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Question Number 122056 by I want to learn more last updated on 13/Nov/20

Commented by 676597498 last updated on 13/Nov/20

your posting iymc questions you will be disqualified i have no option but to report

$$\mathrm{your}\:\mathrm{posting}\:\mathrm{iymc}\:\mathrm{questions} \\ $$$$\mathrm{you}\:\mathrm{will}\:\mathrm{be}\:\mathrm{disqualified} \\ $$$$\mathrm{i}\:\mathrm{have}\:\mathrm{no}\:\mathrm{option}\:\mathrm{but}\:\mathrm{to}\:\mathrm{report} \\ $$

Commented by 676597498 last updated on 13/Nov/20

This guy is posting questions of this competition (iymc.info)

$$\mathrm{This}\:\mathrm{guy}\:\mathrm{is}\:\mathrm{posting}\:\mathrm{questions} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{competition}\:\left(\mathrm{iymc}.\mathrm{info}\right) \\ $$

Answered by mathmax by abdo last updated on 14/Nov/20

we have f_m (x) =(Σ_(n=1) ^∞ e^(xlnn) x^n )^2 ⇒ f_m ^′ (x) =2 (d/dx)(Σ_(n=1) ^∞ x^n e^(xln(n)) )Σ_(n=1) ^∞ x^n e^(xln(n)) =2Σ_(n=1) ^∞ {nx^(n−1) e^(xln(n)) +x^n ln(n)n^x }Σ_(n=1) ^∞ x^n n^x

$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}_{\mathrm{m}} \left(\mathrm{x}\right)\:=\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{e}^{\mathrm{xlnn}} \mathrm{x}^{\mathrm{n}} \right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{f}_{\mathrm{m}} ^{'} \left(\mathrm{x}\right)\:=\mathrm{2}\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:\mathrm{e}^{\mathrm{xln}\left(\mathrm{n}\right)} \right)\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:\mathrm{e}^{\mathrm{xln}\left(\mathrm{n}\right)} \\ $$$$=\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left\{\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{xln}\left(\mathrm{n}\right)} \:+\mathrm{x}^{\mathrm{n}} \mathrm{ln}\left(\mathrm{n}\right)\mathrm{n}^{\mathrm{x}} \right\}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \mathrm{x}^{\mathrm{n}} \:\mathrm{n}^{\mathrm{x}} \\ $$

Commented by I want to learn more last updated on 14/Nov/20

Thanks sir, i appreciate.

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

Commented by mathmax by abdo last updated on 15/Nov/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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