It's the difference between a linear function (just y = x here) and an exponential (y = 12.3. e0.021x). So it reaches 0 twice, where the exponential cuts the linear, and it's asymmetric. What's a quick way of calculating the maximum?
This is the theoretical graph. Because of noise (it's biology!) an actual data point in each set might lie below zero.
Ed: sp
What's this distribution called?

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What's this distribution called?
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Re: What's this distribution called?
So that's y = x  12.3*exp(0.021*x) ?
dy/dx = 1  (12.3*0.021)*exp(0.021*x) = 1  0.2583*exp(0.021*x)
the maximum in y is when the derivative is zero, so
1 = 0.2583*exp(0.021*x)
exp(0.021*x) = 3.8715
0.021*x = ln(3.8715) = 1.3536
x = 1.3536/0.021 = 64.46.
Your maximum looks to be at a slightly larger x; it's important that you repeat these above calculations but at your best numerical precision. (I always tell the students to work with symbols and put the numbers in at the end).
dy/dx = 1  (12.3*0.021)*exp(0.021*x) = 1  0.2583*exp(0.021*x)
the maximum in y is when the derivative is zero, so
1 = 0.2583*exp(0.021*x)
exp(0.021*x) = 3.8715
0.021*x = ln(3.8715) = 1.3536
x = 1.3536/0.021 = 64.46.
Your maximum looks to be at a slightly larger x; it's important that you repeat these above calculations but at your best numerical precision. (I always tell the students to work with symbols and put the numbers in at the end).
molto tricky
Re: What's this distribution called?
Ok if I write it as y = x  A.exp(r.x)
the maximum is at x = ln(A.r) / r
the maximum is at x = ln(A.r) / r
molto tricky
Re: What's this distribution called?
And if your straight line is y=CX+D, i.e. your actual function is
y = Cx+D  A.exp(rx),
then the maximum is at
x = ( ln(C)  ln(A*r) ) / r.
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 Posts: 277
 Joined: Sat Nov 16, 2019 8:18 am
Re: What's this distribution called?
Perfect! Thank you, and thanks to Jaap too.shpalman wrote: ↑Mon Feb 01, 2021 10:23 amSo that's y = x  12.3*exp(0.021*x) ?
dy/dx = 1  (12.3*0.021)*exp(0.021*x) = 1  0.2583*exp(0.021*x)
the maximum in y is when the derivative is zero, so
1 = 0.2583*exp(0.021*x)
exp(0.021*x) = 3.8715
0.021*x = ln(3.8715) = 1.3536
x = 1.3536/0.021 = 64.46.
Your maximum looks to be at a slightly larger x; it's important that you repeat these above calculations but at your best numerical precision. (I always tell the students to work with symbols and put the numbers in at the end).