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The Math Help Thread
- Thread starter -COOLIO-
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Necromanti
Member
I think I fixed the missing parenthesis.
Does this look okay?
B = [1-(1-(p/(k/n)))^2]^k
Does this look okay?
http://www.wolframalpha.com/rayner said:
cpp_is_king
Member
rayner said:
Just to ahow the method:
(3^2^50) mod 7
(3^2 mod 7)^50 mod 7
2^ 50 mod 7
(2^5 mod 7)^10 mod 7
4^10 mod 7
2^5 mod 7
32 mod 7
4
nincompoop
Banned
It's easier to find a power which is congruent to 1 mod 7 first:cpp_is_king said:
3^2 = 2 mod 7
3^6 = (3^2)^3 = 2^3 mod 7 = 1 mod 7
3^96 = (3^6)^16 = 1^16 mod 7 = 1 mod 7
3^100 = (3^96)*(3^4) = 1*3^4 mod 7 = 81 mod 7 = 4.
cpp_is_king
Member
cashman said:
Derivatives and gradients are analagous, so think about what a 0 derivative means and extend that by an extra dimension.
Actually a derivative is really a gradient with just 1 component instead of 2
Al-ibn Kermit
Junior Member
I'm not sure what I did wrong but I got marked down for this problem on my general physics midterm. I asked the TA and he came to the same answer as me. Can anybody else see what they get for this:
My answer was
My answer was
1.1762*10^6 m^2 or which rounded to 1.2 square kilometers.
Total power needed = 50000*10000*.1 = 50 x 10^6 watts/dayAl-ibn Kermit said:
Power produced = 170 * 6 = 1020 watts/day/m^2
Divide to get 49019.6078 m^2.
I played with the numbers a bit and couldn't figure out how you got your answer, so I feel like I might be missing something.
Edit: Actually, your answer seems to be 24 times mine, but I don't see a reason to convert days to hours anywhere in the problem.
Al-ibn Kermit
Junior Member
Therion said:
A watt means 1 joule per second. The way I did it was:
Energy needed: (50,000 people)(10%)(10,000watts)(24hours)(60minutes)(60seconds)= 4,320,000,000,000 joules/day or 4.32*10^12 joules per day
Energy per meter: (170 watts)(6 hours)(60minutes)(60 seconds)= 3,672,000 joules/square meter or 3.672*10^6 joules per day
4,320,000,000,000/3,672,000= 1,176,470.59 square meters of PV cells which I rounded to 1.2 square kiometers.
I'm not sure why you'd convert to joules when you already have watts as a common unit. But anyway your units don't match.Al-ibn Kermit said:
Energy needed: (50,000 people)(10%)(10,000 watts/person/day)(24 hours/day)(60 minutes/hour)(60 seconds/minute)(1 joule/second/watt)= 4.32*10^12 joules/day^2
Energy per meter: (170 watts/hour/m^2)(6 hours/day)(60 minutes/hour)(60 seconds/minute)(1 joule/second/watt)= 3.672*10^6 joules/hour/day/m^2
Divide and your units are m^2 hours/day, so you need to multiply by (1/24 day/hour) to correct it.
Hey everyone, I'm having trouble with this calculus problem. I need to prove that the summation of x as i starts at one and ends at n is equal to n(n+1)/2. I tried looking up how to do it and found basic algebra and induction neither of which I'm suppose to use.
I found a page in my book that has some info but I'm not quite sure what to do.
http://i.imgur.com/TMjoY.jpg
#5 is the theorem i need to prove.
I found a page in my book that has some info but I'm not quite sure what to do.
http://i.imgur.com/TMjoY.jpg
#5 is the theorem i need to prove.
cpp_is_king
Member
Exuro said:
Method 1: write the summation down in order
1 + 2 + ... + n
Rearrange terms
(1+n) + (2 + n-1) + ...
Notice that every single parens adds up to n+1. How many of these groups are there? N/2.
Some profs wont accept that method though.
Method 2: induction
P(1) : 1 = 1*(1+1)/2
Assume P
is trueP(n+1): (n+1)(n+2)/2 = n(n+1)/2 + 2(n+1)/2 = P
+ n + 1But P
= 1+2+...n, so P(n+1) = 1+2+...+n+1I have no idea how you're intended to do this, but here is a way I came up with that uses calculus.Exuro said:
Done Riemann sums? Because that sum is equivalent to the right Riemann sum from 0 to n of f(x)=x partitioned at each integer. That Riemann sum is an overestimate of the integral of x from 0 to n (which is n^2/2). It's also an underestimate of the integral of x+1 from 0 to n (which is (n^2+2n)/2). In fact, the magnitude of error for the two Riemann sums is equal, so the sum you want is the average of them: (n^2+n)/2.
Does that fit at all with where you are in class?
@cpp_is_king: He said he can't use either of those methods.
User Name Here
Member
This site might be of some use: http://www.brightstorm.com/math/
So gaf can anyone help me with some easy algebra 2 work? I need to factor this problem completely.
27x^4y^3-18x^3y^2+9x^2y
I keep screwing this one up. I'm pretty sure I am missing something that is simple.
Edit: I solved if lol so easy I don't know why I didn't see this yesterday.
27x^4y^3-18x^3y^2+9x^2y
I keep screwing this one up. I'm pretty sure I am missing something that is simple.
Edit: I solved if lol so easy I don't know why I didn't see this yesterday.
Rich Uncle Skeleton
Member
There's something weird about your method since you're using the integral of x, and this is usually found in calculus classes when they're first learning about the definition of the Riemann integral -- using the exact summation formula he's trying to prove. So it'd be circular.Therion said:
Exuro, why were those restrictions put on how you should prove it? I could understand not wanting calc students to use induction since most of them haven't properly learned it yet, but what's wrong with the simple algebraic proof?
Rich Uncle Skeleton
Member
Ah OK, I see what you're saying now. I was thrown when you said it used calculus; it doesn't really use any calculus, although it is tangentially related to it.Therion said:
Yes, I should have just called it a proof that might be fitting for a calculus class. To be honest, when I wrote "using calculus" I hadn't yet thought through the proof any further than the words "Riemann sum."Rich Uncle Skeleton said:Ah OK, I see what you're saying now. I was thrown when you said it used calculus; it doesn't really use any calculus, although it is tangentially related to it.
CoffeeJanitor
Member
It's the hardest, though. The first time I dealt with that was in Calc 1...Kind of a kick in the ass.Therion said:
cpp_is_king
Member
CoffeeJanitor said:
Everyone's different, to me its the easiest. I hate numbers, but i got my degree in math
Genesis Knight
Member
Hey guys, any idea of how to start proving that the limit of (1 / (x-1)^2) is infinity as x goes to 1? With the Reciprocal Rule?
Ah, Goya, clever. Thanks!
Ah, Goya, clever. Thanks!
dinazimmerman
Incurious Bastard
Genesis Knight said:
The limit of 1/(x-1) as x goes to 1 is positive infinity, and the limit of a product is the product of the limits.
harriet the spy
Member
cpp_is_king said:Method 1: write the summation down in order
1 + 2 + ... + n
Rearrange terms
(1+n) + (2 + n-1) + ...
Notice that every single parens adds up to n+1. How many of these groups are there? N/2.
Some profs wont accept that method though.
Method 2: induction
P(1) : 1 = 1*(1+1)/2
Assume P
P(n+1): (n+1)(n+2)/2 = n(n+1)/2 + 2(n+1)/2 = P
But P
One other way, which is very similar to your first proof, but both cleaner (what if N/2 is not integer, what does it mean to have a half group?) and completely rigorous, is to add the sum in reverse order to itself. In math terms
S= \sum_{1 <= i <= n} i
S is also equal (easy change of variable) to
S= \sum_{1<= i <= n} (n+1-i)
Therefore 2 S = \sum_{1 <= i <= n} (i +n+1 -i) = \sum_{1 <= i <= n} n+1 = n(n+1)
Thus S=n(n+1)/2
Pictorially (it was a proof that Gauss found when he was four or something)
S = 1 + 2 + 3 ... + n-1 +n
S= n + n-1 + n-2 ... + 2 + 1
2 S =(n+1) + (n+1) .. +(n+1)
Goya said:
Technically, the limit of 1/(x-1) when x goes to 1 depends on which side you approach from - it's either + or - infinity. The square makes the limit well defined to + infinity from whichever side you approach 1.
dinazimmerman
Incurious Bastard
harriet the spy said:
Right! My mistake. =P
Infinity isn't a real number, but it's a valid number on the extended real number line. And if it's implicit we're talking about extended reals, saying the limit of function at a point equals (+/-) infinity makes sense. But yeah, even if we're talking about extended reals, the limit of 1/(x-1) at x=1 doesn't exist, only the right and left limits do.
So to fix my method, you can say that the right (left) limit of 1/(x-1) is + (-) infinity, the right (left) limit of a product is the product of right (left) limits, and (-infinity) * (-infinity) = (+infinity) and (+infinity) * (+infinity) = (+infinity). Since the right and left limits both equal +infinity, the limit must be equal to +infinity. This is kinda pedantic, but it's right.
dinazimmerman
Incurious Bastard
Lonely1 said:Read my edit.. I don't like using the extended real number line since now many of the (mostly algebraic) properties we assume for real numbers are now broken. And fixing them requires us not to use the new reals for anything.
I still don't get what's wrong with extended reals. I find them pretty useful. Read my edit. =P
dinazimmerman
Incurious Bastard
Lonely1 said:
It does have a meaning:
By the fourth line -∞*-∞ = +∞.
The extended reals don't form a field, but they have enough algebraic properties to make them useful.
Goya said:
K.
0=∞-∞=(1+∞
-(∞=1.0=1?
1=∞/∞=(2*∞
/∞=2.1=2?
No, you can assign those properties as long as it doesn't involve ∞+∞ or ∞*∞. They fundamentally breaks the real numbers. So, (-∞*-∞
doesn't has a meaning and can't be defined as you state.dinazimmerman
Incurious Bastard
Lonely1 said:K.
0=∞-∞=(1+∞
0=1?
1=∞/∞=(2*∞
1=2?
No, you can assign those properties as long as it doesn't involve ∞+∞ or ∞*∞. They fundamentally breaks the real numbers. So, (-∞*-∞
Yes, "∞-∞", "0 * (±∞
" and "±∞/±∞" are all meaningless. So? How does that in any way obviate the possibility of assigning a meaning to "-∞ * -∞." I'm not understanding your point. Extending the usual arithmetic operations of R this way is not controversial.The problem if that you are using them to give limits to a function that doesn't has one. There's no way to make f(x)=1/(x-1) a continuous function in 1 using the regular topology, even if you include -∞ and ∞.
ANother example:
Let f:R->R be this function:
f(x) = 1 if x is in Q,
-1 otherwise.
What's the limit of f(x), as x approaches to 1?
Is Answer, is impossible to make this function a continuous one. Is about the same as the other function,and I don't even have to include infinity.
ANother example:
Let f:R->R be this function:
f(x) = 1 if x is in Q,
-1 otherwise.
What's the limit of f(x), as x approaches to 1?
Is Answer, is impossible to make this function a continuous one. Is about the same as the other function,and I don't even have to include infinity.
Lonely1 said:
True the extended real number line doesn't constitute a field anymore, making it not as useful for proofs, but the simplifications it offers are useful as it doesn't lead to inconsistencies. On the proof side, the Hyperreals/Nonstandard Set Theory is, personally speaking, an awesome setting for doing proofs in.
Lonely1 said:
I'm not sure where this ever happens. Do you mean just saying the limit is infinity instead of the limit diverges?
dinazimmerman
Incurious Bastard
This is not an example of saying the limit of a function exists at a point where intuitively it shouldn't. f(x) = 2 * I_Q(x) - 1, where I_Q is the indicator function for the rational numbers, so the same arguments used to argue that I_Q has no limit at any point and is thus nowhere continuous apply also to f. And? I'm pretty sure this is still true when f is defined on the extended real number line.Lonely1 said:
Interestingly, on the real projective line, the limit of 1/x at 0 is ∞ and 1/x is even continuous at 0 if we modify some definitions a bit.
CoffeeJanitor
Member
Hey guys
I am currently doing a lab for physics, and there is a constant we're using for the class (diffusion coefficient). My professor wants us to write down the functional form for this constant...Is a functional form something like
y = mx+b
or
y = ax^2+bx+c
Or do you just say it has a linear. exponential functional form etc. The constant I'm dealing with here seems have variables in both the numerator and the denominator (temperature in numberator, viscosity and radius in denom), so what kind of functional form would that mean?
I am pretty sure he wants temp to be the variable here, as viscosity, radius etc are parameters. Would that mean that this has a linear functional form?
I am currently doing a lab for physics, and there is a constant we're using for the class (diffusion coefficient). My professor wants us to write down the functional form for this constant...Is a functional form something like
y = mx+b
or
y = ax^2+bx+c
Or do you just say it has a linear. exponential functional form etc. The constant I'm dealing with here seems have variables in both the numerator and the denominator (temperature in numberator, viscosity and radius in denom), so what kind of functional form would that mean?
I am pretty sure he wants temp to be the variable here, as viscosity, radius etc are parameters. Would that mean that this has a linear functional form?
dinazimmerman
Incurious Bastard
I think I'm being stupid, but I can't see an easy way to solve this problem:
Anyone have an idea?
EDIT: nvm, figured it out.
Anyone have an idea?
EDIT: nvm, figured it out.
well, G = {a b c}, H1 = {a}, H2 = {b}, H3 = {c}, should do it, right?
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