Remainder term in Taylor series for the complex exponential map

Question

The statement is the following: Let $P_n(z) = 1 + z + \frac{z^2}{2!} + ... + \frac{z^n}{n!}$ be the $n$ first terms of the Taylor series for $e^z$ centered at the origin. Then, prove that $\forall n \in \mathbb{N}$, and $\forall z \in \mathbb{C} : Re(z)< 0$, this inequality holds:

$|e^z - P_n(z)|<|z|^{n+1}$

I am trying to use the integral formula for the remainder:

https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_in_complex_analysis

And then the fact that in the half plane with $Re(z) < 0$, the exponential map has modulus less than one. But this bound is maybe too coarse, because I cannot reach the conclusion. Any help will be much appreciated.

Answer 1

Solution: So I think I finally cracked and it goes like this. I used the result from https://math.stackexchange.com/a/2719893/527701 which was very useful. Let $$P_n(z) = \sum_{k=0}^{n} \frac{z^k}{k!}$$ be the truncation taylor approximation of $exp(z)$ and we are interested in calculating the truncation error i.e. $|e^z-P_n(z)|$. We cannot make use of taylor theorem for reals i.e. the remainder term of the form $R_n(z) = \frac{f^{(n+1)}(\xi)z^{n+1}}{(n+1)!} $ where $\xi$ is between 0 and z because this theorem doesn't hold for complex (as far as I know after my research). This is because in complex analysis, Analytical function are equivalent to Holomorphic function. I.e. if a function is analytic, then in-fact it is infinite time complex differentiable. So I wasn't sure if I can say that there is a radius $r$ such that for $\xi\in B(0,r)$, the above bound holds where $B(0,r)$ is an open ball centred at zero with radius $r$. Moving onto proving this statement.

Proof by Induction:

1. Take $n=0$ and the convention that $P_{-1}(z) = 0$, then using the result that $$e^z - P_n(z) = z\int_0^1 e^{tz} - P_{n-1}(tz) \ dt$$ we can show that $|e^z-1| \leq |z|\int_0^1|e^{tz}|dt = |z|\int_0^1|e^{tRe(z)}|dt \leq |z|e^{Re(z)}\int_0^1dt = |z|e^{Re(z)} \leq |z| $ using the fact that exponential is an increasing function and $Re(z) <0$.

2. Assume this holds for all $n$ i.e. $|e^z - P_{n-1}(z)| \leq \frac{|z|^n}{n!}$ and let us now consider $|e^z - P_{n}(z)|$. $$|e^z - P_n(z)| \leq |z|\int_0^1 \frac{t^{n}|z|^n}{n!}dt = \frac{|z|^{n+1}}{(n+1)!} \leq |z|^{n+1}$$ since $n\in\mathbb{N}$. $\quad \square$

I hope this proof is useful for others. Also please let me know if you spot any mistakes or inconsistency.

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Answer 2

Use $z\int_0^{1} \{e^{tz} -P_{n-1}(tz)\}dt =e^{z} -P_{n}(z)$. A simple induction argument gives the desired inequality. [To begin with $|\int_0^{1}e^{tz}dt| \leq 1$ for $\Re (z) <0$ so $|e^{z} -1| \leq |z|$

Answer 3

Great to see this question already answered! I was about to post a question / comment on exactly the same topic. Silly me, thought my question was very specific, and here it is, beautifully answered already, could've saved myself time thinking about it! I am coming from the physics side and was only vaguely aware of all the maths that deals with remainders of power series, having forgotten most of it.

My solution is as follows. We are talking about the Taylor series for the exponential, $$ \exp(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}, $$ and I am interested in estimating the remainder (using the original notation) $e^z-P_n(z)$ $$ e^z-P_n(z) = \sum_{k=n+1}^{\infty}\frac{z^{k}}{k!}. $$ It is not difficult to verify that the remainder is given by $$ e^{z}-P_{n-1}(z)=\frac{z^{n}}{\left(n-1\right)!}\int_{0}^{\infty}\frac{\exp\left(\frac{zt}{1+t}\right)}{(1+t)^{n+1}}dt $$ for $n\geq1$. For example, one can verify by integrating by parts that $$ \int_{0}^{\infty}\frac{\exp\left(\frac{zt}{1+t}\right)}{(1+t)^{n+1}}dt=\frac{1}{n}+\frac{z}{n}\int_{0}^{\infty}\frac{\exp\left(\frac{zt}{1+t}\right)}{(1+t)^{n+2}}dt, $$ where $n\geq1$. Then, by successively integrating by parts starting with $$ \frac{z^{n}}{\left(n-1\right)!}\int_{0}^{\infty}\frac{\exp\left(\frac{zt}{1+t}\right)}{(1+t)^{n+1}}dt, $$ one recovers the complete power series expansion for $e^{z}-P_{n-1}(z)$.

Now, for $\mathrm{Re}(z)<0$ can write $$ \left|\frac{z^{n}}{\left(n-1\right)!}\int_{0}^{\infty}\frac{\exp\left(\frac{zt}{1+t}\right)}{(1+t)^{n+1}}dt\right|\leq\left|\frac{z^{n}}{\left(n-1\right)!}\right|\int_{0}^{\infty}\frac{\left|\exp\left(\frac{zt}{1+t}\right)\right|}{(1+t)^{n+1}}dt\leq\left|\frac{z^{n}}{\left(n-1\right)!}\right|\int_{0}^{\infty}\frac{dt}{(1+t)^{n+1}}=\frac{\left|z^{n}\right|}{n!}. $$ Of course, the expression for the remainder holds for any $z$, but for $\mathrm{Re}(z)>0$ the exponential $\left|\exp\left(\frac{zt}{1+t}\right)\right|$ cannot be bounded by $1$ for all $0\leq t<\infty$.

By the way, I am sorry I had to present it like this - "It is not difficult...", "It is obvious..." - but I really just engineered this solution by hand, taking the Stieltjes asymptotic series as an example - see the Chapter on asymptotic series in "Advanced Mathematical Methods for Scientists and Engineers" by C.M. Bender and S.A. Orszag, McGraw-Hill (1978) for details.

Cheers!

P.S. For context, this result is quoted in books on quantum field theory when discussing how perturbative expansions (in field theory, but also in quantum mechanics in general) generate asymptotic series that most often do not converge, rather than convergent Taylor series with finite radii of convergence. For example, see "Condensed Matter Field Theory" by Altland and Simons or "Quantum Many-Particle Systems" by Negele and Orland. The following toy example is quoted as a typical ersatz-field-theoretic average: $$ I(g)=\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}\,\exp\left(-\frac{1}{2}x^{2}-gx^{4}\right). $$ Now, the perturbation theory approach would be to expand the integrand in powers of $g$, but such an expansion obviously has zero radius of convergence w.r.t. $g=0$ since the integral diverges for $g<0$. It is then stated then that we are dealing with an asymptotic series, so we should also be mindful of how many terms we wish to include in the expansion, since the error, given by the integral of the remainder of the Taylor series expansion of $\exp(-gx^{4})$), first decreases, and then increases by including more and more powers in the expansion. This is demonstrated by using the result quoted in this discussion, however none of the physics books justify it! - They simply pull it out of their magicians' hats!