Suppose that we have two measured values $x$ and $y$ with maximum absolute errors of $e_x$ and $e_y$.

Is there a formula to find a good upper bound for absolute and relative error of $x^y$?

  • $\begingroup$ very interesting,is there any formula related to this fact?but is this statistical values or what?for exmaple is $e_x$ got by some regression analysis? $\endgroup$ Commented Mar 30, 2013 at 6:52
  • $\begingroup$ No, they are not statistical. Suppose that we have measured $x$ and $y$ by some measurement tools $T_1$ and $T_2$, and these tools have maximum absolute errors of $e_x$ and $e_y$. $\endgroup$ Commented Mar 30, 2013 at 6:57
  • $\begingroup$ If you really know $e_x,e_y$ to be maximal absolute errors, and if you also know the measured values $x_0,y_0$, then you can place each of the true values of $x,y$ in intervals around their observed values, and just compute the maximal errors and relative errors without any fancy formula. Otherwise you might want to use differentials, which really give only approximate error bounds. $\endgroup$
    – coffeemath
    Commented Mar 30, 2013 at 10:33

1 Answer 1


The notation gets confusing, so allow me to set $e_x=\epsilon_x$ and $e_y=\epsilon_y$. The trick here is to observe that $x^y = e^{y \log{x}}$ and Taylor expand as follows:

$$\begin{align}(x+\epsilon_x)^{y+\epsilon_y} &= e^{(y+\epsilon_y)\log{(x+\epsilon_x)}}\\ &=e^{y \log{(x+\epsilon_x)}} e^{\epsilon_y \log{(x+\epsilon_x)}} \\ &= e^{y \log{x} + y \log{(1+\epsilon_x/x)}}\left [1+ \epsilon_y \log{(x+\epsilon_x)}+ O(\epsilon_y^2)\right]\\ &= x^y e^{y[\epsilon_x/x + O(\epsilon_x^2)] } \left [1+ \epsilon_y \log{x} +\epsilon_y \log{(1+\epsilon_x/x)}+ O(\epsilon_y^2)\right]\\ &= x^y \left (1+ \epsilon_x\frac{y}{x} + O(\epsilon_x^2)\right )\left [1+ \epsilon_y \log{x} +\epsilon_y \frac{\epsilon_x}{x}+ O(\epsilon_y^2) + O(\epsilon_x^2 \epsilon_y)\right]\\ &= x^y \left (1 + \epsilon_x\frac{y}{x} + \epsilon_y \log{x} + \left( \frac{1+y \log{x}}{x}\right )\epsilon_x \epsilon_y+ O(\epsilon_y^2) + O(\epsilon_x^2 \epsilon_y) \right) \end{align}$$

If you wish to ignore all second-order error terms, then you have

$$(x+\epsilon_x)^{y+\epsilon_y} \approx x^y \left (1 + \epsilon_x\frac{y}{x} + \epsilon_y \log{x} \right )$$


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