real analysis - Proving uniform continuity of function on a half-open interval whose derivative has a limit at the boundary

I'm given a continuous function $f : (a,b] \rightarrow \mathbb{R}$ for which

• $f'(x)$ exists on $(a,b)$ and


• $\lim_{x \to^+ a} f'(x)$ exists

and asked to prove that $f$ is uniformly continuous.

I am having a bit of trouble with how to show this formally, though I understand the essence of the answer:

• First off, the proof would be immediate if $f$ was defined over the closed interval $[a,b]$ since continuous functions are uniformly continuous on closed domains.




• Second, because the limit of $f'(x)$ exists as $x \to^+ a$, $f$ must be Lipschitz on its domain ($f'$ must be bounded since it is bounded near $a$ and bounded everywhere to the right of $a$ because the domain is closed in that direction).

This is all very well and good for an intuitive answer, but it seems a bit hand wavy. How do I do give a real $\epsilon, \delta$ style argument here? Or is that overkill for this problem?

Thanks.

Answer

Since $f'$ has side limit at $a$, it is bounded on $(a,c)$ for some $c>a$. So $f$ is Lipschitz (and hence uniformly continuous) on $(a,c]$. On the other hand, $f$ is also uniformly continuous on $[c,b]$ by compactness.

Now you can show the following general fact: if a function is uniformly continuous on $(a,c]$ and also on $[c,b]$, then it is uniformly continuous on $(a,b]$. Using this fact and the above observations the proof is finished.

---

PS: This is assuming the side limit of $f'$ exists and is finite, otherwise there are counter-examples.

PPS: $f'$ may still be unbounded on $(a,b]$, you can make counter-examples by adding countably-many disjoint tiny blobs with higher and higher derivatives.

elementary number theory - Is my proof that the square root of all imperfect squares are irrational correct?

I was answering a Quora question about whether $\sqrt{13}$ is irrational or not (link if needed), and I tried to prove that, in fact, the square root of all imperfect squares are irrational.

This is the first proof I have ever attempted, not knowing anything about them before-hand, and I barely know the mathematical symbols, never-mind how to properly set out a proof. So, keeping in mind that I am a complete newbie, can you tell me whether my proof is in fact correct or not, and if it isn't, where I went wrong and how I could improve it next time.

Also, if I chose the wrong symbol, please point out the where the mistake was and what the correct symbol would have been.

Start of Proof

Let's suppose that $n \in \mathbb{N} = \mathbb{Z}^{+}$ is not a perfect square.

This is going to be a proof by contradiction, so we are going to start out by assuming that $\sqrt{n}$ is indeed a rational number, that can be expressed in the irreducible fraction $\frac{A}{B}$ where $A, B \in \mathbb{Z}^{+}$ and $B \neq 1 \because \iff B = 1, \sqrt{n} = A$ which means $n = A^{2}$ which means $n$ is a perfect square.

$\sqrt{n} = \frac{A}{B}$

We can then square both sides to get:

$n = \frac{A^{2}}{B^{2}}$

Since $\frac{A}{B}$ is an irreducible fraction, $A$ and $B$ must not share any factors. When we square a number, we merely repeat its factors, therefore $A^{2}$ and $B^{2}$ must also not share any factors except $1$, making the fraction $\frac{A^{2}}{B^{2}}$ also irreducible.

Bacause it is irreducible, this means $\frac{A^{2}}{B^{2}} \notin \mathbb{Z}^{+} \because B^{2} > B \forall B > 1$ and $ B \in \mathbb{Z}^{+}$ and $B \neq 1$.

Since $n = \frac{A^{2}}{B^{2}}$, this means that $n \notin \mathbb{Z}^{+}$ also.

$\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, \sqrt{n} \neq \frac{A}{B}$

As we had previously defined $n$ to be a positive integer, this is a contradiction. Therefore, our assumption that $\sqrt{n}$ could be expressed as the ratio of two integers was incorrect. Hence $\sqrt{n}$ is irrational $\forall n \in \mathbb{N} = \mathbb{Z}^{+}$ where $n$ is not a perfect square.

$\mathbb{Q.E.D.}$

End of Proof

Thanks for taking the time to read my proof. I would appreciate any and all feedback. As said, I am completely new at this so please show me where I went wrong and how to improve if I did in fact go wrong.

~Edits~:

• Changed the penultimate statement $\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, n \neq \frac{A}{B}$ by adding a radical to the last $n$ that was previously missing: $\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, \sqrt{n} \neq \frac{A}{B}$




• Added a concise contradiction as opposed to ending the proof by simply stating that $\because n \notin \mathbb{Z}^{+}, \sqrt{n} \notin \mathbb{Z}^{+}, \sqrt{n} \neq \frac{A}{B}$ without looping back to the opening when we defined $n$ as an integer.




• Further reinstated why $\frac{A}{B} \notin \mathbb{Z}^{+}$ by adding reasoning that $\because B^{2} > B \forall B > 1$ and $ B \in \mathbb{Z}^{+}$ and $B \neq 1$, along with the fact that $\frac{A^{2}}{B^{2}}$ is irreducible.

Credit to Mathew Daly for helping me improve the summary.

algebra precalculus - Principal value of $(sqrt{2} +sqrt{2}i)^{i+1}$

Wolfram Alpha disagrees with my computation and my first guess is that is because a different branch cut is chosen, but this doesn't seem to be the case after checking, so I'm curious.

I have
\begin{align}

(\sqrt{2} + \sqrt{2}i)^{1+i} &= \exp\big( (1+i) \ln \big( \sqrt{2}(1+i) \big) \big) \\
&= \exp \big( (1+i)\ln \big(2 e^{\frac{\pi i}{4}} \big) \big) \\
&= 2\exp \big( (1+i)\frac{\pi i}{4} \big) \\
&= \frac{2}{e^{\frac{\pi}{4}}}\exp \big( \frac{\pi i}{4} \big) .
\end{align}

Wolfram Alpha however seems to compute an angle of $\theta = 84.7 ^\circ$, as opposed to $\theta = \frac{\pi}{4} = 45 ^\circ$.

The only freedom that we could have would be to pick a different branch $k$: $\ln(e^{\frac{\pi}{4} +2\pi k})$. But this shifts the the angle by $360 ^\circ$, so I'm really puzzled.

How would one reproduce Wolfram Alpha's answer? And more importantly, why do the results disagree? The choice of branch cut doesn't seem to make difference.

Answer

Let's take $\ln 2+\pi i/4$ as the principal logarithm of $\sqrt2(1+i)$.
Then
$$(1+i)\left(\ln 2+\frac{\pi i}4\right)
=\ln2-\frac\pi 4+i\left(\ln2+\frac {\pi}4\right).$$
The "principal value" you seek is the exponential of this.
Note that $\ln2+\pi/4$ radians is about $84.7$ degrees.

number theory - Prove that (integer)-(the sum of it's digits) can be divided by 9

How to prove this:

1. Choose whichever integer you like


2. Subtract from it the sum of it's digits


3. The result can always be divided by 9

For example:

1. I choose 123.


2. The sum of it's digits is 1+2+3=6.


3. 123-6 = 117. And 117/9 = 13.

calculus - Show that function is differentiable but its derivative is discontinuous.

Let $g(x)=x^2\sin(1/x)$, if $x \neq 0$ and $g(0)=0$. If $\{r_i\}$ is the numeration of all rational numbers in $[0,1]$, define
$$
f(x)=\sum_{n=1}^\infty \frac{g(x-r_n)}{n^2}
$$
Show that $f:[0,1] \rightarrow R$ is differentiable in each point over [0,1] but $f'(x)$ is discontinuous over each $r_n$. Is possible that the set of all discontinuous points of $f'$ is precisely $\{r_n\}$?

I'm not seeing how this function is working. I could not even derive it. I need to fix some $ n $ to work? And to see the discontinuity of $ f '(x) $ after that? Can anyone give me any tips? I am not knowing how to work with this exercise and do not know where to start.

Answer

First a stylistic comment: you should use the word "differentiable" in place of "derivable." Second: you should show that $f$ is well-defined on $[0,1]$ so that you can take its derivative (this is easy). We want to consider the difference quotient $\frac{f(x)-f(y)}{x-y}$ and what happens as $x\rightarrow y$ (I will leave it to you to argue why you can interchange sum and limit.. Weierstrass M-test and uniform convergence are your friends). I'll help out with part of the solution.

So we want to evaluate the limit of the difference quotient. To do so, we consider two cases: when $y$ is irrational and when $y$ is rational.

Case 1: $y$ is irrational.

What we have is

$$\lim_{x\rightarrow y}\frac{f(x)-f(y)}{x-y} = \sum_{n=0}^{\infty}\frac{1}{n^2}\lim_{x\rightarrow y}\frac{g(x-r_n)-g(y-r_n)}{x-y}.$$

The important part here is the limit, so I'll focus on that. The limit looks eerily close to a difference quotient (and after a clever introduction of $0$, we see that it is):

$$\lim_{x\rightarrow y}\frac{g(x-r_n)-g(y-r_n)}{x-y} = \lim_{x\rightarrow y}\frac{g(x-r_n)-g(y-r_n)}{(x-r_n)-(y-r_n)}.$$

This is just the derivative of $g$ evaluated at $y-r_n$! Which is equal to $2(y-r_n)\sin\left(\frac{1}{y-r_n}\right)-\cos\left(\frac{1}{y-r_n}\right)$. This is well-defined for all $r_n$ since $y$ is irrational. Since this function is bounded for all $r_n$ and $y$ by the value of $4$, the series converges and so $f'(y)$ is well-defined if $y$ is irrational.

Case 2: $y$ is rational.

If $y$ is a rational in $[0,1]$, then $y=r_k$ for some $k$. Then our difference quotient is

$$\lim_{x\rightarrow r_k}\frac{f(x)-f(r_k)}{x-r_k} = \lim_{x\rightarrow r_k}\left(\frac{1}{k^2}\frac{g(x-r_k)-g(r_k-r_k)}{x-r_k}+\sum_{n\neq k}\frac{1}{n^2}\frac{g(x-r_n)-g(r_k-r_n)}{x-r_k}\right).$$

We could not haphazardly apply the trick from above in this case because we required that $y$ be irrational above (else the denominator in the trigonometric functions will be ill-defined) which is why we split off the term in the series corresponding to $y$ in this case. Notice that the remainder of the series is now susceptible to the trick we did above and the term we pulled out is easy to handle. This gives us:

$$\lim_{x\rightarrow r_k} \frac{f(x)-f(r_k)}{x-r_k} = \lim_{x\rightarrow r_k}\left(\frac{1}{k^2}\frac{g(x-r_k)}{x-r_k} + \sum_{n\neq k}\frac{1}{n^2}\frac{g(x-r_n)-g(r_k-r_n)}{(x-r_n)-(r_k-r_n)}\right).$$

The first part is simply $\frac{1}{k^2}g'(0)$ and the second part is exactly like above.

Do you see how this is also well-defined making $f'$ differentiable everywhere? Can you take it from here?

sequences and series - Bernoulli's representation of Euler's number, i.e $e=lim limits_{xto infty} left(1+frac{1}{x}right)^x $

Possible Duplicates:
Finding the limit of $n/\sqrt[n]{n!}$
How come such different methods result in the same number, $e$?

I've seen this formula several thousand times: $$e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x $$

I know that it was discovered by Bernoulli when he was working with compound interest problems, but I haven't seen the proof anywhere. Does anyone know how to rigorously demonstrate this relationship?

EDIT:
Sorry for my lack of knowledge in this, I'll try to state the question more clearly. How do we prove the following?

$$ \lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x = \sum_{k=0}^{\infty}\frac{1}{k!}$$

Answer

From the binomial theorem

$$\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n {n \choose k} \frac{1}{n^k} = \sum_{k=0}^n \frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{n-k+1}{n}\frac{1}{k!}$$

but as $n \to \infty$, each term in the sum increases towards a limit of $\frac{1}{k!}$, and the number of terms to be summed increases so

$$\left(1+\frac{1}{n}\right)^n \to \sum_{k=0}^\infty \frac{1}{k!}.$$

combinatorics - Prove ${{n+1} choose {m+1}} = sum_{k=m}^{n}{k choose m}$ using a purely combinatorial argument.

Prove ${{n+1} \choose {m+1}} = \sum_{k=m}^{n}{k \choose m}$ using a

purely combinatorial argument.

I don't think I understand how to do a combinatorial proof. I know the left side expands to:

$${{n+1} \choose {m+1}} = \frac{(n+1)!}{(m+1)!(n+1-m-1)!} = \frac{(n+1)!}{(m+1)!(n-m)!}$$

For the right side:

$$\sum_{k=m}^{n}{k \choose m} = {{k+1} \choose {m+1}} = \frac{(k+1)!}{(m+1)!(k+1-m-1)!} = \frac{(k+1)!}{(m+1)!(k-m)!}$$

Which is similar to what I got on the top.

Where do I go from here?

Answer

What you have started there would be an algebraic, rather than a combinatorial argument.

In a typical "combinatorial" argument we look at some set of things, and count the number of element in it in two different ways, yielding two different expressions for the result. Since it is the same things being counted, we now know that the two different expressions have the same value.

In this particular case, it is natural to look at "how many $(m+1)$-element subsets of $\{1,2,3,\ldots,n,n+1\}$ are there?", since that is one definition of the left-hand side $\binom{n+1}{m+1}$. We'd then try to come up with a different way to count those $(m+1)$-element subsets that naturally leads us to conclude that there must be $\sum_{k=m}^{n}\binom km$ of them.

Hint. Consider grouping the subsets according to what the largest element of each subset is.