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We have $f(x,y) = \sec(x+y^2)$

I want to find the first two non-zero terms of $f$ at $(0,0)$ starting by

  1. Taking the first few terms of $\cos x$ centered at zero, $1 - \frac{x^2}{2!} $

  2. Using this to get the first few terms of $\sec x$ centered at zero, $\frac{1}{1 - \frac{x^2}{2!}} = 1+\frac{x^2}{2} + \cdots = 1+\frac{x^2}{2} $ (see comment).

  3. I don't know how to use these to find the first two non-zero terms of $f$, which we are supposed to do.

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You know how to do polynomial long division? –  J. M. Nov 21 '11 at 11:49
    
I know how to do long division, it just didn't give me the results I wanted. I looked it up, and I found that I needed to know the expansion of $\frac{1}{1 - r}$. So the first terms of $\sec x$ are $1 + \frac{x^2}{2}$. There are more terms, but they're dropped off because we only care about the first two. –  brettk Nov 21 '11 at 11:57
    
So now that I have those two parts, I'm not sure how to prove what I want to prove. –  brettk Nov 21 '11 at 11:59

1 Answer 1

up vote 1 down vote accepted

You already have the first few terms of the Taylor expansion of $\sec w$ about $w=0$ (yes, the choice of variable is deliberate).

Everywhere that you see a $w$ in that expansion, write $x+y^2$. Now it depends how many terms you want. But from the wording of the question, it looks as if expanding the $w^2$ term will be enough, and you won't even need the "$y^4$" term.

Alternately, and as a check, look up the formula for the Taylor expansion of a function $f(x,y)$ of two variables. You will need to evaluate some higher partial derivatives, since the first partials are $0$ at $(0,0)$. From the wording of the question, that is not what you are being asked to do. The question is asking you to "recycle" a known expansion to obtain a new one.

Here is a simpler example to illustrate the idea. Suppose you want the Taylor expansion of $\sin(xy^2)$. We could take partial derivatives, but that's doing things the hard way. Just find the Taylor expansion of $\sin w$, and everywhere that you see $w$, substitute $xy^2$.

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I can't see how I missed that, but it was exactly what I needed. –  brettk Nov 21 '11 at 17:08

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