Question

Write out the first four nonzero terms of the Taylor series for f(x) = 5/(1 − x) about 0. Simplify all coefficients.
*my work will be written below shortly*
I just want to know why I got it wrong.

Answer 1

f(x)=5/(1-x) f(0)=5
f'(x)=5/((1-x)^2) f'(0)=5
f''(x)=5/((1-x)^3) f''(0)=10
f'''(x)=5/((1-x)^4) f'''(0)=30
f''''(x)=5/((1-x)^5) f''''(0)=120

Answer 2

5+5x+10x^2/2+30x^3/6+120x^4/24
5+5x+5x^2+5x^3+5x^4

Answer 3

\[\frac{d}{dx}5(1-x)^{-2} \\ =5(-2)(1-x)^{-3}(-1) \\ =10(1-x)^{-3} \\ =\frac{10}{(1-x)^{3}}\]

Answer 4

looks like you did the same thing on the other higher order derivatives

Answer 5

Oh crap I didn't write that in =(, but I did make up for it in f''(0) and f'''(0)...

Answer 6

that is interesting how did you find f''(0) without the correct f''(x) :p

Answer 7

I have it written correctly on the note pad I have currently in m lap haha. It just didn't translate on to here for some reason.

Answer 8

and yeah your evaluations at x=0 look right

Answer 9

Should have been:
f(x)=5/(1-x) f(0)=5
f'(x)=5/((1-x)^2) f'(0)=5
f''(x)=10/((1-x)^3) f''(0)=10
f'''(x)=30/((1-x)^4) f'''(0)=30
f''''(x)=120/((1-x)^5) f''''(0)=120

Answer 10

The only thing I can see wrong in your answer is you have 5 terms instead of 4

Answer 11

well I started at zero because I tried to start at the first term and I got the answer wrong.

Answer 12

5+5x+5x^2+5x^3+5x^4

I was talking instead of doing 5 terms
the question said do 4
so it should be
5+5x+5x^2+5x^3

Answer 13

oh I didn't think of it like that. do you mind working on a different on with me real quick?

Answer 14

i can try

Answer 15

Find the first four nonzero terms of the Taylor series for 1/y about y = −3. Simplify all coefficients.

Answer 16

f''(-3)=-2/27

Answer 17

f'''(-3)=-6/81

Answer 18

oops

Answer 19

but yeah you knew that you had your derivatives right

Answer 20

sorta lol

Answer 21

didn't know you said f'''(x)=-6/x^5
you meant f'''(x)=-6/x^4

Answer 22

f(y)=1/y, f(-3)=-(1/3)
f'(y)=-(1/y^2), f'(-3)=-(1/9)
f''(y)=2/(y^-3), f''(-3)=2/27
f'''(y)=-(6/(y^4)), f'''(-3)=-(6/81)

Answer 23

there is another type-o but I'm just going to shutup now :p

Answer 24

f(y)=1/y, f(-3)=-(1/3)
f'(y)=-(1/y^2), f'(-3)=-(1/9)
f''(y)=2/(y^3), f''(-3)=-2/27
f'''(y)=-(6/(y^4)), f'''(-3)=-(6/81)
there all better

and then it is the plug in time

Answer 25

\[f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2}+f'''(a) \frac{(x-a)^3}{3 \cdot 2 }\]

Answer 26

where a=-3

Answer 27

-1/3+((-1/9)(x+3))+(((-2/27)(x+3)^2)/2)+(((-6/81)(x+3)^3)/6)

Answer 28

and a little simplifying can be done

Answer 29

-1/3+((-1/9)(x+3))+(((-1/27)(x+3)^2))+(((-1/81)(x+3)^3))

Answer 30

looks pretty

Answer 31

pretty ugly :p

Answer 32

but I mean correct

Answer 33

haha I'm going to check and see if I get it right =)

Answer 34

go you!

Answer 35

wrong D=

Answer 36

no way!

Answer 37

\[\frac{-1}{3}-\frac{1}{9}(x+3)-\frac{1}{27}(x+3)^2-\frac{1}{81}(x+3)^3\]
you typed in this?

Answer 38

i mean if so can i see a picture of it

Answer 39

oh darn

Answer 40

oh we are wrong

Answer 41

maybe I should try terms 1-4 excluding 0?

Answer 42

we forgot that we were dealing with x not y

Answer 43

i mean y not x

Answer 44

those dang x's need to be y's I'm sorry

Answer 45

OH! i didn't even notice

Answer 46

yeah and I totally forgot too

Answer 47

Yeah that was it lol

Answer 48

did you still get full credit?

Answer 49

Yes I did thanks

Answer 50

k cool stuff