Using the Chinese population data from 1950 to 1995, let us construct a graph using technology. Before graphing the data though, we must first determine the relevant variables, which are, the year and the population (in millions) of each coinciding year. The parameters are strictly confined to the data for the years 1950 and 1995 in the sense that the data cannot fall below the population number for the year 1950 and cannot fall above the data for the year 1995.
Upon reviewing the graph, we notice that the data appears to increase, but not precisely in a straight line. Going through the trend lines on excel, we find that the trend line that fits the best is the polynomial trend line, which is displayed in the graph down below. If we were to analytically develop one model function to determine if the polynomial trend line is indeed the most accurate fit, I would propose creating a system of equations.
Before jumping to far ahead, we need to make it clear the equation we are going to be analyzing. We will use the equation given to us by the polynomial trend line which is: y= ax2 + bx +c and the reason that we are using this equation is because of the fact that the R2 value is 0.9955. The closer the R2 value is to 1 the better it will fit the graph.
We will rearrange the equation y= ax2 + bx +c so that we can solve for the unknowns which are, the letters a, b and c. To do this, we need to add data to the equation and create three matrices. In order to continue on, let us first add the known values to the equation.
The Essay on Value Of The House Equation 000 Years
In order to create a graph such as the one Ms. Red kin uses to calculate the depreciation of her rental house, first it must be determined which part of the information given is the dependant variable and which is the independent variable. In this case the independent variable is time (in years), and the dependent the value of the house. Next create a graph with the given data, the independent ...
Given y= ax2 + bx +c , we know that the y-values are China’s population in millions and the x-values are the years at which the population is measured. We will use three points to solve this, one from the beginning, one from the middle, and one from the end in order to create the matrices that will then be used to find values for a, b, and c.
So we take the equation y= ax2 + bx +c and plug 1950 (first year) in as the x-value and 554.8 as the y-value and we will do the same with the next two points that will be used.
y= ax2 + bx +c
554.8=a (19502) + b (1950) + c
830.7=a (19702) + b (1970) + c
1120.5=a (19952) + b (1995) + c
Now that we have created our system of equations we can split them up into matrices. Looking at the equations we have come up with, we notice that we have a column of y-values (bold), a column of x-values (blue), and a column of unknowns (purple)
554.8=a (19502) + b (1950) + c
830.7=a (19702) + b (1970) + c
1120.5=a (19952) + b (1995) + c
We can call the y-values matrix ‘B’ or [B] which is going to be a 3×1 matrix, the x-values matrix ‘A’ or [A] which will be a 3×3 matrix, and the unknowns matrix ‘C’ or [C] which is another 3×1 matrix.
[A] = [B] = [C] =
So then we get an equation that looks like this: [B]=[A]*[C]. We can divide [A] from one side of the equation to isolate [C] so, [A]-1*[B]=[C]. We multiply by the inverse of [A]. Multiplying [A]-1*[B] we get the [C] to equal.
Year | 1950| 1955| 1960| 1965| 1970| 1975| 1980| 1985| 1990| 1995| New Population in Millions| 554.8| 620.8| 688.8| 758.7| 830.7|
904.6| 980.6| 1058.6| 1138.5| 1220.5|
Having plugged the original years into the found function we receive the data in the data table above and get a graph that looks like the data is close to the original graph.
We are next shown an equation where the population at time is modeled by: P (t)=
Using a calculator we can run a logistical test on the original data where we get K to equal 1950, L to equal 4.34, and M to equal .0333. We get the data table below.
Year | 1950| 1955| 1960| 1965| 1970| 1975| 1980| 1985| 1990| 1995| New Population in Millions| 537.8| 605.0| 676.5| 751.8| 830.0| 910.0| 991.0| 1071.7| 1151.1| 1228.1|
The Term Paper on Growth Rate Indonesia Population Year
Indonesia is located across the equator and stretch from Sumatra in the west to Iranian Jaya in the east, or from Sabang to Merauke (Dari Sabang Ke Merauke). Its geographic coordinates are 5 00 S, 120 00 E. The total area is 1, 919, 440 sq km. But interestingly only 20% consists of land, the rest is water.The number of islands in the Indonesian archipelago is disputed, but a commonly cited figure ...
The graph created by the logistic function is a very close fit, it almost covers up the original population graph. Even looking at the data table for the function, we see that the data is very close to the original set of data. We know that in polynomial functions, the independent and dependent variables are directly related to each other. In this study the pattern of the original data indicates a continuous gradual incline, so as the years increase so will the population. The logistic function is different in the sense that while it will continue to increase, after awhile it will not increase as much and will begin to slow down until it stops increasing and levels off. The graphs below are to aid in the understanding of the concept.
We are next given a set of data on the population trends in China from the 2008 World Economic Outlook published from the International Monetary Fund (IMF)
Year | 1983| 1992| 1997| 2000| 2003| 2005| 2008|
New Population in Millions| 1030.1| 1171.7| 1236.3| 1267.4| 1292.3| 1307.6| 1327.7|
We will start with the polynomial function and again choose three different points for the above data. So we will get: y= ax2 + bx +c
1030.1=a (19832) + b (1983) + c
1267.4=a (20002) + b (2000) + c
1327.7=a (20082) + b (2008) + c
We can call the y-values matrix ‘B’ or [B] which is going to be a 3×1 matrix, the x-values matrix ‘A’ or [A] which will be a 3×3 matrix, and the unknowns matrix ‘C’ or [C] which is another 3×1 matrix.
[A] = [B] = [C] =
We can divide [A] from one side of the equation to isolate [C] so, [A]-1*[B]=[C]. We multiply by the inverse of [A]. Multiplying [A]-1*[B] we get the [C] to equal .
The Essay on Supply Chain & Logistic Solution
Introduction of Organization The Crystal Group is headquartered in Hong Kong and with the acquisition of the former Martin International becomes one of the largest apparel manufacturers in the world. Our customers are the most important visitors to any of our international clothing operations, – they are not dependent on us, we are dependent on them – they are not an outsider in our ...
Graphing this we get the graph displayed below which we see looks identical to the original to the point where we do not even see the original graph.
Now we run the second function test to see how that one also fits the new data.
Again we have the equation P (t)=
Using a calculator we can run a logistical test on the original data where we get K to equal 1436, L to equal 1.36, and M to equal .0633. We get the data table below.
Year | 1983| 1992| 1997| 2000| 2003| 2005| 2008|
New Population in Millions| 1030.1| 1171.7| 1236.3| 1267.4| 1292.3| 1307.6| 1327.7|
We then take this found data and see how well it fits the original data that we graphed earlier.
As we see the logistic equation fits the original data perfectly. Now let us see how both graphs look together when consolidated into one graph.
As we see here, the data from both the original graphs and their logistic functions line up quite well almost creating one continuous line. The models fit each other with no outliers and creating one solid line on the graph. The same can be seen with the second graph of the original data and the original IMF data and their respective polynomial functions.
In conclusion to this mathematical study, we have found trend lines to graphed data, created system of equations for that data and successfully graphed them against their original graphs. We learned how logistic graphs and polynomial graphs work and how they differ from each other. To create another challenge it would be interesting to try and test more functions. It would be interesting to test parameters and see what would happen if we tested outside of them. It would be a challenge to learn how the logistic function works mathematically to produce the numbers it produces on the calculator.
