Bessel functions are solutions to the Bessel's equation which can be solved by Frobenius method. The solutions obtained by Frobenius method are called Bessel functions. So Bessel functions are particular case of Frobenius solutions.
a0(x1)+a2(x3)+a3(x4)+......=0 u considered all the coefficients to be zero why not taking into consideration that a0(x1)+a2(x3)+a3(x4)+......am(x^m+1)=am+1(x^m+2)+am+2(x^m+3)+am+3(x^m+4)+....an(x^n+1) like a+b=0 if a,b =0 or a= - b
The reason is that the equation a0(x1)+a1(x2)+a2(x3)+......=0 has to hold for all values of x in an open interval. In that case, there is a mathematical theorem called "identity theorem for power series" that proves the coefficients must be zero. A rough idea of why this should be true can be proven by differentiating the equation to make it a0+2a1(x1)+3a2(x2)+...=0 and then setting x=0. Repeating differentiation and setting x=0 shows that all coefficients must be zero. I hope this helps.
You can look at it this way: On the left side of the equation there are some powers of `x` with some coefficients you want to figure out. On the right side of the equation there is 0, which can be thought of as if there was a power series too, but with each power's coefficient being 0 (that is, the first power of `x` appears 0 times, the second power of `x` appears 0 times, etc.). In other words, this: a₀·x⁰ + a₁·x¹ + a₂·x² + a₃·x³ + … = 0 can be understood as this: a₀·x⁰ + a₁·x¹ + a₂·x² + a₃·x³ + … = 0·x⁰ + 0·x¹ + 0·x² + 0·x³ + … so you can now equate the coefficients of corresponding powers of `x` to find out that: a₀ = 0 (the coefficients of x⁰) a₁ = 0 (the coefficients of x¹) a₂ = 0 (the coefficients of x²) a₃ = 0 (the coefficients of x³) … etc. We cannot equate `x` to 0, because then all of its subsequent powers would have to be 0. But this is just the "trivial" solution, i.e., the easiest way to zero everything out. This is boring, because it is a solution for every equation ;J So we're looking for the non-trivial one, in which `x` can be something else than 0 too - an _arbitrary_ value, actually. And the only way for it to work for _any_ value of `x` whatsoever (not just `x=0`), is if the coefficients of all the powers are 0 (that is, none of the powers appears in the equation).
you are the best mathematician i have ever seen Daniel. Thank you so much.
I wish I had find out about your channel at the beginning of the semester 😭 Thank you Daniel
Raul Almaraz same here 4th sem
Same here 😂
We have this in 1st sem
Thank you so much. Your videos really helped me.
Thank you very much!!! I finally understood the method!!! You are awesome!
感謝您的影片 讓我對frobenius了解更深刻
language barrier please
thank you sirrr.....we really appreciate you
ABSOLUTELY BRILLIANT !
Sir you are awesome
Amazing work
Thank you thank you thank you!!!!!
what is the difference between singular and ordinary points...?
What is the difference between Frobenius and Bessel? I'm confused on how to find a difference between this final solution and Bessel's final solution
Bessel functions are solutions to the Bessel's equation which can be solved by Frobenius method. The solutions obtained by Frobenius method are called Bessel functions. So Bessel functions are particular case of Frobenius solutions.
Perfect. thank you
@@daniel_an you are the best
a0(x1)+a2(x3)+a3(x4)+......=0
u considered all the coefficients to be zero
why not taking into consideration that a0(x1)+a2(x3)+a3(x4)+......am(x^m+1)=am+1(x^m+2)+am+2(x^m+3)+am+3(x^m+4)+....an(x^n+1)
like
a+b=0
if a,b =0
or a= - b
The reason is that the equation a0(x1)+a1(x2)+a2(x3)+......=0 has to hold for all values of x in an open interval. In that case, there is a mathematical theorem called "identity theorem for power series" that proves the coefficients must be zero. A rough idea of why this should be true can be proven by differentiating the equation to make it a0+2a1(x1)+3a2(x2)+...=0 and then setting x=0. Repeating differentiation and setting x=0 shows that all coefficients must be zero. I hope this helps.
You can look at it this way:
On the left side of the equation there are some powers of `x` with some coefficients you want to figure out.
On the right side of the equation there is 0, which can be thought of as if there was a power series too, but with each power's coefficient being 0 (that is, the first power of `x` appears 0 times, the second power of `x` appears 0 times, etc.). In other words, this:
a₀·x⁰ + a₁·x¹ + a₂·x² + a₃·x³ + … = 0
can be understood as this:
a₀·x⁰ + a₁·x¹ + a₂·x² + a₃·x³ + … = 0·x⁰ + 0·x¹ + 0·x² + 0·x³ + …
so you can now equate the coefficients of corresponding powers of `x` to find out that:
a₀ = 0 (the coefficients of x⁰)
a₁ = 0 (the coefficients of x¹)
a₂ = 0 (the coefficients of x²)
a₃ = 0 (the coefficients of x³)
…
etc.
We cannot equate `x` to 0, because then all of its subsequent powers would have to be 0. But this is just the "trivial" solution, i.e., the easiest way to zero everything out. This is boring, because it is a solution for every equation ;J So we're looking for the non-trivial one, in which `x` can be something else than 0 too - an _arbitrary_ value, actually. And the only way for it to work for _any_ value of `x` whatsoever (not just `x=0`), is if the coefficients of all the powers are 0 (that is, none of the powers appears in the equation).
How come for this problem you didnt use indicial equation to find r?
zhi zhang It is assumed to have been done beforehand.
this is only wast of time and internet
just because you hate maths, its compulsory for other people
you must be a dumbass