/sci/, I've finished my "sequences and series"

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>>8883202
thats fuckin sick, i didn't realize that happens, gonna look into this now

>>8883202
This vid was pretty good at explaining it. I don't think it needs any explanation though, math is just beautiful that way.

https://www.youtube.com/watch?v=GUtLtRDox3c

Not sure this will be a satisfactory explanation, but it's a generalization of tangent lines. In fact if you take the first two terms of the Taylor series you get the tangent line.

The tangent line is a degree 1 polynomial whose derivative matches the function. In general the kth partial sum of the Taylor series is a degree k polynomial whose first, second, ... and kth derivatives all match the function.

As you take more and more partial sums the approximation gets better and better.

>>8883202
Well, the first trick with this formula is that for any given polynomial f(x), [math]\frac{f^{(n)}(0)}{n!}[/math] just gives you its nth coefficient

So for an arbitrary polynomial p(x), [math]\sum_{i=0}^{\inf} \frac{p^{(n)}(0)}{n!} x^n[/math] just gives you p(x)

Now, if we have p(x)=sin(x), for example, then we just stick [math]sin^{(n)}(0)[/math] in there, and it all works fine.

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Fun math problems. Find the all the zeros for the equation in the image

>>8883202
>>there you go, an almost perfect approximation of your entire original function

Not really. It's just an almost perfect approximation in the neighborhood around x=0.

>>8883202
It only works that way for very special functions. The chance of a generic differentiable function being real analytic on the whole line is small.

>>8883202

Assuming f(x) has the form [math]f(x)=\sum a_n x^n[/math], how do you get the [math]a_n[/math]? Well [math]a_0[/math] is just f(0). What happens when you take the derivative of f(x) and the series?

>>8884061
This.

Just consider that for a function on the reals, both domain and codomain are uncountable, but you consider functions that can be described with the same data as a countable vector space, ∑ a_n x^n.

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>>8883937
I really need to take an analysis class. I am familiar with the concepts, but have not rigorously put it all together in my head.

What's crazy is how similar and different a series expansion is to vectorspace expansions, but are emphatically not vectorspace expansions. That is specifically why it's hard to 100% buy series expansions for me, such that I am always a little suspicious of under what conditions the expansion is valid and if I have satisfied those conditions. WIth vectorspaces, shit's easy.

>>8883202
it only appears to be a "miracle" because the functions that you run into in an introductory calculus class are anomalously well-behaved compared to a "general" function, like >>8884061 said.

https://en.wikipedia.org/wiki/Non-analytic_smooth_function

>>8884694
What do points outside a radius of convergence do? Do they converge to the wrong values, or not converge at all, or some combination of the two?

>>8884708
well if the distance is strictly the radius of convergence, then the series diverges. if the distance is equal to the radius, then it could conceivably diverge or converge. I don't think it can converge to the wrong value, because of this: https://en.wikipedia.org/wiki/Identity_theorem

>>8884708
I'm just going off of memory from my calc class a year ago, but IIRC they usually go to +/- infinity. The graph of this series looks like the normal function within the radius of convergence, then shoots off right at the edge of it, never to be seen again.

>>8884728
>then shoots off right at the edge of it, never to be seen again
That cray cray

>>8883202
>differentiate*
>differentiate*
>differentiate* it again
>differentiate*
>differentiate*
>differentiate*

piggot bump

>>8884708
There are even some functions that outside a certain region will have a higher order taylor expansion be a worse approximation than a lesser order one. I think arctan works like this.

>>8884720
I believe e^(-1/t^2) converges to the wrong value at t = 0