# Question Video: Using the Maclaurin Series of a Function to Find the Value of the Third Derivative at 0 Mathematics • Higher Education

If the Maclaurin series of the function 𝑓 is 𝑓(𝑥) = 3 − (1/2 𝑥) + (5/6 𝑥²) − (11/26 𝑥³) + (21/80 𝑥⁴) + ..., find 𝑓‴(0).

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### Video Transcript

If the Maclaurin series of the function 𝑓 is 𝑓 of 𝑥 is equal to three minus one-half 𝑥 plus five over six times 𝑥 squared minus 11 over 26 times 𝑥 cubed plus 21 over 80 multiplied by 𝑥 to the fourth power and this series continues, find the value of 𝑓 triple prime of zero.

The question gives us the Maclaurin series for our function 𝑓 of 𝑥. We need to use this Maclaurin series to find the value of 𝑓 triple prime of zero. Let’s start by recalling what the Maclaurin series of a function is.

We recall the Maclaurin series for a function 𝑓 of 𝑥 is given by the following power series. 𝑓 of 𝑥 is equal to the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of 𝑥 with respect to 𝑥 evaluated at zero divided by 𝑛 factorial times 𝑥 to the 𝑛th power. And there’s a few things worth pointing out about the Maclaurin series of a function.

First, the Maclaurin series of a function is unique. In other words, any power series representation of a function centered at 𝑥 is equal to zero must be the Maclaurin series of that function. And since the power series representation given to us in the question is centered at zero, it must be the Maclaurin series of our function. But we’re actually told this information anyway.

We need to use this to find 𝑓 triple prime evaluated at zero. Remember, this is the third derivative of 𝑓 with respect to 𝑥 evaluated at 𝑥 is equal to zero. To do this, it will be easier to write our power series for the Maclaurin series of 𝑓 of 𝑥 term by term.

Recall, if we do this and simplify, we get 𝑓 of zero plus 𝑓 prime of zero times 𝑥 plus 𝑓 double prime of zero times 𝑥 squared divided by two factorial plus 𝑓 triple prime of zero times 𝑥 cubed divided by three factorial. And we keep adding terms in this form.

And now we can see that the coefficient of 𝑥 cubed contains 𝑓 triple prime of zero. And we’re told that the coefficient of 𝑥 cubed in the Maclaurin series is negative 11 divided by 26. Since these are both the Maclaurin series for our function 𝑓 of 𝑥, we can just equate the coefficient of 𝑥 cubed in both of these series. So for these to be equal, their coefficients of 𝑥 cubed must be equal. This means 𝑓 triple prime of zero divided by three factorial must be equal to negative 11 divided by 26.

Now, all we need to do is rearrange this equation to find 𝑓 triple prime of 𝑥. We’ll multiply both sides of our equation through by three factorial. To do this, recall, three factorial is equal to three times two times one, which is of course equal to six. This gives us 𝑓 triple prime of zero must be equal to negative 11 times six divided by 26. And if we simplify this expression, we get negative 33 divided by 13.

Therefore, we were able to show if the Maclaurin series of a function 𝑓 of 𝑥 is given by three minus one-half 𝑥 plus five over six 𝑥 squared minus 11 over 26 𝑥 cubed plus 21 over 80 𝑥 to the fourth power and we keep adding terms, then we must have that 𝑓 triple prime evaluated at 𝑥 is equal to zero is given by negative 33 divided by 13.