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Homework Help: Subspace theorem; differential equation for a subspace

  1. Mar 26, 2013 #1
    I can't seem to work out this question because it's so weird The set F of all function from R to R is a vector space given the diffential equation f"(x)+3f'(x)+x^2 f(x) = sin(x) is a subspace of F? Justify your answer

    I know that we have to proof that it's non-empty 0. The zero vector has to be in S 1. if u and v are in U, then u+v lies in S 2. by multiplication scalar k(u)

    So I tried to solve it but it doesn't make sense & this is my soluiton. If anyone know how to solve it please help me! Anything will help

    Check for non-empty f"(0)+3f'(0)+0^2 f(0)= sin(0) =0 therefore the set S is non empty

    Check A1- closed under addition =(f+g)" + 3(f+g)'+ x^2 (f+g) = sin (x) =f" + g" + 3f' + 3g'+ x^2f+ x^2g = sin(x)
    = (f"+3f'+ x^2f) + (g" +3g'+x^2g) - sin (x)

    I do not know what to do with the constant x^2 and sin(x). I even tried to sub (f+g) into them but it doesn't seem right. Also I tried to sub in real number like 1 and the answer doesn't turn into 0 so does that mean it's not a subspace of F?

    Second question: Use subspace theorem to decide whether the following set is a vector space The set V of all real polynomials p of degree at most 2 satisfying p(1) = p (2) i.e poly with the same values at x=1 and x=2

    0) p(x)=0 therefore V is non-empty 1) let t=p+q; t is a polynomial, and for every x∈R it satisfies t(x)=p(x)+q(x). In particular; t(1)=p(1)+q(1) =p(2)+q(2) =t(2) because p,q∈V therefore V is closed under addition 2)Let p be any polynomial in V, let α be any real number, let q=αp (i.e., q(x)=αp(x) for all x∈R), and show that q∈V.

    is the second question look right? please help, thank you!
     
  2. jcsd
  3. Mar 26, 2013 #2

    mfb

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    Check again what is meant by the vector space: It is the space of all functions f. The 0-vector in this vector space is the function f(x)=0 for all x in R.
    Does "f(x)=0 for all x in R" satisfy your differential equation? This is equivalent to the question "is the 0-vector in your potential subspace?"
    The check for addition does not work like that either.

    The second problem is fine.
     
  4. Mar 26, 2013 #3
    It fail the first test for being non-empty ?
    Like if y1+ y2 was a subspace solution. Then in the left hand side we got 2sinx instead of sin x. Is that a good explain? Also how would you do addition for that equation? I saw an example in a text book so I did it like that. Could you explain more please, appreciate :)
    ALSO how can you properly test for non empty ? Since I usually sub in 0 to see the equation equal 0 . Sin 0 =0
     
    Last edited by a moderator: Mar 26, 2013
  5. Mar 26, 2013 #4

    Mark44

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    Let's try to get the terminology right. You're mixing up the words in ways that have no meaning, such as "subspace solution". What you have is a set, the set of solutions to y'' + 3y' + x2y = sin(x).

    First off, does this set contain any members? In other words, are there any solutions to the diff. equation above?

    Is the set closed under addition? If y1 and y2 are solutions to the above diff. equation, is y1 + y2 also a solution?

    Is the set closed under scalar multiplication? If y1 is a solution to the diff. equation, is cy1 also a solution.

    If the answers to all three questions are "yes", then the set is a subspace. If at least one answer was "no" then the set is not a subspace.
     
  6. Mar 26, 2013 #5
    Ok let me get it straight.
    So for the first test for being non-empty, it turn out to be 0 so the first test pass
    Y"+3y'+x^2.y=sinx
    0+3(0)+x^2(0)=sin(0)
    Therefore =0
    Is my working out so far so good?
    For second test, addition
    Let y1+y2 be the solution. If the set was a subspace then y1+y2 be the solution but if we evaluate the left hand side y1+y2 we get 2 sinx not sin x
    Therefore A1 failed and S set is not closed under addition
     
  7. Mar 26, 2013 #6

    Mark44

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    No.
    Is y = 0 a solution of y'' + 3y' + x2y = sin(x)?
    This part is OK.
     
  8. Mar 26, 2013 #7
    So uh it fail the first test as well so does addition right? It's a inhomogenous not homogeneous equation? I'm getting it now, this question is hard because it use differial equation instead of real number so yeah, really sorry!
    But after reading the question again make me thinking again:
    The set of SOLUTIONS to the differential equation f"(x)+3f(x)'+x^2f(x)=sin(x) a subspace of F? Justify your answer

    For question 2/ use the subspace theorem to decide whether following set is a real vector space with usual operations. the set of V of all real poly. p of degree at most 2 satisfying p(1)=p(2) I.e. polynomial with the same value at x=1 and x=2
    I ask my friend about it & show them the way I work out my solution & they say its wrong but u say its right (?)
    They told me u got to work out 10 of the proof A 1-5 and S 1-5. They say u have to since its a vector space , is this necessary?
    Thank you very much for taking ur time answer me! I do me a big big favour :)
    Regards.
     
  9. Mar 26, 2013 #8
    If y=0 mean its a linearly independent but y=sin(x) is a dependent equation. Is that what u mean @Mark44 ?
     
  10. Mar 26, 2013 #9

    Mark44

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    Correct. The set fails both of those axioms.
    The differential equation is nonhomogeneous, right.
    With the work you have done, is this set a subspace of F?
    mfb was the one who said it was right.
    The polynomials of degree at most 2 is already a vector space. If you can establish that the set of polynomials in that space such that p(1) = p(2) is a subspace, then this subspace will be a vector space in its own right. You don't have to verify each of the 10 or so vector space axioms.
     
  11. Mar 26, 2013 #10

    Mark44

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    I have no idea what you're asking. I'm not saying anything about linear independence. All I'm talking about is the function y = 0, whose graph is a horizontal straight line.

    Also, I don't know what you mean when you say that y = sin(x) is a dependent equation.
     
  12. Mar 26, 2013 #11
    LOL I'm confuse so much now...
    subspace theorem:
    1/V≠∅. (V is non-empty.)
    2/If p,q∈V, then p+q∈V. (V is closed under vector addition.)
    3/If p,q∈V and α∈R, then αp∈V. (V is closed under scalar multiplication.

    from my working out i know that test 1 pass because it equal 0
    since f"(0)+3(0)+x^2 (0) = sin(0) =0
    second test fail (addition) because the
    For let y1 and y2 be solutions. If the set of solutions was a subspace, then y1+y2 would be a solution. But if we evaluate the left-hand side at y1+y2, the result is 2sinx, not sinx.
    . therefore it's not a subspace because it's not closed under addition.
    From there I can conclude that's not a a subspace of F

    My question is that how can I SHOW/PROOF that it's not a subspace vector? I already know it's not a subspace of the function. Since the question is justify your answer!

    I do not get you either that's why I'm asking u! idk what you mean by y=0? what does that suppose to mean lol

    If you have a linear homogeneous DE, the set of solutions is a subspace of the space of all functions. If you have a linear non-homogeneous DE, the set of solutions is not a subspace of the space of all functions.
    you mind explain how would you answer this question and justify why it's not a subspace since I'm really confuse atm since this question worth 3 mark and you need to proof it. Really appreciate your time =)
    Cheers.
     
  13. Mar 26, 2013 #12

    Dick

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    Showing that if y1(x) and y2(x) satisfy your equation, then putting y1(x)+y2(x) into your equation gives you 2*sin(x) is already enough to show it's not a vector space. 2*sin(x) is not equal to sin(x). It's not closed under addition.
     
  14. Mar 26, 2013 #13
    hm... Someone explain like that for me as well let y(1) and y(2) be the solution, if the set was a subspace then y(1)+y(2) be the solution but if we evaluate the left hand side Y(1)+y(2) the right hand side is 2*sin(x) <- that's the part where I do not understand why it's 2*sin(x) is it because y(1) and y(2) 1+1=2? so u multiple them by 2 to the other side? ): this question worth 3 mark and I'm really confusing about it ):
    do you mind explain for me, really appreciate your time!
     
  15. Mar 26, 2013 #14

    Dick

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    Define ##D(y(x))=y''(x)+3y'(x)+x^2y(x)##. Your equation is ##D(y(x))=sin(x)##. If ##y_1(x)## and ##y_2(x)## are solutions then ##D(y_1(x))=sin(x)## and ##D(y_2(x))=sin(x)##. If the solution space were a vector space then ##y_1(x)+y_2(x)## would also be a solution, so ##D(y_1(x)+y_2(x))## should be ##sin(x)##. It's not. It's ##2sin(x)##. Since ##D(y_1(x)+y_2(x))=D(y_1(x))+D(y_2(x))##. I thought you had this part.
     
    Last edited: Mar 26, 2013
  16. Mar 26, 2013 #15
    OMG! thank you it's like crystal clear now! you are the life saver. THANK YOU LOTS! Have a nice day!
     
  17. Mar 27, 2013 #16

    Fredrik

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    This is wrong.

    Can you tell us which member of F is the zero vector?

    The set S is defined as the set of all f in F such that ##f''(x)+3f'(x)+x^2f(x)=\sin x## for all x in ℝ. Let's temporarily denote the zero vector of F by z. Do you understand what the statement "z is a member of S" means? (I'm just asking you to use the definition of S to rewrite that statement in a different way). If you don't understand what it means, it will be hard to determine if it's true.

    Is z a member of S? Motivate your answer.


    By the way, someone asked the exact same question the other day. https://www.physicsforums.com/showthread.php?t=680780
     
    Last edited: Mar 27, 2013
  18. Mar 27, 2013 #17
    Well this is an assignment in uni & I guess the person asked is the one in my course as well for engineering. I got the answer and understand the problem
    Our lecture just teach us to test non-empty by sub in 0 to the equation so that exactly what the other person did as me so yeah. It end up that the right hand side only have sin(x) without f(x) so it can't be 0 so its empty and if y1+y2 be the solution if its the scalar then the right hand side should be sin x but it equal 2sinx so its not equal each other . That what I understand so yeah...
     
  19. Mar 27, 2013 #18

    Fredrik

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    Yes, this is correct. Your substitutions in the calculation I quoted weren't. You had substituted 0 for f in two places and 0 for x in two other places. Only the former substitution is correct.

    If z is the zero vector in F, then for all x in ℝ, we have
    $$z''(x)+3z'(x)+x^2z(x)=0+3\cdot 0+x^2\cdot 0=0.$$ Since there exists a real number x such that ##\sin x\neq 0## (for example ##\pi##), it's not true that ##z''(x)+3z'(x)+x^2z(x)=\sin x## for all x.
     
  20. Mar 27, 2013 #19
    Yea that what I understand as well when some people explain to me in the first place but when I explain to my friends they ask sin(0) is 0 so I don't know how to tell them. Further more explanations I was able to notice that sin(x) is by it self like a constant so yeah. The reason x^2 =0 cus of f(x) equal 0 , the question is very tricky ... We usually deal with equations not differential equation like that . Thank you for your help trying to clear thing up :)
     
  21. Mar 27, 2013 #20

    mfb

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    There is absolutely no reason to consider 0 as special value of x here. The equation has to be true for all x (for functions in your set), not just for x=0.
     
  22. Mar 27, 2013 #21

    Mark44

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    Part of the difficulties you're having might be due to English not being your native language, and this might be the reason that many of the things you write make no sense.

    sin(0) = 0 is true, but has nothing to do with this problem. As already noted by others, x = 0 is not relevant here, but the zero function is important (i.e., f(x) = 0 for all x).

    sin(x) is NOT like a constant, so I don't know why you said that.
     
  23. Mar 27, 2013 #22

    Fredrik

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    I think that what you need to tell them is that the members of F are functions, so the zero vector must be a function.

    I think it's useful to be really careful with the notation and terminology. If f is a function, then f(x) is a number, so you shouldn't call f(x) a function. sin is a function, but x is a number, and sin x is a number. (It's still wrong to call sin x a constant, because x doesn't always represent the same number). If we define a function ##s:\mathbb R\to\mathbb R## by ##s(x)=x^2## for all x in ℝ, the differential equation can be written as
    $$f''+3f'+sf=\sin.$$ We can also define a map ##T:F\to F## by $$T(f)=f''+3f'+sf$$ for all ##f\in F##, and write the differential equation as ##T(f)=\sin##. So the set S can be defined by
    $$S=\left\{f\in F\,|\,f''+3f'+sf=\sin\right\}$$ or by
    $$S=\left\{f\in F\,|\,T(f)=\sin\right\}.$$ Note that x isn't even involved! There's no x there to be set equal to 0.

    A differential equation is an equality between functions. Two functions are equal if they have the same domain, and the same value at each point in the domain. So the statement
    $$f''+3f'+sf=\sin.$$ is equivalent to
    $$\text{For all }x\in\mathbb R\text{, we have }(f''+3f'+sf)(x)=\sin x.$$ The "for all" is an essential part of it, and can't be excluded. Sums and products of functions, and products of real numbers and functions, are defined by ##(f+g)(x)=f(x)+g(x)##, ##(fg)(x)=f(x)g(x)## and ##(af)(x)=a(f(x))## for all x in ℝ. So
    $$(f''+3f'+sf)(x) =f''(x)+(3f')(x)+(sf)(x) =f''(x)+3f'(x)+s(x)f(x) =f''(x)+3f'(x)+x^2f(x).$$ So the differential equation is equivalent to
    $$\text{For all }x\in\mathbb R\text{, we have }f''(x)+3f'(x)+x^2f(x)=\sin x.$$ Is the zero vector in F? No, because ##T(0)=0''+30'+g0=0\neq\sin##.

    I don't understand what you mean by these things. They look very wrong.

    Honestly, this problem is extremely easy if you just use the definitions. I think the main reason why you guys are having difficulties is that you still haven't realized that when you prove something, you always have to use the definitions of the terms and notations in the statement that you're trying to prove. It also looks like you and your classmates don't fully understand the concept of "function". Someone who takes a minute to think about how the vector space F is defined, and understands what a function is, will never consider setting x=0.
     
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