Problem 72E

Sine integral function The function is called the sine integral function.

a. Expand the integrand in a Taylor series about 0.

b. Integrate the series to find a Taylor series for Si.

c. Approximate Si (0.5) and Si (1). Use enough terms of the series so the error in the approximation does not exceed 10−3

Solution 72E:

Step 1:

Given , the function “Si(x) = dt” is called the sine integral function.

Integrand is ;

- In this problem we need to expand the integrand in a taylor series about “0”.

We know that the taylor series about “0” is ;

f(x) = f(0) +x ++................

Consider , f(t) = , then f(0) = = 1

, then = 0

= , then = -1, since = = -1

= , then = 0

= , then = 1, since = = 1

……………………………………

Therefore , the taylor series about zero is ;

f(t) = = 1 +t ++................

= 1 - +-................