Kkutie7
  • Kkutie7
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Kkutie7
  • Kkutie7
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
Kkutie7
  • Kkutie7
5+5x+10x^2/2+30x^3/6+120x^4/24 5+5x+5x^2+5x^3+5x^4
freckles
  • freckles
\[\frac{d}{dx}5(1-x)^{-2} \\ =5(-2)(1-x)^{-3}(-1) \\ =10(1-x)^{-3} \\ =\frac{10}{(1-x)^{3}}\]

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freckles
  • freckles
looks like you did the same thing on the other higher order derivatives
Kkutie7
  • Kkutie7
Oh crap I didn't write that in =(, but I did make up for it in f''(0) and f'''(0)...
freckles
  • freckles
that is interesting how did you find f''(0) without the correct f''(x) :p
Kkutie7
  • Kkutie7
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.
freckles
  • freckles
and yeah your evaluations at x=0 look right
Kkutie7
  • Kkutie7
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
freckles
  • freckles
The only thing I can see wrong in your answer is you have 5 terms instead of 4
Kkutie7
  • Kkutie7
well I started at zero because I tried to start at the first term and I got the answer wrong.
freckles
  • freckles
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
Kkutie7
  • Kkutie7
oh I didn't think of it like that. do you mind working on a different on with me real quick?
freckles
  • freckles
i can try
Kkutie7
  • Kkutie7
Find the first four nonzero terms of the Taylor series for 1/y about y = −3. Simplify all coefficients.
freckles
  • freckles
f''(-3)=-2/27
freckles
  • freckles
f'''(-3)=-6/81
Kkutie7
  • Kkutie7
oops
freckles
  • freckles
but yeah you knew that you had your derivatives right
freckles
  • freckles
sorta lol
freckles
  • freckles
didn't know you said f'''(x)=-6/x^5 you meant f'''(x)=-6/x^4
Kkutie7
  • Kkutie7
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)
freckles
  • freckles
there is another type-o but I'm just going to shutup now :p
freckles
  • freckles
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
freckles
  • freckles
\[f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2}+f'''(a) \frac{(x-a)^3}{3 \cdot 2 }\]
freckles
  • freckles
where a=-3
Kkutie7
  • Kkutie7
-1/3+((-1/9)(x+3))+(((-2/27)(x+3)^2)/2)+(((-6/81)(x+3)^3)/6)
freckles
  • freckles
and a little simplifying can be done
Kkutie7
  • Kkutie7
-1/3+((-1/9)(x+3))+(((-1/27)(x+3)^2))+(((-1/81)(x+3)^3))
freckles
  • freckles
looks pretty
freckles
  • freckles
pretty ugly :p
freckles
  • freckles
but I mean correct
Kkutie7
  • Kkutie7
haha I'm going to check and see if I get it right =)
freckles
  • freckles
go you!
Kkutie7
  • Kkutie7
wrong D=
freckles
  • freckles
no way!
freckles
  • freckles
\[\frac{-1}{3}-\frac{1}{9}(x+3)-\frac{1}{27}(x+3)^2-\frac{1}{81}(x+3)^3\] you typed in this?
freckles
  • freckles
i mean if so can i see a picture of it
freckles
  • freckles
oh darn
freckles
  • freckles
oh we are wrong
Kkutie7
  • Kkutie7
maybe I should try terms 1-4 excluding 0?
1 Attachment
freckles
  • freckles
we forgot that we were dealing with x not y
freckles
  • freckles
i mean y not x
freckles
  • freckles
those dang x's need to be y's I'm sorry
Kkutie7
  • Kkutie7
OH! i didn't even notice
freckles
  • freckles
yeah and I totally forgot too
Kkutie7
  • Kkutie7
Yeah that was it lol
freckles
  • freckles
did you still get full credit?
Kkutie7
  • Kkutie7
Yes I did thanks
freckles
  • freckles
k cool stuff

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