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ganeshie8

  • one year ago

show that sum of the vectors drawn from center of a regular polygon to its vertices is 0

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  1. sdfgsdfgs
    • one year ago
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    for regular polygon w even no. of vertices, it is easy to show the sum of vecto5rs from center to vertices is zero by symmetry... it can be proven by symmetry as well for odd no. of vertices but it is less obvious....hmmmm.....

  2. dan815
    • one year ago
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    all the vector sides of an n sided regular polygon centered around the origin can be rewritten was r*e^(2pi*k/n) k is an int from 1 to n and r is the length of each side by this definition we are producing vector from the fact that they are summing to 0

  3. ganeshie8
    • one year ago
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    Consider an octagon for a concrete example with even number of vertices, |dw:1436576807411:dw| How do we use symmetry ?

  4. dan815
    • one year ago
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    |dw:1436576906626:dw|

  5. ganeshie8
    • one year ago
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    @dan815 <nitpicking started> do you mean r*e^(\(\color{red}{i}\)2pi*k/n) ? <nitpickign ended />

  6. dan815
    • one year ago
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    oh yes sry

  7. ganeshie8
    • one year ago
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    so based on that i think the problem translates to proving \[\large \sum\limits_{k=0}^{n-1} e^{i2\pi k/n} = 0\]

  8. dan815
    • one year ago
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    yeah

  9. sdfgsdfgs
    • one year ago
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    |dw:1436577086602:dw|

  10. dan815
    • one year ago
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    what if its odd careful there

  11. ganeshie8
    • one year ago
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    Ahh nice, so we always have opposite vectors for polygons with even number of vertices

  12. sdfgsdfgs
    • one year ago
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    @dan815 it can still be proven by symmetry for odd no. but its less obvious...

  13. dan815
    • one year ago
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    adding vectors to vectors with the same angle of separation between them and the angles add to 360 has to equal 0 another way to say that complex expression in words

  14. geerky42
    • one year ago
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    *

  15. dan815
    • one year ago
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    or also like the outer angles of all regular polygon = 360

  16. dan815
    • one year ago
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    their sum

  17. dan815
    • one year ago
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    |dw:1436577429209:dw|

  18. sdfgsdfgs
    • one year ago
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    |dw:1436577289096:dw|

  19. ganeshie8
    • one year ago
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    Building on dan's method using complex numbers, the vectors can be viewed as solutions to the equation \[\large x^n = r^n\] Clearly the sum of roots of the polynomial \(x^n-r^n\) is \(0\) so i think we're done ?

  20. sdfgsdfgs
    • one year ago
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    @ganeshie8 @dan815 nicely done! :)

  21. dan815
    • one year ago
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    you know kai actually found that equation randomly to generate any n sided regular polygon

  22. ganeshie8
    • one year ago
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    yeah earlier i almost forgot we could use complex numbers here, that equation is same as the equation in vectors because we can treat complex numbers literally as 2D vectors

  23. dan815
    • one year ago
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    right

  24. ganeshie8
    • one year ago
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    ofcourse wid some extra algebra as we cant multiply vectors like we multiply complex numbers

  25. ganeshie8
    • one year ago
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    wonder if there is an useful interpretation dot product/cross product of complex numbers

  26. sdfgsdfgs
    • one year ago
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    btw - just realized that proving the y-componet all vectors in a regular polygon w odd no. of vertices by symmetry will be much less OBVIOUS than I had initially thought! so good that u guys have proven it by using complex nos.! :)

  27. ganeshie8
    • one year ago
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    for even no. of vertices the symmetry argument worked like a charm as we found pairs of opposite vectors !

  28. sdfgsdfgs
    • one year ago
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    by lining up 1 of the vertices in the y-direction, the x-component of all other vectors (except the 1 pointing in y) in an odd-no polygon will equal out as the vectors will be in pair by symmetry. but showing the y-component of the vectors will add to zero is much tricky than i though.

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