An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

Questions for Exam Review


2.Suppose that $Q$ is a refinement of the partition $P$ of the interval $[a, b]$ and that $f$ is a bounded function on $[a, b]$. Show that

$$L(f, P) ≤ L(f, Q) ≤ U(f, Q) ≤ U(f, P)$$.
Derive, as a corollary that $L(f, P_1) ≤ U(f, P_2)$ for all partitions $P_1$ and $P_2$ of $[a, b]$.

3.Define $f$ to be the step function
\[ \left\{
1 & -3 \leq x < 0 \\
-2 & 0\leq x \leq 1\\
\right. \]
Show that $f$ is integrable over $[−3, 1]$ and that $\int_{-3}^{1}f=1$.

4.Suppose that $f_n : [a, b] → R$ is integrable on $[a, b]$ for each $n$ and that $f_n → f$ uniformly on $[a, b]$.
Prove that $f$ is integrable on $[a, b]$.

5.Manipulate the geometric series to show that

$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}= \frac{\pi}{4}$

Note: You’ll need to substitute $x → −x^2$, integrate the result, and plug in the appropriate number. That number is on the boundary of the domain of convergence, though, so you’ll need to use Abel’s theorem to justify the equality.


1 and 2 are fair game for the test.

  1. (h)

The Linearity property of Integration:

Thm. 7.4.2

Assume that $f$ and $g$ are integrable on $[a,b]$. Then

1) $$\int_{a}^{b}Kf=K\int_a^bf$$
for all $K\in\mathbb{R}$


3) If $m\leq f\leq M$, then
$$m(b-a)\leq\int_a^bf\leq M(b-a)$$

4)If $f(x)\leq g(x)$ for all $x\in[a,b]$, then $\int_a^bf<\int_a^bg$ for all $x\in[a,b]$

5)$|f|$ is integrable over$ [a,b] $then $|\int_a^bf|\leq\int_a^b|f|$


Here is the proof for 3:

Let $\epsilon > 0$ and let $P=(-3, - \frac{\epsilon}{3}, 0, 1)$. Then
$$U(f,P) = (1)(3-\frac{\epsilon}{3}) + (1)(\frac{\epsilon}{3}) + (-2)(1) = 3-\frac{\epsilon}{3} + \frac{\epsilon}{3} - 2 = 1$$ and
$$L(f,P) = (1)(3-\frac{\epsilon}{3}) + (-2)(\frac{\epsilon}{3}) + (-2)(1) = 3-\frac{\epsilon}{3} - (2)(\frac{\epsilon}{3}) - 2 = 1 - \epsilon.$$
So, $U(f,P) - L(f, P) = \epsilon.$ So, $f(x)$ is integrable by our criterion of integrability.

Any typos/mistakes you guys see?


Review Question 2. Suppose that $f[a,b] \rightarrow \mathbb{R}$ is bounded, and that $P,Q$ are partitions of $[a,b]$. Then, if $P \subseteq Q$, we have that $L(f,P) \leq L(f,Q) \leq U(f,Q) \leq U(f,P).$

Consider $[x_{k - 1}, x_{k}]$, a sub-interval of $P$, and let $\xi_{j} \in [x_{k - 1}, x_{k}]$. Also, WLOG suppose that $\xi_{1} < \xi_{2} < \dots < \xi_{j}$. We proceed by the PMI. Let $Q_{1} = P\cup\{\xi_{1}\}$, and
$$m_{k} = \sup\{f(x) : x\in[x_{k - 1}, x_{k}]\}.$$
Observe that
m_{k}(x_{k} - x_{k - 1}) &= m_{k}(x_{k} - \xi_{1}) + m_{k}(\xi_{1} - x_{k - 1}) \\
&\leq \inf\{f(x): x\in [\xi_{1}, x_{k}]\} + \inf\{f(x): x\in [x_{k -1},\xi_{1}] \}. \\
It follows that $L(f,P) \leq L(f,Q_{1}).$ So our base case is established. Now assume that $L(f,P) \leq L(f,Q_{s})$, with $Q_{s} = P\cup\{\xi_{1}, \xi_{2}, \dots, \xi_{s}\}$. Then we have that
m_{k}(x_{k} - x_{k - 1}) &= m_{k}(x_{k} - \xi_{s}) + m_{k}(\xi_{s} - \xi_{s - 1}) + \dots + m_{k}(\xi_{1} - x_{k - 1})\\
&\leq \inf\{f(x): x\in[\xi_{s}, \xi_{k}]\} + \inf\{f(x): x\in[\xi_{s - 1}, \xi_{s}]\} + \dots + \inf\{f(x) : x\in [x_{k - 1}, \xi_{1}]\}.\\
Now let $Q_{s + 1} = P\cup\{\xi_{1}, \xi_{2}, \dots, \xi{s}, \xi_{s + 1}\} = Q_{s}\cup\{\xi_{s+1}\}.$ And note that $\xi_{s + 1} \in [\xi_{s}, x_{k}]$. By the same argument as our base case it follows that $L(f,P) \leq L(f,Q_{s + 1})$. Therefore by the PMI it follows that $L(f,P) \leq L(f,Q)$.

If this works it seems like a similar argument would work for $U(f,Q) \leq U(f,P)$. This taken with the fact that for any $P_{1}, P_{2}$, partitions of $[a,b]$ $L(f,P_{1}) < U(f,P_{2})$ should get us what we want.

Once we have this result the corollary is quite easy:
Let $P, P'$ be partitions and let $Q = P\cup P'$. Then we immediately have that $P \subseteq Q$ and $P' \subseteq Q$. It follows that $L(f,P) \leq L(f,Q) \leq U(f,Q) \leq U(f,P')$. Hence $L(f,P) \leq U(f, P')$.


@Ricky_Bobby @Professor.Membrane
Can someone please put up a picture of the work/proof for number 5 that you guys worked on yesterday?


Problem number 5.


Then we plug in $-x^2$


Then we break that up like so.


Then we can integrate this because it is continuous on a compact set.


Thus this implies with an application of Abel's Theorem that



Actually, I'm gonna post my proof of number 5 and would love to get feedback on it.

Let $|c|<1$ and let $I=[-c,c]$. Then,

\sum_{n=0}^{\infty} x^n \rightarrow \frac{1}{1-x}

uniformly for all $x\in I$. Thus, for $-x^2 \in I$ (this is actually equivalent to saying $x \in I$),

\sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} \rightarrow \frac{1}{1+x^2}

uniformly. Because the convergence is uniform, the integral of the series is equal to the series of integrals, so that

\int \sum_{n=0}^{\infty} (-1)^n x^{2n} \;dx = \sum_{n=0}^{\infty} (-1)^n \int x^{2n} dx = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \rightarrow \int \frac{1}{1+x^2} dx = \arctan(x)

uniformly for all $x \in I$. Thus, it converges uniformly on any interval of the form $[-c,c]$ where $|c|<1$. The value $x=1$ is not inside the guaranteed radius of convergence, however, the series $\sum_{n=0}^{\infty} (-1)^n \frac{1}{2n+1}$ converges by the alternating series test. Thus, Abel's Theorem implies that $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$ converges uniformly for all $x\in [0,1]$. Therefore (since every term in the series is a continuous function), the limit must be continuous. We know that the limit is $\arctan(x)$ for all $x \in [-c,c]$, where $|c|<1$, so the only value possible at $x=1$ is

\arctan(1) = \frac{\pi}{4}.


\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}.


What does this come from?



Sorry i was to brief in my comments, because our function is continuous on [0,1], we have that by Thm. 7.2.9 that its is integrable on that interval. Then because the series converges uniformly we know we can perform that operation.


Here is an alternative proof to #4 on the review sheet (@Professor.Membrane suggested a proof that is in the topic Exam I Review). A lot of the same ideas that Caleb has are evident in this proof.

First, I will start by proving a lemma that will help our case:

Let $F(x)$ be bounded on $[a,b]$ and suppose that $F(x) \leq M \forall x \in [a,b].$ Then for any partition, $P$, of $[a,b]$, we have $L(F,P) \leq U(F,P) \leq M(b-a)$ where $M$ is any arbitrary number.

Now let $\epsilon > 0.$ We want to show that $$U(f,P)-L(f,P) = U(f,P) - U(f_{N},P) + U(f_{N},P) - L(f_{N},P) + L(f_{N},P) - L(f,P) \leq \vert U(f,P) - U(f_{N},P) \vert + \vert U(f_{N},P) - L(f_{N},P) \vert + \vert L(f_{N},P) - L(f,P) \vert.$$

Since $(f_{n})$ converges uniformly on $f$, we can choose $N \in \mathbb{N}$ such that $$\vert f_{N}(x) - f(x) \vert < \frac{\epsilon}{3(b-a)}.$$

Since $f_{N}$ is integrable, there exists P, a partition, such that $$ \vert U(f_{N},P) - L(f_{N},P) \vert < \frac{\epsilon}{3}.$$

Now consider $$\vert U(f,P) - U(f_{N},P) \vert = \vert \sum_{k=1}^{n} M_{k}(x_{k}-x_{k-1}) - \sum_{k=1}^{n} M'_{k}(x_{k}-x_{k-1}) \vert = \vert \sum_{k=1}^{n} (M_{k} - M'_{k})(x_{k}-x_{k-1}) \vert = \vert U(f-f_{N},P) \vert < \frac{\epsilon}{3(b-a)}.$$

This is true since we already know that $\vert f_{N}(x)-f(x) \vert < \frac{\epsilon}{3(b-a)}.$

Now using the lemma proved above, let $$M=\frac{\epsilon}{3(b-a)}$$ and let $$F=\vert f( \vert f_{N} - f \vert, P) \vert.$$

Thus, $$L(\vert f_{N}-f \vert,P) \leq U(\vert f_{N} - f \vert, P) \leq \frac{\epsilon}{3(b-a)}(b-a) = \frac{\epsilon}{3}.$$

So, $$U(f,P)-L(f,P) = U(f,P) - U(f_{N},P) + U(f_{N},P) - L(f_{N},P) + L(f_{N},P) - L(f,P) \leq \vert U(f,P) - U(f_{N},P) \vert + \vert U(f_{N},P) - L(f_{N},P) \vert + \vert L(f_{N},P) - L(f,P) \vert = \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.$$

Ok, I am fairly certain this is not completely right and I would love some input. Any ideas?


@ediazloa I think your proof is pretty clear. After reading yours I realized that Abbot defines $\Delta x = (x_{k} - x_{k - 1})$ (and not $(b - a)$.), so good catch. I did basically what you did with that lemma, but started with where I wanted to get and figured out what $M$ needed to be to make everything work.


So, Gaither (melancholy over not having enough analysis to do this semester) and I were discussing an aspect of number 4, and we are both confused about one step:

We are having trouble seeing how $U(f,P) - U(f_n,P) = U(f-f_n,P)$. It doesn't seem to be a true property of the upper (or lower) sums in general; say I have $f_1(x)=x$ and $f_2(x)=-x+1$, and I am interested in the interval $[0,1]$. Let my partition consist of just the endpoints, i.e. $P=\{0,1\}$. Then, $U(f_1,P) = 1$, and $U(f_2,P)=1$, so their difference is zero. But,

$$f_1(x)-f_2(x) = x-(-x+1) = 2x-1,$$

and so it's supremum on the same interval is $1$, thus $U(f_1-f_2,P)=1$.

It could definitely be that we aren't tracking the steps in this proof correctly, so any input would be appreciated!


@Cromer I thought that this may be explained by going back to the definition $$U(f,P) = \sum_{k=1}^{n} M_{k}(x_{k}-x_{k-1}).$$


@Cromer I agree, but you don't need it to do the problem! You just need an inequality.


Since $\left(f_{n}\right) \rightarrow f$ uniformly we can certainly pick $N \in\mathbb{N} \ni$ $$ \left|f_{N}(x) - f(x)\right| \leq \frac{\epsilon}{3(b - a)}.$$ Then for any $n \geq N$ it follows that if we let $M_{k} = \sup\{f(x) : x\in [x_{k - 1}, x_{k}]\}$ and $M'_{k} = \sup\{f_{n}(x) : x\in[x_{k -1}, x_{k}]\}$, then (by the definition of the Supremum) we will have that $$U(f,P) = \sum_{k = 1}^{n} M_{k}(x_{k} - x_{k - 1})$$ and $$U(f_{n},P) = \sum_{k = 1}^{n} M'_{k}(x_{k} - x_{k -1}).$$ Now, this sets up $$|U(f_{n}, P) - U(f,P)| = \left|\sum_{k = 1}^{n} M_{k}(x_{k} - x_{k - 1}) - \sum_{k = 1}^{n} M'_{k}(x_{k} - x_{k -1})\right| = \left|\sum_{k = 1}^{n} (M_{k} - M'_{k})(x_{k} - x_{k -1})\right|. $$ Now $$\sum_{k = 1}^{n} (M_{k} - M'_{k}) \leq \frac{\epsilon}{3(b - a)}$$ by our setup, and $\sum_{k = 1}^{n}(x_{k} - x_{k - 1}) \leq (b - a)$, since each $[x_{k - 1}, x_{k}]$ is a subinterval of $[a , b]$. This taken together yields that $$|U(f_{n},P) - U(f,P)| \leq \frac{\epsilon}{3}$$