Induction on the number of equations

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In summary, the conversation discusses a proof by induction on the number of equations in a differential operator $L$. The base case is for $n=1$ and the inductive hypothesis is that for $n=k$, the system can be reduced to the form $Lx=f \land \psi$, where $\psi$ does not contain any $x$. The inductive step is for $n=k+1$, where the first $k$ equations can be reduced using the inductive hypothesis and the new differential equation $Lx=f$. This system is equivalent to the initial system.
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mathmari
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Hey! :eek:

Let $L$ be a differential operator.

We suppose that we have $n$ equations, that means $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x=f_j}$ and we assume that $\phi$ can be written as $Lx=f \land \psi$, where $\psi$ doesn't contain any $x$.

We prove this by induction on the number of equations, $n$.

  • Base case: For $n=1$ we have one equation, so it is of the form $Lx=f$.
  • Inductive hypothesis: We suppose that it holds for $n=k$, i.e., if $\phi$ contains $k$ equations, then we can reduce it into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
  • Inductive step: We will show that it holds for $n=k+1$, i.e., if we have $k+1$ equations we can reduce this system into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
    From the inductive hypothesis we know that we can reduce the first $k$ equations into the above form. So we have two equations that contain $x$ and its derivatives.

Is this correct so far? (Wondering)

How could we continue? (Wondering)
 
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Is the inductive step as follows? (Wondering) Inductive step: We will show that it holds for $n=k+1$, i.e., if we have $k+1$ equations we can reduce this system into the form $$Lx=f \land \psi \ \ \text{ where } \psi \text{ doesn't contain any } x.$$
From the inductive hypothesis we know that we can reduce the first $k$ equations into the above form. So we have two equations that contain $x$ and its derivatives. We add these two equations and we get a new differential equation $Lx=f$. So the initial system is equivalent to the new differential equation $Lx=f$.
 
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FAQ: Induction on the number of equations

What is induction on the number of equations?

Induction on the number of equations is a mathematical proof technique used to show that a statement is true for all natural numbers. It involves proving a base case and then using a recursive argument to show that if the statement is true for one number, it is also true for the next number.

Why is induction on the number of equations important?

Induction on the number of equations is important because it allows us to prove statements that are true for an infinite number of cases. It is a powerful tool in mathematics and is used to prove many fundamental theorems and properties.

How is induction on the number of equations different from induction on the number of variables?

Induction on the number of equations is a proof technique used to prove statements about the natural numbers, while induction on the number of variables is used to prove statements about mathematical expressions or functions. They are different methods that are used for different types of proofs.

What is the role of the base case in induction on the number of equations?

The base case is the starting point for the induction proof. It is the case for which we directly show that the statement is true. Without a valid base case, the proof cannot continue to the next step and show that the statement is true for all natural numbers.

What are some common mistakes made when using induction on the number of equations?

Some common mistakes when using induction on the number of equations include using the wrong base case, assuming the statement is true without properly proving it, and not considering all possible cases. It is important to carefully follow the steps of the proof and make sure all assumptions and steps are valid.

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