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Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$ I can't have this nice solution
Thank you everyone can help.
|
Using the Taylor series
$$\ln \frac{1+x}{1-x} = \ln (1 + x) - \ln(1 - x) = 2\sum_{k=0}^\infty \frac{x^{2k+1}}{2k+1}, \quad -1 < x < 1;$$
letting $x = 1/3$, we have
$$\ln 2 = 2\sum_{k=0}^\infty \frac{1}{3^{2k+1}(2k+1)}. \tag{1}$$
Using Bernoulli inequality, we have, for all $a > 0$,
$$(2/5)^{2/5} = \frac{1}{a^2}\left(\frac{2}{5}a^5\right)^{2/5}
\le \frac{1}{a^2}\left(1 + \left(\frac{2}{5}a^5 - 1\right)\cdot \frac25\right) = \frac{4}{25}a^3 + \frac{3}{5a^2}$$
with equality if and only if $a^5 = \frac52$.
It suffices to prove that, there exists $a > 0$ such that
$$\frac{4}{25}a^3 + \frac{3}{5a^2} < \ln 2.$$
When $a^5 \approx \frac 52$, from $\frac{4}{25}a^3 + \frac{3}{5a^2} = \frac{4a^5}{25a^2} + \frac{3}{5a^2} \approx \frac{4\cdot (5/2)}{25a^2} + \frac{3}{5a^2} \approx \ln 2$, we have $a^2 \approx \frac{1}{\ln 2} \approx \frac{81}{56} = 1.4464$
and $a \approx 1.2$ where we have used (1) to get $\ln 2 \approx 2\sum_{k=0}^1 \frac{1}{3^{2k+1}(2k+1)} = \frac{56}{81}$. Letting $a = \frac{6}{5}$, it suffices to prove that
$$\frac{4}{25}(6/5)^3 + \frac{3}{5(6/5)^2} < \ln 2.$$
Using (1), we have
$$\ln 2 > 2\sum_{k=0}^5 \frac{1}{3^{2k+1}(2k+1)} = \frac{15757912}{22733865}.$$
It suffices to prove that
$$\frac{4}{25}(6/5)^3 + \frac{3}{5(6/5)^2} < \frac{15757912}{22733865}$$
which is true.
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of $\sin x \cdot \cos x$ I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?
$ \int \sin x \cos x dx $
1) via subsitution $ u = \sin x $
$ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x $
2) via subsitution $ u = \cos x $
$ u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x $
3) using $ \sin 2x = 2 \sin x \cos x $
$ \int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x $
So, we have:
$$ \frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x $$
|
A primitive is unique up to a constant
|
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|
Use the binomial theorem to expand How can we expand this using the binomial theorem?
$(x^2 + 1/x)^7$
|
Let $a = x^{2}$ and $b = \frac{1}{x}$. The binomial is written as $(a + b)^{7}$. Apply Binomial Theorem, so we have:
$$(a + b)^{7} = \dbinom{7}{0} a^{7}b^{7 - 7} + \dbinom{7}{1} a^{6}b^{7 - 6} + \dbinom{7}{2} a^{5}b^{7 - 5} + \dbinom{7}{3} a^{4}b^{7 - 4} + \dbinom{7}{4} a^{3}b^{7 - 3} + \dbinom{7}{5}a^{2}b^{7 - 2} + \dbinom{7}{6}a^{1}b^{7 - 1} + \dbinom{7}{7}a^{0}b^{7 - 0}$$
$$= a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} + b^7$$
Substitute back with $a = x^{2}$ and $b = \frac{1}{x}$, so you get the following answer:
$$(x^{2})^{7} + 7(x^{2})^{6}(\frac{1}{x}) + 21(x^{2})^{5}(\frac{1}{x})^{2} + 35(x^{2})^{4}(\frac{1}{x})^{3} + 35(x^{2})^{3}(\frac{1}{x})^{4} + 21(x^{2})^{2}(\frac{1}{x})^{5} + 7(x^{2})(\frac{1}{x})^{6} + (\frac{1}{x})^7$$
$$= x^{14} + 7x^{11} + 21x^{8} + 35x^{5} + \frac{7}{x^{4}} + 35x^{2} + \frac{21}{x} + \frac{1}{x^7}$$
|
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|
Proving a trig infinite sum using integration How can I prove the following using integration and elementary functions?
Prove that:
$$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2}$$
$0 < \theta < 2\pi$
|
Let,
$$S_1 = \sum_{n=1}^{\infty}\frac{\cos n\theta}{n}\\
S_2 = \sum_{n=1}^{\infty}\frac{\sin n\theta}{n}$$
Then $$S_1 + iS_2 = \sum_{n=1}^{\infty}\frac{\cos(n\theta)+i\sin(n\theta)}{n}=\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}$$
Now, from the Taylor expansion, $\ln (1+x) = x -\frac{x^2}{2}+\frac{x^3}{3} ...$
$$\implies -\ln(1-x) = x+ \frac{x^2}{2}+\frac{x^3}{3} ... = \sum_{n=1}^{\infty}\frac{x^n}{n}$$
$$\begin{align} \therefore S_1+iS_2 &= -\ln(1-e^{i\theta})
\\&=-\ln(1-\cos\theta-i\sin \theta)
\\
&=-\ln(2\sin^2\theta/2 - 2i\sin(\theta/2)\cos(\theta/2))
\\
&=-\ln(2\sin\theta/2)-\ln(\sin\theta/2-i\cos\theta/2)
\\
&=-\ln(2\sin\theta/2)+\ln(\sin\theta/2+i\cos\theta/2)
\\
&=-\ln(2\sin\theta/2)+\ln(e^{i(\pi/2-\theta/2)})
\\
&=-\ln(2\sin\theta/2)+i(\pi/2-\theta/2)
\end{align}
$$
Taking the imaginary part of both sides,
$$S_2 = \frac{\pi}{2} - \frac{\theta}{2}$$
|
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|
How do you calculate $25^{11} \pmod{341}$? How do you calculate $25^{11} \pmod{341}$?
I understand you have to split the exponent into $11 = 1 + 2 + 8$?
|
Note that $ 341 = 11 \cdot 31 $ and hence it is sufficient to fine the residues of $ 25^{11} $ modulo $ 11 $ and $ 31 $ and then apply the Chinese Remainder Theorem.
By Fermat's Little Theorem, $ 25^{10} \equiv 1 \mod 11 \implies 25^{11} \equiv 25 \equiv 3 \mod 11 $.
Also note that $ 25^4 \equiv (-6)^4 \equiv 5^2 \equiv 25 \mod 31 $ and hence $ 25^3 \equiv 1 \mod 31 $ which implies that $ 25^{11} \equiv 25^2 \equiv 5 \mod 31 $.
Therefore, $ 25^{11} \equiv 36 \mod 341 $ after applying the CRT.
|
{
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|
Let M, K, and L be points on line (AB), (BC), and (CA), respectively. Find the maximum area of smallest the three triangles MAL, KBM, and LCK? Let M, K, and L be points on line (AB), (BC), and (CA), respectively. Find the maximum area of smallest the three triangles MAL, KBM, and LCK in respect to ABC?
I try to guess the answer and then prove it, however it doesn't work to randomly guess.
I draw some pictures, but it hardly tells me anything
|
We shall denote the area of a triangle $ABC$ with $S_{ABC}$. We shall prove that $\min(S_{MAL}, S_{KBM}, S_{LCK}) \leq \frac{1}{4}S_{ABC}$.
Without loss of generality assume that $S_{MAL}=\max(S_{MAL}, S_{KBM}, S_{LCK})$. Clearly if $\frac{S_{MAL}}{S_{ABC}} \leq \frac{1}{4}$, we are done. Otherwise consider $\frac{S_{MAL}}{S_{ABC}}=\frac{AM}{AB} \cdot \frac{AL}{AC}>\frac{1}{4}$. Let $\frac{AM}{AB}=x, \frac{AL}{AC}=y, 0 \leq x, y \leq 1$. Then $xy>\frac{1}{4}$. Also, we have $\frac{MB}{AB}=1-x, \frac{LC}{AC}=1-y$. Clearly if any of $K, L, M$ are equal to $A, B$, or $C$, then $\min(S_{MAL}, S_{KBM}, S_{LCK})=0 \leq \frac{1}{4}S_{ABC}$. Thus $0<x, y<1$.
Let $K'$ be on $BC$ s.t. $\frac{BK'}{CK'}=\frac{1-y}{1-x}$. Then $$\frac{S_{K'BM}}{S_{ABC}}=\frac{BK'}{BC} \cdot \frac{BM}{AB}=\frac{(1-x)BK'}{BC}=\frac{(1-y)CK'}{BC}=\frac{CK'}{BC} \cdot \frac{LC}{AC}=\frac{S_{LCK'}}{S_{ABC}}$$
If $BK \leq BK'$, then $$\frac{S_{KBM}}{S_{ABC}}=\frac{BK}{BC} \cdot \frac{BM}{AB} \leq \frac{BK'}{BC} \cdot \frac{BM}{AB}=\frac{S_{K'BM}}{S_{ABC}}$$
Otherwise $BK>BK'$ so $CK<CK'$, so $$\frac{S_{LCK}}{S_{ABC}}=\frac{CK}{BC} \cdot \frac{LC}{AC}<\frac{CK'}{BC} \cdot \frac{LC}{AC}=\frac{S_{LCK'}}{S_{ABC}}=\frac{S_{K'BM}}{S_{ABC}}$$
In any case, we have $$\min(\frac{S_{KBM}}{S_{ABC}}, \frac{S_{LCK'}}{S_{ABC}}) \leq \frac{S_{K'BM}}{S_{ABC}}=\frac{(1-x)BK'}{BC}=\frac{(1-x)(1-y)}{1-x+1-y}$$
It thus suffices to prove that $\frac{(1-x)(1-y)}{1-x+1-y} \leq \frac{1}{4}$, or equivalently, $\frac{1}{1-x}+\frac{1}{1-y} \geq 4$.
Note that $x>xy>\frac{1}{4}$ so $4x-1>0$. Thus
\begin{align}
\frac{1}{1-x}+\frac{1}{1-y}-4 & =\frac{1}{1-x}+\frac{4x}{4x-4xy}-4 \\
& >\frac{1}{1-x}+\frac{4x}{4x-1}-4 \\
& =\frac{(4x-1)+4x(1-x)-4(1-x)(4x-1)}{(1-x)(4x-1)} \\
& =\frac{12x^2-12x+3}{(1-x)(4x-1)} \\
& =\frac{3(2x-1)^2}{(1-x)(4x-1)} \\
& \geq 0
\end{align}
Therefore $\min(S_{MAL}, S_{KBM}, S_{LCK}) \leq \frac{1}{4}S_{ABC}$. Furthermore, equality holds when $M, K, L$ are the midpoints of $AB, BC, AC$ respectively, so $\max(\min(S_{MAL}, S_{KBM}, S_{LCK}))=\frac{1}{4}S_{ABC}$.
|
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|
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$ Please help me to find a closed form for the infinite product
$$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$
where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.
|
Let
$$
f(x)=\prod_{n=0}^\infty\left(1-x^{2^n}\right)^{1/2^n}\tag{1}
$$
and
$$
g(x)=\prod_{n=0}^\infty\left(1+x^{2^n}\right)^{1/2^n}\tag{2}
$$
Then
$$
\begin{align}
f(x)\,g(x)
&=\prod_{n=0}^\infty\left(1-x^{2^{n+1}}\right)^{1/2^n}\\
&=\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{2/2^n}\\
&=\left(\frac{f(x)}{1-x}\right)^2\tag{3}
\end{align}
$$
from which we get
$$
\frac{f(x)}{g(x)}=(1-x)^2\tag{4}
$$
Note that
$$
\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}=\frac{f(x)}{1-x}\tag{5}
$$
and
$$
\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}=\frac{g(x)}{1+x}\tag{6}
$$
Therefore, combining $(4)$, $(5)$, and $(6)$, we get
$$
\frac{\displaystyle\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}}{\displaystyle\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}}=1-x^2\tag{7}
$$
Plug $x=e^{-2}$ into $(7)$ to get
$$
\prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1-e^{-4}\tag{8}
$$
|
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|
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$ I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks
|
Welcome to Math Stack Exchange!
With lab bhattacharjee's helpful hint for the integral, we can rewrite the expression as:
$$\int \left(1 + \frac{(-2x - 5)}{2(x^2 + 1)} + \frac{7}{4(x - 1)} - \frac{3}{4(x + 1)}\right)dx$$
$$= \int 1 dx + \int \frac{(-2x - 5)}{2(x^2 + 1)} dx + \int \frac{7}{4(x - 1)} dx - \int \frac{3}{4(x + 1)} dx$$
$$= \int 1 dx - \int \frac{2x}{2(x^2 + 1)} dx - \int \frac{5}{2(x^2 + 1)} + \int \frac{7}{4(x - 1)} dx - \int \frac{3}{4(x + 1)} dx$$
$$= \int 1 dx - \int \frac{x}{x^2 + 1} dx - \frac{5}{2} \int \frac{1}{x^2 + 1} dx + \frac{7}{4} \int \frac{1}{x - 1} dx - \frac{3}{4} \int \frac{1}{x + 1} dx$$
Next steps should be "direct-cut", meaning that it's really possible to work out the computation one way! :D
*
*For the first integrand, use power rule to obtain $x$.
*For the second integrand, let $u = x^2 + 1 \rightarrow du = 2x dx$.
*For the third integrand, you get $\arctan(x)$. You can find the special integral here.
*For the fourth integrand, let $v = x - 1 \rightarrow dv = dx$.
*For the fifth integrand, let $w = x + 1 \rightarrow dw = dx$.
So we are left off with:
$$x - \frac{5}{2}arctan(x) - \frac{1}{2} \int \frac{du}{u} + \frac{7}{4} \int \frac{dv}{v} - \frac{3}{4} \int \frac{dw}{w}$$
Finally, the expression becomes...
$$x - \frac{5}{2}arctan(x) - \frac{ln(u)}{2} + \frac{7ln(v)}{4} - \frac{3ln(w)}{4} + K \text{ where } K \text{ is an arbitrary constant}$$
$$= x - \frac{5}{2}arctan(x) - \frac{ln(x^2 + 1)}{2} + \frac{7ln(x - 1)}{4} - \frac{3ln(x + 1)}{4} + K$$
The solution for the integral is equivalent to the one in Wolfram Alpha.
|
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|
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr
& = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr
& = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr
& = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$
The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!
|
$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C $$
$$ = \ln(2x + 1)^x + \ln(2x + 1)^\dfrac 12 - x + C $$
$$= \ln({(2x + 1)^x \cdot(2x + 1)^\dfrac 12}) - x + C $$
$$= \ln{(2x + 1)^\dfrac{2x+1}{2} } - x + C $$
$$= \dfrac {1}{2}\cdot (2x+1) \ln{(2x + 1)} - x + C $$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $\omega$ is a complex cube root of unity, show that the following equals null matrix.
If $\omega$ is a complex cube root of unity, show that
$$ \left( \begin{bmatrix}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega \\
\end{bmatrix} + \begin{bmatrix}
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega \\
\omega & \omega^2 & 1 \\
\end{bmatrix} \right)
\begin{bmatrix}
1 \\
\omega \\
\omega^2 \\
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}$$
I tried to solve this and I reduced the L.H.S. to
$$\begin{bmatrix}
2 +2\omega+2\omega^2 \\
2 +2\omega+2\omega^2 \\
2 +2\omega+2\omega^2 \\
\end{bmatrix}
$$ (since $\omega^3=1$)
but couldn't equate it to R.H.S.
Please provide your assistance.
Thank you
|
$$\omega^3=1 \Rightarrow ω^3−1=0 \Rightarrow (ω−1)(ω^2+ω+1)=0 \Rightarrow ω−1=0 \lor ω^2+ω+1= 0$$
Since, $\omega \neq 1 $ as it is complex , $ω^2+ω+1=0$
Hence, LHS = RHS
- Answer originally posted as comment by Git Gud
|
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|
Explain the 1 + 2 + 3 in $ \frac{1 + 1 + 1 + \cdots}{1 + 2 + 3 + \cdots} = \lim_{n \to \infty} \frac{1}{(n+1)/2} $ $$ \frac{1 + 1 + 1 + \cdots}{1 + 2 + 3 + \cdots} = \lim_{n \to \infty} \frac{1}{(n+1)/2} = 0 $$
If $n$ goes to infinity, we can image that a bit by taking a very big number.
Like $1.000.000.000$
$1.000.000.000+1=1.000.000.001$
$1.000.000.001/2=500.000.000,5$
Now how does the $1 + 2 + 3 + ...$ make sense?
|
They're using that $$1+2+3+\dots+n=\frac{n(n+1)}2$$
For example $1+2+3=6=\dfrac{3\cdot 4}{2}$
On the other hand $$\underbrace{1+1+1+\cdots+1}_{n \;\;\rm times}=n$$
Thus $$\frac{1+1+1+\cdots}{1+2+3+\cdots}=\frac{n}{\dfrac{n(n+1)}2}=\frac{2}{n+1}$$
Of course, we're interpreting $$\frac{1+1+1+\cdots}{1+2+3+\cdots}$$ as $$\lim\limits_{n\to\infty}\frac{\sum_{i=1}^n 1}{\sum_{i=1}^n i}$$
If we had, for example $$\lim\limits_{n\to\infty}\frac{\sum_{i=1}^{n^2} 1}{\sum_{i=1}^n i}$$
then we would say that $$\frac{1+1+1+\cdots}{1+2+3+\cdots}=2$$
We can write this more suggestively as $$\frac{{1 + (1 + 1 + 1) + (1 + 1 + 1 + 1 + 1) + \cdots }}{{1 + 2 + 3 + \cdots }} = 2$$ to make clear how many terms we add in each step.
The moral of the story is that you cannot have "voids" in your notation. Writing something like $$\frac{1+1+1+\dots}{1+2+3+\dots}=\text{something}$$
should be better replaced by something informative and clear that really says what is going on.
|
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|
$ e^{At}$ for $A = B^{-1} \lvert \cdots \rvert B $ For a homework problem, I have to compute $ e^{At}$ for
$$ A = B^{-1} \begin{pmatrix}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \end{pmatrix} B$$
I know how to compute the result for $2 \times 2$ matrices where I can calculate the eigenvalues, but this is $3 \times 3$, and I cannot compute eigenvalues, so is there any identity or something which allows computing such exponentials?
Thanks!
|
Joseph is right and $ D^n=\begin{pmatrix}(-1)^n&0&0\\0&2^n&0\\0&0&3^n \end{pmatrix}so (Dt)^n=\begin{pmatrix}(-t)^n&0&0\\0&(2t)^n &0\\0&0&(3t)^n\end{pmatrix}$ so $\left(I+Dt+\frac{1}{2!}(Dt)^2+\frac{1}{3!}(Dt)^3+\cdots\right)=\begin{pmatrix}e^{-t}&0&0\\0&e^{2t}&0\\0&0&e^{3t}\end{pmatrix}$
|
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|
Let $v_1 = (1, 0); v_2 = (1,-1) \space\text{and} \space v_3 = (0, 1).$ I am stuck on the following problem :
Let $v_1 = (1, 0); v_2 = (1,-1) \space\text{and} \space v_3 = (0, 1).$ How many linear transformations
$T \colon \Bbb R^2 \to \Bbb R^2$ are there such that $Tv_1 = v_2; Tv_2 = v_3$ and $Tv_3 = v_1?$ The options are as follows:
(A) $3!$
(B) $3$
(C) $1$
(D) $0$
What I observed that $v_2=v_1-v_3$ and so $T(v_2)=T(v_1)-T(v_3) \implies v_3=v_2-v_1$.
But I do not know how to progress from here. Any idea?
|
Here's a straight forward solution. Every linear transformation $T\colon \mathbb R^2 \longrightarrow \mathbb R^2$ can be represented as a $2\times 2$ matrix, so you want to find $a_1$, $a_2$, $a_3$ and $a_4$ such that
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_1 \\ a_3\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix}\]
and
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ -1\end{pmatrix} = \begin{pmatrix}0 \\ 1\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_1-a_2 \\ a_3-a_4\end{pmatrix} = \begin{pmatrix}0 \\ 1\end{pmatrix}\]
and
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}0 \\ 1\end{pmatrix} = \begin{pmatrix}1 \\ 0\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_2 \\ a_4\end{pmatrix} = \begin{pmatrix} 1 \\ 0\end{pmatrix}\]
Now it's obvious that there are no such matrix, and so the answer is (d), 0.
Actually you've the answer in front of you too. Assume that there is some such $T$. Since $v_2 = v_1+v_3$ we can use the linearity of the transformation to get $v_3 = v_2+v_1$. Combine these two equations to get $v_1 = 0$, which is a contradiction, and again the answer is (d), 0.
|
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|
Prove that a map is Injective How can I prove that $f: (0, \infty) \times (0,\pi) \to \mathbb{R}^2$ where $f(x,y) = (\sinh(x)\sin(y),\cosh(x)\cos(y))$ is injective?
|
Suppose $\cosh a\cos b=\cosh c\cos d$ and $\sinh a\sin b=\sinh c\sin d$. Square the two relations:
\begin{gather}
\cosh^2 a\cos^2 b=\cosh^2 c\cos^2 d\\
\sinh^2 a\sin^2 b=\sinh^2 c\sin^2 d
\end{gather}
Sum and subtract the two:
\begin{gather}
\cosh^2 a + \sinh^2 a = \cosh^2 c + \sinh^2 c\\
\cos^2 b - \sin^2 b = \cos^2 d - \sin^2 d
\end{gather}
Therefore
$$
\cosh 2a = \cosh 2c
$$
which means that $2a=2c$ because $\cosh$ is injective in $(0,\infty)$.
Furthermore
$$
\cos 2b = \cos 2d
$$
that is, $2b=2d$ or $2b=2\pi-2d$, from which
$$
b=d\quad\text{or}\quad b=\pi-d
$$
The second possibility, together with $\cosh a\cos b=\cosh c\cos d$, gives $\cos b=\cos(\pi-b)$, that is $\cos b=0$ and, again, $b=d=\pi/2$.
|
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|
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$ Where do I start to solve a equation for x like the one below?
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
After squaring it, it's too complicated; but there's nothing to factor or to expand?
Ideas?
|
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
$$\implies \sqrt{7x-4}+\sqrt{4x-2}=\sqrt{4x-1}+\sqrt{7x-5} $$
Squaring we get, $$7x-4+4x-2+2\sqrt{(7x-4)(4x-2)}=4x-1+7x-5+2\sqrt{(4x-1)(7x-5)}$$
$$\implies \sqrt{(7x-4)(4x-2)}= \sqrt{(4x-1)(7x-5)}$$
$$\text{Squaring we get, } (7x-4)(4x-2)=(4x-1)(7x-5)$$
$$\text{ On simplification we get, 8-(16+14)x=5-x(20+7)} $$
$$\implies x=1,\text{ which evidently satisfies the given equation. }$$
Generalization:
If $$\sqrt{ax+b}-\sqrt{ax+c}=\sqrt{dx+e}-\sqrt{dx+f} \text{ where }b-c=e-f$$
$$\implies \sqrt{ax+b}+\sqrt{dx+f}=\sqrt{dx+e}+\sqrt{ax+c} $$
Squaring we get, $$ax+b+dx+f+2\sqrt{(ax+b)(dx+f)}=dx+e+ax+c+2\sqrt{(dx+e)(ax+c)} $$
$$\implies \sqrt{(ax+b)(dx+f)}=\sqrt{(dx+e)(ax+c)}\text{ as } b-c=e-f\iff b+f=c+e $$
$$\text{Squaring we get, } (ax+b)(dx+f)=(dx+e)(ax+c)$$
$$\implies x(bd+af-cd-ae)=ec-bf $$
|
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|
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
|
We will use that $\lim\limits_{x\to0}\frac{\sin(x)}{x}=1$, which can be shown geometrically.
First note that
$$
\begin{align}
\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}
&=\frac{\sin^2(x)-x^2}{x^2\sin^2(x)}\\
&=\frac{\sin^2(x)-x^2}{x^4}\left(\frac{\sin(x)}{x}\right)^{-2}\\
&=\frac{\sin(x)-x}{x^3}\left(\frac{\sin(x)}{x}+1\right)\left(\frac{\sin(x)}{x}\right)^{-2}\tag{1}
\end{align}
$$
Next, we have
$$
\begin{align}
\frac{\sin(x/2^{k-1})-2\sin(x/2^k)}{x^3}
&=\frac{2\sin(x/2^k)(\cos(x/2^k)-1)}{x^3}\\
&=\frac{\sin(x/2^k)}{x^3}\frac{2(\cos^2(x/2^k)-1)}{\cos(x/2^k)+1}\\
&=-2^{-3k}\left(\frac{\sin(x/2^k)}{x/2^k}\right)^3\frac2{\cos(x/2^k)+1}\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
\lim_{x\to0}\frac{\sin(x)-x}{x^3}
&=\lim_{x\to0}\sum_{k=1}^\infty2^{k-1}\frac{\sin(x/2^{k-1})-2\sin(x/2^k)}{x^3}\\
&=\lim_{x\to0}\sum_{k=1}^\infty-2^{-2k-1}\left(\frac{\sin(x/2^k)}{x/2^k}\right)^3\frac2{\cos(x/2^k)+1}\\
&=\sum_{k=1}^\infty-2^{-2k-1}\\
&=-\frac16\tag{3}
\end{align}
$$
Plugging $(3)$ into $(1)$, we get
$$
\begin{align}
\lim_{x\to0}\left(\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}\right)
&=-\frac16\cdot2\cdot1^2\\
&=-\frac13\tag{4}
\end{align}
$$
|
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|
If $17 \mid \frac{n^m - 1}{n-1}$ find the values of $n$ where $m$ is even but not divisible by $4$
Let $m, n \in \mathbb{Z}_+$ with $n > 2$, and let $\frac{n^m-1}{n-1}$ be divisible by $17$. Show that either $m$ is even:$ m \equiv 0 \mod 17$ and $n \equiv 1 \mod 17$. Find all possible values of $n$ in the cases when $m$ is even but not divisible by $4$, or divisible by $4$ but not divisible by $8$.
So far, I have done this: First, let's assume that $n \equiv 1 \mod 17$. Also, we know that we can write
$$\frac{n^m-1}{n-1} = \sum_{i = 0}^{m-1} n^i.$$
As $n \equiv 1 \mod 17 \implies n^i \equiv 1 \mod 17$ and so $\sum_{i = 0}^{m-1} n^i \equiv \underbrace{1 + 1 + \cdots + 1} \equiv m$.
Also, we know that $\sum_{i = 0}^{m-1} n^i \equiv 0 \mod 17 \implies m \equiv 0 \mod 17$.
Now, let's assume that $n \not\equiv 1 \mod 17$. Then, $\frac{n^m - 1}{n - 1}$ is divisible by $17$ if $n^m - 1$ is divisible by $17$, i.e $n^m - 1 \equiv 0 \mod 17 \implies n^m \equiv 1 \mod 17$. Fermat's little theorem tells us that $n^{p-1} \equiv 1 \mod p \implies n^{16} \equiv 1 \mod 17$. From here, we have that $m \mid 16$ which tells us that $m$ must be even.
However I am now stuck on the next bit, I'm not sure how to find those values of $n$. Can someone help me please. Thank you.
|
As the first case is perfect, let me start with the second.
As $17$ is prime and hence $\phi(17)=16$
As Primitive Root is a Generator of Reduced Residue System, let's find a primitive root of $17$
$2^4\equiv-1, 2^8\equiv(-1)^2\equiv1\pmod{17} \implies \text{ord}_{17}2=8<\phi(17)$
$3^2\equiv9, 3^4\equiv-4,3^8=(3^4)^2\equiv(-4)^2\equiv-1\pmod {17}$
$\implies \text{ord}_{17}3=16$ i.e., $3$ is a primitive root of $17$
From this, if $d=$ord$_ma,$ ord$_m(a^r)=\frac d{(r,d)}$ where $m,d$ are positive integers and $a,r$ are integers.
So, ord$_{17}(3^n)=\frac{16}{(16,n)}$
Case $1:$ If ord$_{17}(3^n)\equiv2\pmod 4, n=8s$ where $s$ is odd $\implies 3^{8s}=(3^8)^s\equiv(-1)^s\equiv-1\pmod{17}\implies $ord$_{17}(-1)=2$
Case $2:$ If ord$_{17}(3^n)\equiv4\pmod 8, n=4s$ where $s$ is odd
$\implies n\equiv4,12\pmod{16}$
Now, $3^4\equiv-4\pmod{17}$ and $3^{12}=3^8\cdot3^4\equiv(-1)(-4)\equiv4\pmod {17}$
$\implies $ord$_{17}(\pm4)=4$
Case $3:$ If ord$_{17}(3^n)\equiv8\pmod {16}, n=2s$ where $s$ is odd
$\implies n\equiv2,6,10,14\pmod{16}$
|
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|
Telescoping sum of powers
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
\vdots &=& &\vdots & \\
3^3-2^3 &= &3(3^2) &-3(3) &+1\\
2^3-1^3 &= &3(2^2) &-3(2) &+1\\
1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$
Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$
How do this transformation was done?
Thanks!
|
$(x-1)^3=x^3-3x^2+3x-1$; now set $x=n-1$.
|
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|
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$
If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d
Is there any help?
|
If you inspect the system you can see that the solutions for a,b,c,d are all going to have to be small integers. This constrains the system heavily.
I managed to solve the system guessing some small numbers, testing them and tweaking minus signs to get the desired result.
After we have found a,b,c,d the fraction becomes trivial.
Any analytical method to solving this system would have to accommodate high order solutions and would inevitably be very complicated. I would like to see analytical solution but doubt any could be as efficient as trial and error in this case.
|
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|
find the value of $\int \frac {x^2}{({x\sin x+\cos x})^2}\,dx$ In my textbook this question solved in this way:
they take since $$\dfrac{d}{dx}(x\sin x+\cos x)=x\cos x $$ so,
$$\int \dfrac {x^2}{({x\sin x+\cos x})^2}\,dx$$
$$\int \dfrac {x\cos x}{({x\sin x+\cos x})^2}\cdot \dfrac {x}{\cos x}\,dx$$
Is there any other way to do this question? I know substitution, but I have never encountered this type of arrangement on the basis of the differential.
|
Let $u=x\sin{x}+\cos{x}$ so that $du=x\cos{x}~dx$. Note that we chose this since $u$ is composed within another function. Unfortunately, $x\cos{x}$ doesn't quite appear in the integral, but's let's use wishful thinking and pretend that it did. Thus, we can easily solve an integral of the following form:
$$
\int \dfrac {x\cos x}{({x\sin x+\cos x})^2}\,dx = \int \dfrac {1}{u^2}\,dx =\dfrac{u^{-1}}{-1}+K=\dfrac{-1}{x\sin x + \cos x}+K
$$
Now in order to make the above work useful, we multiply by $1$ in a fancy way so that $x\cos{x}$ appears in the integral:
$$\int \dfrac{x^2}{(x\sin{x}+\cos{x})^2} dx = \int \dfrac{x}{\cos{x}} \cdot \dfrac{x\cos{x}}{(x\sin{x}+\cos{x})^2} dx $$
Now we use integration by parts. By pattern matching, this suggests that we let $f=\dfrac{x}{\cos{x}}$ and $dg=\dfrac {x\cos x}{({x\sin x+\cos x})^2}\,dx$. Then $df=\dfrac{x\sin{x}+\cos{x}}{\cos^2{x}}dx$ and $g=\dfrac{-1}{x\sin x + \cos x}$, which yields:
$$ \begin{align*}
\int \dfrac{x^2}{(x\sin{x}+\cos{x})^2} dx &= \int \dfrac{x}{\cos{x}} \cdot \dfrac{x\cos{x}}{(x\sin{x}+\cos{x})^2} dx \\
&= \dfrac{x}{\cos{x}} \cdot \dfrac{-1}{x\sin x + \cos x} - \int \dfrac{-1}{x\sin x + \cos x} \cdot \dfrac{x\sin{x}+\cos{x}}{\cos^2{x}}dx \\
&= \dfrac{-x}{\cos x (x\sin{x}+\cos{x})} + \int \sec^2 {x}~dx \\
&= \dfrac{-x}{\cos x (x\sin{x}+\cos{x})} + \tan{x} + C \\
&= \dfrac{-x}{\cos x (x\sin{x}+\cos{x})} + \dfrac{\sin{x}}{\cos{x}} + C \\
&= \dfrac{-x + \sin x (x\sin{x}+\cos{x})}{\cos x (x\sin{x}+\cos{x})} + C \\
&= \dfrac{-x(\sin^2 x + \cos^2 x) + \sin x (x\sin{x}+\cos{x})}{\cos x (x\sin{x}+\cos{x})} + C \\
&= \dfrac{(-x\sin^2 x -x\cos^2 x) + (x\sin^2{x}+\sin{x}\cos{x})}{\cos x (x\sin{x}+\cos{x})} + C \\
&= \dfrac{\sin{x}\cos{x}-x\cos^2 x}{\cos x (x\sin{x}+\cos{x})} + C \\
&= \dfrac{\cos{x}(\sin{x}-x\cos x)}{\cos x (x\sin{x}+\cos{x})} + C \\
&= \dfrac{\sin{x}-x\cos x}{x\sin{x}+\cos{x}} + C \\
\end {align*} $$
|
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|
A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.
Any hints? Thanks!
|
Here's a kind of ugly way of doing it:
$$\begin{align*}
(12n)^{\log_{4n}(40\sqrt{3})}&=(12n)^{\log_{3n}(45)}\\\\
(3)^{\log_{4n}(40\sqrt{3})}(4n)^{\log_{4n}(40\sqrt{3})}&=(4)^{\log_{3n}(45)}(3n)^{\log_{3n}(45)}\\\\
(3)^{\log_{4n}(40\sqrt{3})}\cdot 40\sqrt{3}&=(4)^{\log_{3n}(45)}\cdot 45\\\\
(3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(4)^{\log_{3n}(45)-\frac{3}{2}}\\\\
(3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(3)^{\log_3(4)\cdot(\log_{3n}(45)-\frac{3}{2})}\\\\
\log_{4n}(40\sqrt{3})-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\
\log_{3n}(45)-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\
\end{align*}$$
Therefore
$$\begin{align*}
\log_{3n}(45)&=\frac{3}{2}\\\\
2025&=(3n)^3\\\\
n^3&=\frac{2025}{27}=75
\end{align*}$$
|
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|
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.
Can you suggest any ideas how to calculate $S$?
|
Write down the function
$$ g(z) = \sum_{n\geq1} \frac{z^n}{n}H_n^2, $$
so that $S=g(-1)$ and $g$ can be reduced to
$$ zg'(z) = \sum_{n\geq1} z^n H_n^2 = h(z). $$
Now, using $H_n = H_{n-1} + \frac1n$ ($n\geq2$), we can get a closed form for $h(z)$:
$$h(z) = z + \sum_{n\geq2}\frac{z^n}{n^2} + \sum_{n\geq 2}z^n H_{n-1}^2 + \sum_{n\geq 2} 2\frac{z^n}{n}H_{n-1}. $$
Now, the first and third sums Mathematica can evaluate itself in closed form (the third one evaluates to the function $p(z)$ below, the first one is $\text{Li}_2(z)-z$), and the middle sum is $z h(z)$.
Substituting this into the expression for $g(z)$, we get
$$g(z) = \int \frac{\text{Li}_2(z) + p(z)}{z(1-z)}\,dz, $$
$$p(z) = -\frac{\pi^2}{3} + 2\log^2(1-z)-2\log(1-z)\log(z)+2\text{Li}_2((1-z)^{-1}) - 2\text{Li}_2(z). $$
Mathematica can also evaluate this integral, giving (up to a constant of integration)
\begin{align}
g(z) &= \frac{1}{3} \left(-2 \log(1-z^3+3 \log(1-z)^2 \log(-z)+\log(-1+z)^2 (\log(-1+z)+3 \log(-z) \right. \\
& \hspace{5mm} \left. -3 \log(z))+\pi ^2 (\log(-z)-2 \log(z))+\log(1-z) \left(\pi^2 - 3 \log(-1+z)^2 \right. \right.\\
& \hspace{5mm} \left.\left. +6 (\log(-1+z)-\log(-z)) \log(z)\right)-6 (\log(-1+z)-\log(z)) \left(\text{Li}_{2}\left(\frac{1}{1-z}\right)-\text{Li}_{2}(z)\right) \right.\\
& \hspace{10mm} \left. -3 \log(1-z) \text{Li}_{2}(z)+3 \text{Li}_{3}(z)\right).
\end{align}
The constant of integration is fixed by requiring $g(0)=0$.
Some care needs to be taken, because the function
is multi-valued, when evaluating $g(-1)$. The answer is
$$ \frac{1}{12}(\pi^2\log2-4(\log 2)^3-9\zeta(3)). $$
|
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}
|
What is $\lim\limits_{n→∞}(\frac{n-x}{n+x})^{n^2}$? What is $$\lim_{n\rightarrow\infty}\left(\frac{n-x}{n+x}\right)^{n^2},$$ where $x$ is a real number. Mathematica tells me the limit is $0$ when I put an exact value for $x$ in (Mathematica is inconclusive if I don't substitute for $x$), but using $f(n)=(n-x)^{n^2}$ and $g(n)=(n+x)^{n^2}$, then $$\lim_{n\rightarrow\infty}f(n)=\lim_{n\rightarrow\infty}g(n)=\infty,$$ and by L'Hopital's rule we have
$$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow\infty}\frac{(s-z)^{-1+s^2} (s+z)^{1-s^2} (s+2 (s-z) \text{Log}[s-z])}{s+2 (s+z) \text{Log}[s+z]} = 1.$$
I'm not sure which to believe - Mathematica or L'Hopital.
|
The interesting form is
$\left(\frac{n-x}{n+x}\right)^{n}$.
The $n^2$ just blows things up.
Taking the log,
$n \ln \frac{n-x}{n+x}
= n \ln \frac{1-x/n}{1+x/n}
= n \big(\ln (1-x/n)- \ln(1+x/n)\big)
$.
Using $\ln(1+z) = z-z^2/2+z^3/3-z^4/4+...$
and
$\ln(1-z) = -z-z^2/2-z^3/3-z^4/4+...$,
and writing $z$ for $x/n$,
this is
$n((-z-z^2/2-z^3/3-z^4/4+...)-(z-z^2/2+z^3/3-z^4/4+...))
= n(-2z-2z^3/3-...)
=-2n(x/n+x^3/(3n^3))
=-2x-2x^3/(3n^2)-...
$.
So $\left(\frac{n-x}{n+x}\right)^{n}
=\exp(-2x-2x^3/(3n^2)-...)
=\exp(-2x)exp( -2x^3/(3n^2)-...)
=\exp(-2x)(1-2x^3/(3n^2) + O(1/n^4))
$.
If you use the $n^2$,
this behaves like
$\exp(-2xn)$
which $\to 0$ if $x > 0$
and $\to \infty$ if $x < 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_{-1}^{1}\frac{\arctan{x}}{1+x}\ln{\left(\frac{1+x^2}{2}\right)}dx$ This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$
where $\arctan{x}=\tan^{-1}{x}$
mark: this integral is my favorite one. Thanks to whoever has nice methods.
I have proved the following:
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$
where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$
I also have got a few by-products
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$
where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$
and
$$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$
and same methods,I have follow integral
$$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$
where $ K $ denotes Catalan's Constant.
|
Here is a solution that only uses complex analysis:
Let $\epsilon$ > 0 and consider the truncated integral
$$ I_{\epsilon} = \int_{-1+\epsilon}^{1} \frac{\arctan x}{x+1} \log\left( \frac{1+x^2}{2} \right) \, dx. $$
By using the formula
$$ \arctan x = \frac{1}{2i} \log \left( \frac{1 + ix}{1 - ix} \right) = \frac{1}{2i} \left\{ \log \left( \frac{1+ix}{\sqrt{2}} \right) - \log \left( \frac{1-ix}{\sqrt{2}} \right) \right\}, $$
it follows that
$$ I_{\epsilon} = \Im \int_{-1+\epsilon}^{1} \frac{1}{x+1} \log^{2} \left( \frac{1+ix}{\sqrt{2}} \right) \, dx. $$
Now let $\omega = e^{i\pi/4}$ and make the change of variable $z = \frac{1+ix}{\sqrt{2}}$ to obtain
$$ I_{\epsilon} = \Im \int_{L_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz, $$
where $L_{\epsilon}$ is the line segment joining from $\bar{\omega}_{\epsilon} := \bar{\omega} + \frac{i\epsilon}{\sqrt{2}}$ to $\omega$. Now we tweak this contour of integration according to the following picture:
That is, we first draw a clockwise circular arc $\gamma_{\epsilon}$ centered at $\bar{\omega}$ joining from $\bar{\omega}_{\epsilon}$ to some points on the unit circle, and draw a counter-clockwise circular arc $\Gamma_{\epsilon}$ joining from the endpoint of $\gamma_{\epsilon}$ to $\omega$. Then
$$ I_{\epsilon} = \Im \int_{\gamma_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz + \Im \int_{\Gamma_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz =: J_{\epsilon} + K_{\epsilon}. $$
It is easy to check that as $\epsilon \to 0^{+}$, the central angle of $\gamma_{\epsilon}$ converges to $\pi / 4$. Since $\gamma_{\epsilon}$ winds $\bar{\omega}$ clockwise, we have
$$ \lim_{\epsilon \to 0^{+}} J_{\epsilon} = \Im \left( -\frac{i \pi}{4} \mathrm{Res}_{z=\bar{\omega}} \frac{\log^2 z}{z - \bar{\omega}} \right) = \frac{3}{2} \frac{\pi^3}{96}. $$
Also, by applying the change of variable $z = e^{i\theta}$,
$$ K_{\epsilon} = -\Re \int_{-\frac{\pi}{4}+o(1)}^{\frac{\pi}{4}} \frac{\theta^2}{1 - \bar{\omega}e^{-i\theta}} \, d\theta = \int_{-\frac{\pi}{4}+o(1)}^{\frac{\pi}{4}} \frac{\theta^2}{2} \, d\theta. $$
Thus taking $\epsilon \to 0^{+}$, we have
$$ \lim_{\epsilon \to 0^{+}} K_{\epsilon} = - \int_{0}^{\frac{\pi}{4}} \theta^2 \, d\theta = - \frac{1}{2} \frac{\pi^3}{96}. $$
Combining these results, we have
$$ \int_{-1}^{1} \frac{\arctan x}{x+1} \log \left( \frac{x^2 + 1}{2} \right) \, dx = \frac{\pi^3}{96}. $$
The same technique shows that
$$ \int_{-1}^{1} \frac{\arctan (t x)}{x+1} \log \left( \frac{1 + x^2 t^2}{1 + t^2} \right) \, dx = \frac{2}{3} \arctan^{3} t, \quad t \in \Bbb{R} .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Product of nilpotent matrices. Let $A$ and $B$ be $n \times n$ complex matrices and
let $[A,B] = AB - BA$.
Question: If $A , B$ and $[A,B]$ are all nilpotent matrices,
is it necessarily true that $\operatorname{trace}(AB) = 0$?
If,in fact, $[A,B] = 0$, then we can take $A$ and $B$ to be strictly upper triangular matrices so that the answer would be yes in this very special case.
|
No. While the conjecture is true for $n=2$ and computer experiments suggest that it may also be true for $n=3$, counterexamples are abundant when $n=4$. Let $A$ be the $4\times4$ Jordan block. The following $B$s are some computer generated random counterexamples that satisfy the condition $B^4=(AB-BA)^4=0$ but $\mathrm{trace}(AB)\neq0$.
\begin{align*}
\pmatrix{0&0&0&0\\ 0&1&1&0\\ -1&-1&-1&0\\ 1&0&0&0},
&\pmatrix{
4& -4& 4& -6\\
2& -2& 2& -3\\
14&-14& 14&-13\\
16&-16& 16&-16
},\\
\\
\pmatrix{
1& 1& 1& 2\\
0& 0& 0& 0\\
1& 2& 1& 2\\
-1& -2& -1& -2\\
},
&\pmatrix{
-1& -1& 0& 0\\
1& 1& 0& 0\\
0& 0& 0& 1\\
1& 1& 0& 0\\
}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/408499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
}
|
Proof of the inequality $(x+y)^n\leq 2^{n-1}(x^n+y^n)$ Can you help me to prove that
$$(x+y)^n\leq 2^{n-1}(x^n+y^n)$$
for $n\ge1$ and $x,y\ge0$.
I tried by induction, but I didn't get a result.
|
Look. I also have tried to do it by induction. It is obvious that it holds for $n=1$ and $n=2$. Assume that it also holds for $n$. Let's prove that inequality for $n+1$:
$$
2^n(x^{n+1} + y^{n+1}) - (x+y)^{n+1} = 2^{n}(x^{n+1} + y^{n+1}) -(x+y)^n(x+y)\geq 2^n(x^{n+1} + y^{n+1}) - 2^{n-1}(x^n + y^n)(x+y) =
2^{n-1}(x^{n+1} + y^{n+1} - x^ny - y^nx) = 2^{n-1}(y(y^n - x^n) + x(x^n - y^n))=
2^{n-1}(y^n-x^n)(y-x) = 2^{n-1}(y-x)^2(y^{n-1}+xy^{n-2}+x^2y^{n-2}+\dots+y^2x^{n-2} + x^{n-1}) \geq 0.
$$
So we have proved that $2^n(x^{n+1} + y^{n+1}) - (x+y)^{n+1} \geq 0$. It is similar to $(x+y)^{n+1} \leq 2^n(x^{n+1} + y^{n+1})$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction that $1^3 + 2^3 + 3^3 + .....+ n^3= \frac{n^2(n+1)^2}{4}$ for all $n\geq1$. Use mathematical induction to prove that $1^3 + 2^3 + 3^3 + .....+ n^3= \frac{n^2(n+1)^2}{4}$ for all $n\geq1$.
Can anyone explain? Because I have no clue where to begin. I mean, I can show that $1^3+ 2^3 +...+ (k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$, but then I don't know where to go. I need further explanation to prove it.
thank you so much for help
Sincerely
|
Hints:
$$1^3=\frac{1^2\cdot2^2}4\;\;\color{green}\checkmark$$
$$1^3+2^3+\ldots+n^3+(n+1)^3\stackrel{\text{Ind. Hyp.}}=\frac{n^2(n+1)^2}4+(n+1)^3=$$
$$=\frac{(n+1)^2}4\left(n^2+4(n+1)\right)=\ldots\ldots\;\;\;\color{green}\checkmark$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/411485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is grateful. Thanks.
|
Applying L'Hopital's Rule, we get
$$
\lim_{x\rightarrow1}{\frac{\frac{1}{2\sqrt{x+1}}+\frac{2x}{2\sqrt{x^2-1}}+\frac{3x^2}{2\sqrt{x^3+1}}}{\frac{1}{2\sqrt{x-1}}+\frac{2x}{2\sqrt{x^2+1}}+\frac{4x^3}{2\sqrt{x^4+1}}}}.
$$
The $2$ in each denominator cancels, and we multiply the numerator and denominator by
$$
\sqrt{x-1}\cdot\sqrt{x^2+1}\cdot\sqrt{x^4+1}
$$
to get
$$
\lim_{x\rightarrow1}{\frac{\left(\sqrt{x-1}\cdot\sqrt{x^2+1}\cdot\sqrt{x^4+1}\right)\frac{1}{\sqrt{x+1}}+\frac{2x}{\sqrt{x^2-1}}+\frac{3x^2}{\sqrt{x^3+1}}}{\sqrt{x^2+1}\sqrt{x^4+1}+2x\sqrt{x-1}\sqrt{x^4+1}-4x^3\sqrt{x-1}\sqrt{x^2+1}}}.
$$
Substituting $x=1$ in the denominator yields $2$, and distributing in the numerator gives us
$$
\frac{1}{2}\lim_{x\rightarrow 1}{\left( \frac{\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x+1}}+\frac{2x\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x^2-1}}-\frac{3x^2\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x^3+1}} \right)}.
$$
The first and third terms go to zero, so factoring the denominator gives us
$$
\frac{1}{2}\lim_{x\rightarrow 1}{\frac{2x\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x-1}\sqrt{x+1}}}.
$$
The $\sqrt{x-1}$ terms cancel, and plugging in $x=1$ gives us
$$
\frac{1}{2}\cdot2\sqrt{2}=\sqrt{2}.
$$
This isn't exactly the simplest or most elegant method, but it works.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit : $\lim\limits_{n\rightarrow\infty}\int_{n}^{n+7}\frac{\sin{x}}{x}\,\mathrm dx$ I have this exercise I don't know how to approach :
Find the limit : $$\lim_{n\rightarrow\infty}\int_{n}^{n+7}\frac{\sin x}{x}\,\mathrm dx$$
I can see that with $n\rightarrow\infty$ the area under the graph of this function becomes really small as $\sin{x} \leq 1$ so $\dfrac{\sin{n}}{n}\rightarrow_{\infty}0$ but can I get something from it?
|
No I don't think that there is a lot that can be got out of it. For example the result is not yet sufficient condition for convergence of the integral from $0$ to $\infty$. However, the Riemann improper integral of the first kind
\begin{equation}
\int_0^\infty \textrm{sinc}(x/\pi) dx
\end{equation}
where $\textrm{sinc}(x) = \Bigg\{ \begin{eqnarray} & \frac{\sin(\pi x)}{\pi x} & , \ \ x \neq 0 \\ &1&, \ \ x = 0 \end{eqnarray}$
converges as we know. Define
\begin{equation}
a_n = \int_0^n \textrm{sinc}(x/\pi) dx \ .
\end{equation}
Now $\{a_n\}_{n=1}^\infty$ is a Cauchy-sequence. The integral under the limit can be expressed as follows.
\begin{eqnarray}
\int_n^{n+7} \frac{\sin(x)}{x} dx & = & \int_n^{n+7} \textrm{sinc}(x/\pi) dx = \int_0^{n+7} \textrm{sinc}(x/\pi) dx - \int_0^n \textrm{sinc}(x/\pi) dx \\
\\
\\
\\
& = & a_{n+7} - a_n \rightarrow 0
\end{eqnarray}
as $n \rightarrow \infty$.
Without the knowledge that the integral from $0$ to $\infty$ converges we can proceed as follows. We first show that $a_n$ is a Cauchy-sequence. We may assume $m \leq n$. Otherwise we can apply $|a_n-a_m| = |a_m-a_n|$ and interchange the order of $m$ and $n$ in the first term of the following calcualtion. We have
\begin{eqnarray}
|a_n-a_m| & = & \bigg| \int_0^n \textrm{sinc}(x/\pi)dx - \int_0^m \textrm{sinc}(x/\pi)dx \bigg| = \bigg| \int_m^n \textrm{sinc}(x/\pi)dx \bigg| \\
& = & \bigg| \int_m^n \frac{\sin(x)}{x}dx \bigg| = \bigg| \bigg|_m^n \frac{-\cos(x)}{x} - \int_m^n -\frac{-\cos(x)}{x^2} dx \bigg| \\
& = & \bigg|-\frac{\cos(n)}{n}+\frac{\cos(m)}{m}-\int_m^n \frac{\cos(x)}{x^2} dx \bigg| \\
& \leq & \bigg|-\frac{\cos(n)}{n} \bigg| + \bigg| \frac{\cos(m)}{m} \bigg| + \bigg| \int_m^n \frac{\cos(x)}{x^2} dx \bigg| \\
& \leq & \bigg|\frac{\cos(n)}{n} \bigg| + \bigg| \frac{\cos(m)}{m} \bigg| + \int_m^n \bigg| \frac{\cos(x)}{x^2} \bigg| dx \\
& \leq & \frac{|\cos(n)|}{|n|} + \frac{|\cos(m)|}{|m|} + \int_m^n \frac{|\cos(x)|}{|x^2|} dx \\
& \leq & \frac{1}{n} + \frac{1}{m} + \int_m^n \frac{1}{x^2} dx \leq \frac{1}{n} + \frac{1}{m} + \bigg|_m^n -\frac{1}{x} \\
& \leq & \frac{1}{n} + \frac{1}{m} - \frac{1}{n} + \frac{1}{m} \leq \frac{1}{n} + \frac{1}{m} + \frac{1}{n} + \frac{1}{m} = \frac{2}{n} + \frac{2}{m} \rightarrow 0
\end{eqnarray}
as $m, \ n \rightarrow \infty$. Hence $a_n$ is a Cauchy-sequence. We can then calculate as above after the definition of $a_n$ and obtain the desired result
\begin{equation}
\lim_{n \rightarrow \infty} \int_n^{n+7} \frac{\sin(x)}{x} dx = 0 \ .
\end{equation}
The existence of the integral from $0$ to $\infty$ can be now established by calculating the distance $|a_M-a|$ where $a$ is the limit of the Cauchy-sequence. For the next integer $m$ from $M$ we have
\begin{equation}
|a_M-a| = |a_M-a_m+a_m-a| \leq |a_M-a_m| + |a_m-a|
\end{equation}
the second term goes to $0$ by definition of limit and the first can be estimated as you thought in your question. Hence
\begin{equation}
a = \lim_{M \rightarrow \infty} a_M = \lim_{m \rightarrow \infty} \int_0^M \textrm{sinc}(x/\pi) dx = \int_0^\infty \textrm{sinc}(x/\pi) dx \ .
\end{equation}
Hopefully you got some perspective to the integration of $\frac{\sin(x)}{x}$.
Then the exercise. We calculate using the previous estimate
\begin{eqnarray}
\Bigg|\int_n^{n+7} \frac{\sin(x)}{x} dx \Bigg| \leq \frac{1}{n+7} + \frac{1}{n} - \frac{1}{n+7} + \frac{1}{n} = \frac{2}{n} \rightarrow 0 \ ,
\end{eqnarray}
as $n \rightarrow \infty$. Note that we succeed to get rid of the $7$ in the question. Another approach is to estimate
\begin{eqnarray}
\sup_{x \in [0,7]} \Bigg|\frac{\sin(x+n)}{x+n} - 0 \Bigg| \leq \sup_{x \in [0,7]} \frac{1}{n} = \frac{1}{n} \rightarrow 0 \ ,
\end{eqnarray}
as $n \rightarrow \infty$. Hence $\frac{\sin(x+n)}{x+n}$ converges uniformly to $0$. Hence
\begin{eqnarray}
\lim_{n \rightarrow \infty} \int_n^{n+7} \frac{\sin(x)}{x} dx = \lim_{n \rightarrow \infty} \int_0^{7} \frac{\sin(x+n)}{x+n} dx = \int_0^7 \lim_{n \rightarrow \infty} \frac{\sin(x+n)}{x+n} dx = \int_0^7 0 dx = 0 \ ,
\end{eqnarray}
where the order of integration and limit is changed by uniform convergence of functions defined on the compact set $[0,7]$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of $\lim\limits_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$ I want to evaluate this limit.
$$\lim_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$$
What I did is:
set $f(x)=x^\frac{1}{3}$
$$f(1)+\int^\infty_0 x^\frac{1}{3} \, dx<f(1)+f(2)+\dots+f(n)<f(n)+\int^\infty_0 x^\frac{1}{3}\,dx$$
how it can help me to evaluate this limit?
I can convert the right\left expressions to limits? or the whole inequality to expression of limit.
I need some advice here,
Thanks!
|
We have
$$\sum_{k=1}^n k^{\frac{1}{3}}\sim_\infty\int_1^nx^{\frac{1}{3}}dx\sim_\infty\frac{3}{4}n^{\frac{4}{3}}$$
so
$$\lim_{x\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)=\frac{3}{4}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
For any positive integer $n$, is it possible to find a nonzero integer $p$ so that $p^2$ is the sum of $i$ nonzero squares for all $1 \leq i \leq n$? I need to prove this result for something I am working on:
For any positive integer $n$, is it possible to find a nonzero integer $p$ so that $p^2$ is the sum of $i$ nonzero squares for all $1 \leq i \leq n$?
What I'd like to do is as follows. First find $a,b,c$ nonzero integers so that $a^2 + b^2 = c^2$. Then find a nonzero integers $d$ and $e$ so that $c^2 + d^2 = e^2$. Then find nonzero integers $f$ and $g$ so that $e^2 + f^2 = g^2$. Note that then
$$
g^2 = e^2 + f^2 = c^2 + d^2 + f^2 = a^2 + b^2 + d^2 + f^2
$$
and I'd like to continue in this way. However, I am wondering if this construction is even possible? Am I always guaranteed that such a string of integers exist? If so, how do I prove this rigorously? Another thought of mine is use Euclid's Formula. For example, let $m = 3$ and $n = 1$. Then
$$
\begin{align*}
a &= 3^2 - 1^2 = 8\\
b &= 2\cdot 3 \cdot 1 = 6\\
c &= 3^2 + 1^2 = 10.
\end{align*}
$$
Furthermore, $c = 2\cdot 5 \cdot 1$, so letting $m = 5$ and $n = 1$ gives
$$
\begin{align*}
d &= 5^2 - 1^2 = 24\\
e &= 5^2 + 1^2 = 26.
\end{align*}
$$
And again, $e = 2\cdot 13 \cdot 1$, so letting $m = 13$ and $n = 1$ gives
$$\begin{align*}
f &= 13^2 - 1^2 = 168\\
g &= 13^2 + 1^2 = 170.
\end{align*}$$
Then
$$\begin{align*}
g^2 = 170^2 = 26^2 + 168^2 = 10^2 + 24^2 + 168^2 = 6^2 + 8^2 + 24^2 + 168^2
\end{align*}
$$
Going one more step we can write $170 = 2\cdot 17 \cdot 5$ so $m = 17$ and $n = 5$ gives
$$\begin{align*}
h &= 17^2 - 5^2 = 264\\
i &= 17^2 + 5^2 = 314,
\end{align*}$$
in which $314 = 2\cdot 157 \cdot 1$. It seems I may be on the right track, but I'm not sure if this is true in general. Maybe it can be proven that this process can be continued arbitrarily given we start with two distinct primes $m$ and $n$ in the first step? Any thoughts or references or anything would be wonderful. Thanks in advance!
|
Yes.
n is the sum of two squres iff all odd $ 3\mod4$ prime factors of n occur an even number of times and every positive integer is the sum of four possibly zero squares. The only positive integers not expressible as the sum of four non-zero squares are $1,3,5,9,11,17,29,41,2\times 4^n,6\times 4^n, 14\times 4^n$. If $i=4k+t$, there exist positive integers expressible as the sum of $k$ squares for $t=1,2,3$, (e.g. $1,2,3$ respectively). For $p^2$ sufficiently large $p^2-1, p^2-2 \text{ and } p^2-3$ can be partitioned as the sum of $k$ positive integers expressible as the sum of four non-zero squares, and hence as the sum of $i$ squares.
|
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|
Given that $x = 4\sin \left( {2y + 6} \right)$ find dy/dx in terms of x My attempt:
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr} $
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr
& {\cos ^2}\left( {2y + 6} \right) + {\sin ^2}\left( {2y + 6} \right) = 1 \cr
& {\cos ^2}\left( {2y + 6} \right) + {\left( {{x \over 4}} \right)^2} = 1 \cr
& {\cos ^2}\left( {2y + 6} \right) = 1 - {{{x^2}} \over {16}} \cr
& \cos \left( {2y + 6} \right) = \sqrt {{{16 - {x^2}} \over {16}}} \cr
& \cos \left( {2y + 6} \right) = {{\sqrt {16 - {x^2}} } \over 4} \cr
& {{dy} \over {dx}} = {1 \over {2\sqrt {16 - {x^2}} }} = 1 \cr} $
Okay I've got it right, but the official answer confuses me, it says:
${{dy} \over {dx}} = {1 \over {8cos\left( {\arcsin \left( {{x \over 4}} \right)} \right)}} = \left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
This is the part i'm struggling to get my head around, although I arrive at the same answer.
Okay
$\arcsin {x \over 4} = 2y + 6$
but how does the answer then go :
$\left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
is there a shortcut or trick I overlooked?
I think I need some sleep, thanks...
|
You got:
$\dfrac{dy}{dx}=\dfrac{1}{8\cos(2y+6)}$
And we have
$x=4\sin(2y+6)\implies y=\dfrac{1}{2}\left(\sin^{-1}\left(\dfrac{x}{4}\right)-6\right)$
Plug that in the top equation:
$\dfrac{dy}{dx}=\dfrac{1}{8\cos\left(\sin^{-1}\left(\dfrac{x}{4}\right)\right)}$
We know $\cos^2x+\sin^2x=1\implies \cos x=\pm\sqrt{1-\sin^2x}$
and therefore we say $\cos(\sin^{-1}x)=\pm\sqrt{1-x^2}$, we have
$\dfrac{dy}{dx}=\pm\dfrac{1}{8\sqrt{1-\dfrac{x^2}{16}}}\implies
\dfrac{dy}{dx}=\pm\dfrac{1}{2\sqrt{16-x^2}}$
|
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|
Finding the closest point to the origin of $y=2\sqrt{\ln(x+3) }$ Given $$y=2\sqrt{\ln(x+3) }$$
How do I determine a (x,y) pair satisfying the above relation which is the closest to the origin (0,0)?
|
To minimize the distance from the origin, minimize the square of the distance from the origin, given by $d^2 = M = x^2+y^2$,
$$M = x^2+y^2=x^2+(2\sqrt{\ln(x+3)})^2 = x^2 + 4\ln(x+3)$$
$$\frac{dM}{dx} = 2x + \frac{4}{x+3} = \frac{2x(x+3) + 4}{x+3} = \frac{2x^2 + 6x + 4}{x+3}.$$
This is only (possibly) equal to zero when the numerator is equal to zero,
$$2x^2 + 6x + 4 = 0 \implies x=-2 \ \ \text{and} \ \ x=-1$$
We know that at least one of these must be where the minimum occurs, so just plug the values of $x$ into the given equation and compute the distance from the origin.
$y(-1) = 2\sqrt{\ln(-1+3)} = 2 \sqrt{\ln(2)} = 1.665 \implies d^2 = (-1)^2 + (1.665)^2 = 2.386$
$y(-2) = 2\sqrt{\ln(-2+3)} = 2 \sqrt{\ln(1)} = 0 \implies d^2 = (-1)^2 + (0)^2 = 1$
So the point nearest to the origin is $(-2,0)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Differential equation Show that the differential equation $$\frac{dy}{dx} =\frac{e^x+x}{\sin y+2}$$ has a solution satisfying $y(0) = \pi$. To do this, separate variables and integrate to get an equation implicitly relating $x$ and $y$.
|
$$
\frac {dy}{dx} = \frac {e^x + x}{\sin y + 2} \\
(\sin y + 2) dy = (e^x + x) dx \\
\int (\sin y + 2) dy = \int (e^x + x) dx \\
-\cos y + 2y = e^x + \frac {x^2}2 + C
$$
Now, substitute $y(0) = \pi$
$$
1 + 2\pi = 1 + C
$$
from which you can find that $C = 2\pi$.
Final answer is
$$
2y - \cos y = e^x + \frac {x^2}2 + 2\pi
$$
is a solution of a given ODE and satisfies $y(0) = \pi$.
|
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|
How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
|
$$\int_0^a f(x)\;dx=2\int_0^\dfrac a2 f(x)\;dx\; if \;f(a-x)=f(x)\;$$
and$\int_0^a f(x)\;dx=0\;\;if \;f(a-x)=-f(x)$
here $f(x)=\sin ^6x\implies f(\pi-x)=\sin ^6(\pi-x)\implies \sin^6x\implies f(x)$
so we can write it as first identity:
$$I=\int_0^\dfrac \pi2 \sin^6 x\;dx$$
from identity:$\int_0^af(x)\;dx=\int_0^af{(a-x)}\;dx$
$$I=\int_0^\dfrac \pi2 \cos^6 x\;dx$$
$$2I=\int_0^\dfrac \pi2 \sin^6 x+\cos ^6 x\;dx$$
$$2I=\int_0^\dfrac \pi2 ({\sin^2 x})^3+({\cos ^2 x})^3\;dx$$
$$2I=\int_0^\dfrac \pi2 (\sin^2 x+\cos ^2 x)({\sin^4 x}+{\cos ^4 x}-\sin^2 x\cos^2 x)\;dx$$
$$2I=\int_0^\dfrac \pi2 ({\sin^4 x}+{\cos ^4 x}+2\sin^2 x\cos^2 x-3\sin^2 x\cos^2 x)\;dx$$
$$2I=\int_0^\dfrac \pi2 ({{\sin^2 x}+{\cos ^2 x}}^2)-3\sin^2 x\cos^2 x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-3\sin^2 x\cos^2 x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34{(2\sin x\cos x)}^2\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34\sin^2 2x\;dx$$
$$2I=\int_0^\dfrac \pi2 1-\dfrac 34\dfrac{(1-\cos 4x)}2\;dx$$
$$2I=\left[x-\dfrac 38(x-\dfrac{\sin 4x}4)\right]_0^\dfrac \pi2 \implies \left[\dfrac {5x}{8}+\dfrac{3\sin 4x}{32}\right]_0^\dfrac \pi2 \implies \left[\dfrac {5\pi}{16}\right]\implies I=\left[\dfrac {5\pi}{32}\right]$$
proof of identity:$$\int_0^{2a} f(x)\;dx=2\int_0^a f(x)\;dx\; if \;f(a-x)=f(x)\;$$
this identity can be prove using this identity:$$\int _0^{2a} f(x)=\int _0^a f(x)\;dx+\int_0^a f(2a-x)\;dx$$
when $f(2a-x)=f(x)$ this become
$$\int _0^{2a} f(x)=\int _0^a f(x)\;dx+\int_0^a f(x)\;dx$$
$$\int _0^{2a} f(x)=2\int_0^a f(x)\;dx$$
|
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|
Showing that $\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$
Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$
Using the substitution $t=\tan\frac{1}{2}x$
$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$
$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$
$=(2-2\sin^2\frac{1}{2}x)\,\mathrm dt$
How do you get this in the form of $t$ instead of $x$ using $\sin A=\dfrac{2t}{1+t^2}$ ?
$$=\int_0^{1/\sqrt3}\frac{2-2\sin^2\frac{1}{2}x}{1-\frac{2t}{1+t^2}}\mathrm dt\,??$$
|
I think you mean
$$\int_0^{\pi/3} \frac{dx}{1-\sin{x}}$$
which may be accomplished using the substitution $t=\tan{\frac{x}{2}}$. Then
$$dt = \frac12 \sec^2{\frac{x}{2}} dx= \frac12 (1+\tan^2{\frac{x}{2}}) dx = \frac12(1+t^2) dx$$
so that $dx = 2 dt/(1+t^2)$ Also
$$t^2=\frac{1-\cos{x}}{1+\cos{x}} \implies \cos{x} = \frac{1-t^2}{1+t^2} \implies \sin{x} = \frac{2 t}{1+t^2} $$
so that
$$1-\sin{x} = \frac{1+t^2-2 t}{1+t^2} = \frac{(1-t)^2}{1+t^2}$$
Then the integral is
$$2 \int_0^{1/\sqrt{3}} \frac{dt}{1+t^2} \frac{1+t^2}{(1-t)^2} = 2 \left [ \frac{1}{1-t}\right]_0^{1/\sqrt{3}} = 1+\sqrt{3} $$
|
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|
Number of solution for $xy +yz + zx = N$ Is there a way to find number of "different" solutions to the equation $xy +yz + zx = N$, given the value of $N$.
Note: $x,y,z$ can have only non-negative values.
|
equation:
$XY+XZ+YZ=N$
Solutions in integers can be written by expanding the number of factorization: $N=ab$
And using solutions of Pell's equation: $p^2-(4k^2+1)s^2=1$
$k$ -some integer which choose on our own.
Solutions can be written:
$X=ap^2+2(ak+b+a)ps+(2(a-2b)k+2b+a)s^2$
$Y=2(ak-b)ps+2(2ak^2+(a+2b)k+b)s^2$
$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$
And more:
$X=-2bp^2+2(k(4b+a)+b)ps-2((4b+2a)k^2+(2b-a)k)s^2$
$Y=-(2b+a)p^2+2(k(4b+a)-b-a)ps-(8bk^2-(4b+2a)k+a)s^2$
$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$X \sim U [-0.5 , 1.5] , Y = X^2$ Given $$ f(x) = \begin{cases} \frac12 &,\ -0.5 \le x\le 1.5\\0 &,\ \mbox{otherwise} \end{cases}$$
find the probability density function of $Y=X^2$.
To solve this I first divided up the pdf of X into three parts:
$$f(x) = \begin{cases} \frac12 &, \ -0.5 \le x\le 0\\\frac12 &, \ 0\le x\le 0.5 \\ \frac12 &,\ 0.5 \le x\le 1.5\\ 0 &, \ \mbox{otherwise} \end{cases}$$
Then applying $g^{-1}(y) = -\sqrt{y}$ for $-0.5 \le x \le 0 $ and $g^{-1}(y) = +\sqrt{y}$ for $ 0 \le x \le 0.5 $ and $ 0.5 \le x \le 1.5 $, I get:
$$ g(y) = \begin{cases} \frac{1}{4\sqrt{y}} &,\ 0 \le x \le 0.25\\ \frac{1}{4\sqrt{y}} &, \ 0\le x\le 0.25 \\ \frac{1}{4\sqrt{y}} &, \ 0.25 \le x \le 2.25 \\ 0 &, \ \mbox{otherwise} \end{cases}$$
Summing up the first two cases I get:
$$ g(y) = \begin{cases} \frac{1}{2\sqrt{y}} &, \ 0 \le x \le 0.25 \\ \frac{1}{4\sqrt{y}} &, \ 0.25 \le x \le 2.25\\ 0 &, \ \mbox{otherwise} \end{cases} $$
Could someone confirm whether my solution is correct or not? And whether my argumentation makes sense? Thanks!
|
It looks fine. However, for future reference, I highly recommend the following method: find the cdf of $X$, find the cdf of $Y$, differentiate the cdf to get back your pdf for $Y$. The steps for this problem would be as follows:
For $x<.5$, the cdf comes out to zero, and for $x>1.5$, it comes out to $1$. For the values in between, we have
$$
cdf_x=P(X<x)=\int_{-0.5}^{x}f(x)\,dx=\frac12(x+0.5)=\frac x2 + 0.25
$$
Now for y, we can say $cdf_y=P(X^2<y)=cdf_x(\sqrt{y})-cdf_x(-\sqrt{y})$. Long story short, we end up with
$$
cdf_y=
\begin{cases}
0 & y<0\\
\sqrt{y} & 0≤y≤0.25\\
0.25+\frac12 \sqrt{y} & 0.25≤y≤2.25\\
1 & y≥2.25
\end{cases}
$$
Differentiating, we find
$$
pdf_y=\frac{d}{dy}cdf_y=
\begin{cases}
0 & y<0\\
\frac1{2\sqrt{y}} & 0≤y≤0.25\\
\frac1{4\sqrt{y}} & 0.25≤y≤2.25\\
0 & y≥2.25
\end{cases}
$$
As you found.
(While in this case the work amounted to the same, this is a favorite method of mine because it works when the way in which you should deal with the multiple inverses isn't necessarily obvious).
|
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|
How to evaluate the trigonometric integral $\int \frac{1}{\cos x+\tan x }dx$
$$\int \dfrac{1}{\cos x+\tan x }dx$$
This can be converted to
$$\int \dfrac{\cos x}{\sin x+\cos^2x}dx$$
But from here I get stuck. Using t substitution will get you into a mess. Are there any tricks which can split the fraction into simpler forms?
|
As commented above the Weirstrass substitution
$$
\begin{equation*}
t=\tan \frac{x}{2}\Leftrightarrow x=2\arctan t,\,dx=\frac{2}{
1+t^{2}}dt,
\end{equation*}
$$
can be applied to the given integral. The integrand becomes a rational
fraction in $t$, because
$$
\begin{equation*}
\cos x=\frac{1-\tan ^{2}\frac{x }{2}}{1+\tan ^{2}\frac{x}{2}}=\frac{1-t^{2}}{1+t^{2}},\,\quad\sin x=\frac{2\tan \frac{x }{2}}{1+\tan ^{2}
\frac{x }{2}}=\frac{2t}{1+t^{2}}.
\end{equation*}
$$
We have
$$
\begin{eqnarray*}
I &=&\int \frac{1}{\cos x+\tan x}\,dx,\qquad t=\tan \frac{x}{2} \\
&=&\int \frac{2}{\left( \frac{1-t^{2}}{1+t^{2}}+\frac{2t}{1-t^{2}}\right)
\left( 1+t^{2}\right) }\,dt \\
&=&\int \frac{2(1-t^{2})}{t^{4}+2t^{3}-2t^{2}+2t+1}\,dt.
\end{eqnarray*}
$$
Since
$$
\begin{equation*}
t^{4}+2t^{3}-2t^{2}+2t+1=\left( t^{2}+(1-\sqrt{5})t+1\right) \left( t^{2}+(1+\sqrt{5})t+1\right),
\end{equation*}
$$
we can expand the integrand into partial fractions
$$
\begin{equation*}
\frac{2(1-t^{2})}{t^{4}+2t^{3}-2t^{2}+2t+1}=\frac{\sqrt{5}}{5}\left( \frac{2t+1+\sqrt{5}}{t^{2}+(1+\sqrt{5})t+1}-\frac{2t+1-\sqrt{5}}{t^{2}+(1-\sqrt{5})t+1}\right).
\end{equation*}
$$
Consequently,
$$
\begin{equation*}
I=\frac{\sqrt{5}}{5}\int \frac{2t+1+\sqrt{5}}{t^{2}+(1+\sqrt{5})t+1}\,dt-
\frac{\sqrt{5}}{5}\int \frac{2t+1-\sqrt{5}}{t^{2}+(1-\sqrt{5})t+1}\,dt.
\end{equation*}
$$
Set $u(t)=t^{2}+(1+\sqrt{5})t+1,v(t)=t^{2}+(1-\sqrt{5})t+1$. Then
$$
\begin{eqnarray*}
I &=&\frac{\sqrt{5}}{5}\int \frac{u^{\prime }(t)}{u(t)}\,dt-\frac{\sqrt{5}}{5}\int \frac{v^{\prime }(t)}{v(t)}\,dt \\
&=&\frac{\sqrt{5}}{5}\log \left\vert u(t)\right\vert -\frac{\sqrt{5}}{5}\log
\left\vert v(t)\right\vert +C \\
&=&\frac{\sqrt{5}}{5}\log \left\vert \frac{t^{2}+(1+\sqrt{5})t+1}{t^{2}+(1-\sqrt{5})t+1}\right\vert +C \\
&=&\frac{\sqrt{5}}{5}\log \left\vert \frac{\tan ^{2}\frac{x}{2}+(1+\sqrt{5})\tan \frac{x}{2}+1}{\tan ^{2}\frac{x}{2}+(1-\sqrt{5})\tan \frac{x}{2}+1}\right\vert +C.
\end{eqnarray*}
$$
|
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|
Binomial expansion of expression with numerator and denominator both linear equations of x How can we expand the following by the binomial expansion, upto the term including $x^3$? That'll be 4 terms.
This the expression to be expanded: $\sqrt{2+x\over1-x}$
I understand how to do the numerator and denominator individually. Now this is what I'm doing - having expanded the denominator (using the standard expansion formula for $(1+x)^{-1}$), do I now simply need to multiply this expansion once with the numerator $(2+x)$? I'm not getting the correct answer, but is this the correct method?
|
We will use the expansion
$\sqrt{1+x} = 1+x/2+x^2(1/2)(-1/2)/2 + x^3(1/2)(-1/2)(-3/2)/6 + ...
= 1+x/2-x^2/8+x^3/8+...
$
where "..." means "terms of higher order than $x^3$"
both in this expansion and in the math below.
Note: I am doing the following math
off the top of my head
as I am entering it,
so the chances for error are
decidedly nonzero.
However, the form of the result should be correct.
Because we are only interested in terms up to order $x^3$,
whenever a term of higher order occurs,
it is dropped and subsumed into
the "..." part.
For those who know the "big-$O$" notation,
the "+..." could also be written as
$+O(x^4)$.
$\begin{align}
\sqrt{2+x\over1-x}
&=\sqrt{(2+x)(1+x+x^2+x^3+...)}\\
&=\sqrt{2+2x+2x^2+2x^3+...+x+x^2+x^3+...}\\
&=\sqrt{2+3x+3x^2+3x^3+...}\\
&=\sqrt{2}\sqrt{1+3x/2+3x^2/2+3x^3/2+...}\\
&=\sqrt{2}(1+(3x/2+3x^2/2+3x^3/2)/2
-(3x/2+3x^2/2)^2/8
+(3x/2)^3/8+...)\\
&=\sqrt{2}(1+3x/4+3x^2/4+3x^3/4
-(3x/2)^2(1+x)^2/8
+27x^3/64+...)\\
&=\sqrt{2}(1+3x/4+3x^2/4+3x^3/4
-(9x^2/32)(1+2x+x^2)
+27x^3/64+...)\\
&=\sqrt{2}(1+3x/4+3x^2/4+3x^3/4
-9x^2/32-9x^3/16
+27x^3/64+...)\\
&=\sqrt{2}(1+3x/4+x^2(3/4-9/32)+x^3(3/4-9/16+27/64))+...\\
&=\sqrt{2}(1+3x/4+x^2((3*8-9)/32)+x^3((3*16-9*4+27)/64))+...\\
&=\sqrt{2}(1+3x/4+15x^2/32+39x^3/64)+...\\
\end{align}
$
This is why computer algebra systems came to be.
|
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|
problem with partial fraction decomposition I want to do partial fraction decomposition on the following rational function:
$$\frac{1}{x^2(1+x^2)^3}$$
So I proceed as follows:
$$\begin{align}
\frac{1}{x^2(1+x^2)^3} &= \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} + \frac{Ex + F}{(1 + x^2)^2} + \frac{Gx + H}{(1 + x^2)^3} \\
1 = &Ax(1+x^2)^3 + B(1+x^2)^3 + (Cx + D)x^2(1+x^2)^2 \\
& + (Ex + F)x^2(1+x^2) + (Gx + H)(x^2) \\
1 = &(A+C)x^7 + (B+D)x^6 + (3A+2C+E)x^5 + (3B+2D)x^4 \\
& + (3A+C+E+G)x^3 + (3B+D+H)x^2 +(A)x + (B)\\
\end{align}$$
But from this, $B=1$, $B=-D$, and $3B+2D=0$, which is an inconsistent system. What am I doing wrong?
|
Your final expression is almost correct; you're just forgetting the $F$. The coefficient of $x^4$ should be:
$$
3B + 2D + F
$$
and the coefficient of $x^2$ should be:
$$
3B + D + F + H
$$
|
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|
Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here? I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was given we have $\sin x=\frac{1}{3\cos x}$ and $\cos x=\frac{1}{3\sin x}$ .
Next, $(\sin^2 x+\cos^2 x)^3=1=\sin^6 x+\cos^6 x+3\sin^2 x \cos x+3\cos^2 x \sin x$ .
Now we substitute what we found above from the given:
$\sin^6 x+\cos^6+\sin x +\cos x=1$
$\sin^6 x+\cos^6=1-(\sin x +\cos x)$
$\sin^6 x+\cos^6=1-\sqrt {\frac 53}$
Not only is this not positive, but this is not even a rational number. What did I do wrong? Thanks.
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Should be $(\sin^2 x+\cos^2 x)^3=1=\sin^6 x+\cos^6 x+3\sin^4 x \cos^2 x+3\cos^4 x \sin^2 x$
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|
System of three equations in three variables? Fibonacci apparently found some solutions to this problem:
Find rational solutions of:
$$x+y+z+x^2=u^2$$
$$x+y+z+x^2+y^2=v^2$$
$$x+y+z+x^2+y^2+z^2=w^2$$
How would you find solutions to this using the mathematics available in Fibonaccis's time? (of course by this I mostly mean without using calculus, series, and modern maths. Also please exclude modular arithmetic notation if possible.) I was able to find little bits of information by adding and subtracting equations, such as $z^2=w^2-v^2$, $y^2=v^2-u^2$, and $y^2+z^2=w^2-u^2$, but I really do not know what to do. Thanks.
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This is not a full answer in that not all solutions are described. But the discussion yields two infinite parametrized families of solutions. And the methods could possibly be studied longer to find more families, and possibly parametrize all solutions. As proof that this works before you invest in studying it, check that the solution it predicts at the end is valid.
There is a known trick for parametrizing rational points on quadratic surfaces, that I think extends to hypersurfaces.
Take the first equation. $(x,y,z,u)=(0,0,0,0)$ is a rational solution. Suppose $(X,Y,Z,U)$ is a different rational solution. Then the line connecting these two points in $4$-space is parametrized by $(x,y,z,u)=t(X,Y,Z,U)$. This line intersects the surface $x+y+z+x^2=u^2$ in precisely two places, since the intersection is found by solving for $t$ in $tX+tY+t^2Z^2=t^2U^2$. One solution is clearly given by $t=0$, and the other is given by $t=\frac{X+Y}{U^2-Z^2}$. Now since the line is parametrized by rational numbers, the intersection of this line with the plane $u=1$ has all rational coordinates: $(a,b,c,1)$. We can solve for $t$ to bring the fourth coordinate to $1$, and have $t=1/U$. So $$\begin{align}a&=X/U\\b&=Y/U\\c&=Z/U\end{align}$$
This establishes a map from rational points on $x+y+z+x^2=u^2$ to rational points on $u=1$. But this map is reversible. Take any rational triple $(a,b,c,1)$ and consider the line connecting this point to $(0,0,0,0)$. This line is parametrized by $(x,y,z,u)=s(a,b,c,1)$, and intersects $x+y+z+x^2=u^2$ in two places. To find both, we substitute: $as+bs+cs+a^2s^2=s^2$, and along with $s=0$, the other solution is with $s=\frac{a+b+c}{1-a^2}$.
So rational solutions to your first equation are given by $$\begin{align}x&=a\frac{a+b+c}{1-a^2}\\y&=b\frac{a+b+c}{1-a^2}\\z&=c\frac{a+b+c}{1-a^2}\\u&=\frac{a+b+c}{1-a^2}\end{align}$$ where $a,b,c$ are any triple of rationals excluding $a=\pm1$.
One infinite family of solutions to the system arises out of this if we take $b=c=0$: $(x,y,z,u,v,w)=\left(\frac{a^2}{1-a^2},0,0,\frac{a}{1-a^2},\pm\frac{a}{1-a^2},\pm\frac{a}{1-a^2}\right)$.
We can see what happens if we throw these into the next equation.
$$\frac{(a+b+c)^2}{1-a^2}+(a^2+b^2)\left(\frac{a+b+c}{1-a^2}\right)^2=v^2$$
Unfortunately this equation is degree 6:
$$(1+b^2)(a+b+c)^2=v^2(1-a^2)^2$$
So trying to proceed as before but this time in $(a,b,c,v)$-space won't work. Lines will not be guaranteed to intersect the surface at two points, which is a crucial element of what we did above.
If we are merely hunting families of solutions, and give up (for now) on finding all solutions, then it would help to have $1+b^2$ be a square. That is, to have $1+b^2=d^2$. We can do this by finding any primitive Pythagorean triple $(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$ and dividing by one of the left terms. Say we choose the second term, so that for integers $m$ and $n$, we have $$\begin{align}b&=\frac{m^2-n^2}{2mn}\\d&=\frac{m^2+n^2}{2mn}\end{align}$$ Now the earlier equation reduces to $$d(a+b+c)=v(1-a^2)$$
If we take $c=0$ (implying $z=0$) then we have another family of solutions to the system that arises out of this. Taking $m,n$ to be free nonzero integers, $a$ a free rational not equal to $1$, we have $$(x,y,z,u,v,w)=\left(a\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},\frac{m^2-n^2}{2mn}\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},0,\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},\frac{m^2+n^2}{2mn}\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2},\pm\frac{m^2+n^2}{2mn}\frac{a+\frac{m^2-n^2}{2mn}}{1-a^2}\right)$$
For example, $m=1$, $n=2$, $a=3/5$ yields $(-9/64, 45/256,0,-15/64, -75/256,75/256)$.
It seems reasonable that some other family could be worked out this way that does not demand $z=0$.
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|
How to understand and create quaternions? I have to multiply two quaternions to calculate a so called spherical linear interpolation between two $R^3$ coordinate systems within the interval $t = [0, 1]$.
I understand how to do the calculation of quaternions basicly works and how to do the slerp. There is a lot of literature available.
But I don't know how to get started: How to create initial quaternions from given coordinate system axis and angles? I basicly fail to understand the meaning of quaternions I guess.
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Okay, I think I got it. The quaternion is created like this:
$$q = \begin{pmatrix} a_x \cdot \sin{\frac{\alpha}{2}} \\ a_y \cdot \sin{\frac{\alpha}{2}} \\ a_z \cdot \sin{\frac{\alpha}{2}} \\ \cos{\frac{\alpha}{2}} \end{pmatrix}$$
I have one quaternion for each axis. Each axis is a vector (with three coordinates) and is aligned along a certain angle. That's six quaternions in total for two coordinate systems:
$$q_{x1} = \begin{pmatrix} \sin{\frac{\alpha_{x1}}{2}} \\ 0 \\ 0 \\ \cos{\frac{\alpha_{x1}}{2}} \end{pmatrix}, q_{y1} = \begin{pmatrix} 0 \\ \sin{\frac{\alpha_{y1}}{2}} \\ 0 \\ \cos{\frac{\alpha_{y1}}{2}} \end{pmatrix}, q_{z1} = \begin{pmatrix} 0 \\ 0 \\ \sin{\frac{\alpha_{z1}}{2}} \\ \cos{\frac{\alpha_{z1}}{2}} \end{pmatrix} $$
and
$$q_{x2} = \begin{pmatrix} \sin{\frac{\alpha_{x2}}{2}} \\ 0 \\ 0 \\ \cos{\frac{\alpha_{x2}}{2}} \end{pmatrix}, q_{y2} = \begin{pmatrix} 0 \\ \sin{\frac{\alpha_{y2}}{2}} \\ 0 \\ \cos{\frac{\alpha_{y2}}{2}} \end{pmatrix}, q_{z2} = \begin{pmatrix} 0 \\ 0 \\ \sin{\frac{\alpha_{z2}}{2}} \\ \cos{\frac{\alpha_{z2}}{2}} \end{pmatrix} $$
The quaternions above are normalized unit quaternions and already simplified, because for the x-axis $a_x = 1$, $a_y = 0$ and $a_z = 0$, and so on: $ a_x^2 + a_y^2 + a_z^2 = 1$.
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|
Factor $x^4 - 11x^2y^2 + y^4$ This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (examples of my working below):
*
*U-Substitution.
*Guess and Check.
*Reversing the question (multiplying the answer out).
Here is my working for each case so far.
(1) U-Substitution.
I tried a simpler case via u-substitution.
Let $u = x^2$ and $v = y^2$. Then $x^4 - 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u - v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.
(2) Guess and Check.
To obtain the first term in $x^4 - 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:
$(x^2 \pm y^2)(x^2 \pm y^2)$.
But I'm at a loss as to how to get the final solution's middle term from guessing and checking.
(3) Reversing the question
Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.
The original answer:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2) = [(x^2 - 3xy) - y^2][(x^2 + 3xy) - y^2]$
$= (x^2 - 3xy)(x^2 + 3xy) + (x^2 - 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$
$= [(x^2)^2 - (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 - 3y^3x] + [y^4]$
$= x^4 - 9x^2y^2 - y^2x^2 + 3y^3x - y^2x^2 - 3y^3x + y^4$
The $3y^3x$ terms cancel out, and we are left with:
$x^4 - 9x^2y^2 - 2x^2y^2 + y^4 = x^4 - 11x^2y^2 + y^4$, which is the original question.
The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x - 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x - 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).
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Let $x^2=u$ and $y^2=v$, then expression becomes $u^2-11uv+v^2$. Let the roots of the quadratic equation $u^2-11uv+v^2=0$ be $\alpha v,\beta v$, then, $\alpha+\beta=11$ and $\alpha\beta=1$ . You can check easily that $\alpha,\beta>0$, so $\sqrt{\alpha},\sqrt{\beta}\in \Bbb R$
therefore, $$x^4-11x^2y^2+y^4=u^2-11uv+v^2=(u-\alpha v)(u-\beta v)=(x^2-\alpha y^2)(x^2-\beta y^2)$$ $$=(x-\sqrt{\alpha}y)(x+\sqrt{\alpha}y)(x-\sqrt{\beta}y)(x+\sqrt{\beta}y)$$ $$=(x-\sqrt{\alpha}y)(x+\sqrt{\beta}y).(x+\sqrt{\alpha}y)(x-\sqrt{\beta}y)\tag{Rearraging}$$ $$=(x^2+(\sqrt{\beta}-\sqrt{\alpha})xy-\sqrt{\alpha \beta}y^2))(x^2-(\sqrt{\beta}-\sqrt{\alpha})xy-\sqrt{\alpha \beta}y^2))\tag{1}$$
Now, WLOG, let $\beta>\alpha$, then $\sqrt{\beta}-\sqrt{\alpha}=\sqrt{\alpha+\beta-2\sqrt{\alpha\beta}}=3$ as $\alpha\beta=1,\alpha+\beta=11$
Therefore, Putting it in $(1)$ gives, $$x^4-11x^2y^2+y^4=(x^2+3xy-y^2)(x^2-3xy-y^2)$$
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|
Limit on the expression containing sides of a triangle To find the bounds of the expression $\frac{(a+b+c)^2}{ab+bc+ca}$, when a ,b, c are the sides of the triangle.
I could disintegrate the given expression as $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} + 2$$ and in case of equilateral triangle, the limit is 3.
Now how to proceed further?
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(Without using the condition that $a, b, c$ are sides of a triangle) We know that
$$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0 $$
This implies that $a^2 + b^2 + c^2 \geq ab + bc + ca$. Hence $\frac{a^2+b^2+c^2} { ab+bc+ca} \geq 1$, so the minimum of the initial expression is 3.
To find the maximum, you will need to use the triangle inequality, which states that $a + b \geq c $.
It might be easier for you to try and find what the limiting case is.
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|
$ 1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$ can be written in the form $\frac{a}{b}$, find $a+b$. The solution to the equation
$$1 + \sqrt{1 - \sqrt{x^4-x^2}} = x$$
can be written in the form $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers. Find $a+b$.
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First isolate the square root on one side:
$$\sqrt{1-\sqrt{x^4-x^2}}=x-1\;.$$
Now square:
$$1-\sqrt{x^4-x^2}=(x-1)^2=x^2-2x+1\;.$$
Then $-\sqrt{x^4-x^2}=x^2-2x$, $\sqrt{x^4-x^2}=2x-x^2$, and $x^4-x^2=(2x-x^2)^2=x^4-4x^3+4x^2$. Simplification yields the equation $4x^3-5x^2=0$, or $x^2(4x-5)=0$. This is easy to solve. Be careful, though: you need to check each solution back in the original equation, because the squaring steps may have introduced extraneous solutions.
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For real numbers $x$ and $y$, show that $\frac{x^2 + y^2}{4} < e^{x+y-2} $
Show that for $x$, $y$ real numbers, $0<x$ , $0<y$
$$\left(\frac{x^2 + y^2}{4}\right) < e^{x+y-2}. $$
Someone can help me with this please...
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When $x+y$ is fixed $x^2+y^2=(x+y)^2-2xy$ takes maximal value when $xy=0.$ So it is enough to prove our inequality when one variable is $0.$ In this case, it can be reduced to $\frac{x^2}{4}\le e^{x-2}$ or $e^z\ge 1+z+\frac{z^2}{4}$ which immediately follows from the fact that $e^z\ge 1+z+\frac{z^2}{2}.$
Alternative way to present solution: Let $x+y=t$
$$\frac{x^2+y^2}{4}=\frac{(x+y)^2-2xy}{4}\le \frac{t^2}{4}\le e^{t-2}$$
and the rest is the same as above.
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|
Applonius Circle/ Ford Circle / Infinite GP / Circle Packing
All the smaller circles are mutually tangent and continue to infinity. What is sum of radii of all the smaller circles?
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This is similar to the Pappus chain in the arbelos and can be answered with circle inversion the same way as that problem is analyzed at Cut The Knot.
Inversion of the figure through a circle centered at $A$ through $C$ (dashed red) takes the semicircles on $AB$ and $AC$ to parallel lines perpendicular to $AC$. The images of all of the inscribed circles are tangent to both lines and so are all congruent.
Given $AB=k, BC=1$, the radius of inversion is $AC=k+1$. Hence the image of $B$ satisfies
$$
AB\cdot AB' = AC^2 = (k+1)^2\\
AB' = \frac{(k+1)^2}{k}
$$
and the common radius of the image circles is
$$
r' = CM=\frac{AB'-AC}{2} = \frac{(k+1)^2-k(k+1)}{2k} = \frac{k+1}{2k}
$$
Let the inscribed circles have centers at $O_1,O_2,\ldots$ with radii $r_1,r_2,\ldots$, and their images have centers at $O_1', O_2',\ldots$ with common radius $r'$. Then since all image circles are tangent along the line $MO_1'$ it is immediate that $MO_n' = (2n-1)r'$.
Take a common tangent to the $n$th circle and its image meeting them at $T_n$ and $T_n'$ respectively. Then $AT_nO_n$ is similar to $AT_n'O_n'$ and hence
$$
r_n = r' \frac{AT_n}{AT_n'}
$$
but $T_n$ and $T_n'$ are also inverses and so satisfy
$$
AT_n \cdot AT_n' = AC^2 = (k+1)^2
$$
and by the Pythagorean theorem
$$
(AT_n')^2 + (r')^2 = (AO_n')^2 = AM^2+(MO_n'^2) \\
(AT_n')^2 = -\left(\frac{k+1}{2k}\right)^2+\left(\frac{(k+1)(2k+1)}{2k}\right)^2+\left(\frac{(2n-1)(k+1)}{2k}\right)^2 \\
(AT_n')^2 = \frac{(k+1)^2}{4k^2}\left(-1+(2k+1)^2+(2n-1)^2\right)
$$
and putting it all together
$$
\begin{align}
r_n &= r' \frac{AT_n}{AT_n'} \\
&= \frac{k+1}{2k} \frac{(k+1)^2}{(AT_n')^2} \\
&= \frac{2k(k+1)}{4k(k+1)+(2n-1)^2}
\end{align}
$$
Then with help from WolframAlpha the sum of the radii is
$$
\begin{align}
\sum_{n=1}^\infty r_n &= 2k(k+1) \sum_{n=1}^\infty \frac{1}{(2n-1)^2+4k(k+1)} \\
&= 2k(k+1) \frac{\pi \tanh(\pi \sqrt{4k(k+1)}/2)}{4\sqrt{4k(k+1)}}\\
&= \frac{\pi}{4}\sqrt{k(k+1)} \tanh(\pi\sqrt{k(k+1)})
\end{align}
$$
QED.
Let $P_1$ be the foot of the perpendicular from $O_1$ to $AC$. As a check we can observe that the semicircle on $AP_1$ that is the inverse of the ray $MO_1'$ (show green dashed in the diagram) passes through the points of tangency between the inscribed circles and near their centers, so is an approximation (and lower bound) for twice the sum of the radii.
Since $AP_1O_1$ is similar to $AMO_1'$
$$
AP_1 = AM\frac{r_1}{r'} = \frac{2k(k+1)}{2k+1}
$$
and half the arc length gives the bound
$$
\sum_{n=1}^\infty r_n > \frac{\pi AP_1}{4} = \frac{\pi k(k+1)}{4k+2}
$$
for example, when $k=50$ this bound gives $\sum r_n>39.6587\cdots$, whereas the exact value from the expression above is $\sum r_n=39.6606\cdots$.
|
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|
Probability of randomly generated quadratic equation having equal roots Could any one help me to solve this problem?
Given that the coefficients of the equation $ax^2+bx+c=0$ are selected by throwing an unbiased die, we need to find what is the probability of the equation having equal roots.
Thank you for Hints.
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Hint: The roots are the same iff $b^2-4ac=0$ so find the number of triples $(a,b,c)$ with each of $a,b,c$ in $\{1,2,3,4,5,6\}$ for which this equation holds, then divide by $6^3$.
OP has asked about finding the $a,b,c$: First from $b^2=4ac$ we see that $b$ must be even, so that it is one of $2,4,6$
$b=2$ (so that $b^2=4=4\cdot 1$) implies $ac=1$ giving the single pair $(1,1)$ for $(a,c).$
$b=4$ (so that $b^2=16=4\cdot 4$) implies $ac=4$ giving the three pairs $(1,4),(2,2),(4,1)$ for $(a,c)$
Finally $b=6$ (so that $b^2=36=4\cdot 9$ implies $ac=9$ giving one pair $(3,3)$ for $(a,b)$. [we can't have $(1,9),(9,1)$ since dice only go to $6$]
So the list of triples $(a,b,c)$ for which $b^2=4ac$ is
$$(1,2,1),(1,4,4),(2,4,2),(4,4,1),(3,6,3).$$
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|
Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$ I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$
This is similar to the series $$\frac{1}{4} + \frac{1\cdot 3}{4 \cdot 6} + \frac{1\cdot3\cdot5}{4\cdot6\cdot8} + \dotsb,$$which one can show is convergent by Raabe's test, as follows: $$\frac{a_{n+1}}{a_n}= \frac{2n-1}{2n+2}= \frac{2n + 2 - 3}{2n+2}=1 - \frac{3}{2(n+1)}.$$
However, in the series I am looking at, $\frac{a_{n+1}}{a_n}=\frac{(2n-1)^2}{(2n)^2}= 1 - \frac{4n-1}{4n^2}$, so Raabe's test doesn't work (this calculation also happens to show that the ratio and root tests will not work).
Any ideas for this one? It seems the only thing left is comparison...
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Since
$$
\prod_{k=1}^n\frac{2k-1}{2k}\times\prod_{k=1}^n\frac{2k}{2k+1}=\frac1{2n+1}
$$
and
$$
\begin{align}
1
&\ge\left.\prod_{k=1}^n\frac{2k-1}{2k}\middle/\prod_{k=1}^n\frac{2k}{2k+1}\right.\\
&=\prod_{k=1}^n\frac{4k^2-1}{4k^2}\\
&\ge\prod_{k=1}^\infty\frac{4k^2-1}{4k^2}\\
&=\prod_{k=1}^\infty\left(1+\frac1{4k^2-1}\right)^{-1}\\
&\ge\exp\left(-\sum_{k=1}^\infty\frac1{4k^2-1}\right)\\
&=\exp\left(-\frac12\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\right)\\
&=e^{-1/2}
\end{align}
$$
we have that
$$
\frac{e^{-1/2}}{2n+1}\le\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2\le\frac1{2n+1}
$$
Thus, since $a_n\ge\dfrac{e^{-1/2}}{2n+1}$, the series diverges.
|
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|
Integral of $\int \frac{dx}{\sqrt{x^2 -9}}$ $$\int \frac{dx}{\sqrt{x^2 -9}}$$
$x = 3 \sec \theta \implies dx = 3 \sec\theta \tan\theta d\theta$
$$\begin{align} \int \frac{dx}{\sqrt{x^2 -9}} & = \frac{1}{3}\int \frac{3 \sec\theta \tan\theta d\theta}{\tan\theta} \\ \\ & = \int \sec\theta d\theta \\ \\ & = \ln | \sec\theta + \tan\theta| + c\end{align}$$
$x = 3\sec \theta \implies \sec\theta = \frac {x}{3}$
$\tan\theta = \frac{\sqrt{x^2 - 9}}{x}$
I have have confused x with 3 but I cannot get the proper answer which is
$$\ln | x + \sqrt{x^2 - 9}| + c$$
I always get $\dfrac{x}{3}$ or $\dfrac{\sqrt{x^2 - 9}}{x}$ or some variation of that, I can't eliminate them to get their answer.
|
$\theta=arc\sec\frac{x}{3}\implies \sec\theta=\frac{x}{3}$ and then $\tan\theta=\frac{\sqrt{x^2-9}}{3}$ so, integral would be $\ln\left(\frac{x}{3}+\frac{\sqrt{x^2-9}}{3}\right)+c=\ln(x+\sqrt{x^2-9})+k(=-\ln 3+c)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$y =f(x) =(ax^2 + bx +c)/(dx^2+ex+f)$ We have to find the conditions for this it takes all real values. $$
y=f(x)=\frac{ax^2+bx+c}{dx^2+ex+f}
$$
We have to find the conditions for this it takes all real values.
MY solution
One approach is to equate it to y and for a quadratic of x and put discriminant greater than equal to 0..That is very lengthy.Is their any better method..?
|
$$f\left( x \right) =\frac { a{ x }^{ 2 }+bx+c }{ d{ x }^{ 2 }+ex+f } =\frac { a{ x }^{ 2 }+bx+c }{ d\left( { x }^{ 2 }+\frac { e }{ d } x+\frac { f }{ d } \right) } =\frac { a{ x }^{ 2 }+bx+c }{ d\left( { x }^{ 2 }+\frac { e }{ d } x+\frac { e^{ 2 } }{ 4{ d }^{ 2 } } -\frac { e^{ 2 } }{ 4{ d }^{ 2 } } +\frac { f }{ d } \right) } =\\ =\frac { a{ x }^{ 2 }+bx+c }{ d{ \left( x+\frac { e }{ 2d } \right) }^{ 2 }+\frac { 4fd-{ e }^{ 2 } }{ 4{ d } } } $$
so range is $$R\setminus \left\{ -\frac { e }{ 2d } \right\} \\ $$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to calculate $\cos(6^\circ)$? Do you know any method to calculate $\cos(6^\circ)$ ?
I tried lots of trigonometric equations, but not found any suitable one for this problem.
|
I'm going to use the value of $\cos 18°=\frac{1}{4}\sqrt{10+2\sqrt{5}}$ obtained in this question.
$\sin^2 18°=1-\left(\frac{1}{4}\sqrt{10+2\sqrt{5}}\right)^2=1-\frac{10+2\sqrt{5}}{16}=\frac{6-2\sqrt{5}}{16}$ so $\sin 18°=\frac{1}{4}\sqrt{6-2\sqrt{5}}$
$\sin 36°=2\cos 18°\sin 18°=\frac{1}{4}\sqrt{10-2\sqrt{5}}$
$\cos 36°=\sqrt{1-\sin^2 36°}=\frac{1}{4}(1+\sqrt{5})$
$\cos 6°=\cos(36°-30°)=\cos 36°\cos 30°+\sin 36°\sin 30°=\frac{1}{4}\sqrt{7+\sqrt{5}+\sqrt{30+6\sqrt{5}}}$
|
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|
How to find the smallest positive integer $K$ such that $(K -\lfloor\frac{K}{2}\rfloor + 1)(\lfloor\frac{K}{2}\rfloor + 1) \geq N$ I am writing a program and I would need an explicit formula for the following:
The smallest positive integer $K$ such that:
$$\left(K - \left\lfloor\frac{K}{2}\right\rfloor + 1\right)\left(\left\lfloor\frac{K}{2}\right\rfloor + 1\right) \geq N$$
where $N$ is a given integer $> 1$.
I tried with
$$K = \left\lfloor 2(\sqrt{N} - 1)\right\rfloor,$$
but it does not seem really correct in general. Can you explain how I can properly get the correct formula?
|
Say we have a positive integer $K$ such that
$$\left(K - \left\lfloor\frac{K}{2}\right\rfloor + 1\right)\left(\left\lfloor\frac{K}{2}\right\rfloor + 1\right) \geqslant N.\tag{1}$$
Now, if $K = 2m$ is even, $(1)$ means $$(m+1)^2 \geqslant N \iff m \geqslant \sqrt{N} - 1 \iff K \geqslant 2(\sqrt{N}-1),$$ and since $K$ is an integer, that is equivalent to $K \geqslant \lceil 2(\sqrt{N}-1)\rceil$.
If $K = 2m-1$ is odd, $(1)$ means that $$(m+1)m \geqslant N \iff 4m^2 + 4m + 1 > 4N \iff (2m+1) > 2\sqrt{N} \iff K > 2(\sqrt{N}-1),$$ and since $K$ is an integer, that implies $K \geqslant \lceil 2(\sqrt{N}-1)\rceil$.
So whether $K$ is even or odd, $(1)$ implies that $K \geqslant \lceil 2(\sqrt{N}-1)\rceil$, and it only remains to show that
$$S := \lceil 2(\sqrt{N}-1)\rceil\tag{2}$$
satisfies $(1)$ (with $K$ substituted by $S$). From $(2)$ follows
$$2(\sqrt{N}-1) \leqslant S \iff \sqrt{N}-1 \leqslant \frac{S}{2} .$$
If $S$ is even, we immediately obtain $\left(\frac{S}{2}+1\right)^2 \geqslant N$, and if $S$ is odd, we obtain
$$\left(\frac{S+1}{2}+1\right)\left(\frac{S-1}{2}+1\right) \geqslant \left(\sqrt{N}+\frac12\right)\left(\sqrt{N}-\frac12\right) = N - \frac14.$$
But the left hand side is an integer, hence $\left(\frac{S+1}{2}+1\right)\left(\frac{S-1}{2}+1\right) \geqslant N.$
|
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|
Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
|
This is akin particularly to the methods described by Elias , Sujaan Kunalan and nbubis :
The sum of a complex number and its conjugate produce a pure real number,
$$ z \ + \ \overline{z} \ = \ 2a \ = \ 2a \ \cdot \ cis (0) = \ 2a \ \cdot \ cis (2 \pi) \ , $$
$ \ cis( \theta ) \ $ being the abbreivated form of $ \ \cos \theta \ + \ i \sin \theta \ \ . $ DeMoivre's Theorem then gives us
$$ z^3 \ = \ r^3 \cdot \ cis(3 \theta) \ = \ 2a \ \cdot \ cis (2 \pi) , $$
with $ \ r \ $ being the modulus of $ \ z \ $ and $ \ \theta \ $ its argument. We thus obtain $ \ r^3 = 2a \ $ and the "trigonometric" equation $ \ cis(3 \theta) \ = \ cis (2 \pi) , $
which is solved (within the "principal circle", $ \ 0 \ \le \ \theta \ < \ 2 \pi \ $ ) by $ \ \theta \ = \ \frac{0 + 2k \pi}{3} \ , \ \text{for} \ k = 0,1,2 , $
which is described by what is also called DeMoivre's Theorem for roots.
EDIT: Oops, need to take this a bit further.
So there are three apparent solutions,
$$z \ = \ (2a)^{1/3} \ , \ (2a)^{1/3} \cdot cis(\frac{2\pi}{3}) \ , \ \text{and} \ (2a)^{1/3} \cdot cis(\frac{4\pi}{3}) \ . $$
For all three, however, we have $ \ z \ + \ \overline{z} \ = \ 2 \cdot (2a)^{1/3} \ , $ which must also equal $ \ 2a \ , $ which is real. Hence, $ \ (2a)^{1/3} \ = \ a \ \ \Rightarrow \ \ a^{1/3} \ \cdot \ (a^{2/3} - 2^{1/3}) = 0 \ , $ which is solved by $ \ a = 0 \ , \ \sqrt{2} \ , \ \text{and} \ -\sqrt{2} \ , $
as found by other posters. So there are three solutions are pure real numbers.
FURTHER EDIT: Hmmm, not quite done, on thinking about this a little more. We've established moduli for solutions, but we should also look at the results for $ \ (2a)^{1/3} \cdot cis(\frac{2\pi}{3}) \ , \ \text{and} \ (2a)^{1/3} \cdot cis(\frac{4\pi}{3}) \ . $
For $ \ a = \sqrt{2} \ , \ z \ = \ \sqrt{2} \cdot cis(\frac{2\pi}{3}) \ \ \text{and} \ \sqrt{2} \cdot cis(\frac{4\pi}{3}) \ $, with $ \ \overline{z} \ = \ \sqrt{2} \cdot cis(\frac{4\pi}{3}) \ \ \text{and} \ \sqrt{2} \cdot cis(\frac{2\pi}{3}) \ , $ respectively, so
$ \ z + \overline{z} \ = \ -2 \sqrt{2} \ , $ but $ \ z^3 \ = \ ( \sqrt{2})^3 \ \cdot \ cis(3 \ \cdot \frac{2 \pi}{3}) \ = \ 2 \sqrt{2} \ , \ $ and likewise for the argument $ \frac{4 \pi}{3} \ . $
The same thing happens for $ \ a = -\sqrt{2} \ , $ since we have $ \ z \ = \ -\sqrt{2} \cdot cis(\frac{2\pi}{3}) \ = \ \sqrt{2} \cdot cis(\frac{5\pi}{3}) \ , $ for which $ \ \overline{z} \ = \ \sqrt{2} \cdot cis(\frac{\pi}{3}) \ $ so,
$ \ z + \overline{z} \ = \ 2 \sqrt{2} \ \ , $ while
$$z^3 \ = \ ( -\sqrt{2})^3 \ \cdot \ cis(3 \ \cdot \frac{2 \pi}{3}) \ = \ (-2 \sqrt{2}) \cdot cis(2 \pi) \ = \ (-2 \sqrt{2}) \cdot (+1) \ = \ -2 \sqrt{2} \ , $$
and similarly for $ \ z \ = \ -\sqrt{2} \cdot cis(\frac{4\pi}{3}) \ = \ \sqrt{2} \cdot cis(\frac{\pi}{3}) \ . $
So it would seem that all of the complex values really are extraneous (as alex.jordan points out).
|
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|
$a, b, c, d$ are positive integers， $a-c|a b+c d$, and then $a-c|a d+b c$
$a, b, c, d$ are positive integers， $a-c|a b+c d$, and then $a-c|a d+b c$
proof: really easy when use $a b+c d-(a d+b c)$
however my first thought is, $a-c| a b+c d+k(a-c)$, and set some $k$ to prove, failed.
question1 : is this method could be done? any other methods?
question2 : And is there any relations between the question and determinant/matrix
$$\begin{array}{cc} a & d \\-c & b \end{array}$$
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As lab gave you the $k$ which works for question 1, here is the answer to question 2:
$$\det \begin{pmatrix} a & d \\-c& b \end{pmatrix} = \det \begin{pmatrix} a & b \\-c & d \end{pmatrix}+\det \begin{pmatrix} a & d-b \\-c & b-d \end{pmatrix}$$
$$ = \det \begin{pmatrix} a & b \\-c & d \end{pmatrix}+(b-d)\det \begin{pmatrix} a & -1 \\-c & 1\end{pmatrix} = \det \begin{pmatrix} a & b \\-c & d \end{pmatrix}+(b-d)\det \begin{pmatrix} a+c & 0 \\-c & 1\end{pmatrix}
$$
The solutions is pretty artificial though...
|
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|
How to solve these two trigonometric equations? How to solve these two trigonometric equations :
$$\sin y \sin(2x+y)=0$$
$$\sin x \sin(x+2y)=0.$$
I know one set of solution will be $(0,0)$. What will be the other set ?
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A possibility is $\sin y=0$, which gives $y=k\pi$ (integer $k$); substituting in the second one gives
$$
\sin x\sin(x+2k\pi)=0
$$
so this implies $\sin x=0$. Similarly, $\sin x=0$ implies $\sin y=0$. Thus we can reduce to the case
$$\begin{cases}
\sin(2x+y)=0\\
\sin(x+2y)=0
\end{cases}
$$
From this you have
$$
\begin{cases}
2x+y=a\pi\\
x+2y=b\pi
\end{cases}
$$
(where $a$ and $b$ are any integers).
This is a linear system, that can be solved as
$$
x=\frac{2a-b}{3}\pi,\quad
y=\frac{2b-a}{3}\pi
\qquad(a,b\text{ any integers}).
$$
This formula encompasses also the case $\sin x=\sin y=0$, because given arbitrary integers $h$ and $k$ one can express
$$
\begin{cases}
h=\frac{2a-b}{3}\\[1ex]
k=\frac{2b-a}{3}
\end{cases}
$$
for suitable integers $a$ and $b$, namely $a=2h+k$ and $b=h+2k$.
Therefore the general solution can be written as
$$
\boxed{\displaystyle x=\frac{2a-b}{3}\pi,\quad y=\frac{2b-a}{3}\pi,
\qquad a,b\text{ arbitrary integers}}.
$$
Noticing that $2a-b\equiv 2b-a\pmod{3}$, we can express the solutions in a different form, as three families:
\begin{align}
&1. & x &= h\pi, & y &= k\pi\\[1ex]
&2. & x &= \frac{\pi}{3}+h\pi, & y &= \frac{\pi}{3}+k\pi\\[1ex]
&3. & x &= \frac{2\pi}{3}+h\pi, & y &= \frac{2\pi}{3}+k\pi
\end{align}
again for arbitrary integers $h$ and $k$.
|
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|
Are there infinitely many rational outputs for sin(x) and cos(x)? I know this may be a dumb question but I know that it is possible for $\sin(x)$ to take on rational values like $0$, $1$, and $\frac {1}{2}$ and so forth, but can it equal any other rational values? What about $\cos(x)$?
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There are infinitely many primitive Pythagorean triples, that is, triples $(a,b,c)$ of positive integers such that $a$, $b$, and $c$ are positive integers $\gt 1$ such that $a^2+b^2=c^2$.
Any such triple determines a right triangle. The sines and cosines of the two non-right angles are the rationals $\frac{a}{c}$ and $\frac{b}{c}$. So there are infinitely many angles between $0$ and $\frac{\pi}{2}$ such that $\sin x$ and $\cos x$ are both rational.
You are undoubtedly familiar with the triples $(3,4,5)$ and $(5,12,13)$. There are infinitely many more. The Wikipedia article linked to above gives a detailed description.
We can already get infinitely many examples by letting $n$ be any integer $\gt 1$, and setting $a=n^2-1$, $b=2n$, and $c=n^2+1$. Make the right-triangle $ABC$, with the right angle at $C$, and $a,b,c$ as above.
Note that $\triangle ABC$ really is a right-triangle, since $(n^2-1)^2+(2n)^2=(n^2+1)^2$.
Let $x=\angle A$. Then $\sin x=\frac{n^2-1}{n^2+1}$ and $\cos x=\frac{2n}{n^2+1}$. Thus both are rational.
|
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|
Partial fractions of $\frac{-5x+19}{(x-1/2)(x+1/3)}$ Alright, I need to find the partial fractions for the expression above. I have tried writing this as $$\frac{a}{x-1/2}+\frac{b}{x+1/3}$$ but the results give me $a=25.8$ and $b=-20.8$, which are slightly wrong because they give me $5x+19$ instead of $-5x+19$. Can you please help? Thanks a lot
|
Either you've started from the wrong equations, or you've made a calculation mistake (don't worry, happens to all of us sometimes). When you write
$$\frac{a}{x-\frac{1}{2}}+\frac{b}{x+\frac{1}{3}}=\frac{-5x+19}{(x-\frac{1}{2})(x+\frac{1}{3})}$$
you get
$$\frac{a(x+\frac{1}{3})}{(x-\frac{1}{2})(x+\frac{1}{3})}+\frac{b(x-\frac{1}{2})}{(x-\frac{1}{2})(x+\frac{1}{3})}=\frac{ax+bx+\frac{a}{3}-\frac{b}{2}}{(x-\frac{1}{2})(x+\frac{1}{3})}=\frac{-5x+19}{(x-\frac{1}{2})(x+\frac{1}{3})}$$
so that the correct equations to solve are
$$\begin{align*}
a+b&=-5\\[0.1in]
\tfrac{a}{3}-\tfrac{b}{2}&=19
\end{align*}$$
|
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|
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
I have had a few ideas about this:
If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$
We also know that $\tan(\alpha +\beta) = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
Then we can write $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
I have tried rearranging $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$ but it has not been helpful.
I also thought if we let $\alpha = \beta$ then I could write $\tan(\alpha+ \alpha) = 1$
(does this also mean $\tan(2\alpha) = 1$?)
then: $\tan(\alpha + \alpha) = \dfrac{\tan\alpha + \tan\alpha}{1- \tan\alpha\tan\alpha}$
which gives: $1 = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$
Anyway these are my thoughts so far, any hints would be really appreciated.
|
from where OP left his step:$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$$
$$\implies{1- \tan\alpha\tan\beta}=\tan\alpha + \tan\beta$$
$$\implies \tan\alpha + \tan\beta+\tan\alpha\tan\beta=1$$
add 1 to both sides
$$\implies\tan\alpha + \tan\beta+\tan\alpha\tan\beta+1=2$$
$$\implies1+\tan\alpha + \tan\beta+\tan\alpha\tan\beta=2$$
factor the above equation
$$\implies1(1+\tan\alpha) + \tan\beta(1+\tan\alpha)=2$$
$$\implies(1+\tan\alpha)(1+\tan\beta)=2$$
Hence proven
|
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|
Stuck at Evaluating the Riemann-Stieltjes Integral Let $$\alpha(x) = \left\lbrace \begin{array}{cc} 0 & x=0\\ \dfrac{1}{2^n} & \dfrac{1}{3^n} < x \leq \dfrac{1}{3^{n-1}}\quad n=1,2,...\end{array}\right.$$
Evaluate $$\int_{0}^{1}{x\mathrm{d}\alpha(x)}$$
Attempt: I dont know how to put this formally, but I know it is similar to problem of evaluating the R-S integral where $\alpha(x)$ is the jump function.
So If I evaluate the function $\alpha(x)$ for some values of $n$ then the function looks like this $$\alpha(x) = \left\lbrace \begin{array}{cc} 0 & x=0\\ \vdots & \vdots \\\dfrac{1}{8} & \dfrac{1}{27} < x \leq \dfrac{1}{9} \\ \dfrac{1}{4} & \dfrac{1}{9} < x \leq \dfrac{1}{3} \\ \dfrac{1}{2} & \dfrac{1}{3} < x \leq 1 \end{array}\right.$$ So $f(x)=x$ is continuous and $\alpha(x)$ is monotonically increasing. Hence the answer should be $$\int_{0}^{1}{x \mathrm{d}\alpha(x)} = \sum_{n=1}^{\infty}{\dfrac{1}{3^n}\dfrac{1}{2^{n+1}}}=\dfrac{1}{2}\sum_{n=1}^{\infty}{\left(\dfrac{1}{6}\right)^n}=\dfrac{1}{10}$$
So my questions are:
(a) Is the answer correct? If yes, then how can I write this formally?
(b) If the answer is wrong, what direction can you provide me to get the right answer?
Thanks!
|
I think your answer is quite right. I'm going to outline an alternative, wimpy approach using integration by parts:
$$\int_0^1 x \, d\alpha(x) = [x \, \alpha(x)]_0^1 - \int_0^1 dx \, \alpha(x)$$
The integrated term is $1/2$, and the resulting integral is straightforward:
$$\int_0^1 x \, d\alpha(x) = \frac12 - \sum_{n=1}^{\infty} \frac{1}{2^n} \left ( \frac{1}{3^{n-1}}-\frac{1}{3^n}\right) = \frac12 - 2 \sum_{n=1}^{\infty} \frac{1}{6^n} = \frac{1}{10}$$
|
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|
If $x,y,z \in \Bbb{R}$ such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then show that $x,y,z \in [2/3,2]$. If $x,y,z\in\mathbb{R}$ such that $$x+y+z=4,\quad x^2+y^2+z^2=6;$$then show that the each of $x,y,z$ lie in the closed interval $[2/3,2]$.
I have been able to solve using $2(y^2+z^2)\geq(y+z)^2$.
Is there any another method to solve it.
|
My Solution:: Given $x+y = 4-z$ and $x^2+y^2=6-z^2$.
Now using the Cauchy-Schwarz inequality, we get $(x^2+y^2)\cdot (1^2+1^2)\geq (x+y)^2$
So we get $(12-2z^2)\geq (4-z)^2\Rightarrow 12-2z^2\geq 16+z^2-8z\Rightarrow 3z^2-8z+4\leq 0$
So $\displaystyle (3z^2-6z)-(2z-4)\leq 0\Rightarrow 3z(z-2)-2(z-2)\leq 0\Rightarrow \frac{2}{3}\leq z \leq 2$
Now $x+y+z=4$ and $x^2+y^2+z^2=6$ are symmetrical expressions on $x,y,z$
So we get $\displaystyle \frac{2}{3}\leq x,y,z\leq 2\Rightarrow x,y,z \in \left[\frac{2}{3}\;,2\right]$
|
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|
Find $F_{n}$ in : $F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$ I'm stuck with the question for a while : Find in $$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
the element $F_{n}$ .
Placing $n-1$ instead on $n$ results in :
$$F_{n-1} +2F_{n-2} + ... + (n-1+1)\cdot F_{0} = 3^{n-1}$$
$$ F_{n-1} +2F_{n-2} + ... + n\cdot F_{0} = 3^{n-1} $$
subtracting both :
$$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} -(F_{n-1} +2F_{n-2} + ... + n\cdot F_{0} )
=3^{n} - 3^{n-1}
$$
But that doesn't help much . Any ideas ?
Thanks
|
Generating functions are your friends.
With experience, you will recognize that sum
as a convolution.
All that follows is standard generating function manipulation.
See the book "generatingfunctionology"
available free online
at
http://www.math.upenn.edu/~wilf/DownldGF.html.
Note: This may seem unnecessarily complicated,
but sometimes generating functions
are the most direct way to get results.
Let
$A(x) = \sum_{n=0}^{\infty} F_n x^n
$,
$B(x) = \sum_{n=0}^{\infty} (n+1) x^n
$,
and
$C(x) = A(x) B(x)
$.
$\begin{align}
C(x)
&= A(x) B(x)\\
&= \left(\sum_{n=0}^{\infty} F_n x^n\right) \left( \sum_{n=0}^{\infty} (n+1) x^n\right)\\
&= \sum_{i=0}^{\infty} \sum_{j-0}^{\infty} (j+1)F_i x^{i+j}\\
&= \sum_{n=0}^{\infty}\sum_{i=0}^{n} (n-i+1)F_i x^n
\quad\text{ Setting }i+j=n\text{, so } j=n-i\\
&= \sum_{n=0}^{\infty} x^n\sum_{i=0}^{n} (n-i+1)F_i\\
\end{align}
$
By assumption,
$\sum_{i=0}^{n} (n-i+1)F_i = 3^n$.
So
$C(x) = A(x)B(x)
=\sum_{n=0}^{\infty}3^n x^n
=\sum_{n=0}^{\infty} (3x)^n
= \dfrac{1}{1-3x}
$.
If we can find $B(x)$,
we can get $A(x) = C(x)/B(x)$.
$\begin{align}
B(x)
&= \sum_{n=0}^{\infty} (n+1) x^n\\
&= \sum_{n=0}^{\infty} (x^{n+1})'\\
&= \left(\sum_{n=0}^{\infty} x^{n+1}\right)'\\
&= \left(\dfrac{x}{1-x}\right)'\\
&= \dfrac{(1-x)+x}{(1-x)^2}\\
&= \dfrac{1}{(1-x)^2}\\
\end{align}
$
so
$\begin{align}
A(x)
&= \dfrac{1}{1-3x}\big/\dfrac{1}{(1-x)^2}\\
&= \dfrac{(1-x)^2}{1-3x}\\
&= (1-2x+x^2)\sum_{n=0}^{\infty} 3^n x^n\\
&= \sum_{n=0}^{\infty} 3^nx^n
-\sum_{n=0}^{\infty} 2\ 3^nx^{n+1}
+\sum_{n=0}^{\infty} 3^nx^{n+2}\\
&= \sum_{n=0}^{\infty} 3^nx^n
-\sum_{n=1}^{\infty} 2\ 3^{n-1}x^{n}
+\sum_{n=2}^{\infty} 3^{n-2}x^{n}\\
&= 1+3x+\sum_{n=2}^{\infty} 3^nx^n
-2x-\sum_{n=2}^{\infty} 2\ 3^{n-1}x^{n}
+\sum_{n=2}^{\infty} 3^{n-2}x^{n}\\
&= 1+x +\sum_{n=2}^{\infty} x^n(3^n-2\ 3^{n-1}+3^{n-2})
\quad(*)\text{ We can stop here to get } F_n\\
&= 1+x +\sum_{n=2}^{\infty} x^n 3^n(1-2/3+1/9))
\quad\text{ Continuing on anyway to get } A(x)\\
&= 1+x +(4/9)\sum_{n=2}^{\infty} x^n 3^n\\
&= 1+x +(4/9)\dfrac{9x^2}{1-3x}\\
&= \dfrac{(1+x)(1-3x)+4x^2}{1-3x}\\
&= \dfrac{1-2x+x^2}{1-3x}\\
&= \dfrac{(1-x)^2}{1-3x}\\
\end{align}
$
From the line labeled "$(*)$",
$F_n
= 3^n-2\ 3^{n-1}+3^{n-2}
= 3^{n-2}(9-6+1)
= 4\ 3^{n-2}
$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Functional equation: $R(1/x)/x^2 = R(x) $ The following can be shown without much hassle.
Suppose $R$ is a rational function satisfying the following functional equation.
\begin{align}
\frac{1}{x^2} R\left( \frac{1}{x} \right) = R(x) \qquad \forall \: x \in \mathbb{R} \backslash \{0\}
\tag{1}
\end{align}
Then
$$
\int_0^\infty R(x) \cdot \log x \,\mathrm{d}x = 0
$$
if it converges.
The proof is done by splitting the integral from $0$ to $1$ and $1$ to $\infty$. And then map $[1,\infty) \to [0,1]$ by $u \mapsto 1/x$. There are many rational functions satisfying this equation. A few examples below
$$
R(x) = \frac{1}{x}\,,\ \frac{x}{(1+x^2)^2} \,,\ \frac{1}{x^2 + x + 1}\,,\ \frac{1}{x^2+1}
$$
My question is: Can one find groups or families of functions satisfying $(1)$?
|
If
$\frac{1}{x^2} R\left( \frac{1}{x} \right) = R(x)$,
$x^2 R(x) = R(1/x)$.
If $R(x) = A(x)/B(x)$,
where $A$ and $B$ are relatively prime polynomials
of respective degrees $n$ and $m$,
$A(1/x)=a(x)/x^n$ and
$B(1/x) = b(x)/x^m$,
where $a(x)$ and $b(x)$
are the reciprocal polynomials of
$A$ and $B$, respectively.
Then
$\begin{align}
x^2 A(x)/B(x)
&= A(1/x)/B(1/x)\\
&= (a(x)/x^n)/(b(x)/x^m)\\
&= (a(x)/b(x))x^{m-n}\\
\end{align}
$
so
$x^{2+n} A(x)b(x)=a(x)B(x)x^m $.
If the degrees of $a(x)$
and $b(x)$ are the same
as $A$ and $B$
(i.e., the constant terms of $A$ and $B$ are non-zero),
the left-hand side has degree
$2+2n+m$
and the right-hand side has degree
$n+2m$,
so
$2+2n+m=2m+n$,
or
$m = n+2$.
In this case,
$A(x)b(x)=a(x)B(x) $.
If, in addition,
$A(x) = a(x) = 1$,
so $n=0$ and $m=2$,
$B(x) = b(x)$,
so $B$ is a symmetric polynomial of degree $2$.
Conversely,
if $R(x) = 1/B(x)$
where $B$ is a quadratic symmetric polynomial,
$R(1/x) = 1/B(1/x) = 1/(B(x)/x^2)
=x^2/B(x) = x^2R(x)$.
I will leave it at this for now,
and possibly work out later
the cases when
$a$ and $b$ are not the same degree as
$A$ and $B$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Express $f'''_{xxx} and f'''_{yyy}$ in terms of $f'''_{uuu} and f'''_{vvv}$. Let $f(x,y)\in C^3(\mathbf{R}^2)$ and let $u=x+y$ and $v=y$.
Express $f'''_{xxx} and f'''_{yyy}$ in terms of $f'''_{uuu} and f'''_{vvv}$.
I'm supposed to use the chain rule, how do I go about?
Thanks!
Alexander
|
If $f$ is of the form $f(x_1,\ldots,x_n)$ I will use the notation $D_if$ for the partial derivativeof $f$ w.r.t. the $i$-th variable. Similarly, $D_{k,i}f=D_kD_if$ denotes the partial derivative of $D_if$ w.r.t. the $k$-th derivative.
Then $$\begin{align}D_1\left(f(x+y,y)\right)&=D_1f(x+y,y)D_1(x+y)+D_2f(x+y,y)D_1(y)\\&=D_1f(x+y,y)\end{align}$$
$$\begin{align} D_2 (f(x+y,y))&=D_1f(x+y,y)D_2(x+y)+D_2f(x+y,y)D_2(y)\\&=D_1f(x+y,y)+D_2f(x+y,y)\end{align}$$
Continuing, $$\begin{align} D_{1,1}(f(x+y,y))&=D_1(D_1(f(x+y,y))\\&=D_1(D_1f(x+y,y))\\&=D_{1,1}f(x+y,y)D_1(x+y)+D_{2,1}f(x+y,y)D_1(y)\\&=D_{1,1}f(x+y,y)\end{align}$$
$$\begin{align} D_{2,2}(f(x+y,y))&=D_2(D_2(f(x+y,y))\\&=D_2(D_1f(x+y,y))+D_2(D_1f(x+y,y))\\
&={{D_{1,1}}f\left( {x + y,y} \right){D_2}\left( {x + y} \right) + {D_{2,1}}f\left( {x + y,y} \right){D_2}\left( {x + y} \right)}\\
&+{D_{1,2}}f(x + y,y){D_2}\left( {x + y} \right) + {D_{2,2}}f(x + y,y){D_2}\left( y \right)\\&={D_{1,1}}f\left( {x + y,y} \right) + {D_{2,1}}f\left( {x + y,y} \right) + {D_{1,2}}f(x + y,y) + {D_{2,2}}f(x + y,y)\\&= {D_{1,1}}f\left( {x + y,y} \right) + 2{D_{1,2}}f(x + y,y) + {D_{2,2}}f(x + y,y)\end{align}$$
...and so on.
Note that $D_2(f(x+y,y))$ here denotes the partial derivative of the function composite function $f\circ g$, where $g(x,y)=(x+y,y)$, in parenthesis, while $D_2f(x+y,y)$ denotes the partial derivative $D_2f$ evaluated at $(x+y,y)$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Improper integral that converges for all $x$ in $ \mathbb{R}$
Let $f(x)$ be defined by the improper integral:
$$f(x)= \int_{0}^{\infty} \cos\left(\frac{t^3}{3} + \frac{x^2 t^2}{2} + xt\right)dt.$$
Show that this improper integral converges for all $x \in\mathbb{R}$.
How do you start this question? Do I evaluate the integral first, and then see if it satisfies all $x \in\mathbb{R}$?
|
Let $x\in\mathbb R$, then there exists an $N>0$ such that $t^2+x^2t+x>0$ for all $t>N$. Then, if $M>N$: you have that $$\int_N^M\cos\left(\frac{t^3}{3}+\frac{x^2t^2}{2}+xt\right)\,dt=\int_N^M\left(\sin\left(\frac{t^3}{3}+\frac{x^2t^2}{2}+xt\right)\right)'\frac{1}{t^2+x^2t+x}\,dt=\left[\sin\left(\frac{t^3}{3}+\frac{x^2t^2}{2}+xt\right)\frac{1}{t^2+x^2t+x}\right]_N^M+\int_N^M\sin\left(\frac{t^3}{3}+\frac{x^2t^2}{2}+xt\right)\frac{2t+x^2}{(t^2+x^2t+x)^2}\,dt.$$ The first term converges as $M\to\infty$, and the last integrand is bounded by $$\left|\frac{2t+x^2}{(t^2+x^2t+x)^2}\right|\leq \frac{ct}{t^4}=\frac{c}{t^3},$$ where $c$ is a positive constant, so it converges as $M\to\infty$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Condition for collinearity of points $(a, a^3), (b, b^3), and (c, c^3)$ The following is a statement I have been trying to prove (while solving problem 1.4.26 in Algorithms (4th edition) by Robert Sedgewick).
Show that three points $(a, a^3), (b, b^3), and
(c, c^3)$ are collinear if and only if $a + b + c = 0$.
I am aware that the rule for collinearity of three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is -
$x_1(y_2- y_3 ) + x_2(y_3 - y_1 ) + x_3(y_1 - y_2 ) = 0$
OR
$(x_2 - x_1) : (x_3 - x_2) = (y_2 - y_1) : (y_3 - y_2)$
But I cannot factor out the term $(a+b+c)$ when I apply either of these rules to the points $(a, a^3), (b, b^3), and (c, c^3)$.
Can someone help me with this proof?
|
Given three distinct points in ${\mathbb{R}^2}$ $P1=(a,a^3)$, $P2=(b,b^3)$ and $P3=(c,c^3)$ for $a,b,c \in \mathbb{R}$, we define ${\vec u},{\vec v} \in \mathbb{R}^2$ such that
$$\vec u = P2-P1=\left( {\begin{array}{*{20}{c}}
{{b^3} - {a^3}} \\
{b - a}
\end{array}} \right), \vec v = P3-P1=\left( {\begin{array}{*{20}{c}}
{{c^3} - {a^3}} \\
{c - a}
\end{array}} \right)$$
Now, $P1, P2, P3$ are collinear if and only if $||\vec u \times \vec v||=0$, we will show that
$$\begin{gathered}
||\vec u \times \vec v|| = ({b^3} - {a^3})(c - a) - ({c^3} - {a^3})(b - a) \\
{\text{ }} = (b - a)({b^2} + ba + {a^2})(c - a) - (c - a)({c^2} + ca + {a^2})(b - a) \\
{\text{ }} = (b - a)(c - a)({b^2} + ba + {a^2} - {c^2} - ca - {a^2}) \\
{\text{ }} = (b - a)(c - a)({b^2} + ba - ca - {c^2}) \\
= (b - a)(c - a)[(b - c)(b + c) + a(b - c)] \\
= (b - a)(c - a)[(b - c)(b + c + a)] \\
= (b - a)(c - a)(b - c)(a + b + c) \\
\end{gathered} $$
since $a+b+c=0$ and $b\ne a, c \ne a, b \ne c$ because $P1,P2,P3$ are distinct, we have $||\vec u \times \vec v||= (b - a)(c - a)(b - c)(a + b + c) = 0$ if and only if $a+b+c=0$.
|
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|
Simplifying a Rational Expression How do you simplify the following expression:
$$\frac{x^3-27}{x^2+x-6}$$
Thanks for the help. Haven't seen this stuff since high school and I'm trying to help my younger sister out.
|
The general idea would be to factor, though there aren't any common factors here:
$$\frac{x^3-27}{x^2+x-6} = \frac{(x-3)(x^2+3x+9)}{(x+3)(x-2)}$$
However, if that + in the denominator is the wrong sign, then this can be reduced, assuming $x\neq3$:
$$\frac{x^3-27}{x^2-x-6} = \frac{(x-3)(x^2+3x+9)}{(x-3)(x+2)}=\frac{x^2+3x+9}{x+2}$$
Or one could flip the sign in the numerator for the 27, assume $x\neq-3$ and then:
$$\frac{x^3+27}{x^2+x-6} = \frac{(x+3)(x^2-3x+9)}{(x+3)(x-2)}=\frac{x^2-3x+9}{x-2}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding volume using triple integrals. Use a triple integral to find the volume of the solid: The solid enclosed by the cylinder $$x^2+y^2=9$$ and the planes $$y+z=5$$ and $$z=1$$
This is how I started solving the problem, but the way I was solving it lead me to 0, which is incorrect. $$\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_{1}^{5-y}dzdxdy=\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\left(4-y\right)dxdy=\int_{-3}^3\left[4x-xy\right]_{-\sqrt{9-y^2}}^\sqrt{9-y^2}dy= {8\int_{-3}^3{\sqrt{9-y^2}}dy}-2\int_{-3}^3y{\sqrt{9-y^2}}dy$$
If this is wrong, then that would explain why I'm stuck. If this is correct so far, that's good news, but the bad news is that I'm still stuck. If someone could help me out, that would be wonderful, thanks!
|
Ok. So you have the triple integral:
$$\begin{align}
\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_1^{5-y} \;dz\;dx\;dy
&= \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}4-y\;dx\;dy \\
&=\int_{-3}^34x-xy\Bigg|_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\;dy \\
&=\int_{-3}^38\sqrt{9-y^2}-2y\sqrt{9-y^2}\;dy \\
&= 8\int_{-3}^33\sqrt{1-\left(\frac{y}{3}\right)^2}\;dy-2\int_{-3}^3y\sqrt{9-y^2}\;dy
\end{align}$$
Now, I'm going to break this up. For the left-hand integral, we must use trig-substitution. Let $\cos(t) = \frac{y}{3}$. This implies that $dy = -3\sin(t)\;dt$. The limits of integration change as well, to $t=\arccos\left(\frac{-3}{3}\right) = \pi$ to $t = \arccos\left(\frac{3}{3}\right) = 0$.
So, the integral becomes:
$$\begin{align}
24\int_\pi^0\sqrt{1-\cos^2(t)}(-3\sin t)\;dt &= -72\int_\pi^0\sin^2(t)\;dt\\
&=72\int_0^\pi\frac{1}{2}-\frac{\cos(2t)}{2}\;dt\\
&=36\int_0^\pi1-\cos(2t)\;dt\\
&=36\left(t - \frac{\sin(2t)}{2}\right)\Bigg|_0^\pi\\
&=\boxed{36\pi}
\end{align}$$
Now, for the left-hand integral, we apply $u$-substitution. If we set $u = 9-y^2$, then $du = -2y\;dy$. The limits are transformed to $u = 9-(-3)^2 = 0$ to $u = 9-(3)^2 = 0$
So, the integral becomes:
$$\begin{align}
-2\int_{-3}^3y\sqrt{9-y^2}\;dy &= \int_0^0\sqrt{u}\;du\\
&= \boxed{0}
\end{align}$$
Well, that wasn't exciting. :)
So, putting it all together, we end up with:
$$V = 36\pi + 0 = \boxed{36\pi}$$
|
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|
Confused about sum of G.P. formula The sum of $n$ terms of the G.P. $a, ar, ar^2, ar^3, \ldots, ar^{n-1}$ is given by $a\dfrac{r^n-1}{r-1}$.
Now consider these two progressions:
1) $r^2, r^4, r^6, ..., r^{2n}$
2) $r, r^3, r^5, ..., r^{2n-1}$
Both of these have $n$ terms. Therefore, $S_{n1} = r^2 \dfrac{(r^2)^n-1}{r-1} = r^2 \dfrac{r^{2n}-1}{r-1}$. Also, $S_{n2} = r \dfrac{r^{2n}-1}{r-1}$. It seems that no matter what $r$ be, the sum of even powers will be $r$ times the sum of odd powers. However, this is not true as a simple example shows:
Consider $r = 0.5$. Now, $r^2 + r^4 + r^6 = 0.328125$, while $r + r^3 = 0.625$, and $ 0.328125 \neq 0.625 \times 0.5$
Where is the error?
|
The last term of the second series is $r^{2n-1}=r\cdot r^{2n-2}=r\cdot (r^2)^{n-1}$, and the ratio of consecutive terms is $r^2$, not $r$, so the sum formula should be
$$S_{2n}=r\cdot\frac{r^{2n}-1}{r^2-1}\;.$$
The ratio in the first series is also $r^2$, so the sum formula yields
$$S_{1n}=r^2\cdot\frac{r^{2n}-1}{r^2-1}\;.$$
This says that $S_{1n}=rS_{2n}$, which is clearly correct, since
$$r^2+r^4+\ldots+r^{2n}=r\left(r+r^3+\ldots+r^{2n-1}\right)\;.$$
Your example doesn’t match your formulas: for $n=3$ you should be comparing $r^2+r^4+r^6$ with $r+r^3+r^5$, not with $r+r^3$.
|
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|
$\frac1a+\frac1b+\frac1c=0 \implies a^2+b^2+c^2=(a+b+c)^2$? How to prove that $a^2+b^2+c^2=(a+b+c)^2$ given that $\frac1a+\frac1b+\frac1c=0$?
|
Expand the right side:
$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc.$$
From the supplementary condition we have
$$\frac{1}{a}+\frac{1}{b} + \frac{1}{c} = 0$$
Or
$$\frac{ab+ac+bc}{abc} = 0.$$
Therefore $ab+ac+bc = 0$ and the result follows. (None of $a,b,c$ can be $0$ else their inverses would be undefined and so the supplementary condition would be ill-posed.)
|
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|
Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that
360 | a2(a2-1)(a2-4)
I attempted it in the following manner.
a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)
The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will have a factor of 4
One of them will have a factor of 3
One of them will have a factor of 2
=> 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.
Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated
Thank you
|
We have $360 = 8 \times 9 \times 5$, so it will suffice to check that $8,9,5|a^2(a^2-1)(a^2-4)$.
For $9$, note that $a^2$ is congruent to either $1$ or $0$ modulo $3$, so both $a(a^2-1)$ and $a(a^2-4)$ are divisible by $3$, and consequently $9|a^2(a^2-1)(a^2-4)$.
Likewise, $a^2$ is congruent to either $1$ or $0$ or $4$ modulo $5$, so again the divisibility $5|a^2(a^2-1)(a^2-4)$ holds.
Finally, we look at divisibility by $8$. If $a$ is even, then $4|a^2$ and $4|a^2 -4$, so you have the divisibility $8|a^2(a^2-1)(a^2-4)$ (even with $16$ in place of $8$). If $a$ is odd, then $a-1,a+1$ are even and one of them is divisible by $4$, so $a^2-1$ is divisible by $8$ and the sought divisibility holds.
As a general rule, if you have a problem like (i.e. "Show that $360|f(a)$ for all $a$") this and don't want to think too much, you can first writhe $360$ as a product of powers of primes (nothing special about $360$ here, of course), and then check that if $P$ is a highest power of a prime dividing $360$ (i.e. $P=8,9,5$ here) you always have $P|f(a)$. This second check can be done by brute force: just check that $P|f(0),f(1),\dots,f(P-1)$ - this requires just the amount of computation that can be done by hand.
|
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|
How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$
How to compute
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$
My Working :
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$
Is it correct
|
Magic answer:
Let $f(n) =\dfrac{n(n-1)}{n^2-n+1}$. Then show $f(n+1) = \dfrac{n(n+1)}{n^2+n+1}$ and thus $$\frac{f(n)}{f(n+1)} = \frac{n(n-1)(n^2+n+1)}{n(n+1)(n^2-n+1)} = \frac{n^3-1}{n^3+1}$$
(I call this a "magic answer" just because most of the other answers here give you reasons for how you would see this, while I just plop in an $f$ that works, as if by magic. This is really the same argument as the others, just distilled to a minimalist essence.)
|
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"url": "https://math.stackexchange.com/questions/462082",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Reduction Integration Question ($\int_0^1{\frac{x^n}{\sqrt{x+1}} dx}$) What integration by parts things do I use for a reduction formula for $\int_0^1{\frac{x^n}{\sqrt{x+1}} dx}$?
I have tried many different ways of obtaining it without success.
|
$$I_n=\int\frac{x^ndx}{\sqrt{x+1}}=\int\frac{x^{n-1}(1+x-1)dx}{\sqrt{x+1}} $$
$$=\int x^{n-1}\sqrt{x+1}dx-\int\frac{x^{n-1}dx}{\sqrt{x+1}}=\int x^{n-1}\sqrt{x+1}dx-I_{n-1}$$
Again, $$\int x^{n-1}\sqrt{x+1}dx= \sqrt{x+1}\frac{x^n}n-\int \frac{x^n}{2n\sqrt{x+1}}dx=\sqrt{x+1}\frac{x^n}n-\frac{I_n}{2n}$$
$$\implies I_n=\sqrt{x+1}\frac{x^n}n-\frac{I_n}{2n}-I_{n-1}$$
|
{
"language": "en",
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|
Identifying the nature of the eigenvalues I wish somebody could help me in this one. We have to choose one of the $4$ options.
Let $a,b,c$ be positive real numbers such that $b^2+c^2<a<1$. Consider the $3 \times 3$ matrix
$$A=\begin{bmatrix}
1 & b & c \\
b & a & 0 \\
c & 0 & 1 \\
\end{bmatrix}.$$
*
*All the eigenvalues of $A$ are negative real numbers.
*All the eigenvalues of $A$ are positive real numbers.
*$A$ can have a positive as well as a negative eigenvalue.
*Eigenvalues of $A$ can be non-real complex numbers.
Now, $\det(A-\lambda I)=0$, so
$$\begin{vmatrix}
1-\lambda & b & c \\
b & a-\lambda & 0 \\
c & 0 & 1-\lambda \\
\end{vmatrix}=0$$
$\implies(1-\lambda)(a-a\lambda -\lambda +\lambda^2)-b(b-b\lambda)-c(ac-c\lambda)=0$
$\implies a-a\lambda -\lambda +\lambda^2-a\lambda+a\lambda^2 +\lambda^2 -\lambda^3-b^2+b^2\lambda-ac^2+c^2\lambda=0$
$\implies-\lambda^3+\lambda^2(2+a)+\lambda(-2a-1+b^2+c^2)+a-b^2-ac^2=0$
I am stuck here, don't know how to proceed. Need your help, please.
|
Let $\mathbf{x}=[x_1\quad x_2\quad x_3]^{T}$ is an arbitrary vector in $\mathbb{R}^3$. Then,\begin{align} \mathbf{x^T}A\mathbf{x}=&[x_1\quad x_2\quad x_3]\begin{bmatrix}
1 & b & c\\
b & a & 0\\
c & 0 & 1\\
\end{bmatrix}\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} \\
\ =& x_1^2+x_3^2+ax_2^2+2bx_1x_2+2cx_1x_3\\
\ =& (x_1+bx_2+cx_3)^2+ax_2^2+x_3^2-(bx_2+cx_3)^2\\
\ =& (x_1+bx_2+cx_3)^2+(a-b^2)x_2^2+(1-c^2)x_3^2-2bcx_2x_3\\
\ >& (x_1+bx_2+cx_3)^2+c^2x_2^2+b^2x_3^2-2bcx_2x_3\\
\ =& (x_1+bx_2+cx_3)^2+(cx_2-bx_3)^2\ge 0
\end{align}
Hence $A$ is positive definite and also it is symmetric $\Rightarrow $ all the eigenvalues of $A$ are positive and real. So option $(2)$ is the correct one.
|
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|
If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$
Thus
($\cos^2 \theta −\sin^2 \theta)=x$ ,
so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?
|
$$
\cos^6\theta-\sin^6\theta = \left ( \cos^2 \theta\right )^3 - \left (\sin^2 \theta \right )^3 = \\
= \left( \cos^2 \theta - \sin^2 \theta\right ) \left(\cos^4 \theta + \sin^2\theta \cos^2\theta + \sin^4\theta \right ) = \\
= x \left ( \cos^4 \theta - 2\cos^2\theta\sin^2\theta + \sin^4 \theta + 3 \cos^2\theta\sin^2\theta\right ) = \\
= x \left( \left(\cos^2\theta - \sin^2 \theta \right )^2 + \frac 34 \sin^22\theta\right ) = x \left( x^2 + \frac 34 \left( 1-\cos^2 2\theta\right )\right ) = \\
= x \left(x^2 + \frac 34 (1-x^2) \right ) = \frac x4 \left(x^2+3 \right )
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solution to cubic inequality How do I find the solutions to this equation? $$(3x^3)-(14x^2)-(5x) \leq 0$$
|
Step 1. Find the roots of the cubic equation $3x^3-14x^2-5x=0$.
$$
\begin{align*}
x(3x^2-14x-5)&=0\\
x(3x+1)(x-5)&=0\\
x&=0\quad\text{or}\\
x&=-\frac{1}{3}\quad\text{or}\\
x&=5
\end{align*}
$$
Step 2. Draw a sign table for $P=3x^3-14x^2-5x$.
Step 3: From your sign table, you can see that for $x\in(-\infty,-\frac{1}{3}]\cup[0,5]$, $P\leq0$.
|
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|
For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well known 1983 IMO Problem 6 is the following:
IMO 1983 #6. Let $a$, $b$ and $c$ be the lengths of the sides of a triangle.
Prove that $$a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0. $$
See: here.
A lot of people have discussed this problem. So this problem (IMO 1983) has a lot of nice methods.
Now I found for quadrilaterals a similar inequality. Are there some nice methods for inequality $(\star)$?
|
WLOG $a = \max(a,b,c,d)$
Let $t = \frac{1}{2} (b+c+d-a) \ge 0$ because $a$,$b$,$c$,$d$ are sides of a quadrilateral
Then decreasing $(a,b,c,d)$ simultaneously by $t$ reduces the desired expression
And $a-t = (b-t)+(c-t)+(d-t)$
Thus it suffices to minimize the expression when $a=b+c+d$, which reduces to:
$
\begin{align}
&(b+c+d) b^2 (b-c) + b c^2 (c-d) - c d^2 (b+c) + d (b+c+d)^2 (c+d) \\
& = \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) - c d^2 (b+c) + d (b+c+d) (b+c+d) (c+d) \\
& \ge \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) - c d^2 (b+c) + \Big( d (b+c) d c + d (b+c) (b+c) c \Big) \\
& \ge \left( b^4 - b^2 c^2 + b^3 d - b^2 c d \right) + \left( b c^3 - b c^2 d \right) + b (b+c) c d \\
& \ge b^4 - b^2 c^2 + b c^3 \\
& \ge 0 \quad\text{because} \quad \frac{1}{3} ( b^4 + 2 b c^3 ) \ge b^2 c^2 \quad \text{by AM-GM}
\end{align}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?
How do I deal with square roots inside the inequality?
|
Let $f(x) = \sqrt{x^2+1}+\sqrt{(1-x)^2+1}$. As $x$ and $y$ are non-negative integers, we only consider $x \in [0, 1]$.
Note that $f(x)$ is symmetric about $x = \frac{1}{2}$ and
$$f'(x) = \frac{x\sqrt{(1-x)^2+1} + (x - 1)\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}}.$$
For $0 < x < \frac{1}{2}$, $x < 1 - x$ so $\sqrt{x^2+1} > \sqrt{(1-x)^2+1}$. Therefore, $$f'(x) = \frac{x\sqrt{(1-x)^2+1} + (x - 1)\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}} < \frac{\frac{1}{2}\sqrt{(1-x)^2+1} - \frac{1}{2}\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{(1-x)^2+1}} < 0.$$ So $f$ is decreasing on $[0, \frac{1}{2})$ and as $f$ is symmetric about $x = \frac{1}{2}$, $f$ is increasing on $(\frac{1}{2}, 1]$. Therefore $f$ attains its minimum value at $x = \frac{1}{2}$ which is $$f\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2}\right)^2+1} + \sqrt{\left(1-\frac{1}{2}\right)^2+1} = \sqrt{\frac{5}{4}} + \sqrt{\frac{5}{4}} = \frac{1}{2}\sqrt{5} + \frac{1}{2}\sqrt{5} = \sqrt{5}.$$ Hence, $f(x) \geq \sqrt{5}$ for $x \in [0, 1]$. Setting $y = 1 -x$ we have $\sqrt{x^2+1} + \sqrt{y^2+1} \geq \sqrt{5}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Double Integral Question on unit square Hints on solving following double integral will be appreciated.
$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$
|
\begin{align}
\frac{x^{2} - y^{2}}{(x^{2}+y^{2})^{2}}
&=
{x^{2} + y^{2} -2y^{2} \over \left(x^{2} + y^{2}\right)^{2}}
=
{1 \over \left(x^{2} + y^{2}\right)^{2}}\,\left\lbrack%
{\partial y \over \partial y}\,\left(x^{2} + y^{2}\right)
-
{\partial\left(x^{2} + y^{2}\right) \over \partial y}\,y
\right\rbrack
\\[3mm]&=
{\partial \over \partial y}\left(y \over x^{2} + y^{2}\right)\,,
\qquad\qquad
\left(x, y\right) \not= \left(0, 0\right)
\\[5mm]&
\end{align}
\begin{align}
\lim_{\epsilon \to 0^{+}}\int_\epsilon^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x
&=
\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{1}\left.{y \over x^{2} + y^{2}}\right\vert_{\,y\ =\ 0}^{\,y\ =\ 1}\
{\rm d}x
=
\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{1}{{\rm d}x \over x^{2} + 1}
\\[3mm]&=
\arctan\left(1\right)
=
{\large{\pi \over 4}}
\end{align}
It seems to be the Mathematica package trick !!!: It's is equivalent to exclude $\left\lbrace \left(0,y\right)\ \ni\ y \in \left(0, 1\right) \right\rbrace$ and takes a limit after integration. It's like this:
\begin{align}
\int_{\epsilon}^{1}
\left.{y \over x^{2} + y^{2}}\right\vert_{\,y\ =\ 0}^{\,y\ =\ 1}\ {\rm d}x
=
\int_{\epsilon}^{1}{1 \over x^{2} + 1}\,{\rm d}x
=
{\pi \over 4} - \arctan\left(\epsilon\right) \to {\pi \over 4}
\end{align}
If you "exclude" both $x = 0$ and $y = 0$ you get:
$$
{\pi \over 4} - \arctan\left(\epsilon\right)-
\arctan\left(1 \over \epsilon\right) + \arctan\left(1\right) \to 0
$$
which is quite obvious ( the integral changes sign when $x \leftrightarrow y$ ).
The problem is related to the singular behavior of the 2D-Green function of the Laplacian operator:
$\left.\nabla^{2}\ln\left(\rho\right)\right\vert_{\rho\ \not=\ 0} = 0$ but its integral around $\vec{\rho} = \vec{0}$ is $\not= 0$.
$\rho \equiv x\,\hat{x} + y\,\hat{y}$. It's common in 2D-Electrostatic.
For example
\begin{align}
\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x
&=
\int_{S}\nabla\times\left({y \over x^{2} + y^{2}}\,\hat{x}\right)\cdot\hat{z}\,
{\rm d}x\,{\rm d}y
=
\oint\left({y \over x^{2} + y^{2}}\,\hat{x}\right)\cdot{\rm d}\vec{\rho}
\\[3mm]&=
\int_{1}^{0}{1 \over x^{2} + 1}\,\left(-{\rm d}x\right)
=
{\large{\pi \over 4}}
\end{align}
|
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|
proof for the Ramanujan's formula ? I found this formula in a textbook in which the proof to the formula was not given
Ramanujam's formula
$$\sqrt{1 +n\sqrt{1 +(n+1)\sqrt{1 + (n+2)\sqrt{1 + (n+3)\sqrt{1 +....\infty}}}}} = n+1$$
Its a great equation andhow do you prove this. its a bit difficult for me and tried different methods to solve this. Iam an undergraduate and I want you to elaborate the method of solving if its complex.
|
For all $x > 0$ and $m \in \mathbb{N}$, define $\varphi_m(x)$ by:
$$(x+1)\varphi_m(x) = \begin{cases}1,&\text{ for } m = 0\\
\\
\sqrt{1+x\sqrt{1+(x+1)\sqrt{\cdots\sqrt{1+(x+m)}}}},&\text{ otherwise. }\end{cases}$$
It is clear for $m > 0$, these functions satisfy the recurrence relations:
$$\begin{align}
&(x+1)^2\varphi_{m}(x)^2 = 1 + x(x+2)\varphi_{m-1}(x+1)\\
\iff &\varphi_{m}(x)^2 = \varphi_{m-1}(x+1) + \frac{1 - \varphi_{m-1}(x+1)}{(x+1)^2}
\end{align}
$$
Notice for any $x > 0, y \in [0,1]$, we have $y + \frac{1-y}{(x+1)^2} \in [y,1]$. This implies
$$\varphi_{m-1}(x+1) \le \varphi_{m}(x)^2 \le 1
\quad\iff\quad
\varphi_{m-1}(x+1)^{2^{-1}} \le \varphi_{m}(x) \le 1$$
Repeat apply this to $\varphi_{m-2}(x+2), \varphi_{m-3}(x+3),\ldots$, we get:
$$\begin{align}
&\varphi_{0}(x+m)^{2^{-m}} \le \varphi_{1}(x+m-1)^{2^{-(m-1)}} \le \cdots \le \varphi_{m}(x) \le 1\\
\implies & \left(\frac{1}{x+m+1}\right)^{2^{-m}} \le \varphi_m(x) \le 1\tag{*1}
\end{align}$$
Notice for fixed $x$, the limit of L.H.S of $(*1)$ goes to $1$ as $m \to \infty$. As a consequence:
$$\sqrt{1+x\sqrt{1+(x+1)\sqrt{1 + \cdots}}} = (x+1)\lim_{m\to\infty}\varphi_m(x) = x + 1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability in coin toss Suppose that you have a fair coin. You start with $\$0$. You win $\$1$ each time you get a head and loose $\$1$ each time you get tails. Calculate the probability of getting $\$2$ without getting below $\$0$ at any time.
|
Consider $P_i$ the probability to obtain $\$2$ when you have $\$i$, then we look for $P_0$. So a recurrence can be
$$P_{i}=\frac{1}{2}P_{i-1} + \frac{1}{2}P_{i+1}$$
with $P_2=1$ and $P_{-1}=0$ because we do not allow a score below of $\$0$. Now
$$P_0=\frac{1}{2}P_{-1} + \frac{1}{2}P_{1}=\frac{1}{2}P_{1},$$
$$P_1=\frac{1}{2}P_{0} + \frac{1}{2}P_{2}=\frac{1}{2}P_{0} + \frac{1}{2}$$
therefore
$$P_0=\frac{1}{2}P_{1}=\frac{1}{2}\left(\frac{1}{2}P_{0} + \frac{1}{2}\right)$$
and $P_0=\frac{1}{3}$.
|
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|
$a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$ $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$
I have tried to use AM-GM inequality, but get no result as follows:
$$a+\sqrt{ab}+\sqrt[3]{abc}\leq a+\frac{a+b}{2}+\frac{a+b+c}{3}$$
|
$a+\sqrt{ab}+\sqrt[3]{abc}=a+\frac{\sqrt{4ab}}{2}+\frac{\sqrt[3]{64abc}}{4}\le a+\frac{a+4b}{4}+\frac{a+4b+16c}{12}=\frac{4(a+b+c)}{3}=28$
|
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|
Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$? I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.
Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?
My proof is kinda simple. Let's work in mod $2$. We'll have:
$\pm 1 \pm 2 \pm \cdots \pm 100 \equiv 101 \mod 2$
but since $+1 \equiv -1 \mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:
$(1 + 0 + \cdots + 1 + 0 \equiv 50\times 1 \equiv 0) \not\equiv (101\equiv 1) \mod 2$ which is contradictory.
Therefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.
Now is there a combinatorial proof of that fact except what I have in mind?
|
Replacing 100 with $n$
and using Brian M. Scott's solution,
we want a partition of
$\{1, 2, ..., n+1\}$
into two sets with equal sums.
The sum is
$\frac{(n+1)(n+2)}{2}$,
and if $n=4k$,
this is
$(4k+1)(2k+1)$
which is odd
and therefore impossible.
If $n = 4k+1$,
this is
$(2k+1)(4k+3)$
which is also odd,
and therefore impossible.
If $n = 4k+2$,
this is
$(4k+3)(2k+2)$,
so it is not ruled out,
and each sum must be
$(4k+3)(k+1)$.
if $n = 4k+3$,
this is
$(2k+2)(4k+5)$
which is also not ruled out,
and each sum must be
$(k+1)(4k+5)$.
Now I'll try to find a solution
for the not impossible cases.
(I am working these out as I enter them.)
For the $n=4k+2$ case,
the sum must be
$(4k+3)(k+1)
=(4k+4-1)(k+1)
=4(k+1)^2-(k+1)
=(2k+2)^2-(k+1)
$.
The square there suggests,
to me,
the formula for
the sum of consecutive odd numbers
$1+3+...+(2m-1)=m^2$,
so $1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is odd,
remove it from the sum
so it is
$(2k+2)^2-(k+1)$.
If $k+1$ is even,
both $1$ and $k$ are odd,
so remove them from the sum.
In either case, we have the desired partition.
For the $n=4k+3$ case,
the sum must be
$(4k+5)(k+1)
=(4k+4+1)(k+1)
=4(k+1)^2+(k+1)
=(2k+2)^2+(k+1)
$.
Again,
$1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is even,
add it to the sum
so it is
$(2k+2)^2+(k+1)$.
If $k+1$ is odd,
$k+2$ is even,
so remove $1$
and add $k+2$ to the sum.
In either case, we have the desired partition.
I do not know if these partitions
are unique.
|
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|
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
|
O method of differences, so powerful and yet so despised...
The $n$th term of the series to be computed is
$$
\prod_{k=1}^n\frac{2k+1}{2k+4}=\frac4{2n+4}\prod_{k=1}^n\frac{2k+1}{2k+2}=4(1-a_{n+1})\prod_{k=1}^na_k
$$ where $$a_k=\frac{2k+1}{2k+2}$$
By telescoping, each partial sum of the series is
$$
\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{2k+1}{2k+4}=4-4\prod_{k=1}^Na_k
$$
Since $1-a_k\sim1/(2k)$, the product $\prod\limits_na_n$ diverges to $0$ hence the sum of the full series is $4$.
$$
\sum_{n=0}^\infty\prod_{k=1}^n\frac{2k+1}{2k+4}=4
$$
More generally, for every $(a,b)$ such that $a>-1$ and $b>a+1$,
$$\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{k+a}{k+b}=\frac{b}{b-a-1}\left(1-\prod_{k=1}^Na_k\right)$$ where $$a_k=\frac{k+a}{k+b-1}$$
hence, if furthermore $b\leqslant 2+a$, then the product $\prod\limits_na_n$ diverges to $0$ hence
$$\sum_{n=0}^\infty\prod_{k=1}^n\frac{k+a}{k+b}=\frac{b}{b-a-1}$$
The question above asks about the case $$a=1/2\qquad b=2$$ which fits these conditions.
Edit: Another exact formula for the partial sums, equivalent to the one above, is
$$
\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{2k+1}{2k+4}=4-4\cdot\frac1{4^N}{2N+1\choose N}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
}
|
The number of solutions to $\frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n,x,y,z\in\mathbb N$ Denote
$$g(n)=\{\{x,y,z\}\mid \frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n,x,y,z\in\mathbb N\},$$
$$h(n)=\{\{x,y,z\}\mid \frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n,1\leq x\leq y\leq z,x,y,z\in\mathbb N\},$$
let $f(n)=|g(n)|$ be the number of members of $g(n)$.
For example, $h(3)=\{\{2,3,6\},\{2,4,4\},\{3,3,3\}\},f(3)=6+3+1=10.$
Since $\{n,n,n\}$ is a solution to $\frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n$, it's easy to see that $f(k)\equiv 1\pmod 3,\forall k\in \mathbb N.$
Question: I find that
$$f(3k)\equiv 0,f(4k+2)\equiv 0,f(6k\pm1)\equiv1 \pmod 2,\forall k\in \mathbb N.$$
I wonder how to prove them?
Edit: I find that $f(n)$ has the same parity to the number of solutions to $\frac{1}x+\frac{2}y=\frac{3}n,$ I think I have got it now.
|
Since I already know how to prove them, I write a proof here now.
It's easy to see that in $h(n)$,
(1)if $x,y,z$ are distinct, then $x,y,z$ add $6$ to $f(n)$,
(2)if just two of them are equal, add $3$ to $f(n)$,
(3)if $x=y=z$, then they add $1$ to $f(n)$.
Since $6$ is even, case (1) didn't change the parity of $f(n)$. Hence $f(n)$ has the same parity of the number of solutions to $\frac{1}x+\frac{2}y=\frac{3}n.$ This is $(3x-n)(3y-2n)=2n^2,$ let $r(n)$ be the number of solutions to this equation.
If $n=3m,$ then $(x-m)(y-2m)=2m^2,$ hence $f(n)\equiv r(n)=d(2m^2)\equiv 0\pmod 2.$
If $n=6m+1$, then $3x-n=a,3y-2n=b,$
$$f(n)\equiv r(n)=\sum_{\substack{ab=2n^2\\a\equiv -n\equiv 2\pmod 3\\b\equiv -2n\equiv 1\pmod 3}}1=\frac{1}2d(2n^2)=d(n^2)\equiv 1\pmod 2.$$
The same to $n=6m-1.$
If $n=4m+2,$ then if $3\mid n$, we get $2\mid f(n),$ too. If $3\not \mid n$, then $2n^2\equiv -1\pmod3,f(n)\equiv r(n)=\dfrac{1}2d(2n^2)=2d((2m+1)^2)\equiv 0\pmod 2.$
Now we get a little more:
$$f(n) \equiv
\begin{cases}
\dfrac{1}2d(2n^2), & 3\not\mid n \\
0, & 3\mid n \\
\end{cases} \pmod 2 $$
|
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"url": "https://math.stackexchange.com/questions/481049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to prove :$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$ For sequence $u_n$ satisfing : $$\begin{cases} u_1=\sqrt{2015}\\ u_{n+1}=u_n^2-2\end{cases}$$
How to prove : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$
|
Here I use its properties as an additive telescoping series:
$$\left(\frac{u_n^2-2}{u_1\cdots u_n}\right)^2=\frac{u_n^4}{(u_1\cdots u_n)^2}-\frac{4u_n^2}{(u_1\cdots u_n)^2}+\frac{4}{(u_1\cdots u_n)^2}$$
$$=\frac{u_n^2}{(u_1\cdots u_{n-1})^2}-\frac{4}{(u_1\cdots u_{n-1})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
Substitute again:
$$=\left(\frac{u_{n-1}^2}{(u_1\cdots u_{n-2})^2}-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_{n-1})^2} \right)-\frac{4}{(u_1\cdots u_{n-1})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
$$=\frac{u_{n-1}^2}{(u_1\cdots u_{n-2})^2}-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
And again:
$$=\left(\frac{u_{n-2}^2}{(u_1\cdots u_{n-3})^2}-\frac{4}{(u_1\cdots u_{n-3})^2}+\frac{4}{(u_1\cdots u_{n-2})^2} \right)-\frac{4}{(u_1\cdots u_{n-2})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
$$=\frac{u_{n-2}^2}{(u_1\cdots u_{n-3})^2}-\frac{4}{(u_1\cdots u_{n-3})^2}+\frac{4}{(u_1\cdots u_n)^2}$$
Reapeat this process ad infinitum (noting that $\lim_{n \rightarrow \infty}u_n= \infty$):
$$=\frac{u_{2}^2}{(u_1)^2}-\frac{4}{(u_1)^2}+\frac{4}{(u_1\cdots u_n)^2}$$
And take the limit $n \rightarrow \infty$:
$$=\frac{(2015-2)^2}{2015}-\frac{4}{2015}+0$$
$$=\frac{2015^2-4\cdot 2015 +4}{2015}-\frac{4}{2015}$$
$$=2015-4=2011$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/481900",
"timestamp": "2023-03-29T00:00:00",
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|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
We can use Parseval's theorem!
If $F$ is the Fourier transform of $f$, and $G$ is the Fourier transform of $g$, then
$$
\int_{-\infty}^\infty\overline{f(t)}g(t)\,dt=
\frac{1}{2\pi}\int_{-\infty}^\infty\overline{F(x)}G(x)\,dx.
$$
We have the Fourier transform pairs
*
*$\displaystyle e^{-|t|}\longmapsto\frac{2}{1+\omega^2}$
*$\displaystyle \frac{d}{dt}f(t) \longmapsto i\omega F(\omega)$
It is clear from those that
$$
f(t)=\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\longmapsto
\frac{ix}{1+x^2}=F(x),
$$
and conveniently $\overline{F(x)}F(x)=\frac{x^2}{(1+x^2)^2}$ is the integrand that we want.
Using this and Parseval's theorem, we see that
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
2\pi\int_{-\infty}^\infty\left(\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\right)^2dt,
$$
and since the integrand is even, one can write
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
\pi\int_0^\infty\left(\frac{d}{dt}\left(e^{-t}\right)\right)^2dt=
\pi\int_0^\infty e^{-2t}dt=\frac{\pi}{2}.
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \pmod 5$
Then I added the two equations:
$4x + 3y \equiv 0 \pmod{5}$
$-4x - 2y \equiv 4 \pmod{5}$
This simplifies to $y \equiv 4 \pmod{5}$.
I then plug this into the first equation: $4x + 3(4) = 0 \pmod{5}$
Wrong work:
Thus, $x = 3$.
But when I plug the values into the first equation, I get $2(3) + 4 \not\equiv 3 \pmod{5}$.
What am I doing wrong?
EDIT:
Revised work:
$x = -3 \pmod{5} = 2 \pmod{5}$.
Now when I plug the values into the first equation, I get $2(2) + 4 \equiv 8 \pmod{5} \equiv 3 \pmod{5}$.
|
Sign error on substitution, it should be $x\equiv -3\pmod{5}$.
You had $4x+(3)(4)\equiv 0$, that is, $4(x+3)\equiv 0$. From this we get $x+3\equiv 0$, so $x\equiv -3\pmod{4}$.
Negative numbers are sometimes troublesome, so we may wish to rewrite as $x\equiv 2\pmod{5}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/484467",
"timestamp": "2023-03-29T00:00:00",
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|
Computing Infinite Continued Fractions I am looking for "tricks" used to compute infinite continued fractions.
For example, $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}$$ is the golden ratio since if we denote it by $x$, then we have $$x=1+\frac{1}{x},$$ which simplifies to $$x^2-x-1=0$$
Are there any other (different/elegant) examples of ways to compute infinite continued fractions?
|
This Infinite Continued Fractions can be written in many different way.
$$ \frac{1}{1} ;\frac{1}{1+\frac{1}{1}};\frac{1}{1+ \frac{1}{1+\frac{1}{1}}};\frac{1}{1+ \frac{1}{1+ \frac{1}{1+\frac{1}{1}}}} ... $$
like this:
$$ \frac{1}{1} ;\frac{1}{2};\frac{3}{2};\frac{5}{3};\frac{8}{5};\frac{13}{8};\frac{21}{13}... $$
there is a Infinite series: $$\frac{13}{8} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2n + 1)!}{(n + 2)!n!(4)^{2n+3}}$$
the limit of this sequance is $\phi=\frac{1 + \sqrt{5}}{2} = 1.680339887...$
also $\phi=\frac{1 + \sqrt{5}}{2}$ is a root of $\phi^{2}-\phi-1=0$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Olympiad number theory problem I found this problem in previous problems of the olympiads of my country
If $t^2+n^2=r^2$, where $t$ has $3$ positive divisors, $n$ has $30$ positive divisors and $t,n,r$ are natural numbers, find the sum of all the possible values of $t$
I did gave it a try but only solved it partially, and I really doubt this is the best way to do it.
My progress:
First, $t$ must be of the form $p^2$ where $p$ is prime
Then, we use pythagorean triplets
Case 1:
$$t=p^2=2ab$$
$$n=a^2-b^2$$
Therfore, $p$ must be $2$, therefore $a=2, b=1$, but this does not satisfy $n$ having $30$ divisors.
Case 2:
$$t=p^2=a^2-b^2=(a-b)(a+b)$$
$$n=2ab$$
Since $p$ is prime, we get $a=b+1$, therefore
$$p^2=2b+1$$
$$n=2(b+1)b$$
If $n$ has $30$ divisors, it must be of one of the forms
$$q^{29},r^{14}q,r^{9}q^2,r^4q^5,r^4q^2s$$
Where $r,q,s$ are different primes
Since $p$ is odd, $p^2=8k+1=2b+1$, $b=4k$ for some integer $k$
Case 2.1
$$8k(4k+1)=r^{29}\implies 4k(4k+1)=2^{28}$$
That is clearly impossible
Case 2.2
$$8k(8k+1)=r^{14}q\implies r=2$$
$$4k(4k+1)=2^{13}q$$
$$\implies 4k=2^{13} \implies q=4k+1=2^{13}+1=3×2731$$
Case 2.3
$$8k(4k+1)=r^9q^2\implies r=2$$
$$8k(4k+1)=2^9q^2$$
$$4k(4k+1)=2^8q^2$$
$$\implies 4k=2^8\implies q^2=4k+1=256+1=257$$
Case 2.4
$$8k(4k+1)=r^4q^5$$
Case 2.4.1($r=2$)
$$4k(4k+1)=2^3q^5$$
$$\implies 4k=8\implies q^5=4k+1=8+1=9$$
Case 2.4.2($q=2$)
$$4k(4k+1)=r^42^4$$
$$\implies 4k=16\implies r^4=4k+1=16+1=17$$
Case 2.5
$$8k(4k+1)=r^4q^2s\implies r=2$$
$$8k(4k+1)=16q^2s$$
$$8k(8k+2)=32q^2s$$
$$(p-1)(p+1)=32q^2s$$
$$p^2-1=32q^2s$$
Where all $p,q,s$ are odd primes
The question now is: how do I proceed from here? Was there an easier way to solve this?
|
Let us start as you did. I will assume that $t$, $n$ and $r$ are coprime. Then $t=a^2-b^2$ and $n=2ab$. See the answer of André Nicolas for the case when $t$, $n$ and $r$ are not coprime.
We get that
\begin{align*}
p^2 &= 2b+1,\\
n &= 2(b+1)b.
\end{align*}
Therefore,
$$n = \frac{p^4-1}{2} = (p-1) \times (p+1) \times \frac{p^2+1}{2}.$$
Note that every prime number $a \neq 2$ divides at most one of the terms $(p-1)$, $(p+1)$, and $\frac{p^2+1}{2}$.
If neither of these terms is a power of 2, then $n$ has at least $4$ prime factors: 2, a prime divisor of $p-1$ (other than 2), a prime divisor of $p+1$ (other than 2), a prime divisor of $\frac{p^2+1}{2}$. This is impossible since $n$ has 30 divisors.
We conclude that one of the numbers $p-1$, $p+1$, $\frac{p^2+1}{2}$ is a power of 2. Note that $\frac{p^2+1}{2}$ is odd. Thus either $p-1$ or $p+1$ is a power of 2.
Note that then $n$ has 3 prime factors: 2, a prime divisor of either $p-1$ or $p+1$ (other than 2), a prime divisor of $\frac{p^2+1}{2}$. So it must be the case that $n=r^4 q^2 s$. Since $n = (p^4-1)/2$ is divisible by 8, we have that $r=2$.
Consider two cases
*
*$p-1$ is a power of 2. Observe that $p\neq 3$. Then $p+1$ is divisible by 2 but not by 4. We have $p - 1 = 8$ and $p=9$. Then $n=3280$. But $3280$ has 20 factors.
*$p+1$ is a power of $2$. Then $p-1$ is divisible by 2 but not by 4. We have $p + 1 = 8$ and $p=7$. Then $n = 1200$. We check that 1200 has 30 factors.
Answer: $t=49$, $n =1200$ and $r=1201$.
|
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|
Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.
|
We can use the inequalities between quadratic mean (a.k.a. square root mean), arithmetic mean, geometric mean and harmonic mean
$$\sqrt{\frac{x^2+y^2}2} \ge \frac{x+y}2 \ge \sqrt{xy} \ge \frac2{\frac1x+\frac1y}.$$
These inequalities are true for any $x,y>0$. The equality holds if and only if $x=y$. They can be generalized for more than two variables (also proofs for two variables are easier). If you are not familiar with these inequalities, you can find a lot of material on them, just try to search for some reasonable queries like this one or this one.
You want to minimize $f(a,b)=\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2$. Notice that
$$f(a,b)=\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2 = a^2+b^2 + \frac1{a^2}+\frac1{b^2} +4.$$
So minimizing $g(a,b)=a^2+b^2 + \frac1{a^2}+\frac1{b^2}$ is an equivalent problem.
Using QM-AM inequality we get
$$\frac{a^2+b^2}2 \ge \left(\frac{a+b}2\right)^2 = \frac14$$
which implies $$a^2+b^2 \ge \frac12.\tag{1}$$
If we combine HM-GM and AM-GM we get
$$\frac2{\frac1{a^2}+\frac1{b^2}} \le \sqrt{a^2b^2} = ab \le \left(\frac{a+b}2\right)^2 = \frac14,$$
which is equivalent to
$$\frac1{a^2}+\frac1{b^2} \ge 8.\tag{2}$$
By adding the inequalities $(1)$ and $(2)$ we get
$$g(a,b) = a^2+b^2 + \frac1{a^2}+\frac1{b^2} \ge 8+\frac12$$
and
$$f(a,b)=g(a,b)+4 \ge 12+\frac12 = \frac{25}2.$$
The inequalities $(1)$ and $(2)$ can be also interpreted geometrically.
Minimizing $a^2+b^2$ for $a+b=1$ is simply finding the point on the line $a+b=1$ which is closest to the origin. If you draw the picture, you immediately see that it is the point given by $a=b=\frac12$.
We also want some geometrical insight into minimizing $\frac1{a^2}+\frac1{b^2}$ for $a+b=1$. Let us transform this problem a bit.
If we denote $x=\frac1a$, then $y=\frac1b=\frac1{1-a}=\frac1{1-\frac1x}=\frac{x}{x-1}$. It is not difficult to see that this is a hyperbole with the asymptotes $x=1$ and $y=1$. And we want to minimize $\frac1{a^2}+\frac1{b^2}=x^2+y^2$, which means that we want to find the closest point to the origin on this hyperbole. (To be more precise, only in the part of the hyperbole in the first quadrant, since $x,y>0$.) Here is a plot from WolframAlpha.
Again we see from the picture that the closest point will be the one with $x=y$. (Which gives us $x=y=2$ and $a=b=\frac12$.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)} \right|$$ converges.
i tried to check for convergence with the 2 criterias a) decreasing and b) zero sequence but yes then i tried to transform the equotation to
$$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n(n+1)} $$
the first fraction is "ok" - the 2nd one i did partial fraction decomposition and finally got $$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n} - \frac{1}{n+1} $$
i then tried to see a pattern by find out the first sequences but im not sure if i'm on the right track.
|
\begin{align}
?
&\equiv
\sum\limits_{n = 1}^{\infty}\left(-1\right)^n \frac{2n + 3}{n\left(n + 1\right)}
=
\sum_{n = 1}^{\infty}\left\lbrack%
{4n + 3 \over 2n\left(2n + 1\right)}
-
{4n + 1 \over \left(2n - 1\right)\left(2n\right)}
\right\rbrack
\\[3mm]&=
\sum_{n = 1}^{\infty}\left\lbrack%
{1 \over n}
+
{1 \over 2n\left(2n + 1\right)}
-
{1 \over n}
-
{3 \over \left(2n - 1\right)\left(2n\right)}
\right\rbrack
\\[3mm]&=
{1 \over 4}
\sum_{n = 1}^{\infty}\left\lbrack%
{1 \over n\left(n + 1/2\right)}
-
{3 \over n\left(n - 1/2\right)}
\right\rbrack
=
{1 \over 4}
\sum_{n = 0}^{\infty}\left\lbrack%
{1 \over \left(n + 1\right)\left(n + 3/2\right)}
-
{3 \over \left(n + 1\right)\left(n + 1/2\right)}
\right\rbrack
\\[3mm]&=
{1 \over 4}\left\lbrack%
{\Psi\left(3/2\right) - \Psi\left(1\right) \over 3/2 - 1}
-
3\,{\Psi\left(1\right) - \Psi\left(1/2\right) \over 1 - 1/2}
\right\rbrack
=
{1 \over 2}\left\lbrack%
\Psi\left(3 \over 2\right) - 4\Psi\left(1\right) + 3\Psi\left(1 \over 2\right)
\right\rbrack
\\[3mm]&=
{1 \over 2}\left\lbrace%
\left\lbrack\Psi\left(1 \over 2\right) + 2\right\rbrack- 4\Psi\left(1\right) + 3\Psi\left(1 \over 2\right)
\right\rbrace
=
1
+
2\left\lbrack\Psi\left(1 \over 2\right) - \Psi\left(1\right)\right\rbrack
\\[3mm]&=
1
+
2\left\lbrace%
\left\lbrack-\gamma - 2\ln\left(2\right)\right\rbrack
-
\left(-\gamma\right)
\right\rbrace
\end{align}
$$
\begin{array}{|c|}\hline\\
{\large%
?
\equiv
\sum\limits_{n = 1}^{\infty}\left(-1\right)^n \frac{2n + 3}{n\left(n + 1\right)}
=
\color{#ff0000}{1 - 4\ln\left(2\right)}}
\\ \\ \hline
\end{array}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/487695",
"timestamp": "2023-03-29T00:00:00",
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drawing balls from an urn (conditional probability) Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black?
I encountered this problem in my AoPS textbook, as a complete newcomer to conditional probability. Is the answer 1/2? I would really appreciate a solution, as I have no way to check my work.
|
The answer is the Probability that the first ball is red and the second ball is black, divided by the Probability that the second ball
is black.
$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$
P(1st ball = Red and 2nd ball = Black )
$=\frac{1}{2}[ (\frac{2}{6}* \frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) ] = \frac{7}{60}$
P(Second ball is BLACK) = $\frac{1}{2}[ (Urn A's Redball*\frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) + ( \frac{3}{6}* \frac{1}{2}*\frac{2}{5} ) ] = \frac{1}{2}[ (1*\frac{1}{2}*\frac{3}{6}) + ( \frac{3}{6}* \frac{1}{2}*\frac{3}{5} ) + ( \frac{3}{6}* \frac{1}{2}*\frac{2}{5} ) ] = \frac{1}{4}$
Therefore,
$\frac{P(\text{First Ball is RED AND Second Ball is BLACK) }}{P(\text{Second Ball is BLACK) }}$ = $\frac{\frac{7}{60}}{\frac{1}{4}} = \frac{7}{15}$
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|
Bessel function to $\sin(kr)$ $J_{\frac{1}{2}}(kr)=\frac{\sqrt{\frac{2}{\pi }} \text{Sin}[\text{kr}]}{\sqrt{\text{kr}}})$ This can be easily obtained by Mathematica,
How to do the details?
|
One way is to use the series definition of the Bessel function together with the duplication formula for the Gamma function.
For $x \geq 0$ we have
$$
\begin{align}
J_{1/2}(x) &= \sqrt{\frac{x}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+1/2)} \left(\frac{x}{2}\right)^{2n} \\
&= \sqrt{\frac{2}{x}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+1/2)} \left(\frac{x}{2}\right)^{2n+1}.
\end{align}
$$
Now
$$
\Gamma(n+1)\Gamma(n+1+1/2) = 2^{-1-2n} \sqrt{\pi} \,\Gamma(2n+2)
$$
by the duplication formula, so
$$
J_{1/2}(x) = \sqrt{\frac{2}{\pi x}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma(2n+2)} x^{2n+1} = \sqrt{\frac{2}{\pi x}} \sin x.
$$
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|
Find all positive integers $n$ for which $1 + 5a_n.a_{n + 1}$ is a perfect square. The sequence $a_1, a_2, \ldots $ is defined by the initial conditions $$a_1 = 20; \quad a_2 = 30$$ and the recursion
$$a_{n+2} = 3a_{n+1} - a_n$$ and
for $n \geq 1$. Find all positive integers $n$ for which $1 + 5a_n * a_{n+1}$ is a perfect square.
I could only find the $n$-th term and don't know how to proceed further.pls help
|
The only such $n$ is $n=3$, with
$$
1 + 5 a_3 a_4 = 1 + 5 \cdot 70 \cdot 180 = 63001 = 251^2.
$$
Let $b_n = a_n/10 = 2, 3, 7, 18, 47, \ldots$ for $n=1,2,3,4,5,\ldots$ .
These are sums of consecutive odd-order Fibonacci numbers:
$2 = 1+1$ (with the first $1$ being $F_{-1}$),
$3 = 1+2$, $7 = 2+5$, $18 = 5+13$, $47 = 13+34$, etc. by induction.
It soon follows that $b_n b_{n+1} = 5 F^2 + 1$ where $F$ is the
Fibonacci number common to $b_n$ and $b_{n+1}$:
$$
2\cdot 3 = 5 \cdot 1^2 + 1,\phantom{M}
3\cdot 7 = 5 \cdot 2^2 + 1,\phantom{M}
7\cdot 18 = 5 \cdot 5^2 + 1,\phantom{M}
18\cdot 47 = 5 \cdot 13^2 + 1,
$$
etc.
So we're looking to make
$$
1 + 5 a_n a_{n+1} = 1 + 500 b_n b_{n+1} = 2500 F^2 + 501
$$
a square, and it's easy to see that $F = 5$ is the only positive integer
that makes this happen even without the hypothesis that $F$ be
a Fibonacci number. (For instance, if $2500 F^2 + 501 = y^2$ with $y>0$,
we may factor $501 = y^2 - 2500F^2 = (y-50F) (y+50F)$,
or bound $y$ between $50F$ and $50F+1$ once $F>5$,
or use the technique I described in
this Mathoverflow answer.) Therefore $n=3$ is the unique answer as claimed.
|
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|
find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you
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Factor a $\tan(x)$ from the denominator:
$$\frac{\tan\tan(x) - \tan\sin(x)}{\tan(x)-\sin(x)} = \frac{\frac{\tan\tan(x)}{\tan(x)} - \cos(x)\frac{\tan\sin(x)}{\sin(x)}}{1-\cos(x)}.$$
The quotients in the numerator go to $1$, so the numerator goes to $1-\cos(x)$. This suggests cancellation.
After two derivatives, the denominator of this quotient will go to $1$, so one strategy is L'Hopital's rule. (I don't have time to compute it but maybe I'll have time later today.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $3^n \ge n^3$ by induction Yep, prove $3^n \ge n^3$, $n \in \mathbb{N}$.
I can do this myself, but can't figure out any kind of "beautiful" way to do it.
The way I do it is:
Assume $3^n \ge n^3$
Now,
$(n+1)^3 = n^3 + 3n^2 + 3n + 1$,
and $\forall{} n \ge 3$,
$3n^2 \le n^3, \,\, 3n + 1 \le n^3$
Which finally gives $(n+1)^3 \le 3n^3 \le 3^{n+1}$ by our assumption.
Now just test by hand for n=1,2,3 and the rest follows by induction.
Anyone got anything simpler?
|
Here is another argument, but it's not necessarily simpler than yours:
1) If $n=1$, $3^1=3\ge1=1^3$; if $n=2$, $3^2=9\ge8=2^3$; and if $n=3$, $3^3=3^3$.
2) Now assume that $3^n\ge n^3$ for some integer $n\ge3$.
Then $\displaystyle\frac{1}{n}\le\frac{1}{3}$, so $\displaystyle\frac{(n+1)^3}{n^3}=\big(1+\frac{1}{n}\big)^3\le\big(\frac{4}{3}\big)^3=\frac{64}{27}\le3$, so
$\;\;\;\;\;\;\;\;3^{n+1}=3(3^n)\ge3n^3\ge(n+1)^3$.
|
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|
Contest math problem algebra proof Let $r, s$ be integers and let
$$a = (2011)^2 + (2011)r + s$$ and
$$b = (2012)^2 + (2012)r + s$$
Show that there exists an integer $c$ with $c^2 + rc + s = ab$.
Can anyone help me with this?
|
Choose r = 2 and s = 1
Then $2011^2 + 2.2011 + 1 = a$
and $2012^2 + 2.2012 + 1 = b$
$a = (2011 + 1)^2$
$b = (2012 + 1)^2$
$ab = (2012.2013)^2$
Let 2012 = u
$ab = (u(u+1))^2$
$ab = u^4 + 2u^3 + u^2$
Let $c = au^2+bu+l$
$ab = (au^2+bu+l)^2 + (au^2 + bu +l)*2 + 1$
$a^2u^4 + b^2u^2+l^2+2(au^2bu+2bul+alu^2)+2au^2+2bu+2c+1$
$a^2u^4+2abu^3+u^2(b^2+2al+2a)+(2bl+2b)u+l^2+2l+1$
Expanding the expression
$a^2 = 1 > a = 1$
$2ab = 2 > b = 1$
$b^2 + 2al + 2a = 1 > l = -1$
Thus $c = (u^2 + u -1)$ which is an integer as u is an integer
Goes to prove that $ab = c^2 + c*2 + 1$ holds true for an integer c.
Thanks
Satish
|
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.