State and prove Taylor’s theorem.
Answer
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Hint: We will here first give the statement of Taylor’s theorem which is given as:
If f(x) be a polynomial function in R then there exists a value $a\in R$ where f is differentiable ‘n’ number of times then f(x) can be expanded as:
$f\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{f'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+...\dfrac{{{f}^{\left( n \right)}}\left( x \right)}{n!}{{\left( x-a \right)}^{n}}$
Then, we will assume a polynomial function in variable x with degree ‘n’ given as:
$f\left( x \right)={{A}_{0}}+\left( x-a \right){{A}_{1}}+{{\left( x-a \right)}^{2}}{{A}_{3}}+{{\left( x-a \right)}^{3}}{{A}_{4}}+...+{{\left( x-a \right)}^{n}}{{A}_{n+1}}$
Then, we will put x=a+h in this function as a result we will obtain a function f(a+h). Then we will differentiate f(a+h) w.r.t. x n times and then we will put h=0 in f(a+h) and all its derivatives as a result of which we will obtain all the values of all the constants written in the expansion of f(a+h). Then we will put the values of those constants in f(a+h) and then we will put the value of h as back to x-a. Hence, our theorem will be proved.
Complete step-by-step solution
Now, we need to state and prove Taylor's theorem. For, this we will first give its statement.
Taylor’s theorem is stated as:
“If f(x) be a polynomial function in R then there exists a value $a\in R$ where f is differentiable ‘n’ number of times then f(x) can be expanded as:
$f\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{f'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+...\dfrac{{{f}^{\left( n \right)}}\left( x \right)}{n!}{{\left( x-a \right)}^{n}}$”
Now, we will give its proof.
To prove this theorem, we will assume a polynomial function f in variable ‘x’ defined as:
$f\left( x \right)={{A}_{0}}+\left( x-a \right){{A}_{1}}+{{\left( x-a \right)}^{2}}{{A}_{3}}+{{\left( x-a \right)}^{3}}{{A}_{4}}+...+{{\left( x-a \right)}^{n}}{{A}_{n+1}}$
Now, let be given as:
$x=a+h$
Thus, we get:
$x-a=h$
Thus, we can write f(x) as:
\[\begin{align}
& f\left( x \right)={{A}_{0}}+\left( x-a \right){{A}_{1}}+{{\left( x-a \right)}^{2}}{{A}_{3}}+{{\left( x-a \right)}^{3}}{{A}_{4}}+...+{{\left( x-a \right)}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a+h \right)={{A}_{0}}+h{{A}_{1}}+{{h}^{2}}{{A}_{3}}+{{h}^{3}}{{A}_{4}}+...+{{h}^{n}}{{A}_{n+1}} \\
\end{align}\]
Now, we will differentiate this function with respect to ‘h’ n times.
First derivative of f is given as:
$\begin{align}
& f'\left( a+h \right)=0+{{A}_{1}}+2h{{A}_{3}}+3{{h}^{2}}{{A}_{4}}+...+n{{h}^{n-1}}{{A}_{n+1}} \\
& \Rightarrow f'\left( a+h \right)={{A}_{1}}+2h{{A}_{3}}+3{{h}^{2}}{{A}_{4}}+...+n{{h}^{n-1}}{{A}_{n+1}} \\
\end{align}$
Now, second derivative of f is given as:
$\begin{align}
& f''\left( a+h \right)=0+2{{A}_{3}}+3.2h{{A}_{4}}+...+n.\left( n-1 \right){{h}^{n-2}}{{A}_{n+1}} \\
& \Rightarrow f''\left( a+h \right)=2{{A}_{3}}+6h{{A}_{4}}+...+n.\left( n-1 \right){{h}^{n-2}}{{A}_{n+1}} \\
\end{align}$
Now, the third derivative of f is given as:
$\begin{align}
& f'''\left( a+h \right)=0+6{{A}_{4}}+...+n.\left( n-1 \right)\left( n-2 \right){{h}^{n-3}}{{A}_{n+1}} \\
& \Rightarrow f'''\left( a+h \right)=6{{A}_{4}}+...+n.\left( n-1 \right)\left( n-2 \right){{h}^{n-3}}{{A}_{n+1}} \\
\end{align}$
If we follow this trend of derivatives, we can see that the ${{n}^{th}}$ derivative of f will be given as:
${{f}^{\left( n \right)}}\left( a+h \right)=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).....2.1{{A}_{n+1}}$
Now, we know that $n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)....2.1=n!$
Hence, we can write the ${{n}^{th}}$ derivative of f as:
${{f}^{\left( n \right)}}\left( a+h \right)=n!{{A}_{n+1}}$
Now, we will put h=0 in f(a+h)
Putting h=0 in f(a+h) we get:
\[\begin{align}
& f\left( a+h \right)={{A}_{0}}+h{{A}_{1}}+{{h}^{2}}{{A}_{3}}+{{h}^{3}}{{A}_{4}}+...+{{h}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a+0 \right)={{A}_{0}}+\left( 0 \right){{A}_{1}}+{{\left( 0 \right)}^{2}}{{A}_{3}}+{{\left( 0 \right)}^{3}}{{A}_{4}}+...+{{\left( 0 \right)}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a \right)={{A}_{0}} \\
& \Rightarrow {{A}_{0}}=f\left( a \right) \\
\end{align}\]
Now, we will put h=0 in all the derivatives to find the values of the different constants.
Putting h=0 in first derivative of f we get:
\[\begin{align}
& f'\left( a+h \right)={{A}_{1}}+2h{{A}_{3}}+3{{h}^{2}}{{A}_{4}}+4{{h}^{3}}{{A}_{5}}+...+n{{h}^{n-1}}{{A}_{n+1}} \\
& \Rightarrow f'\left( a+0 \right)={{A}_{1}}+2\left( 0 \right){{A}_{3}}+3{{\left( 0 \right)}^{2}}{{A}_{4}}+4{{\left( 0 \right)}^{3}}{{A}_{5}}+...+n{{\left( 0 \right)}^{n-1}}{{A}_{n+1}} \\
& \Rightarrow f'\left( a \right)={{A}_{1}} \\
& \Rightarrow {{A}_{1}}=f'\left( a \right) \\
\end{align}\]
Now, putting h=0 in the second derivative of f we get:
$\begin{align}
& f''\left( a+h \right)=2{{A}_{3}}+6h{{A}_{4}}+12{{h}^{2}}{{A}_{5}}+...+n.\left( n-1 \right){{h}^{n-2}}{{A}_{n+1}} \\
& \Rightarrow f''\left( a+0 \right)=2{{A}_{3}}+6\left( 0 \right){{A}_{4}}+12{{\left( 0 \right)}^{2}}{{A}_{5}}+...+n.\left( n-1 \right){{\left( 0 \right)}^{n-2}}{{A}_{n+1}} \\
& \Rightarrow f''\left( a \right)=2{{A}_{3}} \\
& \Rightarrow {{A}_{3}}=\dfrac{f''\left( a \right)}{2} \\
\end{align}$
Now, we can write 2 as:
$2=2.1$
And we know that: $2.1=2!$
Hence, we can say that:
$\begin{align}
& {{A}_{3}}=\dfrac{f''\left( a \right)}{2} \\
& \Rightarrow {{A}_{3}}=\dfrac{f''\left( a \right)}{2!} \\
\end{align}$
Now, putting h=0 in third derivative of f we get:
$\begin{align}
& f'''\left( a+h \right)=6{{A}_{4}}+24h{{A}_{5}}+...+n.\left( n-1 \right)\left( n-2 \right){{h}^{n-3}}{{A}_{n+1}} \\
& \Rightarrow f'''\left( a+0 \right)=6{{A}_{4}}+24\left( 0 \right){{A}_{5}}+...+n.\left( n-1 \right)\left( n-2 \right){{\left( 0 \right)}^{n-3}}{{A}_{n+1}} \\
& \Rightarrow f'''\left( a \right)=6{{A}_{4}} \\
& \Rightarrow {{A}_{4}}=\dfrac{f'''\left( a \right)}{6} \\
\end{align}$
Now, we can write 6 as:
$6=3.2.1$
And we know that: $3.2.1=3!$
Hence, we can say that:
$\begin{align}
& {{A}_{4}}=\dfrac{f'''\left( a \right)}{6} \\
& \Rightarrow {{A}_{4}}=\dfrac{f'''\left( a \right)}{3!} \\
\end{align}$
Now again, if we follow this trend, by putting h=0 in ${{n}^{th}}$ derivative of f we will get:
$\begin{align}
& {{f}^{\left( n \right)}}\left( a+h \right)=n!{{A}_{n+1}} \\
& \Rightarrow {{f}^{\left( n \right)}}\left( a+0 \right)=n!{{A}_{n+1}} \\
& \Rightarrow {{f}^{\left( n \right)}}\left( a \right)=n!{{A}_{n+1}} \\
& \Rightarrow {{A}_{n+1}}=\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
\end{align}$
Now, we will put the values of the constants obtained in f(a+h).
Putting the values of the constants obtained in f(a+h) we get:
\[\begin{align}
& f\left( a+h \right)={{A}_{0}}+h{{A}_{1}}+{{h}^{2}}{{A}_{3}}+{{h}^{3}}{{A}_{4}}+...+{{h}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a+h \right)=f\left( a \right)+hf'\left( a \right)+{{h}^{2}}\dfrac{f''\left( a \right)}{2!}+{{h}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{h}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
\end{align}\]
Now, as we obtained above, h=x-a
Thus, by putting this value in the above mentioned value of f(a+h), we get:
\[\begin{align}
& f\left( a+h \right)=f\left( a \right)+hf'\left( a \right)+{{h}^{2}}\dfrac{f''\left( a \right)}{2!}+{{h}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{h}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
& \Rightarrow f\left( a+x-a \right)=f\left( a \right)+\left( x-a \right)f'\left( a \right)+{{\left( x-a \right)}^{2}}\dfrac{f''\left( a \right)}{2!}+{{\left( x-a \right)}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{\left( x-a \right)}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
& \Rightarrow f\left( x \right)=f\left( a \right)+\left( x-a \right)f'\left( a \right)+{{\left( x-a \right)}^{2}}\dfrac{f''\left( a \right)}{2!}+{{\left( x-a \right)}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{\left( x-a \right)}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
\end{align}\]
As mentioned above, this is Taylor’s theorem.
Hence, Taylor’s theorem is proved.
Note: If we don’t assume h to be equal to x-a, it will still be the expansion of Taylor’s theorem and will still be counted as Taylor’s series. When we put h=x-a, it is just a special case of Taylor’s theorem which is very widely used. Also, calculate the values of all the constants very carefully because it is the most important step of the proof. Any mistake that will result in the wrong solution.
If f(x) be a polynomial function in R then there exists a value $a\in R$ where f is differentiable ‘n’ number of times then f(x) can be expanded as:
$f\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{f'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+...\dfrac{{{f}^{\left( n \right)}}\left( x \right)}{n!}{{\left( x-a \right)}^{n}}$
Then, we will assume a polynomial function in variable x with degree ‘n’ given as:
$f\left( x \right)={{A}_{0}}+\left( x-a \right){{A}_{1}}+{{\left( x-a \right)}^{2}}{{A}_{3}}+{{\left( x-a \right)}^{3}}{{A}_{4}}+...+{{\left( x-a \right)}^{n}}{{A}_{n+1}}$
Then, we will put x=a+h in this function as a result we will obtain a function f(a+h). Then we will differentiate f(a+h) w.r.t. x n times and then we will put h=0 in f(a+h) and all its derivatives as a result of which we will obtain all the values of all the constants written in the expansion of f(a+h). Then we will put the values of those constants in f(a+h) and then we will put the value of h as back to x-a. Hence, our theorem will be proved.
Complete step-by-step solution
Now, we need to state and prove Taylor's theorem. For, this we will first give its statement.
Taylor’s theorem is stated as:
“If f(x) be a polynomial function in R then there exists a value $a\in R$ where f is differentiable ‘n’ number of times then f(x) can be expanded as:
$f\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{f'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+...\dfrac{{{f}^{\left( n \right)}}\left( x \right)}{n!}{{\left( x-a \right)}^{n}}$”
Now, we will give its proof.
To prove this theorem, we will assume a polynomial function f in variable ‘x’ defined as:
$f\left( x \right)={{A}_{0}}+\left( x-a \right){{A}_{1}}+{{\left( x-a \right)}^{2}}{{A}_{3}}+{{\left( x-a \right)}^{3}}{{A}_{4}}+...+{{\left( x-a \right)}^{n}}{{A}_{n+1}}$
Now, let be given as:
$x=a+h$
Thus, we get:
$x-a=h$
Thus, we can write f(x) as:
\[\begin{align}
& f\left( x \right)={{A}_{0}}+\left( x-a \right){{A}_{1}}+{{\left( x-a \right)}^{2}}{{A}_{3}}+{{\left( x-a \right)}^{3}}{{A}_{4}}+...+{{\left( x-a \right)}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a+h \right)={{A}_{0}}+h{{A}_{1}}+{{h}^{2}}{{A}_{3}}+{{h}^{3}}{{A}_{4}}+...+{{h}^{n}}{{A}_{n+1}} \\
\end{align}\]
Now, we will differentiate this function with respect to ‘h’ n times.
First derivative of f is given as:
$\begin{align}
& f'\left( a+h \right)=0+{{A}_{1}}+2h{{A}_{3}}+3{{h}^{2}}{{A}_{4}}+...+n{{h}^{n-1}}{{A}_{n+1}} \\
& \Rightarrow f'\left( a+h \right)={{A}_{1}}+2h{{A}_{3}}+3{{h}^{2}}{{A}_{4}}+...+n{{h}^{n-1}}{{A}_{n+1}} \\
\end{align}$
Now, second derivative of f is given as:
$\begin{align}
& f''\left( a+h \right)=0+2{{A}_{3}}+3.2h{{A}_{4}}+...+n.\left( n-1 \right){{h}^{n-2}}{{A}_{n+1}} \\
& \Rightarrow f''\left( a+h \right)=2{{A}_{3}}+6h{{A}_{4}}+...+n.\left( n-1 \right){{h}^{n-2}}{{A}_{n+1}} \\
\end{align}$
Now, the third derivative of f is given as:
$\begin{align}
& f'''\left( a+h \right)=0+6{{A}_{4}}+...+n.\left( n-1 \right)\left( n-2 \right){{h}^{n-3}}{{A}_{n+1}} \\
& \Rightarrow f'''\left( a+h \right)=6{{A}_{4}}+...+n.\left( n-1 \right)\left( n-2 \right){{h}^{n-3}}{{A}_{n+1}} \\
\end{align}$
If we follow this trend of derivatives, we can see that the ${{n}^{th}}$ derivative of f will be given as:
${{f}^{\left( n \right)}}\left( a+h \right)=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right).....2.1{{A}_{n+1}}$
Now, we know that $n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)....2.1=n!$
Hence, we can write the ${{n}^{th}}$ derivative of f as:
${{f}^{\left( n \right)}}\left( a+h \right)=n!{{A}_{n+1}}$
Now, we will put h=0 in f(a+h)
Putting h=0 in f(a+h) we get:
\[\begin{align}
& f\left( a+h \right)={{A}_{0}}+h{{A}_{1}}+{{h}^{2}}{{A}_{3}}+{{h}^{3}}{{A}_{4}}+...+{{h}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a+0 \right)={{A}_{0}}+\left( 0 \right){{A}_{1}}+{{\left( 0 \right)}^{2}}{{A}_{3}}+{{\left( 0 \right)}^{3}}{{A}_{4}}+...+{{\left( 0 \right)}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a \right)={{A}_{0}} \\
& \Rightarrow {{A}_{0}}=f\left( a \right) \\
\end{align}\]
Now, we will put h=0 in all the derivatives to find the values of the different constants.
Putting h=0 in first derivative of f we get:
\[\begin{align}
& f'\left( a+h \right)={{A}_{1}}+2h{{A}_{3}}+3{{h}^{2}}{{A}_{4}}+4{{h}^{3}}{{A}_{5}}+...+n{{h}^{n-1}}{{A}_{n+1}} \\
& \Rightarrow f'\left( a+0 \right)={{A}_{1}}+2\left( 0 \right){{A}_{3}}+3{{\left( 0 \right)}^{2}}{{A}_{4}}+4{{\left( 0 \right)}^{3}}{{A}_{5}}+...+n{{\left( 0 \right)}^{n-1}}{{A}_{n+1}} \\
& \Rightarrow f'\left( a \right)={{A}_{1}} \\
& \Rightarrow {{A}_{1}}=f'\left( a \right) \\
\end{align}\]
Now, putting h=0 in the second derivative of f we get:
$\begin{align}
& f''\left( a+h \right)=2{{A}_{3}}+6h{{A}_{4}}+12{{h}^{2}}{{A}_{5}}+...+n.\left( n-1 \right){{h}^{n-2}}{{A}_{n+1}} \\
& \Rightarrow f''\left( a+0 \right)=2{{A}_{3}}+6\left( 0 \right){{A}_{4}}+12{{\left( 0 \right)}^{2}}{{A}_{5}}+...+n.\left( n-1 \right){{\left( 0 \right)}^{n-2}}{{A}_{n+1}} \\
& \Rightarrow f''\left( a \right)=2{{A}_{3}} \\
& \Rightarrow {{A}_{3}}=\dfrac{f''\left( a \right)}{2} \\
\end{align}$
Now, we can write 2 as:
$2=2.1$
And we know that: $2.1=2!$
Hence, we can say that:
$\begin{align}
& {{A}_{3}}=\dfrac{f''\left( a \right)}{2} \\
& \Rightarrow {{A}_{3}}=\dfrac{f''\left( a \right)}{2!} \\
\end{align}$
Now, putting h=0 in third derivative of f we get:
$\begin{align}
& f'''\left( a+h \right)=6{{A}_{4}}+24h{{A}_{5}}+...+n.\left( n-1 \right)\left( n-2 \right){{h}^{n-3}}{{A}_{n+1}} \\
& \Rightarrow f'''\left( a+0 \right)=6{{A}_{4}}+24\left( 0 \right){{A}_{5}}+...+n.\left( n-1 \right)\left( n-2 \right){{\left( 0 \right)}^{n-3}}{{A}_{n+1}} \\
& \Rightarrow f'''\left( a \right)=6{{A}_{4}} \\
& \Rightarrow {{A}_{4}}=\dfrac{f'''\left( a \right)}{6} \\
\end{align}$
Now, we can write 6 as:
$6=3.2.1$
And we know that: $3.2.1=3!$
Hence, we can say that:
$\begin{align}
& {{A}_{4}}=\dfrac{f'''\left( a \right)}{6} \\
& \Rightarrow {{A}_{4}}=\dfrac{f'''\left( a \right)}{3!} \\
\end{align}$
Now again, if we follow this trend, by putting h=0 in ${{n}^{th}}$ derivative of f we will get:
$\begin{align}
& {{f}^{\left( n \right)}}\left( a+h \right)=n!{{A}_{n+1}} \\
& \Rightarrow {{f}^{\left( n \right)}}\left( a+0 \right)=n!{{A}_{n+1}} \\
& \Rightarrow {{f}^{\left( n \right)}}\left( a \right)=n!{{A}_{n+1}} \\
& \Rightarrow {{A}_{n+1}}=\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
\end{align}$
Now, we will put the values of the constants obtained in f(a+h).
Putting the values of the constants obtained in f(a+h) we get:
\[\begin{align}
& f\left( a+h \right)={{A}_{0}}+h{{A}_{1}}+{{h}^{2}}{{A}_{3}}+{{h}^{3}}{{A}_{4}}+...+{{h}^{n}}{{A}_{n+1}} \\
& \Rightarrow f\left( a+h \right)=f\left( a \right)+hf'\left( a \right)+{{h}^{2}}\dfrac{f''\left( a \right)}{2!}+{{h}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{h}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
\end{align}\]
Now, as we obtained above, h=x-a
Thus, by putting this value in the above mentioned value of f(a+h), we get:
\[\begin{align}
& f\left( a+h \right)=f\left( a \right)+hf'\left( a \right)+{{h}^{2}}\dfrac{f''\left( a \right)}{2!}+{{h}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{h}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
& \Rightarrow f\left( a+x-a \right)=f\left( a \right)+\left( x-a \right)f'\left( a \right)+{{\left( x-a \right)}^{2}}\dfrac{f''\left( a \right)}{2!}+{{\left( x-a \right)}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{\left( x-a \right)}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
& \Rightarrow f\left( x \right)=f\left( a \right)+\left( x-a \right)f'\left( a \right)+{{\left( x-a \right)}^{2}}\dfrac{f''\left( a \right)}{2!}+{{\left( x-a \right)}^{3}}\dfrac{f'''\left( a \right)}{3!}+...+{{\left( x-a \right)}^{n}}\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!} \\
\end{align}\]
As mentioned above, this is Taylor’s theorem.
Hence, Taylor’s theorem is proved.
Note: If we don’t assume h to be equal to x-a, it will still be the expansion of Taylor’s theorem and will still be counted as Taylor’s series. When we put h=x-a, it is just a special case of Taylor’s theorem which is very widely used. Also, calculate the values of all the constants very carefully because it is the most important step of the proof. Any mistake that will result in the wrong solution.
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