MEPS1996
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In my textbook i have read that g(a) +g'(a)x/1! + g''(a)x^2/2! + g'''(a)x^3/3!..
is the taylor expansion of g(a+x). Isn't this just the maclaurin expansion?
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ztibor
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(Original post by MEPS1996)
In my textbook i have read that g(a) +g'(a)x/1! + g''(a)x^2/2! + g'''(a)x^3/3!..
is the taylor expansion of g(a+x). Isn't this just the maclaurin expansion?
NO.

The Taylor expansion around a is

\displaystyle g(a)+\frac{g'(a)}{1!}(x-a)+\frac{g''(a)}{2!}(x-a)^2+\frac{g'''(a)}{3!}(x-a)^3+...

For a=0 it is known as Maclaurin series
THat is

\displaystyle g(0)+\frac{g'(0)}{1!}\cdot x+\frac{g''(0)}{2!}\cdot x^2+\frac{g'''(0)}{3!}\cdot x^3+...

Can you write me which one from the above that you have read in textbook?
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MEPS1996
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(Original post by ztibor)
NO.

The Taylor expansion around a is

\displaystyle g(a)+\frac{g'(a)}{1!}(x-a)+\frac{g''(a)}{2!}(x-a)^2+\frac{g'''(a)}{3!}(x-a)^3+...

For a=0 it is known as Maclaurin series
THat is

\displaystyle g(0)+\frac{g'(0)}{1!}\cdot x+\frac{g''(0)}{2!}\cdot x^2+\frac{g'''(0)}{3!}\cdot x^3+...
but it says it is the taylor expansion of g(a+x). If you were to find the maclaurin series of g(a+x), ie. differentiate it then put x=0 into the differential and use these in the coefficients, you would get g(a) + g'(a)/1!x +g''(a)/2!x^2.... So it seems as though the maclaurin series and taylor series of g(a+x) are the same...
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(Original post by MEPS1996)
but it says it is the taylor expansion of g(a+x). If you were to find the maclaurin series of g(a+x), ie. differentiate it then put x=0 into the differential and use these in the coefficients, you would get g(a) + g'(a)/1!x +g''(a)/2!x^2.... So it seems as though the maclaurin series and taylor series of g(a+x) are the same...
The Maclaurin series is just a special case of the Taylor series when a=0. It's only used for points around 0, because it's not such a good approximation away from there and we use the Taylor expansion instead.

Simply put, The Taylor series is the general case and the Maclaurin series is just a unique case of Taylor.

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MEPS1996
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#5
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(Original post by majmuh24)
The Maclaurin series is just a special case of the Taylor series when a=0. It's only used for points around 0, because it's not such a good approximation away from there and we use the Taylor expansion instead.

Simply put, The Taylor series is the general case and the Maclaurin series is just a unique case of Taylor.

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its just ive read that the taylor expansion of g(a+x) is: g'(a)/1!x +g''(a)/2!x^2....
This seems to be what you would get if you were to find the maclaurin expansion of g(a+x).
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(Original post by MEPS1996)
its just ive read that the taylor expansion of g(a+x) is: g'(a)/1!x +g''(a)/2!x^2....
This seems to be what you would get if you were to find the maclaurin expansion of g(a+x).
It's the other way round, but what you're saying is correct in a way.

The Taylor expansion is to do with f(x+a), the Maclaurin expansion is when a=0.

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davros
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(Original post by MEPS1996)
its just ive read that the taylor expansion of g(a+x) is: g'(a)/1!x +g''(a)/2!x^2....
This seems to be what you would get if you were to find the maclaurin expansion of g(a+x).
The usual form of the Taylor expansion is the one given by ztibor. If you put y = x+a in your version, so x = y-a, you basically get a rearrangement of the same thing.
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MEPS1996
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(Original post by davros)
The usual form of the Taylor expansion is the one given by ztibor. If you put y = x+a in your version, so x = y-a, you basically get a rearrangement of the same thing.
but its exactly the same as the macluarin expansion of g(a+x)...
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ztibor
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(Original post by MEPS1996)
but its exactly the same as the macluarin expansion of g(a+x)...
The Macalurin series of f(x) is an approximation for f(0) at x=0 in form of series of infinitely
many terms.
So there is need for nth order differentiability of f(x) at x=0 and not at x=a or y=x+a (except a=0)
and we calculate the derivative for x=0 so the will be powers of (x-0) in the series.

For x=a (a=/=0) it will be the Taylor series so that the derivative is calculates in x=a e.g. g'"(a)
but then there are powers of (x-a) in the series e.g (x-a)^3

In your post you gave the form so that for the derivatives you used the x=a substitution (which is g(x) around a), but at the powers you gave powers of x, that is (x-0) (which is g(x) around 0).

This is a contradiction except a=0;
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MEPS1996
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(Original post by ztibor)
The Macalurin series of f(x) is an approximation for f(0) at x=0 in form of series of infinitely
many terms.
So there is need for nth order differentiability of f(x) at x=0 and not at x=a or y=x+a (except a=0)
and we calculate the derivative for x=0 so the will be powers of (x-0) in the series.

For x=a (a=/=0) it will be the Taylor series so that the derivative is calculates in x=a e.g. g'"(a)
but then there are powers of (x-a) in the series e.g (x-a)^3

In your post you gave the form so that for the derivatives you used the x=a substitution (which is g(x) around a), but at the powers you gave powers of x, that is (x-0) (which is g(x) around 0).

This is a contradiction except a=0;
thanks a lot
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