Math Problem (Taylor Series)?
Let f be the function given by f(x)= e^(-2x^2) (read as e raised to the negative 2 x squared)
(a) Find the first four nonzero terms and the general term of the power seriews for f(x) about x=0.
(b) Find the interval of convergence of the power seriews for f(x) about x = 0. Show the analysis that leads to your conclusion.
(c) Let g be the function given by the sum of the first four nonzero terms of the power seriews for f(x) about x=0. Show that l f(x) - g(x) l < 0.02 for -0.6 less than or equal to x less than or equal to 0.6.
Update:Need help on this! It is pretty hard I think you use Alternating Series Estimation Theorem but I don't know how to
Comments
(a) Since e^x = 1 + x + x^2/2 + x^3/3! + ... + x^n/n! + ...
replacing x with -2x^2 yields
e^(-2x^2) = 1 + (-2x^2) + (-2x^2)^2/2 + (-2x^2)^3/3! + ... + (-2x^2)^n/n! + ...
= 1 - 2x^2 + 2x^4 - 4x^6/3 + ... + (-2)^n x^(2n) / n! + ...
(b) Use the ratio test.
r = lim(n-->infinity) |a(n+1) / a(n)|
= lim(n-->infinity) |[(-2)^(n+1) x^(2n+2) / (n+1)!] / [(-2)^n x^(2n) / n!]|
= x^2 * lim(n-->infinity) 2/(n+1)
= 0.
Since |r| = 0 < 1 for all x, the series converges for all x.
==> The interval of convergence is (-infinity, infinity).
(c) Since this series is alternating, the error after 4 terms is bounded by the fifth term.
i.e., Error <= |(-2)^5 x^10 / 5!| = 32 |x|^10 / 120.
Since, we're estimating the error for |x| <= 0.6,
Error <= 32 |x|^10 / 120 <= 32 (0.6)^10 / 120 = .00161... < 0.02.
I hope this helps!
Taylor series is attempting to "wager" what a function sounds like with the help of making use of the values of the function and its derivatives at a close-by factor. occasion : in the experience that your function grow to be f(x)=x^2 ( the sq. of a quantity ) and you had to discover the value at 0.01.. you would be able to desire to artwork with the regularly occurring values at 0 as : (0.01)^2=f(0.01)=f(0)+(0.01)f'(0)+[(0.... right here f is x^2 so f' is 3x and f'' is two and f''' etc are all 0.. Plug them in and you will discover the answer particularly on the brink of (0.01)^2=0.0001