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I am evaluating this formula, it's real value is 0. but maple can't further simplify it?

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- #1

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I am evaluating this formula, it's real value is 0. but maple can't further simplify it?

- #2

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- #3

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How can (√π - √π) not be zero? We are not solving a quadratic equation here, so ##\sqrt{\pi} > 0##.

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How can (√π - √π) not be zero?

If you select a branch at random each time you write a [itex]\sqrt{}[/itex].

It may be that Maple thinks that each [itex]\pi[/itex] is a separate variable, whose values are not necessarily equal.

- #5

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Probably because (√π - √π ) isn'tnecessarilyzero. It depends which of the two square roots you take.

To elaborate on @DrClaude's comment, the square root of a positive real number is by convention the principal square root of that number. i.e., the positive square root.How can (√π - √π) not be zero? We are not solving a quadratic equation here, so ##\sqrt π>0##.

Although the equation ##x^2 - 4 = 0## has two solutions -- x = 2 or x = -2, it is an error to say that ##\sqrt 4 = \pm 2##.

- #6

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Try enteringI am evaluating this formula, it's real value is 0. but maple can't further simplify it?

`Pi`

instead of `pi`

. For Maple, the first one is the famous mathematical constant, the second one is an undefined variable.

- #7

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Conventions are one thing, but what's coded up in the Maple code is another. What's your explanation for why it didn't replace (√π - √π ) in equation 4 of the OP with 0?To elaborate on @DrClaude's comment, the square root of a positive real number is by convention the principal square root of that number. i.e., the positive square root.

Although the equation ##x^2 - 4 = 0## has two solutions -- x = 2 or x = -2, it is an error to say that ##\sqrt 4 = \pm 2##.

- #8

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##\sqrt \pi## in Maple must be a positive number. I'd be very surprised if something as basic as that was coded wrongly.Conventions are one thing, but what's coded up in the Maple code is another.

- #9

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It is a triviality, nothing else. And it certainly does not have anything to do with branches of square roots.

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No idea, but I'm also puzzled why Maple added two expressions whose denominators were both ##\sqrt \pi x^2## to get a single expression with a denominator of ##\sqrt \pi x^2 \sqrt \pi##.Conventions are one thing, but what's coded up in the Maple code is another. What's your explanation for why it didn't replace (√π - √π ) in equation 4 of the OP with 0?

However, I think @S.G. Janssens has hit the nail on the head with his advice to use Pi rather than pi. The Maple documentation backs up this advice - https://www.maplesoft.com/support/help/maple/view.aspx?path=initialconstants.

- #11

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I agree that's worth a try. @ztdep , can you try replacing pi^(1/2) with Pi^(1/2) in your definition of v and see if that fixes it?However, I think @S.G. Janssens has hit the nail on the head with his advice to use Pi rather than pi. The Maple documentation backs up this advice - https://www.maplesoft.com/support/help/maple/view.aspx?path=initialconstants.

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