Answer to Question #297345 in Calculus for Tokyo

Question

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

terms of an appropriate Taylor’s series.

Answer 1

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

Taylor series is given as

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

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

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

$\implies$ $a=9$ and $h=1$

Since we are asked to find the square root

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

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

Substituting the value of $a$ we have

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

$f'(a)={1\over 2}a^{{1\over 2}-1}={1\over 2}a^{-{1\over 2}}={{1\over 2}\over a^{1\over2}}={1\over 2a^{1\over 2}}$

Substituting the value of $a$ we have

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

$f''(a)=(-{1\over 2})({1\over 2})a^{-{1\over 2}-1}=-{1\over 4}a^{-{3\over 2}}={-{1\over 4}\over a^{3\over 2}}=-{1\over 4a^{3\over 2}}$

Substituting the value of $a$ we have

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

$f'''(a)=({-{3\over 2}})({-{1\over 4}})a^{-{3\over 2}-1}={3\over 8}a^{-{5\over 2}}={{3\over 8}\over a^{5\over 2}}={3\over 8a^{5\over 2}}$

Substituting the value of $a$ we have

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

From Taylor series method, we have

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

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

But

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

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

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

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

$\sqrt{10}=5.10471$

$\sqrt{10}=5.1047$

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