Question

How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate `f(1/3)` given `int e^(t^2) dt` from [0,x]?

Answer 1

The approximation is .346.

Explanation:

The Maclaurin series for `e^t=1+t+(t^2)/(2!)+(t^3)/(3!)+...`. So if we plugin t^2 for t to find the series for `e^(t^2)` we get `1+t^2+((t^2)^2)/(2!)+...`. Now we take the antiderivative of this series to find the antiderivative of `inte^(t^2)` and we get `t+t^3/(3*2!)+...`. Finally plug in x=`1/3` to the first two terms and you get `1/3+1/3*((1/3)^4)` which equals `1/3+1/81`, which simplied is `28/81` or if you prefer (and I do) .345 truncated and .356 rounded.

Answer 2

The first two non-zero terms are; `x` and `1/3x^3`

The general maclaurin series is:

`f(x) = x + 1/3x^3 + 1/10x^5 + 1/42x^7 + 1/216x^9 + ...`

The approximation for `f(1/3)` is:

• One Term: `\ \ \ f(1/3) = 0.333333` (6 dp)
• Two Terms: `\ \ f(1/3) = 0.345679` (6 dp)
• Three Terms: `f(1/3) = 0.346091` (6 dp)

For error comparison, using numerical methods we obtain

`f(1/3) =0.346102`

Explanation:

We have;

`f(x) = int_0^x \ e^(t^2) \ dt`

We could derive the Maclaurin series from first principles, using

`f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ...`

But in this case we can start with the series for `e^t` which is probably the most well known Maclaurin Series:

`e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... + x^n/(n!) + ...`

If we substitute `x=t^2` then we can quickly get the Maclaurin Series for the integrand, thus:

`e^(t^2) = 1 + (t^2) + (t^2)^2/(2!) + (t^2)^3/(3!) + (t^2)^4/(4!) + ...`
`\ \ \ \ = 1 + t^2 + t^4/(2!) + t^6/(3!) + t^8/(4!) + ...`

So we we can now write the integral function as follows;

`f(x) = int_0^x \ e^(t^2) \ dt`
`" " = int_0^x \ {1 + t^2 + t^4/(2!) + t^6/(3!) + t^8/(4!) + ... } \ dt`

If we integrate term by term then we get:

`f(x) = [t + 1/3t^4 + 1/5t^5/(2!) + 1/7t^7/(3!) + 1/9t^9/(4!) + ... ]_0^x`
`" " = (x + 1/3x^3 + 1/5x^5/(2!) + 1/7x^7/(3!) + 1/9x^9/(4!) + ... ) - 0`
`" " = x + 1/3x^3 + 1/5x^5/(2) + 1/7x^7/(6) + 1/9x^9/(24) + ...`
`" " = x + 1/3x^3 + 1/10x^5 + 1/42x^7 + 1/216x^9 + ...`

So onto the approximation, and truncating to `6` dp:

If we use one term we have:

`f(1/3) = 1/3`
`" " = 0.333333` (6 dp)

Using two terms we get:

`f(1/3) = 1/3 + 1/3(1/3)^3`
`" " = 1/3 + 1/81`
`" " = 28/81`
`" " = 0.345679` (6 dp)

Using three terms we get:

`f(1/3) = 1/3 + 1/81 + 1/10(1/3)^5`
`" " = 28/81 + 1/2430`
`" " = 841/2430`
`" " = 0.346091` (6 dp)

For error comparison, using numerical methods we obtain

`f(1/3) =0.346102` (6 dp)