Expert: Paul Klarreich - 9/16/2007

Question
Taylor Series for sin(x^2) about x=0 up
to and including terms of order 0(x^8) and use the result to
approximate
I= intergral from 0 to 1 of (sin(x^2))/x dx"


Answer
Questioner:   Kate
Category:  Advanced Math
Private:  No
 
Subject:  Taylor Series
Question:  Taylor Series for sin(x^2) about x=0 up
to and including terms of order 0(x^8) and use the result to approximate
I= integral from 0 to 1 of (sin(x^2))/x dx"
.........................................
Hi, Kate,

I have not worked it all out, but this approach seems to work ok for small values of x, which is, after all, the idea of the Taylor expansion.

The following expansion is known, or at least, fairly easy to derive:

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WARNING: USE COURIER FONT TO VIEW THIS
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            y^3     y^5
sin y = y - ----- + ----- - ....
            3!      5!

Now substitute  y = x^2 and you have:

                x^6     x^10
sin x^2 = x^2 - ----- + ----- - ....
                 3!      5!

..................................
Now for your integral:

{1  sin(x^2)
|   -------- dx
}0     x

take those three terms, divide each term by x:

    x^5     x^9
x - ----- + ----
     3!      5!
Integrate it:

x^2    x^6     x^10
--- - ----- + -----
2     6 3!   10 5!

from 0 to 1, gives:

1      1       1
--- - ---- + -----
2     36     1200
 1800 - 100 + 3
= ---------------
      3600
  1703
= ------ = 0.4731
  3600