Sections 11.10: p. 789, #2, 12, 48

#2a: The series has negative slope at x=1, but the graph is increasing.

#2b: While the function value may be about 2.8 per the series, and the slope is positive as is the coefficient 0.5, the second derivative of the series is 1.5, which suggests that the graph should be concave up -- not concave down, as the graph indicates.

#12:

f[x_]:=Cosh[x]
fPoly[x_]:=1+0/1*x+1/2*x^2+0/Factorial[3]*x^3+1/Factorial[4]*x^4
n=5
fSeries[x]:=(∑_(i=0)^n)((x^(2i))/(Factorial[2i]))
Plot[{f[x],fPoly[x],fSeries[x]},{x,-2,2}]

The ratio test shows that the radius of convergence is infinite.

#48: We take the difference of the series for e^x and 1, and divide by 1: so we get

f[x_]:=(Exp[x]-1)/x
fPoly[x_]:=((1+x+x^2/Factorial[2]+x^3/Factorial[3]+x^4/Factorial[4])-1)/x
n=8
fSeries[x_]:=(∑_(i=1)^n)((x^(i-1))/(Factorial[i]))
Plot[{f[x],fPoly[x],fSeries[x]},{x,-2,2}]

Now we need to integrate f: and so we get

int[x_]:=(∑_(i=1)^n)((x^i)/(i*Factorial[i]))
Plot[int[x],{x,0,2}]