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From what I know the Taylor's series has to:

  • have a radius of convergence, for example 1/(1-x), only converges when x is between -1 and 1. but what if you want to do something with the value x=17, does that mean you cant do anything with the Taylor series then? --> not useful

  • infinitely often differentiable. How many functions are infinitely often differentiable?

  • limit as n goes to infinity of the remainder needs to equal 0. What if this is not the case? How many/little functions satisfy this condition?

my thing is this: a function like 1/1-x is quite easy to work with imo. Since I don't know a lot about Taylor series yet, Taylor series seem like a complication of something that's supposed to be easy, which i hate. obviously not a lot of things in math are useful, but complicating something that's easy seems unnecessary to me, unless its useful, something which I don't know yet

Martin Sleziak
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    "Taylor series seem like a complication of something thats supposed to be easy" No, it is exactly the opposite. The fundamental strategy of calculus is to take a complicated nonlinear function and approximate it locally with a linear function. Taylor series take this strategy a step further and approximate by a quadratic function or a cubic function, etc -- more complicated than linear, but still much simpler than a general nonlinear function. Very often, complicated calculations are greatly simplified by such an approximation. Look at the derivation of $E = mc^2$, for example. – littleO Mar 29 '20 at 11:46
  • A good example to think about is the second derivative test for checking whether a stationary point of $f$ is a local maximum or a local minimum. We approximate $f$ locally by a quadratic function, and quadratic functions are simple enough that it's easy to see if we are looking at a local maximum or a local minimum. – littleO Mar 29 '20 at 11:50
  • Okay thanks. And do most functions have taylor series? If taylor series apply to a small set of functions, then the usefullness is still kinda lost –  Mar 29 '20 at 12:12
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    It's a good question, but there are enough smooth functions out there that this strategy turns out to be extremely useful. It might be helpful to think in terms of Taylor's theorem with remainder. A function only has to be differentiable to be approximated locally by a linear function, or twice differentiable to be approximated locally by a quadratic function, etc. – littleO Mar 29 '20 at 12:19

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In essence, the Taylor Series allows us to translate between polynomials, which are really nice to work with, and sometimes really complex functions that are nearly impossible to compute directly.

For example, how do you compute $\sin(x)$? How would a calculator do that? That's a function that is really hard to compute directly—geometrically, it represents the height as you go around the circle starting at $(1,0$) moving $x$ radians counter-clockwise. But you obviously won't pull out a ruler to measure that. We need a better way. There come the Taylor Series. We have a very useful approximation that becomes an equality as the limit approaches infinity:

$$\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}...$$

Now it's much easier to compute. And in fact, this is how calculators often compute functions like sine and cosine! Just using a few terms of this series gets us very close to the true value of sine. Similarly, how do we compute $\pi$? We're not generally measuring circles—rather, we use approximations including the Taylor Series to get us as approximate as we want—that's the idea of the remainder with the limit as $n\rightarrow \infty$. We want our approximation to get better and better as we carry it out to more terms, and in-fact the approximation gets infinitely close to the true value and becomes exact as $n$ approaches infinity.

And you're right about $\frac{1}{1-x}$, you probably would never need to use the Taylor approximation for that—but how about the other way around? What is the sum $1+x+x^2+x^3+x^4+...$ all the way to infinity? How can we sum an infinite number of terms? Well the Taylor Series tells us that this infinite sum is in fact equal to $\frac{1}{1-x}$. That's precisely how we get the sum of the infinite series. This tells us important things unrelated to pure math like the money multiplier used to compute how much money banks should keep. And those types of problems appear everywhere, in economics to physics to chemistry, etc. (And to answer your question about radius of convergence, it should make intuitive sense here now— what happens to the sum of $1+x+x^2+x^3+x^4+...$ if $|x| \geq 1$?)

The Taylor Series has "restrictions", yes, but that's what make them great. They allow us to work with really complex functions by simplifying them down to making an approximation. Most things in higher level math are very hard to compute directly (like differential equations, which can help epidemiologists model a pandemic, or compute rocket trajectories), so we rely on important approximations to get us close enough to the answer, which allow us to literally reach to the moon and beyond.

rb612
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  • Okay thanks. And do most functions have taylor series? If taylor series apply to a small set of functions, then the usefullness is still kinda lost –  Mar 29 '20 at 12:13
  • @SteezO- we require that we have this idea of “smoothness”—infinite differentiability at the point where we’re expanding — and this is often the case for functions we care about. Think all things that don’t very abruptly change. But there are other approximation methods too which have even more lax requirements but are more advanced to perform. Taylor approximation is like “how are things changing right at this moment?” And uses that to generalize to how things will change over time. – rb612 Mar 29 '20 at 12:19
  • Also, if you’ve ever wondered how root finders (solving for zeros of a function) often work or how we find max or min of functions using a computer—the theory is grounded in? You guessed it—the Taylor Series. – rb612 Mar 29 '20 at 12:25