smokeandmirrors
Banned
Good point, forgot about that. Thanks.
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The Math Help Thread
- Thread starter -COOLIO-
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Good point, forgot about that. Thanks.
GremlinFool
Member
Putting language in layman's term != good math writing. You sacrifice precision and clarity unless the language follows clearly and concisely from proofing conventions.
There's a learning curve involved in acclimating to mathematical writing, but it's a necessary skill to do any upper-level theoretical work.
Sine of Mathness!
i is complex
Treefingers
Member
This is true.
Re: spans, I don't quite remember how those work. It's been a while
If I get some spare time today I'll try to refresh my memory though.I understand, but I have a hard time learning most types of maths without some sort of basic idea first. If I can't put it in laymans terms then I generally don't grasp the concept.
Spanning means that the vectors you have allow you to get any other vector in your set. Think of a regular Cartesian graph: if you consider the x and y axis to be vectors(1,0) and (0,1), then using vector arithmetic allows you to get any vector in that 2d space.
For example, if you want to find (1,2) it's just (1,2) = 1(1,0) + 2(0,1)
This means that to span n dimensions, we need n linearly independent vectors(since if we had less than n, not every vector in our n dimension could be represented)
What this means is any vector in n space, can be written as a linear combination of the spanning vectors.
So to show if vectors span a set, all you need to do is check to make sure that you have enough linearly independent vectors, you just need n of them, n being the dimension of the space.
Here's two examples:
(2,3) and (1,0). We want to make sure that any vector (x,y) can be written as a linear combination of those vectors. ie (x,y) = c1(2,3) + c2(1,0). So we check if they are linearly independent.
a(2,3) + b(1,0) = (0,0)
so 2a + b = 0 and 3a = 0, gives a=b=0, so we see these are linearly independent. Since we have 2 linearly independent vectors, and are in a 2 dimension space, these do span.
Now consider (2,3), (1,0) and (4,6). Note that (2,3) and (4,6) are linearly dependent, since 2(2,3) = (4,6). But from the above example, we see we still have 2 linearly independent vectors, either (1,0) and (2,3) or (1,0) and (4,6), so this set still spans.
So say I choose a vector at random (5,9). I can show that this vector is a linear combination of our spanning vectors.
(5,9) = 3(2,3) - (1,0) + 0(4,6)
If you have any specific questions you need help with that would be easier, because I don't know how much you have covered.
Orbis Tabula
Member
Unless you guys are all really pale, I vote The Tan Gents. Although, maybe it would still be funny if you were all really pale.
Depends on what you know about cosets, but assuming you've proven that different cosets are disjoint, you can just look at where the identity is.
To prove it's a group, you have to consider all four group properties. (Closure, Associativity, identity and inverse) Since all cosets are distinct, meaning if aH and bH share an element, then aH = bH. This means that the only coset of H that is a group must be H itself, since that's the only way it will have the identity: H is a subgroup of G and therefore H must have the identity. (since any cosets with the identity are equal to each other, since they share an element)
Perfect Cha0s
Member
Hi guys, I have a differential equations problem that seems pretty simple, but looking at how the person's paper was graded that I'm comparing to, I'm not too sure if my approach is correct or not.
The problem is:
A bacteria colony grows at an exponential rate, doubling every 5 hours. If a culture starts with 23,000 bacteria, and removes bacteria at a constant 3000 bacteria an hour, how long until the culture is depleted of bacteria?
I started this by setting up a general form "dB/dt = rate in - rate out", where the B variable refers to bacteria, the variable t is time, in hours, the rate in is the exponential growth of the culture, and the rate out is the 3000 per hour out. Defining the rate out, is I think, easy - it would be 3000*t, right?
The rate in is where I think I may be having issues: since it doubles every 5 hours, I used a B(t)=B(initial)*e^kt, where:
B(t) = 2
B(initial) = 1
t = 5
k is the constant being solved for
Solving this, found that k = 0.1386, which I plugged back into the "rate in - rate out = dB/dt" equation from earlier, making
dB/dt = [23000*e^0.1386t] - 3000t
Multiply both sides by dt to separate variables, and integrate, yielding:
B = [(23000/0.1386)*e^0.1386t] - 1500t^2
However, when I graph this function, it never seems to go to 0. I'm sure this is a simple mistake on my behalf. Thanks guys.
EDIT: I think I just realized what I did wrong, which was pretty much all of it.
Thanks anyway guys.
The problem is:
A bacteria colony grows at an exponential rate, doubling every 5 hours. If a culture starts with 23,000 bacteria, and removes bacteria at a constant 3000 bacteria an hour, how long until the culture is depleted of bacteria?
I started this by setting up a general form "dB/dt = rate in - rate out", where the B variable refers to bacteria, the variable t is time, in hours, the rate in is the exponential growth of the culture, and the rate out is the 3000 per hour out. Defining the rate out, is I think, easy - it would be 3000*t, right?
The rate in is where I think I may be having issues: since it doubles every 5 hours, I used a B(t)=B(initial)*e^kt, where:
B(t) = 2
B(initial) = 1
t = 5
k is the constant being solved for
Solving this, found that k = 0.1386, which I plugged back into the "rate in - rate out = dB/dt" equation from earlier, making
dB/dt = [23000*e^0.1386t] - 3000t
Multiply both sides by dt to separate variables, and integrate, yielding:
B = [(23000/0.1386)*e^0.1386t] - 1500t^2
However, when I graph this function, it never seems to go to 0. I'm sure this is a simple mistake on my behalf. Thanks guys.
EDIT: I think I just realized what I did wrong, which was pretty much all of it.
Thanks anyway guys.Partial Gamification
Banned
In the initial conditions, you have t=5 and B(t)=2. Is the B(t)=2 just denoting the doubling every hour, or are you using that in your calculation?
For just the growth, B(5)= population*2^5
I think this is clear but just to state the question, its asking for t such that B(t) = 0
the problem might be with how you found k; but I'll give this a go in a little bit if no one else has stepped in to help us out.
Perfect Cha0s
Member
Yes, essentially I used that as a way of saying that the population was double at t = 5.
Also, I think one of my issues is that because it is already a derivative, that is, dB/dt, I should be using 2000 and not 2000t for my "rate out". I'm still wrapping my head around properly setting up my "rate in".
Partial Gamification
Banned
okay: "doubling every five hours," not every hour, I misread.
B(0)= 23,000
B(5) = 46,000 - 3,000*5 = 31,000
B(t) = B(0)*e^(k*t)
31,000 = 23,000*e^(k*5)
ln(31000/21000)/5 = k
k = 0.0597 (approx.)
edit: I'm hoping for confirmation/corrections from the wiser and less distracted members.
edit2: For your integration, its an indefinite integral right? Are you adding the constant (+C) and figuring this out given the initial conditions?
edit3: something is wrong with the setup... I get an increasing function.
DEAD RABBIT
Member
Nevermind. Misread the problem.
Partial Gamification
Banned
really what I did was have k -> (-1)*k {exponential decay}
edit: I have feeling that I have errored in the setup. * dB/dt *
Perfect Cha0s
Member
really what I did was have k -> (-1)*k {exponential decay}
edit: I have feeling that I have errored in the setup. * dB/dt *
My brain is fried - I need to head to bed.
If it's of any help, though, the instructor made a note that said
"dB/dt = kB - 2000"
on a student's paper that didn't get the problem right and the problem hints at modifying the standard dB/dt = kB model. But I'm off to sleep. I appreciate the help.
Partial Gamification
Banned
No worries, "you don't use it you lose it," and it been too long for me.
youngwerther
Banned
Find the derivative of the function using the definition of the derivative.
f(x)=(x^2-1)/(2x-3)
Question -
Why does my calculus teacher hate me?
I fully do not expect anyone to do that, but if anyone would like, I'm hopelessly stuck on the algebra for this. It's just a huge mess. Doing this a full week after he already taught us the differentiation formulas.
f(x)=(x^2-1)/(2x-3)
Question -
Why does my calculus teacher hate me?
I fully do not expect anyone to do that, but if anyone would like, I'm hopelessly stuck on the algebra for this. It's just a huge mess. Doing this a full week after he already taught us the differentiation formulas.
Partial Gamification
Banned
Note how I re-wrote g/h as g*h^(-1) to employ the product rule. You will likely be asked to use the Quotient Rule but sometimes the product rules is easier, or good for an answer verification.
youngwerther
Banned
Yeah but for this one I'm not allowed to use those. I'm supposed to solve it "using the definition of the derivative" which creates a huge mess of tedious algebra. No credit if I use product or quotient rule.
I'm just going to leave it blank.
I'm just going to leave it blank.
Partial Gamification
Banned
oh shoot, its just setting up the limits.
edit: I don't want this to be a crutch for you (i.e. don't rely on this for answers) but Wolfram Alpha might help make these problem less intimidating.
Partial Gamification
Banned
edit2: the fifth line is my brain just looping, it is not needed.
here is the setup, Y.W.
edit:
Most excellent, thanks for sharing!Great thread, I am in Pre-Cal(126) at my local Community College so this can come in handy. I am posting this also as a heads up for anyone who needs a graphing calculator in a pinch. It has been a godsend for me, and saves me a few bucks
Desmos
Here's my rough work. It was pretty standard algebra, where were you running into trouble?
The amount of water in a tank t minutes after it has started to drain is given by
W=200(t-13)^2
(a) At what rate is the water running out at the end of 9 minutes?
(b) What is the average rate of which the water flows out during the first 9 minutes?
W=200(t-13)^2
(a) At what rate is the water running out at the end of 9 minutes?
(b) What is the average rate of which the water flows out during the first 9 minutes?
youngwerther
Banned
I have no idea.
I set it up correctly, at least.
The derivative of cy^-6 would be c-6y^-7, right?
I set it up correctly, at least.
The derivative of cy^-6 would be c-6y^-7, right?
youngwerther
Banned
Differentiate the function
y=x^5/3 - x^2/3
I got (5/3)x^2/3 - (2/3)x^(-1/3)
that's different than what wolfram alpha gives.
I'm allowed to use the rules on this one.
y=x^5/3 - x^2/3
I got (5/3)x^2/3 - (2/3)x^(-1/3)
that's different than what wolfram alpha gives.
I'm allowed to use the rules on this one.
harriet the spy
Member
Wolfram alpha is showing off and factoring by x^(-1/3).
(5/3)x^2/3 - (2/3)x^(-1/3) = (5/3)x(1-1/3) - (2/3)x^(-1/3)= 1/(3*x^(1/3)) * (5x-2)
You got it right.
youngwerther
Banned
Literally all of my derivatives are unsimplified because I can't do algebra with these fucked up exponents.
I'm fucked.
I'm fucked.
Take the derivative with respect to t, to get W` = dW/dt. This is the rate of water running out with respect to time. This derivative will use the chain rule but since the derivative of (t-13) is just 1 this is easy.
dW/dt=(2)200(t-13) = 400(t-13)
So for part a, just plug in t=9 to find the rate of change at 9 minutes.
dW/dt=(400)(9-13)=-1600
Now for part b we want the average. We find the value at each endpoint and the length between them,so we just find (W at t=9) minus (W at t=0) divided by the time between these two points: 9-0)
So using the formula given, we know at t=9, W is 3200. At t=0 W is 33800
(3200 -33800 ) /9 = -3400
And we are done.
Partial Gamification
Banned
Don't let the operations intimate you. A small percent of people rapidly understand this material, different learning styles and instructional methods.
If:
function = (expression)^exponent
then:
derivative(function) = (exponent)*[derivative(expression)]*(expression)^(exponent-1)
False hope can't float a boat, but don't let your worries tie you down. Just show as much work as you are able, as neatly as possible, and take it one-step-at-a-time. Break down those nasty functions into smaller functions. Good luck!
Thanks a lot. I was mostly stuck on the average rate of change part of the formula. I guess the formula can be:
F(b)-F(a)/b-a
Thanks for the help on the last problem I posted. Here's another:
Prove that the converse of Lagrange's Theorem is true if G=Z_n. That is, prove that if k|n, then there is a subgroup of Z_n of order k.
I was trying to use the idea of a theorem in my book that says, "Fix n in the integers, n>1, and let k be in the integers. The element [k] in Z_n generates a group of order n/d, where gcd(k,n)=d. Moreover, <[k]>=<[d]>." That is really the only train of thought I could invoke for this problem. Thanks for the help.
Prove that the converse of Lagrange's Theorem is true if G=Z_n. That is, prove that if k|n, then there is a subgroup of Z_n of order k.
I was trying to use the idea of a theorem in my book that says, "Fix n in the integers, n>1, and let k be in the integers. The element [k] in Z_n generates a group of order n/d, where gcd(k,n)=d. Moreover, <[k]>=<[d]>." That is really the only train of thought I could invoke for this problem. Thanks for the help.
Hi,
It really depends on what theorems you already covered and if you're allowed to use them without reproving them. I'm 99% sure by the time you see Lagrange theorem you will already have covered the major theorems for cyclic groups...
We know that Z_n is a finite cyclic group, so we can apply a corollary of the fundamental theorem of cyclic groups, and I believe that's what you were referring to above.
(If you wikipedia "fundamental theorem of cyclic groups" you will see it, it's too hard to retype)
So using that theorem for Z_n, it states that if k divides n, the set <n/k> is the unique subgroup with order k of Z_n.
And this is exactly what your question is asking to prove, so that's all there is to it.
Tell me if this wasn't clear or explicit enough and I can provide more info.
Partial Gamification
Banned
The converse of Lagrange's Theorem is also know as Cauchey's Theorem; note there are more complicated tools required to prove this for non-abelian groups.
edit: when k: prime ...?
edit: when k: prime ...?
The converse of Lagrange is always true for finite abelian groups, due to the fundamental theorem of cyclic groups, so it doesn't matter if K is prime. However if it wasn't an abelian group, you could also use Sylow's 1st theorem, so k would need to be a prime power.
Partial Gamification
Banned
Yeah I saw a couple methods for both cases; proofs / analysis is a major weakness, I need to find a good online lecture series.
Himynameischris
Member
Hey guys I'm working on a study guide for an exam for a business class and I'm stuck on these two questions, didn't know where else to post them, can anyone help? I don't need just the answers but why that's the answer, thanks for any help 
52) While glancing over the sensitivity report, you note that the stitching labor has a shadow price of $10 and a lower limit of 24 hours with an upper limit of 36 hours. If your original right hand value for stitching labor was 30 hours, you know that:
A. the next worker that offers to work an extra 8 hours should receive at least $80.
B. you can send someone home 6 hours early and still pay them the $60 they would have earned while on the clock.
C. you would lose $80 if one of your workers missed an entire 8 hour shift.
D. you would be willing pay up to $60 for someone to work another 6 hours.
16) Assume oligopoly firms are profit maximizers, they do not form a cartel, and they take other firms' production levels as given. Then in equilibrium the output effect __________.
a) must dominate the price effect. b. must be smaller than the price effect. c. must balance with the price effect. d. can be larger or smaller than the price effect
52) While glancing over the sensitivity report, you note that the stitching labor has a shadow price of $10 and a lower limit of 24 hours with an upper limit of 36 hours. If your original right hand value for stitching labor was 30 hours, you know that:
A. the next worker that offers to work an extra 8 hours should receive at least $80.
B. you can send someone home 6 hours early and still pay them the $60 they would have earned while on the clock.
C. you would lose $80 if one of your workers missed an entire 8 hour shift.
D. you would be willing pay up to $60 for someone to work another 6 hours.
16) Assume oligopoly firms are profit maximizers, they do not form a cartel, and they take other firms' production levels as given. Then in equilibrium the output effect __________.
a) must dominate the price effect. b. must be smaller than the price effect. c. must balance with the price effect. d. can be larger or smaller than the price effect
I've got a pretty simple number theory problem, but I'm getting stuck on the proof.
Given a,b are integers, and gcd(a,4)=2 and gcd(b,4)=2, find gcd(a+b,4) and prove your answer.
What I have so far:
2|a and 2|b implies 2|a+b implies 2=ua+vb
Then I kind of get lost from there. I know the answer is 4. In words, I can explain that a and b are both even, and neither is a multiple of 4, otherwise their gcds with 4 would be 4. The sum of any two even numbers that are not multiples of 4 will be a multiple of 4. Therefore, the gcd of that sum and 4 is 4. How do I make that proofy?
Given a,b are integers, and gcd(a,4)=2 and gcd(b,4)=2, find gcd(a+b,4) and prove your answer.
What I have so far:
2|a and 2|b implies 2|a+b implies 2=ua+vb
Then I kind of get lost from there. I know the answer is 4. In words, I can explain that a and b are both even, and neither is a multiple of 4, otherwise their gcds with 4 would be 4. The sum of any two even numbers that are not multiples of 4 will be a multiple of 4. Therefore, the gcd of that sum and 4 is 4. How do I make that proofy?
cpp_is_king
Member
Hint: factor 2 out of a and b. is the second term even or odd? What does this mean for a+b?
Thanks for the replies. I can't write it in a modulo congruency because we hadn't yet covered mod arithmetic when this problem was assigned. (My prof actually marked something wrong on a quiz when my friend used Euclid's lemma to find the gcd of two numbers because that wasn't covered until the next section of the book.)
Here's what I ended up doing:
gcd(a,4)=2 implies a=4n+2, since it must be even and not a multiple of 4
gcd(b,4)=2 implies b=4m+2, similarly
gcd(a+b,4) = gcd(4(m+n+1),4) = 4
I thought of this as I was talking out loud a few minutes after I posted. It seems mathematically sound to me.
Here's what I ended up doing:
gcd(a,4)=2 implies a=4n+2, since it must be even and not a multiple of 4
gcd(b,4)=2 implies b=4m+2, similarly
gcd(a+b,4) = gcd(4(m+n+1),4) = 4
I thought of this as I was talking out loud a few minutes after I posted. It seems mathematically sound to me.
cpp_is_king
Member
Yea that's pretty much equivalent to the hint I gave. Basically, a = 2k where k is odd, b = 2j where j is odd. a+b = 2(k+j), and since k and j are both odd, k+j is even, so k+j=2p, so a+b = 4p
Alright so I'm still having some issues with subspaces. I have a few problems where I need to determine if S is a subspace.
A is a particular vector in R_2x2. B is a vector that exists in R_2x2.
BA = 0
AB =/= BA
AB+B = 0
I want to make sure I understand this. B would represent every vector in the subspace and I need to prove it false if possible? The second problem is easy enough as B would include the zero vector which makes the claim false correct?
I'm not sure how to go about the first or third. For the first All I can think of is that A would have to be the zero vector to make it true. Do I just need to do addition with a C vector and mutliply by a scalar?
So for the first I would say C is another vector that is in the subset, and CA = 0 so adding them makes BA + CA = 0 which checks the addition axiom? Then multiplication by a scalar so aBA = a0 = 0?
A is a particular vector in R_2x2. B is a vector that exists in R_2x2.
BA = 0
AB =/= BA
AB+B = 0
I want to make sure I understand this. B would represent every vector in the subspace and I need to prove it false if possible? The second problem is easy enough as B would include the zero vector which makes the claim false correct?
I'm not sure how to go about the first or third. For the first All I can think of is that A would have to be the zero vector to make it true. Do I just need to do addition with a C vector and mutliply by a scalar?
So for the first I would say C is another vector that is in the subset, and CA = 0 so adding them makes BA + CA = 0 which checks the addition axiom? Then multiplication by a scalar so aBA = a0 = 0?
youngwerther
Banned
Need help with a derivative.
It's an easy one but I got stuck and I want to see how it is worked out
f(x) = (sqrt(7))/(t^5)
I used quotient rule and got
(sqrt(7)*(5t^4) - (t^5)(1/2 * 7^(-1/2)))/(t^10)
It's an easy one but I got stuck and I want to see how it is worked out
f(x) = (sqrt(7))/(t^5)
I used quotient rule and got
(sqrt(7)*(5t^4) - (t^5)(1/2 * 7^(-1/2)))/(t^10)
Quotient rule is a bit overkill. Just raise the denominator into the numerator and use the power rule. Set it up like this
sqrt(7)*t^-5
You're treating 7 as a variable when it is a constant. You don't need to use the power rule on sqrt(7). You only need to differentiate the variable t.
You are right for 2, since B can be any vector in R_2x2 fulfilling AB =/= BA we see that B can't be zero, so No subspace.
What you want to do for these questions is check the three conditions of a subspace until you find one that doesn't fit, or until all three are verified. for number 1, we see that it is a subspace due to the following:
Note for this, it says A is a particular vector, and B is whatever, so we keep A constant and play with different B's. Then we just try to put our result in the form given to see if it's fulfilled.
To show it has the identity, just note that if B is the identity, then 0A=0, so this shows 0 is in S.
To show for cB in S for B in S is easy. We have BA=0, so (cB)A=c(BA)=c0=0
So that part is done.
(Here I just called B1 = X and B2 = Y to make the typing easier)
To show for X,Y in S, that we get X+Y in S note that:
|x1 x2||a1 a2| = |0 0|
|x3 x4||a3 a4| = |0 0|
And
|y1 y2||a1 a2| = |0 0|
|y3 y4||a3 a4| = |0 0|
So we need to check if
|x1 + y1 x2 + y2||a1 a2| = |0 0|
|x3 + y3 x4 + y4||a3 a4| = |0 0|
I'll do the first one for you. From above we know that
x1a1 + x2a3 = 0 and y1a1 + y2 a3 =0
so checking, we want (x1 + y1)a1 + (x2 + y2)a3 = 0
But using the fact that x1a1 + x2a3 = 0 and y1a1 + y2 a3 =0, we see this is true.
So we showed all three properties of a subspace, so S is a subspace
Number 3:
Is origin in? Yes since if B = 0, A0 + 0 = 0
For B in S is cB in S?
Yes, since A(cB) + cB = cAB + cB = c(AB+B) = c0 = 0
For X and Y in S, is X + Y in S?
Yes, since we have
AX + X = 0
AY + Y = 0
Need to check if A(X+Y) + (X+Y) = 0
But it clearly does since A(X+Y) + (X+Y) = AX + AY + X + Y = AX + X + AY + Y = 0 + 0 = 0
All three shown, so it's a subspace.
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