Answer to Question #220077 in Calculus for Nick

Question

Calculate the approximate value of √26 to four decimal places using Taylor series method.

Answer 1

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 $26$ in the following form:

$26=25+1$ Since we know square root of $25$

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

$f(26)=f(25+1)$

$\implies$ $a=25$ and $h=1$

Since we are asked to find the square root

$\implies f(a)=a^{1\over 2} \implies f(a+1)=\sqrt {26}$

Substituting the value of $a$ we have

$f(a)=25^{1\over 2} =5$

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

Substituting the value of $a$ we have

$f'(a)={1\over 2(25)^{1\over 2}}=0.1$

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

Substituting the value of $a$ we have

$f''(a)=-{1\over 4(25)^{3\over 2}}=-0.002$

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

Substituting the value of $a$ we have

$f'''(a)={3\over 8(25)^{5\over 2}}=0.00012$

From Taylor series method, we have

$\sqrt {26}=f(a+1)=f(a)+1.f'(a)+{1^2\over 2!}f''(a)+{1^3\over 3!}f'''(a)+...$

$\implies \sqrt {26}= f(a+1)=f(a)+f'(a)+{f''(a)\over 2}+{f'''(a)\over 3!}+...$

But

$f(a)=5$ , $f'(a)=0.1$ , $f''(a)=-0.002$ , $f'''(a)=0.00012$

$\implies \sqrt {26}= f(a+1)=5+0.1+{(-0.002)\over 2}+{0.00012\over 3!}=5.09902$

$\therefore$ $\sqrt{26}=5.0990$ correct to four decimal places