Need Math Help?

[Rici]

You should try drawing them all on your GR. It will help a lot.
This is what I got:

y=p means that p is a horizontal line. Now try to figure out for what values of p it has 4 solutions on the [0 , 2 pi] domain.

 

[Vermanubis]

There is a 25 character password on a computer. The characters can be any letter and any digit and the password is not case sensitive. You are given a hint about the password, the hint is "No character can be repeated in a row, for example F5EE has 2 E's in a row. However T6EO4E is fine as the 2 E's are not in a row".

How many possible combinations are there?

If we discount the hint, the answer is 36^25. With the hint, I dont know how to do it. Any help?

I could be totally, totally wrong, since I'm still new to combinatorics myself, but...

If sequential repetition is disallowed, then that's 36^2 = 1296 excluded combinations. From there, we wanna consider how many of the possible characters we're allowed to choose. Since it's 36 characters and we have 25 to choose, the formula for repetitive combinations would be (n+r-1)!/r!(n-1)! = (36+25-1)!/25!(36-1)! = 60!/875! - 1296. Just calculate all the possible combinations, then factor out the disallowed ones.

Again, could be horribly wrong though.

 

[Jim Morrison]

36 possible characters for each letter and a 25-character password with 60!/875! - 1296 possibilities would be a negative number. Your first calculation of 36^2 is already wrong, it needs to be 36^25 to start with.
Get out your combinatorics book again, Verm.

 

[Vermanubis]

36 possible characters for each letter and a 25-character password with 60!/875! - 1296 possibilities would be a negative number. Your first calculation of 36^2 is already wrong, it needs to be 36^25 to start with.
Get out your combinatorics book again, Verm.

I totally blew the formula up. I was trying to calculate a combination rather than permutation.

36^25 minus the number of possibilities that include sequential repetition, right?

 

[Jim Morrison]

What Rici said on the last page (or this one, dunno your pages) is the right way to calculate this.

 

[Vermanubis]

Rici, would you mind explaining how you calculated the restriction in the permutation?

 

[Rici]

I didn't calculate anything really. I don't know anything about statistics yet, but I tried to look at it logically. The only restriction is that no character can be repeated right after the original one.

So the first character has 36 possibilities. Then, because the second character has to be something different than the first one, it has 35 possibilities. And the third character has to be something else than the second one so, again, it has 35 possibilities. And the fourth has to be different than the third. This goes on till the end, meaning you'll get 36 * 35^24 if you have 25 characters.

 

[Corpsecreate]

Yeah I think that is correct. Nice thinking there Rici!

 

[#HBC | Gorf]

Just checked my grades and saw my grade in this class. Never done this good in class

Course:

Advanced Placement Calculus AB

Teacher:

Gordito's teacher


Student:

Gordito


Assignments


1

Quiz #15: Derivative of the natural log function

Oct 31

A


2

Quiz #16: Derivatives of exponential and logarithmic functions

Nov 2

A


3

Quiz #17: Derivatives of Inverse Trig Functions

Nov 4

A


4

Chapter 4 Test (3 Grades)

Nov 8

A


5

Quiz #18: Function Analysis Part One

Nov 15

A


6

Quiz #19: Function Analysis part 1

Nov 17


B


7

Homework #1

Nov 17

A


GPA

3.89

A

But now we're getting ready to learn about integrals, so question: Is that crap hard? If so, can I get a briefing of it? I'm expecting to be out of school next class and I don't wanna miss the lesson :x

 

[Corpsecreate]

Integration is the opposite of differentiation and is a little harder. It can get very complicated but you wont be given anything super complicated.

Integration gives you the area under a function instead of the Slope like differentiation gives.

Example:

y = x^2

The derivative of x^2 is 2x. The Integral of 2x is x^2 + C. The C is a constant and we have to add this because it will get lost with differentation. For example:

y = x^2 + 3
y = x^2 - 12

The derivative of both of these is 2x because the constant at the end becomes 0. So, whenever you integrate, you dont know what constant was there to begin with so you must always have a "+ C" at the end.

 

[#HBC | Gorf]

So I was already performing integrals in my head. Coolio.

So when y = 2x + 3, integral of that is y = x^2 + 3x + C?

Alright coolio, that's probably all that we're gonna get in the lesson if I know my teacher (plus things like radicals and trig and general expansion of that concept lol).

 

[Rici]

I dont know what kind of level youre being taught at, but it can get pretty damn hard later on, when you get sin- and tan-subtistutions and stuff.

But yeah, all in all it's just knowing the basic rules and how to apply them.

 

[Elyssa Xey Hexen]

Seems like your doing very well at the basic intro. level into calculus.

So here's integration in a theoretical nutshell. You are going to describe something (like the area of a rectangle). The area of a rectangle is length multiplied by its width--or on a coordinate axis, its the "change in x" multiplied by the "change in y". Now, say you sum up any number of these rectangles. You will come up with some larger overall area.

These rectangles can estimate the area of some awkward shape if they become small enough to reduce the amount of error. So, we say that as the area of any one rectangle becomes infinitely small and the amount of rectangles we add goes to infinity, we will approximate the area of the awkward shape.

That whole process is simplified into the expression of an integral. There are simple integral rules like in doing derivatives, but there are also many integrals you cannot do without alternative methods such as U substitution, integration by parts, trigonometric substitution, changing coordinate systems, etc. Even then, there are some integrals that simply cannot be evaluated.

Overall, you will simply be expected to understand the basic ideas behind integration and to memorize some simple integral rules. The rest is getting the skill to recognize what you have to do in order "take" the integral your given.

 

[Hot_ArmS]

Oh man I remember when I was taking calc, tapion was the one who explained integrals and rectangles and riemanns sum to me, calc class was awesome

 

[Kason Birdman]

Heyro. Got some physics stuff on light and waves n' ****.

Two sources are vibrating in phase, and set up waves in a ripple tank. A point P on the second nodal line is 12.0 cm from source A and 20.0 cm from source B. When the sources are started it takes 2.0 seconds for the first wave to reach the edge of the tank, 30 cm from the source. Find the velocity, wavelength and frequency of the wave.

Whatta ya think?

 

[AlphaDelfa]

hey, got a few questions from physics
first off,

a camera lens (n=1.52) is coated with a film of magnesium fluoride (n=1.25). What should be the least thickness of the film be to minimize reflected light with a wavelength of 550 nm?

and second question

A thin film of acetone (n=1.25) is floated on a thick glass plate (n=1.50). Plane light waves of variable wavelength are incident normal to the film. when one views the reflected light it is noted that complete destructive interference occurs at 600 nm and constructive interference at 700 nm. Calculate the thickness of the film.

Any help would be appreciated thanks

 

[Corpsecreate]

You can write this to say that A is less than B, B is less than C and C is less than D:

A < B < C < D

Is this:

A < B = C < D

A correct way of writing "A is less than B and C, B and C are equal, B and C is less than D"?

 

[Hot_ArmS]

That's correct

 

[Corpsecreate]

I got this problem here, I know the answer but I dont know the way to get to the solution.

In a class of 15 students, the teacher thinks of a certain number. In the class, there are 15 chairs arranged in order that are labeled from 1 to 15. The teacher tells the students to sit on any chair they want. If a student is sitting on a chair that is a factor of the number that the teacher thought of, then they get a prize.

13 Students received a prize and the 2 students that did not receive a prize were sitting next to each other.

a) Which chairs are the students who did not receive a prize sitting on?
b) What is the number the teacher thought of?

 

[Gates]

The teacher thought of the number 2520 and the students who did not receive prizes sat in chairs 11 and 13. All integers less than 15 are factors of 2520 except for 11 and 13.

[collapse="Proof by Brute Force"]2520/1=2520
2520/2=1260
2520/3=840
2520/4=630
2520/5=504
2520/6=420
2520/7=360
2520/8=315
2520/9=280
2520/10=252
11 is not a factor
2520/12=210
13 is not a factor
2520/14=180
2520/15=168
[/collapse]

 

[Corpsecreate]

I know the answer, but that is incorrect. 11 and 13 are not next to each other. I solved it by a brute force method also.

Your answer can be obtained reasonably easy if you realise that 840 is the smallest number that has the factors 1-8 and and the remainder when divided by 9 is 1/3 which can be removed by multiplying by 3 giving you 2520. This is not the correct answer though as the order of non-factors is important.

 

[Gates]

Oh, I misread and didn't see the part where the two vacant seats had to be next to eachother.

How about 60060? It's divisible by every number from 1 to 15 except 8 and 9 so the two students would be next to eachother.

It seems like the easiest way to prove that the number in question is 60060 would be to say that since the two numbers are consecutive, both n and n+1 must divide 60060, meaning in turn that n(n+1)|60060 (where n is an integer and 0<n<16). Then you could skip the non-obvious factors (1 through 6) and start with n=7, which would tell you that 8 does not divide 60060. If you skip 8 and go to n=9, you'll find that 9 also does not divide 60060.

But the hard part of the problem, which is what I'm guessing you're having trouble with, is how to get to 60060 in the first place. The key here in finding a number (we'll call it x) is that there are two consecutive integers that are not factors of x. My guess is that one way (probably not the ideal way) to show this is to just list the multiples of each number less than 15 in the set {1, 2,..., 15} and try to find where a subset of it that has more than one element and then claim that it must be a factor. Then you can cross off numbers that have one element that are still factors of x and nonconsecutive numbers. You should be left with x=60060 and 8 and 9 as your missing elements.

[collapse="Sample Proof"]Let A={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Let B_n be a subset of A that contains all elements in A divisible by n.
If |B_n|>1, n|x

n=1
B_1=A
|B_1|=15
Therefore 1|x (no ****)

n=2
B_2={2, 4, 6, 8, 10, 12, 14}
|B_2|=7
So 2|x

n=3
B_3={3, 6, 9, 12, 15}
|B_3|=5
So 3|x

n=4
B_4={4, 8, 12}
|B_4|=3
So 4|x

n=5
B_5={5, 10, 15}
|B_5|=3
So 5|x

n=6
Since 2|x and 3|x and gcd(2,3)=1, 2*3|x (i.e. 6|x)

n=7
B_7={7, 14}
|B_7|=2
So 7|x

n=8
B_8={8}
|B_8|=1
So 8 is not necessarily a factor.

n=9
B_9={9}
|B_9|=1
So 9 is not necessarily a factor.

n=10
Since 2|x and 5|x and gcd(2,5)=1, 2*5|x (i.e. 10|x)

n=11
B_11={11}
|B_11|=1
So 11 is not necessarily a factor.

n=12
Since 3|x and 4|x and gcd(3,4)=1, 3*4|x (i.e. 12|x)

n=13
B_13={13}
|B_13|=1
So 13 is not necessarily a factor.

n=14
Since 2|x and 7|x and gcd(2,7)=1, 2*7|x (i.e. 14|x)

n=15
Since 3|x and 5|x and gcd(3,5)=1, 3*5|x (i.e. 15|x)

The remaining numbers that may or may not be factors are 8, 9, 11, and 13. Of these, only 8 and 9 are consecutive, meaning that 11 and 13 are both factors (11|x and 13|x).

Therefore, the smallest value of x=7!*11*13/2=60060
The number that the teacher thought of is 60060 and the two students who did not receive the prize are sitting in chairs number 8 and 9.
QED[/collapse]This is the most elegant way I can think of to prove this. This problem is kind of beastly and nothing but variations on the brute force method come to mind.

(Also, if you knew the number in the first place but just didn't know how to get there, why not just tell us the number? It would have saved me a lot of frivolous calculations...)

 

[Hot_ArmS]

Gates I didn't know you were so smart

 

[Corpsecreate]

Yeah the problem is getting to 60060 in the first place (thats the same answer I got). If you wanted the answer from me you could have just asked. I didn't want to tell you the answer because it would influence the way you think about the problem.

Oh and x=7!*11*13/2 is 360360 not 60060.

 

[Gates]

Yeah the problem is getting to 60060 in the first place (thats the same answer I got). If you wanted the answer from me you could have just asked. I didn't want to tell you the answer because it would influence the way you think about the problem.

Well it's important for me to know if you're having trouble understanding what the answer is in the first place or if you're just having trouble showing how you got to the solution in an elegant way. The explanation I'll end up giving you is the same either way but I can explain it to you faster if you give me the answer because it's faster to reverse-engineer the problem than to do a ton of work myself.

Haven't you ever gotten stuck on a problem and then looked at the back of the book to reverse-engineer it before? lol

Oh and x=7!*11*13/2 is 360360 not 60060.

You're right, it should be x=7!*11*13/12. My bad, I didn't check my work.

 

[Corpsecreate]

Haven't you ever gotten stuck on a problem and then looked at the back of the book to reverse-engineer it before?

I do it all the time but I always try to do the problem without knowing the answer first.

 

[Kal]

Well, this is exciting. I wish I had known about this thread a long time ago. I'm currently studying first-year cryptography on my own. Is anyone familiar with the subject?

 

[Teczer0]

Anyone here good with Linear Algebra by any chance?

More specifically inverse problems and constraints.

 

[Kal]

By "inverse problems" do you mean finding inverse matrices? If so, I can help you out.

 

[Teczer0]

Nah, I don't mean inverting matrices haha.

I'm not very familiar with the problem myself so I'm probably being really unclear about it.

Its more about finding an optimum solution or something, I need to read more into the problem haha.

 

[Kal]

>_>

Well, if you post the problem, I'll do what I can to help. I'm sure, if it's outside my skillset, someone will be able to lend a hand.

 

[Teczer0]

Well, its related to optimization with constraints, linear programming and Second order cone programming.

Something like that I suppose haha, this makes me wish I took more Linear Algebra classes >_>

 

[Hot_ArmS]

I didn't take linear algebra but we did matrix optimization in another class of mine and solved them with matlab

 

[Teczer0]

Is there a simple command to do that?

My entire research is based on matlab so if you know that would help me out just so I can keep in mind.

 

[Kal]

I know there are a few free and open-source convex optimization programs that can solve SOCPs. I'm not very familiar with matlab so I can't help you there, but I found this.

SOCP problem can be solved in MATLAB using the interior-point algorithm implemented in the fmincon function. The SOC constraint can be defined as the non-linear constraint.

 

[Teczer0]

Well at least that solves one of my problems haha thanks

 

[Hot_ArmS]

Yeah we used fmincon, I'll dig up a PDF of what we did in my class and post it here tomorrow

 

[Kole]

EDIT: Never mind, figured it out.

 

[VictoryIsMudkipz]

Might be back here later, since I got Algebra Finals next week before Summer School is out,currently going over graphing ordered pairs to create Triangles, then finding the distace & slope, & point slope form equations
I swear math is giving me cancer

 

[Kal]

Math is such a wonderful subject. I wish this thread were more active. So, in the interest of some activity, I'm going to post the following fun-ish question.

There is a five-digit number A. With a 1 after it, it is three times as large as with a 1 before it. What is the number?

If you post an answer, include a collapse tag to avoid spoiling it.