What's the class title? Unless it's an introductory course that just ultra-skims the very surface of things, it would be weird to cover kinematics and some non-kinematics stuff in just 5 weeks.
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The Math Help Thread
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Physics 1. University level. The kind that requires calculus. Maybe I described it wrong. We're talking about vectors and 1-D and 2-D motion so far. That is easy at the moment. But my guess is that things will become more complicated quickly.
If it follows the usual path, it'll be pure kinematics, with the next class being electricity and magnetism.
Sort of like how in a quarters system Calc 1 is differential and Calc 2 is integral.
triplestation
Member
I'm gonna take calc physics as well
So scared 😹
So scared 😹
Chance Hale
Member
I currently have it set up where I'm talking Calc 1 and Kinematics at the same time this fall(along with Spanish 2 and maybe Chem 2)
Is that going to be a problem or is precalculus enough for the first part of Physics 1?
Is that going to be a problem or is precalculus enough for the first part of Physics 1?
triplestation
Member
No derivatives or integrals in Precalc
Maybe you don't even have to, I hear it's lots of plug and chug pre-derived formulas
Two words are you deriving and integrating your kinematics equations yourself in ur class or just plugging chugging?
Maybe you don't even have to, I hear it's lots of plug and chug pre-derived formulas
Two words are you deriving and integrating your kinematics equations yourself in ur class or just plugging chugging?
We've done a bit of derivatives stuff, but it's just simple power rules. Like, we may have a displacement equation for the x and y component, and we have to differentiate each equation to get the velocity equations for x and y.
Looks good. Two things I'd add are that, until we specify what alpha is, keep it general (treat it as complex) and use Re(alpha).
For the conclusion, I'd mention that x and y are linearly dependent because alpha is not zero (we eliminated the case in which the vector x or y is zero). Gamma can be zero even if x and y are nonzero vectors, so if our alpha also happens to be zero, then we cannot say that x and y are linearly dependent.
Yeah I mentioned most of that (except for alpha !=0, I'll add that in) just didn't include it in the pic because it's not in english. Thanks again man
I am on a waitlist for an online college algebra course after finishing intermediate algebra. If I miss this course, the next available session does not start until the end of September. Does anyone have resources or tips to keep up on the material? The description for college algebra:
I'd like to be in good shape to start the class.
Math-GAF, I'm having trouble understanding a specific Statistics concept. Here's a question from my quiz to provide an example:
Okay, so actually I understand how to get the confidence interval--since this is a beginner's Stats class, we use the graphing calculator to do all that.
Thing is... how do I find the standard deviation for this problem? My Google-fu hasn't yielded many results, since it just tells me "well enter all the data points into the calculator like this", but I... don't have data points, so I'm really lost. I need the standard deviation to calculate the confidence interval. This isn't the first time I've gotten caught on a question similar to this, and I'm getting frustrated.
Any help is greatly appreciated.
Okay, so actually I understand how to get the confidence interval--since this is a beginner's Stats class, we use the graphing calculator to do all that.
Thing is... how do I find the standard deviation for this problem? My Google-fu hasn't yielded many results, since it just tells me "well enter all the data points into the calculator like this", but I... don't have data points, so I'm really lost. I need the standard deviation to calculate the confidence interval. This isn't the first time I've gotten caught on a question similar to this, and I'm getting frustrated.
Any help is greatly appreciated.
I've tried and tried again, looked over notes but can't seem to find a good explanation on what I'm doing wrong with simplifying rational expressions.
I have the idea of what factoring is and what to do I thought but I can't seem to find the right setup that allows me to cancel out enough common factors to get me to the right end.
Can someone tell me which one is the correct factoring step to do or tell me where I'm going wrong? I have this section and multiplying/dividing to finish before a quiz tomorrow.
Example:
I have the idea of what factoring is and what to do I thought but I can't seem to find the right setup that allows me to cancel out enough common factors to get me to the right end.
Can someone tell me which one is the correct factoring step to do or tell me where I'm going wrong? I have this section and multiplying/dividing to finish before a quiz tomorrow.
Example:
there are a couple of things wrong, but you have some of the basics right.
first, you need to be aware that when factorizing, you need two numbers that add up to -27 and multiply to 10. you also know that since 5 is prime, ie it can only be split up into 1 and 5. so the answer is in the form (5x + a)(x + b) where a and b are numbers you need to determine that form -27x and 10 respectively.
now for a and b, you know that a*b = 10. So they must be some factors of 10, ie 1 and 10 or 2 and 5. however, you also know that when they're added together, you get negative 27x or in laymans terms, 5x*b + a*x = -27x. since it's negative, that means a and b must be both negative to create a positive 10.
I'm not entirely sure where you've pulled some of the factorisations from, so it's always important to check you get the same thing when you expand, so that (5x - 25)(x - 2) =/ 5x^2 - 27x + 10.
you got the right answer however!
Stumpokapow
listen to the mad man
I've tried and tried again, looked over notes but can't seem to find a good explanation on what I'm doing wrong with simplifying rational expressions.
I have the idea of what factoring is and what to do I thought but I can't seem to find the right setup that allows me to cancel out enough common factors to get me to the right end.
Can someone tell me which one is the correct factoring step to do or tell me where I'm going wrong? I have this section and multiplying/dividing to finish before a quiz tomorrow.
Example:
Before proceeding, try to multiply out the factors you came up with to see if you can recover the original terms:
5x(x-25)(x-2) gives you 5x^3 - 135x^2 + 250x
5x(x+5)(x-2) gives you 5x^3 + 15x^2 - 50x
(5x-25)(x-2) gives you 5x^2 - 27x + 50
(5x+5)(x-2) gives you 5x^2 - 5x - 10
None of these are your original numerators or denominators, so we can be confident that your answers are not correct. Whenever you do any kind of math ever, you should check your answers. Checking your answer will show you you're incorrect.
I am assuming based on the way you are asking the question, you have not yet learned the quadratic formula. That's fine. We can approach factoring naively.
Lets's start with your denominator:
5x^2 + 3x - 2
This factors to an answer of the form:
(5x + A)(x + B)
We don't know what A and B are yet. We know that A*B = -2.
It is clear that A and B are one of the following combinations:
1,-2
2,-1
-1,2
-2,1
Why? The factors of 2 are 1 and 2, exactly one of them has to negative for the resulting multiplication to be negative. The order doesn't matter at this stage.
We know that Ax + 5Bx = 3x (or simply A + 5B = 3)
Let's try A=1, B=-2
1 + 5(-2) = -9
Let's try A=2, B=-1
2 + 5(-1) = -3
Let's try A=-1, B=2
-1 + 5(2) = 9
Let's try A=-2, B=1
-2 + 5(1) = 3
Found it.
(5x - 2)(x + 1)
Okay, now your numerator:
(5x + A)(x + B)
A*B = 10
A, B are (factors of 10, either both positive or both negative because the resulting multiplication gives you a positive number):
1, 10
2, 5
5, 2
10, 1
-1, -10
-2, -5
-5, -2
-10, -1
Ax + 5Bx = -27X
A + 5B = -27
Try each (we can be sure immediately that it's not any of the both positive ones because the result of the sum has to be negative:
-1 + 5(-10) = -51
-2 + 5(-5) = -27
Ding ding ding. Got it.
Answer: (5x - 2)(x - 5)
Put it all together:
(5x-2)(x-5) / (5x-2)(x-1) = (x-5)/(x-1)
I didn't see the question until today, sorry if my answer is too late. The theorem addresses the question of when does the inverse of a map exist, and is simple to state and understand in 1D with real numbers.
We can generalize the statement to multi-dimensional spaces using partial derivatives, and to infinite-dimensional spaces using the so-called Frechet derivatives.
Inverse Function Theorem
Let y = y(x) be a map from [a, b] to [c, d]. If the map is C^k(a, b) for some k >= 1, and the Jacobian J(x) := (dy/dx)(x) is not zero for all x in (a, b), then the inverse map x = x
exists and is C^k(c, d).In case you are not familiar with the notations, "y = y(x)" means that y is a function of x, and similarly, "x = x
" means that x is a function of y.The C^k(a, b) means that the map y(x) can be differentiated k times in the open interval (a, b), and that the derivatives up to the k-th order are continuous in this interval.
In other words, if the Jacobian is not zero at a point x and for all points in some neighborhood (a, b) about x, then we know that the inverse map is defined for the point y(x) and all the points in some neighborhood (c, d) about y(x).
Moreover, the inverse map is as smooth as the original map in their respective neighborhoods.
Okay GAF, so I want to use Khan Academy to review/fill in gaps in Calculus and some Trigonometry this Summer but I'm getting it set up and I'm wonder about certain things. So far I'm doing practice problems so that the site can determine my mastery of math. Here's the thing though; I know what I need to learn, I want to skip this so I can just start where I know I need to. Is there any way to bypass this so I can just move to the lessons I want?
Also, I'm worried about how it checks your answers. I've worked with online math programs before and I find it absolutely infuriating when they give you a multi-step problem, make you solve it on your own, and don't let you move on until you get it correctly, because usually I end up staring at my computer screen for way too long to find an arithmetic error, or worse, an error in how the site wants to take the answer in. Here's an example of this from a problem I just did:
Okay, so my answer is apparently incorrect. It's a relatively simple problem in terms of material and I probably either made some simplification/arithmetic error (I haven't checked) or I'm just not giving the answer the site wants. I don't really give a shit. What I'm wonder is if these kinds of situations are common when working with Khan Academy? Because I really can't control my temper when I come across them.
So basically I'm wondering if Khan Academy respects my time or not. Because I just want to use it to review some specific material which shouldn't take too long in theory.
Also, I'm worried about how it checks your answers. I've worked with online math programs before and I find it absolutely infuriating when they give you a multi-step problem, make you solve it on your own, and don't let you move on until you get it correctly, because usually I end up staring at my computer screen for way too long to find an arithmetic error, or worse, an error in how the site wants to take the answer in. Here's an example of this from a problem I just did:
Okay, so my answer is apparently incorrect. It's a relatively simple problem in terms of material and I probably either made some simplification/arithmetic error (I haven't checked) or I'm just not giving the answer the site wants. I don't really give a shit. What I'm wonder is if these kinds of situations are common when working with Khan Academy? Because I really can't control my temper when I come across them.
So basically I'm wondering if Khan Academy respects my time or not. Because I just want to use it to review some specific material which shouldn't take too long in theory.
I've never come across problems with the way Khan Academy computes my answers, really. It's always been my own error, and then I check the hints only to find where I've screwed up. It's extremely rare when you're right and it doesn't accept your answer. Been using that place for about 4 years now without issue.
dr3upmushroom
Banned
We're getting into higher derivatives in my Differential Calculus corse. I'm still having a bit of trouble wrapping my head around the concept of doing derivatives with roots of expressions.
for example h(x)=√(x^2+1)
for example h(x)=√(x^2+1)
Stumpokapow
listen to the mad man
Chain rule:
https://www.khanacademy.org/math/di...ivatives/chain_rule/v/chain-rule-introduction
Let h(x) = f(g(x))
Thus in this case g(x) = x^2 + 1 and f(x) = sqrt(x)
So h(x) = sqrt(x^2 + 1)
By the chain rule:
h'(x) = f'(g(x)) g'(x)
To calculate this:
g'(x) = 2x
f'(g(x)) = 1/2sqrt(g(x)) = 1/2sqrt(x^2 + 1)
So:
h'(x) = 2x/2sqrt(x^2 + 1)
h'(x) = x/sqrt(x^2 + 1)
You have to use the chain rule here.
h(f(x))' = h'(f(x)) * f'(x)
In this instance, h(x) = √(x) while f(x) = x^2+1. You basically have to recognize when a given function is a composite (i.e. one function is inside another).
Rewrite h(f(x)) as (x^2+1)^0.5.
Applying the chain rule, we get 0.5*(x^2+1)^-0.5 * 2x.
I didn't see the question until today, sorry if my answer is too late. The theorem addresses the question of when does the inverse of a map exist, and is simple to state and understand in 1D with real numbers.
We can generalize the statement to multi-dimensional spaces using partial derivatives, and to infinite-dimensional spaces using the so-called Frechet derivatives.
Inverse Function Theorem
Let y = y(x) be a map from [a, b] to [c, d]. If the map is C^k(a, b) for some k >= 1, and the Jacobian J(x) := (dy/dx)(x) is not zero for all x in (a, b), then the inverse map x = x
In case you are not familiar with the notations, "y = y(x)" means that y is a function of x, and similarly, "x = x
The C^k(a, b) means that the map y(x) can be differentiated k times in the open interval (a, b), and that the derivatives up to the k-th order are continuous in this interval.
In other words, if the Jacobian is not zero at a point x and for all points in some neighborhood (a, b) about x, then we know that the inverse map is defined for the point y(x) and all the points in some neighborhood (c, d) about y(x).
Moreover, the inverse map is as smooth as the original map in their respective neighborhoods.
Thanks for the detailed explanation. I kind of understand the gist of it, but parts are beyond my understanding.
I'm not quite sure what the jacobian is (change of variables for triple integrals?), nor am I familiar with the term map (I gather it's like an "image" for matrix equations?).
dr3upmushroom
Banned
Thanks, I can get to that point. Sorry I'm working on hw as I typed out that last post, and didn't write everything. I need to get the second derivative and that's where I get confused.
Stumpokapow
listen to the mad man
Please review basic rules of differential calculus and do practice questions to learn how to identify where you can use them. You have tools in your toolbox: power rule, chain rule, quotient rule, identities for trigonometric derivatives.
In this case, the second derivative of your original function is the first derivative of your first derivative, x/sqrt(x^2 + 1). You can get this using the quotient rule and the chain rule. You use the quotient rule when you are taking the derivative of a function that has numerator and denominator functions.
Quotient rule:
Derivative of f(x)/g(x) = f'(x)g(x) - g'(x)f(x) / g(x)^2
f(x) = x
f'(x) = 1
g(x) = sqrt(x^2 + 1) <-- the denominator is also your original function
g'(x) is the first derivative of your original function: x/sqrt(x^2 + 1) <-- you can also get this using the chain rule
Derivative using Quotient Rule:
1sqrt(x^2 + 1) - (x^2/sqrt(x^2 + 1)) / (x^2 + 1)
Use algebra to simplify:
sqrt(x^2 + 1) / (x^2 + 1)^2
My bad, a map and a function mean the same thing. They are a rule that assigns an element x from one set to an element y from another set. We say that y is the image of x under that rule.
(If you have a matrix equation Ax = y, then the vector y is the image of the vector x. We see that matrices are a map, but they are also a special one, in that the map is linear.)
Below, I'll use the word function for clarity.
I gave the expression (dy/dx)(x) a name so that the theorem will sound more intimate and be easier to remember. The expression in 1D happens to be, in general, the Jacobian of the function.
The Jacobian has a nice geometric interpretation and appears whenever we do a change of variable, but let's just go back to your original problem to see how the inverse function theorem can be used.
----
Consider the differential equation y'(x) = f(x), which holds for all x in (a, b).
We assume that the solution y and its derivative y' are continuous functions, i.e. y is C^1(a, b), and that the RHS function f is "nice"---nothing crazy happens. This makes the differential equation well-defined.
But suppose that this equation is too hard to solve for y(x) by integration.
Could we perhaps rewrite the equation in terms of the inverse function x
, solve the new equation for the inverse function, and invert this to get y(x) back?If the function y happens to be invertible at a point x, then we would have the following equation by the definition of the inverse:
y( x
) = y.(The LHS may look confusing, but it just says, if we plug the point y into the inverse function to get the point x and plug that into the original function, we will get the point y back.)
Furthermore, if the function y is invertible for all points near x, then we could differentiate the equation above with respect to y. By the chain rule,
y'( x
) * x' = 1.We now see why we want the Jacobian y'(x) to be not zero. If it isn't zero, then we have a definition for x'
, the derivative of the inverse function, at the point y:x'
= 1 / y'( x ).Let us go back to the differential equation:
y'(x) = f(x), for all x in (a, b).
Assuming f(x) is also not zero in the interval (a, b), we can take the reciprocal of both sides to get,
1 / y'(x) = 1 / f(x), for all x in (a, b).
If the inverse function exists---by now, all the assumptions of the inverse function theorem are met---then,
1 / y'( x
) = 1 / f( x ), for all y in (c, d).Finally, we substitute x'
for 1 / y'( x ) to get,x'
= 1 / f( x ), for all y in (c, d).This is the new differential equation that we would solve to get the inverse function x
. Whether it's easier to solve this equation and to invert the inverse function is another story.Note that, because y' is continuous and the Jacobian is never zero in (a, b), the function y is always increasing or always decreasing.
Hence, we know the values of c and d from the boundary conditions (BCs). They are c = min(y(a), y(b)) and d = max(y(a), y(b)). The BCs for the new differential equation are x(c) = a and x(d) = b.
Uh, what does it mean to give a geometric description of points in a plane?
When I search on google/YouTube, I keep getting things related to Linear Algebra (which we never talked about in my intro course to Linear Algebra).
A = {(x,y) belongs to R^2 | x < 2 and y belongs to (-3,3)}. Would this be an example of a geometric description?
When I search on google/YouTube, I keep getting things related to Linear Algebra (which we never talked about in my intro course to Linear Algebra).
A = {(x,y) belongs to R^2 | x < 2 and y belongs to (-3,3)}. Would this be an example of a geometric description?
ClayKavalier
Banned
I don't need any help on specific problems but was hoping I could get some suggestions for good math books to self-teach.
I want to brush up on statistics, calculus, and linear algebra, and hopefully eventually learn some real analysis. I Google'd suggested book and got the Courant Lecture Notes on Probability Theory, which seems to be highly regarded, but it's basically unusable to me at this point. The second page has stuff like "P* is countably suadditive" which is meaningless to me.
What should I read to get down the basics and make stuff like that intelligible? Also if someone could suggest a good algebra book I'd really appreciate it, I know basic algebra well but I need to study pre-calculus stuff (I can't remember what sine, cosine, tangent, etc. mean or how they applied) and especially matrices, which I've never actually been taught in school.
Thanks!
I want to brush up on statistics, calculus, and linear algebra, and hopefully eventually learn some real analysis. I Google'd suggested book and got the Courant Lecture Notes on Probability Theory, which seems to be highly regarded, but it's basically unusable to me at this point. The second page has stuff like "P* is countably suadditive" which is meaningless to me.
What should I read to get down the basics and make stuff like that intelligible? Also if someone could suggest a good algebra book I'd really appreciate it, I know basic algebra well but I need to study pre-calculus stuff (I can't remember what sine, cosine, tangent, etc. mean or how they applied) and especially matrices, which I've never actually been taught in school.
Thanks!
Stumpokapow
listen to the mad man
My advice would not be to "self-teach", but rather to look for open courses (Coursera, MIT OpenCourseware) and try to get the textbooks they have and follow along with the courses. I have a pretty dim view of auto-didact claims in general. If you need textbook recommendations, for Calculus, the most common undergraduate textbook is James Stewart - Calculus: Early Transcendentals. "Statistics" is pretty broad. For linear algebra, maybe Gilbert Strang - Intro to Linear Algebra. There are so many. For probability, I like Betsekas and Tsitsiklis - Intro to Probability but there are a variety of texts in that field. For studying canonical RVs and sampling distributions you'll probably want to move into a regression or econometrics text. Weisberg - Applied Linear Regression is good; and then maybe Angrist and Pischke - Mostly Harmless Econometrics... and then maybe Gelman and Hill - Data Analysis Using Regression and Multilevel/Hierarchical Modeling. Another good strategy: Pick a university and start pulling syllabi from the Math/Stats department to get a sense for the topics covered and books used.
But, just for some perspective, you're basically asking how to go from junior high school level math to a knowledge of most of the topics people with a BA/BSc in Math cover. The answer is that you're describing a process that even for very bright, driven people will take them 5+ years. It's always tempting when you're embarking on a big journey to try to plan every step, but most people actually give up on the first step, so spending more time planning than doing is an error. Start with algebra and trigonometry, work through everything, and then if you still have appetite after that move into pre-calculus, and then if you're one of the very few that still wants to keep going, start worrying about the next steps. Algebra and Trig you can get anywhere; any high school text, any GED prep text, any SAT prep text. Start with those.
ClayKavalier
Banned
I know algebra well enough up to about matrices, I was never taught those. We had to use matrices in a Macroeconomics class I took two semesters ago and I was able to pass the tests by just memorizing the steps but I had no clue what I was actually doing, like why you're able to set two equations into a two-by-three grid and multiply diagonals to form a new equation, or what the determinant meant.
I took a very basic Calculus course during undergrad, basically just learned how to take derivatives, and had to teach myself integrals in a Micro course. Not that that was too different than a derivative. Also had to teach myself the Lagrangian, which again isn't hard at all but a ton of undergrads seem to struggle with it in tutoring.
I should have stated my background more, I'm one semester away from earning a Masters in Economics. My undergrad program did not emphasize math whatsoever, and the school I'm currently attending has a much stronger focus on math in undergrad and the grad program is more theory driven. I've been able to teach myself everything up to this point (all As so far) but I know I'm a lot weaker at math than anyone else in my cohort.
I still know some math though, definitely more than the average student. I feels like I'm clueless compared to some of my friends but I think I should be able to teach myself some stuff. Next semester my course load is pretty light so my plan was to brush up as much as I can then delay graduation one semester and take some math courses, which will hopefully be doable with the studying I've done.
So yeah, anyway, don't mean to give my life's story but I pick up mathematical concepts pretty quickly and have some background knowledge. Linear Algebra and Real Analysis are the two courses I'd love to have on my transcript, maybe that isn't realistic but that's what I'd like to work my way toward. Thanks for the suggestions, I'll definitely look through some of those open courses. I've taken enough courses where the professor doesn't follow the book closely or just recommends a colleague's book or other nonsense to be wary about checking random syllabi.
I actually just ordered Stang's Linear Algebra book, read a lot of good impressions. I also ordered Spivak's Calculus text which seems to be very highly regarded, hopefully it's not too over my head.
Chance Hale
Member
Started my calculus class yesterday. 3 and half hours, 4 days a week. 50 minutes in, teacher stopped because he couldn't solve a problem on the board.
"Sorry, I don't know the material very well. We'll get there"
"Sorry, I don't know the material very well. We'll get there"
Just a quick matlab question. I am trying to set the following as my objective function in an optimization algorithm:
objective_function= @(x) (sum(i*cos((i+1)*x(1)+i),i=1..5))*(sum(j*cos((j+1)*x(2)+j),j=1..5));
But I get the following error:
The expression to the left of the equals sign is not a valid target for an assignment.
In relation to the i=1..5 and j=1..5
Anyone have any idea why?
objective_function= @(x) (sum(i*cos((i+1)*x(1)+i),i=1..5))*(sum(j*cos((j+1)*x(2)+j),j=1..5));
But I get the following error:
The expression to the left of the equals sign is not a valid target for an assignment.
In relation to the i=1..5 and j=1..5
Anyone have any idea why?
I don't think you are using Matlab's sum routine correctly. What you can do instead is the following:
i = 1 : 5;
objective_function = @(x) sum(i .* cos((i + 1) * x(1) + i)) * sum(i .* cos((i + 1) * x(2) + i));
Note that .* means entry-wise multiplication between the vector i and the vector cos((i + 1) * x(1) + i). The sum routine then adds the entries of the vector that results from the .* operation.
Stumpokapow
listen to the mad man
Heh, your original post definitely did not give the impression you were at the level you're at, so if my response seemed kinda curt, know that it was just because I made totally the wrong set of assumptions about what you were hoping to do. And... it was probably dumb of me to suggest an intro to regression book to you
I'm confident if you've got this far, the kinds of topics you're asking for help with will be no trouble at all for you.Hypertrooper
Member
I do not know if there are good youtube videos about the topic. I personally have not informed myself about Genetic Algorithm, but will do soon. There is a book by Mitchell Melanie 'Introduction to Genetic Algorithm'. Maybe it is worth checking out?
Hypertrooper
Member
Does anyone have a clue about the spectral theorem? I'm preparing for a exam on friday and I am struggling with transformation matrices. ;-(
Hypertrooper, you just have a minor mistake. You should have,
w2 = 1/sqrt(<v2, v2>) * v2,
and this equals to,
1/sqrt(6) * [1 + i ; -2].
Finally,
U = [w1, w2]
In case you weren't aware of this when you solved this problem, for Hermitian matrices, we know that the eigenvectors corresponding to two distinct eigenvalues are orthogonal.
Hence, v1 and v2 are already orthogonal, and we just need to normalize each to get w1 and w2 (i.e. no need to think about Gram-Schmidt process).
I'm so confused right now that I need help...
Suppose we have a non-homogeneous differential equation of the form: a(x)*y'' + b(x)*y' + c(x)*y = F(x).
Let y(x) be any solution to the aforementioned diffy eq, y_1 be any one solution to the diffy eq, and y_2 be a solution to the homogeneous equation a(x)*y'' + b(x)*y' + c(x)*y = 0.
Why is it that y(x) - y_1 is a solution to the homogenous equation while y_1 + y_2 is a solution to the non-homogeneous equation?...
Suppose we have a non-homogeneous differential equation of the form: a(x)*y'' + b(x)*y' + c(x)*y = F(x).
Let y(x) be any solution to the aforementioned diffy eq, y_1 be any one solution to the diffy eq, and y_2 be a solution to the homogeneous equation a(x)*y'' + b(x)*y' + c(x)*y = 0.
Why is it that y(x) - y_1 is a solution to the homogenous equation while y_1 + y_2 is a solution to the non-homogeneous equation?...
FortuneFaded
Member
Look at what happens to the right hand side of the equation. Let a(x)*y'' + b(x)*y' + c(x)*y = f
for simplicity. f(y(x)-y_1) = f(y(x)) - f(y_1) (since differential equations are linear) = F(x) - F(x) = 0. f(y_1 + y_2) = f(y_1) + f(y_2) = F(x) + 0 = F(x).boviscopophobic
Member
Well, your differential equation can be written as L[y] = F, where L is a linear differential operator, so the statements follow in the same way as they would in linear algebra (where the set of all solutions of an inhomogeneous system of linear equations can be expressed in terms of any specific solution and the set of solutions of the corresponding homogeneous system).
Written out explicitly, let's examine the case of y_1 + y_2. We'll suppress the dependence of a, b, c, and F on x for brevity. We have
a * (y_1 + y_2)'' + b * (y_1 + y_2)' + c * (y_1 + y_2)
= [ a * y_1'' + b * y_1' + c * y_1 ] + [ a * y_2'' + b * y_2' + c * y_2 ]
using the linearity of differentiation. But the term in the first set of square brackets is equal to F since y_1 is a solution of the inhomogeneous equation. Similarly, the term in the second set of brackets is equal to 0, because y_2 is a solution of the homogeneous equation. Hence the whole thing is equal to F, and it follows that y_1 + y_2 was actually a solution of the inhomogeneous equation.
Look at what happens to the right hand side of the equation. Let a(x)*y'' + b(x)*y' + c(x)*y = f
Thanks to the both of you for answers.
I didn't realize that I was dealing with linear differential equations (because I thought a, b, and c were dependent on "x", meaning their values can change and are nonconstant).
I also never knew of the following property: for f
, f(y2-y1) = f(y2) - f(y1) where y2 and y1 are linear. Is there some sort of reference for me to find out more of such properties?Thanks to the both of you for answers.
I didn't realize that I was dealing with linear differential equations (because I thought a, b, and c were dependent on "x", meaning their values can change and are nonconstant).
I also never knew of the following property: for f
Kieli, the functions a, b, and c (these are "given," i.e. we know what they are for a problem) do depend on x, but we are solving for the unknown function y(x) that satisfies your differential equation.
It is with regards to the variable y that the differential equation is a linear one.
In the general context (with the specific case of differential equations in parentheses), we say that an (differential) operator L, which takes on/accepts elements (real-valued functions) y, is a linear operator in y (or shortly, "linear in y"), if the following two conditions hold for the operator L:
1. L(y1 + y2) = L(y1) + L(y2) for all y1, y2.
2. L(alpha * y) = alpha * L
for all y and for all scalars (real numbers) alpha.We can show that these two conditions are equivalent to this single condition:
1. L(alpha1 * y1 + alpha2 * y2) = alpha1 * L(y1) + alpha2 * L(y2), for all y1, y2 and for all scalars alpha1, alpha2.
As others have pointed out, you can check that your differential operator,
L
:= a * y'' + b * y' + c * y,is linear in y, using either set of conditions that I just provided.
(It all works because differentiation is a linear operation!)
boviscopophobic
Member
They do depend on x and are possibly non-constant; however, the important thing here is that they do not depend on y.
No, the "linearity" here is a property of f, not of y1 and y2; the equation above follows directly from the definition of linearity. Recall that a mapping f: U->V of vector spaces is said to be linear iffI also never knew of the following property: for f
f(x + y) = f(x) + f
f(a*x) = a*f(x)
whenever a is a scalar and x,y are elements of U. Note that the linearity is more properly considered to be a property of the mapping, not so much the objects that the mapping is operating on. Most functions are not linear, but we often like the ones that are, because they are convenient to work with and behave nicely.
Some examples of linear [functions/mappings/operators]:
1. A real-valued function f: R->R specified by the equation f(x) = m*x, where m is a scalar. Of course, the graph of such a function is a line, which is why it's reasonable to use the word "linear" to describe it.
2. A linear transformation of finite-dimensional vector spaces specified by matrix multiplication: x |-> A*x. Special case: the identity transformation, x |-> x.
3. The differentiation operator d/dx. The set of C^infinity functions (e.g.) forms a vector space, and differentiation is a linear operator on this space (i.e. it maps a function y to some other function dy/dx, and does so in a linear fashion). Exercise: prove this!
No, the "linearity" here is a property of f, not of y1 and y2
f(x + y) = f(x) + f
f(a*x) = a*f(x)
3. The differentiation operator d/dx. The set of C^infinity functions (e.g.) forms a vector space, and differentiation is a linear operator on this space (i.e. it maps a function y to some other function dy/dx, and does so in a linear fashion). Exercise: prove this!
Got it.
I've seen linearity in the context of matrix equations. A(ax+by) = aAx + bAy where A is matrix, a & b are constants, and x & y are vectors. Didn't make the connection 'til you mentioned it now.
For the exercise, I'm not sure how to go about this. I only know the rule that differentiation of sums = sum of the each differentiated term.
I also looked up what C^infinity functions are, and I understand that they can be repeatedly differentiated. It appears that polynomials and exponentials are elements of this set, but I'm not sure what other functions follow suit.
Hypertrooper
Member
Is anyone fit in the subject of growth of functions? I have to take my written exam in Algorithms & Data structures next Wednesday. One of the subject is growth of function with Big-O etc. My question is if anyone could check out my scribble here.
I am allowed to take cheat sheet with me. So if anyone has some good tips what to put in for the growth of function subject, tell me.
I am allowed to take cheat sheet with me. So if anyone has some good tips what to put in for the growth of function subject, tell me.
GiantEnemyGoomba
Member
I'm gonna be taking integral calculus this upcoming semester. I did really well in my differential calc class (we ended with the fundamental theorem of calculus and u-substitution), but that was last fall semester, so I was wondering what I should brush up on before I start school again since it's been more than half a year. I'm guessing pretty much everything, but idk if stuff like related rates or optimization comes up much in integral calc.
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