Linear Algebra-Subspace Functions

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  • #26
Fredrik
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Well at the point where the function touches the y axes so x=0, then no matter what the sign is both f(x)=f(-x)= the same thing. So does it mean that the vertex point at the origin satisfies this?
The 0 you're talking about is just a member of the domain of f. (And what is f anyway?) This has nothing to do with the zero vector of F(ℝ). You really need to look at the definition of F(ℝ).

I also second micromass' recommendation that you should try to think of a few examples of members of F(ℝ), just to make sure that you know what sort of things are considered "vectors" in this problem.
 
  • #27
micromass
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This equality should be g+f=g. f and g are functions. f(x) and g(x) are real numbers.

Or he should have said ##g(x) + f(x) = g(x)## for all x.
 
  • #28
micromass
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FinalStand, I think it's easier if you first think about this:

Maybe you should start by actually giving some examples of elements in the "vector space"?? Right now, the definition is pretty abstract. So can you give two or three examples of elements?

So instead of trying to find the "0" element or trying to prove anything, I would first focus on finding a few examples of elements in your space.
 
  • #29
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What function f is such that, if g is a function, g(x) + f(x) = g(x)?

This equality should be g+f=g.
It makes sense either way. For you to insist that the equality should be as you wrote, is a bit pedantic, IMO.
f and g are functions. f(x) and g(x) are real numbers.
As micromass points out, the equation should hold for all real x. That's what I meant, but my thought that was that the OP would understand what I meant.
 
  • #30
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Where is a good site I can read the definition? Because I skipped all classes on vector spaces and my prof doesn't post notes online. And the prof only took off 0.5 points off for this question.

And I really have no clue how this works. Now I don't even know what the point of (af)(x)=a(f(x)) is
 
  • #32
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Where is a good site I can read the definition?
Don't you have a textbook? Any linear algebra textbook would have a definition of "vector space."
Because I skipped all classes on vector spaces and my prof doesn't post notes online. And the prof only took off 0.5 points off for this question.

And I really have no clue how this works. Now I don't even know what the point of (af)(x)=a(f(x)) is
 
  • #33
Fredrik
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Where is a good site I can read the definition? Because I skipped all classes on vector spaces and my prof doesn't post notes online. And the prof only took off 0.5 points off for this question.

And I really have no clue how this works. Now I don't even know what the point of (af)(x)=a(f(x)) is
The book you're using for this course must include a definition of the term "vector space". You can also look up this definition on Wikipedia. For this problem, you also need the definition of the vector space F(ℝ). The definition is almost certainly this:

Let F(ℝ) be the set of all functions from ℝ into ℝ. Now this just defines the set. To turn this into a vector space, we must define an addition operation and a scalar multiplication operation. The operations that your book/teacher had in mind are almost certainly defined as follows. Addition is defined like this: For all f,g in F(ℝ), we define f+g by
$$(f+g)(x)=f(x)+g(x)$$ for all x in ℝ. Scalar multiplication is defined like this: For all f in F(ℝ) and all a in ℝ, we define af by
$$(af)(x)=a(f(x))$$ for all x in ℝ.

Now you can easily verify that F(ℝ) is a vector space. (Strictly speaking, it's the triple (F(ℝ),addition,scalar multiplication) that should be called a vector space, but it's standard to abuse the terminology by referring to the set F(ℝ) as a vector space). A key step in this verification is to note that there's a z in F(ℝ) such that x+z=z+x=x for all x in F(ℝ). This unique z is denoted by 0 and is called the zero vector.
 
  • #34
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I know that but for f(x)=f(-x) cant we just add the x of the opposite sign then it givesyou 0? And no I did not buy the textbook because I never needed textbooks for math..well until maybe now.
 
  • #35
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I know that but for f(x)=f(-x) cant we just add the x of the opposite sign then it givesyou 0?
No. All this equation tells you is that the graph of f is its own reflection across the y-axis. Many functions satisfy this relationship. Earlier in the thread micromass asked you if you could come up with two or three examples of functions that satisfy the equation above.
And no I did not buy the textbook because I never needed textbooks for math..well until maybe now.
I don't know what to tell you. If you can't be bothered to go to the class, or make the effort to buy the textbook, I don't know how much good we can do for you.
 
  • #37
Fredrik
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I know that
What does "that" refer to? Do you mean that you already knew that F(ℝ) is defined that way? If so, why have you still not realized that the zero vector is a function?

I suspect that there's also something lacking in your understanding of the concept of "function". Do you know e.g. what it takes for two functions to be equal? What do they have to have in common to be equal?

but for f(x)=f(-x) cant we just add the x of the opposite sign then it givesyou 0?
I don't even know what this means. What is f? Are you saying that this holds for all f in F(ℝ)? Are you saying that there exists an f in F(ℝ) such that this holds? Is f some specific member of F(ℝ), and you're saying that this holds for that f? What do you mean by "add the x of the opposite sign"? What do you mean by "gives you 0"? What 0? The number or the vector?

And no I did not buy the textbook because I never needed textbooks for math..well until maybe now.
You obviously need to rethink that strategy. It can't possibly work for very long.
 
  • #38
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Could it be (f+(-f))(x)=0 then f(-x)+ (-f(-x))=0? Can you guys tell me what the answer is so maybe I can work it out?
 
  • #39
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Could it be (f+(-f))(x)=0 then f(-x)+ (-f(-x))=0? Can you guys tell me what the answer is so maybe I can work it out?
You seem to be fixated on this relatively minor point. What you appear to be doing is making random manipulations of this formula that describes which functions are in the set, without understanding what things are actually in the set. For example, is f(x) = x2 + x + 1/2 in W?

Any function in the set must satisfy f(x) = f(-x), but that is only a characteristic of the members of the set, and is not the definition of any of them.
 
  • #40
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no that is not in the set...for example 1 would not work. You guys kept on repeating the same thing and I do not understand that thing. I dont know any thing and I have no idea what thing you are talking about either. I think I need to go drop this course or drop myself down a building and suicide.
 
  • #41
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And it does not touch the x axis therefore it will not work. But I still dont see the point of this. I think the examples and the definitions are irrelevant. So what if it says x+y=x then why do we still need to proof that it is in f(x)=f(-x) ? of course there is an x that f(x+y=x) = f(-x+y=-x) what is the point on giong over the definitions? And whats the point of proofing this? I think it is all rubbish. i gave up on math I will change my major then. Screw myself with a million screw drivers.
 
  • #42
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no that is not in the set
Are you referring to my example of f(x) = x2 + x + 1/2? Why isn't it in the set? Show me your thinking.
...for example 1 would not work.
Actually, g(x) = 1 IS in set W.
You guys kept on repeating the same thing and I do not understand that thing. I dont know any thing and I have no idea what thing you are talking about either. I think I need to go drop this course or drop myself down a building and suicide.

And it does not touch the x axis therefore it will not work.
What does this have to do with anything? There is no requirement that an arbitrary member of set W has to touch the x-axis.
But I still dont see the point of this. I think the examples and the definitions are irrelevant. So what if it says x+y=x then why do we still need to proof that it is in f(x)=f(-x) ? of course there is an x that f(x+y=x) = f(-x+y=-x) what is the point on giong over the definitions? And whats the point of proofing this? I think it is all rubbish. i gave up on math I will change my major then. Screw myself with a million screw drivers.
 
  • #43
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Dropping the class is probably the best course of action for you, based on your obvious lack of interest in the course.
 
  • #44
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Believe it or not I do have a good mark in it. But what is the zero vector? I know that x+y=x. But for f(x)=f(-x) what is a zero vector for this? f(x)+g(x)=f(-x)? so g(x)=0? or does the value x in g(x) is 0? How is 1 in f(x)=f(-x) when your example is : x^2 + x + 1/2? f(1)=2.5 and f(-1)=0.5? clearly f(x)!=f(-x)? Do we have to prove this by induction or what?

In this thing which one is the vector the funciton itself or the value x? x^2+1 is not in the space I know that since f(x)=f(-x) for all x, but 0+1=1 which is not 0. Now, I love math, but these proofs are useless and worthless. I rather die than drop this course actually. But I probably end up being depressed and jump off a building.

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  • #45
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Are you referring to my example of f(x) = x2 + x + 1/2? Why isn't it in the set? Show me your thinking.
Actually, g(x) = 1 IS in set W.


What does this have to do with anything? There is no requirement that an arbitrary member of set W has to touch the x-axis.

Where the hell did you get g(x)!? are you freaken blind? or mentally retarded like me? This is stupid.
 
  • #47
micromass
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If I define the function ##f(x) = x^2 +1##, then does this satisfy ##f(x) = f(-x)## for all ##x##??

If I define the function ##f(x) = x^2 + x +1##, then does this satisfy ##f(x)=f(-x)## for all ##x##?
 
  • #48
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WHy does that matter!? Why are we looking at examples? no the second one does not satisfy all x. it is not a vectorspace at all. :). But how does this matter? I want ot know how to proof that whether 0 is in the space or not.

So (f+0)(x)=f(x)+0(x)=f(x)=f(-x) work?
 
  • #49
micromass
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WHy does that matter!? Why are we looking at examples? no the second one does not satisfy all x. it is not a vectorspace at all. :). But how does this matter? I want ot know how to proof that whether 0 is in the space or not.

So (f+0)(x)=f(x)+0(x)=f(x)=f(-x) work?

No, it doesn't work.

I'm doing examples to get you familiar with the space you're working in.

We are looking for a function ##f## such that ##f+g=g## for all functions ##g##. So ##f(x) + g(x) = g(x)## for all ##g## and ##x##. What function ##f## satisfies this?
 

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