C4 Binomial Expansion using FP2 Maclaurin?

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lecafe88
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Hi all, I just want some clarification as to whether using FP2 methods of solving C4 Binomial expansion is valid. I get how the C4 binomial expansion works and I know the formula without looking at the sheet, but when being asked binomial questions I prefer using the FP2 trick of differentiating the term again and again and sticking in f(0), f'(0)...etc. I get same answers for the questions I have done, but is it true for all functions being asked at C4?

So the Binomial formula for is (1+ nx+ n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3!........)

Which is similar to the Taylor series f(0)+f'(0)x+f''(0)(x^2)/2!+f'''(0)(x^3)/3!

Say we have function, (9+8x)^0.5 or sqrt (9+8x),

The binomial gives (3) (1+ 4x/9 - 8x^2/81) = 3+4x/3-8x^2/27.

Taylor series gives

f(x)=(9+8x)^0.5
f'(x)=4(9+8x)^-0.5
f''(x)=-16(9+8x)^-1.5

And subbing 0s into the derivatives with the appropriate factorial division and x^n terms gives 3+4x/3-8x^2/27 as well.

Is this type of approach valid for all of these question types and will I get credit if I use this method, and is the binomial expansion just a special case of a taylor/maclaurin series?
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brianeverit
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(Original post by lecafe88)
Hi all, I just want some clarification as to whether using FP2 methods of solving C4 Binomial expansion is valid. I get how the C4 binomial expansion works and I know the formula without looking at the sheet, but when being asked binomial questions I prefer using the FP2 trick of differentiating the term again and again and sticking in f(0), f'(0)...etc. I get same answers for the questions I have done, but is it true for all functions being asked at C4?

So the Binomial formula for is (1+ nx+ n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3!........)

Which is similar to the Taylor series f(0)+f'(0)x+f''(0)(x^2)/2!+f'''(0)(x^3)/3!

Say we have function, (9+8x)^0.5 or sqrt (9+8x),

The binomial gives (3) (1+ 4x/9 - 8x^2/81) = 3+4x/3-8x^2/27.

Taylor series gives

f(x)=(9+8x)^0.5
f'(x)=4(9+8x)^-0.5
f''(x)=-16(9+8x)^-1.5

And subbing 0s into the derivatives with the appropriate factorial division and x^n terms gives 3+4x/3-8x^2/27 as well.

Is this type of approach valid for all of these question types and will I get credit if I use this method, and is the binomial expansion just a special case of a taylor/maclaurin series?
I THINK YOU WOULD GET FULL CREDSIT FOR IT AND YES THE bINOMIAL IS JUST AN EXAMPLE OF A TAYLOR SERIES, HOWEVER I WOULD THINK THAT IT IS A LOT QUICKER TO SIMPLY QUOTE THE STANDARD SERIES RATHER THAN USING REPEATED DIFFERENTIATION. I know that is certainly how I would do them.
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lecafe88
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(Original post by brianeverit)
I THINK YOU WOULD GET FULL CREDSIT FOR IT AND YES THE bINOMIAL IS JUST AN EXAMPLE OF A TAYLOR SERIES, HOWEVER I WOULD THINK THAT IT IS A LOT QUICKER TO SIMPLY QUOTE THE STANDARD SERIES RATHER THAN USING REPEATED DIFFERENTIATION. I know that is certainly how I would do them.
Thanks for the reply. I have trouble using a calculator especially fractions as I don't know how to input fractions properly, to do the square of (9/8) for example I usually do 9x9/8/8. That's plenty of places to trip up on for a binomial question, so I'd rather do differentiation and sticking 0s in.
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username1560589
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(Original post by lecafe88)
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Yes, this approach works for all binomial expansion questions. You can easily prove it by setting f(x) = (a+x)^n.

Binomial expansion is a special case of Maclaurin's series. It is quicker to just use binomial expansion though rather than differentiating.

You will get credit if you provide enough explanation of what you are dong to the examiner.
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davros
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(Original post by lecafe88)
Thanks for the reply. I have trouble using a calculator especially fractions as I don't know how to input fractions properly, to do the square of (9/8) for example I usually do 9x9/8/8. That's plenty of places to trip up on for a binomial question, so I'd rather do differentiation and sticking 0s in.
You shouldn't need a calculator to work out (9/8)^2 and in fact unless a fraction should have been simplified first you'll typically just be using (a/b)^2 = a^2/b^2 all the time anyway

But the answer to your original question is yes - you'll get the same answer, basically because if you represent a function by a power series then that series is unique in the interval of convergence.
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