Determining A Finite Value of Infinity

[willr12]

okay...if you accept that the sequence
1+2+3+4...=-1/12,
I think I have determined a finite value of infinity.
To find the value of the sums of all natural numbers up to a number, you can use the equation
((x^2)+x)/2.
An example would be 4.
4+3+2+1=10.
((4^2)+4)/2 also equals 10.
following this logic,
((x^2)+x)/2=-1/12
is true for the above sequence. this can be rearranged to
(x^2)+x+1/6
This is the resulting quadratic equation. Using the quadratic formula, one obtains the x intercepts as
(-3+-(sqrt3))/6.
and since x is infinity in this situation (since the highest value is infinity), the x intercepts are the values of infinity in the equation. Therefore, by assigning a value of -1/12 to riemann zeta(-1), you also assign finite values to infinity, approximately
-0.211324...
and
-0.788675...
If there is any faulty reasoning, please remember I'm 15 and I most likely have no clue what I'm talking about.

 

[phinds]

...
I think I have determined a finite value of infinity...

Since this is a contraction in terms, I stopped reading right there.

 

[jbriggs444]

okay...if you accept that the sequence
1+2+3+4...=-1/12,
I think I have determined a finite value of infinity.

http://en.wikipedia.org/wiki/Divergent_series

There is no sum of the integers from one to infinity. The series is divergent. The assignment of the value -1/12 as the "sum" of this series does not correspond to anything which you should regard as being an actual sum.

 

[willr12]

Since this is a contraction in terms, I stopped reading right there.

Most people would stop reading when someone says that 1+2+3...=-1/12, that's counterintuitive but generally accepted...so thanks for the ignorance

 

[willr12]

http://en.wikipedia.org/wiki/Divergent_series

There is no sum of the integers from one to infinity. The series is divergent. The assignment of the value -1/12 as the "sum" of this series does not correspond to anything which you should regard as being an actual sum.

Makes sense

 

[pwsnafu]

okay...if you accept that the sequence
1+2+3+4...=-1/12,

No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12

To find the value of the sums of all natural numbers up to a number, you can use the equation
((x^2)+x)/2.
An example would be 4.
4+3+2+1=10.
((4^2)+4)/2 also equals 10.
following this logic,
((x^2)+x)/2=-1/12
is true for the above sequence.

I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.

and since x is infinity in this situation (since the highest value is infinity), the x intercepts are the values of infinity in the equation.

This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.

If there is any faulty reasoning

Unfortunately, everything.

 

[willr12]

No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12
I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.
This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.
Unfortunately, everything.

Good to know. Thanks for the reply

 

[phinds]

Most people would stop reading when someone says that 1+2+3...=-1/12, that's counterintuitive but generally accepted ...

No, as has been pointed out, it is NOT generally accepted.

 

[willr12]

No, as has been pointed out, it is NOT generally accepted.

It is generally accepted as a representation of Riemann zeta (-1)

 

[pwsnafu]

It is generally accepted as a representation of Riemann zeta (-1)

No you have this backwards.
$\zeta(-1)$ is a possible regularization of $\sum_{n=1}^{\infty} n$.
$\sum_{i=1}^{\infty} n$ is not a representation of $\zeta(-1)$.
Regularization and representation are separate concepts.

 

[willr12]

No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12
I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.
This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.
Unfortunately, everything.

Y-value of the vertex in the resulting quadratic equation is -1/12. Any significance do you think?

No you have this backwards.
$\zeta(-1)$ is a possible regularization of $\sum_{n=1}^{\infty} n$.
$\sum_{i=1}^{\infty} n$ is not a representation of $\zeta(-1)$.
Regularization and representation are separate concepts.

good to know thanks

 

[Dragonfall]

okay...if you accept that the sequence
1+2+3+4...=-1/12

I don't.

 

[Mark44]

This thread has run its course, so I'm closing it.