# How do you find the Maclaurin Series for #f(x) = (x-sin x)/ x^3 #?

with

Also

so

Now,

Ultimately defining

#c_k=1/2 ((2 (k + 1) (2 k + 1))/((2 k + 3)!) - 1/((2 k + 1)!) + 2/((2 k + 2)!))#

we have

The initial coefficients are:

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To find the Maclaurin series for ( f(x) = \frac{x - \sin(x)}{x^3} ), we can use the Taylor series expansion. The Maclaurin series is a special case of the Taylor series where the expansion is centered at ( x = 0 ).

- First, let's find the derivatives of ( f(x) ) up to the desired order.

[ f(x) = \frac{x - \sin(x)}{x^3} ]

- Compute the first few derivatives of ( f(x) ):

[ f'(x) = \frac{3x^2 - x\cos(x) - \sin(x)}{x^4} ]

[ f''(x) = \frac{6x^3 - 6x\cos(x) - 3x^2\sin(x) - 2\cos(x)}{x^5} ]

[ f'''(x) = \frac{24x^4 - 18x^2\sin(x) - 12x\cos(x) + 6x^3\cos(x) - 6x^2\sin(x)}{x^6} ]

- Evaluate each derivative at ( x = 0 ) to find the coefficients:

[ f(0) = 0 ]

[ f'(0) = \frac{1}{3} ]

[ f''(0) = -\frac{1}{3} ]

[ f'''(0) = \frac{1}{30} ]

- Write the Maclaurin series using the coefficients found:

[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots ]

Substitute the coefficients:

[ f(x) = 0 + \frac{1}{3}x - \frac{1}{3 \times 2!}x^2 + \frac{1}{30 \times 3!}x^3 + \cdots ]

This gives us the Maclaurin series for ( f(x) ).

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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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