Taylor & Maclaurin polynomials intro
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Worked example: coefficient in Maclaurin polynomial
- [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.