# Taylor series

Watch
Announcements
Thread starter 5 years ago
#1
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?
0
reply
5 years ago
#2
(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?
I am as confused as you are, about what is the question.
0
reply
Thread starter 5 years ago
#3
(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?
0
reply
5 years ago
#4
(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
0
reply
Thread starter 5 years ago
#5
(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?
0
reply
5 years ago
#6
(Original post by Teddysmith123)
This is FP2 have you learnt chapter 6 yet?
Have I learnt chapter 6 of FP2?

is this relevant?
0
reply
Thread starter 5 years ago
#7
(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.
0
reply
5 years ago
#8
(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 ...
0
reply
Thread starter 5 years ago
#9
(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!
1
reply
5 years ago
#10
(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!
you are welcome
0
reply
X

Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (78)
26.44%
No - I have already returned home (36)
12.2%
No - I plan on travelling outside these dates (59)
20%
No - I'm staying at my term time address over Christmas (31)
10.51%
No - I live at home during term anyway (91)
30.85%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.