Find the solution

$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=2{(k+2)}^{k-1}$

Justine White
2022-04-07
Answered

Find the solution

$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=2{(k+2)}^{k-1}$

You can still ask an expert for help

superpms01wks1

Answered 2022-04-08
Author has **13** answers

Complex analysis comes to the rescue. Using Cauchy differentiation formula:

${f}^{n-1}\left(0\right)=\frac{(n-1)!}{2\pi i}\oint \frac{f\left(z\right)}{{z}^{n}}dz$

Now

$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=\frac{(k-1)!}{2\pi i}\oint {\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}\frac{dt}{{t}^{k}}$

$\frac{(k-1)!}{2\pi i}\oint {\left(\frac{\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}dt$

Now performing the change of variable$t={e}^{u}-1$

$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=\frac{(k-1)!}{2\pi i}\oint {\left(\frac{\mathrm{exp}\left(\frac{u}{2}\right)}{\frac{u}{2}}\right)}^{k}{e}^{u}du$

$={2}^{k}[\frac{(k-1)!}{2\pi i}\oint \frac{\mathrm{exp}\left(u(\frac{k}{2}+1)\right\}\left\{{u}^{k}\right\}du}{}$

$={2}^{k}\underset{u\to 0}{lim}\frac{{d}^{k-1}}{d{u}^{k-1}}{e}^{u(\frac{k}{2}+1)}={2}^{k}{(\frac{k}{2}+1)}^{k-1}=2{(k+2)}^{k-1}$

Now

Now performing the change of variable

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