MHB Differentiation of infinite series

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The discussion focuses on finding the derivative of the infinite series defined by the equation y = √{x + √[y + √(x + ...)]}. It establishes that this infinite series can be simplified to y = √{x + √(2y)}. By squaring both sides, the equation y^2 = x + √(2y) is derived, leading to a polynomial equation after further manipulation. The derivative is then obtained through differentiation, resulting in y' = (y^2 - x) / (2y^2 - 2xy - 1). There is also a suggestion that the topic may be more appropriate for a "Calculus" forum rather than one focused on "Linear and Abstract Algebra."
Ande Yashwanth
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Find derivative of y=✓{x+✓[y+✓(x+...)]}infinite.
Here root comes for total inter terms
 
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Ande Yashwanth said:
Find derivative of y=✓{x+✓[y+✓(x+...)]}infinite.
Here root comes for total inter terms
y= \sqrt{x+ \sqrt{y+ \sqrt{x+ \cdot\cdot\cdot}}} is clearly the same as y= \sqrt{x+ \sqrt{y+ y}}= \sqrt{x+ \sqrt{2y}}.

Squaring both sides, y^2= x+ \sqrt{2y} so that y^2- x= \sqrt{2y} and, squaring again, y^4- 2xy^2+ x^2= 2y or y^4- 2xy^2- 2y+ x^2= 0. Differentiating that with respect to x, 4y^3y'- 2y^2- 4xyy'- 2y'+ 2x= 0. Then 4y^3y'- 4xyy'- 2y'= (4y^2- 4xy- 2)y'= 2y^2- 2x so y'= \frac{y^2- x}{2y^2- 2xy- 1}.

(Shouldn't this be in "Calculus" rather than "Linear and Abstract Algebra"?)
 

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