to form a equation having $x,y,z$ as roots and the sum of roots and $\sum{xy}$ but could not get any idea about the product of roots hence could not move ahead.
| Let we set $x=\frac{4}{3}+u,y=\frac{4}{3}+v,z=\frac{4}{3}+w$. The given constraints translate into:
$$\left\{\begin{array}{rcl}v+w &=& -u \\ v^2+w^2 &=& \frac{2}{3}-u^2\end{array}\right.$$
or:
$$\left\{\begin{array}{rcl}v+w &=& -u \\ v\cdot w &=& u^2-\frac{1}{3},\end{array}\right.$$
so $v,w$ are roots of the polynomial $p(t)=t^2+ut+\left(u^2-\frac{1}{3}\right)$. They are real roots provided that $u^2-4\left(u^2-\frac{1}{3}\right)\geq 0$, so the critical values for $u$ are given by $u=\pm\frac{2}{3}$ and:
$$ x\in\left[\frac{2}{3},2\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the formula for the number of golf balls in a triangular pyramid with $n$ layers.
Toni and her friends are building triangular pyramids with golf balls.
Write a formula for the number of golf balls in a pyramid with n
layers, if a $1$-layer pyramid contains 1 ball, a 2-layer pyramid contains 4
balls, a 3-layer one contains 10 balls, and so on.
What is the formula for this question, and what are the steps involved in deriving it?
| If it is a square pyramid, the length of the side of each level will increase by one each time you go down. Thus the number of balls on each level is $k^2$. Therefore the total number of balls with $n$ levels is $\sum\limits_{k=1}^n k^2$
In simplifying this it becomes the Square Pyramidal Number which is $\frac{n(n+1)(2n+1)}{6}$
If it is a tetrahedron, the length of the side of each triangle on each level will increase by one each time you go down. Thus the number of balls on each level is $T(k)$, the $k^{th}$ triangle number. Thus the total number of balls with $n$ levels is $\sum\limits_{k=1}^n T(k)$
In simplifying this, it becomes the $n^{th}$ Tetrahedral Number, $\frac{n(n+1)(n+2)}{6}$
The derivation of the square pyramidal number is outlined in the wiki article linked. To prove the derivation of the tetrahedral number, first note that $T(k) = \sum\limits_{i=1}^k i = \frac{k(k+1)}{2}$, where $T(k)$ denotes the $k^{th}$ triangle number.
So, the $n^{th}$ tetrahedral number is $\sum\limits_{k=1}^n T(k) = \sum\limits_{k=1}^n \frac{k(k+1)}{2} = \frac{1}{2}\sum\limits_{k=1}^n k^2 + k = \frac{1}{2}(P(n)+T(n))$ where $P(n)$ denotes the $n^{th}$ pyramidal number.
This then simplifies to $\frac{1}{2}(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2})$ which after some algebra simplifies to the formula given earlier.
To find the triangle number, $T(n)$, this is $1+2+3+\dots+(n-1)+n$. At the moment we do not know what the total is as we are coming up with a formula for it. Suppose that we give the total a name, $T(n)$. Then $T(n)=1+2+3+\dots+(n-1)+n$.
We try multiplying it by two to see what happens.
$2T(n)=2(1+2+3+\dots+(n-1)+n)$
$= (1+2+3+\dots+(n-1)+n) +(1+2+3+\dots+(n-1)+n)$
$=(1+2+3+\dots+(n-1)+n) +(n+(n-1)+\dots+3+2+1)$ by reversing the order of the second parenthesis
$=(1+n)+(2+(n-1))+\dots+((n-1)+2)+(n+1)$ by grouping terms together as they appear in the parenthesis.
$=(n+1)+(n+1)+\dots+(n+1) = n(n+1)$
Remembering this was the total for $2T(n)$, we divide by two to get
$T(n)=\frac{n(n+1)}{2}$
| {
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"url": "https://math.stackexchange.com/questions/1148720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number Theory : What are the last three digits of $9^{9^{9^9}}?$ I was doing some basic Number Theory problems and came across this problem and was all thumbs :
Find the last three digits of $9^{9^{9^9}}$
How would I go about solving this problem? I am a newbie.
| $(10-1)^{9^{9^9}}=\sum\limits_{i=0}^{9^{9^9}}\binom{9^{9^9}}{i}10^i(-1)^{9^{9^9}-i}$ of course we only care when $i$ is $0,1$ or $2$. So we need only calculate
$(-1)^{9^{9^9}}+\binom{9^{9^9}}{1}10^1(-1)^{9^{9^9}-1}+\binom{9^{9^9}}{2}10^2(-1)^{9^{9^9}-2}=-1+\color{red}{\binom{9^{9^9}}{1}10}-\color{green}{100\binom{9^{9^9}}{2}}$.
So we need only determine the last two digits of $9^{9^9}$ and the last digit of $\binom{9^{9^9}}{2}$ to conclude.
We first calculate the last digit of $\binom{9^{9^9}}{2}=\frac{9^{9^9}}{1}\frac{9^{9^9}-1}{2}$. The last digit of $9^{9^9}$ is $9$. So the last digit of $9^{9^9}-1$ is $8$. The last digit of $\frac{9^{9^9}-1}{2}$ is $4$ because $9^{9^9}-1$ is a multiple of $4$ since $9\equiv 1\bmod 4$.
Therefore the last digit of $\binom{9^{9^9}}{2}$ is $\color{green}6$.
What are the last two digits of $9^{9^9}$? We use the same trick as before, we want $(10-1)^{9^9}=\sum\limits_{i=0}^{9^9}\binom{9^9}{i}10^i(-1)^{9^9-i}$. But we only care about the first two terms:
$(-1)^{9^9}+\binom{9^9}{1
}10^1(-1)^{9^9-1}=10(9^9)-1$ which ends in $\color{red}{89}$.
Therefore $9^{9^{9^9}}$ is $-1+\color{red}{89(10)}-\color{green}{600}+1000k$
and hence ends in $289$.
| {
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"url": "https://math.stackexchange.com/questions/1149994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Solving $n$ unknowns with $n$ independent equations Is it always possible to solve $n$ independent equations with $n$ unknowns?
Or is it possible to solve the following 3 equations with 3 unknowns?
$$x + y + z = a$$
$$x^2 + y^2 + z^2 = b$$
$$x^3 + y^3 + z^3 = c$$
| Consider $x$ , $y$ and $z$ as the roots of the equation :
$$(X-x)(X-y)(X-z)=X^3+AX^2+BX+C=0$$
$A=-(x+y+z)=-a$
$B=xy+yz+zx=\frac{1}{2}\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{1}{2}(a^2-b)$
$C=-xyz=-\frac{1}{6}\left( (x+y+z)^3-3(x+y+z)(x^2+y^2+z^2)+2(x^3+y^3+z^3) \right)=-\frac{1}{6}(a^3-3ab+2c)$
$x$ , $y$ , $z$ are the roots of the equation :
$$X^3-aX^2+\frac{1}{2}(a^2-b)X-\frac{1}{6}(a^3-3ab+2c)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the tangent of the unit circle is given by the equation $z+a^2\overline{z}= 2a$ Let $a$ and $b$ be complex numbers on the unit circle, i.e. $|a|=|b|=1$
(a) Show that the equation of the tangent to the unit circle at $a$ is given by:
$z+a^2\overline{z}= 2a$.
(b) Use the result of part (a) to show that the intersection of the tangents to the unit circle at $a$ and $b$ is:
$\frac{2ab}{a+b}$.
My thoughts: Given that the point $a$ is on the unit circle we know that $|a|=1$. In addition, we know that from the center of the circle to $a$ will be perpendicular with the tangent. The general formula for proving that two lines are perpendicular is: $(z_3-z_1)=e^{i\theta}(z_2-z_1)$ for some complex numbers $z_1$, $z_2$, and $z_3$.
We could use this equation to find a point on the line that is tangent to the circle and then find the slope of the equation. But, I don't quite understand how this would take the form of $z+a^2\overline{z}= 2a$? Thus, I cannot continue on (a) or start on (b). Any hints/help would be much appreciated!
| (a) Since we know the tangent line and the radius from the origin to $a$ are perpendicular, we can say $$z-a = a(e^{\pi i/2}k) = aki$$ where $k$ is some integer.
Manipulating the equation we get:
\begin{align*}
z-a &= aki \\
z &= a(1+ki) \\
\overline{z} &= \overline{a}(1-ki) \\
a^2\overline{z} &= a^2\overline{a}(1-ki)
\end{align*}
Let's simplify the right side: $$a^2\overline{a}(1-ki) = a\cdot a\overline{a}(1-ki) = a\cdot |a|^2(1-ki) = a(1-ki)$$
Now we can continue:
\begin{align*}
a^2\overline{z} &= a^2\overline{a}(1-ki) \\
a^2\overline{z} &= a(1-ki) \\
a^2\overline{z} &= a-aki \\
a^2\overline{z} &= 2a - a - aki \\
a^2\overline{z} &= 2a - z \\
z+a^2\overline{z} &= 2a
\end{align*}
(b) Let's set $w$ as the intersecting point. Using out solution from part a, we can say: $$w+a^2\overline{w}-2a=0=w+b^2\overline{w}-2b$$
Let's manipulate this equation:
\begin{align*}
w+a^2\overline{w}-2a &= w+b^2\overline{w}-2b \\
a^2\overline{w}-b^2\overline{w} &= 2a-2b \\
\overline{w}(a^2-b^2) & = 2(a-b) \\
\overline{w}(a+b) &= 2 \\
\overline{w} &= \frac{2}{a+b} \\
w &= \frac{2}{\overline{a}+\overline{b}}
\end{align*}
Let's take a step back and figure $\overline{a}$ and $\overline{b}$
\begin{align*}
a\overline{a} = |a|^2 \\
a\overline{a} = 1 \\
\overline{a} = \frac{1}{a}
\end{align*}
Using same logic, we can also derive $\overline{b} = \frac{1}{b}$
Now let's substitute these values in:
\begin{align*}
w &= \frac{2}{\overline{a}+\overline{b}} \\
&= \frac{2}{\frac{1}{a}+\frac{1}{b}} \\
&= \frac{2}{\frac{a+b}{ab}} \\
&= \frac{2ab}{a+b}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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proving trigonometric identity involving 2 arguments $\cos^2\theta=\dfrac{m^2-1}{3}$, $\tan^3\dfrac{\theta}{2}=\tan\alpha$.
How to prove $$\cos^\frac{2}{3}\alpha+\sin^\frac{2}{3}\alpha=\frac{2}{m}^{2/3}. $$ I got the following
$$\tan^\frac{2}{3}\alpha=\frac{1- \sqrt\frac{m^2 - 1}{3}}{1+\sqrt\frac{m^2 - 1}{3}} $$
but i can not proceed further.
| $$\tan\alpha=\tan^3\dfrac\theta2$$
$$\iff\frac{\sin\alpha}{\sin^3\dfrac\theta2}=\frac{\cos\alpha}{\cos^3\dfrac\theta2}=\pm\sqrt{\frac{\sin^2\alpha+\cos^2\alpha}{\sin^6\dfrac\theta2+\cos^6\dfrac\theta2}}$$
$$\iff\frac{\sin\alpha}{\sin^3\dfrac\theta2}=\frac{\cos\alpha}{\cos^3\dfrac\theta2}=\pm\frac1{\sqrt{\left(\sin^2\dfrac\theta2\right)^3+\left(\cos^2\dfrac\theta2\right)^3}}\ \ \ \ (1)$$
Now $\left(\sin^2\dfrac\theta2\right)^3+\left(\cos^2\dfrac\theta2\right)^3=\left(\sin^2\dfrac\theta2+\cos^2\dfrac\theta2\right)^3-3\left(\sin^2\dfrac\theta2+\cos^2\dfrac\theta2\right)\left(\sin^2\dfrac\theta2\cos^2\dfrac\theta2\right)$
$=1-\dfrac34\sin^2\theta=\cdots $
Take $2/3$ power of each sides of $(1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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simplifying and factoring a fraction how i get $\frac{(a+b)^2+(a+c)^2+(b+c)^2}{2}$ from $\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-a)(c-b)}$ assuming that $a\ne b\ne c\ne a$
i tried to make
$$\begin{align}
&\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-a)(c-b)}\\
&=\frac{a^4}{(a-b)(a-c)}-\frac{b^4}{(a-b)(b-c)}+\frac{c^4}{(a-c)(b-c)}\\
&=\frac{a^4(b-c)}{(a-b)(a-c)(b-c)}-\frac{b^4(a-c)}{(a-b)(a-c)(b-c)}+\frac{c^4(a-b)}{(a-b)(a-c)(b-c)}\\
&=\frac{a^4(b-c)-b^4(a-c)+c^4(a-b)}{(a-b)(a-c)(b-c)}\\
&=\frac{a^4b-a^4c-ab^4+b^4c+ac^4-bc^4}{(a-b)(a-c)(b-c)}\\
&=\frac{ab(a^3-b^3)+ac(c^3-a^3)+bc(b^3-c^3)}{(a-b)(a-c)(b-c)}\\
&=\frac{(a-b)(a-c)(b-c)(a^2+ab+ac+b^2+bc+c^2)}{(a-b)(a-c)(b-c)}\\
&=a^2+ab+ac+b^2+bc+c^2\\
&=\frac{2a^2+2ab+2ac+2b^2+2bc+2c^2}{2}\\
&=\frac{a^2+2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2}{2}\\
&=\frac{(a+b)^2+(a+c)^2+(b+c)^2}{2}
\end{align}$$
$$\begin{align}
\small bc(b^3-c^3)+ac(c^3-a^3)+ab(a^3-b^3)&\small =bc(b^3-a^3+a^3-c^3)+ac(c^3-a^3)+ab(a^3-b^3)\\
&\small =bc(b^3-a^3)+bc(a^3-c^3)+ac(c^3-a^3)+ab(a^3-b^3)\\
&\small =-bc(a^3-b^3)+bc(a^3-c^3)-ac(a^3-c^3)+ab(a^3-b^3)\\
&\small =b(a-c)(a^3-b^3)+c(b-a)(a^3-c^3)\\
&\small =b(a-c)(a^3-b^3)-c(a-b)(a^3-c^3)\\
&\small =b(a-b)(a-c)(a^2+ab+b^2)-c(a-b)(a-c)(a^2+ac+c^2)\\
&\small =(a-b)(a-c)[b(a^2+ab+b^2)-c(a^2+ac+c^2)]\\
&\small =(a-b)(a-c)(a^2b+ab^2+b^3-a^2c-ac^2-c^3)\\
&\small =(a-b)(a-c)[a^2(b-c)+(b^3-c^3)+a(b^2-c^2)]\\
&\small =(a-b)(a-c)[a^2(b-c)+(b-c)(b^2+bc+c^2)+a(b-c)(b+c)]\\
&\small =(a-b)(a-c)(b-c)(a^2+ab+ac+b^2+bc+c^2)
\end{align}$$
| Hint:
To make the factor $(a-b)(b-c)(c-a)$ appear, write $$b^3-c^3=b^3-a^3+ a^3-c^3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The area of a triangle is $54\sqrt{6}$ square units. Find the lengths of the sides: $5x,6x,7x$ I realize I can use Heron's formula for this question. I did $54\sqrt{6}=\sqrt{9x\cdot 3x\cdot 4x\cdot 2x}$ but from there I must have done something wrong. Thanks for the help.
| $$54\sqrt{6}=\sqrt{9x\cdot 3x\cdot 4x\cdot 2x}=\sqrt{216x^4}=6\sqrt{6x^4}=6x^2\sqrt{6}$$
$$\iff 54=6x^2\iff x^2=9\iff x=3$$
I used the fact that $\sqrt{x^4}=|x^2|=x^2$, since $x^2\ge 0, \forall x\in\mathbb R$
and $x^2=9\implies x=\pm 3\implies x=3$, since $x$ is a positive number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inductive proof, algebra step I have to prove by induction that
$$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 = {n(n+1)(2n+1)\over 6}$$
Base step: $$1^2 = {1(1+1)(2\bullet1 +1)\over 6}$$
$$1^2= {6\over 6} = 1$$
Then I use this $$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 +(n+1)^2 = {n(n+1)(2n+1)\over 6}+(n+1)^2$$
to show that
$$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 + (n+1)^2 = {(n+1)((n+1)+1)(2(n+1)+1)\over 6}$$
$$=$$
$$1^2+2^2+3^2+ \cdot \cdot \cdot + n^2 + (n+1)^2 = {(n+1)(n+2)(2n+3)\over6}$$
I'm stuck with the algebra part, could someone go through the steps of how
$${n(n+1)(2n+1)\over 6}+(n+1)^2 = {(n+1)(n+2)(2n+3)\over 6}$$
| \begin{align}\frac{n(n+1)(2n+1)}{6}+(n+1)^2 &= \frac{n(n+1)(2n+1)}{6}+\frac{6(n+1)^2}{6}\\
&=\frac{1}{6}\left[(n+1)[n(2n+1)+6(n+1)]\right]\\
&=\frac{1}{6}(n+1)(2n^2+7n+6)\\
&=\frac{1}{6}(n+1)(2n^2+4n+3n+6)\\
&=\frac{1}{6}(n+1)(2n(n+2)+3(n+2))\\
&=\frac{1}{6}(n+1)(n+2)(2n+3)\\
&=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $2^x+3^y=z^2$ in nonnegative integers. So, we are trying to find all the solutions to $2^x + 3^y = z^2$ in nonnegative integers. Here are my insights:
First of all, $z^2$ can be either $0$ or $1$ modulo $3$. If $z^2 = 3k$, then
LHS cannot be divisible by $3$ unless $y = 0$ and $x=2k+1$, and so we have $2^x+1 = z^2$, so one of the solutions is $(3,0,3)$ (but we still need to show that this is the only solution in this case).
Now let $y > 0$, $x=2k$. In this case $2^x \equiv 1\ (\mod\ 3)$ and $z^2 \equiv 2^x (\mod 3)$. We have
$$
3^y = (z-2^k)(z+2^k),
$$
so $(z-2^k)=3^a$ and $(z+2^k)=3^b$, so $3^b-3^a = 2^{k+1}$. If $a > 0$, then
$2^{k+1}$ must be divisible by $3$, but it is not, so suppose $a = 0$.
Now we have
$$
2^{k+1} = 3^b - 1,
$$
Аnd this is where I stuck. It is easy to find two more solutions just iterating over $k$, but I do not know how to prove that there are no others. As far as I understand, we need to find some kind of an upper bound for $k$. Note: I am trying to avoid the use of Catalan's conjecture here.
Can you please help me to show that there are no other solutions to this equation and that $2^x+1=z^2$ has a unique solution?
| $2^x + 1 = z^2$, therefore $z=2m+1$ (m is non-negative integer).
$2^x + 1 = 4m^2 + 4m + 1 \; \Rightarrow \;2^x = 4m(m+1) \; \Rightarrow \; 2^{x-2} = m(m+1)$
$m(m+1)$ is the power of $2$ if and only if m=1 (because $GCD(m,m+1)=1$).
Therefore, 2^x+1 =z^2 has one solution, and it is $x=3$, $z=3$.
But there are other solutions, $x=0$, $y=1$, and $z=2$. Also $x=4$, $y=2$, and $z=5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit with both cosine and sine Evaluate $$\lim_{x \to \frac{\pi}{4}}\frac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}$$ I tried substituting and then series expansion but I don't think it is working out.
| This works just fine with a series expansion, but you have to use the expansion around $x = \pi/4$, not the usual series expansions around $x = 0$. Working out the first few terms of these Taylor series,
$$ \eqalign{
\sin x & = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} (x - \pi/4) - \frac{\sqrt{2}}{4} (x - \pi/4)^2 - \frac{\sqrt{2}}{12} (x - \pi/4)^3 + \frac{\sqrt{2}}{48} (x - \pi/4)^4 + \cdots \text,\\
\text{and }\cos x & = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} (x - \pi/4) - \frac{\sqrt{2}}{4} (x - \pi/4)^2 + \frac{\sqrt{2}}{12} (x - \pi/4)^3 + \frac{\sqrt{2}}{48} (x - \pi/4)^4 - \cdots \text.}
$$
Then the numerator becomes
$$ \sqrt{2} - \cos x - \sin x = \frac{\sqrt{2}}2 (x - \pi/4)^2 - \frac{\sqrt{2}}{24} (x - \pi/4)^4 + \cdots \text, $$
and the denominator becomes
$$ (4x - \pi)^2 = 16(x - \pi/4)^2 \text; $$
now you can drop insignificant terms, cancel factors, and get your answer.
I know that your tag said to avoid L'Hôpital's Rule, but if you notice that this rule works after 2 steps, then series expansion should also work with the most significant terms having degree 2. Just be sure to expand around your limit point.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to take this exponentials Given an expansion of a cumulant function as follows:
$$
\kappa(t) = \frac{t^2}{2} + \frac{\rho_3 t^3}{6\sqrt{n}} + \frac{\rho_4t^4}{24n} +O\left(\frac{1}{n\sqrt{n}}\right), (*)
$$
where $\rho_r=\frac{\kappa_r}{\sigma^r}$ is the standardized cumulant. To obtain the corresponding moment generating function, one only needs to take the exponential of the above formula.
$$
m(t) = e^{\kappa(t)} = \exp\left\{{\frac{t^2}{2} + \frac{\rho_3 t^3}{6\sqrt{n}} + \frac{\rho_4t^4}{24n} +O\left(\frac{1}{n\sqrt{n}}\right)}\right\} = \exp\left\{ \frac{t^2}{2}\left[ 1+\frac{\rho_3t}{3\sqrt{n}}+\frac{\rho_4t^2}{12n} + O\left(\frac{1}{n\sqrt{n}}\right) \right] \right\}.
$$
However, I read the following result:
$$
m(t) = \exp\left(\frac{t^2}{2}\right) \left\{ 1+\frac{\rho_3 t^3}{6\sqrt{n}}+\frac{\rho_4t^4}{24n}+\frac{\rho_3^2t^6}{72n}+O\left(\frac{1}{n\sqrt{n}}\right) \right\}. (**)
$$
I do not see how to get $(**)$ from $(*)$. What trick is used here, please? Thank you!
| You wrote $$m(t) = \exp\left\{ \frac{t^2}{2}\left[ 1+\frac{\rho_3t}{3\sqrt{n}}+\frac{\rho_4t^2}{12n} + O\left(\frac{1}{n\sqrt{n}}\right) \right] \right\}$$ Let us expand it first; so $$m(t)=\exp \Big(\frac{t^2}{2}+\frac{\rho_3t^3}{6\sqrt n}+\frac{\rho_4t^4}{24 n}+\cdots \Big)=\exp \Big(\frac{t^2}{2}\Big)\times\exp \Big(\frac{\rho_3t^3}{6\sqrt n}+\frac{\rho_4t^4}{24 n}+\cdots \Big)$$ Now us Taylor series for the second part $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ and make $$y=\frac{\rho_3t^3}{6\sqrt n}+\frac{\rho_4t^4}{24 n}$$
I am sure that you can take from here.
| {
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"url": "https://math.stackexchange.com/questions/1165709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$ Problem
Assume that $a,b\in\mathbb{R}-\{0\}$ and that $a+b\not=0$. Prove that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.
My Proof
Let's assume that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, then it follows that
$$
\begin{equation}
(a+b)^2-ab=0
\end{equation}
$$
Let $x=a+b$ and $y=ab$. Now $b=x-a$ and so $y=a(x-a)=ax-a^2$. The previous equation can be written as
$$
x^2-y=0
$$
Substituting $y=ax-a^2$ in this equation gives
$$
x^2-ax+a^2=0
$$
The discriminant of this quadratic equation (in $x$) is $-3a^2<0$ and therefore $x=a+b$ has no real solution. This means $a+b\in\mathbb{C}$ and therefore either of $a$ or $b$ or both are not real but this contradicts our assumption that $a,b$ are real numbers. Therefore by contradiction $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.
My Question
Is my proof correct? are there any alternative proofs?
| Assume w.n.l.g. that $a>0$ and $b \in (-a,a]$. Now, if $b>0$,
$$
\frac{1}{a} + \frac{1}{b} > \frac{1}{a} > \frac{1}{a+b}.
$$
On the other hand, if $b<0$,
$$
\frac{1}{a}+\frac{1}{b} \le 0 < \frac{1}{a+b}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Olympiad inequality (Cauchy/AM-GM sort) Given $n$ positive numbers $x_1,\ldots,x_n$ ($n\ge 3$) such that the product $x_1x_2\cdots x_n=1$, show that
$$\dfrac{x_1^8}{(x_1^4+x_2^4)x_2}+\frac{x_2^8}{(x_2^4+x_3^4)x_3}+\cdots+\frac{x_n^8}{(x_n^4+x_1^4)x_1}\ge \frac{n}{2}$$
I am having trouble demonstrating this for even $n=3$, but it seems that once that case is established the method for the general case should be identical. I am sure there is some nice application of AM-GM or Cauchy that will solve it. Notice that in each term we have the numerator of degree $8$, and in the denominator there is one factor of degree $4$ and also a linear factor which is quite an annoyance. I have tried to "homogenize" the linear factor so that it is of degree 4 (thus each fraction of degree 0) by doing some substitutions, but I have been unsuccessful with this.
If anyone can provide a hint (e.g suggest a substitution that may be helpful), I would greatly appreciate it. This problem has been bothering me for a while now.
| You can prove it using the inequalities AM-GM, GM-HM (geometric mean and harmonic mean) and the inequality of ordered sequences:
$$
\sum_{cyc} \frac{x_1^8}{\left(x_1^4+x_2^4\right)x_2}=\sum_{cyc} \frac{x_1^4}{x_2}\left(\frac{x_1^4}{x_1^4+x_2^4}\right)=\sum_{cyc} \frac{x_1^4}{x_2}\left(1-\frac{x_2^4}{x_1^4+x_2^4}\right)=\sum_{cyc} \frac{x_1^4}{x_2}-\frac{1}{x_2}\frac{x_1^4x_2^4}{\left(x_1^4+x_2^4\right)}=\sum_{cyc} \frac{x_1^4}{x_2}-\frac{1}{2x_2}\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}=\sum_{cyc} \frac{x_1^4}{x_2}-\sum_{cyc} \frac{1}{2x_2}\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}
$$
Edit: Sorry, I haven't read that you just want a hint, so stop here if you want to do it yourself!
Due to GM-HM we have: $\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}\le x_1^2x_2^2$ and since $x_1^4, x_2^4,\dots, x_n^4$ and $\frac{1}{x_1},\dots,\frac{1}{x_n}$ are oppositely ordered, we have: $\sum_{cyc} \frac{x_1^4}{x_2}\ge\sum_{cyc} \frac{x_1^4}{x_1}=\sum_{cyc} x_1^3$. Therefore:
$$
\sum_{cyc} \frac{x_1^4}{x_2}-\sum_{cyc} \frac{1}{2x_2}\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}\ge\sum_{cyc} x_1^3-\sum_{cyc} \frac{1}{2x_2}x_1^2x_2^2=\sum_{cyc} x_1^3-\sum_{cyc} \frac{x_1^2x_2}{2}
$$
So it is left to prove that:
$$
\sum_{cyc} x_1^3-\sum_{cyc} \frac{x_1^2x_2}{2}\ge \frac{n}{2}\iff 2\sum_{cyc} x_1^3\ge n+\sum_{cyc} x_1^2x_2
$$
But due to AM-GM we have $\sum_{cyc} x_1^3\ge n\cdot\left(x_1^3x_2^3\cdots x_n^3\right)^{\frac{1}{n}}=n$ and $\sum_{cyc} x_1^3=\sum_{cyc} \frac{2x_1^3+x_2^3}{3}\ge\sum_{cyc} x_1^2x_2$ and adding these two yields the desired result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
| Divide $x^3-y^3$ by $x-y$ as polynomials
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I tried to compute $\int_0^1x^2\sin^2xd x$ and mathematica disagrees with me I tried to compute $\int_0^1x^2\sin^2xd x$ and got $\frac 16-\frac 38\sin2-\frac 14\cos2$. However, Wolfram Mathematica gives the result $\frac 16-\frac 18\sin2-\frac 14\cos2$, namely there is something wrong with the coefficient of $\sin$ in my answer. And here is how I computed it(If you don't have the patience to read it through, please help me do the integral and see if we get the same answer):
$$\int_0^1x^2\sin^2xd x$$
$$=\int^1_0x^2\frac {1-\cos 2x}{2}dx$$
$$=\int^1_0\frac {x^2}2 dx+\int^1_0\frac{-x^2\cos 2x}2dx$$
$$=\frac 12*\frac 13[x^3]^1_{x=1_00}+\int^1_0\frac{-x^2\cos 2x}{2}\frac{d(\sin2x)}{2\cos 2x}$$
$$=\frac 16+\int^1_0\frac{-x^2}{4}d(\sin 2x)$$
$$=\frac 16+((\frac {-x^2}{4}\sin 2x)^1_0-\int^1_0\sin 2xd(\frac {-x^2}{4}))$$
$$=\frac 16+(-\frac 14\sin 2+\frac 14\int^1_0\sin 2x*2xdx)$$
$$\frac 16-\frac 14\sin 2+\frac 12\int^1_0x\sin 2xdx$$
But $$\int^1_0x\sin2xdx=\int^1_0x\sin2x\frac{d\cos2x}{-2\sin 2x}=-\frac 12\int^1_0xd(\cos 2x)=-\frac 12((x\cos 2x)^1_0+\int^1_0-\cos 2x\frac {d(2x)}{2})=-\frac 12(\cos 2+\frac 12(\sin 2x)^1_0)=-\frac12(\cos 2+\frac 12\sin 2)$$, So $$\int^1_0x^2\sin^2xdx=\frac 16-\frac 14\sin 2+\frac 12(-\frac 12)(\cos 2+\frac 12\sin 2)=\frac 16-\frac 14 \sin 2-14\cos 2-\frac 18\sin2=\frac 16-\frac 38\sin 2-\frac 14\cos 2$$
| $$\int^1_0x\sin2xdx=\int^1_0x\sin2x\frac{d\cos2x}{-2\sin 2x}=-\frac 12\int^1_0xd(\cos 2x)=-\frac 12((x\cos 2x)^1_0\color{#C00}{+}\int^1_0-\cos 2x\frac {d(2x)}{2})=-\frac 12(\cos 2\color{#C00}{+}\frac 12(\sin 2x)^1_0)=-\frac12(\cos 2\color{#C00}{+}\frac 12\sin 2)$$
If I'm not mistaken, this should be a minus sign, which gives Mathematica's (correct) result.
EDIT:
$$\int^1_0x\sin2xdx = \left[-\frac12x\cos2x\right]^1_0 - \int^1_0-\frac12\cos2xdx = -\frac12\cos2 - \left[-\frac14\sin2x\right]^1_0 = -\frac12\cos2 + \frac14\sin2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sum_{n=1}^\infty \frac{n}{2^n}$.
Evaluate $$\sum_{n=1}^\infty \frac{n}{2^n}$$
My Work:
$$\sum_{n=1}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty n \left(\frac{1}{2}\right)^n$$
If we denote $f(x) = \sum_{n=1}^\infty nx^n$ then we wish to evaluate $f(1/2)$.
Now, $$\sum_{n=1}^\infty nx^n = x \sum_{n=1}^\infty nx^{n-1} = x\sum_{n=1}^\infty (x^n)' = x\left(\sum_{n=1}^\infty x^n\right)' = x \left(\frac{-x}{1-x}\right)' = \frac{-x}{(1-x)^2}$$
Applying $x=1/2$ we get the wrong result of $-2$.
Where is my mistake?
| Yet another solution (maybe less tricky).
You can express this as a double summation and exchange the order (this is directly related to the integration by parts):
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n} &= \sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{2^n}\\
& = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{2^n} 1_{k \leq n}\\
& = \sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{2^n} 1_{k \leq n}\\
& = \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{2^n}\\
& = \sum_{k=1}^\infty \frac{1}{2^k}\frac{1}{1-\frac{1}{2}}\\
& = 2\sum_{k=1}^\infty \frac{1}{2^k}\\
& = 2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the next term in the sequence. $\frac{7}{3},\frac{35}{6},\frac{121}{12},\frac{335}{36},\ldots $ $\dfrac{7}{3},\dfrac{35}{6},\dfrac{121}{12},\dfrac{335}{36},\ldots $
$\bf\text{Answer}$ given is $\dfrac{865}{48}$
I found that $4^{th}$ differencess of the numbers $7,35,121,335\cdots$ are not constant .
and the second differences of the denominator drastically changes,
$3\quad 6\quad 12\quad 36\quad 48\\~\\
\quad 3\quad 6\quad 24\quad \color{red}{12}$
decimal value is also not showing any pattern.
$\frac{7}{3},\ \frac{35}{6},\ \frac{121}{12},\ \frac{335}{36},\ldots $
$2.33,\ 5.83,\ 10.08,\ 9.33,\ldots $
| Okay so after solving a lot this is how the answer goes..
The series is 2+1/3 , 6-1/6 , 10+1/12 , 14-1/24 , 18+1/48..
Both the integral part and fractional part follow a specific sequence.
Hope This Solved Your Problem..
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
Proving $\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$ Proving $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$$
| $$\sum\frac{x^n}{n(n+1)(n+2)(n+3)}=\sum\frac{x^n}6\left(\frac1n-\frac3{n+1}+\frac3{n+2}-\frac1{n+3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $x^4+3x+20=0$ by Ferrari's method Comparing the equation
$$x^4+3x+20=0$$
With the equation
$$(x^2+\lambda)^2-(mx+n)^2=0$$
we get
$m^2=2\lambda,$
$-2mn=3,$
$n^2=\lambda^2-20$
Now, $4m^2n^2=9\Rightarrow 4(2\lambda)(\lambda^2-20)=9\Rightarrow 8\lambda^3-160\lambda-9=0$.
How can I find easily the values of $\lambda$ from the above equation.
Please suggest me.
| HINT: Use binom expansion
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
Reorder the equation as:
$$(x+y)^3-3xy(x+y)-(x^3+y^3)=0$$
$$8\lambda^3-160\lambda-9=0$$
$$\lambda^3-20\lambda-\frac{9}{8}=0$$
Define:
$x+y=\lambda$,
then solve the equation with 2 unknowns
$$3xy=20$$
$$x^3+y^3=\frac{9}{8}$$
Let me know if you cannot get forward from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the indefinite integral $\int \sqrt{1-2\sqrt{x-x^2}} \ \mathrm dx$? What is
$$\int \sqrt{1-2\sqrt{x-x^2}} \ \mathrm dx$$
I have tried substituting everything and it doesn't seem to be working. Substituting trigonometry doesn't seem to work either.
| Since, once more, Jack D'Aurizio is much faster than myself, let us see what can be done with Taylor built at $x=0$ $$\sqrt{1-2\sqrt{x-x^2}}=1-\sqrt{x}-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{5 x^4}{128}-\frac{7
x^5}{256}+O\left(x^{11/2}\right)$$ So, integrating $$\int_{0}^{a}\sqrt{1-2\sqrt{x-x^2}}\,dx =a-\frac{2
a^{3/2}}{3}-\frac{a^2}{4}-\frac{a^3}{24}-\frac{a^4}{64}-\frac{a^5}{128}-\frac{7
a^6}{1536}+O\left(a^{13/2}\right)$$ For $a=\frac 12$, this gives $\frac{14123}{32768}-\frac{1}{3 \sqrt{2}}\approx 0.195297$ which compares quite well to Jack D'Aurizio's value $(\approx 0.195262)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
integration of $\frac{x}{(x^2-1)^\frac{1}{2}}$ When I use the substitution $u=\cosh x$ to integrate $\frac{x}{(x^2-1)^\frac{1}{2}}$, I get $\frac{1}{2}(x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}})$ but when I check online the answer is $\sqrt{x^2-1}$ which is much tidier. I'm probably being thick, but how can my answer be simplified to $\sqrt{x^2-1}$? and if it can't, is there a reason why I got a different answer to the integral?
| Your answer is correct:
\begin{align}
\frac{1}{2}(x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}})&=\frac{1}{2}(x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}}\times \frac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}})\\
&=\frac{1}{2}(x+\sqrt{x^2-1}-\frac{x-\sqrt{x^2-1}}{x^2-(x^2-1)})\\
&=\frac{1}{2}(x+\sqrt{x^2-1}-x+\sqrt{x^2-1})\\
&=\sqrt{x^2-1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction that $\frac1{n+1}+ \frac1{n+2}+\cdots+\frac1{2n}=1-\frac{1}{2}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}.$ Prove that for any positive integer, $$\frac1{n+1}+ \frac1{n+2}+\cdots+\frac1{2n} = \left(1-\frac1{2}\right)+\left(\frac1{3}-\frac1{4}\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n}\right).$$
I have tried using a proof by induction but do not know how to approach the series.
Note: This identity may be established by a clever algebraic manipulation, as seen here, but I am curious as to how an inductive proof might work.
| For each $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}.
$$
Note that the left-hand side constitutes the first $2n-1$ terms of what is called the alternating harmonic series.
Base step ($n=1$): Notice that the left side of $S(n)$ has denominators which range from $1$ to $2n$, whereas the denominators on the right range from $n+1$ to $2n$. Thus, for $n=1$, the denominators on the left range from $1$ to $2$, whereas on the right, they range from $1+1=2$ to $2\cdot 1=2$; that is, there is only one term on the right. Consequently, $S(1)$ say that $1-\frac{1}{2}=\frac{1}{2}$, and this is true.
Inductive step: For some fixed $k\geq 1$, assume the inductive hypothesis $S(k)$
$$
S(k) : 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2k-1}-\frac{1}{2k}=\frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2k}
$$
to be true. We must then show that $S(k+1)$ follows:
$$
S(k+1) : \underbrace{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2k+1}-\frac{1}{2k+2}}_{\text{LHS}}=\underbrace{\frac{1}{k+2}+\frac{1}{k+3}+\cdots+\frac{1}{2k+2}}_{\text{RHS}}.
$$
Starting with the left-hand side of $S(k+1)$ (and filling two more penultimate terms),
\begin{align}
\text{LHS} &= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2}\\[1em]
&= \frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2}\tag{by $S(k)$}\\[1em]
&= \frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{k+1}-\frac{1}{2k+2}\\[1em]
&= \frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{2}{2k+2}-\frac{1}{2k+2}\\[1em]
&= \frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}\\[1em]
&= \text{RHS},
\end{align}
we see that the right-hand side of $S(k+1)$ follows. This completes the inductive step $S(k)\to S(k+1)$.
Hence, by mathematical induction, for all $n\geq 1, S(n)$ is true. $\blacksquare$
| {
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"url": "https://math.stackexchange.com/questions/1181262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integration by partial fractions I've attempted to integrate the function $$\frac{2x^3 + 5x^2 + 8x + 4}{(x^2 + 2x + 2)^2}$$ using several techniques, but none of them are solving it nicely.
I've tried to slve it by partial fractions:
$$\frac{A + Bx}{x^2 + 2x + 2} + \frac{C + Dx}{(x^2 + 2x + 2)^2},$$ but i get wrong coefficients.
| Let us start with $$\frac{2x^3 + 5x^2 + 8x + 4}{(x^2 + 2x + 2)^2}=\frac{A + Bx}{x^2 + 2x + 2} + \frac{C + Dx}{(x^2 + 2x + 2)^2}$$ Remove the denominator, expand and group the terms. You should arrive to $$2x^3 + 5x^2 + 8x + 4=B x^3+ (A+2 B)x^2+(2 A+2 B+D)x+(2 A+C)$$ which gives the four equations $B=2$,$A+2B=5$ so $A=1$, $2A+2B+D=8$ so $D=2$, $2A+C=4$ so $C=2$.
So we have $$\frac{2x^3 + 5x^2 + 8x + 4}{(x^2 + 2x + 2)^2}=\frac{1 + 2x}{x^2 + 2x + 2} + \frac{2 + 2x}{(x^2 + 2x + 2)^2}=\frac{2x+2-1}{x^2 + 2x + 2} + \frac{2x + 2}{(x^2 + 2x + 2)^2}$$ So, setting $u=x^2+2x+2$, you can notice that $2x+2$ is just $u'$ which make things very simple for two parts and what is basically left is $$\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x+1)^2+1}$$
I am sure that you can take from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve recurrence relation using generating function I'm trying to solve: $a_{n+1}-a_n=n^2$, $n\le0$ , $a_0=1$ using generating functions.
Step 1) Multiply by $x^{n+1}$
$$a_{n+1}x^{n+1}-a_nx^{n+1}=n^2x^{n+1}$$
Step 2) Take the infinite sums
$$\sum_{n\ge0}^{\infty}a_{n+1}x^{n+1}-\sum_{n\ge0}^{\infty}a_nx^{n+1}=\sum_{n\ge0}^{\infty}n^2x^{n+1}$$
Our prof gave us the identity: $$\sum_{n\ge0}^{\infty}n^2x^n= \frac{x+x^2}{{1-x}^3}$$
So I factored out an $x^1$ from my RHS(to use the identity) and simplified the LHS to get:
$$(f(x)-a_0)-xf(x)=x\left(\frac{x+x^2}{{1-x}^3}\right)$$
therefore: $$f(x)=x\left(\frac{x+x^2}{{(1-x)}^4}\right)+\frac{1}{(1-x)}=\frac{x(x+x^2)+(1-x)^3}{(1-x)^4}$$
Step 3) Decompose the function by partial fractions
Saving you all the gory details I got:
$$f(x)=\frac{4}{(1-x)^2}-\frac{5}{(1-x)^3}+\frac{2}{(1-x)^4}$$
Step 4) Finding the coefficient of $x^n$ in each term:
I recognized that each term was a derivative of the the power series $\frac{1}{(1-x)}$ to get:
$$4[x^n]\sum(n+1)(x^n)-5\frac12[x^n]\sum(n+2)(n+1)(x^n)+2\frac13[x^n]\sum(n+3)(n+2)(n+1)(x^n)$$
So that whole thing = $a_n$ which equals:
$$4(n+1)-5\frac12(n+2)(n+1)+2\frac13(n+3)(n+2)(n+1)$$
However the answer is: $a_n=1+[n(n-1)(2n-1)\frac16]$
I tried multiplying out to see if they were the same but they're not(I checked on wolfram too).
Could anyone tell me where I went wrong?
Thank you.
| $\begin{array}\\
A(x)
&=\sum_{n=0}^{\infty} a_n x^n\\
&=a_0+\sum_{n=1}^{\infty} a_n x^n\\
&=a_0+\sum_{n=0}^{\infty} a_{n+1} x^{n+1}\\
&=a_0+\sum_{n=0}^{\infty} (a_{n}+n^2) x^{n+1}\\
&=a_0+x\sum_{n=0}^{\infty} (a_{n}+n^2) x^{n}\\
&=a_0+x\left(\sum_{n=0}^{\infty} a_{n}x^{n}+\sum_{n=0}^{\infty} n^2 x^{n}\right)\\
&=a_0+x\left(A(x)+ \frac{x+x^2}{{1-x}^3}\right)\\
\end{array}
$
Now solve for
$A(x)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
What is the number of full binary trees of height less than $h$ Given a integer $h$
What is $N(h)$ the number of full binary trees of height less than $h$?
For example $N(0)=1,N(1)=2,N(2)=5, N(3)=21$(As pointed by TravisJ in his partial answer) I can't find any expression of $N(h)$ neither a reasonable upper bound.
Edit In a full binary tree (sometimes called proper binary tree) every node other than the leaves has two children.
| You can define $T(h,b)$ as the number of trees of height $h$ with $b$ leaves at the bottom. Your $N(h)$ is then the sum over $b$ of $T(h,b)$ To get a tree of height $h+1$ and $b$ leaves on the bottom, you pick a tree of height $h$ and at least $b/2$ leaves on the bottom, then pick $b/2$ leaves to hang new leaves from. This gives a recurrence $$T(h+1,b)=\sum_{k=b/2}^{2^h}T(h,k){k \choose b/2}$$
We have $T(3,2)=8, T(3,4)=8, T(3,6)=4, T(3,8)=1$, giving the total of $21$ cited by TravisJ. Then $T(4,2)=8 \cdot 2 + 8 \cdot 4 + 4 \cdot 6 + 1 \cdot 8=80, T(4,4)=8\cdot 1+8\cdot 6 + 4 \cdot 15+1\cdot 28=144, T(4,6)=8\cdot 4+4 \cdot 20+1\cdot 56=168,T(4,8)=8\cdot 1+4\cdot 15+1\cdot 70=138,T(4,10)=4\cdot 6+1\cdot 56=80,T(4,12)=4\cdot 1+1\cdot 28=32,T(4,14)=8,T(4,16)=1$ for a total of $N(4)=651$ The sequence is given in OEIS A001699 where it is said to approach $1.5028368...^{(2^n)}$ and some references are given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Limit of a function: $\lim_{x\to \infty} \frac x{\sqrt{1+x^2}}$ I'm having trouble finding $\lim_{x\to \infty} {x\over \sqrt{1+x^2}}$ by using the formal analysis proof i.e. $\forall \epsilon>0$ $\exists N: \left|f(x)-L\right|<\epsilon$ $\forall x>N$. I know the answer is 1 but I can't manage to prove it sufficiently. The main problem is rearranging $\left|f(x)-1\right|$ to get a nice equation of $x$ in terms of $\epsilon$. Some help would be greatly appreciated.
So far I have got to $\left|\frac{x}{\sqrt{1+x^2}}-1\right|<\epsilon$ and now I need $x$ to be in terms of $\epsilon$ in order to find a suitable $N$ for $x$ to be greater than.
| \begin{split}
&\left|\frac{x}{\sqrt{1+x^2}}-1\right|\\
=&\left|\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}\right|\\
=&\left|\frac{(x-\sqrt{1+x^2})(x+\sqrt{1+x^2})}{\sqrt{1+x^2}(x+\sqrt{1+x^2})}\right|\\
=&\left|\frac{1}{1+x^2+x\sqrt{1+x^2}}\right|\\
<&\frac{1}{x^2}
\end{split}
Then we can say that $$\forall \varepsilon>0, \exists N>\sqrt{\frac{1}{\varepsilon}}>0, \forall x>N, |f(x)-1|<\frac{1}{x^2}<\frac{1}{N^2}<\varepsilon$$
Hope this can help you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this right? Difference between evaluating and expressing to cartesian form $z= 1 + \sqrt{3} i $ Am i going right with this?
$$z= 1 + \sqrt{3} i $$
i need to i) evaluate $z^9$ and ii) express in cartesian for $z^5$... Which i'm a bit confused with.
First what i did was find the polar form...
So height angle with opposite side $\sqrt{3}$ and adjacent 1 $$\theta = \frac{\pi}{3}$$
$$r=\sqrt{1^2+\sqrt{3}^2}$$ $$r=\sqrt{1+3}$$ $$r=2$$
Hence polar form is.... $$z=2e^{\frac{\pi}{3}i}$$
Then i did part i. Using de moivre's theorem..... $$z^9=\left(2e^{\frac{\pi}{3}i}\right)^{9}$$ $$z^9=512e^{3\pi i}$$
Now here's where i get confused... Does evaluate mean... i go beyond cartesian form?
$$z^9=512e^{3\pi i}$$ $$z^9=512(\cos{3\pi}+i\sin{3\pi})$$
Which equates to....
$$z^9=512(-1+i0)$$ $$z^9=-512$$
for part ii)
Again using de moivre's theorem $$z^5=\left(2e^{\frac{\pi}{3}i}\right)^{5}$$ $$z^5=32e^{\frac{5\pi}{3} i}$$
$$z^5=32\left(\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}}\right)$$ $$z=32\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$$
| All in all, nicely done! Looks like you have a firm understanding of exponentation, polar and cartesian coordinates. The only thing I can point out is the minor algebra mistake right at the end. Notice that $\cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}$ and $\sin\left(\frac{5\pi}{3}\right) =-\frac{\sqrt{3}}{2}.$ Hence for $z^5$ we have $$32\left(\cos\left(\frac{5\pi}{3}\right)+i\sin\left(\frac{5\pi}{3}\right)\right) = 32\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right) \\ = 16\left(1-i\sqrt{3}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Quadratic equation involving floor function. If equations $x^2-3x+4=0$ and $ 4x^2-2\lfloor3a+b\rfloor x+b=0\space (a,b\space\epsilon\space R) $ have a common root then the complete set of values of $a$ is ?
I have not yet been able to develop an approach to this problem. How do I go about it?
Edit :
I was able to solve and obtain the common root as $ x= \frac{b-16}{2\lfloor3a+b\rfloor -12} $.
After this, how do I obtain the range of $a$?
| Begin first by noting that the roots of $x^2-3x+4$ are complex conjugates, $\alpha_1$ and $\alpha_2$, and that the coefficients in $4x^2-2\lfloor 3a+b\rfloor x + b$ are real. If $\alpha_i$ is a nonreal root of a quadratic with real coefficients, its complex conjugate is also a root.
From this, it follows that $g(x)=4x^2-2\lfloor 3a+b\rfloor x + b$ is some multiple of $f(x)=x^2-3x+4$. Looking at the $x^2$ term, we find then that $4f(x)=g(x)$. Equating the coefficients together, we have then $b=4\cdot 4 = 16$ and $-2\lfloor 3a+b\rfloor = -3\cdot 4$ from which it follows after some arithmetic that $\lfloor 3a\rfloor = -10$.
Thus, the possible pairs of points $(a,b)$ are $\{(a,b)~:~\frac{-10}{3}\leq a < -3, b=16\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$? I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$
I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.
| \begin{align}
\int \dfrac{1}{x^2 \sqrt{x-1}}dx=&\int \frac{1}{\sqrt{x-1}}d\left(\frac{x-1}x\right)\\
= &\ \frac{\sqrt{x-1}}x +\int \frac{ d(\sqrt{x-1})}{x}
= \frac{\sqrt{x-1}}x +\tan^{-1} \sqrt{x-1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Problem in Differential Equation. Find particular solution when x=1 and y=1
MY attempts,
$$xy^2\frac{dy}{dx}-x^3-y^3=0$$
Solving as homogeneous equation,
$$-x^3-y^3+xy^2\frac{dy}{dx}=0$$
Let $$y=xv$$
$$\frac{dy}{dx}=v+x\frac{dy}{dx}$$
$$-x^3-x^3v^3+x^3(x\frac{dv}{dx}+v)v^2=0$$
$$x^3(x\frac{dv}{dx}v^2-1)=0$$
$$\frac{dv}{dx}=\frac{1}{xv^2}$$
$$\frac{dv}{dx}v^2=\frac{1}{x}$$
$$\int \frac{dv}{dx}v^2dx=\int \frac{1}{x}dx$$
$$\frac{v^3}{3}=\ln (x)+c_1$$
$$v=\sqrt[3]{3}\sqrt[3]{ln (x)+c_1}$$
$$y=xv$$
$$y=x\sqrt[3]{3\ln (x)+c_1}$$
When y=1, x=1
$c_1=1$$
Therefore, $$y=x\sqrt[3]{3\ln (x)+1}$$
But the given answer is $v^3=\ln (ex^3)$. I wonder why. Can anybody explain?
| $$\begin{array}{l}
x{y^2}\frac{{dy}}{{dx}} - {x^3} - {y^3} = 0 \Rightarrow \frac{{dy}}{{dx}} - \frac{{{x^2}}}{{{y^2}}} - \frac{y}{x} = 0\\
u = \frac{y}{x} \Rightarrow xu = y \Rightarrow udx + xdu = dy\\
\frac{{xdu}}{{dx}} - \frac{1}{{{u^2}}} = 0 \Rightarrow {u^2}du = \frac{{dx}}{x}\\
\frac{{{u^3}}}{3} = \ln (x) + \ln ({{c^{ - 3}}}) \Rightarrow {u^3} = 3\ln ({{c^{ - 3}}}x) = \ln (c{x^3})\\
{u^3} = \ln (c{x^3})\\
{\left( {\frac{y}{x}} \right)^3} = \ln (c{x^3})\\
x = 1 \wedge y = 1 \Rightarrow \ln (c) = 1 \Rightarrow c = e\\
{\left( {\frac{y}{x}} \right)^3} = \ln (e{x^3})
\end{array}$$
They didn't solve it for $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Related Trigonometric Rates A port and a radar station are $3 \text{ mi}$ apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of $13 \frac{\text{mi}}{\text{hr}}$. Find the rate of change of the tracking angle $\theta$ between the shore and the line between the radar station and the ship at $12:30 \text{ PM}$, assuming the ship maintains its speed and course.
The following graphic represents this scenario:
$\hspace{1.95in}$
Let $\dfrac{ds}{dt} = 13 \frac{\text{mi}}{\text{hr}}$.
The ship is given to move from $A \to C$ at $\dfrac{ds}{dt}$ departing at noon and arriving at $12:30 \text{ PM}$ such that:
$$s = \dfrac{13}{2}$$
Let the segment $\overline{CB} = x$.
By the law of cosines:
$$x = \sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}$$
Then apply the law of sines to find that:
$$\dfrac{\sin{\left( \frac{\pi}{4} \right)}}{x} = \dfrac{\sin{\theta}}{s}$$
Through substitution and manipulation of the equality its found that:
$$\dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} = \sin{\theta}$$
Take the derivative with respect to time of both sides to find that:
$$
\\ \begin{align*}
\\ \frac{d}{dt}\left[ \arcsin \left( \dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} \right) \right] &= \frac{d}{dt}\left[ \theta \right]
\\
\\ \dfrac{\dfrac{1}{\sqrt{2}\sqrt{s^2-3\sqrt{2}s+9}} - \dfrac{s^2}{\sqrt{2}{(s^2-3\sqrt{2}s+9)}^{3/2}}}{\sqrt{1-\dfrac{s^2}{2(s^2-3\sqrt{2}s+9)}}}\cdot\frac{ds}{dt} &= \frac{d\theta}{dt}
\\ \end{align*}
$$
Which Wolfram evaluates to be $-4.5193 \frac{\text{rad}}{\text{hr}}$ as seen here.
This doesn't appear to be correct, however.
I'm uncertain of the error I've made, so if you could explicitly state at what point I've made an error in my approach, that would be very helpful. All that I'm certain of is that this is not the expected answer, but my approach appears proper to me (though it is clearly not).
| hint: Let $u = BC \Rightarrow \cos \theta = \dfrac{3^3+u^2-s^2}{2\cdot 3\cdot u}$, then using quotient rule to find $\theta '$, and also first find $s',u', u, s, \sin \theta$ at $12:30$ pm. To find $u'$ you need to use the law of cosine again using $u =\sqrt{3^2+s^2-2\cdot 3\cdot s\cdot \cos (45^{\circ})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$ I have to simplify and evaluate this :
$$\cot 70^\circ+4\cos 70^\circ$$
On evaluating it, the answer comes out to be $1.732$, or $\sqrt 3$ .
I tried to get everything in $\sin$ and $\cos$, but it doesn't go any further. Any hints?
| First $70=90-20$
We can express all in terms of $\cos(20)$ and use that $\frac{1}{2}=\cos(60)=4\cos^3(20)-3\cos(20)$.
Let's write $x=\cos(20), y=\sin(20)$ to write less.
So, $4x^3-3x-\frac{1}{2}=0$.
We square your expression such that we don't have to write radicals, but we can go without it too if we wanted.
$$\begin{align}
\left(\cot(70)+4\cos(70)\right)^2&=\left(\frac{\cos(90-20)+4\cos(90-20)\sin(90-20)}{\sin(90-20)}\right)^2\\
&=\left(\frac{y+4xy}{x}\right)^2\\
&=y^2\frac{16x^2+8x+1}{x^2}\\&=??\end{align}$$
But
$$\frac{1}{x}=8x^2-6.$$
$$\begin{align}??&=(1-x^2)(16x^2+8x+1)(8x^2-6)^2\\&=-1024 x^8-512 x^7+2496 x^6+1280 x^5-1952 x^4-1056 x^3+444 x^2+288 x+36\end{align}$$
Now divide this polynomial by $4x^3-3x-1/2$, which is zero.
$$??=(-256 x^5-128 x^4+432 x^3+192 x^2-180 x-66)\cdot(4 x^3-3 x-1/2) + 3$$
The remainder gives you the value $3$.
We could have worked too with $3\cdot(70)=270-60$ to get smaller numbers in the coefficients, but well, I already wrote it with $20$. You can try this technique with $\cos(70)$ directly to check that you understood how it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Series Solution of Nonlinear ODE I am trying to solve for the series solution of the following nonlinear ODE: $$y'+y^2=0$$ I am stuck in the following step: $$\sum_{n=1}^\infty a_nnx^{n-1}+\sum_{n=0}^\infty \left ( \sum_{k=0}^n a_ka_{n-k} \right )x^n=0$$ Is it possible to solve the DE this way?
| It might be more efficient to attack the problem with force if the general pattern is not readily seen.
Let
\begin{align}
y(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n}
\end{align}
for which the equation $y' + y^{2} = 0$ leads to
\begin{align}
a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \cdots &= - (a_{0} + a_{1} x + a_{2} x^{2} + \cdots )^{2} \\
&= - [ a_{0}^{2} + (a_{0}a_{1} + a_{1} a_{0}) x + \cdots ].
\end{align}
Equating the coefficients leads to
\begin{align}
a_{1} &= - a_{0}^{2} \\
2 a_{2} &= - (a_{0}a_{1} + a_{1} a_{0})
\end{align}
and so on. What is then determined is the general form of the coefficients being
\begin{align}
(n+1) \, a_{n} = - \sum_{k=0}^{n} a_{k} \, a_{n-k}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral relations in Fricke and Klein Can someone please explain how Fricke and Klein obtain the integral relationa stated at the top of p. 34 in this book? The entire book can be previewed on Google Books. It is an old book and I do not speak German. Also, is there a modern account of this anywhere in the modern literature?
| Using the Weierstrass Normal form, let's try to understand the elliptic integral
$$ \int \frac{dy}{\sqrt{4y^3 - g_2 y - g_3}} $$
In order to reduce the number of parameters, Fricke and Klein do a change of variables $y = \frac{g_2}{g_3}z$. The elliptic curve has two basic periods
$$ \Omega = \int \frac{dz}{\sqrt{4z^3 + g(z+1)}}
\text{ and }
-H = \int \frac{z\,dz}{\sqrt{4z^3 + g(z+1)}}
$$
This new parameter $g$ is basically the $J$-invariant in disguise. $\frac{27 J}{1-J} = g $. Let's just call the denominator $R = \sqrt{\dots}$.
What happens if we change $g$? Let's differentiate under the integral sign:
\begin{eqnarray}
\frac{d\Omega}{dg} &=& - &\int \frac{dz}{2R^3} & - &\int \frac{z \, dz}{2R^3}\\
\frac{dH}{dg} &=& & \int \frac{z \, dz}{2R^3} &+& \int \frac{z^2 \, dz}{2R^3}
\end{eqnarray}
These are called the Picard-Fuchs equations. In modern language, this is an instance of the Gauss-Manin connection.
The integral of any derivative along a closed path on the elliptic curve is going to be zero.
$$ \int d\left(\frac{z^a}{R}\right) =
\int \frac{a z^{a-1}}{R} - \int \frac{z^a(12 z^2 + g)}{2R^3}
= 0 \tag{$\ast$}$$
The first step says the derivative and integral are inverse operations - the fundamental theorem of calculus. For a closed path the two endpoints are the same so the integral is zero. For the second step use the quotient rule for derivatives and the definition of R as the denominator.
The first relation comes from setting $a = 0$ without any changes. The second and third relations start from setting $a = 1,2$ with some extra manipulation.
EDIT Yeah, they fall out of $R^2 = 4z^3 + g(z+1)$ and the definition of the periods $\Omega, H$.
Fricke and Klein solve for the periods and ultimately find the differential equation the period $\Omega$ satisfies in terms of the elliptic curve invariant $J$:
$$ \frac{d^2 \Omega}{dJ^2} +
\frac{1}{J} \cdot \frac{d \Omega}{dJ} + \frac{\frac{31}{144}J- \frac{1}{36}}{J^2(J-1)^2}\Omega = 0$$
When is $\oint \neq 0$ ?
The fundamental theorem of calculus says: $\int_a^b F'(x) \, dx = F(b) - F(a)$ so we integrate along a closed path, $a = b$ so the integral should be $F(b) - F(a) = 0$. Yet,
$$ \int_{|z|=1} \frac{dz}{z} = 2\pi i \neq 0 $$
What goes wrong is that $\frac{1}{z}$ is not holomorphic on $\mathbb{C}$, it is only holomorphic on $\mathbb{C} \backslash \{0\}$. By removing the one point we no longer have a plane, but a punctured plane. The circle $|z| = 1$ moves around this puncture and therefore integrating has a residue.
Another important issue is $\frac{1}{\sqrt{z}}$. The problem here is as we move around the unit circle from $1$ back to itself, the square roots go to $-1$.
$$ \sqrt{1} = \color{blue}{\mathbf{1}} \to \sqrt{i} = \mathbf{\tfrac{\color{blue}{1+i}}{\color{blue}{\sqrt{2}}}} \to \sqrt{-1} = \mathbf{\color{blue}{i}}\to \sqrt{-i} = \mathbf{\tfrac{\color{blue}{1-i}}{\color{blue}{\sqrt{2}}}} \to \sqrt{1} = -\mathbf{\color{green}{1}}$$
Basically every number has two square roots except for $0$, therefore the riemann surface $\{ (z, \pm \sqrt{z}): z \in \mathbb{C} \}$ has branching at $0$.
Using the identity $(\ast)$ with $a=1$ we can solve the 2nd identity.
\begin{eqnarray} \frac{1}{2}\Omega = \frac{1}{2}\int \frac{dz}{R}
= \frac{1}{2}\int \frac{R^2 }{R^3 } dz
&=& \frac{1}{2}\int \frac{4z^3 + g(z+1) }{R^3 } dz \\
&=& \frac{1}{2}\int \frac{4z^3 + g(z+1) + (-4z^3 + g(z+2))}{R^3 } dz \\
&=& 2g \int \frac{ z \, dz}{2R^3 } + 3g \int \frac{ dz}{2R^3 } \end{eqnarray}
Using the identity $(\ast)$ with $a=2$ we can solve the 3rd identity.
\begin{eqnarray} \tfrac{1}{2}H = -\tfrac{1}{2}\int \frac{z}{R} \, dz
= -\frac{1}{2}\int \frac{z R^2}{R^3} \, dz
&=& - \frac{1}{2}\int \frac{z(4z^3 + g(z+1)) }{R^3 } dz \\
&=& -\frac{1}{2}\int \frac{ -g(3z^2 + 4z) + g(z+1)z }{R^3 } dz \\
&=& 2g \int \frac{ z^2 \,dz }{2R^3 }
+ 3g\int \frac{ z \,dz}{2R^3 } \end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:
Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.
I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.
I also figured maybe I should play with it this way:
$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$
But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.
So what am I missing (other than sleep and food)? What route should I take to
prove this?
Thanks in advance!
| It holds true for any $ $ irrational $\,x=\sqrt[3]n,\,$ except $\,n=1\,$ (so $\,x^2+x=-1\in\Bbb Q)$
More generally: $ $ if $\ r\in\Bbb Q\ $ and $\,x=\sqrt[3]r\not\in\Bbb Q\,$ then $\,x^2+x = q\in\Bbb Q\iff r = 1.$
Proof $\,\ qx = x^3+x^2 = r+x^2\ $ so $\ qx-r = x^2 = q-x,\ $ so $\,(\color{#c00}{q\!+\!1})\,x = r+q.$
Therefore $\,\ x\not\in\Bbb Q\,\Rightarrow\,\color{#c00}{q = -1}\,\Rightarrow\, 0 = (\color{#c00}{q\!+\!1})x = r+q = r-1,\ $ thus $\,\ r = 1.$
| {
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"url": "https://math.stackexchange.com/questions/1207829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Inequality with condition $x^2+y^2+z^2=1$. let $x,y,z>0$ such that $x^2+y^2+z^2=1$. Find the minimum of $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$
Is the answer $3\sqrt{3}$ by any chance?
| Apply AM-GM and CS inequalities:$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \geq \dfrac{9}{x+y+z} \geq \dfrac{9}{\sqrt{3}\cdot (x^2+y^2+z^2)}= 3\sqrt{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying An Inverse Tan Function I would like to know how this equality holds.
$$
\tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1} \frac{1}{2n-1} - \tan^{-1} \frac{1}{2n+1}.$$
I was told to use the double angle formula for $\tan \theta$ but I can't seem to show this.
Thank you.
For some context, I was asked to show
$$
\sum_{n=1}^\infty \tan^{-1} \frac{1}{2n^2} = \frac{\pi}{4}
$$
| $$
\tan^{-1} \frac1{2n-1} = \arg (2n-1 + i) \\
\tan^{-1} \frac1{2n+1} = \arg (2n+1 + i)
$$
hence
$$
\tan^{-1} \frac1{2n-1} - \tan^{-1} \frac1{2n+1} = \arg \frac{2n-1+i}{2n+1+i} \\
= \arg(2n-1+i)(2n+1-i) = \arg (4n^2 +2i) =\tan^{-1}\frac1{2n^2} \\
=\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n-1)(2n+1)}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find every n $\in \mathbb{N}$ such that $n+1 \mid n^2+3$ Find every n $\in \mathbb{N}$ such that $n+1 \mid n^2+3$
What I did:
$n+1 \mid n^2+3$ and $n+1 \mid (n+1)^2=n^2+2n+1$
So
$n+1 \mid (n^2+3)-(n^2+2n+1) \Longrightarrow n+1\mid-2(n+1)$
$\Longrightarrow n+1\mid0$
And then
$(n+1)k=0$
No idea what to do next.
| Using polynomial division, write $n^2+3=(n+1)(n-1)+4$. Since $n+1|(n+1)(n-1)$ and $n+1|n^2+3$, it must be that $n+1|(n^2+3-(n+1)(n-1))$, or $n+1|4$. This limits our search to $n=0$, $n=1$, or $n=3$. Simple calculations show that all of these answers work.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expand the function in a Maclaurin series $\ln(5\cos^{3}(x))$ $$\ln(5\cos^{3}(x))$$ Need to be expanded: $x^{4}$
I need to end this problem.
So I laid the beginning of the function.
$$\cos x=1-\frac{x^2}{2!}-\frac{x^4}{4!}+o(x^4)$$
$$5\cos^3(x)=5\left(1-\frac{x^2}{2!}+o(x^4)\right)^3=5-\frac{15x^2}{2!}+\frac{35}{8}x^4+o(x^4).$$
I do not know what to do next.
| If you want to be cute about it, note that $\cos x = \sqrt{1 - \sin^2 x}$, so
$$
\ln(5\cos^3 x)=\ln 5 + 3\ln\cos x=\ln 5 + \frac{3}{2}\ln(1-\sin^2 x)=\ln 5 - \frac{3}{2}\sin^2 x - \frac{3}{4}\sin^4 x + O(x^6).
$$
To get the $x^4$ term, you just need $\sin^2 x \approx (x - x^3/6)^2 \approx x^2 - x^4/3$, so
$$
\ln(5\cos^3 x)\approx \ln 5 - \frac{3}{2}\left(x^2-\frac{1}{3}x^4\right)-\frac{3}{4}x^4=\ln 5 - \frac{3}{2}x^2 - \frac{1}{4}x^4.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding all natural solutions to $2^a+5=b^2$ What are all possible natural number solutions $(a, b)$ to the equation
$$2^a+5=b^2 ?$$
The only solution I've found is
$$2^2+5=3^2.$$
This may be the only solution but I don't have a proof.
| For the first equation we can see that $b$ is odd, since $2^a+5=b^2$ and $2^a$ is always even. So rewriting the equation you have $2^a+4= (b-1)(b+1)$. Is clear that if $a=1$, it doesn't has solution so $a\geq 2$. So
$$4(2^{a-2}+1)= (b-1)(b+1)$$
But $b$ is odd, so one between $b-1$ and $b+1$ is divided by 2 and the other by 4 (they're two consecutive even numbers), so the right side is divided by 8, and the only way for the left side to be divisible by 8 is when $2^{a-2}+1$ is even, that is when $2^{a-2}$ is odd, that only happens when $2^{a-2}=1$, so $a-2=0$, that lets $a=2$, that is the only solution you found.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| $$\int \frac{3\sin(x)+2\cos(x)}{2\sin(x)+ 3\cos(x)} \, dx$$
Multiply top and bottom by $\sec^{3}(x)$
\begin{align}
& \int \frac{3\tan(x)\sec^2(x)+2\sec^2(x)}{2\tan(x)\sec^2(x)+ 3\sec^2(x)} dx \\[10pt]
& \int \frac{(3\tan(x)+2)\sec^2(x)}{(2\tan(x)+ 3)\sec^2(x)} \, dx \\[10pt]
& \int \frac{(3\tan(x)+2)\sec^2(x)}{(2\tan(x)+ 3)(\tan^2(x)+1)} \, dx
\end{align}
Substitute $u = \tan(x)$ and $du = \sec^2(x)dx$
$$\int \frac{3u+2}{(2u+3)(u^2+1)}du$$
Use partial fractions to write
$$\int \left(\frac{5u+12}{13(u^2+1)}-\frac{10}{13(2u+3)}\right) \, du$$
$$\frac{1}{13}\int \left(\frac{5u}{u^2+1} + \frac{12}{u^2+1}-\frac{10}{2u+3}\right) \, du$$
for first term, substitute $s = u^2+1$ and $ds = 2u \, du$
$$\frac{1}{13}\int \left(\frac{5}{2s} + \frac{12}{u^2+1}-\frac{10}{2u+3}\right) \, du$$
the integral of $\frac{1}{s} = \ln s$ and the integral of $\frac{1}{u^2+1} = \tan^{-1}u$ turns it into
$$\frac{1}{13} \left(\frac{5\ln s}{2} + 12 \tan^{-1}u- \int \frac{10}{2u+3} \, du\right)$$
Substitute $p=2u+3$ and $dp = 2 \, du$
$$\frac{1}{13}(\frac{5\ln s}{2} + 12 \tan^{-1}u- \int \frac{5}{p} \, dp)$$
$$\frac{1}{13}(\frac{5\ln s}{2} + 12 \tan^{-1}u- 5\ln p)$$
Substituting back for $s = u^2 +1$ and $p = 2u+3$ and $u=\tan(x)$
$$\frac{1}{13} \left(\frac{5\ln (\tan^2(x)+1)}{2} + 12 \tan^{-1}(\tan(x))- 5\ln{(2\tan x+3)}\right)$$
$$\frac{1}{13}(\frac{5\ln (\sec^2(x))}{2} + 12x- 5\ln{(2\tan x+3)})$$
$$\frac{1}{13}(5\ln (\sec(x)) + 12x- 5\ln{(2\tan x+3)})$$
$$= \frac{1}{13}(12x - 5\ln{(2\sin(x)+3\cos(x))}) + \text{constant}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Residue of 1/sin^3(z) What are the residues of $ \frac{1}{sin^{3}z} $?
From the residue theorem the residues are at
$$\lim_{z \rightarrow z_{0}} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} (z-z_{0})^{n} f(z)$$
$$\lim_{z \rightarrow n\pi} \frac{1}{2!} \frac{d^{2}}{dz^{2}} \frac{(z-n\pi)^{3}}{sin^{3}z}$$
$$=\lim_{z \rightarrow n\pi} \frac{1}{2} (z-n\pi)(3csc^{5}(z)+9cot^{2}(z)csc^{3}(z))-6cot(z)csc^{3}z$$
But what is the $\lim_{z \rightarrow n\pi} \frac{1}{2} (z-n\pi)(3csc^{5}(z)+9cot^{2}(z)csc^{3}(z))-6cot(z)csc^{3}z$?
| I find that with higher-order poles, it is easier to simply compute the residue right from the Laurent expansion.
$$\sin{z} = \sin{n \pi} + \cos{n \pi} (z-n \pi) -\frac1{2!} \sin{n \pi} (z-n \pi)^2 - \frac1{6!} \cos{n \pi} (z-n \pi)^3 + \cdots = (-1)^n (z-n \pi) - \frac16 (-1)^n (z-n \pi)^3 + \cdots$$
Thus,
$$\begin{align}\frac1{\sin^3{z}} &= \frac{(-1)^n}{\left [(z-n \pi) - \frac16 (z-n \pi)^3 + \cdots\right ]^3}\\ &= \frac{(-1)^n}{(z-n \pi)^3} \left [1-\frac16 (z-n \pi)^2 + \cdots\right ]^{-3} \\ &= \frac{(-1)^n}{(z-n \pi)^3} \left [1+\frac12 (z-n \pi)^2 + \cdots\right ]\\ &= \frac{(-1)^n}{(z-n \pi)^3} + \frac{(-1)^n}{2} \frac1{z-n \pi} + \cdots \end{align}$$
where the dots represents higher orders. As the residue is the coefficient of $1/(z-n \pi)$, the residue is $\frac12 (-1)^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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An Eigen Value of $\tiny \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is : Let $a,b,c,d$ be distinct non zero real numbers with $a+b=c+d.$ Then, an eigen value of the matrix $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is :
$(i)~a+c ~~(ii)~a+b~~(iii)~a-b~~(iv)~(b-d$
Attempt:
If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :
$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$
$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix} $
Solving $\det ( A - \lambda I) = 0$ results in a third degree equation which is difficult to solve.
Could anyone tell me of an easier method to solve it. Thank you very much for your help in this regard.
| Thank you all for your replies. While I was just reflecting, I thought of the following solution, so, thought of sharing the following solution :
If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :
$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$
$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix} $
Adding the first two columns and placing it in the first results in no change in the determinant value :
Hence :
$\det ( A- \lambda I) = \det \begin{pmatrix} a+b-\lambda&b&1 \\ c+d-\lambda&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$
$= (a+b-\lambda) \det \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix} = (c+d-\lambda) \det \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$
Hence, $\det ( A- \lambda I) =0 \implies a+b = \lambda = c+d$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\tan \alpha$ if $\sin \alpha + \cos \alpha = \frac{\sqrt{3}-1}{2}$ and $\alpha \in (90^\circ,135^\circ)$ I am stuck with this problem. Any ideas on how to solve this?
| Let $\sin A=\dfrac{\sqrt3}2,\cos A=-\dfrac12$
Clearly, $A=120^\circ$ is a solution
So, we can write $\sin\alpha-\sin A=\cos A-\cos\alpha$
Using Prosthaphaeresis Formulas, $$2\sin\frac{\alpha-A}2\cos\frac{\alpha+A}2=2\sin\frac{\alpha-A}2\sin\frac{\alpha+A}2$$
$$\iff\sin\frac{\alpha-A}2\left[\cos\frac{\alpha+A}2-\sin\frac{\alpha+A}2\right]=0$$
If $\sin\dfrac{\alpha-A}2=0,\dfrac{\alpha-A}2=180^\circ n\iff \alpha=A+360^\circ n$ where $n$ is any integer
Else $\tan\dfrac{\alpha+A}2=1=\tan45^\circ$
$\implies\dfrac{\alpha+A}2=m180^\circ+45^\circ\iff\alpha=360^\circ m+90^\circ-A$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can this be shown: $\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\dots}}} = \sqrt a$? $$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}=\sqrt{a}$$
Just for fun. I would like to read the proof of this if it exists. Any references would be appreciated.
| What we have is the $lim_{n\to\infty}$ $a^{\frac{1}{3}}a^{\frac{1}{9}}a^{\frac{1}{27}}..a^{\frac{1}{3^n}}$.
Using the property $(a^b)(a^c)=a^(b+c)$
We get:
$a^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...\frac{1}{3^n}}$
The exponent is in fact is geometric series which has a value of $(1/(1-(1/3))-1=\frac{1}{2}$ so:
We get your expression to be $a^{\frac{1}{2}}=\sqrt{a}$
| {
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Evaluating $\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}$ How do I evaluate
$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{\dfrac{1}{1-\cos x}}\ ?$$
I tried using the fact that $\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} = \exp\left(\ln\bigg(\frac{\sin x}{x}\right)\frac{1}{1-\cos x}\bigg)$ and then I am already stuck.
| Let us start with $$A=\left(\frac{\sin (x)}{x}\right)^{\frac{1}{1-\cos (x)}}$$ Take logarithms; so $$\log(A)=\frac{1}{1-\cos (x)}\log\left(\frac{\sin (x)}{x}\right)$$ Now, use Taylor series $$\sin(x)\approx x-\frac{x^3}{6}$$ $$\frac{\sin (x)}{x}\approx 1-\frac{x^2}{6}$$ Now consider that,for small $y$, $\log(1+y)\approx y-\frac{y^2}{2}$ So $$\log\left(\frac{\sin (x)}{x}\right)\approx -\frac{x^2}{6}-\frac{x^4}{180}$$ Also $$\cos(x)\approx 1-\frac{x^2}{2}+\frac{x^4}{24}$$ Using all of the above, we then have $$\log(A)\approx \frac{-\frac{x^2}{6}-\frac{x^4}{180}}{\frac{x^2}{2}-\frac{x^4}{24}}= \frac{-\frac{1}{6}-\frac{x^2}{180}}{\frac{1}{2}-\frac{x^2}{24}} \approx -\frac{1}{3}-\frac{7 x^2}{180}$$ which shows the limit of $\log(A)$ and how it is approached.
I am sure that you can take from here.
| {
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Proving that for Fibonacci numbers $a_n \lt (\frac {1 + \sqrt 5} 2)^n$ for $n \ge 1$ I'd like to prove that for Fibonacci numbers $a_n \lt \left(\frac {1 + \sqrt 5} 2\right)^n$ for $n \ge 1$. I suppose it needs induction so, after verifying the trivial case $n=1$, the inductive step needs to be justified: $$a_{n+1} \lt \left(\frac {1 + \sqrt 5} 2\right)^{n+1}.$$
Assume $$a_n \lt \left(\frac {1 + \sqrt 5} 2\right)^n.$$
I can only think of multiplying both sides by $\left(\frac {1 + \sqrt 5} 2\right)$, which yields $$a_n\left(\frac {1 + \sqrt 5} 2\right) \lt \left(\frac {1 + \sqrt 5} 2\right)^{n+1}$$
and then somehow show that $a_{n+1} \lt a_n\left(\frac {1 + \sqrt 5} 2\right)$.
I tried using the fact that $a_{n+1}=a_n + a_{n-1}$ but wasn't able to conclude anything useful. Can you suggest me a way to prove it?
| First note the following (this will show the relevance of Antoine's comment):
$$
\left(\frac{1+\sqrt{5}}{2}\right)^n > \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n > \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n.
$$
The explicit formula for Fibonacci numbers that Antoine's comment is presumably referring to is Binet's formula, which states that, for every $n\geq 0$, that
$$
F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right],
$$
where $F_n$ is the $n$th Fibonacci number. Hence, if you can prove Binet's formula, then you will have certainly proven what you set out to prove because
$$
\left(\frac{1+\sqrt{5}}{2}\right)^n > \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right] = F_n.
$$
Thus, let's prove Binet's formula using induction, and your result will follow as a "trivial" consequence, as Antoine's comment notes.
For every $n\geq 0$, let $A(n)$ denote the statement
$$
A(n) : F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right].
$$
Base step: When $n=0$, the expression on the right-hand side of $A(n)$ is $0$, which is $F_0$. For $n=1$, it is also not difficult to check that each side of $A(n)$ is $1$.
Inductive step ($[A(k-1)\land A(k)]\to A(k+1)$): Suppose that for some fixed $k$, that $A(k-1)$ and $A(k)$ are true. Calculating, (where the second equality below follows from $A(k-1)$ and $A(k)$),
\begin{align}
F_{k+1} &= F_{k-1}+F_k\\[1em]
&= \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\right]+\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k}-\left(\frac{1-\sqrt{5}}{2}\right)^{k}\right]\\[1em]
&= \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left(1+\frac{1+\sqrt{5}}{2}\right)-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\left(1+\frac{1-\sqrt{5}}{2}\right)\right]\\[1em]
&= \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left(\frac{3+\sqrt{5}}{2}\right)-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\left(\frac{3-\sqrt{5}}{2}\right)\right]\\[1em]
&= \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k-1}\left(\frac{1+\sqrt{5}}{2}\right)^2-\left(\frac{1-\sqrt{5}}{2}\right)^{k-1}\left(\frac{1-\sqrt{5}}{2}\right)^2\right]\\[1em]
&= \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{k+1}\right],
\end{align}
so $A(k+1)$ is also true, completing the inductive step.
By mathematical induction, for each $n\geq 0, A(n)$ is true.
Your own claim follows as a result.
| {
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Does the limit exist? (Calculus) Consider the function
$$f(x,y)=\frac{2xy^2\sin^2(y)}{(x^2+y^2)^2}.$$
Does the limit exist when $(x,y)$ tends to $(0,0)$?
| Just see this
$$ \frac{2xy^2 \sin^2(y)}{(x^2+y^2)^2} \sim \frac{2 x y^4}{(x^2+y^2)^2} $$
since $\sin t \sim_{t\sim 0} t$. Then we have
$$ \bigg| \frac{2 x y^4}{(x^2+y^2)^2} \bigg| \leq \frac{2 |x| |y|^4}{(x^2+y^2)^2} \leq \frac{2\sqrt{x^2+y^2} (\sqrt{x^2+y^2})^4}{(x^2+y^2)^2} = 2\sqrt{x^2+y^2} < \epsilon $$
$$ \implies \sqrt{x^2+y^2} < \frac{\epsilon}{2} =\delta. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\triangle ABC\;,\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^0,$ Then find ordered pairs $\left(\angle A,\angle B\right)$
In a $\displaystyle \triangle ABC\;,\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^0,$ Then find the ordered pairs $\left(\angle A,\angle B\right)=$
$\bf{Options}::$ $(i)\; \left(45^0,75^0\right)\;\;\;\; (ii)\; \left(105^0,15^0\right)\;\;\;\; (ii) \; \left(15^0,105^0\right)\;\;\;\; (ii)\;\;\;\; (75^0\;,45^0)$
$\bf{My\; Solution::}$ Using Cosine Rule, $$\displaystyle \cos (60^0)=\frac{a^2+b^2-c^2}{2ab}\Rightarrow a^2+b^2-c^2=ab.$$
Now Divide both side by $b^2\;,$ We get $$\displaystyle \left(\frac{a}{b}\right)^2+1-\left(\frac{c}{b}\right)^2=\left(\frac{a}{b}\right)\;,$$ Where $b\neq 0$
Now put value of $$\displaystyle \frac{a}{b}=2+\sqrt{3}\;,$$ We get $$\displaystyle 7+4\sqrt{3}+1-\left(\frac{c}{b}\right)^2=2+\sqrt{3}.$$
So we get $$\displaystyle \left(\frac{c}{b}\right)^2=6+3\sqrt{3}\Rightarrow \frac{c}{b}=\sqrt{6+3\sqrt{3}}$$
Now How can i solve it, Help required, Thanks
| Instead of cosine law, you can try sine law.
Starting from $\frac{\sin A}{\sin (A + 60^0)} = \frac{2 + \sqrt 3}{1}$, and $\sin (A + 60^0) = \sin A \cos 60^0 + \cos A sin 60^0$
Then, $\sin A = [2 + \sqrt (3)][\sin A .(\frac {1}{2}) + \cos A.(\frac {\sqrt 3}{2})$
∴ $\tan A = -\frac {2 \sqrt 3 + 3}{\sqrt 3}$
∴ $\angle A = 105^0$ (but not using special angle values).
| {
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"timestamp": "2023-03-29T00:00:00",
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Where did I go wrong when doing this integral? This is my integral
$$
\int \frac{ (2x-3)}{(x^3 +10x)}\cdot dx \\
$$
This is my work
$$
\int\frac{2x}{x^3 +10x}\cdot dx-\int\frac{3}{x^3 +10x}\cdot dx\\
$$
Looking at them separately:
$$
\int\frac{2x}{x^3 +10x}\cdot dx \\= 2\int\frac{dx}{x^2 +10}
\\=\frac{2}{10}\int\frac{dx}{(\frac{x}{\sqrt{10}})^2+1}
\\=\frac{2}{10}\arctan(\frac{x}{\sqrt{10}})
$$
and
$$
\int\frac{3}{x^3 +10x}\cdot dx\\
\text{Using partial fractions to split the integrand yields}\\
\frac{3}{10}\int\frac{1}{x}\cdot dx -\frac{3}{\sqrt{10}}\int\frac{1}{x^2+10}\\
=\frac{3}{10}\ln{x} - \frac{3}{10\sqrt{10}}\arctan(\frac{x}{\sqrt{10}})
$$
As my final integral I therefore have:
$$
\frac{2}{10}\arctan(\frac{x}{\sqrt{10}})-\frac{3}{10}\ln{x} + \frac{3}{10\sqrt{10}}\arctan(\frac{x}{\sqrt{10}})
$$
However my textbook gives a radically different answer:
$$
\frac{20\arctan \left(\frac{x}{\sqrt{10}}\right)-3\cdot \sqrt{10}\left(\ln \left(\frac{x}{\sqrt{10}}\right)-\frac{\ln \left(\frac{x^2}{10}+1\right)}{2}\right)}{10\cdot \sqrt{10}}
$$
Could someone tell me where I went wrong? I can't figure it out for the life of me.
Thanks!
| Your partial fraction is incorrect. Use
$$\frac{3}{x(x^2+10)}=\frac{A}{x}+\frac{Bx+C}{x^2+10}$$
to get
$$A=\frac{3}{10}, B= -\frac{3}{10} , C=0.$$
Thus
\begin{align*}
\int\frac{3}{x^3 +10x} \, dx & =\frac{3}{10}\int \frac{1}{x} \,dx - \frac{3}{10}\int \frac{x}{x^2+10} \,dx \\
& = \frac{3}{10} \ln |x| - \frac{3}{20} \ln(x^2+10) + C.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{dx}{(x^2 + 1)^\alpha}$ I couldn't follow a step while reading this answer. Since I do not have enough reputation to post this as a comment, I'm asking a question instead. The answer uses "partial integration" to write this $$ \int \frac{dv}{(v^2 + 1)^\alpha} = \frac{v}{2(\alpha-1)(v^2+ 1)^{\alpha - 1}} + \frac{2\alpha -3}{2\alpha - 2}\int \frac{dv}{(v^2 + 1)^{\alpha -1}} $$
I would like to know what this technique is, and how this equality follows from it.
| "Partial integration" just means integration by parts. The important step here is the writing of the fraction as
$$ \frac{1}{(v^2+1)^{\alpha}} = \frac{1}{(v^2+1)^{\alpha}} - \frac{1}{(v^2+1)^{\alpha-1}} + \frac{1}{(v^2+1)^{\alpha-1}} \\
= -\frac{v^2}{(v^2+1)^{\alpha}} + \frac{1}{(v^2+1)^{\alpha-1}}. $$
Then you integrate the first term by parts, integrating $\frac{v}{(v^2+1)^{\alpha}}$ and differentiating $v$; this gives
$$ \int \left( \frac{1}{(v^2+1)^{\alpha}}-\frac{1}{(v^2+1)^{\alpha-1}} \right) \, dv = -\int \frac{v^2}{(v^2+1)^{\alpha}} \, dv \\
= \frac{v}{2(\alpha-1)(v^2+1)^{\alpha-1}} - \frac{1}{2(\alpha-1)}\int \frac{dv}{(v^2+1)^{\alpha-1}}, $$
and then rearranging gives the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Something wrong at $\int \frac{x^2}{x^2+2x+1}dx$ I have to calculate $$\int \frac{x^2}{x^2+2x+1}dx$$ and I obtain: $$\int \frac{x^2}{x^2+2x+1}dx=\frac{-x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$
but I verify on wolfram and this is equal with: $$x-\frac{1}{x+1}-2\log\left(x+1\right)$$ where did I go wrong?
P.S: Here is all steps: $$\int \:\frac{x^2}{x^2+2x+1}dx=-\frac{x^2}{x+1}+\int \:\frac{2x}{x+1}dx=-\frac{x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$ I am sure that is not wrong, but the form on wolfram seems easily then it.
| setting $$x+1=t$$ then we get the integral $$\int \frac{(t-1)^2}{t^2}dt=\int\left(1-\frac{2}{t}+\frac{1}{t^2}\right)dt$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$
I want to find a quick way of evaluating $$\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$$
This problem appeared on the qualifying round of MIT's 2014 Integration Bee, which leads me to think there should be a shortish way (no more than three minutes by hand) to solve it. Examining the indefinite integral (thanks to WolframAlpha) hasn't particularly helped me:
$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=-\frac{2}{3} \left(\sqrt{x^2+1}-2x\right) \sqrt{x+\sqrt{x^2+1}}+C$$
The bounds on the integral hint at trigonometric substitution, but I got nowhere by trying $x=\tan u$.
I also noticed that we can transform the integral by multiplying $\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}$ in the first square root, but there didn't seem to be anything to do after that either.
| $\bf{My\; Solution::}$ Let $$\displaystyle I = \int \sqrt{x+\sqrt{x^2+1}}dx$$
Now Let $$\left(x+\sqrt{x^2+1}\right)=t^2....................\color{red}\checkmark$$
Then $$\displaystyle \frac{\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}dx = 2tdt$$
So We get $$\displaystyle dx = \frac{2\sqrt{x^2+1}}{t}dt\;, $$ Now for Calculation of $$\sqrt{x^2+1}$$ in terms of $t\;,$ We are Using
$$\displaystyle \left(x+\sqrt{x^2+1}\right)\cdot \left(x-\sqrt{x^2+1}\right) = t^2\cdot \left(x-\sqrt{x^2+1}\right)$$
So We get $$\displaystyle \left(\sqrt{x^2+1}-x\right)=\frac{1}{t^2}..................\color{red}\checkmark$$
So We Get $$\displaystyle \sqrt{x^2+1}dx = \frac{t^4+1}{2t^2}dt$$
So We get $$\displaystyle I = \int\frac{t\cdot 2(t^4+1)}{2t^3}dt = \int \left(t^2+t^{-2}\right) dt = 2t-\frac{1}{t}+\mathcal{C}$$
So We Get $$\displaystyle \int\sqrt{x+\sqrt{x^2+1}}dx = 2\sqrt{x+\sqrt{x^2+1}}+\sqrt{x-\sqrt{x^2+1}}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Triple integral volume by equations I have trouble setting up a triple integral to find volume bound by equations, such as:
$$z = x^2 + 3;\quad y = 3 - x^2;\quad x + y = 2;\quad z = 0.$$
I'm not sure how to figure how to find the boundaries for the $z$-axis. How would I go about setting up the triple integral?
| You have $y=3-x^2$ and $x+y=2$. Set them equal to each other and find the intersection points. Namely, $3-x^2=2-x$ gives $x=\frac{1\pm\sqrt{5}}2$. Our lower $x$-limit is $\frac{1-\sqrt{5}}2$ and our upper $x$-limit is $\frac{1+\sqrt{5}}2$.
Furthermore, $3-x^2 \ge 2-x$ on the interval $[\frac{1-\sqrt{5}}2,\frac{1+\sqrt{5}}2]$. Our lower $y$-limit is $2-x$ and our upper $y$-limit is $3-x^2$.
We also see that our lower $z$-limit is $0$ and our upper $z$-limit is $x^2+3$.
Our volume $V$ is therefore represented by the triple integral $$V=\int_\frac{1-\sqrt{5}}2^\frac{1+\sqrt{5}}2 \int_{2-x}^{3-x^2} \int_0^{x^2+3} dz \, dy \, dx.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Large and small time PDE solution I have the following solution for a PDE
$$
u(x,t)=(2x+4t-10)+2e^{\frac{-1}{2}t}+\sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)e^{\frac{-1}{4}n^2\pi^2t}}{n^3\pi^3(n^2\pi^2-2)}
$$
I want to find the first two terms in the asymptotic expansion of u(x, t) as $t\rightarrow 0$ and $t\rightarrow \infty$(fixed $x$). Any hints or suggestions will be appreciated.
| The sum $\Sigma(\cdots)$ is an asymptotic series as $t \to \infty$, so the first two terms of the expansion of $u(x,t)$ as $t \to \infty$ are simply the two terms which do not decrease exponentially, namely
$$
u(x,t) \sim \underbrace{4t}_\text{first} + \underbrace{2x-10}_\text{second} + \cdots
$$
as $t \to \infty$.
Now, the first two terms of the expansion as $t \to 0$ should be what you get when you replace $e^{-n^2\pi^2t/4}$ and $e^{-t/2}$ by the first two terms of their Taylor series. That is,
$$
e^{-n^2\pi^2t/4} \approx 1 - \frac{n^2\pi^2}{4}t \qquad \text{and} \qquad e^{-t/2} \approx 1 - \frac{1}{2}t,
$$
so we should have
$$
\sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)e^{-n^2\pi^2t/4}}{n^3\pi^3(n^2\pi^2-2)} \approx \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n^3\pi^3(n^2\pi^2-2)} - \frac{1}{4} t \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n\pi(n^2\pi^2-2)}.
$$
We expect this to work since the new series we get after multiplying the summand by $n^2$ still converges, but as far as I know this does not constitute a proof. Anyway, we should therefore have
$$
u(x,t) \sim 2x - 10 + 2 - t + \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n^3\pi^3(n^2\pi^2-2)} - \frac{1}{4} t \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n\pi(n^2\pi^2-2)} + \cdots
$$
or, after grouping terms,
$$
u(x,t) \sim \underbrace{2x - 8 + \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n^3\pi^3(n^2\pi^2-2)}}_\text{first} - \underbrace{\left(1 + \frac{1}{4}\sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n\pi(n^2\pi^2-2)}\right) t}_\text{second} + \cdots
$$
as $t \to 0$.
Proving this could be tricky since the estimate $e^{-n^2\pi^2t/4} \approx 1 - n^2\pi^2t/4$ doesn't hold for large $n$. One can probably proceed by truncating the sum after $\lfloor 1/t \rfloor$ terms and showing that the error accrued isn't too large.
| {
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"timestamp": "2023-03-29T00:00:00",
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