Is it a new type of induction? (Infinitesimal induction) Is this even true?

Question

Suppose we want to prove Euler's Formula with induction for all positive real numbers.
At first this seems baffling, but an idea struck my mind today.

Prove: $$e^{ix}=\cos x+i\sin x \ \ \ \forall \ \ x\geq0$$

For $x=0$, we have $$1=1$$ So the equality holds.
Now let us assume that the given equality holds for some $x=k$.
$$e^{ik}=\cos k+i\sin k$$ Now, this is where I added my "own" axiom. Please answer whether this "axiom" is true or not. Now this equality must hold for $x=k+\Delta k$ also, for some infinitely small positive change $\Delta k$ (infinitesimal).
So $e^{i(k+\Delta k)}=e^{ik}.e^{i\Delta k}=(\cos k+i\sin k)(\cos\Delta k+i\sin\Delta k)$
$$=\cos k\cos\Delta k-\sin k\sin\Delta k+i\sin k\cos\Delta k+i\cos k\sin \Delta k$$ $$=\cos(k+\Delta k)+i\sin(k+\Delta k)$$ So we proved it for $x=k+\Delta k$, and hence it must hold for all $x\geq0$.
So is this approach correct, and is this a new type of induction?

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Most of the comments below are indicating that the proof above is wrong. But we see the Euler's Formula does hold for all $x\geq0$, so can someone give a good counter example where this proof doesn't work. Or in other words, these statements are used to prove a wrong equality.?

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Update: Okay, some comments below are suggesting that Euler's Formula is definitely true. So if we prove somehow that $$e^{i\Delta k}=\cos\Delta k+i\sin\Delta k$$ then? But how can we prove it for infinitesimal? Will the concept of limits be used? Can someone solve this mystery?

Answer 1

Here is an example of your logic being used to prove a false statement. I am just basically going to take your exact proof, except replace the constant $e$ with the constant $2$. The logic is all the exact same, and you should notice that no where in your proof do you ever use any properties of the number $e$, so in theory any number could take its place.

Prove (The false euler equation): $$2^{ix}=\cos x+i\sin x \ \ \ \forall \ \ x\geq0$$

For $x=0$, we have $$1=1$$ So the equality holds.
Now let us assume that the given equality holds for some $x=k$.
$$2^{ik}=\cos k+i\sin k$$ Now, this is where I added my "own" axiom. Please answer whether this "axiom" is true or not. Now this equality must hold for $x=k+\Delta k$ also, for some infinitely small positive change $\Delta k$ (infinitesimal).
So $2^{i(k+\Delta k)}=2^{ik}.2^{i\Delta k}=(\cos k+i\sin k)(\cos\Delta k+i\sin\Delta k)$
$$=\cos k\cos\Delta k-\sin k\sin\Delta k+i\sin k\cos\Delta k+i\cos k\sin \Delta k$$ $$=\cos(k+\Delta k)+i\sin(k+\Delta k)$$ So we proved it for $x=k+\Delta k$, and hence it must hold for all $x\geq0$.

But the statement that we have just "proved" is very clearly false.

Answer 2

An argument along these lines is the following: the standard theory of existence and uniqueness of solutions to ODEs implies that a solution to the ODE $y' = iy$ (where $y$ is a function $\mathbb{R} \to \mathbb{C}$) is uniquely determined by its initial value $y(0)$. Now, $e^{ix}$ and $\cos x + i \sin x$ are both solutions to this ODE, and they both have initial value $y(0) = 1$.

The way in which this comes to look like an inductive argument is when you explicitly solve this equation using Euler's method.

In general, there is a "principle of real induction," and it looks something like this: suppose $P(x)$ is a property of a nonnegative real number $x$ such that

• $P(0)$ is true,
• If $P(x)$ is true, then $P(y)$ is true for all $y$ in some interval $[x, x + \varepsilon)$, where $\varepsilon > 0$.
• If $P(x)$ is true for all $x < y$, then $P(y)$ is true.

Then $P(x)$ is true for all $x$. However, it's unclear to me how easy it is to apply this to your case. (Several variations of this are possible; see, for example, this note by Pete Clark.)

Answer 3

The first (not biggest) flaw in you proof is when you go from $k$ to some $x=k+\Delta k$. You are trying to prove some statement for ALL x. So you'd better proove it for ALL $x$ between $k$ and $x=k+\Delta k$. What if there are some $x$ in between for which the statement does not hold? For example, for each rational number there exists even bigger rational number, but still irrational numbers also exist!

To the bigger flaw now.

Let me "proove" much simplier statement: all real numbers are strictly less than 1.

Step 1: Statement holds for $x$ = 0 and all the smaller numbers.

Step 2: for any $x < 1$ there exists some positive change $dx$ such that the statement holds for all the numbers smaller than $x+dx$. F.e. $dx = (1 - x)/2$ will do.

Step 3. So that's it? For any number which satisfies our condition there exists even bigger number which also satisfies, so all the numbers must satisfy it?

Of course not, I know several numbers that are bigger than 1. (By the way, isn't it "Achilles and the tortoise" paradox?)

Moral of this story is that mathemetical induction isn't so obvious as it looks. It is a very important property of the set of integer numbers that you can use induction on it.

Answer 4

In the added paragraph you ask for an example demonstrating where this kind of logic would give "a proof" of a wrong result. Here is one.

Claim. For all real numbers $x\ge0$ we have $$ e^{ix}=\cos 2x+i\sin 2x. $$

Proof. The formula holds for $x=0$. If we assume that the formula holds for $x=k$ AND that the formula holds for all small positive numbers $\Delta k$ AND that the usual formulas for $e^{x+y}$, $\sin(x+y)$ and $\cos(x+y)$ also hold, then we can do the "inductive step" $$ \begin{aligned} e^{i(k+\Delta k)}&=e^{ik}e^{i\Delta k}\\ &=(\cos 2k+i\sin 2k)(\cos2\Delta k+i\sin2\Delta k)\\ &=\cos (2[k+\Delta k])+i\sin (2[k+\Delta k]). \end{aligned} $$ QED.

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Do you now see where the error in your logic is?

Answer 5

The "induction", as stated, does not quite work, as has already been pointed out. However, some rather similar arguments do work, and are quite useful.

Let me give you an example, which is a very basic model for some useful theorems in differential equations.

Lemma. Let $h \colon \mathbb{R} \to \mathbb{R}$ be differentiable with $h(0) = 0$ and $h'(x) = 0$ for each $x$. Then $h(x) = 0$ for all $x$.

Proof: Take any positive constant $c > 0$; we will prove an easier statement that $|h(x)| \leq |c x|$. If this is done, then (since $c$ was arbitrary) you can conclude that $h(x) = 0$ for any $x$.

So, let me try to apply "infinitesimal induction". Certainly, the claim holds for $x =0$. Suppose now that we have proved the claim for some $x$, and let us take some $\Delta x > 0$ (think of $\Delta x$ as being very small). Now, since $h'(x) = 0$, we certainly have $\lim_{y \to 0} \frac{h(x+y)-h(x)}{x-y} = 0$, so (assuming $\Delta x$ is sufficiently small) we can conclude that $\frac{h(x+\Delta x)-h(x)}{\Delta x} < c$, and consequently $$|h(x+\Delta x) | \leq c|\Delta x| + |h(x)| \leq c|x + \Delta x|.$$ Thus, the claim holds for $x + \Delta x$ as well.

Here is a slight subtlety: $\Delta x$ needed to be small, and how small it is might depend on $x$. For example, if I were trying to "inductively" prove the claim $x < 1$, then for each $x$ for which the claim holds, I could find $\Delta x$ such that the claim still holds for $x + \Delta x$ (e.g. $\Delta x = \frac{1- x}{10}$). However, the claim is blatantly false for $x =1$. Hence, we need some additional closure condition.

Returning to the lemma at hand, recall that we are proving the claim that $h(x) \leq c x$. Let me notice additionally the following simple fact: If the claim holds for some $x_i$ ($i =1,2,\dots$) and $\lim_{i \to \infty} x_i = x$ then also the claim holds for $x$. This is not terribly difficult: $$cx - |h(x)| = \lim_{i \to \infty} cx_i - |h(x_i)| \geq 0,$$ because $ cx_i - |h(x_i)| \geq 0$ for each $i$ by assumption.

Combining the "inductive step" (i.e. claim for $x$ implies claim for $x + \Delta x$, provided $\Delta x $ small enough) together with the "continuity" (i.e. claim for $x_i$ implies claim for $x = \lim_i x_i$), we can indeed conclude that the claim holds for all $x > 0$. $\square$

Now, with the above Lemma, you can try to make your reasoning rigorous. Let $f(x) = e^{ix}$, $g(x) = \cos x + i \sin x$ and $h(x) = |f(x) - g(x)|^2$. Assume further that you know $f'(x) = i f(x)$ (you have to know something specific to $e$.)

Clearly, $h(0) = 0$. Doing some algebraic manipulations, you will find that $h'(x) = 0$ for all $x$ (basically because $f'(x) - g'(x) = i(f(x) - g(x))$. Thus, the "infinitesimal induction" will tell you that $h(x) = 0$ always, which is just what you need.

Also, you may find the concept of connectedness to be very relevant.

Answer 6

My issue with your proposed proof is that you assumed your hypothesis is true for some infinitesimal value.

Answer 7

As many answers have pointed out, the method as provided does not work.

However, one could argue that strong induction does work, in the form of a carefully-established integral.

1. It is true for $x=0$ (trivial)

2. Assume it is true for all $y$ such that $0\leq y<x$

3. Show that it is true for $x$.

$$ \frac{e^{ix}-1}i=\int_0^x e^{iy} dy = \int_0^x \cos y + i\sin y\ dy = \sin x-i(\cos x-1) $$ Rearranging gives $$ e^{ix}-1 = i\sin x + (\cos x-1)\\ e^{ix} = \cos x + i\sin x $$

It's not quite formal as written, and you would have to be careful with the exact form of integral being used when doing it fully formally, but I believe it demonstrates how one might do induction of this sort.

Note that, strictly speaking, the above is not sufficient to prove the claim. There is one additional requirement - prove that there can't be a "largest" value for which the claim is true - that is, show that, if it is true for $x$, then there must exist a $y>x$ such that it is true for $y$. Note that this would be satisfied if the interval for which the claim is true is open on the right.

This is necessary to allow it to push "past" a number, rather than getting stuck at a given $x$ (most notably, it is necessary to be able to get past $0$).

In this case, I believe it is sufficient to show that both $e^{ix}$ and $\cos x+i\sin x$ are analytic functions, as this implies (in conjunction with the above) that there is a non-zero region around any given point for which the two functions converge to the same value. Therefore, if it is true at $x$, then there is a point $y>x$ for which the equality holds for all values $[x,y]$. Of course, analyticity is proven using Cauchy-Riemann equations, and I won't go into that here.

Answer 8

The structure of the argument is basically correct, and can be formalized in Robinson's framework in terms of hyperfinite induction. Namely, the formula is first proved for all hyperinteger multiples of a fixed infinitesimal. Then one applies standard part to prove it for all real values of $x$ using the fact that any real $x$ is infinitely close to a suitable hyperinteger multiple of the fixed infinitesimal.

What is missing of course is the base of the induction (namely Euler's formula for the fixed infinitesimal itself).

To prove the formula by induction, it is not necessary to prove the exact formula $e^{i\Delta k}=\cos\Delta k+i\sin\Delta k$ but rather it is sufficient to prove it up to a higher order term, for example by using second-order approximations for both sides. Of course this would require knowing the beginning of Taylor series for each of the functions involved (though not the whole series). Then when you iterate the induction $\frac{1}{\Delta k}$ times, the error introduced into the calculation will be sufficiently small so that it will be erased when you apply the standard part function, resulting in Euler's formula for all real inputs. Then by the transfer principle it holds on the nose also for hyperreal inputs, and in particular for an infinitesimal input $\Delta k$ you will get the formula $e^{i\Delta k}=\cos\Delta k+i\sin\Delta k$ on the nose.

A concrete realisation of this scheme in the context of the study of small oscillations of the pendulum appears in this 2016 publication in Quantum Studies: Mathematics and Foundations.

Answer 9

I love your cleverness! :)

From an aesthetic point of view, induction is an iterative process and iteration is countable. That works fine if you are proving properties of a set where your elements are countable, but you are applying it to the Reals which is an uncountable set.

Another way to see this, is you have shown your statement to be true for k, and k + dk, and (k+dk) + dk, and ((k+dk)+dk) + dk, and so on. But since dk is infinitesimal, so is N * dk. And so, you may not have expanded your statement from k, to cover the entire reals >= 0. Effectively, you haven't gotten any further than the assumption about k, as the game is played.

So you need to show that your method 'expands' to cover an interval around k.

Why does Yuan's method of induction work, because after all it too is an iterative process and thereby a countable process? well, because his assumptions are basically stating, "if you can expand an interval in a countably infinite way, then you can proceed with an induction like process."

By analogy, Yuan is stating "Assume k is an interval where P holds true. And assume k+dk is an expansion of that interval where P hold true. Then using an inductive process, we can countably expand the interval k to cover what we need to."

I hope that helps.

Answer 10

There are two problems with your proof.

The first is handled by ASKASK's answer: even allowing your rule of infinitesimal induction there is a gap in your proof since you would need to prove "Euler's formula for infinitesimals" in order to get $e^{i\Delta k}=\cos\Delta k+i\sin\Delta k$. You haven't done this, so the proof doesn't work.

However, there's also a problem with your rule of infinitesimal induction. I will suppose that we have a theory of infinitesimal quantities, without going into any great detail of what that theory is. Let us suppose we have a theory in which infinitesimals "behave themselves" with respect to arithmetic, and are smaller than any positive quantity. Then infinitesimal induction, if allowed, would go like this:

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To Prove: All non-negative real numbers are less than 1

Base case: $0 < 1$ (prove this either from the ordered field axioms or from your construction of $\mathbb{R}$, as you prefer).

Suppose for the inductive step that for some real number $k$, $k < 1$. That is, $1-k > 0$. Now, we have by definition of infinitesimals that they are less than all positive reals, so for any infinitesimal $\Delta k$, in particular $\Delta k < 1 - k$ and therefore $k + \Delta k < k + (1 - k) = 1$.

This completes the proof?

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So, are we now going to conclude that all real numbers are less than one? Nope! The rule of inference, should we accept it, would lead to contradictions.

It's difficult to say "why" something isn't true. But without going into details of a theory of infinitesimals, my intuition as to "why" this rule of inference doesn't work is that adding an infinitesimal isn't sufficient to get from one real number to another real number. Therefore this induction on infinitesimals doesn't get us any (non-zero) distance along the real line, and certainly doesn't get us all the way along the real line.

Observe also that, never mind infinitesimals, for every $k < 1$ there is neighbourhood of $k$ with a positive radius, all of whose values are less than one. This is a bigger / stronger result than an infinitesimal neighbourhood of $k$, but it still isn't enough to crawl our way up the real line. All it gives us is an open set with an upper bound. So even adding a positive real number isn't sufficient to get us "past" a limit point, if the numbers we're adding get small enough fast enough.

Qiaochu Yuan's answer describes an effective means of crawling up the real line, but note that $\epsilon$ must be a fixed positive value. Not an infinitesimal, and not dependent on $x$. Then you're good to go. Adding a fixed constant does get from inside a set "past" a limit point of the set.

Answer 11

While the argument in general isn't true, as described in other answers, there is another way how to prove euality of functions. The important condition is that both $e^{ix}$ and $\cos x + i sin x$ are holomorphic functions. The Identity theorem states that

If two holomorphic functions $f$ and $g$ on a domain $D$ agree on a set $S$ which has an accumulation point in $D$ then $f = g$ on all of $D$.

So if you prove that there is a sequence of points $x_i$ such that $lim_{i\to\infty} x_i\in D$, you can use the theorem to prove the equality on the whole domain. And for this you could use the induction principle on $i$.

For example you could set $x_0=0$, $x_1=1$ and $x_n=(x_{n-1}+x_{0})/2$. Then if you prove that the equality holds for $0$, $1$ and that if it holds for $x$ and $y$ it also holds for $(x+y)/2$, the theorem gives you equality everywhere.

Answer 12

For the infinitesimal argument, you could write

$$e^{i\Delta x}\approx1+i\Delta x.$$

Then taking the $k^{th}$ power,

$$e^{ik\Delta x}\approx(1+i\Delta x)^k.$$

And in the limit,

$$\lim_{k\to\infty,k\Delta x=x}e^{ix}=\lim_{k\to\infty}\left(1+i\frac xk\right)^k.$$

Developing the RHS by the binomial formula and taking the limit, you will find the Taylor series for $\cos(x)+i\sin(x)$.

Answer 13

This is not really an answer to this question as posted (I do not discuss the exponential function), but it points out that a certain well-known method might be interpreted as induction over the reals. The idea is that to prove something is true, for say all positive $x$, one could prove it for $x=0$ and then show that the set $T\subseteq\mathbb R$ for which it is true is both open and closed.

Compared to usual induction, when we would like to prove a statement or property $P(n)$, we prove $P(0)$ and then (induction step), show that there is no largest integer $n$ for which the statement is true. Indeed, if $n$ were largest, then $P(n)$ is true, but then $P(n+1)$ must also be true, contradicting that $n$ was largest.

Back to the case of the reals, if $T$ had a largest element $t$, then since $T$ is open we have $t+\varepsilon\in T$ for all small enough $\varepsilon>0$, contradicting that $t$ were largest. Note that $T$ must also be unbounded, since if $\sup T<\infty$ then using that $T$ is closed we would have $\sup T\in T$ so $\sup T$ would be the largest element of $T$, but we already saw that $T$ cannot have a largest element.

The above explains the idea but does not yet constitute a proof that $[0,\infty)\subseteq T$. To make it precise recall that a version of induction for the integers is the so-called complete induction, where (after starting with $P(0)$) one proves that $P(m+1)$ holds whenever $P(n)$ holds for all $n$ in the interval of integers $[0,m]$ (that is, whenever $P(n)$ holds for all $n\in\{0,1,\dots,m\}$). (Restated in a different way: One proves that $P(n+1)$ holds whenever each of $P(0),P(1),\dots,P(n)$ holds.)

For the reals, we could define $s=\sup\{x\ge0:[0,x]\subseteq T\}$, and then we would like to show that $s=\infty$. But if $s<\infty$ then, since $T$ is closed we have $s\in T$, and since $T$ is open we have $s+\varepsilon\in T$ for all small enough $\varepsilon>0$, contradicting the definition of $s$.

To illustrate the above with a real-world-example, my answer was inspired by the following MSE question by Saun Dev and more specifically by this answer by Michael Burr to (for which the OP requested an elaboration ... my answer here may provide this).

So the problem there was to prove that if $f:\mathbb R \to \mathbb R$ is a differentiable function such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ then $f(x)>0,\forall x>0$. The answer I referred to provided the following hint: Consider $x=0$. Since $f'(0)>f(0)=0$, we know $$ \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0}\frac{f(h)}{h}>0. $$ Therefore, for $h$ positive and sufficiently close to $0$, the numerator is positive. Continue with this idea.

So to elaborate, let $T(x)$ be the statement that either $x=0$ or $x>0$ and $f(x)>0$. (Formally $T(x)$ is $(x=0)\lor((x>0)\land(f(x)>0))$.) Let $T=\{x:T(x)\}$. What the hint shows is that $T$ is (relatively) open at $0$ in the interval $[0,\infty)$, that is there is $\varepsilon>0$ such that $T(h)$ holds, whenever $h\in[0,\varepsilon)$. Namely $T(h)$ holds for $h\in(0,\varepsilon)$ since $f(h)>0$ (by the hint), and $T(0)$ holds (trivially).

The induction step here could be split into two parts. First, $T$ is open, if $x\in T$ then we could push it further to the right, $T$ is open, there is $\varepsilon>0$ such that $T(h)$ holds, whenever $h\in[x,x+\varepsilon)$. This is essentially what the hint shows. Using that $f'(x)>f(x)\ge0$ (for $x\ge0$) and $$ \lim_{h\rightarrow x}\frac{f(x+h)-f(x)}{h}=f'(x)>0. $$ we conclude that $f(x+h)>f(x)\ge0$ for $h\in[x,x+\varepsilon)$ (or even for $h\in(x-\varepsilon,x+\varepsilon)$, if $x>0$) for all sufficiently small $\varepsilon>0$.
Second part of the induction step (which resembles complete induction for the integers), if $x>0$ and $T(y)$ holds for all $y\in[0,x)$ then $T(x)$ holds too. Indeed in the example here, since $f'(y)>f(y)>0$ for all $y\in(0,x)$, then $f$ must be strictly increasing on $(0,x)$ and by continuity $f(x)>0$ (and, with more detail, $f(x)>f(y)>0$ for each $y\in(0,x)$).

Answer 14

As Eric Wofsey suggested in the comments, this part is not included in your extended induction assumptions!

$$e^{i \Delta k} = \cos \Delta k + i \sin \Delta k$$

Hence, your proof collapses when you use this!

To get more enlightened, you should note that we always start an induction to be true for a specific value not any value. Suppose that you wanted to prove the Euler formula for non-negative integers

$$e^{in} = \cos n + i \sin n, \qquad \qquad n=0,1,2,...$$

Then you would start by saying that will hold for $n=0$ since $1=1$. Next, if being true for $n=k$ implies that it is also true for $n=k+1$, you could conclude that it is valid for all $n$. You should note that the starting value for $k$ is $0$ not any other value.

But in your proof for positive reals, you are saying that we assume it is true for $x=k$ and $x=k+\Delta k$ where $k$ is a specific real number and $\Delta k$ is some arbitrary real number! Then you prove that it is valid for any real number but it didn't need any proof as you took it granted by your assumptions that the equality holds for any real number!

Answer 15

I think this is what you're trying to do.

First use linear approximation: $e^{i\Delta x} = 1+\left(\frac{d}{dx}e^{i\Delta x}\right)|_0 \Delta x=1+i\Delta x$. $\Delta x$ is infinitesimal so higher-order terms disappear.

Doing linear approximations with sine and cosine too, we get $\cos(\Delta x)=1+ 0\Delta x$ and $\sin(\Delta x) = 0+ \Delta x$.

Therefore $e^{i\Delta x}=\cos(\Delta x)+i\sin(\Delta x)$ holds with $\Delta x$ infinitesimal.

Now to move to the finite case, use DeMoivre's formula $(\cos(k)+i\sin(k))^n=\cos(nk)+i\sin(nk)$ ($n$ an integer). This can be proved with the angle addition formulas by induction on $n$.

\begin{align} e^{ix}&=\lim_{\Delta x\rightarrow 0} (e^{i\Delta x})^{x/\Delta x}\\&=\lim_{\Delta x\rightarrow 0} (\cos(i\Delta x)+i\sin(i\Delta x))^{x/\Delta x}\\&=\lim_{\Delta x\rightarrow 0} \cos(x)+i\sin(x) \end{align}

This is not very rigorous, so here is how to make it rigorous. Instead of using $\Delta x$ infinitesimal, use the identity $e^{i\Delta x}=\cos(\Delta x)+i\sin(\Delta x)+\Delta x^2 \delta(\Delta x)$ where $\delta(\Delta x)\rightarrow 0$ as $\Delta x\rightarrow 0$. Then the limit comes out as

\begin{align} e^{ix}&=\lim_{\Delta x\rightarrow 0} (\cos(i\Delta x)+i\sin(i\Delta x)+\Delta x^2 \delta(\Delta x))^{x/\Delta x}\\&=\lim_{\Delta x\rightarrow 0} \left[(\cos(x)+i\sin(x))+\frac{x}{\Delta x}(\cos(x-\Delta x)+i\sin(x-\Delta x))\Delta x^2\delta(\Delta x)+\cdots\right] \end{align}

by the binomial theorem. All terms after the first will go to zero.