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limit of x*floor(1/x) as x goes to 0 #509

Open rtoy opened 1 month ago

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:20:50 Created by ynrobyvr on 2009-10-29 01:21:05 Original: https://sourceforge.net/p/maxima/bugs/1804

wrong result in calculating limit of x*floor(1/x) as x goes to 0

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:20:51 Created by ynrobyvr on 2009-10-29 01:21:08 Original: https://sourceforge.net/p/maxima/bugs/1804/#70ae

wrong result in calculating limit of x*floor(1/x) as x goes to 0

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rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:20:54 Created by rtoy on 2009-10-29 02:12:44 Original: https://sourceforge.net/p/maxima/bugs/1804/#1eee

I get the noun form back. Not wrong, but could be better. What were you expecting?

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:20:58 Created by crategus on 2009-11-09 16:59:42 Original: https://sourceforge.net/p/maxima/bugs/1804/#61d9

With Maxima 5.19post I get a noun form too. There have been serveal changes the last time to improve limit. Furthermore, I think the limit of the example is not defined. Therefore, it seems to be not wrong to return a noun form.

Setting the status to pending and works for me. Dieter Kaiser

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:21:01 Created by crategus on 2009-11-09 16:59:43 Original: https://sourceforge.net/p/maxima/bugs/1804/#1734

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:21:05 Created by rtoy on 2009-11-09 17:17:50 Original: https://sourceforge.net/p/maxima/bugs/1804/#e184

Isn't the limit 1? Let any x small enough, 1/x = n + e, where n is an integer and e < 1. Then floor(1/x) = n and x*floor(1/x) is n/(n+e) = 1 - e/(n+e). As n gets larger (and x gets smaller), this approaches 1.

Did I make a mistake?

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:21:08 Created by crategus on 2009-11-09 21:11:11 Original: https://sourceforge.net/p/maxima/bugs/1804/#6ebd

Sorry, I have no mathematical proof. I have come to the conclusion because of the follwing:

1. The function floor(x) is discontinuous. 2. The function x*floor(1/x) has an infinite number of points of discontinuity in any infinitesimal intervall when aproching zero. 3. Therefore, the function does not approach a limit.

I could be wrong.

Dieter Kaiser

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:21:12 Created by ynrobyvr on 2009-11-11 02:11:14 Original: https://sourceforge.net/p/maxima/bugs/1804/#0ed2

Answer to rtoy. The limit is 1, your proof is essentially orrect. I prefer the following proof: 1/x-1<floor(1/x)<=1/x, hence for x>0 we have 1-x<xfloor(1/x)<=1. It follows that the limit from the right is 1. The proof for the limit from the left is similar. Answer to crategus: The function does have a limit, namely 0. Your staement 2. is correct but your conclusion 3. is erroneous.

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:21:15 Created by ynrobyvr on 2009-11-11 02:11:26 Original: https://sourceforge.net/p/maxima/bugs/1804/#9770

rtoy commented 1 month ago

Imported from SourceForge on 2024-07-03 00:21:19 Created by crategus on 2009-11-12 09:55:27 Original: https://sourceforge.net/p/maxima/bugs/1804/#78a3