Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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For positive $a$, $b$, $c$ with $a+b+c=1$, show that $(ab+bc+ca) \sum_{cyc}\frac{a}{b^2+b} \geq \frac34$
If $a,b,c > 0$ and $a+b+c = 1$, then prove that
$$\left(\frac{a}{b^2+b}+\frac{b}{c^2+c}+\frac{c}{a^2+a}\right)(ab+bc+ca)\geq\frac{3}{4}$$
It's been more than 35 years since I last touched algebra!!
| Hint: Using generalised Holder’s inequality
$$\sum_{cyc}ab \cdot \sum_{cyc} \frac{a}{b^2+b}\cdot \sum_{cyc} a(b+1) \geqslant \left(\sum_{cyc}a\right)^3=1$$
Now it is enough to show $\sum_{cyc} a(b+1) \leqslant \frac43$ which is easy.
| {
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Rational Decomposition
Find the partial fractions for $\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}$
First, I recognize it's not a proper fraction. I also know the denominator is $(x+3)(3x+1)$. Now every time I try long division to convert to a proper fraction by hand I get $x^3-x^2+1+(\frac{-10x-3}{(x+3)(3x+1)})$, but t... | The first step you need to follow is the long division method, which will for sure give you $$\frac{3x^5+7x^4-7x^3-3x^2+x}{3x^2+10x+3}=\frac{(x^3-x^2)(3x^2+10x+3)+x}{(x+3)(3x+1)}=x^3-x^2+\frac{x}{(x+3)(3x+1)}$$Half work done!! Now we just need to break $\frac{x}{(x+3)(3x+1)}$ into partial fractions. That's easy,$$\fra... | {
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Find the Maximum and Minimum Values of $f(x) = \cos\sqrt{x+1} - \cos\sqrt{x}$ I saw today in a fb forum the following excersice:
Find the range of values of
$f(x)=\cos\sqrt{x+1}-\cos\sqrt{x}$
I have tried to find the mix and max of the function but failed. Any ideas?
| To do an analytic check of the range, and possibly other properties, note you can combine the $2 \cos$ terms into a multiple of one $\sin$ term using several identities from List of trigonometric identities. First, note that $\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$, so you get
$$f(x) = \cos\sqrt{x+1}-\cos\sqrt{x} ... | {
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$|x^2-3x+2 | = mx$ has $x_1, x_2, x_3, x_4 $ $s(m) = \frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} + \frac{1}{{x_3}^2 }+ \frac{1}{{x_4}^2}$ express $s(m)$
$|x^2-3x+2 | = mx$ has $x_1, x_2, x_3, x_4 $ 4 distinct solutions
$s(m) = \frac{1}{{x_1}^2} +\frac{1}{{x_2}^2} + \frac{1}{{x_3}^2 }+ \frac{1}{{x_4}^2}$
Express $s... | $$(x^2-3x+2)^2=m^2x^2$$
$$\implies x^4-6x^3+x^2(9+4-m^2)-12x+4=0$$
$$\iff x^4+(13-m^2)x^2+4=12x+6x^3$$
Replace $x^2=\dfrac1y\implies y=\pm\dfrac1{\sqrt x}$
$$\dfrac{1+(13-m^2)y+4y^2}{y^2}=\pm\dfrac{12y+6}{y^{3/2}}$$
$$(1+(13-m^2)y+4y^2)^2= y(12y+6)^2$$
$$\implies 16y^4+y^3(8(13-m^2)-144)+\cdots+1=0$$
Using Vieta's form... | {
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Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
Prove that
$$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$
I tried to use King's rule and to scale by $2$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} ... | $$I=\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin(2x)})dx =\frac12 \int_0^\frac{\pi}{2} x'\ln(\sin x+\cos x+\sqrt{\sin (2x)})dx$$
$$\overset{IBP}=\frac12 \int_0^\frac{\pi}{2}x\,\frac{\sin x-\cos x}{\sqrt{\sin(2x)}}dx\overset{x=\arctan t}=\frac{1}{2\sqrt 2}\int_0^\infty \frac{\arctan t}{1+t^2}\frac{t-1}{\sqrt t}... | {
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Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$. Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$.
I don't know how to approach this question.
| As $9=3^2, 9^{2x}=(3^2)^{2x}=3^{4x}$
Similarly, $27^{x^2-5}=3^{3x^2-15}$
As $3^{4x}\ne0,$
$$\implies1=\dfrac{3^{3x^2-15}}{3^{4x}}=3^{3x^2-4x-15}$$
Now like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$,
if $\displaystyle u^m=1,$
either $\displaystyle m=0 $
or $\displaystyle u=1$
or $\display... | {
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Find, analytically, the value of the following limit. How would one prove that$$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k-1}}$$ converges (rather slowly) to $\frac {1}{\sqrt{2}}$, which appears obvious from numerical computation.
| Another interesting way to prove the lower bound.
Consider:
$$A=\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k-1}}$$
$$I_k=\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-(4k-1)x^2} e^{-4k x^2 \sqrt{1-1/(2k)}} dx$$
Since $(\sqrt{2k}+\sqrt{2k-1})^2=4 k-1+4 \sqrt{k(k-1/2)... | {
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shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{2... | Another way:
Using coordinate geometry,
Any point on the circle $P(2\cos t,2\sin t)$
Distance of $P$ from the line $$\dfrac{|3(2\cos t)+4(2\sin t)-12|}5$$
Now $-\sqrt{6^2+8^2}\le 6\cos t+8\sin t\le?$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Finding the best solution to an inconsistent system $A\mathbf{u} = \mathbf{b}$. Let $A = \begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}$.
1. Find a "best solution" to the inconsistent system $A\mathbf{u} = \mathbf{b}$.
2. Find the orthogonal projecti... | Actually there are three lines $$-x+y=1, 2x-y=2, x+y=0~~~(1)$$ which make a triangle.
pre-multiply the given inconsistent equation $$AX=B,~~ A=\begin{bmatrix} -1 & 1\\ 2 & 1 \\ 1 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 \\ 2\\ 0 \end{bmatrix} ~~~~(2)$$ by $A^{T}$ to get $$ A^T A X= A^T B$$ which is nothing but $${\cal ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the length of a Curve The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is relatively
early in the book, so I would expect the integration to be easy.
Find the length of the curve:
$$ 9x^2 = 4y^3$$
from $(0,0)$ to $\left(2\sqrt{3},3\right)$.
Answer:
The... | From $9x^2=4y^3$, you can also get $x=\frac23y^{3/2}$. And, if $g(x)=\frac23x^{3/2}$, then\begin{align}\int\sqrt{1+\bigl(g'(x)\bigr)^2}\,\mathrm dx&=\int\sqrt{1+x}\,\mathrm dx\\&=\frac23(1+x)^{3/2}.\end{align}So, the length of the curve is$$\frac23(1+3)^{3/2}-\frac23(1+0)^{3/2}=\frac{16}3-\frac23=\frac{14}3.$$
| {
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"source": "stackexchange",
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Tips on solving $a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$ $$a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
This is true with $a>b>0$, according to Wolfram Alpha, but I am not able to prove this.
I to simplify using the fact that $a^2(a+2b)=a^3+2ba^2>3b^3$,... | COMMENT.-The points $(a,b)=(1,t)$ for $t\ge4$ does not satisfy the inequality.
| {
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How is this expression derived? On Concrete Mathematics (page 7) we have the following:
$L_n = L_{n-1} + n = (\frac{1}{2}(n-1)n + 1 ) + n = \frac{1}{2}n(n+1)+1$
How is the last expression derived?
How do you convert $(\frac{1}{2}(n-1)n + 1 ) + n$ into $\frac{1}{2}n(n+1)+1$ ?
I've tried different things but was not abl... | Multiply out, collect the first-degree terms, and then put $\frac12n$ back outside a parenthesis again:
$$ \begin{align} (\tfrac12 (n-1)n + 1) + n &=
\tfrac12 n^2 - \tfrac12 n + 1 + n \\&=
\tfrac12 n^2 + n - \tfrac12n + 1 \\&=
\tfrac12 n^2 + \tfrac12 n + 1 \\&=
\tfrac12n (n + 1) + 1
\end{align} $$
Or you could also do ... | {
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Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$ Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$
I thought it could be good to use the function $f: ]-\infty, 81] \rightarrow \mathbb{R}$ given by $f(x) = \sqrt{81-x}$
Because this function is of class $C^{\infty}$, we can compute its Taylor... | Since $8^2 = 64$ and $9^2 = 81$ and $81 - 64 = 17$, using $8.9$ as an estimate for $\sqrt {80}$ is more than just a lucky guess. Since
$\quad (8.9)^2 < 80 \text{ and } (9)^2 > 80$
we must have
$\quad 0 \lt \sqrt{80} - 8.9 \lt 0.1$
Define the function
$\tag 1 F(x) = \frac{8.9x + 80}{x+8.9}$
For $x \ne -8.9$ we have the... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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"answer_id": 6
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Normal equation question Let $C = \begin{bmatrix}
1 & -1 & 1 \\
1 & 1 & 3 \\
2 & 0 & 4
\end{bmatrix}$ and let $\mathbf{b} = \begin{bmatrix}
3 \\
5 \\
5
\end{bmatrix}$. Find the set of all solutions $\mathbf{x}$ to the normal equation}
\begin{equation*}
C^TC\mathbf{x} = C^T\mathbf{b}.
\end{equation*}
Hence, or otherwise... | *
*Solve the system with $C^T C x = C^T b$ with the Gaussian elimination (or the row echelon form) --- i.e. do not compute the inverse (which does not exist here)
*If $C^T C$ is singular, it means that you will get either no solution or an infinite number of solutions. The row echelon form should help you determine w... | {
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Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?
$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$
| Hint: After squaring one times we get
$$2\sqrt{4+(y-6)^2}\sqrt{16+(y-3)^2}=25-(y-6)^2-(y-3)^2$$
squaring again and simplfying we get
$$4 \left(46 y^2-450 y+975\right)=0$$
| {
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$\frac{1+m_v}{1+m_u}\leq \frac{1+u^T(M+I)^{-1} u}{1+v^T(M+I)^{-1}v} \leq \frac{1+m_u}{1+m_v}$ if $M$ is positive sym. PD & $u,v$ are $0-1$ vectors? Let $n$ be a positive integer. Let $m_u,m_v \in \{1,...,n-1 \}$.
Let $M$ be a $n \times n$ symmetric positive definite matrix with positive entries.
Let $u$ and $v$ be ve... | It doesn't hold.
Here is a counterexample for the right inequality in the first line.
Let $m_u = m+1$ and $m_v=m$. Let the matrix $M$ be diagonal with $M_{ii} = a$ for all diagonal entries, except for $M_{ii} = b$ for the $(m+1)$-last entry. If positive entries of the matrix are required, all other elements can be fil... | {
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Evaluate $ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx $ $$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = ? $$
Attempt:
$$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = \int \frac{3x^{2}}{\sqrt{x^{3}-6} \sin^{2}\sqrt{x^{3}-6}} dx $$
$$ U = \sqrt{x^{3}-6} $... | $$u=\sqrt{x^3-6},du=\dfrac{3x^2}{2\sqrt{x^3-6}}dx$$
$$2\int\csc^2u=-2\cot u+K$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without usin... | Taking $a = b = c = m$ and $d = m + 1$ gives that $a^2 + b^2 + c^2 + d^2 = 4m^2 + 2m + 1$, implying that the minimum value of the expression is at most $4m^2 + 2m + 1$. The calculus approach you mentioned, or maybe an application of the Cauchy-Schwarz inequality would lead to the fact that taking all the terms equal gi... | {
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Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x... | HINT
\begin{align*}
0\leq x^{2} \leq x^{2} + y^{2} \Longleftrightarrow 0\leq \frac{x^{2}}{x^{2}+y^{2}} \leq 1 \Longleftrightarrow 0\leq \frac{x^{2}y^{2}}{x^{2}+y^{2}}\leq y^{2}
\end{align*}
Then apply the squeeze theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why any square can be written in this form? I have noticed that all squares, at least up to $19 ^ 2$ can be written as: $a^2 = 5k + p$, where $a \in \mathbb{Z}+\neq 1$ and $k \in \mathbb{Z}+$ and $p = \{0,\pm1\}$
Some examples:
$4^2 = 5 \cdot 3 + 1$
$13^2 = 5 \cdot 34 - 1 $
What is the intuitive and formal proof to se... | Say we want to consider $b^2$. We look at $b$ modulo $5$, $b = 5c+d$, where $c$ is an integer and $d$ is one of $0$, $1$, $2$, $3$, or $4$. Then
$$ b^2 = (5c+d)^2 = 25c^2 + 10c d + d^2 \text{.} $$
The part "$25 c^2 + 10 c d$" is a multiple of $5$, so can be collected into $k$. This leaves $d^2$. We show in a ta... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Prove this Ramanujan series
Prove how this series: $\frac{1}{1^3}\cdot\frac{1}{2} + \frac{1}{2^3}\cdot\frac{1}{2^2} + \frac{1}{3^3}\cdot\frac{1}{2^3} + \frac{1}{4^3}\cdot\frac{1}{2^4} +...= \frac{1}{6}(\log 2)^3+\frac{\pi ^2}{12}(\log 2) + (\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+..)$
I don't know how to begin but ... | For $0 < x < 1$ we have
$$
\sum_{k=1}^{\infty}x^k = \frac{x}{1-x}
$$
now calling $S_0 = \frac{x}{1-x}$ we have the relationships
$$
x S_3' = S_2\\
x S_2' = S_1\\
x S_1' = S_0
$$
after integration with null constants we have
$$
\left\{
\begin{array}{rcl}
S_1 & = & -\log (1-x) \\
S_2 & = & -\log (1-x) \log (x)-\text{Li... | {
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"timestamp": "2023-03-29T00:00:00",
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How to Evaluate proper Integral. Recently I stumbled upon an integral and its solution in a physics article but I couldn't understand how it was evaluated.I have plotted the function and it indicates that the value of the integral should be finite.
I want to ensure the answer is correct.Does anybody know how to do it?
... | Assuming $A,B,T>0$,
$$\begin{align}
I=&\int_{0}^{T}\frac{1}{(t(T-t))^{3/2}}\exp\left[-\frac{A}{t}-\frac{B}{T-t}\right]\,dt\\
\stackrel{x=1/t}{=}&\int_{1/T}^{\infty}\frac{x}{(Tx-1)^{3/2}}\exp\left[-Ax-\frac{Bx}{Tx-1}\right]\,dx\\
\stackrel{t=Tx}{=}&\;\frac{1}{T^{2}}\int_{1}^{\infty}\frac{t}{(t-1)^{3/2}}\exp\left[-\frac{... | {
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Value of $(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}$ if $\beta$ is the root of $x^3-x-1=0$
If $\beta$ is the root of the equation $x^3-x-1=0$, find the value of
$$(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}.$$
This is what I tried:
$x=\beta$ is a root of $x^3-x-1=0,$
s... | I assume that $\beta$ is the real root of $x^3-x-1=0$ so that the cube roots are well defined.
Let $\mu = 3\beta^2-4\beta$ and $\nu = 3\beta^2+4\beta+2$. We seek $\tau = \mu^{\frac{1}{3}}+\nu^{\frac{1}{3}}$. As in this question, we have
$$
\tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}}\tau
$$
Hoping that $\mu\nu = \beta-\bet... | {
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"answer_id": 1
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$\cos(x)=-\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$. Calculate $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$. If $\cos(x)= -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ a... | You can observe that
$$
\lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25}
$$
Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$.
Also
$$
\lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41}
$$
and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus
$$
\sin y=\tan y\cos y=-9/41
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I... | Rewrite expression as
$$
\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}
$$
so we have
$$
x^{2014} = Q(x) (x-1)^2 + ax + b.
$$
Now you need to find coefficients $a$ and $b$. Your argument with derivatives should work:
$$
1^{2014} = 1 = a + b
$$
$$
2014\cdot 1^{2013} = 2014 = a.
$$
Hence, $b = -2013$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Find maxima and minima of $f(x) = x \cot x$ So I got a function
$$f(x)= x \cot x $$
I would like to find values of $x$ where $f'(x) = 0$
Applying product rule, we get:
$$f'(x) = \cot x - x \cdot csc^2 x $$
Setting equation to zero
$$\cot x - x \cdot csc^2 x = 0 $$
Now, I will try my best at simplifying equation above... | Consider $g(x) = \cos x \sin x -x$, then $g(0)=0$ and $g(x)$ is monotone because $g'(x) = -2\sin^2(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$ I was screwing around lately in GeoGebra and I realized something.
Draw a $\triangle ABC$, and let the bisectors for $\angle A$ and $\angle C$ meet sides $BC$ and $AB$ at po... | Let $AB = c$, $BC = a$, $AC = b$. We will prove that if $EF = FD$, then $ABC$ is either isosceles or $\angle ABC = 60$.
We will first show that $\frac{DF}{FC} = \frac{c}{a+b}$. First, notice that by the angle bisector theorem, we have that $\frac{DF}{FC} = \frac{BD}{BC} = \frac{BD}{a}$. Now, note that $\frac{BD}{c - BD... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
$\pi \cot\pi z\ne -\frac{1}{z}+\lim_{N\to\infty}\sum_{n=-N}^N\frac{1}{z+n}$ I know that
$$\pi\cot\pi z=\frac{1}{z}+\displaystyle\underbrace{\sum_{n=1}^\infty \left(\frac{1}{z-n}+\frac{1}{z+n}\right)}_{S}.$$
Then
$$\begin{align*}S&=\cdots +\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z-2}+\frac{1}{z+2}+\cdots\\&=\cdots +\frac{1... | Note that$$\sum_{n=-N}^N\frac1{z-n}=\frac1{z-N}+\cdots+\frac1{z-1}+\frac1z+\frac1{z+1}+\cdots+\frac1{z+N},$$instead of$$\sum_{n=-N}^N\frac1{z-n}=\frac1{z-N}+\cdots+\frac1{z-1}+\frac1z+\frac1z+\frac1{z+1}+\cdots+\frac1{z+N},$$which is what you wrote.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Formula for calculating the odds per user of winning in a raffle each player can win once Am trying to write a program which gives each user in a raffle contest their odds so far at winning.
The rules of the game are simple:
A predefined number of tickets are sold for example 1000 each user can at most buy 50 tickets ... | Let $N_i$ be the number of tickets that the $i$-th player bought. Then the total number of possible draws of $m$ tickets is the coefficient of $x^m$ in the polynomial
$$
P(x)=\prod_i\left(1-N_i x\right);
$$
the number of possible draws where the $j$-th player doesn't win is the coefficient of $x^m$ in
$$
Q_j(x)=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7},\cos\frac{5 \pi}{7}$
Prove that solution of the equation $8x^3-4x^2-4x+1= 0$ has roots $\cos\frac{\pi}{7},\cos\frac{3 \pi}{7}\space \text{and}\space \cos\frac{5 \pi}{7}$.
How to even solve it ? I have no idea.
B... | $$\cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7}=\frac{8\sin\frac{\pi}{7}\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}}{8\sin\frac{\pi}{7}}=\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}=-\frac{1}{8}.$$
The rest is a similar and use the Viete's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $5^n - 1$ is divisible by $4$ by mathematical induction. I have done it, but I am not sure that the inductive step is right. Can anybody please clear me about it?
Basic steps as:
Taking $n=1$: $p(1)=5-1=4$.
Inductive hypothesis: Assume the statement is true for $p(k)$. $5^k - 1$ is divisible by $4$.
Ind... | The idea is $$5^{k+1} - 1 = 5\cdot 5^{k} - 1 = 4\cdot(5^k) + (5^k - 1)$$
We have $4\cdot(5^k)$ is obviously divisible by $4$ and $5^k - 1$ is divisible by $4$ by inductive hypothesis
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$ How to prove
$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx\tag{1}$$
Here is how I came up with this relation:
In this solution @Kemono Chen elegantly proved
$$\... | In the post body we have
$$\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{*}$$
and from this solution we have
\begin{align}
f(a)&=\frac12\arctan a\ln(1+a^2)=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n}}{2n+1}a^{2n+1}\\
&=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n+1}}{2n+1}a^{2n+1}+\sum_{n=0}^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\... | Method 3. Translate the origin to the point $(\color{red}{\frac32,-\frac{27}{2}})$ and find the coordinates of the point ${\color{blue}{4\choose -13}}$ in the new system:
$${x'\choose y'}={x\color{red}{-\frac32}\choose y\color{red}{+\frac{27}{2}}}={\color{blue}4-\frac32\choose \color{blue}{-13}+\frac{27}{2}}={\frac52 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Odd convergents of $\sqrt 2$ and Pythagorean triangles with consecutive legs I would like to prove the following:
Take any odd convergent of $\sqrt 2$. The denominator gives the hypotenuse of a triangle; the numerator split into two consecutive integers gives the other two sides.
For example, the first $10$ convergen... | Consecutive Legs
In this answer it is shown that
Theorem: Let $m$ and $n$ be positive integers so that
$$
\begin{align}
&m\gt n\\
&m+n\text{ is odd}\\
&m\text{ and }n\text{ are relatively prime}
\end{align}
$$
Then,
$$
\begin{align}
a &= m^2 - n^2\\
b &= 2mn\\
c &= m^2 + n^2
\end{align}
$$
gives all positive,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it:
https://imgur.c... | Since the left side does not depend on changing signs of our variables, it's enough to prove our inequality for non-negatives $a$ and $b$.
Now, by C-S $$4(a^6+b^6)=2\cdot(1^2+1^2)(a^6+b^6)\geq2(a^3+b^3)^2.$$
Thus, it's enough to prove that:
$$2(a^3+b^3)\geq(a+b)(a^2+b^2)$$ or
$$2(a^2-ab+b^2)\geq a^2+b^2$$ or
$$(a-b)^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solutions to a quadratic given 1 solution in form a+bi I was just really confused as to how I only ended up with 1 of 2 answers for the following question.
Given that $-2+bi$ is a solution of $x^2+ax+(3+a) $ find constants $a$ and $b$ given that they are real.
As soon as I saw that $-2+bi$ was a solution, I immediately... | A brute force approach is to substitute $x=-2+bi$ into $x^2+ax+(3+a)$. Then
\begin{align}
x^2+ax+(3+a)
&= (-2+bi)^2+a(-2+bi)+(3+a) \\
&=(7-a-b^2)+b(a-4)i \\
\end{align}
where I have organized the last expression into real and imaginary parts (since $a,b$ are assumed real). If this last expression is to vanish as desir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
how do you differentiate x^(3/4) using first principle $$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac{3}{4}}-(x)^{\frac{3}{4}}\Bigr)}{h}$$
I understand the process till
$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac34}-(x)^{\frac{3}{4}}\Bigr)}{h} * \frac{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}{\Bigl((x+h)^{\frac{3}{4}}+(... | \begin{align*}
\lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}&=\lim_{h\to0}\frac{h^3+3h^2x+3hx^2}{h}\lim_{h\to0}\frac1{(x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}}\lim_{h\to0}\frac{1}{(x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given $a^4+1=\frac{a^2}{b^2}\left(4b^2-b^4-1\right)$, find $a^4+b^4$
Given $$a^4+1=\frac{a^2}{b^2}\left(4b^2-b^4-1\right)$$
Find $$a^4+b^4$$
My approach:
*
*Verify for different values of $a$ and $b$ to see that question is correct and yes, it turns out to be $2$ always for different values of $a$ and $b$.
*Tr... | It is not true that we always have $a^4+b^4=2$. We can rewrite the equation as
$$
\left( a-\frac{1}{a}\right)^2+\left( b-\frac{1}{b}\right)^2=0.
$$
Now chose $a$ such that $a-1/a=i$ and $b-1/b=1$, i.e., with $a := \sqrt{ - \sqrt{3}i + 1}/\sqrt{2}$ and $b=(\sqrt{5} + 1)/2$. This comes from solving the quadratic equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\d... | Note that: if $a,b,c \in \mathbb{R}$ and $a\ne0$, if $ax^2+bx+c=0$, then $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$2x^2-\sqrt{3}x-1=0$,
solving, we get: $x_{1,2}=\frac{\sqrt{3}\pm\sqrt{3-4\cdot2\cdot(-1)}}{2\cdot2}=\frac{\sqrt{3}\pm\sqrt{11}}{4}$
The required expression, $x_1^2+x_2^2=(\frac{\sqrt{3}+\sqrt{11}}{4})^2+(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{... | Let us try another way:
$xy=1$
$x+y=\dfrac{(3+\sqrt{5})^2+(3-\sqrt {5})^2}{9-5}=\dfrac{2(9+5)}{4}=7$
$x^3+y^3=(x+y)^3-3xy(x+y)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the method to solve this system of equations? $\Huge{\text{Updated:}}$
In this question I got an answer for special non-zero values of $A$ and $B.$ I need to solve these polynomial equations for any arbitrary non-zero coefficients.
$\underline{\text{ I am looking for a solution that does not lead to the solut... | You can still eliminate. From $(1)$ we have
$$
y = \frac{ - 3x^2 - A(u^2 + vz)}{3},
$$
and then from $(3)$ we have
$$
x = \frac{uz( - 2A - 3Bz)}{Az^2 + 3}.
$$
Actually, the case of $Az^2+3=0$ leads to $Bz^3-2=0$.
Otherwise we have only two equations left in three variables $z,u,v$, namely $(2),(4)$ and can take the res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos... | I don't know if this counts as a proof, but following your suggestion, I used the cosine laws to obtain your result.
Suppose you have a triangle ABC as in the figure:
Since the angles $A+B+C=\pi$, these are the internal angles of a general triangle.
Using the law of cosines, you can write:
$a^2=b^2+c^2-2bc\cos A\\b^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 2
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Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | $$x^{81} + x^{49} + x^{25} + x^{9} + x=\left(x^{81}-x\right)+\left(x^{49}-x\right)+\left(x^{25}-x\right)+\left(x^9-x\right)+5x.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
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What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$?
If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:
$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.
What is the correct way... | If $7$ worked we would have
\begin{align}
&\frac{21}{30}<\frac{p}{7}< \frac{22}{30}\\
\iff&\frac{21*7}{30*7}<\frac{30p}{30*7}< \frac{22*7}{30*7}
\end{align}
So $7$ is a solution if and only if there is a multiple of $30$ between $21*7$ and $22*7$. Since $21*7=147$ and $22*7=154$ we have a multiple of $30,$ namely $15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
| By AM_GM we get
\begin{align*}
\frac{2a^4+b^4+c^4}{4} & \geq a^2bc\\
\frac{2b^4+a^4+c^4}{4} & \geq b^2ac\\
\frac{2c^4+b^4+a^4}{4} & \geq c^2ab
\end{align*}
Now add these to get your inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove this inequality via weighted Jensen inequality The following inequality is derived from this difficult olympiad problem.
Suppose $a,b,c$ are three positive real numbers such that $abc = 8$, show that
$$\frac{1}{\sqrt{a + 1}} + \frac{1}{\sqrt{b + 1}} + \frac{1}{\sqrt{c + 1}} < 2$$
My idea: Note that $\sqrt{x}$ is ... | The Contradiction method works!
Let $\frac{1}{\sqrt{a+1}}=p,$ $\frac{1}{\sqrt{b+1}}=q$ and $\frac{1}{\sqrt{c+1}}=r.$
Thus, $\{p,q,r\}\subset(0,1),$ $\frac{(1-p^2)(1-q^2)(1-r^2)}{p^2q^2r^2}=8$ and we need to prove that:
$$p+q+r<2.$$
Indeed, let $p+q+r\geq2,$ $r=kr'$ such that $k>0$ and $p+q+r'=2$.
Thus, $$p+q+kr'\geq2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Determining the radius of convergence of a series that I don't know how to express in sigma Question:
Problem 1: Consider the ODE: $(1+x^3)y'' -6xy = 0$
(a) Compute the first $3$ nonzero terms of power series expansion about $x = 0$ for two linearly independent solutions.
(b) Use the ratio test to determine the radiu... | You should get a coefficient recursion ($a_{-1}=a_{-2}=0$)
$$
x^n: (n+2)(n+1)a_{n+2}+(n-1)(n-2)a_{n-1}-6a_{n-1}=0
\\~\\\iff (n+1)[(n+2)a_{n+2}+(n-4)a_{n-1}]=0
\\~\\\implies
a_{n+3}=-\frac{n-3}{n+3}a_n.
$$
so that $a_2=0$ and thus $a_{3k+2}=0$ for all $k$ and also $a_6=0$ and thus $a_{3k}=0$ for $k\ge 2$.
The remaining... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On the equation $\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x$ I'm trying to solve the equation
$$\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x \tag{1}$$
Attempt: I have rewritten the equation as
\begin{align*}
\sqrt{x^2-9} = \frac{2(x+3)}{\left (x-3 \right )^2}-x &\Leftrightarrow \sqrt{x^2-9} = \frac{2(x+3)-x\left ( x-3 \right )^2}... | Hint:
Let $\sqrt{x^2-9}=3\tan2t\ge0,0\le2t<\dfrac\pi2$
$x=3\sec2t$
$$3(\tan2t+\sec2t)=\dfrac{6(\sec2t+1)}{9(\sec2t-1)^2}$$
$$3(\sin2t+1)=\dfrac{2(1+\cos2t)}{3(1-\cos2t)^2}$$
Set $\tan t=u$
$$\dfrac{9(1+u)^2}{1+u^2}=\dfrac{4(1+u^2)^2}{4(1+u^2)u^4}$$
$$\iff9(1+u)^2u^4=(1+u^2)^2$$
As $u=\tan t\ge0$
$$\implies3(1+u)u^2=1+u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS?
$$\frac{d}{dt}\left(\frac{m \ve... | Leave $m$ out and start with,
$$\frac{\vec{u}\cdot\vec{u}} {\sqrt{1-\frac{u^2}{c^2}}}=\frac{u^2} {\sqrt{1-\frac{u^2}{c^2}}}$$
Take derivatives on both sides,
$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} +\frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}
=\left[ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Convergence at the boundary of $\sum_{n= 1}^\infty\frac{(n!)^3}{(3n)!}z^n $ We are given the complex series of functions $\sum_{n= 1}^\infty\frac{(n!)^3}{(3n)!}z^n $. It is a power series, and we can easily calculate that its disk of convergence is $D(0,27)$. How can we know if the series converges at some point of the... | Take some point on the boundary and then try to figure out whether the series you get by plugging in that point for $z$ converges. In the complex case you will mostly work with estimations then as you have infinitely many boundary points. Lets do it for your example:
Let $z \in \mathbb{C}$ with $\vert z \vert = 27$, i.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding $f\in\mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $\deg(f)\leq 3$. What's wrong with my approach?
I'd like to find a polynomial $f(x) \in \mathbb{Q}[x]$ satisfying
$$f(\sqrt{2}+\sqrt{3})=\sqrt{2}$$
and $\deg(f) \leq 3$.
What I've been trying is the following:
Since $f(\sqrt{2}+\sqrt{3})=\s... | Let $e = g + \sqrt{3}h$, $f = j + \sqrt{3}k$.
$$\begin{eqnarray}(x^2 -2)(ex+f) &=& a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x \\
(g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& a(x^3 + 3\sqrt{3}x^2 + 9x + 3\sqrt{3}) +b(x^2 + 2\sqrt{3}x + 3) + cx + \sqrt{3}c +d - x \\
(g + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How many incongruent solutions are there for $x^{2}\equiv49\:\left(10^{6}\right)$? How many solutions are there for $x^{2}\equiv49\:\left(10^{6}\right)$
?
I can see that $x=\pm7$ are two solutions, and I guess that $x$
is a solution to the given congruence iff $\left(\frac{x}{7}\right)^{2}\equiv1\:\left(10^{6}\right)$.... | Let $d=\gcd(x-7,x+7)$ then $d\mid (x+7)-(x-7)=14$ so $d=2$ or $d=1$ since $$2^6\cdot 5^6\mid (x-7)(x+7)$$ So we have this possibilities:
*
*$x-7=5^6a$ and $x+7 = 2^6b$ where $a,b$ are relatively prime
*$x-7=2\cdot 5^6a$ and $x+7 = 2^5b$ where $a,b$ are relatively prime
*$x+7=2\cdot 5^6a$ and $x-7 = 2^5b$ where $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?
$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$
I used
$$\sin... | The identity can also be proved from: $\quad\cos36° = \cos 72° + {1\over2}$
$\cos 72° + {1\over2}= (2\cos^2 36° - 1)+ {1\over2}\;= \large{\phi^2\over2}-{1\over2} ={\phi \over 2}=\cos36°$
$T= \Large \frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}$
$\Large{1\over T} = {1\over 2\cos 6° \cos 12°} + \normalsize \tan 12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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How to solve $(x^2 - 11x + 29)^{(6x^2 + x - 2)}=1$? This question comes from PURE MATHEMATICS 1 for As and A levels. This question is part of exercise 1 and it has 5 Answers : $1/2, - 2/3, 4, 6 $ and $7$ . The first 2 values $1/2$ and $-
2/3 $ I am able to find but the rest I can't. I get this first 2 values by make 1... | By inspection, $$a^b=1 \Rightarrow a=1\textrm{ or }b=0 \textrm{ or } a=-1 \textrm{ if } b \textrm{ is even}$$
So we can just solve three different cases.
Case 1: $a=1$.
\begin{align}x^2-11x+29&=1\\
x^2-11x+28&=0\\
(x-4)(x-7)&=0\\
x_1&=4\\
x_2&=7\end{align}
Case 2: $b=0$.
\begin{align}
6x^2+x-2&=0\\
(3x+2)(2x-1)&=0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Factoring $3x^2+4x-4=0$ using the quadratic formula The correct answer is $(3x-2)(x+2)$ but I am getting $(x-\frac{2}{3})(x+2)$, why?
My calculations:
$x = \frac{-4\pm\sqrt{16-(-48)}}{6} = \frac{-4\pm8}{6}$, which gives the factors $(x-\frac{4}{6})(x+\frac{12}{6}) = (x-\frac{2}{3})(x+2)$.
| Because if the roots of $ax^2+bx+c$ are $r_1$ and $r_2$, then$$ax^2+bx+c=a(x-r_1)(x-r_2).$$You forgot the $a$.
In your case, you get:\begin{align}3x^2+4x-4&=3\left(x-\frac23\right)(x+2)\\&=(3x-2)(x+2).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Let $t$ be a positive integer such that $2^t = a^b ±1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$? Let $t$ be a positive integer such that $2^t = a^b ±1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$?
I found that $t=3$ i... | As mentioned in comments, this is a special case of Catalan's conjecture, now Mihăilescu's theorem, but this special case can be solved on its own without too much suffering.
If $b$ is even, then by letting $c = a^{b/2}$, we get either $2^t = c^2 + 1$ or $2^t = c^2 - 1$.
*
*The first case is eliminated modulo $4$: ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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limit of a n-square root and series of exponents Again, I'm having trouble with the infinite limits:
$$ (1) .... \lim_{n \to \infty} \sqrt[n]{ a^n+b^n } $$ with $a,b$ positive reals.
and to show if the following series is divergent or convergent
$$ (2) ......\sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n} $$
To be ho... | For the first part: Without loss of generality let $a \ge b$. Then by binomial expansion
$$
\sqrt[n]{ a^n+b^n } = a(1 + (b/a)^n)^{1/n} = a\bigg(1 + \frac{(b/a)^n}{n} + ...\bigg)
$$
Since $b/a <1$ each term after the first vanishes when $n \to \infty$. Hence
$$
\lim_{n \to \infty}\sqrt[n]{ a^n+b^n } = \max(a,b)
$$
For t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}... | $$\sqrt{a(2^{x}-2)+1}=1-2^{x}\tag1$$
First of all, we have to have
$$a(2^{x}-2)+1\ge 0\qquad\text{and}\qquad 1-2^x\ge 0\tag2$$
Under $(2)$, we have
$$\begin{align}(1)&\implies a(2^x-2)+1=(1-2^x)^2
\\\\&\implies a(2^x-2)+1=1-2^{x+1}+2^{2x}
\\\\&\implies a\cdot 2^x-2a=2^{2x}-2\cdot 2^{x}
\\\\&\implies 2^{2x}+2^x(-2-a)+2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be ... | There are two square root of $-i$ namely, $$ \pm\frac {\sqrt 2}{2} (1-i)$$
You may verify it by squaring each one $$ (\pm\frac {\sqrt 2}{2} (1-i))^2= \frac {1}{2}(1-2i-1)=-i$$
One way to find them is of course the polar form of complex number $-i=e^{3i \pi /2}$ as mentioned in Kevin's answer.
The other way is looking ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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On the proof of $2^b-1$ does not divide $2^a+1$.
Proposition. For any integer $a$ and $b$ where $b>2$, we never have $2^b-1|2^a+1$.
WLOG let $b<a$. The proof uses a critical fact that can be stated very simply as follows:
$2^a+1$ leaves a reminder of the form $2^r+1$ when divided by $2^b-1$, where $r\leq b$.
This a... | Claim: If $a,b,r, n\in \mathbb N$ if $a \equiv r \pmod b$ then $n^a \equiv n^r\pmod{n^b -1}$.
Pf: wolog $r < a$ so there is a $k > 0; k\in \mathbb N$ so that $a= kb + r$.
So $n^a = n^{a-b}(n^b-1)+ n^{a-b}\equiv n^{a-b}\pmod{n^b-1}$.
and by induction, for any $a-kb \ge b$
$n^a \equiv n^{a-(k-1)b} = n^{a-kb}(n^b-1) + n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Is $\sum_{n=1}^{\infty} 1 = -\frac{3}{12}$ true? Here is how I derived this...
$$1+2+3+4+...=-\frac{1}{12}$$
$$2+4+6+8+...=2(1+2+3+4+...)=2(-\frac{1}{12})=-\frac{1}{6}$$
Thus $1+3+5+7+...=-\frac{1}{6}-(1+1+1+...) $ because $S_{odd}=S_{even}-(1+1+1+...)$
Also $S_{odd}=(1+2+3+4+...)-(2+4+6+8+...)=-\frac{1}{12}+\frac{1}{... | First of all, you have to understand the following point:
There is no single canonical way of defining 'summing infinitely many things'.
As a result, there are several different notions of infinite sums, some of which are not even compatible to each other. For instance, the followings are some selected summability me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
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A property of Nesbitt's inequality I think it's not new however I think it's interesting :
Let $a,b,c>0$ and $x>0$ then the function :
$$f(x)=\frac{a^x}{b^x+c^x}+\frac{b^x}{a^x+c^x}+\frac{c^x}{b^x+a^x}$$
$f(x)$ is increasing
My first try :
The derivative of $f(x)$ is :
$$f'(x)=\sum_{cyc}\frac{a^x\ln(a)(b^x+c^x)-a... | Alternative solution:
WLOG, assume $a \ge b \ge c = 1$. Denote $a^x = u, b^x = v$. Then $u \ge v \ge 1$.
We have
\begin{align*}
f'(x) &= A \ln a + B\ln b \\
&= A\ln \frac{a}{b} + (A + B) \ln b
\end{align*}
where
\begin{align*}
A &= {\frac {u}{v+1}} - {\frac {vu}{ \left( 1+u \right) ^{2}}} -{\frac {u}{
\left( u+v \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Master Theorem: $T(n) = 3T(\frac{n}{3} − 2) + \frac{n}{2}$ I need to solve this recurrence using the Master Theorem; however, I don't know if its is possible since it doesn't follow the format
$T(n) = 3T(\frac{n}{3} − 2) + \frac{n}{2}$
$a = 3$
$b = 3$?
What happens to the -2 inside the recurrence?
| Here's a start.
In
$T(n) = 3T(\frac{n}{3} − 2) + \frac{n}{2}
$,
replace $n$ by $n+c$.
I will choose $c$
so there is no offset.
$T(n+c)
= 3T(\frac{n+c}{3} − 2) + \frac{n+c}{2}
= 3T(\frac{n}{3}+\frac{c}{3} − 2) + \frac{n+c}{2}
$
so if
$c = c/3-2
$,
or
$c=-3$,
this becomes
$T(n-3)
= 3T(\frac{n}{3}-3) + \frac{n-3}{2}
$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find maximum and minimum of : $\Omega=\sin x\sin y\sin z$ Question :
Let : $x,y,z>0$ and $x+y+z=\frac{π}{2}$
Compute the maximum and minimum of the
following expansion :
$\Omega=\sin x\sin y\sin z$
My attempt :
About maximum :
$\Omega=\sin x\sin y\cos (x+y)=\frac{1}{2}\sin x(\sin (2x+y)-\sin y)$
$≤\sin y(1-\si... | For $\Omega= (\sin x \sin y \sin z)$ by GM-AM
$$(\sin x \sin y \sin z)^{1/3} \le \frac{\sin x+\sin y+ \sin z}{3}\le \sin \frac{(x+y+z)}{3}=\frac{1}{2}.$$ The second inequality is due to Jensens Inequality for $x \in (0,\pi/2).$ as $f''(x)<0.$ So finally
$$0<\sin x \sin y \sin z \le \frac{1}{8},~ x,y,z >0, ~ x+y+z=\pi/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of a limit of a sequence $\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}$ I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$
Here there are my diff... | Hint:
$$\begin{align}
{\sqrt{4n^2+1}-2n\over\sqrt{n^2-1}-n}
&={\sqrt{4n^2+1}-2n\over\sqrt{n^2-1}-n}\cdot{\sqrt{4n^2+1}+2n\over\sqrt{4n^2+1}+2n}\cdot{\sqrt{n^2-1}+n\over\sqrt{n^2-1}+n}\\
&={(4n^2+1)-4n^2\over(n^2-1)-n^2}{\sqrt{n^2-1}+n\over\sqrt{4n^2+1}+2n}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$ Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$
In my attempt I used $\xi = \cos(72k) + i\sin(72k)$ with $k=1,2,3,4$
Where I got $\zeta = \f... | By definition, $\xi^5 = 1$ and we have the factorization $$\prod_{k=0}^4 (z - \xi^k) = z^5 - 1.$$ But since $(z - \xi^0) = (z - 1)$, it follows that $$\prod_{k=1}^4 (z - \xi^k) = \frac{z^5 - 1}{z-1} = 1 + z + z^2 + z^3 + z^4,$$ hence $$1 + \xi + \xi^2 + \xi^3 + \xi^4 = 0.$$ Dividing by $\xi^2$ yields $$\begin{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Geometry - Incircle I of $\triangle$ABC with chord MN intersecting AB at P. If BP = MC, find $\angle$AIC I recently found the following problem:
Circle with center $I$ is inscribed in triangle $ABC$ and touches the sides $A$C and $BC$ in points $M$ and $N$. The line $MN$ intersect the line $AB$ at $P$, as $B$ is betwe... |
Let the side lengths of the triangle ABC be $a$, $b$ and $c$. Apply the sine rule to the triangles PBN and AIC,
$$\frac{\sin\alpha}{\sin\beta} = \frac{BN}{BP} = \frac{BN}{MC}
=\frac {\frac12 (a+c-b)}{{\frac12 (a+b-c)}} =\frac {a+c-b}{{a+b-c}}\tag{1}$$
$$\frac{\sin\alpha}{\sin(180-\beta)} = \frac{AM}{AP}= \frac{AM}{AB+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating function for non-isomorphic regular graphs. Determine a generating function for the number of non-isomorphic (n−2)-regular graphs of order n, for n ≥ 2.
I've been staring at this for hours and can't find a place to start, any help would be appreciated.
| We enumerate the complements, namely non-isomorphic 2-regular graphs.
These are sets of cycles and we find
$$\prod_{k\ge 3} \frac{1}{1-z^k}
= (1-z)(1-z^2) \prod_{k\ge 1} \frac{1}{1-z^k}.$$
The term $1/(1-z^k)$ is the OGF of zero, one, two, three
etc. instances of a cycle of order $k.$
This is the OGF
$$(... | {
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"question_score": "3",
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"answer_id": 0
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Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5... | Consider $z^5-1=0$
So the roots of $$0=\dfrac{z^5-1}{z-1}=z^4+z^3+z^2+z+1$$ are $e^{2i\pi r/5},r=1,2,3,4$
As $z\ne0,$ like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ , divide both sides by $z^2$
Replace $z+\dfrac1z=w\implies w^2=?$ to find $$w^2-2+w+1=0$$ whose roots a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
When is $\small \gcd(2n^7\!+1,{3n^3\!+2})>1,$ i.e. when is $\frac{2n^{7}+1}{3n^{3}+2}$ reducible? How can I ind the values of $n\in \mathbb{N}$ that make the fraction $\frac{2n^{7}+1}{3n^{3}+2}$ reducible ?
I don't know any ideas or hints how I solve this question.
I think we must be writte $2n^{7}+1=k(3n^{3}+2)$ with... | Suppose $p$ is prime.
If $p$ divides $2n^7+1$ & $3n^3+2$
then $p$ divides $2(2n^7+1)-(3n^3+2)=n^3(4n^4-3)$
then $p$ divides $4n^4-3$ ( See Footnote )
then $p$ divides $2(4n^4-3)+3(3n^3+2)= n^3(8n+9)$
then $p$ divides $8n+9$ (See Footnote)
then $p$ divides $9(3n^3+2)-2(8n+9)=n(27n^2-16)$
then $p$ divides $27n^2-16$ (Se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find $x,y,z$ for the given conditions $$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, whic... | Continuing your second method.
Denote: $a=2x,b=5y,c=3z$, then:
$$\begin{cases}4x^2+25y^2+9z^2-10xy-15yz-6zx=0\\
x+y+z=5\end{cases} \Rightarrow \\
\begin{cases}a^2+b^2+c^2-ab-bc-ca=0\\
\frac a2+\frac b5+\frac c3=5\end{cases} \Rightarrow \\
2a^2+2b^2+2c^2-2ab-2bc-2ca=0 \Rightarrow \\
(a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3389836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\... | You correctly computed
$$
(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2
$$
only that this is not what was asked for.
But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore
$$
(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i)
$$
which expands to
$$
(1-ip-q+ir+s)(1+ip-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as
$(0, -7)$; no $x$ intercepts.
I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3... | The graph of the function $f(x) = -3|x - 2| - 1$ has vertex $(2, -1)$ and opens downward since the coefficient of $|x - 2|$ is negative. That tells you that the graph of the function cannot intersect the $x$-axis.
Remember that $|x|$ means the distance of the number $x$ from $0$. Therefore, $|x - 2|$ means the distan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Is this a valid way to prove that a sequence $\{a_n\}$ given by $a_n=\frac{1}{\sqrt{n+3}} + \frac{n-3n^2}{2n^2-1}$ converges? I'll be using the Archimedean Property to prove this.
Proof
Let $\epsilon>0$. Then $\epsilon^2/4>0$.
By the Archimedean Property, $\forall \epsilon^2/4>0, \exists N\in \mathbb{N}$ such that $1/N... | I don't see why you need that
$\epsilon^2/4$ statement.
Once you have shown that
$|a_n+\frac{3}{2}|<\frac{2}{\sqrt{n}}
$,
to make
$|a_n+\frac{3}{2}|
\le \epsilon$,
just choose
$n$ so that
$\frac{2}{\sqrt{n}}
\lt \epsilon$
or
$n > 4/\epsilon^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary
Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary.
My attempt is as follows:-
Let $t=\frac{1}{1+x^2}$, and le... | Hint: Factorizing your equation we get
$$x^4+x^2(2-3a)+1-2a=0$$
You can also write $$at^2-3at+1=0$$ where $t=\frac{1}{1+x^2}>0$
$a=0$ is impossible, so we get by the quadratic formula
$$t_{1,2}=-\frac{3}{2}\pm\sqrt{\frac{9}{4}-\frac{1}{a}}$$
If $$\frac{9}{4}-\frac{1}{a}<0$$ then the roots are imaginary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Taylor series of $\sinh{(x)}$ at $\ln{(2)}$. Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
Equating each derivative at $x = \ln{(2)}$ gives:
\begin{equation*}
\begin{split}
f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f''(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(3)}(\l... | Your computation leads naturally to:
$$\sinh x=\frac54\sum_{n=0}^{\infty} \dfrac{(x-\log2)^{2n+1}}{(2n+1)!}
+\frac34\sum_{n=0}^{\infty} \dfrac{(x-\log2)^{2n}}{(2n)!}\\
=\sum_{n=0}^{\infty} \left(1-\dfrac{(-1)^n}4\right)\dfrac{(x-\log2)^{n}}{n!}$$
The radius of convergence is infinite.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to write this as a multiplication: $4 x^2 y^2 - (x^2 + y^2 - z^2)^2$? I have the following polynomial:
$$ 4 x^2 y^2 - (x^2 + y^2 - z^2)^2$$
(which comes up, for example, in computing the area of a triangle using the cosine law).
I would like to convert this to a product.
Wolfram tells me it's
$$ -(x - y - z) (x ... | Note that you have a binomial formula:
$a^2-b^2=(a-b)(a+b)$
Where $a=2xy$ since $a^2=4x^2y^2$ and $b=x^2+y^2-z^2$ obviously.
This leads to $4x^2y^2-(x^2+y^2-z^2)^2=(2x^2y^2-x^2-y^2+z^2)(2x^2y^2+x^2+y^2-z^2)$
With additional use of binomial formulas you are able to factor this term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Algebraic proof of a combinatoric question (Combinatoric proof is given) I had a IMO training about double counting. Then, there is a problem which I hope there is a combinatoric proof. Here comes the problem:
For every positive integer $n$, let $f\left(n\right)$ be the number of all positive integers with exactly $2... | Here is a slightly different proof that
$$\sum_{k=0}^{\lfloor n/2\rfloor} {n\choose 2k} {2k\choose k}
2^{n-2k} = {2n\choose n}.$$
We observe that
$${n\choose 2k} {2k\choose k}
= \frac{n!}{(n-2k)! \times k! \times k!}
= {n\choose k} {n-k\choose n-2k}.$$
We get for our sum
$$\sum_{k=0}^{\lfloor n/2\rfloor} {n\choose k}
{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
| A similar approach to @Quanto's is to define $I_n:=\int_0^1\frac{x^ndx}{1+x^2}$ so $I_0=\frac{\pi}{4}$ and $I_n+I_{n+2}=\frac{1}{n+1}$, and in particular $I_n-I_{n+4}=\frac{2}{(n+1)(n+3)}$. Hence $I_4=\frac{\pi}{4}-\frac{2}{3}$ and$$I_8+I_4-4(I_5+I_7)+6I_6=\underbrace{-\frac{2}{35}}_{I_8-I_4}+6\cdot\underbrace{\frac15}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
He... | $n>1$;
$a_n=\dfrac{2^n+5^n}{3^n+4^n} \gt$
$\dfrac{5^n}{4^n+4^n}=(1/2)(5/4)^n >$
$(1/2)(1+1/4)^n> (1/2)(1/4)n.$
Used : Binomial expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatri... | You didn't go wrong anywhere. You showed what the problem wanted you to show, and then some.
Note that some times the restrictions given in problems and exercises are necessary, and thus not using them means you're wrong somewhere. So you are right to be sceptical. But in this specific case the restriction $0\leq \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A homeomorphism of the unit disk that cannot be extended to the boundary of its domain I have a problem with the following exercise:
Let $D^2$ be the unit disc and S^1 be the unit circle. Show that the function
$
h: {D^2\setminus{S^1}} \to {D^2\setminus{S^1}} \\
h(re^{it})=
\begin{cases}
0 &\text{if}\, r=0 \\
r ... | Observe
$\quad \frac{2 \pi r}{1-r} = 1 \pi \text{ implies } r = \frac{1}{3}$
$\quad \frac{2 \pi r}{1-r} = 3 \pi \text{ implies } r = \frac{3}{5}$
$\quad \frac{2 \pi r}{1-r} = 5 \pi \text{ implies } r = \frac{5}{7}$
$\quad \frac{2 \pi r}{1-r} = 7 \pi \text{ implies } r = \frac{7}{9}$
etc.
Observe
$\quad \frac{2 \pi r}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\int \frac{\sin x}{1+2\sin x}dx$ calculate:
$$\int \frac{\sin x}{1+2\sin x}dx$$
I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification:
$$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$
and I am not able to calulate that.
| Since $u=\tan\frac{x}{2}$ gives$$\int\frac{dx}{1+2\sin x}=\int\frac{2 du}{(u+2)^2-3}=\frac{-2}{\sqrt{3}}\operatorname{artanh}\frac{u+2}{\sqrt{3}}+C=\frac{-1}{\sqrt{3}}\ln\frac{\sqrt{3}+2+\tan\frac{x}{2}}{\sqrt{3}-2-\tan\frac{x}{2}}+C,$$we have$$\int\frac{\sin xdx}{1+2\sin x}=\frac12x+\frac{1}{2\sqrt{3}}\ln\frac{\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $x$ so that rational function is an integer Find all rational values of $x$ such that $$\frac{x^2-4x+4}{x^2+x-6}$$ is an integer.
How I attempt to solve this: rewrite as $x^2-4x+4=q(x)(x^2+x-6)+r(x)$. If we require that $r(x)$ be an integer then we can get some values of $r(x)$ by solving $x^2+x-6=0$, so that $x=-... | $\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}$, for this to be an integer, $\frac{5}{x+3}$ has to be an integer.
Say, $\frac{5}{x+3}=K$, where $K \in \Bbb Z$. This implies for each $K \in \Bbb Z$, $x=\frac{5}{K}-3$ would make the given expression an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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What is the limit of $\lim_{n\to +\infty} n\left(\frac{e}{\left(1+\frac{1}{n}\right)^n}-1\right)$ I have read this post Solve the following limit: $\lim_{n->\infty} n(\frac{\frac{1}{n!}(\frac{n}{e})^n}{\frac{1}{(n+1)!}(\frac{n+1}{e})^{n+1}}-1)$ but I don't understand how do you get from $\log\left(1+\tfrac{1}{n}\right)... | Let
$$a_n=\left(1+\frac{1}{n}\right)^n\implies \log(a_n)=n \log\left(1+\frac{1}{n}\right)$$ Now, using Taylor series
$$\log(a_n)=n \left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor series
$$a_n=e^{\log(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$ Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$
My attempt is as follows:
As $x>1,y>1$ , so $\lo... | Let $\ln{x}=a$ and $\ln{y}=b$.
Thus, $a$ and $b$ are positives, $$a^2+b^2=2(a+b)$$ and by C-S
$$2(a+b)=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2.$$
Thus, $$a+b\leq4$$ and by AM-GM:
$$2\sqrt{ab}\leq a+b\leq4,$$ which gives $$ab\leq4.$$
Id est, $$x^{\ln{y}}=e^{\ln{x}\ln{y}}=e^{ab}\leq e^4.$$
The equality occurs fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Coefficient of $x^4y^3z^3$ in the expansion of $(5x+y-4z)^{10}$ The coefficient of $x^6y^4$ in the expansion of $(2x-3y)^{10}$ is
$$_{10}C_6 \cdot 2^6 \cdot (-3)^4$$
and as for the coefficient of $x^3y^4z^8$ in the expansion of $(x+y+z)^{15}$ is
$$_{15}C_3 \cdot _{15}C_4 \cdot _{15}C_8 = \frac{15!}{3!4!8!} $$
What wou... | We have that by trinomial expansion
$$(5x+y-4z)^{10} =\ldots+\frac{10!}{4!3!3!} (5x)^4y^3(-4z)^3+\ldots$$
therefore the coefficient for $x^4y^3z^3$ is equal to $-\frac{10!5^34^3}{4!3!3!}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Least value of unit vector $|a+b|^2+|b+c|^2+|c+a|^2$ If $a, b, c$ are unit vectors, then least value of $|a+b|^2+|b+c|^2+|c+a|^2$ will be equal to
(1) 1
(2) 3
(3) 9
(4) 12
If am using the concept $a=c=-b$, I am getting the answer 4, but not matching with options provided
| Writing in terms of inner product, you have that
\begin{align*}
|a+b|^2+|b+c|^2+|c+a|^2
&= 2(|a|^2 + |b|^2 + |c|^2) + 2 (a \cdot b + b \cdot c + c \cdot a) \\
&= 6 + 2 (a \cdot b + b \cdot c + c \cdot a)
\end{align*}
Thus, your problem amounts to minimizing the sum of inner products above. Considering that $a, b$ are f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3415387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Cosine of a $2 \times 2$ non-diagonalisable matrix Given
$$A = \begin{bmatrix}
\pi-1 & 1\\
-1 & \pi+1
\end{bmatrix}$$
I need to calculate its cosine, $\cos(A)$. Typically, I use diagonalisation to approach this type of problems:
$$\cos(A) = P \cos(D) P^{-1}$$
However, in this problem, the eigen... | When working this sort of thing out I usually take powers of the matrix and look for a pattern. In this case I found,
$$ M^0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$ M^1 = \begin{bmatrix} \pi -1 & 1 \\ -1 & \pi+1 \end{bmatrix}$$
$$ M^2 = \begin{bmatrix} \pi^2 - 2\pi & 2\pi \\ -2\pi & \pi^2 + 2\pi\end{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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compute $A^2$ and $A^6$. $A= \begin{bmatrix} -1 & 1 & 1 & -1\\ 1 & -1 & -1 & 1\\ 1 & -1 & -1 & 1\\ -1 & 1 & 1 & -1 \end{bmatrix} $ . Fuzhen Zhang's linear algebra, problem 3.11
$A= \begin{bmatrix}
-1 & 1 & 1 & -1\\
1 & -1 & -1 & 1\\
1 & -1 & -1 & 1\\
-1 & 1 & 1 & -1
\end{bmatrix} $
compute $A^2$ and $A^6$.
The an... | We can observe $$A_4 = \begin{bmatrix}A_2&-A_2\\-A_2&A_2\end{bmatrix}$$
where
$$A_2 = -\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$$
Which we see follow the same pattern (except minus sign). So this is nothing but Kronecker product of $A_2$ on $A_2$. Since it is smaller, we can investigate this $A_2$ more readily and see th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that when x approaches to 1-, the function appraochs to negative infinity Prove that
$$\lim_{x\rightarrow1-}f(x):=\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}=-\infty.$$
Using the basic definition.
This is the proof from the textbook:
Let $M\in R$ and assume M<0 without losing generality. As x converges to 1 from ... | I like to let variables
go to zero.
So, in
$\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}
$
I would let
$x = 1-y$.
Then
$x+2 = 1-y+2
=3-y$
and
$\begin{array}\\
2x^2-3x+1
&=2(1-y)^2-3(1-y)+1\\
&=2(1-2y+y^2)-3+3y+1\\
&=2-4y+2y^2-3+3y+1\\
&=-y+2y^2\\
&=y(-1+2y)\\
\end{array}
$
so
$\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find sum of $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$ I want to find the exact sum of this expression: $\sum_{n=1}^{\infty}\frac{(-1)^n}{n-(-1)^n}$
I've already proved that it converges by condition.
Also, I think that it's sort of telescoping series,
because if I open it I get:
$\frac{-1}{2}+\frac{1}{1}-\frac{1}{... | Hint. Note that if $N$ is even then
$$\begin{align}S_N&=\sum_{n=1}^{N}\frac{(-1)^n}{n-(-1)^n}=
-\frac{1}{2}+\frac{1}{1}-\frac{1}{4}+\frac{1}{3}+\dots-\frac{1}{N}+\frac{1}{N-1}\\&=
\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{N-1}-\frac{1}{N}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n},
\end{align}$$
and
$$S_{N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability of getting a pair of symbols when throwing four dice marked with swords, helmets, and skulls I have 4 dices with those faces display:
1-2-3 = sword,
4-5 = helmet,
6 = skull
When rolling 4 dice, what are the probabilities of getting a PAIR of swords ? helmets ? skulls ?
I tried to solve this by using a regul... | I'm assuming you mean getting the probability of two swords, of two helmets, and of two skulls separately. Here, the individual probabilities are:
$P(Sword) = \frac{3}{6} = \frac{1}{2}$
$P(Helmet) = \frac{2}{6} = \frac{1}{3}$
$P(Skull) = \frac{1}{6}$
Here, we use combination instead of permutation for we do not care ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact. From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\b... | Still not what you want, but uses Joachimsthals notations.
The locus is the midpoint of the two intersection points of $$s=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ and (from $s_1^2=s \cdot s_{11}$) $$(\frac{x(O)x}{a^2}+\frac{y(O)y}{b^2}-1)^2=(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)(\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}-1)$$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_{x\to 0} \cot ^2 (x)-\frac{1}{x^2}$
Evaluate $\lim\limits_{x\to 0} \cot ^2 (x)-\dfrac{1}{x^2}.$
I was thinking of using L'Hôpital's Rule, but things got very, very ugly so I wasn't able to solve it. I know from experimentation that the limit is $-\dfrac{2}{3}$ though.
edit: i do not want to use a tayl... | Unacceptably Easy Answer (Before Question Edit)
$$
\begin{align}
\cot^2(x)
&=\frac{\cos^2(x)}{\sin^2(x)}\\
&=\frac{\left(1-\frac12x^2+O\!\left(x^4\right)\right)^2}{\left(x-\frac16x^3+O\!\left(x^5\right)\right)^2}\\
&=\frac1{x^2}\frac{1-x^2+O\!\left(x^4\right)}{1-\frac13x^2+O\!\left(x^4\right)}\\
&=\frac1{x^2}\left(1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Lindquist Identity:$\big(|b|^{p-2}b-|a|^{p-2}a\big)(b-a)=\frac{|b|^{p-2}+|a|^{p-2}}{2}|b-a|^2+\frac{1}{2}\big(|b|^{p-2}-|a|^{p-2}\big)(|b|^2-|a|^2)$ When investigating the properties of the one-dimensional "French tower" function
$$
g: \mathbb{R} \to \mathbb{R}, \
z \mapsto \begin{cases}
| z |^{p - 2} z, & \text{if } ... | The RHS becomes
$$
| b |^{p} - | b |^{p - 2} a \cdot b - | a |^{p - 2} a \cdot b + | a |^{p}.
$$
The LHS is
\begin{align}
& \frac{1}{2} | b |^{p} - | b |^{p - 2} a \cdot b + \frac{1}{2} | b |^{p - 2} | a |^2 + \frac{1}{2} | a |^{p - 2} | b |^{2} - | a |^{p - 2} a \cdot b + \frac{1}{2} | a |^p \\
+ & \frac{1}{2} | b |... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Coincidence? $\left(\frac 1e\right)^{\frac 1e}\approx \ln 2$ Is it a coincidence that
$$\color{lightgrey}{0.6922\cdots =}\left(\frac 1e\right)^{\frac 1e}\approx \ln 2\color{lightgrey}{=0.6931\cdots}$$
?
| I'm not sure if this helps, but I have found that the Taylor series for these values end up being quite similar.
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\ldots,$$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\ldots\,.$$
$\ln(2)=\ln(1+1)$ and $(\frac{1}{e})^\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Probability Question: Find $a$ and $b$ given expected value and density function The density function of $X$ is given by
\begin{align*}
f(x) =\begin{cases}
a+bx^2, & \text{if}\,\,0\leq x\leq 1\\
0, & \text{otherwise}
\end{cases}
\end{align*}
Once $\textbf{E}(X) = 4.25$, find $a$ and $b$
| Using legitimacy,
$$\int\limits_{0}^{1} f(x) dx = 1$$
$$\int\limits_{0}^{1} (a + b x^{2}) dx = 1$$
$$ \bigg[ax + \frac{bx^{3}}{3}\bigg]_{0}^{1} = 1$$
$$ a + \frac{b}{3} = 1$$
$$ 3a + b = 3$$
Using expectation,
$$\int\limits_{0}^{1} x f(x) dx = 4.25$$
$$\int\limits_{0}^{1} (ax + b x^{3}) dx = 4.25$$
$$ \bigg[\frac{ax^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$ Question: If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$
If we construct $g(x)=f(x)-(x-2)(x-3)(x-4)(x-5)$, then is it possible to find f(5)?
| $f(x)=x+1+c(x-1)(x-2)(x-3)(x-4)$ satisfies the hypothesis for any $c \neq 0$. So it is not possible to find $f(5)$ from the given information.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
taylor series approximations of $\ln x$
Compare the errors for the following methods for computing $\ln 2$. Which one is best?
$1.\; P_{n,1}(2)$
$2.\; -P_{n,1}(0.5)$
$3.\;P_{n,1}(\frac{4}{3})-P_{n,1}(\frac{2}{3})$
The Taylor series for $\ln x$ centred at $a=1$ is given by $P_{n,1}(x) = (x-1)-\frac{(x-1)^2}{2}+\frac{(... | Indeed, there is something wrong in your approach.
The explicit formula for the remainder of Taylor's theorem is given by
$$R_{n,1}(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-1)^{k+1}$$ where $\xi \in (1,x)$. Not by $\dfrac{|(x-1)^{n+1}|}{(n+1)}$ as you mentioned.
Based on that, you can reconsider your inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.