I'm guessing you mean 3/1000 of an inch and not 0.003 thousandths = 0.000003 as that would be too small to see.
It's a tough one. I checked out the size of guitar strings and it seems the thinnest string is typically 0.008 inches.
Feeler gauges are used by engineers to judge gaps in machines (eg. spark plug gap) so you might ask a car mechanic to show you one.
Human hair varies from person to person with an average around 0.004 inches. So you need to find someone with fine hair.
Hope that helps. Happy Christmas!
Oh! Step one turned out to be easy.
Theorem. For the sequence 1, 2, a3, 4, a5, a6, a7, 8, a3^2, 10, ........
a3 = 3.
Suppose a3 = 3 + h (delta takes too long to keep typing )
So it is sufficient to show that
So just let i get bigger until this is true. This causes a contradiction with the increasing sequence rule, so a3 ≤ 3.
I'll leave showing a3 ≥ 3 as an exercise for the reader
Useful link but I'm not paying $12 for the full proof.
Here's what I have done since my last post.
(1) Is the sequence 1, 2, 3, 4, ....... in the same format ?
Starting with the identity
So, yes, it is.
(2) I can simplify the formula thus:
As the log may be any base, why not take base 2, then log(2) = 1. The formula is then
where p = a2
Now, the first constraint => if a2 = p, then a4 = p^2, a8 = p^3 etc.
So we know all sequences must have the form
1, p, a3, p^2, a5, a6, a7, p^3, a3^2, a10, .....
My formula fits that for the power of 2 terms, and gives values for the others that form an increasing sequence.
But, what I failed to justify is why those values for a3, a5, a7, a9, a10, .... must be the only possibilities.
I think it is true but, finding a proof has so far eluded me. I was trying to develop a proof by contradiction. Here is a single case to show the way I'm thinking:
Take the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Could the number at a3 be anything else but 3 ? Could it be 3.1 for example ?
1, 2, 3.1, 4, ....
a9 = 3.1^2 = 9.61
a27 = 3.1^3 = 29.791 > 28, so 3.1 is NOT a possible value for a3.
I think, for any 3 + delta, you will always be able to find a term that is a power of (3 + delta) that causes a sub n to exceed a sub n+1.
Then I'd have to generalise this to show that a second term, a2 = p, and a third term a3 = 3^(p + delta) will always cause a contradiction with the increasing sequence constraint.
I'll have a good look at the link, because it looks like it is along the some lines, and see if that enables me to finish my proof. May be a while though as the Christmas party season approaches.
Can you prove that statement?
I thought post 9 did that. Which bit are you not happy with ?
How does the conclusion help, though?
Well I've given you a general formula for the sequence which I think covers all cases. (Looks like Stefy may disagree. )
The question asked how many such sequences. As p > 1 is the only restriction, I claim there are an infinite number of sequences. Hmmmmmm.
I'll just pick a random p and generate a sequence to see if it works.
Here's a sample of test values. I have calculated the first ten terms of three sequences, with p = 3, then p = 7.03 and finally p = pi.
Then, for each, I calculated a2 x a3, and a2 x a5. The results are shown to the right of the sequence table.
The three a2 x a3 results all gave a6 as expected. The a2 x a5 results all gave a10 as expected.
1, 3, 5.704522, 9, 12.81862, 17.11357, 21.84986, 27, 32.54158, 38.45586. ......
The Generator is
Check the property:
As the sequence is increasing you could have B = 1.
I started with a1 = 1, a2 = 3, a4 = 9, a8 = 27 and then looked for the function that would generate this. That's why the ln(2) jumps in. I think you could drop it from the definition and still get a sequence. And then you could switch to log base something else. And you can switch from '3' to any other positive number > 1
Sorry, I haven't quite made myself clear.
When you do a 2-D representation of a 3_D object, the lengths will only look correct if you are looking square on to them. So both AD and CD are seen foreshortened due to the angle of view.
They both look √2 times something, but are in fact √3 times something.
So you need to scale them up in that ratio. CD appears to be √2.r so you can take it that the true length is √3.r
Now can you finish it ?
Is this what you wanted?
This is the view looking down from above.
This will not show the true lengths of AD and CD as these lines are 'slanting' in this view.
A is the centre of the sphere/cube and D is a corner of the cube.
The length CD appears to be √2.r but the true length will be longer in the ratio AD:AB.
Then you can form an equation for r using AD = AB + BC + CD
Which link did you mean ? I'm seeing:
(1) download "fromDoctoPdf" file converting software
(2) download "Filedropper" software.
I don't see a link to the OP's file.
I'm very wary of clicking links for unknown downloads.
If it's a UK syllabus I can view it straight from their site, for example.