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#1 Re: Help Me ! » Prove Fibonacci Sequence » 2010-01-06 08:32:55

How far have you got, fibbingfibonacci? I have a solution to your problem, you need to use induction.

#2 Re: Help Me ! » grid pattern » 2007-03-23 01:33:25

You see, however you pick your 2 x 2 box, all the numbers will have a relation to eachother according to your grid. Let the first number be called A, then the next number will be A+1, A+2, A+3,... filling up your entire grid.

Now, however you pick your square doesn't matter. Let the number in the first  row and first column (of your square) be called x, then the next one is always x+1. The bottom ones in your square will always be x+10 and x+11, such as

x        x+1
x+10  x+11

Do you see why?

Do you know how to proceed from here? All you have to do know is to set up the expression you stated, multiplying diagonally and finding the difference.

(x+1)*(x+10) - x*(x+11) = ??

Edit: I see Jane already posted, but maybe an explanation will help =P

#4 Re: Help Me ! » help with quadratic sequences please. » 2007-02-28 02:41:08

10+1, 40+4, 100+10, 200+20

Skip the second terms for a while and let us simplify the sequence: 1, 4, 10, 20

It's much easier now! We get: n/6(n+1)(n+2)
But that was the simplified model, let's continue:

--> 5n/3(n+1)(n+2) corresponds to the sequence 10,40,100,200. Now we need to add 1/10 of these:

5n/3(n+1)(n+2)+5n/30(n+1)(n+2) =
= 10n/6(n+1)(n+2)+1n/6(n+1)(n+2) =

= 11n/6(n+1)(n+2)

Inserting n=1,2,3,4 you'll see it works out fine for your sequence smile

#5 Re: Help Me ! » PASCAL RIDDLE - Given to me by my Future Girlfriend ! » 2007-02-17 10:25:34

If 65 67 83 is ascii, it would be ACS. If we relate that to time, the closest I get is the Australian Central Standard time. That would be UTC/GMT + 9:30h. Just throwing something out there... does Australia say anything at all?

Also, I was thinking that 65 67 83 could be rows or something in Pascal's triangle. 65 would be row 6, number 5 in the sequence. But 67 doesn't quite work out.

The one thing that comes to mind is that they represent degrees, minutes and seconds like some sort of location. As with 9 000, I'm clueless.

#6 Re: Help Me ! » help with number sequence » 2007-02-07 09:17:32

x is the position of the number in the sequence. Check it for f(1), f(2), f(4) and f(6) in the sequence, it's correct tongue It doesn't give integers for f(3) and f(5) though. Stanley_Marsh made me want to try it out heh.

#7 Re: Help Me ! » Find the next number... » 2006-12-08 19:34:34

check this out: … &go=Search

The only matching (integer) sequence would be the first of the three, none of the other are followed by either 11,12,13 or 14. But then, the rest are still wrong... If it really is a sequence (can't see it myself), nice find.

#9 Re: This is Cool » Nullity? » 2006-12-07 11:17:06

That symbol, uppercase phi, is the reciprocal of the golden ratio. I think he missed something there big_smile

#10 Help Me ! » radii of a lens » 2006-12-06 12:10:45

Replies: 0

I have an optical system of an equiconvex lens and an object located 0.6m from a screen. The object is 0.05m high and it's image is 0.25m on the screen.

I managed to calculate the focal length to f=0.083m by using the fact that the magnification is -5 and the gaussian lens formula, which appears to be reasonable.

The problem is, how do I find the radii of the lens? hmm

#11 Re: This is Cool » 0.9999....(recurring) = 1? » 2006-12-06 06:14:42

Anthony, take a look at this:

-  0.99999999...
= 0.00000000...

Since the number of zero's is infinite, it is the same as writing 0 as there can't be anything after the infinitely many zero's (just as 1.000... is the same as 1, or 5.5999... = 5.6). Thus, there is no difference between 1 and 0.999... they are the same thing!

As pointed out already, there is no such thing as 0.0000...1. The only thing it proofs is that you don't understand the concept of infinity.

If you still don't believe it, prove us wrong. Otherwise none will take you seriously.

#12 Re: Help Me ! » completed equation now what » 2006-11-27 13:41:37

I got confused reading John's posts (sorry ;P) so I did it by using vectors to check his result and got the exact answer

Which is approximately 1.016, but 1 might be good enough for some people wink This is what I did, maybe someone could check it (I usually don't do this in 2D, and it's really late sleep).



We may easily find two points on the line 3x-4y=2 (meaning: find two solutions to the equation). Just pick a value for x and y so it "fits". Two solutions are Q = (2,1) and P = (6,4), check these. These points are on the line, which we will call L. The direction of this vector can be found through QP = (6-2,4-1) = (4,3). Thus, we can write L in the parametric form


We have the point S = (5,2), the distance to L is given by |SR|, where R is a point on L. SR is given by SR=(2+4t,1+3t)-(5,2). Do you see why? This gives SR=(-3+4t,-1+3t). Also note that SR is perpendicular to QP.
This means that 4(2+4t)+3(1+3t) has to be 0. This gives us t=7/11. Let us insert this value for t into SR. We get SR=1/11(-5,10)
The distance is the absolute value for SR. Let us find out what it is!

|SR|=1/11*√(5²+10²) = 1/11*√125

You can apply this technique in 3D systems as well (with x,y,z). It's pretty good to know smile

#14 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-30 10:18:58

I haven't managed to figure out another solution to this problem, all other options I had in mind doesn't seem to work out with 100% accuracy. You think this is the only solution?

#15 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-24 03:09:22

Dross wrote:
numen wrote:

I'll hold it a bit more though. I'll post my solution on monday if I can remember that wink

Would you mind leaving it until, say, Wednesday? Some members may not be able to get on the forum during weekends/very often, for whatever reason.

In fact, could you e-mail me your solution? (I don't know what the deal is with PMs on this forum... I don't seem to be able to, anyways!)

Sure thing. I'll e-mail my solution soon, hoping that, in return, you'll e-mail me yours tongue Hopefully you can confirm it's correct.

#16 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-23 07:19:38

I really have no idea how else to solve this. Your hint is pretty much how I define my groups, if I understand it correctly. If your solution really is different, I'd like to know it as well.

I'll hold it a bit more though. I'll post my solution on monday if I can remember that wink

#17 Re: Help Me ! » I'm just not sure, Help please » 2006-09-22 11:00:50

If it's a multiple of 3, it's between 50-59, but not all numbers between 50-59 are multiples of 3. Cancel out those numbers. For the rest between 50-59, how many are multiples of 4? Cancel out those who are not multiples of 4. You following so far?

Good. So what can you conclude now? Exactly, the number is not a multiple of 3 because none of the numbers between 50-59 remains. Cancel out all remaining numbers that are multiples of 3.

Remember, it has to be a number between 50-79, because it's the only definitions you've got.

Remove your second criteria, you've used it up already. Continue the same way with your other two criterias smile

Edit: Bah, Rod gave the answer away, lol.

#18 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-21 02:23:54

I think I have a solution now, it wasn't that hard when you think about it. But it'll take quite some time, but you said the time taken was irrelevant smile

#19 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-21 00:07:27

So, basically, a prisoner must be absolutely sure that everyone has visited the central room at least once when he says he wants to go home. But they can only contact eachother through the switch... So they must be clever enough to figure out what n is, right? or perhaps be able to count to 100 through the switch? I guess that's two options.

We might have to group the prisoners somehow before they are being put into the cells. hmm

Is the switch on or off from the start?

#20 Re: Puzzles and Games » 1 million! » 2006-09-15 08:29:53

bah, no fun trying anymore tongue

#22 Re: Help Me ! » tough limit problem » 2006-09-13 03:21:19

It's not even in my calculus textbook. sad

#23 Re: Help Me ! » tough limit problem » 2006-09-12 20:49:49

I don't know what L'hopitals rule is, but here's what I did (hoping that I didn't use it!):

#24 Re: Help Me ! » determinant » 2006-09-11 06:31:26

I can see where you're going, but I was hoping for a better way of doing it. Working with 60 2*2 determinants doesn't sound like a lot of fun =P

#25 Re: Help Me ! » limit problems » 2006-09-10 10:08:19

I'm guessing you use ε on your y-axis, and δ on your x-axis (like we do over here). If so, consider this problem (same principle, I'm just using x² instead):

Show that x² --> 4 when x --> 2.
We need to show that, as soon as we've picked our ε>0, we can find a δ>0 so that |x²-4|<ε when 0<|x-2|<δ. Don't forget the 0 right there!

|x²-4| = |x+2|*|x-2|

Let ε>0. If we demand |x-2|<1 (namely that 1<x<3), we get |x+2|<5. If we also demand that |x-2|<ε/5 (notice why we picked 5 below, and not any other number), we get:

|x²-4| = |x+2|*|x-2|<5*ε/5=ε

Thus, if δ is the least of the numbers 1 and ε/5, we get that |x²-4|<ε when 0<|x-2|<δ. Maybe someone could pull out a graph of the definition to make things easier...

That's all there is, I think. I don't see where you got that ε/3=δ from though?

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