Upper Sums - Generalized expression in terms of "n" - (est. of area under curve)

[Njords]

Upper Sums - Generalized expression in terms of "n" - (est. of area under curve)



Hi, I've been stuck on this question enough time now that my brain has gone to soup. I'm not sure if I've over complicated for myself or am reading the question wrong. If anyone is able to spot where I'm going wrong a nudge in the right direction will be very much appreciated. I've uploaded a picture of one of the way's I've attempted to solve.. there's many other pages but generally am hitting the wall in the same part I feel (assuming my interpretation of what is being asked is what I'm trying to do!) - which i believe to be a generalized expression such that substituting for any n would give the sum of LHS of the rectangles.


Since LHS then I guess we want the x sub (i-1)th term but putting that into the function and expressing as sum of first n integers (pink box with #2) is causing brain.exe to crash.




This was the question:

Consider function F(x)=25-5x on the interval [2,4] with "n" sub intervals. Calculate the Upper Sum giving the answer as an expression of "n" but no summation by using
results for "Important Sums" given from class.

The two "Important Sums" in the notes are outlined and marked 1 & 2 respectively in the attached image.

 

[Harry_the_cat]

Sometimes I find formulae and notation can get in the way of your thoughts.

Think of the rectangles - there are n of them.

The width of each is \(\displaystyle \frac{2}{n}\) as you have established.

The first one has height F(2),
the next has height F(2+\(\displaystyle \frac{2}{n}\)),
the next has height F(2+\(\displaystyle \frac{4}{n}\)),
… ,
the last has height F(4-\(\displaystyle \frac{2}{n}\))

\(\displaystyle U = \frac{2}{n}*\sum (25 -5x)\) … think of this expanded out - the sum bit is going to look like: \(\displaystyle (25 -5*2) + (25 - 5*(2+\frac{2}{n})) + (25 - 5*(2+\frac{4}{n})) \)… and there will be n terms

\(\displaystyle U = \frac{2}{n} * (n*25 - 5*\text{Sum of an AP})\) …. the AP has a=2, d=2/n and n=n

Sum of AP = \(\displaystyle \frac{n}{2}(2a+(n-1)d) = \frac{n}{2}(2*2+(n-1)*\frac{2}{n}) = 3n-1\)

So \(\displaystyle U = \frac{2}{n}*(25n-5(3n-1)) = 20 + \frac{10}{n}\)

No doubt you can do it more eloquently using summation symbols but it gets messy.

 

[Njords]

Sometimes I find formulae and notation can get in the way of your thoughts.

Think of the rectangles - there are n of them.

The width of each is \(\displaystyle \frac{2}{n}\) as you have established.

The first one has height F(2),
the next has height F(2+\(\displaystyle \frac{2}{n}\)),
the next has height F(2+\(\displaystyle \frac{4}{n}\)),
… ,
the last has height F(4-\(\displaystyle \frac{2}{n}\))

\(\displaystyle U = \frac{2}{n}*\sum (25 -5x)\) … think of this expanded out - the sum bit is going to look like: \(\displaystyle (25 -5*2) + (25 - 5*(2+\frac{2}{n})) + (25 - 5*(2+\frac{4}{n})) \)… and there will be n terms

\(\displaystyle U = \frac{2}{n} * (n*25 - 5*\text{Sum of an AP})\) …. the AP has a=2, d=2/n and n=n

Sum of AP = \(\displaystyle \frac{n}{2}(2a+(n-1)d) = \frac{n}{2}(2*2+(n-1)*\frac{2}{n}) = 3n-1\)

So \(\displaystyle U = \frac{2}{n}*(25n-5(3n-1)) = 20 + \frac{10}{n}\)

No doubt you can do it more eloquently using summation symbols but it gets messy.

Hi there! Thanks for taking time to respond.

I believe the intent in the question was to use the summation symbols and leaving in terms of n so that you can arrive at the sum of the rectangles for any n, which you've shown Here thanks!. Also you're not kidding about it getting messy!

I'm probably still a little lost here. Maybe not sure what AP means but I will persist until I smash through this wall. lol.


... thanks again

 

[Deleted member 4993]

Hi there! Thanks for taking time to respond.

I believe the intent in the question was to use the summation symbols and leaving in terms of n so that you can arrive at the sum of the rectangles for any n, which you've shown Here thanks!. Also you're not kidding about it getting messy!

I'm probably still a little lost here. Maybe not sure what AP means but I will persist until I smash through this wall. lol.


... thanks again

Sequences like:

1, 4, 7, 10, .....

Or

a, a + d, a + 2d, a + 3d ......

are said to be in Arithmetic Progression (AP) with constant difference between consecutive terms.

 

[Njords]

Sequences like:

1, 4, 7, 10, .....

Or

a, a + d, a + 2d, a + 3d ......

are said to be in Arithmetic Progression (AP) with constant difference between consecutive terms.

ah.. right-o. thanks.

went away cleared my head.. had another go trying to follow Harry_the_cat's help...

I think I got close but close enough in this instance is not good enough. Ended up with 30 + 10/n which doesn't work.. working out
is included if someone can spot the mistake and let me know.. ta

 

[JeffM]

I must admit to finding the question obscurely phrased. As I read the instruction, you are NOT to give an answer in summation notation, but rather to give a closed form in n. You are to do so by using the standard sums given in your notes. But the question is so awkwardly phrased that it may mean the exact opposite. Thus I am not sure that this answer will be helpful. If it is not, I apologize for wasting your time.

Let's take it one step at a time.

Calculate \(\displaystyle \Delta x = \dfrac{4 - 2}{n} = \dfrac{2}{n}.\)

Calculate the height of the jth rectangle:

\(\displaystyle f(2 + j \Delta x) = 25 - 5 * (2 + j \Delta x) = 25 - 10 - 5j * \dfrac{2}{n} = 15 - \dfrac{10j}{n}.\)

Calculate the area of the jth rectangle.

\(\displaystyle \Delta x * \left (15 - \dfrac{10j}{n} \right ) = \dfrac{2}{n} * \left ( 15 - \dfrac{10}{n} \right ) = \dfrac{30}{n} - \dfrac{20j}{n^2}.\)

Now add them up.

\(\displaystyle \displaystyle \sum_{j=1}^n \left ( \dfrac{30}{n} - \dfrac{20j}{n^2} \right ) = \left ( \sum_{j=1}^n \dfrac{30}{n} \right ) - \left ( \sum_{j=1}^n \dfrac{20j}{n^2} \right ) =\)

\(\displaystyle \displaystyle \dfrac{30}{n}\left ( \sum_{j=1}^n 1\right ) - \dfrac{20}{n^2} \left ( \sum_{j=1}^n j \right ) =\)

\(\displaystyle \dfrac{30}{n} * n - \dfrac{20}{n^2} * \dfrac{n(n + 1)}{2} = 30 - \dfrac{10}{n^2} * (n^2 + n) = 20 - \dfrac{10}{n}.\)

Step by step, it is easier than jumping into the middle of the pool.

EDIT: Pardon the mixed metaphor.

 

[Njords]

I must admit to finding the question obscurely phrased. As I read the instruction, you are NOT to give an answer in summation notation, but rather to give a closed form in n. You are to do so by using the standard sums given in your notes. But the question is so awkwardly phrased that it may mean the exact opposite. Thus I am not sure that this answer will be helpful. If it is not, I apologize for wasting your time.

Let's take it one step at a time.

Calculate \(\displaystyle \Delta x = \dfrac{4 - 2}{n} = \dfrac{2}{n}.\)

Calculate the height of the jth rectangle:

\(\displaystyle f(2 + j \Delta x) = 25 - 5 * (2 + j \Delta x) = 25 - 10 - 5j * \dfrac{2}{n} = 15 - \dfrac{10j}{n}.\)

Calculate the area of the jth rectangle.

\(\displaystyle \Delta x * \left (15 - \dfrac{10j}{n} \right ) = \dfrac{2}{n} * \left ( 15 - \dfrac{10}{n} \right ) = \dfrac{30}{n} - \dfrac{20j}{n^2}.\)

Now add them up.

\(\displaystyle \displaystyle \sum_{j=1}^n \left ( \dfrac{30}{n} - \dfrac{20j}{n^2} \right ) = \left ( \sum_{j=1}^n \dfrac{30}{n} \right ) - \left ( \sum_{j=1}^n \dfrac{20j}{n^2} \right ) =\)

\(\displaystyle \displaystyle \dfrac{30}{n}\left ( \sum_{j=1}^n 1\right ) - \dfrac{20}{n^2} \left ( \sum_{j=1}^n j \right ) =\)

\(\displaystyle \dfrac{30}{n} * n - \dfrac{20}{n^2} * \dfrac{n(n + 1)}{2} = 30 - \dfrac{10}{n^2} * (n^2 + n) = 20 - \dfrac{10}{n}.\)

Step by step, it is easier than jumping into the middle of the pool.

EDIT: Pardon the mixed metaphor.

oh boy! I sure did I make this more complicated that it had to be. Thank you very much to all 3 members for your contributions and help.




rebooting brain.exe now. lol.

 

[Njords]

Since I went straight for the deep end had no choice but to keep kicking and sink or swim... lol.

.. in case anyone still cares, discovered my error and if the use of summation notation is good then I'm happy with this result since it works for any n
up to limit as n tends toward infinity. I'm not opposed to any criticism, quite the opposite, I invite it. Since ego doesn't matter and I'd much rather have a
working understanding of it more than anything. So yeah, please let me know if seems ok or should start again... Cheers good people, the help I've got so far is awesome!