As discussed in class, A(m) A(m) approaches e=2.71828183…… as m increases withou

Question

As discussed in class, A(m) A(m) approaches e=2.71828183…… as m increases without bounds. This exercise illustrates that fact. You'll need to enter numerical values of eA(m) that are correct to 4 digits (after the initial zeros). You can compute them with your calculator, and you may find it convenient to use scientific format. Note how the difference of e e and A(m) A(m) approaches zero, and A(m) A(m) approaches e e as m m grows. If you are curious you can see the first 10,000 digits of e here.

eA(1)= . eA(2

e-A'(1)= e-A(2) =. eA(10)= . eA(100)= . eA(1,000)= . eA(10,000)=

Explanation / Answer

A(m)=(1+(1/m))m

e-A(1) =2.718281828459 -(1+(1/1))1=2.718281828459 -2=0.718281828459 =0.7183

e-A(2) =2.718281828459 -(1+(1/2))2=2.718281828459 -2.25=0.468281828459 =0.4683

e-A(10) =2.718281828459 -(1+(1/10))10=2.718281828459-1.110=2.718281828459-2.5937424601=0.1245

e-A(100) =0.01347

e-A(1000) =0.001358

e-A(10000) =0.0001359

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