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## Finding Taylor polynomial approximations of functions

# Worked example: coefficient in Maclaurin polynomial

AP.CALC:

LIM‑8 (EU)

, LIM‑8.A (LO)

, LIM‑8.A.1 (EK)

, LIM‑8.A.2 (EK)

, LIM‑8.B (LO)

, LIM‑8.B.1 (EK)

## Video transcript

- [Instructor] Nth derivative
of g at x equals zero is given by, so the nth derivative
of g evaluated x equals zero is equal to square root of n
plus seven over n to the third for n is greater than or equal to one. What is the coefficient for
the term containing x squared in the Maclaurin series of g? Let's just think about the
Maclaurin series for g. If I were to have my function g of x. The Maclaurin series, I could
say approximately equal to especially if I'm not gonna
list out all of the terms, is going to be equal to,
well it's going to be equal to g of zero plus g prime of zero times x plus g prime prime of zero divided by, I could say two factorial
but that's just two, times x squared, and that's
about as far as we go. Because we just have to think
about what is the coefficient for the term containing x squared. If they said what's the coefficient for the term containing x to
the third, I would keep going. I'd go g prime. I would take the third
derivative evaluated at zero over three factorial. I could do this as a factorial too, but that just evaluates to two. I could do this as one factorial. I could do this as zero
factorial just so you see it's a consistent idea here. I could, of course, keep on
going, but we just care about, they're just asking us
what is the coefficient for the term containing x squared? They just want us to figure out this. What is this thing right over here? To know that, we need to figure out what is the second derivative
of g evaluated x equals zero? Well they tell us that over here. It's a little bit unconventional where they give us a
formula, a general formula for any derivative
evaluated x equals zero, but that's what they're telling us here. So in this case, the n isn't zero. The n is the derivative we're
taking, and that's going to be our second derivative, so this is, so if I wanted to figure out g. If I am figuring out
the second derivative, and I could write it
like that evaluated zero, or I could write it like this just so the notation is consistent. I could write it like that. The second derivative evaluated
x equals zero is going to be equal to, well our n is two, so this is going to be the square root of two plus seven over
two to the third power. Two plus seven is nine, take
the principle root of that. It's gonna give us positive
three over two to the third which is eight, so this
part right over here is 3/8, so the whole coefficient
is going to be 3/8, that's this numerator,
divided by, divided by two. Which of course is equal
to 3/16, and we're done. They didn't want us to
figure out a couple of terms of this which we could call
the Maclaurin polynomial and nth degree Maclaurin polynomial. They didn't want us to
find the entire, you know, keep going with this series. They just wanted to find
one coefficient right here. The coefficient on the second degree term which we just figured out is 3/16.

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