richyw
  • richyw
a little bit confused about summation rules
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Try reading this: http://psychassessment.com.au/PDF/Chapter%2001.pdf
richyw
  • richyw
ok so I have this formula \[v_{av}=\frac{\sum^n_{i=1}v_i\delta t_i}{\sum^n_{i=1}\delta t_i}\]
richyw
  • richyw
yeah I have looked it up. I would just like some clarification on a problem I am working on

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richyw
  • richyw
I also have that \[\delta t_i=\frac{2z_i}{v_i}\]
richyw
  • richyw
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\[v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}\]
richyw
  • richyw
now I am wondering if I can simplify this any further?
richyw
  • richyw
@TuringTest
TuringTest
  • TuringTest
\(v_i\) and \(z_i\) are all different terms depending on index, so I can't see how you could simplify that. I could be wrong...
richyw
  • richyw
right that's what my thinking was.
TuringTest
  • TuringTest
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?
TuringTest
  • TuringTest
\[\sum_i(v_i)\cdot(\frac{z_i}{v_i})=\sum_iz_i\]is valid
TuringTest
  • TuringTest
as long as it's under the same summation sign
richyw
  • richyw
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.
richyw
  • richyw
or final answer I should say.
TuringTest
  • TuringTest
you are right, that rule wouldn't make sense mathematically either
TuringTest
  • TuringTest
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
Zarkon
  • Zarkon
this is valid... \[\sum_i\left[(v_i)\cdot\left(\frac{z_i}{v_i}\right)\right]=\sum_iz_i\]
richyw
  • richyw
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?
Zarkon
  • Zarkon
unless you have more info...you can't simplify it any more
richyw
  • richyw
thanks

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