Question

How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate `f(1/3)` given `f(x)=int_0^x sin(t^2) dt` ?

Answer 1

`f(1/3) ~~0.01233 \ \ \` (5dp)

Explanation:

The Maclaurin series for `sinx` is given by:

`sinx= sum_(n=0)^oo (-1)^n/((2n+1)!)x^(2n+1)`
`\ \ \ \ \ \ \ = x - x^3/(3!) + x^5/(5!) - x^7/(7!) + x^9/(9!) - ...`

And so we can write the integrand `sin(t^2)` as :

`sint^2 =(t^2) - (t^2)^3/(3!) + (t^2)^5/(5!) - (t^2)^7/(7!) + (t^2)^9/(9!) - ...`
`\ \ \ \ \ \ \ \ =t^2 - t^6/(3!) + t^10/(5!) - t^14/(7!) + t^18/(9!) - ...`

And so expressing the integrand as an infinite power series, our function `f(x)` becomes

`f(x) = int_0^x sin t^2 \ dt`
`\ \ \ \ \ \ \ = int_0^x t^2 - t^6/(3!) + t^10/(5!) - t^14/(7!) + t^18/(9!) - ... \ dt`

Note - this assumes that we can actually replace the integrand with a infinite power series, and the integral still be valid. Mathematical vigour will show that in certain circumstances we can do this, and in fact in this particular case the method is valid.

Now we integrate term by term to get:

`f(x) = [t^3/3 - t^7/(7(3!)) + t^11/(11(5!)) - t^15/(15(7!)) + t^19/(19(9!)) - ... ]_0^x`
`\ \ \ \ \ \ = x^3/3 - x^7/(7(3!)) + x^11/(11(5!)) - x^15/(15(7!)) + x^19/(19(9!)) - ...`

So the first two non-zero terms are:

`f(x) = x^3/3 - x^7/42 + ...`

And so using just these terms, we can approximate `f(1/3)` as

`f(1/3) ~~ (1/3)^3/3 - (1/3)^7/42`
`\ \ \ \ \ \ \ \ \ \ \ =1133/91854`
`\ \ \ \ \ \ \ \ \ \ \ =0.01233479`
`\ \ \ \ \ \ \ \ \ \ \ =0.01233 \ \ \` (5dp)