1. taylor series

find exactly the taylor polynomial of degree 4 for

f(x) = e^(2x)

f(x) = e^2x e^-6
f'(x) = 2e^2x 2e^-6
f"(x) = 4e^2x 4e^-6
f"'(x) = 8e^2x 8e^-6
f""(x) = 16e^2x 16e^-6

right?

so to write it out its like

p4(x, -3) = e^-6 + (x-3)/1!*2e^-6 + (x-3)/2!*4e^-6 and so on is that right or since its -3 should it be
p4(x, -3) = e^-6 + (x+3)/1!*2e^-6 + (x+3)/2!*4e^-6

???

2. Originally Posted by helmszee
find exactly the taylor polynomial of degree 4 for

f(x) = e^(2x)

f(x) = e^2x e^-6
f'(x) = 2e^2x 2e^-6
f"(x) = 4e^2x 4e^-6
f"'(x) = 8e^2x 8e^-6
f""(x) = 16e^2x 16e^-6

right?

so to write it out its like

p4(x, -3) = e^-6 + (x-3)/1!*2e^-6 + (x-3)/2!*4e^-6 and so on is that right or since its -3 should it be
p4(x, -3) = e^-6 + (x+3)/1!*2e^-6 + (x+3)/2!*4e^-6

???
no. you want:

$T_4(x) = f(-3) ~+ ~ \frac {f'(-3)(x + 3)}{1!} ~+~ \frac {f''(-3)(x + 3)^2}{2!}$ $+~ \frac {f'''(-3)(x + 3)^3}{3!} ~+~ \frac {f''''(-3)(x + 3)^4}{4!}$