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]?

2 Answers
Mar 22, 2017

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.

Mar 22, 2017

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)