Why is a Non-Zero Root Used in Newton's Method for Sin(x) = x^2?

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The discussion focuses on using Newton's method to find the positive root of the equation sin(x) = x^2, specifically seeking an approximation to six decimal places. The positive root is identified as approximately 0.876726, with emphasis on why a non-zero root is pursued instead of the trivial solution at x = 0. Participants highlight that the problem aims to find a root that is more relevant for Newton's method, which is effective when starting with an approximation close to the actual root. The suggestion is made to begin with an initial guess of 1 to facilitate the approximation process. This approach underscores the importance of selecting appropriate starting points in numerical methods for better convergence.
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Question:
Use Newtons method to approximate the indicated root of the equation correct to six decimal places.

The positive root of sin(x) = x^2

The answer is ...0.876726

Why did they pick this when the obvious root is 0?

Sin(0) = (0)^2 = 0
 
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There are two roots - they want you to find the other one.
The one that it actually helps to use Newton's method for.
 
Yes, they wouldn't ask you to use Newton's Method if they only wanted the trivial root. Try to start with a first approximation that is near the root. I'd try 1.
 

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