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richyw

  • 2 years ago

a little bit confused about summation rules

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  1. hongkongfuiz1234
    • 2 years ago
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    Try reading this: http://psychassessment.com.au/PDF/Chapter%2001.pdf

  2. richyw
    • 2 years ago
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    ok so I have this formula \[v_{av}=\frac{\sum^n_{i=1}v_i\delta t_i}{\sum^n_{i=1}\delta t_i}\]

  3. richyw
    • 2 years ago
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    yeah I have looked it up. I would just like some clarification on a problem I am working on

  4. richyw
    • 2 years ago
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    I also have that \[\delta t_i=\frac{2z_i}{v_i}\]

  5. richyw
    • 2 years ago
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    so plugging this in I get \[v_{av}=\frac{\sum^n_{i=1}v_i\frac{2z_i}{v_i}}{\sum^n_{i=1}\frac{2z_i}{v_i}}\]factoring out constants\[v_{av}=\frac{\sum^n_{i=1}v_i\frac{z_i}{v_i}}{\sum^n_{i=1}\frac{z_i}{v_i}}\]canceling\[v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}\]

  6. richyw
    • 2 years ago
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    now I am wondering if I can simplify this any further?

  7. richyw
    • 2 years ago
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    @TuringTest

  8. TuringTest
    • 2 years ago
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    \(v_i\) and \(z_i\) are all different terms depending on index, so I can't see how you could simplify that. I could be wrong...

  9. richyw
    • 2 years ago
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    right that's what my thinking was.

  10. TuringTest
    • 2 years ago
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    I mean, what about this?\[v_{av}=\frac{\sum^n_{i=1}\cancel v_i\frac{z_i}{\cancel v_i}}{\sum^n_{i=1}\frac{z_i}{v_i}}\]or am I misreading your notation?

  11. TuringTest
    • 2 years ago
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    \[\sum_i(v_i)\cdot(\frac{z_i}{v_i})=\sum_iz_i\]is valid

  12. TuringTest
    • 2 years ago
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    as long as it's under the same summation sign

  13. richyw
    • 2 years ago
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    hmm no I don't think that is correct. This is a physics problem and the answer does not make sense. v is a velocity and z is a distance so the final sum is a velocity.

  14. richyw
    • 2 years ago
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    or final answer I should say.

  15. TuringTest
    • 2 years ago
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    you are right, that rule wouldn't make sense mathematically either

  16. TuringTest
    • 2 years ago
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    actually the more I thinking about it the more I am thinking it should make sense, but I am obviously unsure... @Zarkon @satellite73 series rules question

  17. Zarkon
    • 2 years ago
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    this is valid... \[\sum_i\left[(v_i)\cdot\left(\frac{z_i}{v_i}\right)\right]=\sum_iz_i\]

  18. richyw
    • 2 years ago
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    right and I did that in the numerator of the fraction to get. \[v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}\] Is there any way to express this as a product of \(z_i\) and \(v_i\) in the numerator?

  19. Zarkon
    • 2 years ago
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    unless you have more info...you can't simplify it any more

  20. richyw
    • 2 years ago
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    thanks

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