Need help, URGENT maths question: (1 Viewer)

InteGrand · May 28, 2016

1008 said:
How would you do such questions? Because usually, the width of your partitions is the same, so you can factor that out, and all you're usually left with is a summation which can be easily figured out using standard arithmetic progression formulas, or formulas of summing the first n squares, cubes etc...That doesn't seem to be the case here...

We would go to the general formula. The nth partial Riemann sum formula is:

$S = \sum_{i= 1}^{n} f\left(x_i ^{*}\right) \Delta x_i$ ,

where

$\noindent we have partitioned the interval as $\mathcal{P} = \left \{a = x_0, x_1, \ldots , x_{n-1},x_n = b\right \}$, $x_i^{*}$ is some point in our $i$th subinterval $\left[x_{i-1}, x_{i}\right]$ and $\Delta x_i$ is the width of this $i$th interval ($a$ and $b$ being the end points of our integration interval and $f$ the function we are integrating.$

InteGrand · May 28, 2016

1008 · May 28, 2016

InteGrand said:
$\noindent For your particular question, our function being integrated is $f(x) = x^{j}$, where $j$ is a fixed positive integer. Our interval of integration is $[1,2]$. The partition is given to us as $\mathcal{P} = \left \{1,q,\ldots ,q^{n}=2\right \}$, where $q = \sqrt [n]{2}$. The $i$th subinterval is $\left[q^{i-1}, q^i\right]$, for $i=1,\ldots ,n$. So the width of the $i$th sub-interval is $\Delta x_i = q^{i}-q^{i-1}$. Also, for the lower Riemann sum, we take $x_i ^{*} = q^{i-1}$ for the $i$th subinterval for each $i$, since our function $f$ is increasing.$

I'm somehow struggling with Riemann sums and Riemann Integrals, so if you could provide a general approach when solving such questions by solving this one, that would be great...I understand the concept and I get the formula, just need a rigorous approach for such questions...

InteGrand · May 28, 2016

1008 said:
The thing is that I'm somehow struggling with Riemann sums and Riemann Integrals, so if you could provide a general approach when solving such questions by solving this one, that would be great...I understand the concept and I get the formula, just need a rigorous approach for such questions...

What have you gotten up to using my previous post?

1008 · May 28, 2016

InteGrand said:
What have you gotten up to using my previous post?

How would you rearrange this to get [2^(j+1)-1]/n ? I'm stumped lol

InteGrand · May 29, 2016

1008 said:
How would you rearrange this to get [2^(j+1)-1]/n ? I'm stumped lol

The Riemann Sum doesn't get rearranged to the answer, it gets to the final answer by taking a limit as n -> oo.

Also, the final answer would have (j+1) on the denominator, not n (remember, n is nothing in the end, it goes away when taking the limit to infinity).

1008 · May 29, 2016

InteGrand said:
The Riemann Sum doesn't get rearranged to the answer, it gets to the final answer by taking a limit as n -> oo.

Also, the final answer would have (j+1) on the denominator, not n (remember, n is nothing in the end, it goes away when taking the limit to infinity).

Sorry, I don't understand it. That's alright, I'll just ask my tutor. Thanks anyway

Meanwhile, I've another question:
$\\\text{Find all real numbers }p\text{ such that } I = \int_{2}^{\infty} \frac{1}{x(lnx)^p} dx\text{ converges}$

InteGrand · May 29, 2016

1008 said:
Sorry, I don't understand it. That's alright, I'll just ask my tutor. Thanks anyway

Meanwhile, I've another question:
$\\\text{Find all real numbers }p\text{ such that } I = \int_{2}^{\infty} \frac{1}{x(lnx)^p} dx\text{ converges}$

It converges iff p > 1.

To see this, note that if you transform the integral by using a substitution of u = ln x, it becomes a regular "p-integral", which converges iff p > 1.

1008 · May 29, 2016

InteGrand said:
It converges iff p ≥ 1.

To see this, note that if you transform the integral by using a substitution of u = ln x, it becomes a regular "p-integral", which converges iff p ≥ 1.

Oh right, so when p>1, this integral would converge according to the p-test... But how would you account for the limits, isn't it a condition of the p-test to have the lower limit to be 1 and upper limit to be infinity. I guess I could just say it doesn't matter if you start taking area from right of 1 or right of 2 because area under the curve 1/x is infinite either way.

kawaiipotato · May 29, 2016

1008 said:
Oh right, so when p>1, this integral would converge according to the p-test... But how would you account for the limits, isn't it a condition of the p-test to have the lower limit to be 1 and upper limit to be infinity. I guess I could just say it doesn't matter if you start taking area from right of 1 or right of 2 because area under the curve 1/x is infinite either way.

I think you can justify it since the lower limit will be ln2 < lne = 1
And we know that the integral from ln2 to 1 of the original integrand has a finite value so we can just split it up to the integral from ln2 to 1 of f(x) + integral from 1 to infty of f(x) and apply the p-test.

1008 · May 29, 2016

kawaiipotato said:
I think you can justify it since the lower limit will be ln2 < lne = 1
And we know that the integral from ln2 to 1 of the original integrand has a finite value so we can just split it up to the integral from ln2 to 1 of f(x) + integral from 1 to infty of f(x) and apply the p-test.

And when making the substitution, how would you account for the upper limit, do we just say:

As u = lnx, u-->infinity as x-->infinity so the upper limit of the integral after the substitution would be infinity

unsw-maths-guru · May 29, 2016

um...you mean it converges iff p>1 !!

Paradoxica · May 29, 2016

unsw-maths-guru said:
um...you mean it converges iff p>1 !!

Wrong, it does not converge when p=1.

1008 · May 29, 2016

unsw-maths-guru said:
um...you mean it converges iff p>1 !!

Paradoxica said:
Wrong, it does not converge when p=1.

Lol I think both of you guys are right.

Paradoxica · May 29, 2016

1008 said:
Lol I think both of you guys are right.

$$\noindent $ \int_2^\infty \frac{\text{d}x}{x \log^{p}{(x)}} $ where $p=1 \\ \int_2^\infty \frac{\text{d}(\text{log}\,x)}{\log{x}} = \log{\log{x}}\big|_2^\infty = \log{\log{\infty}} - \log{\log{2}} = \infty$

leehuan · May 29, 2016

Paradoxica said:
Wrong, it does not converge when p=1.

I don't get it.

"It converges when p>1"

implies it does not converge when p=1
______

Edit: Wait, hang on IG why did you use ≥

InteGrand · May 29, 2016

Oops sorry, I was using ≥ for something else at the time and had that in mind and wrote ≥ by mistake. It should just be >.

InteGrand · May 29, 2016

1008 said:
Oh right, so when p>1, this integral would converge according to the p-test... But how would you account for the limits, isn't it a condition of the p-test to have the lower limit to be 1 and upper limit to be infinity. I guess I could just say it doesn't matter if you start taking area from right of 1 or right of 2 because area under the curve 1/x is infinite either way.

The lower limit (a finite one) doesn't matter for these improper integrals if the function doesn't have unusual behaviour at the lower limit (or somewhere in between that and a different lower lower limit) like a singularity there (i.e. as long as the function's integrable over the interval between two different finite lower limits, you can use either one and the convergence status for the improper integral to oo is unaffected). In particular if the function is continuous, then the convergence status of an improper integral from a to oo doesn't depend on the value of a (provided a is taken to be a real number in the domain and the function's domain extends to +oo).

$\noindent E.g. If we know $\int _1 ^\infty \frac{1}{x^{2}}\text{ d}x$ converges, we automatically know $\int _{\pi} ^\infty \frac{1}{x^2}\text{ d}x$ converges. And the reasoning is basically you're only removing/adding a finite piece. Of course, we can't just say that $\int _{0} ^\infty \frac{1}{x^2}\text{ d}x$ converges, because here, we have a singularity at the lower limit (we can't just straightaway say we're only adding on a finite piece), so further investigation is needed (in fact this'd diverge).$

InteGrand · May 29, 2016

1008 said:
And when making the substitution, how would you account for the upper limit, do we just say:

As u = lnx, u-->infinity as x-->infinity so the upper limit of the integral after the substitution would be infinity

Yeah.

leehuan · May 29, 2016

With the p-test btw, just in terms of the actual boundaries

When we were taught it, he demonstrated it with $\int_1^\infty$

Does it have to be 1, or can it be any value of a provided the function remains continuous? $\int_a^\infty$

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