Question #297345

{F} Calculate the approximate value of 10 to four decimal places by taking the first four


terms of an appropriate Taylor’s series.


1
Expert's answer
2022-02-15T11:25:02-0500

Question contains mistake. We have to approximate the value of 10.\sqrt{10}.



Taylor series is given as

f(a+h)=f(a)+h.f(a)+h22!f(a)+h33!f(a)+...f(a+h)=f(a)+h.f'(a)+{h^2\over 2!}f''(a)+{h^3\over 3!}f'''(a)+...

Let's Rewrite 1010 in the following form:

10=9+110=9+1

Now, let's have a function that will enable us use Taylor series method

f(10)=f(9+1)f(10)=f(9+1)

    \implies a=9a=9 and h=1h=1

Since we are asked to find the square root

    f(a)=a12    f(a+1)=10\implies f(a)=a^{1\over 2} \implies f(a+1)=\sqrt {10}

    f(a)=a12    f(a+1)=10\implies f(a)=a^{1\over 2} \implies f(a+1)=\sqrt {10}


Substituting the value of aa we have

f(a)=912=3f(a)=9^{1\over 2} =3

f(a)=12a121=12a12=12a12=12a12f'(a)={1\over 2}a^{{1\over 2}-1}={1\over 2}a^{-{1\over 2}}={{1\over 2}\over a^{1\over2}}={1\over 2a^{1\over 2}}

Substituting the value of aa we have

f(a)=12(9)12=0.166...f'(a)={1\over 2(9)^{1\over 2}}=0.166...

f(a)=(12)(12)a121=14a32=14a32=14a32f''(a)=(-{1\over 2})({1\over 2})a^{-{1\over 2}-1}=-{1\over 4}a^{-{3\over 2}}={-{1\over 4}\over a^{3\over 2}}=-{1\over 4a^{3\over 2}}

Substituting the value of aa we have

f(a)=14(9)32=0.00925f''(a)=-{1\over 4(9)^{3\over 2}}=0.00925

f(a)=(32)(14)a321=38a52=38a52=38a52f'''(a)=({-{3\over 2}})({-{1\over 4}})a^{-{3\over 2}-1}={3\over 8}a^{-{5\over 2}}={{3\over 8}\over a^{5\over 2}}={3\over 8a^{5\over 2}}

Substituting the value of aa we have

f(a)=38(9)52=0.000514f'''(a)={3\over 8(9)^{5\over 2}}=0.000514

From Taylor series method, we have

10=f(a+1)=f(a)+1.f(a)+122!f(a)+133!f(a)+...\sqrt {10}=f(a+1)=f(a)+1.f'(a)+{1^2\over 2!}f''(a)+{1^3\over 3!}f'''(a)+...

10=f(a+1)=f(a)+1.f(a)+122!f(a)+133!f(a)+...\sqrt {10}=f(a+1)=f(a)+1.f'(a)+{1^2\over 2!}f''(a)+{1^3\over 3!}f'''(a)+...

But

f(a)=3f(a)=3 , f(a)=0.166..f'(a)=0.166.. , f(a)=0.00925f''(a)=0.00925 , f(a)=0.000514f'''(a)=0.000514

    10=f(a+1)=5+0.1+0.009252+0.0005143!=5.10471\implies \sqrt {10}= f(a+1)=5+0.1+{0.00925\over 2}+{0.000514\over 3!}=5.10471

    10=f(a+1)=5+0.1+0.009252+0.0005143!=5.10471\implies \sqrt {10}= f(a+1)=5+0.1+{0.00925\over 2}+{0.000514\over 3!}=5.10471


\therefore 10=5.10471\sqrt{10}=5.10471

10=5.10471\sqrt{10}=5.10471

10=5.1047\sqrt{10}=5.1047

10=5.1047\sqrt{10}=5.1047 correct to four decimal places


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