Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Number of ordered Pairs satisfying $4^m-3^n=1$ Find the Number of ordered Pairs $(m,n)$ of positive integers satisfying $4^m-3^n=1$
Mt try:
Trivially $m=n=1$ satisfies
Let $m \gt 1$
$$4^m-3^n=(1+3)^m-3^n=1$$
$\implies$
$$3\binom{m}{1}+3^2\binom{m}{2}+3^3\binom{m}{3}+\cdots+3^m=3^n$$
Now since LHS is not a power of $3$ and RHS is, this is possible only when $m=1$
Hence the only ordered pair is $(1,1)$
is this the right way?
| $4^{m} - 3^{n} = 1$ or $4^m-1=3^n$
$(2^m-1)(2^m+1)=3^n$ therefore $2^m-1$ and $2^m+1$ are perfect powers of 3
Let $2^m+1=3^x$ and $2^m-1=3^y$
$3^x-3^y=2 (x>y)$ therefore $3^y(3^{x-y} -1) = 3^0.2^1$
Hence $ y=0$ and $3^{x-y} -1 =2$ implying $x=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for $n \geq 5, f_{n}+f_{n-1}-1$ has at least $n+1$ prime factors Question -
Prove that for $n \geq 5, f_{n}+f_{n-1}-1$ has at least $n+1$ prime factors, where $f_{n}=2^{2^{n}}+1$
My proof - I proved it using induction,but i got stucked in base case step,
for $n=5$ we get after lots of factoring that
$f_{5}+f_{4}-1$ = $3 \cdot 7 \cdot 13 \cdot 241 .(2^{16}-2^8+1)$
Now i am not able to factor $(2^{16}-2^8+1)$ , i tried mod $3,7,19,13,9$ but none of them working ...
thankyou
| Base case:
$n=5$
By the question and the comments, we get $$f_5+f_4+1=3\times7\times13\times97\times241\times673$$, so base case finished.
Let $p(n)$ be the number of distinct prime factors of $n$.
Assume the proposition is true for integer $k\ge5$, i.e. $$p(f_k+f_{k-1}+1)=p\Big(2^{2^{k+1}}+2^{2^k}+1\Big)\ge k+1$$. Then we let $K=2^{2^k}$.
So $$p(f_k+f_{k-1}+1)=p\big(K^2+K+1\big)\ge k+1$$
When $n=k+1$,
$$p(f_{k+1}+f_{k}+1)=p\big(K^4+K^2+1\big)= p\Big(\big(K^2+K+1\big) \big(K^2-K+1\big) \Big)$$. As the number of distinct prime factors of $K^2+K+1 $ is larger than $k+1$, also $K^2+K+1$ and $K^2-K+1$ are coprime, so $K^2-K+1$ has at least one prime factor which is not a prime factor of $K^2+K+1$, so $$p(f_{k+1}+f_{k}+1)=p\big(K^4+K^2+1\big)\ge k+1+1=k+2 $$, as desired.
| {
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Find radius of largest circle within ellipse $\frac{x^2}{9} + \frac{y^2}4 = 1$ with their intersection only at $(3,0)$ An ellipse is defined by the equation $$\frac{x^2}{9} + \frac{y^2}4 = 1$$
Compute the radius of the largest circle that is internally tangent to the ellipse at $(3,0),$ and intersects the ellipse only at $(3,0).$
How can I write an equation for the largest circle within this ellipse if its equation is given? Is there a property or theorem I'm missing? Is there another way to solve?
| $$\frac{x^2}{9}+\frac{y^2}{4}=1\implies y'=-\frac{4x}{9y}\implies y''=\frac{4(xy'-y)}{9y^2}$$
The radius of curvature, R i.e. radius of circular arc that best approximates the given ellipse at the point $(3,0)$ will be the radius of largest circle which internally touches the ellipse at point $(3,0)$ only, is given as
$$R=\left|\frac{(1+y'^2)^{3/2}}{y''}\right|=\left|\frac{\left(16x^2+81y^2\right)^{3/2}}{36(4x^2+9y^2)}\right|$$
Substituting $x=3$ & $y=0$ in above equation, radius $R$ of the largest circle touching ellipse internally at a single point $(3,0)$, is given as
$$\color{blue}{R}=\left|\frac{\left(16(3)^2+81(0)^2\right)^{3/2}}{36(4(3)^2+9(0)^2)}\right|=\color{blue}{\frac43}$$
| {
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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.
| We have $6z-x-y=6uvw-9u^3=v^3+w^3$. Hence, $6uwv=9u^3+v^3+w^3$. As in general $\sqrt[3]{x_1x_2x_3} \le \frac{x_1+x_2+x_3}{3}$, we have $6uvw<3\sqrt[3]{9}.uvw \le 9u^3+v^3+w^3$, so there is no solution.
B: Likewise we have $6uvw<3\sqrt[3]{9}.uvw \le u^3+9v^3+w^3$, so there is also no solution.
We conclude that there exists no positive integers $x,y,z$ such that $x$ and $y$ are coprime and fulfill the Diophantine equation $(3z)^3=3(x+y)(3z-x)(3z-y)$. QED
A final remark: $z$ is even. The case that $x$ is even and $y$ is uneven implies that $(3z-x)(3z-y)$ is an even number, hence $(3z)^3$ is even.
| {
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Square roots modulo $pq$ where $p$, $q$ are distinct primes - confusion with quote I have a doubt about the following quote from a book:
The Chinese Remainder Theorem implies that, if $p$ and $q$ are
distinct primes, then $s$ is a square modulo $pq$ if and only if $s$
is a square modulo $p$ and $s$ is a square modulo $q$. In particular,
if $s \equiv x^2 \equiv (x')^2 \pmod p$ where $x \neq x'$, and
likewise $x \equiv y^2 \equiv (y')^2 \pmod q$, then $s$ has exactly
four square roots modulo $pq$, namely,
$$s \equiv (xy)^2 \equiv (x'y)^2 \equiv (xy')^2 \equiv (x'y')^2 \pmod{pq}$$
As an attempt to understand the quote, suppose I want to find the square roots of $11$ modulo $133$. So, I have $s = 11$, and, since $133 = 7 \times 19$, I have $p = 7$ and $q = 19$. I first need to separately find the roots of $11$ modulo $7$ and of $11$ modulo $19$:
*
*To find the roots modulo $7$, solve $x^2 \equiv 11 \pmod 7$. Since $11 \equiv 2^2 \equiv 5^2 \pmod{7}$, the roots are $x = 2$ and $x' = 5$.
*To find the roots modulo $19$, solve $x^2 \equiv 11 \pmod{19}$. Since $11 \equiv 7^2 \equiv 12^2 \pmod{19}$, the roots are $y = 7$ and $y' = 12$.
The quote is saying that $xy$, $x'y$, $xy'$ and $x'y'$ are roots of $s$ modulo $pq$. However, this seems to be untrue in my example:
*
*$xy = 2 \times 7 = 14$, but $11 \not\equiv 14^2 \pmod{133} $.
*$xy' = 2 \times 12 = 24$, but $11 \not\equiv 24^2 \pmod{133} $.
*$x'y = 5 \times 7 = 35$, but $11 \not\equiv 35^2 \pmod{133} $
*$x'y' = 5 \times 12 = 60$, but $11 \not\equiv 60^2 \pmod{133} $
Am I missing something?
| The Chinese remainder theorem must be applied correctly.
In your example, it says that there is exactly one $a \pmod{133}$ such that $a \equiv 2 \pmod{7}$ and $a \equiv 7 \pmod{19}$. See, for example, here.
Since we have $1 = 3 * 19 - 8 * 7$, we put $a = 2 * 3 * 19 - 7 * 8 * 7 = -278$ and find that, indeed $a^2 \equiv 11 \pmod{133}$. An explanation of this calculation, can be found, for example, here.
Similar calculations for the other combinations of roots of $11$ mod $7$ and $19$ produce the following four distinct roots of $11$ mod $133$.
$$a \equiv 121 \pmod{133}$$
$$b \equiv 107 \pmod{133}$$
$$ c \equiv 26 \pmod{133}$$
$$ d \equiv 12 \pmod{133}$$
| {
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Show that $f(x) = |1-x^2|^3$ is a differentiable function. $f: \mathbb{R} \rightarrow \mathbb{R}$
$x \rightarrow |1-x^2|^3$
Show that f is a differentiable function and calculate its derivative.
Check whether f if is a continuous function.
$f'(x_0)=lim_{x\rightarrow x_0} \frac{f(x) - f(x_0}{x-x_0} = lim_{x\rightarrow x_0}\frac{|1-x^2|^3-|1-x_0^2|^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{\sqrt{(1-x^2)^2}^3- \sqrt{(1-x_0^2)^2}^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{\sqrt{(1-x^2)^6}- \sqrt{(1-x_0^2)^6}}{x-x_0} = lim_{x\rightarrow x_0} \frac{(1-x^2)^3- (1-x_0^2)^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{(1-x^2)^3- (1-x_0^2)^3}{x-x_0}$
But I didn't get any further. How to get the $x-x_0$ out? I tried multiplying out the numerator and then using polynomial division but I didn't get a solution.
So I tried calculating the derivative directly, but that doesn't show it generally.
$f(x) = |1-x^2|^3 = (\sqrt{(1-x^2)^2})^3$
$f'(x) \\= 3 \cdot (\sqrt{(1-x^2)^2})^2 \cdot \frac{1}{2} \cdot ((1-x^2)^2)^{-\frac{1}{2}} \cdot 4x\cdot (1-x^2) \\ = 3 \cdot |1-x^2|^2 \cdot \frac{1}{|1-x^2|} \cdot 2x \cdot (1-x^2) \\= |1-x^2| \cdot 6x \cdot (1-x^2)$
I looked at the graph. The graph looks like a "W" but with "soft turns". If I can prove that the function is differentiable in every point then it is a continuous function.
So how do I prove it using this definition: $f'(x_0)=lim_{x\rightarrow x_0} \frac{f(x) - f(x_0}{x-x_0}$ ?
| The function is polynomial everywhere except $x=\pm1,$ therefore it suffices to verify derivability with limits only at $x=1$ and $x=-1.$
At each of these values the limit exists (is equal to $0$), hence the function is differentiable on $\mathbb{R}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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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.
| Your hypothesis is true.
Indeed the solution of the equation:
$$
2\sqrt{10-8x}+\sqrt{9-4x-4\sqrt{1-x^2}}=5
$$
is
$$
x=\frac{2+7\sqrt{46}}{50},\text{ with } \sqrt{1-x^2}=\frac{14-\sqrt{46}}{50}.
$$
Substituting this into the derivative of the distance one obtains:
$$
\left[-\frac{8}{\sqrt{10-8x}}+\frac{2(x-\sqrt{1-x^2})}{\sqrt{1-x^2}\sqrt{9-4x-4\sqrt{1-x^2}}}\right]_{x=\frac{2+7\sqrt{46}}{50}}\\
=-\frac{4(14+\sqrt{46})}{15}+\frac{4(14+\sqrt{46})}{15}=0.
$$
| {
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All integer solutions of $x^3-y^3=2020$.
Find all integer pairs $(x,y)$ satisfying $$x^3-y^3=2020\,.$$
First,
$x^3-y^3=(x-y)(x^2+xy+y^2)=2020$
and $2020=2^2\cdot 5 \cdot 101$.
But what next? Can it be worked out by using modulo? Or how?
Any idea? Thanks in advance.
| Let $d = x - y$. Then we want $x^3 - (x-d)^3 = 2020$, which is a quadratic equation in $x$. The discriminant is $24240d - 3d^4 = 3d (8080 - d^3)$, which is nonnegative only for $d \in [0, 8080^{1/3}]$, and since $d$ is an integer it must be between $0$ and $20$.
Moreover, since $(x-y)^3 = d(x^2 + xy + y^2) = 2020$, $d$ must be a divisor of $2020$. This leaves only $6$ possibilities: $d = 1, 2, 4, 5, 10, 20$. We already had a finite problem to solve before, but this observation reduces the number of cases.
For each value of $d$, the solutions to the quadratic equation have irrational solutions, so there are no integer solutions $(x,y)$.
| {
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Find the value of $\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}}$ . We have:
$\sqrt{(1+0)+\sqrt{(4+1)+\sqrt{(9+2)+\sqrt{(16+3)+\sqrt{(25+4)+\cdots}}}}}.$
Basically I'm not getting any clue at the moment for reducing the infinite nested radicals. Any hint would be helpful. Thanks in advance.
| Note that we have the identity
$$n=\sqrt{(n^2-n-1)+(n+1)}$$
Which we can apply indefinitely to give
\begin{align}
2
&=\sqrt{1+3}\\
&=\sqrt{1+\sqrt{5+4}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+5}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+6}}}}\\
&=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+7}}}}}\\
\end{align}
Note that the $n$th line above differs from the provided expression by an $O(n)$ term in the innermost square root. Due to $n$ square roots this error is reduced to zero as $n\to\infty$.
Edit: As shown above, ignoring some of the first terms gives radical expressions for every natural number. For example
$$3=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\cdots}}}}$$
$$4=\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}$$
| {
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Let $a$ and $b$ be elements of odd order in a group. Show that $a^2$ and $b^2$ commute if and only if $a$ and $b$ commute. I really don't know how to solve this problem. I just know that if $|a|=2k_1+1$ and $|b|=2k_2+1$, then,
$a^{2k_1+1}=e=a^0$ and $b^{2k_2+1}=e=b^0$. Also, if $|G|=n$, then, $2k_1+1,2k_2+1\equiv 0 (\mod n)$.
| You got it the other way round. It's $n \equiv 0 \pmod{2k_1 + 1}$ and $n \equiv 0 \pmod{2k_2 + 1}$.
Suppose $a^2b^2 = b^2a^2$. Then:
\begin{align*}
ab &= a^{2k_1+1}(ab)b^{2k_2+1} \\
&= a^{2k_1+2}b^{2k_2+2} \\
&= (a^2)^{k_1+1}(b^2)^{k_2+1} \\
&= (b^2)^{k_2+1}(a^2)^{k_1+1} \\
&= b^{2k_2+1}(ba)a^{2k_1+1} \\
&= ba
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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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?
| Note that
$$1+z^2+z^4+\cdots = \frac{1}{1-z^2}.$$
Integrating both sides,
$$z+\frac{z^3}{3}+\frac{z^5}{5}+\cdots=\frac 12 (\log(1+z)-\log(1-z)) = \tanh^{-1}(z).$$
Observe that multiplying both sides by $z$ gives
$$\sum_{n=1}^\infty \frac{z^{2n}}{2n-1} = z\tanh^{-1}(z)$$
and multiplying the left hand side by $\frac 1z$ and subtracting the $1$ term gives
$$\sum_{n=1}^\infty \frac{z^{2n}}{2n+1} = -1 + \frac{\tanh^{-1}(z)}{z}.$$
Rewriting $\tanh^{-1}(z)$ as $\ln\left(\frac{1-z}{1+z}\right)$, remembering that $z$ is on the unit circle, we can draw vectors $1+z$ and $1-z$ in the complex plane. Doing some basic geometry, we can see that the angle between these is $\frac \pi 2$, and that the lengths of $1+z$ and $1-z$ are $2 \cos \left(\frac \theta 2 \right)$ and $2 \cos \left( \frac \pi 2 - \frac \theta 2 \right) = 2 \sin \left( \frac \theta 2 \right)$.
So $\frac{1-z}{1+z} = -i \cdot \tan \left(\frac x2 \right)$, and so the $\log$ of this is
$$-\frac{i \pi}{2} + \ln\left(\tan \left(\frac x2 \right)\right)$$
(since $\log(-i) = -\frac{i \pi}{2}$).
From here, everything is easily calculatable.
| {
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Evaluating $\lim_{x\to\infty} \frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}}$ $$\lim_{x\to\infty} \frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}}$$
What I did to solve this problem is here
I tried to multiply by $1/x$ the numerator and denominator and I got a limit of $1$ at the end, but I am not sure if what I got is right.
| $$\frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}} \sim \frac{x-x^{2/3}}{x-x^{4/5}}=\frac{1-x^{-1/3}}{1-x^{-1/5}}$$
So $$L=\lim_{x\to \infty} \frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}} =1$$
| {
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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?
| We'll prove that $\ln\frac{3}{2}<\frac{13}{32},$ for which we'll prove that for any $x\geq1$ the following inequality holds.
$$\ln{x}\leq(x-1)\sqrt[3]{\frac{2}{x^2+x}}.$$
Indeed, let $f(x)=(x-1)\sqrt[3]{\frac{2}{x^2+x}}-\ln{x}.$
Thus,
$$f'(x)=\frac{\sqrt[3]2(x^2+4x+1)-3\sqrt[3]{x(x+1)^4}}{3\sqrt[[3]{(x^2+x)^4}}=\frac{2(x^2+4x+1)^3-27x(x+1)^4}{someting\\positive}=$$
$$=\frac{(2x^2+5x+2)(x-1)^4}{someting\\positive}\geq0,$$
which gives $$f(x)\geq f(1)=0.$$
Thus, $$\ln1.5<0.5\sqrt[3]{\frac{2}{3.75}}=\frac{1}{\sqrt[3]{15}}.$$
Id est, it's enough to prove that:
$$\frac{1}{\sqrt[3]{15}}<\frac{13}{32}$$ or
$$32768<32955$$ and we are done!
| {
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what's the biggest integer that can not be made up by 6, 9, and 20? what's the biggest integer that can not be made up by 6, 9, and 20?
for example, 35 can be expressed as 35 = 20 + 9 + 6
46 can be expressed as 46 = 20 * 2 + 6;
27 can be expressed as 27 = 9 * 3 or 27 = 6 * 3 + 9
so, what's the biggest integer that can not be made up by 6, 9, and 20?
| In simple words: we construct a set of rules for making $n+1$ from $n$ if we already have $n=6a+9b+20c$ (e.g. $n+1=6(a-5)+9(b-1)+20(c+2)$) and then we find maximum $N$ such that we can apply no rule of the set. Thus for every $n>N$ we can make $n+1$ from $n=6a+9b+20c$ (if we can find such $a,b,c$) thus we can have any number (say, by induction), starting with a reachable one.
Rigorously:
Consider $(a,b,c)\in$ $S=\{(2, 1, -1),$ $(-1, 3, -1),$ $(-2, -3, 2),$ $(5, -1, -1),$ $(1, -5, 2),$ $(-5, -1, 2),$ $(-4, 5, -1),$ $(-8, 1, 2)\}$.
For every $(a,b,c)\in S\ \ 6a+9b+20c=1$.
Suppose we have $n=6a+9b+20c$ for some $n> N$ and we want to construct $n+1$.
For which $(a,b,c)$ there does not exist $(d_a,d_b,d_c)\in S$ such, that $(a+d_a,b+d_b,c+d_c)\in \mathbb{Z}_+^3$?
We have $$\begin{cases}c<1\\b<5\\a<8\\
\left[\begin{array}{l}a<2\\b<3\end{array}\right.\\
\left[\begin{array}{l}a<5\\b<1\end{array}\right.
\end{cases}$$
and it can be shown with exhaustive search that maximum of $6a+9b+20c$ we can get with that conditions, is $42=7\cdot 6=6+4\cdot 9$, i.e. for every $(a,b,c)\in \mathbb{Z}_+^3$ if $6a+9b+20c>42$ we have such $(d_a,d_b,d_c)\in S$, that $(a+d_a,b+d_b,c+d_c)\in \mathbb{Z}_+^3$.
So starting from $44=4\cdot 6+20$ we can always add $1$ to a number we have.
$43$ is unreachable can be proven with exhaustive search for $c\le\frac{43}{20},b\le\frac{43}{9},a\le\frac{43}{6}$ that is $3\cdot 5\cdot 8=120$ cases.
Python behind the scene
| {
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Does $\sum_{n=1}^\infty \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}$ have a closed form expression? It's clear that the series converges by comparison with $\sum_{n=1}^\infty \frac{1}{2^{n+1}}$, but beyond that I am not sure what can be said.
| The first step I would suggest is to split the sum into two:
\begin{equation}
\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} = \frac{2^n-(2^{n-1}+1)}{2^{n+1}(2^{n-1}+1)} = \frac{1}{2}\frac{1}{2^{n-1}+1}-\frac{1}{2^{n+1}}.
\end{equation}
Thus the initial sum is simply equal to
\begin{equation}
S = \frac{1}{2}\underbrace{\sum\limits_{n=0}^{\infty}\frac{1}{2^n+1}}_{S_1}-
\underbrace{\sum\limits_{n=1}^{\infty} \frac{1}{2^{n+1}}}_{S_2}.
\end{equation}
Obviously, $S_2 = \frac{1}{2}$. For the first one, WolframAlpha gives
\begin{equation}
S_1 = \sum\limits_{n=0}^{\infty}\frac{1}{2^n+1} = -1 +\frac{1}{\log 2}\psi_{1/2}^{(0)}\left(\frac{-i\pi}{\log(2)}\right).
\end{equation}
Therefore,
\begin{equation}
S = \frac{1}{2\log 2} \psi_{1/2}^{(0)}\left(\frac{-i\pi}{\log(2)}\right)-1.
\end{equation}
| {
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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.
$f^{(k+1)}(x) = f^{(k)'}(x)$
| If$$f^{(k)}(x)=\frac{(-1)^k}{2^k}\prod_{j=1}^{k-1}(2j-1)\frac1{\sqrt{x^{2k-1}}}=\frac{(-1)^k}{2^k}\prod_{j=1}^{k-1}(2j-1)x^{1/2-k},$$then\begin{align}f^{(k+1)}(x)&=\frac{(-1)^k}{2^k}\left(\prod_{j=1}^{k-1}(2j-1)\right)\left(\frac12-k\right)x^{1/2-k-1}\\&=-\frac12\frac{(-1)^k}{2^k}\left(\prod_{j=1}^{k-1}(2j-1)\right)\left(2k-1\right)x^{1/2-(k+1)}\\&=\frac{(-1)^{k+1}}{2^{k+1}}\prod_{j=1}^k(2j-1)\frac1{\sqrt{x^{2(k+1)-1}}}.\end{align}
| {
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Find the equation of the two tangent planes to the sphere $x^2+y^2+z^2-2y-6z+5=0$ which are parallel to the plane $2x+2y-z=0$ Find the equation of the two tangent planes to the sphere $x^2+y^2+z^2-2y-6z+5=0$ which are parallel to the plane $2x+2y-z=0$
My Attempt
We need to find a point which is shortest distance from the plane to the sphere. Let it be $(a,b,c).$ Then Equation of the tangent plane at $(a,b,c)$ is given by the formula
$$2a(x-a)+2(b-1)(y-b)+2(c-3)(z-c)=0$$
Finding the diametrically opposite point of $(a,b,c)$. I can find the equation of another tangent parallel to the plane.
I don't know how to find the shortest distance from the plane to the sphere.
Is there any short method to find the tangents?
| Instead complete the square
$$x^2+(y-1)^2+(z-3)^2 = 5$$
then take the gradient
$$\langle x, y-1, z-3 \rangle = \lambda\langle 2, 2, -1\rangle \implies x = y-1 = \frac{3-z}{2}$$
which means
$$x^2 + x^2 + 4x^2 = 6x^2 = 5 \implies x = \pm \sqrt{\frac{5}{6}}$$
This gives you your two points once you plug in for $y$ and $z$.
| {
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What is the Solution to this sum $\sum \limits_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})(\frac{1}{2})^n$ what is the value of this series $\sum \limits_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})(\frac{1}{2})^n$ ?
Anything what's solid and that i got so far is only
$\sum \limits_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})(\frac{1}{2})^n$ = $\sum \limits_{n=1}^{\infty}(1-i^{n^2+n})(\frac{1}{2})^n$
and that the right part reminds me of the geometric series
| Note that$$(1-(-1)^{\frac{n(n+1)}{2}})=\begin{cases}0 & \text{ if } n \equiv 0,3 \pmod{4} \\ 2 & \text{ if } n \equiv 1,2 \pmod{4}\end{cases}$$
Thus
\begin{align*}
\sum_{n=1}^{\infty}(1-(-1)^{\frac{n(n+1)}{2}})\left(\frac{1}{2}\right)^n&=\sum_{\substack{n=1 \\{\small n \equiv 0 \pmod{4}}}}^{\infty}0+\sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}+\sum_{\substack{n=1 \\{\small n \equiv 2 \pmod{4}}}}^{\infty}+\sum_{\substack{n=1 \\{\small n \equiv 3 \pmod{4}}}}^{\infty}0\\
&= \sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}2 \left(\frac{1}{2}\right)^n+\sum_{\substack{n=1 \\{\small n \equiv 2 \pmod{4}}}}^{\infty}2 \left(\frac{1}{2}\right)^n\\
&= 2\left(\sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}\frac{1}{2^n}+\sum_{\substack{n=1 \\{\small n \equiv 2 \pmod{4}}}}^{\infty} \frac{1}{2^n}\right)\\
\end{align*}
Now we just have to focus on these geometric series. For example,
\begin{align*}
\sum_{\substack{n=1 \\{\small n \equiv 1 \pmod{4}}}}^{\infty}\frac{1}{2^n}&=\frac{1}{2}+\frac{1}{2^5}+\frac{1}{2^9}+\dotsb\\
&=\frac{\frac{1}{2}}{1-\frac{1}{2^4}}\\
&=\frac{8}{15}.
\end{align*}
Hopefully you can complete now.
| {
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Show that $\operatorname{sinc}(x-y) = \sum_{n \in \mathbb{Z}}\operatorname{sinc}(x+n)\operatorname{sinc}(y+n)$ My problem is:
Show that for every $x,y \in \mathbb{R}$
$$\operatorname{sinc}(x-y) = \sum_{n \in \mathbb{Z}}\operatorname{sinc}(x+n)\operatorname{sinc}(y+n)$$
Here is what I did so far:
Considering the function $g(t) = f(t + t_{0})$ I can show that the Fourier Transform is given by $\hat{g}(w) = e^{-iwt_{0}} \hat{f}(w)$.
Therefore considering $f(t) = sinc(t)$ I can show that
$$\hat{f}(w) = \begin{cases}
1, & |w| \leq \pi \\
0, & |w| \gt \pi
\end{cases}
$$
So I can use Shannon's Theorem, where $\Delta t = 1$ and write
$$ f(t) = g(t-t_{0}) = \sum_{n \in \mathbb{Z}} \operatorname{sinc}(t-t_{0}-n)sinc(t_{0} + n) $$
Now thinking $t$ as $x$ and having an $t_{0}$ as $y$ for every $x$ I can write
$$ sinc(x-y) = \sum_{n \in \mathbb{Z}} \operatorname{sinc}(x-y-n)\operatorname{sinc}(y + n) $$
Which is close, but not the same to what I should prove.
Can anybody point where I have made any mistake or help me with the final steps?
Thanks in advance.
| For $n\in \mathbb{Z}$, we have
\begin{align}
\sin \pi (x+n) \cdot \sin \pi (y+n)
&= \frac{1}{2}[\cos \pi(x-y) - \cos \pi(x+y+2n)]\\
&= \frac{1}{2}[\cos \pi(x-y) - \cos \pi(x+y)]\\
&= \sin \pi x \cdot \sin \pi y.
\end{align}
Also, we have
\begin{align}
\sum_{n\in \mathbb{Z}} \frac{1}{(x+n)(y+n)}
&= \frac{1}{xy} + \sum_{n=1}^\infty \left(\frac{1}{(x+n)(y+n)} + \frac{1}{(x-n)(y-n)}\right)\\
&= \frac{1}{xy} + \frac{1}{x-y}\sum_{n=1}^\infty \left(\frac{2y}{y^2-n^2} - \frac{2x}{x^2-n^2}\right)\\
&= \frac{1}{xy} + \frac{1}{x-y}\sum_{n=1}^\infty \frac{2y}{y^2-n^2} - \frac{1}{x-y}\sum_{n=1}^\infty \frac{2x}{x^2-n^2}\\
&= \frac{1}{xy} + \frac{1}{x-y}\left(\pi \cot \pi y - \frac{1}{y} \right)
- \frac{1}{x-y}\left(\pi \cot \pi x - \frac{1}{x} \right)\\
&= \frac{\pi \cot \pi y - \pi \cot \pi x}{x-y}
\end{align}
where we have used the following identity
$$\pi \cot \pi z = \frac{1}{z} + 2z \sum_{n=1}^\infty \frac{1}{z^2-n^2}.$$
See: https://mathworld.wolfram.com/Cotangent.html, or https://people.reed.edu/~jerry/311/cotan.pdf, or Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$
Thus, we have
\begin{align}
\sum_{n\in \mathbb{Z}} \frac{\sin \pi(x+n) \cdot \sin \pi(y+n)}{\pi^2(x+n)(y+n)}
&= \frac{1}{\pi^2} \sin \pi x \cdot \sin \pi y \cdot \sum_{n\in \mathbb{Z}} \frac{1}{(x+n)(y+n)}\\
&= \frac{1}{\pi^2} \sin \pi x \cdot \sin \pi y \cdot \frac{\pi \cot \pi y - \pi \cot \pi x}{x-y}\\
&= \frac{\sin \pi x \cdot \cos \pi y - \cos \pi x \cdot \sin \pi y}{\pi (x-y)}\\
&= \frac{\sin \pi (x-y)}{\pi (x-y)}.
\end{align}
(Q. E. D.)
| {
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Flux of $(x,y,z^2)$ through $\{(x,y,z): 2(x^2+y^2)Let $$D=\{(x,y,z): 2(x^2+y^2)<z<3\}$$
and the vector field $$F(x,y,z)=(x,y,z^2)$$
I wanto to compute the flux of $F$ through $\partial D$
Applying the divergence theorem, I should compute
$$\int_D2+2z \ dxdydz$$
To evaluate the integral I pass to cylindrical coordinates but I have some problems at this point. We have $$2r^2<z<3,$$ $$0<r<\sqrt{3/2}$$ so then it becomes
$$\int_D2z \ dxdydz=2\pi\int_0^{\sqrt{3/2}}r \ dr\int_{2r^2}^3 \ z \ dz=2\pi(\sqrt{3/2}9/2-\int_0^{\sqrt{3/2}}2r^5 \ dr)=2\pi(\sqrt{3/2}9/2-\sqrt{3/2}^6/3)$$$$+$$$$\int_D2dz=2\pi(\int_0^{\sqrt{3/2}}6r-4r^2\ dr)$$
But this does not look good at all. So maybe I'm doing something
wrong?
If I try computing the flux directly I first have to compute the boundary, which I am not sure about
$$\partial D=\{(x,y,z):x^2+y^2=\frac{z}{2}, z<3\} \cup \{(x,y,z):x^2+y^2\leq \frac{3}{2}, \ z=0\}=B_1 \cup B_2$$
I can parametrize $B_1$ as $$\phi:(u,v)\mapsto (u,v,2(u^2+v^2))$$ so that $$\phi_u \land \phi_v= (-4u,-4v,1)$$ and gives, for $B_1,$ the integral
$$\int_{2u^2+2v^2<3} \langle F(\phi(u,v)),\phi_u \land \phi_v \rangle dudv= \int_{2u^2+2v^2<3}-4(u^3+v^3)+4(u^2+v^2)^2 \ dudv$$ which is very ugly.
| By using the divergence theorem, we have to evaluate
$$\begin{align}2\pi\int_{r=0}^{\sqrt{3/2}}\int_{z=2r^2}^3(2+2z)r\,dz dr&=2\pi\int_{r=0}^{\sqrt{3/2}}(15r-4r^3-4r^5) dr\\&=
\left[\frac{15r^2}{2}-r^4-\frac{2r^6}{3}\right]_{r=0}^{\sqrt{3/2}}=\frac{27\pi}{2}.
\end{align}$$
As regards the direct computation, the outward flux through $B_1$ is
$$\begin{align}\int_{2u^2+2v^2<3}4(u^2+v^2)-4(u^2+v^2)^2 \ dudv&=2\pi\int_{0}^{\sqrt{3/2}}(4r^2-4r^4)r\,dr\\
&=\left[r^4-\frac{2r^6}{3}\right]_{r=0}^{\sqrt{3/2}}=0
\end{align}$$
whereas the outward flux through $B_2$ is
$$\int_{z=3,r\leq \sqrt{3/2}}z^2 r\,dr=9\cdot \frac{3\pi}{2}=\frac{27\pi}{2}.$$
Their sum is again $\frac{27\pi}{2}$.
| {
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probability with two variables - Taking balls out of the basket There's 3 balls in the basket - White, red and black. Three people chose one ball after the other with return.
$X$ is the various colors that got chosen.
$Y$ is the number of people that chose white.
I need to calculate $P(X+Y≤3|X-Y≥1)$.
So I got that I need to start from that getting each ball is probability of $1/3$, since it's with return.
I'm getting confused with how Can I compute Y? Can I say it's sort of uniform distribution and maybe to try by that? Meaning, that the expected value is $(a-b)/2$?
| We can make a probability table of joint outcomes of $(X, Y)$. Note $X \in \{1, 2, 3\}$, and $Y \in \{0, 1, 2, 3\}$. For $X = 1$, there are only the three outcomes $$(w,w,w), (r,r,r), (b,b,b).$$ For $X = 2$, we have $\binom{3}{2}(2^3 - 2) = 18$ outcomes:
$$(r, r, w), (r, r, b), (r, w, r), (r, w, w), (r, b, r), (r, b, b), \\
(w, r, r), (w, r, w), (w, w, r), (w, w, b), (w, b, w), (w, b, b), \\
(b, r, r), (b, r, b), (b, w, w), (b, w, b), (b, b, r), (b, b, w).$$
For $X = 3$, we have $3! = 6$ outcomes, which are the permutations of $r, b, w$ in some order. The total is $3^3 = 27$.
When $X = 1$, we have either $Y = 3$ with probability $1/3$ or $Y = 0$ with probability $2/3$.
When $X = 2$, we have $Y = 0$, $Y = 1$, $Y = 2$ each with probability $1/3$.
When $X = 3$, we have $Y = 1$ with probability $1$.
So we have
$$\begin{array}{c|cccc}
\Pr[X = x, Y = y] & 0 & 1 & 2 & 3 \\
\hline
1 & \frac{2}{27} & 0 & 0 & \frac{1}{27} \\
2 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 \\
3 & 0 & \frac{2}{9} & 0 & 0
\end{array}$$
The rest is simply conditioning. Select those outcomes for which $X - Y \ge 1$, and among those, tabulate the probabilities for which $X + Y \le 3$. then divide by the sum of the probabilities that you considered.
| {
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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?.
| Note that $\sqrt{g^2(x)}=|g(x)|$
$$L=\lim_{x\to 0} \frac{\sin^2 ax}{x\sqrt{1-\cos ax}}= \lim_{x\to 0} \frac{\sin^2 ax}{x \sqrt{2}~|\sin (ax/2)|} =\lim_{x\to 0} \frac{a^2x^2}{x|ax/2|\sqrt{2}}$$
So $$RL=\lim_{x \to 0^+} f(x)=a\sqrt{2}$$
and$$LL=\lim_{x \to 0^-} f(x)=-a \sqrt{2}$$
| {
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Percentage value higher than 100% I have two values A= 3.8620E+00 B = 1.4396E+00
According to this post, to calculate how much A is higher than B in percentage we do this:
((A-B)/B)*100 = ((3.8620E+00 - 1.4396E+00)/1.4396E+00)*100 = 168.2690%
Does this mean the value A is 168.2690% higher than B?
Does this calculation is correct?
When the percentage is higher than 100%, what does that exactly mean?
| Lets do an easier example. Suppose $B = 100$ and suppose $A = 104$. Then $A$ is $4$ more than $B$. $4$ is $4\%$ of $100$. so $A$ is $4\%$ of $B$ higher than $B$. ANother way of saying this is $A$ is $104\%$ of $B$ which is $4\%$ higher than $100\%$ of $B$, which would be $B$.
And if $C = 160$ then $C$ is $60$ more than $B$. And $60$ is $60\%$ of $B$. So $C$ is $60\%$ of $B$ higher than $B$. And so $C$ is $160\%$ of $B$ which is $60\%$ more than $100\%$ of $B$.
And if $D = 200$ then $D$ is $100$ more than $B$. And $100$ is $100\%$ of $B$ so $D$ is $100\%$ higher than $B$. And $D$ is $200\%$ of $B$ which is $100\%$ more than $100\%$.
And if $E = 280$ then $E$ is $180$ more than $B$. And $180$ is $180\%$ of $B$ so $E$ is $180\%$ higher than $B$. And $E$ is $280\%$ of $B$ which is $180\%$ more than $100\%$.
| {
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"url": "https://math.stackexchange.com/questions/3731001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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An interesting property of nested radicals I found a beautiful form of the sum of square roots.
$a,b,c\in\mathbb{R}$
\begin{eqnarray*}
A &:=& abc\\
B &:=& a+b+c\\
C &:=& ab+bc+ca
\end{eqnarray*}
$$\sqrt{a}+\sqrt{b}+\sqrt{c} = \sqrt{A^0B+2\sqrt{A^0C+2\sqrt{A^1B+2\sqrt{A^2C+2\sqrt{A^5B+2\sqrt{A^{10}C}}}}}}\ \ \cdots\cdots$$
This property is based on the next relation.
$$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 = a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
$$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2 = ab+bc+ca+2(\sqrt{ab^2c}+\sqrt{abc^2}+\sqrt{a^2bc})$$
$$(\sqrt{ab^2c}+\sqrt{abc^2}+\sqrt{a^2bc})^2 = ab^2c+abc^2+a^2bc+2(\sqrt{a^2b^3c^3}+\sqrt{a^3b^2c^3}+\sqrt{a^3b^3c^2})$$
$$\vdots$$
In this way, three sums of square roots generate three sums of square roots. Therefore, it's a recursive structure.
$$$$
The same property can be confirmed for the cube roots.
$a,b\in\mathbb{R}$
\begin{eqnarray*}
A &:=& ab\\
B &:=& a+b
\end{eqnarray*}
$$\sqrt[3]{a}+\sqrt[3]{b} = \sqrt[3]{A^0B+3\sqrt[3]{A^1B+3\sqrt[3]{A^4B+3\sqrt[3]{A^{13}B+3\sqrt[3]{A^{40}B+3\sqrt[3]{A^{121}B}}}}}}\ \ \cdots\cdots$$
Please let me know if you have any interesting information related to these.
| I don't know why, but it seems that the exponent of A in the case of the cube root is in this sequenceA003462.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Sum of distances of a variable point from two fixed points. Given that $A\equiv(4,2)$ and $B\equiv(2,4)$, find a point $P$ on the line $3x+2y+10=0$ such that $PA+PB$ is minimum.
My attempt:
In $\triangle PAB$,
$$PA+PB\ge AB$$ Hence, the minimum value should be $AB=2\sqrt 2$. But in the solution point $P$ is given as $\displaystyle(-\frac{14}{5}, -\frac{4}{5})$ from which $PA+PB$ comes out to be $10\sqrt 2$.
Where am I going wrong? Help is requested!
| If you draw the graph, you will understand Michael's argument. To find the coordinate of $A^\prime$ quickly, you can use the formula for the distance from a point $(a,b)$ to line $Ax+By+C=0$ as follows:
$$d=\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}.$$
Suppose the coordinate of $A^\prime$ is $(a, b)$, then
$$\frac{-(3a+2b+10)}{\sqrt{3^2+2^2}}=\frac{3(4)+2(2)+10}{\sqrt{3^2+2^2}}.$$
The reason of adding a negative sign to the left-hand side is because $A^\prime$ lies under the line $3x+2y+10=0$.
To solve $a$ and $b$, you need to add another equation from the fact that $AA^\prime$ is perpendicular to the line $3x+2y+10=0$.
$$\frac{b-2}{a-4}\cdot \frac{-3}{2}=-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
find all $n$ such that $\varphi(\sigma(2^n)) = 2^n$ Problem: Find all positive integers $n$ such that $\varphi(\sigma(2^n)) = 2^n$, where $\varphi(n)$ is Euler's totient function and $\sigma(n)$ is the sum of all divisors of $n$.
I know that $\sigma(2^n) = 1+2+2^2+2^3+\dots+2^n = 2^{n+1}-1$, so we only need to find all $n$ such that $\varphi(2^{n+1}-1) = 2^n$. Trying out a few $n$, we find that $n=1,3,7,15,31$ work. Not sure how to prove this though. Any answers?
| Taking $n=15$ for example, we have $\sigma(2^{15})=2^{16}-1=(2+1)(2^2+1)(2^4+1)(2^8+1)=3\cdot 5 \cdot 17 \cdot 257$ with all the factors prime. We know that for $p$ prime $\varphi(p)=p-1$, so $\varphi(2^{16}-1)=2\cdot 2^2\cdot 2^4\cdot 2^8=2^{15}$. You should show that this factorization works in all the cases you cite. This works again for $n=31$, but not for $n=63$ because $2^{32}+1=4294967297 = 641×6700417$ is not prime. $n=63$ will then not be a solution. If a number has any prime factors not of the form $2^k+1$ the multiplicative nature of $\varphi$ will ensure that it will have some odd factors, so these are all there are.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the following limit: $\lim\limits_{x\to\infty}(12x^2-2-6x\sqrt{3x^2-2})$ How can I evaluate following limit $$\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)$$
I got stuck. I also can't see application of L'Hospital rule here. Can someone please help me solve this limit? Thanks
| You can say:
$$\lim_{x \to \infty}(12x^2-2−6x\sqrt{3x^2−2})=\lim_{x \to \infty}\bigl(12x^2-2−6x\sqrt{x^2(3-{2\over x^2})} \ \bigr)=\lim_{x \to \infty}\bigl(12x^2-2−6x^2\sqrt{3-{2\over x^2}} \ \bigr)=\lim_{x \to \infty}\bigl(-2+x^2(−6\sqrt{3-{2\over x^2}}+12) \ \bigr)$$
Because $\ \lim_{x \to \infty}\sqrt{3-{2\over x^2}}=\sqrt{3}<2 \ $ so $\ \lim_{x \to \infty}−6\sqrt{3-{2\over x^2}}+12 = 12-6\sqrt{3}>0$.
Then we know that $$\lim_{x \to \infty}\bigl(-2+x^2(−6\sqrt{3-{2\over x^2}}+12) \ \bigr) = \lim_{x \to \infty}(-2+x^2(12-6\sqrt{3}))=-2+\infty = \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
} |
Approximating $f(n,k,c) = \prod_{i=0}^{k-1} \frac{n-ic}{n-i}.$ Problem:
For integers $n>k$ and real $c\in [0,1]$, let
$$f(n,k,c) = \prod_{i=0}^{k-1} \frac{n-ic}{n-i}.$$
How well can we approximate $f(n,k,c)$ for fixed $c$ and sufficiently large $n,k$? I'm particularly interested when $c=1/2$.
Motivation:
If we have $k$ i.i.d. random variables, $X_i$, which each uniformly the integers in $[1,n]$, then by a basic counting argument, we get that the probability they are all distinct is $n!/(n-k)! n^{-k}$.
Another way to see this is to note that, given $X_1, \dots X_i$ are all distinct from one another, the probability $X_{i+1}$ is also distinct from these first $i$ elements is $(n-i)/n$. Thus the probability they are exactly distinct is:
$$ \prod_{i=0}^{k-1} \frac{n-i}n = n!/(n-k)! n^{-k}.$$
I had similar problem where instead I want to calculate:
$$ \prod_{i=0}^{k-1} \frac{n-ic}n $$ for values $c \approx 1/2$.
| It is easy to prove that $\ln \frac{1 - cx}{1-x} \ge (1-c)x$ for $0\le x < 1$. Thus, we have
\begin{align}
f(n,k,c) &= \mathrm{exp}\left(\sum_{i=0}^{k-1} \ln \frac{1 - c\frac{i}{n}}{1-\frac{i}{n}} \right)\\
&\ge \mathrm{exp}\left(\sum_{i=0}^{k-1} (1-c)\frac{i}{n} \right)\\
&= \mathrm{exp}\left( \frac{k(k-1)(1-c)}{2n}\right).
\end{align}
By using $\ln \frac{1 - cx}{1-x} \ge (1-c)x + \frac{1-c^2}{2}x^2$ for $0\le x < 1$, we can get a better lower bound. Omitted.
Update: When $\frac{k}{n}$ is near $1$, the previous lower bounds are not good. Here I give another lower bound.
We have
\begin{align}
&f(n, k, c)\\
=\ & \frac{n^k (n-k)!}{n!}\prod_{i=0}^{k-1} \left(1 - c\frac{i}{n}\right)\\
=\ & \frac{n^k (n-k)!}{n!}\mathrm{exp}\left(\sum_{i=0}^{k-1} \ln \left(1 - c\frac{i}{n}\right)\right)\\
\ge\ & \sqrt{2\pi} \left(1-\frac{k}{n}\right)^{n-k+\frac{1}{2}}\mathrm{e}^{k-1}
\mathrm{exp}\left(\sum_{i=0}^{k-1} \left( -c\frac{i}{n} + (\ln (1-c) + c)\frac{i^2}{n^2} \right)\right)\\
=\ & \sqrt{2\pi} \left(1-\frac{k}{n}\right)^{n-k+\frac{1}{2}}\mathrm{e}^{k-1}
\mathrm{exp}\left(\frac{k(k-1)[(2k-1)\ln(1-c) + (2k-3n-1)c]}{6n^2}\right)
\end{align}
where we have used: i) $\ln (1-cx) \ge -cx + (\ln (1-c) + c)x^2$ for $0\le x \le 1$;
ii) Stirling's formula $\sqrt{2\pi} m^{m+\frac{1}{2}}\mathrm{e}^{-m} \le m! \le \mathrm{e} m^{m+\frac{1}{2}}\mathrm{e}^{-m}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\binom{99}{0}+ \binom{100}{2}+ \binom{99}{3}+ \binom{100}{5}+ \binom{99}{6}+ \binom{100}{8}+ ..+ \binom{100}{98}+\binom{99}{99}$ Hello everyone how can I calculate:
$\binom{99}{0}+ \binom{100}{2}+ \binom{99}{3}+ \binom{100}{5}+ \binom{99}{6}+ \binom{100}{8}+ \cdots+ \binom{100}{98}+\binom{99}{99}$ ?
I tried to mark $\omega = \frac{1}{2}-\frac{\sqrt{3}i}{2}$ and calculate $\binom{99}{0}+ \binom{99}{3}+\binom{99}{6}+ \cdots+\binom{99}{99}$ by $(1+1)^{99} +(1+\omega)^{99} + (1+\overline\omega)^{99}
=\frac{2^{99} +1}{3}$
But I don't know how to continue.
| Hint
$$\binom{100}{k}=\binom{99}{k}+\binom{99}{k-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| Simplify before applying the rule...
$$
\lim_{x\to 0} \frac{\tan x- \sin x}{x^3} = \lim_{x\to 0}\frac{\sin x - \sin x \cos x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0}\frac{1-\cos x}{x^2} = \lim_{x\to 0}\frac{1-\cos x}{x^2}
$$
and then apply the rule
$$
\lim_{x\to 0}\frac{1-\cos x}{x^2}= \lim_{x\to 0}\frac{\sin x}{2x} = \frac 12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Find all $x > 0$ s/t $\sqrt{x} + 1/\sqrt{x}$ and $x^{1/3} + 1/x^{1/3}$ are integers. Why doesn't this get all solutions? This is a math puzzle I've been working on, and I'm not sure why my approach isn't yielding all the solutions for $x$. As in the title, the question is to find all positive reals $x$ where $\sqrt{x} + 1/\sqrt{x}$ and $x^{1/3} + 1/x^{1/3}$ are both integers. Here's my approach--I'm not sure where the error is. Where does the following approach go wrong?
Suppose that $\sqrt{x} + 1/\sqrt{x} = m$ and $x^{1/3} + 1/x^{1/3} = n$, where $m,n \in \mathbb{Z}$. We let $m, n > 0$ since $\sqrt{x} + 1/\sqrt{x} > 0$ and $x^{1/3} + 1/x^{1/3} > 0$ for $x > 0$.
We equivalently have $x - m\sqrt{x} + 1 = 0$ and $x^{2/3} -nx^{1/3} + 1 = 0$. Let $u = \sqrt{x}$, so that we have $u^2 - mu + 1 = 0$ and $u^{4/3} - nu^{2/3} + 1 = 0$. It suffices to find all $u$ satisfying both of these equations.
Applying the quadratic formula, $u = \frac{m \pm \sqrt{m^2 - 4}}{2}$ and $u^{2/3} = \frac{n \pm \sqrt{n^2-4}}{2}$. Squaring the first equation gets $u^2 = (\frac{m \pm \sqrt{m^2 - 4}}{2})^2$.Cubing the second equation gets $u^2 = (\frac{n \pm \sqrt{n^2-4}}{2})^3$.
So there are 4 sign combinations setting these two expressions for $u^2$ equal. Since $m,n$ are integers, I look for integer solutions only, finding only $m = 2, n = 2$ as the viable solution in all 4 cases. I checked this part with WolframAlpha--the general idea was to isolate radicals to one side. If we have a nonzero sum of surds, it ends up being irrational while the other side is rational. That leaves integer solutions to the radicals $\sqrt{m^2-4},\sqrt{n^2-4}$, the only positive solutions to which are $m = 2$ and $n = 2$.
This yields $u = 1$ as the only viable solution. However, the solution states that there are infinitely many such $x$.
| A good way to simplify algebra is to start by getting rid of the radicals. Let $y=\sqrt[6]x$. Then $\sqrt{x}=y^3$ and $\sqrt[3]x=y^2$ and we have to find such $y$ where
$$
y^3+1/y^3 = m
$$
$$
y^2+1/y^2 = n
$$
Further, let $a=y+1/y$. Then $$n = a^2 -2$$ and $$m = a^3-3a = a(a^2-3) = a(n-1)$$
Therefore, $a$ must also be an integer ($a=\frac{m}{n-1}$ is rational and its square is an integer $a^2=n+2$)
Solving $a=y+1/y$ for $y$, we get
$$y=\frac{a\pm\sqrt{a^2-4}}{2}$$ and
$$x=y^6 = \left({\frac{a\pm\sqrt{a^2-4}}{2}}\right)^6$$ for any integer $a\ge2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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How can i evaluate $\int _0^{\infty }\frac{\ln \left(x\right)\sin \left(x\right)}{x^2+1}\:dx\:$ using real methods I've been trying to evaluate this integral for a while now my friend used complex analysis to evaluate this but he got a wrong result, i tried using real methods but i've been stuck. One can probably use differentiating under the integral sign in the following ways,
$$I\left(a\right)=\int _0^{\infty }\frac{\ln \left(x\right)\sin \left(ax\right)}{x^2+1}\:dx$$or$$I\left(a\right)=\int _0^{\infty \:}\frac{x^a\sin \left(x\right)}{x^2+1}\:dx$$
But that seems really complicated, i really have no idea how to proceed, please help me.
| Here's a solid start - let me know if you need for me to expand upon the material provided:
Yes employ the second, but with a slight extension
$$
I(a, t) = \int_0^\infty \frac{x^a \sin(xt)}{x^2 + 1}\:dx
$$
Then if the integral under inspection is $J$, then by the Dominated Convergence Theorem and Leibniz's Integral Rule, we find that
$$
J = \frac{\partial I}{\partial a}\bigg|_{(a,t) =(0, 1)}
$$
Thus, we need to resolve $I(a,t)$. To do so, we employ Fubini's Theorem and take the Laplace Transform with respect to $t$:
\begin{align}
\mathscr{L}\left[I(a,t) \right] &= \mathscr{L}\left[\int_0^\infty \frac{x^a \sin(xt)}{x^2 + 1}\:dx \right] = \int_0^\infty \frac{x^a \mathscr{L}\left[\sin(xt)\right]}{x^2 + 1}\:dx = \int_0^\infty x^a \cdot \frac{x}{s^2 + x^2} \cdot \frac{1}{x^2 + 1}\:dx \\
&= \int_0^\infty \frac{x^{a + 1}}{\left(s^2 + x^2\right)\left(x^2 + 1\right)}\:dx = \int_0^\infty x^{a + 1}\left[\frac{1}{s^2 - 1}\left(\frac{1}{x^2 + 1}- \frac{1}{s^2 + x^2} \right)\right]\:dx \\
&= \frac{1}{s^2 - 1}\left[\int_0^\infty \frac{x^{a + 1}}{x^2 + 1} \:dx - \int_0^\infty \frac{x^{a + 1}}{s^2 + x^2}\:dx \right] = \frac{1}{s^2 - 1}\left[I_1 - I_2\right]
\end{align}
You will observe that both $I_1$ and $I_2$ take the form:
$$
H(b,k,n) = \int_0^\infty \frac{x^k}{x^n + b}\:dx = \frac{1}{n} b^{1 - \frac{k + 1}{n}} \Gamma\left(1 - \frac{k + 1}{n} \right)\Gamma\left( \frac{k + 1}{n} \right)
$$
Where $\Gamma(x)$ is the Gamma Function.
Thus we observe that:
\begin{align}
\mathscr{L}\left[I(a,t) \right] &= \frac{1}{s^2 - 1}\bigg[H\left(1, a+1, 2\right) - H\left(s^2, a+1, 2\right)\bigg] \\
&= \frac{1}{s^2 - 1}\bigg[\frac{1}{2} \cdot 1^{\frac{a + 1 + 1}{2} - 1}\Gamma\left(1 - \frac{a + 1 + 1}{2} \right)\Gamma\left( \frac{a + 1 + 1}{2} \right) - \frac{1}{2} \cdot \left(s^2\right)^{\frac{a + 1 + 1}{2} - 1}\Gamma\left(1 - \frac{a + 1 + 1}{2} \right)\Gamma\left( \frac{a + 1 + 1}{2} \right) \bigg] \\
&= \frac{1}{2\left(s^2 - 1\right)}\Gamma\left(1 - \frac{a + 2}{2} \right)\Gamma\left( \frac{a + 2}{2} \right)\bigg[1 - s^{a} \bigg]
\end{align}
Here as $a$ is to be evaluated at $0$, we may employ Euler's Reflection Formula on the Gamma terms to yield:
\begin{align}
\mathscr{L}\left[I(a,t) \right] &=\frac{1}{2\left(s^2 - 1\right)}\pi\operatorname{cosec}\left(\pi \cdot \frac{a + 2}{2}\right)\bigg[1 - s^{a} \bigg]\\
&=\frac{\pi}{2\left(s^2 - 1\right)}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\bigg[1 - s^{a} \bigg]
\end{align}
We now take the Inverse Laplace Transform:
\begin{align}
I(a,t) &= \mathscr{L}^{-1}\left[ \frac{\pi}{2\left(s^2 - 1\right)}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\bigg[1 - s^{a} \bigg]\right] \\
&= \frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\left[\mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} \right] - \mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} s^{a}\right] \right] \\
&= \frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\left[\sinh(t) - \mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} s^{a}\right] \right]
\end{align}
For the remaining inversion we employ Convolution:
$$
\mathscr{L}^{-1}\left[ F(s)G(s) \right] = \int_0^t f(\tau)g(t - \tau) d\tau
$$
Here let
$$
G(s) = \frac{1}{s^2 - 1} \longrightarrow g(t) = \sinh(t)
$$
And so
$$
F(s) = s^{a} \longrightarrow f(t) = \frac{t^{-(a + 1)}}{\Gamma(-a)}
$$
And so,
\begin{align}
&\mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} s^{a}\right] = \int_0^t \frac{\tau^{-(a + 1)}}{\Gamma(-a)} \sinh(t - \tau)\:d\tau = \frac{1}{\Gamma(-a)} \int_0^t \tau^{-(a + 1)}\sinh(t - \tau)\:d\tau \\
&= \frac{1}{\Gamma(-a)} \int_0^t \tau^{-(a + 1)}\bigg[\sinh(t)\cosh(\tau) - \cosh(t)\sinh(\tau) \bigg]\:d\tau \\
&=\frac{1}{\Gamma(-a)} \left[ \sinh(t)\int_0^t \tau^{-(a + 1)}\cosh(\tau)\:d\tau - \cosh(t)\int_0^t \tau^{-(a + 1)}\sinh(\tau)\:d\tau \right]
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Using sum-to-product formula to solve $\sin(2\theta)+\sin(4\theta)=0$ Trying to use the sum-to-product formula to solve $\sin(2\theta)+\sin(4\theta)=0$ over the interval $[0,2\pi)$, but I'm missing solutions.
$$\sin(2\theta)+\sin(4\theta)=0$$
Apply sum-to-product formula:
$$2\sin\left(\frac{2\theta+4\theta}{2}\right)\cos\left(\frac{2\theta-4\theta}{2}\right)=0$$
$$2\sin(3\theta)\cos(-\theta)=0$$
By odd-even identities: $\cos(-\theta)=\cos(\theta)$
$$2\sin(3\theta)\cos(\theta)=0$$
$$\sin(3\theta)\cos(\theta)=0$$
By the zero-product property
$\sin(3\theta)=0$ or $\cos(\theta)=0$
Then solving for theta gives: $\theta=0, \frac{\pi}{2}, \frac{3\pi}{2}, \pi$.
However, there are missing solutions $\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
A solution online used double angle identities instead:
$$\sin(2\theta)+\sin(4\theta)=0$$
$$\sin(2\theta)+\sin(2*2\theta)=0$$
Apply double angle identity for: $\sin(2*2\theta)$
$$\sin(2\theta)+2\sin(2\theta)\cos(2\theta)=0$$
Factor out $\sin(2\theta)$
$$\sin(2\theta)*[1+2\cos(2\theta)]=0$$
Apply double angle identities:
$\cos(2\theta)= 1-2\sin^2(\theta)$
$\sin(2\theta)= 2\sin(\theta)\cos(\theta)$
$$2\sin(\theta)\cos(\theta)*[1+2(1-2\sin^2(\theta))]=0$$
$$2\sin(\theta)\cos(\theta)*[-4\sin^2(\theta)+3]=0$$
By the zero-product property
$2\sin(\theta)\cos(\theta)=0$ or $-4\sin^2(\theta)+3=0$
Which further simplifies to
$\sin(\theta)=0$, $\cos(\theta)=0$, or $-4\sin^2(\theta)+3=0$
Solving for theta now gives all possible solutions over $[0, 2\pi)$.
My questions are:
(1) Can the sum-to-product formula be used to solve this equation?
(2) If so, why were solutions missing when using the sum-to-product formula but not the double angle identities? What was I doing incorrectly?
| This is an excellent way to proceed with this problem, and the reduction to $\sin(3\theta)\cos(\theta)=0$ is great; this implies that $\sin(3\theta)=0$ or $\cos(\theta)=0$.
*
*The solutions to $\cos(\theta)=0$ are $\theta = \dots,\frac\pi2,\frac{3\pi}2,\dots$.
*The solutions to $\sin(\alpha)=0$ are $\alpha = \dots, 0, \pi, 2\pi, \dots$. But we have $\sin(3\theta)=0$, and so the solutions are $3\theta = \dots, 0, \pi, 2\pi, \dots$, which is the same as $\theta=\dots,0,\frac\pi3,\frac{2\pi}3,\pi,\frac{4\pi}3,\frac{5\pi}3,\dots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Evaluate $\int_0^1 \arctan^3 x\,dx$ I don't want to use the Fourier series. My work \begin{align}J&=\int_0^1 \arctan^3 x\,dx\\
&=[x\arctan^3 x]_0^1 -3\int_0^1
\frac{x\arctan^2 x}{1+x^2}\,dx\\
&=\frac{\pi^3 }{64}-3\int_0^1 \frac{x\arctan^2
x}{1+x^2}\,dx\\
&= \frac{\pi^3
}{64}-\frac{3}{2}\left[\ln(1+x^2)\arctan^2
x\right]_0^1 +3\int_0^1 \frac{\ln(1+x^2)\arctan
x}{1+x^2}\,dx\\
&=\frac{\pi^3 }{64}-\frac{3\pi^2\ln
2}{32}+3\int_0^1 \frac{\ln(1+x^2)\arctan
x}{1+x^2}\,dx\\
\end{align}
How to continue?
| I am not sure if this counts as a solution using Fourier series, but let me present my solution anyway: By the substitution $x=\tan\theta$, we get
$$ J = \int_{0}^{\frac{\pi}{4}} \theta^3 \sec^2\theta \, \mathrm{d}\theta. $$
In order to compute this integral, we will utilize the following regularized expansion:
$$ \sec^2\theta = \frac{4e^{2it}}{(1+e^{2it})^2} = \lim_{r \uparrow 1} \frac{4r e^{2it}}{(1+r e^{2it})^2} = 4 \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} n r^n e^{2in\theta} $$
Plugging this back to $J$, we can interchange the order of limit and integration by the uniform convergence, whence we get
\begin{align*}
J
&= 4 \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} n r^n \int_{0}^{\frac{\pi}{4}} \theta^3 e^{2in\theta} \, \mathrm{d}\theta.
\end{align*}
Now by the integration by parts,
$$ \int_{0}^{\frac{\pi}{4}} \theta^3 e^{2in\theta} \, \mathrm{d}\theta
= -\frac{3i^n}{8n^4} + \frac{3}{8n^4} + \frac{3\pi i^{n+1}}{16n^3} + \frac{3\pi^2 i^n}{64n^2} - \frac{\pi^3 i^{n+1}}{128n}. $$
Plugging this and taking limit,
\begin{align*}
J
&= \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} r^n \left( -\frac{3i^n}{2n^3} + \frac{3}{2n^3} + \frac{3\pi i^{n+1}}{4n^2} + \frac{3\pi^2 i^n}{16n} - \frac{\pi^3 i^{n+1}}{32} \right) \\
&= \sum_{n=1}^{\infty} (-1)^{n-1} \left( -\frac{3i^n}{2n^3} + \frac{3}{2n^3} + \frac{3\pi i^{n+1}}{4n^2} + \frac{3\pi^2 i^n}{16n} \right) + \frac{\pi^3}{64}(1-i) \\
&= - \frac{3}{2} \left( \frac{1}{2^3} - \frac{1}{4^3} + \frac{1}{6^3} - \dots \right)
+ \frac{3}{2} \left( \frac{1}{1^3} - \frac{1}{2^3} + \frac{1}{3^3} - \dots \right) \\
&\quad - \frac{3\pi}{4} \left( \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \dots \right)
+ \frac{3\pi^2}{16} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \dots \right)
+ \frac{\pi^3}{64} \\
&\quad + \underbrace{\text{[imaginary term]}}_{=0}.
\end{align*}
Simplifying this, we get
$$ J = -\frac{3\pi G}{4} + \frac{63\zeta(3)}{64} + \frac{\pi^3}{64} + \frac{3 \pi^2 \log 2}{32}, $$
where $G$ is the Catalan's constant.
| {
"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": 0
} |
If $S=1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+...$ what is the closest integer to $e^S$ If $S=1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+...$ what is the closest integer to $e^S$
I thought that this series could be represented as $\ln{2}$ but it is $\ln{4}$ somehow? Any suggestions please send.
| If your sum is really (there are certainly lots of other ways to go on!):
$\begin{align*}
S_n
&= 1 + \frac{1}{2} + \frac{1}{3}
+ \left(\frac{1}{4} - 1\right) + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}
+ \left(\frac{1}{8} - \frac{1}{2}\right) + \dotsm
+ \left(\frac{1}{12} - \frac{1}{3}\right) + \dotsm \\
&= \sum_{1 \le k \le n} \frac{1}{k}
- \sum_{1 \le k \le \lfloor n / 4 \rfloor} \frac{1}{k}
\end{align*}$
In terms of harmonic numbers:
$\begin{align*}
H_n
&= \sum_{1 \le k \le n} \frac{1}{k}
\end{align*}$
your sum is just:
$\begin{align*}
S_n
&= H_n - H_{\lfloor n / 4 \rfloor}
\end{align*}$
then, as hinted in comments, you can use the (rather crude) bound on the harmonic numbers $\ln n \le H_n \le \ln n + 1$:
$\begin{align*}
\ln n - (\ln \lfloor n / 4 \rfloor + 1)
&\le S_n
\le \ln n + 1 - \ln \lfloor n / 4 \rfloor
\end{align*}$
This tells you that, disregarding the floors, we have approximately:
$\begin{align*}
\ln 4 - 1
&\le S_n
\le \ln 4 + 1
\end{align*}$
Thus $e^{S_n}$ is between $4 / e = 1.4715$ and $4 e = 10.873$. Better bounds on $H_n$ give sharper values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do I find the integer solutions that satisfy $xyz = 288$ and $xy + xz + yz = 144$?
Find all integers $x$, $y$, and $z$ such that $$xyz = 288$$ and $$xy + xz + yz = 144\,.$$
I did this using brute force, where $$288 = 12 \times 24 = 12 \times 6 \times 4$$ and found that these set of integers satisfy the equation. How do I solve this without using brute force?
| We have $288 = 2^53^2.$ Let $x=2^a3^r,$ $y=2^b3^s,$ and $z=2^c3^t$. Then $a+b+c = 5$ and $r+s+t=2$. Since $r,s,$ and $t$ are non-negative integers, one of them must be $0$, say $t=0$. From the equation
$$xy+xz+yz = 144,$$
we see that if a prime number divides any of the variables, it must divide at least one of the others. This forces $r =s=1$ and we must have
$$x=2^a3, y=2^b3, z=2^c.$$
Similarly, if one of the variables is divisible by $8$, then the product of the other two variables is also divisible by $8$, but then $a+b+c\geq 6$, which is too big. This forces $1\leq a,b,c \leq 2.$
So either $c=1$ or $c=2.$ If $c=1$, then $a=b=2$ and $x=y=12$ and $z=2$, which doesn't satisfy the second equation.
If $c=2$, then $a=2, b=1$ (or vv.) and we have $x=12$, $y=6$, $z=4$ which is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Given positive real numbers $a$, $b$, $c$, $d$, $e$ with $\sum_{\text{cyc}}\,\frac{1}{4+a}=1$, prove that $\sum_{\text{cyc}}\,\frac{a}{4+a^2}\le1$.
Let $a, b, c, d, e$ be positive real numbers such that $$\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.$$
| We need to prove that $$1-\sum_{cyc}\frac{a}{4+a^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{a}{4+a^2}+3\left(\frac{1}{5}-\frac{1}{4+a}\right)\right)\geq0$$ or
$$\sum_{cyc}\frac{a^3-a^2-a+1}{(4+a^2)(4+a)}\geq0.$$
Now, by AM-GM $$a^3+\frac{1}{2}\geq\frac{3}{2}a^2$$ and $$\frac{1}{2}a^2+\frac{1}{2}\geq a,$$ which after summing gives $$a^3-a^2-a+1\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Showing that $a$, $b$, $c$, $d$ are in geometric progression iff $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$
If the real numbers $a$, $b$, $c$, $d$ are in geometric progression, show that
$$
\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}
$$
Prove that the converse also holds.
The simplest way I could think of is making/assuming a general G.P
Let $r$ be the common ratio and $a$ be the first term, then
$$b=a r, c=a r^{2}, d=a r^{3}$$
after multiplying a lot of terms, many times, (skipping significant steps here so it is more readable)
$\begin{aligned} \mathrm{LHS} &=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}\right)\left(a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\right) \\ &=a^{2}\left(1+r^{2}+r^{4}\right) \cdot a^{2} r^{2}\left(1+r^{2}+r^{4}\right) \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} \\ &=\left(a^{2} r+a^{2} r^{3}+a^{2} r^{5}\right)^{2} \\ &=\left(a \cdot a r+a r \cdot a r^{2}+a r^{2} \cdot a r^{3}\right)^{2} \\ &=(a b+b c+c d)^{2}=R H S \end{aligned}$
Once after expanding each term, and getting no where, I realised that I had already gotten the answer, all I had to do was take the terms inside the and then I had the RHS. But it all was so tedious and took multiple attempts. Though, this method guarantees that converse holds,
Can it be done more elegantly?
Edit: I am in highschool (and just a little more interested in math) So I don't have knowledge about the much spoken Cauchy's identity.
| Left side is
$$(a^2 + (ar)^2 + (ar^2)^2)((ar)^2 + (ar^2)^2 + (ar^3)^2) = a^4 r^2 (1 + r^2 + r^4)^2$$
Right side is
$$ (a^2 r + a^2 r^3 + a^2 r^5)^2 = a^4 r^2 (1+r^2 + r^4)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Given matrix $A$, find matrix $X$ such that $e^X = A$ Given the following matrix
$$A = \begin{bmatrix} -2 & 2 & 1\\ 2 & -3 & -2\\ -5 & 6 & 4\end{bmatrix}$$
how can we show that there exists a complex matrix $X$ such that $e^X = A$.
I have struggled to find the information about workaround the problem. However, the determinant of matrix $A$ is non-zero and its eigenvalues are $-1$ and $1$.
| The Jordan form of $A$ is given by $$J = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
with $A = PJP^{-1}$ where $P = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 2\end{bmatrix}$.
First we can try to find $\log J$ blockwise. Searching for the logarithm of $\begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix}$ again as some Jordan block, if we set $f(x) = e^{-x}$ then
$$\exp\left(\begin{bmatrix} i\pi & -1 \\ 0 & i\pi\end{bmatrix}\right) = f\left(\begin{bmatrix} -i\pi & 1 \\ 0 & -i\pi\end{bmatrix}\right) = \begin{bmatrix} f(-i\pi) & f'(-i\pi) \\ 0 & f(-i\pi)\end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 0 & -1\end{bmatrix}$$
so we can set (clearly it's not unique)
$$\log J = \begin{bmatrix} i\pi & -1 & 0 \\ 0 & i\pi & 0 \\ 0 & 0 & 0\end{bmatrix}$$
and finally we get
$$\log A = P(\log J) P^{-1} = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 0 & 0 \\ 1 & -2 & -1\end{bmatrix}+i\pi \begin{bmatrix} 1 & 0 & 0 \\ -1 & 2 & 1 \\ 2 & -2 & -1\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $\left(\begin{smallmatrix}0&0&1\\1&0&0\\ 0&1&0\end{smallmatrix}\right)$ diagonalizable over $\mathbb{Z}_2$? Is $A= \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}$ diagonalizable over $\mathbb{Z}_2$?
I tried two approaches and got two different answers so I was hoping someone could point me to a flaw in my reasoning:
First approach:
The minimal polynomial for $A$ is easily found to be $m(x) =x^3-1$ which is the same as $x-1$ over $\mathbb{Z}_2$. Since the minimal polynomial decomposes into distinct linear factors it must be that $A$ is diagonalizable over $\mathbb{Z}_2$.
Second approach:
It follows from the minimal polynomial that $1$ is the only eigenvalue of $A$ . The eigenvector equation is
$\begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y\\ z \end{pmatrix} = \begin{pmatrix} z \\ x\\ y \end{pmatrix} = 1 \times \begin{pmatrix} x \\ y\\ z \end{pmatrix} $ and the only solution is $\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}$.
But $\mathbb{Z}_2^3$ has dimension $3$, so there is no basis for $\mathbb{Z}_2^3$ consisting of eigenvectors for $A$. $A$ can't be diagonalized over $\mathbb{Z}_2$.
What went wrong? Many thanks!
| No, $x^3-1$ is not equal to $x-1$ over $\Bbb Z_2$, although the corresponding polynomial functions are equal indeed. On the other hand, $x^3-1=(x-1)(x^2+x+1)$ and $x^2+x+1$ is irreducible of $\Bbb Z_2$. Therefore, yes, your matrix is not diagonalizable over $\Bbb Z_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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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)$$
But I am unable to convert it into integral form (riemann sum to integral ) as the expression $\frac{k}{n}$ is present instead of $\frac{1}{n}$. Can someone please help me in solving this question. Thanks in advance.
| By your work $$\lim_{n\to +\infty}\left(\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)^2\cdots\left(1+\frac{n^2}{n^2}\right)^n\right)^{\frac{1}{n^2}}=$$
$$=e^{\int\limits_0^1x\ln(1+x^2)dx}=e^{\frac{1}{2}\int\limits_0^1\ln(1+x)dx}=e^{\frac{1}{2}\left((1+x)\ln(1+x)-x\right)|_0^1}=e^{\ln2-\frac{1}{2}}=\frac{2}{\sqrt e}.$$
Can you end it now?
I used your work and $$\lim_{n \to \infty}\sum_{k=1}^n\frac{k}{n}\ln\left(1+\left(\frac{k}{n}\right)^2\right)\frac{1}{n}=\int\limits_0^1x\ln(1+x^2)dx.$$
Since $$\lim_{n\rightarrow+\infty}\left(\frac{2}{\sqrt{e}}\right)^n=+\infty,$$ we see that starting limit does not exist.
| {
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"url": "https://math.stackexchange.com/questions/3753035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $y=f(x)=\frac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$. Question: If $y=f(x)=\dfrac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$.
We have $y=\dfrac{3\left(\frac{3x-5}{2x-m}\right)-5}{2\left(\frac{3x-5}{2x-m}\right)-m} $
How can I find $m$? It is given than $m=3$.
| Now, $$\dfrac{3\left(\frac{3x-5}{2x-m}\right)-5}{2\left(\frac{3x-5}{2x-m}\right)-m}=x$$ or
$$\frac{3(3x-5)-5(2x-m)}{2(3x-5)-m(2x-m)}=x$$ or
$$\frac{-x+5m-15}{x(6-2m)+m^2-10}=x$$ or $$-x+5m-15=x^2(6-2m)+x(m^2-10).$$
We need $$5m-15=0,$$ $$6-2m=0$$ and $$m^2-10=-1,$$ which gives $m=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem about differential of a smooth map Problem:
We have a map $ f: \mathbb{R}^3\rightarrow M(2,R)$
$$f(x,y,z)=exp \begin{pmatrix}
x & y+z \\
y-z & -x \\
\end{pmatrix}$$
Here, we denote the set of all 2-dimensional real square matrices as $M(2,R)$.
Find the points where the differential of $f$ is not injective and draw the graph of this set.
What I have done:
(You can ignore the following part)
Let's denote $\begin{pmatrix} x & y+z \\ y-z & -x \\ \end{pmatrix}$ as $A$.
Then $A^2 = \begin{pmatrix} x^2+y^2-z^2 & 0 \\ 0 & x^2+y^2-z^2 \\ \end{pmatrix} = (x^2+y^2-z^2)I$
Then, we can get the value of $A^n:$
$$
\begin{array}{c|c|c}
&\\\hline
A^0&I&\\
A^1&&A\\
A^2&(x^2+y^2-z^2)I&\\
A^3&&(x^2+y^2-z^2)A\\
A^4&(x^2+y^2-z^2)^2I&\\
A^5&&(x^2+y^2-z^2)^2A\\
\cdots&\cdots&\cdots
\end{array}
$$
It is easy for us to calculate:
$$\begin{align}exp A
& = I+ \color{red}{A}+\frac{A^2}{2!}+\color{red}{\frac{A^3}{3!}}+\frac{A^4}{4!}+\color{red}{\frac{A^5}{5!}}+\cdots \\
& = I (1 + \frac{(x^2+y^2-z^2)}{2!} + \frac{(x^2+y^2-z^4)^2}{4!} +\cdots )+ \color{red}{A} ( \color{red}{1}+ \color{red}{\frac{(x^2+y^2-z^2)}{3!}} + \color{red}{\frac{(x^2+y^2-z^2)^2}{5!}} +\cdots ) \\
& = I (1 + \frac{(x^2+y^2-z^2)}{2!} + \frac{(x^2+y^2-z^4)^2}{4!} +\cdots )\\
& + \color{red}{\frac{A}{\sqrt{x^2+y^2-z^2}}} ( \color{red}{\sqrt{x^2+y^2-z^2}}+ \color{red}{\frac{(x^2+y^2-z^2)^\frac{3}{2}}{3!}} + \color{red}{\frac{(x^2+y^2-z^2)^\frac{5}{2}}{5!}} +\cdots )
\end{align}$$
When $x^2+y^2-z^2>0$,
$$\begin{align}exp A
& = cosh\sqrt{x^2+y^2-z^2}I + \color{red}{\frac{A}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} \\
& = \begin{pmatrix}
cosh\sqrt{x^2+y^2-z^2} + \color{red}{\frac{x}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} & \color{red}{\frac{y+z}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} \\
\color{red}{\frac{y-z}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} & cosh\sqrt{x^2+y^2-z^2} - \color{red}{\frac{x}{\sqrt{x^2+y^2-z^2}} sinh \sqrt{x^2+y^2-z^2}} \\
\end{pmatrix}
\end{align} \tag{1}$$
Similarly, We can calculate that when $x^2+y^2-z^2<0$,
$$exp A= \begin{pmatrix}
cos\sqrt{z^2-(x^2+y^2)} + \color{red}{\frac{x}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}} & \color{red}{\frac{y+z}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}} \\
\color{red}{\frac{y-z}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}} & cos\sqrt{z^2-(x^2+y^2)} - \color{red}{\frac{x}{\sqrt{z^2-(x^2+y^2)}} sin \sqrt{z^2-(x^2+y^2)}} \\
\end{pmatrix} \tag{2}$$
When $x^2+y^2-z^2=0$, $ exp A = \begin{pmatrix} 1+x & y+z\\ y-z & 1-x\end{pmatrix} \tag{3}$
The standard way to solve the problem may be to identify $M(2,R)$ as $\mathbb{R}^4:$
$$\begin{pmatrix} x & z \\ y & w\end{pmatrix} \mapsto (x,y,z,w)$$
then calculate the differential for (1) (2) and (3) and decide when their rank$<3$.
But I found it is hard to calculate. So is to find under which condition rank $<3$.
Any one can give me some ideas or which books deal with related topics?
| You can write $f = g \circ h$ where
$$\begin{array}{l|rcl}
g : & M(2, \mathbb R) & \longrightarrow & M(2, \mathbb R) \\
& A & \longmapsto & \exp A \end{array}$$
and
$$\begin{array}{l|rcl}
h : & \mathbb R^3 & \longrightarrow & M(2, \mathbb R) \\
& (x,y,z) & \longmapsto & \begin{pmatrix}
x & y+z \\
y-z & -x \\
\end{pmatrix} \end{array}$$
According to the chain rule
$$f^\prime(x,y,z) = g^\prime(h(x,y,z)) \cdot h^\prime(x,y,z)$$ and $f^\prime$ is injective if and only if both $g^\prime(h(x,y,z))$ and $h^\prime(x,y,z)$ are injective. As $h$ is linear, its derivative is equal to itself at each point of $\mathbb R^3$. Also $g^\prime(A)= \exp A$.
The characteristic polynomial of $h(x,y,z)$ is $\mu_{h(x,y,z)}(X)= X^2 -(x^2+y^2-z^2)$. We need to deal with 3 cases, depending on the sign of $r(x,y,z) = x^2+y^2-z^2$.
Case 1: $x^2+y^2-z^2 = 0$
In that case, $\mu_{h(x,y,z)}(X)= X^2$, $0$ is an eigenvalue of $h$ (and $h^\prime$) and $f^\prime$ is not injective.
Case 2: $x^2+y^2-z^2 \gt 0$
In that case, $\mu_{h(x,y,z)}(X)$ has two distinct non vanishing roots and $h$ is diagonalizable. $g^\prime(h(x,y,z))$ is also diagonalizable and its eigenvalues are the exponential of the one of $h$. Hence, are also distinct and non zero. $g^\prime(h(x,y,z))$ is injective as $h^\prime(x,y,z)$ and finally $f^\prime(x,y,z)$ is injective too.
Case 3: $x^2+y^2-z^2 \lt 0$
$\mu_{h(x,y,z)}(X)$ has two distinct non vanishing pure imaginary roots $\sqrt{z^2 - x^2-y^2} i, -\sqrt{z^2 - x^2-y^2} i$ and $h$ is again diagonalizable. $g^\prime(h(x,y,z))$ is also diagonalizable and its eigenvalues are the exponential of the ones of $h$. Hence, are also distinct. $f^\prime$ is also injective in that case.
Finally, $f^\prime$ is not injective if and only if $(x,y,z)$ belongs to the cone $x^2+y^2 = z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The sides of the triangle are $a, b, c$. Find the distance between the projections. The sides of the triangle are $\space a, b, c. \space$ Find the distance between the projections of the vertex onto the bisectors of the external angles at the other two vertices.
I've got that there are two right triangles that create the quadrangles. Then this quadrangle can describe a circle and it turns out that there some angles are equal but it didn’t give me anything.
|
Here $BF$ and $CG$ are the external angle bisectors of $\angle ABC=\angle B$ and $\angle ACB=\angle C$, so $\angle ABF = \dfrac12 (\pi - \angle B) = \dfrac{\pi}2-\dfrac{\angle B}2$ and $\angle ACG = \dfrac{\pi}2 - \dfrac{\angle C}2$.
$\Delta AFB, \Delta AGC$ being right angled triangles, we have $\angle BAF=\dfrac{\angle B}2, \ \angle CAG = \dfrac{\angle C}2$, $$AB\sin(\angle ABF) = AF , \ AC\sin(\angle ACG) = AG \\ AF = c\sin\left(\dfrac{\pi}2-\dfrac{\angle B}2\right) = c\cos\left(\dfrac{\angle B}2\right), \ AG =b\cos\left(\dfrac{\angle C}2\right) $$
and $\angle FAG = \angle FAB + \angle BAC + \angle CAG = \dfrac{\angle B}2 +\angle A + \dfrac{\angle C}2 = \dfrac{\angle A}2 + \dfrac{\angle A +\angle B+\angle C}2 = \dfrac{\angle A}2+\dfrac{\pi}2$
Having $$AF=c\cos\left(\dfrac{\angle B}2\right), AG=b\cos\left(\dfrac{\angle C}2\right), \angle FAG = \dfrac{\angle A}2+\dfrac{\pi}2$$ we can use the half-angle formulae in terms of sides which gives $$AF=c\sqrt{\dfrac{s(s-b)}{ac}}, AG=b\sqrt{\dfrac{s(s-c)}{ab}}, \ \text{ where } s=\dfrac{a+b+c}2$$ and apply the cosine rule in $\Delta AFG$ to find the length of $FG$.
Incidentally, the cosine rule calculations are not that bad: $$FG^2=b^2\dfrac{s(s-c)}{ab} + c^2\dfrac{s(s-b)}{ac} -2bc\sqrt{\dfrac{s(s-c)}{ab}}\sqrt{\dfrac{s(s-b)}{ac}}\cos\left(\dfrac{\angle A}2 +\dfrac{\pi}2\right) \\= \dfrac{s}{a}(sb-bc+sc-bc) + \dfrac{2bcs}{a}\sqrt{\dfrac{(s-b)(s-c)}{bc}}\sin\left(\dfrac{\angle A}2\right) \\ = \dfrac{s}a \left(sb+sc-2bc +2bc\sqrt{\dfrac{(s-b)(s-c)}{bc}}\sqrt{\dfrac{(s-b)(s-c)}{bc}}\right) \\ = \dfrac{s}a(sb+sc-2bc +2(s-b)(s-c)) = \dfrac{s}a(2s^2-(b+c)s)=\dfrac{s^2}a(2s-(b+c))\\ =\dfrac{s^2a}a=s^2 \implies FG = s = \dfrac{a+b+c}2$$
This now makes me feel that there can be nice ways of arriving at this length. For a purely geometric approach, my mind thinks of splitters and cleavers at the moment.
| {
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"url": "https://math.stackexchange.com/questions/3756865",
"timestamp": "2023-03-29T00:00:00",
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Integrate: $\int \frac{x}{\left(x^2-4x-13\right)^2}dx$.
Integrate:
$$\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}$$
Is this working out correct? I'm not really sure how WolframAlpha works, so I didn't check it on there.
| Here is an alternative method to integrate as follows
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
$$=\int\frac12 \frac{(2x-4)+4}{\left(x^2-4x-13\right)^2}dx$$
$$=\frac12\int \frac{2x-4}{\left(x^2-4x-13\right)^2}dx+\frac12\int\frac{4}{\left(x^2-4x-13\right)^2}dx$$
$$=\frac12\int \frac{d(x^2-4x-13)}{\left(x^2-4x-13\right)^2}+2\int\frac{d(x-2)}{\left((x-2\right)^2-17)^2}$$
using reduction formula: $\color{blue}{\int \frac{dt}{(t^2+a)^n}=\frac{t}{2(n-1)a(t^2+a)^{n-1}}+\frac{2n-3}{2(n-1)a}\int\frac{dt}{(t^2+a)^{n-1}}} $,
$$=\frac12 \frac{-1}{\left(x^2-4x-13\right)}+2\left(\frac{(x-2)}{2(-17)((x-2)^2-17)}+\frac{1}{2(-17)}\int \frac{d(x-2)}{(x-2)^2-17}\right)$$
using standard formula: $\color{blue}{\int \frac{dt}{t^2-a^2}=\frac{1}{2a}\ln\left|\frac{t-a}{t+a}\right|}$,
$$=-\frac{1}{2\left(x^2-4x-13\right)}-\frac{(x-2)}{17(x^2-4x-13)}-\frac{1}{34\sqrt{17}}\ln\left|\frac{x-2-\sqrt{17}}{x-2+\sqrt{17}}\right|+C $$
$$=-\frac{2x+13}{34(x^2-4x-13)}-\frac{1}{34\sqrt{17}}\ln\left|\frac{x-2-\sqrt{17}}{x-2+\sqrt{17}}\right|+C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factor $x^5-5x^3+4x$ I am trying to factor$x^5-5x^3+4x$ so that I can find the roots. I know from the answers section that the roots are where $x = 0, 1, -1, 2$ and $-2$.
I'm stuck, here's as far as I got:
$$
x^5-5x^3+4x =
x(x^4-5x^2+4)
$$
Let $u = x^2$ and just focus on the term on the right (drop the first $x$ for now):
$$x^4-5x^2+4 = u^2-5u+4x.$$
Master term is $a \times c = 1 \times 4 = 4$.
Seeking a pair of numbers that sum to the middle term $-5$ and whose product is $4$:
\begin{align}
1 \times -4 &= -4, &\text{sum} &= -3 \\
4 \times -1 &= -4, &\text{sum} &= 3 \\
2 \times -2 &= -4, &\text{sum} &= 0
\end{align}
???
I'm not sure how to proceed since I cannot find a pair of numbers that satisfy the condition.
Have I gone wrong somewhere? How can I factor $u^2-5u+4x$?
| $$x^5-5x^3+4x=x(x^4-5x^2+4)$$$$=x(x^4-4x^2-x^2+4)$$$$=x(x^2(x^2-4)-(x^2-4))$$$$=x((x^2-4)(x^2-1))$$
$$=x(x+2)(x-2)(x+1)(x-1)$$
| {
"language": "en",
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Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $2a^3+27c=9ab$
Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $$2a^3+27c=9ab$$
So far, I let the roots of $x^3+ax^2+bx+c=0$ be $r_1, r_2,$ and $r_3$. $r_1=r_2-d$ and $r_3=r_2+d$ because they form an arithmetic sequence with $d$ being the difference. the sum of the roots is $-a$. So, $r_2=-a/3$. We can let the product of the roots be $-c$. So, $(r_2-d)(r_2)(r_2+d)=-c$. Plugging in $r_2=-a/3$ we get $(-a/3-d)(-a/3)(-a/3+d)$. How do I continue with this method?
EDIT: I used hamam_abdallah's hint to get $\frac{-a^3}{27} + \frac{ad^2}{3} = -c$ what do i do after applying vieta's formulas?
| Put $$a=3A$$ The equality to prove $$2a^3+27c=9ab$$ becomes
$$\boxed{2A^3+c=Ab}$$
As you said $r_2=\frac{-a}{3}= -A$ is a root of
$$x^3+3Ax^2+bx+c=0$$ then
$$(-A)^3+3A(-A)^2+b(-A)+c=0$$
$$\iff \; -A^3+3A^3+c=Ab$$
$$\iff \; \boxed{2A^3+c=Ab}$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
So since $n^3-n=(n-1)n(n+1)$, we have that $3\vert(n^3-n).$
Also since we have that $n$ is odd we can say that $n=2k+1$, for some $k \in \mathbb{Z^+}.$
This implies that $(n+1) = 2k +2 = 2(k+1)$, hence $2\vert(n+1).$ Similar argument can be made for $(n-1)$.
Now $24 = 2 \cdot 3 \cdot 4$ so essentially I'm only missing the part where I would have to show that $4$ divides some of the terms. How can I find that?
| Let n be an odd positive integer, so $n=2k+1$ so $$n^3-n=(n-1)n(n+1)\\
=(2k+1-1)(2k+1)(2k+2)\\=2k(2k+1)(2k+2)\\=4k(k+1)(2k+1)\\=4\underbrace{k(k+1)}_{even=2q}(2k+1)\\=4(2q)(2k+1)$$
| {
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Cubes as the sum of odd integers It is well known that
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$4^3=13+15+17+19$
$5^3=21+23+25+27+29$
and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)
We will consider the array
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
&&&&\ddots
\end{matrix}
\end{align*}
and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$.
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
\vdots \\
A_{(i-1)1}&...&A_{(i-1)(i-1)}\\
A_{i1}&...&A_{ij} &...&A_{ii}
\end{matrix}
\end{align*}
It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration.
\begin{align*}
A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\
&=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\
&=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\
&...\\
&=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\
&=(i-1)i-1.
\end{align*}
Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.
Let me know if there is any clarification necessary!
| The OP's claim can be summarized by saying that $n^3$ can be expressed as
the sum of $n$ consecutive odd numbers starting with $n(n+1) - (2n - 1)$ and ending with $n(n+1) - 1$.
This is easy to show by a direct proof, using a result for $n^2$ that
is well-known and easy to prove by induction.
Take the series of odd numbers in reverse order, from largest to
smallest. Then we are saying that:
\begin{eqnarray}
n^3 & = & \big[(n^2 + n - 1) + (n^2 + n - 3) + \ldots + (n^2 + n - (2n - 1))\big] \\
& = & n (n^2 + n) - [1 + 3 + \ldots + 2n - 1]
\end{eqnarray}
This in turn gives:
\begin{equation}
n^2 = 1 + 3 + \ldots + 2n - 1
\end{equation}
which is an exercise in elementary proof by induction.
| {
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Solve for $x$, $(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)} $ $$(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)} $$
Solve for $x$
My Work:
Take ln at the both sides:
$$(2x+6)\ln(1+2i)=(x+1)\ln(-11-2i)$$
and$$\ln(1+2i)=\ln(\sqrt{5})+i\arctan(2)$$
$$\ln(-11-2i)=\ln(5\sqrt{5})+i\left[\arctan\left(\frac{2}{11}\right)+\pi\right]\\=\ln(5)+\ln(\sqrt{5})+i\left[\arctan\left(\frac{2}{11}\right)+\pi\right]$$
And things get dirty here because of $\arctan$ and $\ln$ ...
But
$$(2x+6)\ln(1+2i)=(x+1)\ln(-11-2i)$$ from here
$$2x+6=(x+1)\dfrac{\ln(-11-2i)}{\ln(1+2i)}$$
then
$$x\left(2-\dfrac{\ln(-11-2i)}{\ln(1+2i)}\right)=-6+\dfrac{\ln(-11-2i)}{\ln(1+2i)}$$
then
$$x=\dfrac{-6+\dfrac{\ln(-11-2i)}{\ln(1+2i)}}{2-\dfrac{\ln(-11-2i)}{\ln(1+2i)}}$$
But specifically I cannot find x here.
Any elegant way to solve it, or not elegant but proper, any hint any help, would be appreciated. Thank you.
| $$(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)}$$ is equivalent to $$(1+2i)^{(2x+6)}=(1+2i)^{3(x+1)}$$ or, assuming $(1+2i)^{(2x+3)}\neq 0,$ $$(1+2i)^{3}=(1+2i)^{x}.$$
One solution is $x=3.$
I do not think that other solutions would be trivial to find.
| {
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"timestamp": "2023-03-29T00:00:00",
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Distance between set and point, confused of partial derivatives. Let $H = \{(x,y,z)\ \in \mathbb{R}^{3}: x^2+y^2 - z^2 + 4 = 0$
Compute the shortest distance between H and point $p=(2,4,0)$.
I am a bit confused because I tried a direct approach.
$$ x^2+y^2 + 4 = z^2$$
Let $D(H,p) = \sqrt{(2-x)^{2}+(4-y)^{2} + x^{2} + y^{2}+4}$
So I tried compute $$\frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
$$\frac{\partial D}{\partial y} = (\sqrt{2} (-2 + y))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
It seems not nice to compare with zero.
Do you have another idea?
| A fraction is zero when its numerator is zero and its denominator is not. If your fraction is $0/0$, apply l'Hopital's rule until you can resolve its value.
You wish to study
$$ \frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2} \text{.} $$
This fraction is zero if $\sqrt{2}(-1+x) = 0$, so when $x = 1$, as long as $\left. \sqrt{12 - 2 x + x^2 - 4 y + y^2} \right|_{x=1}$ is not zero. Since
$$ \left. \sqrt{12 - 2 x + x^2 - 4 y + y^2} \right|_{x=1} = \sqrt{11-4y+y^2} $$
Since the discriminant of that quadratic in $y$ is $-28 < 0$, the denominator is not simultaneously zero for any real value of $y$. (The roots of the quadratic in $y$ are $y = 2 \pm \mathrm{i}\,\sqrt{7}$, neither of which is a real number.) Therefore, $\partial D / \partial x = 0$ for $x = 1$ and all $y$, i.e., the set $\{1\} \times \Bbb{R}$ in $\Bbb{R}^2$.
Applying the same analysis to $\partial D / \partial y$, the numerator is zero when $y = 2$ and the denominator for that specialization of $y$ is $\sqrt{8-2x+x^2}$. Again, the discriminant of this quadratic is negative, so it has no real roots and $\partial D/\partial y$ is zero at every point of $\Bbb{R} \times \{2\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\le 1$, prove that $(1+a^2)(1+b^2)(1+c^2)\ge 125$.
QUESTION: Let $a,b,c$ be positive real numbers such that
$$\cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1$$
Prove that $$(1+a^2)(1+b^2)(1+c^2)\ge 125$$ When does equality hold?
MY APPROACH: Firstly, let's try to squeeze out all the information we can from what is given. $$\frac{(1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b)}{(1+a)(1+b)(1+c)}≤1$$ Multiplying this out, we get $$3+2(a+b+c)+(ab+bc+ca)≤1+(a+b+c)+(ab+bc+ca)+abc$$ $$\implies 2+(a+b+c)≤abc$$ Also, since $$1≥\sum_{cyc}\frac{1}{1+a}$$ Therefore by AM-GM,
$$1≥\sum_{cyc}\frac{3}{\sqrt[3]{(1+a)(1+b)(1+c)}}$$
$$\implies (1+a)(1+b)(1+c)≥27$$
That's all I ended up in.. At first, I thought Hölder's inequality could be employed, but that too requires the sum of the powers to be $=1$.. and that is not going to be useful in $(1+a^2)(1+b^2)(1+c^2)$ , since here the sum of powers add up to $3$..
I don't know what to do next.. Any help will be much appreciated..
| Another way.
Since by AM-GM $$\prod_{cyc}\frac{a}{1+a}=\prod_{cyc}\left(1-\frac{1}{1+a}\right)\geq\prod_{cyc}\left(\frac{1}{1+b}+\frac{1}{1+c}\right)\geq$$
$$\geq\prod_{cyc}\frac{2}{\sqrt{(1+b)(1+c)}}=\frac{8}{\prod\limits_{cyc}(1+a)},$$ we see that $$abc\geq8$$ and by Holder we obtain:
$$\prod_{cyc}(a^2+1)\geq\left(\sqrt[3]{a^2b^2c^2}+1\right)^3\geq(4+1)^3=125.$$
| {
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"url": "https://math.stackexchange.com/questions/3763446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that the following power series satisfies this functional equation $f\left(\frac{2x}{1+x^2}\right)=(1+x^2)\,f(x)$.
Show that the following power series satisfies this functional equation $$f\left(\dfrac{2x}{1+x^2}\right)=(1+x^2)f(x)\,,$$
where the series given is $$f(x)= 1+\dfrac{1}{3}x^2+\dfrac{1}{5}x^4+\dfrac{1}{7}x^6+ \cdots\,.$$
I can painstakingly get a relation between the derivatives such as $3f^{(2)}(0) = 2f(0)$ but I was hoping for a better approach.
I would appreciate it if someone could give me a hint. I prefer hints to complete solutions.
| This is a proof that the identity holds in $\mathbb{K}[\![x]\!]$ for an arbitrary base field $\mathbb{K}$ of characteristic $0$, where it is not possible to write $$f(x)=\dfrac{1}{\color{red}2x}\,\ln\left(\dfrac{1+x}{1-x}\right)$$ (although we can technically define $\ln(1+x)$, $\ln(1-x)$, and $\ln\left(\dfrac{1+x}{1-x}\right)$ as power series in $\mathbb{K}[\![x]\!]$). Note that in my comment under the question, I forgot a factor $2$.
Since $f(x)=\displaystyle\sum_{k=0}^\infty\,\frac{x^{2k}}{2k+1}$, we get
$$g(x):=\frac{1}{1+x^2}\,f\left(\frac{2x}{1+x^2}\right)=\frac{1}{1+x^2}\,\sum_{k=0}^{\infty}\,\frac{1}{(2k+1)}\,\left(\frac{2x}{1+x^2}\right)^{2k}\,.$$
Therefore,
$$g(x)=\sum_{k=0}^\infty\,\frac{2^{2k}\,x^{2k}}{2k+1}\,(1+x^2)^{-2k-1}=\sum_{k=0}^\infty\,\frac{2^{2k}\,x^{2k}}{2k+1}\,\sum_{r=0}^\infty\,\binom{-2k-1}{r}\,x^{2r}\,.$$
Since $\displaystyle\binom{-m}{n}=(-1)^n\,\binom{m+n-1}{n}$, we get
$$g(x)=\sum_{k=0}^\infty\,\frac{2^{2k}\,x^{2k}}{2k+1}\,\sum_{r=0}^\infty\,(-1)^r\,\binom{2k+r}{r}\,x^{2r}\,.$$
That is,
$$g(x)=\sum_{k=0}^\infty\,\sum_{r=0}^\infty\,\frac{(-1)^r\,2^{2k}}{2k+1}\,\binom{2k+r}{r}\,x^{2(k+r)}\,.$$
Let $s:=k+r$. Then,
$$g(x)=\sum_{s=0}^\infty\,x^{2s}\,\sum_{k=0}^s\,\frac{(-1)^{s-k}\,2^{2k}}{2k+1}\,\binom{s+k}{s-k}\,.$$
In order to prove that $g(x)=f(x)$, we need to show that
$$\frac{1}{2s+1}=\sum_{k=0}^s\,\frac{(-1)^{s-k}\,2^{2k}}{2k+1}\,\binom{s+k}{s-k}\tag{*}$$
for all $s=0,1,2,\ldots$.
However, we are in luck. The equation (*) is an equality for rational numbers, which is the prime field of $\mathbb{K}$. Therefore, we can simply prove (*) by using the result when $\mathbb{K}=\mathbb{R}$. Note that $f(x)=\dfrac{1}{2x}\ln\left(\dfrac{1+x}{1-x}\right)$ for $x\in\mathbb{R}$ such that $0<|x|<1$. Since
$$f\left(\frac{2x}{1+x^2}\right)=\frac{1+x^2}{4x}\,\ln\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)=\frac{1+x^2}{4x}\,\ln\left(\frac{1+2x+x^2}{1-2x+x^2}\right)\,,$$
therefore
$$\begin{align}f\left(\frac{2x}{1+x^2}\right)&=(1+x^2)\,\left(\frac{1}{4x}\,\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)\right)
\\&=(1+x^2)\,\left(\frac{1}{4x}\cdot 2\ln\left(\frac{1+x}{1-x}\right)\right)
\\&=(1+x^2)\,\left(\frac{1}{2x}\,\ln\left(\frac{1+x}{1-x}\right)\right)
\\&=(1+x^2)\,f(x)\,.\end{align}$$
Thus, (*) holds in $\mathbb{R}$, whence (*) is an equality of rational numbers. Consequently, (*) is true in any field $\mathbb{K}$ of characteristic $0$. Therefore, the identity
$$f\left(\frac{2x}{1+x^2}\right)=(1+x^2)\,f(x)$$
holds in $\mathbb{K}[\![x]\!]$ for any field $\mathbb{K}$ of characteristic $0$.
Remark. I think there has to be a direct combinatorial or algebraic way to prove (*). My proof of (*) is in a very roundabout manner.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differential equation, modulus signs in solution? Question: Find the equation of the curve with gradient $\frac{dy}{dx}=\frac{y+1}{x^2-1}$ that passes through $(-3,1)$.
So I integrated both sides with respect to $x$ which gave me
$\ln{|y+1|}=\frac{1}{2}\ln{|x-1|}-\frac{1}{2}\ln{|x+1|}+c$
Given the point I found $c=\ln{\sqrt{2}}$
So at this point I figure the best way this equation can tidy up is $|y+1|=\sqrt{\frac{2|x-1|}{|x+1|}}$
I'm having trouble getting my head around what I can do with the equation at this point. The textbook gives an answer of $(y+1)^2(x+1)=2(x-1)$ but where have the modulus signs gone? Surely changing them to brackets gives just one possible curve (ie. where $y>-1$ and $x>1$)? And can you just square both sides of the equation like this? ($y=x$ and $y^2=x^2$ aren't the same curve)
Can anyone help me out with perhaps a little intuition here, and maybe some advice with how the solution should be presented here? Thanks in advance.
| $$\frac{dy}{dx}=\frac{y+1}{x^2-1}$$
$$\ln{|y+1|}=\frac{1}{2}\ln{|x-1|}-\frac{1}{2}\ln{|x+1|}+c\qquad\text{is OK.}$$
$y(-3)=1 \implies c=\frac12\ln{\sqrt{2}}\quad$ but not $c=\ln{\sqrt{2}}$ .
$$2\ln{|y+1|}=\ln{|x-1|}-\ln{|x+1|}+\ln(2)$$
$$(y+1)^2=2\left|\frac{x-1}{x+1}\right|$$
Then one have to study the function $f(x)=\frac{x-1}{x+1}$ which must be positive. We show that :
$$f(x)>0 \quad\text{if}\quad \left(x<0\quad\text{or}\quad x>2 \right)$$
$$f(x)<0 \quad\text{if}\quad 0<x<2$$
Thus
$$(y+1)^2=2\frac{x-1}{x+1}\quad\text{if}\quad \left(x<0\quad\text{or}\quad x>2 \right) $$
$$(y+1)^2=-2\frac{x-1}{x+1}\quad\text{if}\quad 0<x<2 $$
The specified point $(x=-3\:;\:y=1)$ belongs to the branche :
$$(y+1)^2(x+1)=2(x-1)\qquad x<0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing characteristic polynomial of $4 \times 4$ matrix Given the following matrix
$$A = \begin{bmatrix} 1 & -3 & 0 & 3 \\ -2 & -6 & 0 & 13 \\ 0 & -3 & 1 & 3 \\ -1 & -4 & 0 & 8\end{bmatrix}$$
I need to calculate $|\lambda I - A|$.
I get to a very complicated determinant. Is there an easier way? The answer says it’s $(\lambda - 1)^4$.
| Hint:
First expand along the 3rd column:
$$\lambda I-A= \begin{vmatrix}
\lambda-1 &3&0&-3 \\ 2&\lambda+6& 0&-13 \\ 0&3&\lambda-1&-3\\ 1&4&0&\lambda-8
\end{vmatrix}=(\lambda-1)\begin{vmatrix}
\lambda-1 &3&-3 \\ 2&\lambda+6& -13 \\ 1&4&\lambda-8
\end{vmatrix}$$
Now, either you apply directly Sarrus' rule, or you simplify the 3rd order determinant (we'll denote it $D$) with elementary operations to make a determinant with more zeros:
$$D=\begin{vmatrix}
\lambda-1 &3&0 \\ 2&\lambda+6&\lambda-7 \\ 1&4&\lambda-4
\end{vmatrix}=\begin{vmatrix}
\lambda-1 &3&0 \\ 0&\lambda-2 &1-\lambda\\ 1&0&\lambda-4
\end{vmatrix}.$$
| {
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If $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$; Find out the no of intersecting point Question from an entrance paper
If $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$ then
A. $f$ and $g$ agree at no point
B. $f$ and $g$ agree at exactly one point
C. $f$ and $g$ agree at exactly two point
D. $f$ and $g$ agree at more then two point
My approach
$f(x)=x^{2}$
$g(x)=x \sin x+\cos x$
$\Rightarrow f(x)=g(x)$
$\Rightarrow x^{2}=x \sin x+\cos x$
$\Rightarrow x^{2}-x \sin x-\cos x=0$
$\Rightarrow F(x)=0$
$F(0)=-1$ & $F(\pi)=\pi^{2}+1 >0 $
$F^{\prime}(x)=2 x-x \cos x$
$F^{\prime}(x)=x(2-\cos x)$
Here $2-\cos x$ is greater than $0$ and $x(2-\cos x)$ will be also greater than 0 for $x>0$
$\Rightarrow F(x)$ is strictly increasing function.
So I can conclude there will be exactly two point on which they will meet.
$\boxed{\text{*My question is how to do this without calculus*}}$
Edit $1$:- I had thought like this. This a eq of power $2$,so It must have two roots. These two roots may be real or complex conjugate.Now how to distinguish in general whether it would have real or imaginary roots for all even powered polynomials
ps Dont give solutions which is similar to mine.
| Solution without calculus
If $x \ge \frac{\pi}{2}$, we have $g(x) \le \sqrt{x^2 + 1} < x^2$,
and thus $f(x) = g(x)$ has no real solution on $[\frac{\pi}{2}, \infty)$.
Also, $0$ is not a solution of $f(x) = g(x)$.
Since $f(0) < g(0)$ and $f(\frac{\pi}{2}) > g(\frac{\pi}{2})$, by continuity,
$f(x)=g(x)$ has at least one real solution on $(0, \frac{\pi}{2})$.
Let us prove that $f(x)=g(x)$ has exactly one real solution on $(0, \frac{\pi}{2})$.
Suppose there exist $0 < s < t < \frac{\pi}{2}$ such that $f(s) = g(s)$ and $f(t) = g(t)$. We have
\begin{align}
\frac{s^2}{\sqrt{s^2+1}} &= \frac{s\sin s + \cos s}{\sqrt{s^2+1}} = \cos (s - \arctan s), \\
\frac{t^2}{\sqrt{t^2+1}} &= \frac{t\sin t + \cos t}{\sqrt{t^2+1}} = \cos (t - \arctan t).
\end{align}
Since $\frac{t^2}{\sqrt{t^2+1}} - \frac{s^2}{\sqrt{s^2+1}} =
\sqrt{t^2+1} - \frac{1}{\sqrt{t^2 + 1}} -
\sqrt{s^2+1} + \frac{1}{\sqrt{s^2 + 1}} > 0$, we have
$$\cos (s - \arctan s) < \cos (t - \arctan t)$$
which, when combined with
$t - \arctan t \in (0, \frac{\pi}{2})$ and $s - \arctan s \in (0, \frac{\pi}{2})$, results in
$$s - \arctan s > t - \arctan t.$$
Thus, we have $\arctan t - \arctan s > t - s$ and $\tan (\arctan t - \arctan s) > \tan (t - s)$
that is $\frac{t - s}{1 + ts} > \tan (t-s)$ which, when combined with $\tan (t-s) > t-s$, results in
$\frac{t - s}{1 + ts} > t-s$ which is impossible. The desired result follows.
Thus, $f(x) = g(x)$ has exactly one real solution on $[0, \infty)$.
Since $f(x)$ and $g(x)$ are both even, $f(x) = g(x)$ has exactly two real solutions on $(-\infty, \infty)$.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed?
One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $\dfrac{163}4$. Which integer was removed?
Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1
I was hoping if someone could spot the flaw in my working for this question.
Attempt.
I began by letting the integer that was removed be $x$.
Then: $$\frac{1 + 2 + \cdots + (x-1) + (x+1) +\cdots + n} {n-1} = \frac{163}{4}$$
There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.
These are equal to $\frac{x(x-1)}{2}$ and $\frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:
$$\frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = \frac{163}{4}$$
which reduces to:
$$\frac{n^2 + n - 2x}{2(n-1)} = \frac {163}{4}$$
And then:
$$2(n^2 + n -2x) = 163(n-1)$$
At first I thought you could consider factors, as 163 was prime then:
$n-1 = 2$ giving $n = 3$ and $n^2 + n - 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn't our positive integer.
I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help
| The average of $1,2,3,\ldots,n$ is the number halfway between the endpoints, $(n+1)/2,$ so the sum is $n(n+1)/2.$ Omitting $x$ from among $1,2,3,\ldots,n,$ we get the sum $n(n+1)/2-x.$
Thus the average of $1,2,3,\ldots,n$ must be a weighted average of $\big( n(n+1)/2-x\big)/(n-1)$ and $x,$ with respective weights $(n-1)/n$ and $1/n.$
$$
\frac{n-1} n \left( \frac{n(n+1)/2} {n-1} - \frac x {n-1} \right) + \frac 1 n\cdot x = \frac{n+1} 2
$$
Therefore we have:
\begin{align}
& \frac{n-1} n \cdot \frac{163} 4 + \frac x n = \frac{n+1} 2 \\[8pt]
& \frac{n(n+1)}{2(n-1)} - \frac x {n-1} = \frac{163} 4
\end{align}
So we get a system of two equations that is quadratic in $n$ and linear in $x.$ I'd try solving for one of those two in terms of the other and then substituting and solving the remaining equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding largest delta value given epsilon for delta-epsilon limit The formal definition of the limit is that $\lim_{x\to c} f(x)=L$ if and only if for any $\varepsilon >0$ there exists a $\delta >0$ such that $|x-c| <\delta \rightarrow |f(x)-L| < \varepsilon$. Use this definition for $\lim_{x \to 5} (x^2 -15x+50)$ to find the largest value for delta that satisfies epsilon equal to one.
Source: FAMAT State Convention 2019, Limits&Derivatives #29
Attempted solution: I set $|x^2-15x+50 - 0| < 1$ and solved for $x$, obtaining the following inequalities: $\frac{15-\sqrt{29}}{2} < x < \frac{15+\sqrt{29}}{2}$ and $x<\frac{15-\sqrt{21}}{2} $ or $ x > \frac{15+\sqrt{21}}{2}$
Next, I tried getting this in the form $|x-c| <\delta$ by changing each inequality as follows:
$\frac{5-\sqrt{29}}{2} < x-5 < \frac{5+\sqrt{29}}{2}$ and $x-5<\frac{5-\sqrt{21}}{2} $ or $ x-5 > \frac{5+\sqrt{21}}{2}$
I don't know how to translate this result to $\delta$ because it does not follow a neat $-\delta<x-c<\delta \rightarrow |x-c| <\delta$
Answer provided by the competition:
$\frac{5-\sqrt{21}}{2}$
Edit: See comment to where I'm currently at
| I don't agree with the provided answer:
$|(x^2-15x+50) - 0| < 1$ is satisfied on two intervals: $(\frac{5-\sqrt{29}}{2}, \frac{5-\sqrt{21}}{2})$ and $(\frac{5+\sqrt{21}}{2}, \frac{5+\sqrt{29}}{2})$. Only the second interval contains $5.$ Now, $|\frac{5+\sqrt{21}}{2} - 5| = |\frac{-5+\sqrt{21}}{2}| = \frac{5-\sqrt{21}}{2} \approx 0.2087$ and $|\frac{5+\sqrt{29}}{2} - 5| = |\frac{-5+\sqrt{29}}{2}| = \frac{\sqrt{29}-5}{2} \approx 0.1926.$ We need to take the least of these as our largest possible $\delta,$ i.e. $\frac{\sqrt{29}-5}{2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$
Let $a,b,c>0$ and such $a+b+c=3$,show that
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\ge 1\tag{1}$$
I tried using Holder's inequality to solve it:
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\sum b\sum \sqrt{a^3+8}\ge (a+b+c)^3$$
But the following is not right
$$\sum\sqrt{a^3+8}\le 9$$
so please help me prove $(1)$
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}=\sum_{cyc}\frac{2a^3}{b\cdot2\sqrt{(a+2)(a^2-2a+4)}}\geq\sum_{cyc}\frac{2a^3}{b(a+2+a^2-2a+4)}=$$
$$=\sum_{cyc}\frac{2a^4}{ab(a^2-a+6)}\geq\frac{2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}ab(a^2-a+6)}=$$
$$=\frac{6(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(3a^3b-a^2b(a+b+c)+2ab(a+b+c)^2)}=\frac{6(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^3b+2a^3c+3a^2b^2+9a^2bc)}$$ and it's enough to prove that:
$$\frac{6(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^3b+2a^3c+3a^2b^2+9a^2bc)}\geq1,$$
which is true by SOS and Tangent Line methods:
$$6(a^2+b^2+c^2)^2-\sum\limits_{cyc}(4a^3b+2a^3c+3a^2b^2+9a^2bc)=$$
$$\sum\limits_{cyc}(6a^4-4a^3b-2a^3c+9a^2b^2-9a^2bc)=$$
$$=2\sum_{cyc}(3a^4-2a^3b-ab^3)+\frac{9}{2}\sum_{cyc}(c^2a^2-2c^2ab+c^2b^2)=$$
$$=2\sum_{cyc}a(a-b)(3a^2+ab+b^2)+\frac{9}{2}\sum_{cyc}c^2(a-b)^2=$$
$$=2\sum_{cyc}\left((a-b)(3a^3+a^2b+ab^2)-\frac{5}{4}(a^4-b^4)\right)+\frac{9}{2}\sum_{cyc}c^2(a-b)^2=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2(7a^2+6ab+5b^2)+\frac{9}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
| {
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How to obtain the sum of the series $\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}$? How to prove what follows?
$$\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\frac{2^{\frac{1}{3}}}{3}\ln\left(\frac{\sqrt{2^{\frac{2}{3}}+2^{\frac{1}{3}}+1}}{2^{\frac{1}{3}}-1}\right)+\frac{\sqrt[3]{2}}{3}\arctan\left(\frac{2^{\frac{2}{3}}+1}{\sqrt{3}}\right)-\frac{2^{\frac{1}{3}}\pi}{6\sqrt{3}}$$
My attempt:
$$\sum_{n=0}^{\infty}\frac{1}{2^n(3n+1)}=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}|_{x=1}$$
We put $$S(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}\implies S^{'}(x)=\sum_{n=0}^{\infty}\frac{x^{3n}}{2^n(3n+1)}$$
$$S^{'}(x)=\sum_{n=0}^{\infty}\frac{(\frac{x^3}{2})^n}{3n+1}=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(1-\frac{3n}{3n+1})=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n\frac{3n}{3n+1}=\frac{1}{2-\frac{x^3}{2}}-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})=\alpha-\beta$$
Where
$$\beta=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})$$
So $$\beta=?$$
Waiting for your help to find a beta or prove equal above.
| Hint. One may prove that
$$
\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\sum_{n=0}^{\infty}\int_0^1\frac{x^{3n}}{2^{n}}dx=\int_0^1\sum_{n=0}^{\infty}\left(\frac{x^3}{2}\right)^{n}dx=\int_0^1\frac{2}{2-x^3}\:dx
$$
Hope you can take it from here.
| {
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Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ I need help with the following question:
Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$
My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^2+18x+95\equiv 3x^2+7x+7\equiv 0\pmod {11} \\ 3x^2+18x+95 \equiv 3x^2+5x+4\equiv 0\pmod {13}$$
I don't know how to solve those equations and I don't know how to combine it to the big solution for the real question (I know about the CRT, but I didn't realy understood how to use it, I'd love help with this).
thanks in advance
| Let's take one equation
\begin{align*}
3x^2+7x+7 & \equiv 0 \pmod{11}\\
4(3x^2+7x+7) & \equiv 4(0) \pmod{11}\\
x^2+28x+28 & \equiv 0 \pmod{11}&& (\because 4(3) \equiv 1 \pmod{11})\\
x^2+6x+6 & \equiv 0 \pmod{11}&& (\because 28 \equiv 6 \pmod{11})\\
(x+3)^2-3 & \equiv 0 \pmod{11}\\
(x+3)^2-5^2 & \equiv 0 \pmod{11}&& (\because 5^2 \equiv 3 \pmod{11})\\
(x-2)(x+8) & \equiv 0 \pmod{11}
\end{align*}
Since $11$ is prime so if $11 | ab$, then $11$ divides at least one of them, so we get
$$x\equiv 2 \pmod{11} \quad \text{ or } \quad x \equiv -8 \equiv 3\pmod{11}.$$
Likewise (you can work this out yourself)
$$3x^2+5x+4 \equiv 0 \pmod{13} \implies x\equiv 2 \pmod{13} \, \text{ or } \, x \equiv \color{blue}{b}\pmod{13}. $$
So we have the following situation
\begin{align*}
x&\equiv 2 \pmod{11} & x&\equiv 2 \pmod{11} & x&\equiv 3 \pmod{11} & x&\equiv 3 \pmod{11}\\
x&\equiv 2 \pmod{13} & x&\equiv \color{blue}{b} \pmod{13} & x&\equiv \color{blue}{b} \pmod{13} & x&\equiv 2 \pmod{13}
\end{align*}
Now use CRT (hopefully you know how to apply it to simple systems like these) to solve these systems.
For example the last system
\begin{align*}
x & \equiv 3 \pmod{11}\\
x & \equiv 2 \pmod{13}
\end{align*}
yields
$$x \equiv 3(13)(6)+2(11)(6) \equiv \color{red}{80} \pmod{143}. $$
Likewise you will get a total of $\color{red}{4}$ incongruent solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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INMO : Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers Let $a$ and $b$ be two positive rational numbers such that $\sqrt[3] {a} + \sqrt[3]{b}$ is also a rational number. Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers.
My first response was, isn't is obvious? but then I tried..
if $a=b$ then we prove it easily . So suppose $a\ne b$.
Let $s=\sqrt[3]{a} + \sqrt[3]{b}$.
then $s^3 = a + b + 3 s \sqrt[3]{a} \sqrt[3]{b}$
then the product $p=\sqrt[3]{a} \sqrt[3]{b}$ is a rational number.
But what should I do next ?
| we have $\sqrt[3]{a}+\sqrt[3]{b}$ is a rational number
and so is their product.(which you found)
Now, squaring $\sqrt[3]{a}+\sqrt[3]{b}$ we can get,
$\sqrt[2/3]{b}+\sqrt[2/3]{a}$ is also rational(using the fact that $\sqrt[3]{ab} \in \mathbb Q$
Similarly,squaring $\sqrt[2/3]{b}+\sqrt[2/3]{a}$
we can get, $\sqrt[4/3]{b}+\sqrt[4/3]{a}=a*\sqrt[3]{a}+b*\sqrt[3]{b} \in \mathbb Q$
Now,multiply $\sqrt[3]{a}+\sqrt[3]{b}$ by a and subtract it from $a*\sqrt[3]{a}+b*\sqrt[3]{b}$
to get the desired result.
| {
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"url": "https://math.stackexchange.com/questions/3775351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve : $2^{x}+2^{y}=12 ; x+y=5, x,y\in \mathbb{R}$ How to solve $2^{x}+2^{y}=12 ; x+y=5, x,y\in \mathbb{R}$?
Attempt:
$2^{x}+2^{y}=12\implies 2^{5-y}+2^y=12\implies 2^y=4,8\implies y=2,3$
| $$2^x+2^y=2^{5-y}+2^y=\frac{2^5}{2^y}+2^y=12 $$
$$2^5+2^{2y}=12\cdot2^y \Longleftrightarrow 2^{2y}-12\cdot2^y+2^5=0 $$
$$2^{2y}-12\cdot2^y+2^5=(2^y-2^3)(2^y-2^2) = 0$$
$$2^y=2^3\Longleftrightarrow y=3 $$
$$2^y=2^2 \Longleftrightarrow y=2 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove distance from foci on an ellipse is equal to twice the semi-major axis (for specific ellipse) Prove that for any point (x,y) on the conic, the sum of the distances to the two foci is always twice the semi-major axis.
I know that this can be proven in general for all ellipses but the practice question specifically asks for this to be proven for $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1. I feel like I'm really close but I've managed to math myself into a corner somehow.
Let the foci ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0) be denoted as F and F'. Let the point on the conic be denoted P(x,y). We are required to show PF + PF' = 2a. In this case, since a = 3, 2a = 6.
PF = $\sqrt{(x-\sqrt{5})^2 + y^2}$ and PF' = $\sqrt{(x+\sqrt{5})^2 + y^2}$
By rearranging the equation for the ellipse, we get y$^2$ = 4 - $\frac{4}{9}$x$^2$.
Substitute this into PF and PF' to get:
PF = $\sqrt{(x-\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 - 2\sqrt{5}x + 9}$ = $\sqrt{(x - \frac{9\sqrt{5}}{5})^2}$ = x - $\frac{9\sqrt{5}}{5}$
PF' = $\sqrt{(x+\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ = x + $\frac{9\sqrt{5}}{5}$
Therefore PF + PF' = 2x
And then I got stuck
| For such a question, the realm of complex numbers helps a lot.
In the complex plane, an ellipse is described by :
$$|z-a| + |z-b|=c$$
where z represents each point on the ellipse, (a,b) is a tuple of complex numbers (which may or may not be real), c is an arbitrary constant $\geq |a-b|$
Also note, (a,b) are the focii of the ellipse as well and c is the major axis.
Coming on to the question, just describe the given equation in the form of complex numbers, the result it direct.
So, the given ellipse is $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1 and
the focii ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0).
Describe this as
$$|z-\sqrt{5}| + |z-\sqrt{-5}|=6$$
This directly means that the sum of distance of any point $z$ on the ellipse is equal to the major axis.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why am I getting derivative of $y = 1/x$ function as $0$? I was finding the derivative of the function: $y = 1/x$. I did the followed steps:
\begin{align*}
\frac{\frac{1}{x+dx} - \frac{1}{x}}{dx} &=\left(\frac{1}{x+dx} - \frac{1}{x} \right) \frac{1}{dx} \\
&= \frac{1}{x dx + (dx)^2} - \frac{1}{x dx}.
\end{align*}
Since, $(dx)^2$ would be extremely small, I removed it, so
$$\frac{1}{x dx} - \frac{1}{x dx}$$
which is equal to zero.
why am I getting the derivative of $y = 1/x$ as $0$?
| $$\frac{\,d}{\,dx} \frac{1}{x}=\lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}= \lim_{h\rightarrow 0}\frac{x-(x+h)}{hx(x+h)}=\lim_{h\rightarrow 0} -\frac{1}{x(x+h)}=-\frac{1}{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$.
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .
What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?
| Here is the second thing I thought of (@iam_agf has the other). Add $1$ to all the terms and set $a+b+c+d=E$ and you get$$\frac Ea=\frac Eb=\frac Ec=\frac Ed=r+1$$ from which either $E=0$ or $a=b=c=d$
In the first case $r=-1$ and in the second $r=3$
Just putting $a+b+c+d=E$, which should be an early thing to try, also gets to the same place.
| {
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"timestamp": "2023-03-29T00:00:00",
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sum of this series: $\sum_{n=1}^{\infty}(-1)^{n-1}(\frac{1}{4n-3}+\frac{1}{4n-1})$ $$\sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{1}{4n-3}+\frac{1}{4n-1}\right)$$
What I did
$$\sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{1}{4n-3}+\frac{1}{4n-1}\right)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{4n-3}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{4n-1} $$
$$=\sum_{n=1}^{\infty} (-1)^{n-1}\int_0^{x}t^{4n-4}dt+ \sum_{n=1}^{\infty} (-1)^{n-1}\int_0^{x}t^{4n-2}\mathrm{d}t $$
$$=\int_0^{x}\frac{1+t^{2}}{1+t^4}dt$$
$$=\frac{1}{\sqrt{2}}\arctan{\frac{x-1/x}{\sqrt{2}}}+\frac{\pi}{2\sqrt{2}}$$
According to the Abel’s limit theorem ,the sum is $\frac{\pi}{2\sqrt{2}}$,but the answer is $-\frac{(\sqrt{2}-1) \pi}{2 (\sqrt{2}- 2)}$.
| A different approach
Apply the linearity (since the sums are convergent by alternating series test) $$\sum_{n\geq 1} (-1)^{n-1}\left(\frac{1}{4n-3} +\frac{1}{4n-1}\right)=\sum_{n\geq 1} \frac{(-1)^{n-1}}{4n-3}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{4n-1}=S_1+S_2$$ We part sum into even and odd parity. That is $$S_1 = \sum_{n\geq 1}\left(\frac{(-1)^{2n-1}}{4(2n)-3}+\frac{(-1)^{2n-1-1}}{4(2n-1)-3}\right)=\sum_{n\geq 1} \left(\frac{1}{8n-7}-\frac{1}{8n-3}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{8n+1}-\frac{1}{8n+5}\right)=\color{red}{\frac{1}{8}\left(\psi_0\left(\frac{5}{8}\right)-\psi_0\left(\frac{1}{8}\right)\right)}$$ Similarly, $$S_2=\sum_{n\geq 1}\left(\frac{1}{8n-5}-\frac{1}{8n-1}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{8n+3}-\frac{1}{8n+7}\right)=\frac{1}{8}\left(\psi_0\left(\frac{7}{8}\right)-\psi_0\left(\frac{3}{8}\right)\right)$$ There we used the classical result $$\sum_{n=0}^{\infty}\left(\frac{1}{an+b}-\frac{1}{an+d}\right)=\color{red}{\frac{1}{a}\left(\psi_0\left(\frac{d}{a}\right)-\psi_0\left(\frac{b}{a}\right)\right)},\; \; a,b,d>0$$ Adding $S_1$ and $S_2$ and using the reflection formula we have $$S_1+S_2=\frac{\pi}{8}\left(\cot\frac{3\pi}{8}+\cot\frac{\pi}{8}\right)=\frac{\pi}{8}\left(\sqrt 2- 1+\sqrt 2+1\right)=\frac{\pi}{2\sqrt 2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2}$ into $\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab}$ I got two algebraic expression, the first should be simplified, with some manipulation I think, to be the second. It's similar to binomial expansion, even though coefficients are a mess.
$$\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2} \tag 1$$
$$\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab} \tag 2$$
It's not exactly a problem related to my studies, but something I'm interested in. I remember something similar, when we add/subtract some terms till we got full expansion or so, but I couldn't do anything. Actually, I'm not sure if this problem is correct, so if there's a proof for this problem as invalid one, it's welcome.
| The first thing I notice is that $b^4 + a^4 -2a^2b^2$ is $(a^2 - b^2)^2$ and $a^2 + b^2 + 2ab= (a+b)^2$.
I can factor $(a^2-b^2)$ as $(a+b)(a-b)$ so
$(a^2 -b^2)^2 = (a-b)^2(a+b)^2$.
And if we have two fractions equal to each other:
$\frac {????????}{(a-b)^2(a+b)^2} = \frac {!!!!!!!!!}{(a+b)^2}$ that has to mean
$\frac {?????????}{(a-b)^2(a+b)^2} = \frac {!!!!!!!!!}{(a+b)^2}= \frac{!!!!!!!!!\cdot (a-b)^2}{(a-b)^2(a+b)^2}$
So if these are equal we will have to be able to factor $(a-b)^2$ out the first numerator.
So let's try to do that:
$\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2}= $
$\frac {(b^5 - a^5)-5(b^3a^2-b^2a^3)}{(b^2 - a^2)^2} = $
$\frac {(b-a)(b^4+b^3a + b^2 a^2+ ba^3 + a^4) - 5b^2a^2(b-a)}{(b-a)^2(b+a)^2}=$ (cancel one of the $(b-a)$ is:
$\frac {b^4 +b^3a + b^2a^2 + ba^3 + a^4 - 5b^2a^2}{(b-a)(b+a)^2} =$
$\frac {b^4 + b^3a -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$ ... now we were told that the fractions were equal so that means $b-a$ factors out the numerator even if it isn't obvious.
$\frac {b^3(b-a)+b^3a + b^3a -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^3a -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)+2b^2a^2 -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)-2ba^3 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)- ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)- a^3(b-a)-a^4 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)- a^3(b-a)}{(b-a)(b+a)^2}=$
$\frac {b^3 + 2b^2a -2ba^2 - a^3}{(b+a)^2}=$ and that's it, we can rearrange the the terms
$\frac {b^3-a^3-2ba^2 +2b^2a}{b^2 + a^2 + 2ab}$
| {
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How can I study the convergence of the improper integral $\int_{2}^{\infty} \frac{\arctan(x+1)+x}{2^x+3^x}\, \mathrm dx$? I need to study the convergence of the following improper integral:
$$\int_{2}^{\infty} \dfrac{\arctan(x+1)+x}{2^x+3^x}\, \mathrm dx$$
I did the following:
$$ -\dfrac{\pi}{2} < \arctan(x+1) < \dfrac{\pi}{2} \\
\implies -\dfrac{\pi}{2} + x < \arctan(x+1) +x < \dfrac{\pi}{2} +x \\
\implies \dfrac{-\dfrac{\pi}{2} + x}{2^x+3^x} < \dfrac{\arctan(x+1) +x}{2^x+3^x} < \dfrac{\dfrac{\pi}{2} + x}{2^x+3^x} \\
$$
I planned to integrate the inequality and then using the comparison criterion to proof its convergence. However, the idea did not work for me.
| Just observe that RHS is less than $$\frac{\frac{\pi}{2}+x}{3^x}$$Since $$\int_2^\infty\frac{1}{3^x} \,dx$$clearly converges, it is enough to show that$$\int_2^\infty\frac{x}{3^x} \,dx$$ converges which can be done using limit comparison test (with $g(x)=\frac{1}{e^x}$).
| {
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Two methods are giving two different answers to the this differential equation : $\frac{dy}{dx}=\frac{1}{2} \frac{d(\sin ^{-1}(f(x))}{dx}$
$f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,\ |x|>1$
If $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$ and
$y(\sqrt{3})=\frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to :
Options:
$1. \quad-\frac{\pi}{6}\\
2. \qquad \frac{2 \pi}{3}\\
3. \qquad \frac{5 \pi}{6}\\
4. \qquad\frac{\pi}{3}$
Now I am getting two Answers in two methods. Can anyone tell me which method is wrong and why?
Method - 1 $f(x) = [\sin(\tan^{-1}x) + \sin(\cot^{-1}x) ]^2 -1 $.
Let $\tan^{-1}x = \theta $. So $f(x) = [\sin(\theta) + \sin(\frac{\pi}{2} - \theta) ]^2 -1 = \sin 2\theta$.
So $f(x) = \sin 2\theta = \frac{2\tan \theta}{1+ \tan^2 \theta} = \frac{2x}{1+x^2}\tag 1$.
Now $\frac{d}{dx} \sin^{-1} \frac{2x}{1+x^2} = \frac{2(1-x^2)}{\sqrt{(1-x^2)^2}(1+x^2)} = \frac{2(1-x^2)}{(1+x^2)(x^2 -1)}$ [Since $|x| > 1$].
Now $\frac{dy}{dx} = \frac{1}{2}\frac{d(\sin^{-1}f(x)}{dx}$ . So $\frac{dy}{dx} = \frac{-1}{1+x^2}$. So $y= - \tan^{-1} x + C$. Now as $y(\sqrt 3) = \frac{\pi}{6}$ , $C = \frac{\pi }{2}$.
So $y = -\tan^{-1} x + \frac{\pi}{2}$.
So $ \displaystyle y(-\sqrt 3) = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}$
Method -2 - $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$. So $y = \frac{\sin^{-1}f(x)}{2 } + C$.
Now $f(x) = \sin (2\tan^{-1} x)$.
SO $\displaystyle y = \frac{\sin^{-1}(\sin (2\tan^{-1} x))}{2 } + C$ . Now as $y(\sqrt 3) = \frac{\pi}{6}$ , $C = 0$.
So $\displaystyle y = \frac{\sin^{-1}(\sin (2\tan^{-1} x))}{2 } $ .
So $ \displaystyle y(-\sqrt 3) = \frac{-\pi}{6}$
I am really confused. Why I am getting two answers? Can anyone please help me out?
| There is no way to predict the value of $y(-\sqrt{3})$ by knowing the solution over $(1,\infty)$.
The problem is ill posed. It is like asking for the value of $y(-1)$ if $y'=1/x$ for $x\ne0$, with $y(1)=0$. Any value for $y(-1)$ can be chosen.
Indeed, there is no reason for the constant of integration to be the same over $(0,\infty)$ and $(-\infty,0)$ in this case or over $(1,\infty)$ and $(-\infty,-1)$ in your case.
By the way, we have
$$
f(x)=\begin{cases}
\frac{\pi}{2}-\arctan x & x>1 \\[6px]
-\frac{\pi}{2}-\arctan x & x<-1
\end{cases}
$$
as witnessed by https://www.desmos.com/calculator/6akvm0e78h
Therefore the differential equation is
$$
y'=-\frac{1}{1+x^2}
$$
and so
$$
y=\begin{cases}
a-\arctan x & x>1 \\[6px]
b-\arctan x & x<-1
\end{cases}
$$
You can determine $a$ by plugging in $a-\arctan\sqrt{3}=\pi/6$, so $a=\pi/2$. However, this does not impose any condition on $b$.
If the instructor wants you to use $b=a$, then the answer would be
$$
\frac{\pi}{2}-\arctan(-\sqrt{3})=\frac{\pi}{2}+\frac{\pi}{3}=\frac{5\pi}{6}
$$
but there is no mathematical justification for this and the instructor is wrong.
| {
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Evaluate $\operatorname{lim}_{n\to\infty} \int_0^\infty \frac{n^2x}{1+x^2} e^{-n^2x^2} dx$ Evaluate the following limit for $a=0$ and for $a>0$
$$\lim_{n\to\infty} \int_a^\infty \frac{n^2x}{1+x^2} e^{-n^2x^2} dx.$$
For $a>0$, I can use dominated convergence theorem as $\frac{n^2x}{1+x^2} e^{-n^2x^2}\leq \frac{1}{x(1+x^2)}\in L^1(a,\infty)$, then the limit can be put into the integral to get the limit as $0$. But I'm not sure how to approach the case $a=0$. Can you please give me some hint?
| Just for your curiosity.
$$\frac{n^2x}{1+x^2} e^{-n^2x^2}=e^{n^2}n^2 x\frac{e^{-n^2(1+x^2)}} {1+x^2 }$$
$$I=\int\frac{n^2x}{1+x^2} e^{-n^2x^2}\,dx=\frac 12e^{n^2}n^2\int \frac{e^{-n^2(1+x^2)}} {1+x^2 }\,d(1+x^2)$$ I suppose that the change of variable $t=1+x^2$ is clear and it makes
$$I=\frac{1}{2} e^{n^2} n^2 \,\text{Ei}\left(-n^2 \left(1+x^2\right)\right)$$ where appears the exponential integral function.
$$J=\int_0^\infty\frac{n^2x}{1+x^2} e^{-n^2x^2}\,dx=\frac{1}{2} e^{n^2} n^2\, \Gamma \left(0,n^2\right)$$ the expansion of which being
$$J=\frac{1}{2}-\frac{1}{2 n^2}+O\left(\frac{1}{n^4}\right)$$
| {
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
| Note that $\sqrt{-1}=i$$$x^2-8x+17=0 \implies x=\frac{8\pm \sqrt{64-68}}{2}=\frac{8+\sqrt{-4}}{2}=4\pm i$$
So (A) option is correct.
| {
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If $\text{ }\big(x-\frac{1}x\big)=i\sqrt{2}$. Then compute $\bigg(x^{2187}-\frac{1}{x^{2187}}\bigg)$. Here $i=\sqrt{-1}$
QUESTION: If $\text{ }\big(x-\frac{1}x\big)=i\sqrt{2}$ , $\text{ }$then compute $$\bigg(x^{2187}-\frac{1}{x^{2187}}\bigg)$$ Here $i=\sqrt{-1}$ .
MY ANSWER: I have done it using the Quadratic formula and De Moivre's Theorem. Let me write down my working before I propose my doubt.. Here's how I did it..
Solving the equation we get $$x^2-(i\sqrt{2})x-1=0$$ $$\implies x=\frac{i\sqrt{2} \pm \sqrt{(i\sqrt{2})^2+4}}{2} $$ $$\implies x=\frac{i\sqrt{2}\pm\sqrt{2}}{2}$$ Take $x=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=e^{\frac{i\pi}4}$
Now we know that $2187=(273\times8)+3$
$$\therefore x^{2187}=e^{2187\times \frac{i\pi}4}=e^{(273\times 2\pi + \frac{3\pi}4)i}=e^{\frac{{3\pi}}{4}i}=\frac{i-1}{\sqrt{2}}$$
$$\therefore x^{2187}-\frac{1}{x^{2187}}= \frac{i-1}{\sqrt{2}}-\frac{\sqrt{2}}{i-1}$$ $$=\frac{(i-1)^2-2}{(i-1)\sqrt{2}}$$ $$=\frac{2}{\sqrt{2}}\frac{(1+i)}{(1-i)}$$ $$=\frac{\sqrt{2}}{2} (1+i)^2$$ $$=\boxed{\sqrt{2}i}$$
Now my first question is that, the quadratic relation gave us two different values for $x$. One with which I have worked out to reach the answer of $\sqrt {2}i$ and the other, $\big(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\big)$ which I had left behind. Now working with that I find that the angle turns out to be $\frac{\pi}{10}$ and stuff becomes much more complicated after that. The official answer to this one is $\sqrt{2}i$ (which matches with what I have found out).
My doubt is why do we not consider the other value of $x$ ?
And is there any alternative (preferably simpler) method(s) to solve this one?
Thank You so much for your help and support.. :)
| In fact, it is easy to verify that both values of $x$ yield the same result. For the whole problem, you just need De Moivre's formula twice (two lines of paper without explanation).
For $x=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$, you have shown the answer is $i\sqrt 2$.
Now, let $x=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i=\cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4}$. Using the De Moivre's formula and the fact that
$$z-\frac{1}{z}=2i\sin(\arg(z))$$
you get
$$x^{2187}-\frac{1}{x^{2187}} = x^3-\frac{1}{x^3} = 2i\sin\frac{9\pi}{4}=i\sqrt 2$$ Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Where does the integration end? I'm new to integrals. I'm solving $$ \int \frac{1}{2x^2+6}$$ but I get a wrong answer: $$ \frac{1}{6}\arctan\frac{x}{\sqrt3}$$
The correct answer should be: $$ \frac{\sqrt3}{6}\arctan\frac{x}{\sqrt3}$$
Here is my full try:
$$ \int \frac{1}{6(\frac{2x^2}{6}+1)} = \int \frac{1}{6(1+(\frac{x}{\sqrt3})^2)} = \frac{1}{6}\arctan\frac{x}{\sqrt3}$$
Can you correct me and give me some source to learn from?
Thanks in advance!
| $$\int \frac{1}{2x^2+6}dx = \frac{1}{2}\int \frac{1}{x^2+3}dx$$
$$x = \sqrt{3}\tan{\theta}\Rightarrow dx = \sqrt{3}\sec^2{\theta}d\theta$$
Plugging our substitution back into the integral yields
$$\frac{\sqrt{3}}{2}\int \frac{\sec^2{\theta}}{3\tan^2{\theta}+3}d\theta = \frac{\sqrt{3}}{6}\int \frac{\sec^2{\theta}}{\sec^2{\theta}}d\theta$$
So we are now left with
$$\frac{\sqrt3}{6}\theta +c$$
Since this is an indefinite integral, we have to write our answer in terms of x. Looking back at our substitution and rearranging for theta, we get to our final answer:
$$\frac{\sqrt3}{6}\tan^{-1}(\frac{x}{\sqrt{3}})+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right| $
How do I find the value of the following expression?
$$
\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right|
$$
I tried writing the numerator as $\tan 40^\circ - \tan80^\circ -\tan20^\circ,$ but then the expression was getting complicated.
| First of all we have (see Morrie's law)
$$
\tan20^\circ\tan40^\circ\tan80^\circ=\sqrt{3}.
$$
The numerator is, setting $x=20^\circ,$
\begin{align}
&\tan40^\circ+\tan100^\circ+\tan160^\circ=\\
&\qquad\qquad=\tan(60^\circ-x)+\tan(120^\circ-x)+\tan(180^\circ-x)=\\
&\qquad\qquad=
\frac{\tan 60^\circ-\tan x}{1+\tan 60^\circ\tan x}+
\frac{\tan120^\circ-\tan x}{1+\tan120^\circ\tan x}+
\frac{\tan180^\circ-\tan x}{1+\tan180^\circ\tan x}=\\
&\qquad\qquad=
\frac{ \sqrt{3}-\tan x}{1+\sqrt{3}\tan x}+
\frac{-\sqrt{3}-\tan x}{1-\sqrt{3}\tan x}-
\tan x=\\
&\qquad\qquad=
\frac{\sqrt{3}\cos x-\sin x}{\cos x+\sqrt{3}\sin x}-
\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}-
\frac{\sin x}{\cos x}=\\
&\qquad\qquad=
-3\cdot\frac{3\sin x\cos^2 x-\sin^3 x}{\cos^3 x-3\sin^2 x\cos x}=\\
&\qquad\qquad=-3\cdot\frac{\sin(3x)}{\cos(3x)}=-3\tan60^\circ=-3\sqrt{3}
\end{align}
So the final result is
$$
\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right|=\left|\frac{-3\sqrt{3}}{\sqrt{3}}\right|=3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find the maximum value of $x^2y$ given constraints Find the maximum value of $${ x }^{ 2 }y$$ subject to the constraint $$x+y+\sqrt { 2{ x }^{ 2 }+2xy+3{ y }^{ 2 } } =k$$ where k is a constant.
I tried it by substituting value of x and then differentiating w.r.t $x$ but not able to proceed further.
| $$g(x,y)=x+y+\sqrt { 2{ x }^{ 2 }+2xy+3{ y }^{ 2 } }= x+y + Q(x,y) = k \tag1$$
for the object function
$$f(x,y)=x^2y $$
By Lagrange multiplier method
$$\dfrac{g_x}{g_y}= \dfrac{f_x}{f_y} $$
$$\dfrac{2y}{x}=1+\dfrac{(4x+2y)}{2Q}=1+\dfrac{(2x+6y)}{2Q}$$
simplifying
$$Q(2y-x)= 2x^2-xy-6y^2$$
has a common factor $ (2y-x)$ to cancel
$$=(x-2y)(2x+3y)\rightarrow Q=(2x+3y)$$
Squaring
$$ Q^2=4x^2+12 x y+9y^2=2x^2+10 xy+6y^2$$
simplifying to find roots of quadratic
$$x^2+5 x y+3 y^2=0;\quad \dfrac{y}{x}=\dfrac{-5\pm \sqrt{13}}{6}; $$
which are a pair of straight lines.
If $(p,q)$ are these roots say $ y=px,\;y=qx\;$ Plug in the first of two roots into (1)
$$ k = x( 1+p+ \sqrt{2+2p+3p^2}) =C x\;$$ say, then the maximum value is
$$x^2y=x^3\cdot\dfrac{y}{x}=p k^3/C^3=\dfrac{pk^3}{( 1+p+ \sqrt{2+2p+3p^2})^3}$$
where
$$p=\dfrac{\sqrt{13}-5}{6}$$
The minimum value can be found by $y=qx$ in a similar fashion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given two real numbers $x,y$ so that $x^{2}+y^{2}+xy+4=4y+3x$. Prove that $3\left(x^{3}-y^{3}\right)+20x^{2}+2xy+5y^{2}+39x\leq 100$.
Given two real numbers $x, y$ so that $x^{2}+ y^{2}+ xy+ 4= 4y+ 3x$. Prove that
$$3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\leq 100$$
I used derivative and Wolfram|Alpha but only the minimum value found
$$\min\{3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\}\Leftrightarrow \left\{\begin{matrix} x\cong 0.0320241\\ y\cong 2.16078\\ \left ( z\cong 2.19235 \right ) \end{matrix}\right.$$
where
*
*$z$ is a root of $472z^{3}- 449z^{2}- 689z- 1305= 0$,
*$y$ is a root of $27468y- 11800z^{2}+ 17833z- 41733= 0$,
*$x$ is a root of $27468x+ 3304z^{2}- 11167z+ 7722= 0$.
Why is unsuccessful ? Here are two examples of my claim.
| From the condition we obtain:
$$y^2+(x-4)y+x^2-3x+4=0,$$ which gives $$(x-4)^2-4(x^2-3x+4)\geq0$$ or $$0\leq x\leq\frac{4}{3}.$$
By the similar way we obtain: $$1\leq y\leq\frac{7}{3},$$ which gives:
$$3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\leq$$
$$\leq\frac{64}{9}-3y^3+\frac{320}{9}+\frac{8}{3}y+5y^2+52.$$
Thus, it's enough to prove that:
$$\frac{64}{9}-3y^3+\frac{320}{9}+\frac{8}{3}y+5y^2+52\leq100$$ or
$$9y^3-15y^2-8y+16\geq0$$ or
$$9y^3-24y^2+16y+9y^2-24y+16\geq0$$ or
$$(3y-4)^2(y+1)\geq0$$ and we are done!
The equality occurs for $x=y=\frac{4}{3},$ which says that $100$ is a maximal such value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Fourier transform of Hermite polynomial via generating function I've spent a lot of times trying to show that
$$
\mathcal{F}[e^{-x^2/2} G(x,t)] = e^{-k^2/2} G(k, -it)
$$
with $G(x,t)$ being the generating function of Hermite polynomial,
$$
G(x,t) = e^{2tx - t^2}
$$
My attempt
We want to show that
\begin{align*}
\mathcal{F}[e^{-x^2/2} G(x,t)]
&= e^{-k^2/2} G(k, -it) \\
&= e^{-k^2/2} e^{-2itk+t^2} \\
&= e^{-\frac{k^2}{2} - 2itk + t^2}
\end{align*}
Computing the Fourier transform:
\begin{align*}
\mathcal{F}[e^{-x^2/2} G(x,t)]
&= \mathcal{F}[e^{-x^2/2} e^{2tx - t^2}] \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/2} e^{2tx - t^2} e^{-ikx} dx
\end{align*}
We know that
\begin{align*}
\sqrt{2\pi}
&= \sqrt{\frac{\pi}{1/2}} \\
&= \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2} dx
\end{align*}
I've been trying to factor the exponential isolating something like $-\frac{1}{2}x^2$.
In fact I should get something like $e^{-\frac{1}{2}x^2} e^{-\frac{k^2}{2} + 2itk + t^2}$ in the integrant so I can get rid of the $\frac{1}{\sqrt{2\pi}}$ and get the desired result but I can't do it.
| We can use the following relation,
\begin{equation*}
\int_{-\infty}^{\infty} e^{-ax^2 + 2bx} dx = \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{a}}
\end{equation*}
Therefore,
\begin{align*}
\mathcal{F}[e^{-x^2/2} G(x,t)]
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx - t^2} e^{-ikx} dx \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} e^{-t^2} dx \\
&= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} dx
\end{align*}
We can rewrite the exponential argument as a 2nd degree polynomial tor the exponential to fit the above relation,
\begin{align*}
-\frac{x^2}{2} + 2tx - ikx
&= -\frac{1}{2}x^2 + (2t - ik)x \\
&= -\frac{1}{2}x^2 + 2(t - \frac{ik}{2})x
\end{align*}
We have $a = \frac{1}{2}$ et $b = t-\frac{ik}{2}$.
Therefore,
\begin{align*}
\mathcal{F}[e^{-x^2/2} G(x,t)]
&= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} dx \\
&= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2 + 2(t - \frac{ik}{2})x} dx \\
&= e^{-t^2} \frac{1}{\sqrt{2\pi}} \sqrt{2\pi} e^{2(t-\frac{ik}{2})^2} \\
&= e^{-t^2} e^{2(t^2 - 2\frac{ikt}{2} - \frac{k^2}{4})} \\
&= e^{-t^2} e^{2t^2} e^{-2ikt} e^{-\frac{k^2}{2}} \\
&= e^{-\frac{k^2}{2}} e^{-2ikt} e^{t^2} \\
&= e^{-\frac{k^2}{2}} G(k, -it)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Distributing $5$ different balls to $4$ different persons We have to find the number of ways of distributing $5$ different balls to $4$ different persons.
Clearly, the answer is $4^5$ as each ball can be given to any of the $4$ persons. However, I wanted to calculate it using a different method.
I assumed that $a$ balls are given to first person, $b$ to second , $c$ to third and $d$ to fourth person. So we have that $a+b+c+d=5$ where $0 \leq a,b,c,d \leq 5$.
But counting the solutions to the above equation assumes that balls are identical. So I tried to find the number of distributions each permutation of $(a,b,c,d)$ produces. That will be $\displaystyle \binom{5}{a}\cdot \binom{5-a}{b}\cdot \binom{5-a-b}{c} $ which equals $\dfrac{5!}{a!b!c!d!}$.
So, now we need to sum this value over all $a,b,c,d$ satisfying $a+b+c+d=5$. Now there are $\displaystyle \binom{8}{3}=56$ solutions to the equation. So there will be $56$ terms in that summation. So how do we do that?
| Work backwards:
Suppose you already gave out $a$ balls to the first person and $b$ balls to the second. You now have $\binom{5-a-b}{c}$ ways of giving $c$ balls to the third person and the remaining balls to the fourth person. Summing this up for all possible value of $c$ gives: $\binom{5-a-b}{0}+\binom{5-a-b}{1}+...+\binom{5-a-b}{5-a-b}=2^{5-a-b}$.
Suppose you gave out $a$ balls to the first person. You now have $\binom{5-a}{b}$ ways of giving $b$ balls to the second person, and for each of those you are left with $2^{5-a-b}$ ways of giving out balls to the remaining two people. Summing this up for all values of $b$ gives:
$\binom{5-a}{0}2^{5-a}+\binom{5-a}{1}2^{4-a}+...+\binom{5-a}{5-1}2^{0}=3^{5-a}$.
Lastly, summing for $a$, you have $\binom{5}{0}3^5+\binom{5}{1}3^4+\binom{5}{2}3^3+\binom{5}{3}3^2+\binom{5}{4}3^1+\binom{5}{5}3^0=4^5$.
This uses the identity:
$$\sum_{k=0}^X\binom{X}{k}Y^k=\sum_{k=0}^X\binom{X}{X-k}Y^k=(Y+1)^X$$
This identity comes from the fact that this sum is what you get when you expand the binomial $(1+Y)^X$
| {
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"url": "https://math.stackexchange.com/questions/3805292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\sum_{cyc} \sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\ge 2+\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ For $a,b,c\geq 0$, no two of which are $0$, prove that:
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\dfrac{a^2+b^2+c^2}{ab+bc+ca}}$$
This inequality actually came up as an accident when I tried to combine 2 known results, and after many testings on computer it still remains true, but there's still no original proof yet. Hope everyone enjoy and have some good ideas for it.
Here's that 2 known results:
$$\dfrac{a^2+b^2+c^2}{ab+bc+ca}\geq \prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)$$
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)}$$
The second one can be proved by direct Karamata's inequality, but it may also inspire some ideas for the original one too.
See the following links:
https://artofproblemsolving.com/community/u410204h2218857p16854913
https://artofproblemsolving.com/community/c6h487722p5781880
https://artofproblemsolving.com/community/u414514h2240506p17302184
| It's just comment.
I think, this inequality is very interesting.
The following way does not help.
By using the Ji Chen's lemma: https://artofproblemsolving.com/community/c6h194103
it's enough to prove three inequalities:
*
*$$\sum_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\geq2+\frac{a^2+b^2+c^2}{ab+ac+bc}$$
2.$$\sum_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\left(\frac{a}{b+c}+\frac{c}{a+b}\right)\geq1+\frac{2(a^2+b^2+c^2)}{ab+ac+bc}$$ and
3.$$\prod_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\geq\frac{a^2+b^2+c^2}{ab+ac+bc}.$$
The first it's just $$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)\geq0,$$ which is true by Muirhed.
The second is true by Muirhead again:
$$\sum_{sym}\left(a^7b-a^5b^2+a^5b^2c-a^4b^3c+\frac{1}{2}a^6bc-\frac{1}{2}a^3b^3c^2\right)\geq0,$$ but the third is wrong!
It's equivalent to:
$$-abc\sum_{sym}(a^4b-a^3b^2)\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$ .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a \ge 3\sum a$$
or $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
| There are many SOS!
My SOS, first is same as Mr. Mike$:$
\begin{align*} \sum \frac{a^3}{bc} -\sum a &=\dfrac{1}{2} \sum {\dfrac { \left( {a}^{2}+ab+ac+{b}^{2}+bc \right) \left( a-b
\right) ^{2}}{bca}}\\&=\dfrac{1}{4}\sum {\frac { \left( 3\,a+4\,b \right) \left( a-b \right) ^{2}}{bc}}+
\dfrac{1}{4}\sum{\frac {a \left( a+b-2\,c \right) ^{2}}{bc}}\\&
=\sum{\dfrac { \left( a+b \right) \left( a-b \right) ^{2}}{ab}}+\dfrac{1}{6}\sum{
\dfrac { \left( 2\,{a}^{2}-{b}^{2}-{c}^{2}+2\,bc-ab-ac \right) ^{2}}{bc
a}}
\\&=\dfrac{1}{7}\sum {\dfrac { \left( a-b \right) ^{2} \left( 5\,a+8\,c \right) }{ac}}+ \dfrac{2}{7}\sum {\dfrac { \left( {a}^{2}-2\,ab+bc \right) ^{2}}{bca}}+\\&\quad
+{\frac {5}{ 42}}\sum{\frac { \left( 2\,{a}^{2}-{b}^{2}-{c}^{2}+2\,bc-ab-ac \right) ^ {2}}{bca}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the limit $\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)$
Find the limit $$\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin
x}\right)$$
\begin{align}
\lim_{x\to 0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)&=\lim_{x\to 0}\left(\frac{\sin x-\arcsin x}{\sin x\times \arcsin x}\right)\\
&=\lim_{x\to 0}\frac{x}{\arcsin x}
\cdot\frac{x}{\sin x}\cdot \frac{\sin x-\arcsin x}{x^2}
\end{align}
Here's where I'm stuck. I know from L'Hopital's Rule that $\frac{x}{\sin x}\to 1$, but what about $\frac{x}{\arcsin x}$? Or is there any other way than using L'Hopital to solve this question?
| $$\frac{1}{\arcsin{x}}-\frac{1}{\sin{x}}=\frac{1}{x+\frac{x^3}{6}+...}-\frac{1}{x-\frac{x^3}{6}+...}\rightarrow0.$$
Your way also helps because $$\frac{\sin{x}-\arcsin{x}}{x^2}\rightarrow\frac{\cos{x}-\frac{1}{\sqrt{1-x^2}}}{2x}\rightarrow\frac{-\sin{x}-\frac{x}{\sqrt{(1-x^2)^3}}}{2}\rightarrow0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $2^x = \sqrt{3^x}+1$ using logarithm. I have problem solving the equation
$$2^x = \sqrt{3^x} +1$$
for $x$ using logarithm. I know the only answer is $2$ which can be proven using graphs or derivatives,etc or by dividing the two sides by $2^x$ which gives the sum of $\sin 60°$ to the power of $x$ and $\cos 60°$ to the power of $x$ equal to $1$, concluding $x=2$.
I'm looking for a way to solve it using logarithm which is not easy because of the "$1$" in one side of the equation.
| Here is my approach.We will solve the equation for $x>0$. The case $x \leq 0$ obviously gives us no solution because:
$$ 2^{-x}=\frac{1}{2^x} \leq 1 < 1 + 3^{-\frac{x}{2}}. $$
Then for the case $x > 0$. We define:
$$f(x):= 2^x - 3^{\frac{x}{2}} - 1, ~x \in \mathbb{R}$$
Hence: $$\frac{df}{dx} = \ln 2 \cdot 2^x - \frac{\ln 3}{2}\cdot 3^{\frac{x}{2}}=\frac{1}{2}\cdot \left(\ln 4 \cdot 2^x - \ln 3\cdot \sqrt{3}^{~x} \right)>0, \forall x > 0$$
Hence $f(x)$ is increasing for all $x>0$ and therefore $f(x)$ meet x-axis at only one point and that is where $f(x)=0$. Easily find that $x=2$ satisfy $f(x)=0$ and that is our only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\tan x + \tan y = 4$ and $\cos x + \cos y = 1/5$, find $\tan(x+y)$.
If $\tan x + \tan y = 4$ and $\cos x + \cos y = 1/5$, find $\tan(x+y)$.
Well, from the first condition, we get
$$\tan x + \tan y = \frac{\sin(x+y)}{\cos x \cos y}=4 \implies \sin(x+y)=4\cos x \cos y$$
Then,
$$\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{4}{1-\frac{\sin x \cos y}{\cos x \cos y}}=\frac{4}{1-\frac{4\sin x \cos y}{\sin(x+y)}}$$
But in this way, I couldn't get use of the second condition. Actually, I squared the first condition and replaced $\tan^2$ with $\sec^2-1$. Then, squared the second condition to find something useful about $\sec^2$'s. Yet, nothing that works.
Therefore, started substituting $\cos x = a$ and $\cos y = b$ to get
$$
\begin{cases}
\cfrac{\pm\sqrt{1-a^2}}{a} + \cfrac{\pm\sqrt{1-b^2}}{b} = 4 \\
a + b = \cfrac{1}{5}
\end{cases}
$$
Nonetheless, this system doesn't seem to be as easy as this problem might be given as a multiple-choice problem (2-3 minutes for solving).
Any help is appreciated.
| I show my attempts: I wrote down the system
\begin{align}
&\frac{s_1}{c_1}+\frac{s_2}{c_2}=4,\\
&c_1+c_2=1/5,\\
&c_1^2+s_1^2=1,\\
&c_2^2+s_2^2=1
\end{align}
solved with Wolfram Mathematica and obtained $8$ solution, but by symmetry they are only $4$. Excluded a complex solution, I have three solutions.
Substituted in
$$
\frac{\frac{s_1}{c_1}+\frac{s_2}{c_2}}{1-\frac{s_1s_2}{c_1c_2}}
$$
I get the three values
$$
(-1.47878, 0.0385584, 0.209877).
$$
Furthermore, if it's a multiple solution test, with proposed results $r_1,\ldots,r_n,$ then you can proceed backward: find the product of tangents as
$$
p=1-\frac{4}{r_i}
$$
then given sum $s$ of tangents and their product, find the values of tangents $t_1,t_2$ solving
$$
t^2-s t+p=0
$$
and finally calculate the sum of cosines as
$$
\cos(\arctan t_1)+\cos(\arctan t_2)
$$
that should give $1/5$ for a correct solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving a system of two equations in polar coordinates Below is a problem I did. I believe the answer is right. Is it? However, I am not sure my reasoning is correct. I am also interested in comments about my style.
Problem:
Find the points of intersection of the follow two pairs of curves.
\begin{align*}
r &= a(1 + \cos \theta) \\
r &= a( 1 - \sin \theta )
\end{align*}
Answer:
\begin{align*}
a \left(1 + \cos \theta \right) &= a \left( 1 - \sin \theta \right) \\
1 + \cos \theta &= 1 - \sin \theta \\
\cos^2 \theta &= \sin^2 \theta = 1 - \cos^2 \theta \\
2 \cos^2 \theta &= 1 \\
\cos \theta &= \pm \frac{1}{ \sqrt{2}}
\end{align*}
Consider $\theta = \frac{\pi}{4}$ as a solution. This corresponds to an $r$ value
of $a\left( 1 + \frac{ \sqrt{2}}{2} \right)$. However, this value does not satisfy the second
equation so it must discarded. Now, we consider $\theta = \frac{3\pi}{4}$. For the first equation, I get:
$$ r = a \left( 1 - \frac{1}{ \sqrt{2}} \right) $$.
For the second equation, I get:
$$ r = a \left( 1 - \frac{1}{ \sqrt{2}} \right) $$
Hence, one of the points of intersection is:
$\left( a\left( 1 + \frac{ \sqrt{2}}{2} \right), \frac{ 3 \pi }{4} \right)$
Now consider $\theta = \frac{5\pi}{4}$ as a solution. In this case, I have:
\begin{align*}
a(1 + \cos\left( \frac{5\pi}{4} \right) ) &= a( 1 - \frac{ \sqrt{2}}{2} ) \\
a(1 - \sin\left( \frac{5\pi}{4} \right) ) &= a( 1 - \frac{ \sqrt{2}}{2} )
\end{align*}
Hence, one of the points of intersection is:
$$ \left( a( 1 - \frac{ \sqrt{2}}{2} ) , \frac{5\pi}{4} \right) $$
Now, we need to consider $\theta = - \frac{5\pi}{4} $
In this case, I have:
\begin{align*}
a(1 + \cos\left( - \frac{5\pi}{4} \right) ) &= a( 1 - \frac{ \sqrt{2}}{2} ) \\
a(1 - \sin\left( - \frac{5\pi}{4} \right) ) &= a( 1 + \frac{ \sqrt{2}}{2} )
\end{align*}
Hence $\theta = - \frac{5\pi}{4}$ is not a solution of the original equation. We picked this solution when we squared both sides.
Now I claim that $(0,a)$ and $(0,b)$ are the same point for all real numbers $a$ and $b$. Hence,
we have to consider the fact that both equations go through the orgin. Therefore, I claim that
$(0,0)$ is the third point of intersection even though $(0,0)$ is not on either curve.
| In general if $a=0$ (as you noted) the curves reduce to a dot at the origin. Otherwise a simpler approach would use $sin\theta =-cos\theta $ leading to $\theta=\frac{3\pi}{4}+n\pi$, for all integers $n$.
Your calculation has a serious error. $sin(\frac{5\pi}{4})$ has the wrong sign.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\frac{1}{5}$ For $a,b,c\geqslant 0.$ Prove$:$
$$\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\dfrac{1}{5}$$
I found an AM-GM proof.
Since $$P+\frac{1}{5}\geqslant 0\Leftrightarrow 6\,{a}^{3}+6\,{b}^{3}+8\,{a}^{2}c-2\,a{c}^{2}+8\,{b}^{2}c-2\,b{c}^{2}-19\,abc+3\,{a}^{2}b+3\,a{b}^{2}+{c}^{3} \geqslant 0$$
And by AM-GM$:$
$$2\,a{c}^{2}\leqslant 6{a}^{3}+\frac49{c}^{3},$$
$$2\,b{c}^{2}\leqslant 6{b }^{3}+\frac49{c}^{3},$$
$$19\,abc\leqslant \frac19{c}^{3}+3a{b}^{2}+3{a}^{2}b+8 \,{a}^{2}c+8\,{b}^{2}c.$$
So we are done!
Is there another nice proof$?$ Thanks for a real lot!
| For the fun : with algebra.
Consider that you look for the minimum value of function
$$f=\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3}$$ with $c=1$ as
@Macavity explained. Computing the derivatives and simplify, we end with the two equations
$$3 a (a+4) b+2 a (a+2)-3 b^3-8 b^2-3 b-1=0\tag1$$
$$-3 a^3-8 a^2+3 a (b (b+4)-1)+2 b (b+2)-1=0\tag2$$ Use $(2)$ to compute $b$ (it is just a quadratic). So
$$b=\frac{\sqrt{9 a^4+30 a^3+61 a^2+33 a+6}-6 a-2}{3 a+2}$$ Pluf in $(1)$ and get a monster. But after a few suarin steps, the only acceptable solution is $a=\frac 13$ so $b=\frac 13$ too and $f_{min}=-\frac 15$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Integrate $f(x)=\frac{1}{1+(x-1/x)^2}$ I came across this rather simple looking function
$$
f(x)=\frac{1}{1+\left(x-\frac{1}{x}\right)^2}
$$
which has a nice looking plot with two peaks symmetric around the vertical axis, whereas each peak is not symmetric by itself. It also turns out to be difficult to integrate by hand, so I have two questions about it:
*
*What kind of function is it? does it have a name?
*How to actually integrate it? For example, if I know how the function behaves for small and large $x$, i.e.:
for very large $x$:
$$
\int\frac{1}{1+x^2} dx = \arctan(x) + C
$$
and for $x\sim 0$:
$$
\int\frac{1}{1+1/x^2} dx = x-\arctan(x) + C
$$
would it be possible to use this information to somehow get to them (at least one-sided) integral of the original function?
| $$I=\int \frac{dx}{1+(x-1/x)^2}=\int\frac{x^2}{x^4-x^2+1}dx =\frac{1}{2} \int \left( \frac{1-\frac{1}{x^2}}{x^2+1/x^2-1} + \frac{1+\frac{1}{x^2}}{x^2+1/x^2-1}\right)dx$$
$$\implies I(x)=\frac{1}{2} \int \left( \frac{1-\frac{1}{x^2}}{(x+1/x)^2-3} + \frac{1+\frac{1}{x^2}}{(x-1/x)^2+1}\right)dx$$
Let $x+1/x=u$ and $x-1/x=v$ in the first and second integrals, respectivelt, then
$$I(x)=\frac{1}{2}\left (\int \frac{du}{u^2-3}+ \int \frac{dv}{v^2+1}\right).$$
$$I(x)=\frac{1}{2} \left( \frac{1}{2\sqrt{3}} \ln \frac{u-\sqrt{3}}{u+\sqrt{3}}+ \tan^{-1}v \right) +C.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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When to use polar substitution? I have solved the problem below, which gave me this question: How do i know when to use polar substitution?
If, instead of polar substitution, I directly set (x, y) = (0,0), I get division by zero. Therefore, I will instead use polar substitution. Is there anything else I should think about?
The problem.
How can the funtion be defined at the origin so that it becoms continuous at all points of the xy-plane?
$$f(x,y) = \frac{x^2+y^2-x^3y^3}{x^2+y^2}, \ (x,y)\neq(0,0)$$
My solution.
$$\text{Polar substitution: } x=r\cos\theta,\ y=r\sin\theta,\ x^2+y^2=r^2\\
\begin{align}
&\lim_{r\to0}\frac{r^2-r^3\cos^3\theta\, r^3\sin^3\theta}{r^2}\\
&\Rightarrow\lim_{r\to0} \frac{r^2(1-r\cos^3\theta\, r^3\sin^3\theta r)}{r^2(1)}\\
&\Rightarrow\lim_{r\rightarrow0} \frac{r^2(1-r\cos^3\theta\, r^3\sin^3\theta)}{r^2(1)}\\
&\Rightarrow\lim_{r\rightarrow0} \frac{1-r\cos^3\theta\, r^3\sin^3\theta}{1}\\
&\Rightarrow\lim_{r\rightarrow0} \frac{1-0\cos^3\theta\, 0^3\sin^3\theta}{1} \Rightarrow f(0,0)=1\\
\end{align}
$$
| If the polar substitution bounds the function between functions of $r$ that have the same limit, you can prove the intermediate function has the same limit by the squeeze theorem. In this case,$$f=1-r^4\cos^3\theta\sin^3\theta=1-\tfrac18r^4\sin^32\theta$$is bound between $1\mp\tfrac18r^4$. But even the looser, more obvious bounds $1\mp r^4$ prove the limit is $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this identity using Jacobi's Triple Product Identity Using Jacobi's Triple Product Identity prove that\begin{align*}
\prod_{n \geq 1}\left(1-x^{n}\right)^{6}=&\ \frac{1}{2}\left\{\prod_{n \geq 1}\left(1+x^{2 n-1}\right)^{2}\left(1-x^{2 n}\right)\times\left(1+4 x \frac{d}{d x}\right) 2 \prod_{n>1}\left(1+x^{2 n}\right)^{2}\left(1-x^{2 n}\right)\right.\\
& \left.-2 \prod_{n>1}\left(1+x^{2 n}\right)^{2}\left(1-x^{2 n}\right) \times 4 x \frac{d}{d x} \prod_{n \geqslant 1}\left(1+x^{2 n-1}\right)^{2}\left(1-x^{2 n}\right)\right\}
\end{align*}
I found this identity in the A Simple proof of Jacobi's Four Square Theorem. I have given a picture. See that after the step
\begin{align*}\prod_{n\geq 1}(1-x^n)^6\ & =\frac{1}{2}\Bigg\{ \sum_{s=-\infty}^{\infty}x^{s^2} \sum_{r=-\infty}^{\infty} (2r+1)^2x^{r^2+r}-\sum_{r=-\infty}^{\infty}x^{r^2+r}\sum_{r=-\infty}^{\infty}(2s)^2x^{s^2}
\Bigg\}\\ & = \frac{1}{2}\Bigg\{ \sum_{s=-\infty}^{\infty}x^{s^2} \times\bigg ( 1+4x\frac{d}{dx}\bigg)\sum_{r=-\infty}^{\infty} x^{r^2+r}-\sum_{r=-\infty}^{\infty}x^{r^2+r}\times 4x\frac{d}{dx}\sum_{r=-\infty}^{\infty}x^{s^2}
\Bigg\}\end{align*}
How did they got the identity using Jacobi's Triple Product Identity
| The Jacobi Triple Product identity states that $$\sum_{n\in\mathbb{Z}} z^nx^{n^2}=\prod_{n=1}^{\infty} (1-x^{2n})(1+zx^{2n-1})(1+z^{-1}x^{2n-1})\tag{1}$$ for all complex numbers $x, z$ such that $z\neq 0,|x|<1$.
Putting $z=1$ we get $$\sum_{n\in\mathbb {Z}} x^{n^2}=\prod_{n=1}^{\infty} (1+x^{2n-1})^2(1-x^{2n})\tag{2}$$ Putting $z=x$ in $(1) $ we get $$\sum_{n\in\mathbb {Z}} x^{n^2+n}=\prod_{n=1}^{\infty} (1+x^{2n})(1+x^{2n-2})(1-x^{2n})$$ Note that the factor $(1+x^{2n-2})$ equals $2$ if $n=1$ and thus we can write above equation as $$\sum_{n\in\mathbb{Z}} x^{n^2+n}=2\prod_{n=1}^{\infty} (1+x^{2n})^2(1-x^{2n})\tag{3}$$ The paper of M D Hirschhorn uses the identities $(2),(3)$ in their derivation. Let me know if there is anything more which is troubling you in that proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with a, b, c about finding minimal and maximal value Find the minimal and maximal value (if they exist) of ${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(a+c)}{a^2+c^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}}$ if are non-negative real numbers, such that at least two of them are positive.
My attempts for the case where all variables are positive:
I tried applying AM-GM on ${\sqrt{a(b+c)}} $ and etc. and then applying $b^2+c^2=>(b+c)^2/2$,but the inequality I received was false. I also tried rewriting $a(b+c)/(b^2+c^2)=(b/a+c/a)/((b/a)^2+(c/a)^2)$ and etc. and then letting $a/b=x$, $b/c=y$ and $c/a=z$ and etc but I was stuck from there. I also tried applying Cauchy Schwarz by squaring the whole expression.
| We can prove that $2$ is a minimum value also by AM-GM:
$$\sum_{cyc}\sqrt{\frac{a(b+c)}{b^2+c^2}}=\sum_{cyc}\frac{2a(b+c)}{2\sqrt{a(b+c)\cdot(b^2+c^2)}}\geq\sum_{cyc}\frac{2a(b+c)}{a(b+c)+b^2+c^2}.$$
Id est, it's enough to prove that:
$$\sum_{cyc}\frac{a(b+c)}{a(b+c)+b^2+c^2}\geq1$$ or
$$abc\sum_{cyc}\left(a^3+2a^2b+2a^2c+\frac{1}{3}abc\right)\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\frac{2a^5 - 5a^4 + 2a^3 - 8a^2}{a^2 + 1}$ Suppose $a$ is a root of $x^2 + 3x - 1.$ Find $\frac{2a^5 - 5a^4 + 2a^3 - 8a^2}{a^2 + 1}.$
I was thinking of factoring the fraction a bit first, than letting $a^2 = 1 - 3a.$ However, that leads nowhere.
| Another way.
For $a^2+3a-1=0$ and $a=\frac{-3+\sqrt{13}}{2}$ we obtain:
$$(a^2-1)^2=9a^2$$ or
$$(a^2+1)^2=13a^2$$ or
$$a^2+1=\sqrt{13}a,$$ which gives: $$\frac{2a^5-5a^4+2a^3-8a^2}{a^2+1}=$$
$$=\tfrac{2a^5+6a^4-2a^3-11a^4-33a^3+11a^2+37a^3+111a^2-37a-130a^2+37a}{\sqrt{13}a}=$$
$$=\tfrac{37-130a}{\sqrt{13}}=\tfrac{37-65(-3+\sqrt{13})}{\sqrt{13}}=\tfrac{232-65\sqrt{13}}{\sqrt{13}}.$$
For $a=\frac{-3-\sqrt{13}}{2}$ we obtain: $$a^2+1=-\sqrt{13}a,$$ which gives
$$\frac{2a^5-5a^4+2a^3-8a^2}{a^2+1}=\tfrac{-232-65\sqrt{13}}{\sqrt{13}}.$$
| {
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In a triangle ABC, $\begin {vmatrix} a&b&c \\b&c&a \\c&a&b \end {vmatrix}=0$, then find $\sin A\sin B+\sin B\sin C+\sin C\sin A$ From the given determinant,
$$a^3+b^3+c^3-3abc=0$$
Which implies $a+b+c=0$ or $a^2+b^2+c^2 -ab-bc-ac=0$
Since the former isn’t possible,
$$a^2+b^2+c^2=ab+bc+ac$$
$$\sin A\sin B+\sin B\sin C +\sin C\sin A=\sin^2A +\sin^2B+\sin ^2C$$
The answer for this is either $\cos^2A+\cos^2 B+\cos^2C$ or one of numbers $0, 1, \frac 94$
How do I proceed from here?
| The sides $a,b,c$ of the triangle satisfy
$$ a^2+b^2+c^2=ab+bc+ac \\
\begin{matrix} \implies \dfrac{1}{2}(2a^2+2b^2+2c^2-2ab-2bc-2ac) & =0 \\
(a-b)^2+(b-c)^2+(c-a)^2 &=0
\end{matrix}$$
This straightaway means that $a=b=c$ (i.e. and equilateral triangle) otherwise the inequality cannot hold.
Therefore, $\sin A\sin B+\sin B\sin C+\sin C\sin A=\dfrac{3}{4}*3=\dfrac{9}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor series of $\frac{x}{e^x-e^{-x}}$ My textbook says that (when $x$ approaches $0$):
$$\frac{x}{e^x-e^{-x}}=\frac{1}{2}-\frac{x^2}{12}+\cdots$$
It is also said that the result can be deduced by using $e^x=1+x+x^2/2+\cdots$.
But when I sub it in directly:
$$\begin{aligned}\frac{x}{e^x-e^{-x}}&=\frac{x}{(1+x+x^2/2+x^3/6\cdots)-(1-x+x^2/2-x^3/6+\cdots)}\\&=\frac{x}{2x+x^3/3+\cdots}\end{aligned}$$
I cannot get the right expansion. So what is the right approach?
| You were on the right track, you could do
$$\frac{x}{2x+x^3/3+o(x^3)}\\=\frac1{2+x^2/3+o(x^2)}\\=\frac12\frac1{1-(-x^2/6)}+o(x^2)\\=\frac12\sum_{k=0}^\infty \frac{(-x^2)^n}{6^n} + o(x^2) \\= \frac12-\frac{x^2}{12}+o(x^2). $$
A question from the comments on the third line: it follows from the identity
$$\left|\frac{1}{A+h} - \frac{1}A \right|= \frac{|h|}{|A||A+h|} $$
Here, we have $A=2+x^2/3\ge 2$. If $h$ is small, say $|h|<1$, then
$|A+h|\ge |A|-|h|\ge 1$. Thus
$$ \left|\frac{1}{A+h} - \frac{1}A \right|\le |h|\times \frac12\times \frac11.$$
Consequently,
$$ \frac{1}{A+h} = \frac{1}A + O(h). $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the arc length of the curve $y = \sqrt{x-x^2}+\sin^{-1}(\sqrt{x})$ Compute the arc length of the curve $y = \sqrt{x-x^2}+\sin^{-1}(\sqrt{x})$ from for $0 \leq x \leq 1$
This problem is pretty brutal! I'd appreciate if somebody could hold my hand through this integral and really lay out the details for me... I've been struggling with it for awhile now and can't get it down!!
Basically we know that: $$L = \int_0^1 \sqrt{1+(\frac{dy}{dx})^2}$$
Where $$\frac{dy}{dx}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}$$
If somebody could help me simplify and integrate this that would be great... Thank so you much!!
| $\frac{dy}{dx}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{x-x^2}}=\frac{1-x}{\sqrt{x-x^2}}=\frac{1-x}{\sqrt{1-x}\sqrt x}=\sqrt {x^{-1}-1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $
But can anyone attack this problem in a different & more standard way?
| We have that
$$H_N=\sum_{k=1}^{N} \frac{1}k=\ln N+\gamma+\frac1{2N}+O\left(\frac1{N^2}\right)$$
then
$$\sum_{k=1}^{2N} \frac{(-1)^{k+1}}k=H_{2N}-H_{N}=\log {2}-\frac1{2N}+O\left(\frac1{N^2}\right)$$
and
$$\sum_{k=n}^{2N} \frac{(-1)^{k+1}}k=\sum_{k=1}^{2N} \frac{(-1)^{k+1}}k-\sum_{k=1}^{n-1} \frac{(-1)^{k+1}}k=$$
$$=-\frac1{2N}+O\left(\frac1{N^2}\right)+\frac1{2(n-1)}+O\left(\frac1{n^2}\right) \sim \frac1{2(n-1)}+O\left(\frac1{n^2}\right)$$
then
$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )\sim \frac n{2(n-1)}+O\left(\frac1{n}\right) \to \frac12$$
| {
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"url": "https://math.stackexchange.com/questions/3834291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
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} |
limit of multivariable function as x,y approach to infinity can i solve this limit using polar coordinate?
$$\lim_{(x,y)\to\infty} \frac{x^2+y^2}{x^2+(y-1)^2}=$$
$$\frac{r^2}{r^2-2r\sin\theta +1}=\frac{1}{1-\frac{2\sin\theta}{r}+\frac{1}{r^2}}=1$$
| Yes of course your solution is fine, as an alternative by $x=u$ and $y-1=v$
$$ \frac{x^2+y^2}{x^2+(y-1)^2}=\frac{u^2+(v+1)^2}{u^2+v^2} =1+\frac{1}{u^2+v^2}+\frac{2v}{u^2+v^2} \to 1$$
for the latter, in order to avoid cases, we can use that by AM-QM
$$\frac{2|v|}{u^2+v^2} \le \frac{2(|u|+|v|)}{u^2+v^2} \le \frac{2\sqrt 2}{\sqrt{u^2+v^2}} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.
| $$\dfrac{a\cdot\dfrac{x^2}a+3b\cdot\dfrac xb+c\cdot\dfrac1{cx}}{a+3b+c}\ge\sqrt[a+b+c]{\left(\dfrac{x^2}a\right)^a\cdot\left(\dfrac xb\right)^{3b}\cdot\left(\dfrac1{cx}\right)^c}$$
The coefficient of $x$ under radical is $$2a+3b-c$$
Let us set $2a+3b-c=0$
$$\implies\dfrac{a\cdot\dfrac{x^2}a+3b\cdot\dfrac xb+(2a+3b)\cdot\dfrac1{(2a+3b)x}}{a+3b+(2a+3b)}\ge\sqrt[a+b+(2a+3b)]{\left(\dfrac{x^2}a\right)^a\cdot\left(\dfrac xb\right)^{3b}\cdot\left(\dfrac1{(2a+3b)x}\right)^{2a+3b}}$$
$$\implies\dfrac{x^2+3x+\dfrac1x}{3(a+2b)}\ge (a^{-a}b^{-3b}(2a+3b)^{-(2a+3b)})^{1/(3(a+2b))}$$
WLOG $a=b=1$
$$\implies\dfrac{x^2+3x+\dfrac1x}{9}\ge (5^{-5})^{1/9}$$
It is sufficient to show $$\dfrac9{5^{-5/9}}\ge\dfrac{15}4\iff12\ge5^{1-5/9}\iff12^9\ge5^4$$ which is obvious
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Convergence of $\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )$ I am trying to find the convergence of the following series:
$$\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )$$
I am stuck trying out different tests but none of them seem to give me an answer. What do you suggest that I should try, and what are the identities or series that I can use to compare this series to so I could maybe solve it like that?
How should I approach finding the convergence of trigonometric series in general and what should I be careful of?
| $$
\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )
$$
The general term can be rewritten as
$$
n\tan \left( \frac{\pi}{2^{n+1}}\right ) = n\frac{\pi}{2^{n+1}}\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}}
$$
Use the $n$-root test
$$
\left(n\tan \left( \frac{\pi}{2^{n+1}}\right )\right)^{\frac{1}{n}}
= \left( n\frac{\pi}{2^{n+1}}\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}} \right)^{\frac{1}{n}}
= n^{\frac{1}{n}}\left(\frac{\pi}{2^{n+1}}\right)^{\frac{1}{n}}\left(\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}} \right)^{\frac{1}{n}} \\
$$
take the limt you get
$$
\lim_{n\to \infty} \left[n^{\frac{1}{n}}\left(\frac{\pi}{2^{n+1}}\right)^{\frac{1}{n}}\left(\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}} \right)^{\frac{1}{n}}\right] = 1\cdot {\frac{1}{2}} \cdot 1 = {\frac{1}{2}} < 1
$$
since the limits of the individaul factors exist, then the limit is the their product. The last one can be found by examining the limit of its logarithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016) I was just doing the following question:
Find a prime number which divides the number $A=14^7+14^2+1$.
I solved it by finding the result which is $A=105413504+196+1=105413701$ and then trying out all prime numbers till I found that 211 divides it. However, obviously this is extremely tedious. I hence looked at the solution which says that $x^7+x^2+1=(x^2+x+1)(x(x-1)(x^3+1)+1)$ and from here by saying that $x=14$ we get the solution. However I can't seem to think of how to intuitively turn $x^7+x^2+1$ into $(x^2+x+1)(x(x-1)(x^3+1)+1)$. I realize that from the question it is obvious to go looking for factors of A and hence trying to factorize $14^7+14^2+1$, but I can't work out how to go about factorizing it, what are the steps which you need to take in order to factorize a given polynomial. Could you please explain to me how to go about factorizing such an expression and how to intuitively think of each step?
| Another way:
$$x^7+x^2+1=x^7-x^4+x^4+x^2+1=(x^2+x+1)(x^4(x-1)+x^2-x+1)=$$
$$=(x^2+x+1)(x^5-x^4+x^2-x+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
What is complex number z if $z^8+16z^4+256=0$? So far, I have set y to equal to $z^4$ and used the quadratic equation to solve $y = -8+8\sqrt{3}i$ or $-8-8\sqrt{3}i$. How do I determine the 8 different values of $z$?
| As you have mentioned, the correct approach is to let $y=z^4$. Then we get
$$
y^2+16y+256=0
$$
Using the quadratic formula, the solutions are
$$
y=-8\pm8i\sqrt{3}
$$
All we need to do know is solve the equations $z^4=-8+i\sqrt{3}$ and $z^4=-8-i\sqrt{3}$. We can do this by writing the two complex numbers in the polar-coordinate form. As a reminder, I will write down how to derive the polar-coordinate form here:
$$
a+bi=r\text{cis}\theta=r(\cos\theta+i\sin\theta)=re^{i\theta}
$$
The magnitude of $-8\pm8i\sqrt{3}$ is $\sqrt{8^2+(8\sqrt{3})^2}=\sqrt{64+64(3)}=\sqrt{256}=16$. Therefore, $r=16$. The angle $\theta$ depends on whether we are considering $-8+8i\sqrt{3}$ or $-8-8i\sqrt{3}$. In the former case, $\theta=120$ degrees. Then, by symmetry, we can find out $\theta$ in the latter case. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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