Read LA 4.3
1) Main Points
4.3 continues with the concept of using vectors to approximate solutions. The book gives a real life example dealing with approximating a constant that works for two specimens. To calculate this approximation we can use vectors like we did in 4.2 to, knowing that the approximate vector will be distanced from the desired vector by r, a residual vector. We know that for the closest approximation the new multiplied vector and the residual vector will be perpendicular meaning their dot product will be equal to zero.
In equations we are dealing with mx + r = y and x dot r = 0. We can solve this by taking the dot product of both sides of mx + r = y, this way we can find m which is the best approximation for a constant of the original problem. This approximation is where lines of best fit come from.
4.3.2 Applies this method to a more complicated situation. if you are given a table of numbers, these as we have learned can be written in vector form. With this more complicated way of thinking about more three dimensional vectors we can conclude that mu + bv + r = s, and u dot r = 0 and v dot r = 0. This can be solved by taking the dot product of u and the equation, then v and the equation. This gives a solvable system of equations which gives the values of m and b needed to make the closest estimate.
This method is called the fundamental problem of linear modeling and is recapped very well in the end of the LA packet.
2) Challenges
I feel pretty good on the two dimensional stuff, but when it gets the vectors with more elements I foresee some difficulty. Overall though I understand the concept I just need to make sure I print the packet and review/apply the concepts
3) Reflections
I now finally understand where a line of best fit comes from, which is very useful in statistical modeling. Right on!
Monday, November 24, 2008
Wednesday, November 19, 2008
Vector projection
Read LA 4.1 - 4.2.
1) Main Points
4.1 continues on our understanding of finding the solution to a linear combination by finding the solution of a vector equation. This chapter acknowledges that sometimes when solving a system of equations there is not a solution, so it introduces the concept of an approximate solution. The approximate solution is best understood by using vectors and the dot product. The dot product(review) looks like this:
(2,3) dot (4,2) = (2*4) + (3*4) = 8 + 12 = 20
The most important thing to remember about the dot product is that:
If the dot product of two vector is 0 then the vectors are Perpendicular (Orthogonal)
Because of this dot product we can use it and the Pythagorean theorem to learn a lot about the relationships of vectors. The length of vector u is equal to the square root of (u dot u), this is derived from the Pythagorean theorem. Furthermore, with our knowledge of the law of cosines we see that the dot product of u and u is equal to (length of u) * (length of u) * Cos(theta). This means that by using the dot product we can find the angle difference of two vectors.
With this knowledge we see that often we have two vectors (ie. u =(2,1) and a=(6,8)) that are not multiples of each other, meaning that we can't multiply the first vector by anything to get the second vector. We want to know in this case, whats the multiple of the first vector that will bring it to be the closest the second. We know that the vectors will be the most similar when the line that passes between the multiplied vector and the desired vector is parallel with the multiplied one. We denote this vector r and it is called the residual vector. According to the dot product u dot r = 0 because they are perpendicular. We want r to be short as possible which means we want it perpendicular. xu + r = a. Because of this we can solve for x to find the multiple that will make the vector closest to the desired one. To do this we set equal:
u dot (xu + r) = a dot r. And knowing that u dot r = 0 because they are perpendicular we solve for x and get in this case x = 4.
2) Challenges
The major challenged I see with this is with the dot product and how to use it algebraically. When solving for x the book distributed u with parenthesis as if it was a multiplication sign, although they are different. I need more clarification on how to use them in equations to avoid mistakes. Other than that it will be important to understand the equations to be able to apply them correctly.
3) Reflections
This seems to be pretty useful considering often there are not solutions to equations, vectors are an easier way to approximate solutions. I hope to see how to apply this vector knowledge to real world stuff.
1) Main Points
4.1 continues on our understanding of finding the solution to a linear combination by finding the solution of a vector equation. This chapter acknowledges that sometimes when solving a system of equations there is not a solution, so it introduces the concept of an approximate solution. The approximate solution is best understood by using vectors and the dot product. The dot product(review) looks like this:
(2,3) dot (4,2) = (2*4) + (3*4) = 8 + 12 = 20
The most important thing to remember about the dot product is that:
If the dot product of two vector is 0 then the vectors are Perpendicular (Orthogonal)
Because of this dot product we can use it and the Pythagorean theorem to learn a lot about the relationships of vectors. The length of vector u is equal to the square root of (u dot u), this is derived from the Pythagorean theorem. Furthermore, with our knowledge of the law of cosines we see that the dot product of u and u is equal to (length of u) * (length of u) * Cos(theta). This means that by using the dot product we can find the angle difference of two vectors.
With this knowledge we see that often we have two vectors (ie. u =(2,1) and a=(6,8)) that are not multiples of each other, meaning that we can't multiply the first vector by anything to get the second vector. We want to know in this case, whats the multiple of the first vector that will bring it to be the closest the second. We know that the vectors will be the most similar when the line that passes between the multiplied vector and the desired vector is parallel with the multiplied one. We denote this vector r and it is called the residual vector. According to the dot product u dot r = 0 because they are perpendicular. We want r to be short as possible which means we want it perpendicular. xu + r = a. Because of this we can solve for x to find the multiple that will make the vector closest to the desired one. To do this we set equal:
u dot (xu + r) = a dot r. And knowing that u dot r = 0 because they are perpendicular we solve for x and get in this case x = 4.
2) Challenges
The major challenged I see with this is with the dot product and how to use it algebraically. When solving for x the book distributed u with parenthesis as if it was a multiplication sign, although they are different. I need more clarification on how to use them in equations to avoid mistakes. Other than that it will be important to understand the equations to be able to apply them correctly.
3) Reflections
This seems to be pretty useful considering often there are not solutions to equations, vectors are an easier way to approximate solutions. I hope to see how to apply this vector knowledge to real world stuff.
Monday, November 17, 2008
Linear combinations, linear independence, span
Read LA 1.3, 2.0 - 2.3, 3.0 - 3.3
1) Main Points
Example 1.3 reminds us that we can combine three vectors (or more) by using the parallelogram or 'tip to tail' method to find the resultant. Also, by using the concept of linear combination we can combine multiples of vectors. Chapter 2 continues with this concept and asks how to find linear a combination that equals a different desired vector. We can do this by using x and y as variables in an equation. If we have the vectors (2,3),(-1,1) and (0,5) and we want the first two together to equal the third we can write it like this:
x(2,3) + y(-1,1) = (0,5)
At this step it is possible to test solutions for x and y but a more efficient way is to solve for an answer like so:
The previous equation shown above is equal to:
(2x,3x) + (-y,y) = (0,5)
Because vectors add across (remember they are usually written vertically) this is equal to:
(2x-y,3x+y) = (0,5)
This gives us a system of equations like so:
2x-y=0
3x+y=5
This can be solved which gives us x=1, y=2 as our solution to the problem above. This means that if we multiply the first vector by 1 and add it to the second vector multiplied by 2 it will be equal to the third.
It is also possible, using algebra to go backwards from a system of equations to three vectors so make sure you know how to do this. A problem with this method however is that sometimes two vectors will be in the same direction and therefor can't be expressed by using a system of equations because the lines may not intersect. In this case they are most easily thought of by thinking of what variables give the desired vector because the linear set up is easier to think about. Creating a system of equations to find a solution is applicable in dimensions beyond 2, however they will produce a system of however many dimensions there are. Lists of vectors can be listed as matrices. This allows us to rewrite vector equations as matrices:
For example:
x(2,3) + y(-1,1) = (0,5)
can be
(2,3, -1,1)(x,y) = (0,5)
(this doesn't come across well because I can't write it vertically, the point is that matrices can consolidate the equation). This means Ax = b, A is the matrix of vectors, x is the variables matrix and b is the goal.
Chapter 3 introduces a 'span" which is a list of all the linear combinations that can be made from the given vectors.
(6,2) is the span of vectors (2,1),(1,1) because 4(2,1) -2(1,1) = (6,2)
A list of vectors is linearly independent if no vector on the list is a linear combination of the other vectors. Otherwise the list is linearly dependent.
2) Challenges
I followed these examples pretty well until the concept of a "span". I don't understand span and I think I need to know more about what a list of linear combinations looks like. Other than that I can see some difficulties in writing vectors with matrices in mind. This to me seems like it could be a little confusing. Also it seems a little difficult to go backwards from a system of equations, this is something I should practice.
3) Reflections
I remember using vectors a lot in physics, they can really account for forces acting in different (or the same) directions. I think linear algebra will seem more applicable when we learn about where to actually use it.
1) Main Points
Example 1.3 reminds us that we can combine three vectors (or more) by using the parallelogram or 'tip to tail' method to find the resultant. Also, by using the concept of linear combination we can combine multiples of vectors. Chapter 2 continues with this concept and asks how to find linear a combination that equals a different desired vector. We can do this by using x and y as variables in an equation. If we have the vectors (2,3),(-1,1) and (0,5) and we want the first two together to equal the third we can write it like this:
x(2,3) + y(-1,1) = (0,5)
At this step it is possible to test solutions for x and y but a more efficient way is to solve for an answer like so:
The previous equation shown above is equal to:
(2x,3x) + (-y,y) = (0,5)
Because vectors add across (remember they are usually written vertically) this is equal to:
(2x-y,3x+y) = (0,5)
This gives us a system of equations like so:
2x-y=0
3x+y=5
This can be solved which gives us x=1, y=2 as our solution to the problem above. This means that if we multiply the first vector by 1 and add it to the second vector multiplied by 2 it will be equal to the third.
It is also possible, using algebra to go backwards from a system of equations to three vectors so make sure you know how to do this. A problem with this method however is that sometimes two vectors will be in the same direction and therefor can't be expressed by using a system of equations because the lines may not intersect. In this case they are most easily thought of by thinking of what variables give the desired vector because the linear set up is easier to think about. Creating a system of equations to find a solution is applicable in dimensions beyond 2, however they will produce a system of however many dimensions there are. Lists of vectors can be listed as matrices. This allows us to rewrite vector equations as matrices:
For example:
x(2,3) + y(-1,1) = (0,5)
can be
(2,3, -1,1)(x,y) = (0,5)
(this doesn't come across well because I can't write it vertically, the point is that matrices can consolidate the equation). This means Ax = b, A is the matrix of vectors, x is the variables matrix and b is the goal.
Chapter 3 introduces a 'span" which is a list of all the linear combinations that can be made from the given vectors.
(6,2) is the span of vectors (2,1),(1,1) because 4(2,1) -2(1,1) = (6,2)
A list of vectors is linearly independent if no vector on the list is a linear combination of the other vectors. Otherwise the list is linearly dependent.
2) Challenges
I followed these examples pretty well until the concept of a "span". I don't understand span and I think I need to know more about what a list of linear combinations looks like. Other than that I can see some difficulties in writing vectors with matrices in mind. This to me seems like it could be a little confusing. Also it seems a little difficult to go backwards from a system of equations, this is something I should practice.
3) Reflections
I remember using vectors a lot in physics, they can really account for forces acting in different (or the same) directions. I think linear algebra will seem more applicable when we learn about where to actually use it.
Monday, November 10, 2008
Catch up / SIR and the spread of disease
10.7
1) main points
Chapter 10.7 works with modeling the spread of disease. The book uses a model called the SIR model which stands for S-susceptible, I-Infected and R-recovered. In differential equation form this comes out to dS/dt= - (Rate susceptibles get sick) = -aSI. We use the constant a, which is proportional to S and I because we assume that if there are more susceptible people around or more infected people around there will be more encounters. It is also negative because we assume that the rate people get sick will eventually decrease. But we must also account for the infected people that are removed from so we add create a differential equation pertaining to I:
dI/dt = (rate susceptibles get sick) - (rate infecteds get sick) = [aSI - bI].
Next we assume that because the recovered are no longer susceptible they increase so we have:
dR/dt = bI
We know that the total population is not changing so S+I+R= Pop so therefore once we know S and I we can calculate for R, so we are only concerned with dS/dt and dR/dt.
We know that a is a measure of how infectious the disease is and with the knowledge that if I=1 and S=762 we see that dS/dt = apprx. -2. With this info we can solve for a and we get a= .0026
When we use a phase plane to plot the information we see how I increases until S= 192 which is the called the threshold population and is the maximum amount of susceptible people to not have an epidemic. In general this number is b/a. This means that the epidemic could have been avoided if all but 192 people were vaccinated
2) Challenges
This model incorporates everything we have learned about differential equations and put it together so a good knowledge of everything is required. My direct understanding of phase planes/slope curves is minimal so sometimes it is hard for me to follow exactly how they come to graph but I do feel like I can adequately interpret them.
3) Reflections
This obviously has amazing real world applications, especially from the preventative standpoint because it can give a relative quota of vaccines to give to a population. I hope to explore this in a bit more detail because it seems like a really useful in regard to disease prevention.
1) main points
Chapter 10.7 works with modeling the spread of disease. The book uses a model called the SIR model which stands for S-susceptible, I-Infected and R-recovered. In differential equation form this comes out to dS/dt= - (Rate susceptibles get sick) = -aSI. We use the constant a, which is proportional to S and I because we assume that if there are more susceptible people around or more infected people around there will be more encounters. It is also negative because we assume that the rate people get sick will eventually decrease. But we must also account for the infected people that are removed from so we add create a differential equation pertaining to I:
dI/dt = (rate susceptibles get sick) - (rate infecteds get sick) = [aSI - bI].
Next we assume that because the recovered are no longer susceptible they increase so we have:
dR/dt = bI
We know that the total population is not changing so S+I+R= Pop so therefore once we know S and I we can calculate for R, so we are only concerned with dS/dt and dR/dt.
We know that a is a measure of how infectious the disease is and with the knowledge that if I=1 and S=762 we see that dS/dt = apprx. -2. With this info we can solve for a and we get a= .0026
When we use a phase plane to plot the information we see how I increases until S= 192 which is the called the threshold population and is the maximum amount of susceptible people to not have an epidemic. In general this number is b/a. This means that the epidemic could have been avoided if all but 192 people were vaccinated
2) Challenges
This model incorporates everything we have learned about differential equations and put it together so a good knowledge of everything is required. My direct understanding of phase planes/slope curves is minimal so sometimes it is hard for me to follow exactly how they come to graph but I do feel like I can adequately interpret them.
3) Reflections
This obviously has amazing real world applications, especially from the preventative standpoint because it can give a relative quota of vaccines to give to a population. I hope to explore this in a bit more detail because it seems like a really useful in regard to disease prevention.
Wednesday, November 5, 2008
11/6: Interacting systems: modeling, phase plane, and trajectories
10.6
1) Main Points
Chapter 10.6 continues exploring differential equations but this time with two interacting systems. The book uses the example of the relationship between robins and worms. Independently the differential equations regarding their populations are dw/dt = aw (positive because with no robins they increase) and dr/dt=-br (negative because with no food the robins decrease). In relation to each other though these equation look like: dw/dt=aw - Effects of robins on worms and dr/dt = -br + Effects of worms on robins. With these considered as variables with constants the equations come out to be (dw/dt = aw - cwr) and (dr/dt = -br + kwr). We look at this first assuming the constants a=b=c=k=1. We want to see both graphs over time but first it is easier to plot (w,r) the relationship between the two populations. This graph is called a phase plane and the point is called a phase trajectory. Using the chain rule we know that (dr/dt = dr/dw * dw/dt) so to find dr/dw we solve and get (dr/dw =(-r+wr)/(w-wr)). When w and r both equal 1 we see that their differential equations = 0 which means this is an equilibrium and the populations don't change. If we graph this on a slope field we see that the derivatives at the different points create a closed curve. This means that if you plug in a certain number of worms and robins the derivative will point in a certain direction that if continued will lead to the same starting values of w and r.
2) Challenges
This chapter is a little difficult for me because I am unfamiliar with slope fields. I understand this concept graphically but when it comes to using differential equations I fee like it will be hard for me to connect the relationships between the two variables. Think of a graph as the two variables can be misleading. I am used to using time as the x axis so it is very important with these problem to consider the variables represented.
3) Reflections
This concept has obvious applications for real life. Populations often affect each other and their populations can change periodically overtime.
1) Main Points
Chapter 10.6 continues exploring differential equations but this time with two interacting systems. The book uses the example of the relationship between robins and worms. Independently the differential equations regarding their populations are dw/dt = aw (positive because with no robins they increase) and dr/dt=-br (negative because with no food the robins decrease). In relation to each other though these equation look like: dw/dt=aw - Effects of robins on worms and dr/dt = -br + Effects of worms on robins. With these considered as variables with constants the equations come out to be (dw/dt = aw - cwr) and (dr/dt = -br + kwr). We look at this first assuming the constants a=b=c=k=1. We want to see both graphs over time but first it is easier to plot (w,r) the relationship between the two populations. This graph is called a phase plane and the point is called a phase trajectory. Using the chain rule we know that (dr/dt = dr/dw * dw/dt) so to find dr/dw we solve and get (dr/dw =(-r+wr)/(w-wr)). When w and r both equal 1 we see that their differential equations = 0 which means this is an equilibrium and the populations don't change. If we graph this on a slope field we see that the derivatives at the different points create a closed curve. This means that if you plug in a certain number of worms and robins the derivative will point in a certain direction that if continued will lead to the same starting values of w and r.
2) Challenges
This chapter is a little difficult for me because I am unfamiliar with slope fields. I understand this concept graphically but when it comes to using differential equations I fee like it will be hard for me to connect the relationships between the two variables. Think of a graph as the two variables can be misleading. I am used to using time as the x axis so it is very important with these problem to consider the variables represented.
3) Reflections
This concept has obvious applications for real life. Populations often affect each other and their populations can change periodically overtime.
Monday, November 3, 2008
Exponential growth and decay / Applications of ODE, equilibria, and stability
10.4 and 10.5
1) Main points
Chapter 10.4 begins to apply our knowledge of differential equations to exponential functions. We know that the solution to the differential equation: dy/dx = y must be a function whose derivative is the same as the function because the derivative = y. We know that the derivative of e^t = e^t so y = e^t must be a possible solution. But we must also account for the constants involved so the family of solutions is y = Ce^kt. It is also important to remember that in the differential equation y can be multiplied by a constant k so: { dy/dx = ky }. This is k is equal to the k in y = Ce^kt because when you take the derivative of y = Ce^kt the constant C is ignored but the constant k matters.
Example:
This means that a differential equation such as dy/dx = .05y has the general solution of y = Ce^kt and k = .05. To find the particular solution you plug in the given constraints, in this example they are y=50 and t=0. So The particular solution of this differential equation would be 50 = Ce^(.05)(0) which gives us C=50 therefore the solution is y=50e^(.05)(t).
The main idea of this chapter is approaching an exponential growth/decay problem from the top down by using what is given about the rate of change to find the actual function.
Chapter 10.5 works off of 10.4 but instead of starting with the differential equation dy/dx = ky it works from the equation dy/dx = k(y-A). We find that the general solution for dy/dt = k(y-A0 is (y = A + Ce^kt). The first example of the chapter introduces a function in which there is an equilibrium solution meaning the function increases or decrease to approach a certain value that creates an equilibrium. This value is the value of the independent variable that makes the differential function equal to zero. If a small change in the initial conditions of a function makes it veer towards the equilibrium solution then it is called a stable equilibrium solution. If it veers away it is unstable.
2) Challenges
When using the dy/dx (Libnitz) notation it is really important to notice which variable (in this case y or x) is represented in the differential equation. For example dy/dx = 2y is different from dy/dx = 2x. The book's examples often jump to the conclusion that from a differential equation such as dP/dt = .02p, C = P0. This comes from the assumption that when t = 0 That the population will be at the initial amount, this information plugged back into the equation gives C = P0. This is an important thing for me to remember. In general it is hard for me to think of these examples backwards. This chapter is centered around using differential equations to find the functions so it is important to distinguish what the different constants mean.
3) Reflections
These concepts allow us to tackle more complicated real life examples. Equilibriums are seen very often in medicine and blood concentration. I think we will find differential equations really useful in analyzing real world things especially because it is often easier to calculate a rate of change than to find an actual function.
1) Main points
Chapter 10.4 begins to apply our knowledge of differential equations to exponential functions. We know that the solution to the differential equation: dy/dx = y must be a function whose derivative is the same as the function because the derivative = y. We know that the derivative of e^t = e^t so y = e^t must be a possible solution. But we must also account for the constants involved so the family of solutions is y = Ce^kt. It is also important to remember that in the differential equation y can be multiplied by a constant k so: { dy/dx = ky }. This is k is equal to the k in y = Ce^kt because when you take the derivative of y = Ce^kt the constant C is ignored but the constant k matters.
Example:
This means that a differential equation such as dy/dx = .05y has the general solution of y = Ce^kt and k = .05. To find the particular solution you plug in the given constraints, in this example they are y=50 and t=0. So The particular solution of this differential equation would be 50 = Ce^(.05)(0) which gives us C=50 therefore the solution is y=50e^(.05)(t).
The main idea of this chapter is approaching an exponential growth/decay problem from the top down by using what is given about the rate of change to find the actual function.
Chapter 10.5 works off of 10.4 but instead of starting with the differential equation dy/dx = ky it works from the equation dy/dx = k(y-A). We find that the general solution for dy/dt = k(y-A0 is (y = A + Ce^kt). The first example of the chapter introduces a function in which there is an equilibrium solution meaning the function increases or decrease to approach a certain value that creates an equilibrium. This value is the value of the independent variable that makes the differential function equal to zero. If a small change in the initial conditions of a function makes it veer towards the equilibrium solution then it is called a stable equilibrium solution. If it veers away it is unstable.
2) Challenges
When using the dy/dx (Libnitz) notation it is really important to notice which variable (in this case y or x) is represented in the differential equation. For example dy/dx = 2y is different from dy/dx = 2x. The book's examples often jump to the conclusion that from a differential equation such as dP/dt = .02p, C = P0. This comes from the assumption that when t = 0 That the population will be at the initial amount, this information plugged back into the equation gives C = P0. This is an important thing for me to remember. In general it is hard for me to think of these examples backwards. This chapter is centered around using differential equations to find the functions so it is important to distinguish what the different constants mean.
3) Reflections
These concepts allow us to tackle more complicated real life examples. Equilibriums are seen very often in medicine and blood concentration. I think we will find differential equations really useful in analyzing real world things especially because it is often easier to calculate a rate of change than to find an actual function.
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