New answers tagged wolfram-alpha
0
Hint: $n^{k(k+1)/(2n)}=\mathrm e^{k(k+1)\log(n)/(2n)}=1+u+u^2/2+\ldots$ with $u=k(k+1)\log(n)/(2n)$.
This yields the expansion
$$
\frac1{4k^2n^2}\left(1-\frac{(k+1)(k-2)}2\frac{\log n}n+O\left(\left(\frac{\log n}n\right)^2\right)\right).
$$
0
Please be careful, because a Taylor expansion at $\infty$ is not a normal series expansion. For $k>0$ it will surely converge to zero, as $\ln(n)$ grows slower than $n$, and hence the denominator is not bounded, as the logarithm is much slower we can forget him for the asympotics
When we look at $$n^\frac{k(k+1)}{2n}$$ this converges to $1$ because the ...
5
I think the apparent answer to your problem is that you entered in Mathematica:
NIntegrate[pi^2 ... blah ]
... and in Mathematica, Pi is the exact real number $\pi$ = 3.1415 ... but pi is just a name like cat or mouse. Capital letters matter. Wolfram Alpha is more flexible about such things, perhaps because it needs to be. This explains why you would get ...
1
1) The limit clearly exists since $\sin$ is bounded and $xy\rightarrow 0$.
2) In this problem you can. In a problem such as $\lim_{(x,y)\rightarrow (0,0)}\frac{x}{y}$ you run into trouble because you seem to suggest that it's ok to write it as $\lim_{z\rightarrow 0}\frac{z}{z}=1$ and that's wrong. To see why it's wrong, $x/y$ can be arbitrarily large if ...
8
Alpha will give you the same $0.1983453912\cdots$ result with $\ \operatorname{sn}(0.2,0.5^2)$
Mathematica and the old and excellent reference Abramowitz&Stegun use the 'parameter' $m=k^2$ instead of the 'modulus' $k$ used by your code, Maple, MATLAB and many other online references like Dlmf and MathWorld which is used when you click at the right on ...
2
Well, here are WolframAlpha's steps:
Ultimately, it's just a matter of applying the multi-variate chain rule (shown in the WA output) together with the fact that the partial derivatives of the binomial can be expressed in terms of the digamma function. Specifically:
$$\frac{\partial}{\partial x} \binom{x}{y}=\binom{x}{y} (\psi ...
Top 50 recent answers are included
