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]?

Question

Elijah Abram · Accepted Answer

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 #

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)

Charles Anctil · Answer

The approximation is .346.

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.

William Abdalla · Answer

undefined To find the first two nonzero terms in Maclaurin's formula for a function $ f(x) $ and use it to approximate $ f(1/3) $, we need to first find the Maclaurin series for the function and then evaluate it at $ x = 1/3 $.

1. **Finding the Maclaurin series:**  
The Maclaurin series for $ e^{t^2} $ is given by:

$$ e^{t^2} = \sum_{n=0}^{\infty} \frac{t^{2n}}{n!} $$

2. **Integrating to find $ f(x) $:**  
Now, integrate $ e^{t^2} $ from $ t = 0 $ to $ t = x $ to find $ f(x) $:

$$ f(x) = \int_{0}^{x} e^{t^2} \, dt $$

$$ f(x) = \int_{0}^{x} \sum_{n=0}^{\infty} \frac{t^{2n}}{n!} \, dt $$

$$ f(x) = \sum_{n=0}^{\infty} \frac{1}{n!} \int_{0}^{x} t^{2n} \, dt $$

$$ f(x) = \sum_{n=0}^{\infty} \frac{1}{n!} \cdot \frac{x^{2n+1}}{2n+1} $$

3. **Approximating $ f(1/3) $:**  
To approximate $ f(1/3) $, we substitute $ x = 1/3 $ into the series expansion of $ f(x) $ and truncate the series after the second term:

$$ f(1/3) \approx \frac{1}{0!} \cdot \frac{(1/3)^{1}}{1} + \frac{1}{1!} \cdot \frac{(1/3)^{3}}{3} $$

$$ f(1/3) \approx \frac{1}{3} + \frac{1}{3!} \cdot \frac{1}{27} $$

$$ f(1/3) \approx \frac{1}{3} + \frac{1}{162} $$

$$ f(1/3) \approx \frac{55}{162} $$

Therefore, the first two nonzero terms in Maclaurin's formula for $ e^{t^2} $ are $ \frac{1}{3} + \frac{1}{162} $, and the approximation of $ f(1/3) $ using these terms is $ \frac{55}{162} $.