Jump to content

Craiger1987

Members
  • Content Count

    7
  • Joined

  • Last visited

Community Reputation

1 Neutral

About Craiger1987

  • Rank
    Lepton

Profile Information

  • Favorite Area of Science
    physics
  1. That's what I mean, all the stuff I've read explains how to simply sum all the integers between the start and stop point using a simple equasion i(I+1)/2, provided the start point is 1. But what if the start point is greater than one? This equasion no longe works. How do you add all the integers between the start and stop point like you could if the start point was 1. Like if the instructions were add everything from 4 to 68 for example. I know I could run it for 1-68 and then subtract the sum of 1-3 to get it but I was wondering if there was a different way to do it that didn't require that additional step.
  2. Yeah I meant sigma, sorry so with sigma if the number below I is one and the top 10 it's just saying 1+2+3...+9+10 and this can be solved for with i(i+1)/2 instead of manually going through all the numbers. But let's say the bottom number we're something other than one, like a 4, the equasion no longer works for summing the integers. Is there a different equasion for when the beginning number is greater than one?
  3. Hi, another question for you fine folks. Just getting into the use of sigma and that thing is super handy. And I know if I wanted to solve for say 1 through 10 I can just use i(i+1)/2. But what if the index number is greater than one? The equation doesn't work anymore. Is there a modification or different equasion to use? Thanks
  4. Lol, kind of sounds it doesn't it? Idk if it's the same in everything but at least in the work that I do we make a distinction between accuracy and precision. Accuracy is relative, if I want to be within say .5 of an exact point something that is .3 is considered accurate, but precision is exacting.
  5. That is really interesting I'm gonna look that up and do some reading
  6. Thanks that helps a lot we never got past geometry in school, which looking back is a shame, and for some reason I never really applied myself heavily in math class back then, I always focused my attention on science, chemistry etc not quite realizing just how much easier life would be once I got into the heavier stuff if I was proficienct with more advanced mathmatics. Didn't quite grasp the gravity of 'math is everything' back then lol. But, now that I'm going back over the base concepts I learned in school on my own I'm actually enjoying it quite a lot and once I'm looking forward to working my way up to the more advanced maths. Thanks again and I'm sure I'll be here asking questions as I go.
  7. Hi all, this is my first post here and I wanted to run something by people who are more knowledgeable than I. In the past couple days I've decided that I need to teach myself more advanced mathmatics as it is of great importance to my studies in physics, astronomy and biology. In school we never went past algebra 1 and geometry and that has proven to be a hindrance. So I decided to brush up on the basic math that I learned in school and then move on to more advanced maths. I decided to make up a bunch of problems to solve while waiting for my books to arrive and I ran into my old nemesis, finding square roots of large numbers. I was taught prime factorization, which didn't work for this number, and the Babylonian Method, which can yield accurate results depending on the accuracy of your initial estimate and how many times you're willing to run the equation. What I wanted was an equation that used simple math and a known number instead of an assumed estimate (that must be high in the case of the Babylonian Method) that didn't have to be high or low that would yeald an equally/greater accurate estimate of the square as does the BM, and would only have to be run once. I realize this simple mathmatics compared to what is normally discussed here, so I apologize, and also please forgive me if my terminology isn't quite correct. What I came up with was √S≈((S/2)/(e/2)+e)/2 in this S is the number we are trying to find the sqaure of and e is the closest perfect square regardless of weather it is high or low. For exaple, using 1,863 the number I was initially trying to find the sqaure of, we know that the closest perfect sqaure is 43. So it would look like this √1,863≈((1863/2)/(43/2)+43)/2 √1,863≈((931.5)/(21.5)+43)/2 √1,863≈(43.325+43)/2 √1,863≈ 86.325/2 √1,863≈ 43.163 Actual square of 1863=43.1624 so the estimate it yielded was very close. I've run a bunch of numbers through it both large and small and so far it appears to give better estimates than the BM (depending on the accuracy of your initial high estimate and number of times you run it through the equation) and without the requirement that the estimate is high, and also we don't have to make an initial estimate ourselves as the starting estimate is fixed for us by the nature of the equation. What do you guys think? Is there something I've missed or perhaps is there an easier equation like it that I am not familiar with? Thanks! Sorry left out a couple words in my haste lol. E is the closest perfect square root.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.