# Math Help - Incredibly Hard MALTHUS Problem Help Needed

1. ## Incredibly Hard MALTHUS Problem Help Needed

I am here once again because I have no idea what I am doing. I am forced into taking this math class to graduate and we have this assignment which I can't figure out. I would be forever grateful for any help possible. The question is stated:

Using Malthus's sequence for unchecked population increase, what would the world's population be in 1900 and in 2000? First write the terms of the sequence. The T(zero) term, which is 1, represents the pop. in the year 1800. So under 1, write 1800 and the worlds population in 1800. Under the rest of the terms in the sequence write the appropriate years and the corresponding populations according to Malthus's sequence. Find the terms in the sequence corresponding to the years 1900 and 2000. You should have the unchecked population numbers for those years.

Then it says "If there was sufficient food for all the people in 1800, how many people would this sequence indicate enough food for in 1900 and in 2000? Use the sequence from the above problem and procedure . The T(zero) term which is 1, represents both the population in 1800 and the food population, since we are assuming enough food for the entire world population at that time.

the last part: "Does this sequence indicate enough food for the number you found above for the year 2000 World Population?"

Oh lord....shoot me now.

2. The population would grow according to a geometric series. The exact numbers would depend on the rate of growth (some percent per year, or common ratio).

The resources would grow according to an arithmetic series - linear growth. Again, the exact numbers would depend on the rate of growth (some number per year, or common difference).

The formulas are fairly straight forward. But without knowing the rates of growth, this is an impossible question. Did your text or teacher give you any rates you can use for these?

For the geometric (population) series, the $n^{th}$ term (n=1 fpr 1800, n=101 for the year 1900, and n=201 for the year 2000) is $u_n$. This can be found with the following formula: $u_n = u_1 * r^{n-1}$ where $u_1$ is the first term (1 in this case) and n is the number of the term you want to find. In 1900, for example, the population $u_{101}$ is: $1*r^{100}$ or just $r^{100}$. If the population is growing at a rate of 2% per year, for example, then r=1.02. If the population is growing at a rate of 5% per year, then r=1.05. If the population is doubling each year (growing at a rate of 100% per year), then r=2.

For the arithmatic (resource) series, the $n^{th}$ term (n=1 for 1800, n=101 for the year 1900, and n=201 for the year 2000) is $u_n$. This can be found with the following formula: $u_n = u_1 +d(n-1)$ where $u_1$ is the first term (1 in this case) and n is the number of the term you want to find. In 1900, for example, the population that could be fed is $u_{101}$ is: $1*d(100)$ or just $100d$. If the amount of resources is increasing by 4 per year, then d=4. If the amount of resources is increasing by 6 per year, then d=6. And so on.