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 $n! = n(n - 1)(n - 2)(n - 3).......(2)(1)$