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[irunshow]

Solved Thanks guys

 

[Sethric]

(a) Your first four terms are right, but the nth term is not. Your f^n(a) will change that. And it is important to have that part right, because you need it for:
(b) Once you have your general formula for the Taylor at x = 2, you need to determine for what values of x it will converge. Remember back to the days of convergent series, when you applied the ratio/root/etc test to determine convergent. One of those will be quite useful here. And the requirements for that test will give you a requirement on x, essentially a radius and interval of convergence. Try that out and let us know what you get. We can help you more once you get there.

 

[irunshow]

Thanks Sethric

I got the nth to = (-1)^n (1/n!)(x-2)^n. is this one correct?

Then i used the ratio test to get this lim n->∞ = abs((-x+2)/(n+1)) so R = ∞ and I = ∞

Is this correct?
Thanks

 

[Sethric]

I got the nth to = (-1)^n (1/n!)(x-2)^n. is this one correct?

Close, but not quite. You can plug in n=2 to see that it does not match up with your series. I recommend looking at the pattern formed by the f^n(a) part of the Taylor series. What is f^n(a) for a particular n?

 

[irunshow]

Actually,

How do I find abs(an +1/an)? Like I am not sure how to find the an term to fit into the series to check for radius of convergenceEdit:

nth to = (-1)^n (1/2^(n+1))(x-2)^n

 

[Sethric]

Also the ratio test requires that the ratio of the absolute value of your n+1-term to your n-term be less than 1. That's where the inequality is coming from.

 

[Sethric]

Ah, the edit.

Edit:

nth to = (-1)^n (1/2^(n+1))(x-2)^n

This is correct. Now apply the ratio test to this.

 

[irunshow]

I got abs(-(x-2)) < 2

So R = 2
And I = (-4,0)Is that abs(-(x-2)) correct? and I use the function centered at -2? or is it +2? If it is +2, then I= (-2,2).

Thanks

EDIT:

Nvm I got
I= (-2,2)
R=2

 

[Sethric]

The absolute value drops the negative sign. So |x-2| < 2. This gives an interval centered at x = 2 with a radius of 2.

 

[irunshow]

Thanks sethric,

I did f(x) = -1/x^2

and i got:

(-1/4) + (1/4)(x-2) - (3/16)(x-2)^2 + (1/8)(x-2)^3 +...+

But I can't seem to find the nth pattern. Is there a formula I can plug into find it?

 

[Char. Limit]

Just remember that the derivative of the nth term of your series for 1/x will be the nth term for your series -1/x^2.

 

[irunshow]

Ic okay thanks Char.limit I will try to work it out and post the answer on there

Edit: so if f(x) = -1/x^2

Then the series representation will have (f(x))' somewhere in it?

 

[Sethric]

Hrmm, is it -1/x^2, or 1/x^2. You have one in the first post, one in the last.

Either way, you know the formula comes from the $$\frac{f^n(a)}{n!}$$ part. Try writing out the each term without actually multiplying through, and without reducing fractions. What are your first four coefficients that way?

edit - Yeah, using char limit's idea is better, saving you a lot of work.

 

[irunshow]

I see. Thanks but I don't really understand char's limit idea. Can someone explain please?

it is -1/x^2 btw

 

[Sethric]

What Char limit is saying is that

$$\frac{-1}{x^2} = \frac{d}{dx} \frac{1}{x}$$

Since both of your series are centered at 2, you can just take a derivative of the first series to find the second. It would make the coefficients easier to find.

 

[irunshow]

I got the first 4 quotients to be:

(-1/4) , (1/4), -(6/32), (24/192), -(120/1536)

I noticed the denominators all divisible by 4.

Also can anyone solve this pattern 1,2,6,24,120?

 

[irunshow]

Ic okay thanks Sethric.

Can I take the dervative of the representations of the series for 1/x to find out what the represenation of the series fo -1/x^2 is?

 

[Char. Limit]

Ic okay thanks Sethric.

Can I take the dervative of the representations of the series for 1/x to find out what the represenation of the series fo -1/x^2 is?

Given, as I believe you have proven, that:

$$\frac{1}{x} = \sum_{n=0}^\infty \frac{\left(-1\right)^n \left(x-2\right)^n}{2^{n+1}}$$

Then the derivative of 1/x is:

$$-\frac{1}{x^2} = \frac{d}{dx} \sum_{n=0}^\infty \frac{\left(-1\right)^n \left(x-2\right)^n}{2^{n+1}}$$

Then just apply the sum of derivatives rule. (Hint: The derivative of a sum is the sum of the derivatives)

 

[Sethric]

*edit* Ooh, sniped, and with much nicer TeX

 

[irunshow]

Guys I got the representation to =

(-1)^n+1 (x-2)^n (1/2^(n+2)) (n+1)

Is this correct?

But I didnt use the sum of the derviates rule. I just added n+1 to first representation and multiplied it by n+1 as well

 

[Char. Limit]

Guys I got the representation to =

(-1)^n+1 (x-2)^n (1/2^(n+2)) (n+1)

Is this correct?

But I didnt use the sum of the derviates rule. I just added n+1 to first representation and multiplied it by n+1 as well

All right, that works. (And you actually did, by assuming that this is true:

$$\frac{d}{dx} \sum_{n=0}^\infty f(x,n) = \sum_{n=0}^\infty \frac{d}{dx} f(x,n)$$

Now, you have a series for -1/x^2. To make it for 1/x^2 you just need to adjust the exponent on your (-1)^(n+1) value.

 

[irunshow]

Thanks guys =)

But how does the sum of the derivates work?

do I do d/dx (-1)^n + d/dx (x-2)^n divided by d/dx (2^(n+1))? and add them together?

Don't I get something like n (-1)^(n-1) for the first term? how does that change into the correct represataion?

 

[Char. Limit]

Thanks guys =)

But how does the sum of the derivates work?

do I do d/dx (-1)^n + d/dx (x-2)^n divided by d/dx (2^(n+1))? and add them together?

Don't I get something like n (-1)^(n-1) for the first term? how does that change into the correct represataion?

No no, what you had was correct for -1/x^2. To make it 1/x^2, all you had to do was adjust your (-1)^(n+1) index to either (-1)^n or (-1)^(n+2).

 

[irunshow]

Thanks char.

I got the radius of convergence to =

lim n-> ∞ of abs( (x-2)(n+2) / (n+1) ) < 2lim n-> ∞ of (n+2)/(n+1) = 1
so

R= 2
I = (-2/2)

Just as a general rule, does the derivative of a series have the same R and I if a is the same?Thanks everyone for the help