Taylor expansion of $\sqrt{x+a}$

I happen to read a book, in which it states that,

if $P_n(x)$ is the $n+1$ th Taylor polynomial (that is, $P_n(x)=a_0+a_1x+\cdots+a_{n}x^n$) of the function $\sqrt{x+a}$ $(a>0)$at $x=0$, then there is a polynomial $Q_n(x)$ such that

$$P_n^2(x)=x+a+x^{n+1}Q_n(x).$$

I have tried to calculate for small $n$, it shows the claim is right.

But I cannot find an easy way to prove it, or understand the result clearly. And is there a more general result for other functions, like $\sqrt[3]{x+a}$?

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You know the binomial theorem? – J. M. Aug 29 '11 at 14:46
@J. M.: I think that's unnecessarily specific -- this is true simply because of the degree of the error term. – joriki Aug 29 '11 at 14:53
I was thinking about how s/he'd derive $Q$, but you're spot on, @joriki. (meta: Why'd you delete your answer?) – J. M. Aug 29 '11 at 14:55
Yes, but I can't prove that coefficients of terms $x^k$($1<k\le n$) are zero. – NGY Aug 29 '11 at 14:57
@J. M.: Because in trying to make it look nicer I'd swapped the roles of $\sqrt{x+a}$ and $P_n(x)$ without noticing that squaring would then leave a term with $\sqrt{x+a}$, so I deleted it while fixing that. – joriki Aug 29 '11 at 14:57

1 Answer

Since $P_n$ is the $(n+1)$-th Taylor polynomial of the function, the error term is of order $x^{n+1}$:

$$\sqrt{x+a}=P_n(x)+R(x)$$

with $R(x)\in O(x^{n+1})$. Squaring that yields

$$x+a=P_n(x)^2+2P_n(x)R(x)+R(x)^2=P_n(x)^2+S(x)$$

with $S(x)\in O(x^{n+1})$. But $S(x)$ is the difference of two polynomials, and thus itself a polynomial, so if it's in $O(x^{n+1})$ it must be of the form $-x^{n+1}Q_n(x)$ with $Q_n(x)$ a polynomial.

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Thanks for your answer. I have not tried this way before(I just tried to calculate the coefficients of terms of $P_n(x)^2-x-a$, but it does not work). – NGY Aug 29 '11 at 15:09