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#176 Re: Help Me ! » Yeses and Nos » 2006-12-24 06:21:03
Pretty good problem.
Let the n be the number of the questions.
In general we'll have two cases:
1. If n is odd, then we can't have (YES=NO) case. So in this case, the answer is 2^(n-1).
2. If n is even, we'll have for the number of (YES=NO) cases:
and the result will be:
.
#177 Re: Help Me ! » A spooky discovery » 2006-12-24 05:34:14
Yes, the same.
Now, i'll uncover my results.
Let
If we use the first greater or equal to the number square indeed, then
and
So: (removing this absolute value for x>y and x>1)
Let us define the error function:
Here are 2 plots of it (fig 1 and 2).
As we see the error is 0 when n is a perfect sqare, and it grows bigger when n=k^2-q for small q.
So in the worst case
But
So we can assume
and we'll get an upper bound for the error function:
And, after some simplification:
And this estimate is very good!
(figures 3 and 4 showing a plot with the error function and the bound).
#178 Re: Help Me ! » A spooky discovery » 2006-12-24 03:49:48
About Mathsy post - good point.
I suppose the best will be to use the first greater or equal to the number exact square.
And the approximation is really good.
As you can see form the image :
which is true for all x,y.
I'll make some tests to see the efficiency of the algoritm.
#179 Re: Help Me ! » A spooky discovery » 2006-12-23 10:28:49
Just "powering" the equation:
...
for b,c =/=0:
So if c<<b, then the approximation will be better.
Here's a plot:
#180 Re: Help Me ! » matices » 2006-12-23 09:32:49
The system has not solution indeed.
#181 Re: Help Me ! » matrix » 2006-12-17 02:52:26
that's a vector - row, it's (8x1) matrix and 8 dimentional vector
#182 Re: Help Me ! » need help (ALGEBRAIC EXPRESSION) » 2006-12-17 02:50:26
<--what about this? Should be EQUALITY, not expression.
#183 Re: Help Me ! » Help » 2006-12-17 02:48:05
#184 Re: Help Me ! » is this right way » 2006-12-17 02:45:12
just having fun!
#185 Re: Help Me ! » Hard(ish) Geometry Problem 3 » 2006-12-09 18:55:16
First I thought the same, but the proof works for every m,n.
If E bisects AD, then E is (the center of the bisectors) in ΔABD, Let BD intersects with AC in K. Then BK=KD and AK is a bisector of the triangle. Thus AP=2PK. Whe we disect ΔBCD in the same way, we have: KC=3PK, SO AC/AP=(AP+PK+KC)/AP=6PK/2PK=3=2+1=2+n/m.
#186 Re: Help Me ! » summation? » 2006-12-09 04:45:27
#187 Re: Help Me ! » Equation ????? » 2006-12-09 04:41:02
This is an equation.
Function is for example:
#188 Re: Help Me ! » Summation » 2006-12-09 04:32:30
You can use the general formula to obtain a formula for intervals:
#189 Re: Help Me ! » Need help factoring this! » 2006-12-09 04:27:56
And another - the degrees of x are only even. So you can plug y=x^2 and the degree of the equation will become form 6 to 3, which is some helpful simplification.
#190 Re: Help Me ! » is this right way » 2006-12-09 04:18:11
LQ, right. Just a little mistake:
#191 Re: Help Me ! » any maths genuises? help plz! simple question. » 2006-12-09 04:14:37
Some good explanation:
http://en.wikipedia.org/wiki/Scientific_notation
#192 Re: Help Me ! » ????? » 2006-12-09 04:08:14
You mean the determinant?
#193 Re: Maths Is Fun - Suggestions and Comments » Error 404!?! » 2006-12-09 04:05:20
Here's the source of the page I got:
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<ADDRESS>Apache/1.3.33 Server at www.mathsisfun.com Port 80</ADDRESS>
</BODY></HTML>#194 Maths Is Fun - Suggestions and Comments » Error 404!?! » 2006-12-09 04:01:25
- krassi_holmz
- Replies: 16
I was posting in the Help section, when some screen saying that there's error 404 and I do not have access to /forum/post on this server on port 80??? 
#195 Re: Help Me ! » Mathbook work format » 2006-12-09 03:54:18
In my opinion you should use something like Proof: or Solution: to distinguish the problem from the solution.
I agree with Rob and mathsy. And if you split the pages you will have not only drawing problems. Sometimes we must work with HUGE equations and you just couldn't write them in a splited row.
#196 Re: Help Me ! » Hard(ish) Geometry Problem 3 » 2006-12-09 03:45:43
Sorry for delaymant now it's edited.
#197 Re: Help Me ! » Hard(ish) Geometry Problem 3 » 2006-12-09 02:55:55
#198 Re: Help Me ! » any maths genuises? help plz! simple question. » 2006-12-09 02:42:13
-1999=-0.1999*10^4.
If you want:
-1999=-0.1999E4
-1999=-(11111001111) in binary
#199 Re: Help Me ! » Coin Toss probability question » 2006-11-26 03:24:42
And, using induction, we will easily prove the interesting fact:
which is according to the pi man's observation.
I'm beginning to think that it will be easier to count combinations without 2 heads than these with two heads.
Once again - combinatorics are COOL!
#200 Re: Help Me ! » Coin Toss probability question » 2006-11-26 03:14:29
Many combinatorial problems yeld to fibonacci numbers.
It's good start to use some reuirrent relation.
Let
Q(n) be the total number of all combinations when a coin is tossed n times,
and
G(n) - the number of combinations in which there are at least 2 consecutive tosses.A combination of this kind we'll call correct.
Q(n) = 2^n (that's easy, because every combination of tosses can be represented as binary string :1 for heads ).
Now notice
G(1)=1;
G(2)=1;
Let compute G(n+1) , n>=2:
1.
If the first is 0, then we'll get
0xxxx... <--n x's
So this will be a correct combination, iff xxxx... is a correct combination.
So we have that the number of correct combinations of n+1 coins, when the first is null, is
G(n).
2. If the first number is 1, we get 2 cases:
2.1. If the second number is one, then we have:
11xxxx... <--n-1 x's,
which is always correct, so for every combination xxx...
In this case we have Q(n-1) correct combinations.
2.2. If the second number is 0:
10xxxxx <-- n-1 x's
As in point 1, this will be correct, iff xxx... is correct, so in this case we have G(n-1) correct cases.
And now:
For example
110
011
111
G(3)=G(2)+G(1)+2^1=0+1+2=3.
1011
1100
1110
1101
1111
0110
0111
0011
G(4)=G(3)+G(2)+2^2=1+3+4=8;
I hope I've lighted a little bit this problem.