(2¹ºº)^t = exp(t*ln(2¹ºº)), so the factor you call ‘c’ is not a constant multiplier for the integral; in fact, that combination of constants doesn’t even show up. The (approximated) integral is actually b∫t²*exp(-at)dt, where a = 100*ln(2) and b = 120. Evaluating this from 0 to T produces the expression: (2b/a³)*(1 - exp(-aT)*(1 + aT + ½(aT)²)).
These factors of exp(-aT) show up when evaluating the integral to T<∞. (Obviously, when T → ∞, the integral converges to (2b/a³).) For a ~ O(5) or higher, then, the entire total utility is found in 10 years, within double precision. That corresponds to a ≈ 7*ln(2). I think this indicates that the model may not be a good approximation of reality. Also, for slower subjective time (a < ln(2) ≈ 0.693), the percentage of total utility found in 10 years drops. For a = 0.1*ln(2), it’s only 3.33%.
Also, you defined either ‘x’ or your linear density function incorrectly. If you want x to be stars/ly^2, the density function should be ρ = x(1 - r/50000). If you do all of the calculations symbolically and don’t plug in values until the end, the equation for total utility as a function of time (before discounting) is exact, not approximate.
Edit: Actually, I made an error when referencing slower subjective time above. That would be a < 0, not a < ln(2). The model doesn’t make sense in this case, because the integral b∫t²*exp(-at)dt diverges for a < 0.
Edit again: Nope, I lost track of the 0.1 factor in 2^(0.1Y). Disregard previous statements about what constitutes slower subjective time; it’s not really relevant anyway.
(2¹ºº)^t = exp(tln(2¹ºº)), so the factor you call ‘c’ is not a constant multiplier for the integral; in fact, that combination of constants doesn’t even show up. The (approximated) integral is actually b∫t²exp(-at)dt, where a = 100ln(2) and b = 120. Evaluating this from 0 to T produces the expression: (2b/a³)(1 - exp(-aT)*(1 + aT + ½(aT)²)).
You are correct! However, you note yourself that “For a ~ O(5) or higher, then, the entire total utility is found in 10 years, within double precision.” So the result does depend on subjective time, which is what I had expected.
This is important—but it still doesn’t change the conclusion, that rational expected utility-maximizers operating in this framework don’t care about the future.
I’m grateful to you for finding the flaw in my math, and am upvoting you on that basis. But I don’t think you should say, “I found an error, therefore I downvote your whole post” instead of “Here is a correction to your work on this interesting and important topic; your conclusion is now only valid under these conditions.”
A general comment to everybody: Most of you are in the habit of downvoting a post if you find a single flaw in it. This is stupid. You should upvote a post if it illuminates an important topic or makes you realize something important. Einstein didn’t say, “Newton’s law is inaccurate. Downvoted.” Voting that way discourages people from ever posting on difficult topics.
Also, you defined either ‘x’ or your linear density function incorrectly. If you want x to be stars/ly^2, the density function should be ρ = x(1 - r/50000). If you do all of the calculations symbolically and don’t plug in values until the end, the equation for total utility as a function of time (before discounting) is exact, not approximate.
No; x is the density as a function of r, and it varies linearly from a maximum at r=0, to zero at r=50,000. The way I wrote it is the only possible function satisfying that.
I’m grateful to you for finding the flaw in my math, and am upvoting you on that basis. But I don’t think you should say, “I found an error, therefore I downvote your whole post” instead of “Here is a correction to your work on this interesting and important topic; your conclusion is now only valid under these conditions.”
A general comment to everybody: Most of you are in the habit of downvoting a post if you find a single flaw in it. This is stupid. You should upvote a post if it illuminates an important topic or makes you realize something important. Einstein didn’t say, “Newton’s law is inaccurate. Downvoted.” Voting that way discourages people from ever posting on difficult topics.
I downvoted not simply because there was a math error, but because of your aggressive comments about downvoting this topic without having arguments against your math, which was in fact faulty. When you spend a significant portion of the post chastising people about downvoting, you should take a little more time to make sure your arguments are as ironclad as you think they are.
The point is not to discourage people from posting on difficult topics; it’s to discourage unwarranted arrogance.
No; x is the density as a function of r, and it varies linearly from a maximum at r=0, to zero at r=50,000. The way I wrote it is the only possible function satisfying that.
You define x as having units of stars/ly^2. Because you’re approximating the galaxy in two dimensions, the general density function should also have units of stars/ly^2. You wrote this density function: x(50000-r), which has units of stars/ly. I wrote the density function ρ = x(1 - r/50000) which has the correct units and fits the boundary conditions you described: ρ(r=0) = x, and ρ(r=50000) = 0.
In your post, you end up redefining x when you solve for it from your density function, so it does not affect the final result. However, in a model that’s supposed to be physically motivated, this is sloppy.
Downvoted, because your math is wrong.
(2¹ºº)^t = exp(t*ln(2¹ºº)), so the factor you call ‘c’ is not a constant multiplier for the integral; in fact, that combination of constants doesn’t even show up. The (approximated) integral is actually b∫t²*exp(-at)dt, where a = 100*ln(2) and b = 120. Evaluating this from 0 to T produces the expression: (2b/a³)*(1 - exp(-aT)*(1 + aT + ½(aT)²)).
These factors of exp(-aT) show up when evaluating the integral to T<∞. (Obviously, when T → ∞, the integral converges to (2b/a³).) For a ~ O(5) or higher, then, the entire total utility is found in 10 years, within double precision. That corresponds to a ≈ 7*ln(2). I think this indicates that the model may not be a good approximation of reality. Also, for slower subjective time (a < ln(2) ≈ 0.693), the percentage of total utility found in 10 years drops. For a = 0.1*ln(2), it’s only 3.33%.
Also, you defined either ‘x’ or your linear density function incorrectly. If you want x to be stars/ly^2, the density function should be ρ = x(1 - r/50000). If you do all of the calculations symbolically and don’t plug in values until the end, the equation for total utility as a function of time (before discounting) is exact, not approximate.
Edit: Actually, I made an error when referencing slower subjective time above. That would be a < 0, not a < ln(2). The model doesn’t make sense in this case, because the integral b∫t²*exp(-at)dt diverges for a < 0.
Edit again: Nope, I lost track of the 0.1 factor in 2^(0.1Y). Disregard previous statements about what constitutes slower subjective time; it’s not really relevant anyway.
You are correct! However, you note yourself that “For a ~ O(5) or higher, then, the entire total utility is found in 10 years, within double precision.” So the result does depend on subjective time, which is what I had expected.
This is important—but it still doesn’t change the conclusion, that rational expected utility-maximizers operating in this framework don’t care about the future.
I’m grateful to you for finding the flaw in my math, and am upvoting you on that basis. But I don’t think you should say, “I found an error, therefore I downvote your whole post” instead of “Here is a correction to your work on this interesting and important topic; your conclusion is now only valid under these conditions.”
A general comment to everybody: Most of you are in the habit of downvoting a post if you find a single flaw in it. This is stupid. You should upvote a post if it illuminates an important topic or makes you realize something important. Einstein didn’t say, “Newton’s law is inaccurate. Downvoted.” Voting that way discourages people from ever posting on difficult topics.
No; x is the density as a function of r, and it varies linearly from a maximum at r=0, to zero at r=50,000. The way I wrote it is the only possible function satisfying that.
I downvoted not simply because there was a math error, but because of your aggressive comments about downvoting this topic without having arguments against your math, which was in fact faulty. When you spend a significant portion of the post chastising people about downvoting, you should take a little more time to make sure your arguments are as ironclad as you think they are.
The point is not to discourage people from posting on difficult topics; it’s to discourage unwarranted arrogance.
You define x as having units of stars/ly^2. Because you’re approximating the galaxy in two dimensions, the general density function should also have units of stars/ly^2. You wrote this density function: x(50000-r), which has units of stars/ly. I wrote the density function ρ = x(1 - r/50000) which has the correct units and fits the boundary conditions you described: ρ(r=0) = x, and ρ(r=50000) = 0.
In your post, you end up redefining x when you solve for it from your density function, so it does not affect the final result. However, in a model that’s supposed to be physically motivated, this is sloppy.