Question

Find the Taylor series about 0 for each of the functions below. Give the first three non-zero terms for (1+2x)^4

Answer 1

i tried expanding it out and making it 16x^4+32x^3+24x^2+8x+1 and taking the derivatives but it didnt work

Answer 2

two ways to do this :-
1) find derivatives using chain rule directly
2) use the known series (1+x)^n

Answer 3

\(f(x) = f(0) + x f'(0) + \frac{x^2}{2!}f^{''}(0) + ....\)

Answer 4

lets try the first method maybe..
find f', f"...

Answer 5

f'(x)=64x^3+96x^2+48x+8

Answer 6

f''(x)=192x^2+192x+48

Answer 7

hey u knw chain rule right ?

Answer 8

yes

Answer 9

4(1+2x)^3*2 right? or something like that

Answer 10

\(f(x) = (1+2x)^4\)

\(f'(x) = 4(1+2x)^3*2 = 8(1+2x)^3\)

\(f''(x) = 8*3(1+2x)^2*2 = 48(1+2x)^2\)

Answer 11

\(f(0) = (1+2*0)^4 = 1\)

\(f'(0) = 8(1+2*0)^3 = 8\)

\(f''(0) = 48(1+2*0)^2 = 48\)

Answer 12

okay!

Answer 13

plug them in the series

Answer 14

1+8x+48x^2/2! yay!!!

Answer 15

wat we're doing is the maclaurin series; hope u knw it...
taylor series @ 0 becomes maclaurin series

Answer 16

can u help me with one more and then i am soooo going to bed!
yeah my professor doesnt teach very well so I'm struggling this semester :/

Answer 17

finding the first 4 terms of sqrt(1-7x)

Answer 18

i know the first term is 1:)

Answer 19

great start :)

Answer 20

ahhh yeauuuh ima pro! lol

Answer 21

lets find f(0), f'(0), f"(0)...

\(f(x) = \sqrt{1-7x}\)

\(f'(x) = \frac{-7}{2\sqrt{1-7x}}\)

\(f^{''}(x) = \frac{-49}{4(1-7x)^{3/2}}\)

\(f^{'''}(x) = \frac{-1029}{8(1-7x)^{5/2}}\)

Answer 22

lets finish this first,
there is a much easier way than this, but lets finish this first ok

Answer 23

\(f(0) = \sqrt{1-7*0} = 1\)

\(f'(0) = \frac{-7}{2\sqrt{1-7*0}} = \frac{-7}{2}\)

\(f^{''}(0) = \frac{-49}{4(1-7*0)^{3/2}} = \frac{-49}{4}\)

\(f^{'''}(0) = \frac{-1029}{8(1-7*0)^{5/2}} = \frac{-1029}{8}\)

Answer 24

plug them in the series

Answer 25

\(f(x) = f(0) + x f'(0) + \frac{x^2}{2!}f^{''}(0) + ....\)

\(f(x) = 1 -\frac{7}{2}x - \frac{49}{8}x^2 - \frac{1029}{48}x^3 - .... \)

Answer 26

yay i got it:)) thank you so much seriously uve helped so much!

Answer 27

np :) im sure u deserve some good sleep after so much hard work... gnite :)