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How to show the expansions \begin{gather*} e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}\\ e^{iy} = 1 + iy - \frac{1}{2}y^2 + \frac{\mu_2 |y|^3}{3!} \end{gather*} where $y \in \mathbb{R}$ and $|\mu_1| \leq 1$ and $|\mu_2| \leq 1$ for all $y$?

I attempts to consider the remainder of Taylor expansion of one real-variable function: $e^{iy} = \cos y + i\sin y$, and their Taylor expansions with Lagrange remainders are \begin{gather*} \cos y = 1 - \frac{y^2}{2}\cos (\theta_1 y) y^2\\ \sin y = y - \frac{1}{2}\sin (\theta_2 y) y^2 \end{gather*} So we may put $\mu_1 = -[\cos (\theta_1 y) + i \sin (\theta_2 y)]$. But here we only have $|\mu_1| \leq 2$, which does not satisfy the requirement...

Any help will be appreciated!

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    $\begingroup$ Sorry but every cosine plus i times sine of a real number has modulus at most 1. $\endgroup$ – Did Jul 31 '14 at 9:25
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    $\begingroup$ @Did: $\cos(0) + i \sin(\pi/2) = 1+i$ has modulus $\sqrt{2}$... The two arguments of $\cos$ and $\sin$ are not the same here! $\endgroup$ – Najib Idrissi Jul 31 '14 at 10:14
  • $\begingroup$ @NajibIdrissi Good catch (stupid me missed that the two arguments could be different, probably because I had in mind the correct approach, working with exp(iy)...). Thanks. $\endgroup$ – Did Jul 31 '14 at 10:56
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Taylor's theorem with integral remainder says that $$f(x)=j_a^nf\>(x)+R_n(x),\qquad R_n(x)=\int_a^xf^{n+1}(t){(x-t)^n\over n!}\ dt\ .$$ Applying this to $f(x):=e^{ix}$ with $a=0$ we find that $$e^{ix}=\sum_{k=0}^n {(ix)^k\over k!}+R_n(x)$$ with $$R_n(x)={i^{n+1}\over n!}\int_0^x e^{it}(x-t)^n\ dt\ .$$ From this we immediately obtain the estimate $$\bigl|R_n(x)\bigr|\leq {|x|^{n+1}\over (n+1)!}\qquad(x\in{\mathbb R})\ .$$

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You made a mistake in your second expansion. The correct one is \begin{align*} \cos y = 1 - \cos(\theta_1y)\frac{y^2}{2}, && \sin y = y - \sin(\theta_2 y)\frac{y^3}{6} \end{align*} then \begin{align*} e^{i y} &= \cos y + i \sin y \\ & = 1 + i y - \cos(\theta_1y)\frac{y^2}{2}- i\sin(\theta_2 y)\frac{y^3}{6} \end{align*} which is closer to your answer ;)

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  • $\begingroup$ @math major's expansion for $\sin$ is correct; he is using $k=1$ instead of $k=2,$ which is what you are using (following the convention at Wikipedia->Taylor's theorem->Explicit formulas for the remainder->"Lagrange form of the remainder": en.wikipedia.org/wiki/…). $\endgroup$ – xFioraMstr18 Jan 31 at 21:14

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