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I have a function where

$$ y = m\sin\Bigl(\frac{x^{1.1}+30}{0.7d}\Bigr)^{2.2} $$ where $m = 45$ and $d = 120$ (constants)

I would like the turning point at the top of the sin curve to peak at 0.75d (90). I would like the starting and finishing values to remain the same.

The current function is plotted in blue. I would like a line similar to the red one. Feel free to ditch the sin function, if another method would be better

imagen

Thanks for any help.

jimjim
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tedioustortoise
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  • Try with $a x^\alpha$, adjusting $a$ so that $a x^\alpha=x^{1.1}$ at the endpoint. –  May 03 '20 at 14:24
  • If you're hoping to have some sort of periodicity, you might just play around with summing $\sin(x), \sin(2x), \sin(3x), \ldots$ in various proportions; for instance, $\sin(x)-\sin(2x)$ looks a lot like what you want on $[0,\pi]$ (although the peak is at 2/3 of the length for that one) – Milo Brandt May 03 '20 at 15:11
  • Thanks Milo, I agree with keeping some sort of periodicity. How could I keep where my max values, which constant m is currently doing – tedioustortoise May 03 '20 at 17:07

1 Answers1

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Consider the curve $$y=\frac{(x-4)^4}{12}-\frac{kx^2}{2}+cx+d,$$ where the constants are given by $d=5-\frac{90^4}{12}$ and $$240=2×30^4-12k(120^2)+24(120c)+24d,\\2M=2d+180c-8100k,$$ with $M$ being the maximum value you want the function to attain.

Allawonder
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  • Thanks. I understand points (1) and (2), I am struggling to grapple with the mathematics of your final expression. Would you mind writing it in the format of y = ... . Furthermore, wouldn’t an (x-90)^2 function lose all its periodicity. I.e. a deceleration to the turning point and acceleration from the turning point and max gradient to be found between turning points. – tedioustortoise May 03 '20 at 17:03
  • @Ollie But you didn't include periodicity among the properties you wanted. In fact, the original function isn't periodic at all. – Allawonder May 03 '20 at 18:20
  • @Ollie The final expression is already in the form $y=\cdots.$ – Allawonder May 03 '20 at 18:22
  • Maybe I have the wrong word for it. I would like still like the qualities of the deceleration and acceleration away from the graph turning points, which is seen in sine, cosine styled functions. Parabolic. functions do not have this quality as they are not trigonometric. – tedioustortoise May 03 '20 at 18:28
  • I’m sorry, but I don’t understand this expression: “with ,>0 a, A>0 and −8100=(0) A−8100=g(0) and −900=(120)A−900=g(120)and A being the maximum value” – tedioustortoise May 03 '20 at 18:30
  • @Ollie Are you talking about the way the curve bends downwards, then upwards, then downwards again? – Allawonder May 03 '20 at 18:30
  • I suppose so. What I mean is that the curve doesn’t decelerate and accelerate at a constant rate. Differentiating the curve leaves another trigonometric function, rather than a constant. – tedioustortoise May 03 '20 at 18:33
  • @Ollie Oh, what I meant is that $a$ and $A$ are both positive numbers chosen so that the value of the function when $x=0$ is equal to $A-8100$ and the value at the endpoint where $x=120$ is equal to $A-900.$ For this to happen we must have $g(120)=g(0)+7200.$ – Allawonder May 03 '20 at 18:33
  • @Ollie Well, I thought you said the sine function could be ditched? If all you want is any curve that bends like the given curve above, then that can be done in many ways. It need not be a sine curve. But of course, a parabola doesn't bend that way. I didn't know you wanted such a condition to be fulfilled as well. I'll look into it. – Allawonder May 03 '20 at 18:36
  • @Ollie I've got a curve that may satisfy all requirements so far. However, I need to clarify: Should the difference $g(120)-g(0)=5$ as is implied in the graph above? – Allawonder May 03 '20 at 18:42
  • yes please. thank you – tedioustortoise May 03 '20 at 18:44
  • @Ollie Hump! Then it will not work. I was thinking of $$y=\frac{M}{1+(x-90)^2},$$ but I will look further into this problem! – Allawonder May 03 '20 at 18:47
  • Thanks for your help. If there is a solution which is roughly correct, I would be interested ub that too. Thank you – tedioustortoise May 03 '20 at 18:52
  • @Ollie OK, I think I've found a curve which satisfies all your conditions so far (except for implicit ones we may not know yet). So consider the curve $$y=\frac{(x-4)^4}{12}-\frac{kx^2}{2}+cx+d,$$ where the constants are given by $d=5-\frac{90^4}{12}$ and $$240=2×30^4-12k(120^2)+24(120c)+24d,,2M=2d+180c-8100k,$$ and $M$ is the maximum value you want the function to attain. – Allawonder May 03 '20 at 19:21
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    Many thanks for this. – tedioustortoise May 04 '20 at 08:39
  • @Ollie Welcome. Glad it helped! – Allawonder May 04 '20 at 15:33