Question

How do you find the Taylor expansion of\[\sqrt x \]at $ x = 1 $ ?

Answer 1

Hint: In order to calculate the Taylor polynomial of the $ f(x) $ which is centred at $ x = a = 1 $ ,use the formula of Taylor polynomial $ {T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}} $ ,up to the value of $ n = \infty $ but for simplicity we find terms up to $ n = 3 $ .Find the derivative of the function up to the order 3 using the property of $ \dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} $ and chaining rule .Putting all the values back into the formula will give your required result.

Complete step-by-step answer:
We are given a function \[f(x) = \sqrt x \]
Let’s have a look into the Taylor formula which we are going to use,
 $ {T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}} $
According to our question, Taylor Polynomial for $ f $ is centred at $ x = a = 1 $
In order to calculate the Taylor expansion of $ f $ ,Since we have not given any information about upto which term we have to expand ,so we have to extend summation up to the term having $ n = \infty $
But as we know that it is not actually possible to go up to $ n = \infty $ .SO for the simplicity we will define the starting 3 terms of the expansion
If we closely look the formula, we conclude that the each term in the formula requires:
1.A derivative of order (i). The $ i $ above the f looks like an exponent but its is not .It is actually representing the order of the derivative.
2.Putting the value of $ a $ in the derivatives.
3.Then Dividing every term by the factorial number $ i! $
4.Finally Multiplying by $ (x - a) $ raised to the power $ i $ .
Let’s Expand our formula up to $ n = 3 $ ,we get
 $
  {T_3}(x) = f(a) + \dfrac{{{f^{(1)}}(a)}}{{1!}}{(x - a)^1} + \dfrac{{{f^{(2)}}(a)}}{{2!}}{(x - a)^2} + \dfrac{{{f^{(3)}}(a)}}{{3!}}{(x - a)^3}....... \\
  {T_3}(x) = f(1) + \dfrac{{{f^{(1)}}(1)}}{{1!}}{(x - 1)^1} + \dfrac{{{f^{(2)}}(1)}}{{2!}}{(x - 1)^2} + \dfrac{{{f^{(3)}}(1)}}{{3!}}{(x - 1)^3}........... \\
  $ ----------(1)
Finding the vales of $ {f^{(1)}}(1) $ , $ {f^{(2)}}(1) $ , $ {f^{(3)}}(1) $ , $ f(1) $
 $ f(x) = \sqrt x \Rightarrow f(1) = \sqrt 1 = 1 $
Let’s find out the derivatives with respect to x up to order 3, using the property of derivative that $ \dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} $ .
1st derivative:
 $ f'(x) = {f^1}(x) = \dfrac{1}{2}{x^{ - \dfrac{1}{2}}}\,\,\,\,\,\,\, \Rightarrow {f^1}(1) = \dfrac{1}{2} $
2nd derivative:
 $ f''(x) = {f^2}(x) = - \dfrac{1}{4}{x^{ - \dfrac{3}{2}}}\,\,\,\,\,\,\,\, \Rightarrow {f^2}(1) = - \dfrac{1}{4} $
3rd derivative:
 $ f'''(x) = {f^3}(x) = \dfrac{3}{8}{x^{ - \dfrac{5}{2}}}\,\,\,\,\,\, \Rightarrow {f^3}(1) = \dfrac{3}{8} $

Putting vales of $ {f^{(1)}}(1) $ , $ {f^{(2)}}(1) $ , $ {f^{(3)}}(1) $ , $ f(1) $ all together in our original formula i.e. equation (1),we get
 $
  {T_3}(x) = f(1) + \dfrac{{{f^{(1)}}(1)}}{{1!}}{(x - 1)^1} + \dfrac{{{f^{(2)}}(1)}}{{2!}}{(x - 1)^2} + \dfrac{{{f^{(3)}}(1)}}{{3!}}{(x - 1)^3}......... \\
   = 1 + \dfrac{1}{2}{(x - 1)^1} + \dfrac{{ - 1}}{4}{(x - 1)^2} + \dfrac{3}{8}{(x - 1)^3}........... \\
   = 1 + \dfrac{1}{2}{(x - 1)^1} - \dfrac{1}{4}{(x - 1)^2} + \dfrac{3}{8}{(x - 1)^3}......... \\
  $
Therefore, we have successfully found out the Taylor expansion of a given function.
So, the correct answer is “ $ 1 + \dfrac{1}{2}{(x - 1)^1} - \dfrac{1}{4}{(x - 1)^2} + \dfrac{3}{8}{(x - 1)^3} $ ”.

Note: $ {T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}} $
 $ \dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x} $
 $ \dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} $
Factorial: The continued product of first n natural numbers is called the “n factorial “and denoted by $ n! $ .
1.Don’t forget to cross-check your answer at least once.
2.Meaning of Zero factorial is senseless to define it as the product of integers from 1 to zero. So, we define it as $ 0! = 1 $ .
3.Factorial of any number can calculated as $ n! = n(n - 1)(n - 2)(n - 3).......(2)(1) $