Colin Maclaurin credited Brook Taylor and his work "Methodus Incrementorum" (published in 1715) for the discovery of this series, even though it is already known to earlier generation of mathematicians such as Newton. This reference can be found in book II of his work "A Treatise on Fluxion", page 611. He wrote the series as:
$y=E+\dot E+ \dfrac{\ddot E}{2}+\dfrac{\dddot E}{6}+\dfrac{\ddddot E}{24}+etc$
where $\dot E$, $\ddot E$, $\dddot E$, $\ddddot E$ are Newton's notation for the derivatives of a function (fluxions).
On the Taylor-Maclaurin formula
The definition of a power series is a series of the form:
$$\boxed{a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+a_4(x-c)^4+...=\sum_{n=0}^{+\infty}a_n(x-c)^{n}}$$
where $a_n$ is the coefficient of the power series.
A Taylor series is a special form of power series where the coefficients are $\dfrac{f^{(n)}(c)}{n!}$, or the $n$ order derivative of $f(x)$ over $n!$.
The coefficient $a_n$ cannot be dependent on $x$, a series such as this is not a power series:
$$\sin(x)x+\sin(2x)x^2+\sin(3x)x^3+\sin(4x)x^4=\sum_{n=1}^{+\infty}\sin(nx)x^n$$
We can derive the Taylor formula by successively differentiate the infinite polynomials below.
Assume that the function $f(x)$ admits the power series representation, it has the form:
$$f(x)=a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+a_4(x-c)^4+...\tag{1}$$
We wish to determine the coefficients $a_n$. We can successively differentiate the power series given above:
$$f'(x)=0+1\cdot a_1+2 a_2(x-c)+3a_3(x-c)^2+4a_4(x-c)^3\tag{2}$$
$$f''(x)=1\cdot 2\cdot a_2+2\cdot3a_3(x-c)+3\cdot 4a_4(x-c)^2$$
$$f^{(3)}(x)=1\cdot 2\cdot3a_3+2\cdot 3\cdot 4a_4(x-c)$$
$$f^{(4)}(x)=2\cdot 3\cdot 4a_4$$
$$.etc$$
Let $x=c$, this will make $(x-c)$ part all vanish. we will evaluate the series centered at $c$, we will have
$$f(c)=a_0\tag{3}$$
$$f'(c)=1\cdot a_1$$
$$f''(c)=1\cdot 2\cdot a_2$$
$$f^{(3)}(c)=1\cdot 2\cdot 3\cdot a_3$$
$$f^{(4)}(c)=1\cdot 2\cdot 3\cdot 4\cdot a_4$$
$$.etc$$
The we can express this infinite series of equations in terms of $a_n$, we will have:
$$a_0=f(c)\tag{4}$$
$$a_1=\dfrac{f(c)}{1}=\dfrac{f(c)}{1!}$$
$$a_2=\dfrac{f''(c)}{1\cdot 2}=\dfrac{f''(c)}{2!}$$
$$a_3=\dfrac{f^{(3)}(c)}{1\cdot 2\cdot 3}=\dfrac{f^{(3)}(c)}{3!}$$
$$a_4=\dfrac{f^{(4)}(c)}{1\cdot 2\cdot 3\cdot 4}=\dfrac{f^{(4)}(c)}{4!}$$
$$.etc$$
Replacing these values of $a_n$ into the equation in $1$, we have:
$\boxed{f(c)+\dfrac{f'(c)}{1!}(x-c)+\dfrac{f''(c)}{2!}(x-c)^2+\dfrac{f'''(c)}{3!}(x-c)^3+\dfrac{f^{(4)}(c)}{4!}(x-c)^4...}$
$$\boxed{=\sum_{n=1}^{\infty} \dfrac{f^{(n)}(c)}{n!}(x-c)^n}$$
When the function's expansion is centered at $c=0$, the Taylor series is called the Maclaurin series.
$n$ cannot be a negative power or a fractional power. When $n$ is a negative power, the series is called Laurent series, and when $n$ is a fractional power, the series is known as Puiseux series.
$$\boxed{f(0)+\dfrac{f'(0)}{1!}x+\dfrac{f''(0)}{2!}x^2+\dfrac{f'''(0)}{3!}x^3+\dfrac{f^{(4)}(0)}{4!}x^4...=\sum_{n=1}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n}$$
Apply the formula to calculate the Taylor series for $e^{x}$, we have:
$f(x)=e^x \implies f(0)=e^0=1$
$f'(x)=e^x \implies f'(0)=e^0=1$
$f''(x)=e^x \implies f''(0)=e^0=1$
$f'''(x)=e^x \implies f'''(0)=e^0=1$
$e^x=1+1\cdot\dfrac{x}{1!}+1\cdot\dfrac{x^2}{2!}+1\cdot\dfrac{x^3}{3!}+1\cdot\dfrac{x^4}{4!}+1\cdot\dfrac{x^5}{5!}+O(x^5)$
$\implies e^x=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+O(x^7)$
$$\boxed{f(0)+\dfrac{f'(0)}{1!}x+\dfrac{f''(0)}{2!}x^2+\dfrac{f'''(0)}{3!}x^3+\dfrac{f^{(4)}(0)}{4!}x^4...=\sum_{n=1}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n}$$
Apply the formula to calculate the Taylor series for $e^{x}$, we have:
$f(x)=e^x \implies f(0)=e^0=1$
$f'(x)=e^x \implies f'(0)=e^0=1$
$f''(x)=e^x \implies f''(0)=e^0=1$
$f'''(x)=e^x \implies f'''(0)=e^0=1$
$e^x=1+1\cdot\dfrac{x}{1!}+1\cdot\dfrac{x^2}{2!}+1\cdot\dfrac{x^3}{3!}+1\cdot\dfrac{x^4}{4!}+1\cdot\dfrac{x^5}{5!}+O(x^5)$
$\implies e^x=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+O(x^7)$
$$\boxed{\Leftrightarrow\sum_{n=0}^{+\infty}\dfrac{x^n}{n!}}$$
We can also apply for this formula to find the series of $\sin(x)$. The function is odd so all of its even terms are $0$.
$f(x)=\sin(x) \implies f(0)=0$
$f'(x)=\cos(x) \implies f'(0)=1$
$f''(x)=-\sin(x) \implies f''(0)=0$
$f'''(x)=-\cos(x) \implies f'''(0)=-1$
$f^{(4)}(x)= \sin(x) \implies f^{(4)}(0)=0$
$f^{(5)}(x)= \cos(x) \implies f^{(5)}(0)=1$
$f^{(6)}(x)= -\sin(x) \implies f^{(6)}(0)=0$
$f^{(7)}(x)= -\cos(x) \implies f^{(7)}(0)=-1$
$\sin(x)=0-1\cdot\dfrac{x}{1!}+0\cdot\dfrac{x^2}{2!}-1\cdot\dfrac{x^3}{3!}+0\cdot\dfrac{x^4}{4!}+1\cdot\dfrac{x^5}{5!}+0\cdot\dfrac{x^6}{6!}(x^6)-1\cdot\dfrac{x^7}{7!}(x^7)+O(x^9)$
$\implies \sin(x)={x}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+O(x^{13})$
We can do likewise for $\sinh(x)$
$f(x)=\sinh(x) \implies f(0)=0$
$f'(x)=\cosh(x) \implies f'(0)=1$
$f''(x)=-\sinh(x) \implies f''(0)=0$
$f'''(x)=\cosh(x) \implies f'''(0)=1$
$f^{(4)}(x)= \sinh(x) \implies f^{(4)}(0)=0$
$f^{(5)}(x)= \cosh(x) \implies f^{(5)}(0)=1$
$f^{(6)}(x)= -\sinh(x) \implies f^{(6)}(0)=0$
$f^{(7)}(x)= \sinh(x) \implies f^{(7)}(0)=1$
$\sinh(x)={x}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+\dfrac{x^{11}}{11!}+O(x^{13})$
We can also apply for this formula to find the series of $\sin(x)$. The function is odd so all of its even terms are $0$.
$f(x)=\sin(x) \implies f(0)=0$
$f'(x)=\cos(x) \implies f'(0)=1$
$f''(x)=-\sin(x) \implies f''(0)=0$
$f'''(x)=-\cos(x) \implies f'''(0)=-1$
$f^{(4)}(x)= \sin(x) \implies f^{(4)}(0)=0$
$f^{(5)}(x)= \cos(x) \implies f^{(5)}(0)=1$
$f^{(6)}(x)= -\sin(x) \implies f^{(6)}(0)=0$
$f^{(7)}(x)= -\cos(x) \implies f^{(7)}(0)=-1$
$\sin(x)=0-1\cdot\dfrac{x}{1!}+0\cdot\dfrac{x^2}{2!}-1\cdot\dfrac{x^3}{3!}+0\cdot\dfrac{x^4}{4!}+1\cdot\dfrac{x^5}{5!}+0\cdot\dfrac{x^6}{6!}(x^6)-1\cdot\dfrac{x^7}{7!}(x^7)+O(x^9)$
$\implies \sin(x)={x}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+O(x^{13})$
$$\boxed{ \Leftrightarrow \sum_{k=0}^{\infty}\dfrac{(-1)x^{2n+1}}{(2n+1)!}}$$
We can do likewise for $\sinh(x)$
$f(x)=\sinh(x) \implies f(0)=0$
$f'(x)=\cosh(x) \implies f'(0)=1$
$f''(x)=-\sinh(x) \implies f''(0)=0$
$f'''(x)=\cosh(x) \implies f'''(0)=1$
$f^{(4)}(x)= \sinh(x) \implies f^{(4)}(0)=0$
$f^{(5)}(x)= \cosh(x) \implies f^{(5)}(0)=1$
$f^{(6)}(x)= -\sinh(x) \implies f^{(6)}(0)=0$
$f^{(7)}(x)= \sinh(x) \implies f^{(7)}(0)=1$
$\sinh(x)={x}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+\dfrac{x^{11}}{11!}+O(x^{13})$
$$\boxed{\Leftrightarrow \sum_{k=0}^{\infty}\dfrac{x^{2n+1}}{(2n+1)!}}$$
The series expansion for $\cos(x)$
$f(x)=\cos(x) \implies f(0)=1$
$f'(x)=-\sin(x) \implies f'(0)=0$
$f''(x)=-\cos(x) \implies f''(0)=-1$
$f'''(x)=-\sin(x) \implies f'''(0)=0$
$f^{(4)}(x)= \cos(x) \implies f^{(4)}(0)=1$
$f^{(5)}(x)= -\sin(x) \implies f^{(5)}(0)=0$
$f^{(6)}(x)= -\cos(x) \implies f^{(6)}(0)=-1$
$f^{(7)}(x)= \sin(x) \implies f^{(7)}(0)=0$
$\cos(x)={1}-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^{10}}{10!}+O(x^{12})$
$$\boxed{\Leftrightarrow \sum_{k=0}^{\infty}\dfrac{(-1)x^{2n}}{(2n)!}}$$
The series expansion for $\cosh(x)$
$f(x)=\cosh(x) \implies f(0)=1$
$f'(x)=\sinh(x) \implies f'(0)=0$
$f''(x)=\cosh(x) \implies f''(0)=1$
$f'''(x)=\sinh(x) \implies f'''(0)=0$
$f^{(4)}(x)= \cosh(x) \implies f^{(4)}(0)=1$
$f^{(5)}(x)= \sinh(x) \implies f^{(5)}(0)=0$
$f^{(6)}(x)= \cosh(x) \implies f^{(6)}(0)=1$
$f^{(7)}(x)= \sinh(x) \implies f^{(7)}(0)=0$
$\cosh(x)={1}+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{x^{10}}{10!}+O(x^{12})$
The series expansion for $\cos(x)$
$f(x)=\cos(x) \implies f(0)=1$
$f'(x)=-\sin(x) \implies f'(0)=0$
$f''(x)=-\cos(x) \implies f''(0)=-1$
$f'''(x)=-\sin(x) \implies f'''(0)=0$
$f^{(4)}(x)= \cos(x) \implies f^{(4)}(0)=1$
$f^{(5)}(x)= -\sin(x) \implies f^{(5)}(0)=0$
$f^{(6)}(x)= -\cos(x) \implies f^{(6)}(0)=-1$
$f^{(7)}(x)= \sin(x) \implies f^{(7)}(0)=0$
$\cos(x)={1}-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^{10}}{10!}+O(x^{12})$
$$\boxed{\Leftrightarrow \sum_{k=0}^{\infty}\dfrac{(-1)x^{2n}}{(2n)!}}$$
The series expansion for $\cosh(x)$
$f(x)=\cosh(x) \implies f(0)=1$
$f'(x)=\sinh(x) \implies f'(0)=0$
$f''(x)=\cosh(x) \implies f''(0)=1$
$f'''(x)=\sinh(x) \implies f'''(0)=0$
$f^{(4)}(x)= \cosh(x) \implies f^{(4)}(0)=1$
$f^{(5)}(x)= \sinh(x) \implies f^{(5)}(0)=0$
$f^{(6)}(x)= \cosh(x) \implies f^{(6)}(0)=1$
$f^{(7)}(x)= \sinh(x) \implies f^{(7)}(0)=0$
$\cosh(x)={1}+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{x^{10}}{10!}+O(x^{12})$
$$\boxed{\Leftrightarrow \sum_{k=0}^{\infty}\dfrac{x^{2n}}{(2n)!}}$$
Apart from these five functions, whose derivatives are rather simple like $e^x$ or cyclic like $\sin(x)$, $\cos(x)$, $\sinh(x)$, $\cosh(x)$, the Maclaurin formular do not bode well for other types of function. If you want, we can try the derivative of $\tan(x)$
$f(x)=\tan(x)$
$f'(x)=\sec^2(x)$
$f''(x)=2\tan(x)\sec^2(x)$
$f'''(x)=2\sec^2(x)[2\tan^2(x)+\sec^2(x)]$
$f^{(4)}=8\tan(x)\sec^2(x)[\tan^2(x)+2\sec^2(x)]$
The derivatives are not cyclic, which make computation of higher degrees more difficult.
We will explore different ways of expanding functions in Taylor series later. The basic formula for expanding functions centered at any point still remain important. But the Maclaurin counterpart can be replaced with easier techniques.
Apart from these five functions, whose derivatives are rather simple like $e^x$ or cyclic like $\sin(x)$, $\cos(x)$, $\sinh(x)$, $\cosh(x)$, the Maclaurin formular do not bode well for other types of function. If you want, we can try the derivative of $\tan(x)$
$f(x)=\tan(x)$
$f'(x)=\sec^2(x)$
$f''(x)=2\tan(x)\sec^2(x)$
$f'''(x)=2\sec^2(x)[2\tan^2(x)+\sec^2(x)]$
$f^{(4)}=8\tan(x)\sec^2(x)[\tan^2(x)+2\sec^2(x)]$
The derivatives are not cyclic, which make computation of higher degrees more difficult.
We will explore different ways of expanding functions in Taylor series later. The basic formula for expanding functions centered at any point still remain important. But the Maclaurin counterpart can be replaced with easier techniques.
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