Suppose you are looking for a function. All you know about the function is that

Question

Suppose you are looking for a function. All you know about the function is that it's infinitely differentiable everywhere, always positive, f(0)=1 and f '(x)=2xf(x) for all x.

a. Use only properties of power series to find a power series representation of for f(x). Do not use any other functions you know. Give your power series in sigma notation form.

b. now find a short formula for f(x) in a different way - solve the initial value problem given above.

c. After completing part (a) and (b), explain why these two representations of f(x) agree.

Explanation / Answer

a) take f(x)= sum(n>=0) a(n) x^n , then f'(x)=sum(n>=1)a(n)nx^(n-1)

so 2xf(x)=f'(x) gives : sum (n>=0)2 a(n) x^(n+1) - sum(n>=1) na(n)x^(n-1) = 0

which can be rewritten in :

-a(1) + sum (n>=1) (2a(n-1)-(n+1)a(n+1)) x^n = 0

so a(1)=0 and a(n+1)=2a(n-1)/(n+1) for n>=1, then a(2n+1)=0 and a(2n+2)=a(2n)/(n+1), which leads by immediate induction to a(2n)=1/n! since a(0)=1

So f(x) = sum (n>=0) x^(2n)/n!

b)

f'(x) = 2x f(x) => f'(x)/f(x) = 2x => ln(f(x))=x^2+const=> f(x)=e^(x^2) since f(0)=1

c)

By replacing x by x^2 in the Taylor expansion of e^x=sum(n>=0) x^n/n!.

we see the 2 solutions are the same, and moreover the radius of convergence of the series is infinite, so it is valid everywhere.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.