MWresearch

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I don't understand your concern  there are problems for the general case. However for this case [latex]x^2\ge 0[/latex] for all real x, so the expression ([latex]ln(x^2)=2ln(x)[/latex]) holds, because if x is negative, that information is lost after squaring.
Except you're not considering the arbitrary coefficient "a" and still completely ignoring that I keep referring to possible positive and negative values in general, not solely x^2. What if I had [math]x^{2}[/math]? Then all values would be negative. There should be one general equation that explains all outputs of log(x) that tells me what the complex numbers are for negative arguments along with their continuum of 2*pi*n solutions resulting from it's trigonometric relation through Euler's identity as well as simultaneously giving me only real output values when the argument is positive, no constraints like x>0 or x<0 for any argument. Basically, an extension of the log(x) function which by hand one could derive the correct output values all real values of x, not only positive values and not only negative values, all values.
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[latex]ln(x^2)=2ln(x) [/latex] for real x.
And that explains the complex result of a negative argument in addition to the real values for all positive and negative values of the argument in one formula how?????
I don't know how many times I have to explain it but I'll do it until you get it. The formula shouldn't restrict anything to only positive values which abs does. Somewhere out there, there's a single extended form of log(x) that explains not only what's happening when you move the exponent but what the complex constant is added for negative values in the formula that simultaneously accounts for both the positive values of the argument yielding only real numbers while the negative values of the argument yield complex numbers and that's what this is all about. Every logarithmic rule that people have written so far only works for either only positive inputs or only negative inputs, but, it should be both for the continuum of results.
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A negative value is easily dealt with as I think you might have been getting at in the OP
[latex]
ln (x) = ln (1 \cdot x) = ln (e^{\pi \cdot i} \cdot x) = ln(e^{\pi \cdot i}) + ln(x) = \pi \cdot i + ln(x)
[/latex]
I tried explaining it and instead of taking my advice on what I'm asking about you decided to act inappropriate and then on top of it you still create the exact same issue but for negative numbers. No separation, there should be one formula for a set of all real numbers if not all complex numbers z, a formula where not only can both positive and negative numbers correspond to ln(x) and ln(x) but also give the +2pi*i*n for all other possible solutions that result from log(x)'s trigonometric relation.
3 
Exactly what the title says. Not an easy function to work with, for some reason it doesn't have a known indefinite integral even though it seems like it almost should. The answer seems to have something to do with the polygamma function though the exact process of differentiating the gamma function seems nonexistent on the internet.
0 
[latex]ln \left( \frac{a \cdot x^j}{b \cdot y^k} \right) = ln(a)  ln(b) + j \cdot ln(x) k \cdot ln(y)[/latex]
The above variables are positive.
Ok so you keep saying x>0, but that's exactly what my issue is. I don't care about those basic rules for positive real numbers, the problem is what's going on that makes the result such that the above statements aren't true for negative numbers. Clearly, factoring out exponents for positive numbers is just "pattern recognition," not the whole story.
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That doesn't really explain anything, that's just a reiteration of basic basic basic logarithm rules that most people here learned years and years and years ago. What I'm talking about is, what the hell is going on when that happens? And that has to at least be related to complex analysis and Euler's Identity because of the constant left over in the negative portion of the domain of the argument of log(a*x). I'm looking for a formula that works for all arguments of log(x).
2 
Just what the title says. Say for instance I have [math]ln(x^2/3)[/math]. I want to be able to turn that into [math]2ln(x...[/math]But, I'm not sure exactly what's happening to a number or what the exact operation is when you take an exponent out of the argument of [math]log(x)[/math]. So, I'm not sure exactly what the result is if I assume [math]ln(x^2/3)=2ln(x/sqrt(3))[/math] because every time I try to do something like that, I get some complex constant left over. It "seems" like when you move exponents out of the argument of a logarithm, you're exponentiation everything in the argument by 1/exponentyouwanttogetridof and for some random reason that exponent then gets moved to the outside of the log operator, but then again, there's always a constant left over when I do that. I guess I could limit it to [math]x*a>0[/math] for [math]log(x*a)[/math] but I'd prefer to have the whole picture of whatever that constant comes from like [math]log(e^(ip*a)))[/math] or whatever.
Btw the exponent command for the [math tags are completely broken and useless.
0 
I'm not solving a differential equation with an unknown function but it seems it should be similar. What I mean is I want to solve for an unknown operator that has some arbitrary transformation of terms and variables. If for instance I had x^5 and I wanted to find some operator that made x^5 turn into x^4, I should get ^4/5. But let's say I didn't know that property of exponents as a given, how would I mathematically investigate the apparent transformation of variables to find that operation that has the properties of turning x^5 into x^4?
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The answer, which, I'm surprised no one else came up with since there's supposedly people here who know a lot of math is that it doesn't disappear, there are complex and imaginary solutions of cos(x) at multiple points not visible in the real plane and it was simply not showing them since cos(x) generally only resides in the real plane.
2 
Yes. But again where in actual practice would the alternator change direction do you think? Because, again, I know of no application in practice where that happens. And again, there is no application where you can't avoid it mechanically. Please answer this so I don't keep asking.
You seem pretty intent on your erroneous assumptions, so why would I bother explaining it? You already think you know the answer, obviously nothing I say to you is going to make the discussion go in a positive direction. If you want to know the answer just use your imagination, there's infinite possibilities for physical circumstance, there's all sorts of actions that aren't perfectly unidirectional or that one does not want to restrict to one direction. You're basically just saying "I could be wrong, but I assume I can't possibly be wrong," no point arguing with someone like that.
I am not too much experienced with alternators, but in normal electric motors, the faster spinning, the higher voltage.
So if it's exactly the same with alternators, which is what I suppose so.
Spinning too fast (if there is nothing like regulator), could cause damage of electronics connected directly to it, in certain extreme situations.(Google shows plentiful of pages after searching for "alternator producing too much voltage", so it's real issue).
In such situations, voltage threshold triggered branch with zener diodes could be used, f.e.
Place fuse in horizontal branch instead of resistor.
Zener diodes f.e. 15 V.
If voltage is exceeding 15 V, current will flow through zener branch, instead of normal branch, and fuse will blow up, disconnecting entire circuit.
Such (or other) security elements should be used in circuits, regardless whether current/voltage regulators are present.
They might fail, and you need to be prepared for it in advance.
Single zener diode costs here $0.05.
Well, I am dealing somewhat with lower power, but on the other hand there is a chance the rotation could spike up in frequency and maintain that frequency in a certain sense, even when alternating in different directions. Why is it better to use a fuse that could blow up instead of a charge regulator? Just more cost effective?
2 
Where's the damn off button on you? How is it possible for your existence to be that big of a of time and resources? I don't give a crap what you think, if you don't have answers, no one cares, go ruin another website and let the legitimate grown up scientists do the discussing.
If you don't want to answer questions, don't waste everyone's time, your incessant trolling is exactly what loses credibility for this site.
2 
This is all about curve fitting. There are methods of deciding how good various functions fit some data. I am not really up on these methods.
It's not that simple though because as I pointed out, more than one curve can perfectly fit data points and yield a 100% correlation. There has to be some way to assume certain parameters, such as that it can't be a polynomial or cos(x) when there was no previous sign of those operators.
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https://en.wikipedia.org/wiki/Inverse_function
@OP: I read what you wrote, based on the fact that you didn't understand why "the domain restricted to [0, pi]" it led me to conclude you weren't aware that you may have asked Wolfram Alpha for something other than what you thought you did.
Honestly, you need to lose the attitude. People help those who are pleasant to deal with. This forum is for the discussion of Science, my dredging up memories of something I ran into over a decade ago is an optional extra.
If you still want the why the imaginary parts didn't appear then the answer is on the Wiki. I thought my response summed it up fairly succinctly but Wiki certainly has the answer if you feel the need to investigate further.
When in the hell did I state I didn't understand why the domain was restricted? Obviously you didn't read it. I merely pointed it out as something to consider, nowhere did I ask a question about the domain being restricted.
Not only that but the words "complex" and "imaginary" didn't occur a single time in your entire "solution to everything" article, but being pleasant obviously solved nothing because the only way to get any god damn attention around here is to shakes things up, otherwise my question would have been answered ages ago. No question answered? No reason to be pleasant. If it were different, that would be nice, but since the staff refused to answer the question I'm going to get someone else to or get enough attention until one of them does. Unfortunately for you, that person definitely isn't you. If there was actually a vertical line test failure in the domain [0,1] from taking the inverse of arccos(x) then obviously no function would be displayed at all in any way, it wouldn't force only the real component of the complex function to be displayed unless there was a complex function to begin with. But as it clearly shows with arccos(x) there are instances where the horizontal AND vertical line test creates complex values which in a normal circumstance should still exist when reflecting over the x axis. Not only that, but cos(z) in wolfram isn't even restricted by Re(z) when displaying cos(z), which means it's not an issue of software refusing to show those complex values, they simply disappear.
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I gave you it. For your poor attitude, which pervades in much of your questions. Nobody owes you anything and not everyone sees a question as the OP intended. Cest la vie.
Obviously I don't give a crap what you think, I care about answers. If you have the audacity to call yourself any remote authority on any science or math, you better have the damn capacity to answer questions. Doesn't matter about anyone owing anything, it matters if this site can actually do what it set out to do.
Furthermore, the fact that you care about something as irrelevant as points is disturbing, this site needs to redefine who it calls "senior."
Here's how it works: people like me ask questions and legitimate scientists answer them because it makes them look good or feel validated. Put it on your resume, put it on a scholarship, I don't care, but that's the only reason for this site's existence.
4 
Is this what you wanted?
http://www.wolframalpha.com/input/?i=1%2F%28+arccos%28x%29%29
"Inverse of" would refer to swapping your range and domain around.
That's not at all what I was referring to, did you even read anything I said or just glance at the title?
I find it rude that I write all that and yet you can't give me the courtesy of reading it before posting.
LOOK at the wolfram alpha link. There's red lines. Those red lines are part of arccos(x). Those red lines can be represented more intuitively by the equation (i*ln(i*x+sqrt(1x^2)). Those red lines are some form of i*arccosh(i*x). Those red lines disappear when you take the inverse of arccos(x). Those red lines are not displayed when you plot the function cos(x). It is likely not a display error since obviously wolfram recognizes imaginary and complex values.
I never bothered to give anyone a 1 mark before but Jesus that's frustrating when I have to hold your hand to explain something you could easily understand yourself if you just read it.
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Is there a way to create special parameters? Like if I have the points (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) (7,7)...and there's no indication of a complicated polynomial anywhere, is some complicated polynomial really more likely than just y=x? Is there some way to rule out certain conditions or assume it's not a polynomial and work with that assumption?
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Wow I can't believe no one familiar with math is answered this, it's definitely not some millennium $1,000,000 problem, it has a real answer. Where do the imaginary hyperbolas go when I take the inverse of arccos(x) (that's right, ARCcos(x), NOT cos(x)).?
Right here http://www.wolframalpha.com/input/?i=arccos%28x%29
See? Past 1 and 1, the equation turns into something like i*arccosh(x)+pi. You can very very very very very very clearly see those red, imaginary logarithms or hyperbolas and see the clear blue arccos(x) on [1,1] that we're all familiar with. BUT, now look what happens when I take the inverse of arccos(x) http://www.wolframalpha.com/input/?i=inverse+of+arccos%28x%29
Not only is the domain restricted to [0, pi] but where did those red imaginary logarithms/hyperbolas go???
2 
https://isc.carma.newcastle.edu.au/standard
Try that...
It is the closest I know to what you are looking for.
That's very close to what I was looking for, thanks.
And if anyone else has any other tools like that though feel free to suggest them.
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Nope  I used Firefox for all my dealings with SF.n either via windows 10 or Linux depending where I am. A bit of empirical research might not go amiss in some posts before claiming an institutional bias. We do have adverts to defray expenses  but the site usage is not exclusive to Google Chrome
To be honest I have every right to be frustrated, after all the complaints of glitches they still won't even fix the most basic problems. But, let's not stray too far off topic.
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Just what the title says. Say I have some random equation like 2^x, then I have some random equation that I tested and it gives me the points (1,2), (2,4), (3,8), (4,16), (5,32), (6,64), (7,128) but there's no proof it works for (8, 256) other than by testing it to see it works for x=8, if I assume the curve is continuous, is there any sort of weak proof that says it's the equation 2^x? Every taylor series should be unique, yet for any given equation that was derived purely from integers, I could use some kind of a*cos(bx)+c that generates a coefficient of "1" for any given integer. For instance, the equation y=2^x generates (1,2), (2,4), (3,8), (4,16), (5,32), (6,64), (7,128). However, the equation y=(2^x)*cos(2*pi*x) generates (1,2), (2,4), (3,8), (4,16), (5,32), (6,64), (7,128) as well, so how do I deal with this? So if I assume there's no trig involved, there's nothing else that could perfectly match another continuous operator at each regular interval but is completely inaccurate at interpolated x values right? Or no?? So is there some weak proof that works if I assume no trig functions are involved?
1 
I guess my first instinct is to wonder why you're searching for a tool like this. Basically, it looks like a tool to 'help' numerology people. "Like, ZOMG people, cos(tan(exp(J1(pi)))) shares 4 of the digits of my birthday!!!!1!!"
The reason no specific tool exists that you're looking for, basically, is that it isn't really all that interesting.
If you really want a tool like this, learn some programming like python or java or similar and make your own. I don't think it would be very hard to code up. You just need a language that can do some higher precision calculations + some string manipulations and you are good to go.
So, again, I am curious why you're looking for this. The results would be little more than curiosities to professionals, I would think.I don't have time to learn python so if it's actually as simple as you say it is, then there should be hundreds of tools like that. Doesn't matter if it's interesting, there's literally an app where all you do is roll toilet paper, I'm sure there's some mathematical help tool. In fact, there's an app that's literally just a red orb and yet people bought it, so being boring especially in math is no excuse.
There is the Inverse Symbolic Calculator;
http://isc.carma.newcastle.edu.au/standardCalcThat link said "404 not found." All I'm looking for is a tool where I punch in a number like 0.7071 and it says "sqrt(2)/2 is 0.7071067811865..." where I set certain parameters like "only use functions ^1/x, tan(x), cos(x), ^x, ln(x), only use numbers 110" or something like that.
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Not that you'd know but that's definitely not what I'm looking for, I' not looking for some little loophole to a perfect sawtooth wave, I'm looking for something that's completely different from a sawtooth wave but happens to coincidentally bear some geometric resemblance to it, I was only using sawtooth wave as an example of the general shape, I definitely wasn't looking for any sort of "exact" sawtooth wave. After doing some research, the only thing I can really say is that it has something to do with a relaxation oscillator or Van Der Pol Oscillator (wait only google chrome works for scienceforums???/scienceforums is sponsored by google (but wait doesn't that make it biased)????????????). Basically I want something that looks nearly like that relaxed van der pol but only in the form of y=f(x) and I don't know how to make it to grow steep first and then release gradually as opposed to how the waves look in the links which is that they gradually build up THEN release quickly/steeply.
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Well I'm here specifically because wolfram alpha doesn't remotely do what I need it to. What I'm trying to do is something like "test .535... for coefficients of e^(n/52)..." at which point it would give me a list of numbers that yield minimum of .535... in the number for different values of n in e^(52/n), likely expanding to .535939048103850580... or some other irrational number.
0 
I don't know what it's called but I'm trying to find a general formula a cosine function that's very similar to a regular cosine function except it's not symmetric about every local minimum, instead it's lopsided. It's like a sawtooth/ramp wave but it's not a triangle and it doesn't have vertical lines, it's just as if a sin(x) got squished on one side of every cycle. Like for any given cycle, it's really steep slope on one side, then after the maximu/minimum it's a really shallow and low slope until the next cycle.
Once again this site is horribly glitched and it's not even letting me post links anymore, so I guess the only picture I can relate it to is a chi distribution. Imagine a chi distribution that keeps repeating and that's basically the type of sine wave I'm trying to find a formula for.
0
How do you differentiate the incomplete gamma function?
in Analysis and Calculus
Posted · Edited by MWresearch
The first problem is d/dx*gamma(x)=gamma(x)*polygamma_0(x) and the polygamma function is already defined in terms of the original gamma function and the derivative of the gamma function, the logic is completely circular, all anyone is saying by writing that is that the derivitive of the gamma function is the derivative of the log of the gamma function divided by the gamma function, it explains nothing about what the actual polygamma function is and what combination of operators it performs on an input and thus there is no clear way to differentiate it. It's like saying that the integral of (2/sqrt(pi))e^(x^2) is the error function when the error function is only defined as the integral of (2/sqrt(pi))e^(x^2), again, the logic is circular and absolutely zero known functions are represented in closed form after taking the integral of the Gaussian curve. The polygamma function appears to be the exact same type of ambiguous problem except with derivatives instead of integrals. Why can't they just admit they don't know the derivative? I might as well say the derivative of the gamma function is spaghetti where spaghetti = the derivative of the gamma function, saying that solves nothing and shows nothing about taking the actual derivative of the function because the logic is simply circular, so if I don't know what the derivative is, how am I suppose to differentiate it?
And THEN holding the input of the gamma function fixed while varying the boundary is suppose to somehow be x^(t1)e^(x) except there's absolutely no known elementary function that you can use to evaluate the boundaries of the lower incomplete (or upper) gamma function at x and infinity to verify that result after integrating the original function, again, completely circular logic that solves nothing. If they don't know of any such function or combination of operators that can accurately describe the integral of the original function in closed form they should simply admit it and stop wasting everyone's time. But since they won't tell anyone and make up this circular logic crap, no outside viewer can have any way of knowing.