Wikipedia:Reference desk/Archives/Mathematics/2013 April 26

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April 26

i don't understand this (derivative)

hello I am having trouble understanding this derivative. this is taken from my notes:

y=2x^3(3-x^3)^4

derivative=(2x^3)[4(3-x^3)^3(-3x^2)]+(3-x^3)^4(6x^2)

=-24x^5(3-x^3)^3+6x^2(3-x^3)^4

take out common factor

6x^2(3-x^3)^3[-4x^3+3-x^3]

=6x^2(3-x^3)(3-5x^3)

I don't understand the "take out common factor" part. Where does the -4x^3 come from? How is there a (3-5x^3) in the answer? help is appreciated, thanksBananas9821 (talk) 17:22, 26 April 2013 (UTC)

The "common factor" is 6x^2(3-x^3)^3. [-24x^5(3-x^3)^3] / [6x^2(3-x^3)^3] = -4x^3. The last expression 6x^2(3-x^3)(3-5x^3) should be 6x^2(3-x^3)^3(3-5x^3). PrimeHunter (talk) 17:52, 26 April 2013 (UTC)

wow that was very easy and i feel dumb now. thanks man!Bananas9821 (talk) 18:14, 26 April 2013 (UTC)

A Function and a Constant Related to e

$\sum _{n=0}^{\infty }{\ {\frac {x^{n}}{n{\color {Red}!}}}}=e^{x}\ ,\qquad \qquad \sum _{n=0}^{\infty }{\ {\frac {x^{n}}{n^{\color {Red}n}}}}=f(x)\ ,\qquad \qquad \sum _{n=0}^{\infty }{\ {\frac {n{\color {Red}!}}{n^{\color {Red}n}}}}=\int _{0}^{\infty }{\frac {f(x)}{e^{x}}}\ dx$

In the above, we've used the fact that $\lim _{n\to 0}{n^{n}}=1.$

— Do either the function f(x) or the constant $\sum _{n=0}^{\infty }{\ {\frac {n!}{n^{n}}}}\approx 2.88$ possess a closed form ?

— Do any of them have a name ?

— Do they present any other interesting properties ?

Any information would be greatly appreciated at this point. Thank you! — 79.113.241.166 (talk) 23:45, 26 April 2013 (UTC)

Mathematics Stack Exchange gets questions like this sometimes e.g. [1]. I don't believe it has a name, it just good for exercises on convergent series. Dmcq (talk) 08:09, 27 April 2013 (UTC)

By the way if you like that you might also like Sophomore's dream. Dmcq (talk) 15:04, 27 April 2013 (UTC)

${\begin{cases}n!<n^{n}<(n!)^{2}\\\\\sum {y(n)>\int {y(x)\ dx}}\end{cases}}=>\ {\begin{cases}{\begin{aligned}\underbrace {\int _{0}^{\infty }{{\frac {1}{x^{x}}}\ dx}} _{\approx \ 2}\ <\ \underbrace {\int _{0}^{\infty }{{\frac {1}{x!}}\ dx}} _{\approx \ 2.27}\ &<\ \underbrace {\int _{0}^{\infty }{{\frac {x!}{x^{x}}}\ dx}} _{\approx \ 2.5}\ <\ \underbrace {\int _{0}^{\infty }{{\frac {1}{\Gamma (x)}}\ dx}} _{F.-R.\ \approx \ 2.8}\\\\\underbrace {\sum _{n=0}^{\infty }{\ {\frac {1}{n^{n}}}}} _{\approx \ 2.3}\quad \ <\quad \underbrace {\sum _{n=0}^{\infty }{\ {\frac {1}{n!}}}} _{{\color {Red}e}\ \approx \ 2.72}\ &<\ \underbrace {\sum _{n=0}^{\infty }{\ {\frac {n!}{n^{n}}}}} _{\approx \ 2.88}\end{aligned}}\end{cases}}$

We also know that

$\underbrace {\int _{0}^{\infty }{{\frac {1}{\Gamma (x)}}\ dx}} _{F.-R.}\ -\ \underbrace {\sum _{n=0}^{\infty }{\frac {1}{\Gamma (n)}}} _{e}\ =\ \int _{0}^{\infty }{\frac {e^{-x}}{\pi ^{2}+\ln ^{2}x}}\ dx$

and I was wondering if similar relations and identities exist for any of the above constants as well... I was also wondering about the nature of f(x) : on one hand, it grows stronger than any other polynomial function; on the other hand, it is signifficantly lesser than the exponential function. Is there any alternate expression for it, or some relation, similar to Sterling's, that might help us better understand its value and behaviour ? — 79.113.231.109 (talk) 17:41, 27 April 2013 (UTC)

Since you got all the series expansion coefficents of f(x), you can compute may integrals involving f(x) and also evaluate summations over the zeroes of f(z) in the complex plane. You can also use Ramanujan's master theorem to compute such integrals if you can find an analytic continuation of the series coefficients. Count Iblis (talk) 22:59, 28 April 2013 (UTC)