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Trigonometric inequality $\sin^{4}x+\cos^{4}x>a$, for $a\in [0,1]$ Solve the inequality:
$\sin^{4}x+\cos^{4}x>a$, for $a\in [0,1]$
My attempt:
$(\sin^{2} x+\cos^{2}x)^2-2\sin^{2}x\cos^2{x}>a$
$\Rightarrow 1-2\sin^{2}x\cos^{2}x>a$
$\Rightarrow 1-2\sin^{2}x(1-\sin^{2}x)>a$
$\Rightarrow 2\sin^{4}x-2\sin^{2}x+1>a$
$t:=\sin^{2}x$
$0\leq 2t^{2}-2t+1\leq 1$
$1)$ $2t^{2}-2t\leq 0$
$\Rightarrow t\in [0,1] \\$
$2)$ $2t^{2}-2t+1\geq 0$
$\Rightarrow t\in \Bbb R$
From $1)$ and $2)$ $\Rightarrow t\in [0,1]$
$\Rightarrow 0\leq \sin^{2}x \leq 1$
$1)$ $\sin^{2}x\leq 1 \Rightarrow -1\leq \sin(x)\leq 1 \Rightarrow x\in \Bbb R$
$2)$ $\sin^{2}x\geq 0 \Rightarrow x\in \Bbb R$
So, $x\in \Bbb R$.
Is this correct?
| Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$
$$\sin^4x+\cos^4x=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}4=\dfrac{2+2\cos^22x}4=\dfrac{2+1+\cos4x}4$$
Now $-1\le\cos4x\le1$ for real $x$
Alternatively,
$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x=1-\dfrac{\sin^22x}2=1-\dfrac{1-\cos4x}4$$
| {
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"url": "https://math.stackexchange.com/questions/1982619",
"timestamp": "2023-03-29T00:00:00",
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Proving that I cannot factor a cubic polynomial with Algebra alone How do I prove a given polynomial is not factor-able using only radicals, basic arithmetic operations, and real, rational numbers? For example, how can I prove that
$$0=x^3-3x+1$$
Has no solution with the given restraints?
Persistently, I want an algebraic (pre-calc. type of algebra) proof.
An idea I've had is to assume it does have one solution expressible with the given and then disproving using contradictions, but it seems be fruitless.
| So we have that $$x^3-3x+1=0 \tag1$$
We can rewrite this as:
$$\begin{align} & x^3+3x^2+3x+1=3x^2+6x \\
\implies & (x+1)^3=3x(x+2)=3\{(x+1)-1\}\{(x+1)+1\}=3(x+1)^2+2(x+1)-1 \\
\implies & (x+1)^3-3(x+1)^2-2(x+1)+1=0 \\
\implies & (x+1)^3-3(x+1)^2+3(x+1)-1=5(x+1)-2 \\
\implies & (x+1-1)^3=5(x+1)-2 \\
\implies & x^3=5x+3 \\
\implies & x^3-5x-3=0 \tag2\end{align} $$
Hence, we see that we can re-arrange $(1)$ to get $(2)$.
Now, $(1)-(2) \implies 5x+3-3x+1=0 \implies x=-2$
So $x=-2$ is a common solution of both $(1)$ and $(2)$. But on verification, we see that $x=-2$ does not satisfy either equation.
This is a clear implication that $(1)$ has no solution in $\mathbb{Q}$.
On application of Cardan's Theorem, we get that the equation, being a cubic one has all $3$ solutions in $\mathbb{R-Q}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the 6 lines can be represented by the following equation: $(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$
Question: Figure $3$ shows six lines passing through the origin. The lines are separated by equal angles. Some
exact values of $\tan(t)$ are given in Table $1$.
$(i)$ Show that the lines can be represented by the following equation:
$$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$
$(ii)$ Find an equation for a hyperbola that does not cross any of the six lines in Figure $3$, giving
reasons for your answer.
I'm just really stuck , how do I even start this question! My approach has been this:
Let $y=mx+c$ for an equation of any line since all lines pass through the origin $(0,0)$ then $y=mx$ and because $m=\tan(t)$ we have the equation of any of these lines is $y=\tan(t)x$
And since there are $6$ lines passing through the origin there are $12$ sub divisions which means the graph is separated into $12$ parts and the angle between each of the parts will be $\frac{2\pi}{12}=\frac{\pi}{6}$
But I am confused , how should I continue? Am I even on the right track?
| Part (i)
See diagram above.
Equations are:
$$\begin{align}\color{red}{x=\pm y \qquad\Rightarrow x^2-y^2=0}\\
\color{blue}{x=\pm y\tan\frac{\pi}{12} \qquad \Rightarrow x^2-y^2\tan^2\frac{\pi}{12}=0}\\
\color{green}{y=\pm x\tan\frac{\pi}{12} \qquad\Rightarrow x^2-y^2\big / \tan^2\frac{\pi}{12}=0}
\end{align}$$
Multiplying the three equations and taking note that
$$\tan\frac{\pi}{12}=2-\sqrt{3}\\
\tan^2\frac{\pi}{12}=7-4\sqrt{3}\\
\frac 1{\tan^2\frac{\pi}{12}}=7+4\sqrt{3}$$
we have:
$$\begin{align}
\color{red}{\bigg(x^2-y^2\bigg)}
\color{blue}{\bigg(x^2-y^2\tan^2\frac{\pi}{12}\bigg)}
\color{green}{\bigg(x^2-y^2\big / \tan^2\frac{\pi}{12}\bigg)}&=0\\
\color{red}{\bigg(x^2-y^2\bigg)}
\color{blue}{\bigg(x^2-(7-4\sqrt{3})y^2\bigg)}
\color{green}{\bigg(x^2-(7+4\sqrt{3})y^2\bigg)}&=0\quad\blacksquare\\
\end{align}$$
Part (ii)
A family of three hyperbola pairs (in alternating spaces between asymptotes) which do not cross any of the six lines is given by setting LHS of the above equation to a constant, i.e.
$$
\bigg(x^2-y^2\bigg)
\bigg(x^2-(7-4\sqrt{3})y^2\bigg)
\bigg(x^2-(7+4\sqrt{3})y^2\bigg)
=c
$$
where $c$ is a constant.
Changing the sign of $c$ places the family of hyperbolas in the alternate set of inter-asymptote spaces.
For only one pair of hyperbola, try, e.g.
$$\bigg(x-y\bigg)\left(x-\frac y{\tan(\pi/12)}\right)=-d$$
where $d$ is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is the sum of the roots of $x^3+9=12x$ zero? I found the roots of the equation to be
$$x=3,\frac{\sqrt{21}}{2}-\frac{3}{2},-\frac{\sqrt{21}}{2}-\frac{3}{2}$$
Algebraically, why is the sum of these roots equal to zero?
| Consider the following cubic form:
$$ax^2+bx^2+cx+d=0$$
For your equation:
$$1\cdot x^3+0\cdot x^2-12\cdot x+9=0$$
Using Vieta's formulas, the sum of roots is equal to $-\frac{b}{a}$.
$$-\frac{0}{1}=0$$
Thus the sum of roots is $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A difficult question on representing a square by a sum of 4 (or 5 ) squares The question above is not about the decomposition of a square $$m^2 = a^2 + b^2 + c^2 + d^2 $$ as a sum of 4 or more squares. We know it is always possible and we also have algorithms to do that.
My question is about going backward, that is if we only have a partial decomposition of $$m^2$$ as a sum of 4 (or 5) squares , can we rebuild $$m^2$$ If we have say 3 out of 4 or 4 out of 5 squares, can we reconstruct the square?
| Not uniquely.
Since for instance, given $a=3, b=5, c=11$
$18^2= 3^2+5^2+11^2+13^2$
$78^2= 3^2+5^2+11^2+77^2$
As we can write $(m+d)(m-d) = a^2+b^2+c^2$
there are as many solutions for $m,d$ as there factorisations of $a^2+b^2+c^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Linear programming vertex proof I want to prove that $x$ is vertex of $\{x\in \mathbb{R}^n\ \colon Ax\leqslant b, \ x\geqslant 0\}$ iff $(x,b-Ax)$ is the vertex of $\{(x,u)\in \mathbb{R}^{n+m}\ \colon Ax+u=b, \ x\geqslant 0, \ u\geqslant 0\}$. Here $A$ is a $m\times n$ matrix.
Definition is: $x \in X$ is called a vertex of a set $X$ if $x$ can't be represented as $x=\lambda y+(1-\lambda)z$, where $y,z\in X$, $y\neq z$ and $\lambda\in (0,1)$.
Any ideas?
| Denote the first polyhedral as $P_1$ and the second polyhedral be $P_2$.
Part 1:
Suppose $x \in P_1$ is not a vertex of $P_1$, then $\exists y,z \in P_1$ such that $$x=\lambda y+(1-\lambda)z$$
check that $\begin{pmatrix} x \\ b-Ax \end{pmatrix}\in P_2$
and we have $$\begin{pmatrix} x \\ b-Ax \end{pmatrix}=\lambda \begin{pmatrix}y \\b-Ay \end{pmatrix}+(1-\lambda )\begin{pmatrix}z \\b-Az \end{pmatrix}$$
Notice that $y \in P_1 \implies \begin{pmatrix}y \\b-Ay \end{pmatrix} \in P_2$ and $z \in P_1 \implies \begin{pmatrix}z \\b-Az \end{pmatrix} \in P_2.$
Hence $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is not a vertex of $P_2$.
Part 2:
Conversely, if $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is not a vertex of $P_2$, then $\exists \begin{pmatrix} y \\ u_u \end{pmatrix}$, $\begin{pmatrix} z \\ u_z \end{pmatrix} \in P_2$.
From definition of $P_2$, we can deduce that $u_y=b-Ay$ and $u_z=b-Az$ and we have
$$\begin{pmatrix} x \\ b-Ax \end{pmatrix}=\lambda \begin{pmatrix}y \\b-Ay \end{pmatrix}+(1-\lambda )\begin{pmatrix}z \\b-Az \end{pmatrix}$$
and we have $$x=\lambda y+(1-\lambda ) z$$
we have to check that $y,z \in P_1$, are you able to prove this?
If you can achieve this, you have proven that $x$ is not a vertex of $P_1$.
Summary:
$x \in P_1$ is not a vertex of $P_1$ if and of if $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is not a vertex of $P_2$
$x \in P_1$ is a vertex of $P_1$ if and of if $\begin{pmatrix} x \\ b-Ax \end{pmatrix} \in P_2$ is a vertex of $P_2$
| {
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"timestamp": "2023-03-29T00:00:00",
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quadratic congruence using Newton-Raphson Method from calculus Question:
Find two solutions of $x^2 + 2x + 2 \equiv 0 \bmod 5^3$
My attempt:
*
*rewrite the equation or simplify (optional step): $x^2 + 2x + 2 = (x+1)^2 + 1 \equiv 0 \bmod 5^3$
*$x = 1$ then: $f(1) = 5$ and $5 \equiv 0 \bmod 5$
*$f'(x) = 2(x+1)$ then: $f'(1) = 4$
*solution $= 1 - \frac{5}{4}$
*using extended euclidean algorithm: $4^{-1} \equiv 1 \bmod 125 \to 4x+125y = 1 \to x=-31$ (test: $125 * (1) + 4 * (-31) = 1$)
*$1-\frac{5}{4} = 1-5*\frac{1}{4} \equiv 1-5*(-31) \bmod 125 \equiv 1+155 \bmod 125 \equiv 156 \bmod 125 = 31$
*But $f(31) = 1025$ and $1025 \bmod 125 = 25 \neq 0$
$31$ is not a correct answer. What am I missing?
| Use the quadratic formula.
$x_{1,2} = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{-2 \pm \sqrt{4-8}}{2} \pmod{125} = \dfrac{-2 \pm \sqrt{121}}{2} \pmod{125} = \dfrac{-2 \pm 11}{2} \pmod{125} = 2^{-1} \times (9, -13) \pmod{125} = 2^{-1} \times (9, 112) \pmod{125} = 63 \times (9, 112) \pmod{125}= (67, 56)$.
Note that the division is a modular inverse!
You can easily test these values, $x = 67$ and $x = 56$ in the original equivalence and make sure the LHS equals the RHS.
You can refer to these handy notes and more notes.
| {
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Remainder in polynomial division The remainder when $x^{50}$ is divided by $(x-3)(x+2)$ is of the form $ax + b$. Find the units digit of $a$.
I tried to tackle the problem using the polynomial remainder theorem but got stuck as the divisor is a quadratic expression.
| Hint
$x^{50}=(x-3)(x+2)q(x)+ax+b$ (Division algorithm)
Now, take $x=3$ and $x=-2$ to get two equations. Two variables and two equations. Hope you get it.
This is the complete problem. (Use the hint and try yourself first)
Taking $x=3$,
$3^{50}=3a+b,$
Taking $x=-2$
$(-2)^{50}=2^{50}=-2a+b$
Subtracting, we get
$3^{50}-2^{50}=5a\implies 9^{25}-4^{25}=5a$
Firstly, observe that the LHS is divisible by $(9-4)=5$(Why?). So, you get an integer value of $a$. (Just for a check)
$\therefore a= 9^{24}+9^{23}\cdot 4+ 9^{22}\cdot 4^2+...+4^{24}$
Now, you may use modular arithmetic.
$a\equiv 1-4+6-4+6-4+...+6 \equiv 1+12\times 2\equiv 5\pmod{10}$ (Why?)
Hope you get it.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $z^4=-4$ using the exponential form method, and with the answers written as $z = x + yi$? I know how to algebraically solve equations like this and find square roots of complex numbers, but how would you solve equations like that with the $n$-th power of $z$ being greater than $2$?
| In exponential form, $z^4 = -4 = 4 \text{ cis } \pi$, where $\text{cis } \theta$ is shorthand for $e^{i\theta} = \cos \theta + i \sin \theta$. Then $z_0 = \sqrt{2} \text{ cis } \frac{\pi}{4}$ (since $\left(\sqrt{2}\right)^4 = 4$), and $z_k$ = $\sqrt{2} \text{ cis } \frac{(1+2k)\pi}{4}$, for $k = 1, 2, 3$.
In rectangular form, let $z = a+bi$ with real $a$ and $b$. Then $z^4 = a^4+4a^3bi-6a^2b^2-4ab^3i+b^4$, which gives
$$
4a^3b-4ab^3 = 0
$$
and
$$
a^4-6a^2b^2+b^4 = -4
$$
Well, $a \not= 0$ and $b \not= 0$, since either $a$ or $b$ being zero would force $z^4$ to be a non-negative real value, so we can divide the first equation by $4ab$ to get
$$
a^2-b^2 = 0
$$
or $a^2 = b^2$. Then the second equation becomes
$$
-4a^4 = -4
$$
or $a^4 = 1$, and $a = \pm 1$, meaning $b = \pm 1$.
| {
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$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ implies $a_n/n^2$ converges? Let $a_0=\pi$, $a_1=\pi^2$, $a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ How can we prove that $a_n/n^2$ converges?
It is easy to see that
$$a_n-a_1=\sum_{k=1}^{n-1}\frac{2a_{k-1}}{k+1}\geq 2a_0\sum_{k=1}^{n-1}\frac{1}{k+1}\to\infty.$$
Then how to do?
| Let $(b_n)=(\frac {a_n} {n^2})$, we have that
\begin{align}
b_{n+1}-b_n&=\frac {n^2a_{n+1}-(n+1)^2a_n}{n^2(n+1)^2}\\
&=\frac {n^2(a_{n+1}-a_n)-(2n+1)a_n}{n^2(n+1)^2}\\
&=\frac {\frac {2n^2}{n+1}a_{n-1}-(2n+1)a_n}{n^2(n+1)^2}\\
&=\frac {2n^2a_{n-1}-(2n+1)(n+1)a_n}{n^2(n+1)^3}\\
&=\frac {2n^2(a_{n-1}-a_n)-(3n+1)a_n}{n^2(n+1)^3}\\
&\le0
\end{align}
Since $(b_n)\geq0$ and $(b_n)$ is decreasing, $(b_n)$ is convergent.
| {
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Then sum of digits in $M\times N$
If $\bf{M = \underbrace{777777777777777777777}_{99-times}}$ and $\bf{N = \underbrace{999999999999999999999}_{77-times}},$ Then sum of digits in $M\times N$
$\bf{My\; Try::}$ We can write $$M = 7(1+10+10^2+\cdots \cdots +10^{98}) = \frac{7}{9}(10^{99}-1)$$
and write $$N = 9(1+10+10^2+\cdots \cdots +10^{76}) = \frac{9}{9}(10^{77}-1) = (10^{77}-1)$$
So $$M\times N = \frac{7}{9}(10^{99}-1)\times (10^{77}-1)=\frac{7}{9}\left[10^{196}-10^{77}-10^{99}+1\right]$$
Now how can i solve it after that, Help required, Thanks
| Note that $N=10^{77}-1$, hence $MN=$
$\underbrace{7\dots7}_{99}\cdot(10^{77}-1)=$
$\underbrace{7\dots7}_{99}\cdot10^{77}-\underbrace{7\dots7}_{99}=$
$\underbrace{7\dots7}_{99}\underbrace{0\dots0}_{77}-\underbrace{7\dots7}_{99}=$
$\underbrace{7\dots7}_{76}6\underbrace{9\dots9}_{22}\underbrace{2\dots2}_{76}3$
Hence the digit-sum of $MN$ is $7\cdot76+6+9\cdot22+2\cdot76+3=891$.
| {
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Limit at negative infinity of a function with a radical $$\lim_{x \to -∞}{\sqrt{x^2 -x} + x}.$$
First I rationalized, to get
$$\frac{x^2-x-x^2}{\sqrt{x^2-x}-x}$$
Then I wanted to factor the whole thing by the largest power(x) in the denominator, to get:
$$\frac{-x/x}{1/x(\sqrt{x^2-x}-x)}$$
After simplifying:
$$\frac{-1}{\sqrt{1-1/x}-1}$$
And evaluating at negative infinity:
$$\frac{-1}{1-0-1}$$
But this is incorrect as the limit would not exist. Where did I go wrong? Am I right in my procedure?
Thanks
| $$\lim _{ x\to -∞ }{ \sqrt { x^{ 2 }-x } +x } =\lim _{ x\to -∞ }{ \frac { \left( \sqrt { x^{ 2 }-x } +x \right) \left( \sqrt { x^{ 2 }-x } -x \right) }{ \left( \sqrt { x^{ 2 }-x } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \sqrt { x^{ 2 }-x } -x \right) } } =\\ =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \left| x \right| \sqrt { 1-\frac { 1 }{ x } } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( -x\sqrt { 1-\frac { 1 }{ x } } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ -x\left( \sqrt { 1-\frac { 1 }{ x } } +1 \right) } } =\\ =\lim _{ x\to -∞ }{ \frac { 1 }{ \left( \sqrt { 1-\frac { 1 }{ x } } +1 \right) } } =\frac { 1 }{ 2 } $$
| {
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Compute the sum of binomial probability I have a sum that is $$\sum_{k=\lfloor{\frac{n}{2}+1}\rfloor}^n\binom{n}{k}p^n$$
I try $n=3$ and $n=4$, I got
$$\sum_{k=\lfloor{\frac{3}{2}+1}\rfloor}^3\binom{n}{k}p^3=4p^3$$
$$\sum_{k=\lfloor{\frac{4}{2}+1}\rfloor}^4\binom{n}{k}p^4=5p^4$$
How do I do it for all $n$?
| Hint. Note that $\lfloor{\frac{n}{2}\pm 1}\rfloor=\lfloor{\frac{n}{2}}\rfloor\pm 1$, and $\binom{n}{k}=\binom{n}{n-k}$ for $k=0,\dots,n$.
Hence
$$2^n=\sum_{k=0}^n\binom{n}{k}=\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor-1}\binom{n}{k}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{k}\\
=\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor-1}\binom{n}{n-k}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{k}\\
\\
=\sum_{j=n-\lfloor{\frac{n}{2}}\rfloor+1}^{n}\binom{n}{j}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{k}.$$
Can you take it from here?
| {
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Does series $(\cosh\frac{1}{n} - 1)^p$ converge? Given: series $\sum^\infty_{n=1}(\cosh\frac{1}{n} - 1)^p$.
Question: Consider different $p$ and find out if it converges.
I tried to represent $\cosh$ as $\frac{e^x - e^{-x}}{2}$. What I got as $n$-th member:
$$\left(\frac{\sqrt[n]{e} + \frac{1}{\sqrt[n]{e}} - 2}{2}\right)^p$$
Notice that degree next to $e$ approaches to $0$ as n approaches to $\infty$:
$$\lim_{n\to\infty} 1/n = 0 \text{ and } \lim_{n\to\infty} -1/n = 0$$
So that I can use Maclaurin series for $e^x$ and get:
$$1 + \frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots + \frac{1}{k!(n+1)^k} + \ldots + 1 - \frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots + (-1)^k\frac{1}{k!(n+1)^k} + \ldots - 2 = \{\forall k: k \text{ is even}\} \frac{2}{2(n+1)^2} + \ldots + \frac{2}{k!(n+1)^k} + \ldots$$
So put it all together:
$$\left(\frac{1}{2(n+1)^2} + \ldots + \frac{2}{k!(n+1)^k} + \ldots\right)^p \forall k: k \text{ is even}$$
As I know, the sum of convergent series is sum of their sums. But how to prove it strictly for infinite number of series? Induction, started from two series? Is my idea even right?
| I would directly use the Taylor polynomial for $\cosh x$ and some basic comparisons. We have
$$\cosh x-1=\frac{x^2}2+R_2(x)=\frac{x^2}2+O(x^3)$$
We thus have that $(\frac1{2n^2}+cn^{-3})^p\le(\cosh\frac1n-1)^p\le(\frac1{2n^2}+Cn^{-3})^p$ for some constants $c,C$. By comparison with $n^{-2p}$, this is summable if and only if $p>\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$
Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$
I set up a triple integral with the bounds of the inmost as $x^2 + y^2 - 2$ to $30 - x^2 - y^2$. The two outer integrals both had the bounds from $-4$ to $4$. When I solved it I got $1024$ as the volume, but this isn't correct. Can someone please show me the steps to finding the bounds, and the correct answer?
| Set $z = z$ in the two equations to find the circle of intersection:
$x^2 + y^2 - 2 = 30 - x^2 - y^2\\
2x^2 +2y^2 = 32\\
x^2+ y^2 = 16$
$\iiint \;dz\;dy\;dx\\
\iint z\;dy\;dx$
$\int_{-4}^{4}\int_{-\sqrt{16-x^2}}^{\sqrt {16-x^2}} 32 - 2x^2- 2y^2\;dy\;dx$
Now we could keep cranking through that in Cartesian, but I think a conversion to cylindrical coordinates will make this easier.
$x = r \cos\theta\\
y = r \sin \theta\\
dy\;dx = r\;dr\;d\theta$
$\int_0^{2\pi}\int_{0}^{4} 2(16 -r^2)r \;dr\;d\theta\\
u = (16-r^2), du = -2r \;dr\\
\int_0^{2\pi}\int_{16}^{0} -u \;du\;d\theta\\
\int_0^{2\pi} 16^2 \;d\theta\\
512 \pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to relate the $|z^3+z^{-3}|$ with $|z+z^{-1}|$? I got stuck on this seemingly simple question:
If $z$ is a complex number satisfying $|z^3+z^{-3}| \le 2$, then the maximum possible value of $|z+z^{-1}|$ is:
(A) $2$
(B) $2^{1/3}$
(C) $2\sqrt 2$
(D) $1$
Using the AM-GM inequality, I showed that $|z|^3 + |z|^{-3} \ge 2$ and so I got:
$$ |z^3+z^{-3}| \le 2 \le |z|^3 + |z|^{-3}$$
I don't inderstand how I can remove the cubic power, and bring any of this in terms of $|z|$ and $|z|^{-1}$. What am I missing?
Thank you.
| $(z+z^{-1})^3=z^3 + 3z + 3z^{-1}+z^{-3}$ therefore $|z+z^{-1}|^3 \le |z^3 + z^{-3}| + 3 |z+z^{-1}| \le 2 + 3|z+z^{-1}|$ and using notation $|z+z^{-1}| = x$ we have:
$x^3 \le 2 + 3x$ equivalent $(x-2)(x+1)^2 \le 0$ therefore $|z+z^{-1}| \le 2 $ is the answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Value of $\sin(\pi/18)$ by nested radicals I recently asked a question about how to find value of $\sin(\pi/18)$ and I understand that there is no expression for $\sin(\pi/18)$ that uses the ordinary arithmetical operations. but I know that we have exact value by nested radicals. I want to know how we get this value ?
http://mathworld.wolfram.com/TrigonometryAnglesPi18.html
|
Theorem 1: For $a_1=\sqrt a$, $a_2=\sqrt{a-\sqrt{a}}, a_3=\sqrt{a-\sqrt{a+\sqrt a}},\ldots,$
We have $a_n$ as$$\lim_{n\rightarrow\infty}a_n=\frac {A-1}6+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt3}\right)\tag1$$where $A=\sqrt{4a-7}$.
Proof: I present to you the proof for the theorem.
Start with $$x^2=y+a,\qquad y^2=z+a\qquad z^2=x+a\tag{2.1}$$
Then we can factor the resulting $8$ degree polynomial into factors $$x^3+\frac 12x^2\left\{1+\sqrt{4a-7}\right\}-\frac 12x\left\{2a+1-\sqrt{4a-7}\right\}+\frac 12\left\{a-2-a\sqrt{4a-7}\right\}\tag{2.2}$$ and $$x^3+\frac 12x^2\left\{1-\sqrt{4a-7}\right\}-\frac 12x\left\{2a+1+\sqrt{4a-7}\right\}+\frac 12\left\{a-2+a\sqrt{4a-7}\right\}\tag{2.3}$$. For brevity, we can let $A=\sqrt{4a-7}$ and use the identity$$\sin^3\theta-\frac 34\sin\theta+\frac 14\sin3\theta=0$$
We first consider $(2.2)$. Substituting $x=s-\frac {A+1}{6}$ and recalling that $A^2=4a-7$, we find that$$s^3+\left(\frac {A-4a}3\right)s+\frac {12a-14-A-8Aa}{27}=0\tag{2.4}$$
And similarly, setting $x=s-\frac {1-A}6$ in $(2.3)$, we get$$s^3+\left(\frac {-A-4a}3\right)s+\frac {12a-14+A+8Aa}{27}=0\tag{2.5}$$
Which we notice is the same as $(2.4)$, but with $A$ replaced with $-A$. Next, setting $s=\frac 23t\sqrt{4a-A}$ and $s=\frac 23t\sqrt{4a+A}$ in $(2.4)$ and $(2.5)$ respectively, we deduce$$\begin{align*} & t^3-\frac 34t+\frac {12a-14-A-8Aa}{8(4a-A)^{3/2}}=0\\ & t^3-\frac 34t+\frac {12a-14+A+8Aa}{8(4a+A)^{3/2}}=0\end{align*}\tag{2.6}$$
Setting $t=\sin\theta$ and using the identity, we get the solutions to $(2.6)$. Simplifying and more simplifying, we get Theorem 1.
Simply plugging in $a=2$ gives you $2\sin\frac {\pi}{18}$ and dividing both sides by two, you get your first equation.$$\frac 12\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\ldots}}}}=\sin\frac {\pi}{18}\tag3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
$$41-4\cdot17=1$$
So the inverse of $17$ is $-4$.
That is, the inverse of $17$ is $37$
Am I right?
| Modular inverses can be rotely computed by the extended Euclidean algorithm, as well as other less-known methods methods that are sometimes simpler for small numbers. A few are below.
Fiirst we consider Gauss's algorithm, which scales the (top & bottom) of the fraction to make the bottom smaller when reduced mod $41$, e.g. $\,2\cdot 17\equiv -7\,$ below (all congruences are mod $41)$
Gauss's algorithm: $\,\ \color{#0a0}{\dfrac{1}{17}}\equiv \dfrac{2}{34}\equiv \dfrac{2}{-7}\equiv \dfrac{-12}{42}\equiv \dfrac{\color{#c00}{-12}}1$
Ext. Euclid in fractions: $\,\ \dfrac{1}{17}\equiv \dfrac{-2}{7}\equiv \dfrac{5}3\equiv\dfrac{\color{#c00}{-12}}1$
Factoring: $\,\ \color{#0a0}{\color{#0a0}{\dfrac{1}{17}}}\equiv \dfrac{42}{17}\equiv 6\cdot \dfrac{7}{17}\equiv 6\cdot\dfrac{-34}{17}\equiv 6(-2)\equiv\color{#c00}{-12}$
Therefore $\ 17x \equiv 14\,\Rightarrow\, x\equiv (\color{#0a0}{1/17})14 \equiv(\color{#c00}{-12})14\equiv -4(3\cdot 14)\equiv -4$
Alternatively we can compute $\,14/17\,$ directly using factoring as above
namely $\,\ {\rm mod}\ 41\!:\,\ \dfrac{14}{17}\equiv 2\cdot \dfrac{7}{17}\equiv 2\cdot\dfrac{-34}{17}\equiv 2(-2)\equiv -4$
Beware $ $ Modular fraction arithmetic is valid only for fractions with denominator coprime to the modulus. See here for further discussion.
| {
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"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Practice test - simplifying expressions another question I am stuck on in a practice test.
The question is $$\frac{5^{2011} - 5^{2009} +24}{ 5^{2009} +1}$$
Can you cancel out the $5^{2009}$ or not?
| Hint:
$$
\begin{align}
\frac{5^{2011} - 5^{2009} +24}{ 5^{2009} +1} & = \frac{5^2(5^{2009}+1) - 5^2 - (5^{2009} + 1) + 1 +24}{ 5^{2009} +1} \\
& = 25 - 1 + \frac{\;\;\cdots\;\;}{5^{2009} +1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculating ...5(5+4(4+3(3+2(2+1(1))))) I've been puzzling over this question for a while now, and I've finally decided to turn to the StackExchange community in order to get an answer.
How would one determine the value of the expression ...5(5+4(4+3(3+2(2+1(1)))))) to n, assuming that in this case n = 5?
Cheers!
| This can be written as:
$$5\cdot 5 + 5\cdot 4\cdot 4 + 5\cdot 4\cdot 3\cdot3 + 5\cdot 4\cdot 3\cdot 2\cdot 2+5!\cdot 1\\
=\frac{5!}{4!}\cdot 5 + \frac{5!}{3!}\cdot 4 + \frac{5!}{2!}\cdot 3 + \frac{5!}{1!}\cdot 2+\frac{5!}{0!}\cdot 1$$
Then the general formula is:
$$\sum_{k=0}^{n-1}\frac{n!}{k!}(k+1) = n!\sum_{k=0}^{n-1}\frac{k+1}{k!}$$
This likely does not have a closed form, but we can say that:
$$\sum_{k=0}^{n-1}\frac{k+1}{k!} = \frac{1}{(n-1)!}+2\sum_{k=0}^{n-2}\frac{1}{k!} $$
So:
$$\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{k+1}{k!}=2e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Four distinct integers a,b,c,d are in A.P. If $a^2+b^2+c^2=d$, then find$ a+b+c+d$. Four distinct integers a,b,c,d are in arithmetic progression. If $a^2+b^2+c^2=d$, then find $a+b+c+d$.
My attempt:-
| As you did, let $a=A-x,b=A$.
Then, we have
$$2x^2\color{red}{-}2x+3A^2-A=0,$$
i.e.
$$x=\frac{1\pm\sqrt{-6A^2+2A+1}}{2}$$
Since $x$ has to be an integer, let $-6A^2+2A+1=m^2$ to have
$$6m^2+\left(6A-1\right)^2=7$$
So, $7-6m^2\ge 0\implies m=0,\pm 1$ giving $(m,A)=(\pm 1,0)$.
Since $a,b,c,d$ are distinct,
$$a+b+c+d=4A+2x=0+2=\color{red}{2}$$
where $(a,b,c,d)=(-1,0,1,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Upper and lower bound of $\sum\limits_{n=1}^{m}\sin(nx)$? According to a post in Desmos, the $\sum\limits_{n=1}^{m}\sin(nx)$ is squeezed between $\frac{1}{2}\cot\left(\frac{x}{4}\right)$ and $-\frac{1}{2}\tan\left(\frac{x}{4}\right)$ where
$$\frac{1}{2}\cot\left(\frac{x}{4}\right)<\sum_{n=1}^{m}\sin(nx)<-\frac{1}{2}\tan\left(\frac{x}{4}\right)$$
From $4n\pi<x<(4n+2)\pi$ where $n\in\mathbb{N}$
$$-\frac{1}{2}\tan\left(\frac{x}{4}\right)<\sin(nx)<\frac{1}{2}\cot\left(\frac{x}{4}\right)$$
From $\left(4n+2\right)\pi<x<4n\pi$
I want to find how to get the upper and lower bounds. I know the partial sum is the following
$$\sum_{n=1}^{m}\sin(nx)=\frac{\sin\left(\frac{mx}{2}\right)\sin\left(\frac{(m+1)x}{2}\right)}{\sin(\frac{x}{2})}$$
The solution should be obvious but I do not know how to manipulate the partial sum to find the bounds. How do I find the upper and lower bounds? Could we use the partial sum or do we need another technique?
| Note that $\sin(a)\sin(b)=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)$.
So your partial sum is also:
$$f_n(x)=\sum_{m=1}^{n}\sin mx =\frac{1}{2} \frac{\cos\frac{x}{2}-\cos\frac{2n+1}{2}x}{\sin\frac{x}2}$$
When $x\in (0,2\pi)$, this means that:
$$f_n(x)\leq \frac{1}{2}\frac{\cos\frac{x}2+1}{\sin\frac{x}{2}}=\frac{1}{2}\cot\frac{x}{4}$$
And similarly, $$f_n(x)\geq \frac{1}{2}\frac{\cos\frac{x}{2}-1}{\sin \frac{x}2}=-\frac12\tan\frac{x}{4}$$
This is only true when $\sin\frac{x}{2}>0$. The inequalities are reversed when $\sin\frac{x}{2}<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solving System of Linear Equations with LU Decomposition of $4 \times 3$ matrix The following is all confirmed to be true:
Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$
U =
$
\begin{bmatrix}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
$
L =
$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-2 & 0 & 1 & 0\\
-1 & -1 & -1 & 0\\
\end{bmatrix}
$
Okay so using that I need to solve the following system:
$
x_2 - 2x_3 = 0 \\
-x_1 + 2x_2 - x_3 = -2 \\
2x_1 -4x_2 + 3x_3 = 5 \\
x_1 - 3x_2 + 2x_3 = 1
$
So step one is solving $Ly = b$, where $y = Ux$
So that is:
$
y_1 = 0\\
y_2 = -2\\
-2y_1 + y_3 = 5 \\
-y_1 - y_2 -y_3 = 1 \\
$
How can we find $y_3$ in the last two equations? Because,
$
-2(0) + y_3 = 5 \\
-(0) - (-2) - y_3 = 1 \\
$
So in the second to last equation $y_3 = 5$, but in the last equation $y_3 = 1$. Very confused.
| Multiplying the first row of $L$ with the first column of $U$ gives us $-1$, which is not the $(1,1)$ entry of $A$. Hence, you have made a mistake in computation of LU decomposition.
There seems to be a missing permutation matrix being involved in your computation.
Your procedure to solve the linear system upon computing the LU decomposition is correct.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Necklace problem with Burnside's lemma
How many necklaces can be made with two red beads, two green
beads, and two violet beads?
I'm trying to solve this with Burnside's Lemma however I'm stuck. Here's what I've got so far:
Let $S=\{$necklace of length 6 with 2 read, 2 green and 2 violet$\}$, $|S|=\frac{6!}{2!2!2!}=90$, let $G=\mathbb{D}_{6}$, $|G|=2*6=12$.
Now let $g\in G$, $g=e\Rightarrow |fix(g)|=|S|=90$ since $e$ sends any $s$ back to itself.
For when $g\neq e$, if $g$ is any one of the rotations by $\frac{\pi}{6}$ clockwise, what should I do from here? I intuitively know that not all rotations sends $s$ back to itself and that different configuration of the color beads also plays a role, yet I don't have a clear route to follow.
And what about flips?
Thanks for any help!
| I upvoted the first answer but I would like to show how to compute the
cycle index $Z(D_6)$ of the dihedral group $D_6$ and apply the
Burnside lemma and the Polya Enumeration Theorem to this problem.
Observe that the OEIS uses the convention of refering to rotational
symmetry as a necklace and as a bracelet when reflectional symmetry is
included, so we are working with a bracelet here.
We need to enumerate and factor the twelve permutations that
contribute to $Z(D_6).$
There is the identity, which contributes
$$a_1^6.$$
The two rotations by a distance of one and five contribute
$$2 a_6.$$
The two rotations by a distance of two or four contribute
$$2 a_3^2.$$
Finally the rotation by a distance of three contributes
$$a_2^3.$$
There are three reflections about an axis passing through opposite
vertices, giving
$$3 a_1^2 a_2^2$$
and three reflections about an axis passing through the midpoints of
opposite edges, giving
$$3 a_2^3.$$
This finally yields the cycle index
$$Z(D_6) = \frac{1}{12}
\left(a_1^6 + 2a_6 + 2a_3^2 + 3a_1^2 a_2^2 + 4a_2^3\right).$$
Do the Burnside calculation first. We have three colors and two
instances of each. The colors must be constant on the cycles. We now
proceed to count these. We get for $a_1^6$ the contribution ${6\choose
2,2,2}.$ There are no candidates for $a_6$ (we do not have six
instances of a color). We do not have three instances either, so zero
for $a_3^2.$ For $a_1^2 a_2^2$ we must choose a pair of colors for the
two-cycles, giving $3\times 2\times {3\choose 2}.$ Finally for $a_2^3$
we get six permutations of the three colors. This yields
$$\frac{1}{12}
\left({6\choose 2,2,2} + 3\times 2\times {3\choose 2}
+ 4\times 6\right)
\\ = \frac{1}{12} (90 + 18 + 24) = 11.$$
On the other hand Polya says that we need the coefficient
$$[A^2 B^2 C^2] Z(D_6)(A+B+C)$$ which is
$$[A^2 B^2 C^2]\frac{1}{12}
\left((A+B+C)^6 + 2(A^6+B^6+C^6) +
2(A^3 + B^3 + C^3)^2 \\+ 3(A+B+C)^2(A^2+B^2+C^2)^2
+ 4(A^2+B^2+C^2)^3\right).$$
Now we may drop the terms that cannot possibly produce multiples of
$A^2 B^2 C^2$ which leaves
$$[A^2 B^2 C^2]\frac{1}{12}
\left((A+B+C)^6
\\ + 3(A+B+C)^2 (A^2+B^2+C^2)^2
+ 4(A^2+B^2+C^2)^3\right).$$
Doing the coefficient extraction then yields
$$\frac{1}{12}
\left({6\choose 2,2,2}
+ 3 \times 3 \times 2 + 4{3\choose 1,1,1}\right) = 11.$$
Here we have used the observation that we must choose a square from
the first term of $(A+B+C)^2 (A^2+B^2+C^2)^2$ and then choose the two
remaining squares from the second factor, which may be done in two
ways.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Show that $\lim\limits_{n\to\infty}\sqrt [n] {n^{2}+n+1}=1$ Proof. Note that $1\leq \sqrt [n] {n^2+n+1}\leq \sqrt [n] {3n^{2}}$. But,
$\lim\limits_{n\to\infty}\sqrt [n] {3n^{2}} =1$. Therefore, by the squeeze lemma, we have $\lim\limits_{n\to\infty} \sqrt [n]{n^2+n+1} =1$
My question is: How did $\lim\limits_{n\to\infty} \sqrt [n] {3n^2} =1$ be? So, how can it prove as primitive methods? We cannot use L'hopital rule, rules of derivative, exp functions.
| We can write $\sqrt[n]{3n^2} = (\sqrt{3}n)^\frac{2}{n} = e^\frac{2\ln(\sqrt{3}n)}{n}$. Note that for $n > 1$ we have $0 < \ln(n) < n$. If $n > 1$ then $\sqrt{\sqrt{3}n} > 1$ (since $\sqrt{3}$ > 1). Thus we have for $n > 1$
\begin{align}
0 < \ln(\sqrt{\sqrt{3}n}) < \sqrt{\sqrt{3}n}\\
0 < 2\ln(\sqrt{\sqrt{3}n}) < 2\sqrt{\sqrt{3}n}\\
0 < \ln(\sqrt{3}n) < 2\sqrt{\sqrt{3}n}\\
0 < \frac{\ln(\sqrt{3}n)}{n} <\frac{2\sqrt{\sqrt{3}} \sqrt{n}}{\sqrt{n} \sqrt{n}} = \frac{2\sqrt[4]{3}}{\sqrt{n}}
\end{align}
Since $\lim_{n \to \infty} \frac{2\sqrt[4]{3}}{\sqrt{n}} = 0$ we have by the squeeze theorem $\lim_{n \to \infty} \frac{\ln(\sqrt{3}n)}{n} = 0$. Thus we have
\begin{equation}
\lim_{n\to\infty} \sqrt[n]{3n^2} = \lim_{n \to \infty} e^\frac{2\ln(\sqrt{3}n)}{n} = e^{2\lim_{n\to\infty} \frac{\ln(\sqrt{3}n)}{n}} = e^{2(0)} = e^0 = 1
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
\hline
x^2 & \\
\hline
& 3 \\
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
\hline
x^2 & 1x \\
\hline
3x& 3 \\
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
$$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
| Had never heard of the box method before I saw you use it!
When you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.
Upon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3)$.
This works for a general polynomial of degree $n$. If a $n$-degree polynomial has roots $\lambda_1, \cdots, \lambda_n$, then the polynomial is equal to $(x - \lambda_1)\cdots(x - \lambda_n)$
Of course we not always have access to all the polynomial's roots at once, but we can partially factorise and work our way through that. Let us go through the polynomial $p = x^4 + 3x^3 - x^2 - 3x$.
First obvious thing is that $\lambda_1 = 0$ and thus $p = x^4 + 3x^3 - x^2 - 3x = x(x^3 + 3x^2 - x - 3)$.
Now comes the tricky part. Finding the roots of $p_1 = x^3 + 3x^2 - x - 3$. What I always start by doing is trying some small numbers. Guessing $\lambda_2 = 1$ turns out to be fine and thus we can factor $p_1$. Now there is some polynomial $p_2$ of degree 2 that multiplied by $(x - \lambda_2) = (x - 1)$ gives $x^3 + 3x^2 - x - 3$. To calculate such polynomial I refer you to Ruffini's rule, which is just a faster way to factorize a polynomial when you know one root. Then we get $p_2 = x^2 + 4x + 3$ which was what you factorized above. Since this is a 2nd degree polynomial, we could keep trying to guess roots, use your box method, or using the quadratic formula to find its roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
How should this fraction involving powers be solved
$$\sqrt { 5 } \cdot { \left( \frac { 5 }{ 4 } \right) }^{ \frac { 1 }{ 2 } }$$
All I know is that this can be written as
$\sqrt { 5 } \cdot { \left( \frac { 5 }{ { 2 }^{ 2 } } \right) }^{ \frac { 1 }{ 2 } }$
Any Ideas?
| $\sqrt { 5 } { \left( \frac { \sqrt5 }{ \sqrt4 } \right)}=$${\left( \frac { \sqrt {5*5} }{ 2 } \right)}$=${\left( \frac { \sqrt {25} }{ 2 } \right)}$=$\frac 52$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{n=1}^\infty \frac{i^n}{n}$ converge? $$\sum_{n=1}^\infty \frac{i^n}{n}$$
I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and look at series of this form or how to look at $i^n$ in an infinite series?
| Since $i^n = i^{n+4}$,
each 4 consecutive terms
is of the form
$\begin{array}\\
\frac{i}{4m+1}+\frac{-1}{4m+2}+\frac{-i}{4m+3}+\frac{1}{4m+4}
&=\frac{i}{4m+1}+\frac{-1}{4m+2}+\frac{-i}{4m+3}+\frac{1}{4m+4}\\
&=i(\frac{1}{4m+1}-\frac{1}{4m+3})-\frac{1}{4m+2}+\frac{1}{4m+4}\\
&=i(\frac{(4m+3)-(4m+1)}{(4m+1)(4m+3)})-(\frac{(4m+4)-(4m+2)}{(4m+2)(4m+4)})\\
&=i(\frac{2}{(4m+1)(4m+3)})-(\frac{2}{(4m+2)(4m+4)})\\
\end{array}
$
Since each of the
real and imaginary parts
are less than
$\frac{2}{(4m+1)^2}$,
and the sum of these converges
absolutely,
the overall sum converges
conditionally.
To write this out explicitly,
this shows that
$\begin{array}\\
|\sum_{k=1}^{4n} \dfrac{i^k}{k}|
&=|\sum_{m=1}^n (i(\dfrac{2}{(4m+1)(4m+3)})-(\dfrac{2}{(4m+2)(4m+4)})|\\
&\le\sum_{m=1}^n |(\dfrac{2}{(4m+1)(4m+3)})|+\sum_{m=1}^n|(\dfrac{2}{(4m+2)(4m+4)})|\\
&\lt\sum_{m=1}^n |(\dfrac{2}{(4m+1)^2})|+\sum_{m=1}^n|(\dfrac{2}{(4m+2)^2})|\\
&=\sum_{m=1}^n 2(\dfrac{1}{(4m+1)^2}+\dfrac{1}{(4m+2)^2})\\
\end{array}
$
and this sum converges.
There are at most
3 additional terms in any sum
beyond the $4n$
and these contribute
at most
$\dfrac{3}{4n}$
which goes to zero,
so they do not affect
the convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help me solve a combinatorial problem Consider a ordered set consisting of $N$ integers. It is called beautiful if for every integer $x$ in the set, either $x-1$ or $x+1$ is present in the set.
How many beautiful sets can be created using the numbers $1$ to $M$ of size $K$ for some arbitrary $M$ and $K$?
The number of sets created using the numbers from $1$ to $6$ of size $4$ is 6. Ex: (1, 2, 3, 4), (2,3,4,5),(3,4,5,6),(1,2,4,5),(1,2,5,6) ,(2,3,5,6).
| We want to count the number of strings of length $M+1$ consisting of $M-K+1$ atoms that look like
$$
\underbrace{\ 0\ }_1,\underbrace{110}_{x^2},\underbrace{1110}_{x^3},\underbrace{11110}_{x^4},\dots
$$
That is, there are $M-K+1$ zeros and $K$ ones. Since the atoms all end with a zero, we've added one to the number of zeros and one to the length the string.
As seen by the monomials under the atoms, we want to count the coefficient of $x^K$ in $\left(1+x^2+x^3+x^4+x^5+\dots\right)^{M-K+1}$. Since
$$
\begin{align}
1+x^2+x^3+x^4+x^5+\dots
&=\frac1{1-x}-x\\
&=\frac{1-x+x^2}{1-x}\\
&=\frac{1+x^3}{1-x^2}
\end{align}
$$
what we are looking for is
$$
\bbox[5px,border:2px solid #C0A000]{\left[x^K\right]\left(\frac{1+x^3}{1-x^2}\right)^{M-K+1}}
$$
To compute this coefficient, we use the Binomial Theorem
$$
\begin{align}
\left(\frac{1+x^3}{1-x^2}\right)^{M-K+1}
&=\sum_{a=0}^{M-K+1}\binom{M{-}K{+}1}{a}x^{3a}\sum_{b=0}^\infty(-1)^b\binom{-M{+}K{-}1}{b}x^{2b}\\
&=\sum_{a=0}^{M-K+1}\binom{M-K+1}{a}x^{3a}\sum_{b=0}^\infty\binom{M{-}K{+}b}{b}x^{2b}
\end{align}
$$
Therefore,
$$
\begin{align}
\left[x^K\right]\left(\frac{1+x^3}{1-x^2}\right)^{M-K+1}
&=\bbox[5px,border:2px solid #C0A000]{\sum_{3a+2b=K}\binom{M{-}K{+}1}{a}\binom{M{-}K{+}b}{b}}\\
&=\left\{\begin{array}{}
\displaystyle\sum_{j=0}^{\lfloor K/6\rfloor}\binom{M{-}K{+}1}{2j}\binom{M{-}\frac K2{-}3j}{\frac K2-3j}&\text{if $K$ is even}\\
\displaystyle\sum_{j=0}^{\lfloor(K-3)/6\rfloor}\binom{M{-}K{+}1}{2j+1}\binom{M{-}\frac{K+3}2{-}3j}{\frac{K-3}2-3j}&\text{if $K$ is odd}
\end{array}\right.
\end{align}
$$
To check this with the example given, $6-4+1=3$, so consider
$$
\left(\frac{1+x^3}{1-x^2}\right)^3=1+3x^2+3x^3+6x^4+9x^5+\dots
$$
Note that the coefficient of $x^4$ is $6$.
Also, since $K$ is even,
$$
\binom{6-4+1}{0}\binom{6-2}{2}=6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many minutes does Train $1$ take for the entire trip?
Two trains simultaneously depart towards each other on one route of Washington's Metro line. Train $1$ departs from Point A on one end and Train $2$ departs from Point B on the other end. At the instant the trains pass each other, it takes Train $1$ another $16$ minutes to reach the point B and it takes Train $2$ another $9$ minutes to reach Point A. How many minutes does Train $1$ take for the entire trip?
Let the speed of Train $1$ be $s_1$ and let the speed of Train $2$ be $s_2$. Then we have $$s_1t+16s_1 = s_2t+9s_2,$$ which gives $t(s_1-s_2) = 9s_2-16s_1$ and so $t = \dfrac{9s_2-16s_1}{s_1-s_2}$. Thus, $t+16 = -\dfrac{5s_2}{s_1-s_2}$.
It doesn't seem possible to get the answer from here, but the answer is $28$. What did I do wrong?
| So you let the speed of train $1$ be $s_1$ and train $2$ be $s_2$. Let us call the distance between $A$ and $B$ as $x$. Let the distance between their common meeting point $C$ and $A$ be $y$, so that $BC$ is then $x-y$.
Then,
1) It is clear that train $2$ takes $7$ minutes less than train $1$. So one equation would be $\frac{x}{s_1} = \frac{x}{s_2} + 7$.
2) The time taken for train $2$ to travel $x-y$ is the same as time taken for train $1$ to travel $y$. So then $\frac{y}{s_1} = \frac{x-y}{s_2}$, or that $ys_2=(x-y)s_1$ or that $y(s_2+s_1) = x$.
3)Train $2$ travels distance $y$ in nine minutes, so $\frac{y}{s_2} = 9$.
4)Train $1$ travels distance $x-y$ in sixteen minutes, so $\frac{x-y}{s_1} = 16$.
From the third point, $y = 9s_2$, and from the fourth, $x-y = 16s_1$, so that $x=16s_1+9s_2$.
Also, from the second point, $9s_2(s_2+s_1) = x$.
From $x=16s_1+9s_2$, we insert this in the first equation to see that $16+\frac{9s_2}{s_1} = 9+\frac{16s_1}{s_2} + 7$, so that $16s_1^2 = 9s_2^2$. Hence, $\frac{s_1}{s_2} = \frac 34$.
Now, substitute for $s_2$ in $x = 16s_1 + 9s_2$. We know that $3s_2 = 4s_1$ ,so that $9s_2 = 12s_1$. Hence, $x = 16s_1 + 12s_1 = 28s_1$. Hence, $\frac{x}{s_1} = 28$.
This means that train $1$ travels the entire distance in $28$ minutes. Train $2$ would do it in $21$ minutes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle
If $a$, $b$, and $c$ are the three sides of a triangle, then
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
lies in what interval?
| The problem is Nesbitt's inequality. (Source: A.M. Nesbitt, Problem 15114, Educational Times (2) 3, (1903), 37-38.)
Lower Bound: Let's say $S= \dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b} $. $S+3= \dfrac{a}{b+c}+1 + \dfrac{b}{c+a}+1 + \dfrac{c}{a+b}+1=(a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right)$. By aritmethic-harmonic mean inequality we can find that $$ (a+b+c)\left( \dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right) \ge \dfrac92$$ Therefore $S\ge \dfrac32$.Equality holds when $a=b=c$.
For other solutions of first part: Nesbitt's Inequality
Upper Bound:
1. Method: By triangle inequality $b+c<a, c+a<b, a+b<c$ and hence $b+c>\dfrac{1}{2}(a+b+c)$, $c+a>\dfrac{1}{2}(a+b+c)$, $a+b>\dfrac{1}{2}(a+b+c)$. We have
$$ S = \dfrac{a}{b+c} + \dfrac{b}{c+a}+\dfrac{c}{a+b} < \dfrac{2a}{a+b+c} + \dfrac{2b}{a+b+c} + \dfrac{2c}{a+b+c} = \dfrac{2(a+b+c)}{a+b+c} = 2 $$
If $a=b$ and $c \to 0$ degenere triangle form, then $S=2$. Hence for all non-degenere triangles $S<2$.
2. Method: Without loss of generality, we assume that $a\leq b \leq c$. Thus,
$$ \dfrac{a}{b+c}\leq \dfrac{a}{a+c}, \quad \dfrac{b}{a+c}\leq \dfrac{c}{a+c}, \quad c < a+b $$
and therefore $$S < \dfrac{a}{a+c} + \dfrac{c}{a+c} + \dfrac{a+b}{a+b} = \dfrac{a+c}{a+c} + 1 = 2 .$$
So, $S<2$.
In summary, we obtain that $$ \dfrac32 \le S <2 .$$
Also this interval is best.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Bernoulli numbers. Today I read a short article about Bernoulli numbers. This article give the way to calculate Bernoulli numbers.
$\displaystyle \frac{x}{e^{x}-1} = f(x)$
$\displaystyle f(x)(e^{x}-1) = x = \sum_{n = 0}^{\infty}B_{n} \frac{x^{n}}{n!} \sum_{k = 1}^{\infty} \frac{x^{k}}{k!} = \sum_{n=1}^{\infty}x^{n} \sum_{k+m = n}^{\infty}\frac{B_{m}}{k!m!} = \sum_{n=1}^{\infty} \frac{x^{n}}{n!} \sum_{m=0}^{n-1} \binom{n}{m} B_{m}$
Now we could calculate Bernoulli numbers for different $n$. But I don't understand the step with $k+m = n$. Please help me with it!
| In slow motion
$$
\begin{array}{l}
\sum\limits_{0\, \le \,n} {B_{\,n} \frac{{x^{\,n} }}{{n!}}} \sum\limits_{1\, \le \,k} {\frac{{x^{\,k} }}{{k!}}} = \sum\limits_{\begin{array}{*{20}c}
{0\, \le \,n} \\
{1\, \le \,k} \\
\end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}} = \sum\limits_{\begin{array}{*{20}c}
{0\, \le \,n} \\
{n + 1\, \le \,n + k} \\
\end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}} = \\
= \sum\limits_{\begin{array}{*{20}c}
{0\, \le \,n} \\
{n + 1\, \le \,q} \\
\end{array}} {B_{\,n} \frac{{x^{\,q} }}{{n!\left( {q - n} \right)!}}} = \sum\limits_{\begin{array}{*{20}c}
{1\, \le \,q} \\
{0\, \le \,n\, \le \,q - 1} \\
\end{array}} {B_{\,n} \frac{{x^{\,q} }}{{n!\left( {q - n} \right)!}}} = \\
= \sum\limits_{\begin{array}{*{20}c}
{1\, \le \,q} \\
{0\, \le \,n\, \le \,q - 1} \\
\end{array}} {B_{\,n} \frac{{q!}}{{n!\left( {q - n} \right)!}}\frac{{x^{\,q} }}{{q!}}} = \sum\limits_{1\, \le \,q} {\left( {\sum\limits_{0\, \le \,n\, \le \,q - 1} {B_{\,n} \left( \begin{array}{l}
q \\
n \\
\end{array} \right)} } \right)\frac{{x^{\,q} }}{{q!}}} = \\
= \sum\limits_{1\, \le \,n} {\left( {\sum\limits_{0\, \le \,m\, \le \,n - 1} {B_{\,m} \left( \begin{array}{l}
n \\
m \\
\end{array} \right)} } \right)\frac{{x^{\,n} }}{{n!}}} \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula? To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ , but I do not see how we can get the antiderivate of $\ \frac{1-x^2}{(x^2+1)^2}\ $ either without any guess (If we know this antiderivate , we can express $\frac{1}{(x^2+1)^2}$ as a linear combination of $\ f(x):=\frac{1}{x^2+1}\ $ and $\ g(x):=\frac{1-x^2}{(x^2+1)^2}\ $ , namely $\ \frac{1}{(x^2+1)^2}=\frac{f(x)+g(x)}{2}\ $)
How can I calculate $\int \frac{dx}{(x^2+1)^2}$ only by using intgration by parts and the substitution rule as well as other basic integration rules ? I am looking for a solution not containing a guess or the iteration formula.
| We have
$$\int\frac{dx}{(1+x^2)^2}=\int \frac{1+x^2-x^2}{(1+x^2)^2}dx=$$
$$\arctan(x)+\frac{1}{2}\int x\frac{-2x}{(1+x^2)^2}dx=$$
$$\arctan(x)+\frac{1}{2}\left(\left[x \frac{1}{1+x^2}\right]-\int \frac{1}{1+x^2}\right)=$$
$$\frac{1}{2}\arctan(x)+\frac{x}{ 2(1+x^2) }+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x$ in $\log_4{(x+4)} \le \log_2{(2x+5)}$
I would like to solve $$\log_4{(x+4)} \le \log_2{(2x+5)}$$ for $x$.
I did:
$$\log_4{(x+4)} \le \log_2{(2x+5)} \Leftrightarrow \frac{\log(x+4)}{\log(4)} \le \frac{\log(2x+5)}{\log(2)}\\ \Leftrightarrow \frac{\log(x+4)}{2\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftrightarrow \frac{1}{2} \cdot \frac{\log(x+4)}{\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftrightarrow\ ???$$
What do I do next?
| Hint: Use that $\log(2)>0$ and that $t\mapsto \log(t)$ and $t\mapsto e^t$ are monotonic functions to continue:
\begin{align}
\frac{1}{2} \cdot \frac{\log(x+4)}{\log(2)} \le \frac{\log(2x+5)}{\log(2)} &\iff \frac{1}{2}\log(x+4) \le \log(2x+5) \\ &\iff \log(\sqrt{x+4}) \le \log(2x+5) \\ &\iff \sqrt{x+4} \leq 2x+5\\ &\iff \ldots\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate the exact value of $\sin\frac{11\pi}{8}$ Calculate the exact value of $\sin\frac{11\pi}{8}$.
The formula $\sin^2x=\frac12(1–\cos2x)$ may be helpful.
I was thinking of using the Angle-Sum and -Difference Identity:
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
For instance: $\alpha=\frac{20\pi}{8}$ and $\beta=\frac{9\pi}{8}$
Am I on the right track?
| As $\dfrac{11\pi}8=\pi+\dfrac{3\pi}8$ which lies in the third Quadrant,
$\sin<0,\cos <0,\tan>0$
Now $\tan\dfrac{11\pi}8=\cdots=\tan\dfrac{3\pi}8$
Using $\tan A=\dfrac{1-\cos2A}{\sin2A},\tan\dfrac{3\pi}8=\dfrac{1-\cos\dfrac{3\pi}4}{\sin\dfrac{3\pi}4}$
$\sin\dfrac{3\pi}4=\sin\left(\pi-\dfrac\pi4\right)=\sin\dfrac\pi4=?$
$\cos\dfrac{3\pi}4=\cos\left(\pi-\dfrac\pi4\right)=-\cos\dfrac\pi4=?$
$$\dfrac{\sin\dfrac{11\pi}8}{\sqrt2+1}=\dfrac{\cos\dfrac{11\pi}8}1=-\sqrt{\dfrac{\sin^2\dfrac{11\pi}8+\cos^2\dfrac{11\pi}8}{(\sqrt2+1)^2+1^2}}=-\dfrac1{\sqrt{2\sqrt2(\sqrt2+1)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Number of distinct sequences of length 10, containing at least 5 consecutive As or at least 5 consecutive Bs Im stuck on this question, asking for the number of distinct sequences of length 10, which contain at least 5 consecutive As or 5 consecutive Bs (for example ABABBBBBBA should be counted, as should ABBBBAAAAA). I know there's 1024 possible strings, but I'm unsure how to do the calculation without explicitly writing code to sum all the posibilites. How do I go about approaching this question without writing down every possibility?
| The generating function for the number of strings with at most $4$ As in a row and $4$ Bs in a row is
$$
\begin{align}
g(x)
&=\overbrace{\ \frac{1-x^5}{1-x}\ }^\text{$0$-$4$ A}\overbrace{\frac{1\vphantom{x^5}}{1-\underbrace{\frac{x-x^5}{1-x}}_\text{$1$-$4$ B}\underbrace{\frac{x-x^5}{1-x}}_\text{$1$-$4$ A}}}^\text{$0$+}\overbrace{\ \frac{1-x^5}{1-x}\ }^\text{$0$-$4$ B}\\
&=\frac{1-x^5}{1-2x+x^5}\\
&=1+2x+4x^2+8x^3+16x^4+30x^5+\sum_{n=6}^\infty a_nx^n
\end{align}
$$
where $a_n=2a_{n-1}-a_{n-5}$ for $n\ge6$. This recurrence follows from the $1-2x+x^5$ in the denominator. Computing several more terms, we get
$$
\sum_{n=6}^{10} a_nx^n=58x^6+112x^7+216x^8+416x^9+802x^{10}
$$
Since there are a total of $2^{10}=1024$ strings of $10$ As or Bs, we get that there are $1024-802=\bbox[5px,border:2px solid #C0A000]{222}$ strings of $10$ As or Bs with at least $5$ As in a row or $5$ Bs in a row.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find $2^{n-1}\prod_{k = 1}^{n-1}{\left(\cos\theta - \cos\left( \frac{k\pi}{n} \right)\right)}$ Evaluate:
$$2^{n-1} \left(\cos\theta - \cos\left( \frac{\pi}{n} \right)\right)\left(\cos\theta - \cos\left( \frac{\pi}{n} \right)\right)...\left(\cos\theta - \cos\left( \frac{(n-1)\pi}{n} \right)\right)$$
The answer is
\begin{align} \frac{\sin(n\theta)}{sin\theta} \end{align}
| I will use multiple angle formula to solve the problem:
$$\sin(mx) = 2^{m-1}\prod_{k=0}^{m-1} \sin\left(x+\frac{k\pi}{m}\right)$$
Now, to solve our problem,
\begin{align}
&2^{n-1} \prod_{k=1}^{n-1} \left( \cos \theta - \cos \left( \frac{k\pi}{n}\right) \right)\\ & = 2^{n-1} \prod_{k=1}^{n-1} (-2) \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}\right) \label{1}\tag{1}\\
&= 2^{2n-2}\prod_{k=1}^{n-1} (-1) \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}\right) \label{2}\tag{2}\\
&= 2^{2n-2}\prod_{k=1}^{n-1} \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \left( -\sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}\right)\right) \label{3}\tag{3}\\
&= 2^{2n-2}\prod_{k=1}^{n-1} \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}+\pi\right) \label{4}\tag{4}\\
&= 2^{2n-2}\prod_{k=1}^{n-1} \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}+ \frac{(2n-k)\pi}{2n}\right) \label{5}\tag{5}\\
&= \frac{2^{2n-2}\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)\prod_{k=1}^{n-1} \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}+ \frac{(2n-k)\pi}{2n}\right)}{\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{6}\tag{6}\\
&= \frac{2^{2n-2}\prod_{k=0}^{2n-1} \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) }{\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{7}\tag{7}\\
&= \frac{2^{2n-1}\prod_{k=0}^{2n-1} \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) }{2\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{8}\tag{8}\\
&= \frac{\sin((2n)\frac{\theta}{2})}{2\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{9}\tag{9}\\
&= \frac{\sin((n\theta)}{2\sin \left(\frac{\theta}{2} \right) \cos\left( \frac{\theta}{2}\right)} \label{10}\tag{10}\\
&= \frac{\sin((n\theta)}{\sin\theta} \label{11}\tag{11}\\
\end{align}
where in $(\ref{1})$, we use $\cos A - \cos B = -2 \sin \left( \frac{A+B}{2}\right) \sin \left( \frac{A-B}{2}\right)$
in $(\ref{2})$, I just take out the $2$ from the product.
in $(\ref{4})$, we use $\sin(A+\frac{\pi}{2})=\cos(A)$
in $(\ref{6})$, I just multiply top and bottom by the same term.
in $(\ref{7})$, I just rearrange the product terms.
in $(\ref{9})$, I use multiple angle formula.
and finally, I use double angle formula in the denominator.
| {
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"timestamp": "2023-03-29T00:00:00",
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The sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically How could we show that the sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically?
When we take the quotient $\frac{a_n}{a_{n+1}}$ we get $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}$. Howcan we conclude that this quotient is $\geq 1$ ?
| $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}>1 \iff 2n^3+n^2>(n^2-2n+1)(2n+3) \iff 2n^3+n^2>2n^3+3n^2-4n^2-6n+2n+3 \iff 2n^2+4n-3>0 \iff (\sqrt{2}n+\sqrt{2})^2-5>0 \iff (\sqrt{2}n+\sqrt{2})^2>5$
Since the last inequality true and we have only biconditionals, it follows that the first inequality, which is precisely what we want, is correct. Therefore the method is basically using what we want to prove. However, keep in mind that you cannot use $\implies$ for once because then you cannot work backwards.
| {
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Show $f(x,y)=(\frac{x}{|x|+|y|},\frac{y}{|x|+|y|})$ it bjective and continuous and its inverse is continuous I'm working through Renzo's Math 490 Introduction to Topology, http://www.math.colostate.edu/~renzo/teaching/Topology10/Notes.pdf, and on page 16, Example 1.8.4, it states:
that a topologist cannot tell the difference between a circle $S^1
=\{(x,y) \in \mathbb{R}^2\ |\ x^2+y^2=1\}$ and a square $T=\{(x,y)\in \mathbb{R}^2 \ |\ |x|+|y|=1\}$ as there is a function $f: S^1\rightarrow
T$ defined by $f(x,y)=(\frac{x}{|x|+|y|},\frac{y}{|x|+|y|})$.
The book states that is is continuous and bijective, and that its inverse, $f^{-1}(x,y)=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})$ is also continuous.
How do I know the function is bijective and continuous, and that the inverse is continuous?
| Injectivity: $$f(x,y) = f(a,b)\Rightarrow \dfrac{x^2+y^2}{(|x|+|y|)^2} = \dfrac{a^2+b^2}{(|a|+|b|)^2}.$$
But $x^2+y^2=a^2+b^2=1$, so $|xy| = |ab|$ or $x^2y^2 = a^2b^2\Rightarrow a^2+b^2-x^2-\dfrac{a^2b^2}{x^2}=0\Rightarrow (x^2-a^2)(x^2-b^2)=0,$ when $x\neq 0.$ If $|x|=|a|$, then $|y| = |b|$ and $\dfrac{x}{|x|+|y|} = \dfrac{x}{|a|+|b|}$. So it follows that $x=a$. If $|x| = |b|$, then similarly $x=a$, again.
Surjectivity: Given $(x,y)\in T$, you can see that $f(\dfrac{x}{\sqrt{x^2+(1-x)^2}}, \dfrac{1-x}{\sqrt{x^2+(1-x)^2}}) = (x,y)$.
Finally, the open mapping theorem implies that $f^{-1}$ is continuous.
| {
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Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$
Show that $g(x) = \log_2x$ for $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$
In other words, simplify $g(x)=\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12)$ into $\log_2x$
I did:
$$\log_2(2x+x^2) + \log_4(36) - \log_2(6x+12) \\
= \log_2(2x+x^2) + \log_{2^2}(36) - \log_2(6x+12) \\
= \log_2(2x+x^2) + \log_2(\sqrt{36}) - \log_2(6x+12) \\
= \log_2{(\frac{(2x+x^2)6}{6x+12})} \\
= \log_2(\frac{12x+6x^2}{6x+12}) \\
= \log_2(\frac{2+x}{12}) \\
= \log_2(\frac{1}{6}+\frac{x}{12}) \\
= \log_2(\frac{12+6x}{72}) \\
= \log_2(12+6x) - \log_2(72)
= ???$$
What do I do next?
| Hint:
$$12x+6x^2=6x(x+2)$$
$$6x+12=6(x+2)$$
| {
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Elementary Algebra The exercise is to solve this equation: $$ \frac{2}{x} + \frac{3}{y} = 1,\quad x, y \in \mathbb{Z}^{*}$$
I tried as follows:
$$ \frac{2}{x} = 1 - \frac{3}{y} \implies \frac{2}{x} = \frac{y-3}{y} \implies x = \frac{2y}{y-3} \in \mathbb{Z}^{*}, \quad y \neq 3 $$
$$ \frac{3}{y} = 1 - \frac{2}{x} \implies \frac{3}{y} = \frac{x-2}{x} \implies y = \frac{3x}{x-2} \in \mathbb{Z}^{*}, \quad x \neq 2 $$
But how further I have no idea.
| Note that if $x,y\in \mathbb{Z}$ and $x\not=0$, $x\not=2$,
$$y = \frac{3x}{x-2} =3+\frac{6}{x-2}\Leftrightarrow \mbox{ $(x-2)$ divides 6}\Leftrightarrow (x-2)\in\{\pm 1, 2,\pm 3,\pm 6\}\\
\Leftrightarrow x\in\{-4,-1,1,3,4,5,8\}.$$
Hence the equation has $7$ solutions:
$$(-4, 2),\;
(-1, 1),\;
(1, -3),\;
(3, 9),\;
(4, 6),\;
(5, 5),\;
(8, 4).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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proof "$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$" I want to proof:
$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$
I have changed the given eqation $1 = cos^2(x)+sin^2(x)$
$\to$ $cos^2(x) = 1- sin^2(x)$
Then another given eqation:
$cos(2x) = cos^2(x)-sin^2(x)$ $\to$ $-sin^2(x)=cos(2x)-cos^2(x)$
After that I have put those equtions together:
$$cos^2(x)=1-sin^2(x)$$
$$cos^2(x)=1+cos(2x)-cos^2(x)$$
So I have the 1+ {...} structure. How can I go on ?
Or is that the wrong way to proof the equation?
I would really appreciate some hints.
| Usually the easiest way to proof trigonometric identities is to use complex exponentials. Here we have $\cos(x) = \frac{1}{2}(e^{ix}+e^{-ix})$ and thus:
$$ \cos^2(\frac{x}{2})
= \frac{1}{4}(e^{i\frac{x}{2}}+e^{-i\frac{x}{2}})^2
= \frac{1}{4}(e^{ix} + 2+ e^{-ix}) = \frac{1}{2}+\frac{1}{2}\cos(x) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the Units Digit to 7 to the 2945 How would I go about finding the Units Digit to 7^(2945)?
I know that:
7^0 = 1
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
...
7^9 = 40353607
| when u are interested only in the unit digit you can multiply step by step like u did in your post but you can forget the other digits.
So you have $$7^2 = 49$$ however you forget what was it exactly and u only remember that $$7^2 = x\cdot 10 + 9 $$
you multiply again
$$7^3=7^2 \cdot 7 = 70x+63 = (7x+6) \cdot 10+3$$
yet again you only remembering that it looked like
$$7^3=x_3 \cdot 10 +3$$
as you can see you dont have to remember $7^9 = 40353607$, $7^9 =x_9 \cdot 10 +7$ will be enougth
so now we can repeat you what u did but make it simpler. We can note that
*
*if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 7
*if the $7^n$ ends with $7$ then $7^{n+1}$ ends with 9
*if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 3
*if the $7^n$ ends with $1$ then $7^{n+1}$ ends with 1
so the sequence of units digits will be $(1,7,9,3,1,7,3,9,\ldots)$
now you want 2945th element of that sequence. The value repeats itself evey 4 steps.$$a_n=a_{n-4}$$
and also
$$a_n=a_{n-20}$$
$$a_n=a_{n-100}$$
so u can say that:
$$a_{2945}=a_{45}=a_5=a_1=7^1=7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I prove this using mathematical induction? \begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}
Using the mathematical induction step I arrive at this :
\begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}
And I don't see any other way to continue except to divide
\begin{align}
n+1(n+3)
\end{align}
into
\begin{align}
n(n+2)+something
\end{align}
and substitute it with the beginning of the fraction.
But that doesn't get me anywhere.
| You have missed some parenthesis.
Write the induction step as:
$$
1\cdot 3 +2\cdot 4+\cdots +n(n+2)+(n+1)(n+3)=
$$
$$
=\frac{n(n+1)(2n+7)}{6}+(n+1)(n+3)=
$$
$$
=\frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}=\frac{(n+1)(2n^2+13n+18)}{6}=
$$
$$
=\frac{(n+1)(2n^2+4n+9n+18)}{6}=\frac{(n+1)[(2n(n+2)+9(n+2)]}{6}=
$$
$$
=\frac{(n+1)(n+2)(2n+9)}{6}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex solutions of $z^3 + 8 = 0$ (help with resolution method) How do I solve the equation $z^3 + 8 = 0 *($z$ being a complex number)* using polar coordinates?
These are the first lines of a solution method I saw:
$$z^3 + 8 = 0$$
$$\Longleftrightarrow z^3 =-8=8e^{i\pi+ 2 \pi in}$$ ($n$ is a natural number)
Could you explain why is $-8$ equal to that exponential multiplied by $8$?
Thank you so much for your help
| $$
z^3 = -8
$$
In polar coordinates you have $z = r e^{i\theta} = r(\cos\theta+i\sin\theta),$ where $\theta$ is real and $r$ is real and nonnegative.
A basic fact you need here is that $z^n = r^n(\cos(n\theta) + i\sin(n\theta)).$ If $n=3$ that says $z^3 = r^3(\cos(3\theta)+i\sin(3\theta)).$
Thus $-8 = 8\cdot( -1 + (i\cdot0) ),$ so you need $r^3=8$ and $\cos(3\theta)=-1$ and $\sin(3\theta)=0.$
You have $\cos=-1$ and $\sin=0$ at $180^\circ$ or $\pi$ radians. One-third of that is $60^\circ$ or $\pi/3$ radians. At that point you have
$$
\cos60^\circ = \frac 1 2 \quad\text{and}\quad\sin60^\circ = \frac{\sqrt 3} 2.
$$
Thus you have $z = 2\left(\dfrac 1 2 + i\dfrac{\sqrt3} 2\right) = 1 + i\sqrt 3.$
However, sine and cosine are periodic with a period of $360^\circ$, so $\theta\mapsto(\cos(3\theta)+i\sin(3\theta))$ is periodic with a period of $120^\circ$. Thus adding $360^\circ/3$ or $120^\circ$ gives you another solution:
$$
2(\cos180^\circ+i\sin180^\circ) = 2(-1+0i) = -2.
$$
Then add another $120^\circ$ to get a third solution:
$$
2(\cos300^\circ + i\sin300^\circ) = 2\left( \frac 1 2 - i\frac{\sqrt 3} 2 \right) = 1 - i\sqrt3.
$$
Another $120^\circ$ brings you back to where you started.
Algebra alone will tell you there cannot be more than three solutions, as follows. Suppose you find one solution to $z^3+8=0$. Call it $a$. Then
$$
z^3 + 8 = (z-a)(\cdots\cdots\cdots).
$$
The other factor besides $(z-a)$ can be found by long division, and must be a $2$nd-degree polyonomial. Then we find another solution -- call it $b$. Then we have $z^3+8 = (z-a)(z-b)(\cdots\cdots).$ A third solution gives us $(z-a)(z-b)(z-c)(\cdots\cdots).$ But the last factor must be constant since otherwise we'd go up to higher than $3$rd degree.
(The specifics: Once you know $z=-2$ is a solution, you have
$$
z^3+8 = (z-2)(\cdots\cdots\cdots).
$$
Long division yields
$$
z^3+8 = (z-2)(z^2+2z+4).
$$
Then solve $z^2+2z+4=0$ by the usual methods used for quadratic equations and get $z = 1 \pm i\sqrt 3,$ agreeing with what we got by polar coordinates.
| {
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Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| One may recall the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ multiplying by $x$ and making $n \to +\infty$ in $(2)$, using $|x|<1$, gives
$$
\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}. \tag3
$$ Then put $x=\dfrac12$ in $(3)$.
| {
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$\text{lcm}(a,b)+7\gcd (a,b) = a^2+b^2$
Find all positive integers such that $\text{lcm}(a,b)+7\gcd (a,b) = a^2+b^2$.
I thought about setting $a = ck$ and $b = dk$, where $\gcd(c,d) = 1$, which makes our expression become $$cdk+7k = (ck)^2+(dk)^2.$$ Equivalently, $$cd+7 = k(c^2+d^2).$$ How do we continue from here or is there an easier way?
| We must find all pairs of coprime positive $c,d$ such that $c^2+d^2| cd+7$.
( we can let $c\leq d$ to ease calculations)
Notice that $c^2+d^2\geq 2cd$, so we must have $cd\leq 7$.
The only pairs of coprime positive integers $c,d$ with $cd\leq 7$ are:
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(2,3)$
Of course $1+d^2> d+7$ for $d\geq 3$.
So we must only check $(1,1),(1,2),(2,3)$ and it turns out that only $(1,1)$ and $(2,3)$ work ( the first gives $k=4$ and the second $k=1$).
So the set of solutions is:
$(4,4),(2,3)$ and $(3,2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you solve the equation $\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0$? $$
\frac{1}{4}(8^{2x+1})+5\cdot8^x - 3 = 0
$$
My work:
$t = 8^x$
$\frac{1}{4}(t^{2+1})+5t - 3 = 0$
Then I don’t know how to do it, but I think I did it wrong.
| $\frac{1}{4}(t^2+1) +5t - 3 = 0$
$(t^2+1) + 20t - 12= 0$
$ t^2 + 20t - 11= 0$
Find it's roots using suitable method.
Here are some methods
And for this question you have to use quadratic formula.
Edit :
You have two different equation in your question. According to other solution is -
$\frac{1}{4}(8^{2x}\cdot8) +5\cdot8^x - 3 = 0$
$2\cdot8^{2x} +5\cdot8^x - 3 = 0$
Put $8^x = t$
$2t^2 + 5t - 3 = 0$
$2t^2 + 6t - 1t - 3 = 0$
$2t(t + 3) - 1(t + 3) = 0$
$(t + 3) (2t - 1) = 0$
$t = -3$ and $ t = \frac{1}{2}$
So $8^x = \frac{1}{2}$
$2^{3x} = 2^{-1}$
On equating powers,
$3x = -1$
$x = \frac{-1}{3}$
| {
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Mathematical Induction and Divisibility $7\mid 11^n-4^n$ Prove $11^{n} - 4^{n}$ is divisible by 7 for all natural numbers n.
I know this holds for the base case of n = 1. Am I supposed to use $11^{n + 1} - 4^{n + 1}$ for the induction step?
Any help would be greatly appreciated!
Thank you in advance!
| Another way to to do it without induction is to note that the mod cycles for powers of $11$ and $4$ modulo congruent to $7$ correspond to each other.
$$\begin{equation}\begin{aligned}11 &\equiv 4 \pmod 7\\ 11^2 &\equiv 2 \pmod 7\\ 11^3 &\equiv 1 \pmod 7 \\ 11^4 &\equiv 4 \pmod 7 \\ &\vdots\end{aligned}\end{equation}$$
$$\begin{equation}\begin{aligned}4 &\equiv 4 \pmod 7\\ 4^2 &\equiv 2 \pmod 7\\ 4^3 &\equiv 1 \pmod 7 \\ 4^4 &\equiv 4 \pmod 7 \\ &\vdots\end{aligned}\end{equation}$$
So for any $n$, $11^n \equiv 4^n \equiv r \pmod 7$, so $11^n - 4^n \equiv r - r \pmod 7 \implies 11^n - 4^n \equiv 0 \pmod 7$ as desired.
| {
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Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$
Vasc's LCF Theorem does not help here. Also I tried MV method, but without success.
| Here is a solution that relies on an uninspiring chain of single-variable inequalities of the form $\frac{x^2}{x^3+2}\leq \alpha+\beta\ln(x)$ on some interval for various nonnegative real $\alpha$ and $\beta$.
Define $f(x)=\frac{x^2}{x^3+2}$. Note first that $f(x)$ is maximized when
$$f'(x)=\frac{2x(x^3+2)-3x^2(x^2)}{(x^3+2)^2}=\frac{x(4-x^3)}{(x^3+2)^2}=0,$$
i.e. when $x=2^{2/3}$, at which point $f(x)=\frac{2^{1/3}}3$. If $$\min(f(a),f(b),f(c),f(d))\leq\frac43-2^{1/3},$$
then without loss of generality let $f(a)$ be less than this bound; then
$$f(a)+f(b)+f(c)+f(d)\leq \left(\frac43-2^{1/3}\right)+\frac{2^{1/3}}3+\frac{2^{1/3}}3+\frac{2^{1/3}}3=\frac43.$$
So, we may assume that
$$f(a),f(b),f(c),f(d)\geq \frac43-2^{1/3}>0.07;$$
this implies $a,b,c,d>0.37$.
Now, define $g(x)=\frac13(1+\ln(x))$. One may check that $f(x)\leq g(x)$ for all $x>0.6$. So, if $a,b,c,d>0.6$, then
$$f(a)+f(b)+f(c)+f(d)\leq \frac43+\frac13\ln(abcd)=\frac43;$$
otherwise, without loss of generality let $a<0.6$; now $0.37<a<0.6$.
Now, define $h(x)=0.36+0.17\ln(x)$; note that $f(x)<h(x)$ for all $x>0.13$. We thus have
\begin{align*}
f(a)+f(b)+f(c)+f(d)&\leq f(a)+1.08+0.17\ln(bcd)\\
&=1.08+\frac{a^2}{a^3+2}-0.17\ln(a).
\end{align*}
One may verify that the one-variable function
$$j(a)=1.08+\frac{a^2}{a^3+2}-0.17\ln(a)$$
is maximized on the interval $[0.37,0.6]$ at $a=0.6$, where its value is $1.3293<\frac 43$. The inequality is thus proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Prove $|\cos z|^2=\cos^2x+\cosh^2y$ That what I tried
$$
\cos z= \cos (x+iy)= \cos x \cos iy- \sin x \sin iy=\cos x \cosh y-i\sin x \sinh y\\
|\cos z|^2= \cos^2(x) \cosh^2(y)+\sin^2(x)\sinh^2(y)\\
\sinh^2(y)=\cosh^2(y)-1
$$
Then
$$
\cos^2(x) \cosh^2(y)+\sin^2(x) \cosh^2 (y)-\sin^2 (x)= \cosh^2 (y)-\sin^2 (x)
$$
What did I do wrong ?
| The formula you want to prove is wrong: for $z=0$, we have $\lvert\cos0\rvert^2=1$, whereas $\cos^20+\cosh^20=2$.
\begin{align}
\lvert\cos z\rvert&=
\cos^2x\cosh^2y+\sin^2x\sinh^2y\\
&=\cos^2x\sinh^2y+\cos^2x+\sin^2x\sinh^2y\\
&=\cos^2x+\color{red}{\sinh^2y}
\end{align}
Your formula is correct as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the smallest n for inequality involving factorial, $n! > 10^6 $, what is $n$? How to find the smallest n for inequality involving factorial, $n! > 10^6 $, what is n?
I solved it using calculator, that is the answer is $n \approx 9$. How to solve this inequality?
Smallest $n$ giving $n! > 10^6$ ?
Edit
inequality relation
| $$
\begin{align*}
9!&=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\\
&=(2\cdot5)(3\cdot7)(4\cdot6)(8\cdot9)\\
&<10\cdot 30\cdot 30\cdot 80\\
&<8\cdot10^5\\
&<10^6.
\end{align*}
$$
and
$$
\begin{align*}
10!&=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\\
&=(2\cdot5)(3\cdot7)(4\cdot6)(8\cdot9)\cdot10\\
&>10\cdot 20\cdot 20\cdot 70\cdot 10\\
&=28\cdot10^5\\
&>2\cdot10^6\\
&>10^6.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Work and time problem . People leaving work one by one and at the same time reducing their rate of doing work.
Question: A group of $30$ people can complete a job working for $10$ hours a day in
$15$ days. The group starts a work. But at the end of every day, starting
from the first day, one person leaves the group and the remaining people
work for $20$ minutes less on the next day.
On which day will the work be completed?
Answer: $29$th day.
Attempted solution:
large version
large version
Please help me solve this problem.Please tell me where I have gone wrong. Whether in concept or in calculations....Thank you..
| Needed amount of work:
$$
W
= 30 \cdot 10 \frac{\text{h}}{\text{d}} \cdot 15 \text{d}
= 4500 \text{h}
$$
Rate of work per person over time:
$$
r(t) = 10 \frac{\text{h}}{\text{d}}
$$
Active people:
\begin{align}
n(1) &= 30 \\
n(t) &= 30 - (t - 1) = 31 - t
\end{align}
Work done until day $t$, where the group in total works $1/3$ hour less each day:
\begin{align}
W(t)
&= \sum_{k=1}^t
\left(
n(k) \, r(k)- \frac{k-1}{3} \frac{\text{h}}{\text{d}}
\right) \, 1 \text{d} \\
&= \sum_{k=1}^t
\left( (31 - k) 10 - \frac{1}{3} k + \frac{1}{3} \right) \text{h} \\
&= \sum_{k=1}^t
\left(
\frac{931}{3} - \frac{31}{3} k \right) \text{h} \\
&=
\left(
\frac{931}{3} \sum_{k=1}^t 1 - \frac{31}{3} \sum_{k=1}^t k \right) \text{h} \\
&=
\left(
\frac{931}{3} t - \frac{31}{3} \frac{t^2+t}{2} \right) \text{h} \\
&=
\left( \frac{1831}{6} t - \frac{31}{6} t^2 \right) \text{h} \\
\end{align}
We can now solve the quadratic equation:
$$
-\frac{4500 \cdot 6}{31}
= t^2 - \frac{1831}{31} t
= \left( t - \frac{1831}{62} \right)^2 - \left( \frac{1831}{62} \right)^2 \iff \\
t = \frac{1831 \pm\sqrt{1831^2 - 4500 \cdot 6 \cdot 4 \cdot 31}}{62} \\
= \frac{1831 \pm \sqrt{4561}}{62} \in \{ 28.44, 30.62 \}
$$
So only $t= 28.44$ makes sense (we have $30$ people, which are all gone after $30$ days), thus we need $29$ days to achieve the $4500$ hours of work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m?
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any help appreciated.
| $$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$$
$$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5+lx^3+2lx^2+3lx+mx^2+2mx+3m}{x^2+2x+3})=10$$
$$\lim\limits_{x \to \infty}(\frac{(1+l)x^3+(4+2l+m)x^2+(3+l+2m)x+(5+3m)}{x^2+2x+3})=10$$
Note that as the limit exists, the coefficient of $x^3$ has to be $0$, so we get $l=-1$
And also the ratio of the leading coefficient ($x^2$)$=10$, so we get $4+2l+m=10$ and thus $m=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Finding out the limit: $\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}$ Find out the limit of the sum $\lim \frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}$ as $n$ tends to $\infty.$
| Since $x-x^2$ has a peak at $x=\frac{1}{2}$ and decreases monotonically to $0$ as $x \downarrow 0$ or $x \uparrow 1$, we see $\frac{1}{n}\sum_{k=0}^n \sqrt{k(n-k)} = \sum_{k=1}^{n-1} \sqrt{\frac{k}{n}(1-\frac{k}{n})} \ge \sum_{k=1}^{n-1} \sqrt{\frac{1}{n}(1-\frac{1}{n})} = (n-1)\sqrt{\frac{1}{n}-\frac{1}{n^2}} = \sqrt{\frac{(n-1)^2}{n}-\frac{(n-1)^2}{n^2}} \sim \sqrt{n}$ which diverges
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factoring a polynomial of degree 5 I was looking at the solutions to a question about BIBO stability. The transfer function was
$$\dfrac{s^2+1}{s^5+2s^4+4s^3+7s^2+3s+5}=\dfrac{1}{s^3+2s^2+3s+5}$$
So the lecturer factorised the bottom polynomial and cancelled accordingly, but it isn't explained in the solutions. Is there a general method for factorising polynomials of degree 5?
| There exists as simple method to find quadratic factors of monic polynomials of any degree that doesn't involve anything more than straightforward high school level algebra. For 5th degrees polynomials this works as follows. We want to factorize the function:
$$f(x) = x^5 + a x^4 + b x^3 + c x^2 + d x + e$$
where the coefficients are integers. Suppose this has a quadratic factor $g(x) = x^2 -p x - q$, then reducing $f(x)$ modulo $g(x)$ yields:
$$\begin{split}
&\left(a p^3+2 a p q+b p^2+b q+c p+d+p^4+3 p^2 q+q^2\right) x\\
& + a p^2 q+a q^2+b p q+c q+e+p^3 q+2 p q^2
\end{split}$$
We want to choose $p$ and $q$ such that this becomes identical to zero. Let's denote the coefficient of $x$ as $A_1$ and the constant term as $A_2$. Since we know that $q$ will have to divide $e$, which will limit the possible values that $q$ can take, we choose to eliminate $p$ in favor of $q$. We can do this easily by setting $A_1$ and $A_2$ equal to zero and then eliminating the highest powers of $p$ moving on to lower and lower powers until $p$ has been completely eliminated.
Using $A_2=0$ you can express $p^3$ into lower powers of $p$ and $q$. Multiplying that expression by $p$, yields an expression for $p^4$, which you can substitute in the equation $A_1 = 0$, this then not only eliminates the $p^4$ term, it also eliminates the $p^3$ term there. We find the equation $A_3 = 0$ with:
$$A_3 = -e p + d q + b q^2 + a p q^2 + p^2 q^2 + q^3$$
We can then use this equation to eliminate $p^3$ and $p^2$ from $A_2$. This yields the equation $A_4 = 0$ where:
$$A_4 = e^2 p - d e q - b e q^2 - a e p q^2 - d p q^3 + c q^4 + a q^5 + p q^5$$
Since this is linear in $p$ we can easily express $p$ in terms of $q$:
$$p = -q\frac{a q^4+c q^3 -b e q-d e }{q^5-d q^3 -a e q^2+e^2}\tag{1}$$
Substituting this in one of the other equations above will yield a polynomial equation for $q$. The intermediary equation where $p^3$ was eliminated from $A_3$ is the most suitable, it will yield the lowest degree non-trivial equation. This yields:
$$
\begin{split}
&q^{10} +b q^9 + (a c - d)q^8 + \left(c^2 + a^2 d - 2 b d - a e\right) q^7 + \left(a c d - d^2 + a^3 e - 3 a b e + c e\right)q^6\\
& + \left(b d^2 + a^2 c e - 2 b c e - 2 a d e + 2 e^2\right)q^{5}+ \left(d^3 + a b d e - 3 c d e - a^2 e^2 + b e^2\right)q^{4}\\
&+ \left(a d^2 e + b^2 e^2 - 2 a c e^2 - d e^2\right)q^{3}+ \left(b d e^2 - a e^3\right)q^{2}+c e^3 q + e^4 = 0
\end{split}\tag{2}
$$
Since $q$ is an integer the rational root theorem applies, but we already knew that $q$ has to divide $e$, so one can just check the possibilities, any solution can be plugged into Eq. (1) to see of this yields an integer $p$.
Plugging in the coefficients into Eq. (2) yields:
$$q^{10}+ 4 q^9+ 11 q^8 + 27 q^7- 12 q^6 - 114 q^5- 168 q^4 - 285 q^3 + 50 q^2 + 875 q +625=0$$
and we see that $q = -1$ is the only solution. Eq. (1) for the given polynomial becomes:
$$p = -q\frac{2 q^4+7 q^3-20 q-15}{q^5-3 q^3-10 q^2+25}$$
and we see that $p = 0$ for $q = -1$, so the quadratic factor is $x^2 - p x - q = x^2 + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
finding value of indefinite integration of $\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$ finding value of indefinite integration $\displaystyle \int\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$
$\displaystyle \int\frac{y^5(y^2-1)+y(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2+1-2)}{y^{10}+1}dy$
$\displaystyle =\int\frac{y^5+y}{y^8-y^6+y^4-y^2+1}-2\int\frac{y^5+y}{y^{10}+1}dy$
i wan,t be able to proceed after that, could some help me with this
| hint: put $u = y^2$, and note that $(y^5 + y)dy = y(y^4+1)dy = \dfrac{u^2+1}{2}du$. Would this enable you to continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Functional equation $(x + y)f(f(x)y) = x^2 f(f(x) + f(y))$ Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that
$$(x + y)f(f(x)y) = x^2 f(f(x) + f(y))
\mbox{, for all } x,y\in\mathbb{R}^+.$$
I tried out various substitutions such as $x=y$, $x=1$, $x+y=x^2$, and nothin' works.
Thanks in advance for your reply.
| \enspace \frac{f(af(b))}{a}=\frac{f(af(a))}{a} = (3) = \frac{f(2f(a))}{2} = \frac{f(2f(b))}{2} = (3) = \frac{f(bf(b))}{b}=\frac{f(bf(a))}{b}\\
B: \enspace (2) \implies \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \\
A \enspace and \enspace B: \enspace \implies \frac{a}{b} \cdot \frac{f(bf(a))}{b^2} = \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \implies a^3=b^3 \implies a = b \enspace \square \\
$$
Now we prove $ (1) $ has no solutions because $ f(x) <0 $ for some $ x $ .
Let: $ x > 1 $.
$$
(1) \implies \frac{f(f(x)(x^2-x))}{x^2} = \frac{f(f(x) + f(x^2-x))}{x+ x^2 -x}\\
f(f(x)(x^2-x)) = f(f(x) + f(x^2-x)) \implies \text{ with injectivity Lemma 1 } \implies \\
f(x)(x^2-x) = f(x) + f(x^2-x) \implies x^2-x=\frac{f(x)}{f(x)} + \frac{ f(x^2-x)}{f(x)} \implies \\
x^2-x -1 = \frac{f(x^2-x)}{ f(x)}
$$
We see that $ x^2-x -1 < 0 $ for $ 1 <x < \frac{1+\sqrt{5}}{2} $ ( $ \implies f(x^2-x) < 0 \lor f(x) < 0 $ ) and conclude $ (1) $ has no solutions in the range $f:\mathbb{R}^+ \to \mathbb{R}^+$
$ \enspace \square $
(Note that : $ \frac{1+\sqrt{5}}{2} $ is the famous golden ratio ).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Simplify $\frac{1}{\sqrt{-1+\sqrt{2}}}$ In an exercise I got as solution $\frac{1}{\sqrt{-1+\sqrt{2}}}$, it now holds that $\frac{1}{\sqrt{-1+\sqrt{2}}} = \sqrt{1+\sqrt{2}}$. But I really don't see how you can manipulate the left hand side to become the right hand side (quite shameful as a mathematician, I must admit).
Does anyone know how to do it?
| You are given the fraction: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} \tag1$$
Square it: $$\bigg(\frac{1}{\sqrt{-1 + \sqrt{2}}}\bigg)^2 = \frac{1^2}{\sqrt{-1 + \sqrt{2}}^2} = \frac{1}{-1 + \sqrt{2}} \tag2$$
Multiply by $1$: $$\frac{1}{-1 + \sqrt{2}}\times 1 = \frac{1}{-1 + \sqrt{2}}\times \frac{-1 - \sqrt{2}}{-1 - \sqrt{2}} = \frac{-1 - \sqrt{2}}{-1} \tag3$$
Now simplify (divide both terms in the numerator by $-1$): $$\frac{-1}{-1} - \frac{\sqrt{2}}{-1} = 1 - (-\sqrt{2}) = 1 + \sqrt{2} \tag4$$ And now finally, to cancel out the first square that we did in $(2)$, square root the $RHS$: $$\sqrt{1 + \sqrt{2}} \tag5$$
And there is you answer. You should end up with the following equation below: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} = \sqrt{1 + \sqrt{2}} \tag{$\checkmark$}$$
EDIT:
The reason we make $1 = (-1 - \sqrt{2})/(-1 - \sqrt{2})$ is because $-1 - \sqrt{2}$ is the conjugate of $-1 + \sqrt{2}$. So pursuant to what @GeorgeN.Missailidis said, we did not need to square the fraction first, because all we could do was multiply it by $(\sqrt{-1 - \sqrt{2}})/(\sqrt{-1 - \sqrt{2}})$ since $\sqrt{-1 - \sqrt{2}}$ is the conjugate of $\sqrt{-1 + \sqrt{2}}$ anyway, but it is neater to cancel out the major roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Solve system of equations with square roots and fractions $$
\begin{cases}
\left(\frac{1+\sqrt{3}}{2}\right)^2x+\left(\frac{1+\sqrt{3}}{2}\right)y+1=0\\
\left(\frac{1-\sqrt{3}}{2}\right)^2x+\left(\frac{1-\sqrt{3}}{2}\right)y+1=0
\end{cases}
$$
I tried setting the equations equal to each other and substituting variables but those methods just made the equation more conplicated.
| swetting $$a=\frac{1+\sqrt{3}}{2}$$ and $$b=\frac{1-\sqrt{3}}{2}$$ then we have to solve
$$a^2x+ay=-1$$
$$b^2x+by=-1$$
from the first equation we get
$$y=-\frac{1+a^2x}{a}$$
plugging this in the second equation we get
$$a^2bx-b(1+a^2x)=-1$$ thus
$$x=\frac{b-1}{ab^2-a^2b}$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Power Series involving Double Factorials This comes from my partial solution to another question. I need to find a closed form for the following summation
$$\sum_{n=1}^{\infty}\frac{n!}{(2n+1)!!}n^x$$
Where $x$ is a fixed integer
This appears similar to many arcsine series and formulas
Particular values:
$$\begin{array}{c|c|c|}
& 0 & 1 & 2 & 3 & 4 \\ \hline
x & \frac{\pi}{2}-1 & \frac{0\pi}{2}+1 & \frac{\pi}{2}+1 & \frac{3\pi}{2}+5 & \frac{16\pi}{2}+25\\ \hline
\end{array}$$
$$\begin{array}{c|c|c|}
& 5 & 6 & 7 & 8 \\ \hline
x & \frac{105\pi}{2}+165 & \frac{841\pi}{2}+1321 & \frac{7938\pi}{2}+12469 & \frac{86311 π}{2}+135577\\ \hline
\end{array}$$
Update 1: @RobertIsrael found a closed form for this sum's Exponential Generating Function, but I have been unable to find any expression for the Maclaurin series.
Update 2: Grant B. has found a simple recurrence, but a complete closed form has not been found
| Why just even powers of $n$? Let
$$ F(j) = \sum_{n=1}^\infty \frac{n! n^j}{(2n+1)!!} = \sum_{n=1}^\infty \frac{2^{-1-n} \sqrt{\pi} n! n^j}{\Gamma(n+3/2)} $$
The exponential generating function is
$$\eqalign{E(z) &= \sum_{j=0}^\infty \frac{F(j)}{j!} z^j\cr
&= \sum_{n=1}^\infty \frac{2^{-1-n} \sqrt{\pi} n! e^{zn}}{\Gamma(n+3/2)}\cr
&= -1 + \frac{2\; e^{-z/2}}{\sqrt{2-e^z}} \arcsin\left(e^{z/2}/\sqrt{2}\right)}$$
EDIT: The last line could be obtained (in hindsight) as follows. As you know, (for $|x| < 1$)
$$\arcsin(x) = \sum _{k=0}^{\infty }{\frac { \left( 2\,k \right) !\,{4}^{-k}{x}^{2\,k
+1}}{ \left( k! \right) ^{2} \left( 2\,k+1 \right) }}
$$
Taking $x = e^{z/2}/\sqrt{2}$, we want to multiply this by
$$ \frac{2 e^{-z/2}}{\sqrt{2-e^z}} = \frac{1}{x \sqrt{1-x^2}} = \sum_{k=0}^\infty \frac{(2k)!\, 4^{-k} x^{2k-1}}{(k!)^2}$$
The Cauchy product is
$$ \sum_{n=0}^\infty \sum_{k=0}^n \frac{(2k)! 4^{-k}}{(k!)^2} \frac{(2(n-k))! 4^{-n+k}}{((n-k))!^2(2n-2k+1)} x^{2n}$$
And it turns out that
$$ \sum _{k=0}^{n}{\frac { \left( 2\,k \right) !\, \left( 2\,n-2\,k
\right) !}{ \left( k! \right) ^{2} \left( \left( n-k \right) !
\right) ^{2}(2n-2k+1)}}={\frac {\sqrt {\pi}{4}^{n}n!}{2\,\Gamma
\left( n+3/2 \right) }}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Computing $16^{74} \bmod 65$ I have the following problem. I need to calculate $16^{74} \bmod 65$. Is there a way to do this without a calculator?
Thank you for your help in advance!
EDIT (M.S.): Notice that this question is linked as an example in one of the answers to the FAQ post: How do I compute $a^b\,\bmod c$ by hand?
| Repeatedly raising to as small powers as possible and immediately reducing modulo $65$ we have
$$(16)^{74} = (16^2)^{37} = (256)^{37} = (260 - 4)^{37} \equiv (-4)^{37} = - 4^{37} = -4 \cdot 4^{36} = -4 \cdot (4^4)^9 = \\
-4 \cdot (256)^9 = -4 \cdot (260 - 4)^9 \equiv -4 \cdot (-4)^9 = 4^{10} = 4 \cdot (4^3)^3 = 4 \cdot (64)^3 = 4 \cdot (65 - 1)^3 \equiv \\
4 \cdot (-1)^3 = -4 \equiv 61 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
Square root positive/negative What is the square root of $9$? Is it always $\pm 3$ or just positive $3$?
Trying to find the solution set of this equation : $x-3 = \sqrt{x+3}$
I want to understand the concept of square root to solve the problem.
Thanks
| There are two numbers $x$ such that $x^2=9$ : $x=3$ and $x=-3$, but with the symbol $\sqrt{9}$ we indicate only the positive one, so $\sqrt{9}=3$.
In other words, the symbol $\sqrt{x}$ indicates, by definition, the positive number $y$ such that $y^2=x$.
This implies that if we search the solution of the equation $x-3=\sqrt{x+3}$, than we must have $x-3\ge0$ and, if squaring we find a solution such that $x-3<0$ this is not a solution of the given equation.
We can say that the equation $x-3=\sqrt{x+3}$ is equivalent to the system:
$$
\begin{cases}
x-3\ge0\\(x-3)^2=x+3
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Similar triangles: the shortest side of triangle $B$ measures $6$. Find the other two sides. Two triangles are similar. Triangle $A$ has sides measuring $3$, $4$, and $5$. The shortest side on triangle $B$ measures $6$. What are the lengths of the other two sides of triangle $B$?
| Let $x,y$ be respectively the lengths of the medium and longest sides of triangle "$B$". Then similarity of triangles implies: $\dfrac{3}{6} = \dfrac{4}{x} = \dfrac{5}{y}\implies x = 8, y = 10.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral $\int \frac{dx}{\sin^2 (x) + \sin(2x)}$ We have to do the following integration
$$\int \frac{dx}{\sin^2 (x) + \sin(2x)}$$
I tried as
In the numerator, $1 = \sin^2(x)+\sin(2x)+\cos^2(x)-\sin(2x)$
But now how to proceed?
| \begin{align}
\int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{\sec^2(x)}{\sin^2(x)\sec^2(x)+2\sin(x)\cos(x)\sec^2(x)}dx\\
&=\int\frac{\sec^2(x)}{\tan^2(x)+2\tan(x)}dx
\end{align}
Letting $u:=\tan(x)$, then $du=\sec^2(x)dx$ gives
\begin{align}
\int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{du}{u^2+2u}\\
&=\int\frac{du}{(u+1)^2-1}
\end{align}
Letting $z:=u+1$, then $dz = du$ gives
\begin{align}
\int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{dz}{z^2-1}\\
&=-\text{artanh }(z)+C\\
&=-\text{artanh }(u+1)+C\\
\end{align}
Reversing the final substitution gives
$$\int\frac{dx}{\sin^2(x)+\sin(2x)} = -\text{artanh }(\tan(x)+1)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Is this a subspace with the given eigenvectors as its basis vectors? Given the eigenvectors:
$$\begin{pmatrix} 2 \\ 3 \end{pmatrix}, \begin{pmatrix}4 \\ 5\end{pmatrix}$$
And this linear combination equation
$$ \begin{pmatrix} 2 \\ 4 \end{pmatrix} = c_1 \begin{pmatrix} 2 \\ 3 \end{pmatrix}+c_2\begin{pmatrix}4 \\ 5\end{pmatrix}$$
Solving the equation gives: $$ c_1 = 3, c_2 = -1 $$
Which as far as I know could be written as the vector:
$$[v]_u = \begin{pmatrix}3 \\ -1 \end{pmatrix}$$
My question is, in this transition from the vector: \begin{pmatrix}2 \\ 4 \end{pmatrix} to this vector \begin{pmatrix}3 \\ -1 \end{pmatrix}
Is the latter vector $(3, -1)$ in a subspace consisting of the mentioned eigenvectors in the linear combination as the basis vectors for this subspace? I can't help but notice how similar it is to this linear combination in an euclidian space with it's basis vectors:
$$\begin{pmatrix} 2 \\ 4 \end{pmatrix} = 2 \begin{pmatrix}1 \\ 0 \end{pmatrix}+4 \begin{pmatrix}0 \\ 1 \end{pmatrix}$$
EDIT:
Let me rephrase the question to a more general one. If we have a linear combination $$ c_1 v_1+c_2v_2 $$
Could one say that this linear combination would create a subspace with the two vectors $v_1$ and $v_2$ as it's basis vectors?
| Note that $\frac{1}{2} \left[ 3 \begin{pmatrix} 4 \\ 5 \end{pmatrix} - 5 \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right] = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, so the space with basis vectors $\Bigg \lbrace \begin{pmatrix} 4 \\ 5 \end{pmatrix} , \begin{pmatrix} 2 \\ 3 \end{pmatrix} \Bigg\rbrace$ is in fact $\mathbb{R}^2$. Hence every vector will be in the subspace with basis vectors the eigenvectors.
In general, $n$ linearly independent vectors will form a basis for $\mathbb{R}^n$, and in this case of $n = 2$ you do indeed have linearly independent vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{-2}+u^{-1}) \right|_1^{1+\sqrt{5}}={1\over 2}$$
Prove $(1)$ using an alternative method other than substitution method.
| Set $\frac{3-\sqrt{5}x}{1+\sqrt{5}x}=u$, we have $\frac{-4\sqrt{5}}
{(1+\sqrt{5}x)^2} dx=du$ thus
$$\int_{0}^{1} {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx=\frac{1}{-4\sqrt{5}}\int_{3}^{\sqrt{5}-2}u\,du=\frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Deriving a Series Representation of an Integral (Riemann Zeta Function) It is known that
$$\int_0^\infty\frac{x^n}{e^x-1}dx=n!\zeta(n+1)$$
for integer values of $n$ (this is also generalises, but that's not important for this question). I have also had a look at how to arrive at this expression, starting from the series representation of the Riemann Zeta function.
However, just out of interest, I've tried to see if I could manage to derive the series representation for the case that sparked my interest in the first place $(n=3)$ on my own - that is, working backwards from this formula to the series that initially defines the Riemann Zeta function. I did manage to arrive at a series representation, but it's not the correct series.
Here's what I did:
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\int_0^\infty\frac{x^3(e^x+1)}{e^{2x}-1}dx\\
&=\int_0^\infty\frac{x^3}{e^{2x}-1}dx+\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx\\
&=\frac{1}{16}\int_0^\infty\frac{x^3}{e^x-1}dx+\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx
\end{align}
Subtracting the first term on the RHS, we have
\begin{align}
\frac{15}{16}\int_0^\infty\frac{x^3}{e^x-1}dx=\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx
\end{align}
Multiplying across the fraction, we have
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{16}{15}\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx\\
&=\frac{1}{15}\int_0^\infty\frac{x^3e^{\frac{x}{2}}}{e^x-1}dx
\end{align}
Letting $z:=e^x$, then $x=\log z\implies dx=\frac{dz}{z}$.
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\int_1^\infty\frac{\log^3z\sqrt{z}}{z(z-1)}dz
\end{align}
Letting $u:=\frac{1}{z}$, we have $z=\frac{1}{u}\implies dz=-\frac{du}{u^2}$, $u(1)=1$ and $u(z\rightarrow\infty)=0$. Thus
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\int_0^1\frac{\log^3\left(\frac{1}{u}\right)\sqrt{u}}{u^2}\frac{1}{1-\frac{1}{u}}du\\
&=-\frac{1}{15}\int_0^1\frac{\log^3u\sqrt{u}}{u}\frac{1}{u-1}du\\
&=\frac{1}{15}\int_0^1\frac{\log^3u}{\sqrt{u}}\frac{1}{1-u}du\\
&=\frac{1}{15}\int_0^1\frac{\log^3u}{\sqrt{u}}\sum_{k=0}^\infty u^k du\\
&=\frac{1}{15}\sum_{k=0}^\infty\int_0^1u^{k-\frac{1}{2}}\log^3u du
\end{align}
Letting $t:=\log u$, then $u=e^t\implies du=e^tdt$, $t(u\rightarrow 0)=-\infty$ and $t(1)=0$.
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\sum_{k=0}^\infty\int_{-\infty}^0 t^3e^{\left(k+\frac{1}{2}\right)t}dt
\end{align}
Letting $p:=-t$, then $dt=-dp$, $p(t\rightarrow -\infty)=\infty$ and $p(0)=0$.
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\sum_{k=0}^\infty\int_0^\infty p^3e^{-\left(k+\frac{1}{2}\right)p}dp
\end{align}
Letting $s:=\left(k+\frac{1}{2}\right)p$, then $dp=\frac{ds}{k+\frac{1}{2}}$. The bounds of integration are unchanged, and thus
\begin{align}
\int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\sum_{k=0}^\infty\frac{1}{\left(k+\frac{1}{2}\right)^4}\int_0^\infty s^3e^{-s}ds\\
&=\frac{\Gamma(4)}{15}\sum_{k=0}^\infty\frac{1}{\left(k+\frac{1}{2}\right)^4}\\
&=\frac{2}{5}\sum_{k=0}^\infty\frac{1}{\left(k+\frac{1}{2}\right)^4}
\end{align}
EDIT: Apparently, this is simply a convoluted way of deriving the correct answer, as the series evaluates to $\frac{\pi^4}{6}$, making the final expression equal to the known value of
$$\int_0^\infty\frac{x^3}{e^x-1}dx = 6\sum_{k=1}^\infty\frac{1}{k^4} = \frac{\pi^4}{15}$$
Many thanks to Simple Art for verifying this result!
| Your so called proof is very much unorthodox. A much shorter proof can be found in this post and it basically goes as follows:
$$\begin{align}\int_0^\infty\frac{x^k}{e^x-1}\ dx&=\int_0^\infty\frac{x^ke^{-x}}{1-e^{-x}}\ dx\\&=\int_0^\infty x^ke^{-x}\sum_{n=0}^\infty e^{-nx}\ dx\\&=\int_0^\infty\sum_{n=1}^\infty x^ke^{-nx}\ dx\\&=\int_0^\infty\sum_{n=1}^\infty\frac1{n^{s+1}}y^{k+1}e^{-y}\ dy\tag{$y=nx$}\\&=\sum_{n=1}^\infty\frac1{n^{s+1}}\int_0^\infty y^{k+1}e^{-y}\ dy\\&=\zeta(s+1)\Gamma(s+1)\end{align}$$
The last line comes by definition. And when $s\in\mathbb N$, $\Gamma(s+1)=s!$
Around the middle part, where you do substitutions, there appears to be missing factors (see comments).
In essence, I think the above is a much simpler derivation that uses the exact same geometric expansion, just in a far simpler way.
As you have edited your question, I give you the relief that your solution is correct. Particularly, scroll down quite a ways and you will find a nice surprise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)=\sin\left(\frac{C}{2}\right)$$
But I have no idea how to proceed.
| Hint:
Prosthaphaeresis Formulas: $\cos A+\cos B=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2$
Double angle formula: $1-\cos C=2\sin^2\dfrac C2$
$\dfrac{A+B}2=\dfrac\pi2-\dfrac C2\implies\cos\dfrac{A+B}2=\sin\dfrac C2$
As $0<c<\pi, \sin\dfrac C2\ne0$
So, we have $\cos\dfrac{A-B}2=2\cos\dfrac{A+B}2$
Expand $\cos\dfrac{A\pm B}2$ and divide both sides by $\cos\dfrac A2\cos\dfrac B2$
Finally, we have $\tan\dfrac A2=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}$ where $2s=a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$$
$$\therefore \bbox[ #FFFDD0, 10px, Border:2px solid #DC143C]{p = {C\over \sqrt{A^2 + B^2}}, \cos \alpha = {-A\over \sqrt{A^2 + B^2}},\sin\alpha = {-B\over \sqrt{A^2 + B^2}}} $$
I did not get the $(3)$ part. Where does $\displaystyle{\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}$ come from ?
| $$
{p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B}
\\
\implies \frac{p^2}{C^2}=\frac{\cos^2(\alpha)}{A^2}=\frac{\sin^2(\alpha)}{B^2}
$$
If
$$
\frac{a}{b}=\frac{c}{d}=k
\\
\implies a=bk, c=dk
$$
So the value of :
$$
\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k(b+d)}{b+d}=k=\frac{a}{b}=\frac{c}{d}
$$
Simply put, in a proportion of two or more equal ratios, the sum of the antecedents divided by the sum of the consequents is equal to any one of the ratios.
In the same way :
$$
\frac{p^2}{C^2}=\frac{\cos^2(\alpha)}{A^2}=\frac{\sin^2(\alpha)}{B^2}={{\sin^2 \alpha + \cos^2 \alpha}\over {A^2 + B^2}}
$$
Hence :
$$
\frac{p}{C}={\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A combinatorial identity with binomial coefficients and floor function. Show that $$ \sum_{m=0}^{2k+1}{2^{m}\binom{n}{m}\binom{n-m}{\bigl\lfloor \frac{2k+1-m}{2} \bigr\rfloor}}=\binom{2n+1}{2k+1} $$
I tried expending the sum and using induction, but could not complete the induction step; I tried using Pascal's identity to obtain $$ \binom{n-m}{\bigl\lfloor \frac{2k+3-m}{2} \bigr\rfloor}=\binom{n-m+1}{\bigl\lfloor \frac{2k+3-m}{2} \bigr\rfloor}-\binom{n-m}{\bigl\lfloor \frac{2k+1-m}{2} \bigr\rfloor} $$ but couldn't find other identities to complete the induction step. Searching Gould's Combinatorial Identities brought up nothing, though I could easily had missed something useful there.
I'm looking to complete the induction step, but would also like to find an algebraic or a combinatorial proof. I would like to avoid using trigonometric and root functions, but would like to use complex numbers in cartesian form.
| Suppose we seek to prove that
$$\sum_{k=0}^{2m+1} {n\choose k} 2^k
{n-k\choose \lfloor (2m+1-k)/2 \rfloor}
= {2n+1\choose 2m+1}.$$
Observe that from first principles we have that
$${n\choose \lfloor q/2 \rfloor} = {n\choose n-\lfloor q/2 \rfloor} =
\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^n}{w^{n+1}}
\left(1+z
+wz^2+wz^3
+w^2z^4+w^2z^5+\cdots\right)
\; dw \; dz.$$
This simplifies to
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}}
\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^n}{w^{n+1}}
\left(\frac{1}{1-wz^2}+z\frac{1}{1-wz^2}\right)
\; dw \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1+z}{z^{q+1}}
\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^n}{w^{n+1}}
\frac{1}{1-wz^2}
\; dw \; dz.$$
This correctly enforces the range as the reader is invited to verify
and we may extend $k$ beyond $2m+1,$ getting for the sum
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1+z}{z^{2m+2}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^n}{w^{n+1}}
\frac{1}{1-wz^2}
\sum_{k\ge 0} {n\choose k} 2^k z^k \frac{w^k}{(1+w)^k}
\; dw \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1+z}{z^{2m+2}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^n}{w^{n+1}}
\frac{1}{1-wz^2}
\left(1+\frac{2wz}{1+w}\right)^n
\; dw \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1+z}{z^{2m+2}}
\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w+2wz)^n}{w^{n+1}}
\frac{1}{1-wz^2}
\; dw \; dz.$$
Extracting the inner coefficient now yields
$$\sum_{q=0}^n {n\choose q} (1+2z)^q z^{2n-2q}
= z^{2n} \sum_{q=0}^n {n\choose q} (1+2z)^q z^{-2q}
\\= z^{2n} \left(1+\frac{1+2z}{z^2}\right)^n
= (1+z)^{2n}.$$
We thus get from the outer coefficient
$$\frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{(1+z)^{2n+1}}{z^{2m+2}} \; dz$$
which is
$$\bbox[5px,border:2px solid #00A000]{
{2n+1\choose 2m+1}}$$
as claimed. I do believe this is an instructive exercise.
| {
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"url": "https://math.stackexchange.com/questions/2087559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Dividing rational expression? Just to clarify, I do not want anyone to do the problem for me. I just need someone to explain this stuff to me because I'm completely lost, and I would really appreciate it if you could take the time to do so.
In the book I've been reading they have this example:
$$\frac{y^3-1}{y^2-1} \div \frac{y^2 + y + 1}{y^2 + 2y + 1}$$
So I understand the basic concept: you flip the fraction so that you're multiplying, then factor everything on top and everything on the bottom, then you cancel things that are the same on top and on the bottom. What I don't understand is, when they factor the numerator, they get rid of a "2" and I'm not really sure how they did that:
$$\frac{y^3 - 1}{y^2 - 1} \div \frac{y^2 + 2y + 1}{y^2 + y + 1}$$
Factor:
$$(y - 1)(y + 1)(y + 1)(y^2 + y + 1) \div (y + 1)(y - 1)(y^2 + y + 1)$$
Cancel things out:
$$y + 1$$
So the problem I have is that they have $(y^3 - 1)(y^2 + 2y + 1)$ on top, but somehow they factor it out to: $(y - 1)(y + 1)(y + 1)(y^2 + y + 1)$; how is that possible? I get that $(y^3 - 1)$ factors out to:
$$(y - 1)(y + 1)(y + 1)$$
but how did they get rid of the "2" in $(y^2 + 2y + 1)$ so that it's $(y^2 + y + 1)$?
| $(y^3-1)/(y^2-1) ÷ (y^2 + y + 1)/(y^2 + 2y + 1)=$
$(y^3-1)/(y^2-1)\times (y^2 +2 y + 1)/(y^2 + y + 1)=$
$\frac {(y^3 -1)(y^2+2y+1)}{(y^2-1)(y^2+y+1)}=M$
I think you've followed and figured all that on your own.
Now comes.... experience, intuition and elbow grease.
Does $y^3 - 1$ factor? I know it does because I've been doing this stuff for 35 years. If it does then it will be of the form $(ay^2 + by + c)(dy + e)$ where $ad = 1$ and $ce = -1$ and ... some mess of inside stuff will cancel out. (Actually $ady^3 + (bd + ea)y^2 + (be+ cd)y + ec = x^3 - 1$ so $ad = 1; bd+ea = 0; be+cd=0; ec=-1$).
First guess is $e =\pm 1; c= \mp1; a = 1;d = 1$ and we get $(y^2 + by \mp 1)(y \pm 1) = y^3 + by^2 \mp y \pm y^2 \pm by - 1= y^3 +(b \pm 1)y^2 + (\pm b \mp 1)y -1 = y^3 - 1$ so $b \pm 1 = 0$ and $b - 1 = 0$ so $b = 1$ and $y^3 - 1 = (y-1)(y^2 + y + 1)$.
Indeed, we should probably learn and put it in a arsenel of tricks that $(x-1)(x^n + x^{n-1} + ...... + x + 1) = x^{n+1} + (1-1)x^{n-1} + ..... + (1-1)x -1 = x^{n+1} - 1$ is a good bit of factoring that we will use a lot in the future.
So $y^3 - 1 = (y-1)(y^2 + y+ 1)$ and we have:
$M =\frac{(y^3 - 1)(y^2+2y+1)}{(y^2 - 1)(y^2+ y + 1)} = \frac{(y-1)(y^2+y+1)(y^2 + 2y + 1)}{(y^2- 1)(y^2 + y+1)} = \frac{(y-1)(y^2 + 2y + 1)}{y^2 - 1}$
Now does $y^2 - 1$ factor. Well, above we say $(x-1)(x^n + .... + 1) = x^{n+1} -1$ so $y^2 - 1 = (y-1)(y+1)$. We can check that: $(y-1)(y+1) = y^2 - y + y - 1 = y^2-1$.
So
$M = \frac{(y-1)(y^2 + 2y + 1)}{y^2 - 1}=$
$\frac{(y-1)(y^2 + 2y + 1)}{(y-1)(y+1)} =\frac{(y^2 + 2y + 1)}{(y+1)}$
Now the question is: does $y^2 + 2y + 1$ factor? Well, yes... $y^2 + 2y + 1 = (y+1)^2$
So $M = \frac{y^2 + 2y + 1}{y+1} = \frac{(y+1)^2}{y+1} = y+1$. and.... we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Integral $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ We have to evaluate the following integral:
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I tried this:
I multiplied both the numerator and denominator by $\sec x$
And substituted $\tan x = t$.
But after that I got stuck.
The book where this is taken from gives the following as the answer: $$\ln(1+t)-\frac14\ln(1+t^4)+\frac1{2\sqrt2}\ln\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}-\frac12\tan^{-1}t^2+c$$ where $t=\sqrt{\cot x}$
| $$\displaystyle\int\frac{\sqrt{\sin x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}\;=\;\int\frac{dx}{1+\sqrt{\cot x}},\\
\displaystyle\cot x=z^{2} \;\;\Rightarrow\;\; -(1+z^{2})dx=2zdz,\\
\displaystyle\int\frac{dx}{1+\sqrt{\cot x}} \;=\; -\int\frac{2zdz}{(1+z^{2})(1+z)} \;=\; -2\int\frac{z(z-1)dz}{z^{4}-1}.\\
\displaystyle\frac{z}{z^{4}-1}\;=\;\frac{1}{2}\left( \frac{z}{z^{2}-1}-\frac{z}{z^{2}+1} \right),\\
\displaystyle\frac{z^{2}}{z^{4}-1}\;=\;\frac{1}{2}(\frac{1}{z^{2}-1}+\frac{1}{z^{2}+1}),\\
\displaystyle\frac{z^{2}-z}{z^{4}-1}=\frac{1}{2}(\frac{1-z}{z^{2}-1})+\frac{1}{2}(\frac{1}{z^{2}+1}+\frac{z}{z^{2}+1})=-\frac{1}{2(z+1)}+\frac{1}{2}(\frac{1}{z^{2}+1}+\frac{z}{z^{2}+1}).\\
\displaystyle-2\int\frac{z(z-1)dz}{z^{4}-1}\;=\;\ln\left| z+1 \right|-\arctan{z}-\ln\sqrt{z^{2}+1}+c,\\
\displaystyle\int\frac{\sqrt{\sin x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}\;=\;\ln\left(\frac{\sqrt{\cot{x}}+1}{\sqrt{\cot{x}+1}} \right)-\arctan{\sqrt{\cot{x}}}+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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} |
How is $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1$? I wondered, why this:
$$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$$
is equal to $\sqrt{2}-1$.
Can anyone explain me, why this is equal? :/
| \begin{align}
\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} &= \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\cdot\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}-1}}\\
&=\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{2-1}}\\
&=\sqrt{\left(\sqrt{2}-1\right)^2}\\\\
&=\sqrt{2}-1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Probability of selecting a ball from a box. Suppose we have $12$ identical balls. Think of the balls as having distinct ID numbers, $1-12$. We have $3$ identical boxes, each of which can contain exactly $4$ balls. One after the other the balls are thrown at random into one of the three boxes.
Suppose Mr. A will buy one ball from each box, so he will buy a total of three balls from the three boxes. Habit of Mr. A is to buy that ball from the box which ID is smallest of all other balls in that box.
For example, the randomly thrown balls in a box have ID $4$, $8$, $9$, and $12$. So Mr. A will buy the ball with ID $4$ from that box.
Then Mr. A goes to another box and here the balls have ID $1$, $3$, $6$, and $11$. He will buy the ball with ID $1$ from that box.
At last, Mr. A goes to the last box and here the balls have ID $2$, $5$, $7$, and $10$. He will buy the ball with ID $2$ from that box.
My question is:
What is the probability that Mr. A will buy the ball with ID $i$, where $i=1,2,\ldots,12$?
What I think is that we need to consider the following two things to calculate the probability:
$(1)$ the position of a ball in the box. Mr. A will buy the ball with ID number $2$ if and only if the box does not contain the ball with ID number $1$.
$(2)$ If Mr. A buy a ball with smallest ID from the first box in front of him, then are there $11$ balls remaining to buy another two balls or are there $8$ balls (excluding all $4$ balls from the first box)?
| For $1\lt j\lt k$, there are $3!\binom{12-k}{3}\binom{8-j}{3}$ ways to buy $1,j,k$.
$3!$ due to the arrangements of $1,j,k$. $\binom{12-k}{3}$ due to the number of ways to choose $3$ balls greater than $k$. $\binom{8-j}{3}$ due to the number of ways to choose $3$ balls greater than $j$ from those left over from choosing $3$ greater than $k$.
The number of ways for $n$ to be the greatest is
$$
\begin{align}
3!\binom{12-n}{3}\sum_{j=2}^{n-1}\binom{8-j}{3}
&=3!\binom{12-n}{3}\left[\binom{7}{4}-\binom{9-n}{4}\right]\\
&=3!\binom{7}{4}\binom{12-n}{3}-3!\binom{7}{4}\binom{12-n}{7}
\end{align}
$$
The number of ways for $n$ to be the middle is
$$
\begin{align}
3!\binom{8-n}{3}\sum_{k=n+1}^9\binom{12-k}{3}
&=3!\binom{8-n}{3}\binom{12-n}{4}\\
&=3!\binom{7}{4}\binom{12-n}{7}
\end{align}
$$
Since the total number of arrangements is $\frac{12!}{(4!)^3}$, and $\frac{3!\binom{7}{4}}{\frac{12!}{(4!)^3}}=\frac1{165}$, the probability of buying the ball labeled $n$ would be
$$
p(n)=\frac1{165}\binom{12-n}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$.
Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.
*Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k}
&=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\
&= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\
\end{align}$$
Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed
$$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\
&\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\
&\small= o\left(\frac{2^n}n \right)
\end{align}$$
Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$
This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).
| Let
$$ S_n=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}. \tag{1}$$
Clearly
$$ S_n=\sum_{k=1}^n\frac{n}{k}\frac1{2^{n-k}}\ge \sum_{k=1}^n\frac1{2^{n-k}}=2-\frac{1}{2^n}. $$
Define
\begin{eqnarray}
f_n(x)=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}\left(\frac12x\right)^k
\end{eqnarray}
and then $f_n(0)=0,f_n(2)=S_n$ and
$$ f_n'(x)=\frac{n}{2^n}\sum_{k=1}^nx^{k-1}=\frac{n}{2^n}\frac{1-x^n}{1-x}.$$
So
\begin{eqnarray}
S_n&=&\frac{n}{2^n}\int_0^2\frac{1-x^n}{1-x}dx\\
&=&\frac{n}{2^n}\int_0^1\frac{1-x^n}{1-x}dx+\frac{n}{2^n}\int_1^2\frac{x^n-1}{x-1}dx.
\end{eqnarray}
Since
$$ \int_0^1\frac{1-x^n}{1-x}dx=\ln n+\gamma+o(1),$$
one
$$ \frac{n}{2^n}\int_0^1\frac{1-x^n}{1-x}dx=o(1). $$
Noting that if $x\in[1,2]$, then $x-1\ge1$ and $x^n-1\le x^n$, one has
$$ S_n\le o(1)+\frac{n}{2^n}\int_1^2\frac{x^n-1}{x-1}dx\le o(1)+\frac{n}{2^n}\int_1^2x^ndx=o(1)+\frac{n}{2^n}\frac{1}{n+1}(2^{n+1}-1).\tag{2}$$
From (1) and (2), one has
$$ 2-\frac{1}{2^n}\le S_n\le o(1)+\frac{n}{2^n}\frac{1}{n+1}(2^{n+1}-1).$$
Letting $n\to\infty$, one has
$$ \lim_{n\to\infty}S_n=2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| Hint: We have $$\tan 7.5^\circ =\frac {\sin 7.5^\circ}{\cos 7.5^\circ} =\frac {2\sin^2 7.5^\circ}{2\cos 7.5^\circ \sin 7.5^\circ} =\frac {1-\cos 15^\circ}{\sin 15^\circ} =\frac {1- \cos (45^\circ -30^\circ)}{\sin (45^\circ -30^\circ)} $$
Can you take it from here? Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of multivariable $\frac{x^2+y^2}{ x^4+y^4}$ Find the limit
$$\lim_{(x, y) \to (0,0)} \frac {x^2+y^2 }{x^4+y^4 }$$
This limit does not exists since when convert it into polar we get $\frac {1 }{ r^2 (1-2 \sin^2 \theta \cos^2 \theta)}$ which is one over zero right?
| plugging $$x=\frac{1}{\sqrt{n}},y=\frac{1}{n^2}$$ in the term above we get
$${\frac { \left( n+1 \right) \left( {n}^{2}-n+1 \right) {n}^{4}}{
\left( {n}^{2}+1 \right) \left( {n}^{4}-{n}^{2}+1 \right) }}
$$ and the limit is $$+\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Points $E$, $F$ and $G$ lies on median of triangle $ABC$. Determine the ratio $|GF|:|EF|$ In a given triangle $ABC$ lets choose inside side $AC$ points $P$, $Q$ and inside side $BC$ points $R$,$S$ so, that $|AP|=|PQ|=|QC|$, $|BR|=|RS|=|SC|$
Next denote point $E$ as intersection of diagonals in trapezoid $ABRP$, point $F$ as intersection of diagonals in trapezoid $PRSQ$ and point $G$ as intersection of diagonals in trapezoid $ABSQ$. Points $E$, $F$ and $G$ lies on median of triangle $ABC$ from vertex $C$. Determinate the ratio $|GF|:|EF|$ for any triangle.
Triangle sketch:
| First off, let's label the intersections of the median with $QS$, $PR$, and $AB$ as $I$, $H$, and $K$, respectively. Note that by the three parallels theorem, $CI = IH = HK$, so let's set the length of $HK = x$.
Note also that triangles $CQS$, $CPR$, and $CAB$ are similar, and using the appropriate ratio of similarities, we can find that $QS = \frac{1}{3} AB$ and $PR = \frac{2}{3} AB$. So $QS = \frac{1}{2} PR$.
Let's consider now trapezoid $APRB$. Using angles, we can conclude that triangles $EPR$ and $EBA$ are similar. So then using that $PR = \frac{2}{3} AB$, we find that $HE: EK = 2:3$. Thus $HE = \frac{2}{5}x$.
Using a similar method for trapezoid $PQSR$, we can conlcude that $HF = \frac{2}{3}x$ and $IF = \frac{1}{3}x$, so $EF = HE + HF = \frac{16}{15}x$. And then from trapezoid $AQSB$, we can conclude that $IG = \frac{1}{2}x$, so $GF = IG - IF = \frac{1}{6}x$.
Thus $GF:EF = \frac{1}{6} : \frac{16}{15}$, or more succinctly, $5:32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$.
My answer is $k<1$ or $k<\frac{1}{5}$, but the answer sheet says $k<\frac{1}{5}$ or $k>1$.
What have I done wrong? Please help.
What I've done
$(1-3k)^2-4(k^2)>0$
$1-6k+9k^2-4k^2>0$
$\frac{6 \pm 4}{10}>k$
$k<1$ or $k<\frac{1}{5}$
| The discriminant of $$x^2+x(1-3k)+k^2=0$$
$$(1-3k)^2-4k^2=1-6k+5k^2=5\left(k-\dfrac15\right)(k-1)$$ which needs to be $\ge0$
Now for $(y-a)(y-b)\ge0$ with $a\le b$
If $y-a\ge0\iff y\ge a;$ we need $y-b\ge0\iff y\ge b\implies y\ge$ max$(a,b)=b$
What if $y-a\le0?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating an expression given values of symmetric polynomials Evaluate $\dfrac x{yz} + \dfrac y {xz} + \dfrac z y$
Given, $z+y+x=4, \qquad xyz=-60, \qquad xy+xz+yz=-17$
How do we do this? I found a common denominator, and substituted it for $-60$, but I am unaware of how to proceed.
Someone already asked the question but there is no useful answer. These are the possible answers:
A. $4/17$
B. $−5/6$
C. $17/60$
D. $−33/60$
E. $33/60$
| Assuming there is a typo.
We have to find -
\begin{align}
& \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \\[10pt]
= {} & \frac{x^2 + y^2 + z^2}{xyz} \tag 1
\end{align}
Also $x + y + z = 4$.
Squaring both sides,
$$x^2 + y^2 + z^2 + 2(xy + yz + zx) = 16$$
$$x^2 + y^2 + z^2 + 2(-17) = 16$$
$$x^2 + y^2 + z^2 = 16 + 34$$
$$x^2 + y^2 + z^2 = 50$$
Put this in equation $(1)$
$$= \frac{50}{-60}$$
$$= \frac{-5}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Maximum of $\frac{1+3a^2}{(a^2+1)^2}$ What is the maximum value of
$\displaystyle{{1 + 3a^{2} \over \left(a^{2} + 1\right)^{2}}}$, given that $a$ is a real number, and for what values of $a$ does it occur ?.
| Let $x=a^2\ge0$. We then have to look at
$$\frac{1+3x}{(x+1)^2}$$
upon differentiating and setting equal to $0$, we get
$$\frac{3(x+1)^2-2(x+1)(1+3x)}{(x+1)^4}\\\implies3(x+1)^2=2(x+1)(1+3x)\\\implies3(x+1)=2(1+3x)\\\implies x=\frac13$$
Thus, the relative maxima (or minima) occurs at
$$a=\pm\sqrt x=\pm\frac{\sqrt3}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How was this equation derived?
Does anyone know how the equation in red was derived?
| First part:
$$\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2=\left(\cos^2\frac\theta4-\sin^2\frac\theta4\right)^2-4\sin^2\frac\theta4\cos^2\frac\theta4.$$
Second part:
$$\cos^2\frac\theta4=\frac1{1+\tan^2\frac\theta4}=\frac1{1+t^2},\\
\sin^2\frac\theta4=\tan^2\frac\theta4\cos^2\frac\theta4=\frac{t^2}{1+t^2}.$$
But $t=\tan\frac\theta4$, not $\cot\frac\theta4$ nor $\arctan\frac\theta4$ !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 +1 -\frac{(a^2+b^2)^2}{(2ab)^2} = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 = \frac{(a^2+b^2)^2}{(2ab)^2}-1 \\
z= \frac{a^2 +b^2}{2ab} \pm \sqrt{\frac{(a^2+b^2)^2}{(2ab)^2}-1} \ , \\
$$
which doesn't nearly suggest the simple end result.
I'm sure it will roll out after reorganizing terms, but I'm interested in an alternative approach which more naturally suggests the end result of $(z-a/b)(z-b/a)$, reflecting the word-use of straightforward in the text.
Context.
Using contour integration to evaluate $I = \int_{0}^{2\pi} \frac{\cos(2\theta)}{a^2+b^2-2ab\cos(\theta)}d\theta$
| Note that
$$\frac{a^2+b^2}{ab}=\frac ab+\frac ba$$
and
$$\frac ab\cdot\frac ba=1$$
Remember that, in a polynomial of second degree, the coefficient of $z$ is the opposite of sum of the roots and the costant term is their product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions of $ 2- \frac{1}{z} = z^* $ in $\mathbb{C}$ I am asked to show that the only solution of the equation $ 2- \frac{1}{z} = z^* $ is $z=1$.
I have tried this a few ways the two where i get closest are as follows.
$\frac{2z-1}{z}=z^*$
$0=z^* z -2z +1$
$\left|z^2 \right| -2z +1=0 = \left|z \right|^2 -2z +1 $
id really like to say $(\left| z \right| -1)^2=0 $ or even better $(z -1)^2=0 $
ive tried a few other things but i always end up with $ \left| z \right| $ instead of $z$
| $$\left|z^2 \right| -2z +1=0 \\ \to \sqrt{x^2+y^2} ^2 -2(x+iy) +1=0\\
x^2+y^2-2x+1-i(2y)=0 $$ I t is better to say $$x^2+y^2-2x+1-i(2y)=0 +i0 \to \\
\begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+y^2=0\\-2y=0 & \to y=0 \end{cases}\\
\begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+0^2=0 \to\\-2y=0 & \to y=0 \end{cases}\\(x-1)^2=0 \to x=1$$
so
$$z=x+iy=1+0y
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt{(a-1)^2+(b+\sqrt{3})^2}
\end{equation}
Is this the right way to go? Should I solve for b or a, and in that case what does that tell me about possible solutions? Any help would be appreciated.
Update:
Solved the equation for $y$, got
\begin{equation}
y=\frac{x}{\sqrt{3}}
\end{equation}
Does this mean that the equation is satisfied for all complex numbers on the line $y=\frac{x}{\sqrt{3}}$?
| Yes, this is a correct way to approach the problem. By squaring each side of your last line and simplifying, one obtains:
$$ 4a - 4\sqrt{3}b = 0 \Rightarrow a = \sqrt{3} b.$$
Thus the set that you are looking for consists of all the complex numbers $z = a + bi$ such that $a = \sqrt{3}b$, which in the complex plane is a straight line.
This was to be expected, because the equation can be rewritten as
$$ |z - (-1 + i\sqrt{3})| = |z - (1 - i\sqrt{3})|,$$
from where the problem can be solved geometrically: you are looking for points of the complex plane at an equal distance from $(-1 + i\sqrt{3})$ and $(1-i\sqrt{3})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
A binomial identity from counting permutations with cycles of length two. Let $n\ge 1$ be an integer. Consider the following sequence:
where $\left\{l_j\right\}_{j=1}^n$ are indices.By analyzing all possible decompositions of the sequence above into distinct pairs we have discovered the following identity:
\begin{equation}
n!\sum\limits_{j=0}^{\lfloor \frac{n}{2} \rfloor}\frac{(2j-1)!!}{2^j j!} \binom{n}{2 j} = (2n-1)!!
\end{equation}
Note that the $j$th term in the sum is just equal to the number of decompositions into pairs where there are exactly $j$ pairs where the first and the last element of thew pair are both zeros.
Now, my question would be is there some alternative way of proving this identity, maybe using analytical methods?
| Suppose we seek to prove that
$$n! \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{(2q-1)!!}{2^q q!} {n\choose 2q} = (2n-1)!!,$$
which is
$$n! \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{2^q q!} \frac{(2q)!}{2^q q!}
{n\choose 2q} = \frac{(2n)!}{2^n n!}$$
or
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{4^q} {2q\choose q} {n\choose 2q} =
\frac{1}{2^n} {2n\choose n}.$$
Now we have
$${2q\choose q} {n\choose 2q} =
\frac{n!}{q!\times q!\times (n-2q)!}
= {n\choose q} {n-q\choose q}.$$
The LHS becomes
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{4^q} {n\choose q} {n-q\choose n-2q}$$
and we introduce
$${n-q\choose n-2q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2q+1}} (1+z)^{n-q} \; dz$$
This controls the range and vanishes when $2q\gt n$ so we may extend
$q$ to $n$, getting for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\sum_{q=0}^n \frac{1}{4^q} {n\choose q} \frac{z^{2q}}{(1+z)^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n}
\left(1+\frac{z^2}{4(1+z)}\right)^n
\; dz
\\ = \frac{1}{4^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(z+2)^{2n}\; dz.$$
Extracting coefficients we find
$$\frac{1}{4^n} {2n\choose n} 2^n$$ which is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2^n} {2n\choose n}}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Basel's problem: $5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)$ Golden ratio:$\phi$
$$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)\tag1$$
We'll known Basel's problem
$$\sum_{k=1}^{\infty}{1\over k^2}=\zeta(2)\tag2$$
How do I transform $(1)$ to $(2)?$
Extra information:
$$\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}=1\tag3$$
Nothing come to my mind! Any hints?
| You want to show that
$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}
=\zeta(2)\tag1
$.
The function you need
is the dilogarithm,
defined by
$L_2(z)
=\sum_{k=1}^{\infty}\dfrac{z^k}{k^2}
$.
There is much useful info here:
https://en.wikipedia.org/wiki/Spence's_function
So you want to show that
$5(L_2(\phi^{-1})-L_2(\phi^{-2}))
=\zeta(2)
$.
Since
$\phi
=\dfrac{1+\sqrt{5}}{2}
$,
$\phi-1
=\dfrac{\sqrt{5}-1}{2}
$,
$1-\phi
=\dfrac{1-\sqrt{5}}{2}
=\dfrac{-1}{\phi}
$
and
$\dfrac{1}{\phi^2}
=\dfrac{(1-\sqrt{5})^2}{4}
=\dfrac{1-2\sqrt{5}+5}{4}
=\dfrac{3-\sqrt{5}}{2}
$,
this is equivalent to
$5(L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2}))
=\zeta(2)
$.
According to the Wikipedia page above,
$L_2(\dfrac{\sqrt{5}-1}{2})
=\dfrac{\pi^2}{10}-\ln^2(\dfrac{\sqrt{5}-1}{2})
$
and
$L_2(\dfrac{3-\sqrt{5}}{2}))
=\dfrac{\pi^2}{15}-\ln^2(\dfrac{\sqrt{5}-1}{2})
$.
Subtracting these,
the $\ln^2$ terms cancel out
and we get
$\begin{array}\\
L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2}))
&=\dfrac{\pi^2}{10}-\dfrac{\pi^2}{15}\\
&=\dfrac{\pi^2}{30}\\
\text{so}\\
5(L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2})))
&=\dfrac{\pi^2}{6}\\
&=\zeta(2)\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Check Convergence of the series: $\sum_{n=1}^\infty {\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}}$ I have to check the convergence of this series:
$$\sum_{n=1}^\infty {\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}}$$
Which is equal to $$\sum_{n=1}^\infty \frac{\sqrt{n+1}}{\sqrt{n}} - 1$$
What can I do here to check whether this series convergences or not? Thank you very much.
| Write the partial sum as
$$s_m = \sum_{n=1}^{m}\left( \sqrt{1+\frac{1}{n}}-1\right)$$
For $n\ge 1$ we have the inequality
$$\left(\frac{1}{n}+1\right)-\left(-\frac{1}{8 n^2}+\frac{1}{2 n}+1\right)^2=\frac{8 n-1}{64 n^4}>0$$
whence by taking the square root
$$\sqrt{\frac{1}{n}+1}>-\frac{1}{8 n^2}+\frac{1}{2 n}+1\\
\sqrt{\frac{1}{n}+1}-1>-\frac{1}{8 n^2}+\frac{1}{2 n}
$$
Hence
$$s_m \gt \sum_{n=1}^{m}\left( -\frac{1}{8 n^2}+\frac{1}{2 n}\right)\\
\gt \frac{1}{2}H_m-\frac{\pi^2}{48}$$
Summary: $s_m$ diverges logarithmically like the harmonic number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find all integers $(x, y)$ such that $1 + 2^x + 2^{2x + 1} = y^2$ Find all integers $$(x, y)$$ such that $$1 + 2^x + 2^{2x + 1} = y^2$$
So I basically used $$ f(x) = 1 + 2^x + 2^{2x + 1} = y^2$$ and created a table from 0 to 20. I got two pairs of integers: $$(0, \pm2)$$ and $$(4, \pm23)$$
I want to know if there's another method or if there are any other pairs of integers, or a generalisation.
| $1 + 2^x + 2^{2x + 1} = y^2 \\\implies 2^x(2^{x+1}+1) = (y+1)(y-1)$
$x<-1$ gives a non-integer LHS (no solutions)
$x=-1$ gives LHS $= 1$ with no solutions for $y$
$x=0$ gives LHS $= 3$ and $y=\pm 2$
For $x>0$, $y$ is odd so put $y=2k+1$ and $2^x(2^{x+1}+1) = (2k+2)(2k) = 4k(k+1)$ which is divisible by $8$ so $x\ge 3$ and $2^{x-2}(2^{x+1}+1) = k(k+1)$.
Clearly we cannot have $k=2^{x-2}$ or $k+1=2^{x-2}$ so we need $(2^{x+1}+1)$ to split into (odd) factors $r,s$ such that $2^{x-2}r = s\pm 1$.
Then $|r|<8$ otherwise $2^{x-2}|r| > (2^{x+1}+1)$. Also $2^{x-2}r^2 = sr\pm r$ so $|r|=3$ is the only viable choice and $2^{x-2}9 = (2^{x+1}+1) \pm 3$ gives $2^{x-2} = 4$ i.e. $x=4$ (and $y=\pm 23$) as the only other solution.
In summary: the only solutions $(x,y)$ are $(0,\pm 2)$ and $(4,\pm 23)$, those you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $z_1^2+z_2^2$ is real, $z_1(z_1^2-3z_2^2)=2$, and $z_2(3z_1^2-z_2^2)=11$, then find $(z_1^2+z_2^2)^2$
If $z_1$ and $z_2$ are complex numbers such that $z_1^2+z_2^2 \in\mathbb R$ and $$z_1(z_1^2-3z_2^2)=2,\qquad z_2(3z_1^2-z_2^2)=11,$$ then find the value of $(z_1^2+z_2^2)^2$. Given answer is $25$.
I have tried many things but I am not getting the answer.
I subtracted two equations to observe that $11/z_2 - 2/z_1$ must be real. Also if $z_1=x_1+iy_1$ and $z_2=x_2+i y_2$, then using the fact that $z_1^2+z_2^2 \in\mathbb R$, we get $x_1 y_1+x_2y_2=0$ but I am not able to compile these results to get the desired value.
| I suggest that we square both equations.
$$z_1^2(z_1^2-3z_2^2)^2=4\\
z_2^2(3z_1^2-z_2^2)^2=121$$
Now add them together and simplify.
$$125=z_1^2(z_1^2-3z_2^2)^2+z_2^2(3z_1^2-z_2^2)^2=z_1^2(z_1^4-6z_1^2z_2^2+9z_2^4)+z_2^2(9z_1^4-6z_1^2z_2^2+z_2^4)=\\
=z_1^6+3z_1^4z_2^2+3z_1^2z_2^4+z_2^6=(z_1^2+z_2^2)^3$$
The rest is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
How to solve $\lim_{x\rightarrow a} (2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})}$ Can someone help me to calculate the following limit?
$$\lim_{x\to a} \Big(2-\frac{x}{a}\Big)^{\tan\dfrac{\pi x}{2a}}$$
Thank you.
| Consider $\lim_{x\rightarrow a} e^{\ln((2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})})}=\lim_{x\rightarrow a} e^{\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}$.
Clearly it suffices to calculate the limit $$\lim_{x\rightarrow a} {\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}.$$
We can do this by using L'Hopitals rule.
\begin{eqnarray}
\lim_{x\rightarrow a} {\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}
&=& \lim_{x\rightarrow a} \frac{\sin(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}{\cos(\frac{\pi x}{2a})}\\
& \overset{\mathrm{H}}{=}& \lim_{x\rightarrow a} \frac{\frac{\pi}{2a}\cos(\frac{\pi x}{2a})\ln(2-\frac{x}{a})-\frac{1}{a}\frac{1}{2-\frac{x}{a}}\sin(\frac{\pi x}{2a})}{-\frac{\pi}{2a}\sin(\frac{\pi x}{2a})}\\
&=& \frac{2}{\pi}.
\end{eqnarray}
Hence the desired limit is $e^{\frac{2}{\pi}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$ If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$.
My Attempt:
$$2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$$
$$2f(x)+3f(\frac {1}{x})=4x + \frac {6}{x}$$
At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method...
| You can solve the following system.
$2f(x)+3f\left(\frac{1}{x}\right)=\frac{4x^2+6}{x}$ and $2f\left(\frac{1}{x}\right)+3f(x)=\frac{\frac{4}{x^2}+6}{\frac{1}{x}}$,
which gives $f(x)=2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $x$ positive integer such that $x^4+x^3+x^2+x+1$ is a perfect square
Find the number of positive integers $x$ for which $x^4+x^3+x^2+x+1$ is a perfect square.
My attempts:
Let $x^4+x^3+x^2+x+1=k^2\implies (x+1)^2(x^2-x+1)=(k-x)(k+x)$
I analysed this a bit found $x=0$ as one which satisfy all condition, how do I find other, please help, try to continue this further, if any other elegant method then add that too.
| Note that the problem is equivalent to finding integer solutions to $$4y^2=4x^4+4x^3+4x^2+4x+4$$
Now proceed to note that if $x>3$, we can find $$(2x^2+x)^2=4x^4+4x^3+x^2 < 4x^4+4x^3+4x^2+4x+4=4y^2$$
And $$4y^2 < 4x^4+4x^3+5x^2+2x+1=(2x^2+x+1)^2 $$
Since $4x^4+4x^3+4x^2+4x+4$ is stuck between squares of two consecutive numbers, it cannot be a square itself, which is a contradiction.
Thus we have that if $x$ is a positive integer it must be less than, or equal to, $3$. Trial and error gives us $x=3, y=11 $ is a valid solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadratic form. so how do I use the synthetic division
| HINT
You can write:
$$x^4-6x^3+16x^2-25x+10= (x^2-2x+k)(x^2+bx+c)+x+a=\\
=x^4+(b-2)x^3+(c-2b+k)x^2+(-2c+bk+1)x+(a+kc)$$
So,
$$b-2=-6→b=-4\\
c-2b+k=16\\
-2c+bk+1=-25\\
a+kc=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using the cosine over cosine in square root I got up to
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
| Fill in details:
$$r:=\tan\left(-\frac{5\pi}{16}\right)=\tan\frac{11\pi}{16}\implies$$
$$s:=\tan\frac{11\pi}8=\tan\left(2\cdot\frac{11\pi}{16}\right)=\frac{2\tan\frac{11\pi}{16}}{1-\tan^2\frac{11\pi}{16}}=\frac{2r}{1-r^2}$$
and
$$-1=\tan\frac{11\pi}4=\tan\left(2\cdot\frac{11\pi}8\right)=\frac{2s}{1-s^2}\implies$$
$$s^2-2s-1=0\implies s_{1,2}=1\pm\sqrt2$$
and now
$$sr^2+2r-s=0\implies r_{1,2}=\frac{-2\pm\sqrt{4+4s^2}}{2s}=\frac{-1\pm\sqrt{1+s^2}}{s}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Binoharmonic summation in sequence and series Please consider this bino-harmonic summation:-
nC0/a + nC0/a+d + nC0/ a+2d. . . . . . .nCn/a+nd
I tried a lot but could not get anything any way all of it was failure..Now I am beginning to think that I am a real dumbass ,please help me solve this..
| I believe you set out to compute
$$
\frac1a\binom{n}{0} - \frac{1}{a+d}\binom{n}{1} + \cdots + (-1)^n\frac{1}{a + nd}\binom{n}{n}.
$$
Consider the binomial expansion of $x^{a-1}(1-x^d)^n$:
$$
x^{a-1}(1-x^d)^n = x^{a-1}\left( \binom{n}{0} - \binom{n}{1}x^d + \cdots + (-1)^n\binom{n}{n}x^{nd} \right) \\
= \binom{n}{0}x^{a-1} - \binom{n}{1}x^{a+d-1} + \cdots + (-1)^n\binom{n}{n}x^{a+nd-1}
$$
Integration w.r.t $x$,
$$
\int x^{a-1}(1-x^d)^n dx = \binom{n}{0}\frac{x^a}{a} - \binom{n}{1}\frac{x^{a+d}}{a+d} + \cdots + (-1)^n\binom{n}{n}\frac{x^{a+nd}}{a+nd} + c
$$
Then, the series that you wanted to compute is given by
$$
\frac1a\binom{n}{0} - \frac{1}{a+d}\binom{n}{1} + \cdots + (-1)^n\frac{1}{a + nd}\binom{n}{n} = \int_0^1 x^{a-1}(1-x^d)^n dx.
$$
Let $x^d = u$. Then $\frac{\text{du}}{\text{dx}} = dx^{d-1} \implies \text{dx} = \frac{\text{du}}{dx^{d-1}}$. The integral gets transformed to
$$
\frac1d\int_0^1 x^{a-d}(1-x^d)^n du = \frac1d\int_0^1 u^{a/d-1}(1-u)^n du
$$
This integral is computed using the Gamma function as follows (See answer to this post):
$$
\frac1d\int_0^1 u^{a/d-1}(1-u)^n du = \frac1d\frac{\Gamma(a/d) \Gamma(n + 1)}{\Gamma(a/d + n + 1)}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$
Is this a correct usage of the chain rule?
| Since it's easy to graph the function with some software (also online) it should be clear that the derivative cannot be everywhere positive and the graph also suggests the function is not differentiable at $-1$ and $1$.
Set $g(x)=\dfrac{2x}{1+x^2}$; then
$$
f(x)=\arcsin\dfrac{2x}{1+x^2}=\arcsin(g(x))
$$
and, by the chain rule,
$$
f'(x)=\frac{1}{\sqrt{1-(g(x))^2}}g'(x)
$$
Now divide et impera:
*
*$\displaystyle 1-(g(x))^2=\frac{(1+x^2)^2-4x^2}{(1+x^2)^2}
=\frac{(1-x^2)^2}{(1+x^2)^2}$
*$\displaystyle \frac{1}{\sqrt{1-(g(x))^2}}=\frac{1+x^2}{|1-x^2|}$
*$\displaystyle g'(x)=2\frac{1+x^2-x\cdot2x}{(1+x^2)^2}$
Therefore
$$
f'(x)=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}=
\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
The derivative does exist at $-1$ and $1$, as it's easy to check with the limits from the left and from the right.
By the way, this shows that
$$
f(x)=\begin{cases}
c_--2\arctan x & \text{if $x<-1$} \\
c_0+2\arctan x & \text{if $-1\le x\le 1$}\\
c_+-2\arctan x & \text{if $x>1$}
\end{cases}
$$
Since
$$
\lim_{x\to-\infty}f(x)=0,\qquad
f(0)=0,\qquad
\lim_{x\to\infty}f(x)=0
$$
we can conclude that
$$
c_-=-\pi,\qquad c_0=0,\qquad c_+=\pi
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
How to factor $x^3-1$ with mod $3$ when I factor $x^3-1$ with $\mod 3$ on maple I get the answer $(x+2)^3$ and I was just wondering what the steps were to get to this solution.
| \begin{align*}
x^3 - 1 &= (x - 1)(x^2 + x + 1)\\[6pt]
&\equiv (x + 2)(x^2 + 4x + 4) \pmod{3}\\[6pt]
&\equiv (x + 2)(x + 2)^2 \pmod{3}\\[6pt]
&\equiv (x + 2)^3 \pmod{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
maximum value of expression $6bc+6abc+2ab+3ac$ If $a,b,c>0$ and $a+2b+3c=15,$ then finding maximum value of $6bc+6abc+2ab+3ac$ is
with the help of AM - GM inequality
$4ab\leq (a+b)^2$ and $4bc \leq (b+c)^2$ and $\displaystyle 4ca \leq (c+a)^2$
and $27(abc)\leq (a+b+c)^3$
could some help me, thanks
| Let $a=5x$, $2b=5y$ and $3c=5z$.
Hence, $x+y+z=3$ and by AM-GM $bc+6abc+2ab+3ac=$
$$=25(xy+xz+yz)+ 125xyz\leq 25 \frac{(x+y+z)^2}{3} + 125 \left(\frac{x+y+z}{3}\right)^3=200.$$
The equality occurs for $x=y=z=1$, which says that the answer is $200$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.