Proving the Uniqueness of Positive Solution of $x(x+1)(x+2)\cdots(x+2020)-1=0$

• MHB
• anemone
In summary, the equation $x(x+1)(x+2)\cdots(x+2020)-1=0$ has exactly one positive solution $x_0$, and this solution satisfies $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$ by using the Intermediate Value Theorem and mathematical induction.
anemone
Gold Member
MHB
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Show that the equation $x(x+1)(x+2)\cdots(x+2020)-1=0$ has exactly one positive solution $x_0$ and prove that this solution $x_0$ satisfies $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$.

To show that the equation $x(x+1)(x+2)\cdots(x+2020)-1=0$ has exactly one positive solution $x_0$, we can use the Intermediate Value Theorem. This theorem states that if $f$ is a continuous function on an interval $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one value $c \in (a,b)$ such that $f(c)=0$.

In this case, we can let $f(x)=x(x+1)(x+2)\cdots(x+2020)-1$ and consider the interval $[1,2020]$. We can see that $f(1)=-1$ and $f(2020)=2020!-1>0$, so by the Intermediate Value Theorem, there exists at least one value $x_0 \in (1,2020)$ such that $f(x_0)=0$. This means that the equation has at least one positive solution.

To prove that this solution $x_0$ satisfies $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$, we can use mathematical induction.

First, we will show that $\dfrac{1}{2020!+10}<1$. This can be proven by noting that $2020!+10>2020!$ and $\dfrac{1}{2020!}$ is a very small number, so $\dfrac{1}{2020!+10}<\dfrac{1}{2020!}<1$.

Next, we will show that if $\dfrac{1}{2020!+10}<x_0<\dfrac{1}{2020!+6}$, then $\dfrac{1}{2020!+10}<x_0+1<\dfrac{1}{2020!+6}$. This can be done by multiplying both sides of the inequality by $x_0+1$ and using the fact that $x_0(x_0+1)(x_0+2)\cdots(x_0+2020)=1$.

1. What is the equation that needs to be proven?

The equation that needs to be proven is x(x+1)(x+2)...(x+2020)-1=0.

2. What is the significance of proving the uniqueness of the positive solution?

Proving the uniqueness of the positive solution is important because it confirms that there is only one value of x that satisfies the given equation, making it easier to solve and analyze.

3. How can one prove the uniqueness of the positive solution?

One way to prove the uniqueness of the positive solution is by using the Intermediate Value Theorem, which states that if a continuous function has opposite signs at two points, then there must be at least one root in between those two points.

4. Can the uniqueness of the positive solution be proven algebraically?

Yes, the uniqueness of the positive solution can also be proven algebraically by showing that there is only one value of x that satisfies the given equation and that this solution is positive.

5. Why is it important to specify that the solution must be positive?

Specifying that the solution must be positive is important because it narrows down the possible solutions and makes the equation easier to solve. It also ensures that the solution is within the desired range and avoids any extraneous solutions.

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