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August 5

A good calculator for Mac OS X

I'm looking for a free advanced calculator for Mac OS X like Microsoft Encart Math, but lets me define my own functions. Grapher is nice for making graphs, but not for displaying calculations. I would like one that can display a lot of decimal places. --Melab±1 ☎ 01:49, 5 August 2011 (UTC)[reply]

The dc (computer program) and bc programming language available in a terminal unix prompt are both powerful command line programs. Probably more powerful than anything with a gui, they can calculate pi to 1000 places.--Salix (talk): 23:34, 5 August 2011 (UTC)[reply]

653xy divisible by 80

If 653xy is divisibile by 80 then the value of x+y is — Preceding unsigned comment added by 49.138.219.184 (talk) 17:08, 5 August 2011 (UTC)[reply]

One answer is zero. (Why?) Another answer is 18. (Why?) There are many more possibilities too (infinitely many, in fact). You need more information to get a unique answer. —Bkell (talk) 19:15, 5 August 2011 (UTC)[reply]

I must be reading the question differently to Bkell. I think there is a single solution. I think that solution will be the only number between 65300 and 65400 to divide by 80 - work this out first - then add the last two digits together. Grandiose (me, talk, contribs) 19:23, 5 August 2011 (UTC)[reply]

This is probably a test to see if the person has learned old tricks. First, 80 is a multiple of 10. Anything times 80 will end with a zero. So, we know that y=0. We want to know if 653x is divisible by 8. If the last 3 digits are divisible by 8, then the whole number is. So, we need a number 53x that is divisible by 8. We have the choices 530, 531, 532, etc... It turns out that 536 is divisible by 8. So, x=6. Then, x+y = 6+0 = 6. I would have made it a little more difficult in appearance by giving a number like 637456284756292653xy (since only the last 2 digits before the x matter). -- kainaw ™ 19:47, 5 August 2011 (UTC)[reply]

Why are you assuming

653xy=653*100+x*10+y

? The standard meaning of this notation is

653xy=653*x*y

.--Antendren (talk) 22:19, 5 August 2011 (UTC)[reply]

Ah, yes, that is how I read the question:

653\times x\times y

. Of course, if x and y stand for digits in a 5-digit number, as other answerers have interpreted it, then there is probably a unique answer; I didn't think of that nonstandard interpretation when I read the question. —Bkell (talk) 22:54, 5 August 2011 (UTC)[reply]

It's a recreational mathematics convention, I think, except that I would have expected capital A and B rather than lowercase x and y. --Trovatore (talk) 22:57, 5 August 2011 (UTC)[reply]

I initially thought it was multiplication, which implies that the question is flawed. So, I assumed that my assumption was wrong and went with another assumption. -- kainaw ™ 12:40, 8 August 2011 (UTC)[reply]

A Simple PDE

Given the PDE $u_{t}+au_{x}=0$ , u(x,0)=f(x) and u(0,t)=g(t) are reasonable conditions (with the domain being the first quadrant in the x-t plane, i.e. $x,t\geq 0$ ). By inspection, I see that u(x,t)=F(x-at) for any differentiable F is a valid solution (and I think this is the most general solution too...all solutions will be of this form). My questions is now what about the given conditions? I was told by my teacher that a solution exists for any f and g but I disagree. I think that you need a certain kind of compatibility between f and g. Any old pair of f and g is not going to work. Here is my argument.
$u(x,0)=f(x)=F(x)$ so f must be the same as F.
$u(0,t)=g(t)=F(-at)$ so g(t)=f(-at).
Does this argument make sense? Who is right? At the very least you need f(0)=g(0), right? So that the two conditions agree on the origin. We don't want a discontinuity there. I think you still need this compatibility even if we don't care about analytic solutions and use FDS to solve it numerically for example. What do you guys think? Thanks! 128.138.138.122 (talk) 19:54, 5 August 2011 (UTC)[reply]

Your argument is correct but you reach the incorrect conclusion, at least when a>0. The last statement, g(t)=f(-at), only makes sense at t=0 since the domain of f and g are t>=0 and x>=0. For t<>0 at least one side of the equation is undefined. F(x)=f(x) only defines F for x>=0, F(x) needs to be defined in terms of g when x<=0 and the values have to match, i.e. f(0)=g(0), at x=0. It's not too hard to show any solution is of the form F(x-at): Restrict u to the line x-at=c. Eliminate one of the variables and calculate the derivative wrt the other one, the original PDE implies the derivative is 0. In other words u is constant on the lines x-at=c, so u is some function of x-at only.--RDBury (talk) 14:18, 6 August 2011 (UTC)[reply]