The long way of writing it out #2

Calculus Level pending

$\delta(m) = \lim_{x \to \infty} \sqrt{\left( 1+\frac{2m^2}{x} \right) ^{\frac{x}{m}}} \times \left( \sum_{n=0} ^ {\infty} \left( \int_0 ^ {\infty} t^ne^{-t}dt \right)^{-1} \right)^m$

Given $\delta(m)$ , express $\delta' \left(2\ln(p)+\frac{\log(11)}{\log(e)} \right)$ in terms of $p$

242p^4

\frac{\pi^2}{p}

2e^{4p+1}

p

1 solution

A Former Brilliant Member
Jun 11, 2020

The first term of the product is $\sqrt {e^{2m}}=e^m$ . The second term is

$\left (\displaystyle \sum_{n=0}^\infty \frac{1}{n!}\right ) ^m=e^m$ .

Hence the given expression becomes $\delta (m)=e^{2m}$ . So $\delta'(m) =2e^{2m}$ .

Hence $\delta' \left (2\ln p+\frac{\log 11}{\log e}\right ) =2e^{\ln (121p^4)}=\boxed {242p^4}$ .

The long way of writing it out #2

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