1.
f(x)=(1+x)21f′(x)=21(1+x)−21f′′(x)=−41(1+x)−23f′′′(x)=83(1+x)−25Taylorseries,f(x)=∑n=03n!fn(x0)(x−x0)nf(x)=f(x0)+1!f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+3!f′′′(x0)(x−x0)3f(x)=f(0)+1!f′(0)(x−0)+2!f′′(0)(x−0)2+3!f′′′(0)(x−0)3f(x)=f(0)+1!f′(0)(0)+2!f′′(0)(0)2+3!f′′′(0)(0)3f(x)=1+21x−81x2+483x3
2.
Given, (1.1)1/2.
Actual value of (1.1)1/2=1.048808848. (9 decimal places)
f(0.1)=(1.1)21f(0.1)=1+21(0.1)−81(0.1)2+483(0.1)3f(0.1)=1.048812500
Error=∣∣approx−exact∣∣=∣∣1.0488125−1.048808848∣∣=0.000003652
3.
Approximate ∫00.1(1+x)21dx ,
∫00.1f(x)dx=∫00.1(1+21x−81x2+483x3)dx=0.102459896
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