Is this formula true Let f(t)=\sum\limits_{k=0}^{\infty}a_{k}P_{k}(x) then a_{k} = \frac{1}{p}\int\limits_{t_{1}}^{t_{2}}f(t)w(t)P_{k}(t)dt where w(t) is weight function [t_{1};t_{2}] is interval P(t) is orthogonal polynomial which is our new base for polynomial The value of p is determined by orthogonality of given polynomial
Clear, concise, and correct, a rarity! Seriously thank you so much for this.
Galt padha raha hai ye aadmi
I apologize, but it's not clear because he omits steps of derivation, at n =2 the steps of derivation should be written clear
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Thnx bro...U got me up with 6 marks
Sir tq for ur teaching I understand clearly
Very clear explanation! Thank you
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Very good explanation. Thank you so much Sir.❤
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Sir u r doing a great job keep it up🖒
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Keep watching
How is the derivative at 2:17 done like this?
aata hi nahi h isse toh btaega kya
I like very much!! If the function f(x) is delta of dirac, how can i expresse that in terms of legenfre polynoms? ThanKs from Brazil!!
THIS IS SO EASY!
THANK YOU SIRRR!!!
Thank You Sir
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thnx a lots monsieur
thank u sir, very understandable
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you are a good man, thank you.
Thank u sirr🥰
We put 4also if power is 4
Thank you Pajeet.
amazing video. thanks
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Sir how to find scalar potential any video is there please reply me
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Good helpful
Thank you
Is this formula true
Let f(t)=\sum\limits_{k=0}^{\infty}a_{k}P_{k}(x)
then a_{k} = \frac{1}{p}\int\limits_{t_{1}}^{t_{2}}f(t)w(t)P_{k}(t)dt
where w(t) is weight function
[t_{1};t_{2}] is interval
P(t) is orthogonal polynomial which is our new base for polynomial
The value of p is determined by orthogonality of given polynomial
ur the best
simple things, you have to tried to teach in simple manner...BTW nice..
Thank u sir❤
very very thank you. sir
Most welcome
How to solve p2(x)=
Thanks sir
TQ sir
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Nice sir
Sir in the final step it's 8/3 or 7/3 please tell me
-(7/3)
2/3 -3 = -7/3
Sir thora dhire samjhaeye please
Supr. tq sir.
nice
sir po(o)= one kaise Hoga niche n! = o hoke infinity NHI hoga
0! = 1 not 0
Kya h bhai
To aai ra bia re mg
you assume a lot