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I would like to consider the following sum:

where each is a non-zero vector in and denotes the usual Euclidean norm on I would like to know under which conditions this sum converges (dependent on and the dimension ).If then this sum converges only if (You can see this just by replacing with ). But what about for other values of ? For instance, does the seriesconverge in any dimension ?In the case where the condition lets us replace at most one of in the sum, but from that point, it doesn't seem like there is anything obvious you can do (mainly due to the term appearing in the denominator, if one chooses to replace ). You can obtain other expressions involving terms like by using the triangle inequality, but I don't know what to do with the resulting sums.One idea I've had is to break up each and consider regions where for some and see if this yields anything in terms of zeta functions (for normed arguments).A remark: for anyone interested in lattice point problems, I have managed to show that

where is the -norm of the remainder in the Gauss circle problem (and its generalisations).*Last edited by zetafunc (2016-11-18 22:51:56)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi:

Do you wish this to be done by computer, if not, it belongs in the highly acclaimed Euler Avenue.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Could you put it into the form of a command?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I never did get the hang of Boole, George or the M command.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Why does this yield "Indeterminate"?

`Sum[1/Abs[a] Boole[a != 0], {a, -10, 10}]`

Similarly, why does the same thing happen here?

```
f[a_] := (Abs[a])^(-1)
Sum[f[a] If[a == 0, 0, 1] , {a, -10, 10}]
```

*Last edited by zetafunc (2016-11-19 21:09:16)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

For the first one:

1/Abs[a] Boole[a != 0]

When a = 0 you are multiplying (1/0) by 0, which is indeterminate.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

The first way that comes to mind would be:

`Sum[1/Abs[a], {a, Range[-10, -1]}] + Sum[1/Abs[a], {a, Range[1, 10]}]`

Or even:

`2 Sum[1/a, {a, 1,10}]`

Or you could define a piecewise function which are like the one we use in math.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Then, the sum I would like to investigate for now is:

```
f[a_] := Piecewise[{{(Abs[a])^(-1), a != 0}, {0, a = 0}}]
g[b_] := Piecewise[{{(Abs[b])^(-1), b != 0}, {0, b = 0}}]
h[c_] := Piecewise[{{(Abs[c])^(-1), c != 0}, {0, c = 0}}]
Sum[f[a]*g[b]*h[c] Boole[a + b + c == 0], {a, -10000, 10000}, {b, -10000,
10000}, {c, -10000, 10000}]
```

where the limits can be changed.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Yes, I would change the limits because 20000^3 = 8 000 000 000 000 which is going to take a long time.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Even that is around 8 million iterations, not a trivial amount.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I do not know yet. For one thing it is okay to define sin(x)/x at x = 0 as one in a piecewise function because sin(0)/0 is what they call a removable singularity. But is it okay to define 1 / 0 as 0 as you have done in your piecewise functions? See you later, I have to get dressed and head out.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Unfortunately I have to get dressed and head out, see you later.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Something strange is happening here. The sum

```
f[a_] := Piecewise[{{(Abs[a])^(-1), a != 0}, {0, a = 0}}]
g[b_] := Piecewise[{{(Abs[b])^(-1), b != 0}, {0, b = 0}}]
h[c_] := Piecewise[{{(Abs[c])^(-1), c != 0}, {0, c = 0}}]
NSum[f[a]*g[b]*h[c] Boole[a + b + c == 0], {a, -10, 10}, {b, -10,
10}, {c, -10, 10}]
```

yields 9.96439. However, changing the limits to -15, 15 gives a bunch of error messages, followed by a ridiculously large number which cannot possibly be correct (of order 10^10). It works fine up to -14,14 but at that point it started spitting out nonsense (-16,16 gives a negative number, which is impossible because the summand is positive).

*Last edited by zetafunc (2016-11-19 22:58:11)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Do you already know the exact answer?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You have some syntax errors.

(a = 0 assigns 0 to a, a == 0 tests whether a is 0. You want the test.)

Start a new session and then try changing your a = 0 , b = 0, c = 0 to a == 0, b == 0, c == 0.

Then change NSum to Sum when using small values. But you will still get an answer close to 10.

Based on your code I would say getting an answer of around 2 as you suggest is not correct.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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I did not suggest the answer would be around 2, I said that it would diverge for d = 2. I corrected the syntax. I get an answer close to 10 for smaller values though.

EDIT: Changed NSum to Sum and it worked well, even with -50,50 for all the ranges. Trying -100,100 now.

*Last edited by zetafunc (2016-11-20 08:11:16)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Please let me see your code.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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