SFNQuestions Posted April 17, 2017 Share Posted April 17, 2017 (edited) If I have a given function that is defined as an integral, like of something that's clearly hard to work with like the gamma function or the exponential integral or the cosine integral or the error function or etc, is there a way to define an inverse of that function in a similar form that's NOT a cop-out dy integral? Not some dy integral that forces me to switch to integrating along the y-axis for no reason that solves absolutely nothing and never will, but something that starts as f(x) = integral(g(x))dx and hopefully takes the form of f-1(x) = integral g^-1(x)dx or something similar, something that is still defined on the domain of x. In fact, you know what, let's take an example: arctangent. The arctangent function can be described as the integral of 1/sqrt(1+x^2) which after complex analysis or trig substitution we know actually takes the form of a complicated complex-valued logarithm containing a polynomial. But now, let's say I wanted to define the actual regular tangent function just starting out with the integral definition of the arctangent function (which for verification purposes we know takes the form of a sum/quotient of real/imaginary exponents)... Edited April 17, 2017 by SFNQuestions Link to comment Share on other sites More sharing options...

fiveworlds Posted April 17, 2017 Share Posted April 17, 2017 Not some dy integral that forces me to switch to integrating along the y-axis for no reason that solves absolutely nothing and never will, but something that starts as f(x) = integral(g(x))dx and hopefully takes the form of f-1(x) = integral g^-1(x)dx or something similar, something that is still defined on the domain of x. We define x as an array of values under which we evaluate the function. Then we reverse the array. Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 17, 2017 Author Share Posted April 17, 2017 We define x as an array of values under which we evaluate the function. Then we reverse the array. This does not help anyone. The correct response would be to start looking at the inverse function theorem which definitely has something to do with the problem at hand. -2 Link to comment Share on other sites More sharing options...

fiveworlds Posted April 19, 2017 Share Posted April 19, 2017 I'm not here to teach you why one-way functions don't have an inverse. The only way to reverse a one-way function back to the state it was in originally is to remember what state it was in originally. For example the undo button uses arrays. There is no inverse for x^0 etc. Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 19, 2017 Author Share Posted April 19, 2017 (edited) I'm not here to teach you why one-way functions don't have an inverse. The only way to reverse a one-way function back to the state it was in originally is to remember what state it was in originally. For example the undo button uses arrays. There is no inverse for x^0 etc. Not only is that a useless tautology which once again illustrates my point, but most known integral functions absolutely have an inverse over a large domain even if it's only known by it's taylor series, so now what you're saying is also completely irrelevant, it's as if you're just here to troll. If you're not going to take the time to answer the question accurately, don't waste the world's time with your post. Edited April 19, 2017 by SFNQuestions -1 Link to comment Share on other sites More sharing options...

John Cuthber Posted April 19, 2017 Share Posted April 19, 2017 ... it's as if you're just here to troll. If you're not going to take the time to answer the question accurately, don't waste the world's time with your post. The correct answer to the question of the OP is either yes, no or sometimes. None of your posts has included any actual answer to the problem- just pointless grumbling about fiveworld's incomplete answer. You might want to read the part of your post which I have quoted above, and then look in the mirror. Link to comment Share on other sites More sharing options...

imatfaal Posted April 19, 2017 Share Posted April 19, 2017 ! Moderator Note post from the OP which was entirely off-topic hidden. Please back to the mathematics. Link to comment Share on other sites More sharing options...

swansont Posted April 19, 2017 Share Posted April 19, 2017 ! Moderator Note Other posts hidden, too, as they were also OT Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 19, 2017 Author Share Posted April 19, 2017 (edited) The correct answer to the question of the OP is either yes, no or sometimes. Incorrect. In math an science, the answer isn't some arbitrary whim, an answer needs to include evidence and reasoning or it isn't credible and therefore is unusable. None of your posts has included any actual answer to the problem- just pointless grumbling about fiveworld's incomplete answer. You might want to read the part of your post which I have quoted above, and then look in the mirror. Off topic from mathematics, and I will continue to report that content until it is censored as per the rules of the site. Edited April 19, 2017 by SFNQuestions Link to comment Share on other sites More sharing options...

swansont Posted April 19, 2017 Share Posted April 19, 2017 Incorrect. In math an science, the answer isn't some arbitrary whim, an answer needs to include evidence and reasoning or it isn't credible and therefore is unusable. That's simply not true. When the question is broad enough, the answer can be "sometimes". Even given that "most known integral functions absolutely have an inverse over a large domain", that does not say that all of them do. So it would seem that "sometimes" is a perfectly reasonable, and correct, answer. (There are a number of examples in science, as well, where that would be an appropriate response) Link to comment Share on other sites More sharing options...

John Cuthber Posted April 19, 2017 Share Posted April 19, 2017 Incorrect. In math an science, the answer isn't some arbitrary whim, an answer needs to include evidence and reasoning or it isn't credible and therefore is unusable. Off topic from mathematics, and I will continue to report that content until it is censored as per the rules of the site. I didn't say it was an arbitrary whim. You asked about "something that's clearly hard to work with like the gamma function or the exponential integral or the cosine integral or the error function or etc, " Imagine that someone could provide an explicit inverse for the error function, but could prove that no such function exists for the Gamma function. In that case the answer to your question "Is there a way to invert a function defined by an integral?" would be "sometimes"; there is sometimes a way to invert such a function. If you are unhappy with the idea that maths sometimes says " It's impossible to tell" then I have bad news for you https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems You are correct- the other things you said - including the comments to fiveworlds- were off topic. Feel free to report yourself. Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 19, 2017 Author Share Posted April 19, 2017 (edited) That's simply not true. It is true unless it is completely common knowledge that it's already been proven. When the question is broad enough, Nope, when the answer is well-known enough, that's the rule, not broad enough. If you ask a broad question like "what's the inverse of a second degree polynomial?" "sometimes where x=0" isn't going to suffice, you need the quadratic formula at the very, most basic east. I didn't say it was an arbitrary whim. And you don't need to, it's very clear that it is. You asked about "something that's clearly hard to work with like the gamma function or the exponential integral or the cosine integral or the error function or etc, " I also specifically asked about a "function that is defined as an integral" and specifically gave a specific example to work through, "In fact, you know what, let's take an example: arctangent. The arctangent function can be described as the integral of 1/sqrt(1+x^2) which after complex analysis or trig substitution we know actually takes the form of a complicated complex-valued logarithm containing a polynomial. But now, let's say I wanted to define the actual regular tangent function just starting out with the integral definition of the arctangent function (which for verification purposes we know takes the form of a sum/quotient of real/imaginary exponents)..." would be "sometimes"; there is sometimes a way to invert such a function. If it's a function that is continuous over some interval and you isolate a branch point, which is the only time you could ask for an inverse, the answer is never ever "sometimes" or "no." If you are unhappy with the idea that maths sometimes says " It's impossible to tell" then I have bad news for you Good thing it's not impossible to tell then. Even the gamma function with 100% certainty has an inverse over any branch point you choose, at the very least over the interval that it is monotonically increasing, it is simply the basic, common, average case that the inverse it is not currently defined in terms of elementary functions or possibly just known techniques, outside of a taylor series. You are correct- the other things you said - including the comments to fiveworlds- were off topic. This comment of yours is off topic. Edited April 19, 2017 by SFNQuestions -1 Link to comment Share on other sites More sharing options...

studiot Posted April 19, 2017 Share Posted April 19, 2017 Your original question is too general to supply a specific answer and there is no universal answer available. So here are some possible lines of attack. Most functions defined by integrals have alternative definitions which may be more easily invertible. Every integral has a counterpart differential equation so using that may be fruitful in some cases. But there are pitfalls to inversion for instance the sine function maps the entire number line to the interval (-1, 1). So there will be no inverse at all for numbers outside this range. Bessel functions cannot be expressed in terms of simpler functions so must be inverted from tables or some other numeric procedure. Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 19, 2017 Author Share Posted April 19, 2017 (edited) Your original question is too general to supply a specific answer and there is no universal answer available. I garuntee you there is a technique that you can attempt to apply to the situation I provided that can also work in similar situations. If I choose to apply the technique to some random problem and I find the result is not expressible in currently defined functions, that's my problem, not yours. Most functions defined by integrals have alternative definitions which may be more easily invertible. From what I have seen, it is exceptionally rare to get an integral definition that somehow turns out to be nice and easy to work with in some alternate definition. Bessel functions cannot be expressed in terms of simpler functions so must be inverted from tables or some other numeric procedure. In terms of functions you know of, or, have cared to define. What the question and example asks for the technique to show how it's done to verify a result, or in other words, any amount of evidence to verify any of the "answers" thus far. Edited April 19, 2017 by SFNQuestions Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted April 20, 2017 Share Posted April 20, 2017 ! Moderator Note Here's a cool idea: if you think something is off-topic, report it and let us handle it. Link to comment Share on other sites More sharing options...

swansont Posted April 20, 2017 Share Posted April 20, 2017 It is true unless it is completely common knowledge that it's already been proven. Nope, when the answer is well-known enough, that's the rule, not broad enough. If you ask a broad question like "what's the inverse of a second degree polynomial?" "sometimes where x=0" isn't going to suffice, you need the quadratic formula at the very, most basic east. But you asked an even broader question. If I asked "is momentum conserved?" a yes or no answer will be wrong. Because momentum conservation depends on the condition of whether or not there is a net external force. You asked about an arbitrary function and did not restrict it to polynomials. The answer is going to be "it depends on the function" Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 20, 2017 Author Share Posted April 20, 2017 (edited) But you asked an even broader question. If I asked "is momentum conserved?" a yes or no answer will be wrong. Because momentum conservation depends on the condition of whether or not there is a net external force. You asked about an arbitrary function and did not restrict it to polynomials. The answer is going to be "it depends on the function" No the answer to momentum will depend on a vector analysis of the situation and whether the system is opened or closed. Like I said, every function can be inverted to some degree if the right domain is chosen, whether or not it happens over a particular domain isn't your problem, the topic you derailed is in reference to an example where we already know what the outcome should be. Edited April 20, 2017 by SFNQuestions Link to comment Share on other sites More sharing options...

John Cuthber Posted April 20, 2017 Share Posted April 20, 2017 (edited) Just for kicks and giggles, I'm going to define a function f ( r ) such that f ( r ) is the area of a circle with radius r. And, I'm going to define it in terms of an integralThe details are here http://www.analyzemath.com/calculus/Integrals/area_circle.html (other such sites are also available) It turns out that f ( r ) = pi r^2 Now, can I find an inverse for this function which is defined in terms of an integral (of course it's an integral- its the area of a curve)Well, yes I more or less can ( if we decide that either r in always >0 or that a negative radius is permitted as an alternative to a positive one) I can calculate the radius ( r ) of a circle, given the area (a) r=(a/pi) ^0.5 So, in this (trivial) case I can find an inverse fro a function that is defined in terms of an integral. (It's entirely possible that someone better at maths than I cam could find more exciting examples but this one will do) OK, That was easy, so I will now try something a bit more complex. I will define a function, g, of two variables x and y such that g(x,y) = x+y,x+y Sorry for my clumsy explanation- what I mean is the equivalent of multiplication by the matrix 1,1 1,1 That matrix has the not very rare property of having no inverse. Now imagine that I define a further function h(x,y) in such a way that it depends on an integral of g(x,y). It doesn't have to be anything clever. The integral of g(x,y) dy,dx over some range will do nicely. Now I contend that since the calculation of the intermediate i.e. g(x,y) is not invertible, the integral of that function also isn't invertible. So, in this case it is not possible to invert a function defined in terms of an integral. So, the answer to the question "Is there a way to invert a function defined by an integral?" is Sometimes You can if it's f(x), but you can't if it's h(x,y) Incidentally, the choice of a function in two variables makes Fiveworld's solution a lot more interesting- it would need a 2 D array, but not all 2 d arrays have an inverse. If you had thought about what he said- rather than discounting it rudely, you would have got to at least part of the answer. (It's possible that someone is going to point out that I'm playing fast and loose with the definition of "function" here by sticking arrays into it. OK, fair cop, but the OP started it by introducing integrals) Edited April 20, 2017 by John Cuthber Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 20, 2017 Author Share Posted April 20, 2017 (edited) Just for kicks and giggles, I'm going to define a function f ( r ) such that f ( r ) is the area of a circle with radius r. And, I'm going to define it in terms of an integral The details are here http://www.analyzemath.com/calculus/Integrals/area_circle.html (other such sites are also available) It turns out that f ( r ) = pi r^2 Now, can I find an inverse for this function which is defined in terms of an integral (of course it's an integral- its the area of a curve) Well, yes I more or less can ( if we decide that either r in always >0 or that a negative radius is permitted as an alternative to a positive one) I can calculate the radius ( r ) of a circle, given the area (a) r=(a/pi) ^0.5 So, in this (trivial) case I can find an inverse fro a function that is defined in terms of an integral. (It's entirely possible that someone better at maths than I cam could find more exciting examples but this one will do) OK, That was easy, so I will now try something a bit more complex. I will define a function, g, of two variables x and y such that g(x,y) = x+y,x+y Sorry for my clumsy explanation- what I mean is the equivalent of multiplication by the matrix 1,1 1,1 That matrix has the not very rare property of having no inverse. Now imagine that I define a further function h(x,y) in such a way that it depends on an integral of g(x,y). It doesn't have to be anything clever. The integral of g(x,y) dy,dx over some range will do nicely. Now I contend that since the calculation of the intermediate i.e. g(x,y) is not invertible, the integral of that function also isn't invertible. So, in this case it is not possible to invert a function defined in terms of an integral. So, the answer to the question "Is there a way to invert a function defined by an integral?" is Sometimes You can if it's f(x), but you can't if it's h(x,y) Incidentally, the choice of a function in two variables makes Fiveworld's solution a lot more interesting- it would need a 2 D array, but not all 2 d arrays have an inverse. If you had thought about what he said- rather than discounting it rudely, you would have got to at least part of the answer. (It's possible that someone is going to point out that I'm playing fast and loose with the definition of "function" here by sticking arrays into it. OK, fair cop, but the OP started it by introducing integrals) Your failure to find in inverse comes purely from your choice to refrain from defining a branch, something that allows any function to be invertible and isn't limited to functional analysis but is applicable to linear algebra as well. Not only that, but this is not the example I provided. Anything that you can possibly identify in any way as any kind of continuous curve or line is invertible over that specific segment, it is only a matter of whether or not you apply the proper techniques to find it in terms of identified functions. In your own limited capabilities, y=x^2 shouldn't actually have an inverse, and yet every actual accredited mathematician in the world agrees that http://www.wolframalpha.com/input/?i=inverse+y%3Dx%5E2 And when it doesn't know the answer, it doesn't say "never going to be an answer, no one will ever know just give up on math" it says "no results in terms of *standard* mathematical functions" because accredited mathematicians know that a function is invertible if you apply the technique of choosing the proper domain and range, it is only the case that we may lack the techniques to explicitly identify it. Edited April 20, 2017 by SFNQuestions Link to comment Share on other sites More sharing options...

John Cuthber Posted April 20, 2017 Share Posted April 20, 2017 Your failure to find in inverse comes purely from your choice to refrain from defining a branch, something that allows any function to be invertible. Not only that, but this is not the example I provided. Feel free to define one for me. The function has no inverse. If you want a simpler example, try k(x) =integral of ax .dx find the inverse where a=0 I know it's not one of the examples you specified. But you have edited your original post to change the goalposts so it's now impossible to tell. Why did you do that? Do you realise it looks like intellectual dishonesty? Just to clarify, what you said was not very specific "something that's clearly hard to work with like the gamma function or the exponential integral or the cosine integral or the error function or etc, " And the example I chose was hard to work with- impossibly so. 1 Link to comment Share on other sites More sharing options...

studiot Posted April 20, 2017 Share Posted April 20, 2017 +1 John for showing more patience than I have to spare. Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 21, 2017 Author Share Posted April 21, 2017 (edited) Feel free to define one for me. The function has no inverse. If you want a simpler example, try k(x) =integral of ax .dx find the inverse where a=0 The x+y function that appears to just be a plane? Do you think that's not invertible in any way? When a=0, the integral gives you a constant line which when inverted gives you simply a vertical line, and to make it functional it only needs to be equivalent to a translated Dirac delta function or limit the range to a finite number. Literally anything that you can recognize as any kind of curve or shape can be flipped over the line y=x, that's really all you need to ask yourself, the rest is simply a matter of domain/range restrictions and whether or not it's in terms of known functions. Matrices are their own group, it has it's own rules. The definition for a bisection or an inversion or invertible does not necessarily have all the same applications and versatility as in functional analysis. Is there any particular reason you continue to completely evade the specific example I provided wherein the end result is already known but it's complex enough to illustrate the application? Honestly this site is so behind, stackexchange and physicsforums answer things so much quicker and with more detail compared to this site and without the cavalier attitude of the academically disappointing precedent set by the staff. I know it's not one of the examples you specified. But you have edited your original post to change the goalposts so it's now impossible to tell. Why did you do that? Do you realise it looks like intellectual dishonesty? As an allegedly accredited staff member you should be able to view the edit history, and if you did you'd see I added the example only shortly after the post in order to give context and a more identifiable goal. Do you realize this looks like abuse and thus a lack of integrity? +1 John for showing more patience than I have to spare. Uh, no one's forcing you to be here, as a sentient being, your frustration is your own choice. If you don't have the expertise or experience for making progress with this particular technique, go ahead and do something else. Edited April 21, 2017 by SFNQuestions -2 Link to comment Share on other sites More sharing options...

John Cuthber Posted April 21, 2017 Share Posted April 21, 2017 (edited) As an allegedly accredited staff member you should be able to view the edit history, and if you did you'd see I added the example only shortly after the post in order to give context and a more identifiable goal. Do you realize this looks like abuse and thus a lack of integrity? Just plain wrong, though I take your point about the timing of the edit.. I'm not a staff member. Any point on the plain gets mapped to a line. or, with the integral every point gets mapped to the same number. Since it's an infinity to one mapping, it's not invertible. If I say (k(x)= 0 can you tell me what x is? Edited April 21, 2017 by John Cuthber Link to comment Share on other sites More sharing options...

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