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Integration as a Linear Transformation

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Skjermbilde_2012_03_10_kl_1_40_59_PM.png

    3. The attempt at a solution
    I: P(R) → P(R) such that I(a0+a1x + ... + anxn) = 0 + a0x + (a1x2)/2 + ... + (an xn+1)/(n+1)

    Clearly this is just integration such that c = 0. It is easily shown that integration is a linear transformation, so I conclude that I is a linear transformation.

    However, in calculus we do not usually require that c = 0. Yet integration, if it is a linear transformation, must map the zero vector of the domain to the zero vector of the codomain. Let f(x) = a0 + a1x. Then F(x) = a0x + (a1x2)/2 + c. A polynomial is zero iff all of its coefficients are zero. If we decide that c is not necessarily zero, doesn't this iff fail? That is, is integration a linear transformation only when we stipulate that c=0?
     
  2. jcsd
  3. Mar 10, 2012 #2

    Fredrik

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    Let U,V be vector spaces. A function T:U→V is said to be linear if T(ax+by)=aTx+bTy for all scalars a,b and all vectors x,y.

    If U=V=ℝ, then the T defined by T(x)=ax+b is clearly not linear, unless b=0. (When U=V=ℝ, both the "scalars" and the "vectors" are members of ℝ). You're working with a different space, and I didn't even read your problem carefully, but my point is just that you need to use the definition of "linear" above, instead of your intuition about what a "line" is. That's why I'm mentioning that the only functions from ℝ to ℝ that are linear are the ones whose graphs are straight lines through the origin.
     
  4. Mar 10, 2012 #3
    And doesn't integration satisfy your definition of linear?

    ∫(f(x) + g(x)) dx = ∫f(x)dx + ∫g(x)dx, for two functions, f,g, in P(R)
    ∫r(f(x))dx = r∫f(x)dx, for some r in R

    Is that not true? And isn't it true that linear transformations must map the zero vector of the domain to the zero vector of the codomain? Maybe I'm misunderstanding.
     
  5. Mar 10, 2012 #4

    HallsofIvy

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    Note that the statement of the problem specifically said "Let F(x) be the polynomial with F(0)= 0 such that F'= f".

    If that condition "with F(0)= 0" were not there, this "theorem" would not be true.
     
  6. Mar 10, 2012 #5
    Does that indicate that integration without the condition "F(0) = 0" is not a linear transformation?
     
  7. Mar 10, 2012 #6

    Fredrik

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    Perhaps you can tell us. :smile: If f,g are arbitrary polynomials, a,b are arbitrary real numbers, and I denotes the map ##f\mapsto F##, what is I(af+bg)?
     
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