if y^(1/m) + y^..... (Calculus)

Question

If y^(1/m) + y^(-1/m) = 2x, prove (x^2 -1)yn+2 + (2n+1)yn+1 + (n^2-m^2)yn =0

Comments

What is the significance of n in the question? Is it a term in a series or an exponent, for example? What is the relation between m and n, since it doesn't appear in the definition of 2x to the expression for y?

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Well m is a normal term or variable in the series whereas n is subscript to y.... If you still have a doubt I can send a pic of the question to you....

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A picture may help because it's not clear what the series is. In other words, what is y subscript n, and is x a substitute for terms in the series, or is it the sum of terms, or whatever.

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here it is

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I see a difference in the central term: y_n+1 has been replaced by xy_n+1, but I still haven't got a definition of the series for y. Nevertheless, from the information you have now provided I may be able to work out what the series and general term is.

Answer 1

From the definition of x, x^2-1=(1/4)(y^(2/m)+2+y^-(2/m)-1=(1/4)(y^(2/m)-2+y^-(2/m))=(1/4)(y^(1m)-y^-(1/m))^2.

We don't have a definition of any series for y, but if we put n=1 we have to prove specifically:

(1/4)(y^(1m)-y^-(1/m))^2y3+3(1/2)(y^(1/m)+y^-(1/m)y2+(1-m^2)y1=0, where y1, y2 and y3 are the first three terms of the unknown series.

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