Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $a, b \in \mathbb{R}$ such that the polynomial $f(x)=x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$ has a triple root. I am given the following polynomial:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$$
with $a, b \in \mathbb{R}$.
I have to find $a$ and $b$ such that the given polynomial has a triple root.
I know that if a polynomial has a triple root $\alpha$ then we have:
$$f(\alpha) = 0$$
$$f'(\alpha) = 0$$
$$f''(\alpha) = 0$$
And in previous exercises (where the degree of the polynomial was $4$, not $5$) I could use $f''(\alpha) = 0$ to find $2$ $\alpha$'s and find $a$ and $b$ for each $\alpha$. I could do that because the second derivative of a $5$th degree polynomial is a quadratic so I could find the $\alpha$'s. Here, it's different. We have:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b$$
$$f'(x) = 5x^4+28x^3+57x^2+52x+a$$
$$f''(x) = 20x^3+84x^2+114x+52$$
And when I try to solve $f''(\alpha) = 0$ I get:
$$20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$$
And I don't know how to find the $\alpha$'s. So, what approach should I use to solve this exercise?
| We'll equate
$\tag 1 (x-u)^3 \,(x^2+vx+w)$
to
$\tag 2 x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b$
The coefficient of $x^4$ in the expansion of $\text{(1)}$ is $v - 3u$. Equating coefficients, $v = 7 + 3u$. So now we want to equate $\text{(2)}$ with
$\tag 3 (x-u)^3 \,(x^2+(7+3u)x+w)$
Equating the coefficients of $x^3$ from $\text{(3)}$ with $\text{(2)}$ now gives
$\quad -6u^2 - 21u + w = 19 \; \text{ iff } \; w = 19 +6u^2 + 21u$
Eliminating $w$,
$\tag 4 (x-u)^3 \,(x^2+(7+3u)x+ 19 +6u^2 + 21u)$
Equating the coefficients of $x^2$ from $\text{(4)}$ with $\text{(2)}$ now gives
$\quad - 10 u^3 - 42 u^2 - 57 u = 26$
So we need to solve
$\tag 5 10 u^3 + 42 u^2 + 57 u - 26 = 0$
Using the rational root theorem you get that $u = -2$.
So $v = 7 + 3u = 1$ and $w = 19 +6u^2 + 21u = 1$ and expanding
$\quad (x+2)^3 \,(x^2+x+1)$
gives
$\tag {ANS} x^5 + 7 x^4 + 19 x^3 + 26 x^2 + 20 x + 8$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the challenging $\int_0^{\infty } \frac{\sin (x)}{\sqrt{x} \left(\cos ^2(x)+1\right)} \, dx$ in terms of Legendre Chi function How to prove $$\int_0^{\infty } \frac{\sin (x)}{\sqrt{x} \left(\cos ^2(x)+1\right)} \, dx=
\sqrt{2\pi} \sum _{k=1}^{\infty } \frac{(-1)^{k-1} \left(\sqrt{2}-1\right)^{2 k-1}}{\sqrt{2 k-1}}$$
Any help will be appreciated.
Update: Using @uniquesailor's hint the problem is solved. Indeed, set $b=3-2 \sqrt{2}$ and make use of $\frac{1}{\cos ^2(x)+1}=\frac{2}{3 \left(\frac{1}{3} \cos (2 x)+1\right)}$, one may broke the integrand into Fourier series based on his Poisson type formula. Then, upon using Fresnel's result $\int_0^{\infty } \frac{\sin (x)}{\sqrt{x}} \, dx=\sqrt{\frac{\pi }{2}}$ and the trigonometric identity $2 \sin (x) \cos (2 n x)=\sin ((2 n+1) x)-\sin ((2 n-1) x)$, the integral is transformed to RHS after rearranging. According to Benidict RHS is also equivalent to $-i \sqrt{2\pi} \chi_{\frac{1}{2}}(i (\sqrt{2}-1))$.
| It can be proved by applying this relationship:
$$2\sum_{n=0}^{\infty} \beta^n cos(nt)-1=\frac{1-\beta^2}{1+\beta^2}\frac{1}{1-\frac{2\beta}{1+\beta^2}cos(t)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Theorems of Ceva and Menelaus
In the parallelogram $ABCD$ $M\in AB, AM:MB=2:7$ and $N\in BC, BN:NC=4:5$. If $DM\cap AN=O$, calculate $AO:ON;DO:OM$.
For the first ratio let $BC \cap DM=Q$. By Menelaus' theorem $\dfrac{AO}{ON}\cdot\dfrac{NQ}{QB}\cdot\dfrac{BM}{MA}=1 \Rightarrow \dfrac{AO}{ON}=\dfrac{BQ}{QN}\cdot\dfrac{AM}{BM}$. If we look at the similar triangles $\triangle BMQ \sim AMD \Rightarrow \dfrac{BQ}{AD}=\dfrac{BQ}{BC}=\dfrac{BM}{AM}=\dfrac{7}{2}$ or $BQ=\dfrac{7}{2}BC$. Now, $\dfrac{AO}{ON}=\dfrac{BQ}{QN}\cdot\dfrac{AM}{BM}=\dfrac{\dfrac{7}{2}BC}{BQ+BN}\cdot\dfrac{2}{7}=\dfrac{\dfrac{7}{2}BC}{\dfrac{7}{2}BC+\dfrac{4}{9}BC}\cdot\dfrac{2}{7}=\dfrac{63}{71}\cdot\dfrac{2}{7}=\dfrac{18}{71}$. Thales theorem tells us that
$\dfrac{AO}{ON}=\dfrac{DO}{OQ}=\dfrac{18}{71}$.
How to find $DO:OM$?
| Let $QB= t$. By Thales: $${DC\over MB} = {QC\over QB} \implies {9\over 7} = {t+9\over t}\implies t=63/2$$
so again by Thales:
$${AO\over ON} = {AD\over QN} = {9\over t+4} ={18\over 71}$$
Let $CP = x$. Again by Thales: $${x\over x+9}={CP\over DP} = {CN\over AD} = {5\over 9}\implies x=45/4$$
so
$${DO\over OM} = {DP\over AM} = {9+x\over 2}={81\over 8 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570490",
"timestamp": "2023-03-29T00:00:00",
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System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$
Solve over reals:
$$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$
I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:
$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$
Here, there are two cases: all three variables are greater than $1$ or only one.
If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.
How can I prove there are no solutions when only variable, say $x$, is greater than $1$?
| Since only $x>1$
Then $x> y$
Using the first equation,
$x^3=y(y-1)$
For the case $0\leq y <1$, the RHS is negative,
So $y<0$
Taking the third equation,
$z^3=x^2-x$
The RHS is always positive hence $0<z<1$
Taking the second equation,
$y^3=z^2-z$
Differentiate RHS to find stationary point,
$2z-1=0$
$z=1/2$,
$-1/4\leq y^3 <0$
Using the first equation again,
$x^3=y(y-1)$
Differentiating RHS to find min/max for $x$
$y=-1/4$,
$x^3=5/16$, which contradicts that $x>1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Combinatorics: Number of ways to distribute 12 balls in group, with given conditions
The number of ways in which 12 identical balls can be grouped in four marked non empty sets A,B,C,D such that n(A) < n(B) is ?
The answer given is 70.
This is how I solved it:-
Now to distribute 12 identical balls to 4 different group such that no group is empty, we have $a+b+c+d = 12$, $$^{12-1}C_{4-1}$$ ways but we cannot have $a=b$ (as $a$ and $b$ are variables and interchangeable n(A) < n(b) and n(A) > n(b) mean the same), which also means that all $a, b, c,d$ should not be equal (as $a,b,c,d$ are variables and interchangeable) i.e. we need to exclude cases of $a=b=c=d$, but there is only one case that is possible which is $a=b=c=d=3$,
thus total ways $=$ $$^{12-1}C_{4-1}-1$$ but that equals 164 but answer given is 70. What am I doing wrong? How will you solve the question?
| Consider $b=a+x_1$ where $x_1\ge1$
Eqn becomesm $2a+x_1+c+d=12$ and $,a, x_1, c, d\ge1$. So we need to find number of positive integral solutions of the eqn
for a=1 we get $x_1+c+d=10$ Number of solutions=C(9,2)=36
for a=2 we get $x_1+c+d=8$ Number of solutions=C(7,2)=21
for a=3 we get $x_1+c+d=6$ Number of solutions=C(5,2)=10
for a=4 we get $x_1+c+d=4$ Number of solutions=C(3,2)=3
for a=5 we get $x_1+c+d=2$ Number of solutions=0
Total number os solutions=70
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $2^n$ always a sum of exactly $n$ primes for $n \geq 2$? For example:
$2^2=4=2+2$
$2^3=8=3+3+2$
$2^4=16=5+5+3+3$
$2^5=32=17+7+3+3+2$
$2^6=64=17+17+17+7+3+3$
How much further can this go?
| This is a very difficult problem but luckily previous results allow us to solve it (thanks to @Luke_Collins). The Weak Goldbach Conjecture states that every odd number $n$ (greater than or equal to $7$) is equal to the sum of three prime numbers. This implies that every even number greater than or equal to $10$ is equal to the sum of $4$ primes. Then consider $n\geq 4$
$$2^n=[2^n-2(n-4)]+2(n-4)$$
We can write $2(n-4)=2+2+\dots+2$ as the sum of $n-4$ $2$'s. Since $2^n-2(n-4)\geq 10$ (and even) for all $n\geq 4$, we are assured that it can be written as the sum of $4$ primes. Since the cases $n\in\{1,2,3\}$ are easy to check, every power $2^n$ can be written as the sum of $n$ primes.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
solving $|2x+1|-|5x-2|\geq1$ Solve:
$|2x+1|-|5x-2|\geq1$
$\Rightarrow |2x+1|\geq1+|5x-2|$
Then I squared the inequality
$\Rightarrow 4x^2-4x+1\geq1+2|5x-2|+25x^2-20x+4$
$\Rightarrow-21x^2+16x-2|5x-2|-4\geq0$
Then I separated into two cases
The first case
$$\begin{cases}-21x^2+16x-10x+4-4\geq0...(a)\\x\geq0...(b) \end{cases}
$$
From (a) we get that
$x\in[0,\frac{2}{7}]$
$(a)\cap(b)\\
\Rightarrow x\in[0,\frac{2}{7}]$
The second case
$$\begin{cases}-21x^2+16x+10-4-4\geq0...(c)\\x<0...(d)\end{cases}$$
From (c) we get that
$x\in[\frac{4}{7},\frac{2}{3}]$
$(c)\cap(d)\\
\Rightarrow x\in\emptyset$
So the solutions to the initial inequality should be $x\in [0,\frac{2}{7}]$
But wolfram said the solutions were $[\frac{2}{7},\frac{2}{3}]$
Where did I make the mistake???
| What about separating it by cases?
Let us consider that $x\geq 2/5$. Then we have
\begin{align*}
|2x + 1| - |5x - 2| = (2x + 1) - (5x - 2) = 3 - 3x \geq 1 \Longleftrightarrow 3x \leq 2 \Longleftrightarrow x\leq 2/3
\end{align*}
Hence the first solution set is given by $S_{1} = [2/5,2/3]$.
We shall then solve the inequation when $-1/2\leq x \leq 2/5$. In this case, we get
\begin{align*}
|2x + 1| - |5x - 2| = (2x + 1) + (5x - 2) = 7x - 1 \geq 1 \Longleftrightarrow 7x \geq 2 \Longleftrightarrow x\geq 2/7
\end{align*}
Consequently, the second solution set is given by $S_{2} = [2/7,2/5]$.
Finally, we have the case when $x\leq -1/2$, from whence we obtain
\begin{align*}
|2x + 1| - |5x - 2| = -(2x + 1) + (5x - 2) = 3x - 3 \geq 1 \Longleftrightarrow 3x \geq 4 \Longleftrightarrow x \geq 4/3
\end{align*}
Therefore the third solution set is given by $S_{3} = \varnothing$.
Gathering all the solution sets, it results that $S = S_{1}\cup S_{2}\cup S_{3} = [2/7,2/3]$, which coincides with the solution proposed by wolfram alpha.
Hopefully it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving the inequality using am gm theorem I was asked to prove that
$a^3+b^3 \le (a^2+b^2)(a^4+b^4)$
I expanded the rhs and used am gm and got $a^6+a^2b^4+a^4b^2+b^6 \ge 4a^3b^3$ and struck.i think I lack intuition or I don't have enough experience.Any hints
| Maybe you want to prove: $(a^3+b^3)^2 \le (a^2+b^2)(a^4+b^4)$. In this case, it reduces to: $2a^3b^3 \le a^2b^4+a^4b^2$, but this is true because $a^2b^4+a^4b^2 - 2a^3b^3 = (ab^2-a^2b)^2 \ge 0$ clearly true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find the derivative of $\sqrt{x+2} -x$ using limit definition? So the function is $f(x) = \sqrt{x+2} -x$, and I keep hitting dead ends trying to solve it using the definition of derivative. If anyone can help, it would be greatly appreciated!
| $\require{cancel}$Note that
$$f(x)=\sqrt{x+2}-x$$
$$f(x+h)=\sqrt{x+h+2}-x-h$$
$$f(x+h)-f(x)=\sqrt{x+h+2}-\sqrt{x+2}-h$$
So we have
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-h}{h}$$
$$= \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}-1$$
$$= \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}\cdot \frac{\sqrt{x+h+2} + \sqrt{x+2}} {\sqrt{x+h+2}+\sqrt{x+2}} - 1$$
$$=\frac{(\cancel{x}+h+\cancel{2})-(\cancel{x}+\cancel{2})}{h(\sqrt{x+h+2} + \sqrt{x+2})} - 1$$
$$=\frac{\cancel{h}}{\cancel{h}(\sqrt{x+h+2} + \sqrt{x+2})} - 1$$
$$= \frac{1}{\sqrt{x+h+2} + \sqrt{x+2}} - 1$$
Let $h\to 0$ to get
$$\boxed{f’(x) = \frac{1}{2\sqrt{x+2}}-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the sum of the following infinite series? $$
\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \cdots
$$
So basically I separated it into two series
where:
one of them is $\left(\frac{1}{3}\right)^n$
And I use geometric series formula to find that this series equals $\frac{1}{2}$.
But I can't figure out the series of the other one.
Apparently the answer for the series combined is: $\frac{5}{8}$
What is the other series?
| Note that when you group adjacent terms, $\frac{1}{3} + \frac{2}{9} = \frac{5}{9}$. Therefore, the sum of the series becomes:
$$\frac{5}{9} + \frac{1}{9} \cdot \frac{5}{9} + \left(\frac{1}{9} \right)^2 \cdot \frac{5}{9} + \cdots$$
$$= \frac{\frac{5}{9}}{1 - \frac{1}{9}}$$
$$= \frac{5/9}{8/9} = \frac{5}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$
Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$.
My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies \left(\frac{\sin^4x}{a}+\frac{\cos^4x}{b}\right)^2=\frac{1}{(a+b)^2} \\ \implies \frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}-\sin^2x\cos^2x\left(\frac{\sin^2x}{a}-\frac{\cos^2x}{b}\right)^2=\frac{1}{(a+b)^2}.$$
Therefore, if we can prove that $$\frac{\sin^2x}{a}-\frac{\cos^2x}{b}=0,$$ then we are done. But, I am not able to prove the same.
| By Cauchy-- Schwarz
$$1=\left( \frac{\sin^2(x)}{\sqrt{a}}\cdot \sqrt{a}+ \frac{\cos^2(x)}{\sqrt{b}}\cdot \sqrt{b} \right)^2 \leq \left( \frac{\sin^4x}{a}+\frac{\cos^4x}{b} \right)(a+b)=1$$
Therefore, we have equality in Cauchy-Schwarz, and hence
$$\frac{\frac{\sin^2(x)}{\sqrt{a}}}{ \sqrt{a}}=\frac{ \frac{\cos^2(x)}{\sqrt{b}}}{ \sqrt{b} }$$
which is the identity you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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how to prove this inequality with $a,b,c\in [1,3]$ let $a,b,c\in [1,3]$,show that
$$3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{45}{a+b+c}\ge 16\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)$$
I had found this simaler problem,https://artofproblemsolving.com/community/c6h615194
I can't prove this it
| WLOG, assume that $c = \max(a,b,c) = 3$. The inequality becomes
$$\frac{3}{a} - \frac{16}{a+3} + \frac{3}{b} - \frac{16}{b+3} + 1 - \frac{16}{a+b} + \frac{45}{a+b+3} \ge 0.$$
Let $f(x) = \frac{3}{x} - \frac{16}{x+3}$.
We have $f''(x) = \frac{6}{x^3} - \frac{32}{(x+3)^3}$.
For $x\in [1,3]$, we have $(x+3)^3/x^3 = (1 + 3/x)^3 \ge 2^3 > 32/6$
and hence $f''(x) > 0$.
Thus, $f(x)$ is convex on $[1,3]$.
We have
$f(a) + f(b) \ge 2 f(\frac{a+b}{2})$. It suffices to prove that
$$\frac{12}{a+b} - \frac{64}{a+b+6} + 1 - \frac{16}{a+b} + \frac{45}{a+b+3} \ge 0$$
or
$$\frac{(a+b-2)(a+b-6)^2}{(a+b)(a+b+3)(a+b+6)}\ge 0.$$
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$ Evaluate the following integral :
$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$
I was tried use change variable ,
If I use $x=y^2$ integral becomes :
$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$
From here I have one idea the derivative under sing integral but I got I difficult integration :
$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$
I already to see you hints or solution!
| Using the principal branch of the logarithm, we have
$$ \begin{align} I &= \int_{0}^{\infty} \frac{\ln(1+t^{4})}{\sqrt{t}(1+t)} \, \mathrm dt = 2\int_{0}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx = \int_{-\infty}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{7} \int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx + \int_{-\infty}^{\infty} \frac{\ln \left(1 \, {\color{red}{+}} \, xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \right)\\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx+ \int_{\infty}^{-\infty} \frac{\ln \left(1 - ue^{i \pi(2n+1)/8}\right)}{1+u^2} \, (- \mathrm du) \right) \\&= 2 \sum_{n=0}^{3} \int_{-\infty}^{\infty} \frac{\ln \left(1 - xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ & \stackrel{(1)}= 2\sum_{n=0}^{3} 2 \pi i \operatorname{Res} \left[ \frac{\ln \left(1-ze^{i \pi(2n+1)/8}\right)}{1+z^2}, i \right] \\ &= 2\sum_{n=0}^{3} 2 \pi i \, \frac{\ln \left(1-ie^{i \pi(2n+1)/8}\right)}{2i} \\ &= 2\pi \sum_{n=0}^{3} \ln \left(1-ie^{i \pi(2n+1)/8}\right) \\ &= 2 \pi \ln \left[ \left(1-ie^{i \pi/8} \right) \left(1-ie^{3 \pi i/8} \right) \left(1+ie^{-3 \pi i/8} \right) \left(1+ie^{- i \pi/8} \right) \right]\\ &= 2 \pi \ln \left[\left(2+ 2 \sin \left(\frac{\pi}{8} \right) \right) \left(2+ 2 \sin \left(\frac{3\pi}{8} \right) \right)\right] \\ &= 2 \pi \ln \left(4 +2 \sqrt{2-\sqrt{2}} +2 \sqrt{2+\sqrt{2}} + \sqrt{2} \right) \\ &= 2 \pi \ln \left(4 + 2 \sqrt{4 +2\sqrt{2}} + \sqrt{2}\right) \\ &= 2 \pi \ln \left(\left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right)^{2}\right) \\ &= 4 \pi \ln \left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right) \end{align}$$
$(1)$ The branch cut for $\ln \left(1-ze^{i \pi(2n+1)/8}\right) $ is in the lower half of the complex plane.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Show that for $n\geqslant1,\ \displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}$
Show that for $n\geqslant1,$ $$\displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}.$$
This is an exercise from the Preliminaries of David Burton's Elementary Number Theory.
I started by simplifying the $\text{RHS}$ into something that could enable me to use mathematical induction.
$$
\begin{align*}
\text{RHS} &= \dfrac{\displaystyle\sum_{k=1}^{n}(2k-1)}{\displaystyle\sum_{k=1}^{n}2k}\cdot2^{2n}\\
&=\frac{n(n+1)-n}{n(n+1)}\cdot2^{2n}\\
&=\frac{2^{2n}n}{n+1}.
\end{align*}
$$
To be proved is that
$$\binom{2n}{n}=\frac{2^{2n}n}{n+1}.$$
To get $\binom{2n}{n}$ on the $\text{RHS}$, I did the following.
$$
\begin{align*}
2^{2n}&=(1+1)^{2n}\\
&=\binom{2n}{0}+\binom{2n}{1}+\binom{2n}{2}+\cdots+\binom{2n}{n}+\cdots+\binom{2n}{2n}.
\end{align*}
$$
From $\displaystyle \binom{n}{k}=\binom{n}{n-k}$, it follows that
$$\begin{align*}
2^{2n}=\left[\binom{2n}{0}+\binom{2n}{2n}\right]+\left[\binom{2n}{1}+\binom{2n}{2n-1}\right]+\cdots+\binom{2n}{n},
\end{align*}$$
but I can't proceed any further.
| Here's a direct proof:
\begin{align}
\binom{2n}{n} &= \frac{(2n)!}{n!^2}\\
&= \frac{\prod_{k=1}^{2n} k}{\prod_{k=1}^n k^2} \\
&= \frac{\left(\prod_{k=1}^n 2k\right)\left(\prod_{k=1}^n (2k-1)\right)}{\prod_{k=1}^n k^2} \\
&= 2^n\prod_{k=1}^n \frac{k (2k-1)}{k^2} \\
&= 2^n\prod_{k=1}^n \frac{2k-1}{k} \\
&= 2^{2n}\prod_{k=1}^n \frac{2k-1}{2k}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to find where two curves have a common value? Suppose I have;
$$f(x) = x^2 - 42x + 364$$
$$g(y) = y^2 - 35y + 364$$
Computing out values I find they have a common value
$$f(6) = g(8) = g(27) = f(36) = 148$$
Is there a way of finding these values?
Working through based on received help...
(1) $$f(x) = g(y)$$
(2) $$f(x) - g(y) = x^2 - 42x + 364 - y^2 + 35y - 364 = 0$$
(3) $$x^2 - 42x - y^2 + 35y = 0$$
Complete the squares...
(4) $$(x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4}$$
(5) $$4 ((x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4} )$$
(6) $$(4x^2 - 168x + 42^2) - (4y^2 - 140y + 35^2) = 42^2 - 35^2$$
(7) $$(2x - 42)^2 - (2y - 35)^2 = 539$$
Considering (7) as Fermat's factorization method;
(8) $$a^2 - b^2 = c$$
(9) $$(a + b)(a - b) = 539$$
(10) $$(2x - 42 + 2y - 35)(2x - 42 - 2y + 35) = 539$$
(11) $$(2x + 2y - 77)(2x - 2y - 7) = 539$$
Then, considering (11) as a simple product of two terms gives;
(12) $$d * e = 539$$
(13) $$(2x - 2y - 7) = d$$
(14) $$(2x + 2y - 77) = 539/d$$
| To find the common values $(x,y)$, for which $f(x)=g(y)$ you have to equate the two functions, and obtain: $$f(x)=g(y)\leftrightarrow x^2-42x+364=y^2-35y+364\leftrightarrow x^2-y^2-42x+35y=0$$
In the left sides, there is the equation of an hyperbola. Obviously, if $(x,y) \in R$, then there are infinite many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Need help to understand how to deduce the value of $\cos^2\displaystyle\big(\frac{\pi}{5}\big)+\cos^2\big(\frac{2\pi}{5}\big)$ I have some trouble understanding the steps provided in the book on solving the last part of the question.
For context purposes (and to provide the full picture), the whole question (with solution) is as follows:
Prove de Moivre's theorem for positive integer exponents.
(I have no problem with this)
Using de Moivre's theorem, show that $\sin5\theta=a \sin^5\theta+b \cos^2\theta\sin^3\theta+c\cos^4\theta\sin\theta$ , where $a,b$ and $c$ are integers to be determined. (No issue with this either)
Express $\displaystyle\frac{\sin5\theta}{\sin\theta}$ in terms of $\cos\theta$, where $\theta$ is not a multiple of $\pi$. (Still no problem)
Hence, find the roots of the equation $16x^4-12x^2+1=0$ in trigonometric form. (I understand this part)
Deduce the value of $\displaystyle\cos^2\Big(\frac{\pi}{5}\Big)+\cos^2\Big(\frac{2\pi}{5}\Big)$. (I don't understand the working given by the book)
My question is, since it is found that $\displaystyle\cos^2\theta=\frac{3\pm\sqrt5}{8}$, how to tell that $\displaystyle\cos^2\frac{\pi}{5}=\frac{3+\sqrt5}{8}$ and how is it known that $\displaystyle\cos^2\frac{2\pi}{5}=\frac{3-\sqrt5}{8}$?
| Since what you asked seems to be answered in comments, here is an alternative way in case you are interested, it uses Euler's formula $$e^{it}=\cos t +i \sin t.\tag{*}$$
First notice that (*) implies $\cos t = \frac{1}{2}(e^{it}+e^{-it})$. Now put $z=e^{i\frac{\pi}{5}}$, and notice that $z^5=e^{i\pi}=-1$. Subsequently, we get $z^{-2}=z^{3}/z^5=-z^3$, and similarly $z^{-4}=z/z^5=-z$.
Then finally we have
\begin{align}
\cos^2 \frac{\pi}{5}+\cos^2 \frac{2\pi}{5}&= \left(\frac{1}{2}(e^{i\frac{\pi}{5}}+e^{-i\frac{\pi}{5}})\right)^2+\left(\frac{1}{2}(e^{i\frac{2\pi}{5}}+e^{-i\frac{2\pi}{5}})\right)^2\\
&=\frac{1}{4}((z+z^{-1})^2+(z^2+z^{-2})^2)\\
&=\frac{1}{4}(z^2+2+z^{-2}+z^4+2+z^{-4})\\
&=\frac{1}{4}(z^2+2-z^3+z^4+2-z)\\
&=\frac{1}{4}(3+1-z+z^2-z^3+z^4)\\
&=\frac{1}{4}\left(3+\frac{z^5+1}{z+1}\right)\\
&=\frac{3}{4}
\end{align}
where we have used $(z+1)(1-z+z^2-z^3+z^4)=z^5+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Domain of definition I need to find the domain of definition of the following functions:
$$
f(x)=\arccos\left(1/(x^2+1)\right), \quad g(x)=\arcsin\left(2x\sqrt{1-x^2}\right).
$$
For $f$, we have to find $x$ such that $-1\leq 1/(x^2+1)\leq 1$. I do this:
$$\left|\frac{1}{x^2+1}\right|\leq 1 \Longleftrightarrow x^2+1\geq 1\Longleftrightarrow |x|\geq0.$$
Does this mean that $D_f=\mathbb{R}$?
For $g$:
$$\left|2x\sqrt{1-x^2}\right|\leq 1 \Longleftrightarrow 4x^2(1-x^2)\leq 1 \Longleftrightarrow -4x^4+4x^2-1\leq 0.$$
I find two solutions $-1/\sqrt2$ and $1/\sqrt{2}$, and the condition $1-x^2\geq 0$ that is $x\in [-1,1]$. So for me the domain is $(-\infty,-1/\sqrt{2}]\cup [1/\sqrt{2},+\infty) \cap [-1,1]$, hence $[1/\sqrt{2},1]$, but I am not sure.
Can you help me please ?
| Yes, the domain of $f$ is all of $\mathbb{R}$. For $g$ you need to solve the system
$$
\begin{cases}
1-x^2 \geq 0 \\
-1 \leq 2x\sqrt{1-x^2} \leq 1.
\end{cases}
$$
The first one gives $-1 \leq x \leq 1$. You can split the second getting two inequalities. Take $2x\sqrt{1-x^2}\leq 1$ first. Note that for $x\leq 0$ this is always true, so you can focus on $0<x\leq 1$. Square of both sides gives $4x^2(1-x^2) \leq 1$, namely $4x^4-4x^2+1 = (2x^2-1)^2\geq 0$, which is always true. Now note that the function $f(x) = 2x\sqrt{1-x^2}$ is odd, so solutions to the other inequality are the same. Therefore the final solution is just $|x| \leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a unique matrix $K \in \mathbb R^{2 \times 2}$ such that $KA + B$ is positive definite ($A$, $B$ given below) Is there a unique matrix $K \in \mathbb R^{2 \times 2}$ (up to multiplication by a constant factor) such that
$$K \begin{pmatrix}
0 &-1
\\
-1 & 0
\end{pmatrix} + \begin{pmatrix}
0 &0
\\
0 & -1
\end{pmatrix}
$$
is a positive definite matrix?
| If $K = M\begin{bmatrix} 0 & -1 \\ -1 & 0\end{bmatrix}$ then $$K \begin{bmatrix} 0 & -1 \\ -1 & 0\end{bmatrix} + \begin{bmatrix}0&0\\0&-1\end{bmatrix} = M + \begin{bmatrix}0&0\\0&-1\end{bmatrix}.$$
There are many choices of $M$ that will work, like $\begin{bmatrix} c_1 & 0 \\ 0 & 1 + c_2\end{bmatrix}$ for any positive $c_1, c_2$. This corresponds to $K = \begin{bmatrix} c_1 & 0 \\ 0 & 1 + c_2\end{bmatrix} \begin{bmatrix} 0 & -1 \\ -1 & 0\end{bmatrix}$ which are not scalar multiples of each other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Valid induction? The problem was presented like this:
$A_{2n} =
\begin{bmatrix}
a & 0 & \cdots & 0 & b\\
0 & a & \cdots & b & 0\\
& & \cdots & & \\
0 & b & \cdots & a & 0\\
b & 0 & \cdots & 0 & a
\end{bmatrix}
n \in N, A_{0} = 1
$
Show that $detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}$
I tried to use induction for this case:
Base case:
$n=1$: $detA_{2n} = \begin{vmatrix}a & b\\ b & a\end{vmatrix} = a^{2}-b^{2} = a^{2}-b^{2} \cdot detA_{2n-2} = a^{2}-b^{2} \cdot detA_{0} = a^{2}-b^{2} \cdot 1 = a^{2}-b^{2} $
Hypothesis:
$ detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2} \Rightarrow detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} $
$detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n+2-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n} \Leftrightarrow detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}$
This feels circular, is this the right way to do it?
| $$A_{2(n+1)} =
\begin{bmatrix}
a & 0 & \cdots & 0 & b\\
0 & a & \cdots & b & 0\\
& & \cdots & & \\
0 & b & \cdots & a & 0\\
b & 0 & \cdots & 0 & a
\end{bmatrix}.$$
We will proceed using the appropriate cofactor matrices. First, expand along the first row, note that we have to negate the $(1,2n+2)$-th entry b since $2n+3$ is odd:
$$\det(A_{2(n+1)})=a \det\begin{bmatrix}
a & \cdots & b & 0\\
& & \cdots & & \\
b & \cdots & a & 0\\
0 & \cdots & 0 & a
\end{bmatrix}-b\det\begin{bmatrix}
0 & a & \cdots & b \\
& & \cdots & & \\
0 & b & \cdots & a \\
b & 0 & \cdots & 0
\end{bmatrix}.$$
Here the cofactor matrices are of size (2n+1) by (2n+1). Now we expand the first resulting cofactor matrix along the last row and the second cofactor matrix along the first column. By the same logic as before, we don't have to negate anything in this case:
$$\det(A_{2(n+1)})=a^2\det(A_{2n})-b^2\det(A_{2n})=(a^2-b^2)\det(A_{2n}).$$
Hence we have finished the second part of the induction. Combine it with the base case check to complete the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving a closed form for an integral with nested radicals Is there a simple way to prove the following identity?
$$\int_0^1\sqrt{\frac{u^2-2-2 \sqrt{u^4-u^2+1}}{4 u^6-8 u^4+8 u^2-4}}\mathrm du=\frac{\sqrt{3+2 \sqrt{3}}}{2^{10/3}\pi}\Gamma\left(\frac13\right)^3$$
Context:
This integral came up in trying to evaluate the complete elliptic integral of the first kind $K(m)$ ($m$ is the parameter),
$$K\left(\exp\left(\frac{i\pi}{3}\right)\right)$$
in terms of simpler functions. In particular,
$$K\left(\exp\left(\frac{i\pi}{3}\right)\right)=C\left(1+i \left(2-\sqrt{3}\right)\right)$$
and $C$ is the integral mentioned in the first part.
I was able to show this through an indirect route, but I am hoping my messy method can be easily outdone.
| Here is another approach that doesn’t requires Hypergeometric Functions.
$$I=\int_{0}^{1}\sqrt{\frac{u^2-2-2\sqrt{u^4-u^2+1}}{4u^6-8u^4+8u^2-4}}du=\int_{0}^{1}\underbrace{\sqrt{\frac{2-u^2+2\sqrt{{u^4-u}^2+1}}{4\left(1-u^2\right)\left({u^4-u}^2+1\right)}}}_{u\rightarrow\sqrt{x}}du$$
$$I=\frac{1}{4}\int_{0}^{1}\underbrace{\sqrt{\frac{2-x+2\sqrt{x^2-x+1}}{x\left(1-x\right)\left(x^2-x+1\right)}}}_{x\rightarrow y+\frac{1}{2}}dx=\frac{1}{4}\int_{-\frac{1}{2}}^{\frac{1}{2}}\underbrace{\sqrt{\frac{\frac{3}{2}-y+2\sqrt{y^2+\frac{3}{4}}}{\left(\frac{1}{4}-y^2\right)\left(y^2+\frac{3}{4}\right)}}}_{f(y)}dy$$
$$I=\frac{1}{4}\int_{0}^{\frac{1}{2}}\left(f\left(y\right)+f\left(-y\right)\right)dy=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{0}^{\frac{1}{2}}\underbrace{\sqrt{\frac{\frac{\sqrt3}{2}+\sqrt{y^2+\frac{3}{4}}}{\left(\frac{1}{4}-y^2\right)\left(y^2+\frac{3}{4}\right)}}}_{y=\sqrt{z^2-\frac{3}{4}}}dy$$
$$I=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{\frac{\sqrt3}{2}}^{1}\underbrace{\frac{dz}{\sqrt{\left(1-z^2\right)\left(z-\frac{\sqrt3}{2}\right)}}}_{z=\cos{\left(\theta\right)}}=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{0}^{\frac{\pi}{6}}\underbrace{\frac{dz}{\sqrt{\cos{\left(\theta\right)}-\cos{\left(\frac{\pi}{6}\right)}}}}_{\sin{\left(\frac{\theta}{2}\right)}=\sin{\left(\phi\right)}\sin{\left(\frac{\pi}{12}\right)}}$$
$$I=\frac{\sqrt{2+\sqrt3}}{2}\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-\sin^2{\left(\frac{\pi}{12}\right)\sin^2{\left(\phi\right)}}}}=\frac{\sqrt{2+\sqrt3}}{2}K\left(\sin{\left(\frac{\pi}{12}\right)}\right)$$
$$I=\frac{\sqrt{2+\sqrt3}}{2}\left(\frac{1}{2\ 3^\frac{3}{4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{4}{6}\right)}\right)=\frac{\sqrt{3+2\sqrt3}}{2^\frac{10}{3}\pi}{\Gamma\left(\frac{1}{3}\right)}^3$$
Some explanation:
1) I’ve started factorizing the denominator to check if there was any common term that could be canceled.
2) Then, I applied a couple of substitutions. The first one had the objective of decreasing the degree of the variable, and the second was aiming to eliminate the first-degree term of the polynomial inside the inner root.
3) Then I rewrote the integral exploiting the symmetry of its integration limits, which result in a sum of two square roots ($f(y)+f(-y)$) that could be rewritten after some algebra as a single square root.
4) After that, some other substitutions were applied to make calculations easier. The last one may look kind trick, but it’s intuitive after rewriting the expression using sines of half angle.
5) Finally, a well-known representation of the Complete Elliptic Integral of the First Kind was found, and its value was taken from a table and then the result was simplified using both duplication and reflection formulas of Gamma.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $A,B \in \mathcal{M_2}(\mathbb{R})$ and $A^2+B^2=AB$, does it follow that $A$ and $B$ commute? Let $A,B \in \mathcal{M_2}(\mathbb{R})$ such that $A^2+B^2=AB$. Is it necessary that $AB=BA$?
I could easily show that such matrices have the property that $(AB-BA)^2=O_2$ (this was actually the question I was asked, then I started wondering if it would be true that the matrices actually commute) by considering the matrix $M=A+\epsilon B$ and computing the determinat of $M \cdot \overline{M}$($\epsilon \in \mathbb{C}\setminus \mathbb{R}$ is a cubic root of unity), but that's all I got. I tried to find some counterexamples, but I have a hard time finding any matrices with that property.
EDIT: To see that $(AB-BA)^2=O_2$ we do the following : by direct computations
$$|\det M|^2=\det M \det \overline{M}=\epsilon^2 \det(AB-BA)$$ and this is a real number if and only if $\det(AB-BA)=0$. From the Cayley-Hamilton theorem we now get that $(AB-BA)^2=O_2$.
| Proposition: Let $A,B \in \mathcal{M_2}(\mathbb{R})$ such that $A^2+B^2=AB$. Then $AB=BA$.
Disclaimer: This proof is absolutely awful.
Proof. Note that for every real number $\lambda\in\Bbb{R}$ we have
$$(\lambda A)^2+(\lambda B)^2=\lambda^2(A^2+B^2)=\lambda^2(AB)=(\lambda A)(\lambda B).$$
If $\det A\neq0$ then after an appropriate change of basis we have
$$A=\begin{pmatrix}1&0\\0&\lambda\end{pmatrix}
\qquad\text{ or }\qquad
A=\begin{pmatrix}1&\lambda\\0&1\end{pmatrix}
\qquad\text{ or }\qquad
A=\begin{pmatrix}\lambda&-1\\1&\lambda\end{pmatrix},$$
for some $\lambda\in\Bbb{R}$. Let $B=\tbinom{a\ b}{c\ d}$. We treat the three cases separately:
*
*Plugging $A=\tbinom{1\ 0}{0\ \lambda}$ and $B$ into the equation yields the system of equations
\begin{eqnarray*}
1+a^2+bc&=&a,\\
b(a+d)&=&b,\\
c(a+d)&=&\lambda c,\\
\lambda^2+d^2+bc&=&\lambda d.
\end{eqnarray*}
If $bc=0$ then $a^2-a+1=0$ which is impossible. Then $\lambda=a+d=1$ and so $A$ is the identity matrix, so certainly $AB=BA$.
*Plugging $A=\tbinom{1\ \lambda}{0\ 1}$ and $B$ into the equation yields the system of equations
\begin{eqnarray*}
1+a^2+bc&=&a+\lambda c,\\
2\lambda+b(a+d)&=&b+\lambda d,\\
c(a+d)&=& c,\\
1+d^2+bc&=&d.
\end{eqnarray*}
If $bc=0$ then $d^2-d+1=0$ which is imposible. It follows that $a+d=1$ and hence that $2\lambda=\lambda d$. If $\lambda=0$ then $A$ is the identity matrix so certainly $AB=BA$. Otherwise $d=2$ and hence $bc=-3$ and $a=-1$, which leads to $\lambda c=0$, a contradiction.
*Plugging $A=\tbinom{\lambda\ -1}{1\ \hphantom{-}\lambda}$ and $B$ into the equation yields the system of equations
\begin{eqnarray*}
\lambda^2-1+a^2+bc&=&\lambda a-c,\\
-2\lambda+b(a+d)&=&\lambda b-d,\\
2\lambda+c(a+d)&=&a+\lambda c,\\
\lambda^2-1+d^2+bc&=&b+\lambda d.
\end{eqnarray*}
Adding the second and third, and subtracting the last from the first, yields
$$(b+c)(a+d)=\lambda(b+c)+(a-d)
\qquad\text{ and }\qquad
a^2-d^2=\lambda(a-d)-(b+c).$$
Isolating $a-d$ from the former and $b+c$ from the latter and plugging it in shows that
$$a-d=(b+c)(a+d-\lambda)=(a-d)(\lambda-a-d)(a+d-\lambda)=-(a+d-\lambda)^2(a-d),$$
which shows that $a=d$. It follows that $b=-c$ and we are left with the system
\begin{eqnarray*}
\lambda^2-1+a^2-b^2&=&\lambda a-b,\\
-2\lambda+2ab&=&\lambda b-a.
\end{eqnarray*}
The latter shows that $a(2b+1)=\lambda(b+2)$, and multiplying the former by $(b+2)^2$ and substituting and cleaning up yields
$$3a^2(b^2-b+1)=(b+2)^2(b^2-b+1),$$
where $b^2-b+1\neq0$ because $b\in\Bbb{R}$. It follows that $b=-2\pm\sqrt{3}a$ and $\lambda=2a\mp\sqrt{3}$ correspondingly. Then
$$A=\begin{pmatrix}
2a\mp\sqrt{3}&-1\\
1&2a\mp\sqrt{3}
\end{pmatrix}
\qquad\text{ and }\qquad
B=\begin{pmatrix}
a&-2\pm\sqrt{3}{a}\\
2\mp\sqrt{3}{a}&a
\end{pmatrix},$$
and a routine check verifies that again $AB=BA$.
Finally, if $\det A=0$ then after an appropriate change of basis we have
$$A=\begin{pmatrix}\lambda &0\\0&0\end{pmatrix}
\qquad\text{ or }\qquad
A=\begin{pmatrix}0&\lambda\\0&0\end{pmatrix},$$
for some $\lambda\in\Bbb{R}$, where the proposition is trivial if $\lambda=0$. If $\lambda\neq0$ a routine check as before shows that for the first form there is no corresponding matrix $B$ satisfying the identity, and for the second form we see that $B$ must be of the form
$$B=\begin{pmatrix}0&\mu\\0&0\end{pmatrix},$$
for some $\mu\in\Bbb{R}$, which shows that $AB=BA$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
| By C-S
$$a^2+b^2+c^2=\frac{1}{3}(1+1+1)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2=\frac{1}{3}.$$
Thus, by C-S again we obtain:
$$\sum_{cyc}\frac{1+a^2}{1-a^2}=\sum_{cyc}\left(1+\frac{2a^2}{1-a^2}\right)\geq3+\frac{2(a+b+c)^2}{\sum\limits_{cyc}(1-a^2)}=$$
$$=3+\frac{2}{3-(a^2+b^2+c^2)}\geq3+\frac{2}{3-\frac{1}{3}}=\frac{15}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $a, b, c>0$, prove $\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b)$
Given $a,b,c>0$, prove that $$\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b).$$
My attempt:
I have that $$\frac{a^4}{a+b}+\frac{c^2(a+b)}{4}\geq a^{2}c$$
$$\frac{b^4}{b+c}+\frac{a^2(b+c)}{4}\geq b^{2}a$$
$$\frac{c^4}{c+a}+\frac{b^2(c+a)}{4}\geq c^{2}b$$
Adding them up yields $$\sum \frac{a^4}{a+b}\geq\frac{3}{4}\sum a^2c-\frac{1}{4}\sum a^2b.$$
So it boils down to proving that $a^2b+b^2c+c^2a\leq ab^2+bc^2+ca^2$, which isn't correct.
Could you help me with this problem?
| Similar to your inequalities, we have
$$ \frac{ 3a^4}{a+b} + \frac{3}{4}c^2 (a+b) \geq 3a^2c $$
$$ \frac{ a^4}{a+b} + \frac{1}{4} b^2(a+b) \geq a^2b $$
$$ \frac{ a^4}{a+b} + \frac{1}{4} a^2(a+b) \geq a^3 $$
Taking the cyclic summation, we get
$$ \sum \frac{ 5a^4 } { a+b} \geq \sum 2a^2 c + \sum \frac{1}{2}a^3 $$
By rearrangement, we know that $ \sum \frac{1}{2}a^3 \geq \sum\frac{1}{2} a^2 c$, thus,
$$ \sum \frac{ a^4 } { a+b} \geq \sum \frac{1}{2} a^2c, $$
The second inequality balances out the $a^2 b$ term which OP had (and is why we have a $3:1$ ratio).
The third inequality balances out the $a^3$ term which the second inequality introduces, and thankfully the other term from the third inequality cancels out with the first.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does $\ \sin \left(\frac{(k\pi)^2}{x+k\pi}\right)\ {\to \sin(x)}\ $ as odd integers $\ k { \to } \infty$? I am investigating $\sin(\frac{1}{x})$ and it's properties out of interest, and this question is related to my investigation.
Initially, I was trying to prove the result in the title for $0<x<2\pi$, but I do think I have shown that for all x, $\ \sin\left(\frac{(k\pi)^2}{x+k\pi}\right)\ {\to \sin(x)}\ $ as $\ k { \to } \infty,\ k\ $ is odd integers only, but just wanted to check my working is correct. I did the following:
\begin{align*}
\sin\left(\frac{(k\pi)^2}{x+k\pi}\right)
& = \sin\left(\frac{(x + k\pi)^2 - (x^2 + 2k\pi x)}{x+k\pi}\right) \\
& = \sin\left(x+k\pi - \frac{x^2 + 2k\pi x}{x+k\pi}\right) \\
& = \sin\left(x+k\pi - (x + \frac{k\pi x}{x+k\pi})\right) \\
& = \sin \left(k\pi - \frac{k\pi x}{x+k\pi}\right)
\end{align*}
Since $k\ $ is odd, this is equal to:
\begin{align*}
& \sin(k\pi)\cos(\ldots) + \cos(k\pi)\sin \left(-\frac{-k\pi x}{x+k\pi}\right) \\
= & \, 0 - \Bigl(-\sin \left(\frac{k\pi x}{x+k\pi}\right)\ \Bigr) \\
= & \, \sin \left(\frac{k\pi x}{x+k\pi}\right) \\
= & \, \sin \left(\frac{k\pi x + x^2 - x^2}{x+k\pi}\right) \\
= & \, \sin \left( x - \frac{x^2}{x+k\pi}\right)
\end{align*}
which, for all $\ x \in \mathbb R, \ { \to } \sin(x)\ $ as $ \ k { \to } \infty$
Firstly, is this all correct? And secondly have I done unnecessary steps or is there is a less convoluted way to get to the result?
Thanks in advance!
| Note also that
\begin{align*}
\sin\left(\frac{(k\pi)^2}{x+k\pi}\right)
& = \sin\left(\frac{(k\pi)^2 - x^2}{x+k\pi} + \frac{x^2}{x+k\pi}\right) \\
& = \sin\left(k\pi-x + \frac{x^2}{x+k\pi}\right) \\
& = \sin\left(x - \frac{x^2}{x+k\pi}\right) \\
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to bring elliptical equation $2x^2+2y^2+3xy-x-y=0$ into canonical form I have this ellipse: $$2x^2+2y^2+3xy-x-y=0$$
Canonical form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
How can I bring my ellipse to that canonical form? It seems like I need some substitution.
| The center of the ellipse is where the gradient vanishes,
$$4x+3y-1=0,\\4y+3x-1=0.$$
Then we center around the solution $\left(\dfrac17,\dfrac17\right)$,
$$2\left(x+\frac17\right)^2+2\left(y+\frac17\right)^2+3\left(x+\frac17\right)\left(y+\frac17\right)-\left(x+\frac17\right)-\left(y+\frac17\right)
\\=2x^2+3xy+2y^2-\frac17=0.$$
In matrix form,
$$z^TAz-\frac17=0$$ where $A=\begin{pmatrix}2&\frac32\\\frac32&2\end{pmatrix}$.
We diagonalize the matrix and find the Eigenvalues $\dfrac12$ and $\dfrac72$. Hence the reduced equation
$$\frac{x^2}2+\frac{7y^2}2=\frac17.$$
| {
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"url": "https://math.stackexchange.com/questions/3617675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$
But I can't progress after this.
| $x^2+4y^2+16z^2=48 \cdots Eq.(1),\;\; xy+4yz+2zx=24 \cdots Eq.(2)$
From (1), $x^2+(2y)^2+(4z)^2=48 \cdots Eq.(3)$
$(2)\times4 \implies 4xy+16zy+8zx=96\cdots Eq.(4) $
$2\times Eq.(3)-Eq.(4)\implies 2[x^2+(2y)^2+(4z)^2]-2[2xy+8zy+4zx]=0 \implies (x-2y)^2+(2y-4z)^2+(x-4z)^2=0
\implies x=2y=4z.$
Now,
$xy+4zy+2zx=24 \implies 3x^2=48\implies x^2=16
$.
Now, $x=4$, then $y=2,z=1$.
$\implies x^2+y^2+z^2=21$.
You will get the same answer even if you take $x=-4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ correctly proved by induction? I have this statement:
Prove by induction that $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ for $n \in \mathbb{N}$
My attempt was:
Base case: $\frac{1}{1} \leq 1$
Assume that $\displaystyle \sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$ is true for some $m \in \mathbb{N}$
To prove: $\displaystyle \sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$
Since $$\sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$$
$$\sum_{k = 1}^m\frac{1}{k^2} + \frac{1}{(m+1)^2}\leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$
$$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$
Now, I'll prove that (1) $$\displaystyle 2 -\frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m + 1}$$
Developing this inequation, gives $$m^2 + 2m + 1 \geq m^2 + 2m$$ getting that (1) is true.
And by transitivity $$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$$ was proved.
But I don't know if this is a valid induction proof and if not, what is the type of this proof?
Thanks in advance.
| Let me show you a different (more direct) way to prove this inequality. Note that $k(k-1)<k^2$ and therefore $$\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}$$ for all $k>1.$
Now for $n>1$ we have $$\sum_{k=1}^n\dfrac{1}{k^2}\le1+\sum_{k=2}^n\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)=1+\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\cdots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right).$$ Simplify the RHS and see it for your self :)
| {
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"timestamp": "2023-03-29T00:00:00",
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If $y = \cos(2\theta)+\sin(2\theta)$ and $x = \cos(2\theta)-\sin(2\theta)$, express $y$ in terms of $x$. If $y = \cos(2\theta)+\sin(2\theta)$ and $x = \cos(2\theta)-\sin(2\theta)$, express $y$ in terms of $x$.
a) $y=\pm \sqrt{2+x^2}$
b) $y=\pm \sqrt{4-x^2}$
c) $y=\pm \sqrt{2-x^2}$
d) $y=\pm \sqrt{2-2x^2}$
I tried to find relations between the 2 but I'm finding it quite difficult and even after subbing them in, I couldn't find anything. Which of the multiple choice is correct? Please help! Thank you!
| Once you observe
$$
y^2 = 1 + 2 \cos 2 \theta \sin 2 \theta\\
x^2 = 1 - 2 \cos 2 \theta \sin 2 \theta
$$
It is clear that $y^2 + x^2 = 2$. Now all you have to prove is the statements above. Note there is nothing special about $2 \theta$ , any angle there and the same works. This is a generalization of $(x-y)^2 + (x+y)^2 = 2(x^2+y^2)$, and if $x^2+y^2$ can be simplified, this expression can be simplified as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find general solution of ODE $(1+t^2)x''+(x')^2+1=0$ I want to solve following ODE:
$$
(1+t^2)x''+(x')^2+1=0
$$
substitute $y = x'$
$$(1+t^2)y' + y^2 + 1 = 0\\
\frac{dy}{dt}=\frac{-(1+y^2)}{1+t^2}\\
\frac{dy}{1+y^2} = \frac{-dt}{1+t^2}
$$
integrating both sides leaves me with
$$
\arctan y = - \arctan t + c
$$
taking $\tan$ from both sides
$$
y = \tan ( - \arctan t + c)\\
\frac{dx}{dt} = \tan ( - \arctan t + c)\\
dx = dt \tan ( - \arctan t + c)
$$
I surely can't integrate that - I must have done mistake somewhere. Could somebody point me where and how to fix it?
| By reassigning the constant
$$\arctan y = \arctan C_1 - \arctan t = \arctan\left(\frac{C_1-t}{1+C_1t}\right)$$
one can take tangent on both sides and integrate getting the solution
$$x = \left(1+\frac{1}{C_1^2}\right)\log\left(1+C_1t\right) - \frac{1}{C_1^2}t + C_2$$
or in the case $C_1 \to \infty$ and $C_1 = 0$ we have
$$x = \log t+C_2$$
$$x = -\frac{1}{2}t^2+C_2$$
respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$y(x+y^3)dx=x(y^3-x)dy$
Solve $y(x+y^3)dx=x(y^3-x)dy$
Attempt: $$\frac{dy}{dx}=\frac{y(x+y^3)}{x(y^3-x)}.$$
Let $y^3=xt.$ Then $$3y^2\frac{dy}{dx} = t+x\frac{dt}{dx}.$$
Therefore
$$\frac{(t+x\frac{dt}{dx})}{(3y^2)} = \frac{t(t+1)}{y^2(t-1)}$$
and hence
$$x\frac{dt}{dx}=\frac{2t^2+4t}{t-1}.$$
Integrating, $$\ln|cx|=\frac{3}{4}\ln|t+2|-\frac{1}{4}\ln|t|.$$
Simplyfing, $$cx^2=\frac{y^3+2x}{y}$$ but the given answer is $y=cx^{1/3}$.
Where have I made a mistake?
| I would like to provide an alternate approach,
Given equation,
$$y(x+y^3)dx=x(y^3-x)dy$$
$$\implies \frac{x+y^3}{y^3-x}=\frac{xdy}{ydx}$$
$$\implies \frac{y^3}{x}=\frac{xdy+ydx}{xdy-ydx}$$
$$\implies y^3 \frac{(xdy-ydx)}{x^2}=\frac{xdy+ydx}{x}$$
$$\implies y^3d(\frac{y}{x})=\frac{d(xy)}{x}$$
$$\implies\frac{y}{x}d(\frac{y}{x})=\frac{d(xy)}{(xy)^2}$$
Integrating, we get
$$\frac{1}{2}\frac{y^2}{x^2}+\frac{1}{xy}=c'$$
If you take $c'=c/2$, it is exactly the same solution you got.
| {
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"url": "https://math.stackexchange.com/questions/3631542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\lim _{x\to \infty }\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right) =1/2$ I don't know how to start. Is it simple algebraic manipulation where,
if, let $a=\sqrt{x+\sqrt{x+\sqrt{x}}} $
and, $b=\sqrt{x}$
the above equation can be manipulated as
$\implies a-b$$.\:\frac{a+b}{a+b}=\frac{a^2-b^2}{a+b}$
giving, $\frac{\sqrt{x+\sqrt{x}}}{\left(\sqrt{x+\sqrt{+x\sqrt{+x}}}+\sqrt{x}\right)\:}$
Now, my mind can't think of any method to solve further.
| $\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\dfrac{\sqrt{x}\sqrt{1+\frac{\sqrt{x}}{x}}}{\sqrt{x}\bigg(\sqrt{1+\dfrac{\sqrt{x}+\sqrt{x}}{x}}+1\bigg)}=\dfrac{\sqrt{1+\frac{\sqrt{x}}{x}}}{\bigg(\sqrt{1+\dfrac{\sqrt{x}+\sqrt{x}}{x}}+1\bigg)}\to\dfrac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy $ f\left(m^{2}+m n\right)=f(m)^{2}+f(m) f(n) $ Question -
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the equation
$$
f\left(m^{2}+m n\right)=f(m)^{2}+f(m) f(n)
$$
for all natural numbers $m, n$
by putting $m=1$ and $f(1)=k$ we get
$f(n+1)=k^2 + kf(n)$
then hint says use $3^2 + 3.1 = 2^2 +2.4$ to get polynomial relation for k..
i am not getting how to use this hint ...i think i am missing some very easy tricks to get at this which i have not lernt yet ...
any help will be appreciated
thankyou
| Now this is my full solution if $0 \notin \mathbb{N}$
$$f(m^2+mn)=f(m)^2+f(m)f(n) \implies P(m,n)$$
Let $f(1)=k$
$$P(1,m) \implies f(m+1)=kf(m)+k^2$$
$$P(1,1) \implies f(2)=kf(1)+k^2=2k^2$$
$$P(1,2) \implies f(3)=kf(2)+k^2=k(2k^2)+k^2=2k^3+k^2$$
$$P(1,3) \implies f(4)=kf(3)+k^2=k(2k^3+k^2)+k^2=2k^4+k^3+k^2$$
and so on, by induction,
$$f(n)=2k^n+k^{n-1}+k^{n-2}+\dots+k^2 \tag{1}$$
for $n \geq3$.
Now $n=6$ in $(1)$ gives
$$f(6)=2k^6+k^5+k^4+k^3+k^2$$
while
$$P(2,1) \implies f(6)=f(2)^2+kf(2)=(2k^2)^2+k(2k^2)=4k^4+2k^3$$
So,
$$2k^6+k^5+k^4+k^3+k^2=4k^4+2k^3 $$
$$\Leftrightarrow 2k^6+k^5-3k^4-k^3+k^2=0$$
and since $k \neq 0$ we can divide by $k^2$:
$$2k^4+k^3-3k^2-k+1=0$$
We can easily find by the rational roots theorem that $k=1$ is the only possible root, and checking back, it works.
Thus, by $(1)$,
$$f(n)=2(1^{n})+\underbrace{1^{n-1}+\dots+1^{2}}_\text{$(n-2)$ terms}=2+n-2=n $$
for all $n \geq 3$. Since $f(1)=k=1$ and $f(2)=2k^2=2$, we can extend the definition:
$$f(n)=n$$
for all $n \in \mathbb{Z^+}$ $\Box$.
I'll make the induction proof of $(1)$ here. Our base case $n=3$ works, now
$$f(n)=2k^n+k^{n-1}+k^{n-2}+\dots+k^2$$
and
$$P(1,m) \implies f(m+1)=kf(m)+k^2$$
So
$$f(n+1)=kf(n)+k^2=k(2k^n+k^{n-1}+k^{n-2}+\dots+k^2)+k^2$$
$$=2k^{n+1}+k^{n}+k^{n-1}+\dots+k^3+k^2$$
So, indeed statement $(1)$ is true for all $n \geq 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$ Prove that $$1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$$ where $n$ is a positive integer.
My book suggests using AM-GM, but I couldn't do it. I just applied AM-GM to the numbers in the LHS, but it looks like I need some more upper bounds.
| By AM-GM, we have:
$ \frac{1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+...+(\frac{1}{n}+...+\frac{1}{n})}{\frac{n(n+1)}{2}}$
$=\frac{1+1+1+...+1}{\frac{n(n+1)}{2}}$, where there are $n$ copies of $1$.
$=\frac{n}{\frac{n(n+1)}{2}}=\frac{2}{n+1}$
$\geq \sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot (\frac{1}{n}...\frac{1}{n})}$
$=\sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n}}$
But $\frac{2}{n+1} \geq \sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n}} \Rightarrow 1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n} \leq (\frac{2}{n+1})^{\frac{n(n+1)}{2}} $, thus concluding our proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $2^{n}-1=\sum\limits_{k=0}^{n-1}2^k$? Recently I discovered that $2^{n}-1=\sum\limits_{k=0}^{n-1} 2^k$. As I don't have a math background, please tell me what this is called and a proof of why this is the case.
| For all positive integer $n$ :$$\begin{align}2^n&=\sum_{k=0}^{n}2^k-\sum_{k=0}^{n-1}2^k&&=(1+2+\cdots+2^{n-1}+2^n)-(1+2+\cdots+2^{n-1})\\[1ex]&=1+\sum_{k=1}^{n}2^k-\sum_{k=0}^{n-1}2^k&&=1+(2+\cdots+2^{n-1}+2^n)-(1+2+\cdots+2^{n-1})\\[1ex]&=1+2\sum_{k=1}^n2^{k-1}-\sum_{k=0}^{n-1}2^k&&=1+2(1+2+\cdots+2^{n-1})-(1+2+\cdots+2^{n-1})\\[1ex]&=1+2\sum_{k=0}^{n-1}2^k-\sum_{k=0}^{n-1}2^k\\[1ex]&=1+\sum_{k=0}^{n-1}2^k&&=1+(1+2+\cdots+2^{n-1})\\[3ex]\therefore\quad\sum_{k=0}^{n-1}2^k&=2^n-1\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum $\text{S} = \sum_{i = 2}^{2008}{\frac{1}{a_i}},$ where $a_1 = \frac{1}{3}$ and $ a_{n + 1} = a_n^2 + a_n.$ Define the sequence $\{a_n\}$ where $n \in \mathbb{Z^+}$ given by $a_1 = \frac{1}{3}$ and $$ a_{n + 1} = a_n^2 + a_n.$$ Let $$\text{S} = \sum_{i = 2}^{2008}{\frac{1}{a_i}},$$
then find $\lfloor S \rfloor$ where $\lfloor X \rfloor$ denotes the greatest integer lesser than or equal to $X$.
P.S.: The obvious approach would be to telescope, but as far as I can see, terms do not cancel at all and the estimation of S becomes cumbersome. I have also tried modifying it by adding $\frac{1}{4}$ to both sides and defining $b_n = a_n + \frac{1}{2}$ gives us $$b_{n + 1} = b_n^2 + \frac{1}{4}$$
but this doesn't help me in any way to estimate S. One can read off that the original sequence is increasing but i am unable to put an upper bound (such as a G.P.) to find [S].
| Start from recurrence relation $a_{n+1} = a_n(a_n+1)$, it is clear if we start from any $a_1 > 0$, $a_n$ will be a strictly increasing sequence.
If for some $N$, we have $a_N = \alpha > 1$, then for all $n \ge N$, we have
$$a_{n+1} = a_{n}(a_{n}+1) \ge a_n(1+\alpha)
\quad\implies\quad \frac{1}{a_{n+1}} \le \frac{1}{a_n(1+\alpha)}$$
This implies for all $k \ge 0$, we have $\displaystyle\;a_{N+k} \le \frac{1}{a_N}\frac{1}{(1+\alpha)^k}$. As a result,
$$\sum_{n=N+1}^\infty \frac{1}{a_n} \le \frac{1}{a_N}\sum_{k=1}^\infty\frac{1}{(1+\alpha)^k}
= \frac{1}{a_N}\frac{\frac{1}{1+\alpha}}{1 - \frac{1}{1+\alpha}} = \frac{1}{a_N\alpha} = \frac{1}{a_N^2}
$$
By brute force, we have
$$(a_1,a_2,a_3,a_4,a_5,\ldots) = (\frac13,\frac49,\frac{52}{81},\frac{6916}{6561},\frac{93206932}{43046721},\ldots)$$
Since $a_n > 1$ start at $n = 4$, we can take $N = 5$. By above argument, we have:
$$5.2182 \sim \sum_{n=2}^5 \frac{1}{a_n}
\le \sum_{n=2}^{2008} \frac{1}{a_n} < \sum_{n=2}^\infty \frac{1}{a_n} \le \sum_{n=2}^5 \frac{1}{a_n} + \frac{1}{a_5^2} \sim 5.4315$$
So the answer is $5$.
| {
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"question_score": "2",
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Prove $\frac1{x^4} < \frac1{x^3} - \frac1{(x+1)^3}$
Prove that for $x \ge 2$, $$\frac1{x^4} < \frac1{x^3} - \frac1{(x+1)^3}.$$
What I have so far is:
$${1\over x^3} - {1\over (x+1)^3} = {(x+1)^3-x^3\over x^3(x+1)^3} = {3x^2+3x+1\over x^3(x+1)^3} > {(x+1)^2\over x^3(x+1)^3} = {1\over x^3(x+1)}.$$
As seen, this won't lead to the correct expression, so could anyone give me any hints on how should I approach the question algebraically?
I've thought of using the graph of ${1\over x^4}$ and using the area-under the graph but I was thinking if there is a trick to solving it via algebraic means.
| $$\frac{3x^2+3x+1}{x^3(x+1)^3}$$
$$\gt\frac{3x^2+3x}{x^3(x+1)^3}$$
$$=\frac{3}{x^2(x+1)^2}$$
$$\gt\frac{1}{x^4}$$
because for $x\ge2$
$$\frac{3}{(x+1)^2}\ge \frac{1}{x^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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System of Equations: Distances from the Torricelli Point to the Vertices of a Triangle I am calculating the length between the Torricelli point and the triangle apex. With the law of cosines I get this system:
$$\left\{\begin{array}{cc}
x^2+y^2+xy&=a^2\,,\\x^2+z^2+xz&=b^2\,,\\z^2+y^2+zy&=c^2\,.\end{array}\right.$$
Please help me solve it.
| Here is some progress. This might be useful if your real task is not about finding the distances from the Torricelli point to the vertices, but the sum of the distances.
Let $ABC$ be a triangle whose internal angles are smaller than $\dfrac{2\pi}{3}$ (which means the Torricelli point is not one of its vertices, otherwise the problem is trivial). Suppose that $a=BC$, $b=CA$, and $c=AB$. The point $O$ is the Torricelli point of the triangle $ABC$. Write $p=OA$, $q=OB$, and $r=OC$. (Note that $x=r$, $y=q$, and $z=p$ in your notations.) Then,
$$a^2=q^2+qr+r^2\,,$$
$$b^2=r^2+rp+p^2\,,$$
and
$$c^2=p^2+pq+q^2\,.$$
Draw an equilateral triangle $BCB'$ such that $B'$ and $A$ are on the opposite sides of $BC$. Suppose that $O'$ is on the line segment $OB'$ such that $O'O=r$. It follows that $O'B'=q$, making
$$p+q+r=AB'=\sqrt{AC^2+B'C^2-2\cdot AC\cdot B'C\cos(\angle ACB')}\,.$$
We have $AC=b$, $B'C=a$, and $\angle ACB'=\angle ACB+\dfrac{\pi}{3}$, whence
$$\cos(\angle ACB')=\cos\left(\angle ACB+\frac{\pi}{3}\right)=\frac{\cos(\angle ACB)-\sqrt{3}\,\sin(\angle ACB)}{2}\,.$$
Since
$$\cos(\angle ACB)=\frac{a^2+b^2-c^2}{2ab}\text{ and }\sin(\angle ACB)=\frac{2\,\Delta}{ab}\,,$$
where $\Delta$ is the area of the triangle $ABC$, we get
$$p+q+r=\sqrt{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}\,\Delta}\,.\tag{*}$$
Because
$$a^2+b^2+c^2=2\,(p+q+r)^2-3\,(rp+pq+qr)\,,$$
we get
$$rp+pq+qr=\dfrac{4}{\sqrt{3}}\,\Delta\,.\tag{#}$$
(This equality can be found using an argument relating the areas of the triangles $BOC$, $COA$, $AOB$, and $ABC$.)
We also have
$$a^2(q-r)+b^2(r-p)+c^2(p-q)=(q^3-r^3)+(r^3-p^3)+(p^3-q^3)=0\,.$$
According to WolframAlpha, the expressions for $p$, $q$, and $r$ are terrible. Here is an expression for $p$:
$$p = \small\sqrt{\frac{-a^6+2b^6+2c^6 + 4 a^4( b^2 +c^2) - 5 a^2( b^4+c^4) +b^2c^2(b^2+c^2)- 3 a^2 b^2 c^2-\sqrt{3\,\Xi_a}}{6\,(a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2)}}\,,$$
where
$$\begin{align}\Xi_a&:=\small-a^{12} + 4 a^10 (b^2+c^2) - 6 a^8( b^4+c^4) - 10 a^8 b^2 c^2 + 4 a^6 (b^6+c^6)\\&\small\ \ \ \ \ + 10 a^6 b^2 c^2(b^2+c^2) - a^4 (b^8+c^8) - 6 a^4 b^2 c^2(b^4+c^4) - 3 a^4 b^4 c^4 \\&\small\ \ \ \ \ \ \ \ \ \ + 2 a^2 b^2 c^2 (b^8+c^6)- b^4 c^4(b^4+c^4) + 2 b^6 c^6\,.\end{align}$$
For $q$ and $r$, the expressions are similar. Apparently,
$$\Xi_a=16\Delta^2(a^2-b^2)^2(a^2-c^2)^2\,.$$
| {
"language": "en",
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"source": "stackexchange",
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Solving Exponential Diophantine Equations Find all positive integers $a, b, c > 1$, all relatively prime with respect to each other such that $b\mid 2^{a} +1, c\mid 2^{b} +1, a\mid 2^{c} +1$. $$$$ It is easy to see that all of $a, b, c$ are odd. Suppose $3\mid a$ then we have $3\mid 2^c+1$ and as $(a, c)=1$ so we have $c=6k+1$ or $c=6k+5$ for some $k$. But $9\mid a$ implies $9\mid 2^c+1$ which is not possible. This is true for all the three $a, b, c$. Hence either all of them are equal to $3$ or have some prime factor other than $3$. Suppose the later one is true. Let $p$ be a smallest prime factor of $b$ other than $3$ then we have $$2^{2a} \equiv 1 \mod p$$ and also we have $$2^{p-1} \equiv 1 \mod p$$
So we have $$2^d \equiv 1 \mod p$$ where $d=\gcd (2a, p-1)$. Now $d \neq 2$ as $p \neq 3$. Also $d \neq 3$ because that would have implied $p=7$ and $3|a$ and looking modulo $7$ we arrive at a contradiction. So let $q$ be a prime factor other than $3$ dividing both $a$ and $p-1$ then we have $q<p$ and also as $a$ divides $2^c+1$ we have $$2^{2c} \equiv 1 \mod q$$ and also $$2^{q-1} \equiv 1 \mod q$$. Again by the same reasoning as above we can show that there is a prime factor $r$ other than $3$ dividing both $c$ and $q-1$ and hence we have $r<q$ and as $c|2^b+1$ we have $$2^{2b} \equiv 1 \mod r$$ and also we have $$2^{r-1} \equiv 1 \mod r$$. Again by the same reasoning we can show that there exists a prime $s$ other than $3$ dividing both $b$ and $r-1$ and hence $s<r<q<p$ and $s$ divides $b$ bot as $p$ was the smallest prime factor other than $3$. So a contradiction and hence there do not exist such positive integers.
$$$$Is My Proof Correct?????
| The reasoning is correct, but not all steps are obvious (to me).
Some remarks:
*
*$3 \mid 2^c+1$ is always true as $c$ is uneven. $2^{2k+1}+1=2.(4)^k+1 \equiv 2.1+1 \equiv 0 \pmod{3}$.
*$2^{6k+r}+1=2^r.(64)^k+1 \equiv 2^r.1+1 \pmod{9}$. Hence $9 \mid 2^{6k+r}+1$ implies $r=3$, and thus $3 \mid c$. Hence, assuming $a=3^n$ implies $a=3$ as $(a,c)=1$, and as $b \mid 2^a+1=9$ we conclude that $b=1$. This implies $c=1$. But $(3,1,1)$ is not a feasible solution.
*$2^a \equiv -1 \pmod{p}$ implies $2^{2a} \equiv 1 \pmod{p}$
*$d=2\gcd(a,(p-1)/2)$. Now $d=6$ implies $2^6 \equiv 1 \pmod{p}$, hence $p=7$. Furthermore, $3 \mid a$ so $a$ is of the form $a=3(2k+1)=6k+3$. Hence $2^a+1=2^{6k+3}+1=$ $8.(2^6)^k+1 \equiv 1.1+1=2 \pmod{7}$. This contradicts with $7 \mid 2^a+1$. We conclude that $d \ge 10$ and $p \ge 11$.
I found an excellent paper with several lemma's regarding $2^n+1$. When looking for all (not coprime) solutions, it is remarkable that only a few prime numbers can appear in $a$, $b$ or $c$; see A057719.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a function $z=f(x, y)$ for the double integral The functions are given on the $xy$ plane, the $xz$ plane and the $yz$ plane, respectively, i.e. $y=1-x^2$, $z=1-x^2$ and $z=1-y^2$ for $0 \leq x \leq 1, \ 0 \leq y \leq 1$ and $0 \leq z \leq 1$.
My original purpose was to find the area bounded by the three functions and the lines $x=0, y=0$ and $z=0$.
To do so I tried to find the function $z=f(x, y)$ that allows me to calculate
$$V=\int_{x=0}^{x=1} \int_{y=0}^{y=1-x^2} f(x, y) \ dy \ dx$$
but I couldn't. The best I could find was $f(x,y)=1-x^2-y^2$ which obviously does not match with $y=1-x^2$ when $z=0$.
How can I find it?
| The region $\mathcal{R}$ defined by the given constraints is the intersection of three solids. The boundary surfaces of each solid are a parabolic cylinder and four planes. You can only solve this problem by using two different functions $z=f(x,y)$ and $z=g(x,y)$. There are only two possibilities: $z=1-x^2$ and $z=1-y^2$.
The intersection of the surfaces $z=1-x^2$ and $z=1-y^2$ is the plane $y=x$. The plane $y=x$ intersects the surface $y=1-x^2$ at $x=(\sqrt{5}-1)/2 = \phi$. Define the points $A,B,C,D$ in the $xy$ plane as $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, and $D=(\phi,\phi)$. The $xy$ domain of the region $\mathcal{R}$ when $z=1-x^2$ is the triangle $T_0$ with the two linear edges $AB$ and $AD$ and the curved edge $y=1-x^2$ for $x\in [\phi,1]$. We could integrate over $T_0$ with one double integral if we integrate with respect to $x$ first, but this would result in some square roots in the integrals. So we break $T_0$ into two triangles $T_1$ and $T_2$. The first triangle $T_1$ is the right triangle with vertices $A$, $D$, and $E=(\phi,0)$. The second triangle $T_2$ is the triangle with linear edges $EB$ and $ED$ and the curved edge of $T_0$. Computing the double integrals over $T_1$ and $T_2$ is easy:
$$
\begin{align}
I_1 &= \iint_{T_1} 1-x^2 \,dA = \int_0^{\phi}(1-x^2)\int_0^x \,dy\,dx \\
&=\int_0^{\phi} x(1-x^2) \, dx = \frac{\phi^2}{2} - \frac{\phi^4}{4}.
\end{align}
$$
and
$$
\begin{align}
I_2 &= \iint_{T_2} 1-x^2 \,dA = \int_{\phi}^1 (1-x^2) \int_0^{1-x^2} dy\,dx \\
&= \int_{\phi}^1 (1-x^2)^2 \,dx = \frac{8}{15} - \phi + \frac{2}{3}\phi^3 - \frac{1}{5}\phi^5.
\end{align}
$$
The $xy$ domain of the region $\mathcal{R}$ when $z=1-y^2$ is the triangle $T_3$ with linear edges $AC$ and $AD$ and curved edge $y=1-x^2$ for $x\in [0,\phi]$. The double integral over $T_3$ is
$$
\begin{align}
I_3 &= \iint_{T_3} 1-y^2 \,dA = \int_0^{\phi}\int_x^{1-x^2} 1-y^2 \,dy\,dx \\
&= \frac{2}{3}\phi - \frac{1}{2}\phi^2 + \frac{1}{12}\phi^4 - \frac{1}{5}\phi^5 + \frac{1}{21}\phi^7.
\end{align}
$$
The total volume is
$$
\begin{align}
V &= I_1+I_2+I_3 = \frac{8}{15} - \frac{1}{3}\phi + \frac{2}{3}\phi^3 - \frac{1}{6}\phi^4 -
\frac{2}{5}\phi^5 + \frac{1}{21}\phi^7 \\
&= \frac{123}{420} + \frac{5}{84}\sqrt{5} \approx 0.425956.
\end{align}
$$
One can easily check this result using Monte Carlo integration with $N$ uniform random samples selected from the unit cube. For each sample inside the region $\mathcal{R}$, add 1 to a variable $S$ that is initialized to 0. The Monte Carlo estimate is $S/N$.
| {
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"If $a+a^3$ is irrational, then $a+a^2$ is also irrational". Given $a\in\mathbb{R}$, define $p=a+a^2$ and $q=a+a^3$. Show that the following statement is false:
"If $q$ is irrational, then $p$ is also irrational".
My approach: I just was lucky to find a counterexample setting $p=1$. Then, I found out that for both $a$ that satisfy the equation $q$ is irrational.
I wonder if someone here could come up with a more neat solution.
| $a^2 +a - p=0$ so $a = \frac{-1\pm\sqrt{1+4p}}2$ and so
$q = (\frac{-1\pm\sqrt{1+4p}}2)^3 + ((\frac{-1\pm\sqrt{1+4p}}2))$
To get a counter example $p$ needs to be rational $q$ does not.
$\sqrt{1+4p}$ will only be rational if $1+4p$ is perfect rational square. It is trivial easy to choose rational $p$ where $4p+1$ is not perfect rational square.
If $\sqrt{1+4p}$ is rational then
$q = (\frac{-1\pm\sqrt{1+4p}}2)^3 + ((\frac{-1\pm\sqrt{1+4p}}2))=$
$r_1 + r_2*\sqrt{1+4p} + r_3*\sqrt{1+4p}^2 + r_4*\sqrt{1+4p}^3=$
$r_1 + r_2*\sqrt{1+4p} + r_3*(1+4p) + r_4*(1+4p)\sqrt{1+4p}=$
$s_1 + s_2\sqrt{1+4p}$ for some $r_1, r_2, r_3, s_1, s_2$.
So long as $s_2\ne 0$ and $1+4p$ is not a perfect square, then $q$ will be irrational.
And $s_2 =0$ is trivially easy to avoid[1].
So there was nothing particularly "lucky" about choosing $p=1$.
[1] $s_2 = r_2 + r_4(1+4p)$
$r_2 = 3(-\frac 12)^2(\pm \frac 12)+1=\pm \frac 38 + 1$ and $r_4=(\pm \frac 12)^3=\pm \frac 18$
So $s_2 = \pm \frac 12 +1 \pm \frac 12p$.
So there are precisely two values of $p$ to avoid.
| {
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Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\frac{1}{\sqrt{2}}\left(\sqrt{46-6\sqrt{5}}-2\sqrt{6-2\sqrt{5}}\right)=$$
$$=\frac{1}{\sqrt{2}}\left(\sqrt{(3\sqrt{5}-1)^2}-2\sqrt{(\sqrt5-1)^2}\right)=\frac{1}{\sqrt{2}}\left(3\sqrt{5}-1-2(\sqrt{5}-1)\right)=$$
$$=\frac{1}{\sqrt{2}}(\sqrt{5}+1)=\frac{1}{\sqrt{2}}\sqrt{6+2\sqrt{5}}=\sqrt{3+\sqrt{5}}.$$
But this way does not work for the starting identity.
How to prove the starting identity without squaring?
Thank you!
| It still works, albeit cumbersome
\begin{align}
& \sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}\\
= & \frac12 \sqrt{\left(\sqrt{1775-425\sqrt{17}} +\sqrt{1207-289\sqrt{17}} \right)^2}
-\frac12 \sqrt{\left(\sqrt{639-153\sqrt{17}} +\sqrt{1207-289\sqrt{17}} \right)^2}\\
= & \frac52 \sqrt{71-17\sqrt{17}} +\frac12\sqrt{1207-289\sqrt{17}}
-\frac32 \sqrt{71-17\sqrt{17}} -\frac12 \sqrt{1207-289\sqrt{17}} \\
=&\left( \frac52-\frac32\right) \sqrt{71-17\sqrt{17}}\\
=& \sqrt{71-17\sqrt{17}}\\
\end{align}
| {
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Solving the system with $x^2 -xy -1=0$ and $2xy-4y^2 +3=0$ I want to solve the following system of quadratic equations:
$$\begin{align}
x^2 -xy -1 &=0 \\ 2xy-4y^2 +3 &=0
\end{align}$$
How can I do it?
My attempt:
$$xy = 1 -x^2$$
$$2x^2=4y^2-1$$
$$x = \sqrt{2y^2 - 0.5}$$
After plugging $x$, I get an equation that I can't solve.
| The given equations
$$\begin{align}
& x^2 -xy = 1\tag1\\ &2xy-4y^2 = -3
\end{align}$$
lead to
$$3(x^2 -xy)+ (2xy-4y^2)=0\implies (x+y)(3x-4y)=0$$
Then, substitute $x+y=0$ and $3x-4y=0 $ into (1) to obtain the
solutions
$$(\pm\frac1{\sqrt2},\mp \frac1{\sqrt2}),\>\>\>
(\pm2,\pm \frac3{2})$$
| {
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show this sum $\sum_{i=0}^{n}(-1)^{i-j}\binom{n+i}{i}\binom{n+k-i}{k-i}\binom{2n}{n+j-i}=\binom{2n}{n}$
Let $n,k$ be positive integers. Show that
$$\sum_{i=0}^{n}(-1)^{i-j}\binom{n+i}{i}\binom{n+k-i}{k-i}\binom{2n}{n+j-i}=\binom{2n}{n}$$
for all $j=0,1,2,\cdots,k$.
I don't know if this is an existing conclusion. After thinking for a long time, I found it difficult to deal with it.
Batominovski's edit:
However, when $k=0$ and $j=0$, the required sum is
$$S=\sum_{i=0}^n(-1)^i\binom{n+i}{i}\binom{n-i}{-i}\binom{2n}{n-i}=\binom{n}{0}\binom{n}{0}\binom{2n}{n}=\binom{2n}{n}.$$
When $k=1$ and $j=0$, the required sum is
$$S=\binom{n}{0}\binom{n+1}{1}\binom{2n}{n}-\binom{n+1}{1}\binom{n}{0}\binom{2n}{n-1}=(n+1)\binom{2n}{n}-(n+1)\binom{2n}{n-1}.$$
Because $\binom{2n}{n-1}=\binom{2n}{n+1}$, we have
$$(n+1)\binom{2n}{n-1}=\binom{n+1}{1}\binom{2n}{n+1}=\binom{n}{1}\binom{2n}{n}=n\binom{2n}{n}.$$
Hence, $S=(n+1)\binom{2n}{n}-n\binom{2n}{n}=\binom{2n}{n}$. When $k=1$ and $j=1$, the required sum is
$$S=-\binom{n}{0}\binom{n+1}{1}\binom{2n}{n+1}+\binom{n+1}{1}\binom{n}{0}\binom{2n}{n}=-(n+1)\binom{2n}{n+1}+(n+1)\binom{2n}{n}.$$
So, similar to the previous case, $S=\binom{2n}{n}$. How to prove this for generalized pairs $(k,j)$? Is there a combinatorial proof?
| Numerical evidence indicates that the upper limit should be $k$ and not
$n$ for this to hold including when $n\lt k.$ So we seek to show that
$$\sum_{q=0}^k (-1)^{q-j} {n+q\choose q}
{n+k-q\choose k-q} {2n\choose n+j-q} = {2n\choose n}$$
where $0\le j\le k.$ The LHS is
$$(-1)^j [w^{n+j}] (1+w)^{2n}
\sum_{q=0}^k (-1)^{q} {n+q\choose q} w^q
{n+k-q\choose k-q}
\\ = (-1)^j [w^{n+j}] (1+w)^{2n}
[z^k] \frac{1}{(1+wz)^{n+1}} \frac{1}{(1-z)^{n+1}}.$$
The inner term is
$$\mathrm{Res}_{z=0} \frac{1}{z^{k+1}}
\frac{1}{(1+wz)^{n+1}} \frac{1}{(1-z)^{n+1}}.$$
Residues sum to zero and the residue at infinity is zero by inspection.
We get for the residue at $z=1$
$$(-1)^{n+1} \mathrm{Res}_{z=1} \frac{1}{z^{k+1}}
\frac{1}{(1+wz)^{n+1}} \frac{1}{(z-1)^{n+1}}
\\ = (-1)^{n+1} \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{k+1}}
\frac{1}{(1+w+w(z-1))^{n+1}} \frac{1}{(z-1)^{n+1}}
\\ = \frac{(-1)^{n+1}}{(1+w)^{n+1}}
\mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{k+1}}
\frac{1}{(1+w(z-1)/(1+w))^{n+1}} \frac{1}{(z-1)^{n+1}}
\\ = \frac{(-1)^{n+1}}{(1+w)^{n+1}}
\sum_{q=0}^n {n+q\choose q} (-1)^q \frac{w^q}{(1+w)^q}
(-1)^{n-q} {k+n-q\choose k}
\\ = - \sum_{q=0}^n
{n+q\choose q} \frac{w^q}{(1+w)^{n+1+q}} {k+n-q\choose k}.$$
Substitute into the coefficient extractor in $w$ to get
$$- (-1)^j \sum_{q=0}^n
{n+q\choose q} {k+n-q\choose k}
[w^{n+j-q}] (1+w)^{n-1-q}.$$
Now with $0\le q\le n-1$ and $j\ge 0$ we have $[w^{n+j-q}]
(1+w)^{n-1-q} = 0.$ This leaves $q=n$ which yields
$$-(-1)^j {2n\choose n} {k\choose k} [w^j] \frac{1}{1+w}
= - {2n\choose n}.$$
This is the claim. We have the result if we can show
that the residue at $z=-1/w$ makes for a zero contribution.
We get
$$\frac{1}{w^{n+1}} \mathrm{Res}_{z=-1/w} \frac{1}{z^{k+1}}
\frac{1}{(z+1/w)^{n+1}} \frac{1}{(1-z)^{n+1}}.$$
This requires
$$\frac{1}{n!}
\left(\frac{1}{z^{k+1}} \frac{1}{(1-z)^{n+1}}\right)^{(n)}
= \frac{1}{n!} \sum_{q=0}^n {n\choose q}
\frac{(-1)^q (k+q)!}{z^{k+1+q} \times k!}
\frac{(n+n-q)!}{(1-z)^{n+1+n-q} \times n!}
\\ = \sum_{q=0}^n {k+q\choose k} (-1)^q \frac{1}{z^{k+1+q}}
{2n-q\choose n} \frac{1}{(1-z)^{2n+1-q}}.$$
Evaluate at $z=-1/w$ and restore the factor in front:
$$\frac{1}{w^{n+1}}
\sum_{q=0}^n {k+q\choose k} (-1)^{k+1} w^{k+1+q}
{2n-q\choose n} \frac{1}{(1+1/w)^{2n+1-q}}.$$
Applying the coefficient extractor in $w$ we get
$$(-1)^j [w^{n+j}] (1+w)^{2n}
\frac{1}{w^{n+1}} w^{k+1+q} \frac{w^{2n+1-q}}{(1+w)^{2n+1-q}}
\\ = (-1)^j [w^{n+j}] (1+w)^{q-1} w^{n+k+1}
= (-1)^j [w^j] (1+w)^{q-1} w^{k+1} = 0$$
because $j\le k.$ This concludes the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find the value of $a$ and $b$? $$
D=d(PA)+d(AB)+d(BQ)=\sqrt{a^2+1^2}+\sqrt{(b-a)^2+2^2}+\sqrt{(4-b)^2+1^2}$$
Given $\frac{\partial D}{\partial a}=0$ and $\frac{\partial D}{\partial b}=0$
find the value of $a$ and $b$
My attempt : i found the answer here but im not getting the answer
My work : $\frac{\partial D}{\partial a}= \frac{2a}{ \sqrt{ 1^2 + a^ 2}} +\frac{-2b+2a}{ \sqrt{ 2^2 + (b-a)^ 2}} $
Now we have $\frac{\partial D}{\partial a}=0$
$\frac{2a}{ \sqrt{ 1^2 + a^ 2}} =\frac{2b-2a}{ \sqrt{ 2^2 + (b-a)^ 2}} $
after that im not able to proceed further
| $$\begin{align}
&\frac{\partial D}{\partial a}= \frac{a}{ \sqrt{ 1^2 + a^ 2}} -\frac{b-a}{ \sqrt{ 2^2 + (b-a)^ 2}}=0\\
&\frac{\partial D}{\partial b}=\frac{b-a}{ \sqrt{ 2^2 + (b-a)^ 2}}-\frac{4-b}{ \sqrt{1^2+ (4-b)^2}} =0
\end{align}
$$
It follows:
$$
\frac{a}{ \sqrt{ 1^2 + a^ 2}}=\frac{4-b}{ \sqrt{1^2+ (4-b)^2}}
\implies a=4-b.
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum and difference formulas for $\csc$ and $\sec$ We all are familiar with the sum and difference formulas for $\sin$ and $\cos$, but is there an analogue for the sum and difference formulas for secant and cosecant? That is,
$$\csc (A\pm B) = ?$$ and $$\sec (A \pm B) = ?$$
I tried a variation of the sum and difference formulas, but they were incorrect. Can it be derived geometrically?
| Note
$\sin^2 A \cos^2 B -\cos^2 A \sin^2 B\\
= \sin^2 A (1-\sin^2 B )-\cos^2 A \sin^2 B\\
= \sin^2 A - (\cos^2 A + \sin^2 A)\sin^2 B
= \sin^2 A - \sin^2 B \\
=\frac12(\cos2B-\cos2 A )\\
$
Thus,
$\csc (A \pm B)
=\frac {1}{\sin A \cos B \pm \cos A \sin B}
=\frac {\sin A \cos B \mp\cos A \sin B}{\sin^2 A \cos^2 B - \cos^2 A \sin^2 B}
= \frac {2\sin(A\mp B)}{\cos2B- \sin2 A }
$
Similarly
$\sec (A \pm B)= \frac {2\cos(A\mp B)}{\cos2B+ \sin2 A }
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3665879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.
Question: If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.
Solution: Observe that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}\\\iff \frac{ab+bc+ca}{abc}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}\\\iff ab+bc+ca\le 3abc+2(a^3+b^3+c^3)\hspace{0.5 cm}(\because abc>0)\\\iff(ab+bc+ca)(a+b+c)-3abc\le 2(a^3+b^3+c^3)\hspace{0.5 cm}(\because a+b+c=1)\\\iff a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\le 2(a^3+b^3+c^3).$$
Thus it is sufficient to prove that $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\le 2(a^3+b^3+c^3).$$
Let us assume WLOG that, $a\le b\le c\implies a^2\le b^2\le c^2.$ Thus by rearrangement inequality, we have $$a^2b+b^2c+c^2a\le a^2a+b^2b+c^2c=a^3+b^3+c^3.\tag 1$$
Again, by rearrangement inequality, we have $$a^2c+b^2a+c^2b\le a^2a+b^2b+c^2c=a^3+b^3+c^3. \tag 2$$
Thus adding $(1)$ and $(2)$, we have $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\le 2(a^3+b^3+c^3),$$ which is our required inequality. Hence, we are done.
Is there a better solution than this one?
| Also, by using the cyclic sum we can use SOS and we can write the solution in one line:
$$3+\tfrac{2(a^3+b^3+c^3)}{abc}-\sum_\text{cyc}\tfrac{1}{a} = \tfrac{\sum\limits_\text{cyc}(2a^3+abc-a^2b-a^2c-abc)}{abc} = \tfrac{\sum\limits_\text{cyc}(a^3-a^2b-ab^2+b^3)}{abc} = \tfrac{\sum\limits_\text{cyc}(a+b)(a-b)^2}{abc}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$ For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$
Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left(
a+b-2\,c \right) ^{2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left(
119\,a+119\,b+30\,c \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2}
\right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right) \left( a+b \right) \left( 16\,{a}^{2}+a
b+16\,{b}^{2} \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right) \left( a+b \right) ^{
2} \Big] \geqq 0$
which is clearly true for $c=\max\{a,b,c\}$
I wish to see another proof without $uvw$! Thanks for a real lot!
You can see also here.
| Here $a,b,c$ replaced for $x,y,z$ for convenience.
For $x,y,z>0$ prove that
\begin{align}
\frac{xy}z+\frac{yz}x+\frac{zx}y+
\frac{81xyz}{4(x+y+z)^2}
&\ge
\tfrac74(x+y+z)
\tag{1}\label{1}
\end{align}
Using Ravi substitution,
\begin{align}
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
,
\end{align}
the inequality \eqref{1} transforms
(omitting lengthy, but straightforward details) into equivalent
inequality
\begin{align}
85\,r^2+32\,r\,R+64\,R^2
-15\,\rho^2
&\ge 0
\tag{2}\label{2}
\end{align}
in terms of the semiperimeter $\rho$,
inradius $r$ and circumradius $R$
of some valid triangle with the side lengths
\begin{align}
a&=y+z
,\quad
b=z+x
,\quad
c=x+y
.
\end{align}
Dividing \eqref{2} by $R^2$, we get
\begin{align}
85\,v^2+32\,v+64
-15\,u^2
&\ge 0
\tag{3}\label{3}
,
\end{align}
where $u=\rho/R$, $v=r/R$.
Now all we have to do is to check \eqref{3}
for all valid triangles, that means for all $v\in[0,\tfrac12]$
and $u(v)\in[u_{\min}(v),u_{\max}(v)]$.
Obviously,
\begin{align}
85\,v^2+32\,v+64
-15\,u^2
&\ge
85\,v^2+32\,v+64
-15\,u_{\max}(v)^2
\tag{4}\label{4}
\end{align}
where
\begin{align}
u_{\max}&=\sqrt{27-(5-v)^2+2\sqrt{(1-2\,v)^3}}
\tag{5}\label{5}
,
\end{align}
and we have \eqref{1} equivalent to
\begin{align}
50v^2-59v+17-15\sqrt{(1-2v)^3}
&\ge 0
\tag{6}\label{6}
,\\
(50v^2-59v+17)^2-15^2((1-2v)^3)
&\ge 0
\tag{7}\label{7}
,\\
(25v-8)^2(1-2v)^2
&\ge 0
\tag{8}\label{8}
,
\end{align}
which holds for all values of $v\in[0,\tfrac12]$,
that is for all valid triangles with the side lengths given by
\eqref{2} and hence,
\eqref{1} is true
for all real $x,y,z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the maximum and minimum value of $f(x)$
$$f(x) = \sin x + \int_{-\frac \pi 2}^{\frac \pi 2} (\sin x + t\cos x)f(t)dt$$
Find the minimum and maximum value of $f(x)$.
My attempt:
Rewrite the functional equation as
$$f(x) = \sin x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) + \cos x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
Then differentiate both sides
$$f'(x) = \cos x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) - \sin x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
for maxima/minima, $f'(x)$ = 0
$$\cos x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) = \sin x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
I got stuck at this point.
| I found a solution. Since
$$f(x) = \sin x \left( 1 + \int_{-\frac \pi 2}^{\frac \pi 2} f(t)dt\right) + \cos x \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt$$
We can rewrite it as $$f(x) = A\sin x + B\cos x$$
This gives us the equations
$$\begin{gather}
A = 1 + \int_{-\frac \pi 2}^{\frac \pi 2}f(t)dt \tag{1} \\
B = \int_{-\frac \pi 2}^{\frac \pi 2} tf(t)dt \tag{2}
\end{gather}$$
Using some integration properties, it is easy to see that $\int_{-\frac \pi 2}^{\frac \pi 2}f(t) = \int_{-\frac \pi 2}^{\frac \pi 2}B\cos t dt$ and $\int_{-\frac \pi 2}^{\frac \pi 2}tf(t)dt = \int_{-\frac \pi 2}^{\frac \pi 2}At\sin t dt$. Evaluating these integrals and substituting in the equations, they simplify to
$$\begin{gather}
A = 1 + 2B \tag{1} \\
B = 2A \tag{2}
\end{gather}$$
Solving this system gives $A = -\frac 13$ and $B = -\frac 23$. Thus
$$f(x) = -\frac 13 \sin x - \frac 23 \cos x$$
The maximum and minimum values of $f(x)$ are $\frac{\sqrt{5}}{3}$ and $-\frac{\sqrt{5}}{3}$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
In one-step induction, why sometimes there are 2 base cases? I noticed in various induction proofs, that it uses 2 base cases (e. g: for $n = 1$ and $2$ the problem trivially holds...)
Why is that? Isn't one base case enough for one-step induction?
| For example.
Prove that:
$$\frac{1}{n+1}+\frac{1}{n+3}+\frac{1}{n+5}+...+\frac{1}{3n-1}\geq\frac{1}{2}$$ for any natural $n$.
For the proof by induction we need to check $n=1$ and $n=2$ because we need a transition from $n$ to $n+2$.
I'll write a full proof and I hope it would be clear.
For $n=1$ we need to check:
$$\frac{1}{2}\geq\frac{1}{2},$$ which is true.
For $n=2$ we need to check:
$$\frac{1}{3}+\frac{1}{5}\geq\frac{1}{2}$$ or
$$\frac{8}{15}\geq\frac{1}{2},$$ which is true.
Now, it's enough to prove that: $$\frac{1}{n+1}+\frac{1}{n+3}+\frac{1}{n+5}+...+\frac{1}{3n-1}\geq\frac{1}{2}\Rightarrow$$
$$\Rightarrow\frac{1}{n+3}+\frac{1}{n+5}+...+\frac{1}{3n-1}+\frac{1}{3n+1}+\frac{1}{3n+3}+\frac{1}{3n+5}\geq\frac{1}{2},$$ for which it's enough to proof that:
$$\frac{1}{2}-\frac{1}{n+1}+\frac{1}{3n+1}+\frac{1}{3n+3}+\frac{1}{3n+5}\geq\frac{1}{2}$$ or
$$\frac{1}{3n+1}+\frac{1}{3n+5}\geq\frac{2}{3(n+1)},$$ which is true by C-S:
$$\frac{1}{3n+1}+\frac{1}{3n+5}\geq\frac{(1+1)^2}{3n+1+3n+5}=\frac{2}{3(n+1)}$$ and by induction we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Need help understanding matrix representations of the symmetric group $S_3$. I have the following map for a representations of $S_3$:
$$e \mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 2) \mapsto
\begin{pmatrix}
0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 3)\mapsto
\begin{pmatrix}
0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0
\end{pmatrix}$$
$$(2\; 3)\mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0
\end{pmatrix}, \quad (1\; 2\; 3) \mapsto
\begin{pmatrix}
0& 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{pmatrix}, \quad (1\; 3\; 2)\mapsto
\begin{pmatrix}
0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0
\end{pmatrix}$$
We can check that any $\sigma \in S_3$ and its image under the map represents the same permutation. For example, consider multiplying the matrix associated with $(2\; 3)$ with the column vector $[a\; b\; c]$:
$$\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0
\end{pmatrix} \begin{pmatrix}
a \\ b \\ c
\end{pmatrix} = \begin{pmatrix}
a\\c\\b
\end{pmatrix}$$
The second and third element in the column vector are interchanged, with the first element remaining fixed. This is the kind of behavior I expected any representation of $S_3$ will be exhibit. However, when I consider the "standard representation" of $S_3$, given as:
$$e \mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 2) \mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1
\end{pmatrix}, \quad (1\; 3)\mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0
\end{pmatrix}$$
$$(2\; 3)\mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 1 & \\ 0 & 1 & -1
\end{pmatrix}, \quad (1\; 2\; 3) \mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1
\end{pmatrix}, \quad (1\; 3\; 2)\mapsto
\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 1 \\ 0& -1 & 0
\end{pmatrix}$$
I don't know how to interpret the result I get from looking at it the way I was looking above, using matrix multiplication. For instance,
$$
\begin{pmatrix}
1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
a\\b\\c\end{pmatrix}= \begin{pmatrix}
a\\-b+c\\c
\end{pmatrix},$$
from which I have no idea what to infer.
I realize I'm not explaining myself very well, but I hope someone can maybe take a stab at answering my question anyway. Thank you very much for your help.
| Your two representations are equivalent, the permutation representation, they just use different bases to represent linear operators as different matrices. The first uses the standard basis.
The permutation representation of $S_3$ is reducible; the span of $(1,1,1)$ is an invariant subspace, as is its complement comprised of $(x,y,z)$ satisfying $x+y+z=0$. Since decomposition into irreducible representations is the primary focus of representation theory, the 2D subspace is what's actually called the "standard representation". In general, the permutation representation of $S_n$ is a direct sum of a 1D trivial subrepresentation and the standard representation of dimension $n-1$, just like for $S_3$.
In the second set of matrices you present, the basis $\{(1,1,1),(1,-1,0),(0,1,-1)\}$ is used instead. To calculate the $2\times2$ part of the matrix for say $(12)$, write out
$$ (12)\left(\color{red}{a}\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\color{blue}{b}\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix}\right)=a\begin{bmatrix}-1\\ \phantom{+}1 \\ \phantom{+}0\end{bmatrix}+b\begin{bmatrix}\phantom{+}1\\ \phantom{+}0\\-1\end{bmatrix} $$
$$ = -a\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+b\left(\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix}\right)=\color{green}{(-a+b)}\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\color{purple}{b}\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix} $$
which matches
$$ \begin{bmatrix} -1 & 1 \\ \phantom{+}0 & 1 \end{bmatrix} \begin{bmatrix} \color{red}{a} \\ \color{blue}{b} \end{bmatrix} = \begin{bmatrix} \color{green}{-a+b} \\ \phantom{+}\color{purple}{b} \end{bmatrix}. $$
(Apologies to the color-blind.)
These $2\times2$ matrices also represent the possible permutations of $\{0,1,\infty\}$ in the Riemann sphere using Mobius transformations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Long division of polynomial I am reading a math history book which says:
"Gregory could have learned in Italy that the area under the curve $y=1 / (1+x^2)$, from
$x=0$ to $x=x$, is arctan x, and a simple long division converts $
1 / (1+x^2)$ to $1 - x^2 + x^4 - x^6+...$"
Can someone please explain how is
$1/(1+x^2) = 1 - x^2 + x^4 - x^6+...$
And
How do you perform long divison of this kind?
Thanks.
| Knowledge of geometric series tells me that
$$\frac1{1-z}=\sum_{n\ge 0}z^n=1+z+z^2+z^3+\ldots\;;$$
if I now set $z=-x^2$, I find that
$$\frac1{1+x^2}=\sum_{n\ge 0}(-x^2)^n=\sum_{n\ge 0}(-1)^nx^{2n}=1-x^2+x^4-x^6+-\ldots\;.$$
That said, it is possible to do the long division, with a little ingenuity:
$$\begin{align*}
\frac1{1+x^2}&=\frac{(1+x^2)-x^2}{1+x^2}\\
&=1-\frac{x^2}{1+x^2}\\
&=1-\frac{(x^2+x^4)-x^4}{1+x^2}\\
&=1-\frac{x^2(1+x^2)-x^4}{1+x^2}\\
&=1-x^2+\frac{x^4}{1+x^2}\\
&=1-x^2+\frac{(x^4+x^6)-x^6}{1+x^2}\\
&=1-x^2+\frac{x^4(1+x^2)-x^6}{1+x^2}\\
&=1-x^2+x^4-\frac{x^6}{1+x^2}\\
&\;\;\vdots
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $a+b+c=0$ find $\frac{a^2+b^2+c^2}{b^2-ca}$ I tried finding the value of $a^2+b^2+c^2$ and it is $-2(ab+bc+ca)$ but that doesn't have any common factor with $b^2-ca$ so that didn't help.
I squared on both sides of $a+b=-c$ to get $a^2+b^2-c^2+2ab=0$. But I didn't proceed any further.
| With $b = \lambda a, c = \mu a$
$$
\frac{1+\lambda^2+\mu^2}{\lambda^2+\mu}=\frac{1+\lambda^2+(1+\lambda)^2}{\lambda^2+\lambda+1} = 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Evaluating $\sum_{r=1}^{2001} f\left(\frac{r}{2002}\right)$, where $f(x)=\frac{4^{x}}{4^{x}+2}$
Let
$
f(x)=\dfrac{4^{x}}{4^{x}+2}
$. Find
$
\sum_{r=1}^{2001} f\left(\frac{r}{2002}\right)
$.
Given,
$$f(x)=\frac{4^{x}}{4^{x}+2}$$
$$
\begin{align}
\sum_{r=1}^{2001} f\left(\frac{r}{2002}\right) &=\sum_{r=1}^{2001} \frac{4^{\frac{r}{2002}}}{4^{\frac{r}{2002}}+2} \\[4pt]
&=\frac{4^{\sum_{r=1}^{2001}\frac{r}{2002}}}{4^{\sum_{r=1}^{2001}\frac{r}{2002}}+2} \\[4pt]
&=\frac{4^{\frac{1}{2002}+\frac{2}{2002}+\cdots \cdot \frac{2001}{2002}}}{4^{\frac{1}{2002}+\frac{2}{2002}+\cdots \cdot \frac{2001}{2002}} +2}
\end{align}$$
Upon simplifying, I am getting
$$\frac{4^{{2001}}}{4^{{2001}}+2}$$
What to do next?
| There is a different way to solve this question.
You need use this identity $\mathit f(x) + f(1-x) = 1$.
Let me prove this first $\pmb :$
$\mathit f(x) = \frac{4^x}{(4^x+2)}----->1 $
Replace $\mathit x\;$ as $\mathit 1-x\\$
$\mathit f(x) = \frac{4^{1-x}}{4^{1-x}+2} = \frac{4\cdot{4^{-x}}}{4\cdot{4^{-x}}+2}$
On simplifying you get $\mathit f(x) = \frac{4}{4+2\cdot{4^x}}$ $\;$ which simplifies as $\mathit = \;\frac{2}{2+{4^x}}----->2$
Add $\mathit equation \;\pmb1$ and $\pmb2\;$ you get the number $\pmb 1$ as the answer.
Now applying this in the summation $$\sum_{i=0}^{2001} \mathit f\left(\frac{r}{2002}\right) $$
Substituting values from $\pmb1$ to $\pmb{2001}$
$\mathit f\left(\frac{1}{2002}\right)$ + $\mathit f\left(\frac{2}{2002}\right)$ + ....... + $\mathit f\left(\frac{1001}{2002}\right)$ + ....... $\mathit f\left(\frac{2000}{2002}\right)$ + $\mathit f\left(\frac{2001}{2002}\right)$
Group the terms $\pmb :$ $\mathit f\left(\frac{1}{2002}\right)$ + $\mathit f\left(\frac{2001}{2002}\right)$ + $\mathit f\left(\frac{2}{2002}\right)$ + $\mathit f\left(\frac{2000}{2002}\right)$ + ........ + $\mathit f\left(\frac{1000}{2002}\right)$ + $\mathit f\left(\frac{1002}{2002}\right)$ + $\mathit f\left(\frac{1001}{2002}\right)$
Now we have grouped $\pmb {2000}$ terms whose sum is $\pmb {1,}$ added $\pmb {2000}$ times.
Finally we have to add the middle term $\mathit f\left(\frac{1001}{2002}\right)$ whose value we can find by substituting $\pmb {\frac{1}{2}}$ in the question $\mathit f(x) = \frac{4^x}{(4^x+2)}$. It's value is $\pmb {\frac{1}{2}}$.
So the final answer is $\mathit 1000 + \frac{1}{2}$ which is $\pmb {\frac{2001}{2}}$
Cheers!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given that $n^4-4n^3+14n^2-20n+10$ is a perfect square, find all integers n that satisfy the condition So, I tried solving that by
$$n^4-4n^3+14n^2-20n+10=x^2\\10=x^2-a^2, a^2=n^4-4n^3+14n^2-20n+10\\10=(x+a)(x-a)$$ but I couldn't find any integers when I solved it
| Let us expand $(x^2+ax+b)^2 = (x^2+ax+b)(x^2+ax+b)$:
$$x^4 + (2a)x^3 + (a^2+2b)x^2 +(2ab)x+b^2$$
We want this to be as close as possible to the given expression, so we need to match the coefficients to the terms in the expansion. Starting from the coefficient after $x^4$ which is $x^3$, when $2a = -4 \Rightarrow a = -2$, we obtain $x^4 - 4x^3 + (4+2b)x^2 - (4b)x + b^2$.
Similarly, when $4+2b=14 \Rightarrow b = 5$, then we have $x^4 - 4x^3 + 14x^2 - 20x + 25$, which is exactly $15$ more than the given expression.
Therefore $x^2 = (n^2 - 2n + 5)^2 - 15$, and you can continue by using the difference of squares method explained in rain1's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Proving $(\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}})\leqq 2$ For $a,b>0.$ Prove$:$ $$(\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}})\leqq 2$$
My proof$:$
Let $a=\frac{x^2}{z}, b=\frac{y^2}{z}$ then $x,y,z>0.$ We need to prove$:$
$$\Big(\frac{x}{\sqrt{z}} +\frac{y}{\sqrt{z}}\Big) \cdot \Big( \frac{\sqrt{z}}{\sqrt{x^2+3y^2}} +\frac{\sqrt{z}}{\sqrt{3x^2+y^2}}\Big) \leqq 2$$
Or $$\Leftrightarrow \frac{1}{\sqrt{x^2+3y^2}} +\frac{1}{\sqrt{3x^2 +y^2}} \leqq \frac{2}{x+y}$$
By C-S we have$:$ $$\text{LHS}^2 \leqq ( \frac{1}{\sqrt{x^2+3y^2}} +\frac{1}{\sqrt{3x^2 +y^2}})^2 \leqq 2 (\frac{1}{x^2+3y^2}+\frac{1}{3x^2+y^2}) \leqq \frac{4}{(x+y)^2} \leqq \text{RHS}^2$$
The last inequality is very easy! But is there another proof$?$
| For example.
Let $a+b=2u\sqrt{ab}.$
thus, by AM-GM $u\geq1$ and we need to prove that
$$\sqrt{a+b+2\sqrt{ab}}\cdot\frac{\sqrt{4a+4b+2\sqrt{3a^2+3b^2+10ab}}}{\sqrt{3a^2+3b^2+10ab}}\leq2$$ or
$$\sqrt{2u+2}\cdot\frac{\sqrt{8u+2\sqrt{12u^2+4}}}{\sqrt{12u^2+4}}\leq2$$
or $$6u^2+2\geq(u+1)(2u+\sqrt{3u^2+1})$$ or
$$4u^2-2u+2\geq(u+1)\sqrt{3u^2+1}$$ or
$$13u^4-22u^3+16u^2-10u+3\geq0$$ or
$$13u^4-13u^3-9u^3+9u^2+7u^2-7u-3u+3\geq0$$ or
$$(u-1)(13u^3-9u^2+7u-3)\geq0,$$ which is obvious.
By this way we can get a best estimation here:
Let $a$ and $b$ be positive numbers and $2-\sqrt3\leq k\leq 2+\sqrt3.$ Prove that:
$$(\sqrt{a}+\sqrt{b})\left(\frac{1}{\sqrt{ka+b}}+\frac{1}{\sqrt{a+kb}}\right)\leq\frac{4}{\sqrt{k+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}),$$
i.e. we need next to show that $(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge (a + b + c)$, but i don't know how to do it.
Any help appreciated
| Consider a point $O$ and the lines $OA$, $OB$, $OC$ of lengths $a,b,c$ respectively such that $\angle AOB=\angle BOC= \angle COA=120^°$. Consider the triangle $\Delta ABC$. using cosine rule you can see that the side lengths of that triangle are $\sqrt{a^2+ab+b^2},\sqrt{b^2+bc+c^2},\sqrt{c^2+ca+a^2}$.
By definition $O$ is the Fermat point of $\Delta ABC$. See the properties of Fermat point and solve this geometric inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Find all sequences that has $\sum_{i=1}^\infty a_i$ converges, where $a_i = \sum_{k=i+1}^\infty a_k^2$.
Find all sequences that has $\sum_{i=1}^\infty a_i$ converges, where $a_i = \sum_{k=i+1}^\infty a_k^2$.
My intuition is that the only sequence of this form is the zero sequence.
Here's what I have so far:
$a_n - a_{n+1} = a_{n+1}^2 \implies a_{n+1} = \sqrt{a_n + \frac{1}{4}} - \frac{1}{2}$, but it doesn't seem to lead me anywhere.
Another line of thought is that if $a_i = 0$ for some $i$, it means that $\sum_{k=i+1}^\infty a_k^2=0$, which means that $a_k = 0$ for $k > i$. This will also mean $a_{i-1} = 0, a_{i-2} = 0, \ldots$, making the whole sequence the zero sequence.
It means that $a_i >0 $ for all $i$, yet $\lim a_i = 0$.
The last line I've tried is $a_1 = a_2^2 + a_3^2 + a_4^2 + \ldots, a_2 = a_3^2 + a_4^2 + \ldots$, so $\sum_{i=1}^\infty a_i = a_2^2 + a_3^2 + a_4^2 + \ldots + a_3^2 + a_4^2 + \ldots = a_2^2 + 2a_3^2 + 3a_4^2 = \sum_{i=2}^\infty (i-1)a_i^2$, which implies a stronger condition of having $ia_i^2 \to 0$. I'm hoping to get a contradiction but it doesn't seem to work.
Python seems to suggest that $(a_n) \approx \frac{1}{n}$ for large $n$.
Any hints?
| Claim: If $a_n > \frac{1}{k}$, then $ a_{n+1} > \frac{1}{k+1}$.
Proof: Verify that for $ k > 0$,
$$ a_{n+1} = \frac{ - 1 + \sqrt{ 1 + 4 a_n } }{2} > \frac{ - 1 + \sqrt{ 1 +\frac{4}{k} } }{2} > \frac{ 1}{k+1}. $$
Corollary: If $ a_1 > \frac{1}{k} $, then $ \sum a_n > \sum \frac{1}{k - 1 + n }$ which diverges.
Hence, the only sequence where $ \sum a_n$ converges is the all-0 sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$\min\left(\sum_{\text{sym}}\frac{(2x^2+y)(4x + y^2)}{\underset{\ne0}{(2x + y - 2)^2}} - 3(\underset{\gt0}{x}+\underset{\gt0}{y})\right)=?$
Given positives $x$ and $y$ such that $2x + y \ne 2$ and $x + 2y \ne 2$, calculate the minimum value of $$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} + \frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} - 3(x + y)$$
We have that $$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} + \frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} - 3(x + y)$$
$$ = \sum_{\text{sym}}\left[\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} - (2x + y)\right] = 2 \cdot \sum_{\text{sym}}\frac{(xy - 2x - y)(xy - 4x - 2y + 2)}{(2x + y - 2)^2}$$
$$ = 2 \cdot \sum_{\text{sym}}\left[\left(\frac{xy - 2x - y}{2x + y - 2}\right)^2 - \frac{xy - 2x - y}{2x + y - 2}\right] \ge 2 \cdot \left[2 \cdot \left(-\frac{1}{4}\right)\right] = -1$$
The equal sign occurs when $\dfrac{xy - 2x - y}{2x + y - 2} = \dfrac{xy - 2y - x}{2y + x - 2} = \dfrac{1}{2}$
$$\iff \dfrac{xy - 2}{2x + y - 2} = \dfrac{xy - 2}{2y + x - 2} = \dfrac{3}{2}$$
$$\implies 2x + y - 2 = 2y + x - 2 \iff x = y$$
Furthermore, $\dfrac{xy - 2}{2x + y - 2} = \dfrac{xy - 2}{2y + x - 2} = \dfrac{3}{2}$
$$\iff 2(xy - 2) = 3(2x + y - 2) = 3(2y + x - 2)$$
$$\implies 2(x^2 - 2) - 3(3x - 2) = 2(y^2 - 2) - 3(3y - 2) = 0$$
$$\implies 2x^2 - 9x + 2 = 2y^2 - 9y + 2 = 0 \implies x = y = \frac{9 + \pm \sqrt{65}}{4}$$
I would like to know whether the above solution is correct and if not, whether there are any other solutions.
Thanks for your attention.
| Solution by Nguyen Van Quy.
First, we have: If $a,b>0,\ a+b\neq 1$ then $$\dfrac{2(a^2+b)(a+b^2)}{(a+b-1)^2} \geq \dfrac{4a+4b-1}{2}.$$ Indeed, inequality equivalent to
$$4(a^3+b^3+a^2b^2+ab)\geq (4a+4b-1)(a+b-1)^2,$$ or
$$4\big[(a+b)^3-3ab(a+b)+a^2b^2+ab\big]\geq
4(a+b)^3-9(a+b)^2+6(a+b)-1,$$ or $$9(a+b)^2-6(a+b)(2ab+1)+(2ab+1)^2\geq
0,$$ or $$(3a+3b-2ab-1)^2\geq 0.$$ Done.
Now replace $(a,b)$ to $\left(x,\dfrac{y}{2}\right)$ and $\left(\dfrac{x}{2}, y\right),$ we get
$$\dfrac{(2x^2+y)(4x+y^2)}{(2x+y-2)^2} \geq 2x+y-\dfrac{1}{2},$$
$$\dfrac{(2y^2+x)(4y+x^2)}{(x+2y-2)^2} \geq 2y+x-\frac{1}{2}.$$
Therefore
$$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} +\frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} \geqslant 3(x + y) -1,$$
or
$$\frac{(2x^2 + y)(4x + y^2)}{(2x + y - 2)^2} +\frac{(2y^2 + x)(4y + x^2)}{(2y + x - 2)^2} - 3(x + y) \geqslant -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ Prove that
$$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$$
Is true for $x$, $y$ and $z$ being positive integers, with $x$ and $y$ being co-prime and $3z<x<y$.
There are 2 cases in this question:
Case 1 - $x$ is odd and $y$ is even (or $x$ is even and $y$ is odd), in this case $z$ is odd.
Case 2 - Both $x$ and $y$ are odd, in this case $z$ is even.
I tried to show some contradiction, we know that for the statement on the RHS to be true, the prime factors must be present in multiples of 3's, however I struggled going any further than that.
Any help would be much appreciated.
| Substituting $t=3z$ and analyzing the function $f(x,y,t)=t^3-3(x+y)(t-x)(t-y)$ we get:
$$f(x,y,t)=t^3-3(x+y)(t-x)(t-y)=t^3-3(x+y)(t^2-t(x+y)+xy)$$
$$f(x,y,t)=t^3-3t^2(x+y)+3t(x+y)^2-(x+y)^3+(x+y)^3-3(x+y)xy$$
$$f(x,y,t)=(t-x-y)^3+(x+y)(x^2+2xy+y^2-3xy)$$
$$f(x,y,t)=(t-x-y)^3+(x+y)(x^2-xy+y^2)=(t-x-y)^3+x^3+y^3$$
Now define $g(x,y,t)=f(-x,-y,t-x-y)=t^3-x^3-y^3$
Let's asume the function $g$ has an integer solution to $g=0$ and let $a,b,c\in\mathbb{N}$ be such a triplet. It follows that $a^3+b^3=c^3$. This contradicts Fermat's Last Theorem and there can not be such triplet solution!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{-1}^{1} [ \frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}] dx $ Evaluate $$\int_{-1}^{1} \left[ \frac{2}{3} x^3 + \frac{2}{3}(2-x^2)^{3/2}\right] dx $$
My attempt :$$ \frac{2}{3} \left[\frac{x^4}{4}\right]_{x=-1}^{x=1} + \frac{2}{3} \left[\frac{(2-x^2)^\frac{-1}{2}}{-1/2}\right]_{x=-1}^{x=1}=0$$
Is its True ?
| sadly, no.
the first part was completely true, although you could have used the property that $x^3$ is an odd function.
$$\frac{2}{3}\int^1_{-1}(2-x^2)^{3/2}dx$$
is a trigonometric integral.
your method ignored the internal $2-x^2$ and did an invalid substitution.
a better substetution is $x = \sqrt{2}\sin(\alpha),dx=\sqrt{2}\cos(\alpha)$ we have to change the bounds too, $\arcsin(\frac{1}{\sqrt{2}})=\frac{\pi}{4},\arcsin(-\frac{1}{\sqrt{2}})=-\frac{\pi}{4}$
$$\frac{2}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}4(1-\sin(\alpha)^2)^{3/2}\cos(\alpha)d\alpha$$
using everyone's favorite trigonometric identity $\cos(\alpha)^2+\sin(\alpha)^2=1$
$$\frac{8}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}(\cos(\alpha)^2)^{3/2}\cos(\alpha)d\alpha$$
$$\frac{8}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}(\cos(\alpha))^4d\alpha$$
using another trigonometric identity $cos(\alpha)^2=\frac{cos(2\alpha)+1}{2}$
$$\frac{8}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{(\cos(2\alpha)+1)^2}{4}d\alpha$$
$$\frac{2}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}[\cos(2\alpha)^2+2\cos(2\alpha)+1]d\alpha$$
using it again:
$$\frac{2}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\cos(4\alpha)+1}{2}+2\cos(2\alpha)+1d\alpha$$
splitting the integral:
$$\frac{1}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\cos(4\alpha)d\alpha+\frac{4}{3}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\cos(2\alpha)d\alpha+\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}1d\alpha$$
all of these integrals are very simple substitutions:
$$\frac{1}{12}[\sin(4\alpha)]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}+\frac{2}{3}[\sin(2\alpha)]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}+[\alpha]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}$$
$$0 + \frac{4}{3}+\frac{\pi}{2}$$
or simply
$$\frac{4}{3}+\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality question. Let $a,b,c>0$ with $a+b+c=1$. Show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{\left(a^3 + b^3 + c^3\right)}{abc}$$
Ohhhkk. So first off,
\begin{align} a^3 + b^3+ c^3 & =a^3 + b^3+ c^3- 3abc +3abc\\
& =\ (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))+3abc\\
& = \ (1-3(ab+bc+ca)) + 3abc \\
\end{align}
Using this the inequality becomes,
$$7 \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 9+ \frac{2}{abc}$$
How do i proceed from here? Was this the right approach? Is there a better one?
| I think the following is a smooth enough.
We need to prove that:
$$(a+b+c)(ab+ac+bc)\leq3abc+2(a^3+b^3+c^2)$$ or
$$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)\geq0,$$ which is obvious.
Your second inequality it's indeed just the first:
$$7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq9+\frac{2}{abc}$$ it's
$$7(a+b+c)(ab+ac+bc)\leq9abc+2(a+b+c)^3$$ or
$$\sum_{cyc}(3abc+2a^3+6a^2b+6a^2c+4abc-7a^2b-7a^2c-7abc)\geq0$$ or
$$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or
$$\sum_{cyc}(a^3-a^2b-ab^2+b^3)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to determine the minimum value of the weighted sum of the distances from two points to a point on a circumference Problem
In a two-dimensional Cartesian coordinate system,
there are two points $A(2, 0)$ and $B(2, 2)$
and a circle $c$ with radius $1$ centered at the origin $O(0, 0)$,
as shown in the figure below.
If $P$ is a point on the circle $c$,
then what is the minimum value of
$$ f = 2\sqrt{2}\lvert{PA}\rvert + \lvert{PB}\rvert? $$
Hypothesis
From my experience,
the solutions to such problems do not seem to be available under elementary form in general,
as indicated by answers to the question Minimize the sum of distances between two point and a circle.
However, when I studied this problem on GeoGebra,
it seems that minimum value is exactly $5$ in this specific situation,
with $P$ located at roughly the position shown below:
I tried to verify my hypothesis as follows.
Since $P$ is located inside $\angle AOB$,
we set its location to $(x, \sqrt{1 - x^2})$ (where $\sqrt{2}/2 < x < 1$).
Therefore,
\begin{align*}
\lvert{PA}\rvert
&= \sqrt{(2 - x)^2 + (1 - x^2)} \\
&= \sqrt{5 - 4x}, \\
\lvert{PB}\rvert
&= \sqrt{(2 - x)^2 + (2 - \sqrt{1 - x^2})^2} \\
&= \sqrt{-4\sqrt{1 - x^2} - 4x + 9}, \\
f
&= 2\sqrt{2} \lvert{PA}\rvert + \lvert{PB}\rvert \\
&= 2\sqrt{2} \sqrt{5 - 4x}
+ \sqrt{-4\sqrt{1 - x^2} - 4x + 9}. \\
\end{align*}
I asked GeoGebra again to plot $f(x)$:
and it seems to confirm my conjecture that
$$\min_{\sqrt{2}/2 < x < 1} f(x) = 5$$
Question
Is my hypothesis correct?
If so, is there a proof of this hypothesis that can be relatively easily by hand
(preferably avoiding, say, the evaluation of $f'(x)$)?
Geometric proofs will be especially appreciated.
| The problem is interesting for sure but I believe that the solution is biased by the fact that $x_A=x_B=y_A$.
I tried to make it more general with $A(x_A,y_A)$ and $B(x_B,0)$ assuming that both points are in the first quadrant. Let $(X,\sqrt{1-X^2})$ the coordinates of point $P$ and assume that we want to minimize
$$f = k\lvert{PA}\rvert + \lvert{PB}\rvert$$
We then have
$$\lvert{PA}\rvert=\sqrt{(X-x_A)^2+\left(\sqrt{1-X^2}-y_A\right)^2}$$
$$ \lvert{PB}\rvert=\sqrt{-2 x_B X+x_B^2+1}$$ which make
$$f=k\sqrt{(X-x_A)^2+\left(\sqrt{1-X^2}-y_A\right)^2}+\sqrt{-2 x_B X+x_B^2+1}$$ to be an hightly nonlinear equation in $X$; this means that, for getting its solution, we need some reasonable estimate.
To generate it, in a preliminary step, let us consider that we want to minimize
$$g= k^2\lvert{PA}\rvert^2 + \lvert{PB}\rvert^2$$ which is more pleasant. Computing its derivative, we have
$$\frac{dg}{dX}=2 k^2 \left(\frac{X y_A}{\sqrt{1-X^2}}-x_A\right)-2 x_B=0\implies X=\frac{k^2 x_A+x_B}{\sqrt{\left(k^2 x_A+x_B\right)^2+k^4 y_A^2}}$$
For the example given in the post, this would give as an estimate $X=\frac{9}{\sqrt{82}}\approx 0.993884$ while the exact solution of the problem is $X=\frac{2+7 \sqrt{46}}{50}\approx 0.989526$.
Back to $f$, computing its derivative, we need to solve for $X$
$$\frac{k X y_A-k \sqrt{1-X^2} x_A}{\sqrt{1-X^2} \sqrt{-2 X x_A+x_A^2-2
\sqrt{1-X^2} y_A+y_A^2+1}}-\frac{x_B}{\sqrt{-2 X x_B+x_B^2+1}}=0$$ which will require a numerical method such as Newton (this will be simple because we already have a good estimate.
For illustration, let us use $x_A=3$, $y_A=4$ and $x_B=5$ and $k=2\sqrt 2$. The preliminary process gives $X_0=\frac{29}{\sqrt{1865}}\approx 0.671519$.
Newton iterates will then be
$$\left(
\begin{array}{ccc}
n & X_n & f(X_n) \\
0 & 0.67151942 & 15.72034369 \\
1 & 0.77667655 & 15.68958161 \\
2 & 0.75966295 & 15.68798018 \\
3 & 0.75880236 & 15.68797655 \\
4 & 0.75880040 & 15.68797655
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$ For $a,b,c$ are reals$.$ Prove$:$ $$P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$$
I found this from Michael Rozenberg's solution. See here.
My proof:
$$P=\frac{1}{16} \, \left( a+b \right) ^{2} \left( a+b-4\,c \right) ^{2}+{\frac {5 \, \left( a+b \right) ^{4}}{14}}$$ $$+{\frac { \left( 3\,{a}^{2}+6\,ab+3\,{ b}^{2}-28\,{c}^{2} \right) ^{2}}{112}}+\frac{3}{8}\, \left( a+b \right) ^{2} \left( a-b \right) ^{2}+\frac{1}{8}\, \left( 2\,c+a+b \right) ^{2} \left( a-b \right) ^{2}$$
I’m looking for an alternative proof. Thanks!
| Let $a+b=2u$ and $ab=v^2$, where $v^2$ can be negative.
Thus, we need to prove that:
$$7c^4-2abc^2-2ab(a+b)c+(a+b)^2(a^2+b^2)\geq0$$ or
$$7c^4-2v^2c^2-4uv^2c+8u^2(2u^2-v^2)\geq0$$ or
$$7c^4+16u^4\geq2v^2(c^2+2uc+4u^2).$$
But, $$c^2+2uc+4u^2=(c+u)^2+3u^2\geq0$$ and $v^2\leq u^2$ it's just $(a-b)^2\geq0.$
Thus, it's enough to prove that
$$7c^4+16u^4\geq2u^2(c^2+2uc+4u^2)$$ or
$$7c^4-2u^2c^2-4u^3c+8u^4\geq0,$$ which is true by AM-GM:
$$7c^4-2u^2c^2-4u^3c+8u^4\geq$$
$$\geq c^4+u^4-2c^2u^2+c^4+3u^4-4u^3c\geq$$
$$\geq4\sqrt[4]{c^4(u^4)^3}-4u^3c=4|u^3c|-4u^3c\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$\sin x = \cos y, \sin y = \cos z, \sin z = \cos x$ For real numbers $x,y,z$ solve the system of equations:
$$\begin{align} \sin x = \cos y,\\
\sin y = \cos z,\\
\sin z = \cos x\end{align}$$
Source: high school olympiads, from a collection of problems for systems of equations, no unusual tricks involved.
So far I found that if we square two equations and use the $\sin^2 x + \cos^2 x=1$ we get $\sin^2 y + \cos^2 z=1$ which yields $\sin^2 y = \sin^2 z$. Is this correct or am I missing something? I don't know how to continue
| I arrived at a different result: (squaring the first two and adding)
$$\sin^2(x)+\sin^2(y)=\cos^2(y)+\cos^2(z)$$
$$\implies 1-\cos^2(x)+\sin^2(y)=\cos^2(y)+\cos^2(z)$$
Now $\cos^2(x)=\sin^2(z)$ from third equation, giving
$$1+\sin^2(y)=\cos^2(y)+(\cos^2(z)+\sin^2(z)) \iff \sin^2(y)=\cos^2(y)$$
$$\implies y= \frac{\pi}{4}+\pi n \implies x=z=\frac{\pi}{4}+\pi n$$
for some integer $n$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in this form: $\sqrt{16x^2 - 3^2}$ so I know to use $x = 3\sec u$:
$$
\begin{align}
& \int \frac{\sqrt{16x^2 - 9}}{x} \, dx \\
= {} & \int \frac{\sqrt{16x^2 - 3^2}}{x} \, dx \\
= {} & \int \frac{\sqrt{16(3\sec u)^2 - 3^2}}{3\sec u} 3\sec u\tan u \, du \\
= {} & \int \frac{(\sqrt{16(3\sec u)^2 - 3^2)}(3\sec u\tan u)}{3\sec u} \, du \\
= {} & \int \sqrt{(16(3\sec u)^2 - 3^2)}(\tan u) \, du
\end{align}
$$
This doesn't seem to make it easy. However, using a calculator online, it suggests I instead use $x = \dfrac{3}{4}\sec{u}$ which simplifies the integral to a crisp $\int 3\tan^2 u \, du$.
My question is, how did the calculator get $a = \dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
| In general, if you have $\sqrt{p x^2\pm q}$
*
*Make the coefficient $x$ equal to $1$ by taking coefficient of $x^2$ out of square root which gives $$\sqrt{px^2\pm q}=\sqrt p\sqrt{x^2\pm \frac{q}{p}}$$
*Above expression: $\sqrt{x^2\pm \frac{q}{p}}$ can be changed into the form: $\sqrt{x^2\pm a^2}$ by equating $a=\sqrt{\dfrac{q}{p}}$
*Substitute $x=a\sec u$ for the form $\sqrt{x^2-a^2}$ and $x=a\tan u$ for the form $\sqrt{x^2+a^2}$
For this case: $$\sqrt{16x^2-9}=\sqrt{16}\sqrt{x^2-\frac{9}{16}}$$ $$\sqrt{x^2-a^2}=\sqrt{x^2-\frac{9}{16}}$$
$$\implies a=\sqrt{\frac{9}{16}}=\frac34$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $ For $a,b,c \in (0,\infty).$ Prove$:$
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $$
My proof by SOS$:$
$$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) \left( a+b \right) \left( b+c \right) \left( c+a \right)\, \cdot \,(\text{LHS}-\text{RHS})$$
$$=\frac{5}{4} abc \sum\limits_{cyc} c^2 (a+b-2c)^2 (a-b)^2 +\frac{1}{4} \sum\limits_{cyc} {c}^{3} \left( 4\,{a}^{2}+3\,ab+4\,{b}^{2} \right) \left( a-b
\right) ^{4}$$
However$,$ it's hard to find this SOS's form without computer.
So I am looking for alternative solution without $uvw.$ Thanks very much!
| We have
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} - 1 - \frac{16abc}{(a+b)(b+c)(a+c)}$$
$$=\sum{\frac{(c-a)^2(c-b)^2}{(c^2+ab)(b+c)(c+a)}}\geqslant 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$ Problem. Let $x, y, z > 0$. Prove that
$$\frac{214x^4}{133x^3 + 81y^3} + \frac{214y^4}{133y^3 + 81z^3} + \frac{214z^4}{133z^3 + 81x^3} \ge x+y+z.$$
It is verified by Mathematica. The inequality holds with equality if $x = y = z$.
When $x = \frac{121}{84}, y = \frac{43}{66}$ and $z = 1$,
$\mathrm{LHS} - \mathrm{RHS} \approx 0.000005327884220$.
It is a stronger version of the inequality in this link:
Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
They can be written as (for $k = \frac{8}{5}$ and $k = \frac{133}{81} \approx 1.641975$, respectively)
$$\frac{x^4}{kx^3+y^3} + \frac{y^4}{ky^3 + z^3} + \frac{z^4}{kz^3 + x^3} \ge \frac{x+y+z}{k+1}.$$
The best constant $k$ is approximately $1.64199$ (see the comment by @Colescu in the link above).
I can prove the inequality of $k = \frac{8}{5}$ by the Buffalo Way.
Several months ago, I tried to prove the inequality of $k = \frac{133}{81}$ by the Buffalo Way without success.
However, I think that the Buffalo Way may work but just I have not found the way.
Any comments and solutions are welcome.
| We can reduce a degree of this inequality.
Indeed, by C-S we obtain:
$$\sum_{cyc}\frac{x^4}{133x^3+81y^3}=\sum_{cyc}\frac{x^4(200x-57y+154z)^2}{(133x^3+81y^3)(200x-57y+154z)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(200x^3-57x^2y+154x^2z)\right)^2}{\sum\limits_{cyc}(133x^3+81y^3)(200x-57y+154z)^2}$$ and it's enough to prove that:
$$214\left(\sum\limits_{cyc}(200x^3-57x^2y+154x^2z)\right)^2\geq(x+y+z)\sum\limits_{cyc}(133x^3+81y^3)(200x-57y+154z)^2,$$ which is true, but BW does not help here and I have no a proof by hand.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For positive $a$, $b$, $c$ with $abc=1$, show $(a + b + c)^3\left(\frac1{a^3+8}+\frac1{b^3+8}+\frac1{c^3+8}\right)\geq9$ Prove that for every $a, b, c > 0$ such that $abc = 1$,
$$ {(a + b + c)} ^ 3 \left(\frac{1}{a ^ 3 + 8} + \frac {1}{b ^ 3 + 8} + \frac {1}{c ^ 3 + 8}\right) \geq 9$$
$$(a + b + c) ^ 3 \left(\frac {1}{a ^ 3 + 8} + \frac {1}{b ^ 3 + 8} + \frac{1}{c ^ 3 + 8})\right) \geq (a + b + c) ^ 3\frac {9}{a ^ 3 + b ^ 3 + c ^ 3 + 24}$$ from Bergstrom's Inequality.
So I need to prove $$(a + b + c) ^ 3 \frac {1}{a ^ 3 + b ^ 3 + c ^ 3 + 24} \geq 1$$ which means $$(a + b + c) ^ 3 \geq a ^ 3 + b ^ 3 + c ^ 3 + 24$$
$$(a + b + c) ^ 3 - a ^ 3 - b ^ 3 - c ^ 3 \geq 24$$
$$ 3(a + b)(b + c)(c + a) \geq 24$$
$$ (a + b)(b + c)(c + a) \geq 8$$
But $$a + b \geq 2\sqrt ab$$
$$b + c \geq 2\sqrt bc$$
$$c + a \geq 2\sqrt ca$$
Multiplying these last 3 inequalities, we get $$(a + b)(b + c)(c + a) \geq 8 \sqrt {a^2b^2c^2} = 8 \sqrt 1 = 8$$
| By AM-GM and C-S:
$$(a+b+c)^3\sum_{cyc}\frac{1}{a^3+8}=(a+b+c)^3\sum_{cyc}\frac{1}{a^3+8abc}=$$
$$=\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)\sum_{cyc}\frac{1}{a^3+8abc}=$$
$$=\sum_{cyc}(a^3+(a+b)(a+c)(b+c))\sum_{cyc}\frac{1}{a^3+8abc}\geq$$
$$\geq\sum_{cyc}(a^3+8abc)\sum_{cyc}\frac{1}{a^3+8abc}\geq9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $ S=\frac{\cos 2x}{1\cdot 3}+\frac{\cos 4x}{3\cdot 5}+\frac{\cos 6x}{5\cdot 7}+\dots=\sum_{n=1}^\infty\frac{\cos (2nx)}{(2n-1)(2n+1)} $
Find a sum of the series:
$$
S=\frac{\cos 2x}{1\cdot 3}+\frac{\cos 4x}{3\cdot 5}+\frac{\cos 6x}{5\cdot 7}+\dots=\sum_{n=1}^\infty\frac{\cos (2nx)}{(2n-1)(2n+1)}
$$
My attempt:
$$
\begin{aligned}
&z=\cos x+i\sin x\\
&S=\frac{1}{2}\text{Re}\sum_{n=1}^\infty\frac{z^{2n}}{2n-1}-\frac{1}{2}\text{Re}\sum_{n=1}^\infty\frac{z^{2n}}{2n+1}
\end{aligned}
$$
But calculating these sums seems a bit difficult to me. Perhaps there is a better approach to this problem?
| $$2S=\sum_{r=1}^\infty\dfrac{\cos2rx}{2r-1}-\sum_{r=1}^\infty\dfrac{\cos2rx}{2r+1}$$
which is real part of $$\sum_{r=1}^\infty\dfrac{(e^{ix})^{2r}}{2r-1}-\sum_{r=1}^\infty\dfrac{(e^{ix})^{2r}}{2r+1}$$
$$=e^{ix}\cdot\sum_{r=1}^\infty\dfrac{(e^{ix})^{2r-1}}{2r-1}-e^{-ix}\cdot\sum_{r=1}^\infty\dfrac{(e^{ix})^{2r+1}}{2r+1}$$
$$=e^{ix}\cdot\ln\dfrac{1-e^{ix}}{1+e^{ix}}-e^{-ix}\left(\ln\dfrac{1-e^{ix}}{1+e^{ix}}-1\right)$$
$$=(e^{ix}-e^{-ix})\left(\ln\dfrac{1-e^{ix}}{1+e^{ix}}\right)+e^{-ix}$$
$$=2i\sin x\left(\ln(-1)+\ln\dfrac{e^{ix/2}-e^{-ix/2}}{e^{ix/2}+e^{-ix/2}}\right)+\cos x-i\sin x$$
$$=2i\sin x\left(\ln(-i)+\ln\tan\dfrac x2\right)+\cos x-i\sin x$$
Now the principal value of $\ln(-1)$ is $-\dfrac{i\pi}2$
| {
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"url": "https://math.stackexchange.com/questions/3710694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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China $1996$ Number theory Problem
Find the smallest positive integer $K$ such that every $K$ -element subset of $\{1,2, \ldots, 50\}$ contains two distinct elements $a, b$ such that $a+b$ divides $a b$
Let $c=g c d(a, b),$ so $a=c a_{1}$ and $b=c b_{1}$.
Therefore, $c a_{1} b_{1}$ is divisible by $a_{1}+b_{1}$
Furthermore, since $g c d\left(a_{1}, b_{1}\right)=1,$ we see that $a_{1}+b_{1}$ is relatively prime to $a_{1}$ and $b_{1},$ s $\left(a_{1}+b_{1}\right) | c$
since $a+b \leq 99 \Rightarrow a_{1}+b_{1} \leq 9 .$
How $a+b \leq 99 \Rightarrow a_{1}+b_{1} \leq 9 .$ ???
| Since $a_1+b_1|c$ then $(a_1+b_1)^2|c(a_1+b_1)=a+b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
| I want to point out that $\frac{13}{32}$ is the value of the Pade $(2,1)$ approximation of $\ln (1+x)$ at $x=\frac{1}{2}$.
In detail, the Pade $(2, 1)$ approximation of $\ln (1+x)$ is $g(x) = \frac{x^2+6x}{6+4x}$.
It is easy to prove that $\frac{x^2+6x}{6+4x} > \ln (1+x)$ for $x > 0$.
Indeed, let $f(x) = \frac{x^2+6x}{6+4x} - \ln(1+x)$. We have
$f'(x) = \frac{x^3}{(3+2x)^2(1+x)} > 0$ for $x > 0$. Also, $f(0) = 0$. The desired result follows.
We have $g(\frac{1}{2}) = \frac{13}{32} > \ln \frac{3}{2}$.
We may find the Pade approximation by hand. We may first try the Pade $(1, 1)$ approximation, second try the Pade $(2, 1)$ approximation,
third try the Pade $(1, 2)$ approximation, and so on, until we find enough approximation.
More details about the Pade $(2, 1)$ approximation:
The Taylor expansion of $\ln(1+x)$ is $x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots$
Let $g(x) = \frac{a_0 + a_1x+ a_2x^2}{1 + b_1x}$. Comparing the coefficients of $x^k$ for $k=0, 1, 2, 3$ of
$$(x - \frac{1}{2}x^2 + \frac{1}{3} x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots)(1 + b_1x)
= a_0 + a_1x+ a_2x^2, $$
we obtain $a_0 = 0, a_1 = 1, a_2 = \frac{1}{6}, b_1 = \frac{2}{3}$. Then, $g(x) = \frac{x^2+6x}{6+4x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Combinatorics - Generating function problem So we have this problem . We want to distribute 20 identical marbles to 3 distinct children so that:
*
*The first child takes at least 4 marbles.
*Every child gets at least 3 marbles.
The given answers where :
*
*$(x^4 + ... + x^{20})(1 + x + ... + x^{20})(1 + x + ... + x^{20})$
*$((1+x+...+x^{20})^{3})$
And in both case the answer is given by the coefficient of $x^{20}$
My question is this:
If the first child takes at least 4 , then the second and the third could never have more 16 so should we change the generating function to: $(x^4 + ... + x^{20})(1 + x + ... + x^{16})(1 + x + ... + x^{16})$ ?
Accordingly, if every child gets at least 3, then it should not have more than 14 because we always need 3+ 3 = 6 for the other two . So i would write $((1+x+...+x^{14})^{3})$ and the correct answer will be given by the coefficient of 20.
Because we care about the coefficient of 20 , we need to make sure that there are no extra things in our function in order for our solution to be correct , if for example we wanted to disturb 10 marbles then because we only need to identify the coefficient of $x^{10}$ then it makes no difference if expand these polynomials. Am I right? Am I missing something?
| For the first problem, the simplest answer is that both generating functions are correct for the desired coefficient. i.e.,
\begin{align*}
&[x^{20}] (x^4+\dotsb+x^{20})(1+\dotsb+x^{20})(1+\dotsb+x^{20})\\
&= [x^{20}] (x^4+\dotsb+x^{20})(1+\dotsb+x^{16})(1+\dotsb+x^{16})
\end{align*}
where $[x^n]$ denotes the coefficient extraction operator.
The reason for this is, for instance, the coefficient of $x^{17}$ is never used from the second product term due to the contribution of terms with power at least $4$ from the first. The difference between our generating functions is that certain cases, such as cases in which the first person selects $4$ marbles and the second selects $17$ are encoded in the first generating function but not in the second. Encoded or not, $20$ marbles cannot be reached from $21$ through selection of marbles in the third term, so $[x^{20}]$ does not count such cases.
In terms of the problem at hand, we note that we can encode more states that won't ultimately affect the extracted coefficient. For much the same reasoning as higher order coefficients like $[x^{17}]$ not being considered when extracting, we could equally well conclude
\begin{align*}[x^{20}] &(x^4+\dotsb+x^{20})(1+\dotsb+x^{16})(1+\dotsb+x^{16})\\
&= [x^{20}] (x^4+\dotsb+x^{a})(1+\dotsb+x^{b})(1+\dotsb+x^{c})
\end{align*}
for $a\geq 20$, $b\geq 16$, and $c\geq 16$.
In fact this holds formally if $a=b=c=\infty$:
\begin{align*}
&[x^{20}] (x^4+\dotsb+x^{20})(1+\dotsb+x^{16})(1+\dotsb+x^{16})\\
&= [x^{20}] (x^4+x^5+\dotsb)(1+x+\dotsb)(1+x+\dotsb)
\end{align*}
This is useful for when we want to actually do the coefficient extraction,
\begin{align*}
&(x^4+x^5+\dotsb)(1+x+\dotsb)(1+x+\dotsb)\\
&= x^4(1+x+\dotsb)^3
= \frac{x^4}{(1-x)^3}
\end{align*}
is more manageable than
\begin{align*}&(x^4+\dotsb+x^{20})(1+\dotsb+x^{16})(1+\dotsb+x^{16})\\
&= x^4(1+\dotsb+x^{16})^3
= \frac{x^4(1-x^{17})^3}{(1-x)^3}.
\end{align*}
On to the actual computation, given that $\frac{1}{(1-x)^n} = \sum_{k=0}^\infty \binom{n+k-1}{n-1} x^k$,
\begin{align*}
[x^{20}] \frac{x^4}{(1-x)^3}
&= [x^{20}] \sum_{k=0}^\infty \binom{k+2}{2} x^{k+4}\\
&= [x^{20}] \sum_{k=4}^\infty \binom{k-2}{2} x^{k}
= \binom{18}{2}.
\end{align*}
For the second problem, I believe they meant $(x^3+\dotsb+x^{20})^3$ as $(1+x+\dotsb+x^{20})^3$ encodes $1\cdot 1\cdot x^{20}$ which is not one of the valid cases.
At least $3$ marbles were taken by each, so you can just look at $(x^3+\dotsb+x^{14})^3$. Another way is to first reserve $3$ marbles per person, and then selecting up to $20-9=11$ marbles for each, which corresponds to the generating function $(x^3)^3(1+\dotsb+x^{11})^3$ (which you'll note is algebraically equivalent). All these generating functions have the same $x^{20}$ coefficient:
\begin{align*}
&\;[x^{20}] (x^3+\dotsb+x^{20})^3\\
= &\;[x^{20}] (x^3+x^4+\dotsb+x^{14})^3\\
= &\;[x^{20}] (x^3+x^4+\dotsb)^3
= [x^{20}] \frac{x^9}{(1-x)^3}\\
= &\;[x^{20}] \sum_{k=0}^\infty \binom{k+2}{2} x^{k+9}
= [x^{20}] \sum_{k=9}^\infty \binom{k-7}{2} x^{k}\\
= &\binom{13}{2}.
\end{align*}
| {
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Are there numbers $N$ such that $N^2=n_1^2+n_2^2+\cdots+n_k^2+1$ In the past few hours I spent some time looking for numbers of the form
$$
N^2=n_1^2+n_2^2+\cdots+n_k^2+1
$$
where $n_i \gt 1 \wedge n_i \neq n_j \wedge k\ge 2$ , and I found none. I searched up to $100$. I wonder if this kind of numbers exist at all...
So I came here for an answer. Can you provide some instances or rules to generate them, or on the contrary they simply can't exist for some reason.
Thanks.
Note that possible numbers are $N^2=n_1^2+n_2^2+1$ or $N^2=n_1^2+n_2^2+n_3^2+1$ etc... we can have any number of squares on the right side...
| $n^2=n1^2+n2^2+n3^2\tag{1}$
Substitute $n1=x, n2=b, n3=c, n=x+k$ to equation $(1).$
We get $$x = 1/2\,{\frac {-{k}^{2}+{b}^{2}+{c}^{2}}{k}}$$
Thus, we get a parametric solution.
$(n,n1,n2,n3) = (k^2+b^2+c^2, -k^2+b^2+c^2, 2bk, 2ck)$
Choose (b,c,k) so that $(n,n1,n2)$ is divisible by $n3$, then
we get the solution of $n^2=n1^2+n2^2+1$.
[n, n1, n2, n3]
[3, 2, 2, 1]
[9, 8, 4, 1]
[19, 18, 6, 1]
[33, 32, 8, 1]
[51, 50, 10, 1]
[73, 72, 12, 1]
[99, 98, 14, 1]
[129, 128, 16, 1]
[163, 162, 18, 1]
After simple algebra, we get a parametric solution below.
$(n,n1,n2,n3)=(1+2p^2, 2p^2, 2p, 1)$
$p$ is arbitrary.
$n^2=n1^2+n2^2+n3^2+n4^2\tag{2}$
Substitute $n1=x, n2=b, n3=c, n4=d, n=x+k$ to equation $(2).$
Thus, we get a parametric solution.
$(n,n1,n2,n3,n4) = (k^2+b^2+c^2+d^2, -k^2+b^2+c^2+d^2, 2bk, 2ck, 2dk)$
Choose (b,c,d,k) so that $(n,n1,n2,n3)$ is divisible by $n4$, then
we get the solution of $n^2=n1^2+n2^2+n3^2+1$.
$(n,n1,n2,n3,n4)=(1+2q^2+2p^2, 2q^2+2p^2, 2q, 2p, 1)$
$p,q$ are arbitrary.
| {
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Prove derivative by induction $f:(0, \infty) \rightarrow \mathbb{R}$
$f(x) = \sqrt{x}$
a) Calculate the first four derivatives
$f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{x}}$
$f''(x) = -\frac{1}{4}\cdot \frac{1}{\sqrt{x^3}}$
$f'''(x) = \frac{3}{8}\cdot \frac{1}{\sqrt{x^5}}$
$f''''(x) = -\frac{15}{16}\cdot \frac{1}{\sqrt{x^7}}$
b) Prove by induction that the following formula holds true:
$f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$
Base Case: $k=1$:
$f'(x) = \frac{(1)}{2}\cdot\prod^{0}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x}} = \frac{1}{2}\cdot\frac{1}{\sqrt{x}}$
Inductive Hypothesis(IH):
Assumption holds true for some k.
Inductive step:
$k \rightarrow k+1$
to show:
$f^{(k+1)}(x) = \frac{(-1)^{k+2}}{2^k+1}\cdot\prod^{k}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k+1}}}$
$\frac{(-1)^{k+2}}{2^k+1}\cdot\prod^{k}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k+1}}} \\
= \frac{(-1)^{k+1}}{2^k}\cdot\frac{-1}{2} \cdot(2k-1)\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}\cdot \frac{1}{\sqrt{x^2}} \\
= [\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}] \cdot\frac{-1}{2}\cdot(2k-1)\frac{1}{\sqrt{x^2}} \\
=^{IH} f^{(k)}(x) \cdot \frac{-1}{2}\cdot(2k-1)\frac{1}{\sqrt{x^2}} \\
= f^{(k)}(x) \cdot (-k+\frac{1}{2})\cdot \frac{1}{|x|} \text{since x $\in$ $(0, \infty)$} \\
= f^{(k)}(x) \cdot [(-k+\frac{1}{2})\cdot \frac{1}{x}]$
This means that to get from one derivative to the next, the factor at the end will be multiplied.
So how do I go on from here?
Usually I would start from the left side. But if I would start here from the left side I would have to transform $f^{(k+1)}(x)$. And I would have to calculate the derivative over the product symbol.
$f^{(k+1)}(x) = f^{(k)'}(x)$
| Hint:
First, I would write the formula in a slightly different way:
$$f^{(k)}(x) = \frac{(-1)^{k\color{red}-1}}{2^k}\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$$
Next, to differentiate $f^{(k)}(x)$, I would use a fractional writing of the exponent of $x$, and use the formula
$$\Bigl(\frac1{x^\alpha}\Bigr)'=-\frac \alpha{x^{\alpha+1}}\quad\text{ with }\quad \alpha=\tfrac{2k-1}2.$$
This yields readily
$$f^{(k+1)}(x)=\frac{(-1)^k}{2^{k+1}}\prod^{k-1}_{j=1}(2j-1)\cdot(2k-1)\ \frac1{x^{\frac{2k-1}2+1}}.$$
| {
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Inequality proof (perhaps inductive?) Came up with this on my own and although it seems true (due to Desmos), I was interested to see a proof of it. I tried an inductive approach myself but unfortunately couldn't come up with anything concrete (just by assuming the statement, proving the base case and fiddling with it).
Prove that
$$ (5^{k})! > 5^{k!} \ \forall \ k \in \mathbb{N} $$
| For $k = 0$, we obtain
$$
\left( 3^k \right)! = 1! = 1 \not> 5 = 5^1 = 5^{0!} = 5^{k!}. \tag{0}
$$
For $k = 1$, we obtain
$$
\left( 3^k \right)! = 3! = 3 \not> 5^1 = 5^{1!} = 5^{k!}. \tag{1}
$$
For $k = 2$, we obtain
$$
\left( 3^k \right)! = 9! > 25 = 5^2 = 5^{2!} = 5^{k!}. \tag{2}
$$
Suppose that $k \in \mathbb{N}$ such that $k \geq 2$ and also
$$
\left( 3^k \right)! > 5^{k!}. \tag{3}
$$
Then we find that
$$
\begin{align}
\left( 3^{k+1} \right)! &= \left( 3 \cdot 3^k \right)! \\
&= \left( 3 \cdot 3^k \right)\left( 3 \cdot 3^k -1 \right) \ldots \left( 3^k +1 \right) \left( \cdot 3^k \right)! \\
&> \left( 3 \cdot 3^k \right)\left( 3 \cdot 3^k -1 \right) \ldots \left( 3^k +1 \right) 5^{k!} \\
&= \left( 3^k + 2 \cdot 3^k \right) \left( 3^k + 2 \cdot 3^k -1 \right) \ldots \left( 3^k + 1 \right) 5^{k!} \\
&> \left( 3^k + 1 \right)^{2 \cdot 3^k} 5^{k!} \\
&> 5^{2 \cdot 3^k } 5^{k!} \\
&> 5^{k+1} 5^{k!} \tag{4} \\
&= 5^{(k+1)k!} \\
&= 5^{(k+1)!}.
\end{align}
$$
In (4) above here we have used the result that
$$
2 \cdot 3^k > k+1
$$
for all $k \in \mathbb{N}$. This result should not be too hard to prove using induction.
Hope this helps.
| {
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Derivative of $p\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}du} $? I found a result,
I have,
$p\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}du} $
$y = - p\left( x \right)\log p\left( x \right) - (1 - p\left( x \right))\log (1 - p\left( x \right))$
I found a solution for the derivative of $y$ at $x=0$ is,
$\frac{{dy\left( 0 \right)}}{{dx}} = \frac{2}{\pi }$
But I could not able to show that. It comes zero for me every time.
| Since you edited the question, I edited my answer.
FIRST PART
We will show that
$$Q'(x) = -\frac{1}{\sqrt{ 2 \pi}} \operatorname{e}^{-\frac{x^2}{2}} $$
The main tool here is the Fundamental theorem of Calculus.
First method
$$
\begin{split}
\frac{\mathrm d}{\mathrm d x} \left[ \frac{1}{\sqrt{ 2 \pi}} \int_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}\,\mathrm du} \right] &= \frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[\int_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}\,\mathrm du}\right]\\
&\overset{t=-u}{=}\frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[\int_{-\infty}^{-x} {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{t^2}}}{2}}}\,\mathrm dt}\right]\\
&=\frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[\int_{-\infty}^{-x} {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{t^2}}}{2}}}\,\mathrm dt}\right]\\
&=-\frac{1}{\sqrt{ 2 \pi}} {{\mathop{\rm e}\nolimits} ^{ - \frac{{{x^2}}}{2}}}\\
\end{split}
$$
Second method (essentially the same)
Let $G(x)$ a primitive of $\mathrm{e}^{-\frac{x^2}{2}}$, i. e. $G'(x) = \mathrm{e}^{-\frac{x^2}{2}}$; therefore
$$
\begin{split}
\frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} \left[ \int_x^\infty {{{\mathop{\rm e}\nolimits} ^{ - \frac{{{u^2}}}{2}}}\,\mathrm du} \right] &= \frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x}\left[G(+\infty)-G(x)\right] \\
&= \frac{1}{\sqrt{ 2 \pi}} \frac{\mathrm d}{\mathrm d x} G(-x)\\
&= -\frac{1}{\sqrt{ 2 \pi}} G'(-x) \\
&= -\frac{1}{\sqrt{ 2 \pi}} \operatorname{e}^{-\frac{x^2}{2}}
\end{split}
$$
Third method
Remembering that $\int_{-\infty}^{+\infty}\operatorname e^{-\frac{u^2}{2}}\, \mathrm du = \sqrt{2\pi}$ we can write
$$
\int_{x}^{+\infty}\operatorname e^{-\frac{u^2}{2}}\, \mathrm du = \int_{-\infty}^{+\infty}\operatorname e^{-\frac{u^2}{2}}\, \mathrm du -\int_{-\infty}^x \operatorname e^{-\frac{u^2}{2}}\, \mathrm du = \sqrt{2\pi} - \int_{-\infty}^x \operatorname e^{-\frac{u^2}{2}}\, \mathrm du
$$
And then apply FTC.
SECOND PART
According to your notation we have
$$C(x) := 2 \log 2 -2h\left(Q\left(\sqrt x\right)\right) $$
If we evaluate $\dot C$, thanks to the chain rule we have
$$
\begin{split}
\dot C (x) &= -2h'\left(Q\left(\sqrt x\right)\right) \cdot Q'\left(\sqrt x\right) \cdot \frac{1}{2 \sqrt x} \\
&= -\log \left[ \frac{Q\left(\sqrt x\right)}{1-Q\left(\sqrt x\right)} \right]\cdot \frac{1}{\sqrt{ 2 \pi}} \operatorname{e}^{-\frac{x}{2}} \cdot \frac{1}{\sqrt x}\\
&= \frac{\operatorname{e}^{-\frac{x}{2}}}{\sqrt{ 2 \pi}} \cdot \frac{\log \left( 1 - Q\left(\sqrt x\right)\right) -\log Q\left(\sqrt x\right) }{\sqrt x}
\end{split}
$$
Since $Q(0)=\frac{1}{2}$, we obtain an indeterminate form $\frac{0}{0}$. Can you handle from here?
THIRD PART
We have to evaluate
$$\lim_{x \to 0}\frac{\log \left( 1 - Q\left(\sqrt x\right)\right) -\log Q\left(\sqrt x\right) }{\sqrt x} \overset{y=\sqrt x}{=} \lim_{y \to 0}\frac{\log \left( 1 - Q(y) \right) -\log Q(y) }{y}$$
to be continued...
| {
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Elementary operation on determinant, but actually basic algebra If $\begin{vmatrix} -1 & a & a \\ b & -1 & b \\ c & c & -1 \end{vmatrix} =0$ then what's the value of $$\frac{1}{1+a}+\frac{1}{1+b} +\frac{1}{1+c}$$
I just expanded the Determinant, to get
$$ab+bc+ac+2abc=1$$
Which further leads to $$\frac{1+a}{a}+\frac{1+b}{b}+ \frac{1+c}{c}= 1+\frac{1}{abc}$$
The solution in the book uses elementary operations, but is it possible to make the "original" equation of determinant in the required form?
This maybe a duplicate question, but I couldn't find it.
| With $$c=\frac{1-ab}{a+b+2ab}$$ we get $$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+\frac{1-ab}{a+b+2ab}}=2$$ after a few simplifications.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration of $\frac{1}{u^4 + (4\zeta^2-2)u^2 + 1}$ I am trying to compute
$$I(\zeta) = \int_{-\infty}^{\infty} \frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1}\, du$$
for positive real $\zeta$. Can anyone help?
I'm way out of practice for integrals except for simple stuff like $\int 1/(1+u^2)\, du = \tan^{-1} u + C$.
Sympy fails on the definite integral and gives me this weird RootSum expression for the indefinite integral:
$$\operatorname{RootSum} {\left(t^{4} \left(4096 \zeta^{8} - 8192 \zeta^{6} + 4096 \zeta^{4}\right) + t^{2} \left(256 \zeta^{6} - 384 \zeta^{4} + 128 \zeta^{2}\right) + 1, \left( t \mapsto t \log{\left (- 512 t^{3} \zeta^{6} + 768 t^{3} \zeta^{4} - 256 t^{3} \zeta^{2} - 32 t \zeta^{4} + 32 t \zeta^{2} - 4 t + u \right )} \right)\right)}$$
Wolfram Alpha gives me the following for the indefinite integral :
$$\begin{align}
& \frac{\frac{1}{a_1}\tan^{-1} \frac{u}{a_1} - \frac{1}{a_2}\tan^{-1} \frac{u}{a_2}}{4\zeta\sqrt{\zeta^2-1}} + C \\
\\
a_1 &= \sqrt{2\zeta^2-2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b-c}\\
a_2 &= \sqrt{2\zeta^2+2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b+c}\\
\end{align}$$
(with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$) but I'm a bit lost how it got there, and then I'm not exactly sure what to do if $\zeta \le 1$ (is the formula still valid?!)
edit: OK, partial fraction expansion is sloooowwwwly coming back to me. It looks like $a_1a_2 = 1$ and $a_1{}^2 + a_2{}^2 = 4\zeta^2-2$, so I guess they used the expansion
$$
\frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
| Note
\begin{align}
& \int_{-\infty}^{\infty}\frac{du}{u^4 + (4\zeta^2-2)u^2 + 1}\\
= & \int_{0}^{\infty}\left( \frac{1+\frac1{u^2}}{u^2+\frac1{u^2} + 4\zeta^2-2} -\frac{1-\frac1{u^2}}{u^2+\frac1{u^2} + 4\zeta^2-2}\right)du\\
= & \int_{0}^{\infty}\left( \frac{d(u-\frac1{u})}{(u-\frac1{u} )^2+ 4\zeta^2} -\frac{d(u+\frac1{u})}{(u+\frac1{u} )^2+ 4\zeta^2-4}\right)\\
= & \int_{-\infty}^{\infty} \frac{dt}{t^2+ 4\zeta^2}- \int_{\infty}^{\infty} \frac{dt}{t^2+ 4\zeta^2-4}\\
=&\frac\pi{2\zeta}-0 =\frac\pi{2\zeta}
\end{align}
| {
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Volume of the solid bounded by $z = 4-x^2$, $y+z=4$, $y=0$ and $z=0$. If I am seeing this problem correctly, when $z=0$, $x = \pm 2$, so $-2 \le x \le 2$.
The $y$ coordinate varies from $0$ to $4$, because when $z=0, y=4$ (the plane $y+z=4$ with $z=0$). So $0 \le y \le 4$.
The $z$ coordinate varies from the plane $z=0$ to the plane $z=4-y$.
Then $0 \le z \le 4-y$.
So the integrals are:
$\displaystyle \int_{-2}^{2} \int_{0}^{4} \int_{0}^{4-y} dzdydx$
Is this correct?
| Note that the $z=f(x,y)$ function is simply the intersection between the parabolic cylinder $z<4-x^2$ and the plane $z<4-y$, and thus the (x,y) domain is split in the intersection of both: in $y=x^2$.
Hence, the integration has two members, one for $y<x^2$ in which $z<4-x^2$:
$$
\int_{-2}^{2}\int_{0}^{x^2}\int_{0}^{4-x^2} \text{dzdydx} = \frac{128}{15}
$$
and for $y>x^2$ in which $z<4-y$:
$$
\int_{-2}^{2}\int_{x^2}^{4}\int_{0}^{4-y} \text{dzdydx} = \frac{256}{15}
$$
The answer is $\frac{128}{15}+\frac{256}{15}=\frac{128}{5}$
| {
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is $f(x,y)=(\sqrt[3]{x} + \sqrt[3]{y})^3$ Differentiable at $P=(0,0)$ $f(x,y)=(\sqrt[3]{x} + \sqrt[3]{y})^3$ and let $P=(0,0)$. the partial derivative at $P$ are $$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(0+h, 0)-f(0,0)}{h}=\frac{(\sqrt[3]{h})^3}{h}=1$$ and $$\frac{\partial f}{\partial y}(x,y)=\lim_{h\to0}\frac{f(0, 0+h)-f(0,0)}{h}=\frac{(\sqrt[3]{h})^3}{h}=1$$ and thus both partial derivative are continuous at $P$ and $f$ is differentiable. But when I derivate directly I get $$\frac{\partial f}{\partial x}(x,y)=\frac{(\sqrt[3]{x} + \sqrt[3]{y})^2}{\sqrt[3]{x^2}}$$ and $$\frac{\partial f}{\partial y}(x,y)=\frac{(\sqrt[3]{x} + \sqrt[3]{y})^2}{\sqrt[3]{y^2}}$$ which are undefined for $P=(0, 0)$ and thus clearly not continuous at $P$. So I'm not sure which derivative is correct and whether $f$ is differentiable or not?!?
| For $f$ to be differentiable at $(0,0)$ there must be a linear form $ax+by$
such that
$$f(x,y)-(ax+by)$$
tends to zero quicker than $\sqrt{x^2+y^2}$ as $(x,y)\to 0$, that is
$$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}\to0.$$
The numbers $a$ and $b$ must be the partial derivatives of $f$ at $(0,0)$, that is $a=1$
and $b=1$ so that
$$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}
=\frac{f(x,y)-x-y}{\sqrt{x^2+y^2}}
=3\frac{\sqrt[3]{x^2}\sqrt[3]y+\sqrt[3]{x}\sqrt[3]{y^2}}{\sqrt{x^2+y^2}}
.$$
But if we take $(x,y)=(t,t)$ we get
$$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}
=\frac{6t}{\sqrt{2}|t|}=\pm3\sqrt2$$
which certainly does not tend to zero as $t\to0$.
Therefore $f$ is not differentiable at $(0,0)$.
| {
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Find idempotent given generator poly and check poynomial by Bezout algorithm I have the cyclic code $C$ of length $8$ and dimension $4$ over $\mathbb{F}_3$ and with check polynomial $$g(x) = (x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^6) = x^4+x^3+x+2$$ where $\alpha \in \mathbb{F}_9$ is a primitive 8-th root of unit over $\mathbb{F}_3$
By polynomial division, I find $$h(x) = x^4+2x^3+x^2+x+1$$
Goal: find the generating idempotent $e(x)$
*
*I know that it's given by the Bezout identity $1 = a(x)g(x) + b(x)h(x)$, so I start dividing $h$ by $g$ (all the division are done in $\mathbb{F}_3$):
$$h = g + (x^3 + x^2 + 2)$$
*
*So now I divide $g$ by the term in brackets:
$$g = x(x^3+x^2+2) + (2x+2)$$
*
*Now I divide $x^3 + x^2 + 2$ by $2x+2$ and I find:
$$x^3 + x^2 + 2 = 2x^2 (2x+2) + 2$$
Now I want to obtain the Bezoùt identity, so I should start from the bottom by writing
$$2 = (x^3 + x^2 + 2) +x^2(2x+2) \quad \star$$
but from the second equation: $2x+2= g +2x(x^3+x^2+2)$
and now I substitute in $\star$:
$$2 = (x^3+x^2+2) + x^2 (g +2x(x^3+x^2+2) )$$
Now, from the first equation: $x^3+x^2+2 = h-g = h+2g$, therefore:
$$2 = h+2g+x^2(g+2x(h+2g))$$
which equals (multiply by 2 both sides):
$$1 = h(2+x^3) + g(1+2x^2+2x^3)$$
Therefore the idempotent $e(x)$ should be: $$e(x) =g(1+2x^2+2x^3)= 2x^7+x^6+2x^5+x^3+x^2+x+2$$ but if I square it I don't $e(x)$ again.
What is wrong?
| The idempotent was calculated correctly. There was a mistake in the verification phase in that the wrong quotient ring was accidentally used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to find the limit of $\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$? The problem is as follows:
A certain tv signal is modeled by the function shown below:
$f(x)=\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$
where $a>0$
Find the $\lim_{x\rightarrow 0^+} f(x)$.
The alternatives given in this problem are as follows:
$\begin{array}{ll}
1.&-a\sqrt{2}\\
2.&\frac{\sqrt{a}}{2}\\
3.&\sqrt{2}\sqrt{a}\\
4.&\sqrt{2}\\
5.&-\sqrt{a}\\
\end{array}$
How exactly should I assess this problem?.
I'm confused about the simbol used in the limit but I think the intended meaning is to find the limit of the function where $x$ approaches to positive?.
Attempting to insert the zero in the function as it is given would yield an infinite value. Thus I thought to reduce the trigonometric expression by doing this:
$f(x)=\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$
$\frac{1-\cos^2 ax}{x\sqrt{1-\cos ax}}\times\frac{\sqrt{1-\cos ax}}{\sqrt{1-\cos ax}}$
$\frac{(1-\cos ax)(1+\cos ax )(\sqrt{1-\cos ax})}{x(1-\cos ax)}$
Simplifying terms in both denominator and numerator it yields
$\frac{(1+\cos ax )(\sqrt{1-\cos ax})}{x}$
By inserting the expression in the numerator inside the square root I'm getting:
$\frac{\sqrt{(1+\cos ax)^2(1-\cos ax})}{x}$
Expanding the whole expression I'm getting:
$\frac{\sqrt{(1^2+2\cos ax+\cos^2ax)(1-\cos ax})}{x}$
$\frac{\sqrt{1+2\cos ax+\cos^2ax-\cos ax-2\cos^2ax-\cos^3 ax}}{x}$
$\frac{\sqrt{1+\cos ax-\cos^2ax-\cos^3 ax}}{x}$
and that's how far I went. What exactly should be done here?. Can someone help me here?.
| Just expand the fraction with $\frac{\sqrt{1+\cos ax}}{\sqrt{1+\cos ax}}$.
So, you get for $a,x >0$
\begin{eqnarray*}\frac{\sin^2ax}{x\sqrt{1-\cos ax}}
& = & \frac{\sin ax}{ax}\cdot \frac{\sin ax}{\sqrt{1-\cos^2 ax}} \cdot a \sqrt{1+\cos ax}\\
& \stackrel{x\to 0^+}{\longrightarrow} & a\sqrt 2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$ Can somebody find an elementary proof of the following identity:
$$
\arcsin ( 1 - 2x) + 2 \arcsin(\sqrt{x}) = \frac\pi2
$$
I noticed it while solving the following integral:
$$ I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} = -2 \sqrt{1 - x}\sqrt{x} + \int \frac{\sqrt{1 - x}}{\sqrt{x}}
$$
where the first equality follows after applying an integration by parts with $f = \sqrt{x}$ and $\mathrm{d} g = 1/\sqrt{1 - x}$. For the sake of simplicity we omit $\mathrm{d}x$ in each integral. Adding the integral to itself:
\begin{align} 2I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} &= -2 \sqrt{1 - x}\sqrt{x} + \int \left(\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1- x}}\right) \\&= -2 \sqrt{1-x}\sqrt{x} + \int\frac{1}{\sqrt{x}\sqrt{1-x}}\end{align}
The last integral on the RHS evaluates to $\arcsin(1 - 2x)$, so $$I = - \sqrt{1-x}\sqrt{x} + \frac{1}{2}\arcsin(1 - 2x) + C.$$
On the other hand, the integral can also be evaluated by applying a $u$-sub with $u = \sqrt{x}$. We find that:
$$
I = 2 \int\frac{u^2}{\sqrt{1 - u^2}} =2 \left(-u \sqrt{1 - u^2} + \int\sqrt{1 - u^2}\right) = - u\sqrt{1 - u^2} + \arcsin u + C_2
$$
So then it follows that $I$ is also equal to $-\sqrt{x}\sqrt{1-x} + \arcsin{\sqrt{x}} + C_2$. Equate the results of the two methods and plug in a random point to find $C - C_2$ and the subsequent identity.
| Here’s an elementary proof. Let $x=\sin^2\phi$. Then, since $\cos(2\phi)=1-2\sin^2\phi$, we have
$$\arcsin(1-2x)=\arcsin(1-2\sin^2\phi)=\arcsin(\cos(2\phi))=\frac{\pi}{2}-2\phi$$
and
$$2\arcsin(\sqrt{x})=2\arcsin(\sin\phi)=2\phi$$
From which the result follows elementarily (without calc).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Using permutations to find the determinant. I came across the following problem
Let matrix A be a $n\times n$ square matrix such that $a_{ij}$ = max{ i , j }. Prove that det(A) = $(-1)^{n-1}n$
I have read the post If $a_{ij}=\max(i,j)$, calculate the determinant of $A$ and understood all the answers adressing the same problem.
I just wanted to see an alternative method which does not use row subtractions.
As, the determinant fromulla $n(-1)^{n-1}$ is reminiscent of $\frac{d}{dx}(x^{n})$ Is there some way we could use differentiation of determinant technique? How would it be if one goes by the formulla of determinant in terms of permutations?
| One alternative approach:
$$
A_{n+1}=
\begin{pmatrix}
1 & 2 & 3 & \cdots &n & n+1 \\
2 & 2 & 3 & \cdots & n & n+1 \\
3&3&3&\cdots &n & n+1\\
\vdots & \vdots &\vdots &\ddots&\vdots & \vdots\\
n & n & n&\cdots& n & n+1 \\
n+1 & n+1 & n+1&\cdots& n+1 & n+1
\end{pmatrix}.
$$
Subtract $\frac{n+1}{n}$ times the second to last row from the last row to get
$$
\begin{pmatrix}
1 & 2 & 3 & \cdots &n & n+1 \\
2 & 2 & 3 & \cdots & n & n+1 \\
3&3&3&\cdots &n & n+1\\
\vdots & \vdots &\vdots &\ddots&\vdots & \vdots\\
n & n & n&\cdots& n & n+1 \\
0 & 0 & 0 &\cdots & 0 & -\frac{n+1}{n}
\end{pmatrix}.
$$
Now, because of the block triangular structure, we have
$$
\det(A_{n+1}) = - \frac{n+1}{n}\det(A_n).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the following limit: $\lim\limits_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$ How can I evaluate following limit $$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=?$$
My first try:
$$\lim_{x\to\infty}(12x^2-2)\to \infty$$
$$\lim_{x\to\infty}(6x\sqrt{3x^2-2})\to \infty$$so
$$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})=\infty-\infty=0.$$
My answer $0$ is correct, but I don't know whether my method is correct.
My second try:
I substituted $3x^2=2\sec^2\theta$
So limit becomes
$$\lim_{x\to\pi/2}(8\sec^2\theta-2-4 \sqrt{3}\sec\theta\tan\theta)$$
I got stuck. I also can't see application of L'Hospital rule here. Can someone please help me solve this limit? Thanks
| $$\lim_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$$
$$=\lim_{x\to\infty}(3x-\sqrt{3x^2-2})^2$$
$$=\lim_{x\to\infty}\frac{(3x-\sqrt{3x^2-2})^2(3x+\sqrt{3x^2-2})^2}{(3x+\sqrt{3x^2-2})^2}$$
$$=\lim_{x\to\infty}\frac{(6x^2+2)^2}{(3x+\sqrt{3x^2-2})^2}$$
$$=\lim_{x\to\infty}\left(\frac{6+\dfrac{2}{x^2}}{\dfrac3x+\dfrac1x\sqrt{3-\dfrac{2}{x^2}}}\right)^2=\color{blue}{\infty}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 5
} |
Finding $\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$ with Fubini's theorem I have to solve the following double integral
$$\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$$
with $A= \left[0,+\infty\right[ \times [0,1].$ So far I've tried to solve it integrating w.r.t. $y$ first.
$$\iint_0^1\frac{dy\,dx}{(1+x^2)(1+x^2 y^2)} = \int_0^\infty\frac{1}{1+x^2}\int_0^1\frac{dy}{1+x^2 y^2} \, dx. $$
I've solved the internal integral by substitution, remembering that $\int\frac{du}{1+u^2}=\arctan u$
Substitution:
$$x^2 y^2= u^2 \to y=\frac{1}{x}u \to dy=\frac{1}{x}du.$$
$$y=0 \to u=0, \qquad y=1 →u=x$$
So:
\begin{align}
& \int_0^∞\frac{1}{1+x^2} \left( \int_0^x\frac{1}{x}\frac{1}{1+u^2}\,du \right) \, dx \\[8pt]
= {} & \int_0^\infty\frac{1}{x}\frac{1}{1+x^2}[\arctan u] \, dx \\[8pt]
= {} & \int_0^\infty\frac{\arctan x}{x(1+x^2)} \, dx.
\end{align}
Now I have to solve this last integral with the Fubini's theorem but I don't know how to do it.
| Solution 1. Let $I$ denote the double integral. Then
\begin{align*}
I
&= \int_0^1 \int_0^\infty \frac{1}{(1+x^2)(1+x^2 y^2)} \, \mathrm{d}x\,\mathrm{d}y \\
&= \int_0^1 \int_0^\infty \frac{1}{1-y^2} \left( \frac{1}{1+x^2} - \frac{y^2}{1+x^2y^2} \right) \, \mathrm{d}x\,\mathrm{d}y \\
&= \int_{0}^{1} \frac{1}{1-y^2} \left( \frac{\pi}{2} - \frac{\pi y}{2} \right) \, \mathrm{d}y \\
&= \frac{\pi}{2} \int_0^1 \frac{1}{1+y} \, \mathrm{d}y \\
&= \frac{\pi}{2} \log 2.
\end{align*}
Solution 2.
\begin{align*}
I
= \int_0^\infty \int_0^1 \frac{1}{(1+x^2)(1+x^2 y^2)} \, \mathrm{d}y \, \mathrm{d}x
= \int_0^\infty \frac{\arctan x}{(1+x^2)x} \, \mathrm{d}x.
\end{align*}
Substituting $x = \tan\theta$, then
\begin{align*}
\require{cancel}
I
&= \int_0^{\frac{\pi}{2}} \frac{\theta}{\tan\theta} \, \mathrm{d}\theta
= \cancel{\left[ \theta \log \sin\theta \right]_0^{\frac{\pi}{2}}} - \int_0^{\frac{\pi}{2}} \log \sin\theta \, \mathrm{d}\theta.
\end{align*}
Regarding the last integral, we observe that the following holds:
$$ I
= -\int_0^{\frac{\pi}{2}} \log \sin\theta \, \mathrm{d}\theta
= -\int_0^{\frac{\pi}{2}} \log \cos\theta \, \mathrm{d}\theta
= -\int_0^{\frac{\pi}{2}} \log \sin(2\theta) \, \mathrm{d}\theta. $$
Using this, we get
$$ 2I
= -\int_0^{\frac{\pi}{2}} \log (\sin\theta\cos\theta) \, \mathrm{d}\theta
= -\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin2\theta}{2}\right) \, \mathrm{d}\theta
= I + \frac{\pi}{2}\log 2. $$
Therefore
$$ I = \frac{\pi}{2}\log 2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic? Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Show that $a_{n}$ converges as $n\to\infty$. What is the limit? Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic?
MY ATTEMPT
The answer to the first question is $a_{n}\to 1/2$ as $n\to\infty$. Indeed,
\begin{align*}
\lim_{n\to\infty}\sqrt{n^{2}+n} - n = \lim_{n\to\infty}\frac{n}{\sqrt{n^{2} + n} + n} = \lim_{n\to\infty}\frac{1}{\sqrt{1 + 1/n} + 1} = \frac{1}{2}
\end{align*}
To test for monotonicity, we can try to study the behavior of the quotient:
\begin{align*}
\frac{a_{n+1}}{a_{n}} & = \frac{\sqrt{(n+1)^{2} + n + 1} - n - 1}{\sqrt{n^{2} + n} - n}\\\\
& = \frac{(n+1)^{2} + n + 1 - (n+1)^{2}}{n^{2} + n - n^{2}}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}\\\\
& = \frac{n+1}{n}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}
\end{align*}
But then I get stuck, because the first factor is greater than one and the second is smaller than one.
Can someone please finish my attempt or provide an alternative approach?
| As you wrote, for $n\ge 1,$
$$a_n=\frac{1}{b_n}$$
with
$$b_n=\sqrt{1+\frac 1n}+1$$
It is clear that $\forall n\ge 1\;\; b_n\ne 0$ and that
$$(b_n) \;\text{ is decreasing}$$
because
$$b_n-b_{n+1}=\sqrt{1+\frac 1n}-\sqrt{1+\frac{1}{n+1}}$$
$$=\frac{ 1 }{n(n+1)(\sqrt{1+\frac 1n}+\sqrt{1+\frac{1}{n+1}})}>0$$
So, $ (a_n) $ is increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?
| Let
$$
f(a,b,c)=\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}
$$Observe $f$ is invariant under permutations: $f(a,b,c)=f(b,c,a)=f(c,a,b)$, etc. Further, observe $f(0,1,2)=f(1,0,-1)=-2$. In other words, $f$ evaluates to $-2$ at $12$ points, and $12$ is greater than the sum of the degrees of the numerator and denominator. Thus $f$ is identically constant (provided no two of $a,b,c$ are equal).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Prove that $\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$ is an integer.
Prove that $$\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$$ is an integer for all pairs of positive integers $a, b$
(American Mathematical Monthly)
My work -
$
v_{p}((3 a+3 b) !(2 a) !(3 b) !(2 b) !)=\sum_{k \geq 1}\left(\left\lfloor\frac{3 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 a}{p^{k}}\right\rfloor+\left\lfloor\frac{3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 b}{p^{k}}\right\rfloor\right)
$
and
$
\begin{array}{l}
v_{p}\left((2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}\right) \\
\quad \quad=\sum_{k \geq 1}\left(\left\lfloor\frac{2 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+2 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+b}{p^{k}}\right\rfloor+\left\lfloor\frac{a}{p^{k}}\right\rfloor+2\left\lfloor\frac{b}{p^{k}}\right\rfloor\right)
\end{array}
$
now
With the substitution $x=\frac{a}{p^{k}}, y=\frac{b}{p^{k}},$ we have to prove that for any nonnegative real numbers $x, y$ we have
$\lfloor 3 x+3 y\rfloor+\lfloor 2 x\rfloor+\lfloor 3 y\rfloor+\lfloor 2 y\rfloor \geq\lfloor 2 x+3 y\rfloor+\lfloor x+2 y\rfloor+\lfloor x+y\rfloor+\lfloor x\rfloor+2\lfloor y\rfloor$
I tried putting $\{x\}+\lfloor x\rfloor=x$ and $\{y\}+\lfloor y\rfloor=y$ and i get things in terms of fractional parts but i am not able to prove after that ....
thankyou
| The left-hand side minus the right-hand side of your inequality is always 0, 1, or 2. The Mathematica plot below mostly proves it. One can cut the work in half by noticing a symmetry $(x, y) \mapsto (-x, -y)$ which induces $L-R \mapsto 2-(L-R)$. The picture strongly suggests that case-by-case reasoning will be messy no matter what. Having the computer check non-negativity for a certain finite collection of points would also be a proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^1 \arctan^3 x\,dx$ I don't want to use the Fourier series. My work \begin{align}J&=\int_0^1 \arctan^3 x\,dx\\
&=[x\arctan^3 x]_0^1 -3\int_0^1
\frac{x\arctan^2 x}{1+x^2}\,dx\\
&=\frac{\pi^3 }{64}-3\int_0^1 \frac{x\arctan^2
x}{1+x^2}\,dx\\
&= \frac{\pi^3
}{64}-\frac{3}{2}\left[\ln(1+x^2)\arctan^2
x\right]_0^1 +3\int_0^1 \frac{\ln(1+x^2)\arctan
x}{1+x^2}\,dx\\
&=\frac{\pi^3 }{64}-\frac{3\pi^2\ln
2}{32}+3\int_0^1 \frac{\ln(1+x^2)\arctan
x}{1+x^2}\,dx\\
\end{align}
How to continue?
| Not an answer, but Wolfram Alpha finds the amazing $$\int (\arctan(x))^3\mathrm{d}x$$ $$=\frac{3}{2}\operatorname{Li}_3(-e^{2i\arctan(x)})-3i\arctan(x)\operatorname{Li}_2(-e^{2i\arctan(x)})+(\arctan(x))^2(x\arctan(x)-i\arctan(x)+3\ln(1+e^{2i\arctan(x)}))+c$$
And the even more beautiful
$$\int_0^1 (\arctan(x))^3\mathrm{d}x=\frac{1}{64}(\pi^2(\pi+\ln(64))+63\zeta(3)-48\pi C)$$
With Catalan's constant $C$ and the Riemann zeta function $\zeta$.
But, continuing from your last line,
$$\int_0^1 \frac{\ln(1+x^2)\arctan(x)}{1+x^2}\mathrm{d}x$$
We can make the substitution $x=\tan\theta, \mathrm{d}x=\sec^2(\theta)\mathrm{d}\theta$ to get
$$\int_0^{\arctan(1)} \theta\ln(\sec^2(\theta))\mathrm{d}\theta$$
Although knowledge of the polylogarithm will still be needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Percent sign after a fraction? Convert to percent Isn't it wrong to write the following with only the percent sign? Instead of $100 \%$?
The change in height as a percentage is
$$
\frac{a - b}{a} \% \tag 1
$$
where $a$ is the initial height and $b$ is the final height.
Because if $a=10$, $b=5$ we have
$$
\frac{10-5}{10}\%=\frac{1}{2} \% = 0.5 \frac{1}{100} = 0.005
\quad \text{what?!}
\tag 2
$$
If we convert a decimal number to percent we multiply it by $100$ and add the percent sign. We have $1\%=\frac{1}{100}$, so with $100 \%$ we multiply the number by $1$, i.e.
\begin{align}
\frac{10-5}{10} \cdot \color{blue}{1}
&=
\frac{10-5}{10} \cdot \color{blue}{100 \%}
=
\frac{5}{10} \cdot
\color{blue}{100 \frac{1}{100}}
=\frac{1}{2} \cdot
\color{blue}{100 \frac{1}{100}} \tag 3
\\
&=0.5\cdot \color{blue}{100 \frac{1}{100}}
=50 \color{blue}{\frac{1}{100}} = 50 \color{blue}{\%} \tag 4
\end{align}
So, shouldn't we instead write $(1)$ as
$$
\frac{a - b}{a} 100 \% \quad ? \tag 5
$$
| Caution: some authors improperly write expressions like $\dfrac{a-b}a\%$ with the meaning that the value should be thought of as percents. In this case, the percent sign should not be taken for a factor. $\dfrac{a-b}a(\%)$ would be safer.
Leaving this on the side, IMO $$\dfrac{a-b}a\%$$ should rarely occur because there are no common reasons to multiply a fraction by one percent (if this really occurred, I would certainly write $0.01$ instead of $\%$). On the opposite, an evaluation like
$$\frac{a-b}a=5\%$$ makes perfect sense.
I also think that $$\dfrac{a-b}a100\%$$ is ugly/superfluous and would puzzle people.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given positive real numbers $a$, $b$, $c$, $d$, $e$ with $\sum_{\text{cyc}}\,\frac{1}{4+a}=1$, prove that $\sum_{\text{cyc}}\,\frac{a}{4+a^2}\le1$.
Let $a, b, c, d, e$ be positive real numbers such that $$\dfrac{1}{4+a} + \dfrac{1}{4+b} +\dfrac{1}{4+c} +\dfrac{1}{4+d} +\dfrac{1}{4+e} = 1.$$ Prove that $$\dfrac{a}{4+a^{2}} + \dfrac{b}{4+b^{2}} +\dfrac{c}{4+c^{2}} +\dfrac{d}{4+d^{2}} +\dfrac{e}{4+e^{2}} \leq 1.$$
My question is how to prove this inequality by using AM-GM inequality?
My solution (using the Chebyshev inequality).
Since $\dfrac{1}{4+a}+\dfrac{1}{4+b}+\dfrac{1}{4+c}+\dfrac{1}{4+d}+\dfrac{1}{4+e} =1,$ we have
$$1 = \dfrac{1}{4+a}+\dfrac{1}{4+b}+\dfrac{1}{4+c}+\dfrac{1}{4+d}+\dfrac{1}{4+e} \geq \dfrac{a}{4+a^2}+\dfrac{b}{4+b^2}+\dfrac{c}{4+c^2}+\dfrac{d}{4+d^2}+\dfrac{e}{4+e^2}$$
$$\Leftrightarrow \dfrac{1-a}{(4+a)(4+a^2)}+\dfrac{1-b}{(4+b)(4+b^2)}+\dfrac{1-c}{(4+c)(4+c^2)}+\dfrac{1-d}{(4+d)(4+d^2)}+\dfrac{1-e}{(4+e)(4+e^2)} \geq 0.$$
Suppose that $a \geq b \geq c \geq d \geq e$. Then, we get
$$\dfrac{1-a}{4+a} \leq \dfrac{1-b}{4+b} \leq \dfrac{1-c}{4+c} \leq \dfrac{1-d}{4+d} \leq \dfrac{1-e}{4+e}.$$
and $$\dfrac{1}{4+a^2} \leq \dfrac{1}{4+b^2} \leq \dfrac{1}{4+c^2} \leq \dfrac{1}{4+d^2} \leq \dfrac{1}{4+e^2}.$$
Applying the Chebyshev inequality, one gets
$$ \sum_{cyc}\dfrac{1-a}{(4+a)(4+a^2)} \geq \dfrac{1}{5} \sum_{cyc}\dfrac{1-a}{4+a}. \sum_{cyc}\dfrac{1}{4+a^2} = \dfrac{1}{5}\sum_{cyc}\dfrac{1}{4+a^2} \sum_{cyc} \left( \dfrac{5}{4+a}-1 \right)=0.$$
| Another way.
By AM-GM $$\sum_{cyc}\frac{a}{4+a^2}\leq\sum_{cyc}\frac{a}{2a+3}.$$
Thus, it's enough to prove that:
$$1-\sum_{cyc}\frac{a}{2a+3}\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{a}{2a+3}\right)\geq0$$ or $$\sum_{cyc}\frac{1-a}{2a+3}\geq0$$ or
$$\sum_{cyc}\left(\frac{1-a}{2a+3}+5\left(\frac{1}{5}-\frac{1}{4+a}\right)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2}{(2a+3)(4+a)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum volume of a trirectangular tetrahedron circumscribing a spherical ball Assume the corner of a room with the floor and two walls, all three planes meeting each other at $90^0$.
Say, the point where all three meet is considered the origin O and you have X, Y and Z axes along the intersection of different two planes.
Now you place a spherical ball of radius r touching the floor and both walls. Then you have another plane over the ball touching it and meeting the X, Y and Z axes resp. at point A, B and C where $OA = OB = 4, OC = 8$.
Now, suppose we adjust the lengths of OA, OB and OC in a way that does not affect the size of the inscribed ball and you manage to minimize the volume of the tetrahedron OABC.
Find the minimum volume of the circumscribing trirectangular tetrahedron OABC thus formed.
If this minimum volume is $V_{min} = m+n\sqrt p,$ where m, n and p are positive integers and p is square free, determine $m + n + p$.
Here is how I attempted it -
Say, the center of the ball is point Q. Then,
$O = (0,0,0), Q = (r,r,r), A = (4,0,0), B = (0,4,0), C = (0,0,8)$.
Equation of plane ABC is $\frac{x}{4} + \frac{y}{4} + \frac{z}{8} = 1 ==> 2x+2y+z-8 = 0$.
Distance of Q (r,r,r) from this plane is r as the plane touches the ball.
Therefore, $\frac{|2r+2r+r-8|}{\sqrt{2^2+2^2+1^2}} = r$.
We get $r = 1, 4$. As both point O (0,0,) and Q (r,r,r) are on the same side of the plane ABC, we find substituting these values in the plane equation that r = 1.
Now as we change the lengths of OA, OB and OC, say OA = a, OB = b, OC = c.
$V = \frac {1}{6}abc$ has to be minimized.
Equation of plane ABC = $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 ==> (bc)x+(ac)y+(ab) - abc = 0$
Radius of the inscribed ball is 1, as solved earlier. Now, is there a simple way to say when the volume will be minimum or do I have to go for Lagrange or AM-GM method? Also, finding $m + n + p$ seems confusing.
Please guide me from here or provide rest of the solution.
| The intercepts of the plane $\frac xa+\frac yb+\frac zc=1$ must satisfy
\begin{align*}
\frac{\left|\displaystyle\sum_{cyc}\frac 1{a}-1\right|}{\sqrt{\displaystyle\sum_{cyc}\frac 1{a^2}}}&=1\\
\left|\displaystyle\sum_{cyc}\frac 1{a}-1\right|^2&=\displaystyle\sum_{cyc}\frac 1{a^2}\\
\left(\displaystyle\sum_{cyc}\frac 1{a}\right)^2+1-2\displaystyle\sum_{cyc}\frac 1{a}&=\displaystyle\sum_{cyc}\frac 1{a^2}\\
\displaystyle\sum_{cyc}\frac 1{a^2}+2\displaystyle\sum_{cyc}\frac 1{ab}+1-2\displaystyle\sum_{cyc}\frac 1{a}&=\displaystyle\sum_{cyc}\frac 1{a^2}\\
\displaystyle\sum_{cyc}\frac 1{a}-\displaystyle\sum_{cyc}\frac 1{ab}&=\frac12\tag{1}\\
abc\displaystyle\sum_{cyc}\frac 1{a}-\displaystyle\sum_{cyc}a&=\frac{abc}2\\
abc&=\frac{\displaystyle\sum_{cyc}a}{\displaystyle\sum_{cyc}\frac 1{a}-\frac 12}\\
\end{align*}
Minimizing $abc$ is easy since the maximum of the denominator and the minimum of the numerator occur simultaneously at equality $a=b=c$. This can be proved by the $AM\ge HM$ inequality as follows
$$\frac{a+b+c}3\ge \frac3{\frac1a+\frac1b+\frac1c}$$
in which $a+b+c$ is maximum when $\frac1a+\frac1b+\frac1c$ is minimum.
So, putting $a=b=c$ in equation $(1)$, we get
\begin{align*}
\frac3a-\frac3{a^2}&=\frac12\\
a^2-6a+6&=0\\
a&=3+\sqrt3&(\because a>2r=2)\\
\end{align*}
Finally, the minimum volume $\frac{abc}6=\frac{(3+\sqrt3)^3}6=9+5\sqrt3\equiv 17.66\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the equation of an hyperbola if the distance between its directrices is $\frac83$ and its eccentricity is $\frac32$ I have this problem:
Find the canonical equation of an hyperbola if the distance between the directrices is $\frac{8}{3}$ and the eccentricity $e=\frac{3}{2}$.
How would you solve it?
This is my try:
The canonical equation of a hyperbola takes the form $x^2/a^2 - y^2/b^2 = 1$, and the foci are at a distance $c > a$ from the origin, and our directrices are located at $a^2/c$, where eccentricity is $\sqrt{a^2+b^2}/a$. If $e = 3/2$, then
\begin{align}
\frac{\sqrt{a^2+b^2}}{a} = \frac32
&\implies \frac32 a = \sqrt{a^2+b^2} \\[4pt]
&\implies \frac94 a^2 = a^2 + b^2 \\[4pt]
&\implies a^2\left(\frac94 - 1\right) = b^2 && (c^2 = a^2 + b^2)
\end{align}
so
\begin{align}
a^2 + a^2\left(\frac94-1\right) = c^2
&\implies c^2 = a^2\left(1 + \frac94 - 1\right) \\[4pt]
&\implies c^2 = \frac94 a^2 \\[4pt]
&\implies c = \frac32 a
\end{align}
so our directrix is located at
$$\frac{a^2}{c} = \frac{a^2}{a\cdot 3/2} = \frac{a}{3/2} = \frac{2}{3}\cdot a,$$
but the distance between directrices is $8/3$, so it's double the distance from the origin, so essentially,
$$ \frac{8}{3} = 2x = 2\cdot \frac{2}{3} \cdot a
\implies \frac{8}{3} = \frac{4}{3} \cdot a
\implies a = 2. $$
And since
$$ c^2 - a^2
= b^2
= \frac{9}{4} \cdot 4^2 - 4^2
= 16 \left( \frac{9}{4} - 1\right)
= 4 \cdot 9 - 16
= 36-16
=20. $$
In total, we have
$$a^2 = 4, \qquad\text{and}\qquad b^2 = 20,$$
which results in the canonical form
$$\frac{x^2}{4} -\frac{y^2}{20} = 1.$$
Also, how would you find the eccentricity of an ellipse if the sides of the square inscribed in it pass through the foci of the ellipse?
My try:
Let’s imagine a generic ellipse, and the square inside has its sides located at $-x$ and $x$, making the side of the square $2x$, and we know that the foci occur at
$$ x = \frac{a^2}{c}
\quad\text{so}\quad
2x = \frac{2a^2}{c},
\qquad\text{(since $c^2 = a^2 + b^2$)},$$
and eccentricity is
$$ \frac{\sqrt{a^2 + b^2}}{a} = \frac{c}{a}, $$
so if $ e = c/a $ and
$$ 2x = \frac{2a^2}{c} = 2a \cdot \frac{a}{c} = 2a\cdot \frac{1}{e} = \frac{2a}{2}
\implies x = \frac{a}{2}
\implies e = \frac{a}{x}
$$
and $x$ would be half the side of the square in this case.
| In the hyperbola exercise, you double-squared the $a$ in your $c^2-a^2 = \cdots = 20$ calculation, effectively multiplying everything by an extra factor of $4$. You should get $9-4=5$, instead.
Here's a clearer path to the solution:
If the hyperbola's transverse semi-axis is $a$, its center-to-focus distance is $c$, and its eccentricity is $e$, then the center-to-directrix distance (call it $d$) is indeed given by $d=a^2/c$, so that (since $e=c/a$) we can write $a = de$.
Since the distance between the directrices is $8/3$, we have $d=4/3$; given the eccentricity $3/2$, we have
$$a = de=\frac43\cdot\frac32=2 \tag{1}$$
Then,
$$c=ae = 2\cdot\frac32=3 \tag{2}$$
and then
$$b^2=c^2-a^2=9-4=5\tag{3}$$
so that, for an origin-centered hyperbola with a horizontal transverse axis, the equation is
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad\to\quad \frac{x^2}{4}-\frac{y^2}{5}=1 \tag{$\star$}$$
(I think the ellipse question should be posted separately, so I won't address it here.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{n\to \infty}\left((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2\cdots(1+\frac{n^2}{n^2})^n\right)^{\frac{1}{n}}$ Calculate the following limit:
$$\lim_{n\to \infty}\left((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2\cdots(1+\frac{n^2}{n^2})^n\right)^{\frac{1}{n}}$$
My attempt:
Let
$$y=\lim_{n\to \infty}\left((1+\frac{1}{n^2})(1+\frac{2^2}{n^2})^2\cdots(1+\frac{n^2}{n^2})^n\right)^{\frac{1}{n}}=\lim_{n\to \infty}\left((1+\frac{1}{n^2})^{\frac{1}{n}}(1+\frac{2^2}{n^2})^{\frac{2}{n}}\cdots(1+\frac{n^2}{n^2})^{\frac{n}{n}}\right)$$
Now, taking logarithm on both sides to get: $$\log y=\lim_{n \to \infty}\sum_{k=1}^n\frac{k}{n}\log(1+(\frac{k}{n})^2)$$
But I am unable to convert it into integral form (riemann sum to integral ) as the expression $\frac{k}{n}$ is present instead of $\frac{1}{n}$. Can someone please help me in solving this question. Thanks in advance.
| Considert the factors
$$\left (1+\frac{k^2}{n^2}\right )^k,\,\, n/2< k \le n.$$
Each of these is greater than
$$\left (1+\frac{(n/2)^2}{n^2}\right )^{n/2} = \left (\frac{5}{4}\right)^{n/2}.$$
There are more than $n/2$ such factors. Thus the product of these factors is at least $(5/4)^{n^2/4}.$ Now take the $n$th root to get $(5/4)^{n/4}\to \infty.$ Thus the limit of the original expression is $\infty.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $y=f(x)=\frac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$. Question: If $y=f(x)=\dfrac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$.
We have $y=\dfrac{3\left(\frac{3x-5}{2x-m}\right)-5}{2\left(\frac{3x-5}{2x-m}\right)-m} $
How can I find $m$? It is given than $m=3$.
| If $y=f(x)=\dfrac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$.
$$f(y) = \frac{3y -5}{2y-m} = x$$
$$2yx -mx = 3y-5$$
$$2yx - 3y +5 = mx$$
$$2\dfrac{3x-5}{2x-m}x-3\dfrac{3x-5}{2x-m} + 5 = mx$$
$$\dfrac{6x^2-10x-9x+15+10x -5m}{2x-m}=mx$$
$$ \dfrac{6x^2 -9x+15 -5m}{2x-m}=mx$$
$$ \dfrac{6x^2 -9x+15 -5m-2x^2m+ m^2x}{2x-m}= 0$$
$$ 6x^2 -9x+15 -5m-2x^2m+ m^2x = 0$$
$$ 3(2x^2 -3x+5) =m(5+2x^2- mx) $$
$$ 3(2x^2 -3x+5) =m(2x^2 -mx+5) $$
You can easily see that
$$m =3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the remainder $1690^{2608} + 2608^{1690}$ when divided by 7? Find the remainder $1690^{2608} + 2608^{1690}$ when divided by 7?
My approach:-
$1690 \equiv 3(\bmod 7)$
$1690^{2} \equiv 2(\bmod 7)$
$1690^{3} \equiv-1 \quad(\mathrm{mod} 7)$[ quite easy to determine , $\frac{2*1690}{7}$..so on]
$\left(1690^{3}\right)^{869} \cdot 1690 \equiv(-1)^{869}1690 \quad(\mathrm{mod} 7)$
$1690^{2608} \equiv -1690 \quad(\mathrm{mod} 7)$....(1)
again for $2608$
$2608 \equiv 4(\bmod 7)$
$2608^{2} \equiv 2(\bmod 7)$
$2608^{3} \equiv1 \quad(\mathrm{mod} 7)$[ quite easy to determine , $\frac{2*2608}{7}$..so on]
$\left(2608^{3}\right)^{563} \cdot 2608 \equiv(1)^{563}2608 \quad(\mathrm{mod} 7)$
$2608^{1690} \equiv 2608 \quad(\mathrm{mod} 7)$...(2)
Now applying property
adding (1) + (2),
$1690^{2608} + 2608^{1690}=918 \quad(\mathrm{mod} 7)$
$\boxed{1690^{2608} + 2608^{1690} \equiv 1 \quad(\mathrm{mod} 7)}$
Is my approach best? or Anyother approach is there comparatively better than it
| Seems fine. Try to use smaller number as fast as you can.
Using Fermat's little theorem,
$$1690^{2608}\equiv 3^{2608}\equiv 3^{6(434)+4} \equiv 3^4 \pmod{7}$$
$$2608^{1690}\equiv 4^{6(281)+4}\equiv 4^4 \pmod{7}$$
$$3^4+4^4 \equiv (-4)^4+4^4 \equiv 2(4^4) \equiv 2^9 \equiv 2^3 \equiv 1 \pmod{7}$$
Remark:
Also, I notice the way you compute $1690^3 \pmod{7}$ is by multiplying $2$ with $1690$ and then get the remainder when you divide by $7$. You don't have to do that. To compute $1690^3$, just multiply $2$ and $3$ and you get $6$ directly. That is once you figure out that $1690 \equiv 3\pmod{n}$, we know that $1690^n \equiv 3^n \mod{7}$, work with $3$ rather than $1690$. In fact, for prime $p$, $gcd(a,p)=1$, $a^n \equiv(a\pmod{p})^{(n \pmod{(p-1)})}\pmod{p}$ can reduce the magnitude of the number that you have to work wth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Looking for an alternative approach to Find the minimum value of the function $f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$ is The minimum value of the function $f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$ is
My work
$$
\begin{aligned}
\text f(x, y) &=4 x^{2}+9 y^{2}-12 x-12 y+14 \\
&=\left(4 x^{2}-12 x+9\right)+\left(9 y^{2}-12 y+4\right)+1 \\
&=(2 x-3)^{2}+(3 y-2)^{2}+1 \geq 1
\end{aligned}
$$
So, minimum value of $\mathrm{f}(\mathrm{x}, \mathrm{y})$ is 1
My question is How can it be done by calculus?
| $$f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$$
$$\frac{\partial f(x,y)}{\partial x}=8x-12$$
$$\frac{\partial f(x,y)}{\partial y}=18y-12$$
Set the partial derivatives equal to $0$; so $x=\frac{3}{2}$ and $y=\frac{2}{3}$
$$f\left(\frac{3}{2},\frac{2}{3}\right)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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In a triangle prove that $\mathrm{cosec}( A) +\mathrm{cosec}(B) +\mathrm{cosec}(C)\le \frac{2\sqrt 3}{9}\left (1+\frac Rr\right)^2$ In a triangle prove that $\mathrm{cosec}( A) +\mathrm{cosec}(B) +\mathrm{cosec}(C)\le \frac{2\sqrt 3}{9}\left (1+\frac Rr\right)^2$
My progress:
$\left (1+\frac Rr\right)^2=\left(\frac{2\prod sin A+\sum sin A}{2\prod sin A}\right)^2\ge \left(\frac{2\prod sin A+\sum sin A sin B}{2\prod sin A}\right)^2\ge\frac{8\prod sin A\sum sin A sin B} {\left(2\prod sin A\right)^2}=2\sum cosec A$
| Let $a=\frac{y+z}{2},$ $b=\frac{x+z}{2}$ and $c=\frac{x+y}{2}$
Thus, in the standard notation we need to prove that:
$$\sum_{cyc}\frac{1}{\sin\alpha}\leq\frac{2}{3\sqrt3}\left(1+\frac{\frac{abc}{4S}}{\frac{2S}{a+b+c}}\right)^2$$ or
$$\sum_{cyc}\frac{1}{\frac{2S}{bc}}\leq\frac{2}{3\sqrt3}\left(1+\frac{2abc(a+b+c)}{16S^2}\right)^2$$ or
$$\frac{ab+ac+bc}{4S}\leq\frac{1}{3\sqrt3}\left(1+\frac{2abc}{\prod\limits_{cyc}(a+b-c)}\right)^2$$ or
$$\frac{\sum\limits_{cyc}(x+y)(x+z)}{4\sqrt{xyz(x+y+z)}}\leq\frac{1}{3\sqrt3}\left(1+\frac{\prod\limits_{cyc}(x+y)}{4xyz}\right)^2$$ or
$$12xyz\sum\limits_{cyc}(x^2+3xy)\cdot\sqrt{\frac{3xyz}{x+y+z}}\leq\left(\sum_{cyc}(x^2y+x^2z+2xyz)\right)^2.$$
But $$\sum_{cyc}(x^2+3xy)\leq\frac{4}{3}(x+y+z)^2$$ and $$\sqrt{3xyz(x+y+z)}\leq xy+xz+yz.$$
Thus, it's enough to prove that
$$16xyz(x+y+z)(xy+xz+yz)\leq\left(\sum_{cyc}(x^2y+x^2z+2xyz)\right)^2.$$
Now, let $\sum\limits_{cyc}(x^2y+x^2z)=6uxyz.$
Thus, by AM-GM $u\geq1$ and we need to prove that
$$16(6u+3)\leq(6u+6)^2$$ or $$(u-1)(3u+1)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate: $\int \frac{x}{\left(x^2-4x-13\right)^2}dx$.
Integrate:
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
Here's my attempt:
I first completed the squares for the denominator:
$$\left(x^2-4x-13\right)^2=(x-2)^2-17 \implies \int \frac{x}{\left(\left(x-2\right)^2-17\right)^2}dx$$
I then used $u$-subsituition:
$$u=x-2 \implies \int \frac{u+2}{\left(u^2-17\right)^2}du = \int \frac{u}{\left(u^2-17\right)^2}du+\int \frac{2}{\left(u^2-17\right)^2}du$$
The first part of the new integral is quite simple:
$$\int \frac{u}{\left(u^2-17\right)^2}du=\frac{-1}{2(u^2-17)}$$
Then I did the second part:
$$\int \frac{2}{\left(u^2-17\right)^2}du = -\frac{1}{2\left(u^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|u+\sqrt{17}\right|-\frac{1}{68\left(u+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|u-\sqrt{17}\right|-\frac{1}{68\left(u-\sqrt{17}\right)}\right) = -\frac{1}{2\left(\left(x-2\right)^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) = -\frac{1}{2\left(x^2-4x-13\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) + C, C \in \mathbb{R}$$
Is this working out correct? I'm not really sure how WolframAlpha works, so I didn't check it on there.
| Hint:
Observe that
$$\left(-\frac u{u^2-a^2}\right)'=\frac{u^2+a^2}{(u^2-a^2)^2}=\frac1{u^2-a^2}+\frac{2a^2}{(u^2-a^2)^2}.$$
Hence
$$\int\dfrac{du}{(u^2-a^2)^2}=-\frac u{2a^2(u^2-a^2)}-\frac1{2a^2}\int\dfrac{du}{u^2-a^2}.$$
The last integral by $\text{artanh}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.