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
The approximation is .346.
Explanation:
The Maclaurin series for
The first two non-zero terms are;
The general maclaurin series is:
# f(x) = x + 1/3x^3 + 1/10x^5 + 1/42x^7 + 1/216x^9 + ... #
The approximation for
- 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^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... + x^n/(n!) + ... #
If we substitute
# 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
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)