Math Is Fun Forum

hi Ganesh,

No cause for concern.  ktesla39 and I have been trying to find a way to protect the forum from spammers.  He has made a copy that he can experiment with without any threat to MIF. Note. He has already found a way to create an Avatar.

I've updated MIF about this.

Bob

Every day I have to remove members who only joined so they can advertise on our site.  It's a lot harder to do if the person uses a guest post rather than a member post so I have configued the guest group so they cannot do this. Sad but necessary or we'd become overwhelmed by rubbish, much of it downright offensive.

It would be wonderful if there was a mechanism that imposed some sort of post restriction on new members, but, as far as I can think, the only way for a new member to qualify would be to make posts, so I'm in a Catch 22 here.  Suggestions welcome.

Bob

Welcome!

I joined in 2010 and I'm still learning

Bob

hi Zach,

That's a reasonable question to ask but, interestingly, no one ever has, before you.

I have a maths degree and have taught maths and computing throughout my career.  I was a head of department for a while.

I only answer questions when I'm reasonably confident I can answer correctly.  But, as with anything you get from the internet, you have to weigh up its accuracy.  I am always willing to add more detail to my posts, so, if you want to consider a proof in more detail, I'll always be willing to do that.

Sometimes, when no one has come up with a response and my own knowledge is nil, I'll 'google it' and provide any link that looks helpful.  There have been occasions when I have no knowledge of what the poster is after but still think I can help in which case a long dialogue may ensue.  As an example, some years back a Canadian asked for help with drilling a hole in a piece of timber. It took several posts to establish that he had the timber mounted in a lathe and could adjust the angles in several ways.  He wanted to drill at a certain angle after he had made two rotations about different axes.  I taught him three dimensional vector geometry and, in particular, how to apply a rotation matrix to coordinates. With that he was able to do the calculations and make his hole on a test piece.  It worked!

There's one problem that was posted years ago which no one could fully answer.  One route to an answer seemes to be via contour integration although there's a big step with the algebra needed first.  I asked my son to help because he is a much better mathematician than me, and he did that missing step with the algebra (in his head!)  I've still got the problem in my subscription list and I hope to complete it one day soon.

Bob

hi Exist,

Welcome to the forum.

I need another proof first; then, it's easy!

Let a = b

then

Factorise:

Cancel the common factor (a-b)

Now to tackle your question:

Let a = b = 1 then

1 + 1 = 1

But 1 + 1 = 2           =>        1 = 2

Now add 1 to each side

1 + 1 = 2 + 1 = 3                   QED.   

He then went on to prove that black = white and got run over crossing the road on a black/white crossing!

Bob

I had my doubts about BBC Basic being able to handle nested function calls, but it worked better than I had hoped:

    5 count = 0
   10 INPUT x
   20 y = FNf(x)
   30 PRINT x,"          ",y
   40
   50
   60 DEF FNf(x)
   65 count = count + 1:PRINT "x="x,"count="count
   70 IF x<0 THEN = -x ELSE  = 0.5*FNf(x-FNf(x-1))

I used it to confirm my table in Excel:

The cells highlighted in yellow I'm fairly sure are correct.

The accuracy may drop off where the number of iterations is large.

Tried x = 3 without much hope. Gave up waiting at count = 100000.

Bob

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