Hey !I need help about Frobenius Method Power Series
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Please show us what you got for y1 and how you got it.
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HallsofIvy
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Do you at least know what a "power series" is? It is an infinite sum of the form \(\displaystyle \sum_{n=0}^{/infty} a_nx^n\). Set y equal to that and calculate the derivatives:
\(\displaystyle y= \sum_{n=0}^{\infty} a_nx^n\)
\(\displaystyle y'= \sum_{n=1}^{\infty} n a_nx^{n-1}\)
\(\displaystyle y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}\).
So \(\displaystyle xy''= \sum_{n=2}^\infty n(n-1)a_nx^{n-1}\)
\(\displaystyle (1+ x)y'= \sum_{n= 1}^\infty n a_nx^{n-1}+ \sum_{n=1}^\infty n a_nx^n\) and
\(\displaystyle 2y= \sum_{n=0}^\infty 2a_n x^n\)
Putting those into the equation we have
\(\displaystyle \sum_{n= 0}^\infty n(n-1)a_nx^{n-1}+ \sum_{n=1}^\infty n a_nx^{n-1}+ \sum_{n=1}^\infty n a_nx^n+ \sum_{n=0}^\infty 2a_nx^n= 0\).
Now we would like to combine all those sums into one sum so that we can set the summand equal to 0. In order to do that, we need to get the same power of x in each and we need to do that by changing the index, n, which will be different in each sum. Those are "dummy" indices- there will be no "n" in the actual sum, so we can change each one as we please.
So in the first sum, \(\displaystyle \sum_{n= 0}^\infty n(n-1)a_nx^{n-1}\), let j= n-1. Then n= j+ 1. When n= 0, j= 1 so that becomes \(\displaystyle \sum_{j=1}^\infty (j+1)Ja_{j+1}x^j\).
In the second sum, \(\displaystyle \sum_{n=1}^\infty n a_nx^{n-1}\), let j= n-1. Then n= j+1. When n= 1, j= 2 so that becomes \(\displaystyle \sum_{j= 2}^\infty (j+1)a_{j+1}x^j\).
in the third sum, \(\displaystyle \sum_{n=1}^\infty n a_nx^n\), let j= n. Then n= j. When n= 1, j= 1 so the sum becomes \(\displaystyle \sum_{j=1}^\infty j a_jx^j\).
in the fourth sum, \(\displaystyle \sum_{n=0}^\infty 2a_nx^n\), let j= n. Then n= j. When n= 0, j= 0 so the sum becomes \(\displaystyle \sum_{j= 0}^\infty 2a_jx^j\).
Now we can use the fact that if a power series is 0 for all x then each coefficient must be 0.
The second sum now starts at j= 2 so we have to handle j= 0 and j= 1 separately. Only the fourth sum has j= 0 so the \(\displaystyle x^0\) term is just \(\displaystyle 2a_0= 0\) so \(\displaystyle a_0= 0\).
The first and third sums start at j= 1 so the first, third, and fourth terms all have a \(\displaystyle x^1\) term. The coefficient is \(\displaystyle 2a_2= a_1+ 2a_1= 2a_2+ 3a_1= 0\) so \(\displaystyle a_2= -(3/2)a_1\).
(We have to get a value for \(\displaystyle a_1\) from the initial value conditions.)
All sums have j= 2 and higher so we can combine all of them. The whole thing, in terms of j, is
\(\displaystyle \sum_{j=1}^\infty (j+1)Ja_{j+1}x^j+ \sum_{j= 2}^\infty (j+1)a_{j+1}x^j+ \sum_{j=1}^\infty j a_jx^j+ \sum_{j= 0}^\infty 2a_jx^j\)
The coefficient of \(\displaystyle x^j\) for \(\displaystyle j\ge 2\) is \(\displaystyle j(j+1)a_{j+1}+ (j+1)a_{j+1}+ a_j+ 2a_j= 0\) so \(\displaystyle (j(j+1)+(j+1))a_{j+1}= (j+1)^2a_{j+1}= -3a_j\) so we have the recursion \(\displaystyle a_{j+1}= -\frac{3}{(j+1)^2}a_j\) for \(\displaystyle j\ge 2\).
(We also have to get a value for \(\displaystyle a_2\) from the initial value conditions.)[/tex][/tex][/tex][/tex]
\(\displaystyle y= \sum_{n=0}^{\infty} a_nx^n\)
\(\displaystyle y'= \sum_{n=1}^{\infty} n a_nx^{n-1}\)
\(\displaystyle y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}\).
So \(\displaystyle xy''= \sum_{n=2}^\infty n(n-1)a_nx^{n-1}\)
\(\displaystyle (1+ x)y'= \sum_{n= 1}^\infty n a_nx^{n-1}+ \sum_{n=1}^\infty n a_nx^n\) and
\(\displaystyle 2y= \sum_{n=0}^\infty 2a_n x^n\)
Putting those into the equation we have
\(\displaystyle \sum_{n= 0}^\infty n(n-1)a_nx^{n-1}+ \sum_{n=1}^\infty n a_nx^{n-1}+ \sum_{n=1}^\infty n a_nx^n+ \sum_{n=0}^\infty 2a_nx^n= 0\).
Now we would like to combine all those sums into one sum so that we can set the summand equal to 0. In order to do that, we need to get the same power of x in each and we need to do that by changing the index, n, which will be different in each sum. Those are "dummy" indices- there will be no "n" in the actual sum, so we can change each one as we please.
So in the first sum, \(\displaystyle \sum_{n= 0}^\infty n(n-1)a_nx^{n-1}\), let j= n-1. Then n= j+ 1. When n= 0, j= 1 so that becomes \(\displaystyle \sum_{j=1}^\infty (j+1)Ja_{j+1}x^j\).
In the second sum, \(\displaystyle \sum_{n=1}^\infty n a_nx^{n-1}\), let j= n-1. Then n= j+1. When n= 1, j= 2 so that becomes \(\displaystyle \sum_{j= 2}^\infty (j+1)a_{j+1}x^j\).
in the third sum, \(\displaystyle \sum_{n=1}^\infty n a_nx^n\), let j= n. Then n= j. When n= 1, j= 1 so the sum becomes \(\displaystyle \sum_{j=1}^\infty j a_jx^j\).
in the fourth sum, \(\displaystyle \sum_{n=0}^\infty 2a_nx^n\), let j= n. Then n= j. When n= 0, j= 0 so the sum becomes \(\displaystyle \sum_{j= 0}^\infty 2a_jx^j\).
Now we can use the fact that if a power series is 0 for all x then each coefficient must be 0.
The second sum now starts at j= 2 so we have to handle j= 0 and j= 1 separately. Only the fourth sum has j= 0 so the \(\displaystyle x^0\) term is just \(\displaystyle 2a_0= 0\) so \(\displaystyle a_0= 0\).
The first and third sums start at j= 1 so the first, third, and fourth terms all have a \(\displaystyle x^1\) term. The coefficient is \(\displaystyle 2a_2= a_1+ 2a_1= 2a_2+ 3a_1= 0\) so \(\displaystyle a_2= -(3/2)a_1\).
(We have to get a value for \(\displaystyle a_1\) from the initial value conditions.)
All sums have j= 2 and higher so we can combine all of them. The whole thing, in terms of j, is
\(\displaystyle \sum_{j=1}^\infty (j+1)Ja_{j+1}x^j+ \sum_{j= 2}^\infty (j+1)a_{j+1}x^j+ \sum_{j=1}^\infty j a_jx^j+ \sum_{j= 0}^\infty 2a_jx^j\)
The coefficient of \(\displaystyle x^j\) for \(\displaystyle j\ge 2\) is \(\displaystyle j(j+1)a_{j+1}+ (j+1)a_{j+1}+ a_j+ 2a_j= 0\) so \(\displaystyle (j(j+1)+(j+1))a_{j+1}= (j+1)^2a_{j+1}= -3a_j\) so we have the recursion \(\displaystyle a_{j+1}= -\frac{3}{(j+1)^2}a_j\) for \(\displaystyle j\ge 2\).
(We also have to get a value for \(\displaystyle a_2\) from the initial value conditions.)[/tex][/tex][/tex][/tex]
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Hey Sir!
Thanks for answer but recursion should be (j+2)/(j+1)^2
HallsofIvy
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I don't know what you mean by "y1(x)" or "y2(x)". You mention them in your first post but even though Subhotosh Kahn asked how you got "y1(x)" you never define them.