Solve Sequence Problem: Estimate World Population 1799-1900

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To estimate the world population from 1799 to 1900, the equation y(n+1) = 1.03y(n) is used, starting with a population of 0.9 billion in 1798. The solution involves solving the difference equation, resulting in y[n] = 0.9 * (1.03)^n. For the year 1900, n is calculated as 103, leading to an estimated population of 18.9 billion. Participants emphasize the importance of allowing others to find solutions independently to maintain engagement in the problem-solving process. The discussion focuses on the methodology rather than providing direct answers.
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K it's been a while and I can't remember how to figure this problem out without going doing tons of work.

Estimating that the world population was 0.9 billion in 1798, use this equation to estimate the population (in billions) in the years 1799, 1800, 1801, 1802, and 1900.

Using the equation y(n+1) = 1.03y(n)

I got all the years except 1900... I don't remember how to find this out the fast way.. please refresh my memory
 
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notme said:
K it's been a while and I can't remember how to figure this problem out without going doing tons of work.

Estimating that the world population was 0.9 billion in 1798, use this equation to estimate the population (in billions) in the years 1799, 1800, 1801, 1802, and 1900.

Using the equation y(n+1) = 1.03y(n)

I got all the years except 1900... I don't remember how to find this out the fast way.. please refresh my memory

It bottles down to solving the difference equation y(n+1)-1.03y(n)=0
solving it we get y[n] = C1 * (1.03)^n C1 is a constant
use initial condition at n =0 we have y(0) = 0.9 (in billions)
so C1 = 0.9
then the solution is y[n] = 0.9 * (1.03)^n in billions
so year 1900 is n= 1900-1798+1 = 103
Then u get y[103] = 18.9 Billion in the year 1900.
 
real10 said:
It bottles down to solving the difference equation y(n+1)-1.03y(n)=0
solving it we get y[n] = C1 * (1.03)^n C1 is a constant
use initial condition at n =0 we have y(0) = 0.9 (in billions)
so C1 = 0.9
then the solution is y[n] = 0.9 * (1.03)^n in billions
so year 1900 is n= 1900-1798+1 = 103
Then u get y[103] = 18.9 Billion in the year 1900.

Don't give the answer, ok? Let the poster find if for him/her self. It takes the fun out of it for them.
 
Dick said:
Don't give the answer, ok? Let the poster find if for him/her self. It takes the fun out of it for them.

sorry u are right... I have avoided this and if u noticed recently I been giving hints and explanations more than solving it to the end (unless required..)
 
not even correct.
 
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