Mathematics: This Will Blow Your Mind

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Rudin
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17 Jul 2015, 10:49 am

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

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Kiriae
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17 Jul 2015, 11:58 am

Infinity, in both cases.
You can add, and add, and add and still add even more - it never ends.

Rudin
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17 Jul 2015, 12:33 pm

Kiriae wrote:

Infinity, in both cases.
You can add, and add, and add and still add even more - it never ends.

Seems correct however it's miraculously not.

1+2+3+4+...=-1/12 and even more miraculously 1^2+2^2+3^2+...=0. It has to do with a formula for Riemann zeta function when the variable is negative.

when s<0. When s is negative and even the sum is 0.

BTW: 1+2+3+...=ζ(-1) , 1^2+2^2+3^2+...=ζ(-2).

Adamantium
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17 Jul 2015, 12:46 pm

Some numberphile vids on this:

Adamantium
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17 Jul 2015, 2:43 pm

Nice write up with satisfying points of clarification at SciAm:

http://blogs.scientificamerican.com/roo ... equal-112/

Quote:

The Numberphile video bothered me because they had the opportunity to talk about what it means to assign a value to an infinite series and explain different ways of doing this. If you already know a little bit about the subject, you can watch the video and a longer related video about the topic and catch tidbits of what's really going on. But the video's "wow" factor comes from the fact that it makes no sense for a bunch of positive numbers to sum up to a negative number if the audience assumes that "sum" means what they think it means.

JJabb
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26 Jul 2015, 6:56 pm

My mind is blown.. also, I LOVE Numberphile. I can watch those videos for hours on end.

slave
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27 Jul 2015, 8:32 pm

JJabb wrote:

My mind is blown.. also, I LOVE Numberphile. I can watch those videos for hours on end.

me 2
I've watched all of them. :nerdy:

Maths is -1/12 cool!

aleclair
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29 Jul 2015, 6:20 pm

The first time I saw this, I thought it was going to be one of those proofs where a false result is proven with the help of division by zero. Looks like I was wrong. Even after completing a math degree it's amazing how much mathematics there still is to learn.

Adamantium
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29 Jul 2015, 6:43 pm

I hope people followed the link to the Scientific American article. It's well worth reading, not least for this:

http://blogs.scientificamerican.com/roo ... equal-112/

ruveyn
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01 Aug 2015, 9:22 pm

Rudin wrote:

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

Rudin
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02 Aug 2015, 7:37 am

ruveyn wrote:

Rudin wrote:

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

Deltaville
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06 Jan 2016, 5:49 am

Rudin wrote:

ruveyn wrote:

Rudin wrote:

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

No these series cannot converge as firstly, they are infinite, and secondly, the Riemann zeta function requires a complex value operator in order to compute. If you use the ratio-limit test, the series is innately divergent.

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Rudin
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07 Jan 2016, 6:03 pm

Deltaville wrote:

Rudin wrote:

ruveyn wrote:

Rudin wrote:

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

They do converge though. One can prove it mathematically.

Earthling
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07 Jan 2016, 6:08 pm

I knew the answer to the first one from Numberphile, but the second one blew my mind. Thanks for sharing Rudin.

Deltaville
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08 Jan 2016, 1:41 am

Rudin wrote:

Deltaville wrote:

Rudin wrote:

ruveyn wrote:

Rudin wrote:

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

They do converge though. One can prove it mathematically.

Your original question is incorrectly expressed. As the convergence/divergence determination is relative to gamma, and you did not mention a complex value operator in your original question, that still postulates a divergent solution in every sense in that word. One thing I learned when I worked for my Honors degree in Math that I have learned is that assume the fewest restrictions as a questions allows and interpret a proposition or explanation in the simplest ideals possible, that is called Occam's Razor.

As long we are not given gamma, the solution will and always will be a divergent series as y>0, XeR, unless otherwise specified of a complex value operator.

I hope this will seal the issue of this thread.

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naturalplastic
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08 Jan 2016, 11:10 am

Rudin wrote:

Deltaville wrote:

Rudin wrote:

ruveyn wrote:

Rudin wrote:

Don't use the internet.

Estimate the value of these sums:

a) 1+2+3+4+...
b)1^2+2^2+3^2+...

The solution involves the Riemann zeta function but tell me what you think.

Neither series converges. If a series converges then its n-th term approaches 0 as n goes to infinity. Clearly this does not happen with either series, hence they do not converge.

That is true, however they do converge due to the Riemann zeta functional equation.

Any convergence or divergence test will make it clear these two series diverge, however they don't.

They do converge though. One can prove it mathematically.

At what point do they converge?

At what point do they stop diverging?

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