147. Taylor’s Theorem

1. Suppose that $f(x)$ is a polynomial of degree $r$. Then $f^{(n)}(x)$ is identically zero when $n>r$, and the theorem leads to the algebraical identity $$f(a+h)=f(a)+h{f}^{\prime }(a)+\frac{h^2}{2!}f”(a)+\cdots +\frac{h^r}{r!}{f}^{(r)}(a).$$

2. By applying the theorem to $f(x)=1/x$, and supposing $x$ and $x+h$ positive, obtain the result $$\frac{1}{x+h}=\frac{1}{x}–\frac{h}{x^2}+\frac{h^2}{x^3}–\cdots +\frac{(-1{)}^{n-1}{h}^{n-1}}{x^n}+\frac{(-1{)}^n{h}^n}{(x+{\theta }_{n}h{)}^{n+1}}.$$

[Since $$\frac{1}{x+h}=\frac{1}{x}–\frac{h}{x^2}+\frac{h^2}{x^3}–\cdots +\frac{(-1{)}^{n-1}{h}^{n-1}}{x^n}+\frac{(-1{)}^n{h}^n}{x^n(x+h)},$$ we can verify the result by showing that $x^n(x+h)$ can be put in the form $(x+{\theta }_{n}h{)}^{n+1}$, or that $x^{n+1}<x^n(x+h)<(x+h{)}^{n+1}$, as is evidently the case.]

3. Obtain the formula $$\begin{array}{c}\sin (x+h)=\sin x+h\cos x–\frac{h^2}{2!}\sin x–h^3\frac{3!}{\cos }x+\dots \\ +(-1{)}^{n-1}\frac{h^{2n-1}}{(2n–1)!}\cos x+(-1{)}^n{h}^{2n}\frac{2n!}{\sin }(x+{\theta }_{2n}h),\end{array}$$ the corresponding formula for $\cos (x+h)$, and similar formulae involving powers of $h$ extending up to $h^{2n+1}$.

4. Show that if $m$ is a positive integer, and $n$ a positive integer not greater than $m$, then $$(x+h{)}^m=x^m+(\genfrac{}{}{0ex}{}{m}{1})x^{m-1}h+\cdots +(\genfrac{}{}{0ex}{}{m}{n–1})x^{m-n+1}{h}^{n-1}+(\genfrac{}{}{0ex}{}{m}{n})(x+{\theta }_{n}h{)}^{m-n}{h}^n.$$ Show also that, if the interval $[x,x+h]$ does not include $x=0$, the formula holds for all real values of $m$ and all positive integral values of $n$; and that, even if $x<0<x+h$ or $x+h<0<x$, the formula still holds if $m–n$ is positive.

5. The formula $f(x+h)=f(x)+h{f}^{\prime }(x+{\theta }_{1}h)$ is not true if $f(x)=1/x$ and $x<0<x+h$. [For $f(x+h)–f(x)>0$ and $h{f}^{\prime }(x+{\theta }_{1}h)=-h/(x+{\theta }_{1}h{)}^2<0$; it is evident that the conditions for the truth of the Mean Value Theorem are not satisfied.]

6. If $x=-a$, $h=2a$, $f(x)=x^{1/3}$, then the equation $$f(x+h)=f(x)+h{f}^{\prime }(x+{\theta }_{1}h)$$ is satisfied by ${\theta }_1=\frac{1}{2}\pm \frac{1}{18}\sqrt{3}$. [This example shows that the result of the theorem may hold even if the conditions under which it was proved are not satisfied.]

7. Newton’s method of approximation to the roots of equations. Let $\xi$ be an approximation to a root of an algebraical equation $f(x)=0$, the actual root being $\xi +h$. Then $$0=f(\xi +h)=f(\xi )+h{f}^{\prime }(\xi )+\frac{1}{2}{h}^{2}f”(\xi +{\theta }_{2}h),$$ so that $$h=-\frac{f(\xi )}{f^{\prime }(\xi )}–\frac{1}{2}{h}^2\frac{f”(\xi +{\theta }_{2}h)}{f^{\prime }(\xi )}.$$

It follows that in general a better approximation than $x=\xi$ is $$x=\xi –\frac{f(\xi )}{f^{\prime }(\xi )}.$$ If the root is a simple root, so that $f^{\prime }(\xi +h)\ne 0$, we can, when $h$ is small enough, find a positive constant $K$ such that $|f^{\prime }(x)|>K$ for all the values of $x$ which we are considering, and then, if $h$ is regarded as of the first order of smallness, $f(\xi )$ is of the first order of smallness, and the error in taking $\xi –\{f(\xi )/f^{\prime }(\xi )\}$ as the root is of the second order.

8. Apply this process to the equation $x^2=2$, taking $\xi =3/2$ as the first approximation. [We find $h=-1/12$, $\xi +h=17/12=1.417\dots$, which is quite a good approximation, in spite of the roughness of the first. If now we repeat the process, taking $\xi =17/12$, we obtain $\xi +h=577/408=1.414215\dots$, which is correct to $5$ places of decimals.]

9. By considering in this way the equation $x^2–1–y=0$, where $y$ is small, show that $\sqrt{1+y}=1+\frac{1}{2}y–\{\frac{1}{4}{y}^2/(2+y)\}$ approximately, the error being of the fourth order.

10. Show that the error in taking the root to be $\xi –(f/f^{\prime })–\frac{1}{2}(f^{2}f”/f^{\prime 3})$, where $\xi$ is the argument of every function, is in general of the third order.

11. The equation $\sin x=\alpha x$, where $\alpha$ is small, has a root nearly equal to $\pi$. Show that $(1–\alpha )\pi$ is a better approximation, and $(1–\alpha +{\alpha }^2)\pi$ a better still. [The method of Exs. 7–10 does not depend on $f(x)=0$ being an algebraical equation, so long as $f^{\prime }$ and $f”$ are continuous.]

12. Show that the limit when $h\to 0$ of the number ${\theta }_n$ which occurs in the general Mean Value Theorem is $1/(n+1)$, provided that $f^{(n+1)}(x)$ is continuous.

[For $f(x+h)$ is equal to each of $$f(x)+\cdots +\frac{h^n}{n!}{f}^{(n)}(x+{\theta }_{n}h),f(x)+\cdots +\frac{h^n}{n!}{f}^{(n)}(x)+\frac{h^{n+1}}{(n+1)!}{f}^{(n+1)}(x+{\theta }_{n+1}h),$$ where ${\theta }_{n+1}$ as well as ${\theta }_n$ lies between $0$ and $1$. Hence $$f^{(n)}(x+{\theta }_{n}h)=f^{(n)}(x)+\frac{h{f}^{(n+1)}(x+{\theta }_{n+1}h)}{n+1}.$$ But if we apply the original Mean Value Theorem to the function $f^{(n)}(x)$, taking ${\theta }_{n}h$ in place of $h$, we find $$f^{(n)}(x+{\theta }_{n}h)=f^{(n)}(x)+{\theta }_{n}h{f}^{(n+1)}(x+\theta {\theta }_{n}h),$$ where $\theta$ also lies between $0$ and $1$. Hence $${\theta }_n{f}^{(n+1)}(x+\theta {\theta }_{n}h)=\frac{f^{(n+1)}(x+{\theta }_{n+1}h)}{n+1},$$ from which the result follows, since $f^{(n+1)}(x+\theta {\theta }_{n}h)$ and $f^{(n+1)}(x+{\theta }_{n+1}h)$ tend to the same limit $f^{(n+1)}(x)$ as $h\to 0$.]

13. Prove that $\{f(x+2h)–2f(x+h)+f(x)\}/h^2\to f”(x)$ as $h\to 0$, provided that $f”(x)$ is continuous. [Use equation (2) of § 147.]

14. Show that, if the $f^{(n)}(x)$ is continuous for $x=0$, then $$f(x)=a_0+a_{1}x+a_2{x}^2+\cdots +(a_n+{ϵ}_x)x^n,$$ where $a_r=f^{(r)}(0)/r!$ and ${ϵ}_x\to 0$ as $x\to 0$.¹

15. Show that if $$a_0+a_{1}x+a_2{x}^2+\cdots +(a_n+{ϵ}_x)x^n=b_0+b_{1}x+b_2{x}^2+\cdots +(b_n+{\eta }_x)x^n,$$ where ${ϵ}_x$ and ${\eta }_x$ tend to zero as $x\to 0$, then $a_0=b_0$, $a_1=b_1$, …, $a_n=b_n$. [Making $x\to 0$ we see that $a_0=b_0$. Now divide by $x$ and afterwards make $x\to 0$. We thus obtain $a_1=b_1$; and this process may be repeated as often as is necessary. It follows that if $f(x)=a_0+a_{1}x+a_2{x}^2+\cdots +(a_n+{ϵ}_x)x^n$, and the first $n$ derivatives of $f(x)$ are continuous, then $a_r=f^{(r)}(0)/r!$.]