JHammond88
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- Apr 23, 2008
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I'm having some considerable trouble solving the following problem, and I sincerely apologize for the ugliness of my formulas...HTML isn't as easy to make pretty as MS Equation:
f(x) = e^(-1/x^2) if x != 0
0 if x = 0
a) Use the definition of a derivative to compute f'(0).
b) Show that f has derivatives of all orders that are defined on RR. I have a hint for this one: I need to show by induction that there is a polynomial P[sub:2pqw4plx]n[/sub:2pqw4plx](x) and a nonnegative integer k[sub:2pqw4plx]n[/sub:2pqw4plx] such that f[sup:2pqw4plx](n)[/sup:2pqw4plx](x) = P[sub:2pqw4plx]n[/sub:2pqw4plx](x)f(x)/x[sup:2pqw4plx]k[sub:2pqw4plx]n[/sub:2pqw4plx][/sup:2pqw4plx] for x != 0. Here, f[sup:2pqw4plx](n)[/sup:2pqw4plx] represents the nth derivative of f.
c) Find a closed expression for P[sub:2pqw4plx]n[/sub:2pqw4plx](x). I have a hint for this one too. My professor is pretty sure that it can be solved by working with P[sub:2pqw4plx]n[/sub:2pqw4plx](x) = g(x)[(d[sup:2pqw4plx]n[/sup:2pqw4plx]/dx[sup:2pqw4plx]n[/sup:2pqw4plx])(h(x,n)] where g and h each represent an identity.
So far, for part (a), I have set y = 1/x[sup:2pqw4plx]2[/sup:2pqw4plx]. I can prove that the lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x[sup:2pqw4plx]2n[/sup:2pqw4plx] goes to 0:
lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x[sup:2pqw4plx]2n[/sup:2pqw4plx] = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]e[sup:2pqw4plx]-1/x^2[/sup:2pqw4plx]/[(x[sup:2pqw4plx]2[/sup:2pqw4plx])[sup:2pqw4plx]n[/sup:2pqw4plx]] = lim[sub:2pqw4plx](y->inf)[/sub:2pqw4plx]y[sup:2pqw4plx]n[/sup:2pqw4plx]/e[sup:2pqw4plx]y[/sup:2pqw4plx]. Using L'Hospital's rule, this = lim[sub:2pqw4plx](y->inf)[/sub:2pqw4plx]n!/e[sup:2pqw4plx]y[/sup:2pqw4plx]. From here, lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x[sup:2pqw4plx]n[/sup:2pqw4plx] = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]x[sup:2pqw4plx]n[/sup:2pqw4plx][f(x)/(x[sup:2pqw4plx]2n[/sup:2pqw4plx])] = 0. Thus, f'(0) = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx][f(x) - f(0)]/(x-0) = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x = 0. So much for (a)
For (b) and (c), however, I am clueless!! I know that the first derivative f'(x) = f(x)(2/x[sup:2pqw4plx]3[/sup:2pqw4plx]), but I don't know how to work the ensuing pattern into the form of f[sup:2pqw4plx](n)[/sup:2pqw4plx](x) = P[sub:2pqw4plx]n[/sub:2pqw4plx](x)f(x)/x[sup:2pqw4plx]k[sub:2pqw4plx]n[/sub:2pqw4plx][/sup:2pqw4plx]. And therefore, I have no way of getting P(n)
Does anyone see what I am missing? Thanks so much for any help!!
f(x) = e^(-1/x^2) if x != 0
0 if x = 0
a) Use the definition of a derivative to compute f'(0).
b) Show that f has derivatives of all orders that are defined on RR. I have a hint for this one: I need to show by induction that there is a polynomial P[sub:2pqw4plx]n[/sub:2pqw4plx](x) and a nonnegative integer k[sub:2pqw4plx]n[/sub:2pqw4plx] such that f[sup:2pqw4plx](n)[/sup:2pqw4plx](x) = P[sub:2pqw4plx]n[/sub:2pqw4plx](x)f(x)/x[sup:2pqw4plx]k[sub:2pqw4plx]n[/sub:2pqw4plx][/sup:2pqw4plx] for x != 0. Here, f[sup:2pqw4plx](n)[/sup:2pqw4plx] represents the nth derivative of f.
c) Find a closed expression for P[sub:2pqw4plx]n[/sub:2pqw4plx](x). I have a hint for this one too. My professor is pretty sure that it can be solved by working with P[sub:2pqw4plx]n[/sub:2pqw4plx](x) = g(x)[(d[sup:2pqw4plx]n[/sup:2pqw4plx]/dx[sup:2pqw4plx]n[/sup:2pqw4plx])(h(x,n)] where g and h each represent an identity.
So far, for part (a), I have set y = 1/x[sup:2pqw4plx]2[/sup:2pqw4plx]. I can prove that the lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x[sup:2pqw4plx]2n[/sup:2pqw4plx] goes to 0:
lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x[sup:2pqw4plx]2n[/sup:2pqw4plx] = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]e[sup:2pqw4plx]-1/x^2[/sup:2pqw4plx]/[(x[sup:2pqw4plx]2[/sup:2pqw4plx])[sup:2pqw4plx]n[/sup:2pqw4plx]] = lim[sub:2pqw4plx](y->inf)[/sub:2pqw4plx]y[sup:2pqw4plx]n[/sup:2pqw4plx]/e[sup:2pqw4plx]y[/sup:2pqw4plx]. Using L'Hospital's rule, this = lim[sub:2pqw4plx](y->inf)[/sub:2pqw4plx]n!/e[sup:2pqw4plx]y[/sup:2pqw4plx]. From here, lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x[sup:2pqw4plx]n[/sup:2pqw4plx] = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]x[sup:2pqw4plx]n[/sup:2pqw4plx][f(x)/(x[sup:2pqw4plx]2n[/sup:2pqw4plx])] = 0. Thus, f'(0) = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx][f(x) - f(0)]/(x-0) = lim[sub:2pqw4plx](x->0)[/sub:2pqw4plx]f(x)/x = 0. So much for (a)
For (b) and (c), however, I am clueless!!
I know that the first derivative f'(x) = f(x)(2/x[sup:2pqw4plx]3[/sup:2pqw4plx]), but I don't know how to work the ensuing pattern into the form of f[sup:2pqw4plx](n)[/sup:2pqw4plx](x) = P[sub:2pqw4plx]n[/sub:2pqw4plx](x)f(x)/x[sup:2pqw4plx]k[sub:2pqw4plx]n[/sub:2pqw4plx][/sup:2pqw4plx]. And therefore, I have no way of getting P(n) Does anyone see what I am missing? Thanks so much for any help!!