Question

Use a power series to approximate the integral 1/(1+x^3) from (2/3) to 0 Use a power series to approximate the integral 1/(1+x^3) from (2/3) to 0 @Mathematics

Answer 1

like taylor series?

Answer 2

A series of the form \[\sum_{n = 0}^{\infty} a _{n}(x-c)^{n}\]

Answer 3

any idea how to start that?

Answer 4

Nope. I know what individually they mean. asubn is a sequence and x is a variable of sort.

An integral is the area under a curve so graphically.

Answer 5

graphically your trying to match a polynomial to the curve of the given equation. you do that by matching derivatives ....

Answer 6

you gather up enough derivatives and you can fit the curve; because derivatives tell you how a curve bends and moves

Answer 7

\[(1+x^3)^{-1}\]
\[-3x^2(1+x^3)^{-2}\]
\[-6x(1+x^3)^{-2}+18x^4(1+x^3)^{-3}\]
\[\begin{array}l-6(1+x^3)^{-2}&+36x^3(1+x^3)^{-3}\\
&+72x^3(1+x^3)^{-3}-162x^6(1+x^3)^{-4}\end{array}\]
but that can get messy

Answer 8

you find values for those at a given "value" in the interval that you are looking at; perferabbly a simple enough value like 0 works since its in your interval

Answer 9

those become the coeffs of your power series

Answer 10

1

0x

0x^2

−6x^3

and so on .....

Answer 11

http://www.wolframalpha.com/input/?i=y+%3D+1-6x%5E3+and+y%3D1%2F%281%2Bx%5E3%29

if you get enough derivatives under your belt, you can see that the curves are starting to match in the interval

Answer 12

i messed up on the coeffs, but the wolf says my poly degrees was good

Answer 13

1-x^3+x^6-x^9+x^12-x^15

Answer 14

http://www.wolframalpha.com/input/?i=y%3D1%2F%281%2Bx%5E3%29+and+y+%3D+1-x%5E3%2Bx%5E6-x%5E9%2Bx%5E12-x%5E15