Hello!

@TeresaLisbon Can I ask more on uniform convergence?

@soupless Hi, I won't be available right now because some urgent work has cropped up, but I'll try and make time. You can drop your question. There are some questions above, I'll try to make sure I can answer them by today evening.

Thank you very much. From what you said earlier, uniform convergence is the convergence of a function such that for every $\epsilon > 0$, there exists $\delta > 0$ such that $|x - y| < \delta \implies |f(x) - f(y)| < \epsilon$.

Now, what I don't get is the part where sequences of functions are introduced. Something like $$\lim\limits_{x \to a} f_{n} = f \text{ uniformly}$$

Can I ask what this means?

Hello, @Koro !

Hi @soupless

@PCMSE We can do that area question by "flipping" the function, I think. The point is to not write $y$ as a function of $x$, but rather $x$ as a function of $y$. Let me show how : first you write , with some collecting of terms and completing the square, $x^2-6xy+9y^2-2x+4y-1 = (x-3y-1)^2 - 2(y+1)$. Now, this equals zero exactly when $x = 3y+1 \pm \sqrt[2(y+1)]$. Therefore, you have two functions $g_i(y)$ here, given by $g_1(y) = 3y+1 + sqrt(2(y+1))$ and $g_2(y) = 3y+1- sqrt(2(y+1))$.

Both functions will be defined in the range [-1,1], and both functions at $y=-1$ have the value $x=-2$. In other words, you have to turn the graph $90$ degrees to see $x$ as a function of $y$, and then calculate the integral of $g_1(y) - g_2(y)$ from $-1$ to $1$, because we are to stop at $y=1$. So I think the answer, is $\int_(-1)^1 g_1(y) - g_2(y) dy$. Let me make sure that this is the case.

This comes out to be 16/3.

I verified this using Desmos.

Here is graphical proof that I got the graph region and the integration right. This, in ONE HOUR of firefighting!

That was a good question.

@Koro Actually, I used orthogonality to create that linear functional on $V$, with nullspace $U$. I forgot that it can be done without it using just basis arguments, so we were definitely on the same page.

I should have asked you if you knew the above fact , because for some reason I thought I should prove it, struggled with an argument with just bases and went for orthogonality. Anyway, it's sorted out thankfully.

@Wolgwang Taking a look now.

Yes, I can understand how to solve the question. Finite differences is actually a very powerful method, let me quickly explain.

So let's imagine you have a polynomial $f(x)$ (with integer/rational coefficients), of degree $n$. You can create the polynomial $g(x) = f(x) - f(x-1)$, which is called the first finite difference of $f$. This is like a "discrete" derivative of $f$.

Now, what you can do is take multiple finite differences. Let's take an example to illustrate. Suppose I start with $f(x) = x^3$. Then, the first finite difference is $f(x) - f(x-1) = x^3 - (x-1)^3 = 3x^2-3x+1$. The second finite difference will be $3x^2-3x+1 - (3(x-1)^2 - 3(x-1)+1) = 6x-6$. The third finite difference will be $6x-6 - (6(x-1) - 6) = 6$, which is a constant polynomial.

Now, there are a few things you can observe about finite differences. The most obvious is that if $f(x)$ is of degree $n$, then the $k$th finite difference of $f(x)$ is a polynomial of degree $n-k$. With $k=n$ you get that the $n$th finite difference is some constant polynomial.

The most important point, however, is that using finite differences, you can actually reconstruct the entire polynomial at many points. Let me quickly explain why.

The finite differences theory is basically an extension of a very simple logic : if I know $f(1)$, and I know $f(2)-f(1)$, then I know $f(2)$. Now, taking this to the next level : if I know $f(1), f(2), f(2)-f(1)$ and $f(3)-f(2)$ then I know $f(3)$. That intuition helps us understand why the first differences are important, and similarly why successive differences are important. Finite differences are often used to guess polynomial patterns for the reason that for any polynomial pattern...

... the finite differences become constant.

I'll get you some useful articles. Here is a proof but also a worked out example and some nice context.

This is Brilliant.org's page on finite differences. In AoPS, the differences are written from top to bottom, here it's written from left to right. Having said that, there's a clear explanation on how finite differences can be used to find polynomial values at integer (or equally spaced) points, which is really useful for competition questions.

Finite differences can be used for polynomial fitting : if you're given f(1),...,f(n) and want to find an n degree polynomial which fits the conditions of $f$, you can use finite differences, get to the constant term and reconstruct the polynomial by using the inverse finite difference, the operation $g(x) = g(x-1)+ f(x)$ will (along with specifying $g(1)$) make $g$ an inverse finite difference of $f$, so building up gets you an answer.

I hope this was helpful. If you want, I'll give you a more detailed demonstration for that particular AoPS problem.

@soupless Ah, ok : I couldn't find our discussion about uniform convergence, but I found something about uniform continuity above. Nevertheless, the point is that when you have a sequence, it's always "there is an N, for all n>N, something happens". This is always the case.

For example, $f_n \to f$ on a domain $A$, uniformly means : for all $\epsilon>0$, there exists $N \in \mathbb N$ such that for all $x \in A$ and $n>N$ we have $|f_n(x) - f(x)| < \epsilon$. So basically, for large enough $n$, $f_n$ can be brought uniformly close to $f$ as much as desired.

It's like, $n$ sort of takes the role of $1/(\delta)$ if you like in the delta-type definitions. $\delta$ being small, is like $n$ being large. Knowing this kind of logic, you can easily transition between a definition for sequences , and a definition for functions.

Teresa Lisbon That is simple and interesting. Thanks :-)

Thank you very much @TeresaLisbon. I finally understood and solved it. Sorry for late response

Was trying to understand and solve

Really a wonderful method @TeresaLisbon. Thank you very much for your time and so much efforts :-)

@TeresaLisbon I am sorry if I could not reply earlier. I get the point now on sequences of functions. What I don't get is the connection between them, if there is one. If I may ask, can you explain it please? Only if you have time.

I could not reply earlier because I fell asleep. Probably because I was lacking some.