Prove the Set of all Odd Functions is a Subspace of a Vector Space

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  • Опубликовано: 4 дек 2024

Комментарии • 19

  • @mefatoum1979
    @mefatoum1979 7 лет назад +7

    thank you,you made it clear :) now i know how to do the same thing for even functions

    • @karanshah9354
      @karanshah9354 6 лет назад +1

      I did that now. It is also a subspace and according to theorem vector space of its own. can you check and confirm

  • @nandithaj7558
    @nandithaj7558 5 лет назад +1

    How do you show that the sum of subspaces of odd and even functions gives a vector space?

  • @rogerrojas6960
    @rogerrojas6960 7 лет назад +9

    Although the zero vector function in the beginning is indeed an odd function, I think your proof showed that it is an even function.

  • @tugaric
    @tugaric 6 лет назад +1

    O(-x) = 0(x) proves that its an even function ... ? To complete the proof you have to establish tge fact that -0(x) = 0(x).

  • @mech_builder7998
    @mech_builder7998 9 лет назад

    In #1 of the first proof, there's a zero vector with (x) immediately after. I'm confused as to what this means. Does it mean 0 times all the x's in the function (and this way the function equals 0)? Furthermore, what does the 0 function mean?

    • @TheMathSorcerer
      @TheMathSorcerer  9 лет назад

      +mech_builder Instead of f, I just called it 0_hat. The 0 indicates that it is the zero function, the hat indicates that it is a vector also. You could have called it anything, say h. The zero function takes every x and sends it to zero, so h(x) = 0 for ALL x, that's all it is:)

  • @youyogee
    @youyogee 4 года назад +1

    If you made 1000 examples I will be happy to watch.....

  • @terryphi
    @terryphi 8 лет назад

    at around 1:00
    should x and y also be elements of V?

    • @terryphi
      @terryphi 8 лет назад

      isn't the definition you gave for a subspace the same as the definition of a linear space?

  • @needlermasta
    @needlermasta 7 лет назад

    Thanks, pretty helpful

  • @OriginalEch3Official
    @OriginalEch3Official 3 года назад

    could you also say: "zero function": R-->{0} ?

  • @raghuhiriyur
    @raghuhiriyur 8 лет назад

    In the case of W={(a,b,c): a is greater than or equal to zero}. if the vector is scaled by negative scalar then W won't be subspace because its components should be greater than or equal to zero.In your case, If I am scaling the function by -1 then cf(-x)= -1 - f(x)= f(x) which does not satisfy the condition for odd function. What am I missing here.

  • @erfanjamshidi3431
    @erfanjamshidi3431 7 лет назад

    you proved all odd functions form subspace , but what about sinx which is an odd function
    sin(x1)+sin(x2)=!! sin(x1+x2)
    clearly the condition CA is not satisfied ,?
    what is the mistake here?

    • @hangduong9608
      @hangduong9608 6 лет назад +2

      yes, all odd functions form subspace, f(x) = sin x which is an odd function and g(x) = x which is an odd function, so (f+g)(x) = - (f+g)(-x). You get the point ? It's about the function, not the x variable.