Thread: Taylor Series

Hello,

I have this problem, that requires us to derive the taylor series based at x = -1 for f(x) from scratch, then find an interval of converge for the series, including endpoint behavior, and finally approximating:
ln((-1 + 2*Pi)/(Pi)) accurate to within 10^-25.

f(x) = ln(x + 2)

Okay, so this is what I have done.

I first found the first 6 derivatives to find a pattern. So they are as follows:

f'(x) = 1/(x+2)
f''(x) = -1/(x+2)^2
f^3(x) = 2/(x+2)^3
f^4(x) = -6(x+2)^4
f^5(x) = 24/(x+2)^5
f^6(x) = -120/(x+2)^6

And so forth.

So then I expanded it to:

1*(x+2)^(-1)
1*(-1)*(x+2)^(-2)
1*(-1)*(-2)*(x+2)^(-3)
1*(-1)*(-2)*(-3)*(x+3)^(-4)

And so forth.

Okay, so the coefficients are factorials. The coefficient for the nth derivative is just:

(+ or -)(n-1)!

Now, the nth derivative can be written as follows:

(+ or -)(n-1)!(x+2)^(-n)

The signs alternate, so I need a factor of (-1)^n or (-1)^(n+1), whichever matches up correctly

For n = 1,2,3,..., the signs are +,-,+,..., therefore I need to use the (-1)^(n+1)

So I have (-1)^(n+1)(n-1)!(x+2)^(-n)

What I really wanted was the nth derivative evaluated at x=-1, so substituted -1 in for x to get:

(-1)^(n+1)(n-1)!(1)^(-n)= (-1)^(n+1)(n-1)!

Now I have to use the Taylor series expansion formula:

S(x) = sum(a_n(x-x_0)^n, where a_n = [S^n(x_0)]/n!

So now I have sum((A_n/n!)(x+1)^n, x, 0, infinity).

A_n is the nth deriv. evaluated at x = -1.

Now what. How do I evaluate the other parts and further expand this if its not done so already. How do I get an approximation accurate to within such a small error bound.

Book has the following result: Sum from n=1 to infinity of [(-1)^(n+1) / n] * (x+1)^n ... I was close just don't really see it. How would I do the remaining parts for accuracy and approximating the function above?

Thanks for your help.

Definition:
A function which is infinitely diffrenciable its "Taylor Series centered at c" is defined as the power series:


You have,

It is centered at
Then, (infinitely diffrenciable for x=2)




In general,

When evaluated at we have,


Thus, its Taylor Series is,

Which simplifies to,

Thus,


Further we can prove that the Lagrange remainder is convergent to 0, thus this power series is a representation for

Originally Posted by ThePerfectHacker

Definition:

Thus,


Further we can prove that the Lagrange remainder is convergent to 0, thus this power series is a representation for

When

RonL

Originally Posted by CaptainBlack

When

RonL

If that did that in front of Cauchy I would be dead.

I continued to derive the Taylor's Series for it using my way...when I last was on here, I was at this point:

So now I have sum((A_n/n!)(x+1)^n, x, 0, infinity).

I substituted what A_n was and simplified.

I ended up with summation(([(-1)^(n+1)]/n)*(x+1)^n, n = 1...infinity) since n = 0 is zero, therefore does not appear.

I think that's the right derivation (or hope, since everything will be messed up).

For part b, it asks to find the interval of convergence for this series, including endpoint behavior.

I decided to use the RATIO TEST to determine this.

So: [[(x+1)^(n+1)]/(n+1)]/[(x+1)^n]/n...which simplifies to:

[n*(x+1)]/(n+1)...now the ratio test says to take the limit of that as
n -> infinity. If the limit is < 1, it converges. If it is > 1, it diverges. If it = 1, it is inconclusive.

I found the limit is x + 1. Oh, I forgot to mention that its the absolute value of that. So, if |x+1| < 1, it converges. If |x+1| > 1, it diverges.

So: will not converge when |-x - 1| > 1 or |x + 1| > 1 or (x < -2 OR x > 0)

Therefore, x needs to be between -2 and 0 for it to converge.

For endpoints, to check convergencejust plug those two points into the original taylor series? And then what?

Now, part c, I have no idea how to approach this part:

"Use your results, along with other ideas, to approximate
ln((-1 + 2*Pi)/(Pi)) accurate to within 10^(-25).

THANKS!