Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

MacLaurins series question help

I need to expand f(x)= (1-x)^-1
I get 1-x+x^2-x^3
Is this the correct answer?

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Original post by K622338

I need to expand f(x)= (1-x)^-1
I get 1-x+x^2-x^3
Is this the correct answer?

Nope. All signs +ve

OP

Original post by RDKGames

Nope. All signs +ve

please can you explain why

Study Forum Helper

Original post by K622338

please can you explain why

Give me a reason why they shouldn't be and I'll tell you why it's wrong.

OP

Original post by RDKGames

Give me a reason why they shouldn't be and I'll tell you why it's wrong.

I did
f(0)= (1-0)^-1 =1
f'(0)= -(1-0)^-2 =-1
f''(0)= 2(1-0)^-3 =2
f'''(0) = -6(1-0)^-4 =-6
So I used the power rule which meant the sign changed for ever derivative I took

Study Forum Helper

Original post by K622338

I did
f(0)= (1-0)^-1 =1
f'(0)= -(1-0)^-2 =-1
f''(0)= 2(1-0)^-3 =2
f'''(0) = -6(1-0)^-4 =-6
So I used the power rule which meant the sign changed for ever derivative I took

f(x) = (1-x)^{-1} \Rightarrow (-1) \cdot (1-x)^{-2} \cdot \dfrac{d}{dx}(1-x) = (1-x)^{-2}

So f'(0)=1

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