So I’m taking Advanced Calculus this semester which is Math 409 at my university. The work is mostly proofs and I suck at it. I’m really want to switch from Applied Math - Computer Science to Computer Science and just get a minor in math, but I’m in too far to switch 
Anyone have some advice at getting better for calculus proofs? It seems like I’m just memorizing someone else’s proof instead of proving things on my own.
The key to proofs is trying to stop looking at things as simply symbols and start looking at things as logic problems and try to word them as such. Sometimes half the problem solves itself when you word it properly.
One mistake I usually see people do in proofs is they tend to fall in love with a single method of proving a feature. Get familiar with using all different kinds of proofs such as proof by contradiction, direct proof, proof by cases, etc.
In my book, capital A and B refer to matricies. so, A belongs to M2R
The proof will still hold if you consider A and B to be the a_ij and b_ij elements of A and B respectively.
These are just some off of the top of my head. When you’re first starting out, you should always be asking why they did that and endevor to uderstand why they did that. Asking professors why they did that is good too. Just model what they and the author do while trying to understand why they did. There are a LOT of gems there if you know where to look … so, learn by experience where you should be looking.
just went through the whole thread, you guys are great.
the easiest way to “learn” how to do proofs is to pay attention in class and/or read your textbook. seriously, most of the proofs that you’ll need to write will be of a form similar to what your professor or TA is showing you in class.
i was always better a number theory and probabilty than adv. calc and linear algebra because the proof parameters and jargon was easier for me to wrap my head around.
April Fools Thoughts:
4/1/14, if taken as 4114 is a palindromic number.
April Fools Day won’t be palindromic for another 100 years
not strictly true, 2040 the date will read: 04/1/40.
yo, is my proffessor of the deep end?
he gave
is y" +2y’ - y = 0 a subset of Vector space V
V= C^2(I)
so, i have to prove axioms 1 and 2 simple enough. u+v=S and ky = S
so i did.
1.u = u" +2u’ - u
v = v" +2v’ - v
then u+v = (u" +2u’ - u)+(v" +2v’ - v) = (u"+v")+2(u’+v’)+(-u-v)
let u+v = w, u’+v’=w’ etc
w" +2w’ - w therefore u, v are elements of that vector space.
he said nope wrong 0 points. why? the point of axiom 1 is to determine that the adition of u and v still falls underneath the vector space, and this is no diffrent. we are still trying to see if u+v is satisfied and if the given is closed under addition.
dude went about it backwards and started here
(u"+v")+2(u’+v’)+(-u-v) = (u" +2u’ - u)+(v" +2v’ - v) = 0+0 = 0
he says that’s how you have to do it. What’s the damn difference, if you start where i start. Both u and v are still elements of the subspace that’s defined, and when added you come to the same function. why is it that in this instance, i have to start backwards, and not start from where i started. IN fact, for other vector spacce problems, had I done what he said to do, it would be incorrect. in both instances u and v are elemnts of the vector space or subspace. is S is a subset of V, then it’s a subspace, therefore an element.
is it because S is an element of V, i have to substitute y with u + v?
When i asked him, he wasn’t clear and he said, well this is how i would start it. but when i told him that for other problems you start the way i did. if you look at vector space section there’s a problem y"+4y = 0
in that instance you start like i did. Is there something that the book and he forgot or aren’t telling me that differentiates subtle properties regarding axiom 1 when applying them to vector spaces or subspace?
I was operating under the assumption that the first zero was omitted. However, I found your observation to be quite clever. Bravo.
I’m not 100% what C^2(I) stands for. Complex numbers? The set of functions with continuous 2nd derivatives? I will assume it is some sort of function space.
It looks like you were trying to prove both of them in one shot. You should do them seperately.
The way you wrote your proof, you need to prove that ku is in the subspace (with k being in the vector space), but you then wrote 1u. You need to use the fact that the n’th derivative of the function ky is equal to k times the nth derivative of y, which is something you’re expected to know from calculus.
As for the addition one, I don’t know if you used something to distinguish the symbol you used for the vector versus what appears to be the same symbol used in the form of the element (iow, did you do something like ((bold)v) or v with an arrow for a vector versus regular v for the variable
Then maybe do something like
((bold)u) + ((bold)v) = (u" +2u’- u)+(v" +2v’-v) = (u"+v")+2(u’+v’)+(-u-v)
=(u"+v")+2(u’+v’)-(u+v) = (u+v)"+2(u+v)’-(u+v) =((bold) u+v) = (0+0) = 0
and we know that 0 is in the subspace, so we showed the sum is equal to an element in the vector space.
I would also point out that neither one of you got ((bold) u+v) in it’s defined form. You both forgot to factor our a minus sign. I see that you tried to use algebra to get the derivatives, but I think your argument would have been stronger if you had used the fact that for differentiable functions f and g, (f’+g’) = (f+g)’, which you may have been thinking, but didn’t explicitly state the form of the theorem here. When in doubt, don’t skip steps.
I’m thinking you didn’t get points because of the above reasons. Are you absolutely certain he insisted on starting with ((bold) u + v) as opposed to ((bold)u) + ((bold)v)? The order is trivial because a+b = c is the same thing as c = a+b. The equals sign nullifies the order the expressions are written in so long as the key steps above are shown. If you agree that all those expressions are equivalent, then changing the order in which they were written doesn’t disprove their logical equivalence. It might make it more confusing to the reader and make it bad practice and I could see you getting penalized that way, and but it doesn’t disprove them.
If he gave you 0 points because you didn’t start with the sum first, then that is ridiculous, but there were other flaws that I pointed out that I think were also taken into consideration.
@"Warrior’s Dreams"
This is what the problem stated.
Let V = C(superscript)2(I), that is the vector space is the complex plane in two dimensions on interval I (this complex plane also tells me that the solution to this differential equation is trigonometric)
S is the subset of V consisting of all solutions that satisfy the differential equation y" +2y’ - y = 0
I also proved that k*u is satisfied, but I didn’t bring it up, beceasue he also marked it as absolutely incorrect, and wanted me to write it a diffrent way, his way, therefore it’s incorrect.
So what I wrote for axiom one
u+v = S, let u and v are elements of V, (or should I have stated that they are elements of the subspace that forms the subset of V, for example, vectors u and v are elements of S)
vector u = u" +2u’ - u = 0 (the u and v that replace y", y’ and y do not have arrows on top)
vector v = v" +2v’ - v = 0
and this should be valid, because axiom 1 states that vector u plus vector v should equal the vector space.
u+v = (u" +2u’ - u)+(v" +2v’ - v) = (u"+v")+2(u’+v’)+(-u-v)
let (u"+v") = w", (u’+v’) = w’, (-u-v) = -w,
(w")+2(w’)+(w), therefore there exists such an element that vectors u and v are elements of the vector space.
my internalized rationilization…Becasue they are closed under addition (this satisfies the parallelogram rule (i use this to understand the whole closed under addition idea better)), it stands that this subspae is a subset of V, therefore it’s an element of V, thus far.
becasue the addition of vectors u and v still results in y"+2w’-w, it stands that the solutions of the differential equation is closed under addition.
for axiom two i wrote
k * vector u = S, where S is an element of S and k is an element of any real number
let vector u = u" +2u’ - u
-> k(u) = k(u" +2u’ - u) = ku" +2ku’ - ku = ku" +2(ku’) - ku = 0
therefore there exists such that u is an element of the subspace and k is an element of all real numbers.
…
therefore the solutions of y" +2y’ - y are a subspace of of the given V.
Good catch on the factoring out, didn’t even notice that.
but no, i didn’t get any points because it wasn’t how he did it.
I talked to him about it and he was being incredibly dense and just opened the book and said look at this example. I then told him that there where other problems from the previous section that only asked to prove axioms 1 and 2, and they did it the way I did it, and he even gave an example in class that did it the way I did. Why was that not incorrect, and he said, well, this is how i’d do it for this one. You can’t add vectors, you’re trying to do this algabreacly.
I told him no, im adding vectors u and v, to prove that there sum is inside the vecotr space, and that they are closed under addition.
He says, you can’t do it like that, I told him why not there’s other problems that are set up similarly and he said lol nope show me proof. Essentially he didn’t give me a reason, and he tried to, but he couldn’t and just said look at book example. Even though the example is just the example.
he told me it has to be like this
(u"+v")+2(u’+v’)+(-u-v) = (u"+v")+(2u’+2v’)-(u+v) = (u" +2u’ - u)+(v" +2v’ - v) = 0+0 = 0
because see this example in the book? And it’s so god damn frustrating. The only reason I can see that it has to be this is because we have to assume that the vector space is closed under addition to begin with, and that the solutions to the differential equation when added equal zero. but by that same token, why am I wrong if i’m still showing that u+v equals zero?
He didn’t give partial credit, and this is the type of instructor that still marks things that are wrong even if you lost all possible points. Had it been those things, he would have marked them, but he didn’t. So i know not that.
At this point, I’m so discouraged in this class. the full physics, chem, and calc sequence never made me feel this stupid and incompetent. I don’t know if it’s me, or him. I feel like I understand what is it where doing, especially regarding axioms 1 and two, but that understanding isn’t doing me any good. But this 25 and all these below 30 percent scores are making me feel shitty.
shit, the examples, aren’t there…anyway, disregading the example part.
do you have any idea as to why this is flat out wrong?
god damn it…
I was just looking over an example
show the differential equation y’+9y=4x^2 is closed under addition.
axiom one states that adding vector u and v should equal our vector space, so for this problem, our vector space
V={y’+9y=4x^2}
so u+v = V, so c
so, in this instance y is an element of V. Usually there’s only one part of the problem that’s an element, and in this case it’s y.
so the vector space is the equation y’+9y, such that our vector space is equal to 4x^2 right?
so if y the element that belongs to our vector space…then is it safe to rationalize that
y is an element of V, and vectors u and v are to be elements of V also, so/
we replace the element that is an element of our defined vector space with the term (u+v), so
in this case
y’+9y -> (u’+v’)+9(u+v), because y is our element, then there should exist such an element that u+v is an element of V…right???
and when worked out, if if u = u’+9u = 4x^2 and v=v’+9v = 4x^2
then
(u’+v’)+9(u+v) = u’+9u + v’+9v = 4x^2+4x^2 =! 4x^2
another example, say matricies would be
V = {x is an element of all real numbers : Ax = 0, where A is a fixed matrix}
in this one, our corresponding element has been defined for us, in this case, it’s X, so
becasue x is our element, there should exist such an element (u+v) which is an element of the vector space, so
Ax -> A(u+v)
So to sum up, for whatever our vector space may be, that which is defined as an element of said vector space, should be replaced by (u+v), since we this is an element we are trying to prove closes under addition.
Oh, S is the SOLUTION SET to that differential equation…
Yea, that makes a BIG difference.
The vectors aren’t of the form y" +2y’ - y = 0, the vectors are the y’s themselves (the functions y(t) that satisfy that diff eq …
The reason it was completely wrong is that you assumed the vectors were that diff eq. Instead, you assume that two vectors u and v satisfy that diff eq. Now you show that the sum u+v satisfies that equation.
So you start by assuming that u+v satisfies that diff eq and through algebra you come to the conclusion that the result is just the sum of two differential equations whose functions that satisfy them are u and v. So it’s consistent with your original assumptions and you can conclude that the space is closed under addition.
Similarly, you plug ku into that differential equation and the k basically factors out because of the reason I stated in a prior post.
I’ve noticed that several times you’ve omitted these little details and that made all the difference in the world. State the problem verbatim and use words to define the symbols you use so we can be on the same page.
So whatever vector is an element of said space, is what we substitute (u+v) or the corresponding term depending on the axiom we do right?
so had the statement said
S is the differential equation y" +2y’ - y = 0, then my original approach would have been correct right? the thing that’s confusing me the most is that the book doesn’t state when and where we substitute u+v, it just states definitions, and then shows 1 example. I’m not comfortable concluding things like this because it costs points on the exam if you are incorrect, and I really don’t trust this professor becuase he contradicts himself all the time. It also bugs me making a conclusion like this without hard evidence, since most of my class mates really don’t understand what’s going on. I want to know the justification.
++++++++++++++++++++++++
so one more time, because y is the vector itself, and is the element that belongs to the space, y itself gets substituted by u+v. so for example
S = {matrix A which is an element of all 2x2 matricies with real numbers such that det(A) = 0}
when we show if this closes under addition
we need to add A+B. but if the question had said instead
S = {the solution x which is an element of all 2x2 matricies with real numbers such that det(A) = 0}
then it would have been A(u+v)
++++++++++++++++++++++++++
One more thing, is it safe to assume that this is how we define the subspace for this particular problem?
S = {y is an element of S such that y" +2y’ - y = 0 on interval I}
The statement
“S is the differential equation y” +2y’ - y = 0"
doesn’t make any sense because an equation is not a set. I honestly think the reason you are making these mistakes is because you aren’t aware of the components of your problem.
You need to determine what the vector space is, what the subset of the elements of that vector space are, how the elements in that set are defined, and the problem statement itself.
To avoid future headaches, just assume that the vector sum u+v is in that set, then use what makes u and v a part of that set, as well as other relevant math knowledge to conclude that your assumption that u+v is in that set is true.
I honestly don’t know about your last question because I don’t know what the problem is actually asking you to do because you gave me an abridged version of it instead of the actual problem itself.
You’re getting symbols confused as well. In the former problem y(t) were your elements, but you need to just stop thinking in terms of symbols for a minute and ask yourself what do these symbols mean in words …
For your first determinant set, you need to start with the sum A+B and use the definition of the set, and the fact that A and B are in that set to determine that the matrix A+B is in that set. A+B is supposed to have a determinant of 0, so if A+B does have a determinant of 0, then that shows A+B is in that set. You will need to use the fact that A and B both have determinant of 0 to demonstrate this.
For your second matrix set, that set is poorly defined. What is x a solution of?
I can see you are trying to generalize what you’ve just learned, but what you really need to be asking yourself is “what does it take to be an element of that set?” That’s why they give you these things in definitions. You need to stop thinking about what order things go in and instead just think logically through it. Examples only get you so far because as soon as they give you a set you’ve never seen before, you don’t know what to do. That’s why you have to use logic to get through it, not examples, not order, just definitions and theorems, because every example you will encounter will use the same idea, even if its manifestation is different. Find the common thread and you won’t have to worry about this kind of problem again.
the last question regards is that how you define the subspace of the problem I am having issues with. S is the subset of V consisting of all solutions that satisfy the differential equation y" +2y’ - y = 0
but that’s what i’m trying to figure out. how and where does this apply. It doesn’t make sense that I’m given a bunch a conditions, and then I have to make them work for whatever problem is given. What is the element, and how does it relate to the axioms and the given definition or statement.
When I’m given problems to solve, and I solve them I’m not learning anything. I’m trying to rationlize what they did,a nd why they did it. And I can’t figure it out.
say I’m given (and this is how the problem is written)
S = {x is an element of IR(superscript 2) such that x = (2k, -3k), where k is an element of any real numbers}
show that S is a subspace of IR superscript 2
what am I supposed to look for?
I know S is defined, and x is an element of any real number(superscript 2) [that is, it belongs to any real number], and our x vector is equal to (2k, -3k) where k is any real scalar.
I know that on the left hand of the definition, we are vaguely describing the subspace and what belongs to what. and on the right hand side we are explicitly stating what equals what and defining any other minuet elements… that’s easy enough. But how am I supposed to apply axiom 1 and 2? What am I looking for?
__
I know vector x = (2k, -3k). and I know that i have to satisfy axiom 1, u+v = S
because our vector x is an element of S, we know that we need to add two vectors which are also elements of S, and the resulting sum should equal S right?
so, u = (2s, -3s) and v = (2t, -3t), where s and t are elements of IR(superscript 2) [these are our two vectors, which are elements of S, hence why we need to add them]
u+v = (2s, -3s) + (2t, -3t) = (2s+2t), (-3s±3t) = 2(s+t), -3(s+t) = (2k, -3k)
s+t = k [we can do this because the sum of s and t will still result in a real number, becuase we defined s and t to be any real number]?
perhaps i’m overthinking it, and trying to rationlize it compeltely wrong.
You are overthinking it. You don’t have to so much make them work as you have to show that they work. You use these theorems because you are trying to find out what sets are vector spaces under what conditions. You need to stop trying to locate a forest by counting its trees and instead locate a forest by using the properties of a forest to determine that.
You don’t really look for anything because everything you need to understand that set is pretty much given to you (with the assumption that you remember stuff from prior classes that are relevant to it). What you did with your (2k, -3k) example is pretty much what you should be doing (except that you forgot the parentheses to show it’s a coordinate), though again, to avoid headaches from your prof, you might want to start with u+v first and show that it can be reduced to u and v. I think you know more than you want to believe and that you are using the fact that you didn’t understand that solution set problem you got a zero on to question everything you know, when if you had read the problem a bit more closely, you would have figured it out.
To use the subspace theorem means to use an established method for proving something is a subspace of a vector space. What you’re supposed to learn is that regardless of how trivial the conditions you are given, if they are a vector space, then they are closed under addition and scalar multiplication. That means that if you can show that ku and u+v are in the set, you’ve demonstrated that the subset is a subspace and thus a vector space.
The first part is to show that it is closed under addition. To do this, you use the first part of the Subspace theorem. This is a litmus test, so to speak. You start out assuming that the sum u+v is in the set. The test is this: If you can show that the vectors u and v are also in the set, then you’ve shown that it’s closed under addition. You do this by showing that the sum u+v can be split into separate elements u and v using the set conditions given to you.
If this has been satisfied, then you show the space is also closed under scalar multiplication. To do this, assume ku is in the set, use the conditions of the set, and if you can show that the element u is in that set according to the set definitions and it can be multiplied by any scalar, then you’ve proven your set is a subspace.
You cannot just say “I have u+v, therefore I have u and v” because that is what you are trying to SHOW.
Similarly, you can’t just say ku is just u times a scalar k because again that is what you are trying to show.
What you are learning is to determine what sets are vector spaces under certain conditions. You need to know that because vector spaces are fundamental to linear algebra. What you’ll eventually do is learn that you can make entire vector spaces from only a couple of vectors (your spanning set), but only if those vectors are linearly independent.
Oops, haven’t been around for a while. This is a month old so maybe you have it all figured out by now. Just in case, my answers or comments are in bold.
I know nobody cares, but since yesterday I’m (almost) officially a mathematician.