Find the MacLaurin polynomial of degree 5 for F(x).?

Since e^t = 1 + t + t^2/2! + ..., letting t = -4x^4 yields

e^(-4x^4) = 1 - 4x^4 + 8x^8 + ...

Hence, integrating term by term yields
∫(x = 0 to t) e^(-4t^4) dt ≈ t - 4t^5/5, by ignoring higher degree terms

Letting t = 0.2 finishes the second part of the question

The Maclaurin polynomial is the Taylor series accepted at x = 0. The maclaurin of cos(x) is a million - x^2/2! + x^4/4! - x^6/6! + ... we would like it to the fourth order, so we would 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 would desire to comprehend ways 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 series, all of us comprehend that the errors interior the approximation is below M*(x-a)^(n+a million) / (n+a million)! the place M is a sure for the n+a million by-fabricated from the function; this is the (n+a million)th by-fabricated from the function<M So thus all of us comprehend that M is a million (because of the fact the (n+a million)th by-fabricated from cos x is the two a sin or a cos, so for confident below a million) and a = 0 (Maclaurin), and x = -0.01, and n =4 so the errors is way less that (-0.01)^5/5! = -8.33333333 × 10^-13 (useful approximation)