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linda

  • 5 years ago

sigma (E) with 3 on top. n=0 on bottom. what does this mean??

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  1. Linda
    • 5 years ago
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    on the side on sigma, it is sin\[\pi\] over 4

  2. anonymous
    • 5 years ago
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    \[\sum_{n=0}^{3}\sin \pi/4\] ?

  3. anonymous
    • 5 years ago
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    Look something like that?

  4. Linda
    • 5 years ago
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    theres an n before the pi symbol, but other than that its right

  5. anonymous
    • 5 years ago
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    \[Ok so \sum_{n=0}^{3}\sin (n*\pi/4)\] So like this?

  6. anonymous
    • 5 years ago
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    Well without the okso at the beginning.

  7. Linda
    • 5 years ago
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    yeah thats right

  8. anonymous
    • 5 years ago
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    The sigma represents a summation. So if it were \[\sum_{n=0}^{3}n\] we start with n=0 plug that value in and keep adding until we reach the limit which is the top value on the sigma, 3.

  9. anonymous
    • 5 years ago
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    So we'd have 0+1+2+3 and then stop.

  10. Linda
    • 5 years ago
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    so when n=0, the whole thing would equal 0 right?

  11. anonymous
    • 5 years ago
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    Yes. You start with whatever n equals under the sigma and plug it into the function to the left of the sigma and then go to the next sequential n value and add it to that result.

  12. Linda
    • 5 years ago
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    so do i just start with n=0 and I get 0, then n=1 I plug it into n*pi over 4 and that equals .785 and so I add that to the zero.... then i go on to n=2 and keep adding on to the .785......

  13. anonymous
    • 5 years ago
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    exactly!

  14. Linda
    • 5 years ago
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    great thanks!

  15. anonymous
    • 5 years ago
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    good method...

  16. anonymous
    • 5 years ago
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    Do u still have doubts @linda ?

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