1
$\begingroup$

I am playing around with using elementary techniques to derive the Taylor Series for $e^x$.

Consider the sequence of integrals $$I_n = \int_0^x t^n e^{-t} dt$$ It can be shown by induction that $$I_n = n! \left ( 1-\frac{1}{e^x}\sum_{k=0}^{n} \frac{x^k}{k!}\right)$$

I want to then consider taking $n\to \infty$ to establish the Taylor Series. This of course relies on $$\lim_{n\to\infty} \frac{I_n(x)}{n!}=0$$ which doesn't seem to easy to show for all $x$.

Any ideas would be fantastic! I am hoping that there is something elementary to use here.

$\endgroup$
4
  • $\begingroup$ An easier way would just be to consider $D(e^t) = e^t$ and just see what that does to the power series coefficient-by-coefficient and you probably need boundary condition that $e^0=1$ to get the dominoes rolling. $\endgroup$ May 12 '20 at 11:24
  • $\begingroup$ You might want to take a look at the Gamma function, using $\int_0^\infty t^n\exp(-t)dt=\Gamma(n+1)=n!$ $\endgroup$ May 12 '20 at 11:25
  • $\begingroup$ Curious method. For $x$ positive $e^{-t}\leq 1$ and you get a very good estimate. For $x$ negative, on the other hand, your method does not give good error estimates (and dividing by $e^x$ is probably not a good idea in that case). $\endgroup$
    – H. H. Rugh
    May 12 '20 at 11:27
  • $\begingroup$ For positive $x$, you can show that $\int_0^x t^n e^{-t} dt \lt \int_0^{\infty} t^n e^{-t} dt = n!$ $\endgroup$
    – Tavish
    May 12 '20 at 11:30
2
$\begingroup$

To finish your proof, take the original integral and bound it like so

$$\left|\frac{I_n}{n!}\right | = \left| \int_0^x \frac{t^n}{n!}e^{-t}\:dt\right | \leq \frac{|x|^{n+1}}{n!} \to 0$$

$\endgroup$
3
  • $\begingroup$ How did you get the last inequality? $\endgroup$ May 12 '20 at 11:36
  • $\begingroup$ @AnInvisibleCarrot $$\int_a^b f(x)\:dx \leq \sup_{x\in[a,b]} f(x) \cdot (b-a)$$ $\endgroup$ May 12 '20 at 11:41
  • $\begingroup$ So wouldn't $|\frac{I_n}{n!}| \leq \frac{ |x|^{n+1} }{n!} \times \max \{ 1 , e^{-x} \} \leq \frac{ |x|^{n+1} e^{|x|}}{n!}$. Your logic does follow though with what I need to do - thank you! $\endgroup$ May 12 '20 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.