3
$\begingroup$

I wish to show that $$ \lim_{n\to \infty} \left(2e^{\frac{it}{\sqrt{n}}} - e^{\frac{2it}{\sqrt{n}}}\right)^n \to e^{t^2}$$

Taylor expanding the inner expression yields $$ 1 + \frac{2-2^2}{2!}\left(\frac{it}{\sqrt{n}}\right)^2 + \cdots + \frac{2-2^k}{k!}\left(\frac{it}{\sqrt{n}}\right)^k + \cdots $$ however I'm not sure this is the way to go? I don't really know how to attack this one. Hints are most appreciated!

$\endgroup$
  • 1
    $\begingroup$ Is $i$ the complex number? Or some constant? $\endgroup$ – Clayton Oct 8 '15 at 12:01
  • $\begingroup$ $i$ is the imaginary unit yes $\endgroup$ – Lionel Ricci Oct 8 '15 at 12:03
  • 1
    $\begingroup$ You've already done the most important step: the inner expression is $ = 1 + t^2/n + o(1/n)$. $\endgroup$ – zhoraster Oct 8 '15 at 12:12
2
$\begingroup$

$$\lim_{n\to \infty} \left(2e^{it/\sqrt{n}} - e^{2it/\sqrt{n}}\right)^n=\lim_{n\to \infty} \left(2\left(\cos\frac{it}{\sqrt{n}}+\sin\frac{it}{\sqrt{n}}\right)-\left(\cos\frac{2it}{\sqrt{n}}+\sin\frac{2it}{\sqrt{n}}\right)\right)^n\\\sim\lim_{n\to\infty}\left(1-\left(\frac{it}{\sqrt{n}}\right)^2-\left(\frac{it}{\sqrt{n}}\right)^3-\frac7{12}\left(\frac{it}{\sqrt{n}}\right)^4...\right)^n=\lim_{n\to\infty}\left(1+\frac{t^2}n+...\right)^n=e^{t^2}\\{\rm as}\;\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x$$

$\endgroup$

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.