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Function given: $$f(\alpha, b,\delta) = \left(a_0 + a_2\cdot(\alpha-\delta)^2 + a_4\cdot(\alpha-\delta)^4\right)\cdot\left(1-\frac{b-1}{2b}\right)\cdot e^{j\cdot \pi} + \left(a_0 + a_2\cdot(\alpha+\delta)^2 + a_4\cdot(\alpha+\delta)^4\right)\cdot\frac{b-1}{2b}$$

  1. Finding the derivative of a function: $$\frac{d f(\alpha, b,\delta)}{d \alpha} = -\frac{\left(12\alpha\delta^2+4\alpha^3\right)\cdot a_4+2\alpha a_2}{b} + \left(4\delta^3+12\alpha^2 \delta \right)\cdot a_4+2\delta \cdot a_2$$

  2. Equate to zero and express $\alpha$:

$$\alpha^3 \cdot (2 \cdot a_4) - \alpha^2 \cdot (6 \cdot \delta \cdot a_4 \cdot b) + \alpha \cdot (6 \cdot \delta^2 \cdot a_4 + a_2) = 2 \cdot \delta^3 \cdot a_4 \cdot b + \delta \cdot a_2 \cdot b$$

Next, I can't find an analytic relation to calculate $\alpha$.

substituting $a_0=1; a_2 = -3/2; a_4 = 7/8; b = 2; \delta = 0.353$ I have to get the three roots of the equation.

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Maybe you can set a condition that the search for the maximum is carried out within the required limits?

Vasili
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Vaisala
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  • I am not sure you calculated the derivative correctly. Is $e^{j\cdot \pi} =-1$? Simplify your expression as much as possible, including removing any constants that won't affect the result. Next, it looks like there are three points where derivative is zero, why do you consider only one root? – Vasili Mar 27 '23 at 16:58
  • you are right, you need to find three extremum points. I double-checked the calculation of the derivative in the Mathcad program, it is correct. I wrote the equation in a more convenient form. I don't know how I can solve this cubic equation – Vaisala Mar 28 '23 at 06:18
  • I plugged the values into an equation solver and I do get three roots, the maximum should be at $x \approx 0.428417$ – Vasili Mar 29 '23 at 12:09
  • Thanks. I also got three real roots, but I didn't understand how you determined that it is the maximum? As I understand it, there are two maxima and one minimum, and the root you have given is just the minimum, or am I wrong? – Vaisala Mar 30 '23 at 04:09

1 Answers1

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Hint.

With

$$f(\alpha, b,\delta) = \left(a_0 + a_2\cdot(\alpha-\delta)^2 + a_4\cdot(\alpha-\delta)^4\right)\cdot\left(1-\frac{b-1}{2b}\right)\cdot (-1) + \left(a_0 + a_2\cdot(\alpha+\delta)^2 + a_4\cdot(\alpha+\delta)^4\right)\cdot\frac{b-1}{2b}$$

introducing the new variables $u = \alpha + \delta, v = \alpha - \delta, w = \frac{b-1}{2b}$ we have

$$ g(u,v,w) = (w-1) \left(a_0+a_2 u^2+a_4 u^4\right)+w \left(a_0+a_2 v^2+a_4 v^4\right) $$

and after substitutions

$$ g_0(u,v) = \frac{1}{4} \left(\frac{7 v^4}{8}-\frac{3 v^2}{2}+1\right)-\frac{3}{4} \left(\frac{7 u^4}{8}-\frac{3 u^2}{2}+1\right) $$

The stationary points are determined by solving

$$ \nabla g_0(u,v) = 0 = \cases{\frac{3}{8} u \left(6-7 u^2\right)\\ \frac{1}{8} v \left(7 v^2-6\right)} $$

with solutions

$$ \left[ \begin{array}{ccc} g_0 & u & v\\ -\frac{37}{56} & 0 & -\sqrt{\frac{6}{7}} \\ -\frac{37}{56} & 0 & \sqrt{\frac{6}{7}} \\ -\frac{1}{2} & 0 & 0 \\ -\frac{5}{28} & -\sqrt{\frac{6}{7}} & -\sqrt{\frac{6}{7}} \\ -\frac{5}{28} & -\sqrt{\frac{6}{7}} & \sqrt{\frac{6}{7}} \\ -\frac{5}{28} & \sqrt{\frac{6}{7}} & -\sqrt{\frac{6}{7}} \\ -\frac{5}{28} & \sqrt{\frac{6}{7}} & \sqrt{\frac{6}{7}} \\ -\frac{1}{56} & -\sqrt{\frac{6}{7}} & 0 \\ -\frac{1}{56} & \sqrt{\frac{6}{7}} & 0 \\ \end{array} \right] $$

Cesareo
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