Get the solution to "Find the solutionlimt→0dk−1dtk−1(t1+tlog⁡1+t)k=2(k+2)k−1"

Justine White

Answered question

2022-04-07

Find the solution

lim_{t \to 0} \frac{d^{k - 1}}{{d t}^{k - 1}} {(\frac{t \sqrt{1 + t}}{\log \sqrt{1 + t}})}^{k} = 2 {(k + 2)}^{k - 1}

superpms01wks1

Beginner2022-04-08Added 13 answers

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

f^{n - 1} (0) = \frac{(n - 1)!}{2 π i} \oint \frac{f (z)}{z^{n}} d z

Now

lim_{t \to 0} \frac{d^{k - 1}}{{d t}^{k - 1}} {(\frac{t \sqrt{1 + t}}{\log \sqrt{1 + t}})}^{k} = \frac{(k - 1)!}{2 π i} \oint {(\frac{t \sqrt{1 + t}}{\log \sqrt{1 + t}})}^{k} \frac{d t}{t^{k}}

\frac{(k - 1)!}{2 π i} \oint {(\frac{\sqrt{1 + t}}{\log \sqrt{1 + t}})}^{k} d t

Now performing the change of variable

t = e^{u} - 1

lim_{t \to 0} \frac{d^{k - 1}}{{d t}^{k - 1}} {(\frac{t \sqrt{1 + t}}{\log \sqrt{1 + t}})}^{k} = \frac{(k - 1)!}{2 π i} \oint {(\frac{\exp (\frac{u}{2})}{\frac{u}{2}})}^{k} e^{u} d u

= 2^{k} [\frac{(k - 1)!}{2 π i} \oint \frac{\exp (u (\frac{k}{2} + 1)} {u^{k}} d u}{}

= 2^{k} lim_{u \to 0} \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}

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