Question

Use Taylor's theorem to find the value of
$\sqrt{25.15}$
​

Answer 1

The value of $\sqrt{25.15}$ by Taylor's theorem is 5.01498.

Step-by-step explanation:

Given: f(a+h) = $\sqrt{25.15}$

To Find: Value of $\sqrt{25.15}$ by Taylor's theorem

Solution:

• Finding the value of $\sqrt{25.15}$ by Taylor's theorem

Consider $f(a+h) = \sqrt{25.15} = \sqrt{25 + 0.15}$ such that a = 25 and h = 0.15. The Taylor's series for $\sqrt{25.15}$ can be written as,

$f (a+h) = \sqrt{25.15} = f(a) + h f'(a) + \dfrac{h^{2}}{2!}f''(a) + \dfrac{h^{3}}{3!}f'''(a) + \ \cdots \cdots$

Since $f(a) = \sqrt{a}$, therefore,

$f (a+h) = \sqrt{25.15} = \sqrt{a} + h\Big(\dfrac{1}{2\sqrt{a}} \Big) + \dfrac{h^{2}}{2!}\Big(\dfrac{-1}{4a^{3/2}} \Big) + \dfrac{h^{3}}{3!}\Big(\dfrac{3}{8a^{5/2}} \Big)+ \ \cdots \cdots$

Substituting the values of 'a' and 'h' in the above equation, we will have,

$\Rightarrow f (a+h) = \sqrt{25.15} = \sqrt{25} + (0.15)\Big(\dfrac{1}{2\sqrt{25}} \Big) - \dfrac{(0.15)^{2}}{2}\Big(\dfrac{1}{4(25)^{3/2}} \Big) + \dfrac{(0.15)^{3}}{6}\Big(\dfrac{3}{8(25)^{5/2}} \Big)+ \ \cdots \cdots$

$\Rightarrow \sqrt{25.15} = 5 + 0.015 - 0.000045 + 8.4375 \times 10^{-9} + \ \cdots \cdots$

$\Rightarrow \sqrt{25.15} \approx 5.01498$

Hence, the value of $\sqrt{25.15}$ by Taylor's theorem is 5.01498.