4

so as the titles states I wan't to proof that no two functions $f_t$ of type $$f_t(x) = \frac{1}{t} \cdot e^{-tx²} \; \text{given that} \; t>0$$ share a point. This is a question from my textbook. However I come to the conclusion that there should be 2 functions should share some point.

$$f_a(x) = \frac{1}{a} \cdot e^{-ax²}$$ $$f_b(x) = \frac{1}{b} \cdot e^{-bx²}$$ Let $f_a(x) = f_b(x)$

$$\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²}$$ $$\frac{1}{a} \cdot e^{a} = \frac{1}{b} \cdot e^{b}$$ $$ae^{-a}=be^{b}$$ In order to proof that the 2 functions don't share a point I would need to proof that a=b This would mean that the function $g(x)=xe^x$ is a one to one function. However this is not true!

Now, all continuous one-one functions have to be monotonic (strictly increasing or decreasing). But we know that this is not the case for $g(x)$ since $\lim_{x \to 0} g(x)=\lim_{x \to +\infty}f(x)=0$ whereas $f(1)>0$

Where is my mistake?

amWhy
  • 209,954
Prankster
  • 567
  • 4
  • 15

4 Answers4

2

Let suppose that for some $c\neq 0$ we have $$\frac{1}{a} \cdot e^{-ac^2} = \frac{1}{b} \cdot e^{-bc^2}\iff \frac{\log b-\log a}{a-b}=c^2<0\ \text{contradiction}$$

2

$\dfrac{1}{a} \cdot e^{-ax²} = \dfrac{1}{b} \cdot e^{-bx²} \implies \dfrac{1}{a} \cdot e^{a} = \dfrac{1}{b} \cdot e^{b}.$

This step is not correct. It should be $$\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²} \implies \frac{b}{a} = e^{(a-b)x^2}.$$ Taking natural log on both sides given $a,b>0$: $$ (a-b)x^2 = \ln b-\ln a. $$ Without loss of generality assuming $a>b>0$, then $$ (a-b)x^2 \geq 0, \;\text{ and }\; \ln b-\ln a < 0. $$ Contradiction.

Shuhao Cao
  • 18,935
1

The question has been amply answered, so we give another proof of non-intersection.

Let $x$ be fixed. Differentiate with respect to $t$. The derivative is negative. So our function, for fixed $x$, is a decreasing function of $t$.

André Nicolas
  • 507,029
0

I don't understand how you go from the $\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²}$ step to the next.

Anyway, here is a hint: \begin{align} \frac{1}{a} \cdot e^{-ax²} &= \frac{1}{b} \cdot e^{-bx²}\\ \frac{b}{a} &= e^{ax^2-bx^2}\\ \text{If } b>a\\ \underbrace{\frac{b}{a}}_{>1}&= e^{\underbrace{(a-b)}_{<0}x^2}\\ \end{align} Contradiction! (The exponent of a negative number can never exceed 1) Can you similarly prove a contradiction for $b<a$

Inquest
  • 6,635