1
$\begingroup$

Could you please give me some hint on how to show (if possible) that the following inequality holds?

$$\bigg|(-1)^m\frac{t^m x^m}{m!}+\sum_{j=m+1}^{\infty}\frac{(-1)^jx^jt^j}{j!}\bigg|\leq \bigg|(-1)^m\frac{t^m x^m}{m!}\bigg|$$

(where $x>0$)

notice that the left side is nothing more than

$$\bigg|e^{-tx}-\sum_{k=0}^{m-1}\frac{(-1)^jx^jt^j}{j!}\bigg|$$

Any hint or advice will be really appreciated.

$\endgroup$
1
$\begingroup$

For $y\geqslant 0$ we have $1\geqslant e^{-y}=\sum_{k=0}^{\infty}(-1)^k y^k/k!$; integrating this $m$ times, we get $$\frac{y^m}{m!}\geqslant\sum_{k=0}^{\infty}\frac{(-1)^k y^{k+m}}{(k+m)!}\underset{k+m=j}{=}(-1)^m\sum_{j=m}^{\infty}(-1)^j\frac{y^j}{j!}.$$

$\endgroup$
0
$\begingroup$

There exist useful estimates for the remainder in Taylor's expansion and that's precisely what you need. Here's your hint: https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder

$\endgroup$

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.