The Student Room Group

Taylor series

Hi I was wondering if anyone could explain how to get the value of A for the taylor series.

So in one example I have (x-1) and the mark scheme says A=1
but then I have (x+1) where A =-1
but then I have tan(x+pi/4) and A = pi/4

is there an actual way to find the A value?:confused:
Reply 1
Original post by Teddysmith123
Hi I was wondering if anyone could explain how to get the value of A for the taylor series.

So in one example I have (x-1) and the mark scheme says A=1
but then I have (x+1) where A =-1
but then I have tan(x+pi/4) and A = pi/4

is there an actual way to find the A value?:confused:


I am as confused as you are, about what is the question.
Original post by TeeEm
I am as confused as you are, about what is the question.



Hi sorry if my first post seemed a bit messy, but when using the taylor expansion we use f(A) instead of f(0) as f(0) is maclaurins series correct?

Now I understand that but when I am given a question like:

find the taylor series expansion of square root of x in ascending powers of (x-1) what value do I use as A and how exactly do I work this out?
Reply 3
Original post by Teddysmith123
Hi sorry if my first post seemed a bit messy, but when using the taylor expansion we use f(A) instead of f(0) as f(0) is maclaurins series correct?

Now I understand that but when I am given a question like:

find the taylor series expansion of square root of x in ascending powers of (x-1) what value do I use as A and how exactly do I work this out?


I am not sure what this means. I assume you are using some kind of formula.

if in powers of (x-1), the expansion is about x=1, in other words it will start converging close to x=1 first and then expand "outwards"

so you evaluate all the derivatives at x=1
Original post by TeeEm
I am not sure what this means. I assume you are using some kind of formula.

if in powers of (x-1), the expansion is about x=1, in other words it will start converging close to x=1 first and then expand "outwards"

so you evaluate all the derivatives at x=1


This is FP2 have you learnt chapter 6 yet?
Reply 5
Original post by Teddysmith123
This is FP2 have you learnt chapter 6 yet?


Have I learnt chapter 6 of FP2?

is this relevant?
Original post by TeeEm
Have I learnt chapter 6 of FP2?

is this relevant?



Yes because it is chapter 6 of FP2 that I am talking about. It would be difficult for you to understand what I'm talking about had you not have studied the topic yourself.
Reply 7
Original post by Teddysmith123
Yes because it is chapter 6 of FP2 that I am talking about. It would be difficult for you to understand what I'm talking about had you not have studied the topic yourself.


I see ...

did you read my last comment ..


if in powers of (x-1), the expansion is about x=1, in other words it will start converging close to x=1 first and then expand "outwards"

so you evaluate all the derivatives at x=1

in other words the value of x which makes the "bracket" zero

so powers of (x-1) require evaluation of derivatives at x=1
so powers of (x-2) require evaluation of derivatives at x=2
so powers of (x+3) require evaluation of derivatives at x=-3
so powers of (x+pi/4) require evaluation of derivatives at x=- pi/4.


PS I have not done chapter 6 in FP2, it was called very different in my days ...
Original post by TeeEm
I see ...

did you read my last comment ..


if in powers of (x-1), the expansion is about x=1, in other words it will start converging close to x=1 first and then expand "outwards"

so you evaluate all the derivatives at x=1

in other words the value of x which makes the "bracket" zero

so powers of (x-1) require evaluation of derivatives at x=1
so powers of (x-2) require evaluation of derivatives at x=2
so powers of (x+3) require evaluation of derivatives at x=-3
so powers of (x+pi/4) require evaluation of derivatives at x=- pi/4.


PS I have not done chapter 6 in FP2, it was called very different in my days ...


Ah ok, I've just read through what you've written and read over the explanation in my textbook and I think I may understand. Thanks anyway for your help! :smile:
Reply 9
Original post by Teddysmith123
Ah ok, I've just read through what you've written and read over the explanation in my textbook and I think I may understand. Thanks anyway for your help! :smile:


you are welcome

Quick Reply

Latest