Question contains mistake. We have to approximate the value of 10.
Taylor series is given as
f(a+h)=f(a)+h.f′(a)+2!h2f′′(a)+3!h3f′′′(a)+...
Let's Rewrite 10 in the following form:
10=9+1
Now, let's have a function that will enable us use Taylor series method
f(10)=f(9+1)
⟹ a=9 and h=1
Since we are asked to find the square root
⟹f(a)=a21⟹f(a+1)=10
⟹f(a)=a21⟹f(a+1)=10
Substituting the value of a we have
f(a)=921=3
f′(a)=21a21−1=21a−21=a2121=2a211
Substituting the value of a we have
f′(a)=2(9)211=0.166...
f′′(a)=(−21)(21)a−21−1=−41a−23=a23−41=−4a231
Substituting the value of a we have
f′′(a)=−4(9)231=0.00925
f′′′(a)=(−23)(−41)a−23−1=83a−25=a2583=8a253
Substituting the value of a we have
f′′′(a)=8(9)253=0.000514
From Taylor series method, we have
10=f(a+1)=f(a)+1.f′(a)+2!12f′′(a)+3!13f′′′(a)+...
10=f(a+1)=f(a)+1.f′(a)+2!12f′′(a)+3!13f′′′(a)+...
But
f(a)=3 , f′(a)=0.166.. , f′′(a)=0.00925 , f′′′(a)=0.000514
⟹10=f(a+1)=5+0.1+20.00925+3!0.000514=5.10471
⟹10=f(a+1)=5+0.1+20.00925+3!0.000514=5.10471
∴ 10=5.10471
10=5.10471
10=5.1047
10=5.1047 correct to four decimal places
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