The Math Thread - Request help with all forms of mathematics

[Sammy56]

DAMNIT! I had the entire solution typed out before you CT and my stupid internet crashed! It's all your fault!

I'll repost mine anyway, since I did it a bit differently.

1) 5x^2 + 25x + 30 ---> (5x+2) (5x+3) (is this right?)
2) 2x^2 + 11x +9
3) 2x^2 + 11x -9

Wow. This has been forever.

1. 5x^2 + 25x + 30
The first thing you have to do is to get the problem in another form so that you can simplify it. These problems are slightly more difficult because there is a number in front of the x^2. So, what you need to do is multiply the first and last number (5 x 30) to get 150. Now, you have to find two numbers that you can multiply to get 150 and add to get 25 (the middle number). These two numbers are 15 and 10 in this case. Now we can rewrite the problem as:

5x^2 +15x + 10x + 30.

Then, separate it:

(5x^2 +15x) + (10x +30)

Now, we factor both sets of parenthesis:

5x(x + 3) + 10(x +3)

You know you are on the right track if you get the same expression in both sets of parenthesis after factoring. Now, you factor out what is still inside the parenthesis (since it should be the same thing) to get:

(x + 3)(5x + 10)

And that is your correct answer. You can then check this answer by using the FOIL (first, outer, inner, last) method.

First: (x)(5x) = 5x^2
Outer: (x)(10) = 10x
Inner: (3)(5x) = 15x
Last: (3)(10) = 30

Add them together to get 5x^2 +10x +15x + 30, which simplifies back to 5x^2 + 25x + 30.

2. 2x^2 + 11x +9

2 x 9 = 18

9 and 2 can be multiplied to get 18 and added to get 11.

2x^2 + 2x + 9x + 9

(2x^2 + 2x) + (9x + 9)

2x(x +1) + 9(x + 1)

(x + 1)(2x + 9) is the correct answer.

3. 2x^2 + 11x -9

This one does not look like it can be simplified. (2 x -9) = -18, but there are no two numbers that can be added to get 11 and multiplied to get -18 that I can see. I believe this one is in it's most simplified form, but it's been awhile since I've done this, so I could be missing something.

5 (x + 3) (x + 2) = (5x +15) (5x + 10)

Wrong. Check it on your calculator. You are violating order of operations. You must multiply the (x+3)(x+2) before you can distribute the 5 or you only distribute the 5 through the first set of parenthesis to get (5x +15)(x+2). If not, you are multiplying that (x+2) by 5 twice.

[Cartesiantheater]

(x + 3)(5x + 10)

And that is your correct answer. You can then check this answer by using the FOIL (first, outer, inner, last) method.

No it isn't. You still have a factor of a 5 in the (5x + 10)

You must factor that five out. The REAL correct answer is, as I already posted:

5(x + 3)(x + 2)


...revenge is a bitch, ain't it?

3) 2x^2 + 11x -9


This one does not look like it can be simplified. (2 x -9) = -18, but there are no two numbers that can be added to get 11 and multiplied to get -18 that I can see. I believe this one is in it's most simplified form, but it's been awhile since I've done this, so I could be missing something.

Yes it can. I just did it. You can use either completing the square or the quadratic formula... well... maybe that's not simplifying it... er... yeah... well, it CAN be solved and written as a product of two binomials! (albeit ones with confusing square roots, etc...)

Wrong. Check it on your calculator. You are violating order of operations. You must multiply the (x+3)(x+2) before you can distribute the 5 or you only distribute the 5 through the first set of parenthesis to get (5x +15)(x+2). If not, you are multiplying that (x+2) by 5 twice.

Yes, but only on my re-multiplying of that one part. The answer I posted IS correct, and YOURS was not because it wasn't fully simplified (as you left a factor of 5 in the second parenthesis).

...again I say...revenge is a bitch, ain't it?


:)

[Sammy56]

Okay. I didn't simplify it completely. You are right. The reason I do that is because that is what they wanted for AP stuff. They wanted two binomials, nothing out front. Stupid rules, I know. I'm just used to them.

However, this:

5 (x + 3) (x + 2) = (5x +15) (5x + 10)

is wrong. You did get the right answer before. I wasn't denying that. Just that your other method right there was TOTALLY wrong. Mine was only partially wrong, so I still win.

maybe that's not simplifying it... er... yeah... well, it CAN be solved and written as a binomial!

The original trinomial is more simplified then your binomial!

Which would you rather have to work with?

2x^2 + 11x -9

or

(x -11/4 + ((193)^(1/2))/4) (x - 11/4 - ((193)^(1/2))/4) ?

You'd have to be an idiot to say the second one.

...again I say...revenge is a bitch, ain't it?

You're the one that should know. Is it?

[Cartesiantheater]

Okay. I didn't simplify it completely. You are right. The reason I do that is because that is what they wanted for AP stuff. They wanted two binomials, nothing out front. Stupid rules, I know. I'm just used to them.

So you get excuses but I don't get mine? I edited my post and all but explained the small error (the error was nothing but haste, because both the binomials I posted indeed ARE in the answer if it is expanded like that- it's just you can only have ONE or the OTHER

for example, you can have EITHER:

(5x +15) (x+2) OR (5x + 10) (x + 3), so as you can see, I was only PARTIALLY wrong as well.)

is wrong. You did get the right answer before. I wasn't denying that. Just that your other method right there was TOTALLY wrong. Mine was only partially wrong, so I still win.

Mine was partially right, as explained above.

The original trinomial is more simplified then your binomial!

Mathematically speaking it is NOT. You have nothing but this form: (x + C)(x + C1), where C and C1 are two constants.

Which would you rather have to work with?

2x^2 + 11x -9

or

(x -11/4 + ((193)^(1/2))/4) (x - 11/4 - ((193)^(1/2))/4) ?

If you actually used your calculator and got the values of those large numbers, I would certainly prefer the later. Just by LOOKING at it you already know where the parabola equals zero. It is certainly easier to use the latter, if the computations are done (via CALCULATOR= no work).

You'd have to be an idiot to say the second one.

Or a certified Math genius

You're the one that should know. Is it?

Well, seems again I have evened the score. Let us cease with this contest. Besides, I think either fireant or probably perfectionist would have our number hear anyway (fireant = engineer in last year of study, perfectionist = bachelors in mathematics [forgot his masters]).

So I shall now concede this debate, in the interest of mathematics, because we are just cluttering the math thread. :bows out of the way:

[Cartesiantheater]

Meet Precious, the overtly adorable poodle...



We catapult Precious into the air at a forty-five degree angle, at an initial velocity of 100 meters per second.



If there is a ten square meter pool of Jello deep enough to save her 1021 meters away, and she doesn't veer left or right, will poor Precious make it to the Jello, or will she become one with the cement, oozing into its crevices?

How high will Precious get before she tumbles down to her fate? (assuming the initial height is zero)




bon appetite

[DontBeAfraid]

I come up shy by about 3 meters.... assuming the pool starts at 1021

of course, she is almost certain to enter the pool after the bounce.

[Cartesiantheater]

I come up shy by about 3 meters....

You're relatively close...

assuming the pool starts at 1021

Yes, it starts at 1021.

of course, she is almost certain to enter the pool after the bounce.

I would have to agree. But would she bounce on the Jello and land on the other side? Or perhaps PEICES of her would?

[DontBeAfraid]

You're relatively close...

Im within a meter.

How high will Precious get before she tumbles down to her fate? (assuming the initial height is zero)

I missed this, she will come screaming in from about 254 meters so... depending on ground conditions varying amounts of her will make it into the pool.

[Cartesiantheater]

Need help with a problem. I know a couple of you can do this. It involves the linear wave equation and proving that a function is a solution to a differential equation:

Linear Wave Equation:

∂^2y/∂x^2 = (1/v^2) ∂^2y/∂t^2

(second partial derivative of y with respect to x equals one divided by velocity squared times the second partial derivative of y with respect to t)

(a.) Show that the function y(x,t) = x^2 + (v^2)(t^2) is a solution to the linear wave equation
(b.) Show that the function in part (a.) can be written as f(x + vt) + g( x -vt), and determine the functional forms for f and g.
(c.) Repeat parts (a.) and (b.) for the function y(x,t) = sin(x) cos (vt)

[FireAnt]

I got part a done. Damn this stuff brings back some horrible memories. It's been about 3 years now since I had to do this stuff, and yes I have forgotten most of it so make sure you check over my results. This is going to seem weird but it gets the answer and kind of makes sense in a funky kinda way.

Start off with the answer (makes it easier)
y=x^2+v^2 *t^2 (1)

Second order partial diff it in terms of x...

∂^2 y/∂ x^2 = 2 (2)

Now second order partial diff eq. (1) in terms of t...

∂^2 y/∂ t^2=2 * v^2 (3)

Use the constant 2 as a algebraical term, rearrange (3) to seperate the 2 and substitute into (2)

∂^2 y/∂ x^2 = 2 = 1/(v^2) * ∂^2 y/∂ t^2

And there you go, you've got the equation you were looking for...

∂^2 y/∂ x^2 = 1/(v^2) * ∂^2 y/∂ t^2

Kind of weird way of showing it but it does definitely answer the question. If you're confused just ask away.

Probably won't be able to do the rest tonight. I'll leave those for the other maths geeks out there.

Oh and on a side note. John, you don't have to post a reply to each and every post. The people who post in this thread are genuinely looking for HELPFUL replies. If you can't answer the question, or atleast try to help them, then don't. Nobody really wants to hear how you can't help them. So please, unless you have a question or an answer, try to stay out of this section. I doubt the bible can show us how to do second order partial differentiation.

[Cartesiantheater]

Thanks FireAnt!

I can't really go over it in depth since I'm at the school library on a "school work only" computer, but I did have time to delete Johnb1's banter.

Looking briefly, your partials are right. I'm a bit lost on the next part, but later I'll look at it more closely.

EDIT- nevermind, I got it. You set 2.) and 3.) equal to each other and then divided the v^2.


Part (b.) is probably the hardest though, from my perspective. I'm lost on how to do that one.

[FireAnt]

The question in Part b is kind of confusing. However, there is a little trick I have picked up for solving stupid questions. Pay attention...

*FireAnt places a blindfold on his head and starts punching random buttons on his keyboard*

This equation should do the trick...

y = (x+vt)^2 /2 + (x-vt)^2 /2

Now to test it. (I sure hope I have done this part right or it could be extremely embarrasing. It's been well over 5 years since I learnt quadratic equations and I tend to forget things.)

y = (x^2+ 2 xvt + v^2 *t^2) / 2 + (x^2 – 2 xvt + v^2 *t^2) / 2

Now just simplify...

y = (2x^2 + 2v^2 * t^2) / 2

And again...

y = x^2 + (v^2) (t^2)

Piece of cake.

The only real way to do questions like this is trial and error. They give you some basic parameters, but the rest is just guessing. I sure hope I have done this right. If not just adjust it till it fits.

I haven't looked at part c yet, but I do know one thing... I hate trig.

[RobsMathStinks]

Simple Question:
Is there a standard symbol for element-by-element multiplication?

Slight elaboration:
I mean for proper type-setting like what might appear in a math text. I don't mean .* or whatever our favorite computer program uses, unless that is actually the standard.

More elaboration:
I've tried ing many times, but the only thing that comes up is a bunch of articles or other posts where people--including me!--seem to have invented their own symbol for it.
Some have equated it to kroenecker product, but I'm virtually certain that's incorrect. I mean:
[1 2 3] ele-by-ele [4 5 6] should be [1*4 2*5 3*6].

Thanks!

[BLOODLINES]

ROB: Apparently your math is better than mine as I have NO CLUE what you are talking about? LOL!

Sorry I was unable to help.

[Cartesiantheater]

Simple Question:
Is there a standard symbol for element-by-element multiplication?

Slight elaboration:
I mean for proper type-setting like what might appear in a math text. I don't mean .* or whatever our favorite computer program uses, unless that is actually the standard.

More elaboration:
I've tried ing many times, but the only thing that comes up is a bunch of articles or other posts where people--including me!--seem to have invented their own symbol for it.
Some have equated it to kroenecker product, but I'm virtually certain that's incorrect. I mean:
[1 2 3] ele-by-ele [4 5 6] should be [1*4 2*5 3*6].

Thanks!

Rob, your math certainly doesn't suck- you wouldn't be asking that question if it did. :P

Set theory and the like is certainly beyond me, but Flynn is a computer scientist, I believe. You should ask Flynn in a PM. Also, maybe Fireant could help (although I wonder if engineers cover that branch of math?)

Good luck

[DontBeAfraid]

although I wonder if engineers cover that branch of math?

Its part of descreet math.... and no, I dont think there is a set standard for it but Im sure there is a common method... which probably involves subscripts...

something like:
Set C_x = A_x * B_x for all integer x between 0 and smallest set length.

[usohypocrisy]

I found a very cool and helpful program, that will show you step by step how to solve math problems (algebra, trig, etc), it's a teaching guide, not a cheat aid. It's called

Algebra Solved.

http://uploading.com/files/61FHZ7TC/....2008.rar.html

[bluenose_ian]

Ok people, who has a maths head, well its been proved we all do but what I want to ask is and i know the answer , just interested who else does.

How would I go about creating a fourth dimension?

its easy.

[Cartesiantheater]

Ok people, who has a maths head, well its been proved we all do but what I want to ask is and i know the answer , just interested who else does.

How would I go about creating a fourth dimension?

its easy.

Uh.... it already exists. It's called time, lol.


But maybe you mean a fourth spatial dimension? Humor us...

[Cartesiantheater]

Simple Question:
Is there a standard symbol for element-by-element multiplication?

Slight elaboration:
I mean for proper type-setting like what might appear in a math text. I don't mean .* or whatever our favorite computer program uses, unless that is actually the standard.

More elaboration:
I've tried ing many times, but the only thing that comes up is a bunch of articles or other posts where people--including me!--seem to have invented their own symbol for it.
Some have equated it to kroenecker product, but I'm virtually certain that's incorrect. I mean:
[1 2 3] ele-by-ele [4 5 6] should be [1*4 2*5 3*6].

Thanks!

Thought I'd bring this one up again. Yeah, Rob is gone, but it's an interesting question, and I don't like questions not being answered. So if this is wrong, I'll bump it again next year ;).





The closest thing I have seen that gives what is asked is the Cartesian product, although it merely describes a relation between two sets where, assuming we're working with the real numbers, every element x in X is assigned to every element y in Y, one element at a time in each set.

The symbol used is just an X.

For example, given two sets A and B, the Cartesian Product between them would do this:

A x B = {(a,b)|a ∈ A and b ∈ B}

In words:

A "times" B equals the set of all possible ordered pairs (a,b) such that the first component a is a member of A and the second component b is a member of set B.

However, just ordered pairs is not enough. I do know that generally when talking about sets, an x is used for multiplication, and when talking about just elements of sets, you'll see a dot often, of just putting the two elements adjacent to each other.

You would have to define your operation, regardless, but I do not believe there is a "standard" notation. At this level of math, what constitutes "multiplication" is subject to the whims of the person describing it anyway. But one way to do it might be this: (I say might because I am barely learning this stuff and so I could be wrong)


Let A and B be two sets whose elements a,b ∈ℝ. The operation f is defined as:

f: S x S → S

where

A x B = {ab| a ∈A and b ∈B}

which in English should mean:

The binary operation f takes the Cartesian product of two sets of the same "universe" and maps them to that same universe (i.e. if you start with two sets that are made of real numbers, you end with a set that is made of real numbers), where given two sets A and B the operation is given by A x B equals all possible sets consisting of a times b where the first component is a member of A and the second component is a member of B.

But as far as I can tell, when you are doing some sort of "multiplication" of two sets, element by element, you would use the Cartesian product in some way or another, which is generally denoted by an x. You would also have to define your multiplication operation, since all the Cartesian product is assure you that whatever operation you are using takes one element of the first set and pairs it with the corresponding element of the second set.




That's my best shot at this point, but I'd put it on a test hoping to at worst get a C. Any others want to try again?