Interval of Convergence: Series f(x)

[rjs123]

Homework Statement

A function f is defined by...

f(x) = \frac{n+1}{3^{n+1}} x^n <br />

a.) find the interval of convergence of the given power series.

b.) Find \lim_{x\rightarrow 0} \frac{f(x) - \frac{1}{3}}{x}

c.) Write the first three nonzero terms and the general term for an infinite series that represents

\int_{0}^{1} f(x)dx

d.) Find the sum of the series determined in part c.

The Attempt at a Solution

a.) i want to use the ratio test, but get stuck...i end up with...\lim_{x\rightarrow \infty} \frac{nx + 2x}{3n+3}

 

[Apphysicist]

Right, so if you rewrite that so it looks a little nicer:

You have the limit AS N GOES TO INFINITY of the absolute value of x*(n+2)/(3n+3).

That's the key here: you're taking the limit for n, seeing what the "final" term looks like compared to the one before it (remember to converge, the terms must tend to zero, thus the ratio of the next term to the one before it must be less than one).

When you take that limit, you have that the absolute value of (x/3) is less than one. So x has to be between -3 and 3, and then you must also check 3 (for if x=3, the ratio is 1, and the ratio test is inconclusive). Do so by plugging 3 in for x in the original series and checking by your other tests to see if it converges or diverges.

 

[rjs123]

Right, so if you rewrite that so it looks a little nicer:

You have the limit AS N GOES TO INFINITY of the absolute value of x*(n+2)/(3n+3).

That's the key here: you're taking the limit for n, seeing what the "final" term looks like compared to the one before it (remember to converge, the terms must tend to zero, thus the ratio of the next term to the one before it must be less than one).

When you take that limit, you have that the absolute value of (x/3) is less than one. So x has to be between -3 and 3, and then you must also check 3 (for if x=3, the ratio is 1, and the ratio test is inconclusive). Do so by plugging 3 in for x in the original series and checking by your other tests to see if it converges or diverges.

i accidentally had as x goes to infinity...not n. Thanks for that catch.

For b, should i plug in x/3 for f(x)? if so i would get 0 as the limit...or should i plug in the whole original f(x) function...which is messy...not sure how to start b.

 

[Apphysicist]

As far as I can tell, (a) refers to f(x) as a series (thus a sum), but since f(x) is not itself a summation, it is more a sequence, a function for discrete values of n (if you plotted f(x) v. n for a fixed value of x, it'd look like a discontinuous function...a bunch of points).

When x=0, the value f(x) is 0 for any n. (0^n=0) You have a 0/0 case, so I would use L'Hopital's rule to take the limit, so you'd get something like:

lim(x-->0) d/dx(f(x)+1/3) divided by d/dx(x)
lim(x-->0) (f ' (x) + 0) divided by 1.

Which seems to me to be 0 because of power rule. (all that n-stuff * n * x^(n-1))

I could have the wrong intuition on that.

 

[Dick]

As far as I can tell, (a) refers to f(x) as a series (thus a sum), but since f(x) is not itself a summation, it is more a sequence, a function for discrete values of n (if you plotted f(x) v. n for a fixed value of x, it'd look like a discontinuous function...a bunch of points).

When x=0, the value f(x) is 0 for any n. (0^n=0) You have a 0/0 case, so I would use L'Hopital's rule to take the limit, so you'd get something like:

lim(x-->0) d/dx(f(x)+1/3) divided by d/dx(x)
lim(x-->0) (f ' (x) + 0) divided by 1.

Which seems to me to be 0 because of power rule. (all that n-stuff * n * x^(n-1))

I could have the wrong intuition on that.

Wrong intuition. The series for f(x) is 1/3+(2/9)*x+(3/27)*x^2+... What's f'(0)? And f(x) only makes sense in the context of these questions if it's supposed to be a series sum. Otherwise how to interpret 'radius of convergence' and 'ratio test'?

 

[rjs123]

Wrong intuition. The series for f(x) is 1/3+(2/9)*x+(3/27)*x^2+... What's f'(0)? And f(x) only makes sense in the context of these questions if it's supposed to be a series sum. Otherwise how to interpret 'radius of convergence' and 'ratio test'?

yes, when you L'HOP...\lim_{x\rightarrow 0} \frac{f(x) - \frac{1}{3}}{x}

it becomes...\lim_{x\rightarrow 0} \frac{f&#039;(x) - 0}{1}

this is where i am stuck...no idea how to take derivative of f(x) for this situation...

 

[Dick]

yes, when you L'HOP...\lim_{x\rightarrow 0} \frac{f(x) - \frac{1}{3}}{x}

it becomes...\lim_{x\rightarrow 0} \frac{f&#039;(x) - 0}{1}

this is where i am stuck...no idea how to take derivative of f(x) for this situation...

Would you say what f(x) actually is? Is it the sum of x^n*(n+1)/3^(n+1) for n from 0 to infinity? I'm guessing it is, but you didn't explicitly indicate any summation in the original post. If so, differentiate the series term by term to get the derivative.

 

[rjs123]

Would you say what f(x) actually is? Is it the sum of x^n*(n+1)/3^(n+1) for n from 0 to infinity? I'm guessing it is, but you didn't explicitly indicate any summation in the original post. If so, differentiate the series term by term to get the derivative.

yes, it is the sum for n from 0 to infinity. The domain of convergence is -3<x<3...i have never taken the derivative of a power series yet so I am unsure of how to do this...is it basically subtracting 1 wherever an n is present?

 

[Dick]

The derivative of a*x^k is a*k*x^(k-1). Every term in the power series has that form, right? Just differentiate every term. Explicitly write three or four terms of the series out to make sure you know what you are doing.

 

[rjs123]

The derivative of a*x^k is a*k*x^(k-1). Every term in the power series has that form, right? Just differentiate every term. Explicitly write three or four terms of the series out to make sure you know what you are doing.

f(x) = \frac{n+1}{3^{n+1}} x^n

then f'(x) would result in the sum of n from 0 to infinity \frac {n+1}{3^{n+1}} * n * x^{n-1}


f'(0) = 0...answer to b would be the limit = 0

this is probably way off though

 

[Dick]

f(x) = \frac{n+1}{3^{n+1}} x^n

then f'(x) would result in the sum of n from 0 to infinity \frac {n+1}{3^{n+1}} * n * x^{n-1}


f'(0) = 0...answer to b would be the limit = 0

this is probably way off though

We are going in circles here. Write out the n=0, n=1 and n=2 terms in the series. Please?

 

[rjs123]

We are going in circles here. Write out the n=0, n=1 and n=2 terms in the series. Please?

when n = 0...\frac{1}{3}

when n = 1...\frac{2}{3^2}x

when n = 2...\frac{3}{3^3}x^2

when n = 3...\frac{4}{3^4}x^3

ok

 

[Dick]

when n = 0...\frac{1}{3}

when n = 1...\frac{2}{3^2}x

when n = 2...\frac{3}{3^3}x^2

when n = 3...\frac{4}{3^4}x^3

ok

Ok! So what is f'(0)? It isn't 0.

 

[rjs123]

Ok! So what is f'(0)? It isn't 0.

it appears \frac{2}{9} would be the answer if i simply added the derivatives of those above terms correct? Everything would equate to 0 except for the 2nd term.

The limit would become -1/9

 

[Dick]

it appears \frac{2}{9} would be the answer if i simply added the derivatives of those above terms correct? Everything would equate to 0 except for the 2nd term.

The limit would become -1/9

Yes, f'(0)=2/9. But how does that show the limit of (f(0)-1/3)/x is -1/9?

 

[rjs123]

Yes, f'(0)=2/9. But how does that show the limit of (f(0)-1/3)/x is -1/9?

i got mixed up...the limit as x approaches 0 is simply 2/9.

For d...the last question...i just use a1/(1-r)...since it is a geometric series...i end up getting 1/2 as a sum.

 

[Dick]

i got mixed up...the limit as x approaches 0 is simply 2/9.

For d...the last question...i just use a1/(1-r)...since it is a geometric series...i end up getting 1/2 as a sum.

I wish you'd show more of your work and not just the answer, but I think 1/2 is right.

 

[rjs123]

For a...finding the interval of convergence... i get here...

\lim_{n\rightarrow \infty} \frac{(n+2)(x)}{(3)(n+1)}

then...should i distribute the top and bottom..and then L'HOP...or just L'HOP right away?

if i distibute i would get nx + 2x/(3n + 3)...then L'HOP would get me 3x/3...or x

the 2nd poster said it was x/3...so I am not sure.

 

[Dick]

For a...finding the interval of convergence... i get here...

\lim_{n\rightarrow \infty} \frac{(n+2)(x)}{(3)(n+1)}

then...should i distribute the top and bottom..and then L'HOP...or just L'HOP right away?

if i distibute i would get nx + 2x/(3n + 3)...then L'HOP would get me 3x/3...or x

the 2nd poster said it was x/3...so I am not sure.

If you are going to use l'Hopital to find that limit you differentiate with respect to n, NOT x. The 2x part is a constant, it's derivative is 0.

 

[jhae2.718]

For a...finding the interval of convergence... i get here...

\lim_{n\rightarrow \infty} \frac{(n+2)(x)}{(3)(n+1)}

then...should i distribute the top and bottom..and then L'HOP...or just L'HOP right away?

if i distibute i would get nx + 2x/(3n + 3)...then L'HOP would get me 3x/3...or x

the 2nd poster said it was x/3...so I am not sure.

I would pull out the constant x/3 from the limit, and then divide through the (n+2)/(n+1) by n, and then take the limit...

 

[rjs123]

i end up getting -3<x<3

I still have to check the endpoints...I just plug -3 and 3 into x for the original equation...but then i get stuck

 

[Dick]

What do you get when you plug x=3 and x=(-3) into the original series?

 

[rjs123]

What do you get when you plug x=3 and x=(-3) into the original series?

i ended up getting that they both diverge...i used the nth term test.

 

[Dick]

i ended up getting that they both diverge...i used the nth term test.

That would be the correct conclusion.