Right-Endpoint-Sum is negative? Approximating Areas

[opus]

Homework Statement

Find ##R_4## for ##g(x)=cos(πx)## on ##[0,1]##

Homework Equations

The Attempt at a Solution

(i) Since we want ##n=4## even sub intervals between ##x=0## and ##x=1##,
intervals are ##[0, \frac{π}{4}]##, ##[\frac{π}{4}, \frac{π}{2}]##, ##[\frac{π}{2}, \frac{3π}{4}]##, ##[\frac{3π}{4}, π]##

(ii) ##\frac{1}{4}\sum_{k=1}^4 cos(πx) = -0.25##
However, the back of the book states that the correct solution is 0.25.

So my question is, do we just take the absolute value of such an answer considering area cannot be negative?

 

[fresh_42]

Homework Statement

Find ##R_4## for ##g(x)=cos(πx)## on ##[0,1]##

Homework Equations

The Attempt at a Solution

(i) Since we want ##n=4## even sub intervals between ##x=0## and ##x=1##,
intervals are ##[0, \frac{π}{4}]##, ##[\frac{π}{4}, \frac{π}{2}]##, ##[\frac{π}{2}, \frac{3π}{4}]##, ##[\frac{3π}{4}, π]##

(ii) ##\frac{1}{4}\sum_{k=1}^4 cos(πx) = -0.25##
However, the back of the book states that the correct solution is 0.25.

So my question is, do we just take the absolute value of such an answer considering area cannot be negative?

What is ##R_4##? I assume that the answer lies there. I mean ##\int_0^1 \cos(\pi x)\,dx = 0## so both values are equally "wrong". I guess it depends on whether you add above or below ##g(x)##.

 

[opus]

Sorry. By ##R_4## I mean the approximate area under a curve when we use the right-most point of the rectangles under the curve.
Please see attached image for a separate example to see what I mean. In terms of actual integral notation, we haven't done that yet for area under a curve, only for antiderivatives, so I am not familiar with how to use that notation in this aspect at this time.

 

[opus]

Here is exact context to the problem (#13).

 

[opus]

Beginning to think this is some kind of stepping stone into doing actual calculations for area under a curve and this method isn't worth much so it's probably unfamiliar. Let me spell out what I have done to get the solution after what I posted.

After getting ##\frac{1}{4}\sum_{k=1}^4 cos(πx)##:

## = \frac{1}{4}\left(cos(\frac{π}{4})+cos(\frac{π}{2})+cos(\frac{3π}{4})+cos(π)\right) = -0.25##

 

[fresh_42]

In this case the result is negative. Those sums are normally used to calculate the integral, which is zero here as both parts are of equal size and opposite sign. If the goal is to calculate the area, then ##|\cos (\pi \cdot x)|## should be used instead of ##g(x)##. But as given, you are right. ##L_4 = + \,0.25##

 

[opus]

Ok great thank you fresh. It's kind of an odd primer for integrals, as in some cases, evaluating the left-most points gives twice the area as evaluating from the right-most points but that is with a very limited ##n##. It's difficult to double check these answers because I don't think they're real integrals so integral calculators won't be of any use. So thanks for clearing that up for me.

 

[fresh_42]

It would be helpful to have an exact definition. Shouldn't you multiply by the interval length, which is ##\frac{\pi}{4}##?

 

[opus]

Yes what I did was multiply by the interval length (which since the interval is ##[0,1]## and n=4, I stated the interval length to be ##\frac{1}{4}##) and by "Sigma Properties", I just brought that out in front of everything since it is a constant.
I've attached the definition for Right-Endpoint Approximation according to my text.

 

[FactChecker]

I agree that you are correct and the book answer is wrong.

One nit-pick: Everywhere that you write the summation, you have 'x' where you should have 'k/4'.

 

[opus]

I agree that you are correct and the book answer is wrong.

One nit-pick: Everywhere that you write the summation, you have 'x' where you should have 'k/4'.

Interesting. What's the reason for that?

 

[FactChecker]

In ##\Sigma_{k=1}^{k=4} \cos(\pi x)##, the variable ##x## is undefined. You want it to take values of 1/4, 2/4, 3/4, 4/4 when ##k=1,2,3,4##, respectively. That is ##\Sigma_{k=1}^{k=4} \cos(\pi \frac k 4)##.

 

[fresh_42]

Interesting. What's the reason for that?

##\frac{1}{4}\sum_{k=1}^{4}\cos(\pi x)=\frac{1}{4}\cdot 4 \cdot \cos(\pi x)=\cos(\pi x)## and ##\frac{1}{4}\sum_{k=1}^{4}\cos(\pi \frac{k}{4})=\frac{1}{4}\cdot \cos(\pi \frac{4}{4})=-0.25##

 

[opus]

So I must have something mixed up. So let me ask, is ##k## an actual value of ##x##, or is it just like a "number of times to do this" thing like an iteration of sorts?

 

[FactChecker]

##k## is used in two ways. It is the counting index in the summation symbol. But it is also used to calculate the input to the cosine function for each term. That is very common.

It is not the "number of times" to iterate in the summation. That is 4, not ##k##. ##k## is the counting index that goes from 1 to 4.

 

[opus]

Ok so then since k is the counting index from 1 to 4, and n = 4, then k is the the number of equal intervals throughout n (assuming regular partition)?

And let me clear this part up also before I confuse myself. Notation is always my weak point:

But it is also used to calculate the input to the cosine function for each term.

Say we are doing a summation and we are on something like ##i=2## where ##n=4##. For the ##i=2##, we aren't necessarily evaluated at ##i=2## but using a left-most or right-most point of the second interval that was evenly divided into ##n##? In other words, just because ##i=2##, it isn't always the case that the ##i## acts like an input ##x## that we plug in ##i## to the function and evaluate it.

 

[fresh_42]

Ok so then since k is the counting index from 1 to 4, and n = 4, then k is the the number of equal intervals throughout n (assuming regular partition)?

And let me clear this part up also before I confuse myself. Notation is always my weak point:

But it is also used to calculate the input to the cosine function for each term.

Say we are doing a summation and we are on something like ##i=2## where ##n=4##. For the ##i=2##, we aren't necessarily evaluated at ##i=2## but using a left-most or right-most point of the second interval that was evenly divided into ##n##? In other words, just because ##i=2##, it isn't always the case that the ##i## acts like an input ##x## that we plug in ##i## to the function and evaluate it.

I'm not quite sure I understood it correctly, it's a bit confusing. What we have is
$$
\underbrace{\frac{1}{n}}_{\text{factor from}\\ \text{arithmetic mean}} \cdot \underbrace{\sum_{i=1}^{i=n}}_{\text{counting index i}}\quad \underbrace{f(i,n)}_{\text{function values to be added;}\\ \text{possibly dependent}\\ \text{on counting variable i}\\ \text{and constants like n or }\pi}
\\
\frac{1}{4} \cdot \underbrace{\sum_{k=1}^4}_{\text{counting k}} \, \underbrace{\cos(\frac{k}{4}\cdot \pi)}_{\text{input k}}
$$
If the values ##v## we add do not depend on "when" we add them, then we have ##\sum_{j=1}^N\,v=N\cdot v## simply "number added times value" as result.
Usually, the values ##v(j)## we add, do depend on "when", i.e. are affected by the counting number, here ##j##. In this case the terms we add vary and we get as result ##\sum_{j=1}^{N}\,v(j)=v(1)+v(2)+\ldots +v(N)##.

What you have written was ##\sum_{k=1}^4\cos(\pi x)##. So the function value ##\cos(\pi x)## was constant! It did not vary, since all that varies is ##k## and the summand ##\cos(\pi x)## has no dependency on ##k##. Thus you had the case ##\sum_{j=1}^N\,v=N\cdot v=\sum_{k=1}^4\,v=4\cdot \cos(\pi x)=4\cos(\pi x)##.

However, you wanted to add ##\cos(\frac{1}{4}\pi)+\cos(\frac{2}{4}\pi)+\cos(\frac{3}{4}\pi)+\cos(\frac{4}{4}\pi)##. Therefore you had to choose the second case, a dependent ##v(k)=\cos(\frac{k}{4}\pi)##. The ##x## had no special meaning in your sum ##\sum_{k=1}^4\cos(\pi x)##, since it did not occur in the summation instruction. From the perspective of the summation ##\sum## it was a constant, just a simple unknown number.

P.S.: I wildly used different summation indices as ##i,j,k## to stress, that they are only valid within a certain summation instruction and which letter is used isn't of any importance. Only rule of course is, not to use more than one per summation. Sometimes we read ##\sum_{i,j} v(i,j)## but this is only an abbreviation for two sums: ##\sum_{i,j}v(i,j)=\sum_i \left(\sum_j v(i,j) \right)##.