Derive coefficients and check validity

After substituting a new expression, transform the convergence condition too. Use radians for small-angle approximations and keep enough terms for any later division or cancellation.

Derive the coefficients before using them

Start with the requested highest power. For a polynomial through $x^n$, find the derivatives at zero through order $n$, then divide each by its factorial. A derivative value is not yet a series coefficient.

Keep three questions separate: can the coefficients be formed, where does the infinite series equal the function, and how accurate is the chosen finite truncation?

A power series has real constant coefficients and ascending powers:

$a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots=\sum_{r=0}^{\infty}a_rx^r.$

Assuming it represents f near zero, differentiate term by term within its interval of convergence. The complete source sequence is:

$f^{\prime}(x)=a_1+2a_2x+3a_3x^2+\cdots+na_nx^{n-1}+\cdots,$

$f^{\prime\prime}(x)=2a_2+3(2)a_3x+4(3)a_4x^2+\cdots+n(n-1)a_nx^{n-2}+\cdots,$

$f^{\prime\prime\prime}(x)=3(2)a_3+4(3)(2)a_4x+5(4)(3)a_5x^2+\cdots+n(n-1)(n-2)a_nx^{n-3}+\cdots.$

After n differentiations the first surviving term is $n!a_nx^0=n!a_n$. Substitution at zero gives the source coefficient table:

Evaluation: $f(0)=a_0$

Coefficient: $a_0=f(0)$

Evaluation: $f^{\prime}(0)=a_1$

Coefficient: $a_1=f^{\prime}(0)$

Evaluation: $f^{\prime\prime}(0)=2!a_2$

Coefficient: $a_2=f^{\prime\prime}(0)/2!$

Evaluation: $f^{\prime\prime\prime}(0)=3!a_3$

Coefficient: $a_3=f^{\prime\prime\prime}(0)/3!$

Evaluation: $f^{(n)}(0)=n!a_n$

Coefficient: $a_n=f^{(n)}(0)/n!$

Putting these coefficients back gives the Maclaurin formula shown above and identified by the source as being in MF27. Maclaurin series are Taylor series centred at zero.

Independent coefficient check. If $f^{\prime\prime}(0)=6$, the coefficient of $x^2$ is $6/2!=3$. Divide by the factorial even when a derivative value looks simple.

Higher derivatives are repeated operations

The full source notation table is:

Order: First

Equivalent derivative notation: $f^{\prime}(x)=f^{(1)}(x)=dy/dx=d(y)/dx$

Order: Second

Equivalent derivative notation: $f^{\prime\prime}(x)=f^{(2)}(x)=d^2y/dx^2=d(dy/dx)/dx$

Order: Third

Equivalent derivative notation: $f^{\prime\prime\prime}(x)=f^{(3)}(x)=d^3y/dx^3=d(d^2y/dx^2)/dx$

Order: Fourth

Equivalent derivative notation: $f^{(4)}(x)=d^4y/dx^4=d(d^3y/dx^3)/dx$

Order: nth

Equivalent derivative notation: $f^{(n)}(x)=d^ny/dx^n=d(d^{n-1}y/dx^{n-1})/dx$: differentiate n times

In general, $d^ny/dx^n$ is not $(dy/dx)^n$ for $n>1$. For example, independently taking $y=x^3$ gives $y^{\prime\prime}=6x$, while $(y^{\prime})^2=9x^4$.

The source also warns that the first-derivative reciprocal rule $dx/dy=1/(dy/dx)$ does not extend by simply taking reciprocals of higher derivatives. The first rule assumes a differentiable local inverse and a nonzero finite derivative.

For $y=x^2$ on $x>0$, independently $d^2x/dy^2=-1/(4x^3)$, whereas $1/(d^2y/dx^2)=1/2$.

Existence of coefficients does not by itself prove equality

The source lists three necessary conditions: derivatives of every order exist, their values at zero are defined, and $f(0)$ is defined. These let us form coefficients, but are not sufficient by themselves to prove that the series equals f near zero.

That equality needs additional justification, such as a Taylor remainder tending to zero. This qualification is separate from the retained source wording.

For example, define $g(0)=0$ and $g(x)=e^{-1/x^2}$ when $x\ne0$. Every derivative at zero is 0, so its Maclaurin series is identically 0, yet $g(x)>0$ away from zero. This counterexample explains why the coefficient derivation starts by assuming a valid power-series representation.