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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    A logistic function has the following formula;

    For the left most equation to be 'easy,' a common number in this series is 1, yes? that means it could be very likely that [x] = 1, and there would be no problem. in fact, if it balances, where there are no negative numbers, this would be very likely.

    Then, the right most equation; this should be [3f 3x] * [1f * 1x -1] = [4f 4x] - 1. seeing as how this is a study of curves, the value is one degree, as that would be such a big curve, yet compressed by many times to produce 'a sized down curve.' this means that the curve can be calculated as one, then multiplied by the other values, and then compressed or expanded - i am not sure - to fit with the rest of the values. this is for a cylinder though.

    For example, if the curve is really thirty degrees, and you have estimated this, then you have a guideline as to how to apply your 'answers.' this would be where the [4f 4x] -1 equals something close to [7], so, you would divide thirty degrees by seven, bringing the values down to be fitted into the one degree curve you worked off of. so, if [1f] = 4 then the answer would be, for [4f], [16f.] if the [x] was 5, then the answer there would be 20, making for a equation of 16 * 20, coming to 36. if you take seven, as we found on the most basic of equations, to multiply by 5 you would get 35, yes? with that minus one, 36 becomes 35! this means that you may reduce all your values to one for this, and, then, just scale to ratio it into working order.
     
  2. Brett Nortje

    Brett Nortje Well-Known Member

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    Heat transfer is where the boiling point of one thing, or progressing to boiling point will influence another. Heat build up is due to friction, where the activity will be where the material is excited due to activity where mass is exerted on other mass, bringing the electrons close together, then rubbing them, making them 'active,' of course. When this happens, the friction leads to orbitals being too close, and, losing protons, leading to a time where the material heats up. This lost proton goes on to make it cooler elsewhere, while the electrons superior numbers will lead to the material gaining in heat, of course.

    Now, there is also a formula for this. It goes like this;

    So, the [r] by the [a] leads to the heat transfer. This equals the [gr] [pr] being the same value, or, closer to what we are looking for, of course. Then, it comes to be that it is evident that this was made by someone that could not express themselves properly, so made such a big formula, so, let's make our own?

    To find the heat transfer rates, we need only take [joules] by the [area] divided by the [density] of the material. This is because the amount of energy is measured in joules, and, the joules that are 'active' are relevant to the [velocity] or [speed] the material is moving at. Then, you would need to find the area of the materials being affected, so you would take the energy, or joules and divide the amount of joules by the area. Then, you would analyse the [speed] or [acceleration] or [velocity] and [density] or [weight] and [mass] to find the time it takes for the two to meet at the same point of heat, of course. Then, you would take the [joules] / [area] * [acceleration] / [weight]. This will give you the time that it takes for the heat to build up to that point.

    To find the heat amount at any other point, you simply divide the total time by the start point time and find it with [calculus], with your angle being divided into [360], yes? This will give you a angle for the joules needed or active, and, therefore the heat.

    Then, to set it up into time, you divide that, instead of by 360, into 60. This will give you minutes of seconds, which is rather obvious, of course.
     
  3. Brett Nortje

    Brett Nortje Well-Known Member

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    Mohr's circle is where we find the stress thresholds of certain materials. This is where we need to know the mass of the material, as that indicates 'density,' and that can be found by finding the collected stress from the mass exerted onto it and that mass with the [mass] of the ['material supporting'], or, in the ['gear'], divided by [360 degrees], for the continuous circulation of the gears, then, like with trig, divided by area supporting the area around the point of stress. This gets continuously less as the area is separated from the region of the stress point, where the two materials meet, and, this can be summed as;

    [mass {a}] / [360] * [45] + [mass {a}] / [mass {b}]. This should be the threshold of the materials, when they come together.
     
  4. Brett Nortje

    Brett Nortje Well-Known Member

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    With holomophic fucntions, we try to find the continuing functions, that is, that the whole function or 'lot of curves' will continue until the whole system is full of them. This is summed up by the formula;

    [​IMG]

    Here we find that [function] = [z^0], which means it must equal zero, yes? Then, [2f] / [1z] would be the answer, as, the answer is found at all that junk on top of the division sign becoming [zs'] cancelled out, yes? THis is because if you observe that [z] - [z^0] / [z] - [z^0] we find the answer is [one], yes?

    So, [1] * [0] / [1], which means it cannot exceed the limits of the 'loop,' being [one], and that this ratio could be one to ten to one hundred or a thousand, yes? it must be [whole] or less, of course.

    Now, if we were to observe that;

    [​IMG]

    [1ay] * [1ay] = [2ay]. / [2ay] + [y] = [1i2a3y]. [1i2a3y] * [1] = [1i2a3y].

    This means that = [1/3 y] [a] = [2/3 y] and [y] = [y]. This means that [i * 2] = [a].
     
  5. Brett Nortje

    Brett Nortje Well-Known Member

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    Chapman Equation;

    [​IMG]

    This is the Chapman equation.

    We obviously need to find the , yes? This would mean that we need to observe [f^4] = [i^4] * [p].

    This means that [p/4] is the answer, and, 3x = [264 + 264 + 264] = 264!
     
  6. Polydectes

    Polydectes Well-Known Member

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    Interesting thread. I learn and use mathematics as I need too I work or school but I quickly forget it because there isn't enough usage of it I'm my daily life. I think most people view it that way. Its always fascinating to come across someone who simply enjoys it. (I'm not assuming you don't use it for work.)
     
  7. Brett Nortje

    Brett Nortje Well-Known Member

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    Heck no, I don't use maths, I didn't even do it at school. I just like to speculate new techniques.
     
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  8. Polydectes

    Polydectes Well-Known Member

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    I'm trying to get into a field where I actually have to use some of this type of math.
     
  9. Brett Nortje

    Brett Nortje Well-Known Member

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    Clairaut's equation;

    This equation is about finding the ever changing system value, or, [f]. This would be for calculating the amount of energy used to get to [f], or, more precisely, how much [f] has to change to stay stable with the engine or building or such device - I am not sure, it does not explain that. However, to keep things brief, you will see this in your textbooks, what it is for, and, I suggest you understand what it is used for before you try to calculate it.

    We find [y] all over the place, and, it is in [y] we find by [x], yes? [1y] / [1x] * [x] + [1y] + [1x] * [f]? This is rather easy now, yes?

    So, we take, for the left hand side, [1y] * [1x] = [1]? Then, to find [f], we take [1f] + [1x] equals, oh, what is missing, [y]! This means [1y] = [2]!
     
  10. Brett Nortje

    Brett Nortje Well-Known Member

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    With trigonometry, the idea that the stress can be worked out easier with this idea of mine amuses me.

    If you were to take the two points of the graph, you could put a circle in the half way point of the two points. This could be scaled to allow for the circle to have the point you are looking for, scaled to the point that it should be, directly bisecting the two points as if they carried on. This would result in the new circle being able to predict the four sections of influence, or, three sections of influence for the circle. These would be 270 degrees between points, with the first or related one being where 'the 135 degrees to either side of the circle,' where the circle will gather stress. This would be where the stress bleeds of, in three sections, as if it were 'a circular triangle.'
     
  11. Brett Nortje

    Brett Nortje Well-Known Member

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    The Fredholm Integral equation is about finding constants in the engine to guide the engineer to find the stress values that the engine can withstand and operate at a constant speed under acceleration and top speed.

    [G / k] / [-f^w] would be the answer, as, that is where all the ones are crossed out. The people that put this theorem together had a lot of constants that they thought should be included, but, they all cancel each other out.
     
  12. Brett Nortje

    Brett Nortje Well-Known Member

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    The Hill differential equation is used to find the differences in the degrees of stress that the engine can handle over various distances.

    [​IMG]

    So, it would be [d4t] + 3 * n * [2nt] + [2m] + [2mt]. Then, we can simplify this by saying everything besides [d4t] + 0, as all that bracket junk is times by zero, the answer is [distance] by [four time].

    The answer is distance divided by four!
     
  13. Brett Nortje

    Brett Nortje Well-Known Member

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    The Ishimori equation is about finding the speed things may turn at without losing power, or, becoming too stressed. This is like where a piston will turn on a train's wheels, and, need to be kept in line with stress bearing materials - the materials need to be strong enough to keep the parts in the vortex, or, cylinder, or area of operation.

    [​IMG] [​IMG]

    As we can see, it comes in two halves, separated by the coma and [1] notes.

    If we were to be totally honest, it is just a lot of multiplication and division, yes? Let's take it on!

    [S / t] = [4s 4d 2x 2y] = [4sd + 2xy] + [duxys] = [4s 14d 2u 4x 4y],

    [2d - 2d =] 1d 1u 4x - [a2] = 8a * [s * x ^ y],

    This means, of course, that we will be taking Speed divided by time times by two, or so, with the 'graph score' being multiplied by itself. In short, we need to take the speed divided by time times two, so it is an average - you know you add two things together and then divide them by two, giving you an average? - then to the power of one. Putting it to the power of one will be where we find the two dimensions of it, but, if we want three dimensions, we merely put it to the power of three, of course, giving direct three dimensional measurements and values for 'the vortex.'
     

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