# Thread: Calc help maclurin polynomials

1. ## Calc help maclurin polynomials

I have to match each polynomial formula with the correct Maclaurin polynomial below. I am confused as to how to do this, so any help would be appreciated.

a) f(x) = (e^-x)-1
b) f(x) = 1/(x+1)
c) f(x) = 1/(1-x)
d) f(x) = (e^x) + sin(x)

1) 1 + x + x^2 + x^3
2) 1 + 2x + (X^2)/(2)
3) -x + (x^2)/(2) - (x^3)/(3)
4) 1 - x + x^2 - x^3

2. Originally Posted by clockingly
I have to match each polynomial formula with the correct Maclaurin polynomial below. I am confused as to how to do this, so any help would be appreciated.

a) f(x) = (e^-x)-1
b) f(x) = 1/(x+1)
c) f(x) = 1/(1-x)
d) f(x) = (e^x) + sin(x)

1) 1 + x + x^2 + x^3
2) 1 + 2x + (X^2)/(2)
3) -x + (x^2)/(2) - (x^3)/(3)
4) 1 - x + x^2 - x^3
The function for the Maclaurin Series is the following sum:
SUM{n=0:infinity} (f^(n)(0)*x^n)/n! ... where f^(n) is the "n'th derivative of f," and n! is "n factorial" = n(n-1)(n-2)*...*(2)(1)

I suspect that sum doesn't make sense to you, so here's the easy way to remember it. We are basically adding all the derivatives of the function in the following way:
f(0) + f'(0)/(1!)*x + f''(0)/(2!)*x^2 + f'''(0)/(3!)*x^3 + ... + f^(n)(0)/(n!)*x^n + ...

All we need in this case is the first 4 terms of the series:
f(0) + f'(0)*x + f''(0)/2*x^2 + f'''(0)/6*x^3

For the 4 functions you've listed, find up to the third derivative and plug those into the above sum. You should see which match with which.

I'll do the first for you:
f(x) = e^(-x) - 1
f'(x) = -e^(-x)
f''(x) = e^(-x)
f'''(x) = -e^(-x)

f(0) = 0
f'(0) = -1
f''(0) = 1
f'''(0) = -1

Let's plug these into the sum to get:
0 + (-1)x + (1)/2*x^2 + (-1)/6*x^3 = -x +1/2*x^2 - 1/6*x^3

This looks like answer 3, except mine is divided by 6. Check to see that number 3 is typed correctly.