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tech / sci.math / Re: For any value of n.

SubjectAuthor
* For any value of n.Dan joyce
+* Re: For any value of n.Mathin3D
|`- Re: For any value of n.Mike Terry
`- Re: For any value of n.Dan joyce

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For any value of n.

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Subject: For any value of n.
From: danj4...@gmail.com (Dan joyce)
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 by: Dan joyce - Thu, 17 Aug 2023 23:54 UTC

n= ((((((sqrt(n + 0.5) )- 1)* 2 + 2)^2 )- 2)/4)

Re: For any value of n.

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Subject: Re: For any value of n.
From: mathi...@gmail.com (Mathin3D)
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 by: Mathin3D - Fri, 18 Aug 2023 00:20 UTC

On Thursday, August 17, 2023 at 7:55:00 PM UTC-4, Dan joyce wrote:
> n= ((((((sqrt(n + 0.5) )- 1)* 2 + 2)^2 )- 2)/4)

It is not an identity. It does have a pair of slns: 0.5, 8.5

Re: For any value of n.

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Subject: Re: For any value of n.
From: danj4...@gmail.com (Dan joyce)
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 by: Dan joyce - Fri, 18 Aug 2023 00:24 UTC

On Thursday, August 17, 2023 at 7:55:00 PM UTC-4, Dan joyce wrote:
> n= ((((((sqrt(n + 0.5) )- 1)* 2 + 2)^2 )- 2)/4)
Or for any value of n=>1
n-1= ((((((sqrt(n - 0.5) )- 1)* 2 + 2)^2 )- 2)/4)

Re: For any value of n.

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From: news.dea...@darjeeling.plus.com (Mike Terry)
Date: Fri, 18 Aug 2023 02:37:12 +0100
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 by: Mike Terry - Fri, 18 Aug 2023 01:37 UTC

On 18/08/2023 01:20, Mathin3D wrote:
> On Thursday, August 17, 2023 at 7:55:00 PM UTC-4, Dan joyce wrote:
>> n= ((((((sqrt(n + 0.5) )- 1)* 2 + 2)^2 )- 2)/4)
>
> It is not an identity. It does have a pair of slns: 0.5, 8.5
>

If n >= -0.5 , we have

((sqrt(n + 0.5) )- 1)* 2 + 2 = 2.sqrt(n + 0.5)

so
((((((sqrt(n + 0.5) )- 1)* 2 + 2)^2 )- 2)/4)
= ((2.sqrt(n + 0.5))^2 - 2) / 4
= (4.(n+0.5) - 2) / 4
= (4n + 2 - 2) / 4
= n.

It's like one of those "think of any number, now double it, now add 17, now blah blah tricks :)

Mike.

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