Question

Help with finding first 4 terms of Taylor series for
F(x)= √(x+1)

Answer 1

Help understanding the process

Answer 2

maybe this will help

http://en.wikipedia.org/wiki/Taylor_series#Definition

set a = 0, then you only need to differentiate the function multiple times plugging in x=0

Answer 3

still not getting it

Answer 4

ok

f(0) = sqrt(1+0) = 1

f'(0) = 1/2(1+0)^-1/2 = 1/2

f''(0) = -1/4 (1+0)^-3/2 = -1/4

f'''(0) = 3/8 (1+0)^-5/2 = 3/8

do you follow?

Answer 5

what about after the 16th?

Answer 6

It gives me 105/32....

Answer 7

haha you lost me .... 16th what?

but you understand how to get the derivatives?

Answer 8

well after I take the 5th derivative I get the coefficent of 105/32 when it should be something/128...

Answer 9

the answer says it should be letter e)

Answer 10

oh ok

f^4(0) = -15/16 (1+0)^-7/2 = -15/16

f^5 (0) = 105/32 (1+0)^-9/2 = 105/32

Answer 11

so what should the series then look like?

Answer 12

once you compute the derivatives you plug them into the formula for taylor series

\[f(x) = 1 + \frac{1}{2} x + \frac{-\frac{1}{4}}{2} x^2 + \frac{\frac{3}{8}}{6} x^3 + \frac{-\frac{15}{16}}{24} x^4 ...\]

Answer 13

oh I see! Thanks a lot. omg you saved me. Do you think you could help with other problems?

Answer 14

sure

Answer 15

It tells me to find f double prime. Not sure how to get started though

Answer 16

actually its f triple prime :)

well from the taylor formula f triple prime is in the 4th term

k=0,1,2,3 so find 4th term of series by plugging in k=3

this gives --> (-1)^3 * x^3/(3+1) = -1/4 x^3

again from taylor formula the 4th term is
\[\frac{f^3 (0)}{6} x^3\]
thus
\[\frac{f^3 (0)}{6} = -\frac{1}{4}\]
so
\[f^3 (0) = -\frac{3}{2}\]

Answer 17

so to find f triple prime, I am basically taking the derivative of -1/4x^3 three times, correct?

Answer 18

seems so :)

Answer 19

okay, I have one last question...
would you be willing to assist?

Answer 20

or anyone?