Math stuff that hasn't been proven

[Char. Limit]

Many recent Calculus texts start by defining
ln(x)= \int_1^x \frac{1}{t}dt
so that (ln(x))'= 1/x follows immediately from the definition,
then defining exp(x) to be the inverse function to ln(x). That greatly simplifies finding the derivative of exp(x). Of course you would still need to show that exp(x)= e^x- that is, that "exp(x)" really is some number to the x-power.

But that's easy. If y= exp(x), then x= ln(y). If x\ne 0, 1= ln(y)/x= ln(y^{1/x}). Going back to the exponential form, y^{1/x}= exp(1) so that y= exp(x)= (exp(1))^x. Define e= exp(1) and we have exp(x)= e^x.

Just wanted to fix something there, as I've been told numerous times never to have the bounds contain the same variable as the integrand.

 

[disregardthat]

It's actually not wrong to use the same variable as bound and as a variable in the integrand. The bad thing about it is the confusion it might generate.

 

[Hurkyl]

I personally don't like that definition since it doesn't show that the natural log is actually a logarithm and it is weird to define a logarithm like that (people will just be confused).

Then they miss the point.

You aren't supposed to learn about logarithms and exponentials from this definition. You were supposed to have already learned in your previous classes the basic algebraic properties of these operations.

The point of this definition is for working out the technical details -- and of the four hooks I know into this set of operations, this one is by far the simplest.

The other two, incidentally, are:

• Define exp(x) via Taylor series
• Define x^y incrementally, first for positive integers y via repeated multiplication, then for positive rational numbers by demonstrating that you can take roots, then for positive real numbers by continuous extension
• Define exp(x) as the solution to a differential equation

 

[dalcde]

You aren't supposed to learn about logarithms and exponentials from this definition. You were supposed to have already learned in your previous classes the basic algebraic properties of these operations.

I'm not saying that you can't learn about logarithms. Looking at that definition, you won't even know that it is a logarithm. Why not define it as a logarithm with base e?

 

[kramer733]

In elementary school or high school, we often use stuff that has never actually been proven (in that class). For example

- Pythagoras' theorem.
- Addition of natural numbers is associative.
- Every number can be uniquely (up to order) decomposed in prime factors.

Accepting such a things really annoyed me, I would always ask why something is true. The answer that most teachers gave me was "can you find an example where it doesn't work," sigh. I had to wait until university to actually see a proof for such a things...

So, were you ever annoyed that something wasn't proven in school?? And what would have liked to see a proof/reason of??

You should've said "Can you prove that it works for EVERYTHING?!"

And yes I hated this with a passion too. Well I only started hating it 2 years ago. Before those 2 years ago, I didn't give a crap about math or school.

 

[TylerH]

In elementary school or high school, we often use stuff that has never actually been proven (in that class). For example

- Pythagoras' theorem.
- Addition of natural numbers is associative.
- Every number can be uniquely (up to order) decomposed in prime factors.

Accepting such a things really annoyed me, I would always ask why something is true. The answer that most teachers gave me was "can you find an example where it doesn't work," sigh. I had to wait until university to actually see a proof for such a things...

So, were you ever annoyed that something wasn't proven in school?? And what would have liked to see a proof/reason of??

I'd like to see a proof for the 2nd and 3rd. They're just true, there is no proof other than: It's true, by definition. Can you even imagine what terrible effect that would have on a high school student asking why? They wouldn't get it, and would likely be offended by its resemblance to how parents tell their kids "just because" when they don't feel like explaining something.

I once had a high school algebra teacher who thought .999... != 1. I would have appreciated a proof of that.

Since we're basically looking back on elementary math with higher math prospective, I think this is related. I really wish my teachers would have taken the time to explain how the associative/commutative/identity/etc. properties of the arithmetic operators are generalized in higher math. The aforementioned teacher started down that road when I asked him why those properties are useful (other than their blatantly obvious application to elementary algebra). But, when I asked him what a set is, he gave up. (Which is rather annoying to think of now. How hard would have been to just say: "a collection of objects/stuff"?)

 

[Tosh5457]

In elementary school or high school, we often use stuff that has never actually been proven (in that class). For example

- Pythagoras' theorem.
- Addition of natural numbers is associative.
- Every number can be uniquely (up to order) decomposed in prime factors.

Accepting such a things really annoyed me, I would always ask why something is true. The answer that most teachers gave me was "can you find an example where it doesn't work," sigh. I had to wait until university to actually see a proof for such a things...

So, were you ever annoyed that something wasn't proven in school?? And what would have liked to see a proof/reason of??

Not really, but I always felt I didn't understand something well if I didn't have at least an intuition of how to deduce some math stuff. What annoyed me were some formulas that we used in physics without deducing them, but that doesn't annoy me anymore, because, who cares?

 

[Hurkyl]

I'm not saying that you can't learn about logarithms. Looking at that definition, you won't even know that it is a logarithm. Why not define it as a logarithm with base e?

I did define it to be the logarithm to the base e.



This process of coming up with the rigorous definition for transcendental functions usually comes after you have already learned the basic facts -- in particular I already know the equation \log x = \int_1^x dt/t.

But, even if I was learning about transcendental functions for the first time, all of the basic properties of \log, \exp, and real exponentiation will be presented around the same time. Pedagogically, choosing one of a myriad of facts to be given the title of "definition" (if any are chosen at all) should be done to simplify the exposition. (unless, of course, you're writing for an audience that reads far too much into the word "definition")


Anyways, to define \log(x) as the inverse of \exp(x) requires you to define \exp(x). To define it as the inverse of e^x requires you to define e and to define x^y.

 

[disregardthat]

Any discussion regarding whether log(x) should be defined as the integral of 1/x or as the inverse function of e^x is pure preference-but a preference based on what? I can't tell.

 

[Hurkyl]

Any discussion regarding whether log(x) should be defined as the integral of 1/x or as the inverse function of e^x is pure preference-but a preference based on what? I can't tell.

The one I'm familiar with is sheer simplicity, as I already mentioned.

If we're operating at this level of rigor and technical detail, then there are five things we need to do:

• Define \log(x)
• Define \exp(x)
• Define e
• Define x^y
• Derive all of the usual identities satisfied by these three operations

By far the simplest programme I know of is to start with \log(x) = \int_1^x dt/t, use change-of-variable to derive \log(xy) = \log(x) + \log(y), define \exp = \log^{-1}, e = \exp(1), x^y = \exp(y \log(x)), and then uses these to derive everything else.

What are the alternatives? I've mentioned the starting points of the main ones I recall. Do you need to see the sketch of how I recall that they proceed?

 

[disregardthat]

Another one (using our knowledge of absolutely converging power series) is to define exp(x) = 1+x+x^2/2 + ..., and define e = exp(1). It is simple algebra + binomial theorem to show that exp(x)exp(y) = exp(x+y). From here it is obvious that exp is positive and increasing, and we can define log = exp^-1, and it follows easily that log(xy) = log(x) + log(y).

At last we define x^y = exp(ylog(x)) for positive x.

(Note that showing that exp(x) = exp(1)^x never came into question here (showing that this coincides with x^n for integer n is trivial)).

It also follows easily from the definition that exp(x)' = exp(x).

EDIT: I can see you have mentioned this before.

 

[JonF]

Yes, but why is the ratio a constant?? That seems nontrivial to me...

I remember asking this question in my high school geometry course and was told “what makes you think it wouldn’t be?”

 

[disregardthat]

pi can be calculated for a circle of radius r, and finding it's a constant independent of r answers the question.

 

[Kindayr]

I'd like to see a proof for the 2nd.

The abstract way that I've learned in class is:

First show that (N, s(n), 1) to be a Peano space.

By the Recursion Theorem we can define for each m\in N the function f_m : N\to N as follows: f_m (1)=s(m), and f_m (s(n))=s(f_m(n)),\, \forall n\in N. We then define a binary operation on N, which we call addition and denote by "+". For each (m,n)\in P\times P, we define m+n to be f_m (n). We also call m+n the sum of m and n.

Note we have then: f_m (1)=s(m)=m+1 and m+(n+1)=f_m (s(n))=s(f_m (n))=(f_m (n))+1=(m+n)+1

(We're almost there)

Claim: k+(m+n)=(k+m)+n,\,\,\forall k,m,n\in N (Associativity of Addition)

Proof. We will use induction on n. Let P(n):=k+(m+n)=(k+m)+n. P(1) is true by definition. So suppose P(n) is true, we must show P(n)\implies P(n+1).

So:
<br /> k+[m+(n+1)]=k+[(m+n)+1]=[k+(m+n)]+1=[(k+m)+n]+1=(k+m)+(n+1).<br />

Where the first, second, and fourth equality are by definition, and the third is the inductive step.
-----

By showing N to be a Peano space, you can prove that induction 'works' on it, and can therefore show the operation of addition to be associative \forall n\in N by the previous proof.

Obviously this way is too abstract and complicated for elementary or high school. So the most intuitive way would be to just explain what induction is and use it.

EDIT: Definition of a Peano space can be found https://www.physicsforums.com/showthread.php?t=518053", (within PF). The proof that there exist's a Peano space can be typed up if anyone is interested, as well as how (N, s(n)=n+1, 1) is a Peano space; which would be sufficient to hold the background to these definitions and claim.

References: https://www.amazon.com/dp/0763727334/?tag=pfamazon01-20

 

[stringy]

If all we are after is some intuition (nothing rigorous) behind the definition

e = \lim_{x\to \infty} \left( 1 + \frac{1}{x}\right)^x,

here's a nice little argument...

Our goal and motivation: find a base a such that d/dx (a^x) = a^x.

From the definition, we have

a^x = \lim_{h\to 0} \frac{ a^{x+h} - a^x}{h}.

We'll let x=1, so

a = \lim_{h\to 0} \frac{ a^{1+h} - a}{h}.

For small h,

a \approx \frac{ a^{1+h} - a}{h},

or

a^{1+h} \approx a(1+h).

But this simplifies to

a \approx (1+h)^{1/h}.

This approximation becomes more and more accurate as h goes to zero. Equivalently, put h=1/x and send x to infinity. We get

a = \lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x.

Put e=a. I would have liked to have seen something like this when I was first learning calculus with the promise that it'll all be made more rigorous later on.

 

[disregardthat]

The problem with this kind of argument is that a^(1+h) is approximately equal to a(1+2h) for small h as well.

 

[stringy]

It's not meant to be an ironclad argument. Of course you can poke holes in it. But you're missing the point. If a Calc 1 student points at the definition and asks, "Why?," you could show them this. If they ask a lot of questions, then it's a good segue into discussing their future enrollment in your university's analysis course.

 

[Dr. Seafood]

I thought that was pretty awesome, stringy.

 

[stringy]

Thank you, but I can't take credit for it. I got that out of a GRE study guide. It was the same guide that started with things like basic analytic geometry, trig identities, and logarithms...and ended with stuff like Lebesgue measure, point-set topology, and group theory. Evidently I remember more from the first half than I do the second. Lol!

 

[HallsofIvy]

pi can be calculated for a circle of radius r, and finding it's a constant independent of r answers the question.

No, it doesn't. The only way to "calculate" pi from a circle is to actually measure the circumference and divide by the diameter. Doing that for any number of circles and finding that you always get the same thing does NOT show that is true for all circles.

 

[disregardthat]

No, it doesn't. The only way to "calculate" pi from a circle is to actually measure the circumference and divide by the diameter. Doing that for any number of circles and finding that you always get the same thing does NOT show that is true for all circles.

What are you talking about? What do you mean by "actually measure"?

Surely we can calculate (mathematically) the circumference by means of mathematical methods for a circle with radius r, must I show it to you?

A circle is, by the way, in this regard, a mathematical object.

 

[Fredrik]

I also find HallsofIvy's complaint a bit strange, but I have one of my own. Edit: Not anymore. See the next few posts after this one.

I would define the length of a differentiable curve C:[a,b]\rightarrow\mathbb R^2 as L(C)=\int_a^b\sqrt{C_1&#039;(t)^2+C_2&#039;(t)}\, dt. A circle around (a,b) with radius r is the range of the curve C defined by \begin{align}
x(t) &=a+r\cos t\\
y(t) &=b+r\sin t\\
C(t) &=(x(t),y(t))
\end{align} for all t\in[0,2\pi). So the circumference of that circle is L(C)=\int_0^{2\pi}\sqrt{x&#039;(t)^2+y&#039;(t)^2}\, dt=r\int_0^{2\pi}dt=2\pi r Yes, we can see that L(C)/2r is independent of a,b and r, but the argument looks circular. You might be able to fix it, but I don't immediately see how.

 

[I like Serena]

Your argument depends on cosine, sine, and pi, whose definitions should not be readily used.

A simple fix is using y(x) = \sqrt{r^2 - x^2}
L(C) = \int_{-r}^{+r} \sqrt{1 + y&#039;(x)^2} dx

This integral measures half the circumference of a circle.
Calculate it and you will find \pi r.

 

[Dr. Seafood]

but the argument looks circular

I see what you did there

 

[Fredrik]

A simple fix is using y(x) = \sqrt{r^2 - x^2}

Cool. That approach looks correct and non-circular. \begin{align}
x(t) &=a+t\\
y(t) &=b+\sqrt{r^2-(x(t)-a)^2}=b+\sqrt{r^2-t^2}
\end{align} We want to show that L(C)/2r is independent of a,b and r.
\begin{align}
\frac{L(C)}{2r} &= \frac{1}{r}\int_{-r}^{r}\sqrt{x'(t)^2+y'(t)^2}\,dt =\frac{1}{r}\int_{-r}^{r}\sqrt{1+\frac{t^2}{r^2-t^2}}\,dt =\frac{1}{r}\int_{-r}^{r}\sqrt{\frac{r^2}{r^2-t^2}}\,dt \\
&=\frac{1}{r}\int_{-r}^{r}\frac{1}{\sqrt{1-\frac{t^2}{r^2}}}\,dt =\left[\text{Variable change: }s=\frac{t}{r}\right] = \int_{-1}^1\frac{1}{\sqrt{1-s^2}} ds
\end{align}That last integral is clearly independent of a,b and r, and Wolfram alpha says that it's equal to \pi.

 

[micromass]

Yes, but to calculate that last integral, we need cosines, sines and pi again. But we need to be careful to use these concepts here.

 

[Fredrik]

We don't need to do the calculation. We just need to see that it's independent of a,b and r.

 

[I like Serena]

Hmm, I think you need to add "a" in the limits of the integrals.
Otherwise that looks correct.

 

[Kindayr]

So would you just define \pi = \int_{-1}^1 \frac{1}{\sqrt{1-s^2}} ds with s=\frac{t}{r} and leave the approximation to someone else?

 

[I like Serena]

Yes, but to calculate that last integral, we need cosines, sines and pi again. But we need to be careful to use these concepts here.

Noooo.
If you want you can define a special function called micromass(x) that happens to be the result of that integral (and that happens to work out to pi).

Oh, and not entirely by coincidence micromass(x) = arcsin(x).

So we found in a mathematically sound argument a definition for arcsin, as well as for pi! Yay!