The DW Math Thread

lizrau

Oh thank you Decus, I never was good at math...

Decus - Dreamweaver

MageFizban - Lost City wrote: »

Yeah, math competition-like questions would be more fun than integrate f(x) =blahbah

I got that integral out of the MIT Integration Bee archives, from 1994 specifically =/

Actually, most of the problems (except the sum and the matrix, which were mine) I grab from AMS and other competition materials...

b:surrender

lizrau wrote: »

Oh thank you Decus, I never was good at math...

You absolutely are; you got the correct answer.

Runemine - Dreamweaver

Couldn't solve my own question with a calculator rofl

http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation Solve that......

SerenityMare - Dreamweaver

Okay, I need help with proving trig identities. x___x


1. Sin ( x + y ) + Sin ( x - y ) = 2 sin x cos y

For this one, I try to expand the left side of the equation but I always end up with 2 sin x and no cos y...

2. tan x + tan ( π - x ) + cot ( π/2 + x ) = tan (2π - x)

π is pie btw.


Thanks for the help! ;__;

Decus - Dreamweaver

SerenityMare - Dreamweaver wrote: »

Okay, I need help with proving trig identities. x___x


1. Sin ( x + y ) + Sin ( x - y ) = 2 sin x cos y

For this one, I try to expand the left side of the equation but I always end up with 2 sin x and no cos y...

2. tan x + tan ( π - x ) + cot ( π/2 + x ) = tan (2π - x)

π is pie btw.


Thanks for the help! ;__;

For number 1, just remember your angle sum/diff identities:

Remember that sin (x +/- y) = sin x cos y +/- cos x sin y

So...

We expand out the two terms:

sin(x + y) + sin (x - y)
(sin x cos y + cos x sin y) + (sin x cos y - cos x sin y)

The cos x sin y terms cancel as they are added and subtracted in the same expression, leaving us with:

sin x cos y + sin x cos y = 2 sin x cos y

And for 2, the trick lies in remembering how shifting graphs works for trig functions:

Remember that the tangent function is odd (meaning f(-x) = -f(x)) and shifting it by pi does nothing to the values since it's period is pi. So tan(pi - x) = - tan x because tan(pi - x) = tan(pi + -x) and that's a shift by pi and a change of sign for the independent variable.

Also remember that cot(pi/2 + x) = - tan x since that's a shift of pi/2 and the cotangent function is, essentially, the tangent function flipped over and shifted by pi/2.

Lastly, we need to remind ourselves that a shift by 2pi does nothing to the tangent function. So tan(2pi - x) = - tan x for similar reasons to what we did above. tan(2pi - x) = tan(2pi + -x) and that's a shift of 2pi and a change of sign for the indep. var. again...

Keeping that all in mind, let's plug junk in:

tan x + tan(pi - x) + cot(pi/2 + x) = tan(2pi - x)

tan x + (- tan x) + (- tan x) = (- tan x)

- tan x = - tan x


Hope that helps, enjoy! b:victory

Decus - Dreamweaver

I gave up with the old problems since no one wanted to do them. I'll pull these straight from competitions from now on, entirely, I promise...

So here's this week's set:

1.

Cafe de la Peche offers three fruit bowls:

Bowl A has 2 apples and one banana;
Bowl B has 4 apples, 2 bananas, and 3 pears;
Bowl C has 2 apples, 1 banana, and 3 pears.

Your doctor tells you to eat exactly:16 apples, 8 bananas, and 6 pears per day.
How many of each type of bowl can you buy so there's no fruit left over?

Find all possible answers, I dare you. Also, the number of bowls must be integers...

2.

At a certain mathematical conference, every pair of mathematicians are either friends or
strangers. At mealtime, every participant eats in one of two large dining rooms. Each
mathematician insists upon eating in a room which contains an even number of his or her
friends. Prove that the number of ways that the mathematicians may be split between
the two rooms is a power of two (i.e., is of the form 2^k for some positive integer k).

3.

A spider, in the top-left-front corner of a 10 x 10 x 10 foot room, sees a big fat fly in the bottom-right-back corner. Describe the shortest path, and the length of the path, that the spider can crawl to get the fly. That's crawl, not jump, fly or spiderman-like-web-shooting!

---

This time, no. 2 is the more challenging one. I pulled that from an old US Math Olympiad question booklet. 1 and 3 are from an old archive of weekly math contest questions that I have compiled for my students.

Have at it!

Decus - Dreamweaver

No takers, at all? I gave you what you wanted

b:cry

krisnda

Not sure what you meant for #1 but... I think

You would need to get:
5 of bowl A,
1 of bowl B
and 1of bowl C
to get exactly:16 apples, 8 bananas, and 6 pears per day.

I could find another solution, but i'm tired, and looking at the other questions (especially #3 ^^).

Decus - Dreamweaver

Krisnda wrote: »

Not sure what you meant for #1 but... I think

You would need to get:
5 of bowl A,
1 of bowl B
and 1of bowl C
to get exactly:16 apples, 8 bananas, and 6 pears per day.

I could find another solution, but i'm tired, and looking at the other questions (especially #3 ^^).

Correct! Yay!

There are actually two more solutions b:avoid

#3 is a favorite of mine, fyi

krisnda

Got the others for #1:

2 of bowl C, and 6 of bowl A

and 2 of bowl B and 4 of bowl A.

I'll let someone else get the tougher ones ^^ lol.

ColdAsLife - Dreamweaver

Krisnda is doomed ... b:shocked

krisnda

Hi Grim, love your siggie ^^.

Ezehc - Lost City

Krisnda wrote: »

Hi Grim, love your siggie ^^.

Decus - Dreamweaver wrote: »

I gave up with the old problems since no one wanted to do them. I'll pull these straight from competitions from now on, entirely, I promise...

So here's this week's set:

1.

Cafe de la Peche offers three fruit bowls:

Bowl A has 2 apples and one banana;
Bowl B has 4 apples, 2 bananas, and 3 pears;
Bowl C has 2 apples, 1 banana, and 3 pears.

Your doctor tells you to eat exactly:16 apples, 8 bananas, and 6 pears per day.
How many of each type of bowl can you buy so there's no fruit left over?

Find all possible answers, I dare you. Also, the number of bowls must be integers...

2.

At a certain mathematical conference, every pair of mathematicians are either friends or
strangers. At mealtime, every participant eats in one of two large dining rooms. Each
mathematician insists upon eating in a room which contains an even number of his or her
friends. Prove that the number of ways that the mathematicians may be split between
the two rooms is a power of two (i.e., is of the form 2^k for some positive integer k).

3.

A spider, in the top-left-front corner of a 10 x 10 x 10 foot room, sees a big fat fly in the bottom-right-back corner. Describe the shortest path, and the length of the path, that the spider can crawl to get the fly. That's crawl, not jump, fly or spiderman-like-web-shooting!

---

This time, no. 2 is the more challenging one. I pulled that from an old US Math Olympiad question booklet. 1 and 3 are from an old archive of weekly math contest questions that I have compiled for my students.

Have at it!

O i just saw this.

number one is just a system of equations which simplifies to A + B = 6 and B + C = 2.

so theres 3 ways to have B and C and u get uhm:

6 0 2
5 1 1
4 2 0

number 3 u just make it flat and draw straight line so it's 10 * sqrt(5).


going to try number 2 now. hm I actually did USAMO twice but that was a long time ago lol. haven't done much math since then.

Ezehc - Lost City

**** too hard

Zanryu - Dreamweaver

I don't even wanna read through all of the pages I didn't read earlier.. Just.. gosh... Math T.T

Esnemyl - Dreamweaver

*waves*
first time on this thread, still here since I poofed? o.o

Got exams coming up.. And I still dont get Histograms in general, or factoirizing (can expand though), never really got hold of them.
Any tips?
I got cookies for helping

Decus - Dreamweaver

Krisnda wrote: »

Got the others for #1:

2 of bowl C, and 6 of bowl A

and 2 of bowl B and 4 of bowl A.

I'll let someone else get the tougher ones ^^ lol.

Yep, nailed it! Way to go Thanks for participating, Krisnda!

Ezehc - Lost City wrote: »

O i just saw this.

number 3 u just make it flat and draw straight line so it's 10 * sqrt(5).


going to try number 2 now. hm I actually did USAMO twice but that was a long time ago lol. haven't done much math since then.

You are correct, nicely done! No. 2 is a toughie ;-;
Cheze I didn't know you had a psy

Esnemyl - Dreamweaver wrote: »

*waves*
first time on this thread, still here since I poofed? o.o

Got exams coming up.. And I still dont get Histograms in general, or factoirizing (can expand though), never really got hold of them.
Any tips?
I got cookies for helping

I am not sure what you are asking as far as histograms are concerned. It's, in its basest parts, a way to represent the distribution of data. You'll put the independent variable (the variable at the start of a process, the thing I control to see a result somewhere else) on the lower axis and the dependent variable (the variable at the end of the process, thing that I observe) on the left (or right, depending on the way you write your histogram) axis.

The thing that separates histograms from other types of bar graphs is that they graph continuous data. What I mean by that is, I graph frequencies of things that are discrete entities (like hair color or dress size) on a bar graph. Histograms instead graph non-discrete data (like weight, height, or time) where the thing being controlled is continuous. It may help to think of a histogram as a collection of frequencies along a continuous axis whereas a bar graph is a collection of frequencies of separate things. (This is a bar graph and this is a histogram.)

Here's a guide for how to make one given some data. It's a little wordy, but the methods are good. Sorry if that seems like a wall of text, but I wasn't sure what bases to cover. Feel free to ask if I have muddled anything as I am still prone to mistakes.

---

Factoring is a big topic... is there anything with which you specifically have an issue? A good way to check is if I give you a few things, which of them you can do, if any? (Btw: There's no shame in not knowing so long as you know on what you should direct your attention.)

3x + 9xy --> ?

x^2 + 4x - 60 --> ?

x^2 - 25 --> ?

x^3 + 3x^2 - 2x - 6 --> ?

There's a tad bit more advanced stuff than this, but I started with that since I am not sure how much you have covered.

Also, hi! Nice to see you back. No need for cookies, I do this purely for fun



EDIT: I just realized I assumed you meant "factoring polynomials!" If you instead were talking about prime factorization for numbers and radicals, that's something else. Stupid me for jumping to conclusions before asking. Normally I'd erase what I wrote earlier but I invested enough effort in it to warrant it staying there till I better understand what you want me to explain

Esnemyl - Dreamweaver

Decus - Dreamweaver wrote: »

EDIT: I just realized I assumed you meant "factoring polynomials!" If you instead were talking about prime factorization for numbers and radicals, that's something else. Stupid me for jumping to conclusions before asking. Normally I'd erase what I wrote earlier but I invested enough effort in it to warrant it staying there till I better understand what you want me to explain

Thanks for the histogram info, but what i meant in factorising..:

i can expand like this, using FOIL
x(2+5)
First xXX= X2(squared)
Outside x X 5 = 5X]
Inside x X 2 = 2X] add these two togewther, makes 7x
Last 2X5= 10

Result: x2+7x+10


The thing with factorising(oppositge of what i jsut displayed there) is, I can sometimes find it hard to find the Highest Common Factor between the numbers..
does this clear things up any mroe o master of math?

Decus - Dreamweaver

Esnemyl - Dreamweaver wrote: »

The thing with factorising(oppositge of what i jsut displayed there) is, I can sometimes find it hard to find the Highest Common Factor between the numbers..
does this clear things up any mroe o master of math?

It does. My hint to you would be to think about what you're given in terms of the coefficients (the numbers multiplied to variables) and the variables.

Take the first example I posted:

3x + 9xy

First, look to the coefficients.

3 and 9 share what factors?

3 -> 1 and 3
9 -> 1, 3, and 9

So 3 is the highest factor shared by both.

Now we look at the variables:

x -> only one x, so just x as a factor
xy -> x and y as factors

Both share an x, so that's the highest factor.

Now we pull 3 and x out from both:

3x / 3x = 1 and 9xy / 3x = 3y

Place the GCF outside of the set of parentheses -> 3x(1 + 3y)

Thus we've found the GCF and factored it out of the expression.

---

Now, when we have a trinomial (three terms) of degree 2 (highest power is 2), the process is a little different. Let's look at the second example I gave:

x^2 + 4x - 60

You almost always factor quadratic expressions like this in this style:
(x + __)(x + __)
The way we fill in the blanks is as such...

We want to find two numbers that add up to the middle coefficient and that multiply to give the rightmost term

x^2 + 4x - 60

I like to start with the latter: What numbers multiply to be -60? (i.e. what are the factors of -60?)

1*60, 2*30, 3*20, 4*15, 5*12, and 6*10 (with one of the numbers being negative in each pair to get -60 instead of pos. 60)

Now which of those pairs can I add/subtract to get 4?

60 +/- 1 isn't = 4 and neither is the second, third, fourth, or fifth pair...

But... (6, 10) works if I do it like this: 10 + -6 = 4 and 10 * -6 = -60

So now we take those two numbers and place them into the sets of parenthesis:

(x + 10)(x + -6) or (x - 6)

If you expand that out, it should give you the original polynomial.

(x + 10)(x - 6) --> x^2 + 10x - 6x - 60 --> x^2 + 4x - 60

It takes practice, but you get better at it by doing. Eventually you just look at expressions and you know the factors you'll need almost intuitively.

---

If none of that was helpful let me know and I'll provide better examples.

Decus - Dreamweaver

Hi mathletes! Been busy with stuff so here's my very late challenges for you this week

As usual, one is a toughie, one is a lightweight, and one is of average difficulty! Enjoy

Primero:

My friend's music fan website started out fast: lots of visitors the first week. But every week, there were fewer and fewer; the second week there was one more visitor than 2/3 of the first week, the 3rd week there were 2 more than 2/3 of the second week, the 4th week there were 3 more than 2/3 of the 3rd week, and so on, until the final week, when there were just 533. How many weeks was the website up, how many visitors did it get the 1st week, and how many total visitors were there?

Segundo:

Can you find a triangle, whose sides are all equal to their opposite angles, in degrees? Prove your answer is unique or prove it's impossible. (Finding the triangle is pretty trivial, fyi)

Tercero:

Find the smallest positive integer n such that if n squares of a 1000x1000 chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.

As usual, all from contest materials... >.>;;

Tally ho!

Mitachi - Dreamweaver

I wrote a trigonometry and geometry exam 3 weeks ago, it was ok except it was a governmental paper. This is why i hated that paper, the questions were weird, most of the time you didn't know what they wanted you to do, there were alot of grammar problems, spelling errors, it was terrible. I passed but i hate just passing in math, i want to do well in it. b:angry

Yorkin - Dreamweaver

I think you should've made a quest line like this one, Dec. I would make PWI challenging again

Seamen - Dreamweaver

Math problem now!!!!!

Decus - Dreamweaver

Seamen - Dreamweaver wrote: »

Math problem now!!!!!

There are three perfectly good ones at the top of this page...

Seamen - Dreamweaver

Sego problem should be an equilateral triangle each side 60 degrees. 60 * 3 is 180 degrees. Too hungry to soothe other ones will do em later

Q____Q - Dreamweaver

Decus - Dreamweaver wrote: »

There are three perfectly good ones at the top of this page...

I looked at them and saw igloos.

Asterelle - Sanctuary_1381265973

Im in UR forum solvin your mathss

1) 10 weeks, with 19,677 first week
2) 60,60,60 triangle. Tried to use law of sines to prove it's only one but got bored
3) n = 1,999

Decus - Dreamweaver

Asterelle - Sanctuary wrote: »

Im in UR forum solvin your mathss

1) 10 weeks, with 19,677 first week
2) 60,60,60 triangle. Tried to use law of sines to prove it's only one but got bored
3) n = 1,999

Oh I always wondered when you'd venture in here. It is an honour to have you participate in my little thread

1) Correct, though you did not include the total, which is 58,100
2) Bingo! (Technically Seamen got there first, though neither of you proved it explicitly, but these problems are all too old for me to bother with details anyway.)
3) Right again!


It's nice to see activity here again, I'll find three more problems and post them later today.

Mitachi - Dreamweaver

Financial Math pl0x

Asterelle - Sanctuary_1381265973

I woke up this morning and realize that x / sin(x) is a monotonically increasing function.... at least in the range (0 ... 180)

That implies that if a/sin(a) = b/sin(b) then a = b
By law of sines a/sin(a) = b/sin(b) = c/sin(c) so then a = b = c.

Since a + b + c = 180 we have a = b = c = 60.