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Summation of product identities

  1. May 12, 2013 #1
    Hi everybody, I am just trying to find a decent identity that relates the sum $$\sum_{k=0}^{n}a_kb_k$$ to another sum such that ##a_k## and ##b_k## aren't together in the same one. If you don't know what I mean, feel free to ask. If you have an answer, please post it. Thanks in advance!
  2. jcsd
  3. May 12, 2013 #2


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    hi eddybob123! :smile:

    that's just aยทb, the inner product (dot product) of two (n+1)-dimensional vectors a and b :wink:
  4. May 12, 2013 #3
    But the a's and the b's aren't vectors, so what is the value of the ''##\cos\theta##"?
  5. May 12, 2013 #4


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    You can always interpret a and b as vectors.

    There is no general formula to express this as the combination of two things, where one thing just depends on all a and the other thing just depends on all b.

    In formulas, there are no (general) functions f,g,h to do this:
    h(F,G)=your result
  6. May 12, 2013 #5
    The individual a's and b's might not be vectors, but we can think of them as components.

    $$\sum_{k=0}^{n}a_{k}b_{k} = \begin{bmatrix} a_0 & a_1 & a_2 & \cdots \end{bmatrix}\begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ \vdots \end{bmatrix} = \langle \vec{a},\vec{b} \rangle $$

    How do we find the cosine of theta, you ask? $$\cos\theta = \frac{\displaystyle \sum_{k=0}^{n}a_{k}b_{k}}{(\sqrt{a_0^2 + a_1^2 + ...})(\sqrt{b_0^2 + b_1^2 + b_2^2 + ...})}$$

    To be completely serious, I am not aware of the kind of general formula you are looking for.
  7. May 12, 2013 #6
    And the magnitudes of the two vectors will be $$\sqrt{a_0{}^2+a_1{}^2+...+a_n{}^2}$$ and $$\sqrt{b_0{}^2+b_1{}^2+...+b_n{}^2}$$Is this right?
  8. May 12, 2013 #7


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