1. We will be in Read Only mode (no new threads, replies, registration) starting at 9:00 EDT for a number of hours as we migrate the forums to upgraded software.

Riemann Sum

Discussion in 'Math' started by Lightfire, Nov 9, 2013.

1. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Good day,

Thank you in advance for your patience and time.

May I once again ask what's wrong in what I did in my work. I know the answer must be 44, but in my work I calculated it as 49. So there's apparently wrong here.

My work is in .pdf file. I am very sorry if you have to have a a PDF reader in order to see the file (which you shouldn't be, my fault).

File size:
191.9 KB
Views:
20
2. WBahn Moderator

Mar 31, 2012
24,562
7,700
You've got several mistakes.

Note that the ratio (k/n) is central to things. k is the present step and n is the number of steps. In essence, (k/n) is the value of x. The values of k and n, in isolation, have no meaning. So as long as k survives, it should always appear to the same order a n does in the denominator in f(x_k). Once you put it in the sum, then n should be one higher in the denominator because the sum is going to essentially produce n in the numerator as it is caries out.

So look at you expression of x_k and then look at the next line where you have f(x_k). Does that make sense to you that you all of a sudden have (1/n) in there by itself?

Then your final line for f(x_k) doesn't agree with the line right before it.

Then when you go to evaluate the sum a 20 magically turns into a 30.

Once you got the wrong answer, did you walk back through it line by line and check your work? You should have caught at least some of these.

File size:
15.7 KB
Views:
21
3. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Questions:

You've got several mistakes.

1. Do you want me to find it myself?

the ratio (k/n) is central to things.

2. What does that mean? What does "things" refers to?

k is the present step

3. What is "present step"? Is that also "the first step"?

number of steps

4. What is steps? Is it the height? Or the number of rectangles?

5. In essence, (k/n) is the value of x.

What "x"?

6. The values of k and n, in isolation, have no meaning.

What does isolation mean?

7. So as long as k survives, it should always appear to the same order a n does in the denominator in f(x_k).

What does "survives" means? What do you mean by the same order? Does that mean so long as "k" is not cancelled or so long as k still does exist, hence survive. Is tha twhat you mean??

Sorry it should only be 33+20k/n+3k^2/n^2?

I'm sorry if I have to ask this. But I am dumb and I cannot go further without knowing this. But if you won't it's Okay.

Last edited: Nov 9, 2013
4. WBahn Moderator

Mar 31, 2012
24,562
7,700
That's certainly the hope, with hints and suggestions on where to look.

Basically, it refers to the basic approach used for the whole thing.

Since k ranges from 1 to n, the ratio (k/n) ranges from 0.0... to 1. This quantity tells you how far into the range of integration you are.

If you want to think of it as the number of rectangles, fine. The "present step" should be interpretted as "present rectangle".

Look at the function you are evaluating. It is a function of x, is it not? Namely,

f(x) = 3x^2 + 2x

In setting up your sum, you are basically shifting it and scaling it so that

x = a + (b-a)(k/n)

So that when k=0 you have x = a (i.e., your lower bound) and when k=n you have x=b (i.e., your upper bound).

Separate from each other.

Basically, yes. You are summing over the index k, so as soon as you perform the sum, the index goes away. It ceases to exist. It does not survive the summation process.

The "order" refers to the magnitude of the exponent. So if you have k^4/n^3, you have k to the 4th order in the numerator and to the 3rd order in the denominator. Within f(x_k), the parameters k and n should always appear as the ratio (k/n).

Depends on where you are referring to. Update your pdf file and let's take a fresh look at things. You will likely get the correct answer this time.

5. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Sorry ... I still can't figure out where I was wrong.

File size:
191.9 KB
Views:
12
6. WBahn Moderator

Mar 31, 2012
24,562
7,700
You have GOT to develop the ability to walk back through your own work and check it line by line. You need to look at what you actually wrote and not what you wanted to write. You need to verify that each step in each line is correct, not just skim over it.  For the rest of your life you will always be making these kinds of mistakes. I don't know anyone that doesn't. I sure as hell do -- I think quite a bit more now than I used to before my stroke. What matters is not that you never make these kinds of mistakes, but that you develop the ability to check your self, catch them, and correct them. Of course, some will always make it through, but the goal is to make THAT a rare event.

File size:
21 KB
Views:
65
File size:
18.3 KB
Views:
59
7. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
I do not want to say final words but all I have to say is this:

If you have "I sure as hell do -- I think quite a bit more now than I used to before my stroke."

Then I'll have : "I think quite a bit more now than I used to after WBahn taught me how to".

Thanks.

8. Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
So here's the final answer:

File size:
191.8 KB
Views:
17
9. WBahn Moderator

Mar 31, 2012
24,562
7,700
Close.

You still have a stray '30' in the third line from the bottom.

In your last line, there is no need to put everything over a common denominator of 6. By inspection the left hand side reduces to 33+10+1.

Lightfire likes this.