Question

completing the table

Answer 1

the first column is
n
second column is : f^(n) x
third column is: f^(n) 0

Answer 2

why isn't it working?

Answer 3

I tried to copy and paste a drawing I drew yesterday and it's not working

Answer 4

:(

Answer 5

ok I'll just write it

Answer 6

n f^(n) x f^(n) (0)
0 1/x
1 -1/x^2
2 1/x^3
3 -1/x^4
4 1/x^5

Answer 7

Taylor series for f(x) = 1/x and a=-3

Answer 8

wouldn't the third column be infinity?

Answer 9

for taylor series of a=-3 u need f^(n) (-3) not f^(n) (0)

Answer 10

here is a table for f(1+x)^-3 in my solution manual with those three column...
it's gonna take me a min to write it. Let me know what you think about that table

Answer 11

ok

Answer 12

n f^(n) x f^(n) (0)
0 (1+x)^-3 1
1 -3(1+x)^-4 -3
2 12(1+x)^-5 12
3 -60(1+x)^-6 -60
4 360(1+x)^-7 360

Answer 13

I don't quite get the third column

Answer 14

is that taylor series for neighborhood of 0 ?

Answer 15

To be quite honest I'm not quite sure...I'm kinda new to Taylor and Maclaurin series

Answer 16

here is the formula that they wrote under it

Answer 17

\[(1+x)^-3= f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3........\]

Answer 18

Is this for Maclaurin

Answer 19

thats right its taylor series for neighborhood of 0

Answer 20

look at this
http://en.wikipedia.org/wiki/Taylor_series

go to Definition

Answer 21

ok so we are taking infinitely many derivatives to make the function go to zero?

Answer 22

:(

Answer 23

not Necessarily go to zero
for example derivatives of 1/x never goes to zero

Answer 24

Lets look at the approximation and convergence example of sin in wikipedia

Answer 25

ok

Answer 26

Pictured on the right is an accurate approximation of sin(x) around the point x = 0

What are we trying to approximate? What does the polynomial of degree 7 do for us?

Answer 27

wait a second

Answer 28

sinx approximately equals that list of polynomials?

Answer 29

oh so if we continue on the right hand with that series we should get sinx?

Answer 30

sorry
numerically --> yes

Answer 31

and What does the polynomial of degree 7 do for us?
sometimes working with polynomials is very easier than other functions like e^x or sin(x) or ln(1+x) ,etc so we approximate this functions by taylor series to make our calculations easier or even possible

Answer 32

i see...so why are we singling out the polynomial of degree 7? wouldn't we need the whole
series to find the approximate value of sinx

Answer 33

that was for example
we use different degrees of polynomials Based on our need

Answer 34

I have a question about that figure in wiki of sinx and the 7th degree polynomial
Are we trying to find a function that traces sinx around the point zero?

Answer 35

sorry if i'm asking the same question over and over

Answer 36

so we are taking derivatives to get a line (function) to trace or original function?

Answer 37

i meant to say "the original function"

Answer 38

yes thats right 'trace'

Answer 39

OMG it's finally coming together. so in the
"List of Maclaurin series of some common functions" in wiki
are we trying to find a function representative of image on the right?

Answer 40

wait a sec

Answer 41

I didn't go to a different page...I just scrolled down and found an image on the right that read
"The real part of the cosine function in the complex plane." The purple one

Answer 42

well thats completely different because its on the complex plane!

Answer 43

look at 4th and 5th figure at that page

Answer 44

yes, ok so that's the approximation of the sin function...just out of curiousity,
and briefly, how does that relate to the complex plane? or is that a completely different
animal and outside of my calc II scope?

Answer 45

I think CalcIII is when i will be introduced to complex planes...that figure just seemed interesting

Answer 46

yes

Answer 47

lol ok :P

back to the function
f(x)= (1+x)^-3

Answer 48

ok

Answer 49

where are u confused about?

Answer 50

in the Taylor series for neighborhood of zero the solution become
\[(1+x)^-3= f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3........\]
\[(1+x)^-3= 1-3x+\frac{4*3}{2!}x^2+\frac{5*4*3}{3!}x^3+........\]
so we're plugging in the third column into the series?

Answer 51

-60=-5*4*3

Answer 52

thats right

Answer 53

now u tell me why the first table for function f=1/x can not be completed for f^n(0)?

Answer 54

because that would be dividing by zero

Answer 55

thats right
because 1/x is not differentiable in the neighborhood of 0

Answer 56

u can complete the table for a=-3

Answer 57

how are we changing the graph? by allowing a=-3
http://www.wolframalpha.com/input/?i=1%2Fx+graph
I just need a visual :\

Answer 58

by the way...you have been completely awesome and patient in helping me understand Taylor series, Can't thank you enought

Answer 59

welcome