Question

Find the Taylor polynomials of orders 0, 1, 2, and 3 generated by f at x = a.
f(x) = 1/x, a = 2

Answer 1

I don't really understand the steps for solving these problems..

Answer 2

f(2) = 1/2
f'(2) = -1/2^2 = -1/4
f'(3) = -1*-2/2^3 = 2/8
f'(4) = -1*-2*-3/16 = -6/16

so the taylor polynomial is
f(x) = 1/2 - 1/4(x-2) + 2/8 (x-4)^2 - 6/16(x-3)^3 + O(4)

something wrong
http://www.wolframalpha.com/input/?i=taylor+expansion+1%2Fx+at+x%3D2

Answer 3

why did you change a from 2 to 3 and 4?

Answer 4

Oh ... forgot to divide by factorial :(

Answer 5

wait could you explain how you got that :/ the process is a bit more important to me at the moment.

Answer 6

the coeffs are generated by the first 3 derivatives of 1/x in this case evaluated at x=a

Answer 7

then plug those into the general setup of the taylor

Answer 8

\[ f(x) = \frac 1/2 - \frac 1/4(x-2 \frac 1{1!} + \frac 2/8 (x-4)^2 \frac 1{2!} - \frac 6/16(x-3)^3 \frac{1}{3!} + O(4) \]
= Same answer was wolframalpha
http://mathworld.wolfram.com/TaylorSeries.html

Gotta go!!

Answer 9

You could just use the series for \(\frac{1}{1-x}\) and replace x with 1-x.

Answer 10

thats the one i was thinking of ... maybe lol

Answer 11

Oh wait, it's supposed to be centered at a=2... nvm.

Answer 12

I'm so confused.. our teacher decided to let a student from a BC calc class come teach our AB class, so the kid taught this to us in about 40 minutes. We're all really lost and I don't have much knowledge of taylor series or the way it works

Answer 13

it works by constructing a polynomial to bend in the same way that the other function does

Answer 14

we see how things bend by looking at their derivatives

Answer 15

whats the derivative of 1/x?

Answer 16

-1/x^2

Answer 17

good, and the second derivative is?

Answer 18

2/x^3

Answer 19

great, and the 3rd derivative, since we want 1,2, and 3

Answer 20

-6/(x^4)

Answer 21

excellent, that gives us 4 numbers that we need to use; evaluate them at x=a=2

Answer 22

so.. we're plugging 2 into all the derivatives?

Answer 23

yes, in order to find out how the thing bends at x=2, we need to plug in x=2

Answer 24

okay so we have 1/2, -1/4, 1/4, and -3/8

Answer 25

i agree with this figures; now the next thing to do is to either try to remember the general setup of a taylor poly, or if you got time, we can see how one takes shape.

Answer 26

the general set up is:
\[T_{poly}=\frac{f^{(0)}(a)}{0!}(x-a)^0+\frac{f^{(1)}(a)}{1!}(x-a)^1+\frac{f^{(2)}(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3...\]

Answer 27

where those numerators are the numbers we determined just a few seconds ago

Answer 28

what happens to the x? (in the (x-a))

Answer 29

nothing, thats apart of the polynomial itself

Answer 30

its like asking what happens to x in: y=2x

Answer 31

ah so we don't have to plug anything in to it right

Answer 32

its the independant variable that receives input in order to graph the thing

Answer 33

well, not "as is" :)

Answer 34

if you wanted to plot the poly, youd have to plug in values for x

Answer 35

hmm. so how does this translate to the answers in my book?
it says:
P0(x) = 1/2
P1(x) = (1/2) - ((x-2)/4)
P2(x) = (1/2) - ((x-2)/4) + (((x-2)^2)/8)
P3(x) = (1/2) - ((x-2)/4) + (((x-2)^2)/8) - (((x-2)^3)/16)

Answer 36

OH

Answer 37

nevermind that question

Answer 38

:)

Answer 39

(I can't do simple math)

Answer 40

Thank you so much for helping me! I really appreciated it!!

Answer 41

youre welcome, and good luck :)