Question

If $f\left( x \right)=\ln \left( 1+2x \right)$ , where $a=2$ and $n=3$ how do you approximate f by a Taylor polynomial with degree n and the number a?

Answer 1

Problems like these are quite simple to solve once we have understood the underlying concepts behind the problem. This particular problem requires a bit of prerequisites which we must have cleared before attempting the problem. We should have a basic as well as an advanced idea of differential and integral calculus, as well as the Taylor's theorem and its various applications to find the series of any given function. The Taylor series is defined as,
$f\left( x-a \right)=f\left( a \right)+\dfrac{\left( x-a \right){{f}^{'}}\left( a \right)}{1!}+\dfrac{{{\left( x-a \right)}^{2}}{{f}^{''}}\left( a \right)}{2!}+\dfrac{{{\left( x-a \right)}^{3}}{{f}^{'''}}\left( a \right)}{3!}+.....$
In this problem we are given the value of $n=3$ , so we need to consider only the first $4$ terms of the above equation. We also need to find the value of the function at ‘x=a’ and then put it in the original equation to find our desired result.

Complete step-by-step answer:
Now we start off with the solution to our given problem by finding the value of the function at ‘x=a’. Since we are given that $a=2$ , so we need to find the value of the function at $x=2$ .
$f\left( 2 \right)=\ln \left( 1+2\times 2 \right)=\ln \left( 5 \right)$
Now we find the first, second and third derivative of the given function,

& {{f}^{'}}\left( a \right)=\dfrac{2}{1+2\times 2}=\dfrac{2}{5} \\\ & {{f}^{''}}\left( a \right)=-\dfrac{4}{{{\left( 1+2\times 2 \right)}^{2}}}=-\dfrac{4}{25} \\\ & {{f}^{'''}}\left( a \right)=\dfrac{16}{{{\left( 1+2\times 2 \right)}^{3}}}=\dfrac{16}{125} \\\ \end{aligned}$$ Now we rewrite the given equation as, $$f\left( x-2 \right)=f\left( 2 \right)+\dfrac{\left( x-2 \right){{f}^{'}}\left( 2 \right)}{1!}+\dfrac{{{\left( x-2 \right)}^{2}}{{f}^{''}}\left( 2 \right)}{2!}+\dfrac{{{\left( x-2 \right)}^{3}}{{f}^{'''}}\left( 2 \right)}{3!}$$ Now we put the values in the respective positions as, $$\begin{aligned} & f\left( x-2 \right)=\ln \left( 5 \right)+\dfrac{\left( x-2 \right)\dfrac{2}{5}}{1!}+\dfrac{{{\left( x-2 \right)}^{2}}\left( -\dfrac{4}{25} \right)}{2!}+\dfrac{{{\left( x-2 \right)}^{3}}\dfrac{16}{125}}{3!} \\\ & \Rightarrow f\left( x-2 \right)=\ln \left( 5 \right)+\left( x-2 \right)\dfrac{2}{5}+{{\left( x-2 \right)}^{2}}\left( -\dfrac{2}{25} \right)+{{\left( x-2 \right)}^{3}}\dfrac{8}{375} \\\ \end{aligned}$$ **Note:** For such problems we need to be through about differential and integral calculus and its various applications in Taylor’s series. We need to be very careful about what is the value of the power ‘n’ and then write the series up to that term only. We also need to consider the value of the constant that is given in this problem which is equal to ‘a’. After finding the values of the derivatives we need to put them in the formed equation to find out our desired answer.