Prove the following for the given expression

Question

Consider the equation $$\frac{\pi ^e}{(x-e)} + \frac{e^{\pi}}{(x-\pi)} + \frac{\pi^{\pi}+e^e}{(x-e-\pi)}=0$$ Prove that

1) This equation has two real roots in $(\pi-e, \pi +e) $

2) This equation has one real root in $(e,\pi)$ and other root in $(\pi, e+\pi)$

My approach : Let $$f(x)=\frac{\pi ^e}{(x-e)} + \frac{e^{\pi}}{(x-\pi)} + \frac{\pi^{\pi}+e^e}{(x-e-\pi)}$$. Then I tried to find the sign of $f(\pi-e)f(\pi +e)$ so as to check for its sign to prove the first part. But this gives me a very tedious equation from which determining the sign is difficult. Moreover $f(\pi +e)$ creates a 0 in the denominator in third term.Any help would be greatly appreciated.

Edit 1: Can anyone prove this using the properties and fundamentals just of polynomials. I know that maths is meant to learn what you don't understand but at current I want to know if this question can be solved using only the fundamentals of algebra and the polynomials without use of calculus

Answer 1

multiply out the term $$\pi^e(x-\pi)(x-e-\pi)+e^\pi(x-e)(x-e-\pi)+(\pi^\pi+e^e)(x-e)(x-\pi)=...$$ ok then consider $$f(x)=\frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^\pi+e^e}{x-e-\pi}$$ and compute $$f(\pi-e)\approx -27.7935443004972883808036072099123560713149915<0$$ and $$\lim_{x \to \pi+e}f(x)=+\infty$$

Answer 2

Let $f(x)=\frac{\pi ^e}{(x-e)} + \frac{e^{\pi}}{(x-\pi)} + \frac{\pi^{\pi}+e^e}{(x-e-\pi)}.$

Thus, $f$ is a continuous function on $(\pi,\pi+e)$ and on $(e,\pi)$ and calculate four limits of $f$ in these bounds.

For example, $$\lim_{x\rightarrow\pi+e^-}f(x)=-\infty$$ and $$\lim_{x\rightarrow\pi^+}f(x)=+\infty$$

Answer 3

Multiplying out the denominator into the numerator we get:

$$f(x)=\pi^e(x-\pi)(x-e-\pi)+e^\pi(x-e)(x-e-\pi)+(\pi^\pi+e^e)(x-e)(x-\pi)$$

I'll show you how to prove the part about $(e,\pi)$, the other proofs are similar. This proof does not use any complex quadratic formula or tough arithmetic calculation.

Take $x_1=e+h$, where $h$ is a very small number $\approx10^{-20}>0$. Compute

$$f(x_1)=\color{green}{\pi^e(e-\pi+h)(-\pi+h)}+\color{blue}{e^\pi(h)(-\pi+h)}+\color{blue}{(\pi^\pi+e^e)(h)(e-\pi+h)}$$

Since $h$ is very very small, we can almost equate $\color{blue}{h\cdot\text{constant value}\approx0}$ and $\color{green}{\text{constant}+h\approx\text{constant}}$. Hence, we get: $$f(x_1)\approx\pi^e(e-\pi)(-\pi)>0$$

Now, take $x_2=\pi-h$. Computing $f(x_2)$ by similar logic, we get: $$f(x_2)=e^\pi(\pi-e)(-e)<0$$

Since $f(x)$ has changed sign in the interval $(e,\pi)$, this implies that it must have crossed the x-axis at some $x_0\in(e,\pi)$ as it is a continuous function. Hence, $f(x_0)=0$ and thus $f(x)$ has a root in $(e,\pi)$.

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For the first proof, you'll need to break the given interval of $(\pi-e,\pi+e)$ at some suitable value of $x'$. Then use the same logic as described above, to prove the existence of one real root each in $(\pi-e,x')$ and $(x',\pi+e)$, for a total of two real roots.

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Can you solve this on your own from here now?

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PS: (since you mentioned you don't have much experience of limits) The technique of taking $h$ is essentially me showing you how to take the Left hand Limits and Right Hand Limits of $f(x)$ without actually mentioning the scary terminology! ;)