So..infinity is a number? What does "closed form" mean?



For math homework, there's a problem that says to find the ratio, fifth term, and the nth term of a sequence: 5, 5^c+1, 5^2c+1, 5^3c+1, ...

I got that a₅ = 5^4c+1 and that a_n = 5^(n-1)c+1.
But what is the ratio? Is it just 5 or is it 5^(n-1)c+1 or something else?



Also, how does S_n = (1/2)(n(2a +(n-1)d)) = n((a₁+a₂)/2) ?
(That is, how does (1/2)(n(2a +(n-1)d)) turn into n((a₁+a₂)/2) ?)
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How does it work that the magnetic field lines points north inside a magnet? What really is a magnetic field? If I remember right it's something about where the north point of the compass would point but..that means if a compass were somehow stuck inside a magnet it would point north. What's actually going on..?

Infinity isn't EXACTLY a number but you can use it like one in many cases.

I've explained "closed form" before. Basically it's a function that returns the same values as a sequence if you pass the natural numbers to it, but it doesn't have any recursion (that is, self-reference) in it.

The ratio is just S_n / S_n-1. For a sequence like this one the ratio is constant across any two consecutive terms. Spoilering the answer:


Without context, I have no idea on Sn = (1/2)(n(2a +(n-1)d)) = n((a1+a2)/2), because I don't know what a, a1, a2, and d are.

What really is a magnetic field?

I mean, you've been studying fields. A magnetic field is a polarized field, just like an electric field. They behave the same way, they're just based on different (but related) underlying forces. "North" in a magnetic field and "positive" in an electric field are analogous.

The compass needle always aligns itself along the field lines, because one end is attracted to the north pole of the magnet and the other end is attracted to the south pole of the magnet. (Of course, that means the pointing end of the compass needle is in fact the compass's SOUTH pole. Don't let that confuse you.)

I thought the pointing end was the compass's north pole..?

Oops, I forgot ._.'''. So basically I can find the sum at any position (n) with closed form and that's what closed form is?

So to find a ratio I just look at one value and the next and if I don't know how much to multiply the first by, I just divide the next from the previous! Although, wouldn't that be 5^c?

I messed up writing it, woops. a = a₁. And 'a' represents the first term in an arithmetic series. d is the difference between two terms in an arithmetic series. S_n in this case is the sum for the arithmetic series. I learned to use those formulas to find the sum of things such as 1, 3, 5, ... up to a specified nth term in class.

Nope, because opposite poles attract. It's the part of the compass LABELED north, but magnetically speaking it's the south pole.

Yup.

Yup.

Um...

5^nc+1 / 5^(n-1)c+1
5^{(nc+1) - ((n-1)c+1)}
5^{nc + 1 + -(nc + -c + 1)}
5^{nc + 1 + -nc + c + -1}
5^{nc + -nc + 1 + -1 + c}
5^c

... yep, you're right, my mistake!

That means that d = a₂ - a₁, so you can substitute that into the expression.

n(2a₁ + (n-1)d)/2 = n((a₁+a₂)/2)
n(2a₁ + (n-1)(a₂ - a₁))/2 = n((a₁+a₂)/2)
2a₁ + (n-1)(a₂ - a₁) = a₁+a₂
2a₁ + (n-1)a₂ - (n-1)a₁ = a₁+a₂
2a₁ - (n-1)a₁ + (n-1)a₂ = a₁+a₂
(2 - n - 1)a₁ + (n-1)a₂ = a₁+a₂
(1 - n)a₁ + (n - 1)a₂ = a₁+a₂
(n - 1)(a₂ - a₁) = a₁+a₂

... by all accounts it doesn't make sense. Did I make a mistake, or did you transcribe something else wrong?

...Woops. It's actually
n(2a₁ + (n-1)d)/2 = n((a₁+a_n)/2)

2a₁ + a₂(n-1) - a₁(n-1) = a₁ + a_n
a₁ - a₁(n-1) + a₂(n-1) = a_n
a₁ - a₁n + a₁ + a₂n - a₂ = a_n
2a₁ - a₁n - a₂ + a₂n = a_n
a₁(2-n) - a₂(1-n) = a_n

It works now, but I'm not sure what it's saying.


Err do you mean the north of the earth is labeled north but it's actually magnetically south?

That is... entirely possible. I don't actually know. If the "north" side of the compass needle points to the south pole of a permanent magnet, then you are correct.

What it's saying is if you know the first term of a sequence, and if you know the (constant) distance between subsequent terms, then you can find any term of the sequence using that last line you derived, and you can find the sum of the sequence at any point using the identity you posted.

It should be noted that this follows from theorems you already know:

Σx = n(n+1)/2 -- the sum of the natural numbers
a(b+c) = ab + ac -- the distributive property of multiplication over addition
a + a + ... + a = na -- the additive definition of multiplication

Why?

Well, consider the sum of the integers:
1 + 2 + ... + n = n(n+1)/2

Our life is going to be much easier if we start with zero, though, so let's substitute (n-1) in place of n:
0 + 1 + ... + (n-1) = (n-1)n/2

Multiply it by d:
d(0 + 1 + ... + n) = d(n-1)n/2

Distribute the d:
0 + d + 2d + ... + nd = d(n-1)n/2

Add a₁ to each term, which is the same as adding n copies of a₁:
a₁ + (a₁+d) + (a₁+2d) + ... + (a₁+nd) = d(n-1)n/2 + n(a₁)

But of course, a₁+d is a₂:
a₁ + a₂ + ... + a_n = d(n-1)n/2 + n(a₁)

Gaussian Quadrature + MATLAB
 

Coda suggested I put this here for easier viewing :P

This problem has to do with Gauss Quadrature, specifically one-point to five-point Gauss-Legendre formulas.

Here's the problem:


And here's my code so far.
I've split it into two parts: the first part of the if statement deals with cases where n is even, and the second where n is odd. (n is the number of points used in the quadrature, and corresponds to the row of both matrices.)

I've tried it against a test code and an example from the text, both where n = 3, and both passed. I tested the example with n = 4, and since Gauss quadrature is accurate to the third degree, and the solutions for n = 3 and n = 4 were the same, I'm reasonably certain that the first half is okay. However, using n = 5 for that example, the answer given didn't match the actual value of the integral, or the solutions for n = 3 and n = 4.

My code fails an assertion (test), but that test is hidden, so I'm not sure which part of my code is messing up. However, since the solution for n = 5 seems wrong, I'm assuming that might be it.

Espy and I discussed this in chat... and failed to find an answer. :x I have been bested! ^^()

As an object is thrown, does it accelerate with anything other than gravity downwards (no air resistance)? That is to say, I throw a baseball cap into the air, and once I release the baseball cap, is the baseball cap accelerating anything because of me or has that ceased?

The answer to your question is F=ma. If you're not touching it, F=0.

Ohh I see. Acceleration needs a force.

Also, if an object were spun in a circle at a constant speed, would there be just a force to the center of the circle or would there be two forces: one to the center of the circle and another to the side?

And an English question: in the above question I used a colon then a question mark. Should I have used parentheses or does it even matter?
Another English question: Between "question mark" and "should," could I haave used a semi-colon?

If the object is spinning at a constant angular velocity, then the net force acting on that object is directed towards the center of rotation. If the object is speeding up, then you have that force as well as a (straight-line) force going straight forward that's causing the object to accelerate.

In practice you still have to apply a force going forward to keep an object spinning due to energy loss from air resistance, friction in the rope, etc. But the net tangential force is zero if it's spinning at a constant rate.

The colon is a fairly weak choice there. You COULD use a dash, a colon, a comma, or parentheses in that context. Each one has a slightly different nuance. Parentheses, for example, deemphasize their contents relative to the sentence, and you could take the contents out without impacting the meaning. A comma suggests that the clause that follows describes the noun before the comma. A colon suggests that the part afterward specifically follows from or equates to the idea introducing it. And a dash would function like parentheses except with stronger emphasis. These aren't hard and fast rules, and my interpretations of them are certainly not the only way to look at them, and you definitely shouldn't try to read too much into a statement based on its choice of punctuation.

A semicolon wouldn't have been wrong in that context but a period is better because the connection between the two thoughts isn't especially strong. Either sentence is strong on its own without need for further information.

Thanks for the responses!

So...
https://i.imgur.com/yxduXUQ.png

Why is theta = -pi/2 the same as x = 0 (my text-book's answer sheet)? Isn't it only the bottom half of x = 0?

I don't think the orange one exists, unless you're also, say, pushing the spinning object towards the right. (Perhaps explain your reasoning behind the orange force?)

In what context?

Thanks for the responses!

I wasn't sure if the orange force would exist or not because I was wondering if there needed to be a force towards the side in order for a ball on a string to swing around in a circle. Apparently that's a fictitious force though..?

I don't have the sheet with me anymore, but I think the problem was: change the polar equation into a coordinate equation.
The polar equation I was given was: theta = -pi/2
(Actually, now that I think about is, is theta = -pi/2 the same as the coordinate equation x = 0 because theta = -pi/2 means that (-n, -pi/2) and (n, -pi/2) are both included?)

Nope, what's propelling the ball in a circle is inertia. And since the ball is swinging around at a constant speed, there's no acceleration in the "around" direction, and hence no force.

And, ick, that's some weird wording. How I understand it, the Cartesian coordinates (0, 1) [x=0, y=1] translates to (pi/2, 1) [theta=pi/2, r=1] in polar, and (0, -1) translates to (-pi/2, 1), so if x=0, theta is either -pi/2 or pi/2.

Thanks for clearing that up!

Err, I thought that polar coordinates were in the form of (r, theta) though?

Regarding the orange force line: The tangential force line is 100% real -- it has a (net) magnitude of zero if it's not accelerating, but it's real. If you try to swing the object faster, then the magnitude will be nonzero until it reaches the new speed. That said, you and Espy are both right in that the force isn't necessary to be there for it to keep spinning, only for it to start spinning.

An interesting exercise is to think about how that force GETS there. When you're spinning the object around on the rope, you're not pushing it from the side in order to get it moving (unless you're playing tetherball) so how is it that you can accelerate it along that axis?

I personally prefer to notate polar coordinates either parametrically -- that is, r=1 θ=π/2 -- or using an angle symbol like 1∠π/2. That said, the standard notation is indeed (r, θ) in 2D and (r, θ, ϕ) in 3D.

Ohh I see. So it's like the moon where the moon just needed to be set with a velocity around the Earth at some point long long ago and now it's just technically falling towards Earth?

Does it get there because..
first the rope is pulled up
the ball gets pulled in the same direction as the rope
then the rope is pulled perpendicular to the ball's motion
then the ball starts going in a circle pattern because of the ball's inertia?
And the direction of velocity of the ball keeps getting changed which is why it can't just get pulled into the center of the circle it's making?

I've not learned notating polar coordinates those ways :o

Pulling the rope perpendicular to the ball's motion will just make it fall in towards the center. Pulling the ball up in the first place is also not putting any force in the direction of the orbit.

I'll give you a hint: It's a little bit of a trick question. It's a question that does indeed have a real answer, but the process of trying to find it will reveal the shortcomings in your model.

The angle symbol notation is used by some graphing calculators. I've not ever seen it used in a textbook. The parametric notation is mostly used when it's not obvious what coordinate system is in use or when you're defining functions in the coordinate space instead of individual points. (In the latter case, you'll often see "r(t) = ..., θ(t) = ..." to specify the coordinate in terms of a parameter.)

Then, is it just that the force comes from the person just pulling on the string to pull on the ball into the direction the person wants the ball to go to once in a while?

Uh whoops, yeah. Been a while since I've done anything with actual polar coords.

More or less, yeah. You can't model it as an object on a fixed pivot with an infinitely thin, infinitely flexible, perfectly inelastic tether. You either have to give the tether some stiffness (at which point you're modeling an object on a rod instead of an object on a string), or you model it more realistically:

When you're spinning an object around on a string, you aren't spinning it around a fixed central pivot. You're pulling it at an angle, applying the force in a circle around the center of rotation. The string isn't parallel to the centripetal force vector or perpendicular to the tangential velocity vector; you're pulling on it to one side in addition to towards the middle.

What about when you swing a thing then stop and let it swing on its own? The centripetal force keeps decreasing this way though.

Also, what's a "polynomial?" I have a review sheet asking me to identify if random expressions are polynomials or not, their degrees, and the "type" of polynomial they are, but I have no idea what a polynomial actually is. Why isn't cosine or sine something a polynomial? What does it mean that the "type" of polynomial is "based on the number of terms?"

If you stop, and you let it keep swinging on its own, then in an ideal system it would just keep moving forever like a planet in orbit. In reality, there's friction that will bleed off that momentum in the form of heat, and it'll spiral towards the center.

A polynomial is just ax^m + bxⁿ + ... + k. Nothing fancier about it than that. The degree of the polynomial is the highest exponent that appears in it. Cosine and sine aren't polynomials because they're not sums of multiples of powers of a single variable. (That said, there are polynomial approximations of those functions, such as Taylor series.) I'm not sure what the "type" of a polynomial is because that's a rather vague term and it could mean several things.

Coda, maybe "type" has to do with second order, third order, etc? Though that doesn't necessary dictate the number of terms, just the maximum number of terms.

Or perhaps, erm, even and odd polynomials?

Thanks for the responses! I dunno what the "type" refers to and still need to ask about that.
So.. if something is ax^3/2 is it a polynomial? What about ax^-1?

Also, how do you find possible imaginary roots of a polynomial? I think I can find not-imaginary ones by having any factor of the constant over any factor of the coefficient of the largest degree, but what about imaginary roots?

Also, when I'm given a problem that tells me to find a polynomial with real coefficients that have a root 2 - i, I know that I have to do something like (x + 2 - i)(x - (2 - i)) but I'm not sure where to go from there. I forgot how to get rid of the i's.

Non-negative integer exponents. Those two examples are not polynomials. A term including x⁰ (that is, a term not including x at all) on the other hand can still be in a polynomial.

You get rid of the i's using i² = -1.

As for finding imaginary roots, we covered that already, you might need to go back a few pages.

Is it reasonable for a car to accelerate at -18 mi/s^2 while it's braking?

(Context: I have a homework problem that asks me to find the acceleration of a car while it's braking: "According to the recent test data, an automobile travels 0.250 mi in 19.9 s starting from rest. The same car, when braking from 60.0 mi/h on dry pavement, stops in 146 ft. Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down or speeding up. (a)Find the acceleration of the car when it is speeding up and when it is braking.
I found 1.51 mi/h^2 while speeding up so I figured maybe braking would be similar.
Tried to use x_f = x_i + v_i*t + .5at^2 and got -(60.0 mi/h)^2/(2*.0277 mi) = 6.50 * 10^4 mi/h^2. I tried to turn it to mi/s^2 because clearly the car isn't braking in a whole hour (still the number's huge?), but it was -18 mi/s^2. I'm not sure where I went wrong on this problem. The car couldn't possibly be accelerating that fast right..? Even earth rotates slower than that, right..?)

Without doing the actual calculation to check your answer, I'm looking at it in reverse. At an acceleration of -18 mi/s^2, it would mean you would stop nearly instantly even going 60 mi/h. So your initial thoughts are correct that the acceleration is too fast. It does still need to travel the 146 ft at least.

Using formulas here http://www.dummies.com/education/sci...-and-velocity/

A speed of 60 mi/h is 88 ft/s.
146 ft = (88 ft/s)x(time) so time to stop is 1.659 seconds.
So then:
1.659 sec = (-88 ft/s)/(acceleration) so acceleration would be -53 ft/s^2, which is a lot less than -18 mi/s^2

Thanks for replying!
(Also I actually used v_f^2 = v_i^2 + 2a(x_f-x_i) oops)

But I'm confused, why is distance = velocity x time ?
Isn't in this case distance = velocity x time + acceleration affecting the velocity during the time x time?

Edit: not confused anymore, thank-you!
Edit2: actually, I'm still confused why I can use the 60.0 m/s as the only speed when speed is changing.

So the part that's confusing you -- and a small mistake Serra made -- is that the formula is specifically x = v_meant. It's not supposed to be the initial speed for the whole distance, it's the average speed.

For uniform acceleration, you should remember this from last semester:
v_mean = (v₀ + v_f)/2.

So Serra's math is off by a factor of 2, because the average speed during braking is not 88 ft/sec, but 44 ft/sec.

The rest of the process is correct:

x = v_meant
146 ft = (44 ft/sec) * t
t = (146 ft) / (44 ft/sec) ~= 3.32 sec

a = Δv/t
a = (-88 ft/sec) / (3.32 sec) -- this one is still 88 because it's the total change in velocity, not the average
a = -26.5 ft/sec²

EDIT: If you want to use v_f² = v_i² + 2a(x_f-x_i) you can, and it gives you the same result:

0 = (88 ft/sec)² + 2a(146 ft)
-7744 ft²/sec² = 2a(146 ft)
-53.0 ft/sec² = 2a
-26.5 ft/sec² = a

EDIT 2: Also your initial answer of 6.50x10⁴ mi/h² was in fact correct:

(65000 mi/h²) / (3600 sec/h) / (3600 sec/h) * (5280 ft/mi) = 26.48 ft/sec²

Dimensional analysis is a wonderful thing. :P

Thanks for catching that Coda. I did miss it.

Also, I need to learn how to write sub and superscript like you XD

Thank-you for explaining! I hadn't realized I was given all the stuff needed to find average velocity :o

Just [ s u b ][/ s u b] and [s u p][/ s u p] xD? (I forgot this existed, and I was lazy)

Also, one more question: for something that uses x_f = x_i + v_it + (1/2)at², if I just try to solve for t I get two answers. Usually one's negative and one's positive, but sometimes both are positive and in either case I don't know what they mean.

(More specifically, I have a problem that says: A boulder takes 1.30 s to fall the last 1/3 of the way down a cliff, and I have to (a)find the height of the cliff and (b)explain what the two solutions to a quadratic meant if I got two solutions and used a quadratic equation.

Along the way, I suppose I had to find the time too. (I didn't figure out how to do the problem myself so I basically looked at the solution and tried to understand it T_T)

Using x_f = x_i + v_it + (1/2)at², I get that
h = (1/2)gt²
(2/3)h = (1/2)g(t-1.30s)² --> h = (3/4)g(t-1.30s)²
(1/2)gt² = (3/4)g(t-1.30s)²

t² = (3/2)(t-1.30s)²
t = .716 s or t = 7.08 s

If I square rooted that above equation, the t = .716 s would disappear, but otherwise I would only know that t must be 7.08 s because .716 s is too short. But what does .716 s mean?)

You will always get two answers for a quadratic (that is, containing x²) equation unless the apex of the parabola is exactly on the x axis.

Do you already understand the physical significance if one value for t is negative? There's a meaningful answer but I don't want to do your homework for you.

If both values for t are in the future, there's a good reason you're confused: it PROBABLY means you've made a mistake somewhere. Either you're not considering the ground to be 0, or you've got the sign for one of the initial parameters wrong (the initial position, the initial velocity, or the acceleration), or you have an error in your computation somewhere, or you have one of the parameters wrong.

In this case, you've got the parameters wrong. Look at the formula more carefully:

x_f = x_i + v_it + (1/2)at²

I'll give you three hints:

1. If the height of the cliff is h, then what's the position of the boulder when you start looking at it?
2. What's the value of v_i?
3. You don't have enough information to solve this all with a single equation.

I solved h to be 246 m based on the time being 7.08 s (since it couldn't have been .716 s because the total time was over 1.30 s).

But what I'm uncertain about is what the .716 s means. Although, I guess maybe .716 s might actually not in the future because when I solved for t, I got it from:
t² = 3/2(t-1.30 s)², and .716 s - 1.30 s < 0 ? Maybe since 7.08 s - 1.30 s is the time it took for the boulder to fall 2/3rds of the way, .716 s - 1.30 s is how long it would have taken the boulder to get to that same 2/3rds mark had the boulder been allowed to fall through the ground 2/3rds(I'm not sure exactly at what point) more then have time be rewound so the boulder would look like it was falling up from through the ground?? (All I remember from class about negative times was something about falling differently...granted this wasn't exactly a negative time, but even if I had considered the ground to be zero, I think the negative numbers would have canceled each other out?)

So yeah, you're still trying to solve it the same way. Try a different approach.

You know that it started at height 3h at time 0, and at time t₁ it's at height h, and at time (t₁ + 1.30). You're assuming that its velocity at time 0 is also 0, and the acceleration is a constant g, because they didn't say otherwise.

How can you plug these values into equations you know to get information you don't know?

Note that you shouldn't get two positive answers if you do this right because the top of the parabola is defined by these initial conditions to be at t=0.