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June 21[edit]

Write naturally in math, automatically translate to TeX[edit]

Is there a product out there where I can write mathematical equations naturally on a surface, and it would automatically convert it to TeX for me? --HappyCamper 01:32, 21 June 2007 (UTC)[reply]

No. Learn to type TeX quickly; that should be more useful. (This unsigned comment added by 129.78.64.102.)

That sounds about right. State-of-the-art PDAs have enough trouble with simple letters and numbers, the mind boggles at what it'd take to get them to understand complex formluas. Black Carrot 04:06, 21 June 2007 (UTC)[reply]

Hmm...nevermind, I just found what I was looking for today. Thanks! --HappyCamper 11:41, 21 June 2007 (UTC)[reply]

Ah, wanna share? —Bromskloss 13:28, 21 June 2007 (UTC)[reply]

"Wanted: math grad student for typing job, $10/hr or ramen+shower." [edited] Tesseran 01:00, 22 June 2007 (UTC)[reply]

dodecagons and hexagons[edit]

How do I calculate the area of a dodecagon and the area of the hexagon if I only know the length of one side, let's say 7cm?Invisiblebug590 06:32, 21 June 2007 (UTC)[reply]

Dodecagon—12 sided figure

Hexgon—6 sided figure

You can't—unless you assume/guarantee something more about the shapes (e.g. that they are regular convex polygons). --CiaPan 07:18, 21 June 2007 (UTC)[reply]

Have you read our dodecagon and hexagon articles? they would seem to answer your question (for regular versions). 213.48.15.234 07:15, 21 June 2007 (UTC)[reply]

P.S. the dodecagon and hexagon are regular polygons.

Invisiblebug590 06:11, 23 June 2007 (UTC)[reply]

See Regular_polygon#Area 67.182.185.143 22:37, 23 June 2007 (UTC)[reply]

Maybe i've turned stupid[edit]

How would you solve $2^{x}+3^{x}=13$ for x?

I can see that the answer's x=2 but i have no idea how to go about getting the solution through manipulation of the equation. 213.48.15.234 07:40, 21 June 2007 (UTC)[reply]

I had enough trouble sleeping before you asked. :) The only thing I know is that you need logarithms to remove the x variables from the exponent. I just don't see how to do it because the "simple" rules of logarithms don't give a simplification for log of the left side of your equation. Yechiel Man 07:29, 21 June 2007 (UTC)[reply]

X=2 is certainly one answer, but I doubt it's the only correct answer. -- JackofOz 07:32, 21 June 2007 (UTC)[reply]

I've tried both logarithms and exponent manipulation, but the old not being able to do much with added exponents of different bases is throwing me off. Jack, I can't see another real answer, maybe there are some imaginary ones. 213.48.15.234 07:36, 21 June 2007 (UTC)[reply]

Since

2^{x}+3^{x}

is a strictly increasing function of

x

, there can be only one solution. —Bromskloss 08:01, 21 June 2007 (UTC)[reply]

\log(a+c)=\log a+\log(1+e^{\log c-\log a})

Assuming Natural Log.

Put a = 2^x and put c=3^x

\log(2^{x}+3^{x})=x\log 2+\log(1+e^{x\log 3-x\log 2})

\log(2^{x}+3^{x})=x\log 2+\log(1+e^{x(\log 3-\log 2)})

Now solve

x\log 2+\log(1+e^{x(\log 3-\log 2)})=\log 13

xF+\log(1+e^{xG})=H

where F = Log 2

where G = Log 3 - Log 2

where H = Log 13

So we have

x+{\frac {\log(1+e^{xG})}{F}}={\frac {H}{F}}

x+{\frac {\log(1+e^{xG})}{\log K}}=L

where e^K=F

where L=H/F

x+\log(1+e^{xG}-K)=L\,

Raise to the e

e^{x}\times (1+e^{xG}-K)=e^{L}\,

e^{x}\times (M+e^{xG})=N\,

where M=1-K

where N=e^L

Now expand

Me^{x}+e^{x}e^{xG}=N\,

e^{x+P}+e^{x(1+G)}=N\,

e^{x+P}+e^{xQ}=N\,

211.28.131.152 07:47, 21 June 2007 (UTC)[reply]

Well, i could do all of that from your first formula, that's exactly the point I got stuck, though :p 213.48.15.234 07:53, 21 June 2007 (UTC)[reply]

shoot he's still going 213.48.15.234 08:11, 21 June 2007 (UTC)[reply]

This transformation seems incorrect to me:

x+{\frac {\log(1+e^{xG})}{\log K}}=L

x+\log(1+e^{xG}-K)=L\,

Logarithmic identity is

{\tfrac {\log A}{\log B}}=\log _{B}A\neq \log {(A-B)}.

--CiaPan 08:34, 21 June 2007 (UTC)[reply]

I believe you're correct there CiaPan, i'm still only a little closer to a solution. :/ 213.48.15.234 09:00, 21 June 2007 (UTC)[reply]

If simple manipulations could solve these kinds of equations, I'm sure Fermat's last theorem would have been proven much earlier. Some related (but much tougher obviously) problems are described in Mihăilescu's theorem. nadav (talk) 09:39, 21 June 2007 (UTC)[reply]

Let me just emphasize these statements:

X=2 is certainly one answer, but I doubt it's the only correct answer. -- JackofOz 07:32, 21 June 2007 (UTC)[reply]
Since $2^{x}+3^{x}$ is a strictly increasing function of $x$ , there can be only one solution. —Bromskloss 08:01, 21 June 2007 (UTC)[reply]

Now, for the complex solutions, $n^{r\cdot \operatorname {cis} (\varphi )}=n^{r\cdot \cos(\varphi )}\cdot \operatorname {cis} (r\cdot \ln(n)\cdot \sin(\varphi ))$ , so we get (for "cis", see this section):

2^{r\cdot \cos(\varphi )}\cdot \cos(r\cdot \ln(2)\cdot \sin(\varphi ))+3^{r\cdot \cos(\varphi )}\cdot \cos(r\cdot \ln(3)\cdot \sin(\varphi ))=13

2^{r\cdot \cos(\varphi )}\cdot \sin(r\cdot \ln(2)\cdot \sin(\varphi ))+3^{r\cdot \cos(\varphi )}\cdot \sin(r\cdot \ln(3)\cdot \sin(\varphi ))=0

Try solving this system, and let me know what you get. -- Jokes Free4Me 09:44, 21 June 2007 (UTC)[reply]

Oh, are we looking for non-real solutions as well? —Bromskloss 09:51, 21 June 2007 (UTC)[reply]

Not really, I was just speculating, however I would like to see if it's possible to find x=2 from the equation. From the answers i've gotten so far, it doesnt look like it is easy to do. 213.48.15.234 09:55, 21 June 2007 (UTC)[reply]

One needs to use a root-finding algorithm.--Patrick 11:31, 21 June 2007 (UTC)[reply]

I haven't worked out the details, but I strongly suspect there are no complex solutions beyond x = 2. Putting x = a+bi, |2^x + 3^x|² is at least (3^a – 2^a)² and at most (3^a + 2^a)²; if 13² is not in this range, we can't have a solution. This implies that 2 ≤ a < 2.61. Examining this range in small increments, with 0 ≤ b < 2π, does not suggest any second solution lurking there. --Lambiam ^Talk 15:54, 21 June 2007 (UTC)[reply]

Your intuition is wrong. I did a quick graph of

2^{x+yi}+3^{x+yi}=13

in the OS X grapher application and got what looked like a periodic curve (but probably isn't), which shouldn't be terribly surprising. Consider the simpler case of

e^{z}=c

for some real

c>1

. Remember that

e^{z}

is a transcendental function and periodic. The functions

2^{z}

and

3^{z}

are similarly shaped and adding them gives you something which is vaguely periodic (I think that the periods will interact in an irrational fashion so there probably does not exist non-zero

w\in \mathbb {C}

such that

2^{z+w}+3^{z+w}=2^{z}+3^{z}

. There are in fact an infinite number of solutions to

2^{z}+3^{z}=13

but only one real solution. Donald Hosek 16:57, 21 June 2007 (UTC)[reply]

My intuition here is mostly useless, but empirically looking at the imaginary part of the sum via a contour plot in Mathematica, with x between -3 and 4, and y between -3 and 3, and one contour line at 0 while sampling on a grid of 3000 points on each axis, shows one contour which strongly seems to lie on the real axis. Zooming the y axis in to between +/- 0.001 supports this hypothesis. This doesn't prove anything, but does suggest only the aforementioned root 2. Baccyak4H (Yak!) 17:28, 21 June 2007 (UTC)[reply]

Have mathematica give you a graph of solutions to

2^{z}+3^{z}=13

. I'm not sure what you're trying to prove with looking at the imaginary part of the graph. I'd spend some time also just looking at graphs of the exponential functions. Look at the real part, imaginary part and modulus (absolute value) to get some sense of what's going on, then you should be able to get some sense of what happens with the function

f(z)=2^{z}+3^{z}

and what the set of solutions to

f(z)=13

should look like. Mathematica should be able to graph those for you as well (you may need to use the

(x,y)

form of the equation I cited above to get it to graph the solutions in a real plane... I'm too old and cheap to know mathematica well). Donald Hosek 17:57, 21 June 2007 (UTC)[reply]

Sorry if I wasn't clear. I did just what you suggested: made a contour plot of Im( 2^(x+i*y) + 3^(x+i*y) ) with x and y being the two axes of the plot. The idea is for the sum to be 13, the imaginary part must be 0. Allowing one contour value at 0 shows where this happens (indeed, the intersection of those contours with those of the real part at value 13 are in principle all of the roots). The only contour which took value zero was apparently coincident with the x axis, meaning y=0, and thus implying all roots indeed are real. Baccyak4H (Yak!) 18:17, 21 June 2007 (UTC)[reply]

Update: This suggests a way to prove it. Ignore the 13 for a moment and try to show that 2^x + 3^x is real implies x is real. Baccyak4H (Yak!) 18:19, 21 June 2007 (UTC)[reply]

Not sure how accurate it is, but Maple turned up no complex solutions. Dlong 18:57, 21 June 2007 (UTC)[reply]

There are 2 complex solutions...somewhere around r = 6.524959 and theta = +- 1.1651783. --HappyCamper 19:51, 21 June 2007 (UTC) <--- update -- actually, I take that back. I think there are an infinite number of solutions. It's easier to see them by solving with r and theta. --HappyCamper 19:55, 21 June 2007 (UTC)[reply]

Good stuff... my efforts didn't look far out enough. Baccyak4H (Yak!) 20:06, 21 June 2007 (UTC)[reply]

If someone could double check these, it would be great. I don't have Mathematica at my disposal. Looks like the next furthest solution is r = 11.3275949 and theta = 1.362440594. --HappyCamper 20:37, 21 June 2007 (UTC)[reply]

Check. Also r = 17.4819... and θ = 1.4525....Baccyak4H (Yak!) 20:48, 21 June 2007 (UTC)[reply]

Check. I get r = 17.48192845163..., θ = 1.452539552949... --HappyCamper 21:24, 21 June 2007 (UTC)[reply]

Come on guys, take a look at the graph of the exponential function. Note that the real values for

e^{z}=c

for

c\in \mathbb {R} ^{+}

are not isolated points, but are actually continuous sets. It shouldn't take a genius to realize that

a^{z}=c

for

a

positive and not 1 will be similar other than scaling (think back to high school algebra for change of base). Now with a bit more effort, you should be able to come up with the idea that the solutions for

2^{z}+3^{z}=c

will also be continuous sets. Donald Hosek 20:55, 21 June 2007 (UTC)[reply]

Whether you meant it to or not, your tone seems like it could offend some people, especially since they're trying their best to help with a problem that resists immediate solution. That aside, though, what you've written is false. If c > 0 is real, for example, the solution set of z such that e^z = c consists of exactly those points of the form Log c + 2kπi for k some integer. This set, as you probably know, is discrete (made up of isolated points), rather than continuous. In fact, the solution set is discrete for any complex c; this is captured by the observation that the exponential map is a covering map (onto the plane minus the origin).

Your second observation is correct: the base e here is not special. Your last claim is also false; the best way to see this in general is a dimension argument. The function that sends z to 2^z + 3^z is an entire function, and in particular a smooth map from the complex plane (a 2-dimensional real manifold) to itself. Then any regular value (here, any regular value of c) has discrete, or 0-dimensional, preimage (solution set). By Sard's theorem (PM), the set of regular values has full measure, which means that if there are any values of c such that the solution set is continuous, those values form a set of measure 0. Thus your assertion holds extremely rarely, if at all. Tesseran 00:59, 22 June 2007 (UTC)[reply]

The dimension argument needs a little more care -- zero-dimensional is not in general synonymous with "discrete". For example the Cantor space is zero-dimensional. But I seem to recall that, if the zeroes of an entire function accumulate, then the function is constantly zero. I think you just figure out what the Taylor series has to be, centered at the accumulation point. That should show that the solution set is discrete, without invoking Sard's theorem. --Trovatore 19:39, 23 June 2007 (UTC)[reply]

In this case, we know that the solution set is a manifold, and a 0-dimensional manifold is discrete (has the discrete topology). As you say, any entire function which vanishes on a set containing a limit point is identically 0, and this implies discreteness immediately. (A much nicer argument.) Tesseran 19:40, 24 June 2007 (UTC)[reply]

Thanks! --Trovatore 22:49, 24 June 2007 (UTC)[reply]

It looks like the solutions a+bi with 2 ≤ a < 2.61 occur with an average frequency for b of once in every interval of length $\pi /((1/ln(2))-(1/ln(3)))=5.9001941$ , that is, twice for every time that the argument of $3^{z}$ overtakes that of $2^{z}$ .--Patrick 08:28, 22 June 2007 (UTC)[reply]

Actually, the 2 ≤ a < 2.61 I quoted from above should be 2 ≤ a < 2.7047.--Patrick 08:39, 22 June 2007 (UTC)[reply]

$2^{x}+3^{x}=2^{2}+3^{2}$

That's all there is to it. The above demonstrates quite clearly how 2 can be the only real solution.

As for imaginary solutions, put x = a + bi and you equate real and imaginary parts and get our original equation. I don't think there are any imaginary solutions.

Fan pressure[edit]

I have a room which in the room a person doing painting and sanding. There are a lot of dust. I want to suct all the dirt particles. Is there any formula to calculate the static pressure that we need to able the fan suct all the dirt particles?

I don't think we can help you practically, other than to advise that you use a vacuum cleaner, but try to clarify the situation. Are the particles floating in the air or sitting on the floor? And how do you know when they have been removed? Yechiel Man 15:12, 21 June 2007 (UTC)[reply]

Any changes in static pressure wont help you, as they will cause no net force on your dust particles. The only way your fan will help is if it causes an air flow strong enough to remove your particles. I agree that a hoover is the best option, though a fan may help ventilate your room to get rid of the smell of paint, as well as the smallest floating dust particles. Good luck. 195.137.96.79 06:27, 22 June 2007 (UTC)[reply]

Be sure to use one fan blowing in and another blowing out on the opposite side of the room, for maximum airflow. Close any other windows, as they would interfere with proper airflow. Also, beware that air saturated with saw dust can explode if exposed to an ignition source. So, avoid smoking or any other activity with an open flame in that room. StuRat 06:45, 22 June 2007 (UTC)[reply]

You may find health and safety regulations could be of help. If you are running a real buisness then there is surly regulations as to the amount of particles in the air, and there may well be standards for type of ventilation you need. Ignore advice about domestic vacuum cleaners which will not be upto the job. --Salix alba (talk) 19:08, 22 June 2007 (UTC)[reply]

Unless it's a domestic setting, and you're just doing a bit of DIY. A vacuum cleaner would be just fine for this. 195.137.96.79 03:40, 24 June 2007 (UTC)[reply]