Use the second degree Maclaurin polynomial for \ln(1+x) to approximate the value of \ln(2)

opatovaL

opatovaL

Answered question

2021-09-07

Use the second degree Maclaurin polynomial for ln(1+x) to approximate the value of ln(2)

Answer & Explanation

Cullen

Cullen

Skilled2021-09-08Added 89 answers

Maclaurin series for a function
f(x) is given by
f(x)=f(0)+f(0)x+f(0)2!x2+
Take f(x)=ln(1+x)
f(x)=11+x=(1+x)1
f(x)=1(1+x)2=1(1+x)2
then f(0)=ln(1+0)=ln(1)=0
f(0)=11+0=1
f(0)=11+02=1
Substitute the values in A
ln(1+x)=0+1×x+(1)×x22!
ln(1+x)=xx22!
is the second degree Maclaurin polynomial for ln(1+x)
Substitute x=1 in ln(1+x)
ln(1+1)=1(1)22!
ln(2)=0.5
The approximate value of ln(2) is 0.5

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