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Which solution of this integral $\int{\frac{x^2+1}{x^4-x^2+1}}dx$ is correct? $$\begin{align}
\int{\frac{x^2+1}{x^4-x^2+1}}dx&=\int{\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}}dx\\
&=\int{\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+1}}dx\\
&=\int{\frac{1}{u^2+1}}du \quad(u=x-\frac{1}{x})\\
&=\arctan(u)+C\\
&=\arctan(x-\frac{1}{x})+C
\end{align}$$
The image of the integral is:
When I check it using wolfram, the result is $\arctan(\frac{x}{1-x^2})+C$
https://www.wolframalpha.com/input/?i2d=true&i=%5Cint%5C%28%7B%29Divide%5BPower%5Bx%2C2%5D%2B1%2CPower%5Bx%2C4%5D-Power%5Bx%2C2%5D%2B1%5D%5C%28%7D%29
It seems Wolfram calculates this integral like this:
$$\begin{align}
\int{\frac{x^2+1}{x^4-x^2+1}}dx&=\int\frac{x^2+1}{(x^2-1)^2+x^2}dx\\
&=\int\frac{\frac{x^2+1}{(x^2-1)^2}}{1+\frac{x^2}{(x^2-1)^2}}dx\\
&=\int{\frac{1}{1+u^2}}du \quad(u=\frac{x}{1-x^2})\\
&=\arctan(u)+C\\
&=\arctan(\frac{x}{1-x^2})+C
\end{align}$$
Which solution is correct?
| Your answer is$$f(x)=\arctan\left(x-\frac1x\right).$$If you differentiate it, you will get indeed$$\dfrac{x^2+1}{x^4-x^2+1}.\tag1$$But $f$ is undefined at $0$. Besides, $\lim_{x\to0^+}f(x)\ne\lim_{x\to0^-}f(x)$.
On the other hand, if$$g(x)=\arctan\left(\frac x{1-x^2}\right),$$then, again, if you differentiate it, you will get indeed $(1)$. But $g$ is undefined at $\pm1$.
A primitive of $(1)$ is$$\begin{array}{rccc}h\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}\arctan\left(x-\frac1x\right)+\frac\pi2&\text{ if }x>0\\0&\text{ if }x=0\\\arctan\left(x-\frac1x\right)-\frac\pi2&\text{ if }x<0.\end{cases}\end{array}$$You can see its graph below:
| {
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Average value of $\frac{x'A^2x}{x'A^3x}$ over surface of $n$-dimensional sphere Suppose $A$ is a diagonal matrix with eigenvalues $1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n}$ and $x$ is drawn from standard Gaussian in $n$ dimensions. In numerical simulations, the following quantity seems to converge to $2$ as $n\rightarrow \infty$
$$z_n=E_{x\sim \mathcal{N}\left(0, I_n\right)}\left[\frac{x^T A^2 x}{x^T A^3 x}\right]$$
Can this be proven or disproven?
$z_n$ can also be written as the following sum
$$z_n=\sum_{i=1}^n i E_{y\sim \mathcal{N}\left(0,A^3\right)}\left[\frac{y_i^2}{\|y\|^2}\right]$$
This quantity can be viewed as the average ratio of quadratic forms $A^2$ and $A^3$ on the surface of $n$-dimensional sphere.
Here's what the distribution looks like for a few values of $n$, means are tending towards $2$
| Not an answer, but wanted to point out an interesting result leading to an identity with the zeta functions. Let $A = \sum_{k\ge 1} X_k^2/k^2$ and $B = \sum_{k \ge 1}X_k^2/k^3$. Then a Taylor series expansion of $f(A, B)$ around $(\mu_A, \mu_B)=\mathbb{E}[(A, B)] = (\zeta(2), \zeta(3))$ is
\begin{align*}
\frac{A}{B} = \frac{\mu_A}{\mu_B} + \begin{pmatrix} 1/\mu_B \\ -\mu_A/\mu_B^2\end{pmatrix}^\intercal \begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix} + \frac{1}{2}\begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix}^\intercal\begin{pmatrix} 0 & -1/\mu_B^2 \\ -1/\mu_B^2 & 2 \mu_A/\mu_B^3 \end{pmatrix}\begin{pmatrix} A - \mu_A \\ B - \mu_B \end{pmatrix} + \cdots
\end{align*}
The first few terms of this expansion are
\begin{align*}
\mathbb{E}\left[\frac{A}{B}\right] &= \frac{\mu_A}{\mu_B} - \frac{\mathbb{E}(A - \mu_A)(B-\mu_B)}{\mu_B^2} + \frac{\mu_A\mathbb{E}(B-\mu_B)^2}{\mu_B^3} + \frac{\mathbb{E}(A-\mu_A)^2(B-\mu_B)}{6\mu_B^3} + \frac{\mathbb{E}(A-\mu_A)(B-\mu_B)^2}{3\mu_B^2} - \frac{\mathbb{E}(B-\mu_B)^3\mu_A}{2\mu_B^4} + \cdots \\
&= \frac{\zeta(2)}{\zeta(3)} - 2 \frac{\zeta(5)}{\zeta(3)^2} + 2 \frac{\zeta(6)\zeta(2)}{\zeta(3)^3} + \cdots
\end{align*}
| {
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How can one solve this without expanding. \begin{array}{l}
\text { If } a+b+c=1, a b+b c+c a=2 \\
\text { and } a b c=3 \text {. What is the value } \\
\text { of } a^{4}+b^{4}+c^{4} \text { ? }
\end{array}
This can be solved by expanding but is there any easy alternative method ?
Here is how I solved :
\begin{array}{l}
(a+b+c)^{2}=1\\
\Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=1\\
\Rightarrow a^{2}+b^{2}+c^{2}=-3 \ldots . .(i)\\
\Rightarrow a b+b c+c a=2 \ldots . \text { (ii) }\\
\text { Squaring of equation (ii), }\\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2\left(a b^{2} c+b c^{2} a+\right.\\
\left.c a^{2} b\right)=4\\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)=4\\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+6=4
\end{array}
\begin{array}{l}
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=-2 \ldots . \text { (iii) }\\
\text { Squaring of equation (i), }\\
\Rightarrow a^{4}+b^{4}+c^{4}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)=9\\
\Rightarrow a^{4}+b^{4}+c^{4}-4=9\\
\Rightarrow a^{4}+b^{4}+c^{4}=13
\end{array}
| $a,b,c$ are roots of $x^3 - x^2 + 2x-3 = 0$
$$a^2+b^2+c^2=(a+b+c)^2 - 2(ab+bc+ac) = 1-2(2)=-3$$
Since
$$x^3 = x^2 -2x+3$$
$$a^3+b^3+c^3 = (a^2+b^2+c^2) - 2(a+b+c) + 9 = -3-2(1)+9=4$$
Since
$$x^4 = x^3 -2x^2+3x$$
$$a^4+b^4+c^4 = 4-2(-3)+3(1)=13$$
| {
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Chapter 2 Exercise 1 Pages 83 Linda J. S. Allen 2010 Exercises for Chapter 2 Exercise 1 Pages 83
textbook: An Introduction to Stochastic Processes with Applications to Biology 2nd Edition Linda J. S. Allen 2010
Link to the textbook
My attempt:
$P=(p_{ij})$ where $p_{ij}=\rm{Prob}\left\{X_{n+1}=i\lvert\ X_{n}=j\right\}$ is The one-step transition probability.
and $P$ is The transition matrix of the DTMC. also The transition matrix P is a stochastic matrix (A nonnegative matrix with the property that each column sum equals one is called a stochastic matrix.)
a stochastic matrix $\iff \begin{cases} \forall i,j\; p_{ij}\geq 0 \\ \forall j;\; \sum_{i=1}^{\infty}p_{ij}=1 \end{cases}$
$P= \begin{pmatrix}
p_{11} & p_{12} & p_{13} \\
p_{21} & p_{22} & p_{23} \\
p_{31} & p_{32} & p_{33}
\end{pmatrix}= \begin{pmatrix}
1 & 0 & 0 \\
\frac{1}{2} & 0 & \frac{1}{3} \\
0 & \frac{1}{3} & 1
\end{pmatrix}$
but I don't feel I am right since the sum of each column doesn't equal 1
Thanks in adavance
| $P= \begin{pmatrix}
p_{11} & p_{12} & p_{13} \\
p_{21} & p_{22} & p_{23} \\
p_{31} & p_{32} & p_{33}
\end{pmatrix}= \begin{pmatrix}
a & 0 & 0 \\
\frac{1}{2} & 0 & \frac{1}{3} \\
0 & b & c
\end{pmatrix}$
Since $\forall j\in [\![1,3]\!];\; \sum_{i=1}^{3}p_{ij}=1$, then it suffices to solve the following system of equations:
$\begin{cases} a+\frac{1}{2}+0=1 \\ 0+0+b=1 \\ 0+\frac{1}{3}+c=1 \end{cases}\implies \begin{cases} a=\frac{1}{2} \\ b=1 \\ c=\frac{2}{3} \end{cases}$
Thus the corresponding stochastic matrix
$$\boxed{P=\begin{pmatrix} \frac12 & 0 & 0 \\ \frac12 & 0 & \frac13 \\ 0 & 1 & \frac23 \end{pmatrix}}$$
| {
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Minimum and maximum points using Lagrange's method F(x,y)=x+2y, G(x,y)=x^2+y^2-1
I want to find the minimum and maximum points using lagrange's method, I tried but I always get lost, I could use a little bit help.
Thank you.
| Define,
$L(x,y, \lambda) = F(x,y) + \lambda G(x,y) = x + 2 y + \lambda (x^2 + y^2 - 1) $
Then, at the extreme points,
$\dfrac{\partial L}{\partial x} = 1 + \lambda (2 x) = 0 $
$\dfrac{\partial L}{\partial y} = 2 + \lambda (2 y) = 0 $
$\dfrac{\partial L}{\partial \lambda} = x^2 + y^2 - 1 = 0$
From the first two equations, it follows that,
$ \lambda = \dfrac{-1}{2x} = \dfrac{-2}{2y} $
Hence,
$2 x = y$
Plug this into the third equation,
$ x^2 + (2 x)^2 = 1 $
$ x = \pm \dfrac{1}{\sqrt{5}} $
$ y = \pm \dfrac{2}{\sqrt{5}} $
Plugging these values into the function $F(x,y)$ we deduce that the maximum is $\sqrt{5}$ and is achieved at $(\dfrac{1}{\sqrt{5}}, \dfrac{2}{\sqrt{5}})$ and the minimum is $-\sqrt{5}$ achieved at $(-\dfrac{1}{\sqrt{5}}, -\dfrac{2}{\sqrt{5}})$.
| {
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Computing local and global minima on Wolfram Alpha Consider $\sqrt{x^2+y^2}+2\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-6x-8y+25}$. I need to find global or local minima of this function, but Wolfram Alpha doesn't seem to find one; the answer is that $1 + 2\sqrt{5}$ is its global minimum
Am I doing something wrong? I use this input for the global minimum, and this input for local minima.
| Using FindMinimum[Sqrt[x^2+y^2]+2Sqrt[x^2+y^2-2x+1]+Sqrt[x^2+y^2-6x-8y+25],{x,y}] locally, I found the following minimum solutions. The output includes the already mentioned solution $5.47214=1+2\sqrt{5}$ and concretely looks as follows:
$$\{5.47214,\{x\to 1,y\to -7.441589669722822*10^{-10}\}\}$$
Hope it helps you to proceed. The plot is:
And here is the plot of the curve given by $\sqrt{x^2-2 x+y^2+1}+\sqrt{x^2-6 x+y^2-8 y+25}+\sqrt{x^2+y^2}-z=0$:
The partial derivate of $2\sqrt{x^2-2 x+y^2+1}+\sqrt{x^2-6 x+y^2-8 y+25}+\sqrt{x^2+y^2}$ for $x$ is:
$\frac{x}{\sqrt{x^2+y^2}}+\frac{2 x-2}{\sqrt{x^2-2 x+y^2+1}}+\frac{2 x-6}{2 \sqrt{x^2-6 x+y^2-8 y+25}}$
with the following contour plot:
and its PDE for $y$ is given by:
$\frac{y}{\sqrt{x^2+y^2}}+\frac{2 y}{\sqrt{x^2-2 x+y^2+1}}+\frac{2 y-8}{2 \sqrt{x^2-6 x+y^2-8 y+25}}$
with the following contour plot:
And finally here is the intersection plot:
followed by the intersection plot where I removed the coefficient 2 in front of the square:
| {
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Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$
$=>S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)}$
$\leq \frac{a^2}{4}.(\frac{1}{2a+b}+\frac{1}{2a+c})+\frac{b^2}{4}.(\frac{1}{2b+c}+\frac{1}{2b+a})+\frac{c^2}{4}.(\frac{1}{2c+a}+\frac{1}{2c+b})$
$=\frac{1}{4}.[a^2.(\frac{1}{2a+b}+\frac{1}{2a+c})+b^2.(\frac{1}{2b+a}+\frac{1}{2b+c})+c^2.(\frac{1}{2c+a}+\frac{1}{2c+b})]$
$\leq \frac{1}{4}.[\frac{a^2}{9}.(\frac{2}{a}+\frac{1}{b}+\frac{2}{a}+\frac{1}{c}) +\frac{b^2}{9}.(\frac{2}{b}+\frac{1}{c}+\frac{2}{b}+\frac{1}{a}) +\frac{c^2}{9}.(\frac{2}{c}+\frac{1}{a}+\frac{2}{c}+\frac{1}{b})]$
$=\frac{1}{36}.[a^2.(\frac{4}{a}+\frac{1}{b}+\frac{1}{c})+b^2.(\frac{1}{a}+\frac{4}{b}+\frac{1}{c})+c^2.(\frac{1}{a}+\frac{1}{b}+\frac{4}{c})]$
$= \frac{1}{36}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) $
We need prove that: $S \le \frac{1}{3}$
$=> S \le \frac{1}{12}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) \le 1 $ <- in there, I don't know how to do that :<
I'm trying to find a solution to continue for me or another solution, Can you help me?
| Let $a\geq b\geq c$.
Thus, $$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{a}{3(a+b+c)}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{3(a+b+c)}\sum_{cyc}\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}\geq$$
$$\geq\frac{1}{3(a+b+c)}\left(\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}-\frac{b(a-b)(b-c)}{(2b+a)(2b+c)}\right)\geq$$
$$\geq\frac{1}{3(a+b+c)}\left(\frac{a(a-b)\frac{a}{b}(b-c)}{(2a+b)(2a+c)}-\frac{b(a-b)(b-c)}{(2b+a)(2b+c)}\right)=$$
$$=\frac{(a-b)(b-c)}{3b(a+b+c)}\left(\frac{a^2}{(2a+b)(2a+c)}-\frac{b^2}{(2b+a)(2b+c)}\right)\geq0$$
because $$a(2b+c)\geq b(2a+c)$$ and $$a(2b+a)\geq b(2a+b).$$
| {
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Remainder Theorem Technique
Determine the remainder when $(x^4-1)(x^2-1)$ is divided by $1 + x + x^2$ (HMMT 2000, Guts Round)
A. Write the division in the form:
$$(x^4-1)(x^2-1)= (1 + x + x^2)Q(x) + R(x)$$
B. Multiply both sides by $x-1$:
$$(x-1)(x^4-1)(x^2-1)= (x^3-1)Q(x) + R(x)(x-1)$$
C. Substitute $x^3=1,x\neq1$, and reduce the resulting equation:
$$(x-1)(x-1)(x^2-1)= R(x)(x-1)$$
D. Divide both sides by $x-1$:
$$R(x)=(x-1)(x^2-1)=x^3 -x - x^2 + 1=-(x^2+x+1)+3=3$$
For someone who knows the method, is it valid to skip Steps B and D, directly substitute $x^3=1,x\neq1$ and use the fact that $x$ is a cube root of unity to get $x^2+x+1=0$.
| Alternate method:
$(x^4 - 1)(x^2 -1)$ has a degree of six, we can write:
$$(x^4 -1)(x^2-1)=Q(x) \left[ 1+x +x^2 \right] + R(x) $$
Evaluate (1) at $\{ \omega, \omega^2 \}$:
$$ \begin{align} R( \omega) &= ( \omega-1)(\omega^2 - 1)= ( \omega -1)( \frac{1}{\omega}-1) \\
R( \omega^2) &= (\omega^2 -1)(\omega-1) = ( \omega -1)( \frac{1}{\omega}-1) \end{align} $$
This means that, $$xR(x) -(x-1)(1-x) = (x - \omega)(x- \omega^2)$$
Rearrange for the answer which is $R(x)=3$.
Note: $(x^4 -1)(x^2-1)$ when divided by $1+x+x^2$ must give a polynomial less than degree two. This follows via the remainder. See Wiki.
| {
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Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)$
$x^4+\dfrac{1}{x^4}=x^4+2+\dfrac{1}{x^4}-2=\left(x^2+\dfrac{1}{x^2}\right)^2-1-1=\left(x^2+\dfrac{1}{x^2}-1\right)\left(x^2+\dfrac{1}{x^2}+1\right)-1$
Because $x^4+\dfrac{1}{x^4}$ is rational number so
$\left(x^2+\dfrac{1}{x^2}-1\right)$ is rational number too and
$x^3+\dfrac{1}{x^3}$ is rational number so
$\left(x+\dfrac{1}{x}\right)$ is rational number.
Am I wrong? Please check my solution, thank you.
| Let $a_k=x^k+\frac1{x^k}$， we have
$\begin{cases}a_1a_4=a_5+a_3\\a_2a_3=a_5+a_1\\a_1a_3=a_4+a_2\end{cases}$
By writing $a_3=u,a_4=v$, we could rewrite the fomula as
$\begin{cases}va_1+0a_2-a_5=u\\a_1-ua_2+a_5=0\\ua_1-a_2+0a_5=v\end{cases}$
Solving the linear equation of $a_1,a_2,a_5$ we could get they're all rational number
| {
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Respective chances that each of $3$ dice is loaded Here's a question from my probability textbook:
There are three dice A, B, C, two of which are true and one is loaded so that in twelve throws it turns up six $3$ times, ace once, and each of the other faces twice. Each of the dice is thrown three times and A turns up $6$, $6$, $1$; B turns up $6$, $5$, $4$; and C turns up $3$, $2$, $1$. What are now the respective chances that A, B or C is loaded?
Here's what I did. The relative proportions of loadedness are$${{\left({3\over{12}}\right)^2\left({1\over12}\right)}\over{\left({3\over{12}}\right)^2\left({1\over12}\right) + 2\left({1\over6}\right)^3}} = {9\over{25}} = {{63}\over{175}}, \quad {{\left({3\over{12}}\right)\left({2\over{12}}\right)^2}\over{\left({3\over{12}}\right)\left({2\over{12}}\right)^2 + 2\left({1\over6}\right)^3}} = {3\over{7}} = {{75}\over{175}}, \quad {{\left({2\over{12}}\right)^2\left({1\over{12}}\right)}\over{\left({2\over{12}}\right)^2\left({1\over{12}}\right) + 2\left({1\over6}\right)^3}} = {1\over{5}} = {{35}\over{175}},$$and so the respective chances are ${{63}\over{173}}$, ${{75}\over{173}}$, ${{35}\over{173}}$.
However, the answer in the back of my book is as follows:
A priori it is equally likely that the loaded die is A, B, or C.
The consequent chances of the observed event are as$${1\over4}\cdot{1\over4}\cdot{1\over{12}}: {1\over4} \cdot {1\over6} \cdot{1\over6} : {1\over6} \cdot {1\over6} \cdot {1\over{12}} = 9 : 12: 4$$The a posteriori chances are ${9\over{25}}$, ${{12}\over{25}}$, ${4\over{25}}$.
I don't really understand why I'm wrong, nor do I understand the solution in the back of the book. Could anyone explain:
*
*Where specifically did I go wrong?
*How did they come up with the answer in the back of the book?
| A calculation such as
$${{\left({3\over{12}}\right)^2\left({1\over12}\right)}\over{\left({3\over{12}}\right)^2\left({1\over12}\right) + 2\left({1\over6}\right)^3}} = {9\over{25}} = {{63}\over{175}}
$$
makes sense when you only roll $A$, it comes up $6,6,1$, and you want to know the probability that $A$ is loaded. Formally, let $H_A, H_B, H_C$ be the three hypotheses "$A$ is loaded", "$B$ is loaded", and "$C$ is loaded". Let $E$ be the evidence we have: that $A$ came up $6,6,1$. Then we would write
$$
\Pr[H_A \mid E] = \frac{\Pr[E \mid H_A] \Pr[H_A]}{\Pr[E\mid H_A] \Pr[H_A] + \Pr[E\mid H_B] \Pr[H_B] + \Pr[E\mid H_C] \Pr[H_C]}
$$
which gives us the calculation you did. (The prior probabilities $\Pr[H_A]$, $\Pr[H_B]$, $\Pr[H_C]$ cancel because they are all $\frac13$.)
In this problem, we didn't just roll $A$; we also rolled $B$ and $C$. Those outcomes are also informative of whether $A$ is loaded (if you imagine that $B$'s rolls were very very suspicious, then that should make us less suspicious of $A$). So we should take $E$ to be the entire set of $9$ rolls. Now we get
*
*$\Pr[E \mid H_A] = \color{blue}{(\frac3{12})^2 (\frac1{12})} \cdot \color{orange}{(\frac16)^3} \cdot \color{purple}{(\frac16)^3} = \frac{3^2 \cdot 1}{12^3 \cdot 6^6}$
*$\Pr[E \mid H_B] = \color{blue}{(\frac16)^3} \cdot \color{orange}{(\frac3{12}) (\frac{2}{12})^2} \cdot \color{purple}{(\frac16)^3} = \frac{3 \cdot 2^2}{12^3 \cdot 6^6}$
*$\Pr[E \mid H_C] = \color{blue}{(\frac16)^3}\cdot \color{orange}{(\frac16)^3} \cdot \color{purple}{(\frac2{12})^2 (\frac1{12})} = \frac{2^2 \cdot 1}{12^3 \cdot 6^6}$
Here, the three colors correspond to the probabilities of the results of $A$, $B$, and $C$ respectively.
If you use these probabilities, you will get answers matching the textbook. (It also takes the shortcut of using odds, which lets us ignore the denominators of $12^3 \cdot 6^6$ and simplify the calculations.)
| {
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Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}
I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?
| Method 1: without using induction
$10^n+3. 4^{n+2}+5=(1+9)^n+3(1+3)^{n+2}+5$
Which after applying binomial expansion, takes the form
$1+3(1+3(n+2))+5+9k$ for some positive integer k. This reduces to $9(n+3)+9k$. Hence, $9$ divides the expression for all $n$.
Method 2: Using induction
Let $P(n)$ denote the statement that $10^n+3.4^{n+2}+5$ be divisible by $9$. Clearly, $P(1)$ is true.
Let $P(n)$ be true.
Then, $P(n+1)= 10^{n+1}+3.4^{n+3}+5= 10^n(9+1)+ 3(3+1).4^{n+2}+5$
which reduces to $10^n+4^{n+2}+5+9.10^n+9.4^{n+2}$ which is nothing but $P(n)$ added to a number divisible by $9$. Hence $P(n+1)$ is also true. The statement $P(n)$ is divisible by $9$ for all $n$.
So yes, induction can be of great use in these problems!
| {
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Why does Solving system of quadratic equations gives extra roots? Consider these system of Equations
\begin{align*}
\begin{cases}
x^2+4x+4=0\\\\
x^2+5x+6=0
\end{cases}
\end{align*}
For solving them
We have
Method 1-
Subtract both equations
So
$-x-2=0$
Hence,
$x=-2$
Method-2
Add both equations
$2x^2+9x+10=0$
After applying quadratic formula,we get
$x=-2$ or $x=-5/2$. But only $x=-2$ satisfies the system of equation.
Why is the $-5/2$ not satisying the system of equations,what is intuition behind the error in method 2?
| HINT
You can factor both polynomials according to your preferred method in order to obtain:
\begin{align*}
\begin{cases}
x^{2} + 4x + 4 = 0\\\\
x^{2} + 5x + 6 = 0
\end{cases} \Longleftrightarrow
\begin{cases}
(x+2)^{2} = 0\\\\
(x+2)(x+3) = 0
\end{cases}
\end{align*}
Can you take it from here?
| {
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Picking two numbers - Calculate the probability We take randomly two numbers without replcement from a box that contains the numbers $1,2,\ldots , 10$.
Calculate the probability of the following events :
a) A = their sum is $11$
b) B = their product is even
c) C = the smaller one is $4$ or $5$
$$$$
I have done the following :
a) To get the sum $11$ we can have the $5$ combinations : $1+10$ , $2+9$ , $3+8$ , $4+7$ , $5+6$
In total there are $10\cdot 9=90$ possible combinations.
So we have the probability $P(A)=\frac{5}{90}=\frac{1}{18}$.
b) The product is even if at least one of the two numbers is even.
So for the first number there are $10$ possible number to take.
In total there are $10\cdot 9=90$ possible combinations.
So we have the probability $P(B)=\frac{5}{90}=\frac{1}{18}$.
c) The smaller number is $4$ or $5$. So do we have to calculate the probability of the union of $E_1$ : the smaller one is $4$ and $E_2$ : the smaller one is $5$ ?
| For $a)$, please note that we could choose any of the $10$ numbers as the first number. Then there is exactly $1$ second number out of remaining $9$ that gives sum of $11$.
So the probability is $\frac{1}{9}$.
In the way you wrote, you should have also counted $10 + 1, 9 + 1$ etc. and the answer should have been $\frac{10}{90}$
For $b)$, only if you pick both numbers odd, you do not get an even number as product. So the probability is,
$1 - \frac{5}{10} \cdot \frac{4}{9}$
$c)$ yes, you have to simply add probabilities for both cases - when the smaller number is $4$ and when the smaller number is $5$.
| {
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Finding three unknowns from three equations Let $a$,$b$ and $c$ be three positive real numbers such that
$$\begin{cases}3a^2+3ab+b^2&=&75\\
b^2+3c^2&=&27\\c^2+ca+a^2&=&16\end{cases}$$
Find the value of $ ab+2bc+3ca$.
My attempt: I observed that $3 . 16+27=75$. Then on replacing $16$ by $c^2+ca+a^2$, $27$ by $b^2+3c^2$ and $75$ by $3a^2+3ab+b^2$, I got $2c^2+ca=ab$.
But after this I am unable to proceed.
Is there a way to proceed from here?
Any constructive hint is appreciated.
|
Please note that
$ (\sqrt3 a)^2 + b^2 - 2 (\sqrt3 a) b \cos 150^0 = 3a^2 + b^2 + 3ab = 75$
$b^2 + (\sqrt3 c)^2 - 2 b (\sqrt3 c) \cos 90^0 = b^2 + 3c^2 = 27$
$ (\sqrt3 a)^2 + (\sqrt3 c)^2 - 2 (\sqrt3 a) (\sqrt3 c) \cos 120^0 = 3a^2 + 3 c^2 + 3 a c = 48$
Angles add to $360^0$ so there is a point $O$ inside $\triangle PQR$ with $OP = \sqrt3 a, OQ = b, OR = \sqrt3 c$ and $PQ = \sqrt{75}, QR = \sqrt{27}$ and $PR = \sqrt{48}$
Next observe that $PQ^2 = QR^2 + PR^2$ which means $\triangle PQR$ is a right triangle.
$ \displaystyle S_{\triangle PQR} = \frac{1}{2} \cdot \sqrt{27} \cdot \sqrt{48} = 18$
But $\displaystyle S_{\triangle PQR} = S_{\triangle POR} + S_{\triangle QOR} + S_{\triangle POQ}$
As we know area of a triangle is $\frac{1}{2} a b \sin \theta$ where $a, b$ are two sides with angle between them being $\theta$.
Adding individual areas we get to,
$\frac{\sqrt3}{4} (ab + 2bc + 3 ac) = 18$
So, $ab + 2bc+3ac = 24 \sqrt3$
| {
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Integration of $\sin^2x$ without using double angle identity of $\cos 2x$ I want to integrate $\sin^2x$ without using the double angle identity of $\cos2x$.
Here's what I tried:
$$
\int \sin^2x dx
= \int \sin x \tan x \cos x dx
$$
Let $u = \sin x$ => $du = \cos x dx$
And if $u = \sin x$, $\tan x = \frac{u}{\sqrt{1-u^2}}$, therefore
$$
=\int u × \frac{u}{\sqrt{1-u^2}}du = \int \frac{u^2}{\sqrt{1-u^2}}du
$$
Now if $t = \sqrt{1 - u^2}$, $2dt = \frac{du}{\sqrt{1-u^2}}$ and $u^2 = 1 - t^2$
$$
= 2\int (1-t^2)dt = 2t - \frac{2t^3}{3}$$
Substituting back $u$ and $\sin x$
$$
= 2\sqrt{1-u^2} - \frac{2}{3}(\sqrt{1 - u^2})^3
$$
$$
= 2\cos x - \frac{2}{3}\cos^3x
$$
But when you differentiate it you get $-2sin^3x$
All the steps seem right to me, why is the answer wrong or what I did is wrong, and is using the double angle formula the only way to integrate it?
| You can avoid using the double angle formula by integrating by parts $$\int\sin^2x\,\mathrm{d}x = -\cos x \sin x + \int \cos^2 x \,\mathrm{d}x = -\cos x \sin x + x - \int \sin^2 x \,\mathrm{d}x. $$
| {
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Does Diophantine equation $1+n+n^2+\dots+n^k=2m^2$ have a solution for $n,k \geq 2$? When studying properties of perfect numbers (specifically this post), I ran into the Diophantine equation
$$
1+n+n^2+\dots+n^k=2m^2, n\geq 2, k \geq 2.
$$
Searching in range $n \leq 10^6$, $k \leq 10^2$ yields no solution. So I wonder if there are any solutions, and if not, is there some elementary reason for that? Or if it can be converted to some known open problem, that would do too...
Some thoughts: Of course if we allow $n=1$ or $k=1$ we could get trivial solutions. We can also see by mod $2$ that both $n$ and $k$ must be odd. Now we could try to solve some small cases such as $k=3$ or $k=5$... So set $k=3$ and try to solve $$1+n+n^2+n^3=2m^2.$$ Left side factors and hence we want to solve $(n+1)(n^2+1)=2m^2$. Now this imples $4 \mid 2m^2$ and so $m$ is even. Also we can see that $4 \nmid n^2+1$ for any integer $n$ (becase $n^2\equiv 0,1 \pmod 4 $). So all powers of $2$ in $2m^2$ except one will divide $n+1$. So let $m=2^t r$ with $2 \nmid r$, then $2m^2=2^{2t+1}r^2$, $2^{2t} \mid n+1$, $2 \mid n^2+1$. So we can put $n=2^{2t}s-1$ with $2 \nmid s$, substitute it back, divide all powers of $2$ and we have the equation
$$
s(2^{4t-1}s^2-2^{2t}s+1)=r^2.
$$
Now the two expressions in the product are coprime, so they both have to be square, and that is farthest I got so far. Also the expression $2^{4t-1}s^2-2^{2t}s+1$ being square has a solution $s=15,t=2$, and I am yet to find another (but of course $s$ is not a square in this case so it does nothing for the original problem).
| Suppose we have $(n,s,x,y)$ with $n,s>1$ odd
$$n^s+1=2x^2\quad\text{and}\quad \frac{n^s-1}{n-1}=y^2,$$
then by a result of Ljunggren, we must have $n=3,s=5$, which gives $x^2=122$. Hence, there are no integral solutions to the system of equations in Theorem $1$ (see below). We conclude that there are no solutions in the positive integers to
$$
\frac{n^{k+1}-1}{n-1}=2m^2
$$
with $n,k\ge 2$.
Theorem 1: Assume there exists a triple $(n,k,m)$ of positive integers with $n,k\ge 2$
$$\frac{n^{k+1}-1}{n-1}=2m^2.$$
Write $k+1=2^rs$ with $s$ odd. Then $r=1$, and there exist positive integers $x,y$ with
$$
n^s+1=2x^2\quad\text{and}\quad \frac{n^s-1}{n-1}=y^2.
$$
Proof: Suppose we have a solution $(n,k,m)$. Write $k+1=2^rs$, then
$$
\frac{n^{2^rs}-1}{n^{s}-1}\cdot \frac{n^s-1}{n-1}=2m^2.
$$
We show that no odd prime divides both factors on the left. Let $p$ be an odd prime, $t=\operatorname{ord}_p\left[n^s-1\right]$, let $g\in\mathbb{Z}$ be a primitive root modulo $p^{t+1}$, and let $\ell\ge 1$ be an integer with $n^s\equiv g^{\ell}\pmod {p^{t+1}}$. Then $\operatorname{ord}_p(\ell)=t-1$. Hence, $\operatorname{ord}_{p}(2^r\ell)=t-1$ and $\operatorname{ord}_p(n^{2^rs}-1)=t$, whence $p$ does not divide $(n^{2^rs}-1)/(n^s-1)$.
By a similar argument, $(n^s-1)/(n-1)$ is odd, which means it must be a perfect square. It follows that there exist positive integers $u,v$ such that $(u,2^r-1,v)$ is a solution (and we can take $u=n^s$). We find that,
$$\prod_{t=0}^{r-1}\left(u^{2^t}+1\right)=\frac{u^{2^r}-1}{u-1}=2v^2.$$
It is very easy to prove that for all integers $t_1\neq t_2$, the greatest common divisor of $u^{2^{t_1}}+1$ and $u^{2^{t_2}}+1$ is a power of $2$. Therefore, each factor in the product on the left is either a perfect square or twice a perfect square.
Assume that $r\ge 2$, then $(u+1)(u^2+1)=u^3+u^2+u+1$ is either a perfect square or twice a perfect square. By Tomita's elliptic curve argument, it cannot be twice a perfect square, so it has to be a perfect square. By the result of Ljunggren, we must have $u=n^s=7$. Because $7^4+1$ is neither a square nor twice a square, we must have $r=2$, so $n=7$ and $k+1=2^rs=4$. However,
$$\frac{7^4-1}{7-1}=400$$
is not twice a perfect square. Therefore, $r=1$. $\square$
| {
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Problem based on area projection in 3-D geometry Considering a Quadrilateral $ABCD$ where $A(0,0,0), B(2,0,2), C(2,2\sqrt 2,2), D(0,2\sqrt2,0)$. Basically I have to find the Area of projection of quadrilateral $ABCD$ on the plane $x+y-z=3$.
I have tried to first find the projection of the points $A,B,C,D$ individually on the plane and then using the projected points find the vectors $\vec{AB}$ and $\vec{BC}$ and then using $|\vec{AB}\cdot \vec{BC}|$ , but I was unable to find the projected points.
Is it the correct approach? If it is not I would highly appreciate a correct approach for the problem.
| As indicated by Intelligenti pauca in the above comments, the best way to go is to find the area of the quadrilateral, then multiply the area found by the cosine of the angle between the two planes which is the same angle between the normals to the planes (or its supplement).
$\begin{equation} \begin{split}
\text{Area} &= \frac{1}{2} ( | AB \times AC | + | AC \times AD | ) \\
&= \frac{1}{2} ( | (2,0,2) \times (2, 2\sqrt{2}, 2) | + | (2, 2 \sqrt{2}, 2) \times (0, 2 \sqrt{2}, 0) | ) \\
&= 2 ( | (- \sqrt{2}, 0, \sqrt{2} ) | + | (-\sqrt{2}, 0, \sqrt{2} ) | )\\
& = 2 ( 2 + 2 ) = 8 \end{split} \end{equation}$
The normal to the plane of the quadrilateral is along $(-1, 0, 1)$ and the normal to the projection plane is along $(1, 1, -1)$, therefore, if $\theta$ is the angle between these two normals, then,
$\cos \theta = \dfrac{ (-1, 0, 1) \cdot (1, 1, -1) }{\sqrt{2}\sqrt{3} } = -\sqrt{\dfrac{2}{3}} $
Hence, the projected area is equal to $8\sqrt{\dfrac{2}{3}}$
| {
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Calculating improper intergal $\displaystyle \int^a_0 \frac{1}{\sqrt{a^2-x^2}} dx$ I tried to solve the improper integral below, but the answer was $\displaystyle \frac{\pi}{2}$. What is wrong with the solution?
$\displaystyle \int^a_0 \frac{1}{\sqrt{a^2-x^2}} dx
= \displaystyle \lim_{p \to a-0} \int^p_0 \frac{1}{\sqrt{a^2-x^2}} dx \\
= \displaystyle \frac{1}{a} \lim_{p \to a-0} \int^p_0 \frac{1}{\sqrt{1-(\frac{x}{a})^2}} dx \\
= \displaystyle \frac{1}{a} \lim_{p \to a-0} \left[ \arcsin{\frac{x}{a}} \right]^p_0 \\
= \displaystyle \frac{\pi}{2a}
$
| Be careful about the change of variable. You need to complete the diferential and change the integration limits. Here $u=\frac{x}{a}$ and $du=\frac{dx}{a}$, then
$\displaystyle \int^a_0 \frac{1}{\sqrt{a^2-x^2}} dx
= \displaystyle \int^a_0 \frac{1}{\sqrt{1-(\frac{x}{a})^2}} \frac{dx}{a} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \int^1_0 \frac{1}{\sqrt{1-(u)^2}} du \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{\pi}{2}.
$
Any doubts?
| {
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Finding a straight line that makes a given angle with a given straight line Question:
Find the equation of the straight line that passes through $(6,7)$ and makes an angle $45^{\circ}$ with the straight line $3x+4y=11$.
My solution (if you want, you can skip to the bottom):
Manipulating the given equation to get it to the slope-intercept form,
$$3x+4y=11...(i)$$
$$\implies 4y=-3x+11$$
$$\implies y=\frac{-3}{4}x+\frac{11}{4}$$
Let, the slope of (i) is $m_1=\frac{-3}{4}$, and the slope of our desired equation is $m_2$. Now, according to the question,
$$\tan(45^{\circ})=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}...(1)$$
$$\implies 1=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}$$
$$\implies \pm \frac{3}{4}+m_2=1-\frac{3}{4}m_2...(ii)$$
Picking positive value from (ii),
$$\frac{3}{4}+m_2=1-\frac{3}{4}m_2$$
$$\implies m_2(1+\frac{3}{4})=1-\frac{3}{4}$$
$$\implies m_2=\frac{1-\frac{3}{4}}{1+\frac{3}{4}}$$
$$\implies m_2=\frac{1}{7}$$
Picking negative value from (ii),
$$-\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -m_2(1-\frac{3}{4})=1+\frac{3}{4}$$
$$\implies m_2=-\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$$
$$\implies m_2=-7$$
Picking $m_2=\frac{1}{7}$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=\frac{1}{7}$$
$$\implies 7y-49=x-6$$
$$\implies -x+7y-43=0$$
$$\implies x-7y+43=0...(iii)$$
Picking $m_2=-7$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=-7$$
$$\implies -7x+42=y-7$$
$$\implies 7x+y-49=0...(iv)$$
Now, we are getting two straight lines as a result. One makes $45^{\circ}$ with $3x+4y=11$ in the counterclockwise direction ($x-7y+43=0$), and the other makes $45^{\circ}$ with $3x+4y=11$ in the clockwise direction ($7x+y-49=0$). So, is $7x+y-49=0$ a valid solution to the above question as it makes $-45^{\circ}$ not $45^{\circ}$?
| I don't see any reason why the clockwise angle would be invalid.
In some contexts, such as trigonometry, we, by convention, write the angle as the counterclockwise angle from the positive $x$-axis. In others, such as true compass bearings, the angle is, again by convention, measured clockwise (from North).
So for this question, with no specified convention, a $45$° angle is a $45$° angle, regardless of whether it is clockwise or counterclockwise. Unless this is from a class or textbook in which you only consider counterclockwise angles, or if the question specifies it, both lines would be a valid solution to the given problem; they pass through the given point and they make a $45°$ angle with the given line.
| {
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Show that the product of the lengths of the perpendiculars drawn from two points to a straight line is $b^2$ Question:
Show that the product of the lengths of the perpendiculars drawn from the points $(\pm c,0)$ to the straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
My attempt:
Let, the length of the perpendicular on drawn upon $bx\cos\theta+ay\sin\theta-ab=0$ from $(c,0)$ is $d_1$ and the length of that drawn upon the same from $(-c,0)$ is $d_2$. Now,
$$d_1=\frac{|bc\cos\theta+0-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1=\frac{|bc\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
Again,
$$d_2=\frac{|-bc\cos\theta-ab|}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}$$
Now,
$$d_1d_2=\frac{|-(bc\cos\theta-ab)(bc\cos\theta+ab)|}{{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2\sin^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2(1-\cos^2\theta)}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{{b^2\cos^2\theta+a^2-a^2\cos^2\theta}}$$
$$\implies d_1d_2=\frac{|a^2b^2-b^2c^2\cos^2\theta|}{a^2-(a^2-b^2)\cos^2\theta}$$
$$\implies d_1d_2=|\frac{b^2(a^2-c^2\cos^2\theta)}{a^2-(a^2-b^2)\cos^2\theta}|$$
[$a^2-(a^2-b^2)\cos^2\theta$ is positive, so bringing it inside the modulus sign doesn't make a difference]
$$\implies d_1d_2=|\frac{b^2(a^2-(a^2-b^2)\cos^2\theta)}{(a^2-(a^2-b^2)\cos^2\theta)}|$$
[Given, $a^2=b^2+c^2\implies c^2=a^2-b^2$]
$$d_1d_2=|b^2|$$(shown?)
The question asked me to show that $d_1d_2=b^2$, but I showed them that $d_1d_2=|b^2|$. So, is there something wrong with my process, or is there something wrong with the question?
| Your answer is correct. Note that the absolute value of a real number $x$ is defined by
$$|x|=\begin{cases}x,\space\text{if}x\geq0 \\ -x,\space\text{if} x<0\end{cases}$$
and since $b^2\geq 0$ for all $b\in\mathbb R$ we have $|b^2|=b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I derive $~\text{opposite}\cdot\sin^{}\left(\theta_{}\right)+\text{adjacent}\cdot\cos^{}\left(\theta_{}\right)=\tan^{}\left(\theta_{}\right)$ Given the below equation .
$$ b \cos^{}\left(\theta_{} \right) = a \cdot \sin^{}\left(\theta_{} \right) $$
I have to derive the below equation .
$$ b \sin^{}\left(\theta_{} \right) + a \cdot \cos^{}\left(\theta_{} \right) = \sqrt{ a ^2 + b^2 } $$
My tries are as below .
$$ \frac{ b }{ a } = \frac{ \sin^{}\left(\theta_{} \right) }{ \cos^{}\left(\theta_{} \right) } $$
$$ \frac{ b }{ a } = \tan^{}\left( \theta_{} \right) $$
$$ \text{adjacent}= a $$
$$ \text{opposite}= b $$
$$ \text{hypotenuse}=\sqrt{ a^2+ b ^2 } $$
$$ b \sin^{}\left(\theta_{} \right) + a \cdot \cos^{}\left(\theta_{} \right) $$
$$ = b \left( \frac{ b \cos^{}\left(\theta_{} \right) }{ a } \right) +a \left( \frac{ a \sin^{}\left(\theta_{} \right) }{ b } \right) $$
$$ = \frac{ b^2\cos^{}\left(\theta_{} \right) }{ a } + \frac{ a ^2 \sin^{}\left(\theta_{} \right) }{ b } $$
I've been got stucked from here .
|
The diagram makes it easier. It is consistent with your given conditions with $\frac ba = \tan \theta$. Note that $c = \sqrt{a^2 +b^2}$ (Pythagoras).
From the diagram, $\sin \theta = \frac bc, \cos \theta = \frac ac$.
So $b\sin \theta + a\cos \theta = \frac {b^2}c + \frac {a^2} c = \frac {c^2}c = c = \sqrt{a^2 + b^2}$ (QED)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the tangent to this equation at origin? $$x(x^2 + y^2) = a(x^2 - y^2)$$
I was trying to find the equation of the tangent as $$(Y-0) = \frac{dy}{dx}(X - 0)$$ where
$$\frac{dy}{dx} = \frac{2ax-3x^2-y^2}{2(x+a)y}$$
So here putting the values of $(x,y)$ as $(0,0)$ makes the derivative non existent.
I am stuck at this point and unable to proceed further.
|
The appearance of an indeterminate implicit derivative at a point on a curve is often an indication of a self-intersection. This is a folium, the best-known example being that of Descartes $ \ ( x^3 + y^3 - 3axy \ = \ 0 ) \ , \ $ which looks like this one rotated counter-clockwise by $ \ 45º \ \ . $ We see that $ \ a \ $ "sets the scale" for the curve, with the "leaf" extending from the origin to $ \ x \ = \ +a \ $ and with a vertical asymptote located at $ \ x \ = \ -a \ \ . $
One way that the self-intersection can be investigated is by "parameterizing" the curve, introducing coordinate functions $ \ x \ = \ x(t) \ \ , \ \ y \ = \ y(t) \ $ which describe the curve in terms of a new independent variable ("parameter") $ \ t \ \ . \ $ (This approach is usually presented in second-semester calculus, so you may not have seen it when you posted your question). Because the curve is symmetric about the $ \ x-$axis, we would like the $ \ x-$coordinate to be given by an even-symmetry function of $ \ t \ . \ $ To avoid having a singularity at the origin, we also wish to have a function for the $ \ y-$coordinate with $ \ y(-\tau) \ = \ y(+\tau) \ = 0 \ $ and $ \ y(0) \ = \ 0 \ \ , \ \ x(0) \ = \ +a \ \ , $ so that the origin is "passed through" twice and the point with $ \ t \ = \ 0 \ $ occurs at the "tip of the leaf"; this would suggest that $ \ y(t) \ $ should have odd symmetry.
We can find such a parameterization if we take $ \ y \ = \ tx \ $ (a "slope parameterization", as this represents a line of slope $ \ t \ $ through the origin) and insert it into the curve equation:
$$ x·(x^2 \ + \ [tx]^2) \ \ = \ \ a· (x^2 \ - \ [tx]^2) \ \ \Rightarrow \ \ x·(1 \ + \ t^2) \ \ = \ \ a \ - \ at^2 \ \ , \ \ x \ \neq \ 0 $$ $$ \Rightarrow \ \ x \ \ = \ \ a·\left(\frac{1 \ - \ t^2}{1 \ + \ t^2} \right) \ \ \Rightarrow \ \ y \ \ = \ \ tx \ \ = \ \ a·\left(\frac{t·(1 \ - \ t^2)}{1 \ + \ t^2} \right) \ \ . $$
We can (for now) generalize these coordinate functions a bit to fit the values of $ \ t \ $ we described above; thus,
$$ x(t) \ \ = \ \ a·\left(\frac{\tau^2 \ - \ t^2}{\tau^2 \ + \ t^2} \right) \ \ , \ \ y(t) \ \ = \ \ -a·t·\left(\frac{\tau^2 \ - \ t^2}{\tau^2 \ + \ t^2} \right) \ \ . $$
The folium is "traced out" with the origin approached "from below" and near the vertical asymptote as $ \ t \ $ approaches $ \ -\tau \ $ from "negative infinity", the "leaf" is covered by $ \ -\tau \ < \ t \ < \ +\tau \ , \ $ then the origin is departed "from above" and near the asymptote as $ \ t \ $ "goes to positive infinity".
We already see that the conflict of derivatives at the origin is resolved by calculating the first derivative at two distinct values of the parameter:
$$ \frac{dy}{dx}|_{x \ = \ 0} \ \ = \ \ \frac{dy/dt}{dx/dt}|_{t \ = \ +\tau} \ \ = \ \ \frac{a·\left( \frac{t^4 \ + \ 4 \tau^2 · t^2 \ - \ \tau^4}{(\tau^2 \ + \ t^2)^2} \right)}{-a·\left(\frac{ 4 \tau^2 · t}{(\tau^2 \ + \ t^2)^2} \right)}|_{t \ = \ +\tau} $$ $$ = \ \ -\frac{ \tau^4 \ + \ 4\tau^4 \ - \ \tau^4}{4·\tau^3} \ \ = \ \ -\tau \ \ , $$
$$ \text{and} \ \ \ \frac{dy/dt}{dx/dt}|_{t \ = \ -\tau} \ \ = \ \ -\frac{ \tau^4 \ + \ 4\tau^4 \ - \ \tau^4}{4·[-\tau^3]} \ \ = \ \ +\tau \ \ . $$
Thus we can account for the two slopes of tangent lines with equal magnitudes and opposite signs at the origin, since the derivative $ \ \frac{dy}{dx} \ $ is the ratio of an even function to an odd function. (We also see that this correctly accounts for the vertical tangent at the end of the folium's "leaf", since the denominator becomes zero for $ \ t \ = \ 0 \ , $ making the derivative undefined.)
We still need to get the "scaling" of the curve set up properly. We can choose any point on the "leaf", say, $ \ x \ = \ \frac{a}{2} \ \ , $ for which
$$ \frac{a}{2}·\left( \ \left[\frac{a}{2} \right]^2 \ + \ y^2 \ \right) \ \ = \ \ a·\left( \ \left[\frac{a}{2} \right]^2 \ - \ y^2 \ \right) $$ $$ \Rightarrow \ \ y^2 \ \ = \ \ \frac{a^3/8}{3a/2} \ \ = \ \ \frac{a^2}{12} \ \ \Rightarrow \ \ y \ \ = \ \ \pm \frac{a}{2 \sqrt3} \ \ . $$
Solving $ \ x(t) \ = \ \frac{a}{2} \ $ gives us $ \ t \ = \ \pm \frac{\tau}{\sqrt3} \ $ and inserting this into the other coordinate function yields
$$ y \left(\pm \frac{\tau}{\sqrt3} \right) \ \ = \ \ -\left(\pm \frac{\tau}{\sqrt3} \right)· \frac{a}{2} \ \ = \ \ \pm \frac{a}{2 \sqrt3} \ \ \Rightarrow \ \ \tau \ \ = \ \ \mp 1 \ \ . $$
So the coordinate functions are "forced" to be $$ x(t) \ \ = \ \ a·\left(\frac{1 \ - \ t^2}{1 \ + \ t^2} \right) \ \ , \ \ y(t) \ \ = \ \ -a·t·\left(\frac{1 \ - \ t^2}{1 \ + \ t^2} \right) \ \ , $$
"tracing out the leaf" over the interval $ \ -1 \ < \ t \ < \ +1 \ . $ The slopes of the two tangent lines at the origin are thus $ \ \frac{dy}{dx}|_{x \ = \ 0} \ = \ \pm \ 1 \ \ . $
$$ \ \ $$
If we follow on Eric Tower's comment on the posted question by writing the two explicit functions which represent the portions of the curve that lie "above and below" the $ x-$axis, we have $ \ x(x^2 + y^2) = a(x^2 - y^2) \ \Rightarrow \ y^2 \ = \ x^2·\left(\frac{a - x}{a + x} \right) \ \ . $ When we perform explicit differentiation on these functions, we obtain
$$ y' \ \ = \ \ \pm \left[ \ \left(\frac{a - x}{a + x} \right)^{1/2} \ + \ \frac12·x·\left(\frac{a + x}{a - x} \right)^{1/2} · \frac{2a}{(a - x)^2} \ \right] \ \ . $$
We needn't fuss with the expression further, as we see that for the two functions,
$$ y'|_{x \ = \ 0} \ \ = \ \ \pm \left[ \ \left(\frac{a - 0}{a + 0} \right)^{1/2} \ + \ \frac12·0·\left(\frac{a + 0}{a - 0} \right)^{1/2} · \frac{2a}{(a - 0)^2} \ \right] \ \ = \ \ \pm \sqrt{\frac{a}{a}} \ \ = \ \ \pm 1 \ \ , $$
without any singularity in the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Is this proof that $2^{23} \equiv 1 \bmod{47}$ correct? By Fermat's little theorem, we have that
$$
2^{46} \equiv 1 \pmod{47}
$$
By writing $2^{23}$ as $\left( 2^{46}\right)^{\frac{1}{2}}$ and knowing that
$$
(a^b) \bmod c = ((a \bmod c)^b)\bmod c
$$
we can conclude that
$$
\left( 2^{46}\right)^{\frac{1}{2}} \bmod{47} =
(2^{46} \bmod{47})^{\frac{1}{2}} \bmod 47 \\
2^{23} \bmod{47}= 1^{\frac{1}{2}} \bmod{47} \\
\therefore 2^{23} \bmod{47} = 1 \bmod{47}
$$
Would this proof be acceptable?
| No, that reasoning is not correct. You are saying: As $2^{46}\equiv 1\bmod 47$ then $2^{23}$ is a square root of $1\bmod 47$. That is true, but there are two possible roots: $\pm1$. So you have either $2^{23}=1$ or $2^{23}=-1$.
Instead you might evaluate $2^{23}\bmod 47$ by square and multiply:
$$ 2^{23} = 2\cdot 2^{22} = 2\cdot 4^{11} = 8 \cdot 4^{10} = 8\cdot 16^5=128\cdot 256^2 \equiv 34\cdot 21^2 = 34\cdot 441 \equiv 34\cdot 18 =612\equiv 1 \bmod 47$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probability of One Event Less Than Probability of Second Event I am having a bit of trouble proving some cases when one probability is smaller than the other probability for all positive integers $a, b$, so some suggestions would be appreciated.
Here is the problem:
For all $a, b \in \mathbb{Z}^+$, if $P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}$ and $P(B) = \dfrac{a^2 + b^2}{a^2 + 2ab + b^2}$, prove that $P(A) < P(B)$.
So I attempted using cases.
Case 1: If $a = b \neq 0$, then we want to show that $P(A) < P(B)$
\begin{align*}
P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b} = \dfrac{2a^2 - 2a}{4a^2 - 2a} < \dfrac{a^2 + b^2}{a^2 + 2ab + b^2} = \dfrac{2a^2}{4a^2} = \dfrac{1}{2} = P(B)
\end{align*}
Thus, this implies that $P(A) < P(B)$ for all $a, b \in \mathbb{Z}^+$
The next two cases are the cases I am having trouble with.
Case 2: If $a > b > 0$, then we want to show that $P(A) < P(B)$.
Case 3: If $b > a > 0$, then we want to show that $P(A) < P(B)$.
I am not sure how to approach cases 2 and 3. But case 3 should follow from case 2. So some assistance would be appreciated. Thanks
| I tried to make $P(A)$ and $P(B)$ comparable to avoid case decisions.
\begin{align}
&P(A)=\frac{(a+b)^2-2ab-a-b}{(a+b)^2-a-b}=1-\frac{2ab}{(a+b)^2-a-b}=1-f(a,b)&&\\ \\
& P(B)=\frac{(a+b)^2-2ab}{(a+b)^2}=1-\frac{2ab}{(a+b)^2}=1-g(a,b)&&
\end{align}
Then we start by comparing the denominators.
\begin{align}
&(a+b)^2-a-b<(a+b)^2 \\ \\ \Longrightarrow & f(a,b)>g(a,b) \\ \\ \Longrightarrow & P(A)<P(B)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proving $\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$
Prove
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$$
Proving right hand side to left hand side:
$$\begin{align}\csc^2x+2\csc x \cot x+\cot^2x
&= \frac{1}{\sin^2x}+\dfrac{2\cos x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x} \tag1 \\[0.8em]
&=\frac{\cos^2x+2\cos x+1}{\sin^2x} \tag2 \\[0.8em]
&=\frac{1-\sin^2x+1+\dfrac{\sin 2x}{\sin x}}{\sin^2x} \tag3 \\[0.8em]
&=\frac{\;\dfrac{2\sin x-\sin^3x+\sin 2x}{\sin x}\;}{\sin^2x} \tag4 \\[0.8em]
&=\frac{2\sin x-\sin^3x+\sin2 x}{\sin^3x} \tag5
\end{align}$$
I could not prove further to the left hand side from here. I would need help. Thank you in advance.
| $$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=$$
$$ \frac {2\sin x +2\sin x \cos x }{2\sin x -2\sin x \cos x}=$$
$$\frac {1 + \cos x }{1 -\cos x}=$$
$$\frac {(1 + \cos x)^2 }{1 -\cos^2 x}=$$
$$\frac { 1+2\cos x +\cos ^2 x}{\sin ^2 x}=$$
$$\csc^2x+2\csc x \cot x+\cot^2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Question about the proof of $\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$ Proof:
Let, $x-a=h\implies x=a+h$
$$\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}$$
$$\lim_{h\to 0}\frac{(a+h)^{n}-a^{n}}{h}$$
$$\lim_{h\to 0}\frac{a^n(1+\frac{h}{a})^{n}-a^{n}}{h}$$
Since $h\to0$, $h$ can be supposed to be less than $a$. So, $|\frac{h}{a}|<1$. Therefore, $(1+\frac{h}{a})^n$ can be expanded with the help of the binomial theorem.
$$\text{rest of the proof...}$$
Questions:
*
*Why do we need to assume that $h$ is smaller than $a$? Does this property not hold if $h$ is greater than or equal to $a$?
*Why does $|\frac{h}{a}|<1$ need to be true for us to expand $(1+\frac{h}{a})^n$ using the binomial theorem?
| Another way.
Use
$x^n-a^n
=(x-a)\sum_{k=0}^{n-1}x^ka^{n-1-k}
$.
Then,
if
$0 < a-c < x < a+c$,
$\begin{array}\\
\dfrac{x^n-a^n}{x-a}
&=\sum_{k=0}^{n-1}x^ka^{n-1-k}\\
\text{so}\\
\dfrac{x^n-a^n}{x-a}
&\lt\sum_{k=0}^{n-1}(a+c)^ka^{n-1-k}\\
&\lt \sum_{k=0}^{n-1}(a+c)^{n-1}\\
&=n(a+c)^{n-1}\\
\text{and}\\
\dfrac{x^n-a^n}{x-a}
&\gt\sum_{k=0}^{n-1}(a-c)^ka^{n-1-k}\\
&\gt \sum_{k=0}^{n-1}(a-c)^{n-1}\\
&=n(a-c)^{n-1}\\
\end{array}
$
I'll work with
the first inequality here.
The second is similar.
If $0 < z < 1$
then
$(1+z)^m
=1+\sum_{j=1}^m \binom{m}{j}z^j
\le 1+\sum_{j=1}^m \binom{m}{j}z
=1+(2^m-1)z
$
so
$(1+z)^m-1
\le (2^m-1)z
$.
Therefore
$\begin{array}\\
\dfrac{x^n-a^n}{x-a}-na^{n-1}
&\lt n(a+c)^{n-1}-na^{n-1}\\
&=na^{n-1}((1+c/a)^{n-1}-1)\\
\text{so if } c<a\\
\dfrac{x^n-a^n}{x-a}-na^{n-1}
&\lt na^{n-1}((1+c/a)^{n-1}-1)\\
&\lt na^{n-1}(2^{n-1}-1)(c/a)\\
\end{array}
$
so if
$c < \dfrac{a\epsilon}{na^{n-1}(2^{n-1}-1)}
$
(remembering that
$x < a+c$)
then
$\dfrac{x^n-a^n}{x-a}-na^{n-1}
\lt \epsilon
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of powers of Bessel function from 0 to infinity Let $J_m(x)$ be the Bessel function of the first kind with order $m$. I was experimenting with the following integral on Wolfram Alpha
$$ \int_0^\infty J_m(x)^4\, dx, $$
and it returns exact value for $m = 1, 2, 3, 4, 5$. Does anyone know if there is an explicit formula for this definite integral for any positive integer $m$?
| If you consider
$$I_m=2\pi \int_0^\infty \Big[J_m(x)\Big]^4\,dx$$ they are given in terms of Meijer G-functions.
Looking at the first
$$I_0=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
1,1,1,1 \\
\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}
\end{array}
\right.\right) \qquad I_1=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
0,1,1,2 \\
\frac{1}{2},\frac{3}{2},-\frac{1}{2},\frac{1}{2}
\end{array}
\right.\right)\qquad I_2=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
-1,1,1,3 \\
\frac{1}{2},\frac{5}{2},-\frac{3}{2},\frac{1}{2}
\end{array}
\right.\right)$$
$$I_3=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
-2,1,1,4 \\
\frac{1}{2},\frac{7}{2},-\frac{5}{2},\frac{1}{2}
\end{array}
\right.\right) \qquad I_4=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
-3,1,1,5 \\
\frac{1}{2},\frac{9}{2},-\frac{7}{2},\frac{1}{2}
\end{array}
\right.\right)\qquad I_5=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
-4,1,1,6 \\
\frac{1}{2},\frac{11}{2},-\frac{9}{2},\frac{1}{2}
\end{array}
\right.\right)$$ and we may conjecture that
$$\color{blue}{I_m=G_{4,4}^{2,2}\left(1\left|
\begin{array}{c}
1-m,1,1,m+1 \\
\frac{1}{2},\frac{2m+1}{2},-\frac{2m-1}{2},\frac{1}{2}
\end{array}
\right.\right)}$$
This has been verified for much higher values of $m$ and the results compared to the results of numerical integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I find this limit without L'Hopital and is my current solution correct? Find $$\lim_{x\rightarrow0} \frac{x^4}{1-2\cos x+\cos^2 x}$$
My solution:
Apply L'Hopital
$$=\lim_{x\rightarrow0} \frac{4x^3}{2\sin x-\sin 2x}$$
Apply L'Hopital again
$$=\lim_{x\rightarrow0} \frac{12x^2}{2\cos x-2\cos 2x}$$
Apply L'Hopital again
$$=\lim_{x\rightarrow0} \frac{24x}{-2\sin x + 4\sin 2x}$$
$$=\lim_{x\rightarrow0} \frac{24}{-2\cos x + 8\cos 2x}=\frac{24}{-2+8}=4$$
Is this correct? How would I approach this problem if I didn't want to use L'Hopital?
| Your result looks right.
Without L'Hospital, here is a hint: $1-2\cos x+\cos^2 x=(1-\cos x)^2=4\sin^4\frac{x}{2}$
This follows from double-angle identity using half-angle: $\cos x=\cos^2\frac{x}{2}-\sin^2 \frac{x}{2}= 1-2\sin^2 \frac{x}{2} \implies 2\sin^2 \frac{x}{2}=1-\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Closed form for infinite sum over unit balls in odd dimensions I stumbled upon the infinite sum
$$
1 + \sum_{k=0}^\infty \frac{(2\pi)^k}{(2k + 1)!!}
= 2 + \frac{2\pi}{3} + \frac{(2\pi)^2}{3\cdot 5} + \frac{(2\pi)^3}{3\cdot 5 \cdot 7} + \dots
$$
(This is actually the sum of the volume of the unit balls in all odd dimensions.)
From computation, it seems that it converges to something around 12.43. I'm wondering if there's a closed form. Any hints?
(The sum of the volumes of the unit balls in all even dimensions is known to be $\exp(\pi)\approx 23.14$. The sum is the same as above, just with even factors in the denominator.)
| Thanks for the comments to @K.defaoite and @Svyatoslav.
From here, we have
$$
\DeclareMathOperator{\erf}{erf}
\sum_{k=0}^\infty \frac{x^k}{(k + 1/2)!}
= \sum_{k=0}^\infty \frac{(2x)^k (2k)!!}{(2k + 1)!}
= \frac{\sqrt{\pi}}{2} \frac{\exp(x)}{\sqrt{x}} \erf{\sqrt{x}},
$$
so
$$
\sum_{k=0}^\infty \frac{(2\pi)^k}{(2k + 1)!!} =
\frac{1}{2} \exp(\pi) \erf(\sqrt{\pi})
$$
which indeed is $12.429316728301792$...
This means that the sum of the volumes of the unit balls in all dimensions is
$$
\begin{align}
\sum_{n=0}^\infty |S_n|
&= \sum_{n=0}^\infty |S_{2n}| + \sum_{n=0}^\infty |S_{2n + 1}|\\
&= \exp(\pi) + \left(1 + \frac{1}{2} \exp(\pi) \erf(\sqrt{\pi})\right)\\
&= 1 + \exp(\pi) \left(1 + \frac{1}{2} \erf(\sqrt{\pi})\right)\\
&\approx 35.57
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve this ODE: $y^{(y(x))}(x)=f(x)$? $$\large{\text{Introduction:}}$$
This question will be partly inspired from:
Evaluation of $$y’=x^y,y’=y^x$$
but what if we made the order of an differential equation equal to the function? Imagine that we had the following linear ordinary differential equation using nth derivative notation. In other words, the $y(x)\,th$ derivative of $y(x)$ is $f(x)$. I would have made $f(x)\to f(x,y)$, but that is too hard to solve. Note that $f(x)$ is any continuous infinitely differentiable function:
$$\frac{d^{y(x)}}{dx^{y(x)}}y(x)=\text D^{y(x)}_x y(x)=f(x)\implies y^{(y)}=f(x)\implies y^{(y)}=f$$
$$\large{\text{Specific Values:}}$$
Here are some examples of points:
$$y^{(y(0))}(0)=f(0), y^{\left(y \left(\frac12\right)\right)} \left(\frac12\right)=f\left(\frac12\right) $$
we could also imagine an inverse relation for $y(x)$ called $y^{-1}(x)$:
$$0=f\left(y^{-1}(0)\right), y^{\left(\frac12\right)} \left(\frac12\right) = f\left(y^{-1}\left(\frac12\right) \right),y’(1)=f\left(y^{-1}(1)\right)$$
$$\large{\text{Problem Statement:}}$$
this presents the problem that if $y(x)\not\in\Bbb Z$, then we need a fractional derivative, but definitions for such derivatives are confusing, so we can use the inverse operation of the fractional derivative which is the fractional integral using the corresponding notation in the bolded link:
$$y^{(y)}=f(x)\implies \text I_x^{y(x)}\text D_x^{y(x)}y(x)=\text I_x^{y(x)}f(x)\implies y+c_1=\text I^y_x f(x)$$
$$\large{\text{Integral Method:}}$$
Now let’s use Cauchy’s Formula for Repeated Integration as the definition for the fractional integration as seen in the bolded link:
$$\,_a\text I^n_x g(x)=\frac 1{Γ(n)}\int_a^xg(t)(x-t)^{n-1}dt\implies y+c_1=\,_a \text I^{y(x)}_x f(x)=\frac{1}{Γ(y(x))}\int_a^x f(t)(x-t)^{y(x)-1} dt\implies \boxed{y!+c_0 Γ(y)=\int_{c_1}^x f(t)(x-t)^{y-1} dt}$$
$$\large{\text{Special Case Solution Using Integral method:}}$$
Let’s now solve the $f(x)=1$ case using the conjectured formula:
$$y^{(y(x))}(x)=1\implies (y(x))!+c_0 Γ(y(x))=\int_{c_1}^x (x-t)^{y(x)-1} dt\implies y!+c_0 Γ(y)=\frac{(x-c_1)^y}y\implies \boxed{y!(y+c_0)-(x-c_1)^y=0}\implies x=\sqrt[y]{y!(y+c_0)}+c_1$$
Here is a complete interactive graph for this conjectured solution.
Another problem is if the fractional derivative has the same definition using $2$ different operator definitions, like this one.
$$\large{\text{Special Case Using Induction:}}$$
Another way for $f(x)=1$ is the following setting all constants of integration to be $0$ via quick induction:
$$y^{(y(x))}(x)=1\mathop\implies^{n=y(x)} y^{(n)}(x)=1\implies y^{(n-1)}=x,y^{(n-2)}=\frac {x^2}{1\cdot2},y^{(n-3)}=\frac{x^3}{3\cdot2\cdot1}\implies y^{(n-k)}=\frac{x^k}{k!}$$
Now let’s set $n=k$:
$$y^{(n-k)}(x)=y^{(n-n)}(x)=y^{(0)}(x)=y(x)=\frac{x^{y(x)}}{(y(x))!}$$
Let’s try to solve for $x$:
$$yx^{-y}=y!$$
but we cannot use the W-Lambert function here. We can also write the conjectured solution as:
$$y^{(y)}=1\mathop\implies^? y\ln(x)-\ln(y)-\ln(y!)=0$$
Using an Inversion theorem will be cumbersome.
Amazingly, we get that the conjectured solution to the differential equation:
$$y^{(y(x))}(x)=1\implies x=\sqrt[y]{yy!}$$
which can be thought of the inverse function of $$y=\sqrt[x]{xx!}$$
graphically shown here
$$\large{\text{Conclusion:}}$$
Is this a correct way to solve $y^{(y)}=f(x)$ and if not, then how? Please correct me and give me feedback!
$$\large{\text{Addendum:}}$$
Also see
Solving $y^{(x)}(x)=ax+b$ in closed form
for a twin question where I was able to find a solution using the following method and assuming all integration constants and $f(x)=1$ for $y(x)\in\Bbb N$:
$$y^{(y(x))}(x)=1=y^{(y)}=1\implies y^{(y-1)}=x+1\implies y^{(x-2)}=\frac{x^2}{2\cdot 1}+x+1$$
Therefore:
$$y^{(y-k)}=\sum_{n=0}^k\frac{x^n}{n!} \implies y^{(y-y)}=y^{(0)}=y(x)=\sum_{n=0}^{y(x)}\frac{x^n}{n!}$$
Here is an interactive graph of the result.
The apparent closed form of the sum may not work also as seen in the question:
$$e^zQ(n,z)=\sum_{m=0}^{n-1}\frac{x^m}{m!},n\in\Bbb N\implies y=\sum_{n=0}^{y} \frac{x^n}{n!}\mathop=^? e^x Q(y+1,x) $$
| Not a complete answer, but a partial answer.
I've asked myself the same question before, but never found a complete solution, so I'd like to share my work on it in the hope that it might help someone. Since nobody has answered yet, I will also share the unfinished one. I hope that's ok.
My Work (the best of it)
Let's assume that $f$ has a Maclaurin Series of the general form $f\left( x \right) = \sum_{k = 0}^{\infty}\left( f_{k} \cdot \frac{x^{k}}{k!} \right)$ (Assumption $1.$) and that $f$ can be written as a Generalized Hypergeometric Function (I will abbreviate "Generalized Hypergeometric Function" in the following with the abbreviation "GHF".) of the form $f\left( x \right) = \operatorname{_{p}F_{q}}\left( \begin{matrix} A_{1},\, A_{2},\, \dots,\, A_{p}\\ B_{1},\, B_{2},\, \dots,\, B_{q} \end{matrix};\, x \right)$ (Assumption $2.$), then we could try the Frobenius Method. But before we do that, I would like to mention the nth derivative formula or nth integral formula for monomials that I used:
$$
\begin{align*}
\operatorname{D}_{x}^{n}~ \left( x^{m} \right) &= \frac{m!}{\left( m - n \right)!} \cdot x^{m - n}\\
\operatorname{I}_{x}^{n}~ \left( x^{m} \right) &= \frac{m!}{\left( m + n \right)!} \cdot x^{m + n}\\
\end{align*}
$$
Rewrite it a bit:
$$
\begin{align*}
\operatorname{D}_{x}^{y\left( x \right)} \left( y\left( x \right) \right) &= f\left( x \right)\\
y\left( x \right) &= \operatorname{I}_{x}^{n}~ \left( f\left( x \right) \right)\\
\end{align*}
$$
Substitute $f\left( x \right) = \sum_{k = 0}^{\infty}\left( f_{k} \cdot \frac{x^{k}}{k!} \right)$:
$$
\begin{align*}
y\left( x \right) &= \operatorname{I}_{x}^{n}~ \left( f\left( x \right) \right)\\
y\left( x \right) &= \operatorname{I}_{x}^{n}~ \left( \sum_{k = 0}^{\infty}\left( f_{k} \cdot \frac{x^{k}}{k!} \right) \right)\\
y\left( x \right) &= \sum_{k = 0}^{\infty}\left( f_{k} \cdot \frac{\frac{k!}{\left( k + y\left( x \right) \right)!} \cdot x^{k + y\left( x \right)}}{k!} \right)\\
y\left( x \right) &= \sum_{k = 0}^{\infty}\left( f_{k} \cdot \frac{x^{k + y\left( x \right)}}{\left( k + y\left( x \right) \right)!} \right)\\
y\left( x \right) &= \sum_{k = 0}^{\infty}\left( \frac{f_{k} \cdot x^{y\left( x \right)}}{\left( k + y\left( x \right) \right)!} \cdot x^{k} \right)\\
\end{align*}
$$
We can rewrite this to the generalized series expansion $\sum_{k = 0}^{\infty}\left( c_{k} \cdot \frac{x^{k}}{k!} \right)$ (General Hypergeometric Series) of the GHF:
$$
\begin{align*}
y\left( x \right) &= \sum_{k = 0}^{\infty}\left( \frac{f_{k} \cdot x^{y\left( x \right)}}{\left( k + y\left( x \right) \right)!} \cdot x^{k} \right)\\
y\left( x \right) &= \sum_{k = 0}^{\infty}\left( \frac{f_{k} \cdot x^{y\left( x \right)} \cdot k!}{\left( k + y\left( x \right) \right)!} \cdot \frac{x^{k}}{k!} \right)\\
y\left( x \right) &= \sum_{k = 0}^{\infty}\left( c_{k} \cdot \frac{x^{k}}{k!} \right)\\
\end{align*}
$$
If we try to simplify this series to a GHF, we can get:
$$
\begin{align*}
\frac{c_{k + 1}}{c_{k}} &= \frac{\frac{f_{k + 1} \cdot x^{y\left( x \right)} \cdot \left( k + 1 \right)!}{\left( k + 1 + y\left( x \right) \right)!}}{\frac{f_{k} \cdot x^{y\left( x \right)} \cdot k!}{\left( k + y\left( x \right) \right)!}}\\
\frac{c_{k + 1}}{c_{k}} &= \frac{f_{k + 1} \cdot x^{y\left( x \right)} \cdot \left( k + 1 \right)! \cdot \left( k + y\left( x \right) \right)!}{f_{k} \cdot x^{y\left( x \right)} \cdot k! \cdot \left( k + 1 + y\left( x \right) \right)!}\\
\frac{c_{k + 1}}{c_{k}} &= \frac{f_{k + 1} \cdot \left( k + 1 \right)}{f_{k} \cdot \left( k + 1 + y\left( x \right) \right)}\\
\frac{c_{k + 1}}{c_{k}} &= \frac{f_{k + 1}}{f_{k}} \cdot \frac{k + 1}{k + 1 + y\left( x \right)}\\
\end{align*}
$$
Via the Assumption $2.$ that $f$ can be represented as a GHF of the form $f\left( x \right) = \operatorname{_{p}F_{q}}\left( \begin{matrix} A_{1},\, A_{2},\, \dots,\, A_{p}\\ B_{1},\, B_{2},\, \dots,\, B_{q} \end{matrix};\, x \right)$ it follows:
$$
\begin{align*}
\frac{f_{k + 1}}{f_{k}} &= \frac{\left( k + A_{1} \right) \cdots \left( k + A_{p} \right)}{\left( k + B_{1} \right) \cdots \left( k + B_{q} \right) \cdot \left( k + 1 \right)}\\
f\left( x \right) &= \operatorname{_{p}F_{q}}\left( \begin{matrix}
A_{1},\, A_{2},\, \dots,\, A_{p}\\ B_{1},\, B_{2},\, \dots,\, B_{q}
\end{matrix};\, x \right)\\
\end{align*}
$$
$$
\begin{align*}
\frac{c_{k + 1}}{c_{k}} &= \frac{f_{k + 1}}{f_{k}} \cdot \frac{k + 1}{k + 1 + y\left( x \right)}\\
\frac{c_{k + 1}}{c_{k}} &= \frac{\left( k + A_{1} \right) \cdots \left( k + A_{p} \right)}{\left( k + B_{1} \right) \cdots \left( k + B_{q} \right) \cdot \left( k + 1 \right)} \cdot \frac{k + 1}{k + 1 + y\left( x \right)}\\
\frac{c_{k + 1}}{c_{k}} &= \frac{\left( k + A_{1} \right) \cdots \left( k + A_{p} \right) \cdot \left( k + 1 \right)}{\left( k + B_{1} \right) \cdots \left( k + B_{q} \right) \cdot \left( k + 1 + y\left( x \right) \right) \cdot \left( k + 1 \right)}\\
\end{align*}
$$
We can combine all of this into a GHF:
$$
\begin{align*}
y\left( x \right) &= \operatorname{_{p + 1}F_{q + 1}}\left( \begin{matrix}
1,\, A_{1},\, A_{2},\, \dots,\, A_{p}\\ y\left( x \right) + 1,\, B_{1},\, B_{2},\, \dots,\, B_{q}
\end{matrix};\, x \right)\\
\end{align*}
$$
Leonard Euler found the following Integral Transformation for this form of the GHF:
$$
\begin{align*}
\operatorname{_{A + 1}F_{B + 1}}\left[ \begin{matrix} c,\, a_{1},\, \dots,\, a_{A}\\ d,\, b_{1},\, \dots,\, b_{B}\\ \end{matrix};\, z \right] &= \frac{\Gamma\left( d \right)}{\Gamma\left( c \right) \cdot \Gamma\left( d - c \right)} \cdot \int_{0}^{1} t^{c - 1} \cdot \left( 1 - t \right)^{d - c - 1}\ \cdot \operatorname{_{A}F_{B}}\left[ \begin{matrix} a_{1},\, \dots,\, a_{A}\\ b_{1},\, \dots,\, b_{B}\\ \end{matrix};\, t \cdot z \right]\, \operatorname{d}t\\
\end{align*}
$$
With this we get:
$$
\begin{align*}
y\left( x \right) &= \frac{\Gamma\left( y\left( x \right) + 1 \right)}{\Gamma\left( 1 \right) \cdot \Gamma\left( y\left( x \right) + 1 - 1 \right)} \cdot \int_{0}^{1} t^{1 - 1} \cdot \left( 1 - t \right)^{y\left( x \right) + 1 - 1 - 1}\ \cdot \operatorname{_{p}F_{q}}\left[ \begin{matrix} A_{1},\, \dots,\, A_{p}\\ B_{1},\, \dots,\, B_{q}\\ \end{matrix};\, t \cdot x \right]\, \operatorname{d}t\\
y\left( x \right) &= \frac{\Gamma\left( y\left( x \right) + 1 \right)}{\Gamma\left( 1 \right) \cdot \Gamma\left( y\left( x \right) \right)} \cdot \int_{0}^{1} \left( 1 - t \right)^{y\left( x \right) - 1}\ \cdot \operatorname{_{p}F_{q}}\left[ \begin{matrix} A_{1},\, \dots,\, A_{p}\\ B_{1},\, \dots,\, B_{q}\\ \end{matrix};\, t \cdot x \right]\, \operatorname{d}t\\
y\left( x \right) &= \frac{y\left( x \right)}{\Gamma\left( 1 \right)} \cdot \int_{0}^{1} \left( 1 - t \right)^{y\left( x \right) - 1}\ \cdot \operatorname{_{p}F_{q}}\left[ \begin{matrix} A_{1},\, \dots,\, A_{p}\\ B_{1},\, \dots,\, B_{q}\\ \end{matrix};\, t \cdot x \right]\, \operatorname{d}t\\
\Gamma\left( 1 \right) &= \int_{0}^{1} \left( 1 - t \right)^{y\left( x \right) - 1}\ \cdot \operatorname{_{p}F_{q}}\left[ \begin{matrix} A_{1},\, \dots,\, A_{p}\\ B_{1},\, \dots,\, B_{q}\\ \end{matrix};\, t \cdot x \right]\, \operatorname{d}t\\
1 &= \int_{0}^{1} \left( 1 - t \right)^{y\left( x \right) - 1}\ \cdot \operatorname{_{p}F_{q}}\left[ \begin{matrix} A_{1},\, \dots,\, A_{p}\\ B_{1},\, \dots,\, B_{q}\\ \end{matrix};\, t \cdot x \right]\, \operatorname{d}t\\
\end{align*}
$$
The best result I got was this:
$$\boxed{1 = \int_{0}^{1} \left( 1 - t \right)^{y\left( x \right) - 1}\ \cdot \operatorname{_{p}F_{q}}\left[ \begin{matrix} A_{1},\, \dots,\, A_{p}\\ B_{1},\, \dots,\, B_{q}\\ \end{matrix};\, t \cdot x \right]\, \operatorname{d}t}$$
I don't know if everything I've done is correct, but I hope it can help or at least give an idea for a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$, then find the value of $\frac{(1+x)^3}{1+x^3}$
Let $x$ be a real number such that $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$. What is the value of $\frac{(1+x)^3}{1+x^3}$?
Of course, one way is that to solve for $x$ from the quadratic $37(1+x)^2=13(1+x^2)$ which gives the value $x=\frac{-37\pm\sqrt{793}}{24}$. Thus we can compute $\frac{(1+x)^3}{1+x^3}=\frac{13}{49}$.
But I am trying to find the result from the equation $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$ without solving for $x$ using some algebra. Here are my attempts to do that:
We have $$\begin{align} & \frac{(1+x)^2}{1+x^2}=\frac{13}{37}\\ &\implies \frac{(1+x)^2}{(1+x)^2-2x}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{(1+x)^3-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+3x(2+x)-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+x(4+x)}=\frac{13}{37}
\end{align}$$
I can't seem to proceed from here.
| Instead of $ \displaystyle \frac{(1+x)^3}{1+x^3}$, I first find $ \displaystyle \frac{1+x^3}{(1+x)^3}$
$ \displaystyle \frac{1+x^3}{(1+x)^3} = \frac{(1+x) (1+x^2 -x)}{(1+x)^3} = 1 - \frac{3x}{(1+x)^2} \tag1$
(as $x \ne -1$)
Now note that $ \displaystyle \frac{1+x^2}{(1+x)^2} = \frac{37}{13} \implies 1 - \frac{2 x}{(1+x)^2} = \frac{37}{13}$
So we know $ \displaystyle \frac{x}{(1+x)^2} = - \frac{12}{13}$ and we can plug into the first equation which gives us $ \displaystyle \frac{49}{13}$. That leads to the answer of $ \displaystyle \frac{13}{49}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to get the maximum value of the sum in two sides in two right triangles? The problem is as follows:
The figure from below shows two triangles intersected on point $E$. Assume $AE=3\,m$ and $ED=1\,m$. Find the maximum value of $AB+EC$
The given choices in my book are as follows:
$\begin{array}{ll}
1.&\sqrt{10}\,m\\
2.&\4\,m\\
3.&\sqrt{5}\,m\\
4.&\4\,m\\
\end{array}$
The official solution for this problem according to my book is shown below:
Let $M=AB+EC$
Notice on the figure:
$AB=3\sin x$
and $EC=\cos x$
$M=3\sin x+\cos x$
$M=\sqrt{10}\left(\frac{3}{\sqrt{10}}\sin x+\frac{1}{\sqrt{10}}\cos x\right)$
$\tan \alpha =\frac{1}{3}$
$M=\sqrt{10}\left(\sin (x+\alpha)\right)$
Also notice that $x$ and $\alpha$ are acute angles.
Therefore:
$0<x+\alpha<\pi$
$\sin (x+\alpha)\leq 1$
Thus the maximum $M$ is $\sqrt{10}$.
There is where it ends the official solution.
But I am confused, where exactly is that alpha that is being talking about?, how did the author came up with the idea of $\sqrt{10}$ and the use of $\frac{1}{3}$.
Could someone please explain me, the logic that the author used and offer an alternate solution, perhaps easier to understand and less to guess what it was meant?.
I appreciate a step by step solution so I can understand.
| Any expression in the form $a\sin x+b\cos x$ can be written as $\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x\right)$. Note that $\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2=1$ which means that we can interpret these two expressions as cosine and sine of some angle $\alpha$. Think of a right triangle with legs $3$ and $1$ and you will understand what $\alpha$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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In $\triangle ABC$ AB=13,BC=14,CA=15 . D,E,F lie on BC ,CA,DE respectively . AD ,DE,AF perpendicular to BC,AC,BF. segment DF =M/N(coprime) .M+N?
In $\triangle ABC$, $AB=13,BC=14,CA=15$. Points $D,E,F$ lie on $BC ,CA,DE$ respectively such that $AD ,DE,AF$ perpendicular to $BC,AC,BF$. Segment $DF$ =$M/N$(coprime). $M+N$ is equal to what ?
What I did -
used herons formula to find area then found $AD$ using it. Using Pythagoras, I found $DC$ as $9$ so $BD$ becomes $5$. Then since $\triangle DBF\sim\triangle ABC$. Then using similarity got $\frac{AB}{DB}=\frac{BC}{FB}=\frac{AC}{DF}$.
So $13/5=15/DF$. So $DF=75/13$. So $M+N=88$.
Where am I going wrong?
Answer shows $21$
|
You already obtained $AD = 12, CD = 9, BD = 5$
In $\triangle ACD, CD:AD:AC = 3:4:5$
As $\angle ADE = \angle ACB$, $\triangle ADE \sim \triangle ACD$. So we obtain,
$DE = \dfrac{36}{5}, AE = \dfrac{48}{5}$
Also as $ABDF$ is cyclic, $\angle ABF = \angle ADE = \angle ACB \implies \triangle ABF \sim \triangle ACD$
So, $AF = \dfrac{52}{5}, EF^2 = AF^2 - AE^2 = 16$
Then $DF = \dfrac{36}{5} - 4 = \dfrac{16}{5} = \dfrac{m}{n}$
$ \therefore m + n = 21$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A-B)\det(A+B)$, even when $A$ and $B$ are not commutative. I am aware of the following identity
$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B)$
When $A = D$ and $B = C$ and when $AB = BA$ the above identity becomes
$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A)\det(A - BA^{-1}B) = \det(A^2 - B^2) = \det(A-B)\det(A+B)$.
However, I couldn't prove this identity for the case where $AB \neq BA$.
EDIT: Based on @Trebor 's suggestion.
I think I could do the following.
$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} =
\det\begin{bmatrix}A & B\\ B-A & A-B\end{bmatrix} = \det(A^2-B^2) = \det(A-B)\det(A+B)$.
| $\begin{pmatrix} A & B \\ B & A \end{pmatrix}\xrightarrow{\text{row1 -= row2}}\begin{pmatrix} A-B & B-A \\ B & A\end{pmatrix}\xrightarrow{\text{col2 += col1}}\begin{pmatrix} A-B & O \\ B & A+B\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove $a^2+b^2+c^2=x^2+y^2+z^2$ given that $a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$ Prove $$a^2+b^2+c^2=x^2+y^2+z^2$$ given that $$a^2+x^2=b^2+y^2=c^2+z^2=\\(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$$ where $a,b,c,x,y,z \in\mathbb R$
A friend forwarded me this problem which recently appeared in a math contest at Costa Rica. I've been working on it but I can't find a definite proof, though I have discovered several facts:
Let $u$ be equal to the value of the equation, then $ab+xy=\frac{-u}{2}$ and $a^2+b^2+c^2+x^2+y^2+z^2=3u$ which implies that $(a+b+c)^2+(x+y+z)^2 = 0 \implies a+b+c=x+y+z=0$.
I also noticed that, because of the equalities, working with complex numbers or vectors would be really tempting, as if $r=a+ix$, $s=b+iy$, $t=c+iz$, then we have $|r|=|s|=|t|=|r+s|=|s+t|=|r+t|$, but I can't find a way to make it useful.
Any ideas/hints/solutions?
| You are very close. Both your approaches point to the fact that $r$, $s$, $t$ form an equilateral triangle (three points on a circle, with center of mass at the origin). Then, if you use the exponential notation, the points are at $Re^{i\phi}$, $Re^{i\phi+2\pi/3}$, $Re^{i\phi+4\pi/3}$. Then it's easy to show that $$\sin^2\phi+\sin^2\left(\phi+\frac{2\pi}3\right)+\sin^2\left(\phi+\frac{4\pi}3\right)=\cos^2\phi+\cos^2\left(\phi+\frac{2\pi}3\right)+\cos^2\left(\phi+\frac{4\pi}3\right)=\frac32$$
Multiply with $R$ and you get your desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Minimize of $x^2+y^2$ subject to $x+y \ge 1$
Consider the problem of minimizing $x^2 +y^2$, subject
to $x + y \ge 1$.
Suppose that you start coordinate descent for this problem at $x = 1$ and $y = 0$.
Discuss why coordinate descent will fail.
The primal problem is equivalent to:
minimize $x^2+y^2$ subject to $-x-y \le -1$.
The lagrangian is: $L(x,y,a) = x^2+y^2-a(x+y-1)$. Then,
$\frac{dL}{dx} = 2x-a=0, x = a/2$
$\frac{dL}{dy} = 2y-a=0, y = a/2$
$\frac{dL}{da} = x+y-1=0$
So, the dual problem is: maximize $a^2/2$ subject to $a=1$. Hence, we get the primal optimal to be $(x,y) = (1/2,1/2)$. Right?
Now if we start at $(1,0)$. Let's solve for $x$.
plug in $y=0$ to get minimze $x^2$ subject to $x \ge 1$. So, $x = 1$.
Then plug in $x=1$ to solve for $y$: minimize $y^2+1$ subject to $y \ge 0$. We will get $y = 0$.
Hence, basically we are stuck at the starting point $(1,0)$.
| First let's observe from the following figure of the gradient fields for the objective function that the optimum occurs at the boundary, not inside the region bounded by the objective function curve (circle).
Now, by Lagrangian, we have $\nabla f = \lambda \nabla g$, where we have the gradient of the objective function $\nabla f=\begin{bmatrix}2x\\ 2y\end{bmatrix}$ and gradient of the constraint function as $\nabla g=\begin{bmatrix}1\\ 1\end{bmatrix}$ and hence $\begin{bmatrix}2x\\ 2y\end{bmatrix}$ $=\lambda\begin{bmatrix}1\\ 1\end{bmatrix}$
Eliminating $\lambda$ from the above system of equations, we get $x=y$ and along with $x^2+y^2=1$, we get $x^2=y^2=\frac{1}{2}$. Notice from the above gradient field that at $x=y=\frac{1}{\sqrt{2}}$ the maximum occurs, whereas $x=y=-\frac{1}{\sqrt{2}}$ is the point where the minimum occurs, but the constraint $x+y\geq 1$ is not satisfied here (hence an infeasible solution).
Now, let's look at the contour plot for the function and the feasible region for the solution, as can be seen from the following figure that minimum occurs at $(0,1)$ and $(1,0)$, where the value of the objective function is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x - \sqrt{ \frac{2}{x} } = 3 $ then $x-\sqrt{2x} = ?$ Please find the value without calculate the value of $x$. I tried to multipy it by $x$, I tried to square it but I still can't find the solution. I tried to make some equation that my help to solve this:
$$x^{2} + \frac{2}{x}-2\sqrt{2x} =9, $$
$$x^{2} - \sqrt{2x} = 3x,$$
$$x^2 - \frac{2}{x} = 3\left(x+\sqrt{\frac{2}{x}}\right).$$
| \begin{align}
&x - \sqrt{\frac{2}{x}} = 3\\
&x\sqrt{x} - \sqrt{2} = 3\sqrt{x}\\
&\sqrt{x}(x-3)=\sqrt{2}\\
&x(x-3)^2=2\\
&(x-2)(x^2-4x+1)=0
\end{align}
Note that $x=2$ is not a solution, so
$$x^2-4x+1=0 \implies x=2+\sqrt{3} \text{ or } x=2-\sqrt{3}$$
The second is not a solution either, so $x= 2+ \sqrt{3}.$
In that case, $x - \sqrt{2x} = 2+ \sqrt{3} - \sqrt{2(2 + \sqrt{3})} = 2 + \sqrt{3} - \sqrt{(\sqrt{3}+1)^2} = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4289510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Urn problem with random halting Consider an urn with $r$ red and $b$ black balls in it. We start drawing balls (without replacement) from it with following rules:
*
*If the drawn ball is red, we stop the draw with probability $p$. Conversely, we continue the draw with probability $1-p$.
*If the drawn ball is black, we stop the draw.
What is the expected number of drawn balls?
For $p=0$ the question was already solved here. It doesn't seem clear to me how to incorporate the random stop into the solution of $p=0$ case.
| You will draw the first ball with probability $1.$
The probability that you stop after drawing the first ball is
$$ \frac{b + pr}{b + r}. $$
The probability that you draw the second ball is
$$ 1 - \frac{b + pr}{b + r} = \frac{qr}{b + r}.$$
The probability that you draw the $n$th ball,
given that you drew the $(n-1)$st ball, is
$$
1 - \frac{b + p(r + 2 - n)}{b + r + 2 - n} = \frac{q(r + 2 - n)}{b + r + 2 - n}.
$$
So if $N$ is the total number of balls drawn,
\begin{align}
\mathbb P(N\geq1) &= 1, \\
\mathbb P(N\geq2) &= \frac{qr}{b + r}, \\
\mathbb P(N\geq3) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1}, \\
\mathbb P(N\geq4) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1}
\cdot \frac{q(r - 2)}{b + r - 2}, \\
\mathbb P(N\geq k) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1}
\cdots \frac{q(r + 2 - k)}{b + r + 2 - k}, \\
\mathbb P(N\geq k + 1) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1}
\cdots \frac{q(r + 1 - k)}{b + r + 1 - k}, \\
\mathbb P(N\geq r + 1) &= \frac{qr}{b + r} \cdot \frac{q(r - 1)}{b + r - 1}
\cdots \frac{q}{b + 1}, \\
\mathbb P(N\geq m) &= 0 \quad\text{if}\quad m \geq r + 2.
\end{align}
The expectation of $N$ is
$$
\mathbb E(N) = \sum_{k=1}^{r+1} \mathbb P(N \geq k).
$$
We can write the probabilities a little more compactly in one of the following forms:
$$
P(N \geq k + 1)
= \frac{r!\, (b+r-k)!\, q^k}{(r-k)!\,(b + r)!}
= \frac{\displaystyle \binom rk q^k}{\displaystyle \binom{b + r}k}
= \frac{r^{(k)} q^k}{(b + r)^{(k)}}
$$
where $a^{(k)} = a(a-1)(a-2)\cdot(a-k+1)$ is the falling factorial.
So for example, substituting $k = j + 1$ in the formula for $\mathbb E(N)$
above, we can write
$$
\mathbb E(N) = \sum_{j=0}^r \mathbb P(N \geq j + 1)
= \sum_{j=0}^r \frac{r!\, (b+r-j)!\, q^j}{(r-j)!\,(b + r)!},
$$
but I have not been able to simplify this or any of the other forms further.
| {
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"url": "https://math.stackexchange.com/questions/4290435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Compute the series $\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}$ Compute the series $$\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}.$$
How do I go about with the index notation, for example to arrange the series instead as $\sum_{n=1}^{\infty}a_n $?
I have tried to simplify the expression as:
$$\begin{align}&\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}=\sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies&\sum_{n=3}^{\infty} \frac{n^4}{n^6} - \sum_{n=3}^{\infty} \frac{1}{n^6} = \sum_{n=3}^{\infty} \frac{(n-1)(n^3+n+1)}{n^6}\\
\implies& \sum_{n=3}^{\infty} \frac{1}{n^2} - \sum_{n=3}^{\infty} \frac{1}{n^6}\end{align}$$
I'm not sure what to do with the index $n=3$ as I know that I can simplify it otherwise as $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{\pi^6}{945}$.
| Just "undo" the extraneous terms.
$$\sum_{n=3}^\infty a_n=\sum_{n=1}^\infty a_n-a_1-a_2.$$
$$\frac{\pi^2}6-\frac54-\frac{\pi^6}{945}+\frac{65}{64}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximizing the cross-sectional area of an isosceles trapezoid with an angle z I have been given the following problem:
A canal must be excavated, which must have a cross section shaped like an isosceles trapezoid. In the cross section, the
bottom line and the two slanting side pieces below the waterline must
together have a length of 150 m.
Determine the angle z so that the water flow can be as large as possible, ie. that the cross-sectional area must be as large as
possible
Image of the isosceles trapezoid
My drawing of the figure with defined sides
First of all, I have divided the trapezoid into one rectangle and two triangles.
In regards to the length of the sides I have said that:
$$150=120+2c\Leftrightarrow 30=2c\Leftrightarrow 15=c$$
I have determined a and b to be:
$$sin(z)=\frac{a}{c} \Leftrightarrow sin(z)\cdot c=a$$
$$cos(z)=\frac{b}{c} \Leftrightarrow cos(z)\cdot c=b$$
so that I can substitute these into the formula for the area of a triangle:
$$A_{triangle}=\frac{1}{2}\cdot a\cdot b=\frac{1}{2} \cdot sin(z) \cdot c \cdot cos(z)\cdot c=\frac{1}{2}c^2 \cdot sin(z) \cdot cos(z)$$
In the same way I have defined the area of the rectangle as:
$$A_{rectangle}=a \cdot 120 = sin(z) \cdot c \cdot 120$$
When I combine the area of the rectangle and the area of the triangle to get the area of the trapezoid I get:
$$A_{trapezoid}=2 \cdot A_{triangle} + A_{rectangle} =\frac{1}{2}c^2 \cdot sin(z) \cdot cos(z)+sin(z) \cdot c \cdot 120$$
Then I find the derivative of the area of the trapezoid:
$$A_{\text {trapezoid }}^{\prime}=\frac{d}{d z} A_{\text {trapezoid }}(z)=\frac{d}{d z}\left(c^{2} \cdot \sin (z) \cdot \cos (z)\right)+\frac{d}{d z}(\sin (z) \cdot c \cdot 120)$$
$$\quad=c^{2} \frac{d}{d z}(\sin (z) \cdot \cos (z))+120 \cdot c \frac{d}{d z}(\sin (z))$$
$$\quad=c^{2} \cdot \cos (z) \cdot \cos (z)+\sin (z) \cdot(-\sin (z))+120 \cdot c \cdot \cos (z)$$
$$\quad=c^{2} \cdot \cos ^{2}(z)-c^{2} \cdot \sin ^{2}(z)+120 \cdot c \cdot \cos (z)$$
Then I let: $cos^2(z)-sin^2(z)=cos(2z)$
At last I try my best to solve for z, but this is as far as I got:
$$\begin{aligned} c^{2} \cdot \cos ^{2}(z)-c^{2} \cdot \sin ^{2}(z)+120 \cdot c \cdot \cos (z)=0 & \Leftrightarrow\end{aligned}$$
$$\begin{aligned} c^{2}\left(\cos ^{2}(z)-\sin ^{2}(z)\right)+120 \cdot c \cdot \cos (z)=0 \Leftrightarrow \end{aligned}$$
$$\begin{aligned}c^{2} \cdot \cos (2 z)+120 \cdot c \cdot \cos (z) &=0 \Leftrightarrow\end{aligned}$$
$$\begin{aligned} c^{2} \cdot \cos (2 z)+120 \cdot c \cdot \cos (z)=0\end{aligned}$$
However, when I try to solve for z using a CAS-tool then $z = 83.03169^\circ$. Although, the answer in my textbook says $z = 96.97^\circ$.
Why is this and what have I done wrong?
Thanks a lot in advance for the help! :)
| There is another way to do this using the pythagorean theorem. The area of a Trapezoid is half the sum of the bases times the perpendicular height. If we assign $x$ to the $b$ dimension in your diagram then:
$$A = (120+x)(15^2-x^2)^{\frac{1}{2}}$$
Maximum area occurs where: $$\frac{dA}{dx} = 0$$
$$(15^2-x^2)^{\frac{1}{2}}+\frac{-2x(120+x)}{2(15^2-x^2)^{\frac{1}{2}}}=0$$
$$(15^2-x^2)^{\frac{1}{2}}=\frac{x(120+x)}{(15^2-x^2)^{\frac{1}{2}}}$$
$$15^2 - x^2 = 120x + x^2$$
$$2x^2+120x -225 = 0$$
Using the quadratic formula: $$x = 1.81981$$
The supplementary angle to angle $z$ is: $$\cos^{-1}(\frac{1.81981}{15}) = 83.03^o$$
$$\text{Angle}\ z = 96.97^o$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+...+x_4 \leq 54, x_1+...+x_5 \leq 90$. Find the maximum of ...
$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+...+x_4 \leq 54, x_1+...+x_5 \leq 90 \text{ for } x_1, ..., x_5 \in R_0^+. \\ \ \\ \text{Find the maximum of } \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}+\sqrt{x_5}.$
Of course, the answer will be $20(x_1=4, x_2=9, x_3=16, x_4=25, x_5=36.)$
My attempt:
\begin{align}
&\text{let } \sum_k \ x_l = x_l+x_{l+1}+x_{l+2}+...+x_{k+l-1}. \ (x_{5n+m}=x_m) \\
&\bigg(\sum_5 \sqrt{x_1} \bigg)^2 \leq 5\bigg(\sum_5 x_1\bigg) \leq 450. \\
&\therefore \bigg( \sum_5 \sqrt{x_1} \bigg) \leq 15\sqrt{2}. \\
\ \\
&\bigg(\sum_4 \sqrt{x_1} \bigg)^2 \leq 4\bigg(\sum_4 x_1 \bigg) \leq 216. \\
&\therefore \bigg( \sum_4 \sqrt{x_1} \bigg) \leq 6\sqrt{6}. \\
\ \\
&\bigg(\sum_3 \sqrt{x_1} \bigg)^2 \leq 3\bigg(\sum_3 x_1 \bigg) \leq 87. \\
&\therefore \bigg(\sum_3 \sqrt{x_1} \bigg) \leq \sqrt{87}. \\
\ \\
&\bigg(\sum_2 \sqrt{x_1} \bigg)^2 \leq 2\bigg(\sum_2 x_1 \bigg) \leq 26. \\
&\therefore \bigg(\sum_2 \sqrt{x_1} \bigg) \leq \sqrt{26}. \\
\ \\
&\bigg(\sqrt{x_1} \bigg)^2 \leq 4. \\
&\therefore \sqrt{x_1} \leq 2. \\
\end{align}
|
Thanks to the hint of Albus Dumbledore:
\begin{align}
&\displaystyle {(10x_5+12x_4+15x_3+20x_2+30x_1)}_{(A)} \bigg(\dfrac 1 {10} + \dfrac 1 {12} + \dfrac 1 {15} + \dfrac 1 {20} + \dfrac 1 {30} \bigg)_{(B)} \geq \Big(\sum \sqrt{x_i}\Big)^2. \\
&A \leq 1200, B = \dfrac 1 3 \\
&\therefore \Big(\sum \sqrt{x_i} \Big)^2 \leq 400, \sum \sqrt{x_i} \leq 20. \\
\ \\
&\text{It is a solution when }x_1=4, x_2=9, x_3=16, x_4=25, x_5=36.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicating at most 2 negative real roots
I end up with following 2 possibilities:-
1)1 positive, 2 negative real roots
2)1 positive, 2 imaginary roots
how to progress further ?
| Given the posted question's precalculus tag, I normally would not have posted the following answer. However, since Siong Thye Goh has given a precalculus answer, I will give an answer that involves derivatives, which is a concept from Calculus (AKA Real Analysis). Note that I am generally ignorant of the precalculus tools available to attack such a problem. Therefore, there may be other precalculus methods of attack besides that given in the answer by Siong Thye Goh.
$f(x)= x^3-x^2-3x-9.$
How many negative real roots does the equation $f(x) = 0$ have ?
The posted question is equivalent to asking how many times $f(x)$ crosses the $x$-axis, for $-\infty < x < 0.$ The behavior of $f(x)$ may be analyzed by analyzing its first and second derivatives, and by noticing that $f(x)$ is a continuous function.
Without belaboring the underlying theorems, what this implies is that if there is an interval $[a,b]$ such that $a < b$, and if there exists $x_1, x_2$ both in $[a,b]$ such that $f(x_1) < 0 < f(x_2)$, then there must exist at least one value $x_0$ in $[a,b]$ such that $f(x_0) = 0$. Note that it doesn't matter whether $x_1 < x_2$ or $x_1 > x_2$.
$f'(x) = 3x^2 - 2x - 3$ and $f''(x) = 6x - 2$.
$\displaystyle f'(x) = 0 \implies ~x ~=
~\left(\frac{1}{6}\right)
~\left[ ~2 ~\pm ~\sqrt{4 + 36} ~\right] ~=
~\left(\frac{1}{3}\right) ~\left[ ~1 ~\pm ~\sqrt{10} ~\right].$
Since the posted question is concerned only about negative roots, attention can be confined to
$\displaystyle (x_3) ~= ~\left(\frac{1}{3}\right) ~\left[ ~1 ~- ~\sqrt{10} ~\right].$
Since $f''(x) < 0$, for all $x < 0$, $f(x)$ is maximized at $x = x_3$, for all $x < 0$.
$\displaystyle (x_3)^3 ~= ~\left(\frac{1}{27}\right)
~\left[ ~1 - 3\sqrt{10} + 30 - 10\sqrt{10} ~\right]
~= ~\left(\frac{1}{27}\right)
~\left[ 31 - 13\sqrt{10} ~\right].$
$\displaystyle (x_3)^2 ~= ~\left(\frac{1}{9}\right)
~\left[ ~1 - 2\sqrt{10} + 10 ~\right]
~= ~\left(\frac{1}{27}\right)
~\left[ 33 - 6\sqrt{10} ~\right].$
$\displaystyle (x_3) ~= ~\left(\frac{1}{3}\right)
~\left[ ~1 - \sqrt{10} ~\right]
~= ~\left(\frac{1}{27}\right)
~\left[ 9 - 9\sqrt{10} ~\right].$
Therefore,
$$f(x_3) = \left(\frac{1}{27}\right) \times
~\left\{ ~\left[ 31 - 13\sqrt{10} ~\right]
~- ~\left[ 33 - 6\sqrt{10} ~\right]
~- ~\left[ 27 - 27\sqrt{10} ~\right]
~- ~\left[ 243 ~\right]
~\right\}.
$$
Simplifying,
$$f(x_3) = \left(\frac{1}{27}\right) \times
~\left[ -272 + 20\sqrt{10} ~\right] ~< ~0.$$
In summary, for $x$ restricted to negative values, $f(x)$ is maximized at $x = x_3$, with $f(x_3) < 0$. Therefore, there is no value of $x < 0$ such that $f(x) \geq 0.$ Therefore, the function $f(x)$ does not cross the $x$-axis in the interval $-\infty < x < 0.$
| {
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"url": "https://math.stackexchange.com/questions/4296303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\cos(2^n)$ diverges Problem:
I have sequence defined as follows:
$$
a_n = \cos(2^n)
$$
I need to show that it diverges.
My progress:
I tried common method here supposing that this sequence has limit $\lim_{n\to\infty}\cos(2^n) = a$ and then trying to get to contradiction with $\lim_{n\to\infty}(\cos(2^{n+1})-\cos(2^n))=0$. But it led me nowhere.
How can I show that this sequence diverges?
| $\textbf{Claim:}$ If $a_{n}$ converges then it converges to either $1$ or $-\frac{1}{2}$.
Proof: Suppose it converges to some $c$ then for each $\epsilon > 0$ $\exists N_{\epsilon} \in \mathbb{N}$ so that if $n \geq N_{\epsilon}$ we have
$$|a_{n+1}-a_{n}| < \epsilon$$
$$|2a_{n}^2-a_{n}-1|<\epsilon$$
This means that $c$ lies an arbitrarily small neighbourhood of one of the roots of the polynomial $2x^2-x-1$ which are $1$ and $-\frac{1}{2}$ respectively as we can take $\epsilon \rightarrow 0$.
$\textbf{Claim:}$ $a_{n}$ does not converge to $-\frac{1}{2}$.
Suppose it converges to $-\frac{1}{2}$ then
$$a_{n+1}+\frac{1}{2} = 2a_{n}^2-1+\frac{1}{2} = 2a_{n}^2-\frac{1}{2} = 2(a_{n}+\frac{1}{2})(a_{n}-\frac{1}{2})$$
Let us choose $M \in \mathbb{N}$ so that for all $n \geq M$ we have $|a_{n}-\frac{1}{2}| \geq \frac{3}{4}$. For $n \geq M$ one has
$$ |a_{n+1}+\frac{1}{2}| \geq \frac{3}{2}|a_{n}+\frac{1}{2}| $$
One must clearly have $a_{M} = \frac{1}{2}$ because otherwise
$\liminf_{k \rightarrow\infty}|a_{k}+\frac{1}{2}| \geq (\frac{3}{2})^{k-M}|a_{M}+\frac{1}{2}| = \infty.$
But if $a_{M} = \frac{1}{2}$ then $2^{m} \equiv \frac{2\pi}{3} \mod 2\pi$ or $2^{m} \equiv \frac{4\pi}{3} \mod 2\pi$; this would imply that $\pi$ is rational which is impossible. This is a contradiction.
$\textbf{Claim:}$ $a_{n}$ does not converge to $1$.
Suppose it converges to $1$ then
$$a_{n+1}-1 = 2a_{n}^2-2 = 2(a_{n}-1)(a_{n}+1)$$
Let us choose $K \in \mathbb{N}$ so that for $n \geq K$ we have $|a_{n}+1| \geq \frac{3}{2}$.
For $n \geq K$ one has
$$|a_{n+1}-1| \geq 3|a_{n}-1|$$
One must have $a_{K} = 1$ because otherwise we have
$$\liminf_{k \rightarrow \infty}|a_{k}-1| \geq 3^{k-K}|a_{K}-1| = \infty$$
But if $a_{K} = 1$ then $2^{K} \equiv 0 \mod 2\pi$; this implies that $\pi$ is rational. This is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all integer solutions of $(a+b+3)^2+2ab=3ab(a+2)(b+2)$ Here is another number theory problem that I am not able to do
Find all integer solutions of $$(a+b+3)^2+2ab=3ab(a+2)(b+2)$$
My attempt$:$
On expanding we get,
$$b^2+4ab+6b+a^2+6a+9=3a^2b^2+6a^2b+6ab^2+12ab$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9=3a^2b^2+6a^2b$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9-3a^2b^2-6a^2b=0$$ or $$\left(1-3b^2-6b\right)a^2+\left(-6b^2-8b+6\right)a+b^2+6b+9=0$$ or $$a=\frac{-\left(-6b^2-8b+6\right)\pm \sqrt{\left(-6b^2-8b+6\right)^2-4\left(1-3b^2-6b\right)\left(b^2+6b+9\right)}}{2\left(1-3b^2-6b\right)}$$ or $$a=\frac{-\left(-6b^2-8b+6\right)\pm \:4\sqrt{3}\left(b+1\right)\sqrt{b\left(b+2\right)}}{2\left(1-3b^2-6b\right)};\quad \:b\ne \:-\frac{3+2\sqrt{3}}{3},\:b\ne \frac{2\sqrt{3}-3}{3}$$ and after further solving $$a=\frac{3b^2+4b-3+2\sqrt{3}b\sqrt{b\left(b+2\right)}+2\sqrt{3}\sqrt{b\left(b+2\right)}}{1-3b^2-6b}$$ and $$a=\frac{3b^2+4b-3-2\sqrt{3}b\sqrt{b\left(b+2\right)}-2\sqrt{3}\sqrt{b\left(b+2\right)}}{1-3b^2-6b}$$ where $b\ne \:-\frac{3+2\sqrt{3}}{3},\:b\ne \frac{2\sqrt{3}-3}{3}$
So, now I can assign some integer values to $a$ and check if that value of $a$ is giving some integer values of $b$ or not. Like if $a=0$ then $b=-3$. But I think it is merely hit and trial because this thing can also apply to the original equation. Any help will be greatly appreciated.
| If $S=a+b$, $P=ab$ we have
$$S^2-6(P-1)S-(3P^2+10P-9)=0\rightarrow$$
$$S_{1,2}=3(P-1)\pm2\sqrt{\Delta}$$
$$\Delta=P(3P-2)$$
Now, we must find $P_i$ such as $\Delta$ be a square: $\Delta=K^2$. That is a difficult. I found only
$P_1=0\rightarrow\Delta=0\rightarrow S=-3$
$P_2=1\rightarrow\Delta=1\rightarrow S=\pm2$
$P_2=9\rightarrow\Delta=225\rightarrow S=24\pm15$
Now, for each pear $(S, P)$ solve $x^2-Sx+P=0$.
I don't know a theorem from the number theory to find all roots of the eq.
$$P(3P-2)=K^2$$
One attempt is
$$3P^2-2P=K^2\rightarrow 2P(P-1)=(K-P)(K+P)$$
but is not easier.
| {
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Find $f(3)$ if $f(f(x))=x^{2}+2$ Let $a,b,f(x),x$ be positive integers such that If $a>b$ then $f(a)>f(b)$ and $f(f(x))=x^{2}+2$ . Find $f(3)$
My approach:
Replacing $x$ with $f(x)$ in the equation gives $f(f(f(x))) = f(x)^2 + 2$, but $f(f(x)) = x^2 + 2$ so $$f(x^2+2) = f(x)^2 + 2$$
how do i proceed after this. Please help. Thanks alot!
| Note that $f(f(1))=3$ so there is some natural number $n$ such that $f(n)=3$.
If $f(1)>3$ then there could be no solution to $f(n)=3$ so we must have $f(1)\in \{1,2,3\}$.
If $f(1)=3$ then we have $3=f(f(1))=f(3)$, a contradiction.
If $f(1)=1$ then we would have $3=f(f(1))=f(1)$, a contradiction.
Thus $f(1)=2$.
It follows that $$f(2)=f(f(1))=1^2+2=3$$ from which we deduce that $$f(3)=f(f(2))=2^2+2=6$$ and we are done.
| {
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proper formula for this geometric series I'm trying to properly write the expression for this geometric series that converges to $\frac{1}{2}$ for all x > 1. It looks like this:
$x = 2, f(x) = \frac{1}{2}$
$x = 3, f(x) = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}$
$x = 4, f(x) = \frac{1}{24} + \frac{2}{24} + \frac{3}{24} + \frac{6}{24} = \frac{1}{2}$
etc.
What is the general equation for this? I can see that the number of actual terms is $1$ term for $x=2$, $2$ terms for $x=3$, $4$ terms for $x=4$, so I figured that the number of terms was $2^{x-2}$. To sum the fractions, the denominator of each term can be $x!$ and the numerator follows the sequence $\{1,2,3,6...\}$ which I can't identify, which is where I'm stuck.
So something like this is... or am I on the wrong track entirely?
$\displaystyle\sum^{2^{x-2}}_{k=1} = \frac{1,2,3,6...}{x!} = \frac{1}{2}$
Any help appreciated. Thanks!
| From a "simple solution that meets the requirements" approach, I would recommend using numerators $x-1$ multiplied by each of the numerators from the sum for $f(x-1)$. So for $x=5$ these additional numerators would be $4,8,12,24$ for the sum as $\frac{1+2+3+6+4+8+12+24}{120}$. Breaking down individual terms could be somewhat tedious in terms of finding a non-recursive term definition. From a value perspective, the recursive presentation might look like
$$f(x)=\frac{f(x-1)}x+\frac{(x-1)f(x-1)}x$$
which is obviously of the form $f(x)=f(x-1)$.
| {
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Let $a, b, c ∈ Z$. Then $ab + ac + bc$ is even if and only if at most one of $a, b$ and $c$ is odd. Is my proof correct? if so, is there something which isn't done nice? I am currently learning proofs on my own and the solutions in my book aren't always complete.
Lemma 1 given: Sum of an odd number of odd numbers is odd.
Lemma 2 given: A product of two integers is odd if both integers are odd.
Proving first implication: $ab+ac+bc$ is even $\Rightarrow$ at most one of $a,b,c$ is odd.
Proof by contradiction:
Suppose that $ab+ac+bc=2k$ and at least 2 of $a,b,c$ are odd.
Case 1: 2 numbers are odd.
WLOG assume that $a$ and $b$ are odd, write $a$ as $2k+1$, $b$ as $2l+1$ and $c$ as $2m$.
$$ab+ac+bc = (2k+1)(2l+1) + (2k+1)(2m) + (2l+1)(2m)$$
$$=(4kl + 2k + 2l + 1) + (4km + 2m) + (4lm + 2m)$$
$$=4kl + 2k + 2l + 4km + 2m + 4lm + 2m + 1$$
$$=2(2kl + k + l + 2km + m + 2lm + m) + 1$$
Since $2kl+k+l+2km+m+2lm+m$ is an integer $ab+ac+bc$ is odd which is a contradiction.
Case 2: three numbers are odd: is a contradiction if we proceed like in case 1.
Proving the second implication: If at most one of $a,b,c,$ is odd $\Rightarrow$ $ab +a c + bc$ is even.
Proof by contradiction:
Suppose that at most one of $a,b,c$ is odd and $ab + ac + bc$ is odd. Since $ab + ac + bc$ is odd,by lemma 1 one or three of the products are odd. By lemma 2 at least two numbers of $a,b,c$ are odd which is a contradiction.
$\blacksquare$
Edit:
Forgot to add second case.
| We'll keep in mind that the sum should be even. We'll also use the fact that $ab + bc + ac = b(a + c) + ac$. This will be for the first implication.
If both $a$ and $c$ are even, then $ac$ must be even and $a + c$ must be even. Having an even sum implies that $b(a + c)$ is even and $b$ can be odd or even as desired.
If both $a$ and $c$ are odd, then $ac$ will be odd and $a + c$ will be even. Having an even sum implies that $b(a + c)$ is odd and regardless of the parity of $b$, $b(a + c)$ will be even. A contradiction.
Therefore, the even sum $ab + bc + ac$ implies that, at most one of $a$, $b$, and $c$ is odd.
| {
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How to solve $3\sec(x)-2\cot(x)>0$ (trigonometric inequality)? I am struggling to find the solution, here is what I already tried (it's for a pre-calculus class so no calculus):
$3\sec(x)-2\cot(x)=0 \rightarrow 3(\frac{1}{\cos(x)})=2\left(\frac{\cos(x)}{\sin(x)}\right) \rightarrow 3\sin(x)=2\cos(x)^2 \rightarrow 3\sqrt{1-\cos(x)^2}=2\cos(x)^2 \rightarrow 9(1-\cos(x)^2)=4\cos(x)^4 \rightarrow 4\cos(x)^4+9\cos(x)^2-9=0$
Then I found the roots ($\pm \sqrt{3}/2$) and the respective $rad$ values for $x$ ($\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6},\frac{7\pi}{6})$, but the problem is that when I graph it I find that the functions are different, although they share the roots (but cosine one has more roots in between!) and some patterns (maximum/minimum values and the roots of the cosine function seem to have some relation to when the original one is positive or negative).
Did I do the transformations right? I though they were supposed to be equal at least on the roots, why are they related in such a weird way?
But anyway, I think my biggest problem right now is that I have no intuition for associating the roots to the original function so that I know when its positive or negative.
Edit: The solution set is for all $\mathbb{R}$.
Edit 2 : Thank you all. So I did it using $\cos(x)^2=1-\sin(x)^2$ and now the roots are right($\frac{\pi}{6},\frac{5\pi}{6}$). My only problem now is associating this with the original function. I don't know how to aproach it. Looking at the graph, the tangent and the secant parts of the function seem to be independent for some reason.
| $ 3 \sec x - 2 \cot x \gt 0 $
Implies
$ \dfrac{3}{\cos x} \gt 2 \dfrac{\cos x }{\sin x} $
Multiply through by $\sin^2 x \cos^2 x $
$ 3 \cos x \sin^2 x \gt 2 \sin x \cos^3 x $
Hence,
$ \cos x \sin x ( 3 \sin x - 2 \cos^2 x ) \gt 0 $
But $\cos^2 x = 1 - \sin^2 x $, so
$ \cos x \sin x (3 \sin x + 2 \sin^2 x - 2 ) \gt 0 $
The quadratic in $\sin x$ factors into $( 2 \sin x - 1 )( \sin x + 2 ) $
Hence, we now have
$ \cos x \sin x (2 \sin x - 1) (\sin x + 2) \gt 0 $
The last term is always positive, so this reduces to
$\cos x \sin x (2 \sin x - 1) \gt 0 $
The zeros of the above function are $0, \dfrac{\pi}{6}, \dfrac{\pi}{2},\dfrac{5 \pi}{6}, \pi, \dfrac{3\pi}{2} $
Considering all three terms, the set where the inequality is satisfied is
$ S = (\dfrac{\pi}{6}, \dfrac{\pi}{2} ) \cup (\dfrac{5 \pi}{6}, \pi ) \cup (\dfrac{3\pi}{2}, 2 \pi) $
| {
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Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less than or equal to $x$.
My try:
Letting $a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
Now knowing that:
$$a-1<[a]\leq a$$ and $$a-\frac{1}{2}<\left[a+\frac{1}{2}\right]\leq a+\frac{1}{2}$$
Adding both the above inequalities and using $(1)$ we get
$$2a-\frac{3}{2}<\frac{9a-5}{4}\leq 2a+\frac{1}{2}$$
So we get $$a \in (-1, 7]$$
Any help here?
| The LHS of this equation is an integer $\,n=\overbrace{\left\lfloor\dfrac{2x+1}3\right\rfloor}^\text{integer}+\overbrace{\left\lfloor\dfrac{4x+5}6\right\rfloor}^\text{integer}=\dfrac{3x-1}2$
We can express $x=\dfrac{2n+1}{3}$ and report in the equation to get $\dfrac 49n+\cdots$ inside the floor values.
This motivates us to set $\,n=9p+r\,$ with $p\in\mathbb Z$ and $r\in\{0,1,2,3,4,5,6,7,8\}$.
Reporting in the equation gives $\quad 9p+r=\left\lfloor 4p+\frac{4r+5}9\right\rfloor+\left\lfloor 4p+\frac{8r+19}{18}\right\rfloor$
And we can get the integer $4p$ out of the floor() function, after simplification we get:
$$p=\left\lfloor \frac{4r+5}9\right\rfloor+\left\lfloor \frac{8r+19}{18}\right\rfloor-r$$
Pluging the $9$ possible values for $r$ gives you as many solutions.
$\begin{array}{|l|ccccccccc|}\hline
r & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
p & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\hline
n & 9 & 10 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
x & \frac{19}3 & \frac{21}3 & \frac{5}3 & \frac{7}3 & \frac{9}3 & \frac{11}3 & \frac{13}3 & \frac{15}3 & \frac{17}3\\\hline
\end{array}$
| {
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$\sum_{k=0}^\infty[\frac{n+2^k}{2^{k+1}}] = ?$ (IMO 1968)
For every $ n \in \mathbb{N} $ evaluate the sum $ \displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right]$ ($[x]$ denotes the greatest integer not exceeding $x$)
This was IMO 1968, 6th problem.
This is a very interesting question I wanted to share, my answer :
|
For every $ n \in \mathbb{N} $ evaluate the sum $ \displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right] $
Now,
$\displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right] = \left[ \dfrac{n+1}{2} \right]+ \left[ \dfrac{n+2}{4}\right] + \left[ \dfrac{n+4}{8}\right]+\cdots+\left[ \dfrac{n+2^k}
{2^{k+1}}\right]\cdots $
$ \because \ \left[n\right] = \left[ \dfrac{n}{2} \right]+ \left[ \dfrac{n+1}{2} \right] \implies \left[ \dfrac{n}{2} +\dfrac{1}{2} \right]= \left[n\right]-\left[ \dfrac{n}{2} \right]\\
\implies \left[ \dfrac{n+2^k}{2^{k+1}} \right] = \left[ \dfrac{n}{2^{k+1}}+\dfrac{1}{2} \right]=\left[ \dfrac{n}{2^k} \right] -\left[ \dfrac{n}{2^{k+1}} \right] $
Now using the above result
$\displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right] = \left(\left[ n \right] - \left[ \dfrac{n}{2} \right] \right) + \left(\left[ \dfrac{n}{2} \right] - \left[ \dfrac{n}{4} \right] \right) + \left(\left[ \dfrac{n}{4} \right] - \left[ \dfrac{n}{8} \right] \right) \cdots $
$\implies \displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right] = \left[ n \right] =
n \ \ (\because n \in \mathbb{N}) $
$\boxed{ \therefore \displaystyle \sum_{k=0}^\infty \left[ \dfrac{n+2^k}{2^{k+1}} \right] = n }$
| {
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How to find the solution of $x^2\equiv 25\pmod{32}$? I am trying to find square root of $57$ modulo $32\times 49$. For that I need to find the solutions of $x^2\equiv 57\pmod{32}$ and $x^2\equiv 57\pmod{49}$ which are $x^2\equiv 25\pmod{32}$ and $x^2\equiv 8 \pmod{49}$. Now I could find the solution of the second one as follows: $x_1=1$ is a solution to $x^2\equiv 8 \pmod{7}$. Let $x_2=x_1+7k=1+7k$ be a solution of
$x^2\equiv 8 \pmod{49}$, so
$$\begin{align*}
(1+7k)^2\equiv 8 \pmod{49}&\implies 1+14k\equiv 8\pmod{49} \\
&\implies 14k\equiv 7 \pmod{49}\\
&\implies 2k\equiv 1 \pmod{7},
\end{align*}$$ thus we can find $k$ as $4$, so $1+7k=29$ and hence $x_2=29$ is a solution of $x^2\equiv 8 \pmod{49}$.
Now for the first equation, $x_1=1$ is a solution of $x^2\equiv 25\pmod{8}$. Now I take $x_2=x_1+8k=1+8k$ to be a solution of $x^2\equiv 25 \pmod{16}$, then I try to proceed in a similar way. But I get stuck because it boils down to finding $k$ such that $2k\equiv 1\pmod{2}$ which has no solution. My question is how to find a solution for $x^2\equiv 25\pmod{32}$?
| Let $32|x^2-25.$ Then $x$ must be odd so let $x=2n+1.$ Then
$32|(x-5)(x+5)=(2n-4)(2n+6)\;$ iff $\;8|(n-2)(n+3).$
Now one of $n-2,n+3$ is odd and the other is even so it is necessary & sufficient that $8|n-2$ or $8|n+3.$ And the rest is easy.
| {
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Epsilon - delta proof $\lim_{x \to \frac{\pi}{4}} \tan(x)=1$ I need to prove the limit using the definition of limit $$\lim_{x\to c}f(x)=L \leftrightarrow \forall \epsilon >0 \hspace{0.5 cm} \exists \delta >0 : 0<|x-c|<\delta \rightarrow |f(x)-L|<\epsilon $$
My attempt
$$|\tan(x)-1|<\epsilon\\ -\epsilon <\tan(x)-1 < \epsilon\\ \tan^{-1}(-\epsilon+1)<x<\tan^{-1}(\epsilon +1)\\ \tan^{-1}(-\epsilon+1)- \frac{\pi}{4}<x-\frac{\pi}{4}<\tan^{-1}(\epsilon +1)-\frac{\pi}{4}$$
I don't know if with this I can give an adequate value for $\delta$. Any advice on how to solve it would be very helpful, thank you.
| Alternative approach:
$\underline{\textbf{Preliminary Results }}$
Result-1
$\displaystyle \lim_{x \to 0} ~\tan(x) = 0.$
Proof:
Attempting to stay within the spirit of the problem,
since the sine function is continuous, I will assume that :
*
*$\displaystyle \lim_{x \to 0} \sin(x) = \sin(0) = 0.$
Edit
If you are given that $~ \displaystyle \lim_{x \to 0}\frac{\sin(x)}{x} = 1, ~$
then the above assumption is a consequence of that.
For $~\epsilon > 0,~$ set $~ \displaystyle \epsilon_1 = \frac{\epsilon}{\sqrt{2}}.$
$\exists ~\delta_1 > 0~$ such that
$~|x| \leq \delta_1 \implies |\sin(x)| < \epsilon_1$.
Also, $~\displaystyle |x| \leq (\pi/4) \implies \frac{1}{\sqrt{2}} \leq \cos(x) \leq 1.$
For $\epsilon > 0,~$ set $~\delta = \min(\delta_1, \pi/4).$
Then $~\displaystyle |x| \leq \delta \implies$
*
*$\displaystyle |\sin(x)| < \frac{\epsilon}{\sqrt{2}}$
*$\displaystyle \frac{1}{\sqrt{2}} \leq |\cos(x)| \leq 1 \implies 1 \leq \frac{1}{\cos(x)} \leq \sqrt{2}.$
Therefore, $~\displaystyle |x| \leq \delta = \min(\delta_1, \pi/4) \implies $
$\displaystyle |\tan(x)| = \left|\frac{\sin(x)}{\cos(x)}\right| < \frac{\epsilon}{\sqrt{2}} \times \sqrt{2} = \epsilon.$
Result-2
$\displaystyle \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}.$
Proof:
See wikipedia trig angle sum identities.
$\underline{\textbf{Main Problem}}$
Given $~\epsilon > 0,~$ set $\epsilon_2 = \min[(1/2), \epsilon/5].$
Invoke Result-1.
Then, there exists a $~\delta_2 ~$ such that
$-\delta_2 < y < \delta_2 \implies |\tan(y)| < \epsilon_2.$
Suppose $-\delta_2 < [x - \pi/4] < \delta_2.$
Set $y = [x - \pi/4] \implies$
*
*$x = (y + \pi/4) \implies \tan(x) = \tan(y + \pi/4)$
*$-\delta_2 < y < \delta_2 \implies -\epsilon_2 < \tan(y) < \epsilon_2.$
Invoke Result-2.
Then
$$\tan(x) = \tan(y + \pi/4) = \frac{\tan(y) + 1}{1 - \tan(y)}.\tag1 $$
In (1) above, let $N$ denote the numerator and let $D$ denote the denominator.
Then $1 - \epsilon_2 < N < 1 + \epsilon_2$
and $1 - \epsilon_2 < D < 1 + \epsilon_2.$
Therefore
$$\frac{1 - \epsilon_2}{1 + \epsilon_2} < \frac{\tan(y) + 1}{1 - \tan(y)} < \frac{1 + \epsilon_2}{1 - \epsilon_2}. \tag2 $$
However, because $~\epsilon_2 = \min[(1/2), \epsilon/5],~$
you have that
*
*$\displaystyle 1 - \epsilon < 1 - 2\epsilon_2 < 1 - \frac{2\epsilon_2}{1 + \epsilon_2} = \frac{1 - \epsilon_2}{1 + \epsilon_2}.$
*$\displaystyle \frac{1 + \epsilon_2}{1 - \epsilon_2} = 1 + \frac{2\epsilon_2}{1 - \epsilon_2} \leq
1 + \frac{2\epsilon_2}{(1/2)} = 1 + 4\epsilon_2 < 1 + \epsilon.$
Therefore $\tan(x) = \tan(y + \pi/4)$ and
$$1 - \epsilon < \tan(y + \pi/4) < 1 + \epsilon.$$
| {
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how to solve this linear system of three equations using Cramer's rule? I have a 3-by-3 matrix,
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
1 & 0 & 1 \\
1 & 1 & -1\\
\end{matrix} \right]$
the known terms are (-6, 2, -5), at the right of "=" symbol.
(1) I've calculated the determinant,
(2) I've used Cramer's rule to find x, y, and z.
but the result isn't correct (the right solutions is (x, y,z) = (1, -5, 1)).
(1) determinant of A,
I used row operations: row 3 <-> row 2. I swapped the rows. (therefore the determinant is $-\det A$).
(2) row 2 <-- $row 2 - row 1$ and I got:
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
0 & -1 & -4 \\
1 & 1 & -1\\
\end{matrix} \right]$
then, I used Laplace in the first column, and I got:
$-\det A$=$\left [ \begin{matrix}
-1 & -4 \\
0 & 1 \\
\end{matrix} \right]$ + $\left [ \begin{matrix}
2 & 3 \\
-1 & -4 \\
\end{matrix} \right]$,
doing algebra here, I get: -1-(-8+3) = -6 but it was -detA, and therefore detA = 6.
I used Cramer's rule, therefore I put the known terms in first column, then in second, and so on.
x = A=$\left [ \begin{matrix}
-6 & 2 & 3 \\
2 & 0 & 1 \\
-5 & 1 & -1\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
x is equal to 3, it should've been equal to 1.
y = A=$\left [ \begin{matrix}
1 & -6 & 3 \\
0 & 2 & 1 \\
1 & -5 & -1\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
-2-6-6-5 = -14-5 = -19 it should have been equal to -5. (Sarrus' rule)
z = A=$\left [ \begin{matrix}
1 & 2 & -6 \\
0 & 0 & 2 \\
1 & 1 & -5\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
4-2 = 2. it should've been equal to 1. (Sarrus' Rule).
| The determinant of$$A_x=\begin{bmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{bmatrix}$$is $6$, which is equal to $\det(A)$. Therefore, $\frac{\det(A_x)}{\det(A)}=1$, which is what you should have got.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove Cayley-Hamilton theorem I have a exercise in my linear algebra textbook:
Let $c_2\lambda^2+c_1\lambda +c_0=0$ be the characteristic equation for the matrix $$A=\begin{pmatrix}1&3\\3&1\end{pmatrix}$$
Prove that $c_2A^2+c_1A +c_0I=0$
This is Cayley-Hamilton theorem.
My solution:
If a root $\lambda$ exists, then it is the eigen value for the eigen vector $\vec{v}$.
If we multiply $c_2A^2+c_1A +c_0I$ with $\vec{v}$ then we get:
$$(c_2\lambda^2+c_1\lambda +c_0)\vec{v}$$
and since the eigen vector is not the zero vector and $c_2\lambda^2+c_1\lambda +c_0=0$ is true, $c_2A^2+c_1A +c_0I=0$ is also true.
Is this enough to prove the theorem and solve the problem?
| You have a specific matrix given:
$$
A=\begin{pmatrix}1&3\\3&1\end{pmatrix}.
$$
Its characteristic equation is given by
$$
p(\lambda):=\det(\lambda I - A) = \begin{vmatrix}\lambda-1&-3\\-3&\lambda-1\end{vmatrix}
= (\lambda-1)^2-(-3)^2 = \lambda^2-2\lambda-8.
$$
Now,
$$
A^2=\begin{pmatrix}1&3\\3&1\end{pmatrix}\begin{pmatrix}1&3\\3&1\end{pmatrix}
= \begin{pmatrix}10&6\\6&10\end{pmatrix}
$$
so
$$
p(A)
= A^2 - 2A - 8I
= \begin{pmatrix}10&6\\6&10\end{pmatrix} - 2 \begin{pmatrix}1&3\\3&1\end{pmatrix} - 8 \begin{pmatrix}1&0\\0&1\end{pmatrix}
= \begin{pmatrix}0&0\\0&0\end{pmatrix}.
$$
| {
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When does this inequality hold? $\sum_{cyc}\frac{a^2+(a+b+d)c}{a^3+3bcd}\ge\frac{(a+b+c+d)^2}{a^3+b^3+c^3+d^3}$
For positive real numbers $a,b,c,d$, prove the inequality $$\frac{a^2+(a+b+d)c}{a^3+3bcd}+\frac{b^2+(a+b+c)d}{b^3+3acd}+\frac{c^2+(b+c+d)a}{c^3+3abd}+\frac{d^2+(a+c+d)b}{d^3+3abc}\ge\frac{(a+b+c+d)^2}{a^3+b^3+c^3+d^3}$$
When does equality hold?
I think that we have to use the $A.M-G.M$ inequality in some way here. I'm pretty sure we should start by simplifying the initial expression in some other way than multiplying.
I've thought about
$$a^3+bcd+bcd+bcd\ge4\sqrt[4]{a^3b^3c^3d^3}=4a^\frac{3}{4}b^\frac{3}{4}c^\frac{3}{4}d^\frac{3}{4}$$
$$a^2+ac+bc+cd\ge4\sqrt[4]{a^3bc^3d}=4a^\frac{3}{4}b^\frac{1}{4}c^\frac{3}{4}d^\frac{1}{4}$$
So
$$\frac{a^2+(a+b+d)c}{a^3+3bcd}=\frac{4a^\frac{3}{4}b^\frac{1}{4}c^\frac{3}{4}d^\frac{1}{4}}{4a^\frac{3}{4}b^\frac{3}{4}c^\frac{3}{4}d^\frac{3}{4}}=\frac{1}{\sqrt{bd}}$$
And we can repeat this for the other 3 expressions and get
$$\frac{2}{\sqrt{bd}}+\frac{2}{\sqrt{ac}}=\frac{a^2+(a+b+d)c}{a^3+3bcd}+\frac{b^2+(a+b+c)d}{b^3+3acd}+\frac{c^2+(b+c+d)a}{c^3+3abd}+\frac{d^2+(a+c+d)b}{d^3+3abc}$$
So now we have to prove
$$\frac{2}{\sqrt{bd}}+\frac{2}{\sqrt{ac}}\ge\frac{(a+b+c+d)^2}{a^3+b^3+c^3+d^3}$$
But I'm not sure how to do this, I think we have to use the $A.M-G.M$ inequality in some way again. Any help is appreciated.
| Instead of solving LHS we will try to make from RHS
so RHS can be written as
$\frac{( a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd)}{( a^3 + b^3 + c^3 + d^3 )}$
which can be written as
$\frac{(a^2+c(a+b+d))}{(a^3+b^3+c^3+d^3)}$ + ..... similarly taking a square and 3 other terms
$a^3+b^3+c^3+d^3 \ge a^3+ (b^3\cdot c^3\cdot d^3)^\frac{1}{3}$
$a^3+b^3+c^3+d^3 \ge a^3 + 3abcd$
$\frac{(a^2+c(a+b+d))}{(a^3+b^3+c^3+d^3)} <= \frac{(a^2+c(a+b+d))}{(a^3+3abcd)}$ .. similarly for the other terms
| {
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How to prove that $ \frac{n+1}{4n^2+3}$ is a Cauchy sequence? By definition, I need to show that for any $\epsilon \gt 0$ there exists $N \in \mathbb{N}$ such that for any $m,n \gt N$:
$ \lvert a_n - a_m \rvert \lt \epsilon$
So I write,
$ |\frac{n+1}{4n^2+3} - \frac{m+1}{4m^2+3} |\leq | \frac{n+1}{4n^2+3}| + |\frac{m+1}{4m^2+3}| = \frac{n+1}{4n^2+3} + \frac{m+1}{4m^2+3} \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
From here I try to show,
$\frac{n+1}{4n^2+3} \lt \frac{\epsilon}{2} \quad \quad \Rightarrow \quad \quad 4\epsilon n^2 -2n +3\epsilon -2 \gt 0$
$n \gt \frac{1}{4\epsilon} + \frac{1}{4\epsilon} \sqrt{1-12\epsilon^2+8\epsilon}$
Can someone please explain to me what should I do from here?
Am I in the right direction at all?
| It's easier to use the fact that, since $n+1\leqslant2n$ and $4n^2+3\geqslant4n^2$, then$$\frac{n+1}{4n^2+3}\leqslant\frac{2n}{4n^2}=\frac1{2n}.$$Now, use the fact that$$\frac1{2n}<\frac\varepsilon2\iff n>\frac1\varepsilon.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to prove fraction inequality proof How can I prove the following inequality?
$$\frac{a^3}{b^3} + \frac{b^3}{c^3} + \frac{c^3}{a^3} \geq \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
I've tried the get common denominators but that doesn't really seem to help since although the denominators are the same, the numerators are hard to compare after doing so.
| Hints (assuming $\,a,b,c\,$ all have the same sign):
*
*$\displaystyle\frac{a^3}{b^3} + \frac{b^3}{c^3} + \frac{c^3}{a^3} \geq \frac{1}{9}\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)^3\,$ by the generalized means inequality;
*$\displaystyle \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3\,$ by AM-GM.
| {
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} |
$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
So for $n=1999$ I get the sum as $2,66,46,67,000$
From this I need to subtract the squares of the even terms twice because subtracting once leaves with only the sum of the squares of the odd nos.
I observed something :
$2^2 + 4^2 + 6^2 + \dots + 1998^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + \dots + (2 \cdot 999)^2$
Therefore to obtain the sum of the square of the even terms, I can take $4$ as common and use the aforementioned formula for $n=999$ and multiply it by $4$.
therefore sum of square of even terms = $1,33,13,34,000$
I need to subtract this sum twice to get the answer, because subtracting once simply leaves me with the sum of the squares of the odd numbers.
The answer is now $1999000$, which still doesn't match the answer key. Can someone explain where I am going wrong ?
| Looking at your work, if you only subtract the sum of the even terms from the sum of all terms, you are only left with the sum of the odd terms, not the alternating sum you have shown above.
Hint: How many times do I need to subtract a term from a summation to get it to go from + to - ?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
For 2 orthogonal vectors, is tan(a)tan(b)=-1?
This is a part of a physics subproblem I was solving where in the solution they casually mentioned:-
$\begin{align}\\
tan(A) \times tan(B) = -1
\end{align}\\$
Without proving it, and proceeded to solve the complete question. I tried using ASS similarity methods but it yielded a totally different result.
I am willing to post my physics question here in case someone thinks this is an XY scenario or I got something wrong! :]
| Use the definition $\tan(x) \triangleq \frac{\sin(x)}{\cos(x)}$ and trigonometric product identities (Google them).
$\begin{align}
\tan(A) \tan(B) &= \frac{\sin(A)}{\cos(A)} \frac{\sin(B)}{\cos(B)}\\
&= \frac{\sin(A) \sin(B)}{\cos(A)\cos(B)}\\
&= \frac{\frac{1}{2} \left(\cos(A - B) - \cos(A + B) \right) }{\frac{1}{2} \left(\cos(A + B) + \cos(A - B) \right)}\\
&= \frac{\frac{1}{2} \left(\cos(A - B) - \cos(90) \right) }{\frac{1}{2} \left(\cos(90) + \cos(A - B) \right)}\\
&= \frac{\frac{1}{2} \left(\cos(A - B) - 0 \right) }{\frac{1}{2} \left(0 + \cos(A - B) \right)}\\
&= 1
\end{align}$
If the angles are taken to be signed angles, then the above proof stills works using the additional fact that $\tan(-B) = - \tan(B)$. This means that angle A is positive/anticlockwise and angle B is negative/clockwise.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
$\int_0^{\frac{\pi}{2}}\ln(\sin^2 x+k^2\cos^2 x)dx$ not by differentiation under the integral? Show that
$$\int_0^{\frac{\pi}{2}}\ln(\sin^2 x+k^2\cos^2 x)dx=\pi\ln \frac{1+k}{2}$$
This is an exercise from Edwards Treatise on Integral Calculus II pg.188.
What solution to this problem can be given ?
In the chapter in Edwards he introduces the technique of substitution $x\mapsto \frac{\pi}{2}-x$. And proves Euler's formulas
$$\int_0^{\frac{\pi}{2}}\ln \sin x dx =\int_0^{\frac{\pi}{2}}\ln \cos x dx=-\frac{\pi}{2}\ln 2$$
The problem can be written as
$$\int_0^{\frac{\pi}{2}}\ln(\tan^2 x+k^2 )dx+2\int_0^{\frac{\pi}{2}}\ln(\cos x)dx=\pi\ln(1+k)-\pi\ln 2$$
Which reduces the question to
$$\int_0^{\frac{\pi}{2}}\ln(\tan^2 x+k^2 )dx=\pi\ln(1+k)$$
What proof can people give of this equation ?
The only success I have had is to differentiate with respect to $k$, and the result is easy to evaluate.
However, it seems that Edwards has at this stage not discussed that method, (although he lists it as a technique, and there is a chapter on it in part I) so is there another solution ? Of course, these old books may be playing by different rules. And the Tripos type questions dont follow any pedagogical pattern. But I guess I really want to know if someone has another solution.
| There is a quite low-level elementary solution. Using some trigonometric identities and change of variable, we obtain
\begin{multline*}
\int\limits_0^{\frac \pi 2} \ln(\sin^2 x + k^2\cos^2x)\,dx = \int\limits_0^{\frac \pi 2} \ln\Big(\frac{1-\cos 2x}2+ k^2\frac{1+\cos 2x}2\Big)\,dx = \\
\frac 12 \int\limits_0^\pi \ln\Big(\frac{1-\cos t}2+ k^2\frac{1+\cos t}2\Big)\,dt = \frac 14 \int\limits_0^{2\pi} \ln\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos t\Big)\,dt.
\end{multline*}
Suppose that $0 < k < 1$. Then we have an integral of a continuous function, which equals to the limit of its Riemann sums:
\begin{multline*}
\frac 14 \int\limits_0^{2\pi} \ln\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos t\Big)\,dt = \frac \pi{2n} \lim\limits_{n\to\infty} \sum\limits_{j=0}^{n-1}\ln\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos \frac{2\pi j}n\Big) =\\ \frac \pi{2n} \lim\limits_{n\to\infty} \ln \prod\limits_{j=0}^{n-1}\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos \frac{2\pi j}n\Big) = \frac \pi{2n} \lim\limits_{n\to\infty} \ln \prod\limits_{j=0}^{n-1}\Big(a^2 - 2ab\cos \frac{2\pi j}n + b^2\Big),
\end{multline*}
where $a = \frac{1+k}{2}$, $b = \frac{1-k}{2}$. Applying the algebraic identity
$$
\prod\limits_{j=0}^{n-1}\Big(a^2 - 2ab\cos \frac{2\pi j}n + b^2\Big) = (a^n - b^n)^2,
$$
we finally get
$$
\frac \pi{2n} \lim\limits_{n\to\infty} \ln (a^n - b^n)^2 = \frac \pi{n} \lim\limits_{n\to\infty} \ln \Big(\Big(\frac{1+k}2\Big)^n - \Big(\frac{1-k}2\Big)^n\Big) = \pi \ln \frac{1+k}2.
$$
| {
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"source": "stackexchange",
"question_score": "5",
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Solving the system $x^2+y^2+x+y=12$, $xy+x+y=-7$ I've been trying to solve this system for well over the past hour.$$x^2+y^2+x+y=12$$
$$xy+x+y=-7$$
I've tried declaring $x$ using $y$ ($x=\frac{-7-y}{y+1}$) and solving from there, but I've gotten to $$y^4+3y^3-9y^2-45y+30=0$$ and I don't see how we can get $y$ from here.
If anyone sees it, I'd appreciate the help. But if you see a simpler solution please do not hesitate to leave a comment and notify me of such.
| $$(x+1)(y+1)=-7+1$$
Let $x+1=a,y+1=b\implies ab=-6$
$$12=x^2+y^2+x+y=(a-1)^2+(b-1)^2+a+b-2=a^2+b^2-(a+b)=(a+b)^2-2(-6)-(a+b)$$
$$\implies(a+b)(a+b-1)=0$$
So, we know $a+b$ and $a,b$
Can you take it home from here?
| {
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"source": "stackexchange",
"question_score": "3",
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Calculate residue of pole with order 5 I am trying to evaluate the residue of $\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}$ at z=0.
Is there a way to do this without having to do the long derivative calculation?
I know we have the formula $Res(f,0)=\lim_{z\to0}\frac{1}{(5-1)!}\frac{\mathrm{d}^{5-1}f}{\mathrm{d}z^{5-1}}z^5f(z)$. But finding the 5th derivative of this function would take a long time and be pretty messy. My first try was to find the residues at the other poles and at infinity since all the residues must sum to zero, but when I tried to do this the residue at infinity ended up being the same as the residue at zero.
I've also considered finding the Laurent expansion but the Partial Fraction Decomposition is also long.
| We will work in the power series ring $\Bbb Q[[z]]$ in the variable $z$ over $\Bbb Q$. Let us denote by $\operatorname{Coeff}_{z^k}\;f$ the coefficient in $z^k$ of some series $f$ in this ring. It may be useful to substitute $w=z^2$, then work in $\Bbb Q[[w]]$, but this is important only to have an easier typing. We may then work modulo $O(z^5)$, respectively modulo $O(w^3)$.
Then the needed residue is:
$$
\begin{aligned}
&\operatorname{Res}_{z=0}
\frac{(z^6-1)^2}{z^5(2z^4-5z^2+2)}
\\
&\qquad= \operatorname{Coeff}_{z^4}
\ \frac 12\cdot \frac{(z^6-1)^2}{z^4-\frac 52z^2+1}\\
&\qquad= \frac 12 \cdot \operatorname{Coeff}_{w^2}
\ \frac{(w^3-1)^2}{w^2-\frac 52w +1}\\
&\qquad= \frac 12 \cdot \operatorname{Coeff}_{w^2}
\ \frac{(w^3-1)^2}{w^2-\frac 52w +1} +O(w^3)\\
&\qquad= \frac 12 \cdot \operatorname{Coeff}_{w^2}
\ \frac{(0-1)^2}{w^2-\frac 52w +1} +O(w^3)\\
&\qquad= \frac 16 \cdot \operatorname{Coeff}_{w^2}
\ \left[\frac 4{1-2w} - \frac 1{1-\frac w2}\right] +O(w^3)\\
&\qquad= \frac 16 \cdot \operatorname{Coeff}_{w^2}
\ \left[\ 4(1 + 2w + 4w^2+\dots) - \left(1+\frac w2+\frac {w^2}4+\dots\right)\ \right] +O(w^3)\\
&\qquad= \frac 16 \cdot \operatorname{Coeff}_{w^2}
\left[\ (4-1) + \left(8-\frac 12\right)w+ \left(16-\frac 14\right)w^2+\dots \ \right]+O(w^3)\\
&\qquad= \frac 16 \cdot \left(16-\frac 14\right) = \color{blue}{\frac{21}8}\ .
\end{aligned}
$$
(The computation was done so that we could finally have the Laurent exansion in zero, but then the numerator $(w^3-1)^2$ has to be considered more carefully.)
Computer check, here sage:
sage: R.<z> = PowerSeriesRing(QQ)
sage: (z^6-1)^2 / z^5 / (2*z^4 - 5*z^2 + 2) + O(z^6)
1/2*z^-5 + 5/4*z^-3 + 21/8*z^-1 + 69/16*z + 261/32*z^3 + 1029/64*z^5 + O(z^6)
(This may be not wanted, but it is just a check, and it may be important to know this can be easily done with computational aid.)
| {
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Limit point of sequence whose general term is $ \frac{1}{1\cdot n} + \frac{1}{2\cdot(n-1)}+\dots+\frac{1}{n\cdot1}$ I need to find the limit point(s) of the sequence whose general term is given by:
$$a_n = \frac{1}{1\cdot n} + \frac{1}{2\cdot (n-1)}+\dots+\frac{1}{n\cdot 1}$$
My observation:
*
*$\frac{1}{1\cdot n} + \frac{1}{2\cdot (n-1)}+\dots+\frac{1}{n\cdot 1} \gt \frac{1}{n\cdot n} + \frac{1}{n\cdot (n-1)}+\dots+\frac{1}{n\cdot 1} \gt \frac{1}{n\cdot n} + \frac{1}{n\cdot n}+\dots+\frac{1}{n\cdot n} = \frac{1}{n}$.
*The series with $a_n$ as the general term is divergent.
Intuitively I feel its a non- monotonic divergent sequence, but I am stuck at how to go further.
Kindly help. Thanks in advance.
| The sequence, $a_n$, can be written
$$\begin{align}
a_n&=\sum_{k=1}^{n} \frac{1}{k(n-k+1)}\\\\
&=\frac1{n+1}\sum_{k=1}^{n}\left(\frac1k-\frac1{k-(n+1)}\right)\\\\
&=\frac2{n+1}\sum_{k=1}^n \frac1k
\end{align}$$
It is easy to show that $a_n$ is monotonically decreasing and bounded below by $0$. So $a_n$ converges. Moreover, from the integral test, we have
$$\sum_{k=1}^n \frac1k \le \log(n)+1$$
Therefore, we see that
$$\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{k(n-k+1)}=0$$
| {
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Convergence of the series with $a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$ I need to determine the convergence of the series whose general term is given by:
$$a_n= \frac{1}{\sqrt{1} \cdot n^2} + \frac{1}{\sqrt{2} \cdot (n-1)^2} +\cdot\cdot\cdot + \frac{1}{\sqrt{n} \cdot 1^2}$$
Observation
*
*$\frac{1}{n^\frac{3}{2}} \lt a_n \lt \sum_{k=1}^n \frac{1}{k^2 }$
*$a_n$ is monotonic decreasing sequence and converges to zero.
This series is bigger than a convergent and smaller than a divergent series by the second observation, if I have not incorrectly done. But its of no help here to determine the convergence. Kindly help, thanks in advance.
| $S_N= \sum_{n=1}^N \sum_{k=1}^n \frac{1}{\sqrt{k}(n+1-k)^2}$
we have
$1\leq k \leq n\leq N$
So
$S_N=\sum_{k=1}^N \sum_{n=k}^N \frac{1}{\sqrt{k}}\frac{1}{(n+1-k)^2}$
$= \sum_{k=1}^N\frac{1}{\sqrt{k}} \left(\sum_{n=k}^N \frac{1}{(n+1-k)^2}\right)$
But
$\sum_{n=k}^N \frac{1}{(n+1-k)^2}\geq \int_{1}^{N-k+2} \frac{1}{x^2}dx
\geq 1-\frac{1}{N-k+2}\geq\frac12$
$S_N\geq \frac{1}{2}\sum_{k=1}^N\frac{1}{\sqrt{k}}$
| {
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For what values of $c$ will $\sum_{n=1}^{\infty}\left(\frac{c}{n} - \frac{1}{n+1}\right)$ converge? (verify solution) For what values of $c$ will $$\sum_{n=1}^{\infty}\left(\frac{c}{n} - \frac{1}{n+1}\right)$$ converge?
The $n$th partial sum is given by:
\begin{align*}s_n&=c+\frac{c-1}{2}+\frac{c-1}{3}+ ... + \frac{c-1}{n-1} - \frac{1}{n}
\\ &=c + (c-1)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}\right)-\frac{1}{n}
\\ & =c + (c-1)\left(\sum_{n=2}^{\infty}\frac{1}{n}\right)-\frac{1}{n}.\end{align*}
Since a constant multiple of a divergent series is a divergent series, the term $(c-1)(\sum_{n=2}^{\infty}\frac{1}{n})$ (and thus the series) will diverge $\iff$ $c \neq 1$.
| You might pave the way differently. We know that if $p>1$ and $$\displaystyle\lim_{n\to\infty}n^pu_n<\infty$$ then $\sum u_n$ converges. This, for the question, is equivalent to have $\displaystyle\lim_{n\to\infty}n^{p-1}(c-1)<\infty$, so it should be $c=1$ if we are looking for the convergence!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For complex $z$, find the value of $z$; $z^2 -2z+( 3-4i) =( 6+3i)$ \begin{array}{l}
\boldsymbol{z^{2} -2z+( 3-4i) =( 6+3i)}\\\\
z^{2} -2z+( 3-4i) =( 6+3i)\\
x^{2} -y^{2} +2xyi-2x-2yi-3-7i=0\\
\left( x^{2} -y^{2} -2x-3\right) +i( 2xy-2y-7) =0\\
x^{2} -y^{2} -2x-3=0\rightarrow ( 1)\\
2xy-2y-7=0\rightarrow ( 2)\\
( 2) \ rewrite\ as\ x=\frac{7+2y}{2y}\\
\\
( 1) \Longrightarrow \frac{( 7+2y)^{2}}{4y^{2}} -y^{2} -2\frac{( 7+2y)}{2y} -3=0\\
4y^{4} +16y^{2} -49=0\\
considering,\ y^{2} =p;\\
4p^{2} +16p-49=0\\
p=-2\mp \frac{\sqrt{65}}{2}\\
\therefore y=\mp \sqrt{-2\mp \frac{\sqrt{65}}{2}}
\end{array}
I got stuck at this point, is this the correct way or have another way to solve this equation. Thank you.
| Yo can do this:
\begin{align*} z^2 -2z+( 3-4i) &= 6+3i
\\ z^2 -2z+1+2-4i & = 6+3i
\\ z^2 -2z+1 & = 4+7i
\\ (z-1)^2 & = 4+7i
\\ z-1 & =(4+7i)^{1/2}\end{align*}
Remember that for complex numbers, the square root has two values. And so
$$z_1=\sqrt{65}e^{\arctan(7/4)i/2},\quad z_2=\sqrt{65}e^{(\arctan(7/4)/2+\pi)i} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$(x-y)(f(f(x)^2)-f(f(y)^2))=(f(x)-f(y))(f(x)^2-f(y)^2$
Find $f: \mathbb{R} \to \mathbb{R}$ which satisfies:
$f\small(0)\normalsize=0, f\small(1)\normalsize=2015. \\
(x-y)(f\small(f(x)^2\small)\normalsize-f\small(f(y)^2\small)\normalsize)=(f\small(x)\normalsize-f\small(y)\normalsize)(f\small(x)^2\normalsize-f\small(y)^2\normalsize)$
My Attempt:
\begin{align}
&f \not\equiv c. \\
\Rightarrow \; & \color{blue} {\exists \ t_k \text{ s.t. } f(t_k) \neq k.} & \tag{1} \label{1} \\
& \color{red} {\exists y_1, y_2 \text{ s.t. } f(y_1) \neq f(y_2).} \tag{2} \label{2} \\
\ \\
&\text{let } f(a)=f(b), f(x) \neq f(a) \text{ by } \color{red}{(\ref 2)}. \\
P(x, a): \; & (x-a)(f\small(f(x)^2)\normalsize-f(f\small(a)^2\normalsize))=(f(x)-f(a))(f(x)^2-f(a)^2). \\
P(x, b): \; & (x-b)(f\small(f(x)^2)\normalsize-f(f\small(b)^2\normalsize))=(f(x)-f(b))(f(x)^2-f(b)^2). \\
& f(x) \neq f(a), f(x) \neq f(b). \\
\therefore \; & (f(x)-f(a))(f(x)^2-f(a)^2) = (f(x)-f(b))(f(x)^2-f(b)^2) \neq 0. \\
\Rightarrow \; & (x-a)(f\small(f(x)^2)\normalsize-f\small(f(a)^2)\normalsize)=(x-b)(f\small(f(x)^2)\normalsize - f\small(f(b)^2)\normalsize). \\
& f\small(f(x)^2)\normalsize - f\small(f(a)^2)\normalsize) = f\small(f(x)^2)\normalsize - f\small(f(b)^2)\normalsize) \ \Rightarrow \ x-a=x-b, a=b. \\
\therefore \; & \color{green}{f(a)=f(b) \Rightarrow a=b.} \label{3} \tag{3}
\end{align}
| Posted on AOPS, and I am posting the arranged solution.
\begin{align}
P(x, y): \; & (x-y)(f(f(x)^2)-f(f(y)^2))=(f(x)-f(y))(f(x)^2-f(y)^2).\\
P(x, 0): \; & xf(f(x)^2)=f(x)^3. \\
\ \\
\therefore \; & yf(f(x)^2)+xf(f(y)^2)=f(x)f(y)(f(x)+f(y)). \text{ (Let this one be $Q(x, y).$)} \\
\ \\
Q(1, 1): \; & f(2015^2)=2015^3. \\
\ \\
Q(1, x): \; & 2015^3x+f(f(x)^2)=2015f(x)(2015+f(x)). \\
Q(x, x): \; & xf(f(x)^2)=f(x)^3. \\
xQ(1, x)-Q(x, x): \; & 2015^3x^2=2015^2xf(x)+2015xf(x)^2-f(x)^3. \\
\ \\
\therefore \; & f(x)^3-(2015x)f(x)^2-(2015^2x)f(x)+2015^3x^2=0. \\
\Rightarrow \; & (f(x)-2015x)(f(x)^2-2015^2x)=0. \\
\ \\
\therefore \; & f(x)=2015x \text{ for } \forall x \leq 0. \ (\because f(x)^2-2015^2x >0.) \\
\ \\
P(x, 0): \; & xf(f(x)^2)=f(x)^3. \\
x<0: \; & xf(2015^2x^2)=2015^3x^3, f(2015^2x^2)=2015^3x^2. \\
\Rightarrow \; & f(x)=2015x \text{ for } \forall x \geq 0. \\
\ \\
\therefore \; & f(x)=2015x.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$? What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$.
Note that $f'(x)=\exp(\frac{1}{2}c^2x^2)c^2x$ and $f''(x)=\exp(\frac{1}{2}c^2x^2)(c^2x)^2+\exp(\frac{1}{2}c^2x^2)c^2$. Then $f'(0)=0$ and $f''(0)=c^2$.
Is that
$$
f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+o(x^2)=1+\frac{1}{2}c^2(\frac{1}{2}c^2x^2)^2?
$$
| Alright, first to simplify the problem you are asking what the Taylor Series expansion of $\exp(ax^2)$ for a constant $a$. In your case, simply take $a=\frac{1}{2}c^2$. But then
$$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\Rightarrow \exp(ax^2)=\sum_{n=0}^\infty \frac{(ax^2)^n}{n!}=\sum_{n=0}^\infty \frac{a^n x^{2n}}{n!}$$
Then for $a=\frac{1}{2}c^2$ this becomes
$$\exp(ax^2)=\sum_{n=0}^\infty \frac{c^{2n} x^{2n}}{2^nn!}=1+\frac{c^2x^2}{2}+\cdots$$
Note that this is different than what you wrote in your question as you erronously stated
$$f''(0)x^2=c^2(\frac{1}{2}c^2x^2)^2$$
when it should have been
$$f''(0)x^2=c^2x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger . I have been asked that which of the two quantities $\sin 28^{\circ}$ and $\tan 21^{\circ}$ is bigger without resorting to calculator.
My Attempt:
I tried taking $f(x)$ to be
$f(x)=\sin 4x-\tan 3x$
$f'(x)=4\cos 4x-3\sec^23x=\cos 4x(4-3\sec^23x\sec 4x)$
but to no avail.
I also tried solving $\tan^2 21^{\circ}-\sin^228^{\circ}=\tan^2 21^{\circ}-\sin^221^{\circ}+\sin^221^{\circ}-\sin^228^{\circ}=\tan^2 21^{\circ}\sin^221^{\circ}+\sin^221^{\circ}-\sin^228^{\circ}$
but again no luck.
There doesn't appear to be a general way of doing this
| We can use some trigonometric identities to help us here. We need to know three things here:
*
*$\sin 2\theta = \frac{2\tan \theta}{1+\tan^2\theta}$
*$\tan 2\theta = \frac{2\tan \theta}{1-\tan^2\theta}$
*$\tan 3\theta = \frac{3\tan\theta-3\tan^3\theta}{1-3\tan^2\theta}$
You were onto something when looking at $\sin4\theta$ and $\tan3\theta$. We can use these definitions to find that $\sin4\theta = \frac{2\tan2\theta}{1+\tan^22\theta} = \frac{2\frac{2\tan \theta}{1-\tan^2\theta}}{1+(\frac{2\tan \theta}{1-\tan^2\theta})^2}$. Simplifying this expression gives us the more manageable $\sin4\theta=\frac{4\tan\theta(1-\tan^2\theta)}{(1+\tan^2\theta)^2}$.
Now we take $\theta$ to be equal to $7^\circ$ or $\frac{7\pi}{180}$. Define $x=\tan\frac{7\pi}{180}$, and look at the following inequality:
*
*$\frac{4x(1-x^2)}{(1+x^2)^2}>\frac{3x-x^3}{1-3x^2}$
This is actually the case assuming that $\sin 28^\circ > \tan 28^\circ$. This expression is equivalent to
*
*$4x(1-x^2)(1-3x^2) > (3x-x^3)(1+x^2)^2$
which in turn is the same as
*
*$12x^5-16x^3+4x>-x^7+x^5+5x^3+3x$
Subtracting the right hand side from the left gives us
*
*$x^7+11x^5-21x^3+x>0$
We can pull out a factor of $x$ to get
*
*$x^6+11x^4-21x^2+1>0$
Something interesting worth noting is that for small $\theta$, $\tan\theta\approx\theta$, and in fact, if $\theta<9^\circ$, the percentage error between the two is less than one percent (https://en.wikipedia.org/wiki/Small-angle_approximation#Error_of_the_approximations). So, we can simply evaluate this expression at $\frac{7\pi}{180}$ instead of $\tan\frac{7\pi}{180}$. If you wish to remain truly calculator-less, I suppose you could further approximate this to $\frac{217}{1800}$ and work it out by hand, but I will be using the expression with pi. Swapping it out for $x$ gives us
*
*$(\frac{7\pi}{180})^6+11(\frac{7\pi}{180})^4-21(\frac{7\pi}{180})^2+1>0$
and running this through WolframAlpha gives us roughly
*
*$0.689>0$
Since this expression is true, it follows that $\sin 28^\circ > \tan 21^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Deriving the formula for $\cos^{-1}x+\cos^{-1}y$ I am trying to prove the following:
$$\cos^{-1}x+\cos^{-1}y=\begin{cases}\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2});&-1\le x,y\le1\text{ and }x+y\ge0\\2\pi-cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2});&-1\le x,y\le1\text{ and }x+y\le0\end{cases}$$
Let $\cos^{-1}x=A, \cos^{-1}y=B.$ Since $-1\le x,y\le1\implies0\le A,B\le\pi\implies0\le A+B\le2\pi$
$\cos(A+B)=\cos A\cos B-\sin A\sin B=xy-\sqrt{1-x^2}\sqrt{1-y^2}$ (since the sine function is positive in $1$st and $2$nd quadrants.)
Therefore, $\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})=\cos^{-1}(\cos(A+B))=\begin{cases}A+B;&0\le A+B\le\pi\implies0\le A,B\le\frac\pi2\\2\pi-(A+B);&\pi\le A+B\le2\pi\implies\frac\pi2\le A,B\le\pi\end{cases}$
From this, how do we conclude about $x+y$?
|
$\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})=\cos^{-1}(\cos(A+B))=\begin{cases}A+B;&0\le A+B\le\pi\implies0\le A,B\le\frac\pi2\\2\pi-(A+B);&\pi\le A+B\le2\pi\implies\frac\pi2\le A,B\le\pi\end{cases}$
$0\leq A+B \leq\pi$
Now, A and B is always greater than $0$,therefore $0\leq A+B$ is always true.
Leaving us with,
$\implies A+B \leq\pi$
$\implies A\leq\pi-B$
$\implies \cos(A)\geq\ \cos(\pi-B)$ (as both LHS and RHS vary from $0$ to $\pi$ where $\cos$ is decreasing.)
$\implies x\geq -y$
$\implies x+y\geq 0$
I will be doing the first case only,leaving the second for you to do yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \textrm{ for any positive real number }a.$$
Again, we start with $$I_1(a)=\int_{0}^{\infty} \frac{d x}{x^{2}+a}= \left[\frac{1}{\sqrt{a}} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)\right]_{0}^{\infty} = \frac{\pi}{2 }a^{-\frac{1}{2} } $$
Then differentiating $I_1(a)$ w.r.t. $a$ by $n-1$ times yields
$$
\int_{0}^{\infty} \frac{(-1)^{n-1}(n-1) !}{\left(x^{2}+a\right)^{n}} d x=\frac{\pi}{2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-3}{2}\right) a^{-\frac{2 n-1}{2}}
$$
Rearranging and simplifying gives $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} =\frac{\pi a^{-\frac{2 n-1}{2}}}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)}
$$
Putting $a=1$ gives the formula of our integral
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\frac{\pi}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)= \frac{\pi}{2^{2 n-1}} \left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
For verification, let’s try $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{10}} &= \frac{\pi}{2^{19}}\left(\begin{array}{c}
18 \\
9
\end{array}\right) =\frac{12155 \pi}{131072} ,
\end{aligned}
$$
which is checked by Wolframalpha .
Are there any other methods to find the formula? Alternate methods are warmly welcome.
Join me if you are interested in creating more formula for those integrals in the form $$
\int_{c}^{d} \frac{f(x)}{\left(x^{m}+1\right)^{n}} d x.
$$
where $m$ and $n$ are natural numbers.
| Denote $I_n = \int_0^\infty \frac1{(1+x^2)^n}dx$, and note that
$$\left( \frac x{(1+x^2)^{n-1}} \right)’
= \frac {2(n-1)}{(1+x^2)^{n}}- \frac {2n-3}{(1+x^2)^{n-1}}
$$
Integrate both sides to get the recursion $I_n= \frac{2n-3}{2(n-1)}I_{n-1} $, along with $I_1=\frac\pi2$. Thus
$$I_n = \frac{(2n-3)!(n-1)\pi }{[2^{n-1} (n-1)!]^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Evaluating $\int\frac{\sec^2(x)}{(4+\tan^2(x))^2}\, dx$ How to solve this integral?$$\int\frac{\sec^2(x)}{(4+\tan^2(x))^2}\, dx$$
I've tried the following:
Starting by substituting $\tan(x) = 2\tan(\theta)\implies \sec^2(x)\ dx= 2\sec^2(\theta)\ d\theta$
$$\implies\int\frac{2\sec^2(\theta)}{(4 + 4\tan^2\theta)^2}\ d\theta\\\implies\int\frac{2\sec^2(\theta)}{16 (\sec^2\theta)^2}\ d\theta\\\implies\int\frac{\cos^2(\theta)}{8} \ d\theta\\\implies\frac\theta{16} + \frac{\sin(2\theta)}{32}+ c$$
Now since, $\tan(x) = 2 \tan\theta\implies \arctan\left(\frac{\tan(x)}{2}\right) = \theta$
So, the final answer to the integral may be,
$$\implies\frac{\arctan\left(\frac{\tan(x)}{2}\right)}{16} + \frac{\sin(2\arctan\left(\frac{\tan(x)}{2}\right))}{32}+ c$$
I checked the answer on wolfram Alpha and found that my answer is not matching with that.
Can anyone tell me what should I do to verify my answer?
| Wolfram|Alpha returns an antiderivative of
$$\int \frac{\sec^2(x)}{\left(4+\tan^2(x)\right)^2} \, dx = \frac{2\sin(2x) - (3\cos(2x)+5) \arctan\left(2\cot(x)\right)}{48\cos(2x) + 80} + C$$
which we can rewrite as
$$-\frac1{16} \arctan\left(2\cot(x)\right) + \frac18 \cdot \frac{\sin(2x)}{3\cos(2x) + 5} + C$$
The antiderivative you found was
$$\int\frac{\sec^2(x)}{\left(4+\tan^2(x)\right)^2} \, dx = \frac1{16} \arctan\left(\frac12 \tan(x)\right) + \frac1{32} \sin\left(2\arctan\left(\frac12\tan(x)\right)\right) + C$$
We have
$$\arctan(x) + \arctan\left(\frac1x\right) = \begin{cases}\frac\pi2 & \text{if }x>0 \\ -\frac\pi2 & \text{if }x<0\end{cases}$$
If we take $2\cot(x)>0$, then
$$\begin{align}
-\frac1{16} \arctan(2\cot(x)) + C &= -\frac1{16} \arctan(2\cot(x)) - \frac\pi2 + C \\
&= \frac1{16} \arctan\left(\frac1{2\cot(x)}\right) + C \\
&= \frac1{16} \arctan\left(\frac12\tan(x)\right) + C
\end{align}$$
so that the first terms "match" (not identical, but they only differ by a constant). A similar conclusion is made by instead assuming $2\cot(x)<0$.
Since $\sin(2x) = 2\cos(x)\sin(x)$, the second term in your solution is
$$\frac1{16} \sin\left(\arctan\left(\frac12\tan(x)\right)\right) \cos\left(\arctan\left(\frac12\tan(x)\right)\right) = \frac18 \cdot \frac{\tan(x)}{4 + \tan^2(x)}$$
(Consider a right triangle with reference angle $\theta$ for which $\tan(\theta)=\frac12\tan(x)$, so the ratio of the legs opposite and adjacent to $\theta$ is $\tan(x)$ to $2$. It follows that this triangle's hypotenuse has length $\sqrt{4+\tan^2(x)}$.)
Now, we have
$$\begin{align}
\frac18\cdot\frac{\tan(x)}{4+\tan^2(x)} &= \frac18 \cdot \frac{\sin(x)\cos(x)}{4\cos^2(x)+\sin^2(x)} \\
&= \frac18 \cdot \frac{\frac12 \sin(2x)}{3\cos^2(x) + 1} \\
&= \frac1{16} \cdot \frac{\sin(2x)}{\frac32(1+\cos(2x))+1} \\
&= \frac18 \cdot \frac{\sin(2x)}{5+3\cos(2x)}
\end{align}$$
and so both solutions also have matching second terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
evaluate the integral involving $e^{-x}$ Evaluate
$$\int_{\ln 2}^{\ln 4} \frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx$$
I made the $u-$sub
$$u=e^{-x}$$
Then my limits become $1/2$ and $1/4$ and
$$du = - e^{-x} dx$$
so rewriting the integral I have
$$-\int_{1/2}^{1/4}\frac{1}{\sqrt{1-u^2}}du=\int_{1/4}^{1/2}\frac{1}{\sqrt{1-u^2}}du=\sin^{-1}(u)\bigg\vert_{1/4}^{1/2}$$
but this is not the answer. The answer is $\sin^{-1}(\frac{\sqrt{15}}{4})-\frac{\pi}{3}.$ I know $\sin^{-1}(1/2)=-\frac{\pi}{3}.$ But don't see how they get the $\sqrt{15}$.
| Notice that :
$$\int_{1/4}^{1/2} \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} = -\left[\cos^{-1} u\right]_{1/4}^{1/2} = \cos^{-1} \dfrac{1}{4} - \cos^{-1} \dfrac{1}{2} = \cos^{-1} \dfrac{1}{4} - \dfrac{\pi}{3}$$
But we know that :
$$\cos^{-1} x = \sin^{-1} \sqrt{1 - x^2}$$
then :
$$\int_{1/4}^{1/2} \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} = \sin^{-1} \sqrt{1 - \dfrac{1}{16}} - \dfrac{\pi}{3} = \sin^{-1} \dfrac{\sqrt{15}}{4} - \dfrac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$ I want to find all solutions to $\frac{a+b}{b} = \frac{a}{a+b}$. I have plugged this into a calculator which tells us that if we solve for $a$, without loss of generality, we obtain
$$a = - \frac{b}{2} \pm \frac{\sqrt{3}}{2}i$$
and vice versa. I am not sure how to show this, however. I have tried a couple of different things. For example,
Method 1:
$$\frac{a}{a+b} = \frac{a+b}{b} \implies (a+b)^2=ab \implies a+b = \pm \sqrt{ab}$$
Method 2:
$$\frac{a}{a+b} \cdot \frac{a-b}{a-b}= \frac{a+b}{b} \cdot \frac{a-b}{a-b} \implies \frac{a(a-b)}{a^2-b^2}=\frac{a^2-b^2}{b(a-b)} \implies \big( a^2-b^2 \big)^2 = ab(a-b)^2$$
But so far neither of these methods are shedding any light for me. Any advice? Thanks in advance.
| You have $(a+b)^2=ab$ or $a^2+ab+b^2=0$. If $a\ne b$, multiply by $(a-b)$, get $a^3-b^3=0$. So $a=b\omega$ where $\omega$ is one of the cube roots of $1$. The possibility $a=b$ does not work, the other two work.
Answer: $b$ is any non-zero number, $a=b\omega$ where $\omega$ is one of the two non-1 cube roots of $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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n-th root in limits I'm having trouble with this kind of series: $$a_n = \sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})$$
I tried to make something like perfect square to simplify those $7$th root junk which is obviously impossible, and as $n$ tends to infinity L'Hopital's rule also can't solve the problem in this form.
The multiple choice:
*
*$$\lim_{n\to\infty}a_n=\frac{9}{7}$$
*$$\lim_{n\to\infty}a_n=+\infty$$
*$$a_n\sim\frac{9}{7}n^{-5/14},\quad n\to+\infty$$
| Rewrite it $$\sqrt{n}\left(\sqrt[7]{n+5} - \sqrt[7]{n-4}\right) = \\ n^{\frac{9}{14}}\left((1+\frac{5}{n})^{\frac{1}{7}}-(1-\frac{4}{n})^{\frac{1}{7}}\right) \\ \approx n^{\frac{9}{14}}\left(1+\dfrac{1}{7}\cdot\dfrac{5}{n} - 1 +\dfrac{1}{7}\cdot\dfrac{4}{n} + o(\dfrac{1}{n})\right) \\ \approx \dfrac{9}{7n^{\frac{5}{14}}}$$. Hence the answer is choice $4$-th down the list.
| {
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Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$ I have this equation to solve:
$$\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$$
Since $\overline{z}\cdot z = |z|^2$ and utilizing the de Moivre's formula this can be simplified to:
$$|z|^3z^4=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
$$|z|^7(\cos{4\alpha} + i\sin{4\alpha})=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
From here I thought just to compare $|z|^7 = 8\sqrt{2}$ and the sine, cosine part.
Is this the correct way to go about it or could it be done using some simpler method?
| Let $z = A e^{i \theta} $ then
$ A^7 e^{i 4 \theta} = 8 \sqrt{2}(e^{-i \dfrac{3 \pi}{10}} )^8 = 2^{7/2} e^{-i \dfrac{12 \pi}{5}} = 2^{7/2} e^{i \dfrac{8 \pi}{5}} $
Hence, $A = \sqrt{2}$, and $\theta = \dfrac{2\pi}{5}$
| {
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Over the field $\mathbb{Z}_2$, does $x^6 + x^3 + 1$ divide $(x^4 + 1)^n + 1$ for some $n \in \mathbb{N}$? I am currently investigating whether the polynomial $x^6 + x^3 + 1 \in \mathbb{Z}_2[x]$ divides $(x^4 + 1)^n + 1$ for some $n \geq 1$.
I think it is impossible, and so far, I've checked in the case where $n$ is a power of 2:
Suppose $n = 2^m$ for some $m \in \mathbb{N}$. In this case, $(x^4 + 1)^n + 1 = x^{4n} + 1 + 1 = x^{4n}$. So if we assume $x^6 + x^3 + 1 \mid x^{4n}$, then $x \in \sqrt{\langle x^6 + x^3 + 1 \rangle} = \langle x^6 + x^3 + 1 \rangle$, which is not possible. (Here, $\sqrt{\cdot}$ denotes the radical of an ideal.)
However, I'm struggling seeing an argument to prove this is impossible when $n$ is not a power of $2$.
| This $n$ exists. Let $F$ be the splitting field of $x^6+x^3+1$ over $\Bbb F_2$. Let $n=|F|-1$. Then for every $y\ne 0$ in $F$ we have $y^n=1=-1$, so $y^n+1=0$. The polynomial $x^6+x^3+1$ has $6$ roots in $F$ and no double roots. Since every root of $x^6+x^3+1$ is a root of $(x^4+1)^n+1$, $x^6+x^3+1$ divides $(x^4+1)^n+1$.
| {
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Finding the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$
Find the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$
My Attempt: Let $z=x+iy$, so, $$\sqrt{x^2+y^2}=\text{max}\{\sqrt{(x-1)^2+y^2},\sqrt{(x+1)^2+y^2}\}$$
Case I: $\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\implies \pm x=x-1\implies x=\frac12$
Case II: $\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2}\implies \pm x=x+1\implies x=-\frac12$
Does that mean if the complex numbers are lying on two lines, either $x=\frac12$ or $x=-\frac12$ then they would satisfy the required equation?
Therefore, the required number of complex numbers is infinite?
But the answer given is zero.
Can we solve this graphically?
| $|z| = max \{ |z-1|, |z+1|\} \implies |z| \ge |z-1|$ and $|z| \ge |z+1|$ . Summing the two equations, we have $|2z| \ge |z-1| + |z+1|$. But triangle inequality states that $|z-1| + |z+1| \ge |2z|$ so $|z-1| + |z+1| = 2|z|$. Because, $|z|$ is one of $|z-1|$ and $|z+1|$, we then deduce that $|z-1| = |z| = |z+1|$. $z$, $z-1$, and $z+1$ are distinct and belong to the horizontal line that pass by $z$. They cannot then be part of the circle of centre 0 and radius $|z|$ because a circle can't have more than 2 intersection points with a line. We conclude then that there's no z that satisfy our last equality.
| {
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$ab+ac+bc=2\quad ,\quad\min(10a^2+10b^2+c^2)=?$
If $ab+ac+bc=2$, find minimum value of $10a^2+10b^2+c^2$
$1)3\qquad\qquad2)4\qquad\qquad3)8\qquad\qquad4)10$
I used AM-GM inequality for three variables:
$$ab+ac+bc\ge3(abc)^{\frac23}\quad\Rightarrow\quad 2\ge3(abc)^{\frac23}$$
$$10a^2+10b^2+c^2\ge 3(10abc)^{\frac23}$$
Hence $10a^2+b^2+c^2\ge2\times 10^{\frac23}$. But it is not in the options.
| Suppose $a,b,c>0$. Let
$$ a=\frac1{\sqrt{10}}r\cos\phi\sin\theta,b=\frac1{\sqrt{10}}r\sin\phi\sin\theta,c=r\cos\theta. $$
Then
$$ 10a^2+10b^2+c^2=r^2$$
and
\begin{eqnarray}
&&ab+ac+bc\\
&=&\frac1{20}r^2\sin(2\phi)\sin^2\theta+\frac1{2\sqrt{10}}r^2\cos\phi\sin(2\theta)+\frac1{2\sqrt{10}}r^2\sin\phi\sin(2\theta)\\
&=&\frac1{20}r^2\sin(2\phi)\sin^2\theta+\frac1{2\sqrt{10}}r^2(\cos\phi+\sin\phi)\sin(2\theta) \tag1\\
&\le&\frac1{20}r^2\sin^2\theta+\frac{\sqrt2}{2\sqrt{10}}r^2\sin(2\theta)\tag2\\
&=&r^2\bigg(\frac1{20}\frac{1-\cos(2\theta)}{2}+\frac{1}{\sqrt{20}}\sin(2\theta)\bigg)\\
&=&\frac{1}{40}r^2+r^2\bigg(-\frac{1}{40}\cos(2\theta)+\frac{1}{\sqrt{20}}\sin(2\theta)\bigg) \tag3\\
&\le&\frac{1}{40}r^2+r^2\sqrt{(\frac{1}{40})^2+\frac{1}{20}}\tag4\\
&=&\frac{1}{4}r^2
\end{eqnarray}
from which one has $r^2\ge8$. So
$$ \min(10a^2+10b^2+c^2)=\min r^2=8.$$
Note:
*
*From (1) to (2),
$$ \sin(2\phi)\le 1, \cos\phi+\sin\phi\le \sqrt2 $$
and "=" holds if and only if $\phi=\frac{\pi}{4}$.
*From (3) to (4),
$$ a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2}$$
is used.
| {
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Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$
Evaluate
$$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$
The solution is $5$.
Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$
We have
$$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$
Two unknowns, two equations, so we should be able to solve for $A+B$ (what we want).
How can I do that?
| Hint 1
$$A^3+B^3=(A+B)((A+B)^2-3AB)$$
Hint 2
$$AB=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391660",
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"source": "stackexchange",
"question_score": "2",
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Maximizing the volume of a cuboid with constraints (lagrange) fails? Given is the following (translated) problem:
You have $12$ meters of wire.
Try to build a wireframe model of a cuboid with sidelengths $x, y, z$ and maximize volume $V(x,y,z)=xyz$.
Show that this is the case iff all sidelengths are equal.
My attempt:
Using Lagrange with the constraint $g(x,y,z)=x+y+z-3$ leads to:
$$\nabla L(x,y,z,\lambda)=\begin{pmatrix}
yz+\lambda \\
xz+\lambda \\
xy +\lambda \\
x+y+z-3
\end{pmatrix} \stackrel{!}{=}0$$
After solving I get:
$$x=y=z=1$$
$$\lambda = -1$$
I calculate the Hessian matrix afterwards and get:
$$H_L=\begin{pmatrix}
0 & z & y & 1\\
z & 0 & x & 1 \\
y & x & 0 & 1 \\
1 & 1 & 1 & 0
\end{pmatrix}$$
Inserting previous values produces this matrix:
$$H_L=\begin{pmatrix}
0 & 1 & 1& 1\\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0
\end{pmatrix}$$
Now the problem:
This matrix has eigenvalues $3$ and $-1$, which should mean it's indefinite and therefore has as saddle point/ no minimum or maximum.
Did I do something wrong or is it not possible to solve this problem with lagrange?
| We have to consider the so-called Bordered Hessian that is the Hessian of the Lagrange function. Here, at the point $(1,1,1,-1)$, it is equal to
$$H_L=\begin{pmatrix}
0 & g_x & g_y& g_z\\
g_x & L_{xx} & L_{xy} & L_{xz} \\
g_y & L_{yx} & L_{yy} & L_{yz} \\
g_z & L_{zx} & L_{zy} & L_{zz}
\end{pmatrix}=\begin{pmatrix}
0 & 1 & 1& 1\\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0
\end{pmatrix}$$
Since we have one constraint, i.e. $m=1$, and there are three variables, a sufficient condition for a local maximum for the constrained problem is that determinant of the $3\times3$ principal minor is positive and the determinant of the entire matrix is negative.
It is easy to verify that both conditions holds:
$$\left|\begin{matrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{matrix}\right|=2>0,\quad\left|\begin{matrix}
0 & 1 & 1& 1\\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0
\end{matrix}\right|=-3<0.$$
and therefore the point $(1,1,1)$ is a maximum for your constrained problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the Arc length of a Polar curve $r=1+\cos(2\theta)$
Given the polar equation, $r=1+\cos(2\theta)$, prove that the length of the curve corresponding to $0 \leq\theta \leq2\pi$ is $8+\frac{4}{\sqrt{3}}\log(2+\sqrt{3})$.
Applying integration to find arc length, $l$, in polar form:
$\int_\alpha^\beta\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta$
$r^2=(1+\cos(2\theta))^2=1+2\cos(2\theta)+\cos^2(2\theta)$,
and, $\frac{dr}{d\theta}=-2\sin(2\theta)\Rightarrow (\frac{dr}{d\theta})^2=4\sin^2(2\theta)=4-4\cos^2(2\theta)$
$\therefore r^2+(\frac{dr}{d\theta})^2=5+2\cos(2\theta)-3\cos^2(2\theta)$
Completing the square, $5+2\cos(2\theta)-3\cos^2(2\theta)=3\left[\frac{16}{9}-(\cos(2\theta)-\frac{1}{3})^2 \right]$
Just doing the integration using (1), but without limits to get a function only, I get:
$\int\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta=\int\sqrt{3\left[\left(\frac{4}{3}\right)^2-(\cos(2\theta)-\frac{1}{3})^2 \right]}d\theta=\sqrt{3}\int\sqrt{\left[\left(\frac{4}{3}\right)^2-(\cos(2\theta)-\frac{1}{3})^2 \right]}d\theta$
The integral has the general form: $\int\sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\arcsin\frac{x}{a}$
Ignoring factor of $\sqrt{3}$ the integral is:
$\int\sqrt{\left[\left(\frac{4}{3}\right)^2-(\cos(2\theta)-\frac{1}{3})^2 \right]}d\theta=\frac{((\cos(2\theta)-\frac{1}{3})\sqrt{(\frac{4}{3})^2-(\cos(2\theta)-\frac{1}{3})^2}}{2}+ \frac{\left(\frac{4}{3}\right)^2}{2}\arcsin\frac{(\cos(2\theta)-\frac{1}{3})}{\frac{4}{3}}$
No log term in the numerical solution.
Another idea I had was to try using $\sqrt{1+\cos(t)}=\frac{\sin(t)}{\sqrt{1-\cos(t)}}$, but this leads to a substitution which doesn't give the log term?
I'm a bit stuck with this one. Most of the other examples I've read, e.g. for $r=1+\cos(\theta)$ seem to be OK. It's the $\cos(2\theta)$ causing the problem for me.
| $r = 1 + \cos2\theta = 2 \cos^2\theta$
$r_{\theta} = - 4 \cos\theta \sin\theta$
$ \displaystyle l = 4 \int_0^{\pi/2}\sqrt{4\cos^4\theta + 16 \cos^2\theta \sin^2\theta} ~ d\theta$
$ \displaystyle l = 8 \int_0^{\pi/2} \cos\theta \sqrt{1 + 3 \sin^2\theta} ~ d\theta$
Now substitute $t = \sin\theta$. Can you take it from here?
| {
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How to apply Stolz–Cesàro theorem in this question? $$\lim _{n \rightarrow \infty}\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}$$
I don't know how to continue after taking the logarithm of it, after that.
This is an exercise to find the limit using Stolz's method.
| Note
\begin{eqnarray}
&&\ln\bigg[\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}\bigg] \\
&=&\sum_{k=2}^n\frac{1}{2^{n-k+1}}\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)\\
&=&\frac{\sum_{k=2}^n2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)}{2^{n+1}}.
\end{eqnarray}
By Stolz's theorem, one has
\begin{eqnarray}
&&\lim_{n\to\infty}\ln\bigg[\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}\bigg] \\
&=&\lim_{n\to\infty}\frac{\sum_{k=2}^n2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)}{2^{n+1}}\\
&=&\lim_{n\to\infty}\frac{\sum_{k=2}^{n+1}2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)-\sum_{k=2}^n2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)}{2^{n+2}-2^{n+1}}\\
&=&\lim_{n\to\infty}\frac{2^{n+1}\ln\bigg(\frac{2^{n}}{2^{n+1}-1}\bigg)}{2^{n+1}}\\
&=&\ln(\frac12)
\end{eqnarray}
and hence
$$\lim _{n \rightarrow \infty}\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}=\frac12.$$
| {
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Product of real quadratic roots If product of roots '$p$' of equation $x^2 – 2ax + 8 – a = 0$ lies between its roots, then maximum
integral value of '$p$' is _____
My approach is as follow
$f\left( x \right) = {x^2} - 2ax + 8 - a$
${D^2} = 4{a^2} - 4\left( {8 - a} \right) \ge 0 \Rightarrow {a^2} + a - 8 \ge 0$
${a^2} + a - 8 \ge 0 \Rightarrow a \in \left( { - \infty ,\frac{{ - 1 - \sqrt {33} }}{2}} \right) \cup \left( {\frac{{ - 1 + \sqrt {33} }}{2},\infty } \right)$
$\alpha + \beta = 2a$ & $\alpha \beta = 8 - a$
$\alpha \beta \in \left( {\alpha ,\beta } \right)$ where $\beta > \alpha $
How do we proceed from hered
| The roots are
$$a\pm \sqrt{a^2-(8-a)}$$
and the product of the roots is $8-a$, so defining $p=8-a$, we wish to maximize $p$ (among integers) subject to constraints
$$ 8-p-\sqrt{(8-p)^2-p}\leq p\leq 8-p+ \sqrt{(8-p)^2-p}.$$
A quick sketch shows optimally the upper constraint must bind so
$$p= 8-p+ \sqrt{(8-p)^2-p}\implies p=5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397707",
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What's the ratio between the segments $\frac{AF.BG}{FG}$ in the figure below? For reference: In the figure below the trapezoid has height $13$ and is inscribed in a circle of radius $15$. Point $E$ is on the minor arc determined by $A$ and $B$, the points $F$ and $G$ intersect with $ED$, $E$C and $AB$. Then the ratio between the segments $\frac{AF \times BG}{FG}$ is?
My progress:
I'm lost in this question...Perhaps using some of the segment $AB$ ratio with the diameter $DC$ ??
$\triangle EFG\sim HJC: \frac{JC}{FG} = \frac{JH}{FE}=\frac{HC}{EG}\\
\triangle BJG \sim \triangle HJC: \frac{JC}{JG}=\frac{JH}{BJ}=\frac{HC}{BG}\\
\triangle KID \sim \triangle KAF: \frac{JC}{FK}=\frac{JH}{AF}=\frac{HC}{AK}$
...???
| Can you see that $~\triangle AFE \sim \triangle CBE~$ and $~\triangle BGE \sim \triangle DAE~$?
That leads to
$~ \displaystyle \frac{BC}{AF} = \frac{CE}{AE}~, \frac{AD}{BG} = \frac{AE}{GE}$
Multiplying, $\displaystyle \frac{BC \times AD}{AF \times BG} = \frac{CE}{GE} \tag1$
As $ \displaystyle \triangle FEG \sim \triangle DEC, ~\frac{CE}{GE} = \frac{CD}{FG}$
Plugging into $(1)$,
$\displaystyle \frac{AF \times BG}{FG} = \frac{BC \times AD}{CD}$
As it is a cyclic trapezium, it must be isosceles and $BC = AD$. Applying Pythagoras,
$OH^2 = OB^2 - BH^2 = 15^2 - 13^2 = 56$
So, $CH = 15 - \sqrt{56}$
$BC \times AD = BC^2 = BH^2 + CH^2 = 13^2 + (15 - \sqrt{56})^2 = 30 (15 - 2 \sqrt{14})$
So the answer is $~(15 - 2 \sqrt{14})$
| {
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Integral on a rotational solid
Compite the integral $$\int_V\frac{4-z^2}{(x^2+y^2)^3}\mathrm dx\mathrm dy\mathrm dz,$$ where $V$ is the solid enclosed by the paraboloids $x^2+y^2=z,x^2+y^2=2z$ and cones $x^2+y^2=(z-2)^2,x^2+y^2=4(z-2)^2.$
My attempt:
$V$ is a rotational solid with $z$ axis as the axis of the rotation. Two cones have picks at the same point above the vertex of the parabolae. I decided to use cyllindrical coordinates: $$\begin{aligned}x&=r\cos\varphi\\y&=r\sin\varphi\\z&=z\\\varphi&\in[0,2\pi].\end{aligned}$$
The Jacobian of the transformation is then $J_\psi(\varphi,r)=r$ and the function under the integral becomes $\frac{4-z^2}{r^6}.$
I thought I should first compute the integral on the solid enclosed by the parabolae and the cone $x^2+y^2=(z-2)^2$ and subtract the integral on the solid enclosed by the parabolae and the cone $x^2+y^2=4(z-z^2).$
Now, I wanted to find the radii of the circles -intersections of the parabolae with cones $x^2+y^2=(z-2)^2$ and $x^2+y^2=4(z-2)^2$ for the first and second integral respectively. Solving $$\begin{cases}x^2+y^2=z\\x^2+y^2=(z-2)^2\end{cases}\\\begin{cases}x^2+y^2=2z\\x^2+y^2=(z-2)^2\end{cases}$$ yields the bounds $r=1,r=2,r=\sqrt{2(3\pm\sqrt 5)}$ and $$\begin{cases}x^2+y^2=z\\x^2+y^2=4(z-2)^2\end{cases}\\\begin{cases}x^2+y^2=2z\\x^2+y^2=4(z-2)^2\end{cases}$$ yields the bounds $r=\sqrt{\frac{17\pm\sqrt{33}}8},\sqrt{2\frac{9\pm\sqrt{17}}4}
$
The first integral I get is
$\int_0^{2\pi}\int_1^{\sqrt{2(3-\sqrt 5)}}\int_{2-r}^r\frac{4-z^2}{r^5}\mathrm dz\mathrm dr\mathrm d\varphi+\int_0^{2\pi}\int_{\sqrt{2(3-\sqrt 5)}}^2\int_{r^2/2}^r\frac{4-z^2}{r^5}\mathrm dz\mathrm dr\mathrm d\varphi+\int_0^{2\pi}\int_2^{\sqrt{2(3+\sqrt 5)}}\int_{r^2/2}^{2+r}\frac{4z-z^2}{r^5}\mathrm dz\mathrm dr\mathrm d\varphi$ and I would analogously write the second integral, but the numbers I got don't seem quite operable. Am I missing something and is there a more efficient method?
| Let $D_-=\{(r,z): r\geq 0, z\leq 2, z\leq r^2\leq 2z, (z-2)^2\leq r^2\leq 4(z-2)^2\}$, and consider the following change of variables: $u=z/r^2$ and $v=(z-2)^2/r^2$, then
$$\left|\det\begin{pmatrix}
\frac{\partial u}{\partial r}& \frac{\partial u}{\partial z}\\
\frac{\partial v}{\partial r}& \frac{\partial v}{\partial z}
\end{pmatrix}\right|=2\frac{4-z^2}{r^5}
.$$
Then
$$\iiint_{V\cap \{z\leq 2\}}\frac{4-z^2}{(x^2+y^2)^3} dx dy dz=2\pi\iint_{D_-}\frac{4-z^2}{r^6}rdrdz=
2\pi\int_{u=1/2}^1\int_{v=1/4}^1\frac{1}{2}dudv=\frac{3\pi}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the inverse of a piecewise function I am struggling with finding the inverse of a piecewise function, namely:
$f(x) = \begin{cases}
x - 1, & 0 \le x < 1\\
2 - x,& 1 < x \le 2 \\
\end{cases} $
For the first case $\\$
$f: y = x - 1$ when $0 \le x < 1 \rightarrow f^{-1}: x = y + 1$ when $-1 \le x < 0$
But I find this because I only have to subtract 1 from the endpoints of the inequality. I don't have a clear intuition how to do this with $2 - x$
| Suppose that $f^{-1}$ there exists, then you have $f(x)=x-1$ when $0\leqslant x<1$ then setting $y=f(x)$ we get $y=x-1$ when $0\leqslant x<1$ then $x=y+1$ when $0\leqslant y+1<1$. Hence $f^{-1}(x)=x+1$ when $-1\leqslant x<0$.Also $f(x)=2-x$ when $0<x\leqslant 2$ then setting $y=f(x)$ we get $y=2-x$ when $1<x\leqslant 2$ then $x=2-y$ when $1<2-y\leqslant 2$. Hence $f^{-1}(x)=2-x$ when $0\leqslant x<1$.
Therefore, the inverse of function $f$ is given by
$$\boxed{f^{-1}: x\mapsto \begin{cases}x+1, \quad \text{if} \quad -1\leqslant x<0,\\ 2-x, \quad \text{if} \quad 0\leqslant x<1 \end{cases}}$$
Now check $(f^{-1}\circ f)(x)=x$ for all $x\in {\rm Dom}(f^{-1})$ and $(f\circ f^{-1})(x)=x$ for all $x\in {\rm Dom}(f)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Asking About Best Upper Bound And lowerBound Find the best lower and upper bounds for
$$\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right),$$
$1)~~$ overall acute-angled $\Delta ABC.$
first of all we know that $\cos(A)-\sin(A) = -\sqrt{2}\sin(A - \frac{\pi}{4})$
$\Pi { -\sqrt{2}\sin(A - \frac{\pi}{4})}=-2\sqrt{2}*\Pi{(\sin(A-\frac{\pi}{4}))}$
$-\frac{1}{2}*\sqrt{2}<\sin(A-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
$-\frac{1}{2}*\sqrt{2}<\sin(B-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
$-\frac{1}{2}*\sqrt{2}<\sin(C-\frac{\pi}{4})<\frac{1}{2}*\sqrt{2}$
Then Lagrange? or there is any other method you all could suggest
Edit:
For Everyone who wanna post or finish using every possible method just write it down here
| $\left(\cos A-\sin A\right)\left(\cos B-\sin B\right)\left(\cos C-\sin C\right) = -2\sqrt{2} \prod_{cyc} (A - \pi / 4)$.
WLOG, $\pi / 2 > A \geq B \geq C > 0$.
Easy to see $\pi /3 \leq A < \pi/2$, $\pi/4 < B \leq A$, and $0 < C \leq B$.
Clearly, upper bound is achieved when $\sin(C - \pi/4) < 0$, i.e., $0 < C < \pi/4$.
Fix $C < \pi/4$, we have $\pi/2 - C/2 \leq A < \pi /2$ and $B = \pi - A -C$.
When $C$ is fixed, the maximum is achieved when $\sin(A - \pi/4)\sin(B-\pi/4)$ is maximzed, where
$$\sin(A - \pi/4)\sin(B-\pi/4) = \sin(A - \pi/4)\sin(3\pi/4 - A - C).$$
Easy to see the above equation is maximized when $A - \pi/4 = 3\pi/4 - A - C = \pi/2 - C/2$, i.e., $A = 3\pi/4 - C/2$ and the maximum value is $\sin^2(\pi/2 - C/2)$.
The next step is to maximize
$$-2\sqrt{2} \sin^2(\pi/2 - C/2)\sin(C-\pi/4),$$
where $0 < C < \pi/4$.
Regarding the lower bound, we shall have $\pi/3 \geq C > \pi/4$ and $\sin(C-\pi/4) > 0$.
This time again we need to maximize $\sin(A - \pi/4)\sin(B-\pi/4)$ in a similar reasoning.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_{-\infty}^\infty \frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$ I recently attempted to evaluate the following integral
$$\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$$
I started by inserting a parameter, $t$
$$F(t)=\int_{-\infty}^\infty\frac{\ln{(tx^4+x^2+t)}}{x^4+1}dx$$
Where F(0) is the following
$$F(0)=\int_{-\infty}^\infty\frac{\ln{(x^2)}}{x^4+1}dx=2\int_0^\infty\frac{\ln{(x^2)}}{x^4+1}dx=4\int_0^\infty\frac{\ln x}{x^4+1}dx$$
We can evaluate this using a common integral from complex analysis and taking the derivative using Leibniz’s rule.
$$\int_0^\infty\frac{x^m}{x^n+1}dx=\frac{1}{m+1}\int_0^\infty\frac{(m+1)x^m}{(x^{m+1})^\frac{n}{m+1}+1}dx=\frac{1}{m+1}\int_0^\infty\frac{du}{x^\frac{n}{m+1}+1}$$
$$=\frac{1}{m+1}\frac{\pi}{\frac{n}{m+1}\sin{\frac{\pi}{\frac{n}{m+1}}}}=\frac{\pi}{n\sin{\frac{\pi(m+1)}{n}}}=\frac{\pi}{n}\csc{\frac{\pi(m+1)}{n}}$$
$$\int_0^\infty\frac{\ln{x}}{x^n+1}dx=\frac{d}{dm}\int_0^\infty\frac{x^m}{x^n+1}dx\Big|_{m=0}$$
$$=\frac{\pi}{n}\frac{d}{dm}\csc{\frac{\pi(m+1)}{n}}\Big|_{m=0}=-\frac{\pi^2}{n^2}\csc{\frac{\pi(m+1)}{n}}\cot{\frac{\pi(m+1)}{n}}\big|_{m=0}=-\frac{\pi^2}{n^2}\csc{\frac{\pi}{n}}\cot{\frac{\pi}{n}}$$
Therefore
$$F(0)=4\int_0^\infty\frac{\ln{x}}{x^4+1}dx=-\frac{\pi^2\sqrt 2}{4}=-\frac{\pi^2}{2\sqrt 2}$$
Now that we found F(0), we can start applying Feynman’s trick.
$$F’(t)=\int_{-\infty}^{\infty}\frac{dx}{tx^4+x^2+t}$$
Using a formula I derived we can continue
$$\int_{-\infty}^\infty\frac{dx}{ax^4+bx^2+c}=\frac{\pi}{\sqrt{c}\sqrt{b+2\sqrt{ac}}}$$
$$F’(t)=\frac{\pi}{\sqrt{t}\sqrt{1+2t}}$$
Integrating both sides
$$F(t)=\pi\sqrt2\ln{(\sqrt{2t}+\sqrt{2t+1})}+C$$
Set $t=0$
$$C=F(0)=-\frac{\pi^2}{2\sqrt2}$$
Therefore
$$F(t)=\pi\sqrt2\ln{(\sqrt{2t}+\sqrt{2t+1})}-\frac{\pi^2}{2\sqrt2}$$
$$I=\pi\sqrt2\ln{(\sqrt2+\sqrt3)}-\frac{\pi^2}{2\sqrt2}$$
WolframAlpha confirms it numerically
I am not satisfied with this solution. I am curious as to what other solutions there might be. How else can we solve this integral?
| \begin{align}J&=\int_{-\infty}^\infty \frac{\ln{(x^4+x^2+1)}}{x^4+1}dx\\
&=2\int_{0}^\infty \frac{\ln{(x^4+x^2+1)}}{x^4+1}dx\\
&=4\int_{0}^\infty \frac{\ln x}{1+x^4}dx+\underbrace{\int_{0}^\infty \frac{\left(1+\frac{1}{x^2}\right)\ln\left(\left(x-\frac{1}{x}\right)^2+3\right)}{\left(x-\frac{1}{x}\right)^2+2}dx}_{u=x-\frac{1}{x}}\\
&=4\int_{0}^\infty \frac{\ln x}{1+x^4}dx+\int_{-\infty}^\infty\frac{\ln(u^2+3)}{u^2+2}du\\
&=4\underbrace{\int_{0}^\infty \frac{\ln x}{1+x^4}dx}_{=A}+2\underbrace{\int_{0}^\infty\frac{\ln(u^2+3)}{u^2+2}du}_{B}\\
\end{align}
Calculation of $B$.
Let \begin{align}a\in[0,1],F(a)&=\int_{0}^\infty\frac{\ln\Big(a(u^2+2)+1\Big)}{u^2+2}du,F(1)=B,F(0)=0\\
F^{\prime}(a)&=\int_0^\infty \frac{1}{1+a(2+x^2)}dx=\left[\frac{\arctan\left(\frac{ax}{\sqrt{a(1+2a)}}\right)}{\sqrt{a(1+2a)}}\right]_0^\infty=\frac{\pi}{2\sqrt{a(1+2a)}}\\
B&=\frac{\pi}{2\sqrt{2}}\left[\ln\left(1+4a+2\sqrt{2}\sqrt{a(1+2a)}\right)\right]_0^1=\boxed{\frac{\pi\ln\left(2\sqrt{6}+5\right)}{2\sqrt{2}}}
\end{align}
Calculation of $A$.
\begin{align}R&=\int_0^\infty\int_0^\infty \frac{\ln(xy)}{(1+x^4)(1+y^4)}dxdy\\
&\overset{u(x)=xy}=\int_0^\infty \int_0^\infty \frac{y^3\ln u}{(1+y^4)(u^4+y^4)}dudy\\
&\frac{1}{4}\int_0^\infty\left[\frac{\ln\left(\frac{y^4+1}{y^4+u^4}\right)}{u^4-1}\right]_{y=0}^{y=\infty}\ln u du\\
&=\int_0^\infty \frac{\ln^2 u}{u^4-1}du\\
&=\frac{1}{2}\underbrace{\int_0^1 \frac{\ln^2 u}{u^2-1}du}_{=-\frac{\pi^3}{8}}-\frac{1}{2}\underbrace{\int_0^1 \frac{\ln^2 u}{1+u^2}du}_{=0}\\
&=\boxed{-\frac{\pi^3}{16}}
\end{align}
On the other hand,
\begin{align}R&=2A\int_0^\infty \frac{1}{1+u^4}du\\
\int_0^\infty \frac{1}{1+u^4}&=\int_0^\infty \frac{1+\frac{1}{u^2}}{\left(u-\frac{1}{u}\right)^2+2 }du\\
&\overset{z=u-\frac{1}{u}}=
\int_{-\infty}^{+\infty}\frac{1}{2+z^2}dz=2\int_0^{+\infty}\frac{1}{2+z^2}dz\\
&=\left[\frac{\arctan\left(\frac{u}{\sqrt{2}}\right)}{\sqrt{2}}\right]_0^\infty=\boxed{\frac{\pi}{2\sqrt{2}}}
\end{align}
Therefore,
\begin{align}A&=\frac{-\frac{\pi^3}{16}}{2\times \frac{\pi}{2\sqrt{2}}}=\boxed{-\frac{\pi^2}{8\sqrt{2}}}\\
J&=4A+2B=\boxed{-\frac{\pi^2}{2\sqrt{2}}+\frac{\pi\ln\left(2\sqrt{6}+5\right)}{\sqrt{2}}}
\end{align}
NB:
I assume,
\begin{align}\int_0^\infty \frac{\ln^2 x}{1+x^2}dx=\frac{\pi^3}{8}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Convergence of double summation of divergent and convergent series I am looking at two series
(a)$$\sum_{n = 1}^\infty \dfrac{1}{n^2 (n + 1)^{2p}}\sum_{m = 1}^n (m + 1)^{2p}$$
and
(b)
$$\sum_{n = 1}^\infty \dfrac{1}{(n + 1)^{2p}}\sum_{m = 1}^n \dfrac{(m + 1)^{2p}}{m^2}$$
Assume $p \in (0, 1]$ for both (a) and (b). I wish to figure out which of the ones is summable and not summable (and for which power this holds).
Suppose $p = 1$ for simplicity, then, the first few terms of (a) are:
$\dfrac{2^2}{2^2} + \dfrac{2^2 + 3^2}{2^2 3^2} + \dfrac{2^2 + 3^2 + 4^2}{3^2 4^2} + \ldots = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{3^2 4} + \dfrac{1}{4^2} + \dfrac{1}{3^2} + \ldots$
Since this seems to converge at similar rate as $1/n^2$, does this indicate that (a) is convergent?
Similarly for (b), the first few terms are,
$\dfrac{2^2}{2^2} + \dfrac{2^2 + 3^2/2^2}{3^2} + \dfrac{2^2 + 3^2/2^2 + 4^2/3^2}{4^2} + \ldots = 1 + 2^2/3^2 + 1/2^2 + 1/4 + 3^2/4^3 + 1/3^2 + \ldots$
For this one I really have no idea what I could use to determine convergence.
| The first series is divergent for all $p>0$, and therefore so is the second, whose terms are larger.
To see this, consider $\alpha>0$, and let's show that $\sum_{k=1}^nk^\alpha\sim Cn^{\alpha+1}$ for some constant $C$.
We have
$$\frac{1}{n^{\alpha+1}}\sum_{k=1}^nk^\alpha=\frac{1}{n}\sum_{k=1}^n\left(\frac kn\right)^\alpha$$
This is a Riemann sum, for the continuous function $f(x)=x^\alpha$, therefore
$$\frac{1}{n^{\alpha+1}}\sum_{k=1}^nk^\alpha\to\int_0^1 x^\alpha\mathrm dx=\frac{1}{\alpha+1}$$
Hence
$$\sum_{k=1}^nk^\alpha\sim\frac{n^{\alpha+1}}{\alpha+1}$$
Hence the term in the first series is equivalent to $\dfrac{n^{2p+1}}{(2p+1)n^2n^{2p}}\sim\dfrac{1}{(2p+1)n}$, and the series is divergent, per the limit comparison test.
For the second series, just note that
$$\sum_{m=1}^n\frac{(m+1)^{2p}}{m^2}\ge\frac{1}{n^2}\sum_{m=1}^n(m+1)^{2p}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x$ I was wondering if it is possible to compute this integral in closed form:
$$
\int_{-\infty}^{\infty} \big(\frac{x^2}{x^2 + a}\big)^2 e^{-x^2/2} \, \mathrm{d}x
$$
I tried making a substitution with $s = x^2$ and also tried evaluating it in Wolfram Alpha, but was not successful.
| This is too long for a comment.
Since, in comments, you already received the answer and a good hint, let me address the more general case of
$$I_n= \int_{-\infty}^{+\infty} \left(\frac{x^2}{x^2+a}\right)^n\, e^{-\frac{x^2}{2}}\,dx=2\int_{0}^{+\infty} \left(\frac{x^2}{x^2+a}\right)^k\, e^{-\frac{x^2}{2}}\,dx$$
The idea is to write
$$\frac{x^2}{x^2+a}=\frac{x^2+a-a}{x^2+a}=1-\frac {a}{x^2+a}$$ and to use the binomial expansion to face known integrals since using Tricomi's confluent hypergeometric function
$$\int_0^\infty \left(a+x^2\right)^{-n}\, e^{-\frac{x^2}{2}}\,dx=\frac 1 {2 ^n}\sqrt{\frac \pi 2 } \, U\left(n,n+\frac{1}{2},\frac{a}{2}\right)$$
They write
$$I_n=\frac{\sqrt{2\pi }} {2^n\,n!}\Bigg[ P_{n}- { \sqrt{\frac{a\pi}2}\,e^{a/2} } \, \text{erfc}\left(\sqrt{\frac{a}{2}}\right)\,Q_{n}\Bigg]$$
where the first polynomials in $a$ are
$$\left(
\begin{array}{ccc}
n & P_n & Q_n \\
0 & 1 & 0 \\
1 & 2 & 2 \\
2 & 4 a+8 & 4 a+12 \\
3 & 6 a^2+54 a+48 & 6 a^2+60 a+90 \\
4 & 8 a^3+160 a^2+696 a+384 & 8 a^3+168 a^2+840 a+840
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the probability density function of the random variable $\frac{1}{X}$? How to find the probability density function of the random variable $\frac{1}{X}$?
Let $X$ be a random variable with pdf
$f_X(x)= \begin{cases} 0 ; x \le 0 \\ \frac{1}{2} ; 0 < x \le 1 \\ \frac{1}{2x^2} ; 1 < x < \infty \end{cases}$
I was trying to avoid using formula. So we see that
$P(Y \le y) = 1 - P(X \le y)$
Case$1$: if $y \le 0$
Then $F(y) = 1 - P[X \le y] = 1$
Case $2:$ if $0 < y \le 1$
Then $F(y) = 1 - P(X \le y) = 1 - (0 + \int_{0}^{y}\frac{1}{2}dy) = 1-\frac{y}{2}$
Case $3:$ if $1 < y < \infty$
Then $F(y) = 1 - P(X \le y) = 1 - (\frac{1}{2} + \int_{1}^{y}\frac{1}{y^2}dy) = \frac{1}{2}+\frac{1}{y}-1$
Then on differentiating we can get the probability density function.
I thibk that the distribution function that i got is not correct . Can someone help me out please
| The PDF of $X$ is given by
$$
f_X(x) = \left\{ \begin{array}{ccc}
0 & \mbox{if} & x \leq 0 \\[2mm]
{1 \over 2} & \mbox{if} & 0 < x < 1 \\[2mm]
{1 \over 2 x^2} & \mbox{if} & 1 < x < \infty \\[2mm]
\end{array} \right. \tag{1}
$$
Thus, $X$ is a positive random variable.
Since $Y = {1 \over X}$, it is immediate that $Y$ is also a positive random variable.
Hence,
$$
F_Y(y)= P(Y \leq y) = 0 \ \ \mbox{for} \ \ y < 0. \tag{2}
$$
Fix $y$ in the interval $0 < y < 1$.
Then ${1 \over y} > 1$.
Now, we find that
$$
F_Y(y) = P(Y \leq y) = P\left( {1 \over X} \leq y \right) =
P\left( X \geq {1 \over y} \right)
$$
which can be evaluated using (1) as
$$
F_Y(y) = \int\limits_{1 \over y}^\infty \ {1 \over 2 x^2} \ dx =
{1 \over 2} \ \left[ - {1 \over x} \right]_{1 \over y}^\infty
$$
or
$$
F_Y(y) = {1 \over 2} \ \left[ 0 + y \right] = {y \over 2}
$$
Thus,
$$
F_Y(y) = {y \over 2} \ \ \mbox{for} \ \ 0 < y < 1. \tag{3}
$$
Next, we fix in the interval $y > 1$.
Then it follows that $0 < {1 \over y} < 1$.
Now,
$$
F_Y(y) = P(Y \leq y) = P\left( {1 \over X} \leq y \right) =
P\left( X \geq {1 \over y} \right) = 1 - P\left( X \leq {1 \over y} \right)
$$
which can be evaluated using (1) as
$$
F_Y(y) = 1 - \int\limits_{0}^{1 \over y} \ {1 \over 2} \ dx =
1 - {1 \over 2} \left[ {1 \over y} - 0 \right] = 1 - {1 \over 2 y}.
$$
Thus,
$$
F_Y(y) = 1 - {1 \over 2 y} \ \ \mbox{for} \ \ y > 1 \tag{4}
$$
Combining the three cases, we find that
$$
F_Y(y) = \left\{ \begin{array}{ccc}
0 & \mbox{if} & y \leq 0 \\[2mm]
{y \over 2} & \mbox{if} & 0 < y < 1 \\[2mm]
1 - {1 \over 2 y} & \mbox{if} & y > 1 \\[2mm]
\end{array} \right. \tag{5}
$$
From (5), we find the PDF of $Y = {1 \over X}$ as
$$
f_Y(y) = F_Y'(y) = \left\{ \begin{array}{ccc}
0 & \mbox{if} & y \leq 0 \\[2mm]
{1 \over 2} & \mbox{if} & 0 < y < 1 \\[2mm]
{1 \over 2 y^2} & \mbox{if} & y > 1 \\[2mm]
\end{array} \right. \tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Can we prove the inequality without opening the parentheses? $(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$$
I tried to use the following inequalities:
$$x^2+y^2+z^2≥xy+yz+xz$$
and The Cauchy–Schwarz inequality:
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
$$x^2+y^2+z^2≥x+y+z$$
I also tried
$$x+y+z ≥3\\ x^2+y^2+z^2≥3\\ xy+yz+xz≥3$$
But, I couldn't make progress.
I want to solve this inequality without expansion. That is, without opening the parentheses. Is this possible?
| Increase the right hand side to $9(x^2+y^2+z^2)$ by switching $xy+yz+zx$ to $x^2+y^2+z^2$ (Cauchy).
Divide the left hand side by $xyz=1$.
This leads to $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$, which is easy to show. Namely, $x+y+z\geq 3\sqrt[3]{xyz}$ by AM-GM and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3/\sqrt[3]{xyz}$ by HM-GM. This also shows that equality holds iff $x=y=z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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${\log _ {x}}8 - {\log _{4x}} 8 = {\log _{2x}} 16$
${\log _ {x}}8 - {\log _{4x}} 8 = {\log _{2x}} 16$
I tried solving this problem by change of base and by $\frac{1}{\log{x}}$, but I really cannot seem to solve it no matter how hard I try.
I could only answer it by substituting $x$ for mcq answers.
Where do I even begin to solve it?
| HINT
As @TheoBendit has mentioned in the comments, I would recommend you to start with noticing
\begin{align*}
\log_{x}(8) - \log_{4x}(8) = \log_{2x}(16) & \Longleftrightarrow \frac{\log_{2}(8)}{\log_{2}(x)} - \frac{\log_{2}(8)}{\log_{2}(4x)} = \frac{\log_{2}(16)}{\log_{2}(2x)}
\end{align*}
where $x > 0$ and $x\not\in\{1,1/2,1/4\}$.
If we let that $y = \log_{2}(x)$, one arrives at equivalent equation:
\begin{align*}
\frac{3}{y} - \frac{3}{2 + y} = \frac{4}{1 + y} & \Longleftrightarrow \frac{6}{y(2 + y)} = \frac{4}{1 + y}\\\\
& \Longleftrightarrow
\begin{cases}
3(1 + y) = 2y(2 + y)\\\\
y(2 + y)(1 + y) \neq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
2y^{2} + y - 3 = 0\\\\
y(2 + y)(1 + y) \neq 0
\end{cases}
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximize $z$ over $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$ Suppose that $x$, $y$, and $z$ are real numbers such that $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$. What is the largest possible value of $z$?
I tried applying Cauchy-Schwarz to get $(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$, but this doesn't say anything. I also tried some different ways to apply Cauchy, but they all didn't do much.
Thanks in advance!!
| Define $r = [x,y,z]^T$, then your objective function using Lagrange multipliers is
$ f(r,\lambda_1, \lambda_2) = c^T r + \lambda_1 (b^T r - 3) + \lambda_2 (r^T r - 6) $
where $c = [0, 0, 1]^T , b = [1, 1, 1]^T $
Differentiating, and equating to zero:
$\nabla_r f = c + \lambda_1 b + 2 \lambda_2 r = 0 \hspace{15pt}(1) $
$f_{\lambda_1} = b^T r - 3 = 0 \hspace{15pt}(2)$
$f_{\lambda_2} = r^T r - 6 = 0 \hspace{15pt}(3)$
From $(1)$ it follows that
$r = -\dfrac{1}{2 \lambda_2} (c + \lambda_1 b ) \hspace{15pt}(4) $
Plugging this into $(2)$ and $(3)$ results in
$ ( b^T c + \lambda_1 b^T b ) + 6 \lambda_2 = 0 \hspace{15pt}(5)$
and
$ ( c + \lambda_1 b)^T (c + \lambda_1 b) = 24 \lambda_2^2 \hspace{15pt} (6)$
Plugging in the values of vectors $c$ and $b$, we get
$ 1 + 3 \lambda_1 + 6 \lambda_2 = 0 \hspace{15pt} (7)$
and
$ 1 + 2 \lambda_1 + 3 \lambda_1^2 = 24 \lambda_2^2 \hspace{15pt}(8) $
Substituting for $\lambda_2$ from $(7)$ into $(8)$
$ 1 + 2 \lambda_1 + 3 \lambda_1^2 = 24 \left( \dfrac{-1 - 3 \lambda_1 }{ 6} \right)^2 \hspace{15pt}(9) $
Aftern simplifying, this becomes
$ 9 \lambda_1^2 + 6 \lambda_1 - 1= 0 $
Therefore, solutions for $\lambda_1, \lambda_2 $ are
$ (\lambda_1, \lambda_2) = ( \dfrac{ - 1- \sqrt{2} }{3} , \dfrac{\sqrt{2}}{6} ) , ( \dfrac{ -1 + \sqrt{2}}{3}, - \dfrac{\sqrt{2}}{6} ) $
The corresponding value for $z$ for the first solution can be computed from equation $(4)$
$r = -\dfrac{1}{2 \lambda_2} (c + \lambda_1 b ) \hspace{15pt}(4) $
$ z_1 = - \dfrac{3}{\sqrt{2}} ( 1 + \dfrac{-1 - \sqrt{2}}{3} ) =1 - \sqrt{2} $
And for the second solution
$ z_2 = \dfrac{3}{\sqrt{2}} ( 1 + \dfrac{-1 + \sqrt{2}}{3} ) = 1 + \sqrt{2}$
Hence, the maximum is $z_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle Find the angle $\alpha$ of a triangle with sides $a,b$ and $c$ for which the equality $$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}$$ holds.
My idea is to use the law of cosines: $$\cos\alpha=\dfrac{b^2+c^2-a^2}{2bc}$$ after simplifying the given equality and plug in something, but this does not seem to be easy.
$$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}\\\dfrac{2a+b+c}{(a+b)(a+c)}=\dfrac{3}{a+b+c}\\(2a+b+c)(a+b+c)=3(a+b)(a+c)$$ Am I missing something?
| Multiplying both sides of the equation by $(a+b)(a+c)(a+b+c)$, we get \begin{align*}
(a+c)(a+b+c)+(a+b)(a+b+c)&=3(a+c)(a+b)\\
a^2 + a b + 2 a c + b c + c^2+a^2 + 2 a b + a c + b^2 + b c&=3 a^2 + 3 a b + 3 a c + 3 b c\\
2 a^2 + 3 a b + 3 a c + b^2 + 2 b c + c^2&=3 a^2 + 3 a b + 3 a c + 3 b c\\
b c &=-a^2+ b^2 + c^2\\
\end{align*}
Hence, we have $\cos\alpha=\frac{1}{2}$. Since we know $0<\alpha<\pi$, we have $\alpha=\boxed{\frac{\pi}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{x\to\infty}\frac{f^{-1}(x)}{\ln(x)}$, where $f(x)=e^x+x^3-x^2+x$
$$f(x)=e^x+x^3-x^2+x$$
What is the following?
$$\lim_{x\to\infty}\frac{f^{-1}(x)}{\ln(x)}$$
Do I need to calculate the inverse of $f(x)$ or is there some other way that this limit can be solved?
| From $f'(x) = e^x + 3 x^2 - 2 x+1$ we deduce that $f'(x)\ge 3x^2 -2 x + 1>0$, hence $f$ is a bijection from ${\mathbb R}$ onto $f({\mathbb R}) = {\mathbb R}$ and we have $f^{-1}(x)\to +\infty$ when $x\to +\infty$. Let $y = f^{-1}(x)$, we have
\begin{equation}
\frac{\ln x}{y} = \frac{\ln(f(y))}{y} = \frac{y + \ln(1 + e^{-y}(y^3 -y^2+y))}{y} = 1 + \frac{1}{y}\ln(1 + e^{-y}(y^3 -y^2+y))
\end{equation}
Hence $\frac{\ln x}{y}\to 1$, hence $\frac{y}{\ln x}\to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the number of ways in which 4 letters can be selected from the 9 letters of the word ALGORITHM, given at least one vowel is included. My solution:
3 cases:
1 vowel:
$6P3 \times 3C1 = 360 $
2 vowels:
$6P2 \times 3C2 \times 2! = 180 $
3 vowels:
$6P1 \times 3! = 36 $
Total number of ways: 360 + 180 + 36 = 576
Am I correct?
| Using the Inclusion-Exclusion principle, we count the number of combinations that use $1$ vowel, subtract those that use $2$, and add those that use all $3$.
The generating function for this is
$$[x^4]:(1-(1-x)^3)\frac{1}{(1-x)^6}$$
where
$$1-(1-x)^3=\binom{3}{1}x-\binom{3}{1}x^2+x^3$$
$$\frac{1}{(1-x)^6}=\binom{5}{5}+\binom{6}{5}x+\binom{7}{5}x^2+\binom{8}{5}x^3+\dots$$
and gives $3\cdot56-3\cdot21+6=111$.
Also, the GF can be re-arranged to give
$$[x^4]:\frac{1}{(1-x)^6}-\frac{(1-x)^3}{(1-x)^6}$$
$$[x^4]:\frac{1}{(1-x)^6}-\frac{1}{(1-x)^3}$$
$$=\binom{9}{5}-\binom{6}{2}$$
$$=111$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to describe Image of matrix function f by eigenvectors given
$$
\textbf{A}=\begin{bmatrix}
4& 1& 2\\
2 & 2 & 4\\
1&1&2
\end{bmatrix}
$$
$f({\bf x})=A{\bf x}, {\bf x}\in R^3$
I know eigenvectors : $v_1=[0,-2,1]^T$, $v_2=[-2,2,1]^T$, $v_3=[2,2,1]^T$
eigenvalues: 0,2,6
the kernel is solution of Ax=0, $x=[0, -2t,t]^T, t \in R$
the image is $[4x_1+x_2+2x_3, 2x_1+2x_2+4x_3, x_1+x_2+2x_3]^T,x_1,x_2,x_3\in R$
but I can't figure how to use $\{v_1,v_2,v_3\}$ to describe image. the kernel = $v_1*t$ or $(v_2+v_3)*t , t\in R$.
| Given that
$$
A=\left[ \begin{array}{ccc}
4 & 1 & 4 \\
2 & 2 & 4 \\
1 & 1 & 2 \\
\end{array} \right]
$$
Note that $A$ has rank two.
(Obviously, the second row is twice the third row. In other words,
$R_2 = 2 R_3$ or that $0 R1 + R_2 - 2 R_3 = 0$. Hence, the rows of $A$
are linearly dependent)
A simple calculation gives the eigenpairs of $A$ as:
$$
\lambda_1 = 0, \ \ \mathbf{v}_1 = \left[ \begin{array}{c}
0 \\
-2 \\
1 \\
\end{array} \right]
$$
$$
\lambda_2 = 2, \ \ \mathbf{v}_2 = \left[ \begin{array}{c}
-2 \\
2 \\
1 \\
\end{array} \right]
$$
$$
\lambda_3 = 6, \ \ \mathbf{v}_1 = \left[ \begin{array}{c}
2 \\
2 \\
1 \\
\end{array} \right]
$$
$A$ has three distinct eigenvalues $0, 2, 6$ with corresponding
eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.
Obviously
$$
\mathcal{B} = \left\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \right \}
$$
forms a basis of $\mathbf{R}^3$.
If $f(\mathbf{x}) = A \mathbf{x}$, then we have
$$
f(\mathbf{x}) = \left[ \begin{array}{c}
4 x_1 + x_2 + 2 x_3 \\
2 x_1 + 2 x_2 + 4 x_3 \\
x_1 + x_2 + 2 x_3
\end{array} \right] = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3
$$
since $\mathcal{B}$ forms a basis of $\mathbf{R}^3$.
A simple calculation gives
$$
c_1 = 0, \ c_2 = - {x_1 \over 2} + {x_2 \over 4} + {x_3 \over 2}, \ \
c_3 = {3 x_1 \over 2} + {3 x_2 \over 4} + {3 x_3 \over 2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $x,y,z$ satisfying $x(y+z-x)=68-2x^2$, $y(z+x-y)=102-2y^2$, $z(x+y-z)=119-2z^2$ Solve for $x,y,z$:
$$x(y+z-x)=68-2x^2$$
$$y(z+x-y)=102-2y^2$$
$$z(x+y-z)=119-2z^2$$
After some manipulation, I obtain
$$xy+xz=68-x^2$$
$$yz+xy=102-y^2$$
$$xz+yz=119-z^2$$
After combining equations, I get
$$y=\frac{-51-x^2+z^2}{x-z}$$
This seems too tedious. Is there a simpler way?
| Your system is equivalent to
\begin{align}
x(y+z+x)&=68\\
y(z+x+y)&=102\\
z(x+y+z)&=119.
\end{align}
Therefore, in effect, you have
\begin{align}
xa&=68\\
ya&=102\\
za&=119\\
a&=x+y+z.
\end{align}
This means that
$$a=\frac{68}{a}+\frac{102}{a}+\frac{119}{a}$$
or $$a^2=68+102+119\implies a=\pm 17,$$
which results in either $x= 4, y=6, z=7$ or $x=-4, y=-6, z=-7.$ (Thanks to paw88789 for pointing this out).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
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