I am using the following function with the intention of getting a skewed cosine the derivative of which decreases as x approaches 1.
A picture might explain better than I can: y = (cos(0.5pix)^1.5)*x
I would like the local maxima between 0 and 1 to be 1 for whatever value I give to the float coeficient and exponent. This would allow me to alter the shape of the curve without affecting the height of the curve
What I have been doing to address this previously is dividing by the local maxima between 0 and 1 everytime I alter the exponent (shape of the curve), however this is not very convenient. It seems like there ought to be a better way.
I thought perhaps computing the derivative might get the old brain gears going and lead me to an answer, but no luck so far.
If anyone has any suggestions I would be very appreciative.
please excuse my rather unorthodox terminology.
Here is a simpler polynomial with fractional exponent instead of a trig function. Function is taken in form $ y={x^2(1-x^m)} \tag1 $
We find critical and maximum values by differentiation
$$ y'(x)=0=2 x -(2+m)x ^{1+m},$$
$$\rightarrow x_{crit}= (\dfrac{2}{2+m})^{\frac{1}{m}}\tag2$$
Plug into (1) to evaluate $y_{max}$
$$ y_{max}= (\dfrac{2}{2+m})^{\frac{2}{m}} \dfrac {m}{2+m} \tag3$$
To normalize divide (1) maximum y-value at (3). Instead of unity introduce $A$ as maximum height for generality's sake.
All curves pass through $(0,0),(1,0)$ on x-axis:
$$y= A\dfrac{x^2(1-x^m)}{(\dfrac{2}{2+m})^{\frac{2}{m}} (\dfrac{m}{m+2})} \tag4$$
The constant height occurs at a variable $x$
$$x_{crit} =(\dfrac{2}{2+m})^{\frac{1}{m}} = f(m),\;y_{max}=A\quad \tag5$$
Graphed for $ A=1, m= (2,4,5,8)$.
EDIT1:
$(\approx 0.61 <x_{crit} <1)$ for $(0<m<\infty)$