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## #1 Re: Help Me ! » Another counting question » 2006-12-22 07:20:44

Sounds like it's just an ambiguous question. When it said "at one end", it probably just meant "at this one end", when you could just as easily interpret it as meaning "an end (either one)". Just goes to show teachers need to be careful in how they phrase their questions.

## #2 Re: Help Me ! » solve for x by cramer's rule » 2006-12-19 07:00:16

Wow, I vaguely remember something about that. That's kind of cool actually - I wonder why that works. Already spent way too long messing around here today though, got to get back to work!

(please ignore my previous post )

## #3 Re: Help Me ! » solve for x by cramer's rule » 2006-12-19 06:26:49

Hi, it's been a while since I've done this kind of thing, and I'm not sure what background you have, so please ask if you don't understand something in my explanation.

First off, in your post you are forming several different matrices and taking their determinants. As far as I know, there is no use for the determinants of the matrices you call Dx, Dy, and Dz, only of the original matrix of a's, b's, and c's.

There are two ways I remember that you can approach the problem. You can try to invert the matrix, or you can use elementary row operations to solve for each of the variables. I'm not sure which of these you've been exposed to. But either way you should try to understand why the determinants are important before you try to use them.

Solve:
x + 2y + x = 0
3x - y - z = 5
5x - 2y - 3z = 12

Hopefully you've seen matrix multiplication. It allows me to rewrite this:
( 1    2    1  )      (x)      (0)
( 3  -1   -1  )   x  (y)  =  (5)
( 5  -2   -3  )       (z)      (12)

If you multiply out the matrices, this one is exactly the same as the three previous equations. So why do we care about the determinants? Because if a determinant is zero then you can multiply by the inverse on both sides. It cancels the matrix on the left, and you are left with:

(x)                      (0)
(y) = (inverse) *  (5)
(z)                       (12)

If the inverse exsits, you can solve the problem by finding the inverse and plugging it into this equation. If it doesn't, you can't solve the problem no matter how you approach it. But the determinant can never be directly used to get the solution like you're trying to do.

Have you been taught how to find inverses of matrices?

## #4 Re: Help Me ! » Can someone help me with these two sums in complex numbers » 2006-12-19 05:59:09

Could you add parentheses please? Otherwise I'm not sure what exactly the questions are. Not sure if I will be able to help until I see where they go.

## #5 Re: Help Me ! » Another Physics Problem » 2006-12-19 05:53:38

fusilli, what you say is true, and if you get the same answer that way, that might be a good way to think about it. I think you're right too Stanley -- at least you're expression for the normal force looks like it matches mine.

I will say that the reason these approaches look simpler though is because you are combining a number of steps into one. As long as you are careful that's fine, but it is much easier to make mistakes that way and it's harder to justify your steps to others. The most organized and convincing way (in a physics teacher's point of view) to answer a problem like this:

1) Draw a free body diagram and label all the relevant forces
2) Write down the equations for the forces and accelerations you want to use
3) Use these equations step by step to show how you get your answer
4) Plug in numbers for your symbols at the end (sure you can do it earlier, but it's much easier to grade and you don't have to use your calculator until the end this way, so there are fewer oportunities for mistakes)

## #6 Re: Help Me ! » Compound interest » 2006-12-17 08:46:09

Let's start with an example. Say you have \$100 in a bank and you will be making a 1% interest rate per year - only added on at the end of every year. How much money will you have in three years?

Well, after one year you will have:

100 * (1.0 + 0.01) = 101 dollars

After two years you will have:
101 * (1.0 + 0.01) = 100 * (1.0 + 0.01) * (1.0 + 0.01)

And after three years you will have:
100 * (1.0 + 0.01)^3 = \$103.03

Does that make sense? If so then let's go to the next step, involving symbols. Let's say instead of getting 1% every year, pertend you get r% interest per year, compounded n times per year. You want to know how much money you make after t years. Then, your compounding will give you

[align=center]

[/align]

Where A-sub-i is the money you start with and A-sub-f is the money you end with. You're multiplying by (1+r/n) each time because you would only get the full r% interest if n was once per year. Now pretend that you wanted the compounding to happen extremely often -- you're always getting some of your interest added back in. This means n is extremely large. There's a formula (which  you can derive with some simple calculus) that

[align=center]

[/align]

If you send n to infinity. So the formula you can use for compounding interest is:
[align=center]

[/align]

Which gets more accurate the larger n is. I think that's how it is anyway, I've never actually had an economics class or anything, so if I have my terminology confused then maybe someone can correct me.

## #7 Re: Help Me ! » One Last Derivative » 2006-12-03 17:12:07

Yeah, that's definitely something you should be sure you're clear on. I would bet \$20 your teacher means the "inverse of csc", not 1/csc, but you should definitely ask.

## #8 Re: Help Me ! » One Last Derivative » 2006-12-03 17:02:01

I must be misreading the problem. It looks to me as though you are writing (for the first part):

[align=center]

[/align]

In this case, by applying the chain rule I get:

[align=center]

[/align]

or

[align=center]

[/align]

[align=center]

[/align]

This is different from what both you and your teacher have, so perhaps I'm missing something. Anyway, csc is 1 over sin, not the inverse sine function, so you have to be careful (just like you can't write tan-1 as cos/sin and evaluate from there in the term on the left you get the wrong answer either way). Off the top of my head I can't remember a good technique for evaluating csc-1, so I'll take the wimpy way out and refer you to a tabulated answer

http://www.efunda.com/math/derivatives/TrigD.cfm

## #9 Re: Help Me ! » Trig Functions Subtitutions in ∫√ (a^2-x^2) , x = asinθ » 2006-12-03 06:12:01

Well, you already showed that you went from x to theta using x=2sin(theta), so you can get theta in terms of x by simply inverting this if you like: theta = sin^-1[(1/2)x]. It's probably a bad idea to do this before integration, however, because you have an easy function to integrate here and the substitution would make it quite tricky. If you're doing an indefinite integral, you can just integrate this and then sub in for x, remembering to put in the unknown constant. If it's definite, you just need to remember to use the theta limits while you have the expression in terms of theta, and the x limits while you have the expression in terms of x. Does that make sense?

Also, you should be careful about your d's. I see that through most of your work you have both D(theta) and d(theta). First off, in introductory calculus, everyone thinks in terms of lower case d's, so you probably should too to avoid confusion. Second, at this level you should always have the same number of d's in your equation as integrals. Each integral gets rid of one d, so if you have two d's and one integral, you will be left with a d in the final state. Then, since d(theta) is infinitely small, your answer is zero! (actually, it would represent a sort of "density" in theta as I understand it). It might be a good idea to go back and try to figure out how you got the second d and where your mistake is. Again, ask if you're confused and I'm sure someone will explain.

## #10 Re: Help Me ! » statistics and probability » 2006-11-30 16:16:13

Definitely true - step 2 is entirely optional. I just find it easier to think about things and explain them in terms of performing thought experiments to see what outcome you'd get.

## #11 Re: Help Me ! » Calculating the CDF (probability) » 2006-11-29 15:45:29

Hi benn,

From looing online, it looks like you're right that the cdf sums the pdf probabilities. Basically, the cdf tells you the fraction of numbers that are less than or equal to some value in your distribution. If you know calculus (if not that's ok too), it represents a definite integral of your pdf from the minimum to your test value.

I'd be very surprised if they expect you to find some function that fits this distribution. Probably if you just make another table where you list the fractions that are less or equal to the values in each bin then that would be good enough.

## #12 Re: Help Me ! » statistics and probability » 2006-11-29 15:35:15

Here's a strategy:
1) First figure out what the probability is of rolling each number on the die (this shouldn't be too tough).
2) Then, predend that you rolled the die a thousand times (or whatever number you want). Figure out how many times you can expect to roll each number based on the probabilities you got in 1. Make sure you don't round away fractions!
3) You can directly use the results of 2 to plug into the formulas for E[x] and V[x], the expected mean and standard deviation.

## #13 Re: Help Me ! » Help with variation =S » 2006-11-19 09:33:08

Intensity represents the number of particles passing through some area in a given amount of time - in your case you're talking about light "particles" coming from a lamp. Let's say you know how many particles there are coming out of your lamp per second. You should also assume these particles fly out uniformly in all directions - this wouldn't make sense for a flashlight, say. You should then be able to answer questions like: "If I put a screen 20m away, how many particles will hit every square centimeter of it per second?"

One way to think about it is in terms of spheres. Assuming no particles are created or destroyed on the way out, at radius 20m the initial particles will be spread out over an area of 4*pi*20^2 - just the surface area of a sphere at your radius. So because the sphere has an area proportional to the square of its radius, the light particles become more diffuse as the square of the distance they travel outwards. And since the energy carried is proportional to the number of particles carrying it, you can say the same thing about the heat energy they deliver.

## #14 Re: Help Me ! » A reversal of the Birthday Problem » 2006-11-06 13:53:52

Intersting. This post covers a lot of different topics in some detail and I am left unsure what exactly it is that you're trying to do. It sounds as though there are a lot of different aspects that you want to look at in the model and this can be overwhelming for potential responders, most of whom aren't experts in statistics and none of whom have probably have read the papers you're referring to.

It would help if you could reduce your post to a simpler question or two if there are some concepts you are struggling with. At this point I'm not sure what the level is of the marterial you're struggling with.

## #15 Re: Help Me ! » I just want a hint (Limits & Calculus) » 2006-11-05 04:11:26

I assume you are talking about taking the derivative of the denominator in using L'Hopital's Rule here. Yes, it is definitely true that you can make the simplification there that sin(x) + xcos(x) = x + x + O(x^2) for x very close to zero ... not quite sure where you're going with this though ...

## #16 Re: Help Me ! » I just want a hint (Limits & Calculus) » 2006-11-03 16:01:53

Actually, I would also recommed the Taylor Expansion approach for this problem. I think it's more intuitive, plus it has the advantage of also being useful in situations where you aren't dealing with 0/0 or inf/inf, but you want to simplify the function without removing the x dependence.

You know:
[align=center]

[/align]
and
[align=center]
[/align]

As can be shown with a Taylor Expansion. When you take the limit as x becomes small, you know the higher order terms become small faster than the lower order terms, so you can ignore terms above x^n to the "nth order" of accuracy.

Expanding to first order still leaves you with 0/0 for this problem (which is why you need two iterations of L'Hopital's Rule), but to second order,

[align=center]

[/align]

In the limit as x->0, which gives you the same answer.

## #17 Re: Help Me ! » I just want a hint (Limits & Calculus) » 2006-11-03 15:47:17

Actually, the rule was developed by a Swiss person, who was paid off by a French person, or so is believed

http://en.wikipedia.org/wiki/L%27hopitals_rule

## #18 Re: Help Me ! » Probabilty » 2006-11-03 15:43:32

If you don't have graphing calculator, there are many tables you can find online which give you the information you need for this, such as the one near the top of this site. If you don't understand how to use this, please ask.

http://www.statsoft.com/textbook/sttable.html

## #19 Re: Help Me ! » Find the area » 2006-11-01 21:48:38

I'll get you started with the basic idea. Generally when you have these absolute value problems you should start by isolating one of the absolute terms, and thinking about just one at a time. In your case, we have:

[align=center]

[/align]

There are two possibilities for x+y: it can either be negative or positive. If it is positive the absolute value changes nothing and we can just take it away like this:

[align=center]

[/align]

Or it could be negative. In that case |x+y| = -(x+y), so you can substitute that in:

[align=center]

[/align]

These two equations describe different sets of points, but when you take them both together you get the same points as in the original equation with two absolute values. So you really can use these two equations in place of the one before. We started with one equation for the points in this graph, and by removing an absolute value we ended up with two equations for two different parts of the points that satisfy this graph. There is now one absolute value left in each equation. You can remove it in the same way as before and each equation will split into two more giving you four total equations that, taken together, specify the same set of points as the one original equation with two absolute values. Does that make sense?

Once you have all four equations try sketching the graph. It should be pretty obvious how to find the area of the result, but please ask if you get stuck.

## #20 Re: Help Me ! » hyperbollic functions » 2006-10-27 11:04:22

I'm pretty sure that if you allow these numbers to be complex you can actually always pick A and B so that this expression equals Asinh(x+B). It looks as though the problem you're trying to solve requires you to keep your constants real. Then depending on the values of a and b you may only be able to get one of the sinh and the cosh expressions to work without having lograthim of a negative number involved.

I recommend this approach: use the exponential expansions for your hyperbolic functions:
[align=center]

[/align]
[align=center]
[/align]

You want to show that there are some values you can pick for A and B such that y=y1 or y=y2. Start with y1 and pick A in terms of a and B such that the positive exponential in x has the correct coefficient, and then see if you can figure out what B needs to be so that the negative exponential in x has the correct coefficient. Your final answer will only be real depending on the relative signs of a and b. You will then need to show that your y2 expression takes care of the cases of a and b that give imaginary results for your y1 answer.

Does that make sense? Let me know if you need more detailed advice and I'll try to get back to it sometime in the next couple of days.

## #21 Re: Help Me ! » Another Physics Problem » 2006-10-25 17:00:07

I was using F-sub-p in place of P in your diagram to represent the horizontal force being applied to the large block. I probably should have stuck with P to be less confusing, but P is often used to represent pressure, so I instinctively avoid that. It is not the same as the normal force F-sub-N being applied between the two blocks. I will be a little more explicit here:

The small block is accelerated by the normal force F-sub-N, so

[align=center]

[/align]

Where the small block has mass m and is being accelerated horizontally with acceleration a. But we know that the big block is also accelerated at exactly a too (otherwise they wouldn't always be touching). So F-sub-p (or "P" the way you were using it) is

[align=center]

[/align]

Since it has to push both masses at acceleration a. If you need a better justification for this let me know and I can give it to you but I'm tired now, so I'll go on to the punchline. I then solved for a and plugged it back into the F-sub-N equation:

[align=center]

[/align]

The stuff inside the {} is just a. Does this all make sense?

## #22 Re: Help Me ! » Simple Calculus » 2006-10-25 16:45:28

A function is discontinuous at some point if there is a break in it at that point - this is mathematically represented by what polylog wrote. If there is a break at a point then you won't be able to find a limit for the function at that point, or in some cases you can but it won't be the same as the function evaluated at the point. Any point at which a function goes to infinity is guaranteed to be discontinuous.

The function listed here actually has two singularities, but they aren't both removable. This definition may not be formally correct, but the way I think about it a singularity is removable if you can "remove" it and make your function continuous by redefining the value of your function at that one point. For example,

[align=center]

[/align]

technically has a singularity at x = 0, but you can "remove it" if you redefine f = 1 only at x = 0. In this case f is continuous at x = 0. You can logically see that it's smooth there, and if you formally evaluate the limit of sin(x)/x at x = 0 you will find it has 1 as its limit (this can be shown either through L'Hopital's rule or a Taylor expansion). A similar substitution can be made at one but not both of the singularities in the function you're trying to solve, but you should try a little more to think this through yourself before we give the answer away. Let us know if you're still stuck!

## #23 Re: Help Me ! » trajectory » 2006-10-22 15:36:17

Sweet, I never got to shoot guns when I was in school.

You are asking some pretty broad questions though. What level of class is this? If you already know a little calculus that would make it a lot easier to explain some of this - you don't need to remember formulas then, you can figure them out yourself! But if not then that's ok too.

Whether you know calculus or not, the key is to not just know the formulas, you should try to really understand them. You're starting with a constant acceleration, g, from gravity. Acceleration is what changes velocity over time, so

[align=center]

[/align]

You start at some initial speed v sub i, and then over time gravity changes your speed more and more as you can see from the right most term. Of course, velocity changes position in the same way, but it's a little more complicated because the velocity isn't constant like acceleration is. Each of the terms should look reasonable though. What you get from a little calculus is:

[align=center]

[/align]

These two formulas can tell you all you need to know about the vertical movement of the projectile. As for the horizontal movement, as long as you can ignore air resistance it will always be going at the same speed. So as long as you treat the up and down movement of the projectile separately from the side to side movement you really only need to work with two formulas for most projectile problems.

But all formulas aside, it helps to clearly picture what you're dealing with (drawing a sketch helps).  All math aside, how far do you think a bullet would go if you shot it at 90 degrees (straight up)? This is a very broad topic so I'm going to stop here for now, but feel free to ask more specific questions.

## #24 Re: Help Me ! » Statistics » 2006-10-22 14:43:23

Hi stormswimmer,

Rather than giving away the answer straight off I'm going to try and guide your thinking. Please ask again if you still need more help after this or if you want us to check your thinking.

First think about what the mean and standard deviation really mean outside of any math. The mean is an average of a bunch of numbers, and the standard deviation represents how spread out the numbers are. So if you added 25 to every number, how much do you think the average of those numbers should change by? And will they become more or less spread out than they were before?

## #25 Re: Help Me ! » Multiplication of Power Series » 2006-10-21 00:06:09

Well, you can multiply the results of what they converge to together rather than the series themselves.  But if you want to work with expressions in series form I don't see how you can get around multiplying things out term by term.

I'm not sure how advanced calculus can help you here. I'm willing to bet though that the theorem just says you're allowed to multiply them out term by term - that doing so doesn't change the answer. This wouldn't actually help you do the multiplication though.