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#101 Re: Help Me ! » Factoring Quadratic Equations » 2014-07-04 05:13:12

Maburo

For 16, C is correct. Great!


For 18, you must also consider the coefficient for

. You need two numbers which multiply to produce
and sum to the middle coefficient for
,
. We can, in fact, find these two numbers! They are
and
because
and
. Then we replace the coefficient of
with the new numbers we found, and we get:

Then factor out
:

Then factor out
:

Then the answer is A


19.   Yes, fractions can be quite tricky. Plus, the hardest part of factoring these quadratics is finding the new coefficients for the middle term. So mix that with fractions, and they can be a little nasty. Anyhow, let's follow the same steps and see what we can come up with. The coefficient in front of

is
and the constant is
, so we need two numbers that multiply to produce
and sum to the middle coefficient,
. Two numbers which have these properties are
and
because
, and
. Again, we then replace the coefficient of the middle term with our new numbers:

Then factor out
from the first two terms, and
from the last two:

Then factor out
:

Thus, the answer is D.


20.   You are correct!

#102 Re: This is Cool » Tracing a Function » 2014-07-04 02:21:52

Maburo

Thank you for sharing, benice. This is cool!

#103 Re: Jai Ganesh's Puzzles » 10 second questions » 2014-07-04 02:20:16

Maburo


#104 Re: Help Me ! » Factoring Quadratic Equations » 2014-07-04 02:14:07

Maburo

Hello demha,

Don't let the double variables confuse you. Double variable quadratics can still be factored the same way.

13.   We need to look for two numbers which multiply to produce

and sum to
. No such real numbers exist, so the answer is C - cannot be factored.


14.   This should be a better example to rid any confusion surrounding double variable quadratics, since this one can be factored. Again, we must look for two numbers which multiply to produce

and sum to
. Two such numbers are
and
because
and
. Once we have found these numbers, we replace the middle term with our new numbers as coefficients:

Then factor terms with k and n:

Now factor out the common factor of
:

And the answer is then D. See, it's no different with two variables!


15.   This one is a little different in that it uses the concept of "difference of squares." We have


We factor out a 2:

The difference of squares basically states that
. In this case,
, so we can write

And thus the answer is E.


16.   I'm sure you can do this one! Try it out.


17.   Remember the difference of squares? We need it to factor this one! We have

. To use the difference of squares we need
. This is possible if we let
and
. Then

or

Thus, the answer is F.


Try doing the rest on your own. I don't see any more with double variables, but if you ever run into them, you know they shouldn't be treated any differently!

#105 Re: Jai Ganesh's Puzzles » 10 second questions » 2014-07-03 03:49:08

Maburo


#106 Re: This is Cool » Blobs » 2014-06-26 06:19:51

Maburo

A similar example - but this time made with a warped ellipse, rather than a warped circle, in the shape of the sine function. It resembles a snake!

Elliptical Sine Snake

#107 Re: This is Cool » Tracing a Function » 2014-06-26 03:56:04

Maburo

Amazing. How do you trace the curve in order to plot the distances corresponding to θ?

#108 Re: This is Cool » Tracing a Function » 2014-06-26 02:11:57

Maburo

That is very interesting. Did you plot both y with θ and x with θ? For example, in your first plot, the blue curve was the distance along the x axis for corresponding values of θ, and the red curve was for distance along y axis?

Thanks for sharing!

#109 Re: This is Cool » Blobs » 2014-06-25 10:05:21

Maburo

Here is an example I made using Desmos:

Blob

#110 This is Cool » Blobs » 2014-06-25 06:51:23

Maburo
Replies: 2

While playing around with functions and graphng, I found an equation for a blob which takes the shape of any function. It is simply the equation for a circle, but modified slightly.

For a blob that takes the shape of a function

, use the equation
where
varies.

Try it out with a graphing calculator capable of using sliders for

, or use a CAS.

For example, you could examine a blob that warps its shape around the curve

by plotting
with
and varrying
.

You can also use it to make a heart around

by plotting
.

#111 Re: This is Cool » Tracing a Function » 2014-06-22 06:45:27

Maburo

Ah, I see. It should be

. This works! Thank you for your help.

There are still many functions you could do this to where the resulting graph will not be solvable with elementary functions. However, they can still be drawn. Are you able to have a computer draw these?

#112 Re: This is Cool » Tracing a Function » 2014-06-22 06:11:30

Maburo

Ah, indeed it is. How are you finding this? And could you do it for

?

#113 Re: This is Cool » Tracing a Function » 2014-06-22 05:56:49

Maburo

How about something simple, like

. Of course, there will be some places where this will be undefined.

#114 Re: This is Cool » Tracing a Function » 2014-06-22 05:50:23

Maburo

Indeed! That's what I am looking for, for an ellipse. Now what if we did this with some other function?

#115 Re: This is Cool » Tracing a Function » 2014-06-22 04:48:59

Maburo

I don't have any images because I do not know how to generate them. I have, however, found a link to a dynamic image of the sine function (and cosine) being generated by tracing the unit circle. What I want to do, though, is put the point on any function or curve and trace out the resulting curve in the same way.

Unit circle and sine function

#116 This is Cool » Tracing a Function » 2014-06-22 03:37:26

Maburo
Replies: 19

Say we have the unit circle and we take any point on it and connect a line from the origin to that point, marking the angle

from the horizontal axis to the line. We then rotate the line around the circle, varying the angle
and plot the height of the point on the circle above the horizontal axis,
. Plotting the value of
for every angle
gives us the sine function,
.

What if, instead of a unit circle, we traced an ellipse? We would make a line from the origin to a point on the ellipse, noting the angle

between the horizontal axis and the line, then varying
and plotting the height above the horizontal axis,
, for corresponding values of x. So, we apply the same process of obtaining the sine function from the unit circle on other shapes.

I have absolutely no background in programming, so I have not been able to have a computer generate these functions for me. I imagine that if we did this to an ellipse, we would get a sine function stretched out along the x-axis.

But what if we did this to any function like a polynomial, rational, exponential, logarithmic, sinusoidal, or even parametric or polar? I'm not sure how to find out what these would look like, as I do not know any programming languages (although I have used Maple a bit).

I can only try to imagine what some of the simpler ones may look like. For example, an ellipse should look like a sine function stretched along the x-axis. The polar function

should look like a cycloid with every other arch missing.

I would like to know if anybody who can program something like this could show me what some graphs generated this way look like? Or maybe there is a way to define these functions so I can graph them myself? If anybody knows anything about curves generated this way, or could show me what some look like, I would be very interested to hear/ see what you have!

#117 Re: This is Cool » Explanation? » 2014-06-11 06:55:53

Maburo

Hello again. Here it is, written with WriteLaTeX. This version has been cleaned up. It is much more organized and includes a lot more explanation than my posts on the forum.

Proof

#118 Re: Jai Ganesh's Puzzles » 10 second questions » 2014-06-09 08:21:03

Maburo

#119 Re: Jai Ganesh's Puzzles » 10 second questions » 2014-06-08 02:34:03

Maburo

#120 Re: This is Cool » Explanation? » 2014-06-07 16:22:41

Maburo

Now that we have shown that the nth difference in our pyramid gives a row of

, and that every row after that is all
, we will show that
can be expressed as an alternating sum:

Remember the definition:

   and   

We can continue this and notice a pattern:

and

So in general:

What we want to prove is that:

To do so, we will prove that

  and then look at the case where
and we already proved that
  when 


Again we use induction:

We must assume that

Then




Expanding:

We see pairs of


and




Now we rewrite our previous expansion as:



Every k has been replaced with k+1, thus proving that

Now for the best part! We shall look at the case where k=n:

becomes
which we proved was equal to

Substituting n for k in our sum:

Now, x is arbitrary; it does not matter where in a sequence of perfect powers of n we begin taking differences, so we will let x=0:


That is my proof smile Let me know if you guys see any errors (maybe I am completely wrong). Also, I apologize if it is a messy proof either due to my lack of LaTeX skills or lack of organization. Off to bed now! Bye.

#121 Re: Puzzles and Games » Next Term in the Sequence » 2014-05-16 05:45:06

Maburo

Those are right, phrontister! That was an interesting way to find how they are related, and not what I had intended, but you are correct nonetheless.

I intended them to be related by

13)     

14)     

15)     

16)     


These ones are all related too

#122 Re: Jai Ganesh's Puzzles » 10 second questions » 2014-05-15 16:55:57

Maburo


#123 Re: Puzzles and Games » Next Term in the Sequence » 2014-05-15 16:30:42

Maburo

You both got it!

9)     

10)     

11)     

12)     

These are all related. Tell me how!

#124 Re: This is Cool » Explanation? » 2014-05-15 15:16:03

Maburo

Here is the proof:

Consider the difference pyramid discussed earlier in the thread. Each row is the difference between consecutive terms of the previous row. We will define a function to describe the differences:

. Here
represents the number of times we have taken the difference,
is a number in a row, and
is the exponent. By this, we can define the first row (the 0th difference) with
. Thus, by our definitions,
. This should make sense because
is the first difference between two consecutive powers of n.

We can expand this:






Now we can apply the same process for further differences:

This is where we use induction: we want prove that the nth difference is equal to n!

To do this we need to assume two things:
First, we assume that the nth difference is, in fact, equal to n!, and we need to assume that after we reach a row of n!, the next difference is 0.

Assume:


Now we prove this for

:






This was just the first part of my proof, showing that the nth difference of perfect powers of n is equal to n!. I will use this to prove that n! can be expressed as an alternating sum. I have to go for now, so I will post that part later. Bye for now! And let me know if I made any errors or did anything improperly. I didn't catch any mistakes, but I'm sure you guys can if I made any.

#125 Re: Introductions » Hello » 2014-05-15 11:37:34

Maburo

Hi Bob. I'm going to the University of Alberta in Edmonton.

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