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I have following function: $a = \dfrac{x\cdot\left(1-\mathrm{e}^{-\frac{yz}{x}}\right)}{z}$

With: $x>0; 0\le y \le 1; z>0$

I am espacially interested in the behaviour of the function for $z$. Looking at graphs and working with numbers it is fairly easy to see that the function is convex and always rising for $z$ in the given value range.

Sadly I am unable to prove it mathemacially. I tried it with the derivations:

$\frac{d}{dz} [f(z)]=\dfrac{y\mathrm{e}^{-\frac{yz}{x}}}{z}-\dfrac{x\cdot\left(1-\mathrm{e}^{-\frac{yz}{x}}\right)}{z^2}$

$\frac{d^2}{dz^2} [f(z)]=\dfrac{\mathrm{e}^{-\frac{yz}{x}}\cdot\left(2x^2\mathrm{e}^\frac{yz}{x}-y^2z^2-2xyz-2x^2\right)}{xz^3}$

Since there are additions/subtractions I cannot show these are always positve or negative. I also tried to use a Hessian Matrix to prove it for the multivariate function, but due to the size of the derivatives I am not finding a proper solution.

Is there an easier way I am missing to proving those characteristics of the function?

Nino
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1 Answers1

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Let us rewrite $a(x,y,z)$ in the following way: $$ a(x,y,z) = \frac{x}{z} (1-e^{-\frac{yz}{x}}) = y \frac{x}{zy} (1-e^{-\frac{yz}{x}}) = y \varphi\left(\frac{yz}{x}\right), $$ where we introduce $$ \varphi(\sigma) := \frac{1-e^{-\sigma}}{\sigma}. $$ Hence, to study the shape of the function $a$, it suffices to study $\varphi$, which is a real-function of a single variable, which you can plot easily. First $$ \varphi'(\sigma) = - \frac{e^{-\sigma}}{\sigma^2} \left( e^{\sigma} - (1+\sigma) \right). $$ From this, you check that $\varphi' \leq 0$ so $\varphi$ is decreasing on $\mathbb{R}$ (there must be a typo in your question). This stems from the very classical inequality $1+x \leq e^x$ valid for any $x \in \mathbb{R}$. Similarly, $$ \varphi''(\sigma) = \frac{e^{-\sigma}}{\sigma^3} \left(2 e^\sigma + \sigma^2 - 2 \sigma -2 \right) $$ so that $\varphi'' \geq 0$ thanks to the same inequality.

Hence, for a fixed $y > 0$ and $x > 0$, as a function of $z$, your function is decreasing on $\mathbb{R}$ and convex.

cs89
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