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 November 7th, 2011, 03:35 AM #1 Newbie   Joined: Jul 2009 Posts: 9 Thanks: 0 a rectifiable path Let f(t) = exp( (-1+i )(t^-1) ) for 0< t <=1 and f(0)=0 show that f is a rectifiable path . " sorry for not using latex "
 November 7th, 2011, 04:19 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: a rectifiable path [color=#000000]Compute, $\int_{0}^{1}\begin{Vmatrix}f'(t)\end{Vmatrix}\ ;dt=\frac{\sqrt{2}}{e}<+\infty$ .[/color]
 November 8th, 2011, 10:51 AM #3 Newbie   Joined: Jul 2009 Posts: 9 Thanks: 0 Re: a rectifiable path Thank you Zardoz for your answer , ^__^ But how can i show that f(t) is of bounded variation
 November 8th, 2011, 02:27 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: a rectifiable path [color=#000000]You have, $f(t)=e^{\frac{-1+i}{t}}\Rightarrow f#39;(t)=\frac{(1-i)e^{\frac{i-1}{t}}}{t^2}$ so $\begin{Vmatrix}f'(x)\end{Vmatrix}=\begin{Vmatr ix}\frac{(1-i)e^{\frac{i-1}{t}}}{t^2}\end{Vmatrix}=\begin{Vmatrix}1-i\end{Vmatrix}\begin{Vmatrix}\frac{e^{\frac{i-1}{t}}}{t^2}\end{Vmatrix}=\sqrt{2}\frac{e^{-\frac{1}{t}}}{||t||^2}$ $\int_{0}^{1}\begin{Vmatrix}f'(x)\end{Vmatrix}\ ;dx=\sqrt{2}\int_0^1\left(e^{-\frac{1}{t}}\right)#39;\;dt=\frac{\sqrt{2}}{e}<\i nfty$ .[/color]

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