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Let $i$ be the complex number $i^2 = -1$. I want to prove that $$ \Biggl| e^{i x} - \sum_{j = 0}^N \frac{(i x)^j}{j!} \Biggr|\leq \frac{ | x |^{N+1} }{(N+1)!} $$ hold for all $x \in \mathbb{R}$. This should follow from some form of Taylor's theorem.

I don't think I can apply the real Taylor's theorem, because it's a complex function, but I also can not apply complex Taylor theorem because I am only considering $x \in \mathbb{R}$, instead of an open subset of $\mathbb{C}$. How can I deduce this? Thank you!

Johnny T.
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    Use induction, integrating the inequality at each step to go to the next step. Can you prove this for $N=0$? – Kavi Rama Murthy Dec 26 '23 at 12:04
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    You can write $e^{ix} = cos(x) + isin(x)$, and split the series to real (even $j$'s) and complex (odd $j$'s), and use the triangular inequality. After that you can apply the real Taylor's theorem for each of the terms. – Amit Dec 26 '23 at 13:20
  • @geetha290krm I am not seeing how to prove it for $N=0$... How do you deal with the case $|x| \geq 1$? – Johnny T. Dec 26 '23 at 15:35
  • I am not posting an answer because this question has been answered before on MSE. – Kavi Rama Murthy Dec 26 '23 at 23:13
  • The answer can be found here: https://math.stackexchange.com/questions/2718800/remainder-term-in-taylor-series-for-the-complex-exponential-map – Johnny T. Dec 31 '23 at 16:57

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The Taylor formula with the integral form tells you that the remainder is $R_n(x)=\int _0^x f^{[N+1]} (t)(x-t)^N {dt \over N!}$. In your case, $\vert f^{[N+1]}(t) \vert =1$ your result follows.

Thomas
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