What IS a Taylor Series?

If you have studied Physics, you may be familiar with the small angle approximation cos θ ≈ 1 - θ²  / 2. This is a surprisingly accurate approximation for small values of theta (try it!), but where does it come from? The idea of Taylor's Theorem is to use algebraic manipulation to find polynomials (i.e. anything of the form a + bx + cx² + dx³...) that approximate non-polynomial functions (e.g. cos x, sin x, e^x etc.)  around a certain point. Quite a mouthful, and perhaps a little overwhelming at first, especially when faced with the seemingly complex formula. Taking a step back, let's consider, if I asked you to find a quadratic that acted as much like cos x as possible around a certain point (say for this example x = 0), how would you try to figure this out? A good place to start here is to make sure the quadratic has the same value as cos x at 0. Plugging 0 into cos x, we find it equal to 1. So if we formalise this new quadratic, we have something of the form f(x) = a + bx + cx² and substituting 0 in we get f(0) = a. We can now comfortably set a equal to 1, and so f(x) = 1 + bx + cx². We've now locked our function into any quadratic that intercepts the y axis at (0,1). Not bad, but not great either. We can do better.

Next you might think a good idea would be if our function had the same gradient as cos x around (0,0), and you'd be right! The derivative of cos x is -sin x, meaning that the gradient of cos x at 0 is -sin(0) = 0. Taking the derivative of our quadratic, we get f ' (x) = b + 2cx, and so f ' (0) = b. Equating the two, we find b = 0, and so f(x) = 1 + cx². So now we have a quadratic with a maximum/minimum at y = 1. A natural continuation would be to take the second derivative and equate, as even though the gradient of cos x at 0 is 0, it is decreasing rapdily around that point. Taking the second derivative of cos x gives us -cos x, and subbing in 0 we get -cos(0) = -1. Doing the same for our quadratic gives f '' (x) = 2c, and so f '' (0) = 2c. Equating we get 2c = -1, and so c = -1/2. Putting this all together we get cos x ≈ 1 - x² / 2. That's it! Getting back to our original approximation is just a matter of changing variables.  Of course you may ask if continuing this process of taking derivatives and equating would yield a more accurate approximation, and you'd be right. This is where the taylor series formula T(x) = f(a)  + f ' (a)(x - a)  + f '' (a)(x - a)²  / 2! + f ⁽³⁾ (a)(x - a)³  / 3! + ... comes into play. Setting a to 0 and evaluating for f(x) = cos x we find it starts exactly the same way as our approximation. And that's basically it! Looking at the (x - a)ⁱ terms might seem confusing at first, but replacing a with 0 we find that the expansion is actually just incrementing powers of x (a polynomial!), and using (x - a) just allows us to treat the point as if it were at x = 0. Finally, the presence of increasing factorials as divisors is a direct result of differentiating the polynomial as we did earlier. To see how this works, take f(x) = x⁵ and try differentiating until the x term is gone. Can you find an alternate way of writing that number?