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Problem $$I(x) = \int_{1}^x \frac{e^t - 1}{t}$$

Find $I'( \sqrt{x} )$.

Solution

We know that $F'(x) = f(x)$ by the fundamental theorem of calculus so $$I'(x) = \frac{e^t -1}{t}$$ And so $$I'( \sqrt{x}) = \frac{ e^{\sqrt{x}} -1 }{ \sqrt{x}}$$


Problem

Find the fourth taylor polynomial of $I(x)$ around $0$ (Guess this makes it a Maclaurin series?)

Taylor series: $$ \sum_{n=o}^\infty \frac{ f^{(n)} (a) }{n!} (x-a)^n$$

So as we know the first derivative, we can compute the first Taylor expansion:

$$\frac{e^t-1}{t} (x)^n$$

Do I just continue like this? Am I doing this whole problem right? I'm just not feeling sure at all!

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  • $\begingroup$ I've had 2 answers but they both got deleted I think? Is everything okay with the question? $\endgroup$ – Paze Apr 12 '14 at 20:00
  • $\begingroup$ I misread the question at first but I have posted what I believe to be a correct solution. $\endgroup$ – Ellya Apr 12 '14 at 20:15
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Here is a start. Using the power series of $e^t$, we have

$$ I(x) = \int_{1}^x \frac{e^t - 1}{t} = \int_{1}^x \sum_{k=1}^{\infty}\frac{t^{k-1}}{k!} dt = \sum_{k=1}^{\infty}\frac{1}{k!}\int_{1}^{x}{t^{k-1}} dt. $$

I think you can finish the problem.

Note: You only need a fourth degree polynomial.

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  • $\begingroup$ Why would I like to turn $e^t$ into the power series? $\endgroup$ – Paze Apr 12 '14 at 20:09
  • $\begingroup$ @Paze: It is easier. $\endgroup$ – Mhenni Benghorbal Apr 12 '14 at 20:11
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Hint: $$I(x)=\int^x_1\dfrac{e^t-1}{t}dt=\int^x_1\sum^\infty_{k=1}\dfrac{t^{k-1}}{k!}dt-\ln(x)= \sum^\infty_{k=1}\dfrac{x^{k}-1}{k\times k!}-\ln(x)$$

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