Gram-Schmidt Orthogonalization and Legendre Polynomials

[@MathTheBeautiful]

Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.

[@piyushsharma5177]

Teachers who are passionate in teaching aint interested in whether you can score better than the guy sitting next to you, rather getting the knowledge and applying it on things that others have yet not touched upon. Great explanation sir.

[@mohammedal-haddad2652]

Absorbing the question is much important than getting the answer. That's the real lesson.

[@mehmetalidemir8380]

If this is how you guys over in USA are getting educated in mathematics, then you can consider yourselfs as lucky. Well made video, thanks for sharing!

[@jeanpi314159]

I don''t think so : who has access to study at MIT ? how much is it ? and how many people earn enough money and get good education in the USA ? statistically a rare minority. USA aren't great in their educationnal system : Cuba is far above and for free, everywherein the country and for everyone. See the statistics about This teacher is great, but he cannot change this system, though.

[@wernhervonbraun4222]

6:45 I always ask myself, Why am I doing this basically on anything I do. Also, it seems like a lyric from Talking head song hahaha.

[@adwaithkj]

had this concept in communications theory class, and I had no clue. This made things clear. Thanks, sir. Beautifully explained

[@skye-chor]

Best explanation on the platform. Thank you from Hong Kong

[@MathTheBeautiful]

Thank you! And yes, you're correct :)

[@fouriertransformationsucks438]

Writing on the blackboard and explaining step by step is ten times better than giving 30 pages of slides and read through it.

[@rohanjoshi3785]

true bro

[@gossipGirlMegan]

This a the most beatiful explaination in the world! Thanks a lot ,professor!

[@MathTheBeautiful]

Thank you! It's the name of the channel!

[@ivanrodionov9724]

Excellent, simple and elegant explanation! Thank you sir!

[@holyshit922]

You can play with other orthogonal polynomials using this approach Inner product for other orthogonal polynomials Chebyshev Polynomials (your transcription from cyrylic of name that guy is terrible and leads to misreading) (p,q) = Int(p(x)*q(x)/sqrt(1-x^2),x=-1..1) With integration by parts we can derive nice reduction formula for Int(x^{n}/sqrt(1-x^2),x=-1..1) Base cases are easy Int(x/sqrt(1-x^2),x=-1..1) = 0 Int(1/sqrt(1-x^2),x=-1..1) = pi , (arc sine) Hermite Polynomials (p,q) = Int(p(x)*q(x)*exp(-x^2),x=-infinity..infinity) With integration by parts we can derive nice reduction formula for Int(x^{n}*exp(-x^2),x=-infinity..infinity) Base case can be reduced to Gamma by substitution and then by reflection formula Laguerre Polynomials (p,q) = Int(p(x)*q(x)*exp(-x),x=0..infinity) This can be expressed as Gamma function of simply factorial

[@louisderouiche2091]

great prof

[@MathTheBeautiful]

Thank you, it means a lot!

[@rishabhgarg9217]

Great series of lectures. But i wanna ask a question How can you say that (1,x,x^2) is a worst basis ? only because they are not orthogonal ? I do agree that they are not orthogonal but you are determining that orthogonality using only that particular definition of inner product. Suppose if we have a different def. of inner product satisfying those 3 properties of inner product then how we can generalize that {1,x,x^2} is a bad basis ? Moreover, we are interested in orthogonal basis only because make decomposition too much easy but in case of polynomial space we can easily decompose a quadratic poly. wrt to {1,x,x^2} basis too much easily. So what's the whole point of evaluating a basis wrt to they are orthogonal or not in this particular case ?

[@MathTheBeautiful]

You are exactly right. Whether a basis is good or bad, or convenient or not, is determined by the applications. For some applications, {1,x,x^2} is the best, for some, it's the worst, for others, it's in-between.

[@rishabhgarg9217]

Much thanks for your reply. Btw i am very much thankful to you. By watching these videos about LA, i haven't only learned Linear algebra from you but a lot of things. You have completely changed my view how and what i think about Maths. You have completely changed my perception about maths. Thanks really thanks for putting these lectures on web. I love to hear about the new upcoming topics you gonna cover in near future. P.S - I am already watching your new set of videos on Vector Calculus.

[@lemmavideos2163]

Thanks for letting me know. I'm glad you're enjoying them. Next are Calculus and Complex Variables.

[@thomasjefferson6225]

you know i didnt realize the beauty of this when my teacher geeked out on it in my applied linear algebra course. He made it seem like this was the coolest thing in the world, along with fourier series and expansions. Oh boy, was I too young mathematically to understand the beauty of this. its taking the already nice basis, and deflating it to an even nicer basis. Its beautiful. Now im curious about convergence properties.

[@benkim7300]

Thanks!

[@semashkosam2843]

Very helpful!

[@braytonsoldevillaquique1791]

The best

[@holyshit922]

def LegendreP(n): coeff = [] coeff.append(1-(n&1)) coeff.append((n&1)) scalingFactor = coeff[0] for k in range(n): coeff.append(((k-n)*(k+n+1))/((k+2)*(k+1))*coeff[k]) scalingFactor += coeff[k+1] coeff.pop() for k in range(len(coeff)): coeff[k] /= scalingFactor return coeff As you can see this function works in O(n) time (or rather in O(n) floating-point operations) but we have to repeat our calculations if we change our mind and want to calculate coefficiens for polynomial of different degree Unfortunately this quite fast way for calculation coefficients does not belong to the algebra

[@MathTheBeautiful]

Thank you for sharing!

[@holyshit922]

This will look better from fractions import Fraction def LegendreP(n): coeff = [] coeff.append(Fraction(1-(n&1))) coeff.append(Fraction(n&1)) scalingFactor = coeff[0] for k in range(n): coeff.append(Fraction(k-n,k+2)*Fraction(k+n+1,k+1)*coeff[k]) scalingFactor += coeff[k+1] coeff.pop() for k in range(len(coeff)): coeff[k] /= scalingFactor return coeff def PrintPoly(coeff,c): s = ''; for k in reversed(range(len(coeff))): if coeff[k] < 0: s += str(coeff[k])+'*'+c+'^'+str(k) elif coeff[k] > 0: s +='+'+ str(coeff[k])+'*'+c+'^'+str(k) if s.startswith('+'): s = s[1:] print(s) n = int(input('Enter degree of polynomial: ')) PrintPoly(LegendreP(k),'x') I wrote also modified Gram-Schmidt orthogonalizaton method but it took O(n^4) floating point operations Here we have O(n) floating-point operations when we need only one polynomial and O(n^2) if we want all polynomials up to degree n

[@quinnquynguyen8129]

He's saying L_2(x) = x^2-1/3, but P_2(x)=1/2*(3x^2-1) so is he missing a time 3/2 in his third answer somehow?

[@estebanlopez1701]

I bet this guy speaks french

[@MathTheBeautiful]

Nope. But I, too, am a faucet expert.

[@estebanlopez1701]

@@MathTheBeautiful jajaja. I said it because of the way you pronounce Legendre. Really nice lecture BTW.

[@jeanpi314159]

Eventhough he cannot speak french, it's easy for a frenchman (as myself) to understand him. His approach ( in its principle) looks like the one I discovered a couple of years ago : connecting things that seem far away, finding out unexpected correspondances . This is unity or universality of maths, but this is what makes maths a poetic and beautiful world.

[@joshsmit779]

I can tell you work out

[@ChristGodinyouItrust]

Are you kidding, you watch this video and your one comment is about how he looks? Wow!

[@snippletrap]

Gaaaay

[@l.p.3584]

I think he made a mistake. Integral for dot product (1, x) using his formula is not 0, but 2. Everything else is OK.

[@shavuklia7731]

He is right. You integrate x over [-1,1]; that yields zero.

[@infernobeatsmusic]

You have to square the minus sign.. the answer is zero.