Question

Though 1 quick question, when: (f^(5))(0) = e^0 = 1 How does that become 5^5?

Answer 1

5^5 e^0 = 5^5

Answer 2

So is f(n) referring to the previous derivative? not just the original?

Answer 3

each derivative of the original function is centered at 0 in order to determine the constants ofthe power series

Answer 4

Is the Maclauren series any different?

Answer 5

\begin{array}
e^{5x}\\
5e^{5x}\\
5*5e^{5x}\\
5*5*5e^{5x}\\
etc
\end{array}

Answer 6

mac and taylor are the same; only one of them uses 0 and the other uses any number

Answer 7

\[e^{5x}=\frac{f^0(0)}{0!}x^0+\frac{f^1(0)}{1!}x^1+\frac{f^2(0)}{2!}x^2+...\]

Answer 8

the other one is just like it except for the "0"

\[e^{5x}=\frac{f^0(a)}{0!}(x-a)^0+\frac{f^1(a)}{1!}(x-a)^1+\frac{f^2(a)}{2!}(x-a)^2+...\]

Answer 9

Ahh, I see

Answer 10

I'm just so nervous over this exam! I'm forgetting this stuff as I'm learning it

Answer 11

And the only difference between a Taylor and a Maclauren is?