Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Simplify $\int \frac{1}{\sqrt{2-x^2}}\, dx$ Somewhere in the provided answer:
$$\int \frac{1}{\sqrt{2-x^2}} dx = \sin^{-1}{\frac{x}{\sqrt{2}}}$$
How did they get that? What I have:
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2(1-\frac{x^2}{2})}} = \frac{1}{\sqrt{2} \sqrt{1-\frac{x^2}{2}}}$$
$$\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}} = \frac{1}{\sqrt{2}} \sin^{-1}{\frac{x}{\sqrt{2}}}$$
So I have an extra $\frac{1}{\sqrt{2}}$ ... I probably had some stupid mistakes?
| I'm going to use a $u$-substitution to make it clearer why this is the case. Let $x = \sqrt{2}\sin \theta$ so $dx = \sqrt{2} \cos \theta d\theta $. We have that
\begin{eqnarray}
\int \dfrac{dx}{\sqrt{2-x^2}} &=& \int\dfrac{\sqrt{2}\cos \theta d\theta}{\sqrt{2 - (\sqrt{2}\sin \theta)^2}} \\
&=& \int\dfrac{\sqrt{2}\cos \theta d\theta}{\sqrt{2 - (\sqrt{2}\sin \theta)^2}} \\
&=& \int\dfrac{\sqrt{2}\cos \theta d\theta}{\sqrt{2 - 2\sin^2 \theta}}\\
&=& \int\dfrac{\cos \theta d\theta}{\sqrt{1 - \sin^2 \theta}} \\
&=& \int\dfrac{\cos \theta d\theta}{\sqrt{\cos^2\theta}} \\
&=& \int\dfrac{\cos \theta d\theta}{cos\theta} \\
&=& \int d\theta = \theta+C.
\end{eqnarray}
Since $x = \sqrt{2}\sin \theta$, we have $\sin \theta = \dfrac{x}{\sqrt{2}}$ and so $\theta = \sin^{-1}\left(\dfrac{x}{\sqrt{2}}\right)$. Therefore
$$
\int \dfrac{dx}{\sqrt{2-x^2}} = \theta+C = \sin^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve trigonometric equation $\tan\theta + \sec\theta =2\cos \theta$
$$\tan\theta + \sec\theta =2\cos \theta,\quad 0\le \theta\le 2\pi$$Find all the possible solutions for the equations.
Multiply both sides by $\sec\theta - \tan \theta$.
$$\implies (\tan\theta + \sec\theta)(\sec\theta - \tan\theta) = (\sec\theta -\tan\theta)2\cos \theta$$
$$\implies 1 = 2 -2\sin \theta$$ $$\implies \sin \theta=\frac12 \implies \theta = \arcsin\frac12$$Such a solution gets me two solutions $\frac{\pi}6$ and $\frac{5\pi}6$. But when I Wolfram it, I am supposed to get one more solution i.e $\frac{3\pi}2$, but at $\frac{3\pi}2$ $\tan \theta$ and $\sec\theta$ aren't defined.
| $$\sec\theta+\tan\theta=2\cos\theta=\frac2{\sec\theta}$$
$$\implies \sec\theta-\tan\theta=\frac1{\sec\theta+\tan\theta}=\frac{\sec\theta}2$$
$$\implies \sec\theta\left(1-\frac12\right)=\tan\theta$$
$$\implies \frac{\tan\theta}{\sec\theta}=\frac12\iff \sin\theta=\frac12=\sin\frac\pi6$$
$$\implies \theta=n\pi+(-1)^n\frac\pi6\text{ where }n \text{ is any integer}$$
If $n$ is even $=2m$(say), $\theta=2m\pi+\frac\pi6$
As $0\le \theta\le2\pi, 0\le2m\pi+\frac\pi6\le2\pi\implies 0\le 12m+1\le12\implies m=0$
Similarly, if $n$ is odd $=2m+1$(say), $\theta=(2m+1)\pi-\frac\pi6\implies m=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge? For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge?
The solution states:
The general term is of the form $u_n(x)=\frac{x^{n-1}}{(2n-1)}$, and hence
$$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{(2n+1)}\cdot\frac{(2n-1)}{|x^{n-1}|}$$
------edit start-------
$$=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$$
------ edit end -------
$$=\frac{(2n-1)}{(2n+1)}|x|$$
clearly
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|$$
My question is:
*
*How do you get from $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ to $u_n(x)=\frac{x^{n-1}}{(2n-1)}$?
*Why do the absolute restriction only apply to the $|x^n|$ and $|x^{n-1}|$ in the next line and not to the rest of the equation?
*and lastly, We go from $=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$ to $=\frac{(2n-1)}{(2n+1)}|x|$ I.E. how does $\frac{|x^n|}{|x^{n-1}|}=|x|$ in the next line? (I understand that once you apply the limits, that $\frac{(2n-1)}{(2n+1)} = 1$)
Edits
I have added a line to the equation that was not there earlier and clarified my last question.
| $$\frac{x^0}{1}+\frac{x^1}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$$
You can see the pattern, in the numerator It's:
$$x^0, x^1, x^2, x^3, \dots, x^{n-1}$$
And in the denominator, it's the odd numbers:
$$1, 3, 5, 7, 9, \dots, 2n-1$$
Therefore, it is:
$$u_n(x)=\frac{x^{n-1}}{(2n-1)}$$
While all odd numbers are positive ($n>0$), you can take them out of the absolute and $2n+1$ produce the same series of number as $2n-1$ which is odd numbers:
$$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{\color{red}{(2n+1)}}\cdot\frac{\color{red}{(2n-1)}}{|x^{n-1}|} = \frac{|x^n|}{|x^{n-1}|}=|\frac{x^n}{x^{n-1}}|=|x|$$
Hence:
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explain this logic to me please $$
\ \frac{3}{(x^2+4)(x^2+9)} = \frac{Ax + B}{(x^2+4)} + \frac{Cx+D}{(x^2+9)}
$$
Instructions say that "we can anticipate that $$ A = C = 0,$$ because neither the numerator nor the denominator involves odd powers of x, whereas nonzero values of A or C would lead to odd degree terms on the right"
I understand what they're saying, but I don't follow the logic. Can someone please explain in layman's terms? Thanks
| The following seems to be in the spirit of the instructions, but precise. (However, the substitution $y=x^2$ is better.)
Let $f(x)$ be the function on the left-hand side. Then $f(-x)=f(x)$ for all $x$, that is, $f(x)$ is an even function. So
$$f(x)=\frac{1}{2}\left(f(x)+f(-x)\right).$$
Now look at the right-hand side. We have
$$\begin{align*}\frac{1}{2}\left(f(x)+f(-x)\right)&=\frac{1}{2}\left(\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+9}+\frac{-Ax+B}{x^2+4}+\frac{-Cx+D}{x^2+9}\right)\\
&=\frac{B}{x^2+4}+\frac{D}{x^2+9}\end{align*}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/140324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find the inverse of a $4\times4$ matrix My matrix looks like this:
$$\left(\begin{array}{rrrr}
1& 1 & 1 & 1\\
1& -1 & 1 & 0\\
1& 1 & 0 & 0\\
1& 0 & 0 & 0
\end{array}\right)$$
The right lower half are all zeros.
Is there a quick way to find an inverse of this matrix?
I have the solution, but I'm unable to find the algorithm to get the inverse.
| You can also do it by inspection. Let $X$ be the inverse, and $\overline{x_i}$ the i-th row.
Then
$$
\left(\begin{array}{rrrr}
1 & 1 & 1 & 1\\
1 & -1 & 1 & 0\\
1 & 1 & 0 & 0\\
1 & 0 & 0 & 0
\end{array}\right)
X
= I
$$
So, considering first the last row, we get
$$ (1 \; \phantom{-}0 \; \phantom{-}0 \; \phantom{-}0) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4} \end{array}\right) = \overline{x_1} = (0 \; \phantom{-}0 \; \phantom{-}0 \; \phantom{-}1) $$
Then, from row 3 we get
$$(1 \; \phantom{-}1 \; \phantom{-}0 \; \phantom{-}0) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4} \end{array}\right)
= \overline{x_1} + \overline{x_2} = (0 \; \phantom{-}0 \; \phantom{-}1 \; \phantom{-}0) \Rightarrow \overline{x_2} = (0 \; \phantom{-}0 \; \phantom{-}1 \; -1)
$$
and
$$(1 \; -1 \; \phantom{-}1 \; \phantom{-}0) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4} \end{array}\right)
= \overline{x_1} - \overline{x_2} + \overline{x_3} = (0 \; \phantom{-}1 \; \phantom{-}0 \; \phantom{-}0) \Rightarrow \overline{x_3} = (0 \; \phantom{-}1 \; \phantom{-}1 \; -2)
$$
$$(1 \; \phantom{-}1 \; \phantom{-}1 \; \phantom{-}1) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4} \end{array}\right)
= \overline{x_1} + \overline{x_2} + \overline{x_3} + \overline{x_4} = (1 \; \phantom{-}0 \; \phantom{-}0 \; \phantom{-}0) \Rightarrow \overline{x_4} = (1 \; -1 \; -2 \; \phantom{-}2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/141936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Point on the bisector and excenter Given a triangle ABC, $\angle BAC = 20^{\circ}, \angle ACB=30^{\circ}$. M is a point inside the triangle such that $\angle MAC=\angle MCA=10^{\circ}$. L is a point on AC (L is between A and C) such that $AL=AB$. If $AM \cap BC =K$, prove that $K$ is the center of the excircle of $\triangle ABL$. Find $\angle AMB$.
Proving that $K$ is the excenter of $\triangle ABL$ is easy. However, I cannot find $\angle AMB$.
| $\hspace{2cm}$
Since $\overline{AB}=\overline{AL}$, $\triangle ABL$ is isosceles. $\overline{AK}$ bisects $\angle BAL$; therefore, $\overline{BL}\perp\overline{AK}$. Let $J$ be the intersection of $\overline{BL}$ and $\overline{AK}$. Drop perpendicular $\overline{KN}$ to $\overline{AB}$ and perpendicular $\overline{KP}$ to $\overline{AC}$.
$\angle ABC=130^\circ$ and $\angle ABJ=80^\circ$; therefore, $\angle JBK=50^\circ$. Being an external angle of $\triangle ABC$, $\angle NBK=50^\circ$. Therefore, $\triangle BJK=\triangle BNK$. Thus, $\overline{NK}=\overline{JK}$. $\triangle ANK=\triangle APK$; therefore
$$
\overline{PK}=\overline{NK}=\overline{JK}
$$
Thus, $K$ is the center of the excircle to $\triangle ABL$ tangent to $\overline{BL}$.
Being an external angle of $\triangle AMC$, $\angle KMC=20^\circ$. Therefore, $\triangle KMC$ is isosceles, giving $\overline{MK}=\overline{KC}$.
Because $\triangle CKP$ is a $30{-}60{-}90$ triangle, we can place $Q$ so that $\triangle CQP$ is also $30{-}60{-}90$ and $\triangle KQC$ is equilateral. Therefore, $\overline{KQ}=\overline{KC}$. Furthermore, $\overline{KP}=\overline{QP}=\frac12\overline{KQ}$. Thus,
$$
\frac{\overline{JK}}{\overline{MK}}=\frac{\overline{KP}}{\overline{KQ}}=\frac12
$$
Therefore, $\overline{MJ}=\overline{JK}$, and $\triangle MBK$ is isosceles. Since $\angle MBJ=\angle JBK=50^\circ$, we have that $\angle BMJ=40^\circ$, which leaves $\angle AMB=140^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Computing $A^{50}$ for a given matrix $A$ $A =\left(
\begin{array}{ccc}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right)$ then what would be $A^{50}$?
For real entries
| the characteristic polynomial is
\[
\chi_A(t) = (1-t)\bigl(t^2 - 1)
\]
so the eigenvalues are $\pm 1$, we have
\[
A - 1 = \begin{pmatrix} 0 & 0 & 0\\\ 1 & -1 & 1 \\\ 0 & 1 & -1 \end{pmatrix}
\]
which has defect $1$ (so $A$ isn't diagonalizable). It holds
\[
(A - 1)^2 = \begin{pmatrix} 0 & 0 & 0\\\ -1 & 2 & -2 \\\ 1 & -2 & 2
\end{pmatrix}
\]
so a basis of $\ker (A-1)^2$ is $\\{(2,1,0), (2,0,-1)\\}$, we have $A \cdot (2,1,0) = (0, 1, 1)$, so $\\{(2,1,0), (0,1,1)\\}$ is the basis of $\ker(A-1)^2$ we will use. It holds
\[
A + 1 = \begin{pmatrix} 2 & 0 & 0\\\ 1 & 1 & 1 \\\ 0 & 1 & 1 \end{pmatrix}
\]
we choose the basis $\{(0,1,-1)\}$ of its kernel. So we let
\[
S := \begin{pmatrix} 2 & 0 & 0\\\ 1 & 1 & 1 \\\ 0 & 1 & -1 \end{pmatrix}
\]
Then
\[
S^{-1}AS = \begin{pmatrix} 1 & 0 & 0\\\ 1 & 1 & 0 \\\ 0 & 0 & -1 \end{pmatrix}
\]
so
\[
S^{-1}A^{50}S = \begin{pmatrix} 1 & 0 & 0\\\ 50 & 1 & 0 \\\ 0 & 0 & -1 \end{pmatrix}
\]
and finally
\[
A^{50} = \begin{pmatrix} 1 & 0 & 0\\\ 25 & 1 & 0 \\\ 25 & 0 & 1 \end{pmatrix}
\]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Quadratic equation Suppose that $a+b = 2m_{1}$ and $ab = 4m_{1}^{2}-3m_{2}$. Why is the quadratic equation $$y^{2}-2m_{1}y+(4m_{1}^{2}-3m_{2})=0$$ instead of $$y^{2}+2m_{1}y+(4m_{1}^{2}-3m_{2})=0$$
In other words, why is it $-2m_{1}y$ instead of $2m_{1}y$ in the second term?
| $a+b = x$ and $ab = y$
$$(a+b)^2 = x^2$$
$$a^2+2ab+b^2 = x^2$$
$$a^2+2ab+(x-a)^2 = x^2$$
$$a^2+2y+x^2-2xa+a^2 =x^2$$
$$2a^2+2y-2xa =x^2-x^2=0$$
$$2(a^2-xa+y) =0$$
Equation $(1)$
$$a^2-xa+y =0$$
if you put $a=x-b$ then
$$(x-b)^2-x(x-b)+y =0$$
$$x^2-2xb+b^2-x^2+xb+y =0$$
Equation $(2)$
$$b^2-xb+y =0$$
As you see Equation $(1)$ and Equation $(2)$ are the same Quadratic equation . $a$ and $b$ are roots of the Quadratic equation. As you see $x$ must be minus $y$ must be plus while writing the quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/144936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2$ I need to prove that
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq (a_1 + a_2 + \ldots + a_k)^2\;,$$ where $a_1, a_2, \dots, a_k$ is some set of reals.
Firstly:
Can I presume without the loss of generality that $a_1 \leq a_2 \leq \ldots \leq a_n$ ?
This is how far I got:
I used the formula $\left \langle a,b \right \rangle \leq |a||b|$:
$$\begin{align*}\left \langle a,1 \right \rangle &\leq |a||1|\\
(a_1 + a_2 + \ldots + a_k) &\leq \sqrt{(a_1^2 + a_2^2 + \ldots + a_k^2)}\sqrt{k}
\end{align*}$$
Square it:
$$(a_1 + a_2 + \ldots + a_k)^2 \leq k(a_1^2 + a_2^2 + \ldots + a_k^2)$$
Now I have to prove that:
$$\frac{k(k+1)}{2}\left(\frac{a_1^2}{k} + \frac{a_2^2}{k-1} + \ldots + \frac{a_k^2}{1}\right) \geq k(a_1^2 + a_2^2 + ... + a_k^2)$$
But I'm not sure how. Any pointers?
| Try considering instead the vectors $(\frac{a_1}{1},...,\frac{a_k}{\sqrt{k}})$ and $(1,...,\sqrt{k})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Eigenvector basis I have this matrix:
$$
\begin{pmatrix}
3 & \sqrt{2} \\
\sqrt{2} & 2 \\
\end{pmatrix}
$$
I found the eigenvalues ( that are $1$ and $4$) but I find it really difficult to find the bases... ($(cI-A)x=0$ )
I think maybe one is $(1,1)$ but I'm not sure..
| The eigenvalues are correct.
To find the eigenvectors, you need to solve systems of linear equations. Namely, to find eigenvectors of $1$, you need to find all vectors $\mathbf{x}$ for which $(A-(1)I)\mathbf{x}=\mathbf{0}$ (this is equivalent to $(I-A)\mathbf{x}=\mathbf{0}$, but it involves changing fewer entries, so it may be less prone to errors).
$$A-I = \left(\begin{array}{cc}
3&\sqrt{2}\\
\sqrt{2}&2\end{array}\right) - \left(\begin{array}{cc}1&0\\0&1\end{array}\right) = \left(\begin{array}{cc}2 & \sqrt{2}\\
\sqrt{2} & 1
\end{array}\right).$$
When is $(A-I)\mathbf{x}=\mathbf{0}$? We need to solve the system
$$\left(\begin{array}{cc}
2 & \sqrt{2}\\
\sqrt{2}&1\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right).$$
At this point, you probably know how to solve systems of linear equations. For example, we can use Gaussian elimination on the matrix:
$$\begin{align*}
\left(\begin{array}{cc}
2 & \sqrt{2}\\
\sqrt{2} & 1
\end{array}\right) &\to \left(\begin{array}{cc}
1 & \frac{\sqrt{2}}{2}\\
\sqrt{2} & 1 \end{array}\right) &\text{(divide first row by }2\text{)}\\
&\to \left(\begin{array}{cc}
1 & \frac{\sqrt{2}}{2}\\
0 & 1 - \sqrt{2}\left(\frac{\sqrt{2}}{2}\right)\end{array}\right) &\text{(subtract }\sqrt{2}\text{ times the first row from row 2)}\\
&=\left(\begin{array}{cc}
1 & \frac{\sqrt{2}}{2}\\
0 & 0
\end{array}\right).
\end{align*}$$
The system has infinitely many solutions (as it should, since there have to be eigenvectors associated to the eigenvalue). They are all vectors $(x_1,x_2)^t$ such that $x_1+\frac{\sqrt{2}}{2}x_2 = 0$. That means that we must have $x_1 = -\frac{\sqrt{2}}{2}x_2$. If we set $x_2=\sqrt{2}$, then $x_1=-1$. So one eigenvector is $(-1,\sqrt{2})^t$.
Now you need to do the same thing with $A-4I$, that is, with the matrix
$$\left(\begin{array}{rr}
-1 & \sqrt{2}\\
\sqrt{2} & -2
\end{array}\right).$$
(No, $(1,1)$ is not an eigenvectors; note that $A(1,1)^t = (3+\sqrt{2},2+\sqrt{2})^t$, which is not a scalar multiple of $(1,1)$)
| {
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"url": "https://math.stackexchange.com/questions/146788",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Probability and distribution function Can some one please tell me how to do it?
Let $(i,j)$ denote the numbers on the top faces when a pair of fair dice is rolled.
Let $X(i,j)=i+j-3$.
Find the range of $X$.
Find the probability distribution function $f_{X}(x)$ of $X$.
Find each of the following
*
*$p(x>11)$
*$p(x<7)$
*$p(x \ge0)$
| The range: The values of $i$ and $j$ range from $1$ to $6$. So the smallest possible value of $i+j$ is $2$, and the largest is $12$. So $i+j-3$ ranges over the integers from $-1$ to $9$.
Thus the range of $X$ is the set of all integers $n$ such that $-1\le n\le 9$. You could write it also as $\{-1,0,1,1,3,4,5,6,7,8,9,\}$.
The probability distribution function: Recall that $f_X(x)=P(X=x)$. We need to specify $f_X(x)$ for all $x$ in the range.
As a start, we find $f_X(-1)=P(X=-1)$. We have $X=-1$ precisely if $i+j=2$, which happens if both dice show a $1$. The probability of this is $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.
Next we find $f_X(0)$. We have $X=0$ precisely if $i+j=3$. This happens if the yellow die shows $1$ and the pink shows $2$, or the other way around. So $f_X(0)=\frac{2}{36}$.
Next we find $f_X(1)$. Note that $X=1$ precisely if $i+j=4$, that is, if two dice give sum $4$. We have probably long known that the sum of two fair dice is $4$ with probability $\frac{3}{36}$.
Let's get slightly more abstract. Let $Y=X+3$. Then $Y=i+j$ is just the sum of two dice, and $X=k$ iff $Y=X+k$. So $f_X(1)=f_Y(4)=\frac{3}{36}$.
Similarly, $f_X(2)=f_Y(5)=\frac{4}{36}$, $f_X(3)=f_Y(6)=\frac{5}{36}$, $f_X(4)=f_Y(7)=\frac{6}{36}$, $f_X(5)=f_Y(8)=\frac{5}{36}$, and so on.
The numbered questions: $1$. We want $P(X\gt 11)$. Since $X=Y-3$, We have $X\gt 11$ precisely if $Y \gt 14$. Of course the sum of two dice cannot be $\gt 14$, and therefore $P(X\gt 11)=0$.
$2$. This one is more work. The most mechanical way to do it is to observe that $X \lt 7$ precisely if $X$ takes on the values $-1$ to $6$. So add up the values of $f_X(k)$ obtained earlier, $k=-1$ to $k=6$.
Or else note that this is the probability that $Y \lt 10$, that is, the probability that the sum $Y$ of two dice is less than $10$. It is easier to find the probability that $Y\ge 10$. This is $\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{6}{36}$. So the probability that $Y \lt 10$ is $\frac{30}{36}$.
$3$. More of the same.
| {
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} |
How to factor the quadratic polynomial $2x^2-5xy-y^2$? How do I factor this polynomial: $2x^2-5xy-y^2$ ?
| EDIT: Your example cannot be factored over the integers, because $25 + 8 = 33$ is not a perfect square. That is, taking $$ a x^2 + b x y + c y^2 = 2 x^2 - 5 x y - y^2, $$ we get
$$ a = 2, \; b = -5, \; c = -1, \; \Delta = b^2 - 4 a c = (-5)^2 - 4 \cdot 2 \cdot (-1)= 25 + 8 = 33, $$
and $33$ is nonnegative but is not a square.
ORIGINAL: Either the one-variable function $$ a x^2 + b x + c $$ or the quadratic form
$$ a x^2 + b x y + c y^2, $$ with integers $a,b,c,$ factor over the rational numbers if and only if the discriminant
$$ \Delta = b^2 - 4 a c $$ is a square.
You are familiar with this because of the quadratic formula
$$ \frac{-b \pm \sqrt \Delta}{2a} $$
which gives the roots $x$ for $ a x^2 + b x + c =0. $
Only a little changes when inserting the letter $y,$ giving $ a x^2 + b x y + c y^2. $ First, if both $a,c$ are $0,$ then we have $bxy$ which is already factored. So, let me show the traditional case, when $a \neq 0.$ Also let
$$ \Delta = \delta^2, $$ say with integer $\delta \geq 0.$
$$ a x^2 + b x y + c y^2 = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + 4 a c y^2 \right) $$
$$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - b^2 y^2 + 4 a c y^2 \right) $$
$$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - (b^2 y^2 - 4 a c y^2) \right) $$
$$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - (b^2 - 4 a c ) y^2 \right) $$
$$ = \frac{1}{4a} \left( 4 a^2 x^2 + 4 a b x y + b^2 y^2 - \Delta y^2 \right) $$
$$ = \frac{1}{4a} \left( (2ax+by)^2 - \Delta y^2 \right) $$
$$ = \frac{1}{4a} \left( (2ax+by)^2 - \delta^2 y^2 \right) $$
$$ = \frac{1}{4a} \left( \; (2ax+by + \delta y) \; (2ax+by - \delta y) \; \right) $$
$$ = \frac{1}{4a} \left( \; (2ax+ (b + \delta) y) \; (2ax+ (b - \delta) y) \; \right) $$
Now, either $b,\delta$ are both even or both odd. Either way, we may absorb a factor of $4$ into
$$ = \frac{1}{a} \left( \; (ax+ \frac{(b + \delta)}{2} y) \; (ax+ \frac{(b - \delta)}{2} y) \; \right) $$
Finally, since
$$ \frac{(b + \delta)}{2} \frac{(b - \delta)}{2} = \frac{b^2 - \Delta}{4} = ac $$
is divisible by $a,$ by unique factorization we may write
$$a = a_1 a_2$$ with $a_1$ dividing the first fraction and $a_2$ the second, thus finally getting
$$ a x^2 + b x y + c y^2 = \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) y \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) y \right) \; $$ in integers.
NOTE, October 3, 2013: we may simply take $$ a_1 = \gcd \left( a,
\frac{(b + \delta)}{2} \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1} $$ without paying attention to any prime factorizations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$ I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.
So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula.
Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.
How would I start with this integration?
| You can use integrate by parts:
\begin{align}
I&=\int_0^1\sqrt{1+\frac{1}{x}}dx=x\sqrt{1+\frac{1}{x}}|_0^1-\int_0^1x\frac{-\frac{1}{x^2}dx}{2\sqrt {1+\frac{1}{x}}}
=\sqrt{x(1+x)}|_0^1+\int_0^1\frac{d\sqrt{x}}{\sqrt {1+x}}\\
&=\sqrt{2}+\ln(\sqrt{x}+\sqrt{x+1})|_0^1
=\sqrt{2}+\ln(1+\sqrt{2})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 7
} |
Integral of $\int \frac{x^2 - 5x + 16}{(2x+1)(x-2)^2}dx$ I am trying to find the integral of this by using integration of rational functions by partial fractions.
$$\int \frac{x^2 - 5x + 16}{(2x+1)(x-2)^2}dx$$
I am not really sure how to start this but the books gives some weird formula to memorize with no explanation of why $\frac {A}{(ax+b)^i}$ and $ \frac {Ax + B}{(ax^2 + bx +c)^j}$
I am not sure at all what this means and there is really no explanation of any of it, I am guessing $i$ is for imaginary number, and $j$ is just a representation of another imaginary number that is no the same as $i$. $A$, $B$ and $C$, I have no idea what that means and I am not familiar with capital letters outside of triangle notation so I am guessing that they are angles of lines for something.
| Write $$\frac{1}{(2x-1)\cdot (x-2)^{2}} = \frac{A}{2x+1} + \frac{Bx + C}{(x-2)^{2}}$$
Once you have written this down it makes the job more easier. Now, the denominator terms cancel and you are left with
\begin{align*}
1 &= A \cdot \bigl(x-2)^{2} + B\cdot x \cdot (2x+1) + C \cdot (2x+1) \\\ 1 &= A \cdot \bigl(x^{2} - 4x +4) + 2Bx^{2} + Bx + 2Cx + C \\\ 1 &= x^{2} \cdot (A + 2B) + x \cdot (-4A + B + 2C) + (4A + C)
\end{align*}
Now comparing both the sides you find that
\begin{align*}
A+2B &=0 \\\ -4A+B+2C &=0 \\\ 4A+C &=1
\end{align*}
From here find the value of $A,B$ and $C$ and try to solve the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\int \frac {\sqrt {x^2 - 4}}{x} dx$ I am trying to find $$\int \frac {\sqrt {x^2 - 4}}{x} dx$$
I make $x = 2 \sec\theta$
$$\int \frac {\sqrt {4(\sec^2 \theta - 1)}}{x} dx$$
$$\int \frac {\sqrt {4\tan^2 \theta}}{x} dx$$
$$\int \frac {2\tan \theta}{x} dx$$
From here I am not too sure what to do but I know I shouldn't have x.
$$\int \frac {2\tan \theta}{2 \sec\theta} dx$$
I also know I shouldn't have dx anymore.
$$dx = 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int \frac {2\tan \theta}{2 \sec\theta} 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int {2\tan^2 \theta} \; \mathrm d\theta$$
$$2\int {\tan^2 \theta} \; \mathrm d\theta$$
I have no idea how to find the integral of $\tan^2 \theta$
So I use Wolfram Alpha:
$$\tan \theta - \theta + c$$
Now I need to replace theta with x.
$$x = 2 \sec\theta$$
With same mathmagics I produce
$$ \frac {x}{2} = \sec \theta$$
$$ \theta = \operatorname {arcsec} \left(\frac{x}{2}\right)$$
$$\tan \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) - \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) + c$$
This is wrong but I am not sure why.
| \begin{aligned}
& \int \frac{\sqrt{x^{2}-4}}{x} d x \\
=& \int \frac{x^{2}-4}{x \sqrt{x^{2}-4}} d x \\
=& \int \frac{x^{2}-4}{x^{2}} d\left(\sqrt{x^{2}-4}\right) \\
=& \sqrt{x^{2}-4}-4\int{\frac{1}{x^{2}} d\left(\sqrt{x^{2}-4}\right)} \\
=& \sqrt{x^{2}-4}-4\int \frac{d\left(\sqrt{x^{2}}-4\right)}{\left(\sqrt{x^{2}-4}\right)^{2}+4}\\
=& \sqrt{x^{2}-4}-2\tan ^{-1}\left(\frac{\sqrt{x^{2}-4}}{2}\right)+C
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Shortest distance between point and line of intersection.
Find the shortest distance between the point $Q(11, 2, -1)$ and the line
of intersection created by the planes
x .$ \left(\begin{array}{cc} 1\\
-1\\ 3\\ \end{array}\right) = 0 ~$ and x
$= a \left(\begin{array}{cc} 2\\ 1\\ 2\\ \end{array}\right)~+ $ $~b
\left(\begin{array}{cc} 3\\ 1\\ -3\\ \end{array}\right) $
To get the direction of the line I took $(1, -1, 3)\times(v_1 \times v_2)$ where $v_1$ and $v_2$ are the directional vectors of the second plane.
$
\left(\begin{array}{cc}
1\\
-1\\
3\\
\end{array}\right)
\times$
$
\left(\begin{array}{cc}
-5\\
12\\
-1\\
\end{array}\right)
=$
$
\left(\begin{array}{cc}
-37\\
-16\\
7\\
\end{array}\right)
$
and the planes go through the origin, so my line is just
$\lambda
\left(\begin{array}{cc}
-37\\
-16\\
7\\
\end{array}\right)
$
If I imagine a triangle created by the line going from the line of the intersection to the point Q as being the hypotenuse, then if I project Q onto the direction $(-37, -16, 7)$, I get the magnitude of the base of the triangle
$\frac{446}{\sqrt{1674}}$
and all I need to do is use Pythagoras to find the last side of my triangle which will be the perpendicular distance from the line to my point. Therefore I get:
Shortest distance = $\sqrt{126 - (\frac{446}{\sqrt{1674}})^2} \approx 2.678$
I have done it a few times now and keep getting this answer; however, my provided answer is $\sqrt{6} \approx 2.449$. Where am I going wrong?
| Your cross product is incorrect. $$\begin{pmatrix} 1 \\ -1 \\ 3\end{pmatrix} \times \begin{pmatrix} -5 \\ 12 \\ -1\end{pmatrix} = \begin{pmatrix} (-1) \times (-1) - 12 \times 3 \\ 3 \times (-5) - 1 \times (-1) \\ 1 \times 12 - (-1) \times (-5)\end{pmatrix} = \begin{pmatrix} -35 \\ -14 \\ 7\end{pmatrix} = 7 \begin{pmatrix} -5 \\ -2 \\ 1\end{pmatrix}$$
As you have correctly observed since both the planes pass through the origin, so does the line of intersection. Hence, if $P$ is the projection of $Q$ on to the line, then $\Delta OPQ$ is a right-angled triangle right-angled at $P$. Hence, the projection of $Q$ onto the line is nothing but $OP$.
Hence, the magnitude of the base of the triangle i.e. $OP$ is $$\dfrac{(11,2,-1) \cdot (-5,-2,1) }{\sqrt{30}} = \dfrac{-55-4-1}{\sqrt{30}} = - \dfrac{60}{\sqrt{30}} = -2 \sqrt{30}$$
Hence, the shortest distance of the point $Q$ to the line is $$PQ = \sqrt{OQ^2 - OP^2} = \sqrt{11^2 + 2^2 +(-1)^2 - \left( -2 \sqrt{30}\right)^2} = \sqrt{126 - 4 \times 30} = \sqrt{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Angle of a triangle inscribed in a square Say we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $\angle DNF$?
I'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed.
|
Firstly We can write
(1) Oklid relation from $\triangle CNB$
$h^2=m(k+x)$
(2) Pisagor relation from $\triangle DPN$
$a^2=(x+k)^2+(x+k+m-h)^2$
(3) Pisagor relation from $\triangle FRN$
$b^2=h^2+k^2$
(4) Pisagor relation from $\triangle DCF$
$c^2=x^2+(x+k+m)^2$
if DNF triange is a right triangle,It must satisfy $a^2+b^2=c^2$.
$(x+k)^2+(x+k+m-h)^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2+(x+k+m)^2-2h(x+k+m)+h^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2-2h(x+k+m)+2h^2+k^2=x^2$
$xk+k^2-h(x+k+m)+h^2=0$
$xk+k^2-h(x+k+m)+m(k+x)=0$
$-h(x+k+m)+(k+m)(k+x)=0$
$\frac{x+k}{x+k+m}=\frac{h}{k+m}$
This result is equal to the rates of thales formula for similar triangles $\triangle CRN \sim \triangle CBE$
Thus $a^2+b^2=c^2$ is correct for $\triangle DNF$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Arc length of $y^3 = x^2$ I am trying to find the arc length of $y^3 = x^2$ and I am suppose to use two formulas, one for in terms of x and one for in terms of y.
At first I need to find (0,0) (1,1) and I start with in terms of x
$$y\prime = \frac{2}{3}x^\frac{-1}{3}$$
$$\left(\frac{2}{3}x^\frac{-1}{3}\right)^2 = \frac{4}{9x^\frac{2}{3}}$$
$$\int_0^1 \sqrt{1 + \left( \frac{2}{3}x^\frac{-1}{3}\right)^2}\, dx$$
$$\int_0^1 \sqrt{1 + \frac{4}{9x^\frac{2}{3}}}\, dx$$
This is undefined at 0 so it is an improper function. I do not think I can continue.
Now in terms of y.
$$x = y^\frac{3}{2}$$
$$x \prime = \frac{3\sqrt{y}}{2}$$
$$ \left(\frac{3\sqrt{y}}{2}\right)^2 = \frac{9y}{4}$$
$$\int_0^1 \sqrt{1+ \frac{9y}{4}}\, dy$$
I have no idea how to factor that out, I have tried many ways but I can not get it.
| For the last integral, let $u=1+\frac{9y}{4}$. Then $du=\frac{9}{4}\,dy$, so $dy=\frac{4}{9}\,du$. The indefinite integral is $\int \frac{9}{4}u^{1/2}\,du$, and it's almost over.
The first integral is a bit more scary. I would rewrite your $\sqrt{1 + \frac{4}{9x^{2/3}}}$ as $\sqrt{\frac{9x^{2/3} + 4}{9x^{2/3}}}$ and then pull the denominator out to conclude that we are interested in the indefinite integral
$$\int \frac{1}{3x^{1/3}}\sqrt{9x^{2/3}+4}\,\,dx$$
Now let $u=9x^{2/3}+4$. Then $du=6x^{-1/3}\,dx=\frac{6}{x^{1/3}}\,dx$. Then the $\frac{1}{3x^{1/3}}\,dx$ term is just $\frac{1}{18}du$.
You are right in viewing the above integral as an improper integral, since there is the appearance of trouble at $0$. In principle, we should integrate from $\epsilon$ to $1$, and find the limit as $\epsilon$ approaches $0$ from the right. In practice, the substitution $u=x^{2/3}$ makes the problem disappear.
Remark: There is a fair bit of difference in appearance of difficulty between the two approaches. But in fact they are just a substitution away from each other. However, the appropriate substitution may not be obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotics for Bell number Concrete Mathematics EXERCISE 9.46
Show that the Bell number $\varpi_n=e^{-1}\sum_{k\ge0}k^n/k!$ of exercise 7.15 is asymptotically equal to
\[ m(n)^ne^{m(n)-n-1/2}/\sqrt{\ln n} \]
where $m(n)\ln m(n) = n-\frac12$, and estimate the relative error in this approximation.
Part of the answer is that (According to the errata, I've edited the answer)
For convenience we write just $m$ instead of $m(n)$. By Stirling's approximation, the maximum value of $k^n/k!$ occurs when $k\approx m\approx n/\ln n$, so we replace $k$ by $m+k$ and find that
\begin{align*}
\ln\frac{(m+k)^n}{(m+k)!}=&n\ln m-m\ln m+m-\frac{\ln 2\pi m}2\\
&-\frac{(m+n)k^2}{2m^2}+O(k^3m^{-2}\log n)+O(1/m)\tag1
\end{align*}
Actually we want to replace $k$ by $\lfloor m\rfloor+k$; this adds a further $O(km^{-1}\log n)$. The tail-exchange method with $|k|\le m^{1/2+\epsilon}$ now allows us to sum on $k$, ...
How can we derive equation (1) (especially when $|k|\le m^{1/2+\epsilon}$)? I try to expand $\ln(m+k)!$ using Stirling's approximation. It gets
\[
\ln(m+k)!=(m+k)\ln(m+k)-(m+k)+\frac12\ln(m+k)+\sigma+O(1/m)
\]
where $e^\sigma=\sqrt{2\pi}$. However, the term $k\ln m$ in
\[
(m+k)\ln(m+k)=(m+k)(\ln m+\ln(1+k/m))=(m+k)\left(\ln m-k/m+O(k/m)^2\right)
\]
never vanishes, and it's $\Omega(1)=\omega\left(k^3m^{-2}\log n\right)$ when $k$ is small.
Any help? Thanks!
| You're doing the right things, as far as I can tell. Maybe you're not using $n = m \log m + 1/2$ or $n/\log n = m + o(m)$ in the right place? (The latter asymptotic comes from $\log n = \log m + \log(\log m + 1/(2m))$ and doing the division of $n$ by $\log n$. The dominant term is $m$.) At any rate, here's the derivation I get.
We need Stirling's approximation
$$\log (m+k)! = (m+k) \log(m+k) - (m+k) + \frac{1}{2}\log(m+k) + \frac{1}{2} \log (2\pi) + O\left(\frac{1}{m}\right)$$
and $$\log(m+k) = \log m + \log\left(1 + \frac{k}{m}\right) = \log m + \frac{k}{m} - \frac{k^2}{2m^2} + O\left(\frac{k^3}{m^3}\right).$$
We have
$$\log \frac{(m+k)^n}{(m+k)!} = n \log (m+k) - \log(m+k)!.$$
Let's take these two terms separately.
\begin{align*}
n \log (m+k) &= n \log m + \frac{nk}{m} - \frac{nk^2}{2m^2} + O\left(\frac{nk^3}{m^3}\right) \\
&= n \log m + k \log m + \frac{k}{2m} - \frac{nk^2}{2m^2} + O\left(\frac{k^3 \log n}{m^2}\right). \tag{1}
\end{align*}
And
\begin{align}
- \log(m+k)! = &-(m+k)\left(\log m + \frac{k}{m} - \frac{k^2}{2m^2} + O\left(\frac{k^3}{m^3}\right)\right) \\
& + (m+k) - \frac{1}{2}\left(\log m + \frac{k}{m} + O\left(\frac{k^2}{m^2}\right)\right) - \frac{1}{2} \log (2\pi) + O\left(\frac{1}{m}\right) \\
=&- (m+k)\log m -k - \frac{k^2}{m} + \frac{k^2}{2m} + m+k - \frac{\log m}{2} - \frac{k}{2m} \\
&- \frac{1}{2} \log (2\pi) + O\left(\frac{k^3}{m^2}\right) + O\left(\frac{1}{m}\right)\\
=& - (m+k) \log m + m - \frac{\log(2 \pi m)}{2} - \frac{k^2}{2m} - \frac{k}{2m} + O\left(\frac{k^3}{m^2}\right) + O\left(\frac{1}{m}\right). \tag{2}
\end{align}
Adding Eqs. (1) and (2), we get
\begin{align}
&\log \frac{(m+k)^n}{(m+k)!} \\
&= n \log m - m \log m + m - \frac{\log(2 \pi m)}{2} - \frac{(m+n)k^2}{2m^2} + O\left(\frac{k^3 \log n}{m^2}\right) + O\left(\frac{1}{m}\right),
\end{align}
which is the expression given by Concrete Mathematics.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$
Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$.
I've been struggling with some of these Galois group questions.
| Let $f=(x^3-1)(x^2-5)\in\mathbb{Q}[x]$. Note that
$$\begin{align}f&=(x-1)(x^2+x+1)(x^2-5)\\ &=(x-1)(x-\zeta_3)(x-\zeta_3^2)(x-\sqrt{5})(x+\sqrt{5})\end{align}$$
where $\zeta_3=\frac{-1+\sqrt{3}}{2}$ is a primitive cube root of unity. Thus, the splitting field of $f$ over $\mathbb{Q}$ is $$K=\mathbb{Q}(\zeta_3,\sqrt{5})=\{a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5}\mid a,b,c,d\in\mathbb{Q}\}.$$
Suppose that $\phi:K\to K$ is an automorphism of $K$ fixing $\mathbb{Q}$. Then
$$\begin{align}\phi(a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5})&=\phi(a)+\phi(b)\phi(\zeta_3)+\phi(c)\phi(\sqrt{5})+\phi(d)\phi(\zeta_3)\phi(\sqrt{5}) \\ &=a+b\phi(\zeta_3)+c\phi(\sqrt{5})+d\phi(\zeta_3)\phi(\sqrt{5})\end{align}$$
so that $\phi$ is determined entirely by the values of $\phi(\zeta_3)$ and $\phi(\sqrt{5})$; that is, if $\psi$ is an automorphism of $K$ such that $\psi(\zeta_3)=\phi(\zeta_3)$ and $\psi(\sqrt{5})=\phi(\sqrt{5})$, then in fact $\psi=\phi$.
Because $\zeta_3^2+\zeta_3+1=0$ and $(\sqrt{5})^2-5=0$, we must have that
$$0=\phi(0)=\phi(\zeta_3^2+\zeta_3+1)=\phi(\zeta_3)^2+\phi(\zeta_3)+1$$
and
$$0=\phi(0)=\phi((\sqrt{5})^2-5)=\phi(\sqrt{5})^2-5$$
so that $\phi(\zeta_3)$ must be one of the two roots of $x^2+x+1$, namely $\zeta_3$ and $\zeta_3^2$, and $\phi(\sqrt{5})$ must be one of the two roots of $x^2-5$, namely $\sqrt{5}$ and $-\sqrt{5}$.
Thus, there are at most four elements of $\operatorname{Gal}(K/\mathbb{Q})$ (there are two choices for where $\zeta_3$ goes, and two choices for where $\sqrt{5}$ goes):
$$\begin{align}
\operatorname{id}_K(\zeta_3)&=\zeta_3 & \operatorname{id}_K(\sqrt{5})&=\sqrt{5}\\
\phi(\zeta_3) &= \zeta_3^2 & \phi(\sqrt{5})&=\sqrt{5}\\
\rho(\zeta_3) &= \zeta_3 & \rho(\sqrt{5})&=-\sqrt{5}\\
\sigma(\zeta_3) &= \zeta_3^2 & \sigma(\sqrt{5})&=-\sqrt{5}\\
\end{align}$$
Check that all four do in fact define valid automorphisms, and see what the group structure on $\operatorname{Gal}(K/\mathbb{Q})$ is by composing them and seeing what comes up. For example, we have that $\phi\circ \rho=\sigma$, because
$$(\phi\circ\rho)(\zeta_3)=\phi(\rho(\zeta_3))=\phi(\zeta_3)=\zeta_3^2=\sigma(\zeta_3)$$
$$(\phi\circ\rho)(\sqrt{5})=\phi(\rho(\sqrt{5}))=\phi(-\sqrt{5})=-\phi(\sqrt{5})=-\sqrt{5}=\sigma(\sqrt{5})$$
Are there any elements of order 4?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
"Ballot numbers" sum up to Catalan numbers Summing certain numbers and comparing the results with OEIS, I found that
$
\sum_{k=1}^n \frac{k^2}{n} \binom{2n-k-1}{n-1} = C_{n+1} - C_{n},
$
where $C_n$ denotes the $n^{\textrm{th}}$ Catalan Number. How can I prove this equation? And is there any combinatorial interpretation?
Some background information: The number $\frac{k}{n} \binom{2n-k-1}{n-1}$ denotes the number of unranked trees of size $n$ with a root degree $k$ (these numbers are known as ballot numbers, see e.g. the book Analytic Combinatorics from Flajolet and Sedgewick, page 68). So one must have $\sum_{k=1}^n \frac{k}{n} \binom{2n-k-1}{n-1} = C_{n}$, since there are $C_n$ many trees of size $n$.
| Notice that:
$$
\sum_{k=1}^n \frac{k^2}{n} \binom{2n-k-1}{n-1} \stackrel{k \to n+1-k}{=} \sum_{k=1}^n \frac{(n+1-k)^2}{n} \binom{n+k-2}{n-1} \stackrel{\binom{a}{b} = \binom{a}{a-b}}{=} \sum_{k=1}^n \frac{(n+1-k)^2}{n} \binom{n+k-2}{k-1} = \sum_{k=0}^{n} \frac{(n-k)^2}{n} \binom{n+k-1}{k}
$$
This this nothing but a convolution of two sequences $a_{k} = \frac{k^2}{n}$ and $b_{k} = \binom{n+k-1}{k}$. Therefore the generating function for the sum $c_n = \sum_{k=0}^{n} a_{n-k} b_k$ is the product of generating functions of $a_k$ and $b_k$:
$$
\sum_{k=0}^\infty x^k a_k = \frac{1}{n} \sum_{k=0}^\infty x^k k^2 = \frac{1}{n} \left( x\frac{\mathrm{d}}{\mathrm{d} x}\right)^2 \sum_{k=0}^\infty x^k = \frac{1}{n} \frac{x(1+x)}{(1-x)^3}
$$
$$
\sum_{k=0}^\infty \binom{n+k-1}{k} x^k = \frac{1}{(1-x)^n}
$$
Thus:
$$
c_n = [x]^n \left( \frac{1}{n} \frac{x(1+x)}{(1-x)^3} \frac{1}{(1-x)^n} \right) = [x]^n \left( \frac{1}{n} \frac{x}{(1-x)^{n+3}} + \frac{1}{n} \frac{x^2}{(1-x)^{n+3}} \right) = \frac{1}{n} \left(\binom{2n+1}{n-1} + \binom{2n}{n-2} \right)
$$
Now, using
$$
\binom{2n}{n-2} = \binom{2n+1}{n-1} - \frac{2n}{n-1} \binom{2n-1}{n-2}
$$
which is easily proven using recurrence relation for factorials comprising binomial coefficients, we get
$$
c_n = \frac{2}{n} \binom{2n+1}{n-1} - \frac{2}{n-1} \binom{2n-1}{n-2} = C_{n+1} - C_{n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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} |
Two sums for $\pi$ I was looking for a simple way to evaluate the integral $\int_0^\infty \frac{\sin x}{x}dx$ ( a belated look at this question). There are symmetries to be exploited, for one thing. So I had an idea, but the idea depends, optimistically, on two expressions for $\pi$ that I cannot prove, and my question is whether anyone sees reasonable proofs of either expression (1, 2 below).
Put
$$\int_0^\infty \frac{\sin x}{x}dx = \int_0^\pi \sum_{k=0}^{k=\infty} \left\{\frac{\sin x}{x + 2 k \pi }-\frac{\sin x }{x+2k\pi+\pi}\right\} \, dx$$
The intuition behind this step is that as we make our way around the unit circle from 0 to $\pi$, for each angle $\alpha$ we have an angle $\alpha + \pi$ for which the value of $\frac{\sin x} {x}$will be smaller and negative. No need to integrate beyond $\pi$, and for angles greater than $2\pi$ we are simply dividing up the original integral's range by adding multiples of $\pi$ to the value of $x$.
WLOG we can form Riemann sums by dividing the range of integration into an even number of equi-angle subranges $dx_n$. For example, if $n = 6$, we have
$$\int_0^\pi \sum_{k=0}^{k=\infty}\sin x \left\{\frac{1}{x + 2 k \pi }-\frac{1 }{x+2k\pi+\pi}\right\} \, dx$$
$$\approx \left\{ \sum_{k=0}^\infty \sum_{n=1}^6 \sin \left( \frac{n\pi}{6}\right)\left(\frac{1}{\pi}\right) \left(\frac{1}{\frac{n}{6} + 2k}-\frac{1}{\frac{n}{6}+2k+1}\right) \right\} \left\{ \frac{\pi}{6} \right\}.$$
Taking this as true, we note that pairs of summands of the Riemann sums symmetric about $\frac{\pi}{2}$ sum to $1$, and the value of the summand at $\frac{\pi}{2}$ is $1/2$. Again, taking this as true, we have that
$$\sum f(x_n) = \frac{1}{2} + 1\cdot \frac{n-2}{2} + \sin\pi\cdot f(\pi) = \frac{n-1}{2},$$
and
$$\sum f(x_n) \, dx_n = \left(\frac{n-1}{2}\right) \cdot \frac{\pi}{n}.$$
So for the approximation above for $n = 6$, we expect a value of about $\frac{5 \pi}{12}$ for high $k$.
In words, we are only counting n - 1 summands, since the last summand at $\pi$ is $0$. Since we divided the interval into $n$ subintervals, we are stuck with the sum above and the conditional limit (conditioned at least on the truth of the two limits below):
$$\int_0^\infty \frac{\sin x}{x} \, dx = \lim_{n = 1}^\infty \frac{n-1}{2}\cdot \frac{\pi}{n} = \frac{\pi}{2}.$$
The problem is this: we need to prove, to begin with, that
$$\sum_{k=0}^\infty \left(\frac{ 1}{\pi/2 + 2\pi k} - \frac{1}{\pi/2 + 2k\pi + \pi}\right) = \frac{1}{2}$$
or
$$ \frac{1}{\pi}\sum \frac{1}{1/2 + 2k}- \frac{1}{1/2 + 2k\pi +1}= \frac{1}{2},$$
that is,
(1) $\sum_{k=0}^\infty \frac{4}{3+16k+16k^2}= \frac{\pi}{2}$.
We must also prove for $0< j < n$ that
(2) $\sin (\frac{j\pi}{n})( \sum_{k=0}^{\infty} (\frac{ 1}{\frac{j\pi}{n} + 2\pi k} - \frac{1}{\frac{j\pi}{n} + 2k\pi + \pi}) + \sum_{k=0}^{\infty} (\frac{ 1}{\frac{(n-j)\pi}{n} + 2\pi k} - \frac{1}{\frac{(n-j)\pi}{n} + 2k\pi + \pi}))= 1$.
or , factoring $\frac{1}{ \pi}$ out of the sum and mutiplying:
$$\sin \left( \frac{j\pi}{n}\right)\left( \sum_{k=0}^\infty \left(\frac{1}{\frac{j}{n} + 2 k} - \frac{1}{\frac{j}{n} + 2k + 1}\right) + \sum_{k=0}^\infty \left( \frac{ 1}{\frac{(n-j)}{n} + 2 k} - \frac{1}{\frac{(n-j)}{n} + 2k + 1}\right)\right)= \pi.$$
(1) might be a special case of something at Wolfram's site, but I didn't spot it. (2) looks messy, but maybe induction?
If this is otherwise correct, I think one could work back from the relations for $\pi$ and establish the value of the integral. It is important not to forget that k starts at 0.
Edit: the following is equivalent to (2) and maybe easier to scan. For $0<r<1$,
$$(2) \sum_{k=0}^{\infty}\frac {2(\sin \pi r)(1+4k+4k^2-r+r^2) }{(1+2k+r)(1+2k-r)(2+2k-r)(2k+r)} = \pi$$
This was derived assuming r rational, but I don't think it matters.
| $$
\sum_{k=0}^N \frac{1}{\alpha + 2 k} = \frac{\Psi(N+1+\alpha/2) - \Psi(\alpha/2)}{2}
$$
where $\Psi(t) = \ln(t) + O(1/t)$ as $t \to \infty$, so
$$\sum_{k=0}^\infty \left(\frac{1}{1/2 + 2 k} - \frac{1}{3/2 + 2 k} \right) = \frac{\Psi(3/4) - \Psi(1/4)}{2} = \frac{\pi}{2}$$
However, I think the evaluation of $\Psi(3/4) - \Psi(1/4)$ may be just as hard as your integral.
EDIT: It turns out that $\Psi$ has the integral form
$$ \Psi(t) = \int_0^\infty \left( e^{-s} - \frac{1}{(s+1)^t} \right) \frac{ds}{s}$$
so that
$$ \Psi(3/4) - \Psi(1/4) = \int_0^\infty \left((s+1)^{-3/4} - (s+1)^{-1/4}\right) \frac{ds}{s} = \left. 4 \arctan((s+1)^{1/4}) \right|_0^\infty = \pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How can I calculate $\lim_{x \to 0} \log(\cos(x))/\log(\cos(3x))$ without l'Hopital? How can I calculate the following limit without using, as Wolfram Alpha does, without using l'Hôpital?
$$
\lim_{x\to 0}\frac{\log\cos x}{\log\cos 3x}
$$
| Another way is via the approximations $\cos(x) = 1-\frac{x^2}{2}+O(x^4)$ and $\log(1+x) = x+O(x^2)$:
$$\begin{align}\frac{\log(\cos(x))}{\log(\cos(3x))} &= \frac{\log(1-\frac{x^2}{2}+O(x^4))}{\log(1-\frac{(3x)^2}{2}+O(x^4))} \\
&= \frac{-\frac{x^2}{2}+O(x^4)}{-\frac{(3x)^2}{2}+O(x^4)} \\
&= \frac{-\frac{x^2}{2}}{-\frac{x^2}{2}}\frac{1+O(x^2)}{9+O(x^2)} \\
&= \frac{1}{9}+O(x^2)\end{align}$$
And now the limit as $x\rightarrow0$ is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Factoring by grouping: $x^4 - y^4 -4x^2 + 4$ Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms.
Thank you.
| $x^4 - y^4 -4x^2 + 4 $
$= (x^4 -4x^2 + 4) - y^4$
$= [(x^2)^2 -2(2x^2) + 2^2] - (y^2)^2$
The expression $(x^2)^2 -2(2x^2) + 2^2$ has the form: $a^2 -2ab + b^2 = (a - b)^2$
where $a=x^2$ and $b=2$
Thus, $(x^2)^2 -2(2x^2) + 2^2 = (x^2 - 2)^2$
Hence
$[(x^2)^2 -2(2x^2) + 2^2] - (y^2)^2$
$= (x^2 - 2)^2 -(y^2)^2$
which is a difference of two squares i.e. has the form: $a^2-b^2 = (a+b)(a-b)$
where $a= x^2-2$ and $ b = y^2$
Thus, $(x^2 - 2)^2 -(y^2)^2 = (x^2 -2 + y^2)(x^2 -2 - y^2)$
$\therefore $ $x^4 - y^4 -4x^2 + 4 = (x^2 -2 + y^2)(x^2 -2 - y^2)$ Answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the indefinite integral of $1/(16x^2+20x+35)$ Here is my steps of finding the integral, the result is wrong but I don't know where I made a mistake or I may used wrong method.
$$
\begin{align*}
\int \frac{dx}{16x^2+20x+35}
&=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{35}{16}} \\
&=\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}} \\
&=\frac{1}{16}\int \frac{dx}{(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2}\\
&=\frac{1}{16}\frac{4}{5}\textstyle\arctan ((x+\frac{\sqrt{10}}{4})\cdot \frac{4}{5}) \\
&=\frac{1}{20}\textstyle\arctan(\frac{4x+\sqrt{10}}{5})
\end{align*}
$$
| Your problem is this step:
$$\frac{1}{16}\int \frac{dx}{x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}}
=\frac{1}{16}\int \frac{dx}{(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2}$$
for which you use this equality:
$$\textstyle x^2+\frac{20}{16}x+\frac{10}{16}+\frac{25}{16}
=(x+\frac{\sqrt{10}}{4})^2+(\frac{5}{4})^2$$
but that's just not true! The right-hand side expands into $x^2 + 2\frac{\sqrt{10}}{4}x + \frac{10}{16} + \frac{25}{16}$: as you can see the $x$ term is wrong, and the square root is unnecessary. Just to remind you, the general rule for completing the square is:
$$\textstyle x^2 + bx + c = (x + \frac{b}{2})^2 + c - \frac{b^2}{4}$$
No square roots anywhere!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Steps to get Inverse of Pentagonal I have solved http://projecteuler.net/problem=44 by getting the inverse equation from Wikipedia http://en.wikipedia.org/wiki/Pentagonal_number:
Pentagonal:
$f(n) = \frac{n(3n - 1)}{2}$
Inverse Pentagonal:
$n = \frac{\sqrt{24f(n) + 1}+1}{6}$
am interested in the steps from Pentagonal equation (quadratic?) to the Inverse.
I note that it is similar to What is the inverse of $f(n)=\frac{n^2+n}{2}$? and I've tried the same strategy:
$f(n) = \frac{n(3n - 1)}{2}$
*6 + 1 on each side
$6f(n) + 1 = 9n^2 -3n + 1$
but this isn't correct because I want:
$6f(n) + 1 = 9n^2 -6n + 1$
to give:
$(3n-1)^2$ on the right hand side
| To imitate the procedure used in the solution of the other problem, starting from
$$2f(n)=3n^2-n,$$
multiply both sides by $12$. We get
$$24f(n)=36n^2-12n.$$
Note that $36n^2-12n=(6n-1)^2-1$. The rest is easy. We get $(6n-1)^2=24f(n)+1$, then $6n-1=\sqrt{24f(n)+1}$, then $6n=\sqrt{24f(n)+1}+1$.
Another way to solve the same problem is to write our equation as
$$3n^2-n-2f(n)=0,$$
and use the Quadratic Formula.
Remark: Look at the quadratic equation $ax^2+bx+c=0$, where $a\ne 0$. Multiply both sides by $4a$. We get the equivalent equation
$$4a^2x^2+4abx+4ac=0.$$
Note that $4a^2x^2+4abx=(2ax+b)^2-b^2.$
So quickly we arrive at the equation
$$(2ax+b)^2=b^2-4ac.$$
From this we conclude that
$$2ax+b=\pm\sqrt{b^2-4ac},$$
and then straightforward algebra yields
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$$
the important Quadratic Formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$ I recently came across this problem
Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }} = 3$
After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.
So my question is
Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.
Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .
If the solution from that book is required I can post it but it is not complete as I mentioned.
Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist.
| $x + 1 = \sqrt {1 + x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{}.....}}}$. Put $x=2$ gives you the solution. For proof see http://zariski.files.wordpress.com/2010/05/sr_nroots.pdf
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/165671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 0
} |
Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead of this step.
| $$x_n=\bigg(1+\frac{1}{n}\bigg)^n\longrightarrow x_{n+1}=\bigg(1+\frac{1}{n+1}\bigg)^{n+1}$$
$$\frac{x_{n+1}}{x_{n}}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}}=\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)=\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)$$
$$=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)≥\bigg(1-\frac{n}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg)$$
$$≥^*\frac{1}{1+\frac{1}{n+1}}\bigg(1+\frac{1}{n+1}\bigg)≥1$$
It means that your sequence is increasing.
≥*: $$(n+2)(n^2+n+1)=(n+2)\bigg((n+1)^2-n\bigg)≥(n+1)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 14,
"answer_id": 3
} |
Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
| By AM-GM one has $\,abc\leq\left(\frac{a+b+c}{3}\right)^3=1$. And with the
Cauchy–Bunyakovsky–Schwarz inequality we obtain
$$\sum_{cyc}\frac{1}{2a^2b+1}=\sum_{cyc}\frac{c^2}{2a^2c^2b+c^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2c+a^2)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab+a^2)}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/168704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
$(\sin\theta+\cos\theta)^2=1+\sin2\theta$ 49) $(\sin\theta+\cos\theta)^2=1+\sin2\theta$
Left Side:
\begin{align*}
(\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta
\end{align*}
This can either be $1$ or I can power reduce it. I don't know.
Right Side:
\begin{align*}
1+\sin2\theta=1+2\sin\theta\cos\theta
\end{align*}
Thank you!
| Your left hand side isn't good: $(a + b)^2 = a^2 + 2ab + b^2$. After that use $\sin^2(\theta) + \cos^2(\theta) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int 1/(1+x^{2n})\,dx$ for an arbitrary positive integer $n$? How to find
$$\int\dfrac{dx}{1+x^{2n}}$$
where $n \in \mathbb N$?
Remark
When $n=1$, the antiderivative is $\tan^{-1}x+C$. But already with $n=2$ this is something much more complicated. Is there a general method?
| I realized after I wrote this up that this is given in one of the papers mentioned by Dave L. Renfro, but I did all this work and the approach is not exactly the same, so here goes.
We wish to evaluate
$$
\int \frac{1}{1+x^n}\ dx.
$$
We will do this by partial fraction decomposition. Note that the roots of $1+x^n$ are the $2n$-th roots of unity that are not $n$-th roots of unity. That is to say $x^{2n}-1=(x^n-1)(x^n+1)$. It follows that the set of roots of $1+x^n$ is
$$
\left\{\exp\left(\frac{(2k-1)\pi i}{n}
\right):0\leq k\leq n-1\right\}.
$$
If we consider the roots (excluding -1 if $n$ is odd) we have that
$$
\left(x-\exp\left(\frac{(2k+1)\pi i}{n}
\right)\right)\left(x-\exp\left(\frac{(2(n-k)-1)\pi i}{n}
\right)\right)=\left(x-\exp\left(\frac{(2k+1)\pi i}{n}
\right)\right)\left(x-\exp\left(\frac{-(2k+1)\pi i}{n}
\right)\right)
$$
$$
=x^2-\left(\exp\left(\frac{(2k+1)\pi i}{n}\right)+\exp\left(\frac{-(2k+1)\pi i}{n}
\right)
\right)x+1=x^2-2\cos\left(\frac{(2k+1)\pi}{n}\right)x+1.
$$
Let $x_k=\frac{(2k+1)\pi}{n}$ and $\alpha_k=\exp((2k+1)\pi i/n)$, then by partial fraction decomposition (for $n$ even) we have that
$$
\frac{1}{1+x^n}=\sum_{k=0}^{n/2-1}\frac{a_kx+b_k}{x^2-2\cos(x_k)x+1}=\sum_{k=0}^{n/2-1}\frac{(a_kx+b_k)\prod_{\overset{j\neq k}{j\neq n-1-k}}(x-\alpha_j)}{1+x^n}=\sum_{k=0}^{n/2-1}\frac{\frac{a_kx+b_k}{x-\alpha_{n-1-k}}\prod_{j\neq k}(x-\alpha_j)}{1+x^n}.
$$
Furthermore
$$
1=\sum_{k=0}^{n/2-1}\frac{a_kx+b_k}{x-\alpha_{k}^{-1}}\prod_{j\neq k}(x-\alpha_j).
$$
If we set $x=\alpha_k$ for $0\leq k\leq n/2$ we obtain
$$
\frac{a_k\alpha_k+b_k}{\alpha_k-\alpha_{k}^{-1}}\prod_{j\neq k}(\alpha_k-\alpha_j)=1.
$$
Note that
$$
\prod_{k=1}^{n-1}(x-\exp(k2\pi i/n))=(1+x+\cdots+x^{n-1})
$$
so
$$
\prod_{k=1}^{n-1}(1-\exp(k2\pi i/n))=n.
$$
Furthermore
$$
\prod_{j\neq k}(\alpha_k-\alpha_j)=\prod_{j\neq k}\alpha_k(1-\frac{\alpha_j}{\alpha_k})=\alpha_k^{n-1}\prod_{k=1}^{n-1}(1-\exp(k2\pi i/n))=n\alpha_k^{n-1}=-n\alpha^{-1}.
$$
So we are left with
$$
\frac{(a_k\alpha_k+b_k)(-n\alpha_k^{-1})}{\alpha_k-\alpha_{k}^{-1}}=1.
$$
and
$$
-n(a_k+\alpha_k^{-1}b_k)=\alpha_k-\alpha_{k}^{-1}=2i\sin(x_k)
$$
implying that
$$
a_k+\cos(x_k)b_k-i\sin(x_k)b_k=-\frac{2i}{n}\sin(x_k).
$$
Hence $b_k=\frac{2}{n}$ and $a_k=-\frac{2}{n}\cos(x_k)$. So for even $n$ we have.
$$
\frac{1}{1+x^n}=-\frac{1}{n}\sum_{k=0}^{n/2-1}\frac{2\cos(x_k)x-2}{x^2-2\cos(x_k)x+1}
$$
If $n$ is odd we have the additional term
$$
\frac{a}{1+x}
$$
and it follows that $a\prod_{\alpha_k\neq 1}(x-\alpha_k)=a(1-x+\cdots-x^{n-2}+x^{n-1})=1$. Setting $x=-1$ we obtain $a=\frac{1}{n}$.
Noticing that
$$
\frac{2\cos(x_k)x-2}{x^2-2\cos(x_k)x+1}=\frac{\cos(x_k)(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}+\frac{2\cos^2(x_k)-2}{(x-\cos(x_k))^2+1-\cos^2(x_k)}
$$
$$
=\frac{\cos(x_k)(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}-2\frac{\sin^2(x_k)}{(x-\cos(x_k))^2+\sin^{2}(x_k)}
$$
$$
=\cos(x_k)\frac{(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}-2\sin(x_k)\frac{\csc(x_k)}{(\frac{x-\cos(x_k)}{\sin(x_k)})^2+1}.
$$
So we have for even $n$
$$
\int\frac{1}{1+x^n}\ dx=-\frac{1}{n}\sum_{k=0}^{n/2-1}\left\{\cos(x_k)\int\frac{(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}\ dx-2\sin(x_k)\int\frac{\csc(x_k)}{(\frac{x-\cos(x_k)}{\sin(x_k)})^2+1}\ dx\right\}
$$
$$
=-\frac{1}{n}\sum_{k=0}^{n/2-1}\left\{\cos(x_k)\log|x^2-2\cos(x_k)x+1|-2\sin(x_k)\arctan\left(\frac{x-\cos(x_k)}{\sin(x_k)}\right)\right\},
$$
and for odd $n$
$$
\int\frac{1}{1+x^n}\ dx=\frac{1}{n}\log|x+1|-\frac{1}{n}\sum_{k=0}^{(n-1)/2-1}\left\{\cos(x_k)\log|x^2-2\cos(x_k)x+1|-2\sin(x_k)\arctan\left(\frac{x-\cos(x_k)}{\sin(x_k)}\right)\right\}
$$
where $x_k=(2k+1)\pi/n$, $n\in\mathbb{Z}_{>0}$.
| {
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"url": "https://math.stackexchange.com/questions/171024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Verify trigonometry equation $\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A$ Sorry for asking so many of these type of questions.
How would I verify the following trigonometry identity:
$$\frac{\sin A+\tan A}{\cot A+\csc A}=\sin A \tan A.$$
My work is
$$\frac{\sin A + \frac{\sin A}{\cos A}}{\frac{\cos A}{\sin A} + \frac{1}{\sin A}}.$$
Do I have to use a common denominator between the sin and tan to solve the identity?
| For what it's worth (at this late date)...
Since cotangent and cosecant are linked by the Pythagorean Identity, the "conjugate factor" method is helpful:
$$
\frac{\sin A + \tan A}{\cot A + \csc A} \cdot \frac{\cot A - \csc A}{\cot A - \csc A} = \frac{(\sin A + \tan A)(\cot A - \csc A)}{\cot^2 A - \csc^2 A}$$
$$= \frac{\sin A \cot A - \sin A \csc A + \tan A \cot A - \tan A \csc A}{-1} = - (\cos A - 1 + 1 - \frac{1}{\cos A} )$$
$$= \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A} = \sin A \tan A .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve for $+r$ ; $A=2\pi r^2+2\pi rh$
Solve for $+r$
$$A=2\pi r^2+2\pi rh$$ Since $2\pi$ is common on both sides of the $+$ so I will take it out
$$A=2\pi (r^2+rh)$$
Now, divide both sides by $2\pi$
$$\dfrac{A}{2\pi}=r^2+rh$$
Then, we can divide by the $h$
$$\dfrac{Ah}{2\pi}=r^2+r$$
Then;
$$0=r^2+r-\dfrac{Ah}{2\pi}$$
Quadratic formula $=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=1,b=1,c=?$ What are the values of $a,b,c$
| No! $\sqrt{r^2 + r} \neq 2r.$
What you have is a polynomial equation $r^2 + r - \dfrac{Ah}{\pi} = 0.$
Since this is a homework, I will only give a hint: the two roots $r_1, r_2$ of quadratic $ax^2 + bx + c = 0$ are $$ r_{1,2} = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complete the square and write in standard form for $9y^2-6y-9-x=0$
Complete the square and write in standard form: $x-a=A(y-b)^2$
$9y^2-6y-9-x=0$
I do not know how to complete the square with $4$ terms. I started off like: $$9y^2-6y-9=x$$
I don't know whether to start trying to complete the square or if I would do this: $$9y^2-6y=x+9$$
And then complete the square. If someone can either tell me the first one or second one that would be very helpful. Thanks!
EDIT:
$$9y^2−6y=x+9 \\
(3y)^2 - 2 \cdot 3 y + 1^2 = x + 9 + 1 \\
(3y - 1)^2 = x + 10 \\
9\left ( y - {1 \over 3}\right )^2 = x + 10$$
| $$ 9y^2−6y=x+9 \\
(3y)^2 - 2 \cdot 3 y + 1^2 = x + 9 + 1 \\
(3y - 1)^2 = x + 10 \\
9\left ( y - {1 \over 3}\right )^2 = x + 10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Verify $\frac{\sin^3A + \cos^3A}{\sin A + \cos A} = 1 - \sin A\cos A$ How can I verify the following trigonometric identity?
$$\frac{\sin^3 A + \cos^3 A}{\sin A+\cos A} = 1-\sin A\cos A.$$
My work so far is
$$\begin{align*}
&\frac{\sin\cos(\sin^2+\cos^2)}{\sin+\cos}\\
&\frac{\sin\cos(1)}{\sin+\cos}
\end{align*}$$
| $$
a^3+b^3 = (a+b)(a^2-ab+b^2).
$$
Use that to factor the numerator.
Also, it is not correct to say that $s^3+c^3 = sc(s^2+c^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$
I have not been sucessful to do this.
| This is an infinite series, but it is not geometric because there is no common ratio.
So, let
$$S = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + \dots + n \cdot \frac{1}{2^n} + \dotsb.$$
Multiplying this equation by 2, we get
$$2S = 1 \cdot 1 + 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} + \dots + n \cdot \frac{1}{2^{n - 1}} + \dotsb.$$
Subtracting these equations, we find
$$S = 1 \cdot 1 + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + \dots + 1 \cdot \frac{1}{2^n} + \dotsb.$$
So, $S=1+\dfrac12+\dfrac14+\dfrac18+\dotsb.$ Even though it didn't begin as one, we've managed to rewrite $S$ as an infinite geometric series. Thus, we may easily find its sum:
$$S = \frac{1}{1 - 1/2} = \boxed{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a number with $3^n$ equal digits is divisible by $3^n$ Prove a number with $3^n$ equal digits is divisible by $3^n$.
My thoughts about the problem are: a number with $3^n$ equal digits $d$ is equal to $d\frac {10^{3^n} - 1} {9}$. We use Lifting The Exponent lemma, or plain induction.
| $10^{3^n}-1$
=$-1+(1+3^2)^{3^n}$
=$-1+ 1+(3^nC_1)(3^2)+(3^nC_2)(3^2)+\cdot\cdot\cdot+(3^2)^{3^n}$
which will be divisible by $3^{n+2}$ if n+4≥n+2 , (n-1+6)≥n+2, $\cdot\cdot\cdot\ , 2.3^n≥n+2$ which is true for n≥0.
So, $\frac{10^{3^n}-1}{9}$ will be divisible by $3^n$ for n≥0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Inequality question: Sum of squares and determining minimum value. Given that $a^2 + b^2 + c^2 = 6\;$, determine the minimum value of $ab + bc + ca\;$.
I know that $a^2 + b^2 + c^2 ≥ ab + bc + ca\;$, so that means $6 ≥ ab + bc + ca\;$, but I don't know where to go from there.
| You need a lower bound for $ab+bc+ca$. The inequality $ab+bc+ca \le a^2+b^2+c^2$ goes in the "wrong" direction. It is useful for producing an upper bound for $ab+bc+ca$.
Use instead the fact that
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=6+2(ab+bc+ca).$$
So to minimize $ab+bc+ca$ we must minimize $(a+b+c)^2$. Clearly $(a+b+c)^2 \ge 0$ always. But we can arrange for $a+b+c$ to be $0$ in infinitely many ways. One example is $a=\sqrt{3}$, $b=-\sqrt{3}$, $c=0$. (Any point where the plane $x+y+z=0$ meets the sphere $x^2+y^2+z^2=6$ will do it.)
So the minimum value of $ab+bc+ca$ under our constraint is $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ We have $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$. I want to expand this as a Laurent series in $z_0=2$ on $\{4<|z-2|<\infty\}$.
The partial decomposition is:
$$f(z)=\frac{1}{4}\frac{1}{z-2}-\frac{1}{4}\frac{1}{z+2}+\frac{1}{6-z}$$
In my reference, they expand $\frac{1}{z+2}$ and $\frac{1}{6-z}$ with the help of the geometric series, but they leave $\frac{1}{z-2}$ as it is. Why?
| $$\frac{1}{z+2}=\frac{1}{(z-2)+4}=\frac{1}{4}\frac{1}{1+\left(\frac{z-2}{4}\right)}=\frac{1}{4}\left(1-\frac{z-2}{4}+\frac{(z-2)^2}{16}-\frac{(z-2)^3}{64}+\cdots\right)$$
$$\frac{1}{6-z}=-\frac{1}{(z-2)-4}=\cdots$$
and now just mimic the first development above of the given fraction in powers of $\,z-2\,$
Pay attention to the fact that the first (and also the second one, btw) development is valid for
$$\left|\frac{z-2}{4}\right|<1\Longleftrightarrow|z-2|<4$$
Once you've done all the above just add the different fractions which form part of $\,f(z)\,$
| {
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"url": "https://math.stackexchange.com/questions/174460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove $\cos 3x =4\cos^3x-3\cos x$ How would I solve the following double angle identity.
$$\cos 3x =4\cos^3x-3\cos x $$
I know $\,\cos 3x = \cos(2x+x)$
So know I have $\,\cos 2x +\cos x \,$ , Which is $\,(2\cos^2x-1)\cos x$
But I am not sure what to do next.
| \begin{eqnarray}
\cos(3x) &=& \cos(2x+x)\\
&=& \cos(2x)\cos x - \sin(2x)\sin x\\
&=& (\cos^2 x - \sin^2 x)\cos x - 2\sin^2 x\cos x\\
&=& \cos^3 x -(1-\cos^2 x)\cos x - 2 (1 -\cos^2 x)\cos x\\
&=& 2 \cos^3 x -\cos x + \cos^3 x -2 \cos x\\
&=& 4 \cos^3 x - 3 \cos x
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
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Could You check whether this expression is nonnegative? I'm trying to determine if $$\bigl(x+y)^4(y+z)^4(z+x)^4 \geq$$
$$8x^2y^2z^2\bigl((x+y)^2 + (y+z)^2\bigr)\bigl((y+z)^2 + (z+x) ^2\bigr)\bigl((z+x)^2 + (x+y)^2\bigr)$$
for $x,y,z>0$.
| Updated Version: Call the given polynomial $f(x,y,z)$. Note that this is a symmetric polynomial--that is, if we permute $x,y,z$ in the polynomial, we end up with the same polynomial--so without loss of generality, we may assume that $x,y\geq z$.
Another nice observation is that every term of the expanded polynomial will have the same total degree--namely $12$--so the polynomial is homogeneous, too. This suggests that we might want to rewrite $x=\alpha z$, $y=\beta z$, where $\alpha,\beta\geq 1$ (since $x,y\geq z>0$), for then each term will have a factor of $z^{12}$, and our $3$-variable problem reduces to the (less daunting) $2$-variable problem of determining whether $$\frac{f(\alpha z,\beta z,z)}{z^{12}}\geq 0$$ for all $\alpha,\beta\geq 1$.
To make our lives slightly easier (so we aren't working with so much at once and everything fits on one line), let's set $$p(x,y,z):=\bigl((x+y)(x+z)(y+z)\bigr)^4$$ and $$q(x,y,z):=8x^2y^2z^2\bigl((x+y)^2+(x+z)^2\bigr)\bigl((x+y)^2+(y+z)^2\bigr)\bigl((x+z)^2+(y+z)^2\bigr),$$ so that $f=p-q$. Now, $$p(\alpha z,\beta z,z)=\bigl((\alpha+\beta)(\alpha+1)(\beta+1)\bigr)^4z^{12}$$ and $$q(\alpha z,\beta z,z)=8\alpha^2\beta^2\bigl((\alpha+\beta)^2+(\alpha+1)^2\bigr)\bigl((\alpha+\beta)^2+(\beta+1)^2\bigr)\bigl((\alpha+1)^2+(\beta+1)^2\bigr)z^{12}.$$
These are still messy, so let's perform another substitution for increased simplicity. Set $a:=\alpha+1$, $b:=\beta+1$, so $$p(\alpha z,\beta z,z)=\bigl((a+b-2)ab\bigr)^4z^{12}$$ and $$q(\alpha z,\beta z,z)=8(a-1)^2(b-1)^2\bigl((a+b-2)^2+a^2\bigr)\bigl((a+b-2)^2+b^2\bigr)(a^2+b^2)z^{12}.$$ Hence, we need to show that $$\bigl((a+b-2)ab\bigr)^4-8(a-1)^2(b-1)^2\bigl((a+b-2)^2+a^2\bigr)\bigl((a+b-2)^2+b^2\bigr)(a^2+b^2)\geq 0$$ for all $a,b\geq 2$.
At this point, though, I'm stuck. Perhaps this will give you (or someone else) an idea on how to get the rest of the way--or perhaps Mathematica will give you what you need. Sorry!
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $x^2 \cdot y^2 + x^2 + y^2 = c^2$ with $x$, $y$, $c \in \mathbb{Z}^+$ I am working on Project Euler 390.
The question is about triangles, and finding the area of a triangle with sides $\sqrt{a^2+1}, \sqrt{b^2+1}$ and $\sqrt{a^2+b^2}$, with $a, b \in \mathbb{Z}$. I have narrowed the problem down to solving the equation
$$x^2 \cdot y^2 + x^2 + y^2 = (2\cdot c)^2 \text{ with } x, y, c \in \mathbb{Z}^+$$
This is not a problem for $c \le 10^6$ (brute force), but I have to calculate up to $10^{10}$.
I would like to know how to solve these kind of equations, without any brute force attack. I have searched for a few days on Google, but the general solutions to the Diophantine equations I found were never appliable to my problem.
Any suggestions are welcome (even the name of this kind of equation), although I would appreciate not being told the answer to the problem.
| Let $y=x+a=>x^2\cdot y^2+x^2+y^2=x^2(x+a)^2+x^2+(x+a)^2=x^4+x^3(2a)+x^2(a^2+2)+x(2a)+a^2$
Let this be equal to $(x^2+px+q)^2$ where p,q are integers, so that $c=x^2+px+q$.
So, $x^4+x^3(2a)+x^2(a^2+2)+x(2a)+a^2=x^4+x^3(2p)+x^2(2q+p^2)+x(2pq)+q^2$
Expanding the RHS and comparing the coefficients of different powers of x,
coefficients of cubic power $=>p=a$
coefficients of square $=>a^2+2=2q+p^2=>q=1$ as $p=a$
coefficients of first degree $=>2pq=2a=>q=1$
constants $=>q^2=a^2=>a=±q=±1=>p=a=±1$
So, $y=x±1$.
The RHS($x^2+px+q$) reduces to $x^2±x+1$ but unfortunately this must be odd as x is even, so can not be equals to 2c.
So, there can be no solution following this approach.
Now, if we arrange $x^4+x^3(2a)+x^2(a^2+2)+x(2a)+a^2=(2c)^2$ as a quadratic equation of a, $a^2(x^2+1)+2a(x^3+x)+x^4+2x^2-4c^2=0$--->(1).
As a=x-y is integer, so the discriminant($D^2$) of (1) must be perfect square.
So,$D^2=(2x^3+2x)^2-4(x^2+1)(x^4+2x^2-4c^2)$
So, D is even=(2E, say)
$=>E^2=(x^3+x)^2-(x^2+1)(x^4+2x^2-4c^2)=4c^2+4c^2x^2-x^4-x^2$
$=>x^4-(4c^2-1)x^2+E^2-4c^2=0$--->(2)
As x is integer, the discriminant($D_1^2$) of (2) must be perfect square.
So, $D_1^2=(4c^2-1)^2-4.1.(E^2-4c^2)$
$=>D_1^2+(2E)^2=(4c^2+1)^2$, clearly $D_1$ is odd.
Now $4c^2+1$ can always be expressed as the sum of two squares(not necessarily in a unique way)=($r^2+s^2$, say), where r,s are integers.
If we take $E=rs =>D_1=r^2-s^2$
So,$x^2=\frac{4c^2-1 ± D_1}{2}=\frac{r^2+s^2-2±(r^2-s^2)}{2}=r^2-1\ or s^2-1$
Now, either I have made some mistake or I doubt the existence of a non-trivial solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Bounding an expression involving digamma function Let $\psi$ be the digamma function. I have a conjecture that
$$\frac ax > \log(x) - \psi(x)$$
holds for all $x > 0$ if (and only if) $a \ge 1$. I do not know how to prove it. Please help.
| Let
$$f(x) = \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x}.$$
Then for $x > 0$, we have
$$f'(x) = \frac{1}{x^2} - \frac{1}{4\sinh^2(x/2)} = \frac{1}{x^2}\left(1 - \left(\frac{x/2}{\sinh (x/2)} \right)^2 \right) > 0. $$
This shows that $f(x)$ is increasing. Also, it is easy to find that $f(0+) = 0$ and $\lim_{x\to\infty} f(x) = \frac{1}{2}$. Thus we find that $0 < f(x) < \frac{1}{2}$ for $0 < x < \infty$ and hence the Laplace transform $\mathcal{L}f(s) = \int_{0}^{\infty} f(x) \, e^{-sx} \; dx$ of $f(x)$ satisfies
$$ 0 < \mathcal{L}f(s) < \int_{0}^{\infty} \frac{1}{2} \, e^{-sx} \; dx = \frac{1}{2s}. $$
But we also have
$$ \begin{align*}
\mathcal{L}f(s)
&= \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x} \right) e^{-sx} \; dx\\
&= \left[ \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \right]_{0}^{\infty} \\
&\qquad + s \int_{0}^{\infty} \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \; dx \\
&= s \int_{0}^{\infty} \left( \frac{x}{2} - \log (sx) + \log s + \log \left(1 - e^{-x} \right) \right) e^{-sx} \; dx \\
&= \frac{1}{2s} + \gamma + \log s - s \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-(n+s)x}}{n} \; dx \\
&= \frac{1}{2s} + \gamma + \log s - \sum_{n=1}^{\infty} \frac{s}{n(n+s)} \\
&= \frac{1}{2s} + \log s - \psi_0(1+s),
\end{align*}$$
where we have used the fact that $\int_{0}^{\infty} e^{-x} \log x \; dx = -\gamma$ and
$$ \psi_0(1+x) = -\gamma + \sum_{n=1}^{\infty} \frac{x}{n(n+x)}. $$
Finally, since $\psi_0(1+s) = \frac{1}{s} + \psi_0 (s)$, we have
$$ \frac{1}{2s} < \log s - \psi_0(s) < \frac{1}{s}. $$
Therefore the inequality holds if $a \geq 1$ and only if $a > \frac{1}{2}$.
Now we prove that $a = 1$ is the sharp bound. We are to calculate the limit
$$\alpha = \lim_{x\to 0} x(\log x - \psi_0(x)).$$
But by our previous calculations,
$$\begin{align*}\alpha
&= \lim_{s\to 0} s(\log s - \psi_0(s)) \\
&= \lim_{s\to 0} s\left( \frac{1}{2s} + \mathcal{L}f(s) \right) \\
&= \frac{1}{2} + \lim_{s\to 0} s \int_{0}^{\infty} f(x) \, e^{-sx} \; dx \\
&= \frac{1}{2} + \lim_{s\to 0} \int_{0}^{\infty} f(u/s) \, e^{-u} \; du \qquad (u = sx) \\
&= \frac{1}{2} + \int_{0}^{\infty} \lim_{s\to 0} f(u/s) \, e^{-u} \; du \\
&= \frac{1}{2} + \int_{0}^{\infty} \frac{1}{2} \, e^{-u} \; du \\
&= 1.
\end{align*}$$
Thus if $\frac{a}{x} > \log x - \psi_0 (x)$ is true for all $x > 0$, then we must have $a \geq \alpha = 1$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Does this Dirichlet series converge to zero? Consider the periodic Dirichlet series that has this iterative definition:
$$\text{a1}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{6}}+...$$
$$\text{a2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1}{\sqrt{6}}+...$$
$$\text{a3}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2}{\sqrt{6}}+...$$
$$\text{a4}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2+a3}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2+a3}{\sqrt{6}}+...$$
$$...$$
continuing this iteration, will it converge to zero?
| The original series is
$$
\begin{eqnarray}
a_1 &=& +\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{5}}+\frac{2}{\sqrt{6}} + \dots \\
&=& \sum_{n=1}^{\infty}\frac{a_n}{\sqrt{n}},
\end{eqnarray}
$$
where $a_n=2\cos\left(\frac{\pi}{3}n\right)=e^{i\pi n/3}+e^{-i\pi n/3}$.
The successive iterates are
$$
a_{k+1}=a_{k} + a_{k}\left(-\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{6}} - \frac{1}{\sqrt{9}} + \ldots\right) = a_{k}\left(1 + \frac{1}{\sqrt{3}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}\right);
$$
i.e., each is equal to the previous value times a fixed constant factor. The constant factor is
$$
1 + \frac{1}{\sqrt{3}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}=1+\frac{(\sqrt{2}-1)\zeta(\frac{1}{2})}{\sqrt{3}}=0.65076...,$$
which is less than $1$, so the iterates converge to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to divide using addition or subtraction We can multiply $a$ and $n$ by adding $a$ a total of $n$ times.
$$ n \times a = a + a + a + \cdots +a$$
Can we define division similarly using only addition or subtraction?
| To divide $60$ by $12$ using subtraction:
$$\begin{align*}
&60-12=48\qquad\text{count }1\\
&48-12=36\qquad\text{count }2\\
&36-12=24\qquad\text{count }3\\
&24-12=12\qquad\text{count }4\\
&12-12=0\qquad\;\text{ count }5\;.
\end{align*}$$
Thus, $60\div 12=5$.
You can even handle remainders:
$$\begin{align*}
&64-12=52\qquad\text{count }1\\
&52-12=40\qquad\text{count }2\\
&40-12=28\qquad\text{count }3\\
&28-12=16\qquad\text{count }4\\
&16-12=4\qquad\;\text{ count }5\;.
\end{align*}$$
$4<12$, so $64\div 12$ is $5$ with a remainder of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 1
} |
Calculate absolute values with unknown constant I am to calculate all $x$ if $f(x) = g(x)$ and if
$$f(x)= |2x+2| + |3-2x|$$
$$g(x)= x + 14$$
How do I mix regular numbers with absolute values in such a sense? I thought I could calculate it like this:
$$|2x+2| + |3-2x| = 3x+2 + 3+2x = 3x+5$$
But then I realised that the end value of either of the absolute values is determined by $x$. It feels like a catch 22: how do I calculate this equation without knowing the value of $x$ until I've calculated it?
| First, partition of range of $x$ into 3 parts:
*
*$x\geq\frac{3}{2}$, in this case, $f(x)=(2x+2)+(2x-3)=4x-1$. Then solve the eqation $4x-1=x+14$ and find that $x=5$.
*$-1\leq x\leq\frac{3}{2}$, then $f(x)=(2x+2)+(3-2x)=5$. Then you can find that the solution of $5=x+14$ is $x=9$, which out of the range $-1\leq x\leq\frac{3}{2}$. Therefore, this is not a solution.
*$x\leq-1$, then $f(x)=(-2x-2)+(3-2x)=1-4x$. The solution of $1-4x=x+14$ is $x=-\frac{13}{5}$, which is in the range.
Therefore, the solution of $f(x)=g(x)$ is $x=5$ and $-\frac{13}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality $2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$ Assume that $a,b,c$ are real numbers from the interval $(\frac{1}{2},1)$. What is the proof that
$$2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$$
holds?
| For the left inequality: Since the sum of any two of $a,b,c$ is greater than 1, one has that
$a+1$, $b+1$, $c + 1 < a + b + c$. So you have
$$\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} > \frac{a+b}{a + b + c} +\frac{b+c}{a + b + c} +\frac{c+a}{a + b + c} $$
$$= \frac{2a + 2b + 2c}{a + b + c}$$
$$ = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods? I have seen the Fresnel integral
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral. I was wondering, however, does anyone know a real analysis method to evaluate this integral?
| Let $u=x^2$, then
$$
\int_0^\infty \sin(u) \frac{\mathrm{d} u}{2 \sqrt{u}}
$$
The real analysis way of evaluating this integral is to consider a parametric family:
$$\begin{eqnarray}
I(\epsilon) &=& \int_0^\infty \frac{\sin(u)}{2 \sqrt{u}} \mathrm{e}^{-\epsilon u} \mathrm{d} u = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\int_0^\infty u^{2n+\frac{1}{2}} \mathrm{e}^{-\epsilon u} \mathrm{d} u \\ &=& \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \Gamma\left(2n+\frac{3}{2}\right) \epsilon^{-\frac{3}{2}-2n} \\
&=& \frac{1}{2 \epsilon^{3/2}} \sum_{n=0}^\infty \left(-\frac{1}{\epsilon^2}\right)^n\frac{\Gamma\left(2n+\frac{3}{2}\right)}{\Gamma\left(2n+2\right)} \\
&\stackrel{\Gamma-\text{duplication}}{=}&\frac{1}{2 \epsilon^{3/2}} \sum_{n=0}^\infty \left(-\frac{1}{\epsilon^2}\right)^n\frac{\Gamma\left(n+\frac{3}{4}\right)\Gamma\left(n+\frac{5}{4}\right)}{\sqrt{2} n! \Gamma\left(n+\frac{3}{2}\right)} \\
&=& \frac{1}{(2 \epsilon)^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} {}_2F_1\left(\frac{3}{4}, \frac{5}{4}; \frac{3}{2}; -\frac{1}{\epsilon^2}\right) \\
&\stackrel{\text{Euler integral}}{=}& \frac{1}{(2 \epsilon)^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} \frac{1}{\operatorname{B}\left(\frac{5}{4}, \frac{3}{2}-\frac{5}{4}\right)} \int_0^1 x^{\frac{5}{4}-1} (1-x)^{\frac{3}{2}-\frac{5}{4} -1} \left(1+\frac{x}{\epsilon^2}\right)^{-3/4} \mathrm{d} x \\
&=& \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} \frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{1}{4}\right)} \int_0^1 x^{\frac{5}{4}-1} (1-x)^{\frac{1}{4} -1} \left(\epsilon^2+x\right)^{-3/4} \mathrm{d} x
\end{eqnarray}
$$
Now we are ready to compute $\lim_{\epsilon \to 0} I(\epsilon)$:
$$\begin{eqnarray}
\lim_{\epsilon \to 0} I(\epsilon) &=& \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \int_0^1 x^{\frac{1}{2}-1} \left(1-x\right)^{\frac{1}{4}-1} \mathrm{d} x = \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)} \\ &=& \frac{1}{2^{3/2}} \Gamma\left(\frac{1}{2}\right) = \frac{1}{2} \sqrt{\frac{\pi}{2}}
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "75",
"answer_count": 8,
"answer_id": 2
} |
Calculating :$((\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ))$ What is the easiest way to calculate :
$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ)) $$
| $\sqrt3+\tan(30^\circ-x)=\sqrt3+\frac{\frac{1}{\sqrt3}-\tan x}{1+ \frac{1}{\sqrt3}\tan x}$
$=\sqrt 3+\frac{1-\sqrt 3 \tan x}{\sqrt 3+\tan x}=\frac{4}{\sqrt 3+\tan x}$
$$\Longrightarrow(\sqrt 3+\tan(30^\circ-x))(\sqrt 3+\tan x)=4$$
Put $x=1,2,....,14^\circ$.
To complete, $x=15^\circ$, $(\sqrt 3+\tan(30^\circ-15^\circ))(\sqrt 3+\tan 15^\circ)=4$
$\implies (\sqrt 3+\tan 15^\circ)^2=4\implies \sqrt 3+\tan 15^\circ=2$ as $\tan 15^\circ>0$.
So, the answer should be $4^{14}\cdot 2=2^{29}$
A little generalization :
Assuming $A≠n\frac{\pi}{2}$(where $n$ is any integer) so that $\cot A$ and $\tan A$ are non-zero finite,
$\cot A+ \tan(A-y)= \cot A+ \frac{\tan A-\tan y}{1+\tan A\tan y}=\frac{\cot A + \tan A}{1+\tan A\tan y}=\frac{\csc^2A}{\cot A + \tan y}$ (multiplying the numerator & the denominator by $\cot A$)
$\implies (\cot A+ \tan(A-y))(\cot A + \tan y)=\csc^2A $ , here $A=30^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Solve $x^2+x^2t^2+2xt^2+t^2=1$ for $x$. Solving
$$x^2+x^2t^2+2xt^2+t^2=1$$
for $x$ yields
$$x=\frac{1-t^2}{t^2+1}.$$
I had to use software because I could not manipulate it algebraically, and sadly the program did not tell me how it reached that solution.
What is the method?
| This equation can be written as $x^2+t^2(x+1)^2=1$ which $\implies t^2=\frac{(1+x)(1-x)}{(1+x)^2}\implies t^2=\frac{(1-x)}{(1+x)}$ (assuming $x\neq -1$) $ \implies t^2(1+x)=1-x\implies t^2+t^2x=1-x\implies x(t^2+1)=1-t^2\implies x=\frac{1-t^2}{1+t^2}$
Here, we have to check whether $x=-1$ satisfies the equation separately. Putting in the equation $x=-1$, it satisfies the equation. Thus $x=-1, \frac{1-t^2}{1+t^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find convergence of integral and compute it Preparing for my calculus exam I found this exercise and I want to see if I'm on the right track
Find out the convergence of the following series and compute it.
$$\int_0^\infty \frac {\,dx} {(x+1)\sqrt {x^2 + x + 1}}$$
To find if the integral is convergent or divergent I'm thinking to compute this limit:
$$\lim_{t\rightarrow\infty} \int_0^t \frac {\,dx} {(x+1)\sqrt {x^2 + x + 1}}$$
And if that limit is a constant then my integral is convergent if not then the integral is divergent and I should not compute it.
Now, to compute the integral so I can compute that limit I'm thinking to apply integration by parts.
That $\frac {1}{x+1}$ from the integral is $\frac {d}{\,dx}\ln(x+1)$.
Any tips and/or corrections?
| If the problem is to examine the convergence alone, just observe that for some constant $C > 0$ we have
$$ \frac{1}{(x+1)\sqrt{x^2+x+1}} \leq C \min\left\{ 1, \frac{1}{x^2}\right\}. \tag{1}$$
Indeed, on $[0, 1]$, continuity of the integrand shows that it is bounded by some constant $C_1$. On $[1, \infty)$, we have
$$ \frac{1}{(x+1)\sqrt{x^2+x+1}} \leq \frac{1}{x \sqrt{x^2}} = \frac{1}{x^2}.$$
Thus for $C = \max \{ C_1, 1 \}$ we have $(1)$. Therefore
$$ \begin{align*}
\int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x^2+x+1}}
& \leq C \int_{0}^{\infty} \min\left\{ 1, \frac{1}{x^2}\right\} \; dx \\
& = C \left( \int_{0}^{1} dx + \int_{1}^{\infty} \frac{1}{x^2} \; dx \right)
= 2C < \infty.
\end{align*} $$
This method is quite general for convergence analysis. When applying this method, we first list all the singularities (the point where the integrand blows up) of the integrand, including points $\pm \infty$ at infinity. For example, if we are dealing with
$$ \int_{0}^{\infty} \frac{dx}{x^{3/2}\sqrt{x + 1}} $$
instead, then our list of singularity will be $\{0, \infty\}$. Then examine the behavior of the integrand near each singularity point $x_0$, by approximating it to familiar functions such as $(x - x_0)^{r}$. In many cases, this information solely determines the convergence behavior of the integral. In our example above, near $x = 0$ the function is approximately $x^{-3/2}$, whose integral near $x = 0$ diverges to infinity. This proves the divergence of the integral above. Rigorous justification of this estimation would be to find a suitable estimation for the integral. In this example, we may argue by
$$ \frac{1}{x^{3/2}\sqrt{x + 1}} \geq \frac{1}{x^{3/2}\sqrt{2}} \quad \text{on} \quad (0, 1]. $$
But in this case, we are asked to find its value. When we succeed in finding its value, then convergence also follows.
We make the substitution $x + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan t$. As $x$ ranges from $0$ to $\infty$, $t$ ranges from $\frac{\pi}{6}$ to $\frac{\pi}{2}$. Also differentiating both sides, we have $ dx = \frac{\sqrt{3}}{2} \sec^2 t \; dt$. Thus
$$ \begin{align*}
\int \frac{dx}{(x+1)\sqrt{x^2 + x + 1}}
&= \int \frac{1}{\left( \frac{1}{2}+\frac{\sqrt{3}}{2} \tan t \right) \left( \frac{\sqrt{3}}{2} \sec t \right)} \cdot \frac{\sqrt{3}}{2} \sec^2 t \; dt \\
&= \int \frac{dt}{\frac{1}{2}\cos t + \frac{\sqrt{3}}{2} \sin t} \\
&= \int \frac{dt}{\sin\left(t+\frac{\pi}{6}\right)}.
\end{align*}$$
This shows that
$$ \int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x^2 + x + 1}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{dt}{\sin\left(t+\frac{\pi}{6}\right)} = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u} $$
for $u = t + \frac{\pi}{6}$. Already it is clear that this improper integral converges, for the integrand is bounded. To find its value, we proceed the calculation.
$$\begin{align*}
\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{du}{\sin u}
&= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{\sin^2 u} \; du
= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{\sin u}{1 - \cos^2 u} \; du \\
&= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ds}{1 - s^2} \qquad (s = \cos u) \\
&= \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2}\left( \frac{1}{1-s} + \frac{1}{1+s} \right) \; ds \\
&= \frac{1}{2}\left[ \log(1+s) - \log(1-s) \right]_{-\frac{1}{2}}^{\frac{1}{2}}
= \log 3.
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Little help with some algebra Note: This isn't homework, I'm skipping ahead of class. Please answer all these equations, I'm deathly stuck on them.
Use the substitution method only please. (Find $x$ and $y$.)
\begin{align}
ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2
\end{align}
$a$ is not equal to $b$ or $-b$
Note - It's $ax$ as the first word, I am worried the latex might mess up there.
Equation 2:
\begin{align}
6x+5y=7x+3y+1=2\left({x+6y-1}\right)
\end{align}
Equation 3:
\begin{align}
\sqrt{2}x+\sqrt{3}y=0
\end{align}
\begin{align}
\sqrt{3}x-\sqrt{8}y=0
\end{align}
Thank you for the help!
Note: Please provide the full solution, the more hints i get the more confused i get...
| Equation 2:
$$6x+5y=7x+3y+1\Rightarrow 0=x-2y+1 \Rightarrow 2y-1=x$$
$$7x+3y+1=2(x+6y-1)\Rightarrow 5x-9y+3=0$$
Substituting the $\,x\,$ from the first equation you get:
$$5(2y-1)-9y+3=0\Rightarrow y-2=0\Rightarrow y=2\Longrightarrow x=2y-1=3$$
Equation 3:
$$\sqrt{3}x-\sqrt{8}y=0$$
Hence $$\sqrt{3}x=\sqrt{8}y$$
So
$$x=\frac{\sqrt{8}}{\sqrt{3}}y$$
Substituting in the second:
$$\sqrt{2}\frac{\sqrt{8}}{\sqrt{3}}y+\sqrt{3}y=0$$
Now you have
$$\left(\sqrt{\frac{16}{3}}+\sqrt{3}\right)y=0\Longrightarrow y=0\Longrightarrow x=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/190420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Inequality. $\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that (using rearrangements inequalities, you can also view this exercise here, exercise number 3.1.8 )
$$\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$$
thanks.
| Observe $a + b = 1 - c$, $a + c = 1 - b$, and $b + c = 1 - a$, so the desired inequality is
$$\frac{ab+c}{1 - c}+\frac{ac+b}{1 - b}+\frac{bc+a}{1 - a} \geq 2$$
Similarly, we substitute $c = 1 - a - b$, $b = 1 - a - c$, and $a = 1 - b - c$ in the numerator, and the desired inequality becomes
$$\frac{ab + 1 - a - b }{1 - c}+\frac{ac+1 - a - c}{1 - b}+\frac{bc+ 1 - b - c}{1 - a} \geq 2$$
This can be rewritten as
$$\frac{(1 - a)(1-b)}{1 - c}+\frac{(1 - a)(1 - c)}{1 - b}+\frac{(1 - b)(1 - c)}{1 - a} \geq 2$$
It's natural to let $A = 1 - a$, $B = 1 - b$, and $C = 1 - c$ here. So we want to show under the condition that $A + B + C = 2$ that we have the following.
$$\frac{AB}{C}+\frac{AC}{B}+\frac{BC}{A} \geq 2 {\hspace 1 in}(*)$$
Without loss of generality, we may assume $A \leq B \leq C$. Then the rearrangement inequality says the left-hand side of $(*)$ is at least as large of what you get by any permutation of the denominator. So you have
$$\frac{AB}{C}+\frac{AC}{B}+\frac{BC}{A} \geq \frac{AB}{A}+\frac{AC}{C}+\frac{BC}{B}$$
$$ = B + A + C$$
$$ = 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality.$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$
Let $a,b,c \gt 0$. Prove that (Using Cauchy-Schwarz) :
$$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$$
I tried to use Cauchy-Schwarz in the following form
$$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz}\leq \sqrt{(A+B+C)(x+y+z)}\tag{1}.$$
I wrote $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}=\frac{\sqrt{2a(c+a)(a+b)}+\sqrt{2b(b+c)(a+b)}+\sqrt{2c(b+c)(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}.$$
and then I applied on $(1)$:
\begin{eqnarray}
A &=& 2a(c+a) &\mbox{and}& x=a+b;\\
B &=& 2b(a+b) &\mbox{and}& y=b+c;\\
C &=& 2c(b+c) &\mbox{and}& z=c+a ,
\end{eqnarray}
but I did not obtain anything. Thanks for your help. :)
| You got it wrong. It should be:
$$
\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}} \leq 3
$$
See my comment on Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$ for a proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Convergent or divergent series examples Suppose $\sum a_n$ is convergent. Is $\sum {{a_n} \over {1+|a_n|}}$ convergent or divergent?
| If $\sum a_n$ converges absolutely, the the answer is affimative. We claim that this is no longer the case for conditional convergence. Note that
$$ \frac{x}{1+|x|} = x - x|x| + O(x^3)$$
near the origin. Now consider the series
$$\sum_{n=1}^{\infty} a_n = \frac{2}{\sqrt{1}} - \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{1}} + \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} + \cdots.$$
This series converges conditionally. Now then we have
$$ \frac{a_{3n-2}}{1+|a_{3n-2}|} + \frac{a_{3n-1}}{1+|a_{3n-1}|} + \frac{a_{3n}}{1+|a_{3n}|} = -\frac{2}{n} + O\left( \frac{1}{n^{3/2}}\right). $$
Therefore the sum $\sum \frac{a_n}{1+|a_n|}$ diverges.
Slightly modifying this argument also generates a conditionally convergent series $\sum a_n$ whose corresponding sum $\sum \frac{a_n}{1+|a_n|}$ also converges, thus the answer is inconclusive.
| {
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"timestamp": "2023-03-29T00:00:00",
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Three inequalities with sums of fractions over two positive integers In a proof, I arrive at three inequalities for all $p,q \geqslant 0$:
\begin{align}
\frac{p+1}{q+1} + \frac{q+1}{p+1} &\geqslant 1 +
\frac{p}{2q+1} + \frac{q}{2p+1} + \frac{1}{p+q+1};\cr
\frac{2q+3}{p+1} + \frac{p+2}{q+2} &\geqslant 2 +
\frac{2q+1}{2p+1} + \frac{2}{p+q+2};\cr
\frac{q+1}{p+1} + \frac{q+2}{p+2} + \frac{p+2}{q+2} +
\frac{p+1}{q+1} &\geqslant 2 + \frac{2p+1}{2q+2} +
\frac{2q+1}{2p+2} + \frac{2}{p+q+2}.
\end{align}
Any idea on how to attack these?
EDIT: Following the pieces of advice in comments, I expanded everything to get rid of the fractions and form bivariate polynomials which must be positive. In the first case, there is an obvious factor $pq$. Since the polynomial is zero if $p=q$ (the inequality is tight), this means that $p-q$ is another factor, yielding $pq(p-q)^2(2p+2q+3) \geqslant 0$. In the case of the second polynomial, there is a trivial factor $q$, and, again $p-q$. I didn't know how to guess the two last factors, but Wolfram|Alpha helped: $q(p-q)(p-q-1)(2p+2q+5)
\geqslant 0$.
The last one is fearsome, although the polynomial must have a factor $p-q$.
| The first one can also be slightly generalized. If $c>=0$ then:
$$
\frac{p+c}{q+c} + \frac{q+c}{p+c} \ge 1 + \frac{p}{2q+c} + \frac{q}{2p+c} + \frac{c}{p+q+c}
$$
The proof is the same as suggested in my comment above.
| {
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"timestamp": "2023-03-29T00:00:00",
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solution to equation $a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$ Does the equation
$$ a \cdot \cos(\theta) - b \cdot \sin(\theta) = c$$
have a closed-form solution for $\theta$? What about the case where $a^2 + b^2 = 1$?
| Using the identity $ sin(x)^2 + cos(x)^2=1 $, we have,
$$a \cdot\, \cos(\theta) - b \cdot \sin(\theta) = c \Rightarrow a\sqrt{1-\sin^2(\theta)}= b \cdot \sin(\theta) + c $$
$$ \Rightarrow a^2\,(1-\sin^2(\theta)) = ( b\sin(\theta) +c)^2 $$
$$ (a^2 + b^2)\sin(\theta)^2 + 2bc \sin(\theta) + (c^2-a^2)=0\,. $$
The above equation is quadratic in $\sin(\theta)$. Solving the equation in $\sin(\theta)$ yields,
$$ \sin(\theta) = \frac{ -2bc \pm \sqrt{4b^2c^2-4(a^2+b^2)(c^2-a^2)}}{2(a^2+b^2)} $$
I believe you know how to find theta from here. Do not forget that $-1\leq \sin(\theta) \leq 1$.
| {
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Limit of quotients with square roots: $\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ You can't use L'Hospital's rule :S
$$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$$
I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
| This is not any different than some of the other solutions, just hopefully a bit easier to follow:
$$
\begin{align}
\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
&=\frac{\color{#C00000}{\sqrt{6-x}-2}}{\color{#00A000}{\sqrt{3-x}-1}}
\frac{\color{#C00000}{\sqrt{6-x}+2}}{\color{#C00000}{\sqrt{6-x}+2}}
\frac{\color{#00A000}{\sqrt{3-x}+1}}{\color{#00A000}{\sqrt{3-x}+1}}\\
&=\color{#C00000}{\frac{(6-x)-4}{\sqrt{6-x}+2}}
\color{#00A000}{\frac{\sqrt{3-x}+1}{(3-x)-1}}\\
&=\frac{\color{#00A000}{\sqrt{3-x}+1}}{\color{#C00000}{\sqrt{6-x}+2}}
\frac{\color{#C00000}{2-x}}{\color{#00A000}{2-x}}\\
&=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\tag{1}
\end{align}
$$
Take $\lim\limits_{x\to2}$ of $(1)$:
$$
\begin{align}
\lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
&=\lim_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\\
&=\frac12\tag{2}
\end{align}
$$
| {
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Trigonometric integral involving trig multiplication $$\int \sin^3(3x)\cos^{-2}(3x)dx$$
Let$u=3x$; then $du=3dx$, so $dx=\dfrac{du}{3}$
$$\dfrac{1}{3}\int \sin^3(u)\cos^{-2}(u)du$$
Expand $\sin^3(u)$ to $\sin^2(u)\sin(u)$
$$\dfrac{1}{3}\int \sin^2(u)\sin(u)\cos^{-2}(u)du$$
Re-write $\sin^2(u)$ to $(1-\cos^2(u))$
$$\dfrac{1}{3}\int (1-\cos^2(u))\cos^{-2}(u)\sin(u)du$$
Multiply $(1-\cos^2(u))$ by $\cos^{-2}(u)$
$$\dfrac{1}{3}\int (\cos^{-2}(u)-1)\sin(u)du$$
Let $v=\cos(u)$; then $dv=-\sin(u)du$, so $-dv=\sin(u)du$
$$-\dfrac{1}{3}\int (v^{-2}-1)dv$$
Integrate
$$-\dfrac{1}{3}(-v^{-1}-v)$$
Re-write in terms of $x$
$$-\dfrac{1}{3}(-\cos^{-1}(3x)-\cos(3x))+C$$
Incorrect answer according to MyMathLab (Pearson Education)
| Your answer is correct. However, note that
$$-\dfrac{1}{3}(-\frac{1}{\cos(3x)}-\cos(3x)) = \dfrac{1}{3} (\cos(3x)+\sec(3x))$$
Differentiating your answer, we have:
$$\dfrac{1}{3} \frac{\text{d}}{\text{d}x} (\cos(3x)+\sec(3x))+C = \dfrac{1}{3} \cdot 3 (-\sin (3x)+\tan (3x) \sec (3x)) = \sin(3x)\tan^2(3x) = \frac{\sin^3 (3x)}{\cos^2 (3x)}$$
So your integral is correct.
Try and avoid the $\sin^{-1}$ notation for $\frac{1}{\sin x}$, as people may confuse with $\arcsin$.
| {
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What is the coefficient of the $x^3$ term in the expansion of $(x^2+x-5)^7$ (See details)? I fail to see a simple way to answer this.
As such, this is my long winded approach:
Using the multinomial theorem,
$$(x_1 + x_2 + \cdots + x_m)^n
= \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
\prod_{1\le t\le m}x_{t}^{k_{t}},$$
we have the specific parameters $m=3$, $n=7$, $x_1=x^2$, $x_2=x$, and $x_3=-5$.
Via the theorem,
$$(x^2+x-5)^7=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}\prod_{1 \le t \le 3}x_{t}^{k_t}=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}x^{2k_1}x^{k_2}(-5)^{k_3}.$$
The coefficient of the $x^3$ term is the summation of the multinomial coefficient multiplied by the $(-5)^{k_3}$ factor evaluated at all the solutions of the equation $2k_1+k_2=3$ where $0\le k_1\le 7$ and $0 \le k_2 \le 7$.
Those values are $(k_1,k_2)=\{(1,1),(0,3)\}.$ Given that $k_1+k_2+k_3=7$, $k_3$ are respectively $5$ and $4$.
Hence, the coefficient of the $x^3$ is
$${7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4.$$
Given that the definition of the multinomial coefficient is
$${n \choose k_1,k_2,\ldots ,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!},$$
$$
\begin{align}
{7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4&=\frac{7!(-5)^5}{1!1!5!}+\frac{7!(-5)^4}{0!3!4!}\\
&=7\cdot 6(-5)^5+\frac{7\cdot 6\cdot 5(-5)^4}{3!}\\
&=-109375
\end{align}$$
This could be atrociously wrong. Either way, I am desperate for a much simpler process. This is ridiculous to do in a timed testing environment without the formulas given.
I would like to see a very simple but also very general way of arriving at the correct answer (preferably without college methods, but I am open to any methods).
What says you, Math.SE?
| Given that you're looking for the coefficient on a small monomial and the exponent is small, we can compute directly.
$$(x^2 + x - 5)^2 = \cdots + 2 x^3 - 9 x^2 - 10 x + 25$$
Rather than go through the usual method of multiplying two polynomials, you could do it term-wise. The only way to produce an $x^3$ is as $x^2 \cdot x$ or $x \cdot x^2$, so that coefficient is $1 \cdot 1 + 1 \cdot 1$, and so forth. Continuing:
$$(x^2 + x - 5)^3 = \cdots - 29 x^3 + 60 x^2 + 75 x - 125 $$
Ah, let me introduce a trick for repeated exponents: rather than multiplying in copies of $(x^2 + x - 5)$ one at a time, we can skip a lot of steps by simply squaring an intermediate value. Squaring the previous equation gives
$$\begin{align*}(x^2 + x - 5)^6 &= \cdots + (2 \cdot (-29) \cdot (-125) + 2 \cdot 60 \cdot 75) x^3
\\ & + (2 \cdot 60 \cdot (-125) + 75^2) x^2 + (2 \cdot 75 \cdot (-125)) x + 125^2
\\&= \cdots + 16250 x^3 - 9375 x^2 - 18750 x + 15625 \end{align*}$$
Finally, when computing the seventh power, we can only compute the coefficient we care about:
$$\begin{align*}(x^2 + x - 5)^7 &= \cdots + (16250 \cdot (-5) + (-9375) \cdot 1 + (-18750) \cdot 1) x^3 + \cdots \\ &= \cdots - 109375 x^3 + \cdots\end{align*}$$
| {
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Sum of Infinite Surds Recently I came across this question. Compute $$\sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}.$$
I know how to do it. By equating the above expression to $x$ (i.e., $x = \sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}$) and $x = \sqrt{10+x}$ and so on. But how is this justified? I mean if we take the expression $\sqrt{10+x}$, literally it is equal to $\sqrt{10+\sqrt{10+\sqrt{100 +\dots}}}$ which alters the original expression. Can someone clarify my doubt?
| $x=\sqrt{10+\sqrt{10^2+\sqrt{10^4+\cdots}}}$
$\implies \frac x{\sqrt{10}}=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ (This is the well-known Golden Ratio)
so, $\frac x{\sqrt{10}}=\frac{ 1+\sqrt 5}{2}$
Alternatively, $ \frac x{\sqrt{10}}=\sqrt{1+\frac x{\sqrt{10}}}$
Squaring we get, $\frac {x^2}{10}=1+\frac x{\sqrt{10}}\implies x^2-\sqrt{10}x-10=0$
$\implies x=\frac{\sqrt 10\pm \sqrt {50}}{2}=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$
or $y=\sqrt{1+y}$ putting $\frac x{\sqrt{10}}=y$
$\implies y^2=1+y\implies y=\frac{1\pm\sqrt5}2\implies x=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$
But $x>0,$ which means $x=\frac{\sqrt{10}(1+\sqrt 5)}{2}$
For the converge, one may check for "Geometric Infinite Surd" here.
| {
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Solving roots of complex number Given $z= \dfrac{i-1}{2}$. Evaluate $z^{1/2}$ and show the roots.
I got $z^{1/2}=\dfrac{1}{2^{1/4}}\left(\cos\dfrac{3\pi}{8} + i\sin\dfrac{3\pi}{8}\right)$
First of all is my $z^{1/2}$ correct? And secondly how to show the roots??
I would be glad and appreciated if someone could help me out.
| Partly correct; what you have found is the principle value. A complete answer should be: $$z^{1/2}=\left(\frac{1}{\sqrt{2}}e^{i\frac{3\pi}{4}+2n\pi}\right)^{1/2}=\frac{1}{2^{1/4}}e^{i\frac{3\pi}{8}+n\pi}=\frac{1}{2^{1/4}}\left(\cos\left(\frac{3\pi}{8}+n\pi\right)+i\sin\left(\frac{3\pi}{8}+n\pi\right)\right)=\pm\frac{1}{2^{1/4}}\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of digits and product of digits is equal (3 digit number) My child got a question in school (grade) that is:
Find biggest and smallest 3 digits number, which has sum of it's digits equal to product of those digits.
Help please since I cannot explain my child this question.
Example would be:
$1+2+3=1\cdot 2\cdot 3$
$3+2+1=3\cdot 2\cdot 1$
Numbers 123 and 321
But this is just example of this, how to solve it as a problem.
| So, we need find digits $a,b,c$ such that $a+b+c=abc$
Clearly, $abc \ne 0$, if $a=0,b+c=0\implies b=c=0$
So, $$\frac{a+b}{ab-1}=c$$ which is integer.
(1)If $ a=3m+1,b=3n+1$ or $a=3m-1,b=3n-1$
$3\mid (ab-1)$, but $3 ∤ (a+b)\implies (ab-1) ∤(a+b)$
(2)If $ a=3m\le 9,b=3n\le 9\implies 1\le m,n\le 3$
$$ \frac{a+b}{ab-1}=\frac{3(m+n)}{9mn-1}$$
$\implies (9mn-1)\mid (m+n)$ as $(9mn-1,3)=1$
But $(m+n)\le 3+3=6$
$\implies (9mn-1)\le 6$ which is clearly impossible as $m,n \ge 1$.
(3)If $ a=3m+1\le 9 ,b=3n-1\le 9\implies 0\le m\le 2, 1\le n\le 3 $,
$$ \frac{a+b}{ab-1}=\frac{3(m+n)}{(3m+1)(3n-1)-1}$$
$\implies (3m+1)(3n-1)-1\mid (m+n)$ as $((3m+1)(3n-1)-1,3)=(3(3mn-m+n)-2,3)=1$
$ (3m+1)(3n-1)-1\le 5\implies (3m+1)(3n-1) \le 6\implies m\le1$ and $n\le 2$
If $m=1\implies 3m+1=4\implies 3n-1\le 1\implies n<1$, but $1\le n\le 3 $
If $m=0$, $$\frac{3(m+n)}{(3m+1)(3n-1)-1} \ becomes \frac{3n}{3n-2}$$
$\implies (3n-2)\mid 3n\implies (3n-2)\mid n$ as $(3n-2,3)=1$
So, $3n-2\le n\implies n\le 1$ ,but $1\le n\le 3$ so, $n=1$
So, $a=1,b=2\implies c=\frac{a+b}{ab-1}=3$
As $a+b+c=abc$ is symmetric $a=3,b=1$ and $a=3,b=2$ will also be solutions corresponding to cases (4) $a=3m,b=3n+1$ and (5) $a=3m,b=3n-1$ respectively.
So, the only solution is $1,2,3$.
Clearly, $321$ is the largest and $123$ is the smallest.
| {
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Need help to simplify the expression involving powers $$\left(1-\frac{\sqrt{3}-i}{2}\right)^{24}$$
somehow this should be equal to :$$\left(2-\sqrt{3}\right)^{12}$$
but I can't see how...
| $$
a=\frac{\sqrt{3}-i}{2} = \cos 30^\circ-i\sin30^\circ.
$$
Look at the triangle whose vertices are $0$, $a$, and $1$. Since the distance from $0$ to $1$ and the distance from $0$ to $a$ are both equal to the radius of the unit circle, the triangle is isosceles. The angle at the center of the circle is $30^\circ$ and the other two angles must be equal to each other. Since they have to add up to $180^\circ$, they must each be half of the remaining $150^\circ$, hence each $75^\circ$.
The short side of the triangle is just $1-a$. Hece $1-a=|1-a|(\cos75^\circ+i\sin75^\circ)$. Now
$$
|1-a|=\left|1-\frac{\sqrt{3}-i}{2}\right| = \left|\frac{2-\sqrt{3}-i}{2}\right| = \frac{2\sqrt{2-\sqrt{3}}}{2} = \sqrt{2-\sqrt{3}}.
$$
Hence
$$
(1-a)^{24} = \left(\sqrt{2-\sqrt{3}}\right)^{24} (\cos(24 \cdot 75^\circ) + i\sin(24 \cdot 75^\circ)) = \left(2-\sqrt{3}\right)^{12}\cdot(1).
$$
($24\cdot75^\circ=1800^\circ = 5\text{ full circles}$, so the cosine is $1$ and the sine is $0$.)
| {
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Trigonometric proof If $4 \tan(\alpha - \beta) = 3 \tan \alpha $, then prove that
$$\tan \beta = \frac{\sin(2 \alpha)}{7 + \cos(2 \alpha)}$$
This is not homework and I've tried everything so I would just like a straight answer thank you in advance.
| $$4\sin(\alpha-\beta)=3\tan\alpha\cos(\alpha-\beta)$$
so
$$4\sin\alpha\cos\beta-4\cos\alpha\sin\beta=3\tan\alpha\cos\alpha\cos\beta+3\tan\alpha\sin\alpha\sin\beta.$$
Divide through by $\cos\beta$ to find
$$4\sin\alpha-4\cos\alpha\tan\beta=3\tan\alpha\cos\alpha+3\tan\alpha\sin\alpha\tan\beta,$$
and solve for $\tan\beta$:
$$\tan\beta=\frac{4\sin\alpha-3\tan\alpha\cos\alpha}{4\cos\alpha+3\tan\alpha\sin\alpha}=\frac{\sin\alpha\,\cos\alpha}{4\cos^2\alpha+3\sin^2\alpha}=\frac{\tfrac{1}{2}\sin(2\alpha)}{3+\cos^2\alpha}.$$
Finally, use that $2\cos^2\alpha=1+\cos(2\alpha)$.
$$\tan\beta=\frac{\sin(2\alpha)}{7+\cos(2\alpha)}.$$
| {
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$\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ I recently met the inequality $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ , where a , b , c are all positive real numbers. I wanted to prove it but had some difficult time , seeing no connection to any known standard inequalities I began to simplify the expression multiplied by $(a+b)(b+c)(c+a)$ , after some simple and elementary but tedious calculations I obtained:-
$$(a+b)(b+c)(c+a) \left(\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b}\right) \\=a(a-c)^2 + b(b-a)^2 + c(c-b)^2$$ , which is obviously non-negative thus proving the inequality. But I am wondering , is there any other way(s) to prove the inequality? Also, what is the shortest way of deriving the above said identity ?
| We can set $a+b=C$ and so on, then minimize
$$ f(A,B,C)=\sum_{cyc}\frac{2C-B}{A}. $$
If we regard $f$ as a function of $B$ and $C$ only, we have that $\frac{\partial f}{\partial B}=0$ implies $(B^2-AC)(2A-C)=0$, and $\frac{\partial f}{\partial C}=0$ implies $(C^2-AB)(2B-A)=0$. So we have four stationary points:
$$ (B/A,C/A)\in\left\{(1,1),(2,4),(2,1/2),(1/2,1/4)\right\}.$$
Without loss of generality we can additionally assume $A\leq B\leq C$, having stationary points for
$$(A,B,C) = (\lambda,\lambda,\lambda)\quad\mbox{or}\quad(\lambda,2\lambda,4\lambda).$$
We can now substitute these values into $f$ to prove the inequality.
| {
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Trigonometry Inequality This is the first time I'm posting here. If you can also tell me how to format this like a pro, I'll be very grateful.
1st question:
Prove the following inequality:
$$0^{\circ} < a, b, c < 180^{\circ}$$
$$\sin a \times \sin b \times \sin c \le \sin\left(\frac{a+b}{2}\right) \times \sin\left(\frac{a+c}{2}\right) \times \sin\left(\frac{a+b}{2}\right)$$
2nd question:
Prove the following inequality:
$$a + b + c = 90^{\circ}$$
$$\sin a \times \sin b \times \sin c \le \frac{1}{8}$$
| First inequality
By Jensen we have
$$\sin\left(\frac{a+b}{2}\right) \geq \frac{\sin(a)+\sin(b)}{2}$$
By AM-GM we get
$$\frac{\sin(a)+\sin(b)}{2} \geq \sqrt{ \sin(a) \sin(b)}$$
Combining we get
$$\sqrt{\sin(a) \sin(b) } \leq \sin\left(\frac{a+b}{2}\right)$$
Similarly you get
$$\sqrt{\sin(a) \sin(c) } \leq \sin\left(\frac{a+c}{2}\right)$$
$$\sqrt{\sin(b) \sin(c) } \leq \sin\left(\frac{b+c}{2}\right)$$
Multiplying them you get the desired inequality.
Second Inequality:
By AM-GM:
$$\sin a \times \sin b \times \sin c \le \left( \frac{\sin(a)+\sin(b)+\sin(c)}{3} \right)^3$$
Now, by Jensen:
$$\frac{\sin(a)+\sin(b)+\sin(c)}{3} \leq \sin(\frac{a+b+c}{3})=\frac{1}{2}$$
Combining the two Yields the desired result.
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Sides of triangle and an altitude Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Let $h$ be the altitude drawn on the side of length $a$ Then is $a^2 + 4h^2 - (b+c)^2$ always negative ?
|
For acute/right angled triangle $h=b\sin C=c\sin B$ and
for obtuse triangle (let $\angle B>\frac \pi 2$), $h=c\sin(\pi-B)=c\sin B$
$4h^2=2h\cdot 2h=(2b\sin C)(2c\sin B)$
$=2bc(\cos(B-C)-\cos(B+C))$
$=2bc(\cos(B-C)+\cos A)$
as $\cos(B+C)=\cos(\pi-A)=-\cos A$
We know, $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$
$$\implies 1+\cos A=\frac{b^2+c^2-a^2}{2bc}+1=\frac{(b+c)^2-a^2}{2bc}$$
$\implies (b+c)^2-a^2=2bc (1+\cos A)$
$4h^2-((b+c)^2-a^2)$
$=2bc(\cos(B-C)+\cos A)-2bc (1+\cos A)$
$=2bc(\cos(B-C)-1)\le0$ as $\cos(B-C)\le 1$ and $bc>0$
| {
"language": "en",
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"question_score": "3",
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} |
$a+b=c \times d$ and $a\times b = c + d$ There is a 'nice' relationship between the integers (1,5) and (2,3) as
$$1+5=2 \times 3;$$
$$1\times 5 = 2 + 3.$$
So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$
$$a\times b = c + d.$$
To find this, $a, b, c, d$ must satisfy $$(a+1)(b+1)=(c+1)(d+1).$$ However, this condition is only necessary but not sufficient.
Any idea?
| There is no other solution for positive integers because if we consider the pairs $(n, x)$ and $(m, y)$ such that $$n+x=m \times y;$$
$$n\times x = m + y$$
We have $x=my-n$ and $m+y=n(my-n)=nmy-n^2$ and $y-nmy=-n^2-m$ and finally $$y=\frac{n^2+m}{nm-1}$$
If we supose that $n$ is a positive integer fixed and $m$ variable this is a hyperbola.
For the case n=1 we have $y(2)=3$ are the only integers points and in $x=my-n=2\times 3-1=5$ we find $(1,5)$ and $(2,3)$ as we are expecting.
If we consider the case $n=2$ then $y(2)=2$ and in $x=my-n=2\times 2-2=2$ we find $(2,2)$ and $(2,2)$ trivially satisfies the condition.
For $n=3$ we have $y(5)=1$ and in $x=my-n=5\times 1-3=2$ we find $(3,2)$ and $(5,1)$.
But for $n>3$ don't exist intenger $m$ such that $y(m)$ is integer, then there is not more solutions for the condition.
| {
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"source": "stackexchange",
"question_score": "8",
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Rudin Series ratio and root test. In Rudins Principles of Mathematical Analysis he says consider the following series
$$\frac 12 + \frac 13 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{2^3} + \frac 1{3^3} + \frac 1{2^4} + \frac 1{3^4} + \cdots$$
for which
$$\liminf \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim \limits_{n \to \infty} \left( \dfrac {2}{3} \right)^n =0, $$
$$\liminf \limits_{n \to \infty} \sqrt[n]{a_n} = \lim \limits_{n \to \infty} \sqrt[2n]{\dfrac{1}{3^n}} = \dfrac{1}{\sqrt{3}}, $$
$$\limsup \limits_{n \to \infty} \sqrt[n]{a_n} = \lim \limits_{n \to \infty} \sqrt[2n]{\dfrac{1}{2^n}} = \dfrac{1}{\sqrt{2}}, $$
$$\limsup \limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} = \lim \limits_{n \to \infty} \dfrac 12\left( \dfrac {3}{2} \right)^n =+\infty, $$
The root test indicates convergence; the ratio test does not apply.
In the book he defines the root and ratios test for the lim sup. I am not exactly sure how he goes from the lim sup to the lim and also why there is a $2n$ (which I assume comes from even terms of the sequence) in the root test. Also why is he checking the lim inf? I believe that my understanding of lim sups and infs are not well developed or I would probably understand what’s going on.
Also how does he get the terms that he is taking the limit of. A nudge in the right direction to figure this out would be much appreciated. Thank you!!
| For definiteness, call the terms of our sequence $a_1,a_2,a_3,\dots$. A similar analysis with minor differences of detail can be made if we call the first term of our sequence $a_0$.
Note that for $n=1,2,3,\dots$ we have $a_{2n-1}=\dfrac{1}{2^n}$ and $a_{2n}=\dfrac{1}{3^n}$.
The $k$-th root of the $k$-th term is "small" when the $k$-th term is a power of $\dfrac{1}{3}$. The $k$-th root of the $k$-th term is "large" when the $k$-th term is a power of $\dfrac{1}{3}$.
More precisely, $\liminf \sqrt[k]{a_k}=\lim\inf \sqrt[2n]{\frac{1}{3^n}}=\dfrac{1}{3}$. For even $k$ the $k$-th root is constant.
Also, $\limsup\sqrt[k]{a_k}=\liminf\sqrt[2n-1]{\dfrac{1}{2^n}}$. But
$$\sqrt[2n-1]{\dfrac{1}{2^n}}=\left(\frac{1}{2^n}\right)^{1/(2n-1)}=\left(\frac{1}{2^n}\right)^{2n/(2n(2n-1))}=\left(\frac{1}{\sqrt{2}}\right)^{2n/(2n-1)}.$$
The expression on the right has limit $\dfrac{1}{\sqrt{2}}$.
That takes care of one of the gaps.
For the Ratio Test, we are interested in the behaviour of $\left|\dfrac{a_{k+1}}{a_k}\right|$.
Let $k$ be odd, say $k=2n-1$. Then $a_k=\dfrac{1}{2^n}$. And $a_{k+1}=a_{2n}=\dfrac{1}{3^n}$. It follows that
$$\frac{a_{k+1}}{a_k}=\frac{a_{2n}}{a_{2n-1}}=\left(\frac{2}{3}\right)^n.$$
This has very pleasant behaviour for large $n$, indeed for any $n$: it is safely under $1$, indeed has limit $0$.
Now let $k$ be even, say $k=2n$. Then $a_k=\dfrac{1}{2^n}$. and $k+1=2n+1$. The $2n+1$-th term of our sequence is $\dfrac{1}{2^{n+1}}$. It follows that in the case $k=2n$ we have
$$\frac{a_{k+1}}{a_k}=\frac{a_{2n+1}}{a_{2n}}=\frac{\frac{1}{2^{n+1}}}{\frac{1}{3^n}}=\frac{1}{2}\left(\frac{3}{2}\right)^n.$$
This unfortunately behaves badly for large $n$: we would like it to be safely under $1$, and it is very much over.
The limit of the ratios $\dfrac{a_{k+1}}{a_k}$ does not exist. The ratios do not (uniformly) blow up, since for $k$ odd, the ratios approach $0$. The ratio behaves very nicely at odd $k$, and very badly at even $k$. So the Ratio Test is inconclusive. The bad behaviour prevents us from concluding convergence. But the good behaviour prevents us from concluding divergence.
| {
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"source": "stackexchange",
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How do I go about simplifying this complex radical? I'm having a difficult time trying to simplify the radical below. When I type the radical to the left into my calculator, I arrive at the correct answer to the right. But I cannot seem to figure out how to simplify the radical on the left by hand. I've googled the heck out of it, but perhaps I'm using an incorrect search phrase, because I cannot find a similar example. Can someone please either point in the right direction or give me a step by step system for simplifying the radical on the left to arrive to the solution on the right?
radical image:
$$\sqrt{\frac{1+\frac{\sqrt{39}}{8}}{2}}=\frac{\sqrt{26}+\sqrt 6}{8}$$
| Method One:
This is another technique for simplifying nested radicals of this form if they can be simplified. It is based on the following binomial formula $a^2+2ab+b^2$. First let's put the nested radical in this form $\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}$. Set $$a^2+b^2=\frac{1}{2}$$ and $$2ab=\frac{\sqrt{39}}{16}$$ corresponding to the each real number underneath the radical sign. The second equation results in $b=\frac{\sqrt{39}}{32a}$. Substitute b into the first equation to arrive at the following equation $a^2+(\frac{\sqrt{39}}{32a})^2=\frac{1}{2}$; $a^2+\frac{39}{1,024a^2}=\frac{1}{2}$. Multiply all terms by $a^2$ and set the equation equal to zero. So, $$a^4-\frac{1}{2}a^2+\frac{39}{1,024}=0$$ Use the quadratic formula to arrive at $a_1^2=\frac{26}{64}$ and $a_2^2=\frac{6}{64}$ or $a_1=\pm\frac{\sqrt26}{8}$ and $a_2=\pm\frac{\sqrt6}{8}$. If we add the two positive roots $a_1$ and $a_2$ and we can see that $$\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}=\frac{\sqrt{26}+\sqrt 6}{8}$$ The equality $\sqrt{\frac{1}{2}-\frac{\sqrt{39}}{16}}$=$\frac{\sqrt{26}-\sqrt 6}{8}$ is computed the same way. If the result of these computations do not denest the nested radical nicely the result will be even for nested than the original nested radical. Nevertheless, the result will still be equal to the original nested radical.
Method Two:
Another similar method where we use the quadratic equation but only to degree two not four is as follows. Set the nested radical as the sum of two square roots so that $$\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}=(\sqrt{a}+\sqrt{b})$$ Then square both sides so that $$\frac{1}{2}+\frac{\sqrt{39}}{16}=a+2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=\frac{1}{2}$$ and set (2) $$2\sqrt{a}\sqrt{b}=\frac{\sqrt{39}}{16}$$ Square both sides of (2) to get $4ab= \frac{{39}}{256}$ and solve for $b$ to get $b=\frac{39}{1,024a}$. Putting $b$ into (1) gives $a+\frac{39}{1,024a}=\frac{1}{2}$. Multiply all terms by $a$ and convert to the quadratic equation $$a^2-\frac{1}{2}a+\frac{39}{1,024}=0$$ Solving the quadratic gives $a=\frac{26}{64}$or $a=\frac{6}{64}$. As a result at the same time $b=\frac{26}{64}$or $b=\frac{6}{64}$. Replacing $a$ and $b$ in the sum or square roots formula above gives $\frac{\sqrt{26}+\sqrt 6}{8}$.
| {
"language": "en",
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"source": "stackexchange",
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find the values of a and b for which the following system is consistent find the values of $a$ and $b$ for which the following system is consistent
$x+ y-z+w=1$
$ax+y+z+w=b$
$3x+2y+aw=1+a$
Note: I have been using gauss elimination, but I got confused when completed "power-1" in the 3rd line
how to move in column 4?
I feel the value that can be entered is a $2$ and $a=1$, but I do not have proof for that
| We have the augmented matrix
$$\left[\begin{array}{cccc|c}
1&1&-1&1&1\\
a&1&1&1&b\\
3&2&0&a&1+a
\end{array}\right]\;.$$
Subtracting $a$ times the first row from the second row and $3$ times the first row from the third row gives us this matrix:
$$\left[\begin{array}{cccc|c}
1&1&-1&1&1\\
0&1-a&1+a&1-a&b-a\\
0&-1&3&a-3&a-2
\end{array}\right]\;.$$
Interchange the last two rows and multiply the new middle row by $-1$:
$$\left[\begin{array}{cccc|c}
1&1&-1&1&1\\
0&1&-3&3-a&2-a\\
0&1-a&1+a&1-a&b-a\\
\end{array}\right]\;.$$
Subtract $1-a$ times the second row from the third row:
$$\left[\begin{array}{cccc|c}
1&1&-1&1&1\\
0&1&-3&3-a&2-a\\
0&0&4-2a&-a^2+3a-2&b-a^2+2a-2\\
\end{array}\right]\;.$$
If $a\ne 2$, then $4-2a\ne 0$, and we can pivot on $4-2a$ to complete the Gaussian elimination, and the system will be consistent. If $a=2$, the matrix is
$$\left[\begin{array}{cccc|c}
1&1&-1&1&1\\
0&1&-3&1&0\\
0&0&0&0&b-2\\
\end{array}\right]\;,$$
which is consistent if and only if $b=2$.
Thus, the system is consistent when $a\ne 2$, and it is consistent when $a=b=2$.
| {
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Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$ I have this inequality that I don't know how to prove. Possibly the inequality between means might be useful. For $n \in \mathbb{N}$:
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$$
| We induct on $n$. For $n=1$, it's trivial.
Now assume it holds for $n=k$, then
$$
\frac{1}{(k+1)+1}+\cdots+\frac{1}{3(k+1)}+\frac{1}{3(k+1)+1}=\frac{1}{k+2}+\cdots+\frac{1}{3k+4}
$$
$$
=(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})-\frac{1}{k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}
$$
Since the first expression is at least $1$ by the inductive hypothesis, it suffices to show
$$
\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\ge \frac{1}{k+1}
$$
But by Cauchy-Schwarz we obtain
$$
(3k+2+3k+3+3k+4)(\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4})\ge9
$$
Which is equivalent to what we want after dividing by $9(k+1)$. So the inductive hypothesis always holds.
We can also use user8268's hint and apply Cauchy or AM-HM directly:
$$
(k+1+\cdots+3k+1)(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})\ge (2k+1)^2
$$
So it suffices to show
$$
(2k+1)^2\ge (k+1+\cdots+3k+1)=(1+\cdots+3k+1)-(1+\cdots+k)
$$
or
$$
4k^2+4k+1\ge \frac{(3k+1)(3k+2)}{2}-\frac{k(k+1)}{2}=\frac{8k^2+8k+2}{2}=4k^2+4k+1
$$
but this is clearly true.
| {
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Writing a number as a linear combination using only coefficients in $\mathbb{Q}$ Can someone please tell me how to write $\frac{1}{7+\sqrt[3]{2}}$
as a linear combination of $1$, $\sqrt[3]{2}$ and $\sqrt[3]{2}^2$using
only coefficients in $\mathbb{Q}$ ?
| Let $\omega = \sqrt[3]{2}$, it has minimal polynomial $\omega^3 - 2$.
Like when you divide complex numbers ($\frac{1}{a + ib} = \frac{a - ib}{a^2 + b^2}$) an important part of this is the norm (recall $N(a + ib) = a^2 + b^2$) but unlike the complex case, the automorphisms of the field don't help (because there are not automorphisms of $\mathbb Q(\omega)$!).
The definition of the norm in the general (non-Galois) case is not by multiplying all the conjugates together but the determinant of the linear operator "multiplication by $a + \omega b + \omega^2 c$". So write [a,b,c] as the vector representation of that number. Multiplication by it is represented by the matrix [a,b,c; 2c,a,b; 2b,2c,a] (not that [1,0,0]*that = [a,b,c]) or $$\left( \begin{array}{ccc}
a & b & c \\
2c & a & b \\
2b & 2c & a \end{array} \right)$$ so we have $N(a + \omega b + \omega^2 c) = \text{det}(\text{that matrix}) = a^3 - 6abc + 2b^3 + 4c^3.$ So dividing by this number is represented by the inverse of that matrix, and we know $\text{det}(M)M^{-1}$ has a nice simple form so let's compute [1,0,0]*det(m)/m = [a^2 - 2*c*b, -b*a + 2*c^2, -c*a + b^2] to find $N(a + \omega b + \omega^2 c)/(a + \omega b + \omega^2 c) = (a^2-2bc) + (-ab + 2c)\omega + (-ac + b^2)\omega^2.$
Putting $a = 7$, $b = 1$, $c = 0$ this gives $$\frac{1}{7 + \omega} = \frac{49 - 7 \omega + \omega^2}{345}.$$
| {
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Quadratic congruence There is question
For each of prime $p$, show that the congruence
$x^2 \equiv1 \pmod {p^a}$
has precisely two solutions.
Continue and show that the congruence
$x^2 \equiv 1 \pmod {2^a} $
has one solution if $a=1$, two solutions if $a=2$, and four solutions if $a \ge 3$.
I don't know how to do. Help please?
| A hint has been given already for odd primes. So let us deal with $2^a$. If $a=1$ or $a=2$, you should be able to verify the assertion, by just calculating.
For practice, we can also do an explicit computation for $a=3$. It is easy to verify that $1^2\equiv 1\pmod{8}$, that $3^2\equiv 1\pmod{8}$, that $5^2\equiv 1\pmod{8}$, and that $7^2\equiv 1\pmod{8}$. And of course if $x$ is even then we cannot have $x^2\equiv 1\pmod{8}$, so there are $4$ solutions modulo $8$.
Now let us look at general $a\ge 3$. Suppose that $x^2\equiv 1\pmod{2^a}$. This can be rewritten as
$$(x-1)(x+1)\equiv 1\pmod{2^a}.$$
Note that $x$ must be odd, so both $x-1$ and $x+1$ are even. If $x-1$ is congruent to $0\pmod{4}$, then $x+1\equiv 2\pmod{4}$. And if $x-1\equiv 2\pmod{4}$, then $x+1\equiv 0\pmod{4}$.
So if $x$ is odd, one of $x-1$ or $x+1$ is divisible by $4$, and the other is divisible by $2$ but by no higher power of $2$.
If $2^a$ divides $(x-1)(x+1)$, where $a\ge 3$, there are $4$ possibilities:
(i) $2^a$ divides $x-1$, that is, $x\equiv 1\pmod{2^a}$. Informally, $x-1$ all by itself contributes enough $2$'s.
(ii) $2^a$ divides $x+1$, that is, $x\equiv -1\pmod{2^a}$. Informally, $x+1$ contributes enough $2$'s.
(iii) $2^{a-1}$ divides $x-1$, but $2^a$ doesn't. Informally, $x-1$ does not quite have enough $2$'s, but $x+1$ chips in with the only $2$ it has. Then $x-1\equiv 2^{a-1}\pmod{2^a}$, that is, $x\equiv 1+2^{a-1}\pmod{2^a}$.
(iv) $2^{a-1}$ divides $x+1$, but $2^a$ doesn't. That gives $x\equiv -1+2^{a-1}\pmod{2^a}$.
It is almost obvious that if $a\ge 3$, these $4$ solutions are incongruent modulo $2^a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Polynomial Orthogonal Complement
Let $V = \mathbb{P^4}$ denote the space of quartic polynomials, with
the $L^2$ inner product $$\langle p,q \rangle = \int^1_{-1}
p(x)q(x)dx.$$ Let $W = \mathbb{P^2}$ be the subspace of quadratic
polynomials. Find a basis for and the dimension of $W^{\perp}$.
The answer is $$t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35};\,\, \dim (W^{\perp}) =2$$
How did they get that?
| Well, for a finite dimensional vector space we have $\dim W + \dim W^\perp = \dim V$ so that covers the dimension part. For the basis of the orthogonal complement, we have
$$\int_{-1}^1 ax^4 + bx^3 + cx^2 + dx + e\ dx = 0$$
$$\int_{-1}^1 x(ax^4 + bx^3 + cx^2 + dx + e)\ dx = 0$$
$$\int_{-1}^1x^2(ax^4 + bx^3 + cx^2 + dx + e)\ dx = 0$$
Because the standard basis of $\mathbb{P}^2$ must satisfy the orthogonality conditions. Therefore we get
$$\frac{a}{5} + \frac{c}{3} + e = 0$$
$$\frac{b}{5} + \frac{d}{3} = 0$$
$$\frac{a}{7} + \frac{c}{5} + \frac{e}{3} = 0$$
Solving this system yields
$$a = \frac{35}{3}e,\ \ b=-\frac{5}{3}d,\ \ c=-10e$$
with two parameters to vary. Your solutions follows by taking $(a=0,\ b=1)$ and $(a=1,\ b=0)$ respectively.
| {
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"source": "stackexchange",
"question_score": "3",
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Definite integrals using u substitution (verification needed) Would someone mind verifying this?
$
\int_{0}^{\ln(\pi + 1)}e^x \sin(e^x - 1) \space dx
$
$
u = e^x - 1
\Rightarrow \frac{du}{dx} = e^x
\Rightarrow du = e^x \space dx
\Rightarrow dx = \frac{1}{e^x} \space du
$
$
\int_{0}^{\ln(\pi + 1)} e^x \sin(u) \frac{1}{e^x} \space du
= \int_{0}^{\ln(\pi + 1)} \sin(u) \space du
= -\cos(u) \space |_{0}^{\ln(\pi + 1)}
= -\cos(e^x - 1) \space |_{0}^{\ln(\pi + 1)}
= [-\cos(e^{\ln(\pi + 1)} - 1)] - [-\cos(e^{0} - 1)]
= [-\cos(\pi + 1-1)] - [-\cos(1 - 1)]
= [-\cos(\pi)] - [-\cos(0)]
= -\cos(\pi) + 1
$
and a second one
$\int_{\frac{\pi^2}{4}}^{\pi^2}\frac{\cos(\sqrt{x})}{\sqrt(x)} \space dx \\
$
$
u = \sqrt{x}
\Rightarrow \frac{du}{dx} = \frac{1}{2 \sqrt{x}}
\Rightarrow du = \frac{1}{2 \sqrt{x}} \space dx
\Rightarrow dx = \frac{du}{\frac{1}{2 \sqrt{x}}} = 2\sqrt{x} \space du
$
$
\int_{\frac{\pi^2}{4}}^{\pi^2} = \frac{\cos(u)}{\sqrt(x)} 2\sqrt{x} \space du
= 2 \int_{\frac{\pi^2}{4}}^{\pi^2} \cos(u) \space du
= 2\sin(\sqrt{x}) \space |_{\frac{\pi^2}{4}}^{\pi^2}
$
$
= 2\sin(\sqrt{x}) \space |_{\frac{\pi^2}{4}}^{\pi^2}
= [2\sin(\sqrt{\pi^2})] - [2\sin(\sqrt{\frac{\pi^2}{4}})]
= [2\sin(\pi)]-[2\sin\frac{\pi}{2}]
= [2(0)]-[2(1)]
= 0 - 2
= -2
$
(Thanks. :))
| You arrived at the correct results in a rather unusual way. Here is how it should be:
$u = e^x- 1\Rightarrow \frac{du}{dx} = e^x\Rightarrow du = e^x \space dx$
and $0\le x\le\ln(\pi + 1)\Rightarrow 0\le u\le \pi$
Therefore,
\begin{gather}
\int_{0}^{\ln(\pi + 1)} e^x \sin(e^x-1) \space dx
= \int_{0}^{\pi} \sin(u) \space du
= -\cos(u) \space |_{0}^{\pi}
= [-\cos(\pi)] - [-\cos(0)]
= 1+1=2
\end{gather}
| {
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Trying to solve $x^2 – 11x = y^2$ knowing that $x$ and $y$ are integers. Once upon a time, I knew how to solve this equation. Yesterday, I tried (and failed) to solve it again. I remember the solutions, but I can't figure out how to find them…
| You can factor the equation by completing the square and clearing denominators and you end up getting
$$
(2x - 11 - 2y)(2x - 11 + 2y) = 121
$$
After this you can consider all possible factorizations of $121$, so you'll have to solve some systems of linear equations.
Added in response to comments
To factor the original equation by completing the square and clearing denominators you can do the following.
$$
\begin{align}
x^2 - 11x = y^2 &\iff x^2 - 11x - y^2 = 0\\
&\iff (x^2 - 11x \quad \quad \quad) - y^2 = 0\\
&\iff \left( x^2 - 11x + \color{red}{ \left( \frac{11}{2} \right)^2} - \color{red}{ \left( \frac{11}{2} \right)^2} \right) - y^2 = 0\\
&\iff \left( x^2 - 11x + \color{red}{ \left( \frac{11}{2} \right)^2} \right) - y^2 = \color{red}{ \left( \frac{11}{2} \right)^2}\\
&\iff \left( x - \frac{11}{2} \right)^2 - y^2 = \frac{121}{4}\\
&\iff \left( \frac{2x - 11}{2} \right)^2 - y^2 = \frac{121}{4}\\
&\iff \frac{1}{4} (2x - 11)^2 - y^2 = \frac{121}{4}\\
&\iff \hspace{-2.0cm}\underbrace{(2x - 11)^2 - 4y^2}_{\text{Difference of squares, factors as $A^2 - B^2 = (A-B)(A+B)$}} \hspace{-2.0cm}= 121\\
&\iff (\underbrace{2x - 11}_{A} -\underbrace{2y}_{B})(2x - 11 + 2y) = 121
\end{align}
$$
The method's name is just completing the square, and you should google for this if you want more information. In fact you can find some online videos where the method is explained, like this one.
Finally, yes, all you have to do after factoring the equation is considering the pairs of equations you mentioned $(2x - 11 -2y, 2x -11 + 2y) = (\pm 11, \pm 11), (\pm 121, \pm 1), (\pm 1, \pm 121)$. Just be careful that for some of these you may not get solutions (I only solved a couple of them so I don't know if all of them give integer solutions).
| {
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"url": "https://math.stackexchange.com/questions/234524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Recurrence relation, Fibonacci numbers $(a)$ Consider the recurrence relation $a_{n+2}a_n = a^2
_{n+1} + 2$ with $a_1 = a_2 = 1$.
$(i)$ Assume that all $a_n$ are integers. Prove that they are all odd and the
integers $a_n$ and $a_{n+1}$ are coprime for $n \in \mathbb N$
$(ii)$ Assume that the set $\{a_n , a_{n+1} , a_{n+2}\}$ is pairwise coprime for $n \in \mathbb N$. Prove that all $a_n$ are integers by induction.
$(b)$ Consider the recurrence relation $a_{n+2}a_n = a^2_{n+1} + 1$ with $a_1 = 1, a_2 = 2$
and compare this sequence to the Fibonacci numbers. What do you find?
Formulate it as a mathematical statement and prove it.
| Here is a hint how to solve (i).
We are given the recurrence $$a_1 = 1,\, a_2 = 1,\, a_{n-1} a_{n+1} = a_n^2 + 2$$
Lemma When $n > 1$, $a_{n} < a_{n+1}$. proof: By induction assume $a_{n-1} < a_n$. $$a_{n+1} = \frac{a_n^2+2}{a_{n-1}} > \frac{a_n^2+2}{a_{n}} = a_n+\frac{2}{a_{n}} > a_n.$$
Corollary When $n > 2$, $a_n > 2$.
Theorem $a_n$ are all integers. proof: We have
*
*$a_{n-1} a_{n+1} = a_{n}^2 + 2$
*$a_{n-2} a_{n} = a_{n-1}^2 + 2$
*$a_{n-3} a_{n-1} = a_{n-2}^2 + 2$
thus $a_{n-1}$ divides $a_{n-2}a_n - 2$ and $a_{n-2}^2+2$ thus it divides their sum: $a_{n-1}|a_{n-2}(a_{n-2}+a_n)$ multiply by $a_n$ to get $a_{n-1}|(a_{n-1}^2+2)(a_{n-2}+a_n)$ but $a_{n-1}\not|(a_{n-1}^2+2)$ [if it did then we would have $a_{n-1}|2$ which is a contradiction] so we deduce $a_{n-1}|a_{n-2}+a_n$. Squaring gives $a_{n-1}|a_{n-2}^2+2a_{n-2}a_n+a_n^2$ which we can rewrite to $a_{n-1}|a_{n-2}^2+2a_{n-1}^2+4+a_n^2$ but we already know $a_{n-1}|a_{n-2}^2+2+2a_{n-1}^2$ [since both $a_{n-2}^2+2$ and $2a_{n-1}^2$ are multiples of $a_{n-1}$] so we conclude that $a_{n-1}|a_{n}^2+2$.
Lemma $a_n$ are odd. proof: Let $o_i \equiv a_i \pmod 2$ this definition is valid due to $a_i$ being integers) then assume by induction $o_i = 1$ for all $i \le n$. Then clearly $o_{n+1} \equiv o_{n-1}^{-1}(o_{n}^2+2) \equiv 1^{-1} (1^2+0) \equiv 1 \pmod 2$.
Lemma For $n > 1$, $a_{n-1}$ and $a_{n}$ are coprime. proof: Suppose they were not, since the sequence is strictly increasing we have $a_{n-1}|a_{n}$ thus $a_{n-1}|a_{n-2}a_{n} = a_{n-1}^2+2$ thus $a_{n-1}|2$ contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Remainder Theorem with polynomial as divisor I'm aware that in remainder theorem you take the divisor and make it equal to zero. For the result of that just plug in x into whatever the polynomial dividend is.
But its different when the divisor is also a polynomial since it has multiple roots:
$$p(x) = \frac{2x^6 - x^4 - 4x^2 - 8}{x^2 + 4x - 3}$$
| $x^2+4x-3$ into $2x^6-x^4-4x^2-8$.
First, look at leading terms. $x^2$ into $2x^6$. That's $2x^4$. Multiply divisor ($x^2+4x-3$) by $2x^4$ (getting $2x^6+8x^5-6x^4$) and subtract from dividend ($2x^6-x^4-4x^2-8$), getting $-8x^5+5x^4-4x^2-8$. Leading terms again: $x^2$ into $-8x^5$ is $-8x^3$. Multiply divisor by $-8x^3$, subtract from $-8x^5+5x^4-4x^2-8$. Write the whole thing out the way you would write out $123$ into $456789$ and you'll see we're just doing division but with polynomials instead of numbers. Keep going until what's left of the dividend has degree less than that of the divisor, that is, less than 2. At that point what's left is the remainder, and the quotient is $2x^4-8x^3+$ whatever other terms you have worked out along the way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to understand why $x^0 = 1$, where $x$ is any real number? Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$
I understand that the logarithm with base $a$ of $x = c$,
tells us that
$$a^c = x$$
and for $c =$ positive; values for $x$ are greater than $1$, and for $c =$ negative; values for $x$ are less than $1$, and for $c = 0$, values for $x$ are...$1$.
So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.
I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".
Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?
| RULE 1
$$\frac{x^m}{x^n} = x^{m-n}$$
thus if m=3 and n=3,
$$\frac{x^3}{x^3} = x^{3-3} = x^0$$
RULE 2
Anything devided by its self is 1, why?
$$\frac{10}{10} = 1,$$
because $\frac{y}{y} = y\left(\frac{1}{y}\right) = \left(\frac{y}{1}\right)\times \left(\frac{1}{y}\right)$ which can be displayed as
$$ \frac{y}{1} \times \frac{1}{y} = \frac{1}{1} = 1$$
here the $y$ values cancel out to become 1
COMBINING THE RULES
so if anything devided by it's self is 1
and $x^0 = x^{m-n} = \frac{x^m}{x^n}$
then
$X^0$ must = 1, as its simply just a way of saying that value devided by itself =?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 18,
"answer_id": 11
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Question on Euler Totient, and finding a function Determine $f(n)$ such that for all $n\geq 1$,
$$\frac{1}{\varphi (n)}=\sum_{d\vert n}\left(\frac{1}{d}\right)f\left(\frac{n}{d}\right)$$
This is not a homework question, just a question I stumbled upon.
I have tried writing $\varphi (n)$ as
$$\varphi (n)=\sum_{d\vert n}^{}{\mu(d) \frac{n}{d}},$$ where $\mu$ is the Möbius function.
I am not sure if this is the right approach, but I was stuck here.
Sincere thanks for any help!
| Assume that $f$ is multiplicative. This will ensure that we can determine $f(n)$ as the product of its values at prime powers $f(p^v).$ We set $f(1)=1.$
The values of $f(p^v)$ can be determined recursively. Start with $f(p)$, which produces the equation $$ \frac{1}{p-1} = f(p) + \frac{1}{p} f(1) $$ or
$$ f(p) = \frac{1}{p} \frac{1}{p-1}.$$
Now claim that $f(p^v) = 0$ when $v\ge 2.$ Reasoning inductively, we find
$$ \frac{1}{p^v-p^{v-1}} = f(p^v) + \frac{1}{p^{v-1}} f(p) + \frac{1}{p^v} f(1)$$
which implies
$$ f(p^v) = \frac{1}{p^v-p^{v-1}} - \frac{1}{p^v} \frac{1}{p-1} - \frac{1}{p^v}
= \frac{p - 1 - (p-1)}{p^{v+1}-p^v} = 0.$$
This shows that
$$f(p^v) = \begin{cases}
1 & \text{if} \quad v=0 \\
\frac{1}{p} \frac{1}{p-1} & \text{if} \quad v=1 \\
0 & \text{otherwise.}
\end{cases}$$
To conclude we now identify this function. It must be zero if the square of a prime divides $n$, and positive otherwise, hence it is a multiple of $\mu^2(n).$ The denominator is simply $n\varphi(n)$, so that the end result is
$$ f(n) = \frac{\mu^2(n)}{n\varphi(n)}.$$
The above process reflects the Dirichlet convolution
$$ f \star \frac{1}{n} = \frac{1}{\varphi}.$$
This would suggest a possibility to compute a closed form of the function $G(s)$ from this post. However we have the Euler product
$$ \sum_{n\ge 1} \frac{1/\varphi(n)}{n^s} =
\prod_p \left( 1 + \frac{1}{p-1} \frac{1}{p^s} +
\frac{1}{p}\frac{1}{p-1} \frac{1}{p^{2s}} +
\frac{1}{p^2}\frac{1}{p-1} \frac{1}{p^{3s}} + \cdots \right)$$
which is
$$ \prod_p \left( 1 + \frac{p}{p} \frac{1}{p-1} \frac{1}{p^s} +
\frac{p}{p^2}\frac{1}{p-1} \frac{1}{p^{2s}} +
\frac{p}{p^3}\frac{1}{p-1} \frac{1}{p^{3s}} + \cdots \right)$$
which yields in turn
$$\prod_p \left( 1 + \frac{p}{p-1} \frac{1/p^{s+1}}{1-1/p^{s+1}} \right)$$
Now we examine the roots and the singularities of this expression as in
$$ 1 + \frac{p}{p-1} \frac{1/z/p}{1-1/z/p}$$
getting $$ z = \frac{1}{p(1-p)},$$
so no closed form appears possible.
| {
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"url": "https://math.stackexchange.com/questions/240513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use residues to evaluate $\int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x$, where $|a|<1$ Use residues to evaluate $$
\int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x
$$
where $|a|<1$.
Try considering the integral of the form
$$
\int_C \frac{\exp(az)}{\cosh(z)}\,\mathrm dz,
$$
where $C$ is the contour given by $y=0,\, y=\pi,\, x=-R,\, x=R$.
| This doesn't use residues until we use
$$
\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}=\pi\csc(\pi z)
$$
which can be proven using residues.
We just expand things in powers of $e^x$:
$$
\begin{align}
&\int_0^\infty\frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x\\
&=\int_0^\infty e^{(a-1)x}\frac{1+e^{-2ax}}{1+e^{-2x}}\,\mathrm{d}x\\
&=\int_0^\infty\left(e^{(a-1)x}-e^{(a-3)x}+e^{(a-5)x}-\dots\right)\,\mathrm{d}x\\
&+\int_0^\infty\left(e^{(-a-1)x}-e^{(-a-3)x}+e^{(-a-5)x}-\dots\right)\,\mathrm{d}x\\
&=\frac1{1-a}-\frac1{3-a}+\frac1{5-a}-\dots\\
&+\frac1{1+a}-\frac1{3+a}+\frac1{5+a}-\dots\\
&=\frac1{a+1}-\frac1{a+3}+\frac1{a+5}-\dots\\
&-\frac1{a-1}+\frac1{a-3}-\frac1{a-5}-\dots\\
&=\frac12\left(\dots+\frac1{\frac{a+1}2-2}-\frac1{\frac{a+1}2-1}+\frac1{\frac{a+1}2}-\frac1{\frac{a+1}2+1}+\frac1{\frac{a+1}2+2}-\dots\right)\\
&=\frac12\pi\csc\left(\pi\frac{a+1}2\right)\\
&=\frac\pi2\sec\left(\frac\pi2a\right)
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that if $x$ is odd then $x^2 -1$ is divisible by $8$.
If $x$ is odd then prove that $x^2-1$ is divisible by $8$.
I start by writing: $x = 2k+1 $ where $k\in\mathbb{N}$.
Then it follows that:
$(2k+1)^2 -1 = 4k^2 +4k + 1 -1 $
Therefore:
$$\frac{4k^2 +4k}{8} = \frac{k(k+1)}{2}$$
At the end part I can see that for what $k$ is, the number on top is divisible by 2. I was expecting the end result to be a number, not a fraction. Or is it "divisible" from the definition $\text{even} = 2k$ that completes the proof?
| Are you familiar with the mod operator?
We want to show
$x^2 - 1 \equiv 0\mod8$
$x^2 \equiv 1\mod8$ for $x = 2k+1$ for some $k$
$(2k+1)^2 \equiv 1 \mod 8$
$4k^2 + 4k + 1 \equiv 1 \mod 8$
$4k^2 + 4k \equiv 0 \mod 8$
$4 (k^2 + 4k) \equiv 0 \mod 8$
This is what we want to show.
If $k$ is odd then $k^2$ is odd. And $k^2 + k$ is even.
If $k$ is even then $k^2 + k$ is even
Hence $k^2 + k$ is always of the form $2a$, And hence
$4 (k^2 + 4k) \equiv 4(2a) \equiv 8a \equiv 0 \mod 8$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a row-reduced echelon matrix. $A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\ 1& -2& 1& 1\end{pmatrix}$.
I should find a row-reduced echelon matrix $R$ which is row equivalent to $A$ and an invertible $3 \times 3$ matrix $P$ such that $R = PA$.
I know that if a matrix is row equivalent to another that means that we can obtain such a matrix by using elementary row operations. But how should i apply this to such question?
| $$A =\begin{pmatrix} \;1& 2& 1& 0\\ \!\!-1 &0 &3 &5\\ \;1& \!\!-2& 1& 1\end{pmatrix}\stackrel{R_2+R_1\,,\,R_3-R1}\longrightarrow\begin{pmatrix} \;1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& \!\!-4& 0& 1\end{pmatrix}\stackrel{R_3+2R_2}\longrightarrow \begin{pmatrix} \;1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& 0& 8& 11\end{pmatrix}$$
Now, for example: to add 1st. row to 2nd one and to substract 1st. one from 3rd. one is to multiply $\,A\,$ from the left by matrix
$$P_1:=\begin{pmatrix}1&0&0\\1&1&0\\\!\!\!-1&0&1\end{pmatrix}$$
Take it from here, and show that each step you get an $invertible$ $\,3\times 3\,$ matrix, so at the end $\,P\,$ is a product of these matrices and, thus, an invertible one.
Added: Thus,
$$P_1A=\begin{pmatrix} \;1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& \!\!-4& 0& 1\end{pmatrix}$$
which is the middle matrix above.
Now you try to generalize the above to ge the matrix $\,P_2\,$ s.t. $\,P_2P_1A\,$ gives us the rightmost matrix above (the already reduced form)
| {
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How to simplify this equation: $1\cdot N + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \dots + N \cdot1$? $$1 (N) + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \cdots + N (1) $$
I need to write it in simplest form?
here 1(N) means 1 multiply by N
| $$
\sum_{k=1}^Nk(N-k+1)=\sum_{k=1}^N kN-\sum_{k=1}^Nk^2+\sum_{k=1}^Nk=N\,\frac{N(N+1)}2-\frac{N(N+1)(2N+1)}6+\frac{N(N+1)}2=\frac{N(N+1)}2\left(N-\frac{2N+1}3+1\right)=\frac{N(N+1)(N+2)}6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/247150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Discrete random variable, joint density question Discrete random variables $X$ and $Y$ have the joint density shown in the table below.
$ \ \ \ \ X \ 1 \ \ \ 2$
$Y$
$1 \ \ \ \ \ \ \frac{1}{10} \ \ \ \frac{2}{10}$
$2 \ \ \ \ \ \ \frac{3}{10} \ \ \ \frac{4}{10}$
What is the conditional density of $Y$ given $X = 1$?
What is the procedure for doing this problem? I have an exam tomorrow and we never covered this in the lectures. I have just found it on an old exam paper which means it could potentially be on tomorrows exam.
| Given $X=1$ we are looking at the first column. The "ratio"in that column is $1$ to $3$.
More formally, we want $\Pr(Y=1|X=1)$ and $\Pr(Y=2|X=1)$. Note that $\Pr (X=1)=\frac{1}{10}+\frac{3}{10}=\frac{4}{10}$.
So
$$\Pr(Y=1|X=1)=\frac{\Pr(X=1 \cap Y=1)}{\Pr(X=1)}=\frac{1/10}{4/10}=\frac{1}{4}.$$
By a similar argument, $\Pr(Y=2|X=1)=\dfrac{3}{4}$.
Basically, we confine attention to the column $Y=1$, and scale up the probabilities in that column so they add to $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of fraction powered by n $$\lim_{n\to\infty} \left({\frac{2n+1}{3n+2}}\right)^{n} = ? $$
Simple method gives an indeterminate expression. Any idea how to think out this case?
| Probably a bit cumbersome but straightforward:
$$
\lim_{n \to \infty} \bigg(\frac{2n +1}{3n+2} \bigg)^n=\lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(\frac{n+\frac{1}{2}}{n+\frac{2}{3}} \bigg)\bigg)^n = \lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(\frac{n+\frac{2}{3}-\frac{1}{6}}{n+\frac{2}{3}}\bigg)\bigg)^n
=\lim_{n \to \infty}\bigg( \frac{2}{3} \cdot \bigg(1-\frac{1}{6n+4} \bigg)\bigg)^n= \lim_{n \to \infty}\bigg( \frac{2}{3} \bigg)^n \lim_{n \to \infty} \bigg(1-\frac{1}{6n+4} \bigg)^{(6n +4-4)\cdot \frac{1}{6}}=0 \cdot e^{-\frac{1}{6}} \cdot 1
= 0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Comparison test Use the comparison test to determine whether or not
$$\sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}$$
converges.
Would I use $\frac{1}{n^\frac{3}{2}}$ for the test and then show,
$$\lim_{x \to \infty} \dfrac{\sqrt{n}}{1+n+n^2}\frac{n^\frac{3}{2}}{1}=\lim_{x \to \infty} \dfrac{n^2}{1+n+n^2} = 1$$
$$\sum\limits_{n=1}^\infty \frac{1}{n^\frac{3}{2}}\Rightarrow \sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}\Rightarrow converges $$
| $$\frac{\sqrt n}{1+n+n^2}\leq \frac{\sqrt n}{n^2}=\frac{1}{n^{3/2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove equation with induction Prove by induction that for every natural number $n\in\mathbb{N}$ and every real number $x\in\mathbb{R}$:
$$(1+x)(1+x^2)(1+x^4)\cdot\,\dots\,\cdot(1+x^{2^{n-1}}) = 1 +x +x^2+\dots+(x^{2^n-1})$$
I proved for $n=1$ but i don't know how to continue with $(n+1)$
THANKS
| I think you made some mistake in the formula. We prove
$$(1+x)(1+x^2)\cdots (1+x^{2^{n-1}}) = 1+x+x^2+x^3+\cdots + x^{2^n-1}.$$
For $n=1$ we get $x+1=x+1$ which holds.
Assume the statement holds for $n$. We prove it holds for $n+1$. So we have
$$(1+x)(1+x^2)\cdots(1+x^{2^{n-1}})(1+x^{2^{n}}) = (1+x+\cdots +x^{2^n-1})(1+x^{2^{n}}),$$
by our induction assumption. But
$$(1+x+\cdots +x^{2^n-1})(1+x^{2^{n}})=1+x+\cdots +x^{2^n-1} + x^{2^n}+x^{2^n+1} \cdots + x^{2^n+2^{n-1}}=1+x+\cdots+x^{2^{n+1}-1},$$
proving the statement.
| {
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What is wrong with my polynomial division? $x^3-2x^2-5x+6$ $$x^3-2x^2-5x+6$$
I want to get the solutions of this. I did a polynomial division.
First, I know that $(x+1)$ is a factor since $1^3-2\cdot1^2-5\cdot1+6 = 0$
So my division goes like this:
$$..........x^2-3x-2$$
$$(x+1)|\overline{x^3-2x^2-5x+6}$$
$$x^3+x^2$$
$$......\overline{0-3x^2}-5x$$
$$........-3x^2-3x$$
$$...............\overline{0-2x}+6$$
$$....................2x-2$$
$$....................\overline{0 + 8}$$
(Sorry for the improvised formatting. Ignore any dots you see there.)
So I get, at the end, the quadratic $x^2-3x-2+8=x^2-3x+6$
However, $\triangle = (-3)^2-4(1)(6)=9-4(6)=9-24=-15$
Therefore, there are no solutions since $\triangle$ is negative.
But I definitely did something wrong, since I do know that the solutions are $1,-2,3$
What did I do wrong?
| The factor is $x-1$ because $+1$ is a root.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $x^4+4$ an irreducible polynomial? We know that $p(x)=x^4-4=(x^2-2)(x^2+2)$ is reducible over $\mathbb{Q}$ even not having roots there.
What about $q(x)=x^4+4\in \mathbb{Q}[x]$? Again, no roots.
| One may use the same version of completing the square that proves that $x+\dfrac1x \ge 2$ when $x>0$:
$$
x+\frac1x = \left(x-2+\frac1x\right)+2 = \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2.
$$
Similarly
$$
x^4+4 = \left( x^4 +4x^2 + 4 \right) - 4x^2 = \left(x^2+2\right)^2 - (2x)^2
$$
then factor that as a difference of two squares.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the greatest value... The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 1281. What is the greatest value can be the sum of the cubes?
| Call the terms $\dfrac{b}{r}$, $b$, and $br$. Square the sum. We get
$$\frac{b^2}{r^2}+b^2+b^2r^2 +2b\left(\frac{b}{r}+b+br\right).$$
Thus $441= 1281+2b(21)$. It follows that $b=-20$. The rest is routine.
If we want to do the rest prettily, use
$$\frac{1}{r^3}+r^3=\left(\frac{1}{r}+r\right)^3-3\left(\frac{1}{r^2}+r^2\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\sum_{n=1}^{\infty } n^{2} \frac{a^n}{(1+a)^{n+1}}$ How to find the sum ?
$$\sum_{n=1}^{\infty } n^{2} \frac{a^n}{(1+a)^{n+1}}$$
| Let $f(x) = \displaystyle \sum_{n=0}^{\infty} x^n = \dfrac1{1-x}$, when $\vert x \vert <1$. Then $$f'(x) = \sum_{n=0}^{\infty} nx^{n-1} = \dfrac1{(1-x)^2}$$
$$g(x) = xf'(x) = \sum_{n=0}^{\infty} nx^{n} = \dfrac{x}{(1-x)^2}$$
$$g'(x) = \sum_{n=0}^{\infty} n^2x^{n-1} = \dfrac{1+x}{(1-x)^3}$$
$$xg'(x) = \sum_{n=0}^{\infty} n^2x^{n} = \dfrac{x(1+x)}{(1-x)^3}$$
Set $x = a/(1+a)$ to get that
$$\sum_{n=0}^{\infty} n^2\dfrac{a^{n}}{(1+a)^n} = \dfrac{\dfrac{a}{1+a}\left(1+\dfrac{a}{1+a}\right)}{\left(1-\dfrac{a}{1+a}\right)^3}=(a+1)(2a^2+a)$$
So the final solution is $2a^2+a$
| {
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"timestamp": "2023-03-29T00:00:00",
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find the points on a curve where the tangent line is horizonta
Find the points on the curve $x^3 + y^3 = 2xy$ where the line tangent to the curve will be horizontal
I know that that this means that the derivative of the curve will be equal to 0. This is what I get: $$\frac{(2Y-3X^2)}{(3Y^2-X)} = 0$$
..and then I'm stuck. Please help.
| So, $2Y=3X^2$
Putting the value of $Y$ in the given equation, $$X^3(27X^3-16)=0\implies X=0,$$ or $X^3=\frac{16}{27}$
But, $X=0\implies Y=0,$ from $2Y=3X^2$
$ \frac{2Y-3X^2}{3Y^2-2X}$ will be $\frac00$ hence undefined.
So, $X\ne0$
$X^3=\frac{16}{27}\implies \frac{3Y^2}{2X}=\frac{3\left(\frac{3X^2}2\right)^2}{2X}=\frac{27}{8}\cdot X^3$ (as $x\ne0$)
$\implies \frac{3Y^2}{2X}=\frac{27}{8}\cdot \frac{16}{27}\ne1\implies 3Y^2-X\ne0$
So, $ \frac{2Y-3X^2}{3Y^2-2X}$ will be $0$ as $2Y=3X^2$ but $3Y^2-2X\ne0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Specific inequality Let $x,y,z$ be different real numbers . Prove that:
$$\frac{x^2y^2+1}{(x-y)^2}+\frac{y^2z^2+1}{(y-z)^2}+\frac{z^2x^2+1}{(x-z)^2} \geq \frac{3}{2}$$
| Hint :
We have Dao Hai Long inequality :
$$\sum \frac{a+b}{a-b}.\frac{b+c}{b-c}=-1$$
So :
$$\sum \frac{(a+b)^2}{(a-b)^2} \geq 2$$
So:
$$\sum \frac{ab}{(a-b)^2} \geq \frac{-1}{4} (1)$$
Then :
$$\sum \frac{1-ab}{a-b}=-1$$
So:
$$\sum \frac{(1-ab)^2}{(a-b)^2} \geq 2 (2)$$
From (1) and (2) we have solution
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all roots of $\,(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$ The question is to find all complex roots of
$$(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$$
and it is meant to be solved by hand.
Is there any quick way to solve this using some trick that I'm not aware of? The solution given by Wolfram used a lot of multiplying out and factoring that would be very difficult to think of or write. Also, just by observation $x=0$ and $x=-6$ are solutions.
| $360 = 2^3 \times3^2 \times 5 = 1 \times 2 \times 3^2 \times 4 \times 5 = (-5) \times (-4) \times (-3)^2 \times (-2) \times (-1)$. We see that $x=0$ and $-6$ are solutions.
Let $(x+3) = y$. Then we get
$$(x+1)(x+2)(x+3)^2(x+4)(x+5) = y^2 (y^2-1)(y^2-4) = 360$$ Note that since $x=0$ and $-6$ are solutions, we have $y = \pm 3$ to be solutions.
Calling $y^2=t$, we then get that
$$t(t-1)(t-4) = 360$$
Since $y=\pm3$ is a solution, $t=9$ is a solution for the above equation.
Hence,
$$t^3 - 5t^2 + 4t = 360 \implies t^3 - 5t^2 + 4t - 360 = (t-9)(t^2+at+b)$$
Hence, $a-9=-5$ and $9b = 360$. This gives us $a = 4$ and $b=40$. Hence, the solutions are
$$t = 9, -2 \pm 6i$$
Hence,
$$y = \pm3 , \pm \sqrt{-2\pm6i}$$
Hence, $$x = 0,-6, -3 \pm \sqrt{-2\pm6i}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$ Could you help me show that
$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$$
| Divide numerator and denominator by $3^{2n}$, yielding:
$$\frac{\frac{n^2}{3^{2n}}+1}{\left(\frac{n^3}{3^n}+1\right)^2}$$
Prove the numerator and denominator both approach $1$ as $n\to\infty$.
This amounts to showing $\frac{n^2}{3^{2n}}\to 0$ and $\frac{n^3}{3^n}\to 0$ as $n\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267322",
"timestamp": "2023-03-29T00:00:00",
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Proving :$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$
Possible Duplicate:
Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?
How to prove $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$
| As $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$$
$$\tan (A+B+C)=\frac{\tan(A+B)+\tan C}{1-\tan(A+B)\tan C}=\frac{\frac{\tan A+\tan B}{1-\tan A\tan B}+\tan C}{1-\tan C\left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)}=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}$$
So, $$A+B+C=\arctan\left(\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\right)+m\pi$$ where $m$ is any integer.
Putting $\tan A=1, \tan B=2,\tan C=3$,
So, $$\arctan 1+m_1\pi+\arctan 2y+m_2\pi+\arctan 3+m_3\pi=\arctan 0+m\pi$$ where $m_i$ are integers.
The general value of
$\arctan 1+\arctan 2+\arctan 3=\arctan 0+(m-m_1-m_2-m_3)\pi$
$=(n+m-m_1-m_2-m_3)\pi=r\pi$ where $n$ is any integer, hence $r=n+m-m_1-m_2-m_3$ is.
As the special value of each of the inverse trigonometric function lies in $\left(0,\frac\pi2\right)$ so their sum will lie in $(0,\frac{3\pi}2).$
Hence, the special value being $0\cdot\pi=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple Calculus Function I am doing a course in calculus and I was given this problem :
Given that $f(x)=3x^4−6x^3+4x^2−7x+3$, evaluate $f(−2)$.
The answer is meant to be 129 according to the tutor, but no matter how many times I try and work it out I can't see how they got that answer.
| Horner's method?
$$3 + x \cdot ( -7 + x \cdot (4 + x \cdot (-6 + 3\cdot x)))=$$
$$=3 + (-2 )\cdot ( -7 + (-2) \cdot (4 + (-2) \cdot (-12)))=$$
$$=3 + (-2) \cdot ( -7 + (-2) \cdot (4 + 24))=$$
$$=3 + (-2) \cdot ( -7 + (-2) \cdot 28)=$$
$$=3 + (-2) \cdot ( -7 - 56)=$$
$$=3 + (-2) \cdot (-63)=$$
$$=129$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
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