Looking over this thread this might be ridiculously basic but I'm not sure if I've done this right.

Known facts: we have 2 parallel lines one which goes through (3, 1) and the other that has the solution 0,8x + 2y -2 = 0

Question: the two lines and the coordinate axes limits an area. How big is it?

Solution: 0.8x + 2y -2 = 0 = (0.8x - 2 = -2y)/-2 = y = -0.4x + 1
y1 = -0.4x + 1
y2 = (3*0.4) + 1 - 0.4x = -0.4x + 2.2
y1 has coordinates (0, 1) and 1/0.4 = 2.5 (2.5, 0)
y2 has coordinates (0, 2.2) and 2.2/0.4 = 5.5 (5.5, 0)

(2.2*5.5)/2 - (1*2.5)/2 = 4.8

Answer: 4.8 units

does this look right?
ty in advance.

The final result could be right, the way you wrote it down is hard to follow though.

Ok. So let me try it again.

First lines known facts are that it goes through (3, 1) and is parallel with second line.
Second lines known facts are 0,8x + 2y - 2 = 0

We start off by calculating what x should be for both lines by calculating the correct formula for the second line: 0,8x + 2y - 2 = 0 should read (0,8x - 2 = - 2y) but since we want a single y we divide both sides with -2 which gives us y = -0.4x + 1

So our second line is y = -0.4x + 1 which tells us our first line is y = -0.4x + 3*0.4 + 1 which is y = known slope (-0.4x) plus x coordinate from known facts(3) times the slope(0.4) [it's actually -3*-0.4] plus y coordinate from known facts (1)

No we have Y1 = -0.4x + 1 and Y2 = -0.4x + 2.2 therefor we also have coordinates from when they break the y axe Y1(0, 1) Y2(0, 2.2). Then we have to figure out where they break the x axe which for Y1 is 1/0.4 = 2.5 and for Y2 2.2/0.4 = 5.5

Now we know Y2 (which is the bigger of the two) breaks the axes at (0, 2.2) and (5.5, 0) which creates a triangle with the area: (5.5*2.2)/2 = 12.1/2 = 6.05
Y1 with coordinates breaking the axes at (0, 1) and (2.5, 0) creates a triangle with the area (1*2.5)/2 = 1.25 and since it's within the bigger triangle, the area that the two lines make should be the rest of Y2s area - Y1s. 6.05 - 1.25 = 4.8

Sorry if it's still messy, not sure how else to explain it.

Okay. Since each term is getting smaller (in an absolute sense) we know that the number that the series converges to will have to be less than or equal to the next number in the series. This allows us to say definitively that b_(n+1) is the greatest possible error/remainder we could have between S_n and S.

Yes.

Thank you.

The bolded doesn't convince me of anything, because you don't say why.

Ace, a piece of advice:

You should write mathematics as if you're teaching someone, as if you're explaining the concept to someone who has never seen it before. Additionally, you should not be guessing when you write things down. If you, yourself, are not convinced, how might others be?

A full proof that the alternating series test works is actually not easy to write down (for a calculus student, but it is a nice exercise for a student in real analysis). But there is a nice picture in Stewart (and elsewhere) that tells the story. You're using Stewart, right?

A version of this picture appears in Stewart. Hopefully my notes are helpful. The number line was constructed 0, S1, S2, ..., and then I put S in at the very end. Hopefully you see how the S_n's squeeze in to "find" S, and then see why the error bound is true.

Noted, and thanks for the drawing; I've seen the similar one in Stewart's calc book as well as the one Bruce posted, but I think it has sunk in now. How is the following:

since S_n + b_(n+1) = S_(n+1) and
since S_(n+1) <= S <= S_n, then
|S-S_n| <= b_(n+1)

Your second line is only true half the time...

Why is the second line ever true? Not saying that it isn't, but I don't see a good justification for it yet.

Aaah. What if I say something like "where (n+1) is an even integer"? I'm guessing there is a more general way to say it mathematically that doesn't require the clarification my argument requires.

Isn't it true if we assume that the picture Wyman posted describes the nature of the relationship between s, s_n, and s_(n+1)?

(and add the bit about (n+1) being an even integer)

But why can we assume that?

The assumption on {b_n} was that it's positive and monotone decreasing to 0.

My argument is a bit hand-wavy, but it provides enough justification for a student in calc 2 imo.

If you want more rigor, you can see any number of analysis texts. The gist of the argument is that WLOG we asume b_1 is positive, and we claim (and show, in the proof) that S_odd is monotonic decreasing and S_even is monotonic increasing, and that moreover both are bounded (the evens above by s1, and the odds below by s2). Hence they both converge. Since the tails differ by b_n, which goes to 0, the limits must be the same. The even and odd partial sums converge to the same limit, so the sequence of partial sums (hence the series) converges to that limit as well (choose max(N_even, N_odd) s.t. |S - S_n| < epsilon for n > N_{e/o}). This shows convergence of series satisfying the criteria of the AST.

Now, S_even increases to S (i.e., S_even < S) and S_odd > S, so
S_2n < S < S_{2n+1} = S_2n + b_{2n+1}, and similar for odds, showing the error bd.

I would even say that the odd elements are bounded by any even element, and vice versa, and thus you can avoid using the theorem on monotonic bounded sequences... but yes, this may be above the level of Calc 2.

My point was not that I didn't think you could prove it - my point was to warn Acemanhattan about handwaviness.

The way I try to speak this argument to my calc students is that we start at 0 and add b_1 to get s_1.

Then we subtract b_2 to get s_2, but since b_2 was less than b_1, we get 0 < s_2 < s_1.

Now we add b_3 to get s_3, but since b_3 is smaller than b_2, we don't get all the way back to s_1, so we get:
0 < s_2 < s_3 < s_1.

Then we subtract b_4 to get s_4, but since b_4 is smaller than b_3, we don't get all the way back to s_2.
0 < s_2 < s_4 < s_3 < s_1.

And so on. If you keep doing this, you'll get

0 < s_2 < s_4 < s_6 < ... < s_{2n} < ... < ... < s_{2n+1} < ... < s_5 < s_3 < s_1. (EQN *)

Notice that if we just look at the evens, we get s2 < s4 < s6 < ...
And if we just look at the odds we get s1 > s3 > s5 > ...

To get that S is wedged in between the S_odd and the S_evens, we can take a leap of faith -- doesn't it seem like it has to be?

Aside, which I don't tell my calc class, which I tried to describe in the last post: Or we can use some facts from analysis, i.e., that a monotone bounded sequence converges. Since s_2 is smaller than all the s_odd (which decrease) and since s_1 is larger than all the s_even (which increase), we know that {s_odd} and {s_even} are monotone bdd, hence they converge. Also, since |s_2n - s_{2n+1}| = b_{2n+1} --> 0, we know that s_odd and s_even must converge to the same limit, which we'll call S.

{s_odd} converges to S means that there exists an N_odd such that for n_odd > N_odd, |S - S_n_odd| < epsilon. Similarly there's an N_even such that for n_even > N_even, |S - S_n_even| < epsilon.

So take the max of N_odd and N_even, call it N, and note that for n > N, |S - S_n| < epsilon. Hence {S_n} --> S. This shows convergence.

Now, looking at (EQN *), we can rewrite, this time including S, as:
0 < s_2 < s_4 < s_6 < ... < s_{2n} < ... < S < ... < s_{2n+1} < ... < s_5 < s_3 < s_1.

And now the error bound becomes clear.

5141 was before I saw 5140.

Forgive me; I've been fairly busy and only popped my head in intermittently, so I forget who is where in terms of level. I did think you were more advanced, but I wasn't sure.

Sorry. But maybe 5141 will help anyway. :shrug:

No worries

Point taken about the handwaviness.

I made the assumption that because of Wyman's knowledge that I am only in...ahem...Calc 3, his graphical proof (though not formally rigorous) could be trusted to describe what was happening in general.

Got an A in my signals class

I'm looking for a chart of the composition of Earth's atmosphere by mass (ie, nitrogen comprises x% of the mass of the atmosphere.) Everything I google gives me the composition by volume. Anyone know a good resource for this?

The ideal gas law and the periodic table. A mole of nitrogen should take up the same volume as a mole of oxygen or argon.

Thanx a lot!

Note that my r1 and r2 are the reciprocal of your roots because they are solutions for z^-1 instead of z. Then since the roots are complex, you can reduce h[n] to sine functions multiplied by (sqrt(2)/2))^n using Euler's identities like

sin(x) = [exp(jx) - exp(-jx)] / 2j.

Depending on what is in your table of z-transform pairs, you could do this directly without a partial fraction expansion. For example, it may have the transform of a 2nd order transfer function like:

1 / [1 - 2r*cos(theta)*z^-1 + r^2*z^-2].

Then you can make the appropriate identifications for r and cos(theta), and recognize that the 1 - z^-1 in the numerator will transform to h[n] - h[n-1]. That comes from the partial fraction expansion though.

When you have 2 complex conjugate roots on a circle, the possible ROCs are actually |z| < |1/r1| for causal and |z| > |1/r1| for anti-causal using my notation, and the other root gives the same 2 possibilities, so they are either inside or outside the circle. In general, there will be a maximum of 3 ROCs for 2 roots, not 4, because it is either inside both, outside both, or the annulus in between. It can't be inside the inner one and outside the outer one (real roots) because the whole transfer function has to converge everywhere in the ROC.

I'm supposed to find an expression for

-ln[sum_y[sum_x[q(x)p(y|x)^(1/(1+rho))]]^(1+rho)]
(the 1+rho power is around all terms in the y summation)

given that q(x) = N(0,P/(2W)) and p(y|x) = N(x,N_o/2)

so I write the terms as Gaussians, but my two main questions are:

1) Won't the q(x) term just = 1 when summed over all x?
2) How do we treat this sum_y when there are no y values?

The homework even tells the solution when rho=1 ==> ((1/2)ln(1+P/(2*N_o*W))

Any direction/ideas? Thanks