SRK Math Thread

[Axl_m4ster]

It doesn’t look like you’ve taken complex variables.

Much like the integers are a subset of the rational numbers, and the rational numbers are a subset of the real numbers, so too are the real numbers a subset of the complex numbers. These are numbers in the form
a+ bi, where a and b are real numbers.

The truth is, you’ve been working with complex numbers your whole life. Except you have only been working with complex numbers in the form a + 0*i, where a is a real number.

The whole a+bi additive notation is a convenience. When you see a+bi, you are essentially talking about a coordinate (a,b). Since you’ve been only working with real numbers, you’ve been working with coordinates in the form (a,0) where again, a is a real number.

Think of the xy-plane. You have an x-axis and a y-axis. Yet you cannot solve an equation as simple as x^2 = -1. There is no real number that satisfies said equation. If however, we accept sqrt(-1) = i as a solution, then this opens up the possibility of solving other equations with numbers that have a negative number inside a square root.

I won’t derive the complex number system here, but it can be shown that the sum of two complex numbers and the product of two complex numbers (where multiplication is defined as (a+bi) (c+di) = (ac-bd) + (bc+ad)i) ) always spits out a complex number. You should plug in b = d = 0 to see that the usual multiplication works with this definition. You can also go here for more info

http://en.wikipedia.org/wiki/Complex_numbers

Anyways, with this system in mind, we can now think about the complex plane. Think of the same xy-plane you’ve always used, except now, ONLY the x-axis is composed of every single real number you’ve ever used. The y-axis does NOT have any real number (although again, points ON this y-axis are points whose coordinates are themselves real numbers) (and I suppose since the x- and y-axes intersect at 0, I supopse they have a common point there)Much like in the real xy-plane, when you are on the x-axis, your y-coordinate is 0 (that is, you deal with points of the form (a,0), where a is a real number ) so too, when you are on the complex plane, and you are on the x-axis, you are dealing with points in the form (a,0). Every single point on this line is a real number. Notice your y-coordinate has always been zero. This is because again, you only had need of real numbers. But now you have discovered complex numbers and so you need to consider what something like (1,1) means in the complex plane. Well, it means: go right one, and up one. This is the complex number 1+1*i, or more colloquially, 1+ i.

the e^i(theta) notation is a representation of a coordinate in polar form. It’s like taking two different routes to get to the same point (-1,0). In rectangular coordinates, you start at the origin, walk one unit to the left, and you’re at (-1,0). But with polar form, you declare points by rotation of a radius. So your point on a graph is contingent on a radius r, and an angle of rotation theta. So, rather than thinking in terms of rectangles, you look at points in terms of where they are on a circle or radius r. Since we’re merely looking at coordinates of thr form (a,b) = (rsin(theta),r cos(theta)) , we use the additive notation and write this point, which lies on said circle of radius 1, as cos(theta) + isin(theta). It can be shown that

e^i(theta) = cos(theta) + isin(theta), so this can be taken as truth.

So basically, this whole notation is a fancy way of saying this:
You can take the polar route to the point (-1,0) as follows: Form a circle of radius 1 at the origin (ie, the point (0,0) ), go out the length of the radius (which is one) and stay right there, then rotate said radius eminating from the origin 180 degrees counterclockwise (i.e. Pi). The point I end up at as I hang onto the very radius that is being rotated is the point (-1,0) in the complex plane, which, informally speaking, can be likened to the point (-1,0) in the real plane.

or to put it another way:

e^(i*pi) = cos(pi) + isin(pi) = -1 + 0i = -1.

From algebra, we then get the well-known identity

e^(i*pi) + 1 = 0.

Exactly!
Exactly what he said.

my brain just exploded

[Warrior_s_Dreams]

As an added bonus:

Notice the definition of multiplication in the complex plane

(a+bi) (c+di) = (ac-bd) + (bc+ad)i.

Let

a = sin(theta1)
b = cos(theta1)
c = sin(theta2)
d = cos(theta2)

Then you get

(a+bi) (c+di) = ( sin(theta1) + icos(theta1) ) (( sin(theta2) + icos(theta2) )

= ( sin(theta1)sin(theta2) - cos(theta1)cos(theta2) ) + i* (cos(theta1)sin(theta2) + sin(theta1)cos(theta2) )

= cos(theta1 + theta2) + i*sin(theta1 + theta2)

which are the trig identities you should know from elementary trig.

[Jnana]

snip

I will pocket those 3 questions since they’re, at the moment, drawing me to even more questions (which is good). I’ll likely be popping back up here with more specific and basic questions starting with: Why does 3 - (-2) = 5? School just tells us to change the (minus) to a + and switch the sign of the right hand term. However, I don’t get why that is accepted. The way I’m looking at it is, you have some quantity (3) of something and you’re diminishing its value or taking away from it by some other quantity (2). But it’s (-2) and I tend to look at (-#)s as…guh, this is kind of where it’s hard to explain. I would like to use something like debt or the example of throwing a ball up against gravity (taking the acceleration of gravity to be positive and the ball travelling at a negative velocity going to a positive velocity as it reaches its max height and back down), but even then I can’t put it into a clear general definition. Now I could suppose a negative quantity is something that is to be subtracted from a relevant quantity, but then that is putting the operation of subtraction into play when I’m only trying to understand the quantity itself. I try to think that (-#)s aren’t things that necessarily ‘exist’ (if that’s a safe word to use in this context), but rather just represent a type of exclusion to some extent. In that case something like 3 + (-2) makes sense, but when something like 3 - (-2) comes up, what exactly is going on for that to equal 5 and not 3 or 1? Maybe it’s the operations that I don’t get? idk. Help. o_O and examples please.

[pedoviejo]

I will pocket those 3 questions since they’re, at the moment, drawing me to even more questions (which is good). I’ll likely be popping back up here with more specific and basic questions starting with: Why does 3 - (-2) = 5? School just tells us to change the (minus) to a + and switch the sign of the right hand term. However, I don’t get why that is accepted. The way I’m looking at it is, you have some quantity (3) of something and you’re diminishing its value or taking away from it by some other quantity (2). But it’s (-2) and I tend to look at (-#)s as…guh, this is kind of where it’s hard to explain. I would like to use something like debt or the example of throwing a ball up against gravity (taking the acceleration of gravity to be positive and the ball travelling at a negative velocity going to a positive velocity as it reaches its max height and back down), but even then I can’t put it into a clear general definition. Now I could suppose a negative quantity is something that is to be subtracted from a relevant quantity, but then that is putting the operation of subtraction into play when I’m only trying to understand the quantity itself. I try to think that (-#)s aren’t things that necessarily ‘exist’ (if that’s a safe word to use in this context), but rather just represent a type of exclusion to some extent. In that case something like 3 + (-2) makes sense, but when something like 3 - (-2) comes up, what exactly is going on for that to equal 5 and not 3 or 1? Maybe it’s the operations that I don’t get? idk. Help. o_O and examples please.

I think of negative sign as a direction.

that is -3 = moving three units to the left on the number line.

the subtraction sign for me simply states, move x amount of numbers in the opposite direction, from whatever number you are moving away from.

so 5 - (-3) = 8

because I start at five, and then move in the opposite direction of the second operator (is that the right word?)

however, im most likely wrong, but that is how calculators work from my understanding.

[Toodles]

1. Why does 3 - (-2) = 5?

Debt example makes it easy.
Your cash in pocket is an asset; the more of it you have, the higher your net worth. If you add more cash, you increase your net worth, so you’re adding a positive thing. If I add $2 to your cash, your net worth goes up $2. 3+ (+2)=5, easy.
Your debt to the back is a liability; the more of it you have, the lower your net worth. If you add more debt, you decrease your net wroth, so you’re adding (!) a negative thing. If you break my lamp and tell me you’ll pay me $4 for it later, your net worth has just gone down $4. 5 + (-4) = 1.
You use a dollar out of your pocket to buy a soda. You’re subtracting (because you’re getting rid of something) a positive thing (because cash in pocket is a positive thing). 1-1=$0, you’re new net worth.

Just a recap before the final part. You had $3 cash, made an easy $2, and then owed me $4 for breaking my stuff. That’s cash, cash, and a debt, or a positive, positive, and a negative. All of that together is a sum, or addition.
(+3) + (+2) + (-4) = 1.
You then got rid of some of your cash, subtracting a positive $1 when you bought the soda.
(+3) + (+2) + (-4) - (+1) = 0. (You still have $4 in your pocket, which is the same amount you owe me, so your net worth is $0).
The point of this is to show how you can add a positive, add a negative, and subtract a positive. Last is subtracting a negative.

Now, your birthday comes up, and your present is I will forgive your debt because Im a cheap bastard. I’m taking away (subtracting) a debt ( a negative). I’m taking away a bad thing.
(+3) + (+2) + (-4) - (+1) - (-4) = 4
You still have that $4 in your pocket, but don’t owe me anymore, so your net worth is $4.

If instead I had just giving you $4 ( added +4) , and NOT forgiven the debt, and you turned around and gave me the money back to pay the debt you owe, you’d be in the exact same situation. That’s why subtracting -4 is the same thing as adding +4.

(I love euler’s formula. It’s so weird.)

[Warrior_s_Dreams]

Why does 3 - (-2) = 5?

There are a couple of ways to see this

1. Geometric

Forget that the symbols you see here are numbers for a second and consider them to be evenly spaced symbols of distance 1 unit apart. Start from the symbol labeled -2 and walk along the line until you get to the symbol 3. You will have traveled 5 units.

Distance between -2 and 3
|-------------------|

|—|---|—|---|—|
-2 -1 0 +1 +2 +3

2. Algebraic

So you’ve probably noticed that when you multiply a number by 1, you end up with that number when you’re done. Similarly, when you add 0 to something, you end up with that number when you’re done. A number with this property is called an identity with respect to that operation. There are two such operations: addition (where 0 is the additive identity) and multiplication (where 1 is the multiplicative identity).

You’ve probably also noticed that if you have a number, say 3, and subtract 3 from 3, you end up with 0, that is, you end up with the additive identity. A number that yields the additive identity 0 under addition is said to be the inverse of a number. So, the inverse of 3 is -3 under addition, since 3 - 3 = 0. Addition also has the property that I can rearrange the position of each of these numbers, and they will still give me the same sum, so this also means that the inverse of -3 is 3 since if I start with the fact that 3 - 3 = 0, and I use the axiom that I can add them in any order, then we have -3 + 3 = 0. In a sense then, there is no such thing as “subtraction,” rather, there is adding the additive inverse of a number to a different number, which need not be the inverse of said number. For example, the additive inverse of 2 is -2, yet I can add -2 to 3 to get 3 - 2 = 1. Since I did not get 0, -2 is not the additive inverse of 3.

Just as subtraction is merely adding inverses of elements to other elements, so to is there no such thing as “division,” but rather, multiplying a number by an multiplicative inverse. For instance, the multiplicative inverse of 2 is 1/2 and the multiplicative inverse of 1/2 is 2 because multiplying each of these numbers yields the multiplicative identity 1 since 2*(1/2) = (1/2)2 = 1. Similarly, the multiplicative inverse of 3 is 1/3, but 2(1/3) = 2/3 =/= 1. So 2 and 1/3 are not inverses of each other since their product does not yield 1, yet this operation is the same as the so called “division.”

So, with this is mind, what number when multiplied by -1 yields 1? It is -1, since (-1)*(-1) = 1. So we have an interesting property: a number that is its own inverse. This is why -1 1 = -1 and (-1)(-1) = 1.

One other thing: It doesnt matter which order we multiply things. So (235) = (23)5 = 2(35) This is called the associativity property of multiplication.

Knowing this, it becomes a routine calculation to determine the sum 3-(-2):

3-(-2)
= 3±(-2) Since I’m adding the inverse of the number 2 to a number called 3
= 3±(-1)2 Since -2 = -(12) = (-1)2
= 3+12 Since -(-1)2 = (-1)(-1)2 = 12
= 3+2 = 5.

[Nando]

hey guys

2 sats question
I can’t figure out how to start these problems
need help cuz the homework wasn’t worded like these are

1. A 2003 survey showed that 4.6 percent of the 250 Americans surveyed had suffered some kind of identity theft in the past 12 months. Construct a 98 percent confidence interval for the true proportion of Americans who had suffered identify theft in the past 12 months.

A random sample of monthly rent paid by 12 college seniors living off campus gave the results below (in dollars). Find a 99 percent confidence interval for μ, assuming that the sample is from a normal population. (35 points)

900 810 770 860 850 790
810 800 890 720 910 640

for 2, would it be
low 742.20
high 882.80?
( i just used mega stat and ran the numbers under descriptive stats… lol… i fucking love that shyt btw…

[Einlanzer]

i need some assistance with an induction proof.

“Prove by induction that n!<n^n for all integers n>1.”

it doesnt seem too difficult, but i dont quite understand how induction proofs work and the tutor is gone…

[Warrior_s_Dreams]

i need some assistance with an induction proof.

“Prove by induction that n!<n^n for all integers n>1.”

it doesnt seem too difficult, but i dont quite understand how induction proofs work and the tutor is gone…

Induction is a two-step process:

1. Show that this case is true for a base case. Since the inequality is definied for n>1, plug in n=2 in the inequality. If it is a true statement, go on to 2)

2. Induction step. Assume now, that the inequality k!< k^k is true.
   You must show that the inequality (k+1)!< (k+1)^(k+1) is also true.

To do this, you need to

1. use the inequality k!< k^k that you assumed to be true
2. use the fact that 1<… .< k-2< k-1< k < k+1
3. start the inequality off with k! < [what?]
4. HINT: How can I rewrite (k+1)! so that I can use 1)?

[Warrior_s_Dreams]

hey guys

2 sats question
I can’t figure out how to start these problems
need help cuz the homework wasn’t worded like these are

1. A 2003 survey showed that 4.6 percent of the 250 Americans surveyed had suffered some kind of identity theft in the past 12 months. Construct a 98 percent confidence interval for the true proportion of Americans who had suffered identify theft in the past 12 months.

A random sample of monthly rent paid by 12 college seniors living off campus gave the results below (in dollars). Find a 99 percent confidence interval for μ, assuming that the sample is from a normal population. (35 points)

900 810 770 860 850 790
810 800 890 720 910 640

for 2, would it be
low 742.20
high 882.80?
( i just used mega stat and ran the numbers under descriptive stats… lol… i fucking love that shyt btw…

If you just memorize examples, you won’t get anywhere, as you are experiencing, because you miss the entire point of the material. You need to be familiar with the theory BEHIND the material because life isn’t going to just say: Theres a problem. Here are the mean, SD, and use a Poisson distribution to figure out the rest.

For 1) Use facts about the binomial random variable to fill in the missing values for finding your CI.

For 2), you can deduce that yourself. Does the calculation match the work you did by hand?

[Nando]

yea
thanks

[Warrior_s_Dreams]

Normally I hear people complaining about math and how it’s hard, which is why it’s odd that I find somebody complaining about the people who complain about math being hard.

http://thebestpageintheuniverse.net/c.cgi?u=math

[xlxlxlxl]

(sqrt(cos(x))cos(400x)+sqrt(abs(x))-0.4)(4-xx)^0.1
google it

[xlxlxlxl]

ok today for the life of me i couldn’t solve this equation
i had to find the zeros to this equation using synthetic division going down the list of possible real numbers which were positive or negative 1,2,4,8
x^4 -2x^3 + x^2 +12x+8

i factored it down to
(x+1)(x^3 -3x^2 +4x+8)

i went down the rest of the list after 1 and could not find another zero to further factor it

Factored completely you get
(x+1)^2 *(x^2 -4x +8)

-1| 1 -3 4 8
…| 0 -1 4 -8

---

…1 -4 8 |0

You have 2 real zeros (-1 and 1) and 2 imaginary because the discriminant of x^2 -4x +8 is negative.

(x^3 -3x^2 +4x+8)

you split the polynomial in two so you have

x^3 -3x^2 and 4x+8, and then factor each one individually.

so x^3 -3x^2 = x^2(x-3) then set it to zero
and 4x+8 = (4x-8) then set it to zero

you can only do this if the first half and the second half are factorable. If they are not, then you are stuck there [(x^3 -3x^2 +4x+8)]

Polynomials do NOT work like that. You CANNOT split it up like that. If x^2 were a factor, there would be neither an x^1 term nor an x^0 term.

What is wrong with this site and editing posts?

[pedoviejo]

Factored completely you get
(x+1)^2 *(x^2 -4x +8)

-1| 1 -3 4 8
…| 0 -1 4 -8

---

…1 -4 8 |0

You have 2 real zeros (-1 and 1) and 2 imaginary because the discriminant of x^2 -4x +8 is negative.

Polynomials do NOT work like that. You CANNOT split it up like that. If x^2 were a factor, there would be neither an x^1 term nor an x^0 term.

What is wrong with this site and editing posts?

I remember that’s how the teacher in Intermediate thought us to do polynomials that looked similar to that. Where you would just split it in half like that. But what was it degree was it. Maybe you know what Im talking about. ill get back to you.

[Havatchu]

Simple rundown.

I was average at math until 7th grade when algebra kicked in. After that went to a Sylvan learning center in the summer, and it helped alot. By the time I was in 9th grade I was making alot more freinds and skating so of course I had to be the cool kid and just fuck off, talk the whole class and get C’s. By the time I started college I had to take remedial math classes which were basically the math classes I had fucked around in for so long. After that I did statistics last spring semester.

I like math, but I have to work on it for HOURS for me to be comfortable with the material. It’s the only subject where I blank out everything I learned if I’m not psychologically prepared for the test. I remember countless times I would learn something just fine, do the homework well, and when it came time to do it again just fucking forget everything.

[Davidstar]

(xlxlxlxl being cool guy)

[media=youtube]5VL_5MX7z94[/media]

[Squirtguns]

Yo, for the first time i’m really appreciative of my math background. After taking Quantum mechanics and Thermodynamics, and finding a real use for calculus, I finally feel like math has a real use besides figuring out how much to tip.

[Ephidel]

Which is more difficult Statistics or Trigonometry?

[ResolutePower]

Which is more difficult Statistics or Trigonometry?

Short answer: stats. But you really can’t even compare the two. You can learn trig in a single chapter of a high school math class. There isn’t a course that’s devoted strictly towards the topic. It’s simply a tool that you may use when solving other types of math problems. On the other hand, stats is an entire field of study.