# 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]?

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 #

We have;

We could derive the Maclaurin series from first principles, using

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

If we integrate term by term then we get:

If we use one term we have:

Using two terms we get:

Using three terms we get:

For error comparison, using numerical methods we obtain

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The approximation is .346.

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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 ).

**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!} ]

**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} ]

**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} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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