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## #1 Re: Help Me ! » area » 2012-09-06 03:43:26

That's the answer I get Julie.

## #2 Re: Help Me ! » combinations » 2012-08-03 07:40:57

First multiply the number of options available in steps 1, 2, and 4.  This gives you 6 meat options X 4 rice/bean options X 8 salsa options (assuming you can choose no salsa) = 192.

Now you need to determine the number of ways of choosing options in step 3.  This looks like

Multiply those together and you get a total of 192 X 32768 = 6,291,456 possible combinations.

## #3 Re: Help Me ! » Local min/max of multivariable function » 2012-07-18 03:14:36

If it wasn't on a boundary point then it would be at a critical point, and we already know that there is only 1 critical point for T(x, y) and that it is a global maximum.

## #4 Re: Help Me ! » Local min/max of multivariable function » 2012-07-18 00:54:07

I think this is right.

First take the partial derivatives of T(x, y) and set them equal to 0:

Which agrees with Bob's answer.  This is the only critical point for T(x, y) in the region x^2 + y^2 <= 4 (indeed, it's the only critical point anywhere).  This means that to find the minimum of T(x, y) in this region we only need to consider the boundary x^2 + y^2 = 4.

Find the critical point:

Now we have just 3 points to check by hand, the boundary cases (-2, 0) and (2, 0) and the critical point (-1/3, +- sqrt{35/9}):

So the minimum temperature is -57/9 and occurs at (-1/3, +- sqrt{35/9})

## #5 Re: This is Cool » Shaking The Foundations Of Mathematics. » 2012-07-16 00:38:03

Can you show the steps you took to get from

to

It's not obvious to me how you do that.

## #6 Re: Help Me ! » Factorial Equation » 2012-07-11 02:56:34

Those solutions have some good ideas, but there are flaws in both of them.

Edit: Rather, the first solution is flawed, and while the second looks correct he skips a lot of steps and makes it hard to follow.

## #7 Re: Help Me ! » Factorial Equation » 2012-07-11 01:28:46

That doesn't really get you anywhere though.  b! + c! is the numerator, b! - 1 is the denominator.  You don't learn much by saying that the numerator has factors that aren't in the denominator.  Consider the example we already know - 3! + 4! = 30, 3! - 1 = 5.  The numerator is divisible by 3 and the denominator is not, but you can still divide them and get another factorial as the quotient.

## #8 Re: Help Me ! » x + 1/x » 2012-07-07 02:40:43

Add the two equations together to get (x + 1/x) + (x - 1/x) = (a + b) + (a - b) => 2x = 2a => x = a.  The rest is easy from there.

## #9 Re: Help Me ! » Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that: » 2012-07-06 08:39:47

Unless I made a mistake somewhere I believe that the inequality is wrong.  If you let a be a very small number (like 10^-10), b = 0.5, and c = 1 - a - b, and p = 0.25 the inequality will fail.

## #10 Re: Help Me ! » Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that: » 2012-07-06 04:46:15

Here's what I've got so far.  Rewrite the LHS to read

The partial derivative with respect to a is

Setting this equal to 0 gives us

You can show from here that a = b = c = 1/3 is a critical point, but I haven't been able to show that it's the only critical point.

## #11 Re: Help Me ! » Factorial Equation » 2012-07-05 07:37:46

All factorials above 1! are even, so that doesn't tell you much.

## #12 Re: Help Me ! » Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that: » 2012-07-03 08:53:40

First show that

is minimal when a = b = c = 1/3.  From there:

## #13 Re: Help Me ! » hey im new and need help with these 2 questions.. » 2012-06-21 03:02:23

Based on the second question he probably wants f'(x)?

For the first question use the chain rule.  Rewrite f(x) like this:

Then

Try the second one on your own, ask if you get stuck or if you don't understand what I did for the first question.

## #14 Re: Help Me ! » Need help with inverse functions » 2012-06-11 07:16:53

The second one is easy enough.

It should be simple enough to solve for x from there.

The first question is a bit trickier, I don't know offhand of a good way of finding the inverse function to that.

Edit: After searching around it seems that the first problem does not have a simple inverse function.  Your only options there are trial and error or some method of numerical approximation.

## #15 Re: Puzzles and Games » Can you suggest six names of great mathematicians quickly? » 2012-06-11 03:06:00

A few others to consider:

Gauss
Riemann
Archimedes

## #16 Re: Help Me ! » Unique Relationship to find prime numbers » 2012-06-06 07:39:03

Indeed, and also readily apparent that all composite odd numbers are covered.

Edit: In fact, you can even simplify it further to (2p + 1)(2q + 1) = 4pq + 2p + 2q + 1, 0 < p, q.

## #17 Re: Help Me ! » Unique Relationship to find prime numbers » 2012-06-06 06:09:04

You can take credit for it, you did the hard part and I'm not interested in publishing or anything like that.

## #18 Re: Help Me ! » Unique Relationship to find prime numbers » 2012-06-06 05:31:54

I just took your basic idea and modified it.  The equations you used, x = 6y + 5 and x = 6y + 7, simply generate all odd numbers except those divisible by 3.  But the equation x = 2y + 1 generates all odd numbers.  This has two benefits: 1) You don't have to check if a number is missing from the list because it's divisible by 3 to make sure it is prime, and 2) You only have one possibility for m and a: m = 2p + 1 and a = 2q + 1.  This lets you generate your list from one formula instead of four.

## #19 Re: Help Me ! » Unique Relationship to find prime numbers » 2012-06-06 05:21:12

It's interesting, but your method could be improved to produce all odd numbers except the primes with a single formula.  Instead of starting from x = 6y + 5 and x = 6y + 7, use x = 2y + 1.  You eventually get a formula of

n = 4qb + 2b + 4pq + 2p + 2q + 1

Where b, p, q are integers and 0 < b, q and 0 <= p

## #20 Re: Help Me ! » probability » 2012-02-01 08:27:14

Interesting.  How do you know that all squares are of the form 7k, 7k+1, 7k+2, 7k+4?  It's not immediately obvious to me why this is true.

## #21 This is Cool » Sum of squares and primes » 2012-02-01 08:24:45

TheDude
Replies: 1

I was working on this problem http://www.mathisfunforum.com/viewtopic.php?pid=200241#p200241 in the Help Me! section and came across something interesting.  It seems that if a number C has prime factors

It also appears that the opposite is mostly true: if C has a prime factor p that cannot be written as the sum of the squares of two integers then C itself cannot be the sum of the squares of two integers.  The exception is cases where p appears as a factor of C an even number of times, in which case x and y will both also have p as a factor exactly half the number of times that C does.

Is this a well known fact that I'm not aware of, or is there a proof of this somewhere?  It's not at all clear to me why this appears to be true.  For example, 2 can be written 1^2 + 1^2 = 2, and 2^2 + 3^2 = 13.  If we take 2*2*13 = 52 we get 4^2 + 6^2 = 16 + 36 = 52.  The factors 2*2*13 have no obvious relationship with 4 and 6, yet we see that they are connected somehow.  And this is not an isolated case, I tested numbers up to 440 and found no exceptions to this fact aside from the already mentioned case.

## #22 Re: Help Me ! » probability » 2012-02-01 07:53:57

For a number to be divisible by 5 its last digit must be 0 or 5.  Make a list of the final digits of the squares of all single digit numbers:

0 ->  0 -> 0
1 ->  1 -> 1
2 ->  4 -> 4
3 ->  9 -> 9
4 -> 16 -> 6
5 -> 25 -> 5
6 -> 36 -> 6
7 -> 49 -> 9
8 -> 64 -> 4
9 -> 81 -> 1

This means that the probability of the last digit of the square of a number being 0 is 10%, 1 is 20%, 4 is 20%, 5 is 10%, 6 is 20%, and 9 is 20%.  You can pair 0's and 5's with other 0's and 5's, 1's with 9's and 4's, 4's with 1's and 6's, 6's with 4's and 9's, and 9's with 1's and 6's.  This means that the probability of x^2 + y^2 being divisible by 5 is (0.2*0.2) + (0.2*0.4) + (0.2*0.4) + (0.2*0.4) + (0.2*0.4) = 0.36 = 36%.

For x^2 + y^2 to be divisible by 7, both x and y must be divisible by 7 (I think, I don't have a proof for this).  That means that there is a 1/7 * 1/7 = 1/49 chance that x^2 + y^2 is divisible by 7.  Multiply that by 36% to get 1/49 * 9/25 = 9/1225.

## #23 Re: Help Me ! » How can I get this equation? » 2012-01-31 15:25:32

I think he means this expression:

At least it ends up working out to the expression he's looking for:

## #24 Re: Help Me ! » Oh,Oh,Oh,merry analysis!!! » 2012-01-25 08:52:49

Sorry to be pedantic, but when base isn't specified isn't log(n) usually base 10 and ln(n) base e?