## richyw 3 years ago a little bit confused about summation rules

1. hongkongfuiz1234

2. richyw

ok so I have this formula $v_{av}=\frac{\sum^n_{i=1}v_i\delta t_i}{\sum^n_{i=1}\delta t_i}$

3. richyw

yeah I have looked it up. I would just like some clarification on a problem I am working on

4. richyw

I also have that $\delta t_i=\frac{2z_i}{v_i}$

5. 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$v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}$

6. richyw

now I am wondering if I can simplify this any further?

7. richyw

@TuringTest

8. 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...

9. richyw

right that's what my thinking was.

10. 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?

11. TuringTest

$\sum_i(v_i)\cdot(\frac{z_i}{v_i})=\sum_iz_i$is valid

12. TuringTest

as long as it's under the same summation sign

13. 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.

14. richyw

or final answer I should say.

15. TuringTest

you are right, that rule wouldn't make sense mathematically either

16. 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

17. Zarkon

this is valid... $\sum_i\left[(v_i)\cdot\left(\frac{z_i}{v_i}\right)\right]=\sum_iz_i$

18. 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?

19. Zarkon