How do i show that this is a $C^{\infty}$ function that cannot be a 1-1 analytic function.
$$ \phi(t) = (g(-t),g(t)) $$
$$ g(t) = \begin{cases} 0 & t \leq 0\\ t e^{-1/t} & t > 0 \end{cases} $$
How do i show that this is $C^{\infty}$
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calculus
real-analysis
ordinary-differential-equations
multivariable-calculus
differential-geometry
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0This function is identically $0$ in the open set $(- \infty ,0)$. What can you conclude about its injectivity? What about its analiticity at $t=0$? – 2017-01-13
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0Sorry i know its not 1-1 there are infinite items that get mapped to the element 0 as for analiticity i had to look up what an analytic function was about 6 hours ago and i cant seem to understand why the Taylor series isnt equal to f(x) close to 0 – 2017-01-13
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0That's because the Taylor series should be $0$ at that point (all left derivatives are $0$!). However, the function is not $0$ on the right of $0$. – 2017-01-13
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0As an aside, you should probably learn some of the important properties of analytic functions (esp. the identity theorem), since otherwise you cannot really understand why the distinction between analytic functions and smooth functions is important. – 2017-01-13
1 Answers
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To show it's $C^\infty,$ all you need to do is show the derivative exists at zero (since it obviously exists elsewhere). For that, just differentiate the $t>0$ part and show that $\lim t\rightarrow 0^+$ is zero no matter how many derivatives you take. This follows since there's always that $e^{-1/t}$ factor, that kills any of the $1/t^p$ terms out front.
To show it's not analytic, you just use the same fact. All of the derivatives are zero at zero. This means that the Taylor series at zero is zero. However the function is nonzero, thus not equal to the taylor series.