Thread: 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.

Please post the whole question.

please see attched question and my answer to part a -

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?

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

Have you solved the exercices? Or? ...

Originally Posted by Siron

Have you solved the exercices? Or? ...

Not yet will have a go again.

This is what I think, for your first exercice you have calculated the Taylor serie about 0 for f (so practically an maclaurin serie).
(this is just an approximation)
Now you have to use this answer and the fact that .
If you say: then you get this for the function .
Do you notice any relation with ? ...

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

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)

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