Algebra question

[gGo]

OK, I know this is a simple question. Working on a problem with summation [MATH]a_n=\sum_{k=n+1}^2n[/MATH] and comparing this sum to the right approximation of the area under [MATH]1/x[/MATH] over the interval [MATH][1,2][/MATH].

The idea is that [MATH]a_n[/MATH] will subdivide [MATH][1,2][/MATH] into [MATH]n[/MATH] subintervals. Normally, to find the endpoints of an Riemann sum we use [MATH]x_k = a+k \cdot \Delta x[/MATH]. In this problem, [MATH]a=1, \Delta x = 1/n[/MATH], and, if n=2, then k starts at 2, which breaks the formula. If [MATH]n=2[/MATH], then [MATH]k = 3, 4[/MATH]

[MATH]x_3 = 1+3/2 = 5/2[/MATH] (should be 3/2)​

[MATH]x_4 = 1+4/2 = 6/2[/MATH] (should be 4/2)​

I am told that the left endpoints are [MATH]\frac{k-1}n[/MATH] and the right endpoints are [MATH]\frac kn[/MATH], which checks out. But I'm curious how we figure that out. The best I can figure is to shift the index, like this:

Let [MATH]m=k-n[/MATH]. Then, for [MATH]n=2[/MATH], we have that [MATH]k=3, 4[/MATH] and [MATH]m=1, 2[/MATH] (that is, starts at 1). Then the right endpoints are [MATH]x_m = a+\frac mn = 1+\frac mn = \frac{n+m}{n} = \frac kn[/MATH].​

Is this right? Is there a better way?

 

[Dr.Peterson]

Working on a problem with summation an=2∑k=n+1n\displaystyle a_n=\sum_{k=n+1}^2n and comparing this sum to the right approximation of the area under 1/x\displaystyle 1/x over the interval [1,2]\displaystyle [1,2].

I don't think you said what you mean: k doesn't appear in the summand. Did you perhaps mean [MATH]a_n=\sum_{k=n+1}^2k[/MATH]? No, that doesn't make sense for n>1.

Please correct the question.

 

[gGo]

The summand should have been [math]\frac 1k[/math]. I can't find how to edit the post (if it's allowed). But I think it doesn't affect my question since I was really asking how to find the formula for the subinterval endpoints.

 

[Dr.Peterson]

The summand should have been [math]\frac 1k[/math]. I can't find how to edit the post (if it's allowed). But I think it doesn't affect my question since I was really asking how to find the formula for the subinterval endpoints.

But that doesn't fix the limits of the summation, which make no sense if n>1; and that is essential to the question. Can you show us an image of the actual problem?

You can only edit during the first half hour, which prevents some confusing situations. But you are encouraged to restate the question as needed, if you find an error. And sometimes you can ask a moderator to edit for you.

 

[gGo]

Oh rats. The top should be 2n. (Ugh). Can I just start a new thread at this point

[math]a_n = \sum_{k=n+1}^{2n} \frac 1k[/math]

 

[Dr.Peterson]

Oh rats. The top should be 2n. (Ugh). Can I just start a new thread at this point

[math]a_n = \sum_{k=n+1}^{2n} \frac 1k[/math]

Ah! And I can even see how that happened ...

Fixed version:

OK, I know this is a simple question. Working on a problem with summation [MATH]a_n = \sum_{k=n+1}^{2n} \frac 1k[/MATH] and comparing this sum to the right approximation of the area under [MATH]1/x[/MATH] over the interval [MATH][1,2][/MATH].

The idea is that [MATH]a_n[/MATH] will subdivide [MATH][1,2][/MATH] into [MATH]n[/MATH] subintervals. Normally, to find the endpoints of an Riemann sum we use [MATH]x_k = a+k \cdot \Delta x[/MATH]. In this problem, [MATH]a=1, \Delta x = 1/n[/MATH], and, if n=2, then k starts at 2, which breaks the formula. If [MATH]n=2[/MATH], then [MATH]k = 3, 4[/MATH]

[MATH]x_3 = 1+3/2 = 5/2[/MATH] (should be 3/2)​

[MATH]x_4 = 1+4/2 = 6/2[/MATH] (should be 4/2)​

I am told that the left endpoints are [MATH]\frac{k-1}n[/MATH] and the right endpoints are [MATH]\frac kn[/MATH], which checks out. But I'm curious how we figure that out. The best I can figure is to shift the index, like this:

Let [MATH]m=k-n[/MATH]. Then, for [MATH]n=2[/MATH], we have that [MATH]k=3, 4[/MATH] and [MATH]m=1, 2[/MATH] (that is, starts at 1). Then the right endpoints are [MATH]x_m = a+\frac mn = 1+\frac mn = \frac{n+m}{n} = \frac kn[/MATH].​

Is this right? Is there a better way?

First, I don't see why you say, "if n=2, then k starts at 2, which breaks the formula." In that case, the formula you quote becomes \(x_k=1+\frac k2\) which nicely gives 1.5 and 2 for k = 1, 2. Of course, the summand can't really be \(\frac{1}{k}\) as you are saying; are you sure you didn't mean \(\frac{1}{1+\frac k2}\)?

I'm starting to see what's going on. You're thinking that the sum you were given is written in the exact form of the Riemann sum; it isn't. What I'm starting to do is to write the actual Riemann sum, since that's what your formula is for. Finish that, and then compare THAT to the sum you were given, and make a variable substitution -- the k in the Riemann sum will not be the k in the given sum, so use a different letter. And maybe that's what you're starting to do toward the end.

Alternatively, you can look at the given summation and work out what the summands will be, and see how they relate to what has to be added for the Riemann sum, without writing the latter at all in terms of the formulas you were given.

 

[gGo]

OK. Really sorry for all these mess ups. I'm also embarrassed enough that if you simply abandon this thread I'd be happy to let it die (or actively try to kill it). But you've invested time, which I appreciate.

The ultimate idea is I want a algebraic way to get the end points. Yes, I'm comparing it to how we calculate Riemann sum endpoints (and yes, I oftenrepeat or confuse indices confused on paper, though they're usually straight in my head). The question might be, why am I comparing it to that. I didn't repeat the whole problem because I'm only curious about that one thing.

The idea of the problem is to compare (no more typos!)

[math]a_n = \sum_{k=n+1}^{2n} \frac 1k[/math]​

to the right approximation of the integral

[math]\int_1^2 \frac 1x dx[/math].​

Now, if I was doing the right-Riemann sum, with [math]n[/math] subints, the endpoints' formula is

[math]x_i = 1+\frac in[/math]​

If 3 subintervals, it checks (obviously)

[math]x_1 = 1+\frac 13 = \frac 43, x_2 = 1+\frac 23 = \frac 53, x_3 = 1+\frac 33 = \frac 63 [/math].​

​

Now, back to a_n. Maybe this picture will help (you can even change n): https://www.desmos.com/calculator/0h5i8upbqi

The goal is to figure out the formula for the end points. Actually, someone else told me the answer: The left endpoints are [math]\frac{k-1}n[/math] and the right end points are [math]\frac kn[/math]. So the real goal is to understand how he got that (you know, in case of harder problems).

We use [math]n=3[/math] subintervals again. In this case, we'll have [math]k=4, 5, 6[/math]. Each subinterval should be [math]\frac 1n[/math], or [math]\frac 13[/math] wide. The left endpoint is 1. Writing this up clearer, I think the way to find the formula for the endpoint is to use the index [math]i[/math] to represent the [math]i[/math]th interval, and then let [math]i=k-n[/math] so that the first subinterval, which has [math]k=n+1[/math] has [math]i=1[/math] (or something like this).

Then the right endpoints workout as [MATH]x_i = 1+\frac in = 1+\frac{k-n]n = \frac kn[/MATH].

The math display has stopped working for me and I'm having difficulty bringing this home, anyway (and it's late at night). So I'm just going to end this post here and either hope for some input or insight.

​

​

​

 

[Dr.Peterson]

I suppose the one thing I'm really unsure of is what your real goal is. Are you trying to figure out how they got the limits for that summation from the Riemann sum, or something else? I think maybe you're trying to figure out what this person who gave you an answer was doing. I'm not sure whether that's a useful thing to do, until I see whether it makes any sense.

As I understand it, no one is claiming that the given sum is a Riemann sum, or that it is something you are supposed to be able to come up with; they are just giving it to you and asking you to compare it to the Riemann sum. Here is how I would solve that problem:

Starting with the given summation, we know that the first term is 1/(n+1), and the last term is 1/(2n). Clearly this is not identical to the Riemann sum.​

​

What is the Riemann sum? As I showed, the way I would do it (and the formula you gave agrees), the n points are 1 + k/n, with k varying from 1 to n. So the terms are 1/(1 + 1/n) = n/(n + 1) through 1/(1 + n/n) = 1/2. [Are you asking where that formula came from?]​

​

How do these compare? Each term of the sum is 1/n times the corresponding term of the Riemann sum. And that answers the question.​

Now, what do you mean by "left and right endpoints"? You say they are (k - 1)/n and k/n; but those don't seem to have anything to do with the problem. Ah, I guess you mean the left and right sides of the kth interval! Yes, that's true for the Riemann sum as we have done it; but it appears that you are trying to do that for the given summation. Am I right? If so, I don't think that makes sense, because the summation doesn't have "endpoints" -- it just has n terms.

As for the summation you were given, the index k there is just what you'd use to find the Riemann sum from 0 to 1 rather than from 1 to 2, with the limits set appropriately to relate it to the Riemann sum you were told to do. That is, that k is n more than what it needs to be for your Riemann sum; and the summand, 1/k, is n times what it would be. They've just done an index substitution and a multiplication on the Riemann sum. But I don't think that's a task you need to do.

Can you perhaps show an image of the actual problem you're working on?

 

[gGo]

I might have to get back to this another day. Thanks again. I might have enough in your previous post, but can't dig into it right now.

And I am trying to avoid loading the full problem or image, since it's homework (not for me, but others, and this particular professor will find out about that!).

Part of the problem says "show that the term [math]a_n[/math] is equal to the right-hand approximation of area under [math]1/x[/math] over [1,2]." The two are equal because the terms [math]a_n[/math] correspond to the function value at the endpoints of the subintervals.

In my solution, I showed they were equal with a series of graphs (the one I linked to above) and said "See? I told you they were equal!". In order to make that graph, someone told me "the left endpoints will be [math]\frac{k-1}n[/math] and the right will be k/n". He seemed to just know this. I want to know how he knew that. That's the ultimate goal. But your right, I might not have needed those formulas. That doesn't kill my intrested in figuring it out.

Also, your comments now have me wondering if there isn't a better way to show that a_n equals the right sum.

 

[Dr.Peterson]

In order to make that graph, someone told me "the left endpoints will be [MATH]\displaystyle \frac{k-1}n[/MATH] and the right will be k/n". He seemed to just know this. I want to know how he knew that.

Okay, I take this to mean that he was talking not about the summation you were given, but the Riemann sum; and that by "endpoints" he means the left and right x-coordinates for the nth rectangle. (One problem I've had here is that you mixed several things together without clearly stating which bit answers which question.)

But that's wrong. The formulas he gave tell you how far the kth rectangle's endpoints are from the starting point, 1! That should be obvious, because the nth end point by this formula is n/n = 1, not 2.

To get his formulas, you can just see that the kth right endpoint is obtained by moving to the right 1/n units, k times, which takes you to k/n. The kth left endpoint is 1/n to the left of that: k/n - 1/n = (k-1)/n.

To find the endpoints in your Riemann sum, you have to add these amounts to 1.

Here's one way to get the correct locations of the endpoints for your problem directly, without using his formulas for relative positions:

You are dividing the interval from 1 to 2 into n equal parts. The width of the interval is 1, so each interval has width 1/n. So the first right endpoint is at 1 + 1/n, and for each point after that, we add 1/n. This is an arithmetic sequence, if you are familiar with that.​

​

The kth right endpoint is reached by starting with 1 + 1/n, and adding 1/n to it (k-1) times: 1 + 1/n + (k-1)1/n = 1 + k/n = (n+k)/n. (Observe that this is the starting point, 1, plus his amount, k/n, as I said.)​

​

The kth left endpoint is 1/n to the left of the kth right endpoint: (n+k)/n - 1/n = (n+k-1)/n. This is equal to 1 + (k-1)/n, which agrees with his formula shifted right one unit.​

This work can be simplified by using 0-based indexing for the points, but that might complicate things too much for you all at once.

 

[gGo]

Ha! You pegged it perfectly.

I knew all those ideas (the indented part of your reply). Since the right-approximation corresponds to the summation, I "mixed metaphors", making it too difficult for me to see the obvious.

TY. [EDIT] ...and good, clear explanation.

 

[gGo]

Is there a better way to prove that this summation is equal to the right-hand approximation of the area under 1/x? That is, better than drawing a few pictures?

 

[Dr.Peterson]

Is there a better way to prove that this summation is equal to the right-hand approximation of the area under 1/x? That is, better than drawing a few pictures?

Better way than what?

Which summation? The summation you were given is NOT equal to the right-hand approximation of the area under 1/x; I showed how it IS related in post #8 (indented part). And I didn't draw any pictures.