Maclaurin polynomial for (e^2x)(sin2x)??!!!?

Question

okay AP Calculus BC question.... how do you get a four term maclaurin polynomial for (e^2x)(sin2x)?!?!?

thank you anyone that can help!!!

Answer 1

Multiply together the Maclaurin Series for e^(2x) and sin(2x).



Using the said series, we have

e^(2x) = 1 + (2x) + (2x)^2/2! + (2x)^3/3! + (2x)^4/4! + ...

.........= 1 + 2x + 2x^2 + (4/3)x^3 + (2/3) x^4 + ...



sin(2x) = (2x) - (2x)^3/3! + (2x)^5/5! - ...

..........= 2x - (4/3) x^3 + (4/15) x^5 - ...



Using the Distributive property yields

e^(2x) sin(2x)

= [1 + 2x + 2x^2 + (4/3)x^3 + (2/3)x^4 + ...] [2x - (4/3)x^3 + (4/15)x^5 - ...]

= 2x + 4x^2 + (4 - 4/3) x^3 + (8/3 - 8/3) x^4 + (4/3 - 8/3 + 4/15) x^5 + ...

= 2x + 4x^2 + (8/3) x^3 - (16/15) x^5 + ...



Double checked with Wolfram Alpha:

http://www.wolframalpha.com/input/?i=taylor+series+e^%282x%29+sin%282x%29



I hope this helps!

Answer 2

The Maclaurin polynomial is the Taylor sequence based at x = 0. The maclaurin of cos(x) is a million - x^2/2! + x^4/4! - x^6/6! + ... we'd like it to the fourth order, so we'd desire to approximate cos x to: cos x ? a million - x^2/2! + x^4/4! So for x = - 0.01, we've cos(-0.01) ? a million - (-0.01)^2/2! + (-0.01)^4/4! ? 0.99995 Now we'd desire to renowned how good of an approximation our answer is. there's a formulation to locate the blunders in the Maclaurin polynomial: as quickly as we approximate a function to the nth order of the Taylor sequence, all of us comprehend that the blunders in the approximation is under M*(x-a)^(n+a million) / (n+a million)! the place M is a certain for the n+a million by-made from the function; this is the (n+a million)th by-made from the function

Answer 3

The Maclaurin polynomial is the Taylor sequence based at x = 0. The maclaurin of cos(x) is a million - x^2/2! + x^4/4! - x^6/6! + ... we want it to the fourth order, so we'd desire to approximate cos x to: cos x ? a million - x^2/2! + x^4/4! So for x = - 0.01, we've cos(-0.01) ? a million - (-0.01)^2/2! + (-0.01)^4/4! ? 0.99995 Now we'd desire to renowned how good of an approximation our answer is. there's a formula to locate the errors interior the Maclaurin polynomial: as quickly as we approximate a function to the nth order of the Taylor sequence, all of us know that the errors interior the approximation is under M*(x-a)^(n+a million) / (n+a million)! the place M is a sure for the n+a million derivative of the function; it is the (n+a million)th derivative of the function