Wikipedia:Reference desk/Archives/Mathematics/2014 April 12

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

April 12

What is the relation between death rates and life expectancy?

"In the nineteenth century twenty-two out of every thousand people died each year, more than 2% of the population– today only 5 out of a thousand people die each year. In the nineteenth century the average lifespan was thirty-six years – today it’s about eighty." - From a review of "The Remedy" on Goodreads by Watchingthewords. 2% of the population is the same as one person out of 50. So if 2% of the population dies every year, then that means that, in any given year, on average, one person out every 50 people will die. If 50 people die every year, then it will take 40 years to kill 1,000 people, which means that your average lifespan is going to be about 40 years. 40 is not that much different from 36, so maybe we are on the right track. But then we look at today's numbers and we get something that doesn't make any sense. If only 5 people out of 1,000 die every year, then it will 200 years to kill 1,000 people, and nobody lives 200 years. I suspect there is something wrong with this picture, but I don't know what. I suppose an expanding population could skew the numbers somewhat, but I don't see how they can skew it that much. Pergelator. — Preceding unsigned comment added by 50.43.12.61 (talk) 00:19, 12 April 2014 (UTC)

Please don't double post. I believe you already got your answer at the original Desk where you posted this Q. StuRat (talk) 00:26, 12 April 2014 (UTC)

I only posted here because I did not get an answer to my question from the science crowd. I got several replies and comments, but no answer. Maybe there is no answer, but I thought that surely someone would have a some kind of model that would relate these two numbers. — Preceding unsigned comment added by 50.43.12.61 (talk) 07:40, 12 April 2014 (UTC)

The first comment there was the correct answer. Population is not static, because it is aging. That only means there has been a recent increase in life expectancy. Imagine a new drug came out that cured all diseases and slowed all aging processes, so people could live for 250 years. Then the death rate would suddenly drop to zero for a while, until those people reached 250. Then they would start dying at the usual rate again. IBE (talk) 13:45, 12 April 2014 (UTC)

You can also construct a simple model to demonstrate the relationship, but you need to make some simplifying assumptions. Those assumptions won't apply to the real world, but let's look at just how the mathematics works. Suppose you have 1000 people. We only care about their age, so lets think of them as 1000 columns on a bar graph. Each year, each column grows by 1. Then exactly 5 of them are chosen at random to die. Those 5 columns go back to zero. Suppose 5 new people are always born, but they start at zero. Now the average age (average height of a column) is the sum of all ages divided by a thousand; call it X. The total number of years is just the sum of all the ages, which is 1000 * X. The total number of years goes up by a thousand when everyone gets a year older. The total number goes down on average by 5 * X when 5 people die. So whenever 5 * X (5 * the average age) equals 1000, it will remain constant. On average, it will only be constant when X is 200, allowing for some random variation (you might not kill average age people). So the average age should fluctuate around 200. I'm pretty sure the life expectancy works out to the same thing, ie 200. But when the average age is below 200, the situation is not static. Further, in the real world, people don't die randomly, they are more likely to die as they get older, so in reality your 0.5% rate is not at all static for a young population. Trust me, it will increase. IBE (talk) 14:40, 12 April 2014 (UTC)

How to Calculate this?

I saw this equation on the page of Riemannian Curvature: $\left.{\frac {d}{ds}}{\frac {d}{dt}}\tau _{sX}^{-1}\tau _{tY}^{-1}\tau _{sX}\tau _{tY}Z\right|_{s=t=0}=(\nabla _{X}\nabla _{Y}-\nabla _{Y}\nabla _{X})Z=R(X,Y)Z$

I am poor at calculation, so could you tell me how to calculate $\left.{\frac {d}{ds}}{\frac {d}{dt}}\tau _{sX}^{-1}\tau _{tY}^{-1}\tau _{sX}\tau _{tY}Z\right|_{s=t=0}$ in detail please? Thank you! — Preceding unsigned comment added by 14.19.152.149 (talk) 11:00, 12 April 2014 (UTC)

The basic idea is that to first order in t, you have $\tau _{tX}=I+t\nabla _{X}$ for any t and X. (Here I is the identity map, and if it is still difficult to assign meaning to this, you can write it in coordinates.) Once you have convinced yourself that it is true, the above formula for the Riemann tensor can be obtained by just keeping the quadratic st term in the expansion:

${\begin{aligned}\tau _{sX}^{-1}\tau _{tY}^{-1}\tau _{sX}\tau _{tY}&=(I-s\nabla _{X})(I-t\nabla _{Y})(I+s\nabla _{X})(I+t\nabla _{Y})\\&=I+st(\nabla _{X}\nabla _{Y}-\nabla _{Y}\nabla _{X})+{\text{higher order terms.}}\end{aligned}}$

--Sławomir Biały (talk) 23:19, 12 April 2014 (UTC)

Write as a decimal the sum of 300 hundreds and 109 hundredth

Write as a decimal the sum of 300 hundreds and 109 hundredth — Preceding unsigned comment added by 213.205.241.243 (talk) 23:03, 12 April 2014 (UTC)

Show us what you've done so far on your homework, and we will be glad to help. StuRat (talk) 23:36, 12 April 2014 (UTC)

Be aware that, as written, the units of the two values to add are different. In order to add them, you have to determine what they are and how to convert from one to the other (or from both to a common base). I'm guessing that one "hundreds" means 100 and that one "hundredth" means 0.01, but maybe not... -- SGBailey (talk) 02:26, 14 April 2014 (UTC)