1. ## Taylor Series about -1

I have a result for part a) which is
1+(3/2*x)+((27/16)*x^2)+((27/16)*x^3)+......
using this and the fact that 1-3x=4-3(x+1) i need to find a taylor series for
g(x) =[16/(1-3x)^2]

I know its by substitution but I just don't know where to start.

I did post part a of this question but I was confident in what I was doing - this I haven't got clue as once I have resolved this I need to check first four terms in the Taylor series by finding first, second and third derivatives.

4. ## Re: Taylor Series about -1

If your answer a is correct (is that sure? Or? ...) than you can do (like you said) a substitution, if you say: x+1=t
Then you get: 16/(4-3t)^2, this is exactly the same as f(x).

EDIT:
I checked, question a seems right, I guess you've used the binomial serie?

5. ## Re: Taylor Series about -1

Originally Posted by Siron
If your answer a is correct (is that sure? Or? ...) than you can do (like you said) a substitution, if you say: x+1=t
Then you get: 16/(4-3t)^2, this is exactly the same as f(x).

EDIT:
I checked, question a seems right, I guess you've used the binomial serie?
Yes I have used binomial expansion that's what we r to use

6. ## Re: Taylor Series about -1

Have you solved the exercices? Or? ...

7. ## Re: Taylor Series about -1

Originally Posted by Siron
Have you solved the exercices? Or? ...
Not yet will have a go again.

8. ## Re: Taylor Series about -1

This is what I think, for your first exercice you have calculated the Taylor serie about 0 for f (so practically an maclaurin serie).
$\displaystyle \frac{16}{(4-3x)^2}=\frac{1}{\left(1-\frac{3}{4}x \right)^2}=1+\frac{3}{2}x+\frac{27}{16}x^2+\frac{2 7}{16}x^3$ (this is just an approximation)
Now you have to use this answer and the fact that $\displaystyle 1-3x=4-3(x+1)$.
If you say: $\displaystyle x+1=t$ then you get this $\displaystyle \frac{16}{(4-3t)^2}$ for the function $\displaystyle g$.
Do you notice any relation with $\displaystyle f$? ...

9. ## Re: Taylor Series about -1

yes I do notice that but I have to find the series about -1 and this is what I don't understand-

10. ## Re: Taylor Series about -1

Originally Posted by vidhi96
yes I do notice that but I have to find the series about -1 and this is what I don't understand-
I know the answer should be= (5/32)=(3x/32)+(27(x+1)^2/256)+(27(x+1)^3/256)

11. ## Re: Taylor Series about -1

I have tried and but I can't seem to get my head around this as this is really a new thing for me- can you please help any further