College maths help.

Discussion in 'Science' started by Brett Nortje, Jan 21, 2017.

  1. Brett Nortje

    Brett Nortje Well-Known Member

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    I have always preferred college maths to high school maths, as it is so much more satisfying understanding it, as, it has directly useful formulas for use in the 'maths using sectors.' this means it is actually more useful than those other formulas, and, as awlays, i am here to try to improve the state of understanding through my own methods.

    So, to find the value of [z] you need to find, for any place you wish to place [z] how it has an equal value to the other places you place [z] as this is calculus, yes?

    This means that f = [x + iy] = z? let's look at the whole suggested example;

    f[z] = [x + iy]^2 = x^2 - y^2 + 2ixy.

    Looking at this we can easily gather that x^2 and y^2 equals x * x = 2x * x = 4x, and the same for [y]. so, from the second equals sign onwards, we find that it means [2i 5x - 5y].

    So, [4x + 4i 4y] = [2i 5x - 5y] and the answer to that is [z]. i suppose the catch is finding where 4i = 2i, which means that [4x + 4y] = [5x - 5y] = 2i = z. this means that [2i] = [z].
     
  2. Brett Nortje

    Brett Nortje Well-Known Member

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    [QUOTEhttps://en.wikipedia.org/wiki/Chapman–Kolmogorov_equation]In mathematics, specifically in the theory of Markovian stochastic processes in probability theory, the Chapman–Kolmogorov equation is an identity relating the joint probability distributions of different sets of coordinates on a stochastic process. The equation was derived independently by both the British mathematician Sydney Chapman and the Russian mathematician Andrey Kolmogorov.[/QUOTE]

    [pi^n...,...,i^n=1 [f^1...F^n]

    So, now the equation is like this;

    p^i3i1 [f^3 | f^1] = p^i3i2 [f^3 | f^2] p^i2i1 [f^1 f^2] df2.

    If we were to observe that [p^32i = 32ip] * [32f] * [8p] * [8f] d * 4f, we would understand it is much easier now, yes? this means that [16p * 16f] = [32i] * [40p] * [300f]. this means that d = 372[ipf] - [pif]196.
     
  3. Brett Nortje

    Brett Nortje Well-Known Member

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  4. Brett Nortje

    Brett Nortje Well-Known Member

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    I find my way of doing maths could be summed up as the process of elimination. if you were to observe that the way things pan out, i usually 'cross things out,' things that end up being one or so, you would agree that one is not worth writing down, of course. this means that the sum may be simplified by eliminating repeats or things that eliminate themselves.

    This is quite common in college maths, as the creators of the maths would list all things, maybe repeating them, to make sure they were right! think of a old man trying to teach the times tables - they would repeat them over and over until the students had the right idea about what the times tables were, yes? this would mean that they drilled it into you, and, you either remembered, you left, or you eventually remembered.

    So, with that in mind, the creators of maths must have gotten it into their heads that you need to take everything and include it. i remember when i was summarizing field theory from einstein, he had so many values in there that it was hell of a hard to work out. then, i came with basic science, being that [mass * acceleration = power] or something, and everybody said it was so easy then.

    Now, when eliminating a few values from the formula, the right way to do things is to look for weakness - do not worry about the order of the sum, just start eliminating as you understand it to mean. you must remember to eliminate in the brackets or outside the brackets though, as in the end all brackets fall away, so you might want to wait with that idea. otherwise it is pretty straight forward to see the sum come to a conclusion, and, then regard [x / y * x / y] as [2x / 2y], which equals [1x / 1y], yes?
     
  5. sdelsolray

    sdelsolray Well-Known Member

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    Do you enjoy talking to yourself?
     
  6. Brett Nortje

    Brett Nortje Well-Known Member

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    the company is good.
     
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  7. Brett Nortje

    Brett Nortje Well-Known Member

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    So, an equation of this kind is about syncing limits as frequently used values - the values may not go above a certain point, but have a pattern. once again, we are back to patterns in maths, where there is a science to everything. well, actually, everything has values we give to them, so, philosophically, we find that there are values in life, as, there is familiarity between people's eye line when they stand up, there is standards when people break bones and find that they take the same amount of time, more or less, to heal, and find similarity between different atoms too, with them weighing the same and having the same melting points, also found to have patterns.

    This equation is no different, as is stated in the quote i supplied. this though is only an equation of the first kind...

    [1] The first kind of fredholm equation. g[t] = a + b k [t,s] f ds.

    This is easy to work out, as there is one t on the left of the equals sign, and, that means that g * t = all that other stuff, yes? of course! now, if you want to get through the right hand side with elimination, then you need to merely see how many [t] there are on the right hand side - there is one, like on the left, so {[g]t = a + b k f ds} = [3s * k * f * d * b + a = gt] = [3s 1k 1f 1d 1b 1t + 1a = gt] = [3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = gt] = [3 + 6 = gt] = [9 = gt] = [9 = 1g - 1t] = [9 - 1 = 1 {/*+-} 1] = [8 = g].

    So, we would say that g = 8, yes?
     
  8. Brett Nortje

    Brett Nortje Well-Known Member

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    If you need to divide [x^2] by [y] you would need to know that it is [1x^2 / 1y], leaving you at an 'unworkable equation.' this means you need to inflate the values, making them more manageable, so, [2x^2 / 2y] / 2?

    This will leave you with [8x / 2y] = [4x / 2y] = [2x / 1y], for example.
     
  9. Brett Nortje

    Brett Nortje Well-Known Member

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    The hill differential equation is about 'movement.' what it comes down to is movement of mass that remains bound, yes? i am sure you already knew that! this you might do in college so listen up, please?

     
  10. dadoalex

    dadoalex Well-Known Member Past Donor

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    Yeah...

    From here on...

    You and the pooch got real cozy.

    [x + iy]^2 = x^2 - y^2 + 2ixy is wrong and should be X^2 + Y^2+ 2IXY.

    this

    x^2 and y^2 equals x * x = 2x * x = 4x,

    Is also flat wrong. X^2 does not equal 2x*x and 2x*x does not equal 4x (it equals 2x^2)
     
  11. Brett Nortje

    Brett Nortje Well-Known Member

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    Sorry if i got that one wrong, and thanks for pointing that out. sometimes i get it mixed.
     
  12. dadoalex

    dadoalex Well-Known Member Past Donor

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    A suggestion:

    Substitution is often easier than working with multiple variables. If you go back to the original

    f = [x + iy] = z

    And substitute Q = iy then the problem becomes

    f=[x+q]=z

    Which, in your later calculations comes to:

    f=x^2 +2xq + q^2

    The iy can be substituted back into the equations after simplification.

    Calculus was my favorite classes (all 24 hours) because of its application to real world problems. All truth in the universe resides within math and science.
     
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  13. Brett Nortje

    Brett Nortje Well-Known Member

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    The ishimori equation is one where we try to find the value of one part of the total spin, a third in other words, of the total spin to be applied to the total spin of an atom. this means we could consider one third negative spin or two thirds positive spin quite easily, or, other such things to do with 'spin.'

    A much easier way to do this is to observe that this is all equal to speed times by time. finding the speed would be down to mass being divided by either a negative spin, or a positive spin times by two, of course. this means, instead of working out the spin of the particle, we could instead deduct the amount of energy being spent as two or minus one, and then weigh the atoms depending on how much mass they have, of course.

    Or, we could look on the two furtherest equals signs, where, if [as / at] equals all that junk, and it is equal to [2a^2s] * [as / ax] ^ [as ay] it is easy to see that it is merely [speed divided by time].
     
  14. mudman

    mudman Well-Known Member

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    Ummm...no.

    He was right (about this part anyway).

    i^2 = -1 which is why it's -y^2 not +y^2.
     
  15. Brett Nortje

    Brett Nortje Well-Known Member

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    Laplace's equations are for finding heat and energy potentials of mass. these can be simplified further by taking the weight regarding the load bearing structure, and, then weight by density to find the ability to conduct these 'energies' or masses by container density and so forth.

     
  16. Brett Nortje

    Brett Nortje Well-Known Member

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    Maurer-Cartan form is used for changing frames or something, but, it reminds me of calculus, of course. the goal of this formula is to find the manifold with variables.

    The goal is to find [g] of course. after you find [g] you need to apply it to [p], but that is something else.

    So, we have a [w^g = L^g-1] then, the rest is multiplying it by volume, of course. this leads to L being a negative number, or [w] being a negative number, yes? [g] divided by [g] means that [L] to the power of [g] * [g^-1] means [L] is also a negative number, in the end, so [w] is a negative number too. This means that [v] is also a negative number, so, on the cross section, they will be in the bottom right and top left sections of the cross section.
     
  17. Brett Nortje

    Brett Nortje Well-Known Member

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    Pell's equation is there to determine different determinates for area of cones, like screws and other highly technical parts of an engine or even the holes in them. it has been simplified to mean this;

    I suppose the objective here is to find [x]? well, the thing is, there are so many equals signs that this will be easy, as we only need to calculate one of the [x]'s in ratios to other [x]'s.

    This means that [x] will be equal to [y] at any given point, merely being the reverse of it, as, every time [x] goes up, so does [y], minus [N], so [x] = [y] divided by [n].

    Being a cylinder, there needs to be balance with the cone, of course, as it is supposed to have a 'opposing form.' where the n comes from, decreases from both of them, as x = y, yes?
     
  18. Brett Nortje

    Brett Nortje Well-Known Member

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    Poisson's equation is where to find the missing values of manifolds. this would be where the holes are covered up, and, charges found to be released by each hole due to stress.

    As we can see, we need to find this 'p squiggle thing.' that means we need to find the d squiggle thing first, being [12 d] * p * [x y z] = f [x y z], so;

    [xyz]^2 / [12d] / [xyz] = p, or, [f] / [x y z] = [p]. this means that [p] = [12d] * [x y z]^2 = f.
     
  19. Brett Nortje

    Brett Nortje Well-Known Member

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    The riccati equation is done like this;

    As you can see, the fanciest squiggle, this "v" character, appears on each side of the equals sign, yes? that means, to find this value, the other stuff doesn't matter, but, rather only the values that affect [v]. v i believe to be volume, by the way.

    So, [minus v], as that looks like a sum sign, equals [-4q / 4q * v] = [1q + 1q] * [v + 4v]. so, [-v] = [1q * v] = [2q * 5v] meaning that v is a minus number, being -3.
     
  20. Brett Nortje

    Brett Nortje Well-Known Member

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    Sine gordon equation is about engines, i presume, or, measuring particles in a atom rotating around, specifically the electron, i would suppose.

    So, applying logic to this, the sum of b equals the square root of, first, one, as, we see a double entry above and below the square root sign, being that "(1 - w^2)" by itself, it would come to one, yes? then, the [1 - v ^2 + v^k] would equal [1 - 4v + 2kv]. that is on the square root side of this new word i don't understand, being 'arc tangent h.'

    Then, the sum of = [1-4v+2kv] square rooted by this 2 arc tangent what what.
     

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