Determine the value of expression (Calc-II)

[alingy1]

Please look picture.
I represented the expression with this integral :

integral of (1-x)^3 from 0 to 1.

I got 0, which makes no sense...

How should I proceed?

 

[alingy1]

Correction:

I represented this using:

sum of (1-i/n)^3(1/n), i=1..infinity.

I expand it, and apply summation rules.

Problem solved?

 

[HallsofIvy]

It took me a moment to understand what you were doing! The problem you gave "Find the integral of (1- x)^3, from 0 to 1" can be done with a simple integration. But what you are doing is using the basic "Riemann sums" definition. Yes, if you divide the interval from 0 to 1 into n parts each part has length 1/n. And, at the right endpoint of the ith interval the function value is (1- i/n)^3 so the "area" of the thin rectangle there is (1- i/n)^3(1/n) and the entire area, and so the integral, is approximated by the sum.

No, i does not go from 1 to infinity- it goes from 1 to n and then, to get the actual integral, you take the limit as n goes to infinity.
You should have
\frac{1}{n}\sum_{i=1}^n \left( 1- 3\frac{i}{n}+ 3\frac{i^2}{n^2}- \frac{i^3}{n^3}\right)
\frac{1}{n}\sum_{i= 1}^n 1- \frac{3}{n^2}\sum_{i= 1}^n i+ \frac{3}{n^3}\sum_{i= 1}^n i^2-\frac{1}{n^4}\sum_{i= 1}^n i^3

 

[Saitama]

Correction:

I represented this using:

sum of (1-i/n)^3(1/n), i=1..infinity.

I expand it, and apply summation rules.

Problem solved?

I don't see how you get that. Do you see that it should be:
$$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3$$

EDIT: Ok, your form is correct too. Now just convert it to a definite integral which is:
$$\int_0^1 (1-x)^3\,dx$$
There is no need to expand it.

 

[alingy1]

Hmm... If I evaluate the integral, I get 1/4. But that answer is just right if I assume n goes to infinity. If am asked to determine the value of this expression, should I not use hallsofivy's last expression and the summation formulas to expand the i's into n?

I think my final answer should contain n's.

Ultimately, the integral is irrelevant in this problem?

 

[SammyS]

...

If am asked to determine the value of this expression, should I not use hallsofivy's last expression and the summation formulas to expand the i's into n?
...

Yes, use Halls's last expression.

Then use the summation formulas, the results of which will contain n, but not i .

 

[alingy1]

What about the integral? Is it "useless" in this specific problem?

 

[SammyS]

What about the integral? Is it "useless" in this specific problem?

I think Halls has the correct interpretation of what's expected. Use the basic "Riemann sums" definition of the integral - taking the limit as n → ∞ .

To check your answer, find the value of the following integral in the usual way, by finding the anti-derivative of the integrand, etc.

$\displaystyle \int_0^1 (1-x)^3\,dx$​

 

[alingy1]

Why would it be take the infinite sum? The question is determine the value of this expression without saying as n goes to infinity.

 

[Saitama]

Why would it be take the infinite sum? The question is determine the value of this expression without saying as n goes to infinity.

We can do that too but this is easily done using definite integrals. If you insist, here's how. I will be using the form I shown before because that doesn't requires expanding the terms as HallsOfIvy has done.

The sum under consideration is:
$$\frac{1}{n}\sum_{i=1}^{n-1} \left(\frac{i}{n}\right)^3=\frac{1}{n^4}\sum_{i=1}^{n-1}i^3$$
Since $\displaystyle \sum_{i=1}^{n-1}i^3=\left(\frac{(n-1)n}{2}\right)^2$, we have the following limit:
$$\begin{aligned}
\lim_{n\rightarrow \infty} \frac{1}{n^4}\left(\frac{(n-1)n}{2}\right)^2 &= \lim_{n\rightarrow \infty} \frac{(n-1)^2}{4n^2}\\
&= \lim_{n\rightarrow \infty} \frac{1}{4}\left(1-\frac{1}{n}\right)^2 \\
&= \boxed{\dfrac{1}{4}}\\
\end{aligned}$$

 

[alingy1]

I'm sorry. Perhaps I was not clear. My question is: the way you just did it supposes that n->infinity. But the original question is : determine the value of this expression in the picture. And in the picture, there is NO indication that n->infinity.
Do you understand what I'm trying to convey?

 

[Saitama]

I'm sorry. Perhaps I was not clear. My question is: the way you just did it supposes that n->infinity. But the original question is : determine the value of this expression in the picture. And in the picture, there is NO indication that n->infinity.
Do you understand what I'm trying to convey?

Ah ok, I thought this was another question related to Riemann sums. In that case, you leave it till the following step:
$$\frac{1}{4}\left(1-\frac{1}{n}\right)^2$$
I hope your issue is addressed now. :)

 

[alingy1]

Excellent! This forum is filled of hardworking people!