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Let $g(x) = e^{-1/x^2}$ for $x$ not equal to zero, and $g(0) = 0$.

a) Please Show that $g^{(n)}(0) = 0$, for all $n = 0,1,2,3,4, \ldots$

Can someone please elaborate on the comments below for this one?

b) Please Show that the Taylor Series for $g$ about 0 agrees with $g$ only at $x = 0$.

I think this would be easy once I have part a, all I have to do is plug in n = 0?

Can someone please show how to do this?

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    In a), do you mean "..., 8, 16, ...", or did you forget the 3? A more usual notation for the $n$-th derivative of $g$ is $g^{(n)}$ (with parentheses around the order). About the question itself: Once you have a), then b) is almost immediate, since the value of the Taylor series everyhwere is immediate from a) and it's a basic property of the exponential function that it can't take that value. For a), have you tried induction?2011-04-08
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    @joriki, The missing parenthesis is my fault. They were present in the original and I forgot to put them back in changing to LateX.2011-04-08
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    @joriki, I forgot the 3, sorry. For a, I have tried induction, but I messed up on the setup.2011-04-08
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    @ Pete, Also, before you say "which is a very standard sort of limit" below, I don't know how to simplify it, the professor skipped l'hopitals rules and such2011-04-08
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    @knucklebumpler: In that case, it's actually my own fault, since I approved your edit and didn't notice :-)2011-04-08
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    @user8917: I added the $3$; usually, if someone points out an error in your question (or answer), it's good style to correct it so that the question can be understood without reading through all the comments.2011-04-08
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    @GEdgar, can you please write out a solution, I am not fully sure I get it2011-04-09

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