# Taylor series

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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?

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?

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#2

(Original post by

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?

**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?

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(Original post by

I am as confused as you are, about what is the question.

**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?

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#4

(Original post by

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?

**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?

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**

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(Original post by

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

if in powers of

so you evaluate all the derivatives

**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**
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#6

(Original post by

This is FP2 have you learnt chapter 6 yet?

**Teddysmith123**)This is FP2 have you learnt chapter 6 yet?

is this relevant?

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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.

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#8

(Original post by

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.

**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.

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 ...

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(Original post by

I see ...

did you read my last comment ..

if in powers of

so you evaluate all the derivatives

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 ...

**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 ...

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#10

(Original post by

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!

**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!

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