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

Solution 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.

Solution 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.)