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#51 Maths Teaching Resources » Tools to play with Venn Diagramm and probabilities? » 2013-02-11 08:07:42

White_Owl
Replies: 2

I have a bunch of problems like this:
P(A) = .1
P(B) = .2
P(AUB) = .3
P(A|B) = ?
Some of the problems are really complicated and consist of multiple events (right now the largest one have five, but more is possible).

Is there any tool which knows how to work with probability functions and can construct a new formula upon the given set of known probabilities?

#52 Re: Help Me ! » integrate x/sqrt(3x+25) » 2013-01-25 05:31:27


Yes, it does look promising.
Thank you very much.

#53 Re: Help Me ! » integrate x/sqrt(3x+25) » 2013-01-24 10:21:44

Ok... Then:


And after substituting back u for x I have:

But now if I differentiate it back:


Where did 9 come from?

#54 Help Me ! » integrate x/sqrt(3x+25) » 2013-01-24 08:23:56

White_Owl
Replies: 5

In my homework for this week I have a question:

If I do a substitution u=3x+25, then I have


Which does not help much.



If I try to find such u that du will eliminate x from numerator, then: du=1/x dx and u have to be ln(x)


Which is also not very helpful...



Finally I run the original integral through Maxima, and it gave me:


Which is very scary and I have no idea how it come to this formula.


So which "u" should I use? How to find such "u"? Or maybe there is another approach I should use? I am completely stuck.

#55 Re: Help Me ! » Distance to the wall of the box » 2012-12-07 05:42:15

John E. Franklin wrote:

minValue(maxValue(dTop,dBottom),maxValue(dLeft,dRight))

Yes, I am allowed to use min() and max() functions, and right now  I actually have an equation similar to the one you showed here. It works perfectly but I still believe there is a way to avoid individual calculations of these distances.
I think, the ray tracing algorithms are dealing with the similar problem. Do we have any experts in 2.5D graphics here? I remember reading some articles on that subject, but it was too many years ago.

#56 Re: Exercises » Bafflers? » 2012-12-06 10:58:18

The more I think about the problem #1 in this thread, the less I believe in the answers given to it in this thread smile
I do not think we can talk about percentage of white marbles in the jar at all. We can talk only about probabilities.

At the beginning we have 50 blue and 50 white.
On a first step we can remove 25 blue, 25 white or any number of blues and white (up to 25). So after putting 25 blue ones back into the jar we can have either:
50b+50w, or 75b+25w, or some numbers in between.
After repeating this process the second time, numbers can become 50b+50w, or 100b+0w, or some numbers in between.

In other words, after 10 steps we can have 50% marbles in the jar are white, or 0% (possibly even after the second step) or any number between 0 and 50.

#57 Help Me ! » Distance to the wall of the box » 2012-12-06 08:53:53

White_Owl
Replies: 4

The problem:
We have a squared box of size w*w.
Inside the box, we have a point at (x,y).
From the point we draw a line at some angle d.
If x, y, d, and w are known, find the distance to the closest wall of the box.

x and y are always positive and less then w (point is always inside the box). The d can be anything.
On the attached image I put two points. I need to know the length of green lines.

From a simple definitions of sin and cos, I have four distances to the four walls of the box
dl = (0 - x) / cos d
db = (0 - y) / sin d
dr = (w - x) / cos d
dt = (w - y) / sin d
From the set {dl, db, dr, dt} remove negative values (those are distances along the same line but in opposite directions).
And the final answer is the minimal from two remaining values.

Looks like the final answer is correct, but solutiona is fairly complicated. Maybe I am not seeing the obvious, but is it possible to reduce this algorithm to a one formula?

#58 Re: Maths Teaching Resources » Graphing Calculator » 2012-07-10 01:28:28

bob bundy, Thank you. I think this will do.

bobbym, Yes, the program would be nice too. Especially if it will be able to export the graph in .tex format...

#59 Maths Teaching Resources » Graphing Calculator » 2012-07-09 07:27:53

White_Owl
Replies: 6

Does anyone know good (and preferably free) graphing calculator which is able to draw graphs of relations, not just functions?
For example I have an equation

. Are there any ready to use applications which can handle it?

#60 Re: Help Me ! » Range of values of x » 2012-07-06 04:34:02

bob bundy wrote:

That's the same interpretation as zetafunc.

And the same equation.

He jumped a few steps but gets to the same solution.

ummmm.... Ok... I see now, were we went into different directions.
Although I still think my solution is more clear.
Well it is more clear for me smile

#61 Re: Help Me ! » Range of values of x » 2012-07-05 09:56:56

And I do not agree with zetafunc.'s solution.
The description "solid cube of side s has a square hole of side x" gives me an image as in the attached picture.

From it, the surface of the remaining solid is: 4*s^2 + 2*(s^2-x^2) + 4*s*x. Here:
4*s^2 - for four untouched sides of the original cube
2*(s^2-x^2) - for face and back of the original cube
4*s*x - for the inner surface of the hole

The volume of the remaining solid is: s^3-s*x^2.

Therefore we have an equation:

From this we can see that if s=6, then x=0. But if s=7, then x=7/5. And with increasing of the s, x also increases. (in fact, for s=10, x=5).
Therefore, the final answer: x in {1, 2, 3, 4, 5}.

#62 Re: Help Me ! » Creating LaTeX documents? » 2012-07-05 08:18:45

I prefer to use TeXstudio: http://texstudio.sourceforge.net/  for actual typing of tex document. It has a lot of features and (very important to me) it looks and works very-very similar on both Linux and Windows.

In addition to the Au101's template I also always have the next two commands before \begin{document} and one command after:

\title{My School\\ My Class\\ Homework \#1}
\author{Name \texttt{(email)}}
\begin{document}
\maketitle

And all my homework files have a package:
\usepackage{tikz}
for all graphics needs. The PGF/TikZ is actually a system which consists of a simple drawing engine and a lot of sub-packages built on top of it. Sometimes I feel that PGF/TikZ is as complicated as TeX itself, but it really is fun to use smile
You can check this website: http://www.texample.net/tikz/examples/ for examples of what can be drawn and how it was done.


Almost always, I also use tablists . It is an addition for standard enumeration environment, which allows to put items into columns. Something like:

\begin{enumerate} % default
\item A
\item B
\end{enumerate}
\begin{tebenum} % from tablists
\item C
\item D
\end{tabenum}

This will produce
1. A
2. B
1. C  2. D

If you need to add pieces of code in some program language (C, Java, Basic, etc ) you can look at the listings package. It generates teletype-like text, with ability to number each line and it can emphasize key-words of the language. It already knows more then 40 languages.


And I would add my thumbs up for: http://en.wikibooks.org/wiki/LaTeX
It will give you the documentation on all "standard" TeX.
For help on packages you have to remember just one command: texdoc <package_name> Use it in the terminal and it will automatically find html/pdf/dvi/anything else which contains documentation for the selected package.
And if you ever forget how to

#63 Re: Help Me ! » g(x) = x^(2/3); g'(0)=? » 2012-06-26 06:43:22

Ok.
Question #1.
According to Stewart's textbook, Calculus Early Transcendentals 7th edition, we can substitute x for an exact value if we are looking for the slope of the tangent at that point.
Chapter 2.7, Pages 144-145.


And Example #2 does exactly that.


Question #2.
Yes, I did a mistake in the final step of calculating the limits. Sorry about that. It should be:


But still, I have
vs
.
Where did 2/3 disappeared?


Question #3.
Actually, I do not see why did you say that domain of the function is just positive numbers? I believe the domain is full R.
After all, we can do :


Not all graphing calculators are smart enough for such function. I am playing with QtiPlot and for "x^(2/3)" it draw the similar image as bob bundy showed. But if I rewrite formula into "(x^2)^(1/3)", I have a more correct graph.
Which still has a strange horizontal part near the zero, which does not disappear when I zoom in.

#64 Help Me ! » g(x) = x^(2/3); g'(0)=? » 2012-06-24 14:04:54

White_Owl
Replies: 8

The problem states: if g(x)=x^{2/3}, show that g'(0) does not exist.

So my answer is:

exists if and only if both left and right limits at the point are equal.

Now lets find left and right limits separately:

From this we can see that left and right limits at point 0 are equal by an absolute value, but have opposite signs; therefore, the

does not exists.


But according to the rules of differentiation

or in this case:


My problem is: how

correlate with
?
They are both supposed to be equal to the slope of tangent line, aren't they? What am I missing?

#65 Re: Help Me ! » Limits of x+sqrt(x^2+2x) » 2012-06-17 09:29:27

Wow!
Thank you, guys.
Sometimes you have to be creative to get an answer. And I always believed math is a precise science... I wonder why smile

#66 Re: Help Me ! » Limits of x+sqrt(x^2+2x) » 2012-06-17 03:57:57

anonimnystefy wrote:

Hi White_Owl

Ok... But if we continue from here:


????

#67 Re: Help Me ! » Limits of x+sqrt(x^2+2x) » 2012-06-17 03:08:01

anonimnystefy wrote:

Where are the other parts of the fraction?

"other" parts?

#68 Re: Help Me ! » Limits of x+sqrt(x^2+2x) » 2012-06-17 02:13:38

Ok... but how this will help?


Or did I make a mistake somewhere?

#69 Help Me ! » Limits of x+sqrt(x^2+2x) » 2012-06-16 12:18:30

White_Owl
Replies: 29

As text book suggested in one of the examples, I tried to multiply by conjugate radical:


After a few iterations I came up with

Which does not lead me anywhere....
So I am starting to think that multiplication by conjugate radical is not a way to go.
But what is the way to go???

By graphing the function I see that


But I have no idea how to prove it.

#70 Re: Help Me ! » Need help with inverse functions » 2012-06-11 07:57:25

Thank you.
I got the second one.

As for the first... Well I thought of using the numerical approximation as a justification for finding

without formally finding the
itself.
Unfortunately, (or luckily) in that case we do not need approximation. The x=0 already gives the y=4. Not much of approximation...

#71 Help Me ! » Need help with inverse functions » 2012-06-11 06:48:09

White_Owl
Replies: 5

Working with Stewart's "Calculus, Early Transcendentals"
Stuck on two problems:

1.6 #17
If

, find
.

The answer here is almost obvious:

.
Just by simple brute force approach (plug various numbers into x and once g(x) become 4, the value of x is the value of

But I do not think that is the approach expected from students...
Unfortunately I do not see how to make an inverse function. Where to start?


1.6 #26
Find inverse of the function:

Same problem here. Where to start? How to take

outside?

#72 Re: Help Me ! » Circle in a 3D space » 2012-05-08 07:31:28

Well, I would not say that the chord is forever fixed smile

OK, let me think out loud:
We can define a plane by three points. Let's say the first one is defined by points (C1, A, B) and the second one by (C2, A, B). So basically I need a formula of a circle which goes through points A and B, and have a center in the point C1.
For the the plane we can use formula ax+by+cz+d=0, so for the plane we have a set:


It would give us a, b, c, and d.
Each point of the circle also is a member of the same plane and at the same time the distance between it and a center is fixed.
So we should add to the previous set of three formulas two more:


And this should give us a circle in any arbitrary plane defined by three points....
Is this correct or did I made a mistake somewhere?

Now the question is how to simplify it into more usable formulas.

#73 Re: Help Me ! » Circle in a 3D space » 2012-05-08 06:33:42

No, what you draw is a moving center of a circle in the same plane as the original circle.

I need something like this. I can draw a circle C1 using


But now I need to draw the circle with a center in C2, but in a different plane. The original plane and the new one intersect along the chord AB.

#74 Help Me ! » Circle in a 3D space » 2012-05-08 05:24:14

White_Owl
Replies: 7

The goal is to draw a circle which turns around one of its chords.
I had an idea to define a plane with a base point on a circle's chord and a normal vector, then turn the vector.... But while I can imagine this graphically I am completely stuck in the algebraic part.
Or am I thinking in the wrong direction completely?

#75 Re: Coder's Corner » Random Latin Square generation » 2011-04-27 06:24:26

And the third approach which always work:

take a Latin Square
randomly exchange two columns or two rows in it
repeat several times

This would generate almost random Latin Square, but it would be isomorphic to the original, so not a really random.
And it is also not a very fast procedure...

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