Okay, thanks but I still don't understand the multi variable summation notation-
What does the i,j,k under the sigma represent and why is there nothing on "top" of the sigma?
Does this notation also assume that:
"the summation is meant to be over all triples i, j, k where...
Well, we just learned that the
lim x -> a
notation simply means that the x value is approaching a, but it isn't a. So it is the value right next to a.
I guess it's just a practical and less formal way of saying the same thing, although simply saying that x minus a is greater than zero...
I can't understand the sum notation shown in Wikipedia or in this article:
http://mathforum.org/library/drmath/view/53159.html
I want to find the sum notation for (a+b+c)^n
however I can't understand the sum notation:
I don't understand the use of brackets or what they mean here and in the...
Can you find the surface area of a sphere without using the cylinder formula?
(I mean, you can end up with the same formula, but you use different reasoning)
Okay, is there a better way of stating my example? Is this okay:
Diminishing returns especially happen when, for example, a man chooses to major in a state college (with state scholarships) only to choose to live off his wife upon completion of the degree. Of course he may have some...
Sorry, I didn't mean to make anyone mad. It could have easily been a man who lives of his wife, I just stated the traditional scenario.
"If that same woman enhances the education (and moral compass) of her children - there is an absolute value to society that should not be diminished - IMO."...
On the what to teach, shouldn't practical things such as cooking, car mechanics, and personal accounting be required for a GED?
Specially since these skills don't require high levels of abstract thinking and can be learned through apprenticeship, the skills can be more easy for younger...
The above method is too tedious, here is a better way:
dy/dx = Q - ay - by^3
dx/dy = \frac{1}{Q - ay - by^3}
dx = \frac{dy}{Q - ay - by^3}
Now integrate both sides (I don't know how to do it) but Wolf math does...
I'm not sure this is right, but it may work!:
dy/dx + (a+b.y^2)y = Q
dy/dx + ay+by^3 = Q
dy/dx = Q - ay - by^3
y = Qx - a\int(ydx-b\int(y^3dx
Now we convert dx into dy and y terms:
dy/dx = Q - ay - by^3
dx = \frac{dy}{Q - ay - by^3}
Bam! Here it is
y = Qx -...
Thanks for the reply, it's cleared up some things and I think I get it now.
Euclid's proof shows that if P + 1 is not prime, then P + 1 can not be divided by a prime number in the finite set of prime numbers since when you divide P + 1 by a factor of P, that factor will end up dividing 1...
Why can't we make a equal some function?
a = -8x
\frac{a}{-8}=\frac{-8x}{-8}
Here the function is globally onto.
I don't understand why you choose y=0 in the discriminant:
Delta = 6^2-4(-8)a
instead of
Delta = (6 - 6y)^2 - 4(a + 8y)(-8 - ay)
We can either replace a for one function...
I don't understand how this part is valid:
"If q is not prime then some prime factor p divides q. This factor p is not on our list: if it were, then it would divide P (since P is the product of every number on the list); but as we know, p divides P + 1 = q. Then p would have to divide the...
I think you can still use the quadratic equation:
a has to be a value that makes this expression always true (since you can't take the square root of negative numbers):
http://latex.codecogs.com/gif.latex?0\leq&space;(6-6y)^2-4(a+8y)(8+ay)
I'm not sure what you do next...
I thought this expression had to be negative for there not to be any zero roots in the denominator.
a> -\frac{9}{8}
Why isn't that ^ the answer? I don't understand what onto and f:R-->R mean.
If you plug in any number less than -9/8 on any graphing calculator, you will get a continuous...
Oh, I think I know what it is know. Isn't it weird how there is an -8 on top where there is an a in the bottom and vise versa? You simply put in a=-8 and the answer is always 1!
((-8)*x^2+6x-8) / ((-8)+6x-8*x^2)=1
-8 is smaller than -9/8 so the calculus was true.
I don't quite get what...
Okay, so I'm just going to treat the top equation as a quadratic, and the bottom equation as a quadratic, and then multiply them together.
Now I replace the first fraction with a,b, and c with a,6, and -8. Then I replace the second fraction with a,b, and c with -8, 6, and a.
Now you...
Okay, I am trying to find a formula for the position of two bodies (or more) in space based on the initial velocity and time.
I'm trying to integrate the equation for gravitational acceleration to find the velocity equation, however, the radius is changing, and the degree (theta) is also...
DArN iT!
I thought I had found a way around it though.
Have you at least tried it this way?
I have another hunch:
and look!
\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{(\frac{L}{N}k)^{(\frac{Lk}{N})}}}
Which is so close to:
\lim_ {N\rightarrow...
But Hey! According to wikipedia:
http://en.wikipedia.org/wiki/The_fundamental_theorem_of_calculus
,this image:
,and this formula:
When you take the derivative of a remain sums, the last sum times dx should be your answer.
In this case, I did not have a dx because I'm...
\lim_{N\rightarrow +\infty}{\frac{1}{N}\sum_{k=1}^L{1^k}}=1
L=Nh (it doesn't let me put two characters on top of the sigma)
It's 1^k right? nvm it always equals 1.
Oh, I think that what I meant was from k=1 or 0 to k=Nh not N on the sigma sign.
So, is what I wrote above right? It...
Based on my friends, AP Stats was really easy, compared with Physics AP (but then again Since Physics AP deals with the application of Calculus to Physics and we were learning Calculus in the same year, we had some blind spots until the end of the year when we learned Calculus).
One of my...
I'm trying to find the average value of x^4 from 0 to L
So, I simply made an infinite number of function values be divided by the number of those function values:
(h)^(4) + (2h)^(4) +...(Nh)^(4)
N
P.S: Nh=L
I factored out the h^4...:
[(1)^(4) + (2)^(4) +...(N)^(4)](h)^(4)
N...
Yeah, you're right. So, how many other ways are there of doing integrals and derivative?
You can ignore the following rant:
"I didn't like the whole "work backwards" process for finding the integrals of function, so I tried to find the average value then multiply that value by the length...
I must have done something wrong. There shouldn't be a -1 next to the 2^L.
You wouldn't happen to know a a formula that would get (2^L)/(log(2)L) or how we can edit the original conditions to remove that -1 in the end.
I have hunch that dividing the original sigma by N+1 instead of N might...
That's Awsome! Thanks for showing that to me Mute.
The following is true, right?:
^{lim}_{N\rightarrow\infty}\frac{1}{N}\sum_{k=0}^N{^{lim}_{h\rightarrow0}(2^h)^k}=\frac{1}{N} 2^h \frac{1-2^{hN}}{1-2^h}=\frac{h}{L}\frac{e^h}{1-e^h}(1-e^{L})=\frac{2^L}{log(2)L}
(Nh=L)
I have a feeling...
Has anyone here heard of discrete calculus? I think that's what I'm trying to do, but the most advanced math course I've taken is Calculus I so I don't understand how you reduced the original Sigma equation to this:
\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}
?
The guy at this website talks...
I'm trying to find the average of an infinite series.
We have an infinite number of subdivisions: N subdivisions
All of the subdivision's lengths are equal and virtually zero: h length
If we multiply the subdivisions (h) by the number of them (N) we should get the length of all of them...
\frac{1}{N}\sum_{k=1}^N{(e^h)^k}=\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}=\frac{e^L}{L}
and
Nh=L
I know that it should equal (e^L)/L but I don't know why.
Here's another situation:
2^(h)+2^(2h)...2^(Nh)
N
Nh=L
(based off of what you had-I don't know if this is right though)...
I'm not sure how you got the right part of that equation (induction?). Could you show me the logic behind it or at least tell me what math course that you take to learn that sort of stuff?
Using whatever you have there, Can you prove that it equals:
e^L
L
?
What was that? I think I had the mute button on. Here's what I meant
e^(h) + e^(2h)... e^(Nh)
N
I want to factor out the h's, so that I have something like:
[e^(1) + e^(2)... e^(N)][something]
N
Well...
I have two limits:
h-> 0
N-> infinity
When I multiply these limits together, I get a constant:
Nh = L
I have the following "thing" (since it's not a function or is it?):
e^(h) + e^(2h) +...e^(Nh)
N
If we can factor out the h from ^ that ^, then we can put it to the side...
I want to know if there is a way to factor out the b from e^(ab) so that:
e^(ab) = e^(a)[f(b)] or e^a + f(b)
or
e^(ab) = some other function where a and b are separated.
Sorry, but I'm curious- What is this? I don't recognize the aT thing or the double absolute value marks. Could you tell me what type of math is this? Is it Calculus III?
I think I'm having a definition problem. If potential energy is the energy that you could potentially use. How can you use energy that you don't have, or energy that is negative? Does it have to do about where E mech is?
So potential energy is the negative work done by a field (or spring)...
I am dealing with polynomials (and other functions) which are continuous and always differentiable.
If g(y) = x and f(x) = y, then I thought that g'(y) = dx/dy, since f(x) = dy/dx
so the integral of dx/dy + C= g(x)
What I think you are saying is g(y) = the integral of (dx/dy)(dy/dx) + C then...
Can you integrate this function (perhaps by using integration by parts)?
I know it depends on what f(x) is, but we can use general notation such as f''(x) or the integral of f(x) so that we can simply plug in for any f(x).
The point of this equation is to solve for x when f(x) is a big...
What you are saying is that the Mechanical Energy is zero at infinity? Then what would happen if the net mechanical energy positive or negative? If I am doing work against gravity, my potential energy is positive and it increases since my work has contributed to the increase in potential energy...
If you ignore the air particles, then the normal force (or what you call n) should be zero (since there isn't any normal force, since we are ignoring internal forces).
If you don't ignore the air particles, then the air particles hit against the falling body (which causes the momentum of the...
I have two physics books that state that
U(r) = -GMm/r
What I don't understand,is how can potential energy be negative?
I've done the integral of GMm/r^2 from infinity to r, but I don't quite get the concept of negative potential energy.
I don't understand why using r=infinity as...
@ homology dy/dx = f'(x)dx Your right, it is false:
dy/dx = f'(x)dx/dx = f'(x)
Still, that doesn't change the question since I originally just ignored dx (because the change in f(x) per dx or that the f'(x) is based on dx)
Well... I found a cool trick to find the derivative of the inverse of functions
What I originally took the derivative of y:
dy/dx = f'(x)dx
Then I flipped the derivative:
dx/dy = 1/f'(x)dx If we solved for y and took the derivative of it with respect to y, you would get the...
I think he means an easy explanation for them. A limit is a value that a function approaches.
For example, you are lifting weights one day and you decide to increase the weights at small increments. Eventually you will approach a maximum amount of weight or what you may call your limit. If...
Yes, I'm trying to integrate that function. Using this method:
u = f'(x) and dv = dy
I want to integrate this function enough times to see a pattern.
While integrating, I keep in mind that f(x) is the function of a polynomial. When integrating it is preferable to derive any f''''''(x)...
If x = the integral of dy/f'(x) what is a x equal to? I used integration by parts. I'm trying to integrate it enough times to find a pattern (using f'''''' notation so I don't get lost). I'm having trouble doing it. If anyone can spot something please share!
Thank you, I was trying to verify if that worked by doing it both ways. I figured out my problem. I had the signs on the v(x)/h(x) derivative rule switched.