Question

Assume that sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(8 x^2) to evaluate the integral from 0 to 0.8 sin(8 x^2)dx
Your answer will be an infinite series. Use the first two terms to estimate its value

Answer 1

is it
\[ \int_{0}^{0.8} \sin (8x^2) dx = \int_{0}^{0.8} (8x^2/1! - (8x^2)^3/3! + (8x^2)^5/5! -(8x^2)^7/7! + ...) dx\]

Answer 2

Does not work !!!

Answer 3

should we wrap it about zero? i can never remember if the mac or the taylow is wrapped at zero

Answer 4

i think it's mac

Answer 5

the derivative of sin(8x^2) are:

sin(8x^2)

16x cos(8x^2)

-16^2 x^2 sin(8x^2)

-16^3 x^3 cos(8x^2)

which at x=0 simply goes zero all over, so we might have to pick a different center

Answer 6

x=pi might work better

Answer 7

or, x = sqrt(pi) lol

Answer 8

well, the mac wraps about zero according to the google

Answer 9

the wolfs got a series expansion about zero, but it doesnt say how

Answer 10

can you post the link??

Answer 11

http://www.wolframalpha.com/input/?i=sin%288x%5E2%29

Answer 12

thanks ... the expansion at zero looks like mine ... but i tested integrating first few terms, and it doesn not work at all.

Answer 13

Oo... some how i got close this time
http://www.wolframalpha.com/input/?i=integrate+%288+x^2-%28256+x^6%29%2F3%2B%284096+x^10%29%2F15-%28131072+x^14%29%2F315%2B%281048576+x^18%29%2F2835-%2833554432+x^22%29%2F155925%29+from+0+to+0.8

http://www.wolframalpha.com/input/?i=integrate+sin%288x^2%29+from+0+to+0.8

Answer 14

first four term does not give close value at all
http://www.wolframalpha.com/input/?i=integrate+%288+x^2-%28256+x^6%29%2F3%2B%284096+x^10%29%2F15-%28131072+x^14%29%2F315%29+from+0+to+0.8

Not even 5 terms
http://www.wolframalpha.com/input/?i=integrate+%288+x^2-%28256+x^6%29%2F3%2B%284096+x^10%29%2F15-%28131072+x^14%29%2F315%2B%281048576+x^18%29%2F2835%29+from+0+to+0.8

Need at lest six terms, but that is fairly complicated :((