June 17

(I'm assuming below rigid particles, rods, elastic collisions etc - the simplest case)

I'm stuck (totally) on a problem. There are two masses (A and B) (each mass m) connected by a rod of length L. For clarity I'll say A is at (0,0), and B at (L,0)

Another point mass C (also mass m) is at (0,-d) and is moving at velocity v in the direction (0,1) ie towards point A.

What happens when A and C collide - I'd expect some sort of rotation of A-B - but I can't get that to make any sense mathematically. It seems that what I've described is impossible (or maybe my stupidity is a factor) Any help appreciated. Thanks87.102.32.96 01:45, 17 June 2007 (UTC)[reply]

The animations in the Elastic collision article might help. I believe what would happen is this: C would move towards A at a constant velocity v until they collided. All the energy of the system would leave C and move into A, which would start A moving along the same line, at velocity v. This would send the bar rotating about its center point, without the center point moving, until B collided with C. The bar would stop moving, and C would start back on its merry way. Black Carrot 03:01, 17 June 2007 (UTC)[reply]

No, that's wrong. The first part is right, but I think the bar would resist rotation somewhat, and some of the system's energy would go into moving the bar forward. In other words, it would rotate about its center point a bit more slowly, and that centerpoint would move forward at a constant velocity. It would, then, miss C and keep moving forever. The article about Moment of inertia might help. Black Carrot 03:05, 17 June 2007 (UTC)[reply]

This one's not too bad if you take it step by step. At the instant that C collides with A, you can ignore the presence of B, because the collision is at right angles to the rod and so the rod can't exert any forces on A. So conservation of momentum and energy mean that C comes to rest and A heads off in the (0,1) direction with velocity v, as Black Carrot says. Now consider the pair of particles A and B. Their centre of mass (call it D) is halfway along the rod, initially at (L/2, 0), but since A is moving with velocity v, D is now moving at velocity v/2 in direction (0,1). By conservation of momentum, D will continue to move at this velocity, since there are no external forces on A or B. However, the rod joining A and B is now rotating with an instantaneous angular velocity w = v/L, and by conservation of angular momentum, will continue to do so (the rod between A and B is now under tension, and supplies the necessary centripetal force). So C has come to rest at the origin. Its original momentum mv is now taken up by the motion of D, with total mass 2m and velocity v/2. The original kinetic energy mv²/2 is now shared between the translational KE of the rod, which is 2m(v/2)²/2, and its rotational KE, Iw²/2, where its moment of inertia is I = 2m(L/2)². Add these together, and it all works out. C stays at rest, and A and B move off, spinning gently with an angular velocity of v/L (in radians/second) around their centre of mass, which itself moves steadily at speed v/2. In fact, A and B describe interlocking cycloids, each coming to rest instantaneously when they reach x coordinate L. --Prophys 10:03, 17 June 2007 (UTC)[reply]

Thanks - much clearer now.

Here's a bit of a math conundrum for you: say I took the number 1, and I added half of 1 (0.5), then I added half of half of 1 (0.25), and continued in this pattern an infinite number of times. Would I ever reach 2, or would you just end up with a number that has an infinite number of decimal places?

Here's the start of the math to explain what I mean a bit more clearly:
1 + 0.5 = 1.5
+ 0.25 = 1.75
+ 0.125 = 1.775
+ 0.0625 = 1.8375
continued...

Thanks for the help. I thought this up one day and I tried it with a calculator but wasn't able to get that far with my 7 digit screen. Taylor 08:13, 17 June 2007 (UTC)[reply]

This is just an example of an infinite series. Take a look at the article for a lot of interesting information. — Kieff | Talk 08:23, 17 June 2007 (UTC)[reply]

Thanks for that. Does this mean that the series would never reach 2, since there would always be a room at the end equivalent to the amount just added? Or does it count as 2 on the account of 0.999...=1?Taylor 09:11, 17 June 2007 (UTC)[reply]

If you summed the first 5 billion terms of that series, you'd end up with a number very close to 2, but not quite! If you took the sum of the first 5 thousand trillion terms, the sum would still be slightly smaller than 2. So, if the series has a finite number of terms, then it will never reach 2. BUT, if the series has an infinite number of terms, then by the same principle as the 0.9999... = 1 statement, it is equal to 2.

You can also consider how far you are from 2. For 1, that is 1. For 1.5, it is 1/2. For 1.75, 1/4. Then 1/8, 1/16, and so on. Each time you halve the distance. Half a positive number is again a positive number, so at each step you will still have a positive distance to 2. But you can get arbitrarily close: the limit of 1, 1/2, 1/4, ... is 0, since you can make the number as small as you want – provided you allow it to remain positive. To give an impression, here is where you would be after 100 steps if the screen size is not a factor:

1.9999999999999999999999999999992111390947789881945882714347172137703267935648909769952297210693359375

This falls short of 2 by the amount of

0.0000000000000000000000000000007888609052210118054117285652827862296732064351090230047702789306640625

See also Zeno's paradoxes.  --Lambiam^Talk 12:48, 17 June 2007 (UTC)[reply]

What you want to do is write the numbers in binary:

${\displaystyle 1_{2}+0.1_{2}+0.01_{2}+0.001_{2}+\cdots =1.111\ldots _{2}\,\!}$

This is, indeed, the binary equivalent of 0.999… in base 10. We must carefully distinguish two possibilities: (1) adding as often as we like, and (2) the infinite sum. For (1), we can get as close to two as we like, but always fall short; for (2), we get exactly two. And for those whose consider such mathematical conundrums silly, I offer the physical reality of quantum tunnelling: A particle approaches an insurmountable obstacle, but if the wall is thin enough we may find the particle on the other side because of quantum uncertainty in its position! What would Zeno make of that? :-) --KSmrq^T 19:36, 17 June 2007 (UTC)[reply]

0.999... = ${\displaystyle {\frac {1+1/2+(1/2)/2+...}{2))}$. So just see the 0.999... article. 202.168.50.40 21:56, 17 June 2007 (UTC)[reply]

It's not a conundrum at all. If you limit yourself to adding up a finite number of terms of the series, then you will obviously never reach 2. Adding n terms will produce a sum of 2-2⁻ⁿ < 2 for all n ∈ N. If, however, you add up an infinite number of terms, the sum will be exactly 2. Confusion over this is simply confusion over the mathematical concept of infinity. Which is, admittedly, a subtle thing; it took mathematicians a long time to be sure they understood it.

Why does in this page [1] say the method won't catch a minimum if x,y,z can be negative? --Taraborn 09:09, 17 June 2007 (UTC)[reply]

The Lagrange multipliers method gives necessary conditions for the global minima and maxima. If the variables are assumed to be nonnegative, then adding the constraint results in a closed, bounded region, and since the target function is continuous, it must have a minimum and a maximum - and they will be found among the points which are "suspected" by the method. If, however, the variables are not assumed to be nonnegative, then the region is unbounded, and there is no guarantee that a minimum or a maximum exists. If, say, there is no minimum, then none of the points suspected by the method will be a minimum. However, the method does not give sufficient conditions, so by itself it cannot tell if any of the points found is actually a minimum. -- Meni Rosenfeld (talk) 10:09, 17 June 2007 (UTC)[reply]

Note also that the advice is specific to the example, not a general rule. 84.239.133.38 14:43, 17 June 2007 (UTC)[reply]

On closer inspection, the example given there is quite misleading. The final result is that the minimum is obtained at 3 points, when in reality it is obtained at infinitely many points (e.g., ${\displaystyle (x,1-x,0)}$ for any ${\displaystyle 0\leq x\leq 1}$). More care needs to be taken when solving a system where the constraints involve both equalities and inequalities. -- Meni Rosenfeld (talk) 15:45, 17 June 2007 (UTC)[reply]

No. They are correct. The minimum is attained at only one point. --Spoon! 17:26, 17 June 2007 (UTC)[reply]

We are clearly not speaking about the same thing. I'm referring to "find the minimum and maximum of ${\displaystyle xyz}$ subject to ${\displaystyle x+y+z=1}$ and ${\displaystyle x,y,z\geq 0}$. -- Meni Rosenfeld (talk) 17:38, 17 June 2007 (UTC)[reply]

Consider the objective function 1+x². It attains a minimum at x = 0, but can grow arbitrarily large so has no maximum. Similarly, the objective function 1−x² attains a maximum at x = 0, but can grow arbitrarily small so has no minimum. Now multiply the objective function by y to give y(1+x²), but add the constraint that y = 1; clearly nothing should change. The first equations, now including Lagrange multiplier, become

${\displaystyle {\begin{aligned}2xy&{}=0\lambda \\1+x^{2}&{}=1\lambda \\y&{}=1\end{aligned))}$

The only solution is x = 0, y = 1, λ = 1 — a minimum; but without bounds we again have no maximum. (Note that the positive value of λ tells us the constraint is pulling us away from a smaller minimum, at y = 0.)

Incidently, the recent question about a maximum on a circle invoked the same idea behind Langrange multipliers; if a component of the gradient is parallel to a direction of allowed movement, we must not be at an extremum. --KSmrq^T 20:49, 17 June 2007 (UTC)[reply]

Thanks for this incidental note -- pleasantly unexpected. Tesseran 09:13, 18 June 2007 (UTC)[reply]

I started off with

${\displaystyle {\frac {\text{dy)){\text{dx))}={\frac {w\cos(\beta )-v\sin(\theta )}{w\sin(\beta )-v\cos(\theta )))}$

Where sin theta = y/sqrt[x^2+y^2] and cos theta = x/sqrt[x^2+y^2]

And I got to this point (using u=y/x):

${\displaystyle {\frac {w\sin(\beta )-{\frac {v}{\sqrt {u^{2}+1)))){w\cos(\beta )-w\sin(\beta )u)){\text{du))={\frac {\text{dx)){x))}$

Basically I want to know if this can be solved?

${\displaystyle {\frac {A-{\frac {B}{\sqrt {u^{2}+1)))){E-Fu)){\text{du))={\frac {\text{dx)){x))}$

220.239.108.132 12:05, 17 June 2007 (UTC)[reply]

Are w, β and v all constants?  --Lambiam^Talk 13:17, 17 June 2007 (UTC)[reply]

Yes they are constants. 220.239.108.132 15:33, 17 June 2007 (UTC)[reply]

What I'd do is switch to polar coordinates, so that

${\displaystyle x=r\cos \theta \,,}$

${\displaystyle y=r\sin \theta \,,}$

in which r is considered a function of θ. Abusing the "prime" notation for the derivate with respect to θ, so that r' = dr/dθ, we have

${\displaystyle {\frac {dy}{dx))={\frac {dy/d\theta }{dx/d\theta ))={\frac {r'\sin \theta +r\cos \theta }{r'\cos \theta -r\sin \theta ))\,.}$

Next, do a change of variable, in which r = exp z, so r' = z' exp z = z'r. Then we can rewrite the last fraction as:

${\displaystyle {\frac {z'r\sin \theta +r\cos \theta }{z'r\cos \theta -r\sin \theta ))={\frac {z'\sin \theta +\cos \theta }{z'\cos \theta -\sin \theta ))\,.}$

This is still the result of rewriting of dy/dx, which is equal to the r.h.s. of your first equation, and so, using A := w cos(β) / v, B := w sin(β) / v,

${\displaystyle {\frac {z'\sin \theta +\cos \theta }{z'\cos \theta -\sin \theta ))={\frac {A-\sin \theta }{B-\cos \theta ))\,.}$

You can solve this for z', leading to an equation of the form

${\displaystyle z'=f(\cos \theta ,\sin \theta )\,,}$

in which f is a rational function. If we can find the antiderivative

${\displaystyle \int f(\cos \theta ,\sin \theta )\,d\theta \,,}$

we're done. This is the point to call t := tan(¹/₂θ) to our aid:

${\displaystyle \cos \theta ={\frac {1-t^{2)){1+t^{2))}\,,}$

${\displaystyle \sin \theta ={\frac {2t}{1+t^{2))}\,,}$

${\displaystyle d\theta ={\frac {2dt}{1+t^{2))}\,,}$

By this change of variable, the indefinite integral becomes

${\displaystyle \int {\frac {2f({\frac {1-t^{2)){1+t^{2))},{\frac {2t}{1+t^{2))})}{1+t^{2))}\,dt\,.}$

The integrand is a rational function of t, and so the antiderivative can be found in the standard way by the method of partial fractions.  --Lambiam^Talk 00:56, 18 June 2007 (UTC)[reply]

I attacked this in a slightly different way, based on my own version of intuition (so it may not be the most elegant method).

• The RHS is, of course, easy to integrate so we ignore that.
• The LHS can be broken into two terms, the first one a simple fraction with a similarly simple integral, the second the bit containing the square root.
• When you have a ${\displaystyle {\sqrt {a^{2}+u^{2))))$, the traditional substitution is ${\displaystyle u=a\tan \theta }$, making use of the identity ${\displaystyle \sec ^{2}\theta =1+\tan ^{2}\theta }$.
• In this case it seems to simplify things significantly, leaving us with an integral of ${\displaystyle {\frac {1}{E\cos \theta -F\sin \theta ))}$, which can be solved via substitution of ${\displaystyle t=\tan {\frac {\theta }{2))}$, and making use of various half-angle identities and partial fractions (although you need to watch out for special cases, such as when the denominator becomes a perfect square).

I haven't pulled out a solution myself, but it certainly looks doable without resulting to any particularly complicated integration techniques, thankfully. Confusing Manifestation 01:33, 18 June 2007 (UTC)[reply]

Hi all, it's me again. I was hoping to get some aid with another complex analysis problem. For reference, here's the statement I have to prove.

Let f be analytic in the the upper half plane and continuous in the closed upper half plane. Assume also that ${\displaystyle |f(z)|\rightarrow 0}$ as ${\displaystyle |z|\rightarrow \infty }$ in the upper half plane. If ${\displaystyle f(z)=u(z)+iv(z)}$, where u and v are real-valued, show that for ${\displaystyle x\in \mathbb {R} }$,

${\displaystyle u(x)=-{\frac {1}{\pi ))\int _{-\infty }^{\infty }{\frac {v(y)}{x-y))\ dy}$ and ${\displaystyle v(x)={\frac {1}{\pi ))\int _{-\infty }^{\infty }{\frac {u(y)}{x-y))\ dy}$.

I wanted to use the Cauchy integral formula using (as my countour) the semicircle centered at the origin with radius R, then send R to infinity and all that jazz. Playing free and loose (i.e., not checking whether or not this is legitmate), I was able to prove the statement I wanted, but each expression was divided by 2. Since f is not analytic ON my entire contour (although it is analytic within it), is there something I'm missing? Any help would be appreciated. –King Bee ^{(τ • γ)} 14:11, 17 June 2007 (UTC)[reply]

I think that's the correct approach. You might have missed that the semicircle contains the point x, while the formulation of Cauchy's integral formula in our article requires the contour to go around the point x. You should use a version of Cauchy's integral formula which allows the contour to go through x, or a version of the residue theorem which allows for poles on the contour. -- Jitse Niesen (talk) 21:04, 17 June 2007 (UTC)[reply]

Yeah, that's what I'm thinking. I wonder if the version I want says something like

${\displaystyle f(\omega )={\frac {1}{\pi i))\int _{C}{\frac {f(z)}{z-\omega ))\ dz}$,

since my function isn't necessarily analytic ON the contour. –King Bee ^{(τ • γ)} 21:10, 17 June 2007 (UTC)[reply]

• For those interested parties, I figured out the problem. I was using the wrong contour. What I want is an "archway" contour, that "jumps" over the given point x in the problem. Thanks for the help, however! –King Bee ^{(τ • γ)} 12:20, 19 June 2007 (UTC)[reply]

How do calculate the power required to perform classic gym lifts like the deadlift in a certain time (1.5sec)? If it was only a matter of P = E/t it would be easy, but the fact that joints are working in different directions makes it difficult. I'm unable to make use of the information in the torque article, which I suppose should be of use. Could anyone give me a few guidelines how to calculate this kind of movements? Thanks in advance, Jack Daw 14:59, 17 June 2007 (UTC)[reply]

This is not really a maths question; it is more a question for the Science section of the Reference desk. If you post the question there, it may help if you could clarify what you mean by "effect". Do you mean power?  --Lambiam^Talk 18:06, 17 June 2007 (UTC)[reply]

I meant power, thanks. The reason I put my question here is that I need only the purely mathematical part to my answer - that is, the proper equations - rather than an explanation of the concepts of force, mass, power and so forth. Still think I should post it in the science section? Jack Daw 14:05, 19 June 2007 (UTC)[reply]

What are your assumptions? If you assume that there is no friction and that the mass of your own body is negligible, then by conservation of energy, the power your muscles output is exactly equal to the power received by the weight. Of course, if you do not assume these, then the power your muscles output is equal to the power received by the weight, plus the power received by the moving body parts, plus the power lost to friction. Do you aspire to calculate these other terms? Doing so requires quite a lot of data. -- Meni Rosenfeld (talk) 21:51, 19 June 2007 (UTC)[reply]

I have been trying to do this problem for a day or 2, and it is just not working. I'm supposed to find the inverse function of ${\displaystyle x^{2}+x}$(assuming ${\displaystyle x=>-1/2}$). I know the basics of inverse functions, but the 2nd x is screwing me up. I have no clue what to do with it. This is the only type of inverse function that I am currently having trouble with.69.49.70.112 15:35, 17 June 2007 (UTC)[reply]

Hint: It has something to do with a square root (I'm sure you know that from a basic parabola). See completing the square. x42bn6 Talk Mess 15:42, 17 June 2007 (UTC)[reply]

As far as I can tell, the function has no inverse, since it is not one to one. Both 0 and -1 get mapped to 0. I'm an idiot, and should really read the whole problem before responding. Follow the hint above, it should help you solve your problem. –King Bee ^{(τ • γ)} 15:59, 17 June 2007 (UTC)[reply]

Ok, I got through the problem I ended up with ${\displaystyle x={\sqrt {y+(1/4)))-(1/2)}$ I checked it out on the graphing calculator and it looks pretty good. Thanks Guys!

It might be useful to note that the domain might change things. If one takes x<=-0.5, then you will have a negative square root. x42bn6 Talk Mess 16:17, 17 June 2007 (UTC)[reply]

²²sir help me on this prolems as i am little bit confused.

• 1)find the two distinct right coset of R in (C+).

if i write a+(c+id) and b+(c+id) are'nt they distinct right coset of R in c for different value of a and b element R

• 2)sir tell me whether this statement is true or false give reason also to prove or disprove it ...........if every proper subgroups of group G is cyclic than group G is cyclic.

sir is'nt it true because even G is proper subgroup of itself and if it is cyclic than it is cyclic.

• 3)sir how should i start to show that M²(R×S)≈ M²(R)×M²(S) where M²(R)and M²(R) represent matrices of second order and R and S are rings.(≈) denotes isomorphism.
• 4)if H intersection K is a normal subgroup of G than H and K are normal subgroups of G where H and K are subgroups of G .............sir in my opinion it is false but what example i should give to disprove
• 5)let G and H be groups of orders P and q where p and q are distinct prime numbers show that any homomorphism from G to H must take any element of G to the identity element of H........... i am unable to think over it neither i am getting any example in book pls help

—Preceding unsigned comment added by 59.176.119.128 (talk • contribs) 16:41, 2007 June 17

You should sign your posts on talk pages, and remember that we don't do your homework for you here, but we can help you.

1) I think you're thinking about this one backwards. You need to find two elements ${\displaystyle z_{1},z_{2}\in \mathbb {C} }$ so that ${\displaystyle z_{1}+\mathbb {R} }$ and ${\displaystyle z_{2}+\mathbb {R} }$ have different elements in them. Concentrate on the imaginary parts of ${\displaystyle z_{1))$ and ${\displaystyle z_{2))$ and you should get it.

2) No group is a proper subgroup of itself. Also, the statement you are trying to prove is false. Now, go find a counterexample.

3) Smells like the Chinese Remainder Theorem to me.

4) I feel like this one is false. Pick one of your favorite pathological groups, and find two non-normal subgroups whose intersection is the trivial subgroup. That should suffice.

5) Smells like the fundamental homomorphism theorem to me. Think about what type of groups G and H are. Assume you have a homomorphism and look at its kernel.

Good luck. –King Bee ^{(τ • γ)} 17:44, 17 June 2007 (UTC)[reply]

How about this. Assume f homomorphism, f(e_G) ≠ e_H, e_G and e_H respective identities of G and H. Consider f(e_G)f(g), g ∈ G. To avoid spoiling all the fun, you can determine f(e_G)f(g) easily from the properties of homomorphisms. What are you forced to conclude? What does this imply? —Preceding unsigned comment added by 129.78.64.102 (talk • contribs)

He is trying to show that f maps all the elements to the identity of H, not just e_G. nadav (talk) 10:41, 18 June 2007 (UTC)[reply]

Ok, I misread. Never mind, you can still adapt the fact to get the necessary result. You need to know that any group of prime order is cyclic, so only one element has order 1. That should be all you need, the rest is piecing it together.

Well, it's a little bit more complicated than that (but not by much). The kernel of a homomorphism is a subgroup of the domain (of that homomorphism). Since H is of prime order, H has no proper nontrivial subgroups (and neither does G). Since ${\displaystyle p\neq q}$, what does this mean about the kernel of any homomorphism from G to H? (Think G/kerφ, which is isomorphic to Imφ....) –King Bee ^{(τ • γ)} 13:19, 18 June 2007 (UTC)[reply]

It can be more complicated, but it doesn't have to be. (This part is to King Bee, not the original poster.) Take G and H finite groups, with g in G. How might you prove (using additional assumptions on g and H if necessary) that this particular element g is sent to the identity by any homomorphism f: G → H? Tesseran 21:58, 18 June 2007 (UTC)[reply]

I suppose I would look at the orders of the elements. =)–King Bee ^{(τ • γ)} 22:43, 18 June 2007 (UTC)[reply]