- #1
phrankle
- 6
- 0
For solving a series solution near a regular singular point with the Frobenius method, why is it that the indices of summation derivatives aren't shifted?
For example, in my textbook and lecture notes
y = [tex]\sum[/tex]A[tex]_{}n[/tex]x[tex]^{}n+r[/tex] from n=0 to infinity
y' = [tex]\sum[/tex](n+r)A[tex]_{}n[/tex]x[tex]^{}n+r-1[/tex] from n=0 to infinity
y'' = [tex]\sum[/tex](n+r)(n+r-1)A[tex]_{}n[/tex]x[tex]^{}n+r-2[/tex] from n=0 to infinity
But shouldn't the index for y' be from n=1 to infinity because it shifts up when you take the derivative of a summation? Shouldn't the index for y'' be from n=2 to infinity?
Thanks.
For example, in my textbook and lecture notes
y = [tex]\sum[/tex]A[tex]_{}n[/tex]x[tex]^{}n+r[/tex] from n=0 to infinity
y' = [tex]\sum[/tex](n+r)A[tex]_{}n[/tex]x[tex]^{}n+r-1[/tex] from n=0 to infinity
y'' = [tex]\sum[/tex](n+r)(n+r-1)A[tex]_{}n[/tex]x[tex]^{}n+r-2[/tex] from n=0 to infinity
But shouldn't the index for y' be from n=1 to infinity because it shifts up when you take the derivative of a summation? Shouldn't the index for y'' be from n=2 to infinity?
Thanks.