Taylor series

A polynomial is the easiest kind of function: you can differentiate it, integrate it and evaluate it with arithmetic alone. A Taylor series is the attempt to replace a hard function with a polynomial that behaves like it.

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$

Match the value at $a$, then the first derivative, then the second, and so on. Each term fixes one more derivative, and matching infinitely many usually recovers the function exactly.

When $a = 0$ it is called a Maclaurin series, which is the case the exam asks for almost every time.

Watch it converge — and stop

At degree $1$ the polynomial is just $x$ — the tangent line at the origin, which tracks the curve to within $0.01$ out to about $\pm 0.4$ and then leaves.

Drag the degree up and each odd term pushes that window outward: degree $3$ holds to about $\pm 1$, degree $7$ to about $\pm 2.5$, degree $11$ to about $\pm 4$. Near the centre the improvement is dramatic.

Watch the edges of the panel while you do it. The worst-error readout covers the whole visible window, and it does not fall steadily — it is about $78$ at degree $3$, about $221$ at degree $7$, then about $67$ at degree $11$. A higher-degree polynomial fits better where it fits and diverges harder where it does not, because its leading term grows faster. The improvement is local, and the readout is honest about that.

Sine’s series converges everywhere, so pushing the degree far enough covers any interval you like — the wild edges retreat as the degree climbs. That is a special property, not the general case, and the next section is the general case.

Those three check the pattern: the derivative of the degree-3 polynomial is the degree-2 polynomial for cosine, exactly as $\sin' = \cos$ demands. A Taylor series differentiates term by term into the series of the derivative, which is one of the most useful facts in the unit.

The four to know

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \quad (|x| < 1)$$

The last verified line is $e^x$’s defining property showing up in its series: the polynomial’s derivative is the polynomial one degree lower, which is the only way a function can be its own derivative.

The fourth series is the geometric series from chapter 10.1 read backwards, and it is the source of most series you will be asked to derive rather than recall — substitute, differentiate or integrate it and you get many others.

The radius of convergence

Sine and $e^x$ converge everywhere. $\tfrac{1}{1-x}$ does not: its series is geometric with ratio $x$, so it converges only for $|x| < 1$.

That number — $1$ here — is the radius of convergence, and outside it the series does not approximate the function badly. It fails to converge at all.

At $x = \tfrac12$ the series gives $2$, which is $\tfrac{1}{1 - 1/2}$. At $x = \tfrac13$ it gives $\tfrac32$, which is $\tfrac{1}{1 - 1/3}$. Both inside the radius, both correct.

At $x = 2$ the terms are $1, 2, 4, 8, \ldots$ — growing without bound, while the function $\tfrac{1}{1-2}$ is a perfectly ordinary $-1$. The function is fine; the series is not.

What series are for

Beyond approximation, they make otherwise impossible integrals possible.

$\int e^{-x^2}dx$ has no elementary antiderivative — chapter 6.3 said so. But substituting $-x^2$ into the series for $e^x$ gives a polynomial, and polynomials always integrate:

$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$

$$\int e^{-x^2}dx = x - \frac{x^3}{3} + \frac{x^5}{10} - \cdots$$

The verified line is the first three terms integrated, and it matches. The integral that had no answer in elementary functions has one as a series — which is how such integrals are actually computed, in this course and outside it.