Example 5.1.1

Here is the power series, which we will truncate to guess radius of convergence:

f[x_,m_]:=(∑_(n=0)^m)(((x-3))^n)

Here is a list of the truncated series, which is not output to save space:

functionlist=Table[f[x,m],{m,1,100,10}];

Here is a plot, which is uninformative:

Plot[Evaluate[functionlist],{x,-10,10}]

Here is a more imformative plot:

Plot[Evaluate[functionlist],{x,0,5},PlotRange->{Automatic,{-10,10}},ImageSize->{400,Automatic}]

Here we can readily see that convergence is only for 2 < x< 4, which is what we determined using the ratio test.

Example 5.1.8

Limit[(1-1/(n+1))^n,n->Infinity]
Limit[(1-1/n)^n,n->Infinity]

Here is the power series, which we will truncate to guess radius of convergence:

f[x_,m_]:=(∑_(n=1)^m)((n! (x^n))/(n^n))

Here is a list of the truncated series, which is not output to save space:

functionlist=Table[f[x,m],{m,1,100,10}];

Here is a plot, which is uninformative:

Plot[Evaluate[functionlist],{x,-10,10}]

Here is a more imformative plot:

Plot[Evaluate[functionlist],{x,-5,5},PlotRange->{Automatic,{-20,20}},ImageSize->{400,Automatic}]

Here we can readily see that convergence is only for -e < x< e, which is what we determined using the ratio test.

Example 5.1.15

f[x_]=(1/(1-x));
x0=0;
Df[x_,n_]:=Derivative[n][f][x]
Table[{n,Df[x,n],Df[x0,n]},{n,0,6}]//TableForm

Here is the power series, which we will truncate to guess radius of convergence:

f2[x_,m_]:=(∑_(n=0)^m)(x^n)

Here is a list of the truncated series, which is not output to save space:

functionlist=Table[f2[x,m],{m,1,20,4}];

Here is a more imformative plot:

Plot[f[x],{x,-2,2},PlotStyle->{Red,Thickness[0.01]},PlotRange->{All,{-20,20}},DisplayFunction->Identity]
Plot[Evaluate[functionlist],{x,-2,2},PlotRange->{Automatic,{-20,20}},ImageSize->{400,Automatic},DisplayFunction->Identity]
Show[%,%%,DisplayFunction->$DisplayFunction]

Here we can readily see that convergence is only good for -1 < x< 1, which is what we determined using the ratio test. The Red line is the original function.

Example 5.1.16

f[x_]=(1/(1-x));
x0=2;
Df[x_,n_]:=Derivative[n][f][x]
Table[{n,Df[x,n],Df[x0,n]},{n,0,6}]//TableForm

Here is the power series, which we will truncate to guess radius of convergence:

f2[x_,m_]:=(∑_(n=0)^m)(((-1))^(n+1))(((x-2))^n)

Here is a list of the truncated series, which is not output to save space:

functionlist=Table[f2[x,m],{m,1,20,4}];

Here is a more imformative plot:

Plot[f[x],{x,0,4},PlotStyle->{Red,Thickness[0.01]},PlotRange->{All,{-20,20}},DisplayFunction->Identity]
Plot[Evaluate[functionlist],{x,0,4},PlotRange->{Automatic,{-20,20}},ImageSize->{400,Automatic},DisplayFunction->Identity]
Show[%,%%,DisplayFunction->$DisplayFunction]

Here we can readily see that convergence is only good for 1 < x< 3, which is what we determined using the ratio test. The Red line is the original function.