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Determine the highest order of contact at $x_0 = 0$

$f(x) = x^2$

$g(x) = \sin x$

My definition of contact is fuzzy. So I took the derivative at each step, and inspected when $f^{(k)}=g^{(k)}=0$

It's clear that $f^{(k)}(0)=0 \forall k$

$g'(0) = 1, g''(0)= 0, g'''(0)=-1, g''''(0)=0, ...$

Is the highest point of contact 2? or $\infty$?

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You have $f(0)=g(0)$, but $f^\prime(0)\neq g^\prime(0)$. "Order of contact" is related to the concept of two curves being tangent to each other. – J. M. Nov 30 '11 at 6:47
Prod @J.M. to turn that into an answer. – Willie Wong Nov 30 '11 at 8:47

To settle this:

As mentioned here, "order of contact" is related to the concept of tangency. For instance, we say that a curve and its tangent line have first-order contact, since the function and first derivative values agree at the point of intersection.

For the functions given in the OP, we know the Maclaurin ($x=0$) expansion $\sin\,x=x-\dfrac{x^3}{3!}+\cdots$; comparing this with $x^2$ shows that at $x=0$, the function values of these two functions agree, but the first derivative values don't. We can thus say that the order of contact is zero.

Put another way: it doesn't matter if the higher derivatives match; the order of contact between $f(x)$ and $g(x)$ is related to the degree of the first nonzero term of $f(x)-g(x)$. The first nonzero term of $\sin\,x-x^2$ is $x$, which is of degree $1$; the order of contact is thus $1-1=0$.

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