My Math Forum Publishers

July 17th, 2017, 04:53 PM   #11
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 Originally Posted by Micrm@ss This is not accurate at all. It should give the same value as the standard methods. You might say the error is small, but in a large program, such small errors will compound quickly. So if your method doesn't give the correct answer, what is its benefit? Is it faster? More stable?
It has an f(x) representation. That being the integral is in the form of a function, it is not a code or a series of sums. It is literally like what you do when to take the integral of x, which spits out (x^2)/2 +c. My method modifies this for power functions.

July 17th, 2017, 04:55 PM   #12
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 Originally Posted by Tkipp It has an f(x) representation. That being the integral is in the form of a function, it is not a code or a series of sums.
That is interesting! So what would the representation of $\int e^{x^2}dx$ be?

July 17th, 2017, 05:02 PM   #13
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 Originally Posted by Micrm@ss That is interesting! So what would the representation of $\int e^{x^2}dx$ be?
I would like to show you but I would like to get it published somewhere first. Maybe an reactional math journal or something. For the moment I just want to share the data runs. For example, that previous table which compared my function against the true values produced by Matlab. Once, again, it is fairly close, although I am still missing something in the divisor.

With that being said, I will say this, the method I came up with does not produce the integral of exp(x^2) instead it is the integral of a function that acts very closely to exp(x^2) that can be integrated. The reason, in my paper, why I call the Equivalent Compound Functions or Mirrors.

Last edited by Tkipp; July 17th, 2017 at 05:05 PM.

July 17th, 2017, 05:07 PM   #14
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 Originally Posted by Tkipp I would like to show but I would like to get it published somewhere first. Maybe an reactional math journal or something. For the moment I just want to share the data runs. For example, that previous table which compared my function against the true values produced by Matlab. Once, again, it is fairly close, although I am still missing something in the divisor. With that being said, I will say this, the method I came up with does not produce the integral of exp(x^2) instead it is the integral of a function that acts very closely to exp(x^2) that can be integrated. The reason, in my paper, why I call the Equivalent Compound Functions or Mirrors.
Ok, so it doesn't give an exact form of $\int e^{x^2}dx$. It also produces values that are not the same as the standard software values. So again, why is your method any good? Is it faster than standard software? Is it stable? Is it more accurate? These will be the main question you will be asked when you publish your method, and it doesn't seem that you know the answer (yet).

July 17th, 2017, 05:56 PM   #15
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 Originally Posted by Micrm@ss Ok, so it doesn't give an exact form of $\int e^{x^2}dx$. It also produces values that are not the same as the standard software values. So again, why is your method any good? Is it faster than standard software? Is it stable? Is it more accurate? These will be the main question you will be asked when you publish your method, and it doesn't seem that you know the answer (yet).
Nearly exact, the closest I have seen for an integral function, and it can be closer than what I have, shown here it just takes some fine tuning. (I probably can make it perfect given the right resources and time.) The big thing is it produces a f(x) function that represents the integral of a power function, and how close it can be depends on the divisor. The produced functions do not touch/use imaginary values, or anything complex and stay within the simple function range.

Not to mention the derivative of my modified integral is very close to the real function as well. All in all, for you to produce a graph now in days of a power function integral you need to wait for the computer program to produce it point by point. This method creates a simple f(x) function that you type in, and produces an near identical graph.

July 17th, 2017, 06:00 PM   #16
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 Originally Posted by Tkipp Nearly exact, the closest I have seen for an integral function, and it can be closer than what I have, shown here it just takes some fine tuning. (I probably can make it perfect given the right resources and time.) The big thing is it produces a f(x) function that represents the integral of a power function, and how close it can be depends on the divisor. The produced functions do not touch/use imaginary values, or anything complex and stay within the simple function range. Not to mention the derivative of my modified integral is very close to the real function as well. All in all, for you to produce a graph now in days of a power function integral you need to wait for the computer program to produce it point by point. This method creates a simple f(x) function that you type in, and produces an near identical graph.
Sure, you seem very convinced. Time to convince a journal of your discovery then. There's nothing more I can say or do for you if you refuse to give specifics of your discovery. I can't even recommend journals since I'd need to know the specifics. All I can say is that you should look at journals which do things similarly to what you did. You did read journal papers right? If only to check your method is new. If so, then just publish in those journals which had similar papers as yours.

 July 17th, 2017, 06:13 PM #17 Senior Member   Joined: Oct 2009 Posts: 841 Thanks: 323 By the way, your approximation to $\int e^{x^2}dx$ had an error of about 0.01, which you claimed to be the best you've seen. Let me already show you a better one: $$\int e^{x^2}dx = \text{sgn}(x) \sqrt{1 - \text{exp}\left(-x^2 \frac{\frac{4}{\pi} + 0.147x^2}{1+0.147x^2}\right)}$$ Accurate to 0.00012. There are even similar functions known accurate to 7 decimal places.
July 18th, 2017, 02:08 PM   #18
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 Originally Posted by Micrm@ss By the way, your approximation to $\int e^{x^2}dx$ had an error of about 0.01, which you claimed to be the best you've seen. Let me already show you a better one: $$\int e^{x^2}dx = \text{sgn}(x) \sqrt{1 - \text{exp}\left(-x^2 \frac{\frac{4}{\pi} + 0.147x^2}{1+0.147x^2}\right)}$$ Accurate to 0.00012. There are even similar functions known accurate to 7 decimal places.
Ohh, you are so right....

One question, though, when did the integral of exp(x^2) from 0 to 5 become 1 (that's what your function produces)? It's been 7.35E9 for a really long time, forever I think. Also, at x=0.1 the integral of exp(x^2) is 0.1003; your great function produces .11246 which is 12% off of the real.

Also, don't bother replying if you are going to give me examples like that.

Last edited by skipjack; July 18th, 2017 at 11:40 PM.

July 18th, 2017, 02:34 PM   #19
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 Originally Posted by Tkipp Ohh, you are so right.... One question, though, when did the integral of exp(x^2) from 0 to 5 become 1 (that's what your function produces)? It's been 7.35E9 for a really long time, forever I think. Also, at x=0.1 the integral of exp(x^2) is 0.1003; your great function produces .11246, which is 12% off of the real. Also, don't bother replying if you are going to give me examples like that.

Correct, my function is describing the error function, which is defined as
$$\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt$$
So it's a trivial modification. My apologies. Not sure if the hostile tone was necessary though.

Last edited by skipjack; July 18th, 2017 at 11:40 PM.

July 18th, 2017, 03:05 PM   #20
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 Originally Posted by Micrm@ss Correct, my function is describing the error function, which is defined as $$\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt$$ So it's a trivial modification. My apologies. Not sure if the hostile tone was necessary though.
You're still wrong.
Using complex functions, the integral of exp(x^2) = ((1/2)sqrt(pi))erfi(x)). Your two functions only describe the error function, not the imaginary error function, nor the other components needed to create the complex integral of exp(x^2). I would recommend you look through the internet more before you spit out random things that you find out there.

I am sure you will try to say something else to cover up that you were wrong earlier.

Last edited by skipjack; July 18th, 2017 at 11:29 PM.

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