Help with the taylor expansion. Maybe I'm just rusty and out of practice but I don't know how to some this problem using the taylor expansion. I have attached a picture of the problem in a comments.

Question

Help with the taylor expansion. Maybe I'm just rusty  and out of  practice but I don't know how to some this problem using the taylor expansion. I have attached a picture of the problem in a comments.

anonymous · Answer

it is part b

amistre64 · Answer

a taylor series matches a polynomial to a function such that their derivatives are equal .. they move the same.

a polynomial has a constant term with an infinite number of variable parts raised to powers .... if we zero out the variable parts, all that is left is the constant term:

$f=c_o+c_1x+c_2x^2+c_3x^3+c_4x^4+...$

$f'=c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4+...$

$f''=2c_2+3.2c_3x+4.3c_4x^2+5.4c_5x^3+6.5c_6x^4...$

$f'''=3.2c_3+4.3.2c_4x+5.4.3c_5x^2+6.5.4c_6x^3...$

notice how the repeated derivatives are forming factorials attached to the constant terms.  when the variable parts are zeroed out we have:

$f=c_o$

$f'=c_1$

$f''=2c_2$

$f'''=3.2c_3$
.....

$f^{(n)} = n!~c_n$

we can now solve for the coefficients of our polynomial representation:

$$c_n=f^{(n)}(0)/n!$$

amistre64 · Answer

on the case of this sqrt(1+x)

(1+0)^(1/2) = 1

1/2  (1+0)^(-1/2) = 1/2

-1/4  (1+0)^(-3/2) = -1/4

-1/4 . -3/2 = 3/8

3/8 . -5/2 = -15/16

-15/16 . -7/2  = 105/32

etc ...  divide by the appropriate n! become the coefficients of the taylor exansion centered at x=0

amistre64 · Answer

the more terms you can gather, the closer you can get to the true value of sqrt(1+x) near x=0

amistre64 · Answer

http://www.wolframalpha.com/input/?i=y%3Dsqrt%281%2Bx%29%2C+y%3D1%2Bx%2F2-x%5E2%2F8%2Bx%5E3%2F16-15x%5E4%2F%2816*4%21%29%2B105x%5E5%2F%2832*5%21%29