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hi zetafunc,

Not sure I have understood your solution. Are you assuming the question was this:

What, in connection with this question, is the next number below?

4, 13, 19, 21, 29, ?

Bob

hi Salingna

Welcome to the forum.

When you are asked to find the product of numbers, this means **multiply** them together.

Looking at the choices, I'm guessing that the questioner wanted you to round off your answer.

Bob

hi faria66

I said suppose he uses x of the first and y of the second, then the cost is 5x + 6y

He sells (x+y) at 7 so his sales come to 7x + 7y

So (5x+6y) times 120/100 = 7x + 7y

From this work out x/y

Took about 1 minute, longer to write down.

Bob

hi HoaGio

How about 64 ?

Thus

Set n = 6 and f(n) = 64

Welcome to the forum. Happy Christmas!

Bob

hi SuperLynx

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.

http://uk.farnell.com/ck-tools/t3525-41 … dp/1518960

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!

Bob

hi MathsisFun,

I think it's a great new venture. Some folk will prefer this to the static page and it may pull in new members.

Happy Christmas!

Bob

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

Bob

hi,

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.

Bob

hi BonaviaFx

I find it helps to switch log base like this

That should help with the first part and a similar trick will help with the others.

Bob

Hi PatternMan,

The key phrase in what you ask is "you could understand what you are doing so well". If you have truly understood, then you'll know you're right. If you're wrong, then you haven't properly understood.

Bob

Stefy wrote:

Can you prove that statement?

I thought post 9 did that. Which bit are you not happy with ?

Agnishom wrote:

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.

Bob

hi Andrey GAP

Welcome to the forum.

Sorry, but I don't understand what you mean by " b in degree a "

What branch of mathematics is this from / Whose theorem ? / Can you give an example ?

Bob

Using

Putting m = 1

for all sequences following the rule.

Now let

then

so the general function must be

Let

All sequences must have this form.

If 0<p<1 then a value for B is not possible.

If p < 0 then B is again not possible.

So p > 1.

Bob

hi BonaviaFx

Use the factor theorem:

If you eliminate c from these you should end up with the required equation.

Bob

I've simplified and generalised the above to:

Take logs

'De-log'

Bob

Here's another:

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

Bob

That's a mighty fine geogebra demonstration Agnishom. I am well impressed!

Stefy wrote:

An n-sphere is the set of all points in n+1 dimensional space

So a 1-sphere is in 2-D space.

Bob

Substitute 1-sphere in my question about 1-circle.

Bob

hi Agnishom,

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 ?

Bob

1-circle

What is this, please ? I googled it and got, as the first hit, a chance to buy one!

Bob

hi Agnishom,

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

Bob

Thanks bobbym, got it ok.

Bob

Oh ha!

Since when has a circle been 1-D ? Try drawing one on a line.

Bob

hi bobbym,

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.

Bob

Sorry to be picky, but a soft rubber ball is 3 dimensional and his shapes are 2-D. Have you got a 2-D soft rubber ball by any chance ?

Bob