Math Is Fun Forum

I've been working on this since you posted it.

If x is negative, it's easy to work out f(x).

When positive each f(x) depends on f(x-1) so I reasoned that it should be possible to create a chain of f(y) where each y is smaller than the one before so that eventually y is negative and we can work that one out and hence all those in the chain.

On paper I kept making errors so I put the whole lot into a spreadsheet with a LOOKUP table with 2 columns, x and f(x)

That spreadsheet is growing and growing as the chains get longer and longer.  MS Excel is not very kind to me.  If I use the lookup for an x that isn't yet in the table, Excel gives me the nearest value anyway rather than saying it cannot do it.  As a result I'm having to check everything by hand anyway. At the moment my table has 26 entries and I think there's some errors.

I'm considering writing a program with a subprocedure that creates the chain. That way I should be able to get correct answers and may have an answer to your puzzle sometime in the next two weeks.

Great puzzle!  Simple?  No it isn't! 

Bob

That's what I got!

Bob

We have a slightly different way of tackling this question. It comes to the same thing, but my way shows how I got two answers.

When I have two triangles that are similar I write them like this:

AEB
ADC

What this means is that A is common to both; D is the enlargement of E and C is the enlargement of B.

The ratio of a side in the larger to the matching side in the smaller will be equal to the ratio of another pair. (equals the enlargement factor) 

ie.  AC/AB = AD/AE.  This leads to your calculation and hence solves for x.

But that assumes that CD is parallel to BE.  It looks right but the question doesn't specifically say so.

A is obviously common but maybe angle ACD = AEB which still leads to similar triangles but with the second answer.

My notation for this case is

ACD
AEB

Now I can pick out the alternative ratios

AC/ AE = AD/AB

and form a second equation for x.

Bob

That's what I did at first.

But the question doesn't say that ADC is an enlargement of AEB, just that the triangles are similar.

So it could be that ACD is an enlargement of AEB. That leads to a second scale factor and hence answer for x.

Important lesson for any maths question with a diagram: Don't make assumptions because of the way the diagram has been drawn.

Bob

Watch https://www.youtube.com/watch?v=GiWL7iKPoLM

Bob

Not roughly correct; but rather, completely correct. The stored gravitational energy at the top is called potential energy and the energy of movement is called kinetic energy.

Bob

Yes, it is very helpful!

But why won't the code command show the word math as white on black. It's bizarre!

Bob

[b]bold text[/b]

Something very odd is happening here.  I can use the square brackets code command to show a command without it being performed.  The above shows how to create bold test using a bbcode command.

But when I change the 'b' to 'math' the command is in black on a black background so it doesn't show.

That makes it very hard to explain what to do to use Latex.

Every line of Latex must start with a square bracket [                 then the word math            then a ] bracket.

At the end of your Latex you must 'switch off' the Latex interpreter by using a similar command but this time /math

Here's and example of what you can then do:

If you click on this expression the underlying code shows but without the math /math command that tells the interpreter what to display.

Hope that makes sense.

Bob

Here's what I do:

6 and 10 both have more than two factors so there are several alteratives to try.

6  10              6   5             3   10         3   5

1   1               1   2             2    1          2   2

By trial I found that 3 times 5 - 2 times 2 gives 11

(3x     )(2x     )

The 3x must multiply the 5 so

(3x   2)(2x   5)

I want + 11 so that fixes the signs

(3x -2)(2x + 5)

If you want to check first if an integer solution is lurking there you can check B^2 - 4AC to see if it has an integer root:

B^2 - 4AC = 121 + 240 = 361. This has root 19 so if I used the quadratic formula a simple 'solution' is available.  That check saves wasting time looking for a factorisation if there isn't one.

Bob

Hello Tony.

Welcome to the forum.

I'm pretty certain it's a German Shepherd.

Bob

Bob

You're welcome.  I enjoyed working it out so you did me a favour by setting the problem.

My 'complete proof' is meant to cover all cases, so if I'm right it does carry on doing this as infinitum.  The problem with proves is, sometimes, they have flaws in them.  Andrew Wiles surprised the mathematical world by coming up with a proof of Fermat's last theorem.  His proof was taken away by other mathematicians for close scrutiny. They did find some points where he had jumped ahead too fast and he went back to the drawing board to clean up the criticisms (which he did!).

The common multiple bit is fairly robust.  Then I had to show the lowest common multiple.  I think it's ok too but maybe others will spot a weakness.  When I was at university we spent a lot of time showing even the most obvious results are generally true so I've had practice.

You have to be careful assuming results just because they work for a few easy numbers. Have a look here:

http://www.mathisfunforum.com/viewtopic.php?id=30690

Another example is the statement N^2 + N + 41 is a prime number generator.  That is if you substitute N = 1, 2, 3 etc
each time you get a prime number.

If you try some N values it look promising. But the statement is FALSE. Can you find a value of N for which it fails?

Has someone already come up with your theorem? Not as far as I know. I tried an internet search and came  up with nothing.  So I suggest you claim it as yours and give it a fancy name.  The history of maths is full of disputed claims (eg. Leibnitz and Newton over calculus) so you'll be in with a select crowd.

But now, can you pull this out of the sphere of 'pure maths' and find a useful application?

Best wishes,

Bob