2(5) − cot [4 arctan 0.2 + (i/2) ln i] − 1

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In summary: It seemed to me that the standard solution is to first analyze the limiting behavior of a specific case, then generalize the results. In this case, the limiting behavior of a series of velocities is a velocity that is ZERO. I figured that the "1/12" thing was the result of some sort of programming error, but that the general idea of the problem was interesting in its own right.I was wrong. It seems that in this case, if the velocities are added in a certain order, the sum can assume a finite value. I'm not sure what is happening here. I'd really like to understand why this is, and what is going on in this case.In summary, the conversation discusses a formula for summing an infinite
  • #1
Jenab2
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Don't ever divide anything by the quantity in the title.

Post your favorite "fancy zeros" here.
 
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  • #2
$$\frac{1}{12} + \sum _{n = 1} ^\infty n$$
 
  • #3
axmls said:
$$\frac{1}{12} + \sum _{n = 1} ^\infty n$$

That should probably be

−1/12 + Σ(2,∞) 1/n⁴

Edit: whoops, no. That doesn't seem quite right, either. I evaluated ten million terms of the sum and came up with −0.0010100996222299347, so

−1/12 + 1/999 + Σ(2,∞) 1/n⁴

seems to be nearer to zero.
 
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  • #4
Jenab2 said:
That should probably be

−1/12 + Σ(2,∞) 1/n⁴

Edit: whoops, no. That doesn't seem quite right, either. I evaluated ten million terms of the sum and came up with −0.0010100996222299347, so

−1/12 + 1/999 + Σ(2,∞) 1/n⁴

seems to be nearer to zero.

Nope, it's written as I intended.
 
  • #5
Unless you specify how divergent sums are to be evaluated, the formula is not well-defined.
Yes there is a specific way that leads to -1/12, but this is by far not the only way to assign finite values to divergent sums.
 
  • #6
mfb said:
Unless you specify how divergent sums are to be evaluated, the formula is not well-defined.
Yes there is a specific way that leads to -1/12, but this is by far not the only way to assign finite values to divergent sums.

It was a somewhat tongue-in-cheek answer, if that wasn't clear.
 
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  • #7
I tried something like this to mess with my maths teacher in senior year once, replaced pi with some weird sums.
 
  • #8
1 / Σ(1,∞) n = 0
1 / { 1/a + 1 / Σ(1,∞) n } = a, a≠0.
 
  • #9
Jenab2 said:
1 / Σ(1,∞) n = 0
1 / { 1/a + 1 / Σ(1,∞) n } = a, a≠0.

See here: https://en.m.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯

There are ways to assign a value to that sum using alternate methods. That was the "tongue-in-cheek" aspect of my response.
 
  • #10
## \begin{Vmatrix} \vec\nabla \times \vec\nabla f \end{Vmatrix} ##
 
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  • #11
axmls said:
See here: https://en.m.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯
There are ways to assign a value to that sum using alternate methods. That was the "tongue-in-cheek" aspect of my response.
Ah. My difficulty in appreciating the assignment was caused by my thinking of scalars in vector terms. Consider velocities in the same direction, classically being added, tail to head:

v₁ + v₂ + v₃ + v₄ + ...

where each velocity is in the direction of the +x axis and the magnitude of the velocities is proportional to the subscript.

How is it that an object, moving through an infinite succession of changes-of-velocity, all of them forward, might end up moving BACKWARD at a speed of 1/12 velocity units?

I'd figured that this was a case of getting a strange result out of an indeterminate form.
 

Related to 2(5) − cot [4 arctan 0.2 + (i/2) ln i] − 1

1. What is the value of the expression 2(5) − cot [4 arctan 0.2 + (i/2) ln i] − 1?

The value of the given expression is undefined, since the natural logarithm of imaginary numbers (such as i) is undefined. Therefore, the entire expression is undefined.

2. What are the possible values of cot [4 arctan 0.2 + (i/2) ln i]?

The possible values of cot [4 arctan 0.2 + (i/2) ln i] range from positive to negative infinity, since the values of arctan 0.2 and ln i can vary between 0 and infinity, resulting in a range of possible values for cot.

3. Can the expression 2(5) − cot [4 arctan 0.2 + (i/2) ln i] − 1 be simplified?

No, the expression cannot be simplified as the values of arctan 0.2 and ln i cannot be simplified further. However, if the imaginary component (i) is removed, the expression can be simplified to 9 - cot[4 arctan 0.2] - 1.

4. How does the value of the expression change if the value of arctan 0.2 is increased?

The value of the expression will also increase, as the value of cot [4 arctan 0.2 + (i/2) ln i] will increase, resulting in a larger value overall.

5. Can this expression be evaluated using a calculator?

Yes, this expression can be evaluated using a calculator, but it is important to use a scientific calculator that can handle complex numbers and has a cot function. Standard calculators may not be able to evaluate this expression accurately.

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