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

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SUMMARY

The discussion centers on using Newton's Method to approximate the positive root of the equation sin(x) = x^2, specifically targeting the root at approximately 0.876726. Participants clarify that the non-zero root is preferred over the trivial root of 0 because it is the more relevant solution for the application of Newton's Method. The recommendation is to start with an initial approximation close to the desired root, with 1 being suggested as a suitable starting point for the iterative process.

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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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