What Are Orthogonal Polynomials? Inner Products on the Space of Functions
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- Опубліковано 14 жов 2024
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Oh my god! Where was this I all these days! You are not just teaching math, you are teaching the historical development of the subject. And that puts so much sense into many seemingly arbitrary "mathy" definitions.
Wish my lecturers were even a little bit more like him - teaching people how to think rather than just giving answers with no explanation. Came here for one little piece of info and took away a whole method
This is so beautiful! I just got here when I googled "orthogonal polynomials". I just wanted to know what it meant that's all. But what I got instead was 10 whole minutes of riveting math. The determination of the integrals just by "looking at it" was so beautiful. I have not studied the forward process that this video is building up off backwards. I can visualize why perpendicular vectors have a zero dot product and understand orthogonality that way. I took linear algebra 3 decades ago, and have forgotten most of it. Watching this video gave me a blood rush and a desire to study linear algebra again, so I can understand this is better context. Cant wait to watch the other videos on this channel.
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Damn, I watch a lot of math videos and this is the first one I have found from this guy. The lucidity and confidence at answering objections is truly remarkable. This is a good teacher!
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You’re a wonderful teacher. It’s very easy to follow your thoughts and you answer all my questions.
Thank you!
This dude's a very inspiring teacher.
He could dismiss ∫(√x)(√x)dx since √x is not a polynomial and the inner product is being defined over the space of polynomials.
Same as Cogito :)
You can define the inner product space on all continuous functions, not just polynomials.
Note: (sqrt(x) )^2 is not really "x" but is "x for all x >= 0"
i.e., the domain is x >= 0 which eans
1) The expression is not even defined on [-1, 1] so it is not in the space we are considered
2) The square IS positive because it is positive on the entire domain.
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Thank you so much for this video. You make it so simple, it's incredible!
Fun fact extending the sqrt(x) debate - let g be a piecewise function that is 0 on the interval [-1, 0) and sqrt(x) on [0, 1]. So now g is a valid function for this inner product (unlike sqrt(x)). But now (g, g) = 1/2, versus the debated (sqrt(x), sqrt(x)) =? 0, so no violation of axiom 3!
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I think this might be more philosophy than math, but one thing I'm a little confused about is the idea that all inner products are "equal." (I know you don't mean this in the mathematical sense of equality) On the one hand, I see why the choice of operation is arbitrary as long as the operation satisfies the conditions. On the other hand, only one definition of "length" on the space of geometric vectors corresponds to the value that we actually measure when we use a ruler. In that sense, there does seem to be a "prefered" definition of length.
Is there something about the nature of measurement that forces us to pick a definition of our "inner product" and did out brains have to pick the "inner product" that it did?
Like, if our brains had defined length with some weird weighted dot product (whatever the brain equivalent of that is) rather than the dot product with equal weights, would our concept of physical length still make sense?
Spot on. *Geometric vectors* - for which the concept of length is primary, as you stated - certainly have a preferred inner product, i.e. a*b*cos(gamma). I refer to it as the "dot product" to distinguish it from the more general inner product.
I think the inner product space definition is missing some conditions.
e.g.
(ca, b) = c(a, b)
and similarly when constant is in 2nd spot
Correct!
fantastic teacher !
Thanks but couldn't an alternative dismissal of sqrt[x] simply be that by definition it isn't a polynomial because of its noninteger exponent and consequently isn't even part of the vector space?
Yes, that's correct. But then the same question could be asked of general functions.
+MathTheBeautiful Does it matter if you multiply the functions first, then take the integral of their product, or the other way around? Sorry, I'm very slow at calculus, and you gloss over these details, thanks.
You must first multiply the functions, THEN integrate.
Is the interval of the integral chosen arbitrarily, in so far as (p, p) > 0?
Sir, we can also dismiss the sqrt(x) as it is multivalued, so there is no one to one relationship. Hence it is not a function. Can this also be another reason to reject the 3rd point?
I don't think so. Positive sqrt(x) is a legitimate function for positive arguments.
Thank you sir
If polynomials form a vector space, would the dual be the set of all possible definite integrals? If so, what would be a basis for the dual space?
The dual is a slightly more complicated question because you would be talking about the dual of an infinite dimensional vectorspace.
In this case the polynomials are a dense subset of a space called L^2. This space Has itself as its dual, and because of the denseness of the set of polynomials the dual of it is also L^2.
L^2 is the "set" of all functions for which the integral of the square of itself is finite.
The basis question is even more complicated because it depends on what you mean by basis. Do you mean a normal basis or a Hilbertbasis?
Actually, googling orthogonal polynomials brought me here.
Why I googled that? Well, if you must know I was doing a linear regression in R. When I moved from a simple regression to a polynomial regression I couldn't make sense of the output. Somewhere I saw "google orthogonal polynomials" and now I am trying to see how they used that in that programming language.
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the sound is weird. maybe because the mic receives the original voice and also the sound of the speakers.
01:30 Can this trick be extended to other intervals under the parabola basing on this fact?
01:48 Does it mean that `y = x` is not a "unit" polynomial?
06:32 OK nevermind ;>
nice one
+MathTheBeautiful The integral of f(x) = 1 from -1 to 1 should be 2, not 0. Why do you say it is 0 at 2:40?
Don't they cancel out essentially because it is from -1 to +1 ,the areas would sum to 0?
Because that is not what is said: It is not the integral of f(x) = 1 from -1 to 1; it is the integral of f(x) = x from -1 to 1.
I don't think the dismissal of (√x,√x) is complete. I agree that the lower limit of the integral is outside the domain of √x, but if you go ahead and evaluate it, the integral of (√x)(√x) from -1 to 1 is 1, because (√x)(√x) = |x|.