July 17th, 2017, 04:53 PM |
#11 | |

Newbie Joined: Jul 2017 From: US Posts: 12 Thanks: 0 | Quote:
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July 17th, 2017, 04:55 PM |
#12 |

Senior Member Joined: Oct 2009 Posts: 406 Thanks: 140 | |

July 17th, 2017, 05:02 PM |
#13 | |

Newbie Joined: Jul 2017 From: US Posts: 12 Thanks: 0 | Quote:
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.
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July 17th, 2017, 05:07 PM |
#14 | |

Senior Member Joined: Oct 2009 Posts: 406 Thanks: 140 | Quote:
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July 17th, 2017, 05:56 PM |
#15 | |

Newbie Joined: Jul 2017 From: US Posts: 12 Thanks: 0 | Quote:
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 | |

Senior Member Joined: Oct 2009 Posts: 406 Thanks: 140 | Quote:
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July 17th, 2017, 06:13 PM |
#17 |

Senior Member Joined: Oct 2009 Posts: 406 Thanks: 140 |
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 | |

Newbie Joined: Jul 2017 From: US Posts: 12 Thanks: 0 | Quote:
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.
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July 18th, 2017, 02:34 PM |
#19 | |

Senior Member Joined: Oct 2009 Posts: 406 Thanks: 140 | Quote:
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.
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July 18th, 2017, 03:05 PM |
#20 | |

Newbie Joined: Jul 2017 From: US Posts: 12 Thanks: 0 | Quote:
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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