Questions and Answers #3 Forms of Linear Equations

Questions and Answers Sheet 3 Forms of Linear Equations Question #1 I am trying to convert part of an equation from its log form into a linear form. Specifically, I am trying to convert 104 log(𝑥) , into 𝑥 4 , but I'm really unsure of how to get from this first stage to the second. My experience with logarithms and exponents is limited, though I believe that 104 log(𝑥) can be re-written as 104 + 10log(𝑥) , but I'm not sure that this helps my plight! Any very basic guidance would be greatly appreciated. Answer: A cool rule about logarithmic functions is that 𝑎 ⋅ log 𝑏 (𝑥) = log 𝑏 (𝑥 𝑎 ) . Fron this, 4 104 log(𝑥) = 10log(𝑥 ) . If we assume that you have a logarithm to the base 10, then 10log10 (𝑥 4) = 𝑥 4. Question #2 I came across the following form of first order linear partial differential equation and I was wondering if there is a name for it? 𝜕𝑔 𝜕𝑓 𝜕𝑥 𝜕𝑦 − 𝜕𝑔 𝜕𝑓 𝜕𝑦 𝜕𝑥 = ℎ(𝑥, 𝑦), where 𝑔(𝑥, 𝑦) as well as ℎ(𝑥, 𝑦) are given and 𝑓(𝑥, 𝑦) is the unknown. Answer: As Cameron Williams said, this is a first-order linear PDE, to which the method of characteristics can be applied. Your equation has a special form thanks to g; in fact, the left hand side of the equation is known as the Poisson bracket of f and g. A consequence of this is that the characteristic curves of the PDE are precisely the level sets 𝑔(𝑥, 𝑦) = 𝑐. Let's see why. Let (𝑥(𝑡), 𝑦(𝑡)) be any solution curve of the ODE 𝑥 ′ (𝑡) = −𝑔𝑦 (𝑥(𝑡), 𝑦(𝑡)) 𝑦 ′ (𝑡) == 𝑔𝑥 (𝑥(𝑡), 𝑦(𝑡)) (I prefer using subscripts for derivatives.) This curve is a level curve of g, because 𝑑 𝑑𝑡 𝑔(𝑥(𝑡), 𝑦(𝑡)) = 𝑥 ′ 𝑔𝑥 + 𝑦 ′ 𝑔𝑦 = −𝑔𝑦 𝑔𝑥 + 𝑔𝑥 𝑔𝑦 = 0 Also, the derivative of unknown function f along this curve is 𝑑 𝑑𝑡 𝑓(𝑥(𝑡), 𝑦(𝑡)) = 𝑥 ′ 𝑓𝑥 + 𝑦 ′ 𝑓𝑦 = −𝑔𝑦 𝑓𝑥 + 𝑔𝑥 𝑓𝑦 = ℎ(𝑥(𝑡), 𝑦(𝑡)) Thus, the problem of solving the PDE splits into parametrizing level curves of g, and then integrating h along those curves. Question #3 I can solve a system of linear equations by adding the constituent equations together or subtracting them from one another (via the so-called "elimination" method), but can one do the same with a system of linear inequalities. In other words, is an operation of the following form valid? 𝑥 > 𝑏, 𝑎 > 𝑏 Therefore 𝑥+𝑎 >𝑦+𝑏 Answer: Yes, it is valid. You can prove it's validness in two steps: First, if 𝑎 < 𝑏 then 𝑎 + 𝑥 < 𝑏 + 𝑥 . Secondly, if 𝑥 < 𝑦 then 0 < 𝑦 − 𝑥 , by adding −𝑥 to both sides. Hence 𝑏 + 𝑥 < 𝑏 + 𝑥 + 𝑦 − 𝑥 = 𝑏 + 𝑦 . Finally, 𝑎 + 𝑥 < 𝑏 + 𝑦 Question #4 I found it "If the row echelon form of a linear equation system has a line with zero entries, then it has more than one solution." Let I 3𝑥1 = 2 II 0𝑥1 = 0 a linear equation system in row echelon form and a line with zero entries. 2 If we solve this equation system, the solution ℒ would be ℒ = {𝑥1 = } . Contradiction. The statement is 3 wrong. This looks way too easy. Is there anything I might forgot about? Answer: Let's start from the linear system 𝐴𝑥 = 𝑏. You can form the matrix [𝐴|𝑏] and perform row reduction to get, say, the row echelon form [𝑈|𝑐]. The system has at least a solution if and only if c is not a pivot column. In this case, the system has a single solution if and only if every column in U is a pivot column independently on the number of zero rows in [𝑈|𝑐]. If the system has more unknowns than equations, then, as soon as it has a solution it will have infinitely many of them, because it's impossible that every column in U is a pivot column. The case when the system has a single solution and there are zero rows in [𝑈|𝑐] is possible when the system has more equations than unknowns. The example you present is indeed one of these. The statement “if the row echelon form has a zero column then the system has no solution” is clearly false. If the system has a solution at all, it will have infinitely many of them, because the unknown corresponding to the zero column can get whatever value you want. Question #5 I had one doubt in matrix form of linear equations Say , we have a system of equations 𝐴𝑥 = 𝑏 such that A is an n\times n matrix and b is a 𝑛 × 1 matrix and so is x then, if we are told that rank(𝐴) = 𝑛 , then, do we need to check that rank(𝐴|𝑏) = 𝑛 or can we say that since b is a linear combination of the component vectors in A then augmenting it in A won't increase the number of linearly independent column vectors and hence the rank(𝐴|𝑏) = 𝑛 and can't ever be 𝑛 + 1 . Answer: I am not sure that I understand you when you write “do we need to check”. Assuming that rank(𝐴) = 𝑛 , then rank(𝐴|𝑏) = 𝑛 if and only if the system 𝐴𝑥 = 𝑏 has a solution. Does this answer your question? Question #6 I have a linear problem with double derivate of space and time, which has Dirichlet boundary condition in (1)2 and Neumann's boundary condition in (1)3 : 𝛿 2𝑢 − 𝑐 2 Δ𝑢 = 𝑓 in Ω 𝛿𝑡 2 𝑢 = 0 on 𝛿Ω or ∇𝑢 ⋅ 𝑛 = 0 on 𝛿 with conditions 𝑢(0, 𝑥) = 𝑢0 (𝑥) in Ω 𝛿𝑢 (0, 𝑥) = 𝑣0 (𝑥) in Ω 𝛿𝑡 Which is the weak form of the problem? My attempt: 𝛿2𝑢 ∫ ( 𝛿𝑡 2 − 𝑐 2 Δ𝑢) ⋅ 𝑣𝑑𝑥 = ∫ 𝑓𝑣 ⋅ 𝑑𝑥 , for ∀𝑣 ∈ 𝑉𝑛 . Handling two different parts separately 𝛿2𝑢 ∫Ω 𝛿𝑡 2 𝑣 𝑑𝑉 + 𝑐 2 [∫Ω ∇𝑢 ⋅ ∇𝑣𝑑𝑉 − ∫𝛿Ω ∇𝑢 ⋅ 𝑛̂ ⋅ 𝑣𝑑𝐴] = ∫ 𝑓𝑣 ⋅ 𝑑𝑥 Answer: I finally got the result which should be right one. By integrating and multiplying Equation (1) with v ∫ 𝑣 𝑢̈ 𝑑𝑥 + 𝑐 2 𝑣Δ𝑢 = ∫ 𝑓𝑣 𝑑𝑥 using Green I we get ∫ 𝑣 𝑢̈ 𝑑𝑥 − ∫𝑐 2 𝑣 𝜕 𝜕𝑢 𝑑𝑆 + ∫ ∇𝑐 2 𝑣 ⋅ ∇𝑢𝑑𝑥 = ∫ 𝑓𝑣 𝑑𝑥. 𝜕𝑛 So the weak form is (𝑣, 𝑢̈ ) + 𝑐 2 (∇𝑣, ∇𝑢) 𝜕𝑢 −𝑐 2 (𝑣, ) = (𝑓, 𝑣). 𝜕𝑛 𝜕Ω Question #7 I have a representation of a linear equation in standard form (𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0) which I am representing as a set of coefficients: a,b,c. I want to normalize these so that any two equations that represent the same line can be compared programatically to see if they're equal just using the co-efficients. For this I need the co-efficients of any two equations representing the same line to be equivalent after normalizing. I normalized the values by dividing all coefficients by √𝑎2 + 𝑏 2 + 𝑐 2 which scales the coefficients to be the same but I don't know how to account for sign changes in this. For example, if I have equations: 2𝑥 ± 4𝑦 + 2 = 0 (represented as 2, -4, 2) and −2𝑥 + 4𝑦 ± 2 = 0 (represented as -2, 4, -2), how can I transform the coefficients to make them equal? Answer: This is the same problem as standardizing projective homogeneous coordinates. You have the right first step, namely, dividing all coordinates by the magnitude. Now you have the problem that a set of coordinates and its negative are equivalent. If the last non zero coordinate is negative, then negate all the coordinates. This gives standardized coordinates. Question #8 I have a set of linear equations, of the form 𝐴𝑥 = 𝑏 . However, the value of b depends on the sign of the solution of x, such that 𝑏𝑖 = 𝑐𝑖 + 𝑑𝑖 if 𝑥𝑖 > 0 , and 𝑏𝑖 = 𝑐𝑖 − 𝑑𝑖 if 𝑥𝑖 < 0 . I could solve this by trying each combination of positiveness of x, but that would require solving 2𝑁 sets of linear equations, which becomes unfeasible quite quickly. Is this solvable analytically? Answer: You can write the system as 𝐴𝑥 = 𝑐 + 𝐸𝑑 where E is a diagonal matrix whose diagonal elements are 𝑒𝑖 = ±1 . This solves to 𝑥 = 𝐴−1 𝑐 + 𝐴−1 𝐸𝑑 . You can solve this while leaving the 𝑒𝑖 as unknowns. Then you have the 𝑥𝑖 expressed as linear combinations of the 𝑒𝑖 , and can quickly run through the various combinations to see which ones have sign(𝑥𝑖 ) = 𝑒𝑖 for all i . Question #9 I have a system of linear equations of the form ∑𝑛𝑗=1 2 𝑥𝑗 𝑘𝑖𝑗 ∈ 𝑍, 𝑖 = 1, … , 𝑚 or equivalently 2𝐾 ⋅ 𝑥⃗ ∈ 𝑍 𝑛 where K is an integer matrix with orthogonal rows and columns, such that the GCD of every row and every column is 1, and 𝑥⃗ is a vector of real numbers. what are the constraints on 𝑥⃗ ? Answer: 1 1 The matrix 𝐾 = ((1 1), (1 1)) satisfies your conditions, but has the solution (𝑥1 , 𝑥2 ) = ( , ). 4 4 Question #10 I have a very simple question: suppose I have two 2D linear equations in general form 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 I'd like to know what's the (intuitive) geometric interpretation of their addition and subtraction (𝑎1 + 𝑎2 )𝑥 + (𝑏1 + 𝑏2 )𝑦 + (𝑐1 + 𝑐2 ) = 0 (𝑎1 − 𝑎2 )𝑥 + (𝑏1 − 𝑏2 )𝑦 + (𝑐1 − 𝑐2 ) = 0 Answer: Consider the lines given by the following equations: 1𝑥1 𝑦 − 1 = 0 (𝐴) 2𝑥 + 3𝑦 − 4 = 0 (𝐵) Their sum (in your sense) has the equation 3𝑥 + 4𝑦 − 5 = 0 (𝐴 + 𝐵) However, that first line is also given by this equation (multiplying-through by -2 doesn't change the solution set): −2𝑥 − 2𝑦 + 2 = 0 (𝐴′ ) but now the sum with (B) is 0𝑥 + 1𝑦 − 2 = 0 (𝐴′ + 𝐵) which represents an entirely different line than the previous sum. Unfortunately, your notion of "addition" for lines is (as we say) "not well-defined", because a line does not have a unique equation that describes it. To salvage the idea, you need to try to find a unique representative equation for a given line. You might consider requiring that the x coefficient must be 1; this would allow us to reject (A') from our 3 addition process, but it would mean re-writing (B) as 1𝑥 + 𝑦 − 2 = 0 prior to the addition process. So, 2 the sum of the lines represented by (A) and (B) is the line represented by 5 2𝑥 + 𝑦 − 3 = 0 2 which you would re-write as 5 3 4 2 1𝑥 + 𝑦 − = 0 if you wanted to add it to another line. That seems to work fine, except ... There's no such representative equation for a horizontal line (since all representative equations have x coefficient 0). Subtraction will always result in a horizontal line by killing the x coefficient! Hmmm ... So, you can't require the x-coefficient to be 1 (or any particular number), because sometimes there isn't an x coefficient; the same goes for the y coefficient ... and also for the constant term. Hmmm ... Question #11 I have been learning that the standard expression of a linear equation in two variables is of the form: 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 while 𝑎 ≠ 0 and 𝑏 ≠ 0 . I want to understand the purpose of the constant 'c' in this equation and where it was derived from, whether it be graphically or algebraically. Answer: Let's start with the fact that we want linear equations in two variables to represent a line. Suppose we didn't have a constant c. If we considered all equations of the form 𝑎𝑥 + 𝑏𝑦 = 0 , we would 𝑎 only get lines through the origin, as follows: we could solve our equation to get 𝑦 = − 𝑥, which always 𝑏 passes through (0,0). Adding the constant c allows us to consider all lines in the plane. Specifically the equation 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 𝑐 𝑐 𝑏 𝑎 0 passes through the points (0, − ) and (− , 0) Question #12 I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of 𝑦 ′ + 𝑃(𝑥)𝑦 = 𝑄(𝑥) Here is the equation: (2𝑥𝑦 + 𝑥 2 + 𝑥 4 )𝑑𝑥 − (1 + 𝑥 2 )𝑑𝑦 = 0 It is not exact since partial derivatives are not equal. Any help would be appreciated. Answer: To put it into the form you requested: −(1 + x 2 )dy + (2 xy + x 2 + x 4 )dx = 0   dy 2 xy + x 2 + x 4 − =0 dx 1 + x2  dy 2x x2 + x4 + (− ) y = dx 1 + x2 1 + x2  dy 2x + (− ) y = x2 2 dx 1+ x Question #13 I have some question to ask you that explain to me and remove my confusion. 1.The book said that we call for those equations linear that have first degree power or general forms of linear equations now my question is that we call for those kind of equations linear that have general form of linear equation or can change to general form or other forms of linear equations form example can we 10 call = 5 linear or square root of (𝑦 − 9) = 10 linear? 𝑥 2.Can those rational and radical equations that can be manipulated to linear forms called linear equations like the examples above? 3.Are all first degree one variable equations linear? Answer: To define linear equations, we need to understand linear functions of a variable. Let us assume a function f:R\rightarrowR , such that 𝑓(𝑎𝑥1 + 𝑏𝑥2 ) = 𝑎𝑓(𝑥1 ) + 𝑏𝑓(𝑥2 ) For all 𝑥1 , 𝑥2 , 𝑎, 𝑏 ∈ 𝑅 Any function with this property is a linear function. Now, we can define a linear equation to be 𝑓(𝑥) = 0 where f is a linear function in x. Now, when you look at functions like 𝑓(𝑥) = 10 𝑥 −5 This is clearly not a linear function, and hence the equation 𝑓(𝑥) = 0 per se is not a linear equation, but with a bit of manipulation, we can define a 𝑔(𝑥) = 10 − 5𝑥 , such that the following is true if x solves g(x) = 0  x solves f(x) = 0 And hence the equations that you have mentioned though are not linear, they are analogous to a linear equation, and hence they share the same solution set. This gives us a tool to solve a non-linear equation using a corresponding transformation to get a linear equation, which is easier to solve Question #14 I have the equation (1 + 𝑦 2 )𝑑𝑥 + (𝑥𝑦 + 1)𝑑𝑦 = 0 . It is a linear differential equation of the first order. Now the problem is that this doesnt have the regular form of this type of equations which is 𝑑𝑦 𝑑𝑥 + 𝑝(𝑥)𝑦 = 𝑞(𝑥) . I know I can divide both sides by dx and then I'll have (1 + 𝑦 2 ) + (𝑥𝑦 + 1)𝑦 ′ = 0 . Still, it should be (𝑥𝑦 + 1) + 𝑦 ′ not (𝑥𝑦 + 1)𝑦 ′ . Answer: If you set 𝑀(𝑥, 𝑦) = (1 + 𝑦 2 ), 𝑁(𝑥, 𝑦) = 𝑥𝑦 + 1 , you see that 𝑀𝑦 = 2𝑦, 𝑁𝑥 = 𝑦 and 𝑀𝑦 −𝑁𝑥 −𝑀 = −𝑦 1+𝑦 2 is a function respect to y. Now think of an integrating factor 𝜇(𝑦) and then make your OE exact. Question #15 I know how to use methods like system of linear equations to show that any vector in 𝑅𝑛 can be expressed in the form of unique linear combinations of a given set of n linearly independent vectors, but how to expand this result to that any given set of exactly n linearly independent vectors must form a basis of 𝑅𝑛 space? How to understand it conceptually? I saw there are similar questions, but the answers in those posts do not really convince me. Thanks in advance. Answer: I'm not sure if this answers your question but in order for a set of vectors to be a basis of a vector space it has to be linear independant and has to span the entire space. Now if you're given a set of n linearly independant vectors in 𝑅𝑛 you only have to show that it spans 𝑅𝑛 . But as you said yourself this is indeed the case since you can express any vector as a linear combination of vectors from your set. If that isn't satisfactory, you could also think about it this way: Suppose you have n linearly independant vectors (𝑣1 , … , 𝑣𝑛 ) and another vector w not in the span of your (𝑣1 , … , 𝑣𝑛 ) . Then (𝑣1 , … , 𝑣𝑛 𝑤) is linear dependant since it's a set of 𝑛 + 1 vectors in a vector space of dimension n. This means there are 𝜇, 𝜆𝑖 ∈ 𝑅, 𝑖 = 1, … , 𝑛 not all 0 with 0 = 𝜇𝑤 + ∑𝑛𝑖=1 𝜆𝑖 𝑣𝑖 . Now 𝜇 ≠ 0 because otherwhise (𝑣1 , … , 𝑣𝑛 ) would be 1 linear dependant. Therefore 𝑤 = − ∑𝑛𝑖=1 𝜆𝑖 𝑣𝑖 but this means w is in the span of (𝑣1 , … , 𝑣𝑛 ) a 𝜇 contradiction. Question #16 I tried researching and found that I can use a system of linear equations and solve by an inverse matrix to find the cubic equation given 4 points which satisfy the function f(x) of the general form 𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 I can also find a cubic of the form 𝑎𝑥 3 + 𝑑 with no 𝑥 2 or x term from 2 points, however I was wondering how one would go about finding a full general form cubic given only the minimum and maximum. Example minimums could be (-1,4) and (2,3) Answer: Well, since the extrema are the roots of the derivate, it would be nice to have the minimum and the maximum (and an initial condition too.) Question #17 I was stuck at numbers 2 and 6. This is a linear ODE. 2. (𝑦 + 1)𝑑𝑥 + (4𝑥 − 𝑦)𝑑𝑦 = 0 Divided it by dx : (𝑦+1)+(4𝑥−𝑦)𝑑𝑦 𝑑𝑥 =0 Transposed the (𝑦 + 1) (4𝑥−𝑦)𝑑𝑦 𝑑𝑥 = −(𝑦 + 1) Divided it by (4𝑥 − 𝑦) 𝑑𝑦 = 2𝑦(𝑥 − 3) 𝑑𝑥 The farthest I can go is: 𝑑𝑥 =( 𝑑𝑦 2𝑥 3𝑥 2 )−( 6𝑦 3𝑥 3 ). The rest of the problems were easily solved but those 2 are very hard to bring to the standard form. Answer: for Nr. 2 is not linear, since it contain 𝑦 ∗ 𝑦 ′ so you can not bring it to standard form but with 𝑑𝑦 𝑑𝑥 you have 𝑑𝑥 𝑑𝑦 + 4 𝑦+1 ∗𝑥 = 𝑦 𝑦+1 Question #18 I was thinking of functions who have differential equations in which the function is 'factorable' from first order differential equation. Eg: Functions satisfying the equation: 𝑑𝑦 = 𝜂(𝑥)𝑦(𝑥) 𝑑𝑥 Example functions: 𝑦 = 𝑒 𝑎𝑥 , 𝑦 = 1 1−𝑥 .. etc Now, I notice that the solution to the differential equation is: 𝑦 = 𝑒 ∫ 𝜂𝑑𝑥 But here is the interesting thing, the above expression has the same form of the function which is used when we solve linear differential equation Answer: 𝑑𝑦 = 𝜂(𝑥)𝑦(𝑥) 𝑑𝑥 First, change of symbol x for t: 𝑑𝑦 = 𝜂(𝑡)𝑦(𝑡) 𝑑𝑡 𝑑𝑦 𝑑𝑡 − 𝜂(𝑡)𝑦(𝑡) = 0 Second, let −𝜂(𝑡) = 𝑝(𝑡) 𝑑𝑦 𝑑𝑡 + 𝑝(𝑡)𝑦(𝑡) = 0 This is the equation from your cited document in the particular case 𝑔(𝑡) = 0: 𝑑𝑦 𝑑𝑡 + 𝑝(𝑡)𝑦(𝑡) = 𝑔(𝑡) So it is not surprizing that you observe the relationship of solutions. Question #19 If a system of linear equations can be represented by the compact form, 𝐴𝑥 = 𝑏 how will a system of nonlinear equations (system of quadratic equations and system of exponential equations) be represented in compact form? Answer: This is way too general. Perhaps something can be said about quadratic equations: you can write them as 𝑥 ⋆ 𝐴𝑥 + 𝑏 ⋆ 𝑥 + 𝑐 = 0,where x is the vector of unknowns, b is a vector and c is a scalar. But a system of such things... That would require 3-dimensional matrices, and it would not be nice. If you think about the degree of freedom: you need two dimensions for all variations of the form 𝑥𝑖 , 𝑥𝑗 , and a third one for the different equations. As for exponential equations, I am not even sure what you mean, and what the degree of freedom would be. I do not think there is a compact form. Question #20 If I get the slope of a line, and one point that is on it, then, are they define exactly ONE line? The point–slope form of linear equations (𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )) need only the slope and the coordinates of that given point. And this equation describes exactly one line. Is it enough? Answer: Yes, that is enough. You can determine unambiguously whether a point is on the line from the equation. There is a problem with vertical lines in this form, however. You just need to use 𝑥 = 𝑥1 for that.