We calculate the derivatives of *f*, evaluate them at 0, and plug them into the formula with the appropriate value of *n* for each part of the problem. Since the function we're looking at is *f*(*x*) = *e*^{x}, all the derivatives are also *e*^{x}, so each derivative evaluated at 0 is
*e*^{0} = 1.
We get then and so on, up to Another way to do this problem would be to find *g*_{5}(*x*) to start, and then cut the polynomial off at the appropriate degrees to get *g*_{1} through *g*_{4}. (a) The first-degree Taylor polynomial for *f*(*x*) at 0 is *g*_{1}(*x*) = 1 + *x*. This is our linear approximation: (b) The second-degree Taylor polynomial for *f*(*x*), centered at 0, is . This function curves more like *e*^{x} but still gets away pretty quickly: (c) The 3rd-degree Taylor polynomial for *f*(*x*) at 0 is .
If we graph the 2nd and 3rd degree polynomials with *f*(*x*) we can see that *g*_{3} sticks a little more closely to *f*(*x*) when *x* > 0: (d) The 4th-degree Taylor polynomial for *f*(*x*) at 0 is . This function gets even closer to *f*(*x*) = *e*^{x} for *x* > 0, and is starting to look a little more like *e*^{x} for *x* below zero, also. (e) The 5th-degree Taylor polynomial for *f*(*x*) at 0 is . Getting even closer... |