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I was trying to do a Taylor expansion, and could not get the result Wolfram alpha gave me.

I tried expanding $(1-x+x^2)^q$ to which Wolfram alpha produced

$$1-qx-(1/2)q(q+1)x^2.$$

I know the Taylor series is

$$f(0)+xf'(x)+x^2\frac{f''(x)}{2!} +...$$

The first term should be $1$, yet when I compute it, I arrive at $1^q$, and I am confused from here. Any advice?

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    Hem, what is $1^q$ ? – 2017-02-09
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    @Peter: I am not asking how he came to this expression, I am asking him what $1^q$ is. – 2017-02-09
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    1^(q) was what I got for the first term, for f(a), the first term in the Taylor expansion. I know that is wrong, and it should 1, but I cannot think of a good reason for why this is the case. – 2017-02-09
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    @YvesDaoust Now, I understand what you mean! – 2017-02-09
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    The formula contains a typo : It has to be "1-qx+(1/2)..." – 2017-02-09

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Hint : The first two derivates of $$f(x)=(1-x+x^2)^q$$ are

$$f'(x)=(2x-1)\cdot q(1-x+x^2)^{q-1}$$

$$f''(x)=(2x-1)^2\cdot q(q-1)(1-x+x^2)^{q-2}+2q(1-x+x^2)^{q-1}$$

Plugging in $0$ gives $f'(0)=-q$ and $f''(0)=q(q-1)+2q=q^2-q+2q=q^2+q$

Note that $1^x=1$ for every real $x$

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    You give everything, except what the OP is querying about :-) – 2017-02-09
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    I have already arrived at those very derivatives, I just can't get my final answer in the form 1-qx + (1/2)q(q+1)x^(2). – 2017-02-09
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    Peter, thank you. I applied this example I created to a different problem involving a Taylor series of dot products and fractions. I arrived at the term I expected. So, when one is solving a Taylor expansion, about x=0, one plugs x=0 into the derivative of the term I am is solving for? If I wanted to expand around x=3, would I solve the problem the same way and plug in for x=3 after I take the derivative for a given term? What does the a term mean in the Taylor series f(x) = f(a) + f'(x)*f(x-a) + ... ? I originally thought I was supposed to set a=0. – 2017-02-09
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    The general formula is $f(x)=f(a)+f'(a)\cdot (x-a)+f''(a)/2\cdot (x-a)^2+\cdots$. In the case $a=0$ we get the Maclaurin-Series $f(x)=f(0)+f'(0)\cdot x+f''(0)/2 \cdot x^2+\cdots$ – 2017-02-09