Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Calculus 2: Power Series Questions?
1. Let P3(x) be the cubic Maclaurin polynomial for the function f defined by f(x) = (e^x)(sin x), x = all real numbers
A. Compute P3(x).
B. Use the result to compute f (1/2) = P3 (1/2)
C. Use Taylor's Theorem to approximate | f(1/2)-P3(1/2) |
2. Let f be the function defined by f(x) = summation of [-1^n)(x^n)]/(5n+1)
A. Find Dom f
B. Compute a power series representation for the function g defined by g(x) = Domain x (f(x))
C. Use the power series representation for f to show that
(23/48) < integral from 1/2 to 0 of f(x)dx < (85/176)
3. Compute integral from 1/2 to 0 of cube root(1+x^2) dx as follows:
A. Find a power series representation for the function f defined by f(x) = cube root(1+x^2), x = all real numbers
B. use the power series representation for f to compute a series that converges to integral from 1/2 to 0 of cube root(1+x^2) dx
C. Use the result to compute bounds for the same equation.
1 Answer
- kbLv 79 years agoFavorite Answer
1A) Using the definition:
f(x) = e^x sin x ==> f(0) = 0
f '(x) = e^x (sin x + cos x) ==> f '(0) = 1
f ''(x) = e^x (2 cos x) ==> f ''(0) = 2
f '''(x) = e^x (2 cos x - 2 sin x) ==> f '''(0) = 2.
So, P₃(x) = 0 + 1x + 2x^2/2! + 2x^3/3! = x + x^2 + x^3/3.
---------
B) Simply let x = 1/2 in the result in (A).
C) Since f ''''(x) = e^x * -4 sin x, the error bound is
|f ''''(c) * (1/2)^4/4!| for some c in (0, 1/2)
= (4e^c sin c) * (1/16) / 4!
= e^c sin c / 96
< e^(1/2) * 1/96
= e^(1/2)/96.
----------------------
2A) Use the Ratio Test.
r = lim(n→∞) |[(-1)^(n+1) x^(n+1) / (5n+7)] / [(-1)^n x^n/(5n+1)]|
..= |x| * lim(n→∞) (5n+1)/(5n+7)
..= |x|.
So, this series converges (at least) for |x| < 1.
Check the endpoints:
x = -1 ==> Σ 1/(5n+1); divergent by Integral Test.
x = 1 ==> Σ (-1)^n/(5n+1); convergent by Alternating Series Test ({1/(5n+1)} is a decreasing sequence which converges to 0).
So, the domain is [-1, 1).
B) The notation is confusing...
C) Since f(x) = Σ(n=0 to ∞) (-1)^n x^n/(5n+1), integrating from 0 to 1/2 yields
∫(x = 0 to 1/2) f(x) dx
= Σ(n=0 to ∞) (-1)^n (1/2)^(n+1)/[(n+1)(5n+1)].
= Σ(n=0 to ∞) (-1)^n / [2^(n+1) * (n+1)(5n+1)]
Note that this is an alternating series; so any sum ending with a + term is an overestimate, and any ending with a - term is an underestimate.
In particular, using the first 2-3 terms:
1/(2 * 1) - 1/(4 * 12) < ∫(x = 0 to 1/2) f(x) dx < 1/(2 * 1) - 1/(4 * 12) + 1/(8 * 33)
==> 23/48 < ∫(x = 0 to 1/2) f(x) dx < 85/176.
----------------
3A) Using the binomial series with exponent 1/3:
(1 + t)^(1/3) = 1 + Σ(n = 1 to ∞) [(1/3)(1/3 - 1)...(1/3 - n + 1)/n!] t^n.
Let t = x^2:
(1 + x^2)^(1/3) = 1 + Σ(n = 1 to ∞) [(1/3)(1/3 - 1)...(1/3 - n + 1)/n!] x^(2n).
Integrate both sides from 0 to 1/2:
∫(x = 0 to 1/2) (1 + x^2)^(1/3) dx
= 1/2 + Σ(n = 1 to ∞) [(1/3)(1/3 - 1)...(1/3 - n + 1)/n!] (1/2)^(2n+1)/(2n+1).
C) What bounds?
I hope this helps!
