Express the limit as a definite integral

[mcastillo356]

TL;DR Summary

Assumed continuity implies integrability, can´t see the Riemann sum in this exercise

Hi, PF, there goes the definition of General Riemann Sum, and later the exercise. Finally one doubt and my attempt:

(i) General Riemann Sums
Let $P=\{x_0,x_1,x_2,\cdots,x_n\}$, where $a=x_0<x_1<x_2<\cdots<x_n=b$, be a partition of $[a,b]$, having norm $||P||=\mbox{max}_{1<i<n}\Delta x_i$. In each subinterval $[x_{i-1},x_i]$ of $P$ pick a point $c_i$ (called a tag). Let $c=(c_1,c_2,\cdots,c_n)$ denote the set of these tags. The sum

$$R(f,P,c)=\sum_{i=1}^n f(c_i)\Delta x_i$$
$$\qquad{=f(c_1)\Delta x_1+f(c_2)\Delta x_2+f(c_3)\Delta x_3+\cdots +f(c_n)\Delta x_n+}$$

is called the Riemann sum of $f$ on $[a,b]$ corresponding to partition $P$ and tags $c$.
(...) For any choice of the tags $c$, the Riemann sum $R(f,P,c)$ satisfies

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Therefore, if $f$ is integrable on $[a,b]$, then its integral is the limit of such Riemann sums, where the limit is taken as the number $n(P)$ of subintervals of $P$ increases to infinity in such a way that the lengths of all the subintervals approach zero. That is,

$$\displaystyle\underset{||P||\to 0}{\lim_{n(P)\to\infty}}{R(f,P,c)}=\int_{a}^{b} f(x) \, dx$$

EXAMPLE 4 Express the limit $$\displaystyle\lim_{n\to\infty}{\sum_{i=1^{n}}{ \displaystyle\frac{2}{n}}\left(1+\displaystyle\frac{2i-1}{n}\right)^{1/3} }$$
as a definite integral.

Solution We want to interpret the sum as a Riemann sum for $f(x)=(1+x)^{1/3}$. The factor $2/n$ suggests that the interval of integration has length $2$ and is partioned into $n$ equal subintervals, each of length $2/n$. Let $c_i=(2i-1)/n$ for $i=1,2,3,\cdots,n$. As $n\to\infty$, $c_1=1/n\to 0$ and $c_n=(2i-1)/n \to 2$. Thus, the interval is $[0,2]$ and the points of the partition are $x_i=\frac(2i)/{n}$. Observe that $x_{i-1}=(2i-2)/n<c_i<2i/ n=x_i$for each $i$, so that the sum is indeed a Riemann sum for $f(x)$ over $[0,2]$. Since $f$ is continous on that interval, it is integrable there, and

$$\displaystyle\lim_{n\to\infty}{\sum_{i=1^{n}}{ \displaystyle\frac{2}{n}}\left(1+\displaystyle\frac{2i-1}{n}\right)^{1/3} }=\int_{0}^{2}{(1+x)^{1/3}} \, dx$$

(iii) Question: how can I state the following equivalence?

$$x_{i-1}=(2i-2)/n<c_i<2i/ n=x_i$$
$$\Longleftrightarrow$$
$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Attempt to solve: I can`t. $c_i$ is a tag set; any partition $P$; but I can say: if $f$ is continous, I can assert the correspondence.

Regards!

 

[andrewkirk]

You didn't define the lower and upper Darboux sums $L(f,P), U(f,P)$ but let's use the standard definitions from here. The relationship follows automatically from the definition since $t_i\in[x_i,x_{i+1}]$ gives us:
$$\inf_{t\in [x_i,x_{i+1}]} f(t) \le f(t_i) \le \sup_{t\in [x_i,x_{i+1}]} f(t)$$
We then sum the inequality over $i$ to get the result.

BTW the definition of the Riemann integral in the OP is no good, as it requires the axiom of choice (because it requires choosing a set of tags in every one of the uncountably infinitely many possible partitions), and Riemann integrability does not need that axiom. See the second definition in this section for an intuitive, workable and robust definition of a Riemann integral.

 

[mcastillo356]

You didn't define the lower and upper Darboux sums $L(f,P), U(f,P)$ but let's use the standard definitions from here. The relationship follows automatically from the definition since $t_i\in[x_i,x_{i+1}]$ gives us:
$$\inf_{t\in [x_i,x_{i+1}]} f(t) \le f(t_i) \le \sup_{t\in [x_i,x_{i+1}]} f(t)$$
We then sum the inequality over $i$ to get the result.

$$L(f,P)=\sum_{i=0}^{n-1}\inf_{t\in [x_{i},x_{i+1}]}=\sum_{i=0}^{n-1}\sup_{t\in [x_{i},x_{i+1}]}=U(f,P)$$

This is my first step, but shouldn't be the last. Working on it.

BTW the definition of the Riemann integral in the OP is no good, as it requires the axiom of choice (because it requires choosing a set of tags in every one of the uncountably infinitely many possible partitions), and Riemann integrability does not need that axiom.

Brilliant remark. Unfortunately, the link to Wikipedia is only a hint to me.

 

[SammyS]

TL;DR Summary: Assumed continuity implies integrability, can´t see the Riemann sum in this exercise

EXAMPLE 4 Express the limit
$$ \displaystyle \lim_{n\to\infty}{\sum_{i=1^{n}}{ \displaystyle\frac{2}{n}}\left(1+\displaystyle\frac{2i-1}{n}\right)^{1/3} }$$

as a definite integral.

You have a typo in your LaTeX for the summation: $\displaystyle \sum_{i=1^{n}}$ .

In particular, I wondered what this meant: $\displaystyle i=1^{n}$

It's simply the result of a brace, $\rbrace$ , in the wrong place. \sum_{i=1^{n}} should be \sum_{i=1}^{n} .

Giving $\displaystyle \quad \quad \sum_{i=1}^{n}$

 

[Mark44]

It's simply the result of a brace, $\rbrace$ , in the wrong place. \sum_{i=1^{n}} should be \sum_{i=1}^{n} .

In fact, braces aren't even needed at all around that upper limit, n. Braces are needed only when the thing (exponent, subscript, fraction numerator or denominator, radicand, etc.) is more than 1 character.
This will work exactly the same \sum_{i=1}^n .

 

[mcastillo356]

Hi, PF
I will quote the textbook to see if I have solved the question, that is, how can I state first $x_{i-1}=\frac{(2i-1)}{n}<c_i<\frac{2i}{n}=x_i$ for each $i$, and observe that the sum is indeed a Riemann sum for $f(x)$ over $[0,2]$:
"Note in Figure 5.13 that $R(f,P,c)$ is a sum of signed areas of rectangles between the $x$-axis and the curve $y=f(x)$. For any choice of the tags $c$, the Riemann sum $R(f,P,c)$ satisfies

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Therefore, if $f$is integrable on $[a,b]$, then its integral is the limit of such Riemann sums, where the limit is taken as the number $n(P)$ of subintervals of $P$ increases to infinity in such a way that the lengths of all the subintervals approach zero. That is,

$$\underset{||P||\to 0}{\lim_{n(P)\to\infty}}R(f,P,c)=\int_a^b\,f(x)dx$$

With my words: this paragraph states that tagged partitions implie, likewise partitions, bounded Riemann sums; thus, integrability. Right?

Figure 5.13

 

[Mark44]

Hi, PF
I will quote the textbook to see if I have solved the question, that is, how can I state first $x_{i-1}=\frac{(2i-1)}{n}<c_i<\frac{2i}{n}=x_i$ for each $i$, and observe that the sum is indeed a Riemann sum for $f(x)$ over $[0,2]$:
"Note in Figure 5.13 that $R(f,P,c)$ is a sum of signed areas of rectangles between the $x$-axis and the curve $y=f(x)$. For any choice of the tags $c$, the Riemann sum $R(f,P,c)$ satisfies

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Therefore, if $f$is integrable on $[a,b]$, then its integral is the limit of such Riemann sums, where the limit is taken as the number $n(P)$ of subintervals of $P$ increases to infinity in such a way that the lengths of all the subintervals approach zero. That is,

$$\underset{||P||\to 0}{\lim_{n(P)\to\infty}}R(f,P,c)=\int_a^b\,f(x)dx$$
With my words: this paragraph states that tagged partitions implie, likewise partitions, bounded Riemann sums; thus, integrability. Right?

What you've written doesn't make sense to me.
The "tags" of a partition are merely the set of points $c_i$ in the various subintervals.

When you say "thus integrability," that's trivially true since the assumption is that "if f is integrable on [a, b]..." This is like saying "if x = 2 <whole bunch of other equations> then x = 2."

 

[mcastillo356]

What you've written doesn't make sense to me.
The "tags" of a partition are merely the set of points $c_i$ in the various subintervals.

Definetely true

When you say "thus integrability," that's trivially true since the assumption is that "if f is integrable on [a, b]..." This is like saying "if x = 2 <whole bunch of other equations> then x = 2."

Yeah, runnig in circles, it's the raw evidence

You didn't define the lower and upper Darboux sums $L(f,P), U(f,P)$ but let's use the standard definitions from here. The relationship follows automatically from the definition since $t_i\in[x_i,x_{i+1}]$ gives us:
$$\inf_{t\in [x_i,x_{i+1}]} f(t) \le f(t_i) \le \sup_{t\in [x_i,x_{i+1}]} f(t)$$
We then sum the inequality over $i$ to get the result.

Is there an equivalence relationship between partition tags and bounded Riemann sums, as claimed in the original post? Quoted above is the way to prove it, if so? If that is the case, could I be given the next step?
Regards!

 

[mcastillo356]

Hi, PF

You didn't define the lower and upper Darboux sums $L(f,P), U(f,P)$ but let's use the standard definitions from here. The relationship follows automatically from the definition since $t_i\in[x_i,x_{i+1}]$ gives us:
$$\inf_{t\in [x_i,x_{i+1}]} f(t) \le f(t_i) \le \sup_{t\in [x_i,x_{i+1}]} f(t)$$
We then sum the inequality over $i$ to get the result.

$$\underset{t\in[x_i,x_{i+1}]}{\inf{f(t)}}\leq f(t_i)\leq\underset{t\in[x_i,x_{i+1}]}{\sup{f(t)}}$$

$$\displaystyle\sum_{t=0}^n{\inf f(t)}\leq\displaystyle\sum_{i=0}^n{f(t_i)}\leq\displaystyle\sum_{t=0}^n{\sup f(t)}$$

$$L(f)=\displaystyle\lim_{t_i \to\infty}{\displaystyle\sum_{i=0}^{\infty}{\inf f(t_i)}}\leq\displaystyle\lim_{t_i \to\infty}{f(t_i)}\leq \displaystyle\lim_{t_i \to\infty}{\displaystyle\sum_{i=0}^{\infty}{\sup f(t_i)}}=U(f)$$

$\therefore$ $f(t)$ is Darboux integrable, where

$$L(f)=\inf\{L(f,P)|P\in{\mathcal P([x_i,x_{i+1}])}\}$$
$$U(f)=\sup\{U(f,P)|P\in{\mathcal P([x_i,x_{i+1}])}\}$$
Right?

 

[mcastillo356]

Not right 😒

 

[andrewkirk]

Hi, PF

$$\underset{t\in[x_i,x_{i+1}]}{\inf{f(t)}}\leq f(t_i)\leq\underset{t\in[x_i,x_{i+1}]}{\sup{f(t)}}$$

$$\displaystyle\sum_{t=0}^n{\inf f(t)}\leq\displaystyle\sum_{i=0}^n{f(t_i)}\leq\displaystyle\sum_{t=0}^n{\sup f(t)}$$

Yes, not right. As I said in post #2, you need to sum the first inequality over $i$. Yet you have summed the three parts differently - over $i$ for the middle part but over $t$ in the two outside parts. Why? You need to sum all parts identically for the inequality to continue to hold.
And the range over which the inf and sup are taken ($t\in [x_i,x_{i+1}]$) has disappeared. Why?

 

[pasmith]

Yes, not right. As I said in post #2, you need to sum the first inequality over $i$. Yet you have summed the three parts differently - over $i$ for the middle part but over $t$ in the two outside parts. Why? You need to sum all parts identically for the inequality to continue to hold.
And the range over which the inf and sup are taken ($t\in [x_i,x_{i+1}]$) has disappeared. Why?

It is also necessary to multiply each term by the width of the corresponding subinterval, $x_{i+1} - x_i$.

 

[mcastillo356]

Hi, PF, let's give another try

$$\inf_{t\in{[x_i,x_{i+1}]}}f(t)\le f(t_i)\le \sup_{t\in{[x_i,x_{i+1}]}}f(t)$$
$$\sum_{t=0}^n \inf_{t\in{[x_i,x_{i+1}]}}f(t)(x_{i+1}-x_i)\le \sum_{t=0}^n f(t)(x_{i+1}-x_i)\le \sum_{t=0}^n \sup_{t\in{[x_i,x_{i+1}]}}f(t)(x_{i+1}-x_i)$$
$$\lim_{||N||\to 0}\left(\sum_{t=0}^n \inf_{t\in{[x_i,x_{i+1}]}}f(t)(x_{i+1}-x_i)\le \sum_{t=0}^n f(t)(x_{i+1}-x_i)\le \sum_{t=0}^n \sup_{t\in{[x_i,x_{i+1}]}}f(t)(x_{i+1}-x_i)\right)$$
$$\Leftrightarrow{L(f)=U(f)}$$
$$\Leftrightarrow{\int_a^b f(t)\,dt}$$
where $a=x_i$, $b=x_{i+1}$. BW!

 

[mcastillo356]

Hi, PF, wrong again

 

[pasmith]

Darboux and Riemann integrability are equivalent: $f$ is Darboux integrable iff it is Riemann integrable, and where the two integrals exist they give the same value. The proof should be given in any decent analysis textbook which covers both integrals, or you can find one here. It requires more than the observation that Riemann sums are bounded by the lower and upper Darboux sums.

 

[mcastillo356]

You didn't define the lower and upper Darboux sums $L(f,P), U(f,P)$ but let's use the standard definitions from here. The relationship follows automatically from the definition since $t_i\in[x_i,x_{i+1}]$ gives us:
$$\inf_{t\in [x_i,x_{i+1}]} f(t) \le f(t_i) \le \sup_{t\in [x_i,x_{i+1}]} f(t)$$
We then sum the inequality over $i$ to get the result.

For each interval $[x_i,x_{i+1}]$ of any partition:
$L(f,P)$ approximates the area in that section under the function by means of a little rectangle of height the lowest value $m_i$ that the function takes in the interval.
$R(f,P,c)$ approximates the area in that section under the function by means of a little rectangle of height the value $f(c_i)$ of the function.
$U(f,P)$ approximates the area in that section under the function by means of a little rectangle of height the higher value $M_i$ that the function takes in the interval.
From the fact that $m_i\le f(c_i)\le M_i$, inmediately it follows
$$L(f,P)\le R(f,P,c)\le U(f,P)$$
If we want to detail more, simply observe that
$$L(f,P)=\sum_{i=0}^{n-1}(x_{i+1}-x_i)m_i\le \sum_{i=0}^{n-1}(x_{i+1}-x_i)f(c_i)=R(f,P,c)\le \sum_{i=0}^{n-1}(x_{i+1}-x_i)M_i=U(f,P)$$
BW, L&P.

 

[andrewkirk]

@mcastillo356 Very good. You have proven that the Riemann sum is bounded below and above by the lower and upper Darboux sums.
Is that as far as you wanted to go or do you want to prove additional results?