- #1

transgalactic

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prove that this function differentiable endles times on x=0 ?? http://img502.imageshack.us/img502/6778/83126617mm0.th.gif

i was told

"once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable"

differentiable around zero part:

[tex]

f'(0)=\lim_{x->0}\frac{e^\frac{-1}{x^2}-0}{x-0} \\

=\lim_{x->0}\frac x{2e^{1/x^2}}

[/tex]

i know that exponentials grow faster then polinomials but the power of the exp is 1/x^2

so the denominator goes to infinity but faster then the numenator so the expression

goes to 0.

power series around zero part:

the power series of e^x

[tex]

g(x)=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^3)\\

[/tex]

i put -1/x^2 instead of x

[tex]

g(x)=e^x=1+\frac{-1}{x^2}+\frac{x^{4}}{2}+\frac{-x^6}{6}+O(-x^6)\\

[/tex]

what now??

i was told

"once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable"

differentiable around zero part:

[tex]

f'(0)=\lim_{x->0}\frac{e^\frac{-1}{x^2}-0}{x-0} \\

=\lim_{x->0}\frac x{2e^{1/x^2}}

[/tex]

i know that exponentials grow faster then polinomials but the power of the exp is 1/x^2

so the denominator goes to infinity but faster then the numenator so the expression

goes to 0.

power series around zero part:

the power series of e^x

[tex]

g(x)=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^3)\\

[/tex]

i put -1/x^2 instead of x

[tex]

g(x)=e^x=1+\frac{-1}{x^2}+\frac{x^{4}}{2}+\frac{-x^6}{6}+O(-x^6)\\

[/tex]

what now??

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