Maan, I'm sad because I don't understand it exactly. Let's see . . .
How much she is still in my life minus k1 times pain yields zero when k1=1 and pain=how much she is still in my life. As she leaves my life, this value becomes zero, so long as k1 is positive. If k1 is negative, then this term always increases.
This term multiplied by one over one plus something decaying, once t is greater than k2. After t surpasses k2, the exponential decay increases for increasing values of d. So this whole value (1/(1+e^((tk2)/d))) approaches 1 as t increases, approaching slower if d is smaller and slower if k2 is larger. So this is a damping factor.
So as time increases, the damping factor thing approaches 1. This makes it effectively go away over time. If k1 is positive, then as she leaves my life the value decreases. If k1 is negative, the value increases. So long as k1 is positive, her leaving my life results in a decreased derivative, meaning pain is decreasing faster! Yay! Less pain for the win.
Ugh, thinking that through was painful. Maybe that has to do with it being 12:30 past midnight. That is, it is 30 minutes past midnight. Goodnight.
0128: "dPain over dt"
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 X is kiss
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0) "How much she's still in my life" = Y. Breikup occurs from t = x to t = x, x << 1 (with time mezhured in days)
1) Giv'n: dP/dt = (Y(t)kP)/(1+e^(Kt)/d)
2) For simplicity, let's choose the units so that the presence/pain proportionality constant k = 1. We may do this since there is no other information provided about the relationship of presence and pain.*
2) Common Kno'lij: dP(0)/dt = (Y(0)P(0))/(1+e^K/d) >> 0 (sharp increase in pain @ breikup)
2a) Y(0) > P(0)
2b) K/d >/> 0
3) CK: P(0) > 0
3a) From this & 2a) follows: Y(0) >> 0
4) CK: dY(0)/dt << 0
5) Is it realistic to assume that d^2P(0)/dt^2 = d^2Y(0)/dt^2 = 0? Ie. that the breikup will be the steepest increase in pain & drop in presence?
6) CK: Y(1) << 1 (assuming it's a real breikup)
6a) From this & 3a) follows that Y must have the shape of an exponential drop around these parts.
7) CK: P(1) >> 0 (as abov)
â€¦
Anyone wanna continue? I'm starting to get bored here.
1) Giv'n: dP/dt = (Y(t)kP)/(1+e^(Kt)/d)
2) For simplicity, let's choose the units so that the presence/pain proportionality constant k = 1. We may do this since there is no other information provided about the relationship of presence and pain.*
2) Common Kno'lij: dP(0)/dt = (Y(0)P(0))/(1+e^K/d) >> 0 (sharp increase in pain @ breikup)
2a) Y(0) > P(0)
2b) K/d >/> 0
3) CK: P(0) > 0
3a) From this & 2a) follows: Y(0) >> 0
4) CK: dY(0)/dt << 0
5) Is it realistic to assume that d^2P(0)/dt^2 = d^2Y(0)/dt^2 = 0? Ie. that the breikup will be the steepest increase in pain & drop in presence?
6) CK: Y(1) << 1 (assuming it's a real breikup)
6a) From this & 3a) follows that Y must have the shape of an exponential drop around these parts.
7) CK: P(1) >> 0 (as abov)
â€¦
Anyone wanna continue? I'm starting to get bored here.
Re: "dPain over dt" Discussion
According to Maple, Pain(t) equals:
[math]Pain \left( t \right) = \left( 1+{e^{{\frac {t{\it k2}}{d}}}} \right) ^
{{\it k1}\,d} \left( {e^{{\frac {t{\it k2}}{d}}}} \right) ^{{\it k1
}\,d}{\it \_C1}+ \left( {e^{{\frac {t{\it k2}}{d}}}} \right) ^{{\it
k1}\,d}G \left( {e^{ \left( t{\it k2} \right) {\it k1}}}+{e^{{\frac {
t+{\it k2}+{\it k1}\,dt{\it k1}\,d{\it k2}}{d}}}} \right) \left( 1+
{e^{{\frac {t{\it k2}}{d}}}} \right) ^{1}{{\it k1}}^{1}[/math]
So if k1 is positive it decreases, and if k1 is negative it increases.
[math]Pain \left( t \right) = \left( 1+{e^{{\frac {t{\it k2}}{d}}}} \right) ^
{{\it k1}\,d} \left( {e^{{\frac {t{\it k2}}{d}}}} \right) ^{{\it k1
}\,d}{\it \_C1}+ \left( {e^{{\frac {t{\it k2}}{d}}}} \right) ^{{\it
k1}\,d}G \left( {e^{ \left( t{\it k2} \right) {\it k1}}}+{e^{{\frac {
t+{\it k2}+{\it k1}\,dt{\it k1}\,d{\it k2}}{d}}}} \right) \left( 1+
{e^{{\frac {t{\it k2}}{d}}}} \right) ^{1}{{\it k1}}^{1}[/math]
So if k1 is positive it decreases, and if k1 is negative it increases.
Re: "dPain over dt" Discussion
Actually it feels more like a graph of both a ln an e value, so while she can be in your life, you will never reach the equalibrium to which there is no pain, so you have two paths that the relationship can take, painful and recurring (e) or painful with an end (ln), so either way you look at it, yes you're boned. But it hurts. But you can have her with you. But it hurts. So you're boned. Mobius strip of the internets strikes again!
I think, without a doubt, I have proven that XKCD is a ghey tranvestite that isalso straight, but a hermaphrodite. Congratulations internets, your mobiusstrip strikes again.
 Freddino18
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Re: 0128: "dPain over dt"
Eight months. Pain has transformed from constant agony to periods of relative normalcy, bookended by depression and interspersed with occasional bouts of agony, at the same intensity as the beginning.
Blenders!
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