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Math.SE seems to be going the stackoverflow way. Pity.


8h
comment Proving this binomial identity $\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$
@darijgrinberg: If $f(x) = \sum_{k=0}^{\infty} a_k x^k$, then the partial sums $\sum_{k=0}^{n} a_k$ are given by the coefficients of $x^n$ in the series expansion of $\dfrac{f(x)}{1-x}$. So we need to find coefficient of $x^n$ in $$ \frac{1}{(1-\frac{x}{2})^{n-1}(1-x)}$$ How to figure that out is a different problem, but Alex's insight definitely helps.
1d
comment Rational vs irrational
The difference between a rational and irrational is irrational :-)
1d
revised Convergent or Divergent? $\sum_{n=1}^\infty\bigl(2^{\frac1{n}}-1\bigr)$
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1d
comment Convergent or Divergent? $\sum_{n=1}^\infty\bigl(2^{\frac1{n}}-1\bigr)$
@DepeHb: Uh. You are right! Thanks for pointing that out. That is easily fixed, fortunately (take $\frac{1}{\sqrt{2}}$ instead)...
2d
comment Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$
You are welcome.
Jan
29
revised Proving that $\sin1 $(radian) is irrational without using Taylor Series Expansion.
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Jan
29
comment Proving that $\sin1 $(radian) is irrational without using Taylor Series Expansion.
btw, if it was $\sin 1^{\circ}$, the proof would be easy: $\sin 45^{\circ}$ is a polynomial in $\sin 1^{\circ}$ with rational (in fact, integer) coefficients! Since $\sin 45^{\circ}$ is irrational...
Jan
29
answered Proving that $\sin1 $(radian) is irrational without using Taylor Series Expansion.
Jan
29
revised Proving that $\sin1 $(radian) is irrational without using Taylor Series Expansion.
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Jan
29
comment Irrational numbers in reality
This is not a mathematical question. Seems like you found a bug around getting close votes by offering a bounty :-) I have flagged for migration to physics.se, so let's see.
Jan
27
comment Can't figure out $O(n \log n)$ divide-and-conquer algorithm
You could implement so it is $\Omega(n \log n)$ instead of $O(n)$ though...
Jan
27
revised Can't figure out $O(n \log n)$ divide-and-conquer algorithm
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Jan
27
comment Can't figure out $O(n \log n)$ divide-and-conquer algorithm
This is the famous maximum sum subarray problem...
Jan
27
answered Can't figure out $O(n \log n)$ divide-and-conquer algorithm
Jan
27
comment Why $\frac{1}{n}\sum_{j=1}^mj^p \asymp\frac{1}{n}m^{p+1}$ as $n\to\infty$?
@Lionville: Ok, good to know you haven't ignored it completely :-)
Jan
27
comment How to proceed with the following integration?
Manish, this site has latex support using MathJax, please use it instead of posting image links. Here is a reference: meta.math.stackexchange.com/questions/5020/…
Jan
27
revised How to proceed with the following integration?
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Jan
27
revised Why $\frac{1}{n}\sum_{j=1}^mj^p \asymp\frac{1}{n}m^{p+1}$ as $n\to\infty$?
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Jan
27
comment Calculating Euler's totient function values.
@Amad27: Once you solve this yourself (based on OohAah's comment), please add an answer and tick that.
Jan
27
comment Calculating $\sum_{k=0}^{n-1}\frac{1}{a+bk^2}$.
A closed form is unlikely, but you can try and use Euler-MacLaurin Summation formula, which is beyond algebra-precalculus.