Build the standard series

Compare logarithm partial sums graphically

The source uses $Y_1=\ln(1+X)$ and successively adds the following curves. Both copies print all five input-screen/display/observation rows:

Source row: 1

Input or selection: $Y_2=X$ (red) with $Y_1$ (blue)

Printed observation: The line does not resemble the logarithm across the shown window.

Source row: 2

Input or selection: Add $Y_3=X-X^2/2$ (black)

Printed observation: It starts to resemble the logarithm.

Source row: 3

Input or selection: Add $Y_4=X-X^2/2+X^3/3$ (magenta)

Printed observation: It resembles the logarithm more closely.

Source row: 4

Input or selection: Add $Y_5=X-X^2/2+X^3/3-X^4/4$ (green)

Printed observation: It resembles the logarithm still more closely.

Source row: 5

Input or selection: Hide the other polynomial graphs, leaving $Y_1$ and $Y_5$

Printed observation: Compare the fourth-degree partial sum directly with the logarithm.

The reconstruction below uses the same inputs and successive selections. Its graph window is illustrative: the source does not specify the original calculator settings. Near zero, even the first line is a local approximation because the logarithm’s derivative there is 1.

The source remark calls the approximation valid only for $-1<x\le1$; this is the infinite series’ representation interval.

Keep two checks separate. A graph comparison does not prove convergence. Being inside the convergence interval also does not guarantee that a finite truncation meets a chosen error tolerance. These distinctions qualify the printed observation.

Examples 1–2 retain the full Teacher factorisation, logarithm law, substitution through $x^6$, transformed inequality and closed-endpoint answer.

Derive the exponential series

Both copies print the complete derivative table and derivation:

Order: 0

Derivative: $f(x)=e^x$

Value at zero: $f(0)=1$

Order: 1

Derivative: $f^{\prime}(x)=e^x$

Value at zero: $f^{\prime}(0)=1$

Order: 2

Derivative: $f^{\prime\prime}(x)=e^x$

Value at zero: $f^{\prime\prime}(0)=1$

Order: 3

Derivative: $f^{\prime\prime\prime}(x)=e^x$

Value at zero: $f^{\prime\prime\prime}(0)=1$

Order: 4

Derivative: $f^{(4)}(x)=e^x$

Value at zero: $f^{(4)}(0)=1$

Every later derivative also equals $e^x$, with value 1 at zero. Substitution into the coefficient formula gives

$e^x=1+1x+\frac1{2!}x^2+\frac1{3!}x^3+\frac1{4!}x^4+\cdots+\frac1{n!}x^n+\cdots$

$=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots+\frac{x^n}{n!}+\cdots.$

The representation is valid for every $x\in\mathbb R$. This does not mean that a low-degree polynomial is accurate at every real input.

Derive the sine series

Both copies show the repeating derivative cycle:

Order: 0

Derivative: $\sin x$

Value at zero: 0

Order: 1

Derivative: $\cos x$

Value at zero: 1

Order: 2

Derivative: $-\sin x$

Value at zero: 0

Order: 3

Derivative: $-\cos x$

Value at zero: −1

Order: 4

Derivative: $\sin x$

Value at zero: 0

Order: 5

Derivative: $\cos x$

Value at zero: 1

Thus

$\sin x=0+1x+\frac0{2!}x^2+\frac{-1}{3!}x^3+\frac0{4!}x^4+\frac1{5!}x^5+\cdots$

$=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,\qquad x\in\mathbb R.$

For x in radians sufficiently small that cubic and higher terms can be neglected, the source gives $\sin x\approx x$. Radians are essential; all-real convergence of the infinite series is a different statement from small-angle accuracy of this linear truncation.

Derive the cosine series

The complete source cycle is:

Order: 0

Derivative: $\cos x$

Value at zero: 1

Order: 1

Derivative: $-\sin x$

Value at zero: 0

Order: 2

Derivative: $-\cos x$

Value at zero: −1

Order: 3

Derivative: $\sin x$

Value at zero: 0

Order: 4

Derivative: $\cos x$

Value at zero: 1

Order: 5

Derivative: $-\sin x$

Value at zero: 0

Hence

$\cos x=1+0x+\frac{-1}{2!}x^2+\frac0{3!}x^3+\frac1{4!}x^4+\frac0{5!}x^5+\cdots$

$=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots,\qquad x\in\mathbb R.$

When x is in radians and cubic and higher powers may be neglected, the source gives $\cos x\approx1-x^2/2$. Independently, the cubic coefficient is zero, so the first omitted nonzero term is $x^4/24$, compared with the sine approximation’s first omitted term $-x^3/6$.

Compare cosine partial sums graphically

The source writes the general term as $(-1)^r x^{2r}/(2r)!$, with the cosine series valid for all real x. Both copies give this five-row calculator comparison:

Source row: 1

Input or selection: $Y_1=\cos X$ (blue), $Y_2=1$ (red)

Printed observation: The constant function does not look like cosine across the shown window.

Source row: 2

Input or selection: Add $Y_3=1-X^2/2!$ (red); $Y_2$ becomes magenta

Printed observation: The quadratic starts to look like cosine.

Source row: 3

Input or selection: Add $Y_4=1-X^2/2!+X^4/4!$ (black)

Printed observation: It looks more like cosine.

Source row: 4

Input or selection: Add $Y_5=1-X^2/2!+X^4/4!-X^6/6!$ (green)

Printed observation: It looks even more like cosine.

Source row: 5

Input or selection: Hide $Y_2,Y_3,Y_4$, leaving cosine and $Y_5$

Printed observation: Compare the final partial sum directly with cosine.

The reconstruction uses an explicitly illustrative common window. Near zero even the constant 1 approximates cosine: its error begins at order $x^2$. Adding successive even powers improves local agreement, but a plot is not a proof of convergence or a guarantee of accuracy at every displayed input. The original screens and their changing colours are retained in the source artwork archive.

The source asks: Can you think of an alternative method to obtain the series expansion of $\cos x$? Its full answer differentiates the sine series:

$$\cos x=\frac{d}{dx}(\sin x)=\frac{d}{dx}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots.$$

Termwise differentiation is justified within the power series’ radius of convergence, which is infinite for sine. Thus this produces cosine’s all-real expansion, rather than a new finite-polynomial identity.

Derive the tangent series

Both copies mark this expansion not in MF27 and print every step through the fifth derivative. Writing derivatives as polynomials in $\tan x$ makes evaluation at zero immediate. Start with $f(x)=\tan x$, so $f(0)=0$.

First derivative:

$$f^{\prime}(x)=\sec^2x=1+\tan^2x,\qquad f^{\prime}(0)=1.$$

Second derivative:

$$f^{\prime\prime}(x)=2\tan x\sec^2x$$ $$=2\tan x(1+\tan^2x)=2\tan x+2\tan^3x,\qquad f^{\prime\prime}(0)=0.$$

Third derivative:

$$f^{\prime\prime\prime}(x)=2\sec^2x+6\tan^2x\sec^2x$$ $$=(2+6\tan^2x)(1+\tan^2x)$$ $$=2+8\tan^2x+6\tan^4x,\qquad f^{\prime\prime\prime}(0)=2.$$

Fourth derivative:

$$f^{(4)}(x)=16\tan x\sec^2x+24\tan^3x\sec^2x$$ $$=(16\tan x+24\tan^3x)(1+\tan^2x)$$ $$=16\tan x+40\tan^3x+24\tan^5x,\qquad f^{(4)}(0)=0.$$

Fifth derivative:

$$f^{(5)}(x)=16\sec^2x+120\tan^2x\sec^2x+120\tan^4x\sec^2x$$ $$=(16+120\tan^2x+120\tan^4x)(1+\tan^2x)$$ $$=16+136\tan^2x+240\tan^4x+120\tan^6x,\qquad f^{(5)}(0)=16.$$

Substitute these values into the Maclaurin formula:

$$\tan x=0+1x+\frac0{2!}x^2+\frac2{3!}x^3+\frac0{4!}x^4+\frac{16}{5!}x^5+\cdots$$ $$=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\cdots=x+\frac{x^3}{3}+\frac{2x^5}{15}+\cdots.$$

Use radians. When x is small enough to neglect cubic and higher powers, the source gives $\tan x\approx x$. Independently, its first omitted nonzero term is $x^3/3$.

The tangent series is a local expansion: unlike sine and cosine, its representation interval is $|x|<\pi/2$, bounded by the nearest poles. The derivative calculation by itself does not prove that interval. Do not apply the all-real statement from the preceding cosine page to tangent.

The source also offers a QR-linked exploration of Maclaurin expansions and validity. This is the destination decoded from the original QR; its current external content has not been reviewed. The original artwork is preserved with the source images.

Know which formulae may be quoted

The source summary lists the binomial, exponential, sine, cosine and logarithm expansions as available in MF27. They may be quoted without proof unless the question asks for their derivation. Tangent is not included in that list. The general terms and representation intervals are:

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^r}{r!}+\cdots,\qquad x\in\mathbb R.$$ $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+\frac{(-1)^r x^{2r+1}}{(2r+1)!}+\cdots,$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+\frac{(-1)^r x^{2r}}{(2r)!}+\cdots.$$

The sine and cosine expansions apply for all real x in radians. The source repeats the small-angle forms $\sin x\approx x$, $\cos x\approx1-x^2/2!$ and $\tan x\approx x$, assuming radians and that cubic and higher powers can be neglected.

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+\frac{(-1)^{r+1}x^r}{r}+\cdots,\qquad -1<x\le1.$$

The binomial expansion is derived next. Its printed $|x|<1$ condition must be distinguished from the unrestricted finite polynomial when the exponent is a positive integer.