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Question: How do you write the taylor series for the function \(f\left( x \right)=\sqrt{x}\) at \(a=16\)?...

How do you write the taylor series for the function f(x)=xf\left( x \right)=\sqrt{x} at a=16a=16?

Explanation

Solution

We start solving the problem by recalling the fact that the taylor series of the function f(x)f\left( x \right) at x=ax=a is defined as f(x)=f(a)+f′(a)×(x−a)1!+f′′(a)×(x−a)22!+.......fn(a)×(x−a)nn!+.....f\left( x \right)=f\left( a \right)+{{f}^{'}}\left( a \right)\times \dfrac{\left( x-a \right)}{1!}+{{f}^{''}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{2}}}{2!}+.......{{f}^{n}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{n}}}{n!}+...... We then use this result for the given function at a=16a=16. We then find the general form of all the derivatives of the function f(x)f\left( x \right) and substitute in the formula. We then make the necessary calculations to get the required answer.

Complete step-by-step solution:
According to the problem, we are asked to find the taylor series for the function f(x)=xf\left( x \right)=\sqrt{x} at a=16a=16 and also the radius of convergence.
We know that the taylor series of the function f(x)f\left( x \right) at x=ax=a is defined as f(x)=f(a)+f′(a)×(x−a)1!+f′′(a)×(x−a)22!+.......fn(a)×(x−a)nn!+.....f\left( x \right)=f\left( a \right)+{{f}^{'}}\left( a \right)\times \dfrac{\left( x-a \right)}{1!}+{{f}^{''}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{2}}}{2!}+.......{{f}^{n}}\left( a \right)\times \dfrac{{{\left( x-a \right)}^{n}}}{n!}+......
Now, let us use this result to find the Taylor series for the given function at a=16a=16.
f(x)=f(16)+f′(16)×(x−16)1!+f′′(16)×(x−16)22!+.......fn(16)×(x−16)nn!+.....f\left( x \right)=f\left( 16 \right)+{{f}^{'}}\left( 16 \right)\times \dfrac{\left( x-16 \right)}{1!}+{{f}^{''}}\left( 16 \right)\times \dfrac{{{\left( x-16 \right)}^{2}}}{2!}+.......{{f}^{n}}\left( 16 \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+..... ---(1).
Now, let us find the derivatives of the functions f(x)f\left( x \right).
We have f(x)=xf\left( x \right)=\sqrt{x}.
⇒f′(x)=d(x12)dx\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d\left( {{x}^{\dfrac{1}{2}}} \right)}{dx} ---(2).
We know that d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}. Let us use this result in equation (2).
⇒f′(x)=12x−12\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}} ---(3).
Now, f′′(x)=d(12x−12)dx{{f}^{''}}\left( x \right)=\dfrac{d\left( \dfrac{1}{2}{{x}^{\dfrac{-1}{2}}} \right)}{dx}.
⇒f′′(x)=−14x−32\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{-1}{4}{{x}^{\dfrac{-3}{2}}} ---(4).
Now, f3(x)=d(−14x−32)dx{{f}^{3}}\left( x \right)=\dfrac{d\left( \dfrac{-1}{4}{{x}^{\dfrac{-3}{2}}} \right)}{dx}.
⇒f3(x)=38x−52\Rightarrow {{f}^{3}}\left( x \right)=\dfrac{3}{8}{{x}^{\dfrac{-5}{2}}} ---(5).
Now, f4(x)=d(38x−52)dx{{f}^{4}}\left( x \right)=\dfrac{d\left( \dfrac{3}{8}{{x}^{\dfrac{-5}{2}}} \right)}{dx}.
⇒f4(x)=−1516x−72\Rightarrow {{f}^{4}}\left( x \right)=\dfrac{-15}{16}{{x}^{\dfrac{-7}{2}}} ---(6).
Similarly, we get fn(x)=(−1)n(2n−1)(2n)x−2n+32{{f}^{n}}\left( x \right)=\dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{x}^{\dfrac{-2n+3}{2}}} ---(7).
Now, let us substitute equations (2), (3), (4), (5), (6), and (7) in equation (1).
f(x)=16+(12(16)−12)×(x−16)1!+(−14(16)−32)×(x−16)22!+.......((−1)n(2n−1)(2n)(16)−2n+32)×(x−16)nn!+.....f\left( x \right)=\sqrt{16}+\left( \dfrac{1}{2}{{\left( 16 \right)}^{\dfrac{-1}{2}}} \right)\times \dfrac{\left( x-16 \right)}{1!}+\left( \dfrac{-1}{4}{{\left( 16 \right)}^{\dfrac{-3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{2}}}{2!}+.......\left( \dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{\left( 16 \right)}^{\dfrac{-2n+3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+......
⇒f(x)=4+(x−16)8−(x−16)2512+.......((−1)n(2n−1)(2n)(16)−2n+32)×(x−16)nn!+.....\Rightarrow f\left( x \right)=4+\dfrac{\left( x-16 \right)}{8}-\dfrac{{{\left( x-16 \right)}^{2}}}{512}+.......\left( \dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{\left( 16 \right)}^{\dfrac{-2n+3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+......
So, we have found the taylor series of the function f(x)f\left( x \right) at x=ax=a as
⇒f(x)=4+(x−16)8−(x−16)2512+.......((−1)n(2n−1)(2n)(16)−2n+32)×(x−16)nn!+.....\Rightarrow f\left( x \right)=4+\dfrac{\left( x-16 \right)}{8}-\dfrac{{{\left( x-16 \right)}^{2}}}{512}+.......\left( \dfrac{{{\left( -1 \right)}^{n}}\left( {{2}^{n}}-1 \right)}{\left( {{2}^{n}} \right)}{{\left( 16 \right)}^{\dfrac{-2n+3}{2}}} \right)\times \dfrac{{{\left( x-16 \right)}^{n}}}{n!}+......

Note: We should perform each step carefully to avoid confusion and calculation mistakes. We should confuse between the variables a and x while solving this type of problem. We can also find the Maclaurin series for the given function f(x)f\left( x \right) by taking a=1a=1. Similarly, we can expect problems to find the Taylor series of the function g(x)=sin⁡3xg\left( x \right)=\sin 3x.