## Expansions and order of contact

I am asking my mentor for my class about this question, but unfortunately the answers I get from her often take forever and do not always clear things up for me, so I hope someone out there in Physics Forum has a good way of explaining this to me. Here goes...

The question has some Mathematica input (Normal[Series] command) and output for the function f(x) = e -xCos(x). Then the question is asked "Mathematica is trying to tell you that, at x = 0, e-x Cos[x] and 1 - x + (x3)/3 - (x4)/6 + (x5)/5 have order of contact ____?"

Now to answer, I went back to what we learned about the definition of an expansion in powers of x was. This definition from the materials is: Given a function f[x], the expansion of f[x] in powers of x is: a[0] + a[1]x + a2[x]2 + ... + a[k]xk, where the numbers a[0], a[1], a[2], ..., a[k] are chosen such that for every positive integer m, the function f[x] and the polynomial a[0] + a[1]x + a[2]x2 + ... + a[m]xm have order of contact m at x = 0.

From this definition, I believed the answer to the question would be "5" because we see that in the expansion given, we have a 5th degree polynomial. Instead, the answer was 4. Can you please explain where my reasoning was faulty, and why the actual answer is 4? My guess is that the (x5)/5 term is actually xm+1making it a 5th degree polynomial where m = 4. Is that correct? Or am I barking up the wrong tree?

 I plugged in $$e^{-x}\cos(x)-[1-x+x^3/3-x^4/6+x^5/5]$$ into wolfram alpha to find the taylor expansion, it was $$-x^5/6-x^7/630+x^8/2520-x^9/22680+O(x^{11})$$ This tells me that $$e^{-x}\cos(x)=1-x+x^3/3-x^4/6+x^5/30+O(x^7)$$ So they agree up to fourth order, but not to fifth order. For instance, since the first order terms are the same, their tangent planes are the same, that is, the contact is at least of order 1.