Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Use the de Moivre theorem to evaluate the following $$\frac{(1+i)(\sqrt{3} + i)^3}{(1-\sqrt{3}i)^{3}} = 1-i$$
What confuses me is how would I do the numerator because I have two expressions.
| $1+i=\sqrt 2 (\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$
$ \sqrt3+i=2(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})) $
$ 1-i\sqrt3=2(\cos(\frac{-\pi}{3})+i\sin(\frac{-\pi}{3})$
Then you can simply apply De Moivre's theorem:
The numerator becomes $8\sqrt2(\cos(\frac{9\pi}{12})+i\sin(\frac{9\pi}{12}))=8\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))$
The denominator becomes $8(\cos(-\pi)+i\sin(-\pi))=-8$
So the fraction is equal to $-\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))=1-i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/263430",
"timestamp": "2023-03-29T00:00:00",
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Compute $\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space$ How would you approach
$$\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space?$$
The way I see here involves Dirichlet kernel. I wonder what else can we do, maybe some easy/elementary approaching ways. Thanks !
| Let $I=\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx$
As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
$\displaystyle I=\int_0^{\frac\pi2} \frac{\sin (2n+1)(\frac\pi2-x)}{\sin (\frac\pi2-x)} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\sin \{n\pi+\frac\pi2-(2n+1)x\}}{\cos x} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ if $n$ is even.
$\displaystyle =-\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ if $n$ is odd.
If $n$ is odd,
$\displaystyle 2I=\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx-\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\sin (2n)x}{\sin x\cos x} dx$
$\displaystyle =2\int_0^{\frac\pi2} \frac{\sin (2n)x}{\sin 2x} dx$
$\displaystyle =2\frac12\int_0^{\pi} \frac{\sin ny}{\sin y} dy$
$\displaystyle =2\int_0^{\frac{\pi}2} \frac{\sin ny}{\sin y} dy$ as $\displaystyle\frac{\sin ny}{\sin y}$ is an even function.
So, $\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx=I=\int_0^{\frac{\pi}2} \frac{\sin nx}{\sin x} dx$ if $n$ is odd.
Similarly, $\displaystyle \int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (n+1)x}{\sin x} dx$ if $n$ is even.
If we put, $2n+1=2013, n=1006$ which is even.
S0, $\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (1007)x}{\sin x} dx$
Now, if we put $2n+1=1007,n=503$ which is odd.
So, $\displaystyle \int_0^{\frac{\pi}2} \frac{\sin (1007)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (503)x}{\sin x} dx$
Now, if $2n+1=503,n=251$
The reduction of $n$ will follow :$2013,1007,503,251,125,63,31,15,7,3,1$
So,$\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin x}{\sin x} dx=\frac{\pi}2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Quick way to iterate multiples of a prime N that are not multiples of primes X, Y, Z, ...? Is there a way to quickly iterate multiples of some prime $N$ while avoiding multiples of blacklisted primes $X$, $Y$, $Z$, ...? By quickly I mean is there a faster way than:
*
*Increment current number by N.
*Check if current number is divisible any each number in blacklist (M checks if there are M primes in the blacklist).
*If so, jump to 1, else return the current number as the next number in the iteration.
My end goal would be an O(1) algorithm that lets me answer queries like, "Give me the 5th number that is a multiple of N but not a multiple of X, Y, or Z." But the above algorithm requires computing the 1-4th numbers first, so it's linear time.
I suspect the answer is no because it seems that if there is a way it would allow speeding up the Sieve of Eratosthenes. Although in some cases the sieve avoids redundant marking by starting at a prime square, it still redundantly marks numbers sometimes. For example the sieve will wastefully mark 60 three times, once for 2, once for 3, and once for 5. You can see this by watching 60 light up three different colors in the linked Wikipedia article's first graphic:
| Up to $kN$ (inclusive) there are $k$ positive integer multiples of $N$, of which $\lfloor k/X \rfloor$ are multiples of $X$, etc. By inclusion-exclusion, the number that escape a "blacklist" of $3$ primes is
$$ k - \left\lfloor \frac{k}{X} \right\rfloor - \left\lfloor \frac{k}{Y} \right\rfloor - \left\lfloor \frac{k}{Z}
\right\rfloor + \left\lfloor \frac{k}{XY} \right\rfloor + \left\lfloor \frac{k}{XZ} \right\rfloor + \left\lfloor \frac{k}{YZ}\right\rfloor - \left\lfloor \frac{k}{XYZ}\right\rfloor $$
Approximate it without the $\lfloor \cdot \rfloor$'s, and you're off by at most $4$, so then adjust...
EDIT: For example, suppose you want the $35$'th positive integer multiple of $N$ (any prime other than $5$, $7$ or $11$) that is not a multiple of $X=5$, $Y=7$ or $Z=11$. Thus you want $k$ so that
$$f(k) = k - \left\lfloor \frac{k}{X} \right\rfloor - \left\lfloor \frac{k}{Y} \right\rfloor - \left\lfloor \frac{k}{Z}
\right\rfloor + \left\lfloor \frac{k}{XY} \right\rfloor + \left\lfloor \frac{k}{XZ} \right\rfloor + \left\lfloor \frac{k}{YZ}\right\rfloor - \left\lfloor \frac{k}{XYZ}\right\rfloor = 35$$
Now
$$\eqalign{k &- \frac{k}{X} - \frac{k}{Y} - \frac{k}{Z}
+ \frac{k}{XY} + \frac{k}{XZ} + \frac{k}{YZ} - \frac{k}{XYZ} \cr
&= k \left(1-\frac{1}{X}\right)\left(1 -\frac{1}{Y}\right) \left( 1 - \frac{1}{Z}\right) = \frac{48 k}{77}}$$
which would be $35$ for $k = 56.14\ldots$. Now $f(56) = 34$ so we try $f(57)$ and find that this is $35$. Thus $57N$ is the multiple of $N$ we are looking for.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x)\,\mathrm dx$ What would you recommend me for the integral below?
$$
\int_{0}^{1}\ln(1 - x)\ln(x)
\ln(1 + x)\,\mathrm dx
$$
For instance, for the version without the last logarithm would work to use Taylor series, but in this case things are a bit more complicated and it doesn't seem to work.
| $$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$
By using the the algebraic identity $ab=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2$ we get:
$$I=\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x^2\right)\:dx}_{t=x^2}-\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)\:dx$$
$$=\underbrace{\frac{1}{16}\int _0^1\frac{\ln \left(t\right)\ln ^2\left(1-t\right)}{\sqrt{t}}\:dt}_{I_1}-\underbrace{\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1-x\right)\:dx}_{I_2}+\frac{1}{2}\underbrace{\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx}_{I}-\underbrace{\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1+x\right)\:dx}_{I_3}$$
$$\frac{1}{2}I=\frac{1}{16}I_1-\frac{1}{4}I_2-\frac{1}{4}I_3$$
$$I_1=\frac{1}{16}\int _0^1\frac{\ln \left(t\right)\ln ^2\left(1-t\right)}{\sqrt{t}}\:dt=\frac{1}{16}\lim_{\alpha\rightarrow 1/2\\\beta\rightarrow 1}\frac{\partial ^3}{\partial \alpha \partial \beta ^2}\text{B}\left(\alpha ,\beta \right)$$
$$=\frac{7}{4}\zeta \left(3\right)-6-\ln ^2\left(2\right)+4\ln \left(2\right)+2\zeta \left(2\right)-\frac{3}{2}\ln \left(2\right)\zeta \left(2\right)$$
$$I_2=-\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x\right)\:dx}_{t=1-x}=-\frac{1}{4}\int _0^1\ln \left(1-t\right)\ln ^2\left(t\right)\:dt=\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)^3}$$
$$=-\frac{1}{2}\zeta \left(3\right)+\frac{3}{2}-\frac{1}{2}\zeta \left(2\right)$$
For this one make use of $\ln ^2\left(1-x\right)=2\sum _{k=1}^{\infty }\left(\frac{H_k}{k}-\frac{1}{k^2}\right)x^k$.
$$I_3=-\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1+x\right)\:dx$$
$$=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k}\int _0^1x^k\ln \left(x\right)\:dx+\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k^2}\int _0^1x^k\ln \left(x\right)\:dx$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k\left(k+1\right)^2}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k^2\left(k+1\right)^2}$$
$$=\frac{1}{16}\zeta \left(3\right)-\frac{1}{4}\zeta \left(2\right)+\frac{1}{2}\ln ^2\left(2\right)+\frac{3}{2}-2\ln \left(2\right)$$
Collecting the results after multiplying by $2$ we have:
$$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$
$$=\frac{21}{8}\zeta \left(3\right)-3\ln \left(2\right)\zeta \left(2\right)+\frac{5}{2}\zeta \left(2\right)-\ln ^2\left(2\right)-6+4\ln \left(2\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Getting the equation of an ellipse using the constant and the foci Find the equation of the ellipse with the foci at (0,3) and (0, -3) for which the constant referred to in the definition is $6\sqrt{3}$
So I'm quite confused with this one, I know the answer is $3x^2+2y^2=54$ through trial and errror, but is there any way to solve it without trial and error?
The constant referred to is the sum of the distances from the foci right? How do I use it? I've been experimenting with how you can get $6 \sqrt3$ from the equation like using distance formula etc. I'm kinda stuck with this one, can someone please explain thanks!
| Let $(x,y)$ be a point of ellipse then from definition
$$\sqrt{(x-0)^2+(y-3)^2}+\sqrt{(x-0)^2+(y+3)^2}=6\sqrt{3}$$
$$\sqrt{x^2+(y-3)^2}+\sqrt{x^2+(y+3)^2}=6\sqrt{3}$$
$$2x^2+2y^2+18+2\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=108$$
$$x^2+y^2+9+\sqrt{(x^2+(y+3)^2)(x^2+(y-3)^2)}=54$$
$$(x^2+y^2-45)^2=(x^2+(y+3)^2)(x^2+(y-3)^2)$$
$$(x^2+y^2)^2-90x^2-90y^2+45^2=x^4+x^2(y+3)^2+x^2(y-3)^2+(y^2-9)^2$$
$$2x^2y^2+y^4-90x^2-90y^2+45^2=x^2(2y^2+18)+y^4-18y^2+81$$
$$-90x^2-90y^2+45^2=18x^2-18y^2+81$$
$$108y^2+72x^2=36\cdot54$$
$$3x^2+2y^2=54$$
| {
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"timestamp": "2023-03-29T00:00:00",
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There is a $3\times 3 $ orthogonal matrix with all non zero entries.? Is the statement true?
There is a $3\times 3 $ real orthogonal matrix with all non zero entries.
for orthogonality, $AA^T=A^TA=I_3$, please give me hint
| Gordon Pall found all rational orthogonal matrices, 3 by 3, in 1940, see PALL_PDF
Given an odd number
$$ n = a^2 + b^2 + c^2 + d^2,$$ the matrix
$$ \frac{1}{n} \begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2(-ad+bc) & 2(ac+bd) \\ 2(ad+bc) & a^2 - b^2 + c^2 - d^2 & 2(-ab+cd) \\ 2(-ac+bd) & 2(ab+cd) & a^2 - b^2 - c^2 + d^2 \end{pmatrix}$$
is orthogonal, and all rational orthogonal matrices can be written in this manner. This is formula (10) on page 755 of Pall's article, the second page of the pdf.
Right, so you want no zeroes in the matrix. Take, for example, $d=0,$ and nonzero $a,b,c$ which do not make a Pythagorean triple in any order. It would be enough to just add on the condition $c^2 \geq 1 + a^2 + b^2.$
| {
"language": "en",
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If $\prod x_k=1$, then $\prod\frac{1+x_k}{2}\leq ( \frac{x_1+\cdots +x_n}{n})^{n-1}$ Despite many attempts, no one at StackOverflow has succeeded in solving that old question about proving a deceptively simple-looking inequality.
I propose now a weaker and slightly simpler inequality (no more squares) which may perhaps be easier to prove.
Let $x_1,x_2, \ldots ,x_n$ be positive numbers whose product is $1$. Prove or find a counterexample :
$$
\prod_{k=1}^n \big(\frac{1+x_k}{2}\big) \leq \bigg( \frac{x_1+x_2+x_3+ \ldots x_n}{n}\bigg)^{n-1}
$$
(this inequality follows from the old one by putting $x_i=a_i^2$ and using Cauchy-Schwarz).
| As mentioned in comment, one can use equal variable method for $f(x) = \log \left(\frac{1+x}{2}\right)$. Note that
$$g(x) = f'(1/x) = \frac{x}{x+1}$$
is strictly concave for $x > 0$. By corollary 1.6, the maximum of LHS is reached when $0 < x_1 \leq 1 \leq x_2 = \cdots = x_n = t$. Substitute $x_1 = \frac{1}{t^{n-1}}$, we have to prove
$$\left(\frac{(n-1)t + \frac{1}{t^{n-1}}}{n}\right)^{n-1} \ge \left(\frac{1+t}{2}\right)^{n-1}\left(\frac{1+\frac{1}{t^{n-1}}}{2}\right)$$
for $t \ge 1$. By AMGM applied to $(n-2)$ $(1+t)/2$, and $(1+t)(1+t^{-(n-1)})/4$, we see that
$$RHS \leq \left(\frac{(n-2)\left(\frac{1+t}{2}\right) + \frac{1+t+t^{-(n-1)} + t^{-(n-2)}}{4}}{n-1}\right)^{n-1}$$
So it suffices to show that
$$\frac{(n-1)t + \frac{1}{t^{n-1}}}{n} \ge \frac{(n-2)\left(\frac{1+t}{2}\right) + \frac{1+t+t^{-(n-1)} + t^{-(n-2)}}{4}}{n-1}$$
Clearing denominator and simplify,
$$\Leftrightarrow 4(n-1)^2 t + 4(n-1)t^{-(n-1)} \ge 2n(n-2)t + 2n(n-2) + n (1+t+t^{-(n-1)} + t^{-(n-2)})$$
$$\Leftrightarrow (2n^2-5n+4)t + (3n-4)t^{-(n-1)} \ge 2n^2 - 3n + nt^{-(n-2)} \hspace{5mm} (*)$$
We prove the last inequality now. AM-GM gives
$$\frac{n(n-2)}{(n-1)} t^{-(n-1)} + \frac{n}{n-1} \ge nt^{-(n-2)}$$
and
$$(2n^2-5n+4)t + \frac{2n^2-5n+4}{n-1}t^{-(n-1)} \ge (2n^2-5n+4)\frac{n}{n-1}$$
(*) follows from adding up the last two inequalities.
| {
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"timestamp": "2023-03-29T00:00:00",
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show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion I am trying to show that
$$
\ln{\Big|\frac{1-x}{1+x}\Big|}
$$
belongs to $L^2({\Bbb{R}})$ but not to $L^1({\Bbb{R}})$ by using it's taylor expansion (this is the entire statement of the problem). More important to me than the solution to this particular problem is to understand the general technique I may use to do this.
Since there are singularities at $\pm1$ I am inclined to think that this needs to be done in three parts. Expanding around $0$ I find:
$$
ln\frac{1-x}{1+x} \approx -2x - \frac{2x^3}{3} - \frac{2x^5}{5} - \frac{2x^7}{7} ...
= \sum_0^{\infty}{\frac{-2}{2n-1}x^{2n-1}}
$$
Which is (please correct me if I am wrong) valid between $\pm1$ because of the singularities at $\pm1$.
With robjohn's suggestion an expansion for all x such that |x| > 1 is given by
$$
\ln{\Big|\frac{1-x}{1+x}\Big|} = \ln{\Big|\frac{\frac{1}{x}-1}{\frac{1}{x}+1}\Big|} = \ln{\Big|\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\Big|}
$$
And since $|x|>1 => |\frac{1}{x}|<1$ we are in the domain of our previous expansion therefore:
$$
ln\frac{1-x}{1+x} \approx -\frac{2}{x} - \frac{2}{3x^3} - \frac{2}{5x^5} - \frac{2}{7x^7} ...
= \sum_0^{\infty}{\frac{-2}{2n-1x^{2n-1}}}
$$
Thank you.
| Hint 1:
For $|x|\lt1$, we have the standard
$$
\log\,\left|\frac{1-x}{1+x}\right|=-2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\dots\right)
$$
For $|x|>1$, we have
$$
\begin{align}
\log\,\left|\frac{1-x}{1+x}\right|
&=\log\,\left|\frac{1/x-1}{1/x+1}\right|\\
&=\log\,\left|\frac{1-1/x}{1+1/x}\right|\\
&=-2\left(\frac1x+\frac1{3x^3}+\frac1{5x^5}+\frac1{7x^7}+\dots\right)
\end{align}
$$
Hint 2:
$\log(x)$ is both $L^1$ and $L^2$ on finite intervals
$$
\int_0^1|\log(x)|\,\mathrm{d}x=\int_0^\infty t\,e^{-t}\,\mathrm{d}t=1
$$
$$
\int_0^1|\log(x)|^2\,\mathrm{d}x=\int_0^\infty t^2e^{-t}\,\mathrm{d}t=2
$$
So the difference must be near $\infty$.
| {
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Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
| As @rlgordonma has let us make use of the substitution $x = e^{-y}$ to get the integral as
$$\int_0^{\infty} dy e^{-y} \left ( \frac{(e^{-y} - (1-y))^2}{y^2 (1-e^{-y})^2} \right ) $$
which can be rewritten as $$\sum_{k=1}^{\infty} k \int_0^{\infty} dy \left ( \frac{(e^{-y} - (1-y))^2}{y^2} \right ) e^{- k y} $$
If we call $$\int_0^{\infty} dy \left ( \frac{(e^{-y} - (1-y))^2}{y^2} \right ) e^{- k y} = I(k)$$ as @rlgordonma has, we get that
$$I(k) = (k+2) \log{ \left( \frac{k (k+2)}{(k+1)^2} \right )} + \frac{1}{k}$$
and we want to hence evaluate $$\sum_{k=1}^{\infty} k I(k).$$
Let us write down the first few terms to see what happens
$$kI(k) = 1 + k(k+2) \log(k) + k(k+2) \log(k+2) - 2 k(k+2) \log(k+1)$$
$$1I(1) = 1 + 3 \log(1) + 3 \log(3) - 6 \log(2)$$
$$2I(2) = 1 + 8 \log(2) + 8 \log(4) - 16 \log(3)$$
$$3I(3) = 1 + 15 \log(3) + 15 \log(5) - 30 \log(4)$$
$$4I(4) = 1 + 24 \log(4) + 24 \log(6) - 48 \log(5)$$
$$5I(5) = 1 + 35 \log(5) + 35 \log(7) - 70 \log(6)$$
We see that $$I(1) +2I(2) +3 I(3) + 4I(4) + 5I(5) = 5 + 2(\log 2 + \log 3 + \log 4 + \log 5) -46 \log 6 + 35 \log 7$$
So we see that if we sum upto $n$ terms, we will get a sum of the form $$n + 2 \log(n!) + (\cdot) \log(n+1) + (\cdot) \log(n+2)$$ and then we can call our good old reliable friend, Stirling, to help us with $\log(n!)$. Let us now proceed along these lines. We get
$$S_n = \sum_{k=1}^n k I(k) = \sum_{k=1}^{n} \left(1 + k(k+2) \log(k) + k(k+2) \log(k+2) - 2 k(k+2) \log(k+1) \right)$$
$$S_n = n + \sum_{k=1}^n \overbrace{\left(k(k+2) + (k-2)k - 2(k-1)(k+1) \right)}^2\log(k)\\
+ ((n-1)(n+1)-2n(n+2)) \log(n+1) + (n(n+2)) \log(n+2)$$
$$S_n = n + 2 \sum_{k=1}^n \log(k) - (n^2 + 4n + 1) \log(n+1) + (n^2 + 2n) \log(n+2)$$
$$\sum_{k=1}^n \log(k) = n \log n - n + \dfrac12 \log(2 \pi) + \dfrac12 \log(n) + \mathcal{O}(1/n) \,\,\,\,\,\, \text{(By Stirling)}$$
Hence,
$$S_n = \overbrace{2 n \log n - n + \log(2 \pi) - (n^2 + 4n + 1) \log(n+1) + (n^2 + 2n) \log(n+2) + \log(n)}^{M_n} + \mathcal{O}(1/n)$$
The asymptotic for $M_n$ can now be simplified further by writing $$\log(n+1) = \log (n) + \log \left(1 + \dfrac1n \right)$$
and
$$\log(n+2) = \log (n) + \log \left(1 + \dfrac2n \right)$$
and using the Taylor series for $\log \left(1 + \dfrac1n \right)$ and $\log \left(1 + \dfrac2n \right)$.
$$M_n = \log(2 \pi) - \dfrac32 - \dfrac2{3n} + \dfrac3{4n^2} - \dfrac{17}{15n^3} + \mathcal{O}\left(\dfrac1{n^4}\right)$$
Now, letting $n \to \infty$ gives us
$$\log(2 \pi) - \dfrac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 7,
"answer_id": 3
} |
Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
| Denote $X=\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$ and $Y=\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$. In fact, if $X\ge 1$, then $Y\ge \sqrt{3}$. ($\sqrt{3}$ is optimal, because when $a=b=c=\frac{1}{\sqrt{3}}$, $X=1$ and $Y=\sqrt{3}$.)
By Cauchy-Schwarz inequality,
$$3(a^2+b^2+c^2)\ge (a+b+c)^2.$$
It follows that
$$Y=\frac{a^2+b^2+c^2}{abc}\ge \frac{(a+b+c)^2}{3abc}=\frac{abc}{3} X^2\ge \frac{abc}{3}.$$
If $abc\ge 3\sqrt{3}$, we are done. Otherwise, by inequality of arithmetic and geometric means,
$$Y\ge 3(abc)^{-\frac{1}{3}}\ge \sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/281966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluate the limit I am hung up on this limit: $\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$
I must be missing something related to dealing with square roots but I can not for the life of me figure out what.
Here is my work so far:
$\displaystyle\lim_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x} = \lim_{x\to0} \frac{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}$
$=\displaystyle \lim_{x\to0} \frac{\sqrt{1+x}^2 + (\sqrt{1+x} + \sqrt{1-x}) - (\sqrt{1+x} + \sqrt{1-x})-\sqrt{1-x}^2}{x(\sqrt{1+x} + \sqrt{1-x})}$
$= \displaystyle \lim_{x\to0} \frac{1+x - 1-x}{\sqrt{1+x} + \sqrt{1-x}}= 0$.
After this I end up with the answer 0, but I know that it should come out to 1. If someone could look over this and see where I am going wrong and point me in the right direction I will be eternally thankful!
|
$$\lim\limits_{x\to0} \frac{\sqrt{1+x} + \sqrt{1-x}}{x}$$
Note that evaluating $\quad \displaystyle \lim \frac{\sqrt{1+x} + \sqrt{1-x}}{x}\;$ as $\;x\to 0\;$ gives you $\;2\;$ in the numerator, and $\;0\;$ in the denominator. So the task that ultimately remains is to evaluate the limits as $\;x \to 0^+\,$ and as $\;x \to 0^-$.
Hint: (prior to post's edit): Regarding your algebraic manipulations, if you are attempting to "simplify" the expression to make the limit more evident: try multiplying numerator and denominator by $\;\sqrt{1+x} - \sqrt{1-x}$, and be careful with algebra!
$$\lim\limits_{x\to0} \frac{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{(\sqrt{1+x} - \sqrt{1-x})x}$$ $$= \lim_{x \to 0} \frac{2x}{(\sqrt{1+x} - \sqrt{1-x})x}$$
$$=\lim_{x \to 0} \frac{2}{\sqrt{1+x} - \sqrt{1-x}}$$
Now be sure to take the limit as $x \to 0^+$ and as $x\to 0^-$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to expand $\tan x$ in Taylor order to $o(x^6)$ I try to expand $\tan x$ in Taylor order to $o(x^6)$, but searching of all 6 derivative in zero (ex. $\tan'(0), \tan''(0)$ and e.t.c.) is very difficult and slow method.
Is there another way to solve the problem?
Any help would be greatly appreciated :)
| Here's the long division method suggested by coffemath.
\begin{align}
\sin(x)
= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{(2n+1)}
&=& \sum_{n=0}^\infty a_{(2n+1)} x^{(2n+1)} \\
\cos(x)
= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{(2n)}
&=& \sum_{n=0}^\infty b_{(2n)} x^{(2n)} \\
\end{align}
\begin{align}
\tan(x) = \frac{\sin(x)}{\cos(x)}
\end{align}
\begin{align}
\sin(x) - \tan(x)\cos(x) &= 0 \\
\sin(x) - truncated(\tan(x))\cos(x) &= remainder\\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) -
\left( 0 \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \right) \\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) -
\left( x \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( \frac{x^3}{3} - \frac{x^5}{30} + \dots \right) \\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) -
\left( x + \frac{x^3}{3} \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( \frac{2x^5}{15} - \frac{4x^7}{315} + \dots \right) \\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) -
\left( x + \frac{x^3}{3} + \frac{2x^5}{15} \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( \frac{17x^7}{315} - \frac{29x^9}{5670} + \dots \right)
\end{align}
\begin{align}
\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \dots = \sum_{n=0}^{\infty}c_{(2n+1)}x^{(2n+1)}
\end{align}
Where the coefficients are given recursively as:
\begin{align}
a_{(2n+1)} &= \frac{(-1)^n}{(2n+1)!} \\
b_{(2n)} &= \frac{(-1)^n}{(2n)!} \\
c_{(2n+1)} &= a_{(2n+1)} - \sum_{k=0}^{n-1} b_{(2n-2k)} c_{(2k+1)}. \quad \blacksquare
\end{align}
This series converges with radius $\frac{\pi}{2}$ around $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
What will be the slope of $BC$? The vertex $A$ of triangle $\triangle ABC$ is $(3,-1)$. The equation of median $BE$ is $$6x+10y-59=0$$ and angle bisector $CF$ is $$x-4y+10=0.$$ Then what is the slope of $BC$?
Let slopes of $AC, CF, BC$ be $m1, m2, m3$ respectively, then
$$(m1-m2)/(1+m1*m2) = (m2-m3)/(1+m2*m3)$$
How to find $m1$?
| Let's begin at point $C$.
As $C \in CF$ it follows that
$$C=(4y_C-10, y_C).\quad (1)$$
As $E$ is midpoint of $AC$ then the coordinates of E are the arithmetic mean of coordinates of $A$ and $C$, as it is shown in equation below:
$$E=(\frac{4y_C-7}{2}, \frac{y_C-1}{2}).\quad (2)$$
We know that $E \in BE$, therefore
$$6x_E+10y_E-59=0. \quad (3)$$
From $(2)$ and $(3)$ we get
$$y_C=5.$$
So we already know the location of three points:
$$A = (3, -1),$$
$$C = (10, 5),$$
$$E = (\frac{13}{2}, 2).$$
Let $K$ such that $\{K\}=CF \cap BE$, then
$$\left \{
\begin{array}{l}
x_K -4 y_K +10 =0\\
6x_K +10y_K -59 =0\\
\end{array}
\right. \Rightarrow K = (4, \frac{7}{2}).$$
We know that the slope of a stragiht line defined by PQ can be calculated by
$$m_{PQ}=\frac{y_P - y_Q}{x_P-x_Q}.$$
Let's then calculate $m_1$ and $m_2$ now:
$$m_1= m_{AC} = m_{CE} = \frac{3}{\frac{7}{2}} = \frac{6}{7} \quad (4)$$
$$m_2= m_{CF} = m_{CK} = \frac{\frac{3}{2}}{6} = \frac{1}{4} \quad (5)$$
Let $\gamma = m(\angle KCE)$ then
$$\tan \gamma = \frac{m_1-m_2}{1+m_1m_2}. \quad (6)$$
From $(4)$, $(5)$ and $(6)$ we get
$$\tan \gamma = \frac{1}{2}$$
As $m(\angle BCK)= m( \angle KCE)$ it follows that
$$ \frac{1}{2} = \frac{m_2-m_3}{1+m_2m_3} \Rightarrow \frac{1}{2} = \frac{\frac{1}{4}-m_3}{1+\frac{1}{4}m_3} \Rightarrow $$
$$\Rightarrow \frac{1}{2} + \frac{1}{8}m_3=\frac{1}{4} - m_3 \Rightarrow m_3= -\frac{2}{9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!
| Not an independent answer but completing the circle start by others.
Several people here have derived the expression:
$$I_n = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{n+1+2k}$$
Using the digamma function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(z)}$
which has the expression and asymptotic expansion:
$$\begin{align}
\psi(x) &= - \gamma -\frac{1}{x} - \sum_{k=1}^{\infty}(\frac{1}{k} - \frac{1}{x+k})\\
&\approx_{ae} \ln{x} - \frac{1}{2x} + \sum_{k=1}^{\infty}\frac{B_{2k}}{2 k\ x^{2k}}
\end{align}
$$
where $B_{k}$ is the k-th Bernoulli number, one can simplify $I_n$ as:
$$\begin{align}
I_n &= \frac{1}{2}( \psi(\frac{n+1}{2})-\psi(\frac{n+1}{4}) - \ln{2})\\
&\approx_{ae} \frac{1}{2}( \frac{1}{n+1} + \sum_{k=1}^{\infty}\frac{4^k(4^k-1)B_{2k}}{2 k(n + 1)^{2k}} )\\
&\approx_{ae} \frac{1}{2}( \frac{1}{n+1}
+ \frac{1}{(n+1)^2}
- \frac{8}{(n+1)^4}
+ \frac{144}{(n+1)^6} - \frac{4352}{(n+1)^8} + .. )
\end{align}
$$
In particular, the leading terms up to $O(\frac{1}{n^4})$ is given by:
$$I_{n} \approx \frac{1}{2}(\frac{1}{n+1} + \frac{1}{(n+1)^2} ) \approx \frac{1}{2}( \frac{1}{n} - \frac{1}{n^3} )$$
agree with @rlgordonma result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
What's the integration of $\int \sin^4 (x)dx$? What's the integration of $\int \sin^4 x \,dx$? I don't see the approach to this question.
I have a issue with this question as well:
$$\int \sin x \cos x (\sin x+\cos x) \,dx.$$
I simplify this to $\sin^2 x \cos x + \sin x \cos^2 x$
and set $u= \sin x$, $du = \cos x \,dx$.
So I got
$$\int u^2 \,du + ???.$$
I don't get the second part now
| Recall the identities
$$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}2$$
and
$$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}2$$
Hence,
$$\sin^4(x) = \left(\dfrac{1-\cos(2x)}2 \right)^2 = \dfrac{1 - 2 \cos(2x) + \cos^2(2x)}4 = \dfrac{1 - 2 \cos(2x) + \dfrac{1+\cos(4x)}2}4$$
Hence,
$$\sin^4(x) = \dfrac{3-4\cos(2x) + \cos(4x)}8$$
Now you should be able to integrate this term by term and obtain an answer.
For the second part, note that $$\int \sin^2(x) \cos(x) dx = \int \sin^2(x) d(\sin(x))$$ and $$\int \cos^2(x) \sin(x) dx = -\int \cos^2(x) d(\cos(x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\frac{1}{1 + a^2 \tan^2x}$? Can you please help me out with evaluating this integral?
$$\int_0^{2\pi} \frac{1}{1 + a^2 \tan^2x}dx$$
| $$
\begin{align}
\int_o^{2\pi}\frac1{1+a^2\tan^2(x)}\mathrm{d}x
&=4\int_0^{\pi/2}\frac1{1+a^2\tan^2(x)}\mathrm{d}x\\
&=4\int_0^\infty\frac1{1+a^2u^2}\mathrm{d}\arctan(u)\\
&=4\int_0^\infty\frac1{1+a^2u^2}\frac1{1+u^2}\mathrm{d}u\\
&=\frac4{a^2-1}\int_0^\infty\left(\frac{a^2}{1+a^2u^2}-\frac1{1+u^2}\right)\mathrm{d}u\\
&=\frac4{a^2-1}\left(\int_0^\infty\frac{a^2}{1+a^2u^2}\mathrm{d}u-\int_0^\infty\frac1{1+u^2}\mathrm{d}u\right)\\
&=\frac4{a^2-1}\left(\int_0^\infty\frac{|a|}{1+u^2}\mathrm{d}u-\int_0^\infty\frac1{1+u^2}\mathrm{d}u\right)\\
&=\frac4{|a|+1}\int_0^\infty\frac1{1+u^2}\mathrm{d}u\\
&=\frac{2\pi}{|a|+1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving a second-order nonlinear ordinary differential equation Let's start with equation with two parameters $$y=a x^b.$$ Then we calculate $y'=a b x^{b-1}$ and solve $a=y/x^b$ from original. Substitute that to the derivative and $$y'=b \frac{y}{x}.$$ Then differentiate again and substitute $b=x y'/y$ and we get $$y''=\frac{y'^2}{y}-\frac{y'}{x}.$$ How to solve this "properly", without knowing where the final form came from?
| Hint: $(d/dx)(y'/y) = (y y'' - (y')^2)/y^2$
Specifically:
$$y y'' - (y')^2 = -\frac{y y'}{x}$$
$$\frac{y y'' - (y')^2}{y^2} = -\frac{y'}{x y}$$
$$\frac{d}{dx} \left ( \frac{y'}{y} \right ) = -\frac{1}{x} \frac{y'}{y}$$
$$\frac{d}{dx} \log{\left ( \frac{y'}{y} \right )} = -\frac{1}{x}$$
$$\log{\left ( \frac{y'}{y} \right )} = -\log{x} + K = \log{\left ( \frac{1}{x} \right )} + K$$
$$\frac{y'}{y} = \frac{d}{dx} \log{y} = \frac{b}{x}$$
$$\log{y} = b \log{x} + K' = \log{x^b} + K'$$
$$y = C x^b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/296534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding $\frac{FC}{EG}$ from $FG+EG=DG,EG+DG=DA=2EC=AF-FG$ If $FG+EG=DG,EG+DG=DA=2EC=AF-FG$ .How to find $\frac{FC}{EG}$
| This is really more an algebraical than a geometrical exercise. Essentially we are given 4 equations in 6 variables. Then we add one more variable $FC$ and exclude $EG$ by introducing the ratio. So the total number of variables remains 6.
Therefore, we must infer the two missing relations from the given geometrical construction. Adding line segments $CG$ and $AG$ and considering pairs of right-angled triangles $\triangle CFG$ and $\triangle GEC$ sharing hipotenusa $CG$ we obtain
$$FC^{2}+FG^{2}=EC^{2}+EG^{2}$$
$$\left(\frac{FC}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{EC}{EG}\right)^{2}+1$$
$$\left(\frac{FC}{EG}\right)^{2}=\left(\frac{EC}{EG}\right)^{2}-\left(\frac{FG}{EG}\right)^{2}+1 \tag{1}$$
Now consider $\triangle GFA$ and $\triangle ADG$ sharing $AG$
$$AF^{2}+FG^{2}=DG^{2}+AD^{2}\tag{2}$$
This is basically where geometry ends and all that follows is a tedious algebraic manipulation. Obviously we want to find $\frac{FG}{EG}$ and $\frac{EC}{EG}$ to use them in $(1)$ so let's divide $(2)$ by $EG$
$$\left(\frac{AF}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{DG}{EG}\right)^{2}+\left(\frac{AD}{EG}\right)^{2}\tag{3}$$
Now it is time to use the given relations to start eliminating the unknowns. Take
$$AD=2EC$$
and substitute in $(3)$
$$\left(\frac{AF}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{DG}{EG}\right)^{2}+4\left(\frac{EC}{EG}\right)^{2}\tag{4}$$
Now combine
$$AF-FG=DG+EG$$
and
$$DG=FG+EG$$
to obtain
$$AF=2FG+2EG$$
Plug this in $(4)$
$$\left(\frac{2FG+2EG}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{FG+EG}{EG}\right)^{2}+4\left(\frac{EC}{EG}\right)^{2}$$
$$3\left(\frac{FG+EG}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=4\left(\frac{EC}{EG}\right)^{2}$$
$$4\left(\frac{FG}{EG}\right)^{2}+6\frac{FG}{EG}+3=4\left(\frac{EC}{EG}\right)^{2}$$
$$\left(\frac{EC}{EG}\right)^{2}-\left(\frac{FG}{EG}\right)^{2}=\frac{3}{2}\frac{FG}{EG}+\frac{3}{4}\tag{5}$$
We have not used one more relation, namely
$$AF=FG+2EC$$
Insert this in $(4)$ together with $DF=FG+EG$ to obtain
$$\left(\frac{FG+2EC}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{FG+EG}{EG}\right)^{2}+4\left(\frac{EC}{EG}\right)^{2}$$
$$\left(\frac{FG}{EG}\right)^{2}+4\frac{FG}{EG}\cdot\frac{EC}{EG}+4\left(\frac{EC}{EG}\right)^{2}+\left(\frac{FG}{EG}\right)^{2}=\left(\frac{FG}{EG}\right)^{2}+2\frac{FG}{EG}+1+4\left(\frac{EC}{EG}\right)^{2}$$
$$\left(\frac{FG}{EG}\right)^{2}+2\frac{FG}{EG}\left(2\frac{EC}{EG}-1\right)-1=0\tag{6}$$
Now $(5)$ and $(6)$ give the system to determine $\frac{FG}{EG}$ and $\frac{EC}{EG}$. I trust WolframAlpha that the only solution that makes sense is
$$\frac{FG}{EG}=\frac{1}{3} \qquad \frac{EC}{EG}=\frac{7}{6}$$
Hence, substituting these values in $(1)$ we obtain
$$\left(\frac{FC}{EG}\right)^{2}=\frac{9}{4}$$
Finally
$$\frac{FC}{EG}=\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the equation $x^2=x \sin x+ \cos x$ true? I came across a problem that says: The equation $x^2=x \sin x+ \cos x$ is true for:
$1.$No real value of $x$
$2.$Exactly one real value of $x$
$3.$Exactly two real values of $x$
$4.$Infinitely many real values of $x$ .
Thanks in advance for your time.
| Consider
$f(x) = x^2-x \sin x- \cos x$.
$f(0) = -1$,
$f(\pi/2) = \pi^2/4 - \pi/2 > 0$,
and
$f(-\pi/2) = f(\pi/2) > 0$,
so $f$ has at least two real roots.
$f(-x) = f(x)$
(since $f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x)
= x^2 - x \sin(x) - \cos(x) = f(x)$,
so we only need look at $x > 0$.
Since $\sin$ and $\cos$ are bounded by $1$,
$f(x) > x^2 - x - 1
> 0$
for $x > 2$ certainly
(you can get the actual root, but we only need a bound).
The only way that case (4) can hold
is if $f$ has an infinite number of roots
in a finite region
(like $\sin(1/x)$
in $(0, 1)$).
However this cannot hold
since $f'(x)$ is bounded.
So the answer seems to be (3).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
A log improper integral Evaluate :
$$\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x$$
I found it can be simplified to
$$\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x$$
I found the exact value in the table of integrals:
$$2\pi\left(\ln ^22+\frac{\pi ^2}{12}\right)$$
Anyone knows how to evaluate this?
| I find a way to get the number using gamma functions, nothing is rigorous.
Consider the integral $I(\beta) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos x)^\beta dx$.
We know:
$$2 \frac{d^2}{d\beta^2} I(\beta) \bigg|_{\beta=0} = 4 \int_{0}^{\frac{\pi}{2}} \ln^2(\cos x) dx$$
is the integral we want. Introduce $u = \frac{1 + \sin x}{2}$, we have:
$$\begin{align}
I(\beta) &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos x)^{\beta-1} d\sin x\\
&= \int_0^1 (4 u (1-u))^{\frac{\beta-1}{2}} d( 2u )\\
&= 2^{\beta} \int_0^1 u^{\frac{\beta+1}{2}-1} (1-u)^{\frac{\beta+1}{2}-1} du\\
&= 2^{\beta} \frac{\Gamma(\frac{\beta+1}{2})^2}{\Gamma(\beta+1)}
\end{align}$$
Using the taylor expansion of various terms at $\beta = 0$,
$$\begin{align}
2^{\beta} &= 1 + \ln(2) \beta + \frac{\ln^2 2}{2}\beta^2 + \,...\\
\Gamma(\frac{\beta+1}{2}) &= \sqrt{\pi} \left(
1 - \frac{\gamma + 2\ln 2}{2} \beta + \frac{\pi^2+2( \gamma + 2\ln 2)^2}{16}\beta^2 + \, ...\right)\\
\Gamma(\beta+1) &= 1 -\gamma \beta + \frac{6\gamma^2 + \pi^2}{12} \beta^2 + \,...
\end{align}
$$
We get:
$$\begin{align}
&I(\beta) = \pi \left( 1 - \ln(2) \beta + \frac{\pi^2 + 12 \ln^2 2}{24}\beta^2 + \,... \right)\\
\implies &2 \frac{d^2}{d\beta^2} I(\beta)\bigg|_{\beta=0} = 2\pi \left( \frac{\pi^2}{12} + \ln^2 2 \right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 2
} |
Prove the identity $1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$ Here is the problem:
$$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
Can you help me prove that this is an trigonometric identity?
| $$\begin{align}
2\cdot \color{blue}{\cos^2\left(45^{\circ}-\frac{x}{2}\right)}&=\\
2\cdot \color{blue}{\frac{1+cos(double angle)}{2}}&=\\
2\cdot \color{blue}{\frac{1+cos(90^{\circ} - x)}{2}}&=\\
1+\cos\left(90^{\circ}-x\right) &= \\
1+\sin(x)&\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How to find the minimal polynomial when given only the characteristic polynomial? I am trying to find all possible Jordan forms of a transformation with Characteristic Polynomial $(x-1)^2(x+2)^2$. How can I find its minimal polynomial? Or do I just assume the $2$ (minimal and characteristic pol.) coincide?
| Actually, in the general case to obtain the Jordan canonical form you need to find monic, non-constant polynomial $d_{i}$ such that $$d_{1} \mid d_{2} \mid \dots \mid d_{n},$$ with $$d_{1} d_{2} \cdots d_{n} = \text{the characteristic polynomial},$$ and then $d_{n}$ is the minimal polynomial. (See Smith normal form.) I am assuming all eigenvalues are in the underlying field.
In this case there aren't many possibilities, it's either $n = 1$ and $d_{1} = (x-1)^2(x+2)^2$, or $n = 2$ and then you have three possibilities
$$
\begin{align}
&d_{1} = (x-1) (x+2) = d_{2}\\
&d_{1} = x-1, d_{2} = (x-1)(x+2)^2\\
&d_{1} = x+2, d_{2} = (x-1)^2(x+2)\\
\end{align}
$$
The corresponding JCF are
$$
\begin{bmatrix}1&1&&\\&1&&\\&&2&1\\&&&2\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&2&\\&&&2\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&2&1\\&&&2\end{bmatrix},
\quad
\begin{bmatrix}1&1&&\\&1&&\\&&2&\\&&&2\end{bmatrix}.
$$
Since $n \le 2$ in this case, the characteristic polynomial and the minimal polynomial together determine the JCF. But for instance if you take the characteristic polynomial to be $(x-1)^4$ and the minimal polynomial to be $(x-1)^{2}$, then there are two possible sequences of $d_{i}$
$$
d_{1} = (x-1)^{2} = d_{2},
\quad\text{and}\quad
d_{1} = d_{2} = (x-1) , d_{3} = (x-1)^{2}.
$$
The corresponding JCF are
$$
\begin{bmatrix}1&1&&\\&1&&\\&&1&1\\&&&1\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&1&1\\&&&1\end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Proof about an inequality We have $a,b,c>0$, prove that:
$$\dfrac {1+\sqrt {3}} {3\sqrt {3}}\left( a^{2}+b^{2}+c^{2}\right) \left( \dfrac {1} {a}+\dfrac {1} {b}+\dfrac {1} {c}\right)\geq a+b+c+\sqrt {a^{2}+b^{2}+c^{2}}$$
| Because the inequality is homogeneous, we can assume that $a^2 + b^2 + c^2 = 1$.
Then by the power mean inequality
$$\frac{a+b+c}{3} \le \sqrt{\frac{a^2 + b^2 + c^2}{3}} = \frac{1}{\sqrt{3}}$$
and
$$\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \le \sqrt{\frac{a^2 + b^2 + c^2}{3}} = \frac{1}{\sqrt{3}}.$$
Thus
$$a+b+c+1 \le 1 + \sqrt{3}= \frac{1 + \sqrt{3}}{3\sqrt{3}} 3\sqrt{3} \le \frac{1 + \sqrt{3}}{3\sqrt{3}} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence Given
$$
y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.
$$
Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck.
| Note that
$$
\begin{align}
\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}}
&=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\
&=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\
&=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\
&\ge\frac{n}{n+1}\left(1+\frac{n+2}{n(n+2)}\right)\\
&=1
\end{align}
$$
Therefore,
$$
\left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1}
$$
Similarly,
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\left(\frac{n+2}{n+1}\right)^{n+1}\left(\frac{n}{n+1}\right)^n\\
&=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\
&=\frac{n+1}{n}\left(1-\frac1{(n+1)^2}\right)^{n+1}\\
&\ge\frac{n+1}{n}\left(1-\frac{n+1}{(n+1)^2}\right)\\
&=1
\end{align}
$$
Therefore,
$$
\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n
$$
Bernoulli's Inequality
In the preceding, we used Bernoulli's Inequality: for all $x\ge-1$ and non-negative integer $n$,
$$
(1+x)^n\ge1+nx
$$
This can be proven by induction:
Note that the inequality above is true for $n=0$.
Suppose that $x\ge-1$ and for a non-negative integer $n$, we have
$$
(1+x)^n-nx\ge1
$$
Then
$$
\begin{align}
(1+x)^{n+1}-(n+1)x
&=(1+x)^n-nx+x(1+x)^n-x\\
&\ge1+x((1+x)^n-1)\\
&\ge1
\end{align}
$$
If $-1\le x\le0$, then both $x$ and $(1+x)^n-1$ are negative. If $x\ge0$, then both both $x$ and $(1+x)^n-1$ are positive. Therefore, if $x\ge-1$, $x((1+x)^n-1)\ge0$. This justifies the last inequality above.
Note that if $x\ne0$ and $n\ge1$, the last inequality is strict. Thus, for $x\ne0$ and $n\ge2$, we have
$$
(1+x)^n\gt1+nx
$$
Negative Exponents
Bernoulli's Inequality is also true for negative integer exponents. That is, for $x\gt-1$ and non-negative $n\in\mathbb{Z}$,
$$
1-nx\le(1+x)^{-n}
$$
Suppose that
$$
(1-nx)(1+x)^n\le1
$$
which is trivially true for $n=0$, and strictly true for $x\ne0$ and $n=1$. Then
$$
\begin{align}
(1-(n+1)x)(1+x)^{n+1}
&=(1-nx)(1+x)^n-(n+1)x^2(1+x)^n\\
&\le1
\end{align}
$$
Therefore, for all non-negative $n\in\mathbb{Z}$,
$$
(1-nx)\le(1+x)^{-n}
$$
where the inequality is strict for $x\ne0$ and $n\ge1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 7,
"answer_id": 1
} |
Integer solution $(x,y,z)$ to $xyz=24$
Find the total number of integer solutions $(x,y,z)\in\mathbb{Z}$ of the equation $xyz=24$.
I have tried $xyz = 2^3 \cdot 3$
My Process:
Factor $x$, $y$, and $z$ as
$$
\begin{cases}
x = 2^{x_1} \cdot 3^{y_1}\\
y = 2^{x_2} \cdot 3^{y_2}\\
z = 2^{x_3} \cdot 3^{y_3}
\end{cases}
$$
Now $2^{x_1+x_2+x_3} \cdot 3^{y_1+y_2+y_3} = 2^3 \cdot 3$, so
$$
\begin{cases}
x_1+x_2+x_3 = 3\\
y_1+y_2+y_3 = 1
\end{cases}
$$
where $x_1,x_2,x_3,y_1,y_2,y_3\geq 0$. Thus the number of solutions for the first is $\binom{5}{2}=10$ and second is $\binom{3}{2}=3$. This gives a total number of positive solutions of $10 \cdot 3 = 30$
How can I calculate the total number of solutions? Is this process correct?
| Your logic is correct for positive solutions. Now you need the sign pattern to be $+++,+--,-+-,--+$, four possibilities, so multiply by four.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
A hard log definite integral: $\int_0^{\pi/4}\ln^3\sin x\,\mathrm dx$ Show that:
$$\int_0^{\pi/4}\ln ^3\sin x\text{d}x=\frac34\text{Im}\left(\text{Polylog}\left(4,i\right)\right)-3\text{Im}\left(\text{Polylog}\left(4,\frac12+\frac{i}{2}\right)\right)-\frac{23}{128}\pi^3\ln 2+\frac32\text{Im}\left(\text{Polylog}\left(3,\frac12+\frac{i}{2}\right)\right)\ln 2-\frac38G\ln^22-\frac5{16}\pi\ln^32-\frac38\pi\zeta\left(3\right)$$
| A related problem. Here is a different form for the solution in terms of the hypergeometric function
$$\int_0^{\frac{\pi}{4}}\ln ^3\sin x\text{d}x = -\frac{\pi\,\ln^3(2)}{32}-\frac{3}{8}\sqrt {2} \left(\ln^2( 2 ) + 4\,\ln
\left( 2 \right) +8 \right)\times$$
$${_7F_6\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\bar{\alpha}, \alpha\, ;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\bar{\beta} , \beta;\,\frac{1}{2}\right )}\sim -6.037581109, $$
where $i=\sqrt{-1},$
$$ \alpha = \frac{3\ln(2)+2+2i }{2\ln(2)}, \quad \beta = \frac{\ln(2)+2+2i }{2\ln(2)}, $$
and $ \bar{\alpha},\bar{\beta} $ are the complex conjugates of $\alpha,\beta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that the difference of two consecutive cubes is never divisible by $3$. Here is my proof:
Let $n \in \Bbb Z$. Then, $n$ is of the form $2k$(even) or $2k + 1$(odd), for some $k \in \Bbb Z$.
Without loss of generality (not sure if I can use this), let $n = 2k$.
Then, $n + 1 = 2k + 1$.
$$\begin{align}
(n + 1)^3 - n^3 & = (2k + 1)^3 - (2k)^3 \\
& = 8k^3 + 12k^2 + 6k + 1 - 8k^3 \\
& = 3(4k^2 + 2k) + 1
\end{align}$$
Let $m = (n + 1)^3 - n^3$.
Then, $m = 1 + 3(4k^2 + 2k) \Rightarrow m \equiv 1 \pmod 3$.
Therefore, $\forall m$, when divided by $3$, there remains a remainder $1$. So $(n + 1)^3 - n^3$ is not divisible by $3$.
Do I also have to prove this for the other case where $n$ is odd, or is this proof sufficient?
| Let, number 1 = x then
number 2 = x+1
(If Difference of cubes is divisible by 3 then if we mod that value by 3 it should give reminder as 0.)
Now, [Difference of cubes] mod 3 ,
= [(x+1)^3 - x^3] mod 3
= [x^3 + 3(x)(1)(x+1) + 1 - x^3] mod 3
= [3(x)(x+1) + 1] mod 3
= 1 (because 3(x)(x+1) is always divisible by 3)
Thus, the reminder will be always 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$ Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$
using Taylor's Theorem?
I am thinking of expanding it about $x=0$ but I got something like
$$f(x) = -x^2 + \frac{x^4}{2} - \dots$$
Is my approach correct? Could you give me some hints/guides here?
Thanks.
| Sure.
$$\ln(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \ldots$$
In particular, the terms alternate sign. If we subtract $x^2$, we are left with
$$\ln(1 + x^2) - x^2 = -\frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \ldots$$
The first term of the sum is negative, and since the sum is a strictly alternating sum whose terms go to $0$ (since $|x| < 1$), the error between a sum of the first $n$ terms and the actual value is bounded by the $n+1$st term. Here, we only need to note that the first term is negative and the second term is smaller in absolute value - thus we conclude that
$$\ln(1 + x^2) - x^2 \leq 0$$
with equality only when $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Solving an equation with fractional powers I was trying to find the maximum value for a function. I took the first derivative and arrived at this horrible expression:
$$ (x^2 + y^2)^\frac{3}{2} - y {\frac{3}{2}}(x^2 + y^2)^{\frac{1}{2}}2y = 0$$
How can I find the extrema by hand?
| Define a new variable:
$$ r = \sqrt{x^2 + y^2} $$
Then it follows:
$$ \begin{array}{rcl}
(x^2 + y^2)^\frac{3}{2} - y \frac{3}{2}(x^2 + y^2)^{\frac{1}{2}}2y & = & 0 \\
(x^2 + y^2)^\frac{3}{2} - 3(x^2 + y^2)^{\frac{1}{2}}y^2 & = & 0 \\
r^3 - 3y^2r & = & 0 \\
r^2 - 3y^2 & = & 0 \\
r^2 & = & 3y^2 \\
x^2 + y^2 & = & 3y^2 \\
x^2 & = & 2y^2 \\
x & = & \pm \sqrt{2}y \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/311570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Taylor expansion of $x^x-1$ around 1 How do I find Taylor expansion of(around 1):
$$f(x)=x^x-1$$
The answer should be:
$$(x-1)+(x-1)^2+\frac 12(x-1)^3+\cdots$$
How the answer was obtained?
| Using the power series for $\log(1+x)$, we get
$$
\begin{align}
(1+x)\log(1+x)
&=(1+x)\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right)\\
&=x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-\dots\tag{1}
\end{align}
$$
Plugging $(1)$ into the power series for $e^x$ yields
$$
(1+x)^{1+x}=1+x+x^2+\frac{x^3}{2}+\frac{x^4}{3}+\frac{x^5}{12}+\frac{3x^6}{40}-\frac{x^7}{120}+\dots\tag{2}
$$
Alternatively, use the binomial theorem to get
$$
(1+x)^{x+1}=1+\tfrac{x+1}{1}x+\tfrac{(x+1)x}{2}x^2+\tfrac{(x+1)x(x-1)}{6}x^3+\dots\tag{3}
$$
which also produces $(2)$.
Plug $x-1$ into $(2)$ and subract $1$ to get the Taylor Series for $x^x-1$ at $x=1$:
$$
x^x-1=(x-1)+(x-1)^2+\frac{(x-1)^3}{2}+\frac{(x-1)^4}{3}+\frac{(x-1)^5}{12}+\dots\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/311903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Nested Summation I have the following nested sum :
$$\sum_{i=1}^{n}\sum_{j=1}^{i}\sum_{k=1}^{j}x = x+1$$
I don't have a clue how to solve this one, can somebody help me?
Thanks in advance.
${}{}$
| I will use the rules found on this page.
First, note that $x$ is not in the base or limit of any of the sums, so it can be "pulled out":
$$x\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j 1= x+1$$
Work from inside to outside, simplifying the sum:
$$\sum_{k=1}^j 1 = j$$
$$\sum_{j=1}^ij = \frac{i(i+1)}{2}=\frac{1}{2}\cdot\left(i^2 + i\right)$$
$$\sum_{i=1}^n\frac{1}{2}\cdot\left(i^2 + i\right) = \frac{1}{2}\left(\sum_{i=1}^ni^2+\sum_{i=1}^ni\right) = \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right)$$
Now we're done, and can plug back in:
$$\frac{x}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right)=x+1$$
Now, solving for $x$:
$$x\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)=1$$
$$
\begin{align*}
x &= \frac{1}{\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)} \\
&= \frac{6}{n^3 + 3 n^2 + 2 n - 6}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/312859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}$$ $$ T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix}
$$
Then the standard matrix for T is...
I'm not exactly sure how to approach this problem.
Could anyone explain how to solve this problem?
| The standard matrix has columns that are the images of the vectors of the standard basis
$$
T \Bigg (\begin{bmatrix}1\\0\\0\end{bmatrix} \Bigg),
\qquad
T \Bigg (\begin{bmatrix} 0\\1\\0 \end{bmatrix} \Bigg),
\qquad
T\Bigg (\begin{bmatrix}0\\0\\1 \end{bmatrix}\Bigg).
\tag{1}
$$
So one approach would be to solve a system of linear equations to write the vectors of the standard basis in terms of your vectors
$$
\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix},
\qquad
\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix},
\qquad
\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix},
$$
and then obtain (1).
Alternatively, note that if $A$ is the standard matrix you are looking for, then
$$
A \cdot
\begin{bmatrix}-2 & 3 & -4\\ 3 &-2&-5 \\ -4&3&5 \\ \end{bmatrix}
=
\begin{bmatrix} 5 & -4 & -6\\ 3 & 6 & -40 \\ 14 & -14 & -2 \\ \end{bmatrix},
$$
and multiply on the right by the inverse of
$$
\begin{bmatrix}-2 & 3 & -4\\ 3 &-2&-5 \\ -4&3&5 \\ \end{bmatrix}.
$$
Spoiler And the matrix $A$ is...
$$\begin{bmatrix}-1& 5& 3\\ 5& 3&-1\\-2& -2 & -4\end{bmatrix}$$
Many Thanks to @MartinSleziak for correcting two misprints in comments below.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/313798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
} |
Value of $\sum_{k=1}^{\infty}\frac{1}{k^2+a^2}$ So my question is to find the value of $\sum_{k=0}^\infty\frac{1}{k^2+1}$ and more generally $\sum_{k=0}^\infty\frac{1}{Q(k)}$
where Q is a quadratic polynomial with no zeroes on the integers.
I can prove it converges, by the comparison test. I think I 've seen such a sum before, and I think it has a hyperbolic function solution. But the way they did it there involved complex analysis, and I'm not very comfortable with that.
| Using the residue theorem, you can show that
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \coth{\pi a}$$
This is equivalent to saying that
$$\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2} \left (\frac{\pi}{a} \coth{\pi a} - \frac{1}{a^2}\right )$$
You can also derive this by considering the Maclurin expansion of $z \coth{z}$:
$$z \coth{z} = 1 + \sum_{k=1}^{\infty} \frac{B_{2 k} (2 z)^{2 k}}{(2 k)!}$$
where $B_{2 k}$ is a Bernoulli number, which also shows up in Riemann zeta functions of even, positive argument:
$$\zeta(2 k) = (-1)^{k+1} \frac{B_{2 k} (2 \pi)^{2 k}}{2 (2 k)!}$$
To evaluate the sum, factor out $n^2$ from the denominator and Taylor expand:
$$\begin{align}\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} &= \frac{1}{n^2} \frac{1}{1+ \frac{a^2}{n^2}}\\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{\infty} (-1)^k \left (\frac{a^2}{n^2}\right )^{k} \\ &=\sum_{k=0}^{\infty} (-1)^k a^{2 k}\sum_{n=1}^{\infty} \frac{1}{n^{2 k+2}} \\ &=\sum_{k=0}^{\infty} (-1)^k a^{2 k} \zeta(2 k+2)\\ &= \frac{1}{2 a^2}\sum_{k=1}^{\infty} \frac{B_{2 k} (2 \pi a)^{2 k}}{(2 k)!} \\ &= \frac{1}{2 a^2} ( \pi a \coth{\pi a} - 1)\\ \end{align}$$
The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
probability proof requiring conditional probability
Consider the experiment where two dice are thrown. Let $A$ be the event that the sum of the two dice is 7. For each $i\in\{1,2,3,4,5,6\}$, let $B_i$ be the event that at least one $i$ is thrown.
(a) Compute $P(A)$ and $P(A\mid B_1)$.
(b) Prove that $P(A\mid B_i)=P(A\mid B_j)$ for all $i$ and $j$.
(c) Since you know that some $B_j$ always occurs, does it make sense that $P(A)\neq P(A\mid B_i)$? (After all, if $E$ is an event with $P(E)=1$, then for any event $F$, $P(F\mid E)=P(F)$. What is going on? Does this seem paradoxical?)
(original image)
I am stuck on this question especially the proof. Can anybody help me out.
| Hint:
$\qquad\qquad\text{Event }A\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & & & & & \bullet \\\hline
2& & & & & \bullet & \\\hline
3& & & & \bullet & & \\\hline
4& & & \bullet & & & \\\hline
5& & \bullet & & & & \\\hline
6& \bullet & & & & & \\\hline
\end{array}$ $\qquad\qquad\text{Event }B_1\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\\hline
2& \bullet & & & & & \\\hline
3& \bullet & & & & & \\\hline
4& \bullet& & & & & \\\hline
5& \bullet & & & & & \\\hline
6& \bullet & & & & & \\\hline
\end{array}$
$\qquad\qquad\text{Event }B_2\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & \bullet & & & & \\\hline
2&\bullet & \bullet & \bullet& \bullet & \bullet & \bullet \\\hline
3& & \bullet & & & & \\\hline
4& & \bullet & & & & \\\hline
5& & \bullet & & & & \\\hline
6& & \bullet & & & & \\\hline
\end{array}$ $\qquad\qquad\text{Event }B_3\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & & \bullet & & & \\\hline
2& & & \bullet & & & \\\hline
3& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\\hline
4& & & \bullet & & & \\\hline
5& & & \bullet & & & \\\hline
6& & & \bullet & & & \\\hline
\end{array}$
$\qquad\qquad\text{Event }B_4\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & & & \bullet & & \\\hline
2& & & & \bullet & & \\\hline
3& & & & \bullet & & \\\hline
4& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\\hline
5& & & & \bullet & & \\\hline
6& & & & \bullet & & \\\hline
\end{array}$ $\qquad\qquad\text{Event }B_5\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & & & & \bullet & \\\hline
2& & & & & \bullet & \\\hline
3& & & & & \bullet & \\\hline
4& & & & & \bullet & \\\hline
5& \bullet & \bullet& \bullet & \bullet & \bullet & \bullet \\\hline
6& & & & &\bullet & \\\hline
\end{array}$
$\qquad\qquad\text{Event }B_6\\\begin{array}{c|c|c|c|c|c|c|c|}
& 1 & 2 & 3 & 4 & 5 & 6\\\hline
1& & & & & & \bullet\\\hline
2& & & & & & \bullet \\\hline
3& & & & & & \bullet \\\hline
4& & & & & & \bullet \\\hline
5& & & & & & \bullet \\\hline
6& \bullet& \bullet & \bullet &\bullet & \bullet & \bullet \\\hline
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim\limits_{n\rightarrow \infty}\int_1^3\frac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and evaluate it. I am trying to show that $\lim_{n\rightarrow \infty}\int_1^3\dfrac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and what its value is. I know that to do this I must show that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow x^{33}$ uniformly on $[1,3]$ and that each $\dfrac{nx^{99}+5}{x^3+nx^{66}}$ is integrable on $[1,3]$ and the rest will follow. I am having a difficult time showing that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow x^{33}$ uniformly on $[1,3]$. So far my proof for uniform convergence is as follows.
Let $\epsilon > 0$, choose $N\in \mathbb{N}$ such that $\left|\frac{4}{N+1}\right|<\epsilon$. Then $n\geq N$ implies
\begin{align*}
\left|\frac{nx^{99}+5}{x^3+nx^{66}}-x^{33}\right|&=\left|\frac{nx^{99}+5-x^{33}(x^3+nx^{66})}{x^3+nx^{66}}\right|\\
&=\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\\
&\leq\left|\frac{4}{n+1}\right|\\
&\leq\left|\frac{4}{N+1}\right|<\epsilon
\end{align*}
and so we have uniform convergence on $[1,3]$.
| You are on the good way, and to complete the proof you can by triangle inequality do:
$$\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\leq \frac{5+x^{36}}{x^3+nx^{66}}\leq\frac{5+3^{36}}{1^3+n1^{66}}= \frac{5+3^{36}}{1+n}\rightarrow0,\quad\forall x\in[1,3]. $$
Hence, $\forall \epsilon>0,$ we can find $N\in\mathbb{N}$ such that $\forall x\in [1,3],\quad\forall n\geq N$ we have $\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\leq \epsilon. $
Now, you can conclude the uniform convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits. Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits.
| If $N=1\times 3\times 5\times \cdots \times 2013$
we need to find $N\pmod{1000}$
Now $1000=8\cdot125$
Now, $1\cdot3\cdot5\cdot7\equiv1\pmod 8$ (Find one generalization here )
So, $(1\cdot3\cdot5\cdot7)\cdots (2001\cdot2003\cdot2005\cdot2007)\equiv1\cdot1\cdots\cdot1\cdot1\equiv1\pmod 8 $
So, $N\equiv 1\cdot2009\cdot2011\cdot2013\pmod 8\equiv9\cdot11\cdot13\equiv1\cdot3\cdot5\equiv 7\pmod 8$
Clearly, $125|N\implies N\equiv0\pmod{125}$ as $125$ is a factor of $N$
Now applying the Chinese Remainder Theorem,
$$N\equiv 0\cdot b_1\cdot\frac{1000}{125}+7\cdot b_2\cdot\frac{1000}8\pmod{1000}$$
where $b_1\cdot\frac{1000}{125}\equiv1\pmod{125}\text{ and } b_2\cdot\frac{1000}8\equiv1\pmod 8$
We don't need to calculate $b_1$ as it's multiplier is $0$
$125b_2\equiv1\pmod8\iff 5b_2\equiv1\pmod8$
Trying multiplication of $5$ with numbers coprime to and $<8,$ we get $b_2\equiv5\pmod 8$
So, $$N\equiv7\cdot5\cdot125\pmod{1000}\equiv(8\cdot4+3)125\equiv3\cdot125\equiv375\pmod{1000}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Range of $z,$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$. Then value of $z$ lie in the interval.
My Try:: Let $x,y,z$ be the roots of the quadratic equation
$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$
Let $xyz = p, $ Then let $f(t) = t^3-3t^2-9t-p.$
Now we will check the extermes value of $f(t)$
$f^{'}(t) = 3t^2-6t-9 = 3(t^2-2t-3)$
$f^{''}(t) = 3(2t-3)$
Now for Max. or Min. , $f^{'}(t)=0$ or $t=-1\;,t=3$
So $f^{''}(-1) = -15=-$(ve)
so Local Max. at $t=-1$
and $f^{''}(3) = 9 = +$(ve)
So Local Min. at $t=3$
Now My question is How can I check Range of $z$ using Graph of cubic equatio using Derivatives
Thanks
| We have $x+y+z = 3$ and $xy+yz+zx = -9$
$xy+z(3-z)=-9$
$\Rightarrow xy+3z-z^2=-9$
$\Rightarrow xy+3z-z^2+9=0$
$\Rightarrow -xy-3z+z^2-9=0$
$\Rightarrow z^2-3z-9=xy$.....(1)
By A.M G.M inequality we have,
$\displaystyle |xy|\le(\frac{x+y}{2})^2=(\frac{3-z}{2})^2$
Using this in 1 we have,
$-(\frac{3-z}{2})^2\le z^2-3z-9\le (\frac{3-z}{2})^2$
$4(z^2-3z-9)\le (z^2-6z+9)$
$3z^2-6z-45\le 0$
$z^2-2z-15\le0$
Evaluate this and the left part and take the intersection of the two answers as your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the remainder. Let $a,b$ be positive integers such that $7$ divide $a^2+b^2$ .How to find the remainder when we divide $ab-1$ by $7$
| Answer is 6
a^2+b^2+2=(a+b)^2-2ab+2
2(ab-1)=(a+b)^2-(a^2+b^2)-2
=49d+7k-2
ab-1=7[k+7d]/2-1=7[k+7d]/2-7+6
Note that both a,b are divisible by 7
Assume if not a=7n+1,7n+2,7n+3,b=7m+1,7m+2,7m+3,etc. and check
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.
Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.
Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.
In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.
Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.
Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.
Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.
We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.
| Suppose $a^2 + b^2 = 3 c^2$
Write $a = 3^p u$, $b = 3^q v$, and $c = 3^r w$,
where $u, v, w$ are all relatively prime to 3.
This does not assume that $a, b, c$ have no common divisor greater than 1.
Assume $p \le q$ (if $p > q$, switch their roles in what follows).
$a^2+b^2 = (3^p u)^2 + (3^q v)^2
= 3^{2p}(u^2+ 3^{2(q-p)}v^2)
$, so an even power of 3 divides $a^2+b^2$.
($u^2+ 3^{2(q-p)}v^2$ has a remainder of 1 or 2 mod 3
depending on if $p < q$ or $p = q$.)
But $3 c^2 = 3 (3^r w)^2
= 3^{2r+1} w^2$,
so an odd power of 3 divides $c^2$.
By unique prime factorization, this is a contradiction.
Note: I wrote this because the assumption
of $a, b, c$ having no common factor is,
to me, either an implicit use of unique factorization
or an infinite descent contradiction based on
the powers of 3 dividing them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Find the value $\sum_{n=1}^{\infty}(e-(1+\frac{1}{n})^n)$ How to find the following series' value?
$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^n\bigg)$$
| Taylor expansion of $\frac {e^{x} - 1} {x}$ around $x = 0$ is $$1 + \frac {x} {2} + \frac {x^2} {3} + \cdots > 1 + \frac {x} {2}.$$ Put $n = 1/x$ and let $n \to \infty$, we have $$n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ By Minkowski inequality, we have $$e - \left(1 + \frac {1} {n}\right)^n \geqslant n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ But $\sum \frac {1} {2n}$ diverges, then so does your series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
| As you probably have already realized the formula
$$
\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2
$$
can be proved using induction.
For $n=1$ you have
$$
1^3 = \left(\frac{1\cdot 2}{2}\right)^2.
$$
Say that the formula holds for $n$ and prove that it holds for $n=1$. So you have/get
$$
\sum_{k=1}^{n+1} k^3 = \left[\sum_{k=1}^n k^3\right] + (n+1)^3 = \frac{n^2(n+1)^2}{2^2} + (n+1)^3.
$$
I will leave you to prove that this right hand side is indeed equal to
$$
\frac{n^2(n+1)^2}{2^2} + (n+1)^3 = \dots = \frac{(n+1)^2(n+2)^2}{2^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 10,
"answer_id": 5
} |
$x\in [0,1]$ and $n\in N$ $\implies$ $\left| \frac{1-2xn}{n^2} \right| \leq \frac{1}{\sqrt n}$ Let $x\in [0,1]$ and $n\in N$. According to the solutions manual this is true:
$$\left| \frac{1-2xn}{n^2} \right| \leq \frac{1}{\sqrt n}$$
How can I see this is true ?
| Since $0 \le x \le 1$,
$0 \ge 1-2xn \ge 1-2n$.
So, $|1-2xn| \le 2n-1$.
We thus want to show that
$\frac{2n-1}{n^2} \le \sqrt{n}$
or $2n-1 \le n^{2.5}$.
Looking at the $2n-1$, I think of
$(n-1)^2 = n^2-2n+1$.
Since $(n-1)^2 \ge 0$,
$0 \le (n-1)^2 = n^2-2n+1$
or $2n-1 \le n^2$
and since $n^2 \le n^{2.5}$ we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equal angles formed by the tangent lines to an ellipse and the lines through the foci. Given an ellipse with foci $F_1, F_2$ and a point $P$. Let $T_1, T_2$ the points of tangency on the ellipse determined by the tangent lines through $P$. Show that $\widehat {T_1 P F_1} = \widehat {T_2 P F_2}$.
| Let the ellipse be:
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have:
$$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$
i.e.,
$$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$
Solving for the point of tangency, the discriminant $\Delta$ must be zero:
$$\Delta$=$4a^4k^2(t-ka)^2-4(b^2+a^2k^2)[a^2(t-ka)^2-a^2b^2] $$
At $\Delta=0$, we get $ 2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$.
Let $k_1$ be $PF_1$'s slope, $k_1=\dfrac{t}{a+c}=\tan A_2$, the $ \angle T_1PF_1=|A_1-A_2|$, now we calculate $\tan(A_1-A_2)$:
$$\tan(A_1-A_2)=\dfrac{\tan A_1-\tan A_2}{1+\tan A_1 \tan A_2}=\dfrac{k-k_1}{1+k k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$$
Since $b^2=a^2-c^2$, we get:
$$ RHS=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$$
Let $k_2$ be line $PF_2$'s slope, then $k_2=\dfrac{t}{a-c}=\tan\angle PF_2T_2$. Since $\angle T_2PF_2=\dfrac{\pi}{2}-\angle PF_2T_2$, so $\tan \angle T_2PF_2=\dfrac{1}{\tan\angle PF_2T_2}=\dfrac{a-c}{t}=\tan(A_2-A_1)$. Since the angles are acute, we have:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Fourier series of function $f(x)=0$ if $-\pi$$f(x) = \begin{cases}0 & \text{if }-\pi<x<0, \\
\sin(x) & \text{if }0<x<\pi.
\end{cases}$$
My attempt:
I went the route of expanding this function with a complex Fourier series.
$$f(x) = \sum_{n=-\infty}^{+\infty} C_{n}e^{inx}$$
$$C_{n} = \frac {1}{2\pi} \int_{0}^{\pi} \frac {e^{ix}-e^{-ix}}{2i} e^{-inx} \,\mathrm dx = \frac {1}{\pi}\left(\frac {1}{1-n^2}\right)$$
because only even $n$ terms survive, odd $n$ are 0
$$
C_0 = \frac {1}{2\pi} \int_{0}^{\pi} \sin(x)\, \mathrm dx = \frac {1}{\pi}
$$
so
$$
f(x) = \frac{1}{\pi} + \frac {1}{\pi} \left(\frac {e^{i2x}}{1-2^2} + \frac {e^{i4x}}{1-4^2}+\frac {e^{i6x}}{1-6^2}+\cdots\right) + \frac {1}{\pi} \left(\frac {e^{-i2x}}{1-2^2} + \frac {e^{-i4x}}{1-4^2}+\frac {e^{-i6x}}{1-6^2}+\cdots\right)
$$
In sine and cosine terms,
$$
f(x) = \frac{1}{\pi} + \frac {2}{\pi} \left(\frac {\cos(2x)}{1-2^2} + \frac {\cos(4x)}{1-4^2}+\frac {\cos(6x)}{1-6^2}+\cdots\right)
$$
But the answer in my book is given as
$$
f(x) = \frac{1}{\pi} + \frac{1}{2} \sin(x)+ \frac {2}{\pi} \left(\frac {\cos(2x)}{2^2-1} + \frac {\cos(4x)}{4^2-1}+\frac {\cos(6x)}{6^2-1}+\dotsb\right)$$
I don't understand how there is a sine term and the denominator of the cosines has $-1$.
| Note that, $C_1$ is a special case and you need to handle separately as
$$ C_1 = \frac {1}{2\pi} \int_{0}^{\pi} \sin(x) e^{-ix} dx \neq 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Matrix 4x4 calculation Let $$A = \begin{pmatrix}
1 & a+3 & 8 & 2 \\
0 & 0 & 2 & 0 \\
0 & a-2 & 5 & 0 \\
0 & 10 & a & a+3
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
w \\
\end{pmatrix} = \begin{pmatrix}
0 \\
7 \\
-2 \\
0 \\
\end{pmatrix}$$
I'm a bit stuck on this question... I need to find values of a for which this matrix has unique solutions.
So far I found that:
$z=\dfrac{7}{2}$
$y= \dfrac{-39}{2(a-2)}$
$w= \dfrac{(390-(7a^2)+14a)}{ (2(a-2)(a+3))}$
Can someone help to solve it completely?
| $$\begin{align}\det A &= 1\cdot\det\left(\begin{array}{ccc}0 & 2 & 0\\a-2 & 3 & 0\\10 & a & a+3\end{array}\right)\quad (\text{expanded down first column})\\ &= 1\cdot(a+3)\cdot\det\left(\begin{array}{cc}0 & 2\\a-2 & 5\end{array}\right)\quad(\text{expanded down third column})\\ &= 1\cdot(a+3)\cdot\bigl(0-2(a-2)\bigr)\\ &= -2(a+3)(a-2).\end{align}$$
You'll have unique solutions if and only if $\det A\neq 0.$
In the comments below, you've come to see that as long as $a\neq 2$ and $a\neq -3$, then there will be a unique solution. Another way to see that is simply to look at the equations you've already generated: $$z=\frac72\tag{1}$$ $$y=\frac{-39}{2(a-2)}\tag{2}$$ $$w=\frac1{a+3}\left(\frac{195}{a-2}-\frac{7a}2\right)\tag{3}$$ Now $(2)$ doesn't even make sense when $a=2$, so in that case, we have no solutions at all. Likewise, $(3)$ makes no sense when $a=-3$, so we've no solutions in that case, either.
Otherwise, $(1)$ through $(3)$ make sense, and plugging the values thus determined into $$x=-(a+3)y-8z-2z\tag{4}$$ gives us the value of $x$ needed for the solution. Since $x,y,z,w$ are determined once we've plugged in our appropriate $a$, then the solution thus determined is unique.
| {
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"source": "stackexchange",
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Variable Exponents Question If $a,b$ and $c$ are different positive integers and $2^a\cdot2^b\cdot2^c =64$ then $2^a+2^b+2^c$=?
This is so far my work: I got $2^a\cdot2^b\cdot2^c=2^6$ then $abc=6$ is this so far in the right track?
| Note that
$2^a\cdot2^b\cdot2^c=2^{a+b+c}$ and $64=2^6$.
You know that $2^a\cdot2^b\cdot2^c=64$, then you have that $2^{a+b+c}=2^6$, and from this you conclude that $a+b+c=6$.
Solution $(a,b,c)= (1,2,3)$, therefore $2^a+2^b+2^c= 2^1+2^2+2^3=2+4+8=14$.
| {
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Having $A_1=a+b+c$,$A_2=a^2+b^2+c^2$, $A_3=a^3+b^3+c^3$ - how to get $a,b,c$? Perhaps I'm just a bit dense at the moment - I've re-read some of my notes from monthes ago concerning elementary symmetric polynomials, and I find that I've no idea how to approach the "inverse" problem: if you have the -say- three values $A_0,A_1,A_2,A_3$ given, and you know
$$ \begin{matrix}
A_0 &=& a^0+b^0+c^0 &= 3 \\
A_1 &=& a^1+b^1+c^1 &= 29 \\
A_2 &=& a^2+b^2+c^2 &= 315 \\
A_3 &=& a^3+b^3+c^3 &= 3653 \\
\end{matrix}$$
- just to give some example values, then
Q: how would I approach the finding of $a,b,c$ by some general path, which one could as well use for problems with more values/variables.
I guess, that would involve somehow a matrix/an eigenvalue-formulation but I don't get an idea for the first step at the moment...
(P.s.: I've no good idea for the tags; please feel free to improve my selection)
| As $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$
$ A_1^2=A_2+2(ab+bc+ca)\implies ab+bc+ca=\frac{A_1^2-A_2}2$
Again, $a^3+b^3+c^3-3abc=(a+b+c)\{a^2+b^2+c^2-(ab+bc+ca)\}=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$
$\implies A_3-3abc=A_1\left(A_1^2-3\frac{A_1^2-A_2}2\right)$
Express $abc$ in terms $A_1,A_2,A_3$
So, the equation whose roots are $a,b,c$ is $x^3-A_1x^2+\frac{A_1^2-A_2}2x-abc=0$
| {
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Generating function for the sequence $1,1,3,3,5,5,7,7,9,9,\ldots$ The generating function for the sequence $\left\{1,1,1,1,...\right\}$ is $$1 + x + x^2 + x^3 ... = \frac{1}{1-x}$$
What is the generating function for the sequence $\left\{1,1,3,3,5,5,7,7,9,9,\dots \right\}$?
This is my attempt:
what we want to do is after the initial $\left\{1,1,1,1,...\right\}$ function we want to add two more functions shifted two to the right such as $\left\{0,0,1,1,1,...\right\}$ and keep adding such function 2 at a time and each new two are shifted 2 more to the right then the previous one. so first would be $\left\{1,1,1,1,\dots \right\}$ then add 2 of $\left\{0,0,1,1,1,1,\dots\right\}$ then add 2 more of $\left\{0,0,0,0,1,1,1,1,\dots\right\}$ and keep doing that. this gives us the function
$$\frac{1}{1-x}+\frac{2x^2}{1-x}+\frac{2x^4}{1-x}+...$$ or $$\frac{1}{1-x}+\frac{2x^{2k}}{1-x}$$ however $i$ do not know where to go from here, how do $i$ finish this problem?
| Here is another approach.
The generating function writes $(1+x)(1+3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots)$ where the second term appears as the derivative of $x+x^3 + x^5 + x^7 + x^9 + \dots$
The later being the odd part of $1+x+x^2 + x^3 + x^4 + \dots$, we have:
$$
x + x^3 + x^5 + x^7 + x^9 + \dots = \frac{1}{2}\left(\frac{1}{1-x}-\frac{1}{1+x}\right)
$$
Hence:
$$
1 + 3x^2 + 5x^4 + 7x^6 + 9x^8 + \dots = \frac{1}{2}\left(\frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}\right)
$$
Finnaly:
$$
\frac{1+x}{2}\left(\frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}\right) = \frac{x^2+1}{(x-1)^2(1+x)}
$$
| {
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"source": "stackexchange",
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$(x-a)(x-b)(x-c)(x-d)=ex$ We can verify that $x=125,162,343$ are the roots of equation $(x-105)(x-210)(x-315)=2584x$.
My question is,Could you find five positive integers $a,b,c,d,e$, which $(x-a)(x-b)(x-c)(x-d)=ex$ has four positive integer roots?
Thanks in advance.
| a. A parametric solution can be given to,
$$(x-a)(x-b)(x-c)(x-d) = ex$$
such that $a,b,c,d$ and all four roots $x_i$ are all positive integers as,
{$a,b,c,d$} = {$2n,\; 2n+1,\; (n+1)(n+3),\;2(n+1)(n+2)$}, and $e = n(n-1)(n+1)(n+2)(2n+3)$
b. In general, this quartic problem can be solved by two quadratics. One must find six positive integers $a,b,c,d,p,q$ that obey,
$(a-p)(b-p)(c-pq)+p(a-p)(1-q)(c-dq)+p(1-q)(b-dq)(c-dq)=0\tag{1}$
a quadratic in $d$, such that,
$abc-pqx(a+b+c+d-p-dq-x)=0\tag{2}$
a quadratic in $x$ has positive integer roots. Then the four roots of,
$$(x-a)(x-b)(x-c)(x-d) = \frac{(p-a)(p-b)(p-c)(p-d)}{p}x\tag{3}$$
are $p,dq,x_1, x_2$ where the $x_i$ are the two roots of (2). If these four roots are integers, then the RHS of (3) obviously is also an integer.
Proof: Factor out $(x-p)$ from (3) and substituting $x = dq$ will yield (1). For the other two roots, eliminate $x$ between (2) and (3), and you will get also get (1).
P.S. Its quite easy for Mathematica to find small $a,b,c,d,p,q$ which led me to the parametrization above. However, one must avoid trivial solutions such as,
$$(a-p)(b-p)(c-p)(d-p)(a-dq)(b-dq)(c-dq)(q-1) = 0$$
| {
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$\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$ Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?
| $x$ can be set to $\sin^22\theta$ as $0<x<\frac 12\implies 0<\sin2\theta<\frac1{\sqrt 2}$
$\implies 0< 2\theta< \frac\pi4$
$1+\sqrt x=1+\sin2\theta=(\cos\theta+\sin\theta)^2\implies \sqrt{1+\sqrt x}=\cos\theta+\sin\theta$
$1-\sqrt x=1-\sin2\theta=(\cos\theta-\sin\theta)^2\implies \sqrt{1+\sqrt x}=\cos\theta-\sin\theta$ as $\cos\theta>\sin\theta>0$ as $0< \theta< \frac\pi8$
$1-x=\cos^2\theta, \sqrt{1-x}=\cos2\theta,$
As $\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta$
$1+\sqrt{1-x}=1+\cos2\theta=2\cos^2\theta, 1-\sqrt{1-x}=1-\cos2\theta=2\sin^2\theta$
$$\text{ Then, }\frac{\sqrt{1+\sqrt x}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt x}+\sqrt{1-\sqrt{1-x}}}$$
$$=\frac{\cos\theta+\sin\theta+\sqrt2\cos\theta }{\cos\theta-\sin\theta+\sqrt2\sin\theta}$$
$$=\frac{(\sqrt2+1)\cos\theta+\sin\theta }{\cos\theta+(\sqrt2-1)\sin\theta}$$
$$=(\sqrt2+1)\cdot\frac{(\sqrt2+1)\cos\theta+\sin\theta }{(\sqrt2+1)\cos\theta+(\sqrt2+1)(\sqrt2-1)\sin\theta}$$
$$=(\sqrt2+1)$$ as $(\sqrt2+1)\cos\theta+\sin\theta\ne 0$ as $\cos\theta>\sin\theta>0$ as $0< \theta< \frac\pi8$
| {
"language": "en",
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$\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$ when $z_k$ solves $z^3+8=0$ The problem in the book is
Compute $\int_0^\infty \frac{dx}{x^3+8}$.
I set up the keyhole contour, apply the residue theorem, and go through the tedious algebra. I get stuck in doing so, but looking at the solution, I see the following remark:
Note then that $\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$.
Here, $z_k$ represents one of the poles of the function $\frac{\log z}{z^3+8}$. What I want to know is how the author makes this claim.
I am trying to compute the residues using the simple formula
$$\textrm{Res}(f(z); z_k) = \frac{A(z_k)}{B'(z_k)}$$
when $z_k$ is a simple pole. In this case, letting $B(z) = z^3+8$, I have the following:
$$z_k = -2, 1\pm\sqrt{3}i.$$
This yields the following residues
$$\begin{align*}
\textrm{Res}\left(\frac{\log z}{z^3+8}; z=-2\right) &= \frac{1}{12}\left(\log 2 + i\pi\right), \\
\textrm{Res}\left(\frac{\log z}{z^3+8}; z=1+\sqrt{3}i\right) &= \frac{2}{12(-1+\sqrt{3}i)}\left(\log 2 + \frac{i\pi}{3}\right), \\
\textrm{Res}\left(\frac{\log z}{z^3+8}; z=1-\sqrt{3}i\right) &= \frac{2}{12(-1-\sqrt{3}i)}\left(\log 2 - \frac{i\pi}{3}\right).
\end{align*}
$$
However, when I sum the results, I get a non-zero real and imaginary part. What am I doing wrong?
| Things are a lot easier if you choose a contour which is a circular wedge shape encompassing a single pole. I chose a wedge angle of $2 \pi/3$ and used the residue theorem with just the one pole at $z=\pi/3$ and got the correct result, $\pi/(6 \sqrt{3})$. Let me explain.
Consider
$$\oint_C \frac{dz}{z^3+8}$$
where $C$ is the contour that goes from $[0,R]$ along the real axis, then along a circular arc of angle $2 \pi/3$ at radius $R$, then along the line $z=t e^{i 2 \pi/3}$ from $t=R$ to $t=0$. The integral along the circular arc vanishes as $2 \pi/(3 R^2)$ as $R \rightarrow \infty$. So we are left with
$$\oint_C \frac{dz}{z^3+8} = e^{i 2 \pi/3} \int_{-\infty}^0 \frac{dt}{t^3+8} + \int_0^{\infty} \frac{dx}{x^3+8}$$
This is equal to i 2 \pi times the residue at the only pole within $C$, namely that at $z=2 e^{i \pi/3}$. I will let you work out the algebra inevaluating this rsidue; the bottom line is that
$$\oint_C \frac{dz}{z^3+8} = (1-e^{i 2 \pi/3}) \int_0^{\infty} \frac{dx}{x^3+8} = \frac{\pi}{6} e^{-i \pi/6}$$
Multiply both sides by $e^{i \pi/6}$, noting that $e^{i \pi/6} - e^{i 5 \pi/6} = \sqrt{3}$. Then it follows that
$$\int_0^{\infty} \frac{dx}{x^3+8} = \frac{\pi}{6 \sqrt{3}}$$
as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The roll of the dice Find the probability of rolling a sum of three or less on two dice.
(Yes, I know this is an "easy" question but I second quess myslef and like second opinions.)
| $$
\begin{array}{c|cccccc} X+Y & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \\
1 & \color\green{2} & \color\green{3} & 4 & 5 & 6 & 7 \\
2 & \color\green{3} & 4 & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\ddots & \vdots & & & & & \vdots
\end{array}
$$
Since each entry in the above table is equally likely:
$$ \Pr\left(X+Y \leq 3 \right) = \frac{3}{6 \times 6} = \frac{1}{12}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$ I'm having trouble proving $$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$$ where $n\in\mathbb{Z}$ and $\theta\in\mathbb{R}$. Can anyone suggest a hint?
| In order to proof
$$\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)}=\frac{\sin\left(\frac{\left(n+1\right)\alpha}{2}\right)\cdot\sin\left(\varphi+\frac{n\alpha}{2}\right)}{\sin \frac{\alpha}{2}}$$
observe
$$\sin \frac{\alpha}{2}\cdot \sin \left(\varphi + k\alpha\right)=\frac{1}{2}\left[\cos \left(\varphi + k\alpha - \frac{\alpha}{2}\right)-\cos \left(\varphi+k\alpha+\frac{\alpha}{2}\right)\right]$$
because $\sin u \cdot \sin v = \frac{1}{2}\left[\cos \left(u-v\right) - \cos \left(u + v\right)\right]$.
Then $\sin \frac{\alpha}{2}\cdot\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)}$ is a telescopic series:
$$\sin \frac{\alpha}{2}\cdot\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)} = \frac{1}{2}\left[\cos \left(\varphi + \frac{\alpha}{2}\right)-\cos \left(\varphi + n\alpha +\frac{\alpha}{2}\right)\right].$$
And use $\sin u \cdot \sin v = \frac{1}{2}\left[\cos \left(u-v\right) - \cos \left(u + v\right)\right]$ again in order to reduce the last expresion.
Also, for the identity
$$\sum_{k=1}^{n}{\cos \left(\varphi + k\alpha \right)}=\frac{\sin\left(\frac{\left(n+1\right)\alpha}{2}\right)\cdot\cos\left(\varphi+\frac{n\alpha}{2}\right)}{\sin \frac{\alpha}{2}}$$
we can use the identity $\cos u \sin v =\frac{1}{2}\left[\sin \left(u+v\right)-\sin \left(u-v\right)\right]$:
$$\cos \left(\varphi + k\alpha\right)\cdot \sin \frac{\alpha}{2}=\frac{1}{2}\left[\sin \left(\varphi + k\alpha + \frac{\alpha}{2}\right) - \sin \left(\varphi + k\alpha - \frac{\alpha}{2}\right)\right].$$
And so on.
| {
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Abstract algebra polynomial problem
Let $F$ be any field and $a,b\in F,\,\,a\neq b$. Find the greatest common divisor of $f(x) = x + a$ and $g(x) = x + b$.
Since the degree of both is $1$, the gcd is $1$ or $f(x)$ or $g(x)$, since $a\neq b$. So $\gcd(f(x),g(x))=1$.
Am I right for the answer and proving?
| Let $a \sim b$ denote that $a$ is associated with $b.$ The only divisors up to associates of $x+a$ are $1$ and $x+a$, and the only divisors of $x+b$ up to associates are $1$ and $x+b.$ Since $a\neq b$, $x+a$ is not associated to $x+b$ so the greatest common divisor is the only common divisor, $1.$
| {
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| A bit late.. but all the calculus and double sums aren't necessary.
$$\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=0}^{\infty}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}+1$$
Hence $$\displaystyle \sum_{n=1}^{\infty}\frac{n}{2^{n}}=2$$
| {
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"question_score": "35",
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Simplify : $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$? Just out of curiosity, is $$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},a \gt0\quad?$$
Thanks
| $$\sqrt{\frac{a}{2}} + \sqrt{\frac a2} = 2 \frac{\sqrt a}{\sqrt 2} = \frac {2}{\sqrt 2} \sqrt{a} = \sqrt 2 \sqrt a = \sqrt{2a} > \sqrt a$$
Added: perhaps your question originates from not remembering that $$(a + b)^2 \not\equiv a^2 + b^2\;?$$ If we do square each side of the proposed inequality (which we can do, without having to worry about whether doing so might change the direction of the inequality, since $a\gt 0$), then note: $$\left(\sqrt{\frac a2} + \sqrt{\frac a2}\right)^2\;\; \geq \; \;(\sqrt a)^2$$ $$ \iff \frac a2 + 2\cdot \frac a2 + \frac a2 \;\;\geq \;\; a $$ $$\iff 2a \geq a\quad \checkmark$$
| {
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$x^4 - y^4 = 2z^2$ has no solution How do I prove that the equation $x^4 - y^4 = 2 z^2$ has no solutions using the fact that the equations $x^4 + y^4 = z^2$ and $x^4 - y^4 = z^2$ have no solutions.
I cant think of a method of reducing the above equation to one of these forms.
| IDENTITY--The identity $(X^4 - Y^4)^2 + 4(XY)^4 = (X^4 + Y^4)^2 $ solves the proposer's question.
The proposer assumes that there is a non-zero triple of integers X,Y,Z
such that $(X^4 - Y^4 ) = 2(Z^2)$. WOLOG we may assume $(X,Y,Z) =1$.
We substitute " $2(Z^2)$ " for $(X^4 -Y^4)$ in the above IDENTITY.
This translates to $(2Z^2)^2 + 4(XY)^4 = (2B)^2$ where $X^4 + Y^4 = 2B$.
( Note that X,Y, are both odd as $X^4 = Y^4 = 2(Z^2)$ by hypothesis. Hence we may write $X^4 + Y^4) = 2B$ for some nonzero integer $B$ in the above IDENTITY.)
Dividing both sides by $4$ yields
$Z^4 + (XY)^4 = B^2$, which Fermat proved is impossible in non-zero integers $Z$,$(XY)$, and $B$ and is the connection that the proposer was seeking.
Many thanks to JW Tanner and Tinlyx for improving the formatting of this proof.
I wanted to share the above IDENTITY equation with the readership because it is an important tool to shorten and simplify Fermat-type equations of exponent $4$ which may or may not have a "$2$" as a coefficient.
The first answer given to this problem above is correct, but this solution is also very elegant, and simpler as well.
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor expansion of $\dfrac{e^x}{\cos x}$ Let $g\left(x\right) = \dfrac{e^x}{\cos x}$ and find the first four terms of its taylor polynomial.
The first four terms are: $$1 + x + x^2 + \frac{2x^3}{3} + \frac{x^4}{2}$$
Why do you have to solve this problem term by term? Why does just dividing the terms of the series work out?
Why doesn't the division of the series of $e^x$ and $\cos (x)$ which gives $\sum \limits_{n=0}^{\infty} \dfrac{x^n (2n)!}{n! (-1)^{n}(x^{2n})} $ give the correct series.
| I hate denominators.
First write ${e^x\over\cos(x)}=a_0+a_1x+a_2x^2+a_3x^3+\cdots$
Then, since I hate denominators, I consider $e^x=\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)$
Let's work from the right hand side, and take an intuitive guess that we only need expand $\cos(x)$ to a third degree polynomial:
$$\begin{align*}
\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots) & = (1-x^2/2+\cdots)(a_0+a_1x+a_2x^2+a_3x^3+\cdots) \\
%& = (a_0+a_1x+a_2x^2+a_3x^3+\cdots)-(x^2/2)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)+\cdots\\
& = a_0+a_1x+(a_2-a_0/2)x^2+(a_3-a_2/2)x^3+\cdots\\
& = e^x\\
& = 1+x+x^2/2+x^3/6
\end{align*}$$
Thus, $a_0=1$, $a_1=1$, $a_2=1$, $a_3=2/3$
$${e^x\over\cos(x)}=1+x+x^2+{2\over3}x^3+\cdots$$
So remember, when dividing power series, "I hate denominators".
| {
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"timestamp": "2023-03-29T00:00:00",
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How to draw an ellipse if a center and 3 arbitrary points on it are given? How to draw an ellipse if a center and 3 arbitrary points on it are given?
| The generally centered 2D ellipse equation is
$$ \begin{pmatrix} x\\y\\1 \end{pmatrix}^\top \begin{bmatrix} C_{11} & C_{12} & 0 \\ C_{12} & C_{22} & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{pmatrix} x\\y\\1 \end{pmatrix} = 0 \\ C_{11} x^2 + C_{22} y^2 + 2 C_{12} x y - 1 = 0$$
Using the three points you can find the three coefficients $C_{11}$, $C_{22}$ and $C_{12}$. First form a 3×3 system using the above equation for the three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$.
$$\begin{bmatrix} x_1^2 & y_1^2 & 2 x_1 y_1 \\x_2^2 & y_2^2 & 2 x_2 y_2 \\ x_3^2 & y_3^2 & 2 x_3 y_3 \end{bmatrix} \begin{pmatrix} C_{11} \\ C_{22} \\ C_{12} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$
The above system is easily solved by inverting the 3×3 matrix.
Once the coefficients are known you can find the angle of incline of the ellipse by
$$ \theta = \frac{1}{2} \arctan \left( \frac{2 C_{12}}{C_{22}-C_{11}} \right) $$
Note the angle is measured clockwise from horizontal
Getting the semi-major and semi-minor axes $a$ and $b$ is a little more involved.
*
*Find $\eta = C_{11} + C_{22}$
*Find $\zeta = \frac{C_{22}-C_{11}}{\cos(2 \theta)}$
*Semi major axis $a= \sqrt{ \frac{2}{\eta - \zeta} }$
*Semi minor axis $b= \sqrt{ \frac{2}{\eta + \zeta} }$
The equation of the ellipse is now
$$ \left( \frac{(b^2-a^2) \cos^2 \theta}{a^2 b^2} + \frac{1}{b^2} \right) x^2 + \left( \frac{(a^2-b^2)\cos^2\theta}{a^2 b^2} + \frac{1}{a^2} \right) y^2 + \frac{2 (a^2 - b^2)\sin\theta\cos\theta}{a^2 b^2} x y - 1 = 0$$
Example
Consider the points $(\sqrt{3}+1,-1)$, $(1,\sqrt{3}-1)$ and $(-1,\frac{5 \sqrt{3}}{13} + \frac{7}{13})$
Find the coefficients $$\begin{pmatrix} C_{11} \\ C_{22} \\ C_{12} \end{pmatrix} = \begin{pmatrix} \frac{5-2\sqrt{3}}{9} \\ \frac{2 \sqrt{3}+5}{9} \\ \frac{2}{9} \end{pmatrix} $$
Find the angle $$\theta = \frac{1}{2} \arctan \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{12}$$
Find the axes:
*
*$\eta = \frac{10}{9}$
*$\zeta = \frac{8}{9}$
*$a^2 = \frac{2}{\frac{2}{9}}= 9$
*$b^2 = \frac{2}{2} = 1$
Find the ellipse equation
$$ \left( \frac{5}{9} - \frac{2 \sqrt{3}}{9} \right) x^2 + \left( \frac{2 \sqrt{3}}{9} + \frac{5}{9} \right) + \frac{4}{9} x y - 1 =0 $$
Confirm results with GeoGebra:
| {
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If $\left |z-3\right |=\left |z+i\right |$, where $z=x+iy$, prove that $3x+y=4$
If $\left |z-3 \right |=\left |z+i\right |$, where $z=x+iy$, prove that $3x+y=4$.
I have got to the point where I have $\left |z \right |= \sqrt{x^2+(y+1)^2} = \sqrt{(x-3)^2+y^2}$
But really don't know where to go from here or if my start is anywhere near right :(
Can someone help me.
| $|z-3|=\sqrt {(x-3)^2+y^2}$
$|z-i|= \sqrt {x^2+(y+1)^2}$
Now ,
$(x-3)^2+y^2=x^2+(y+1)^2$
$x^2+9-6x+y^2=x^2+y^2+2y+1$
$8=6x+2y$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove inequality: $\sum \frac{a^4}{a^3+b^3} \ge \frac{a+b+c}{2}$
Prove inequality with $a,b,c >0$ $$\frac{a^4}{a^3+b^3} +
\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}$$
I tried the inequality: $\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c$, but seem like it doesn't help
| There is an extensive discussion of this inequality here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=23113
I like the sum of squares method the best:
\begin{align}
& \sum_{cyc}{\frac{a^4}{a^3+b^3}}-\frac{1}{2}\sum_{cyc}{a} \\
& =\frac{(a-b)^2(b-c)^2(c-a)^2(ab+ac+bc)^2}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)} \\
& +\sum_{cyc}{\frac{c^2(b-c)^2(2c^3(a^3+b^3)+c^2a^2(a+b)^2+2ca^4b+a^4(a^2+b^2))}{4(a^3+b^3)(b^3+c^3)(c^3+a^3)}} \\
& \geq 0
\end{align}
| {
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} |
$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $ Help me please to prove that:
for any $\triangle ABC$ we have the following inequality: $$\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}+\sin{\frac{C+A}{2}} > \sin{A}+\sin{B}+\sin{C}. $$
It's about convexity ?
thanks :)
| Hint: Eliminate $\angle C$
$A+B+C=180^0 \implies C=180-(C+B)$
Now you just need to show :
$\sin{\frac{A+B}{2}}+\sin{(90-\dfrac{A}{2})}+\sin{(90-\dfrac{B}{2})} \ge \sin{A}+\sin{B}+\sin{(180-(A+B))}$
$\sin{\frac{A+B}{2}} +\cos{\frac{B}{2}}+\cos{\frac{A}{2}} \ge \sin A+ \sin B+ \sin(A+B)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x}{2\cos^2x-1} = \frac{1}{\cos x}$$
$$2\cos{^2}(x) = 2\cos{^2}(x) - 1$$
$$0 = -1 ????$$
| Recall that
$$\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \tan(B)}$$
Hence,
$$\tan(2x) = \dfrac{2\tan(x)}{1-\tan^2(x)} = \tan(x)$$
This implies $\tan(x) = 0 \text{ or }1-\tan^2(x) = 2$. $1-\tan^2(x) = 2$ is not possible if $x \in \mathbb{R}$. Hence, $$\tan(x) = 0 \implies x = n \pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
| Hint: Complete the square in your last expression $2x^2 + 12x + 16$.
Solution:
Completing the square gives
$$2x^2 + 12x + 16 = 2(x^2 + 6x + 4) = 2(x^2 + 2\cdot 3x + 3^2 - 1) = 2(x+3)^2 - 2$$
You already showed that your inequality is equivalent to $2x^2 + 12x + 16 \geq 0$.
Using the result of completing the sqare, we continue: $$2x^2 + 12x + 16 \geq 0 \\ \iff (x+3)^2 - 1 \geq 0 \\ \iff (x+3)^2 \geq 1 \\ \iff \left| x+3\right| \geq 1 \\ \iff \left| x - (-3)\right| \geq 1 \\ \iff \text{the distance between }x\text{ and }-3\text{ is at least }1 \\\iff x \leq -4 \text{ or } x\geq -2.$$
By the fact that $x = -4$ is forbidden, we can also write
$$x < -4 \text{ or }x\geq -2.$$
Comment:
I like your start where you multiply by $(x + 4)^2$ (and not just by $x+4$). In this way, you avoid the case-by-case analysis!
| {
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Prove: $\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ How can I show:
$$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
I've tried multiplying with its "conjugate" but that doesn't seem to help that much.
Thanks!
| Probably you meant
$$\lim _{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
Which can dealt using the identity
$$x^3-y^3=(x-y)(x^2+y^2+xy)$$
now apply it for $x=x^{\frac{1}{3}}, \,y=(x+1)^{\frac{1}{3}}$
so multiply the numerator and denominator with
$x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}$
and you will get:
\begin{align}\lim _{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})&=\lim _{x\to\infty}x^{\frac{2}{3}}\frac{(x+1)^{1/3}-x^{1/3})(x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}) }{x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}}\\
&=\lim _{x\to\infty}x^\frac{2}{3}\frac{1}{x^\frac{2}{3}+(x+1)^\frac{2}{3}+((x+1)x)^\frac{1}{3}},\mathrm{now\,divide\,with \, x^\frac{2}{3}}\\
&=\lim _{x\to\infty}\frac{1}{1+(1+\frac{1}{x})^\frac{2}{3}+\left (\frac{x^2+x}{x^2}\right )^\frac{1}{3}}\\
&=\frac{1}{3}
\end{align}
| {
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Proving that $a_n$ is an integer for every $n$ For every $k\ge1$ integer number if we define the sequence : $a_1,a_2,a_3,...,$ in the form of :$$a_1=2$$
$$a_{n+1}=ka_n+\sqrt{(k^2-1)(a^2_n-4)}$$
For every $n=1,2,3,....$ how to prove that $a_n$ is an integer for every $n$
| $a_{n+1}=ka_n+\sqrt{(k^2-1)(a^2_n-4)}\Rightarrow a_{n+1}-ka_n=\sqrt{(k^2-1)(a^2_n-4)}$
Squaring both sides we have,
$a_{n+1}^2+k^2a_n^2-2ka_{n+1}a_n=k^2a_n^2-4k^2-a_n^2+4$
$\Rightarrow a_{n+1}^2+k^2a_n^2-2ka_{n+1}a_n-k^2a_n^2+4k^2+a_n^2-4=0 $
$\Rightarrow a_{n+1}^2-2ka_{n+1}a_n+4k^2+a_n^2-4=0 \dots (1)$
Replacing $n+1$ by n we have similarly,
$\Rightarrow a_{n}^2-2ka_{n}a_{n-1}+4k^2+a_{n-1}^2-4=0 \dots (2)$
By $(1)-(2)$ we have,
$a_{n+1}^2-a_{n-1}^2-2ka_{n}(a_{n+1}-a_{n-1})=0$
$(a_{n+1}-a_{n-1})(a_{n+1}+a_{n-1}-2ka_n)=0$
$\Rightarrow$ either $a_{n+1}=a_{n-1}$ or $a_{n+1}=2ka_n-a_{n-1}$
Now we use induction,
Hypothesis: $\{a_{k}\}_{k=1}^{n}$ are all integers.
As $a_{n+1}=a_{n-1}$ or $a_{n+1}=2ka_n-a_{n-1}$ so $a_{n+1} $ is also integer.
| {
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$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
| $$f''(x)=2a, 4a=f''(x-1)+f''(x-2)=10, a=2.5$$
$$f'(x)=2ax+b=5x+b, 5(x-1)+5(x-2)+2b$$
$$f'(x-1)+f'(x-2)=10x-2, -15+2b=-2, b=6.5$$
$$f(x)=ax^2+bx+c, 2.5(x-1)^2+6.5(x-1)+2.5(x-2)^2+6.5(x-2)+2c=f(x-1)+f(x-2)=5x^2-2x+9$$
let $x=0$ on both side, $2.5-6.5+10-13+2c=9, c=8$
so $a+b+c=2.5+6.5+8=17$
| {
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Multivariable limit $\lim_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1}$ $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$
According to my textbook the limit equals $2$.
What I have tried:
Using the squeeze theorem:
$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$
I have also tried to use the squeeze theorem with two other equations and obtained different values:
$$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$
$$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$
| The following "trick" can be useful. Let $x=r\cos\theta$, $y=r\sin\theta$. Then we are finding
$$\lim_{r\to 0}\frac{r^2}{\sqrt{r^2+1}-1}.$$
This can be computed using any of the usual one variable techniques.
| {
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"url": "https://math.stackexchange.com/questions/345262",
"timestamp": "2023-03-29T00:00:00",
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Why is $\displaystyle\int^{1/2}_{-1/2} \cos(\tfrac{2}{3} \pi x)\, dx= 0$? Why is
$$\int^{1/2}_{-1/2} \cos\left(\frac{2}{3} \pi x\right)\, dx= 0\qquad ?$$
Because mathematically, we have
$$\sin\left( \frac{2}{3}\pi \right) - \sin\left(-\frac{2}{3}\pi\right) = 2\sin\left(\frac{2}{3}\pi\right)$$ instead.
And what about
$$\int^{1/2}_{-1/2} \sin\left(\frac{2}{3} \pi x\right) \,dx $$
What is this equal to ?
I know this sounds like a very dumb question. But i would like to see different approaches to tackle this problem.
| $$\int\limits_{-1/2}^{1/2}\cos\frac{2}{3}\pi x\,dx=\left.\frac{3}{2\pi}\sin\frac{2}{3}\pi x\right|_{-1/2}^{1/2}=\frac{3}{2\pi}\left(\sin\frac{\pi}{3}-\sin\left(-\frac{\pi}{3}\right)\right)=$$
$$=\frac{3}{\pi}\sin\frac{\pi}{3}=\frac{3\sqrt 3}{2\pi}\neq 0$$
| {
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What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 + \frac{2}{3} (x^2-1)^{-2/3} $$
whereas according to Wolfram Alpha (see alternate form), the correct result is:
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 - \frac{2}{3} (x^2-1)^{-5/3} $$
So apparently, my calculation for $u'v$ is correct, but $uv'$ is wrong. What am I missing here?
| So
$$\begin{align}
(uv)' = u'v + uv' &= \frac{-4}{3}x(x^2 - 1)^{-5/3}\frac{2}{3}x + (x^2 - 1)^{-2/3}\frac{2}{3} \\
&= \frac{-8}{9}x^2(x^2 - 1)^{-5/3} + \frac{2}{3}(x^2 - 1)^{-2/3}\\
&= \frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2(x^2 - 1)^{-5/3} + \frac{\color{red}2}{\color{red}3}(\color{red}x^\color{red}2-1)(x^2-1)^{-5/3} \\
&= \left[\frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2 + \frac{\color{red}2}{\color{red}3}\color{red}x^\color{red}2\right](x^2 - 1)^{-5/3} - \frac{\color{red}2}{\color{red}3}(x^2 - 1)^{-5/3}\\
&= -\frac{2}{9}x^2(x^2 - 1)^{-5/3} - \frac{2}{3}(x^2 - 1)^{-5/3}.
\end{align}
$$
| {
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If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| As Glen has shown, $9\mid \{(a-b)^2+3ab\}$
$\implies 3\mid \{(a-b)^2+3ab\}\implies 3\mid(a-b)^2$
$\implies 3\mid(a-b)\implies 9\mid(a-b)^2$
$\implies 9\mid3ab\implies 3\mid ab$
But if $3$ divides $a,3$ must divide $b$ and conversely as $3\mid(a-b)$
| {
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"url": "https://math.stackexchange.com/questions/348964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
$c$.
I have no idea how to solve these types of problems. Can anybody help me please?
| A number whose digits are all equal and of length $3^n$ is thus of the form $c = \sum\limits_{i=0}^{3^n-1} a \cdot 10^i = a \dfrac {10^{3^n}-1}{10-1}$ by the geometric series.
Since we have to account for the possibility that $a = 1$ we need to show that $3^n \mid \dfrac{10^{3^n}-1}9$, i.e. $3^{n+2} \mid 10^{3^n}-1$.
This we can do by induction: Base case is trivial, suppose $3^{n+2} \mid 10^{3^n}-1$.
Then we have $10^{3^n} \equiv 1 \pmod {3^{n+2}}$, and so: $10^{3^n} \equiv 1 + k 3^{n+2} \pmod {3^{n+3}}$. Now compute $10^{3^{n+1}} = \left(10^{3^n}\right)^3$ modulo $3^{n+3}$.
Addendum: In general, when considering repdigits it is useful to have them in the form obtained through the geometric series. Since the length of the repdigit is in the exponent, and exponents are well-behaved under modulo calculations, it is one of the most informative forms for discovering properties of repdigits; in particular those pertaining to divisibility.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int\limits_0^\infty \frac{\cos(ax)}{\cos(bx)}\frac{1}{1+x^2}dx$ I would like to show that
$$\text{PV}\int_0^\infty \frac{\cos(ax)}{\cos(bx)}\frac{1}{1+x^2}dx = \frac{\pi}{2}\mathrm{sech}(b)$$
using complex analysis. $a$ and $b$ are real numbers and $a \neq b$.
Please give some hints.
| What the question asking for cannot be right!
At least for $0 < a < b$, we have:
$$\begin{align}\operatorname{PV} \int_0^{\infty} \frac{\cos a x}{\cos b x} \frac{dx}{1+x^2}&= \frac12 \operatorname{PV} \int_{-\infty}^{\infty} \frac{\cos a x}{\cos b x}\frac{dx}{1+x^2}
\\&= \frac12 \lim_{\epsilon\to 0+} \Re\left[\int_{-\infty+i\epsilon}^{\infty+i\epsilon} \frac{\cos a z}{\cos b z}\frac{dz}{1+z^2}\right]\tag{*}
\end{align}$$
The last equality is true because at the poles $\pm \frac{(2k-1)\pi}{2 b}, k = 1, 2,\ldots$ of the integrand $\frac{\cos a z}{\cos b z}\frac{1}{1+z^2}$, the residues are all real. Their contribution to the integral is $-\pi i$ times the residues and hence is imaginary.
We can evaluate the integral $(*)$ by completing the contour in upper half plane.
Notice when the $y$ in $z = x + iy$ becomes big,
$\frac{\cos a z}{\cos b z} \sim e^{-(b-a)(y - ix)} \to 0$.
The upper half circle at infinity contributes nothing to the contour integral and we have:
$$\lim_{\epsilon\to 0+}\int_{-\infty+i\epsilon}^{\infty+i\epsilon} \frac{\cos a z}{\cos b z}\frac{dz}{1+z^2}
= 2 \pi i \operatorname{Res}( \frac{\cos a z}{\cos b z}\frac{1}{1+z^2}; z = i )
= 2 \pi i \frac{\cos a i}{\cos b i}\frac{1}{2i} = \pi \frac{\cosh a}{\cosh b}$$
From this, we get:
$$\operatorname{PV} \int_0^{\infty} \frac{\cos a x}{\cos b x} \frac{dx}{1+x^2} = \frac{\pi}{2} \frac{\cosh a}{\cosh b}$$
This is not what the OP asking to show...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How many intersections has the following two curves at the point $(0,0)$? The curves are as follows:
$y^2+yx^2-x^3$ and $y^2-x^5$
| Let $x=z^2,y=z^5$
$\implies (z^5)^2+z^5\cdot z^2-(z^2)^3=0\implies z^6(z^4+z-1)=0$
So, $z$ has $6$ repeated values for $z=0$,
So, $x$ will have $\frac62=3$ repeated values as one value $t^2$ of $x$ corresponds to two values of $z$ namely, $\pm t$
If $x=0,y^2=0\iff y=0$
So,we will have $3$ intersections.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This is where I get stuck. Any help appreciated.
| Let's assume it's true for $k$.
Then $\sum_{i=1}^k\dfrac{1}{i^2}\leq2-\dfrac 1 k$
Let's try $k+1$.
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}=\sum_{i=1}^k\dfrac{1}{i^2}+\dfrac{1}{(k+1)^2}\leq2-\dfrac 1 k+\dfrac{1}{(k+1)^2}$
As $\dfrac{1}{(k+1)^2}-\dfrac 1 k<-\dfrac{1}{k+1}$ ,
$\sum_{i=1}^{k+1}\dfrac{1}{i^2}\leq2-\dfrac{1}{(k+1)^2}$
which shows that it is true for $k+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometry Question - Trihedral angles, planar geometry, spherical geometry
Two rays, $OX$ and $OY$, are drawn in the horizontal plane $\pi$, and the third ray, $OZ$, is drawn in space so that the rays $OX$, $OY$, and $OZ$ form a trihedral angle $OXYZ$. The planar angles of this trihedral angle are $a = \angle ZOY$, $b = \angle ZOX$, and $c = \angle YOX$. Find the angle between the ray $OZ$ and the plane $\pi$.
I was not sure how to calculate this.
| For simplicity, choose $X$, $Y$, $Z$ along their rays such that $|OX|=|OY|=|OZ|=1$. Let $P$ be the foot of the perpendicular dropped from $Z$ into the plane of $\triangle XOY$; then, the angle you seek ---call it "$\theta$"--- is given by
$$\sin\theta=\frac{|PZ|}{|OZ|} = |PZ|$$
We just need a useful expression for $|PZ|$.
Let $Q$ be the foot of the perpendicular dropped from $Z$ onto line $OX$; then, in $\triangle ZOQ$, we have
$$|QZ| = |OZ| \sin \angle ZOX = \sin b$$
Now, $ZQ \perp OX$ by construction; since $ZP \perp \triangle XOY$, we also have $PQ \perp OX$. Thus, in (right) $\triangle ZPQ$, the planar angle at $Q$ matches the dihedral angle along $OX$ (call it "$x$"), and we have
$$\sin\theta = |PZ| = |QZ| \sin \angle ZQP = |QZ| \sin x = \sin b \; \sin x$$
(Had we dropped our perpendicular onto line $OY$, we'd have $\sin\theta = \sin a \; \sin y$. The spherical Law of Sines ---$\frac{\sin a}{\sin x}=\frac{\sin b}{\sin y}=\frac{\sin c}{\sin z}$--- assures us that we have equivalent expressions for $\sin \theta$.)
All that remains is to find $\sin x$ in terms of $a$, $b$, $c$, which is a simple matter of applying the appropriate Spherical Law of Cosines to get
$$\sin\theta = \frac{\sqrt{1+2\cos a \cos b \cos c - \cos^2 a - \cos^2 b - \cos^2 c}}{\sin a}$$
Alternatively (but equivalently), if you know the formula for the volume of a parallelepiped determined by (unit) vectors $\overrightarrow{OX}$, $\overrightarrow{OY}$, $\overrightarrow{OZ}$,
$$\begin{align}
V &= |OX|\;|OY|\;|OZ|\;\sqrt{1+2\cos a \cos b \cos c - \cos^2 a - \cos^2 b - \cos^2 c} \\
&= \sqrt{1+2\cos a \cos b \cos c - \cos^2 a - \cos^2 b - \cos^2 c}
\end{align}$$
then note that that volume is also given by "base-times-height", the product of the area of the parallelogram determined by $OX$ and $OY$, and the length of altitude $PZ$:
$$V = 2|\triangle XOY| \cdot |PZ| = |OX|\;|OY|\;\sin a \cdot h = h \sin a = \sin\theta \; \sin a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Conjugate Ferrers diagrams
Let $\pi=\langle \pi_1,\pi_2,... \rangle , \ \pi_1\ge\pi_2\ge...,$ be a partition of a number and $\pi'=\langle \pi_1',\pi_2',... \rangle$ be a partition conjugated to $\pi$, which means that Ferrers diagram for $\pi'$ is transposed Ferrers diagram for $\pi$. For example partition conjugated to $\langle 4,4,2,1 \rangle$ is $\langle 4,3,2,2 \rangle$.
Prove identities:
*
*$\displaystyle\sum_{i}\left\lceil \frac{\pi_{2i-1}}{2} \right\rceil = \sum_{i}\left\lceil \frac{\pi'_{2i-1}}{2} \right\rceil$
*$\displaystyle\sum_{i}\left\lfloor \frac{\pi_{2i-1}}{2} \right\rfloor = \sum_{i}\left\lceil \frac{\pi'_{2i}}{2} \right\rceil$
*$\displaystyle\sum_{i}\left\lfloor \frac{\pi_{2i}}{2} \right\rfloor = \sum_{i}\left\lfloor \frac{\pi'_{2i}}{2} \right\rfloor$
No idea how to even start. Nice observation is that $\pi_1'$ is the number of the elements in $\pi$ but it gives us nothing I think.
| HINTS: If you take the first, third, fifth, etc. of a set of $n$ elements, you end up taking $\left\lceil\frac{n}2\right\rceil$ elements; if you take the second, fourth, sixth, etc., you end up taking $\left\lfloor\frac{n}2\right\rfloor$ elements
First identity:
$$\begin{array}{l|l}
\begin{array}{ccc}
\pi\\ \hline
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet\\
&
\end{array}&
\begin{array}{ccc}
\pi'\\ \hline
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet\\
\color{red}{\bullet}
\end{array}
\end{array}$$
Second identity:
$$\begin{array}{l|l}
\begin{array}{ccc}
\pi\\ \hline
\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet\\
&
\end{array}&
\begin{array}{ccc}
\pi'\\ \hline
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet&\bullet\\
\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet
\end{array}
\end{array}$$
Third identity:
$$\begin{array}{l|l}
\begin{array}{ccc}
\pi\\ \hline
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet\\
\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}\\
&
\end{array}&
\begin{array}{ccc}
\pi'\\ \hline
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet&\color{red}{\bullet}\\
\bullet&\bullet&\bullet&\bullet\\
\bullet&\color{red}{\bullet}&\bullet\\
\bullet
\end{array}
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a parametrization of a hyperbola who has a fixed signature How do I find parametrization of the hyperbola $x^2-y^2=1$ which is the unit sphere of a quadratic form with signature $(1,-1)?$
The only parametrization that comes to mind is $x=\cosh t,y=\sinh t$. What is a quadratic form of a parametrization?
| $$\text {As }(a^2+b^2)^2=(2ab)^2+(a^2-b^2)^2$$
For $ab\ne 0,$
$$\left(\frac{a^2+b^2}{2ab}\right)^2=1+\left(\frac{a^2-b^2}{2ab}\right)^2\implies (x,y)\text{ can be}\left(\frac{a^2+b^2}{2ab},\frac{a^2-b^2}{2ab}\right)$$
For $a^2-b^2\ne 0\implies a\ne \pm b$
$$\left(\frac{a^2+b^2}{a^2-b^2}\right)^2=1+\left(\frac{2ab}{a^2-b^2}\right)^2\implies (x,y)\text{ can be}\left(\frac{a^2+b^2}{a^2-b^2},\frac{2ab}{a^2-b^2}\right)$$
Try putting $a=b\tan\theta$ or $b\cot\theta$ or $b=a\tanh y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355111",
"timestamp": "2023-03-29T00:00:00",
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Equivalent modulos In my course notes I came onto two examples for finding remainders of huge numbers with fermat's little theorem and need some helping analyzing them.
1) Find the remainder when 5^183 is divided by 11.
I know that 5^10 ≡ 1 mod(11) so (5^(10*18)) = 5^180*5^3.
That ≡ 1(5^3)mod11 but then the examples says 3*5 mod 11 ≡ 4 mod 11. How did 125 mod11 turn to 3*5 mod 11?
2) why is 2^(18*9)*2^14 mod 19 = 2^14 mod18
| For the first question, $5^3=5^2\cdot 5$. But $5^2\equiv 3$.
For the second, unless there is a typo, I suspect the question has shape find the remainder when $a^{2^{68}}$ is divided by $19$. For the calculation, if $a$ is not divisible by $19$, we want to use the fact that $a^{18}\equiv 1\pmod{19}$. So we want to begin by calculating the remainder when $2^{68}$ is divided by $18$. Also, it looks as though the thing you call $68\ast 3$ is meant to be $18\ast 3$.
If the problem is to find the ramainder when $2^{68}$ is divided by $19$, note that by Fermat's Theorem we have $2^{18}\equiv 1\pmod{19}$. Note that $68=3\cdot 18+14$. So $2^{68}=2^{18\cdot 3}2^{14}=(2^{18})^3\cdot 2^{14}\equiv 2^{14}\pmod{19}$.
So now we need to calculate the remainder when $2^{14}$ is divided by $19$. We have $2^{3}\equiv 8\pmod{19}$, so $2^6\equiv 64\equiv 7\pmod{19}$. It follows that $2^{12}\equiv 49\equiv 11\pmod{19}$, and therefore $2^{14}\equiv 44\equiv 6\pmod{19}$. There are a number of other ways to do the calculation.
Remark: There is a more systematic way, called the binary method for exponentiation, which I recommend you look up. All congruences are modulo $19$. We have $2^2\equiv 4$, so $2^4\equiv 16\equiv -3$, so $2^8\equiv 9$, so $2^{12}= 2^8\cdot 2^4\equiv -27\equiv 11$, so $2^{14}=2^{12}\cdot 2^2\equiv 11\cdot 4\equiv 6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the number of triples $(a,b,c)$ with $-47 \leq a,b,c \leq 47$ and $a+b+c >0$
Kate is looking for ordered triples $(a,b,c)$ of distinct integers such that $-47 \leq \; a,b,c \; \leq 47$ and $a+b+c>0$. How many such ordered triples can Kate find?
My first step was to let $a_1=a+47, \; b_1=b+47$ and $c_1=c+47$ so that $a_1+b_1+c_1>141$ and $0 \leq \; a,b,c \; \leq 94$. Do I simply have to solve for all $a_1+b_1+c_1=142, \; 143, \cdots, 282$ and sum these results. Or is there a better way?
You can solve the above sub problems using a stars and bars method http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
| Let $N = 47$ and $X = \{-N,\ldots,N\}$. We have $|X^3| = |X|^3 = (2N+1)^3$.
Let $\mathscr{N}_{???}$ be the number of solutions for any equation ??? involving $(a,b,c) \in X^3$.
Because of symmetry,
$$\mathscr{N}_{a+b+c>0} = \mathscr{N}_{a+b+c<0} \implies \mathscr{N}_{a+b+c>0} = \frac12 \left((2N+1)^3 - \mathscr{N}_{a+b+c=0} \right)$$
For any fixed $c$, the number of solutions for $a + b + c = 0$ is $(2N+1) - |c|$, we have:
$$\mathscr{N}_{a+b+c=0} = \sum_{c = -N}^{N} \left( (2N+1) - |c|\,\right) = (2N+1)^2 -2\frac{N(N+1)}{2} = 3N^2+3N+1$$
and hence
$$\mathscr{N}_{a+b+c>0} = \frac12\left((2N+1)^3 - (3N^2+3N+1)\right) = \frac12 N (8N^2 + 9N + 3) = 425303$$
Method 2
Another way to get the same answer is try to attack a more general question:
Given $n \in \mathbb{Z}$ and $d, L \in \mathbb{Z}_{+}$, what is the
number of integer solutions for $a_1 + a_2 + \cdots + a_d = n$
where all $ 0 \le a_i < L$.
Let $\varphi_{d}(n)$ and $\varphi_{d}(n;L)$ be the number of non-negative integer solutions where $\sum_{i=1}^{d} a_i = n$ and those with further restriction that $a_1,\ldots,a_d < L$. i.e.
$$\begin{align}
\varphi_{d}(n)
&= \left|\,\{ (a_i) \in \mathbb{N}^{d} : \sum_{i=1}^{d} a_i = n \}\,\right|\\
\varphi_{d}(n;L)
&= \left|\,\{ (a_i) \in \mathbb{N}^{d} : \sum_{i=1}^{d} a_i = n \wedge a_1, \ldots a_d < L \}\,\right|
\end{align}$$
By induction, it is not hard to show:
$$\varphi_{d}(n) = \begin{cases}\binom{n+d-1}{d} = \frac{n(n+1)\cdots(n+d-1)}{d!},& n \ge 0\\ \\0, & n < 0\end{cases}$$
To compute $\varphi_{d}(n;L)$, we can start with the set of solutions for $\varphi_{d}(n)$
and then proceed to remove those solutions with at least one $x_i \ge L$. For first approximation, we get:
$$\begin{align}\varphi_{d}(n;L)
&= \varphi_{d}(n) - \sum_{j=1}^{d} \left|\{ (a_i) \in \mathbb{Z}^d : \sum_{i=1}^{d} a_i = n \wedge a_j \ge L \}\right| + \cdots\\
&= \varphi_{d}(n) - d \varphi_{d}(n-L) + \cdots\\
\end{align}$$
However, for those solutions with at least two $a_i \ge L$, this procedure removed it more than once. We need to add back their contribution. For second approximation, we get:
$$\begin{align}\varphi_{d}(n;L)
&= \varphi_{d}(n) - d \varphi_{d}(n-L) + \sum_{1 \le j < k \le d} \left|\{(a_i) \in \mathbb{Z}^d : \sum_{i=1}^{d} a_i = n \wedge a_j, a_k \ge L \}\right| + \cdots\\
&= \varphi_{d}(n) - d \varphi_{d}(n-L) + \binom{d}{2} \varphi_{d}(n-2L) + \cdots
\end{align}$$
For those solutions with at least three $a_i \ge L$, this again add back too much. We can use Inclusion-Exclusion to organize this mess. In the end, we will have:
$$\varphi_{d}(n;L)
= \sum_{k=0}^{d} (-1)^k \binom{d}{k} \varphi_{d}(n - kL)
= \sum_{k=0}^{\min(d,\lfloor\frac{n}{L}\rfloor)}(-1)^k \binom{d}{k} \varphi_{d}(n-kL)
$$
Back to our original problem, it is clear the number of solutions for
$$a + b + c < 0,\quad|a|, |b|, |c| \le N$$
is equal to the number of solutions for $$x + y + z = 3N,\quad 0 \le x, y, z < 2N+1$$
Since $\lfloor\frac{3N}{2N+1}\rfloor = 1$, we get:
$$\begin{align}\mathscr{N}_{a+b+c>0}
&= \mathscr{N}_{a+b+c<0}\\
&= \varphi_3(3N;2N+1)\\
&= \varphi_3(3N) - \binom{3}{1}\varphi_3(3N-(2N+1))\\
&= \varphi_3(3N) - 3\varphi_3(N-1)\\
&= \frac12 N(3N+1)(3N+2)) - \frac12 (N-1)N(N+1)\\
&= \frac12 N(8N^2+9N+3)\\
&= 425303
\end{align}$$
UPDATE
Oops, forget the requirement that $a, b, c$ are distinct. We need to count the
cases where $a, b, c$ fails to be distinct. There are two possible cases:
*
*Case 1, two of $a, b, c$ is the same. There are 3 possible subcases.
For the subcase where $a = b \ne c$, we have $2N(2N+1)$ combination.
$2 \lfloor\frac{N}{2}\rfloor$ of which sums to $0$.
The other 2 subcases give same amount of counting.
*Case 2. $a = b = c$. There are $2N+1$ combination, only 1 of them sums to $0$.
From this, the number of solutions for $a + b + c > 0$ with $a \ne b \ne c$ is given by:
$$\begin{align}
\mathscr{N}^{distinct}_{a+b+c>0} &= \mathscr{N}_{a+b+c>0} - \frac12\left( (6N+1)(2N+1) - 6\lfloor\frac{N}{2}\rfloor - 1\right)\\
&= \frac12 (N-1)N(8N+5) + 3\lfloor\frac{N}{2}\rfloor\\
&= 411930
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving functional equation for generating function Find the functional equation for the generating function whose coefficients satisfy $$
a_n = \sum_{i=1}^{n-1}2^ia_{n-i}, \text{ for } n\ge 2, a_0 = a_1 = 1
$$
This is what I've tried so far:
$$
\begin{align}
g(x) -1 -x &= \sum_{n\ge2} \sum_{i=1}^{n-1} 2^i a_{n-i}\\
&=\sum_{n\ge2} \sum_{i=1}^n 2^i a_{n-i}x^n\\
\end{align}
$$
at which point I'm stuck. How do I proceed?
| We have
\begin{align}
g(x) & = \sum_{n=0}^{\infty} a_n x^n = 1 + x + \sum_{n=2}^{\infty} \sum_{i=1}^{n-1} 2^i a_{n-i}x^n = 1 + x + \sum_{i=1}^{\infty} \sum_{n=i+1}^{\infty} 2^i a_{n-i} x^n\\
& = 1+x+\sum_{i=1}^{\infty} \sum_{k=n-i=1}^{\infty} 2^i a_{k} x^{k+i} = 1+x + \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} (2x)^i a_{k} x^{k}\\
& = 1 + x + \dfrac{2x}{1-2x} \cdot (g(x)-1)
\end{align}
This gives us
$$\dfrac{1-4x}{1-2x} \cdot g(x) = 1+x - \dfrac{2x}{1-2x} = \dfrac{1+x-2x-2x^2-2x}{1-2x}$$
This gives us
$$\color{red}{g(x) = \dfrac{1-3x-2x^2}{1-4x}}$$
Now let us simplify this to get the coefficients,
$$g(x) = \dfrac{1-3x-2x^2}{1-4x} = (1-3x-2x^2) \left(\sum_{k=0}^{\infty}(4x)^k\right) = \sum_{k=0}^{\infty} \left(4^k x^k - 3\cdot 4^k x^{k+1} - 2^{2k+1}x^{k+2}\right)$$
Hence,
\begin{align}
g(x) & = 1 +4x + \sum_{k=0}^{\infty} 4^{k+2} x^{k+2} - 3 x - \sum_{k=0}^{\infty}\left(3 \cdot 4^{k+1} x^{k+2} + 2^{2k+1} x^{k+2} \right)\\
& = 1 + x + \sum_{k=0}^{\infty} 2^{2k+1}x^{k+2} = 1 + x + \sum_{k=2}^{\infty} 2^{2k-3} x^k
\end{align}
Hence, we get that
$$\color{red}{a_k = 2^{2k-3}}$$
Note that we can verify that
$$\sum_{i=1}^{n-1} 2^i a_{n-i} = \sum_{i=1}^{n-2} 2^i 2^{2n-2i-3} + 2^{n-1} \cdot a_{1} = 2^{2n-3} - 2^{n-1} + 2^{n-1} = a_n$$ for $n \geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
limit with $\arctan$ I have to find the limit
and want ask about a hint:
$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$
I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard
| Here is another approach, just compute the Taylor series of $\arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}}$ at the point $x=\infty$
$$ \arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}}= \frac{1}{2}\, \frac{1}{x^{3/2}}-\frac{7}{8}\, \frac{1}{x^{5/2}}+O \left( \frac{1}{x^{7/2}} \right). $$
So the above expansion gives
$$ n^{3/2}(\arctan((n+1)^{\frac{1}{2}}) - \arctan((n)^{\frac{1}{2}})= \frac{1}{2}\, -\frac{7}{8}\, \frac{1}{n}+O \left( \frac{1}{n^{2}} \right)$$
$$ \implies \lim_{n\to \infty} n^{3/2}( \arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}})=\frac{1}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find this $\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2}{2k+1}$ Let $m,n,k\in \mathbb{N}$, and $m,n,k\ge 1,m\neq n\neq k\neq m $, such that
$$
\dfrac{1}{2m+1}+\dfrac{1}{2n+1}=\dfrac{2}{2k+1}
$$
Is there a solution? Or does this not have any solution?
| $$\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2n+1+2m+1}{(2m+1)(2n+1)}=\frac{2(n+m+1)}{(2m+1)(2n+1)}$$
So this is of the form $\frac{2}{2k+1}$ if and only if $$(n+m+1)\mid(2m+1)(2n+1)$$
Now $(2n+1)(2m+1)=4nm + 2(n+m) + 1 = 4nm -1 + 2(n+m+1)$. So we want:
$$(n+m+1)\mid 4nm-1$$
Let $d=n+m+1$. Then $n\equiv -1-m\pmod d$ and therefore:
$$0\equiv 4nm-1\equiv 4(-1-m)m-1=-(2m+1)^2\pmod d$$
So $n+m+1\mid(2m+1)^2$, and, by symmetry, and $n+m+1\mid(2n+1)^2$.
Now assume $n<m$ and fix some $n\geq 1.$ We can find an $m$ if and only if there is a $d\mid (2n+1)^2$ such that $2n+1<d\leq(2n+1)^2$. In that case $m=d-n-1$.
If $(m,n)$ is a pair satisfying this condition and $m>n$ then there is some factoring $(2n+1)^2=PQ$ with $P>Q>1$ and $m=P-n-1$.
So we are basically talking about factorings of odd squares in non-trivial fashion.
The general solution for this is $P=a^2c$ and $Q=b^2c$ with $a>b$, with $a,b,c$ odd. This yields the general solution, with $a,b,c$ odd and $a>b$:
$$2n+1=abc,2m+1=a(2a-b)c,2k+1=bc(2a-b)$$
The base case is when $c=1$ and $a>b$ are odd and relatively prime. Then $n=\frac{ab-1}{2},m=\frac{a(2a-b)-1}{2},k=\frac{b(2a-b)-1}{2},$ and we get: $$\begin{align}\frac{1}{2n+1}+\frac{1}{2m+1}&=\frac{1}{ab}+\frac{1}{a(2a-b)}\\
&=\frac{2a-b+b}{a(2a-b)b}\\
&=\frac{2}{b(2a-b)}\\
&=\frac{2}{2k+1}\end{align}$$
All other solutions can be found by dividing a base solutions by an odd integer.
When $a=2k+1$ and $b=1,$ you get the selected answer:
$$\frac{1}{2k+1}+\frac{1}{(2k+1)(4k+1)}=\frac{2}{4k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Test the convergence of $\sum_{n=1}^{n=\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)$. Test the convergence of
$$
\sum_{n=1}^{\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)
$$
$$
\frac{u_{n+1}}{u_{n}}=\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
| A straightforward computation gives
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}
=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
=\lim_{n\to\infty}\frac{(n+1)^4+2^{n+1}}{2^{n+1}(n+1)^2}\frac{n^2 2^n}{n^4+2^n}
=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\frac{2^n}{2^{n+1}}\frac{(n+1)^4+2^{n+1}}{n^4+2^n}
=\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\lim_{n\to\infty}\frac{1}{2}\frac{\frac{(n+1)^4}{2^{n+1}}+1}{\frac{n^4}{2^{n+1}}+\frac{1}{2}}
=1\cdot\frac{1}{2}\frac{1+1}{0+\frac{1}{2}}=1
$$
Since this limit equals 1 you can't say anything about convergence. But you can apply another approaches. For example cnsider two sums as it was done in Sami Ben Romdhane's answer. I want to emphasize that you can even find exact value for this sum. Indeed, decompose
$$
\sum\limits_{n=1}^\infty\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)=\sum\limits_{n=1}^\infty\frac{n^2}{2^n}+\sum\limits_{n=1}^\infty\frac{1}{n^2}
$$
And see this and this questions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$
find the value
$$\dfrac{b+c}{a}$$
my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $
so$\sqrt{9-8\sin 50^{\circ}}<2$,then $a=1$
then $$\dfrac{b^2}{16}(1-\cos{(2c)})+\dfrac{b}{4}\sin{c}=1-\sin{50^{0}}$$
so $b=4$
then we have $\sin{c}-\cos{(2c)}=-\sin{50^{0}}$
then my question: How can prove this $c$ must equality 10?
Thank you everyone: yesterday,when I go to bed, I have consider this:let $f(c)=\sin{c}-\cos{(2c)}$.then we have $f(10)=\sin{10}-\cos{20}=\cos{80}-\cos{20}=2\sin{\dfrac{80-20}{2}}\sin{\dfrac{80+20}{2}}=\sin{50}$, by other hand, we have $f'(c)=\cos{c}+2\sin{2c}>0,0<c<\dfrac{\pi}{2}$,so if we $f(c)=f(10)$,we must $c=10$
| $$\sin c-\cos(2c)$$
$$\implies \cos(90-c)-\cos(2c)$$
$$\implies -2\sin\Bigg(\dfrac{90+c}2\Bigg) \sin \Bigg(\dfrac{90-3c}2\Bigg)=-\sin(50^{\circ})$$
$$2\sin\theta \sin\phi=\sin 50^{\circ}$$
So, one of the solution comes when one of $\sin \theta^{\circ}$ or $\sin\phi$
is equal to $1/2$ and other is $\sin 50^{\circ}$
So, $$1)\dfrac{90+c}2=50 ;\sin\Bigg(\dfrac{90-3c}2\Bigg)=1/2$$
Or $$2)\dfrac{90-3c}2=50 ; \sin\Bigg(\dfrac{90+c}2\Bigg)=1/2$$
Case $2$ does not hold true , So, from $1$st case $c=10^{\circ}$
And as the solution is unique, it's the one we need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
To find the logarithm of $1728$ to the base $2 \sqrt{3}$
Find the logarithm of: $1728$ to base $2\sqrt{3}$.
Let, $\log_{2\sqrt{3}} 1728 = y$, then
$$\begin{align} (2\sqrt{3})^y &= 1728\\
2^y(\sqrt3)^y &= 1728\\2^y(3^\frac12)^y &= 1728\\2^y(3^\frac y2)
&= 1728\\2^y × 3^\frac y2 &= 2^6 × 3^3 \end{align}$$
What should I do next to find the logarithm of $1728$?
| $$12^3 = 1728 = (2\sqrt 3)^y = (\sqrt 12)^y = (12^\frac 1 2)^y = 12^\frac y 2 $$
$$ 3 = \frac y 2 $$
$$ y = 6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation: $x^2+\frac {1}{x^2}=2^{1-y^2}$. I found the following problem interesting but do not know how to proceed:
Solve the equation: $$x^2+\frac {1}{x^2}=2^{1-y^2}.$$
Can someone point me in the right direction?
| We can proceed with the problem in the following way:
We know that for any non-negative $a,b ; \quad a+b \geq 2 \sqrt {ab}$. So, $x^2 + \frac {1}{x^2} \geq 2 \sqrt {x^2. \frac {1}{x^2}}=2$. Hence the min. value of the left hand side of the equation is $2$. On the other hand ,the max. value of the
right hand side is $2.$ So the given equation is equivalent to $x^2 + \frac {1}{x^2}=2$ and $2^{1-y^2}=2$.Hence solving the equations seperately,we get $x=\pm 1$ and $y=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/362388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Expressing $\cos\theta - \sqrt{3}\sin\theta = r\sin(\theta - \alpha)$ My book explains that $a\cos\theta + b\sin\theta$ is a sine (or cosine) graph with a particular amplitude/shift (i.e. $r\sin(\theta + \alpha)$) and shows me some steps to solve for $r$ and $\alpha$:
$$r\sin(\theta + \alpha) \equiv a\cos\theta + b\sin\theta$$
$$\Rightarrow r\sin\theta\cos\alpha + r\cos\theta\sin\alpha \equiv a\cos\theta + b\sin\theta$$
I see the basic trig identity $\sin(a + b) \equiv \sin a\cos b + \sin b\cos a$ is being used, and the coefficients $a$ and $b$ are easily identified as:
$$r\sin\alpha = a$$
$$r\cos\alpha = b$$
Then the book squares and adds the equations:
$$r^2(\sin^2\alpha + \cos^2\alpha) = a^2 + b^2 \Rightarrow r = \sqrt{a^2 + b^2}$$
I see that the basic identity $\sin^2\alpha + \cos^2\alpha \equiv 1$ is being used here.
All the examples in the book involve $r$ being a positive quantity.
An exercise asks me to express $\cos\theta - \sqrt{3}\sin\theta$ in the form $r\sin(\theta - \alpha)$. I tried this and ran into problems as the squaring and square rooting process only ever produces a positive $r$, but the answer is $-2\sin(\theta - \frac{1}{6}\pi)$.
My workings:
$\cos\theta - \sqrt{3}\sin\theta = r\sin(\theta - \alpha) = r\sin\theta\cos\alpha - r\sin\alpha\cos\theta$
$\Rightarrow r\cos\alpha = -\sqrt{3}$, $r\sin\alpha = -1$
$\Rightarrow r^2\cos^2\alpha = 3$, $r^2\sin^2\alpha = 1 => r^2(\cos^2\alpha + \sin^2\alpha) = 4$
$\Rightarrow r^2 = 4 \Rightarrow r = \pm2$.
My question is how should the process in the book be refined so one knows whether $r$ should be positive or negative?
(sorry if this is a bit long winded for such a basic question but I thought showing what I do and don't know might get me an answer targeted at my simple level!).
| In your specific case you get either
$$\sin\alpha = \frac{1}{2}\text{ and } \cos\alpha=-\frac{\sqrt{3}}{2}\text{ for } r=2$$
or
$$\sin\alpha = -\frac{1}{2}\text{ and } \cos\alpha=\frac{\sqrt{3}}{2}\text{ for } r=-2$$
In the first case you get $\alpha=\frac{5\pi}{6}$, and in the second case you get $\alpha=-\frac{\pi}{6}$. (I use the function $r\sin(\theta + \alpha)$ with a positive sign of $\alpha$). You have to be careful when computing $\alpha$ because you usually cannot simply take the principle branch of the functions $\arcsin(x)$ and $\arccos(x)$. You have to choose a value that satisfies both equations (the one for $\sin(\alpha)$ and the one for $\cos(\alpha)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/363222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the standard matrix of the transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ that corresponds to the reflection through the line Find the standard matrix of the transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ that corresponds to the reflection through the line $x_2=2x_1$ followed by reflection through the line $x_1=3x_2$
I am very confused about this question, could anyone explain how to solve the problem? THANKS!
| First you change coordinates to where your basis is given by $\begin{pmatrix} 1 \\ 2\end{pmatrix}$ and $\begin{pmatrix} 2 \\ -1\end{pmatrix}$.
This is done by multiplying by
$$\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}.$$
Next, you reflect the second coordinate. This is done by multiplying by
$$ \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}.$$
Finally, you change your coordinates back by multiplying by
$$\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}.$$
Thus, your transformation is multiplication by
$$ \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix} = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}\begin{pmatrix} 1 & 2 \\ -2 & 1\end{pmatrix} = \frac{1}{5}\begin{pmatrix}-3 & 4 \\ 4 & 3 \end{pmatrix}.$$
Now try to do the second reflection yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Deriving Closed Form for a Recursion via Generating Functions Consider (1) $a_{n+2} = 2a_{n+1} - a_n + 4n3^n$ with $a_0 = a_1 = 1$.
Using generating functions and setting $A(x) = \sum a_nx^n$ we obtain
$$\begin{align*}&\quad\sum a_{n+2}x^{n+2} = \sum2a_{n+1}x^{n+2} - \sum a_nx^{n+2} + \sum 4n3^nx^{n+2}\\ &\implies [A(x) - a_0 - a_1x] = 2x[A(x)-a_0] - x^2A(x) + \sum_n 4n3^nx^{n+2}\end{align*}$$
Is this correct so far? Is there always a best way to go about rearranging the obtained generating function, or does it vary from problem to problem? Further, is it simpler to use this method here or to instead obtain a particular solution through undetermined coefficients? Any help is much appreciated.
Is it possible to decompose $4n3^nx^{n+2}$ into $x^2$$\sum 4n \times 1/(1-3x)$ and does this help?
| This is how I would approach this. Of all the terms $4n3^n$does not depend on the $a_i$, so I'll keep that one apart. The first thing is to view an expression obtained from the remaining terms as a multiple of $A(x)$. So first separate
$$
a_{n+2}-2a_{n-1}+a_n = 4n3^n.
$$
We can recognise the left hand side as the coefficient of $x^{n+2}$ in $(1-2x+x^2)A(x)$. So if we multiply each displayed recursion relation by $x^{n+2}$ and add up, then we get an equation for $(1-2x+x^2)A(x)$, except for the terms of degree${}\leq1$, which are $1-x$ because $a_0=a_1=1$ and $(1-2x)(1+x)\equiv1-x\pmod{x^2}$. So we get
$$
(1-2x+x^2)A(x) = 1-x+4x^2\sum_{n\geq0}n(3x)^n,
$$
where the term $x^2$ before the summation comes from the fact that $4n3^n$ was multiplied by $x^{n+2}$ rather than by $x^n$.
Now doing the final summation can be done recognising that we can write $n+1=\binom{n+1}n=(-1)^n\binom{-2}n$ and $1=\binom nn=(-1)^n\binom{-1}n$ for all $n\geq0$, so that
$$
\sum_{n\geq0}n(3x)^n
=\sum_{n\geq0}\left(\tbinom{-2}n-\tbinom{-1}n\right)(-3x)^n
=(1-3x)^{-2}-(1-3x)^{-1}
=\frac{3x}{(1-3x)^2}.
$$
Putting everything together gives
$$
A(x)=\frac{1-7x+15x^2+3x^3}{(1-x)^2(1-3x)^2},
$$
where the numerator was obtained as $(1-3x)^2(1-x)+4x^2\times 3x$. Working out a concrete expression for the coefficients of $A(x)$ is now standard by partial fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
complex integral with parametrization of an ellipse I'm having trouble with the following complex integral:
$$\int_{C}^{}\frac{dz}{\sqrt{1 - z^{2}}}$$ where C is a positively oriented ellipse $${x^{2}\over a^{2}} +{ y^{2}\over b^{2}} = 1$$ where $$a^{2} - b^{2} = 1$$
I know z(t) = $a\cos(t) + ib\sin(t)$ but I'm not sure about how to factor after plugging z into the integrand and simplifying using the fact that $$a^{2} - b^{2} = 1$$
| As you say: $z(t) = a\cos t + ib\sin t$ gives a bijection from $[0,2\pi)$ to the ellipse. Given the definition of $z$, we can see that $\operatorname{d}\!z=(-a\sin t + ib\cos t) \, \operatorname{d}\!t$. Hence:
$$\oint_C \frac{\operatorname{d}\!z}{\sqrt{1-z^2}} = \int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{1-(a\cos t + ib\sin t)^2}}\, \operatorname{d}\!t$$
The trick here is to use the fact that $a^2-b^2=1$, or rather $1=a^2-b^2$. Notice that
\begin{array}{ccc}
1-(a\cos t + ib\sin t)^2 &\equiv& 1- a^2\cos^2t +b^2\sin^2t - 2iab\cos t \sin t \\ \\
&\equiv& a^2-b^2 - a^2\cos^2t +b^2\sin^2t - 2iab\cos t \sin t \\ \\
&\equiv& a^2(1-\cos^2t) -b^2(1-\sin^2) - 2iab\cos t \sin t \\ \\
&\equiv& a^2\sin^2t - b^2\cos^2t - 2iab\cos t \sin t \\ \\
&\equiv& (-a\sin t + ib\cos t)^2
\end{array}
It follows that:
$$\oint_C \frac{\operatorname{d}\!z}{\sqrt{1-z^2}} = \int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{(-a\sin t + ib\cos t)^2}}\, \operatorname{d}\!t$$
This integrand has an explicit anti-derivative:
\begin{array}{ccc}
\int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{(-a\sin t + ib\cos t)^2}}\, \operatorname{d}\!t &=& \left[ t \, \frac{-a\sin t + ib\cos t}{\sqrt{(-a\sin t + ib\cos t)^2}} \right]_0^{2\pi} \\ \\
&=& 2\pi i \frac{b}{\sqrt{-b^2}} \\ \\
&=& 2\pi \frac{b}{|b|}
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the Laurent series at the given point and its residue
a) $\displaystyle \frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$
b) $\displaystyle \frac{1}{1-\cos(z)}$ at $z_0=0$
I'm having trouble understanding Laurent series, please help!
| A related problem. For the second one, you can advance as
$$ \frac{1}{1-\cos(z)} = \frac{1}{\frac{z^2}{2!}-\frac{z^4}{2!}+\frac{z^6}{6!}-\ldots }= \frac{1}{\frac{z^2}{2!} \left(1-\frac{2! z^2}{4!}+\frac{2! z^4}{6!}-\ldots \right) } $$
$$ =\frac{2}{z^2} \left(1-\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right) \right)^{-1} =\frac{2}{z^2}( 1-t )^{-1}, $$
where $t=\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right).$
$$\implies \frac{2}{z^2}( 1-t )^{-1}=\frac{2}{z^2}( 1+t+t^2+t^3+\ldots )$$
$$ =\frac{2}{z^2} \left(1+\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right)+\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right)^2 + \dots \right) $$
$$ =\frac{2}{z^2} + \frac{1}{6} + \frac{z^2}{120}+\ldots\,. $$
Clearly, the coefficient of $z^{-1}$ is $0$. Note that, we have used the identity
$$ (1-x)^{-1}=1+x+x^2+x^3+\ldots\,. $$
Different approach: You can use another technique to find the residue without using the Laurent series. See here for details.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express ${x^3 + 2x^2 + 61 \over (x + 3)^2 (x^2 + 4)}$ in partial fractions $$\eqalign{
& {x^3 + 2x^2 + 61 \over (x + 3)^2(x^2 + 4)} \equiv {A \over (x + 3)} + {B \over (x + 3)^2} + {Cx + D \over (x^2 + 4)} \cr
& \equiv {A(x + 3)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x + 3)^2 \over (x + 3)^2(x^2 + 4)} \cr} $$
so:
$$x^3 + 2x^2 + 61 \equiv A(x + 3)(x^2 + 4) + B(x^2 + 4) + (Cx + D)(x + 3)^2$$
Solving for $B$, making $x=-3$:
$$\begin{align}
- 27 + 18 + 61 & = 0 + 13B + 0 \\[6pt]
13B & = 52 \\[6pt]
B & = 4 \end{align} $$
The other three variables are tricky as there is no value that I can see that is capable of eliminating two and leaving one, I've tried to "compare coefficients" but I am left with more than one unknown so I think the only other options available to me is solving simultaneously with 3 variables (something I do not know how to do), have i missed something that could make this easier to write in partial fractions?
Thank you.
| You're doing the problem right. I don't think there's a way of avoiding solving a $3\times 3$ system. However, you can try "nice" values for $x$ to make it a nice system, e.g. $x=0, x=1, x=-1$.
| {
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Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers.
Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$
For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$
and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach.
Is this problem solvable, at least partially?
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?
| Using the fact that: $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ and letting $x=m^{\frac{1}{3}}$ and $y=n^{\frac{1}{3}}$, we get the statement: $$m+n=(m^{\frac{1}{3}}+n^{\frac{1}{3}})(m^\frac{2}{3}-(mn)^{\frac{1}{3}}+n^{\frac{2}{3}})$$Maybe this will help.
| {
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"timestamp": "2023-03-29T00:00:00",
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Verifying a proof that if $x,y,z \geq 0$ and $x+y+z = 1$, then $0 \le xy + yz + zx - 2xyz \le \frac{7}{27}$ I was working some recreational problems from a book (The Art and Craft Of Problem Solving, Zeitz) and came across one from the '84 IMO:
Suppose that $x, y, z$ are non-negative reals, with $x + y + z = 1$. Prove that
$$0 \le xy + yz + zx - 2xyz \le \frac{7}{27}.$$
I'm afraid that I'm here today to ask for a proof check. I ask because I was able to prove the claim without using any sophisticated inequalities, and - although I have checked it myself - I can't help but feel a bit suspicious.
I have posted my solution as an answer below. Of course, I would love it if somebody would give it a careful read.
Shorter, more elegant solutions are also welcome.
| For proving the second inequality, take $ x=a+\frac{1}{3}, y=b+\frac{1}{3}, z=c+\frac{1}{3}$. Since $x+y+z=1$, so $a+b+c=0$. Since $x,y,z\geq 0$, so $a,b,c\geq -\frac{1}{3}$.By simple algebraic manipulation we get,
$$xy+yz+zx-2xyz=\frac{2}{3}(ab+bc+ca-3abc)+\frac{7}{27}$$
We just need to show that $ab+bc+ca-3abc\leq 0$. Since $a+b+c=0$, so $a^2+b^2+c^2=-2(ab+bc+ca)$ and $a^3+b^3+c^3=3abc$. Thus,
$$ab+bc+ca-3abc=-\frac{1}{2}(a^2+b^2+c^2)-(a^3+b^3+c^3)$$
which is equal to $-\frac{1}{2}\lbrace a^2(1+2a)+b^2(1+2b)+c^2(1+2c)\rbrace\leq 0$ .
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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I cannot find the last factor of this expression? I'm supposed to factor $x^8-y^8$ (the exponents are 8 for both if it is too difficult to see) as completely as possible. It is easy to factor this to $(x+y)(x-y)(x^2+y^2)(x^4+y^4)$. However, the book says "(Hint: there are 5 factors. Note that we sat real coefficients, not just integers)". What is this elusive 5th factor? Thanks!
| Another approach: if you know that
$$x^4+1=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$
then you can do as follows:
$$x^4+y^4=y^4\left(\frac{x^4}{y^4}+1\right)=y^4\left(\frac{x^2}{y^2}+\sqrt 2\frac xy+1\right)\left(\frac{x^2}{y^2}-\sqrt 2\frac xy+1\right)=$$
$$=\left(x^2+\sqrt 2xy+y^2\right)\left(x^2-\sqrt 2xy+y^2\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Differential equation with no constants. Is there a way to calculate $\psi$ which is a function of $x$ out of this differential equation:
$$
\frac{d^2 \psi}{d x^2} = x^2 \psi
$$
| $\dfrac{d^2\psi}{dx^2}=x^2\psi$
$\dfrac{d^2\psi}{dx^2}-x^2\psi=0$
Note that this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf.
Let $\psi=e^{-\frac{x^2}{2}}y$ ,
Then $\dfrac{d\psi}{dx}=e^{-\frac{x^2}{2}}\dfrac{dy}{dx}-xe^{-\frac{x^2}{2}}y$
$\dfrac{d^2\psi}{dx^2}=e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}-xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}+\left(x^2e^{-\frac{x^2}{2}}-e^{-\frac{x^2}{2}}\right)y=e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-2xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}+(x^2-1)e^{-\frac{x^2}{2}}y$
$\therefore e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-2xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}+(x^2-1)e^{-\frac{x^2}{2}}y-x^2e^{-\frac{x^2}{2}}y=0$
$e^{-\frac{x^2}{2}}\dfrac{d^2y}{dx^2}-2xe^{-\frac{x^2}{2}}\dfrac{dy}{dx}-e^{-\frac{x^2}{2}}y=0$
$\dfrac{d^2y}{dx^2}-2x\dfrac{dy}{dx}-y=0$
You can apply the procedure in Help on solving an apparently simple differential equation to get $y=c_1\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt}}{\sqrt{t}}dt+c_2\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt}}{\sqrt{t}}dt$
$\therefore\psi=c_1e^{-\frac{x^2}{2}}\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt}}{\sqrt{t}}dt+c_2e^{-\frac{x^2}{2}}\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt}}{\sqrt{t}}dt$
$\psi=c_1\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt-\frac{x^2}{2}}}{\sqrt{t}}dt+c_2\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt-\frac{x^2}{2}}}{\sqrt{t}}dt$
$\psi=C_1\int_0^\infty e^{-\frac{t^2}{4}+xt-\frac{x^2}{2}}~d(\sqrt{t})+C_2\int_0^\infty e^{-\frac{t^2}{4}-xt-\frac{x^2}{2}}~d(\sqrt{t})$
$\psi=C_1\int_0^\infty e^{-\frac{t^4}{4}+xt^2-\frac{x^2}{2}}~dt+C_2\int_0^\infty e^{-\frac{t^4}{4}-xt^2-\frac{x^2}{2}}~dt$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial and distinct roots Find all real m such that $x^3-2x^2-2x+m$ has 3 distinct rational roots.
Source: School exam paper. No idea why it seems so hard
I don't think rational root theorem works, since m is not necessarily integer (clearly it is rational though)
Assuming $m=\frac{r}{s}$, we need to check $sx^3-2sx^2-2sx+r=0$ for all $s,r$ coprime...
| Let $a, b, c$ be the rational roots. Then we have
$$a+b+c=2,\quad ab+bc+ca=-2, \quad m=-abc.$$
Replace $c=2-a-b$,
$$a^2+b^2+ab-2(a+b)-2=0.$$
(If $a=b$, then $a$ is irrational) Thus,
$$\Delta_b=(a-2)^2-4(a^2-2a-2)=-3a^2+4a+12$$
is a square rational.
The rest is easy.
Added after diner. Since
$$-3a^2+4a+12=-3(a+\frac{2}{3})^2+4\times \frac{10}{3}$$
is a square. Let $\frac{3}{2}(a+\frac{2}{3})=\frac{p}{q}$ (suppose $\gcd(p,q)=1$), then
$$-3p^2+30 q^2$$
is a square of integer, say $3r$, thus
$$-p^2+10q^2=3r^2.$$
It is clear that $2|p-r$. If both are even, then $q$ is even too, contradicting $\gcd(p,q)=1$. So $p$ and $r$ both are odd, then $p^2+3r^2=4 \mod(8)$, therefore $q$ is even, hence $10q^2=0 \mod (8)$, this is an contradiction with $p^2+3r^2=10 q^2$.
If I didn't make mistake, Conclusion: there is no $a, b, c$, there is no $m$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^\tau \frac{t\sin(t z)}{z\cos(t z)-\sin(tz)}\text{d}t$ I'm trying to evaluate the following definite integral. Mathematica gives me a complicated expression which I think I can simplify, but I was wondering if there was a "nice" way to evaluate it.
$$\displaystyle\int_0^\tau \frac{t\sin(t z)}{z\cos(t z)-\sin(tz)}\text{d}t$$
| This integral doesn't seem to have an elementary expression, at least for general $\tau$. At least Mathematica and my own results agree on this, as the final answer involves polylogarithm.
Let's represent the integral as:
$$I=\frac{1}{z^2} \int_0^{\tau z} \frac{y \sin y}{z \cos y-\sin y}dy$$
My attempt uses integration by parts with:
$$u=y, \qquad dv=\frac{\sin y}{z \cos y-\sin y}dy$$
Antiderivative for $dv$ was obtained by Mathematica, but it's trivial to prove it by differentiation:
$$v=-\frac{1}{1+z^2} \left(y+z \log |z \cos y-\sin y| \right)$$
Using the integration by parts formula, we obtain:
$$I=-\frac{1}{1+z^2} \times \\ \times \left( \tau^2 +\tau \log |z \cos \tau z-\sin \tau z|-\frac{1}{z^2} \int_0^{\tau z} y dy-\frac{1}{z} \int_0^{\tau z} \log |z \cos y-\sin y| dy \right)$$
To simplify the last integral, we do a series of transformations:
$$ \log |z \cos y-\sin y|=\frac{1}{2} \log (z \cos y-\sin y)^2=$$
$$=\frac{1}{2} \log ((z^2-1) \cos^2 y-2 \sin y \cos y+1)=$$
$$=\frac{1}{2} \log \left(\frac{z^2+1}{2}+ \frac{z^2-1}{2}\cos 2y-z \sin 2y \right)=$$
$$=\frac{1}{2} \log \left(\frac{z^2+1}{2}\right)+\frac{1}{2} \log \left(1+ \frac{z^2-1}{z^2+1}\cos 2y-\frac{2z}{z^2+1} \sin 2y \right)=$$
$$=\frac{1}{2} \log \left(\frac{z^2+1}{2}\right)+\frac{1}{2} \log \left(1+ \cos \left( 2y+\arcsin \frac{2z}{z^2+1} \right)\right)$$
Simplifying and making a substitution in the last integral $x=2y+\arcsin \frac{2z}{z^2+1}$, we obtain:
$$I=-\frac{1}{2(1+z^2)} \left(\tau^2 +\tau \log \frac{ 2(z \cos \tau z-\sin \tau z)^2}{1+z^2}-\frac{1}{2z} \int_a^{2\tau z+a} \log (1+\cos x) dx \right)$$
$$a=\arcsin \frac{2z}{1+z^2}$$
The last integral is well known in it's definite form:
$$\int_0^{\pi/2} \log (1+\cos x) dx=2 G-\frac{\pi}{2} \log 2$$
where $G$ is Catalan's constant. However, the indefinite integral isn't so nice and involves polylogarithm.
| {
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Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:
Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$
I see that $x_n^{\frac{1}{2}}=\frac{2}{3}\frac{5}{6}\frac{9}{10} \ldots..\frac{(n-1)(n+2)}{n(n+1)}$. now not sure what to do next ?Any idea?
| Let $f(n)=\frac{n-1}{n+1}$. Then show $$1-\frac{1}{n(n+1)/2} = \frac{f(n)}{f(n+1)}$$
| {
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Fractions in Questions and Answers
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