find $f(a,b,c,d,e,f)$ definition based on input and output

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0
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I wonder if it will be possible to find an approximated function based on a simulation result.



I have a specific input and I would like it to produce a specific output. Is it possible to construct something like that and if yes so how?



  • $f(28.78, 15.38, 6.58, 1.94, 0.29, 0.02) = 47$

  • $f(22.05, 23.26, 20.20, 12.94, 4.06, 0.48) = 17$

  • $f(20.89, 23.11, 20.83, 14.04, 4.56, 0.57) = 16$

  • $f(13.88, 18.00, 22.72, 22.99, 10.54, 1.89) = 10$

  • $f(9.92, 13.72, 19.24, 28.67, 17.35, 4.07) = 7$

  • $f(2.99, 4.33, 6.86, 12.72, 41.03, 30.08) = 2$

  • $f(1.48, 2.20, 3.56, 6.69, 22.18, 62.88) = 1$

Those numbers were found in a simulation done here.




calculus functional-analysis functions




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  • 1




    6-dimensional function approximation based only on 7 points is going to be insanely inaccurate.
    – lisyarus
    Jul 28 at 13:41










  • @lisyarus I can produce more points. But I would like to understand the process, with an example so I can code it. This is a simulation result
    – Ilya Gazman
    Jul 28 at 13:44














up vote
0
down vote

favorite












I wonder if it will be possible to find an approximated function based on a simulation result.



I have a specific input and I would like it to produce a specific output. Is it possible to construct something like that and if yes so how?



  • $f(28.78, 15.38, 6.58, 1.94, 0.29, 0.02) = 47$

  • $f(22.05, 23.26, 20.20, 12.94, 4.06, 0.48) = 17$

  • $f(20.89, 23.11, 20.83, 14.04, 4.56, 0.57) = 16$

  • $f(13.88, 18.00, 22.72, 22.99, 10.54, 1.89) = 10$

  • $f(9.92, 13.72, 19.24, 28.67, 17.35, 4.07) = 7$

  • $f(2.99, 4.33, 6.86, 12.72, 41.03, 30.08) = 2$

  • $f(1.48, 2.20, 3.56, 6.69, 22.18, 62.88) = 1$

Those numbers were found in a simulation done here.




calculus functional-analysis functions




share|cite|improve this question















  • 1




    6-dimensional function approximation based only on 7 points is going to be insanely inaccurate.
    – lisyarus
    Jul 28 at 13:41










  • @lisyarus I can produce more points. But I would like to understand the process, with an example so I can code it. This is a simulation result
    – Ilya Gazman
    Jul 28 at 13:44












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I wonder if it will be possible to find an approximated function based on a simulation result.



I have a specific input and I would like it to produce a specific output. Is it possible to construct something like that and if yes so how?



  • $f(28.78, 15.38, 6.58, 1.94, 0.29, 0.02) = 47$

  • $f(22.05, 23.26, 20.20, 12.94, 4.06, 0.48) = 17$

  • $f(20.89, 23.11, 20.83, 14.04, 4.56, 0.57) = 16$

  • $f(13.88, 18.00, 22.72, 22.99, 10.54, 1.89) = 10$

  • $f(9.92, 13.72, 19.24, 28.67, 17.35, 4.07) = 7$

  • $f(2.99, 4.33, 6.86, 12.72, 41.03, 30.08) = 2$

  • $f(1.48, 2.20, 3.56, 6.69, 22.18, 62.88) = 1$

Those numbers were found in a simulation done here.




calculus functional-analysis functions




share|cite|improve this question











I wonder if it will be possible to find an approximated function based on a simulation result.



I have a specific input and I would like it to produce a specific output. Is it possible to construct something like that and if yes so how?



  • $f(28.78, 15.38, 6.58, 1.94, 0.29, 0.02) = 47$

  • $f(22.05, 23.26, 20.20, 12.94, 4.06, 0.48) = 17$

  • $f(20.89, 23.11, 20.83, 14.04, 4.56, 0.57) = 16$

  • $f(13.88, 18.00, 22.72, 22.99, 10.54, 1.89) = 10$

  • $f(9.92, 13.72, 19.24, 28.67, 17.35, 4.07) = 7$

  • $f(2.99, 4.33, 6.86, 12.72, 41.03, 30.08) = 2$

  • $f(1.48, 2.20, 3.56, 6.69, 22.18, 62.88) = 1$

Those numbers were found in a simulation done here.





calculus functional-analysis functions





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 13:25









Ilya Gazman

456927




456927







  • 1




    6-dimensional function approximation based only on 7 points is going to be insanely inaccurate.
    – lisyarus
    Jul 28 at 13:41










  • @lisyarus I can produce more points. But I would like to understand the process, with an example so I can code it. This is a simulation result
    – Ilya Gazman
    Jul 28 at 13:44












  • 1




    6-dimensional function approximation based only on 7 points is going to be insanely inaccurate.
    – lisyarus
    Jul 28 at 13:41










  • @lisyarus I can produce more points. But I would like to understand the process, with an example so I can code it. This is a simulation result
    – Ilya Gazman
    Jul 28 at 13:44







1




1




6-dimensional function approximation based only on 7 points is going to be insanely inaccurate.
– lisyarus
Jul 28 at 13:41




6-dimensional function approximation based only on 7 points is going to be insanely inaccurate.
– lisyarus
Jul 28 at 13:41












@lisyarus I can produce more points. But I would like to understand the process, with an example so I can code it. This is a simulation result
– Ilya Gazman
Jul 28 at 13:44




@lisyarus I can produce more points. But I would like to understand the process, with an example so I can code it. This is a simulation result
– Ilya Gazman
Jul 28 at 13:44










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Calling



$$
A_mtimes 7 = left(
beginarrayccccccc
28.78 & 15.38 & 6.58 & 1.94 & 0.29 & 0.02 & 1 \
22.05 & 23.26 & 20.2 & 12.94 & 4.06 & 0.48 & 1 \
20.89 & 23.11 & 20.83 & 14.04 & 4.56 & 0.57 & 1 \
13.88 & 18. & 22.72 & 22.99 & 10.54 & 1.89 & 1 \
9.92 & 13.72 & 19.24 & 28.67 & 17.35 & 4.07 & 1 \
2.99 & 4.33 & 6.86 & 12.72 & 41.03 & 30.08 & 1 \
1.48 & 2.2 & 3.56 & 6.69 & 22.18 & 62.88 & 1 \
cdots & cdots &cdots & cdots &cdots & cdots & 1 \
endarray
right);;
b_7times m = left(
beginarrayc
47 \
17 \
16 \
10 \
7 \
2 \
1 \
cdots
endarray
right)
X_7times 1 = left(
beginarrayc
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
x_6 \
x_7 \
endarray
right)
$$



we can adjust a hyper-plane with coefficients $X$ such that the error



$$
E(X) = ||Acdot X-b||^2
$$



is minimum



This can be obtained by solving



$$
X^* = left((A^topcdot A)^-1cdot Aright)cdot b
$$



NOTE



There are many ways to obtain this kind of data adjusting.






share|cite|improve this answer





















  • Can you please provide more details with a full example. It is still hard for me to code.
    – Ilya Gazman
    Jul 28 at 16:54










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Calling



$$
A_mtimes 7 = left(
beginarrayccccccc
28.78 & 15.38 & 6.58 & 1.94 & 0.29 & 0.02 & 1 \
22.05 & 23.26 & 20.2 & 12.94 & 4.06 & 0.48 & 1 \
20.89 & 23.11 & 20.83 & 14.04 & 4.56 & 0.57 & 1 \
13.88 & 18. & 22.72 & 22.99 & 10.54 & 1.89 & 1 \
9.92 & 13.72 & 19.24 & 28.67 & 17.35 & 4.07 & 1 \
2.99 & 4.33 & 6.86 & 12.72 & 41.03 & 30.08 & 1 \
1.48 & 2.2 & 3.56 & 6.69 & 22.18 & 62.88 & 1 \
cdots & cdots &cdots & cdots &cdots & cdots & 1 \
endarray
right);;
b_7times m = left(
beginarrayc
47 \
17 \
16 \
10 \
7 \
2 \
1 \
cdots
endarray
right)
X_7times 1 = left(
beginarrayc
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
x_6 \
x_7 \
endarray
right)
$$



we can adjust a hyper-plane with coefficients $X$ such that the error



$$
E(X) = ||Acdot X-b||^2
$$



is minimum



This can be obtained by solving



$$
X^* = left((A^topcdot A)^-1cdot Aright)cdot b
$$



NOTE



There are many ways to obtain this kind of data adjusting.






share|cite|improve this answer





















  • Can you please provide more details with a full example. It is still hard for me to code.
    – Ilya Gazman
    Jul 28 at 16:54














up vote
2
down vote



accepted










Calling



$$
A_mtimes 7 = left(
beginarrayccccccc
28.78 & 15.38 & 6.58 & 1.94 & 0.29 & 0.02 & 1 \
22.05 & 23.26 & 20.2 & 12.94 & 4.06 & 0.48 & 1 \
20.89 & 23.11 & 20.83 & 14.04 & 4.56 & 0.57 & 1 \
13.88 & 18. & 22.72 & 22.99 & 10.54 & 1.89 & 1 \
9.92 & 13.72 & 19.24 & 28.67 & 17.35 & 4.07 & 1 \
2.99 & 4.33 & 6.86 & 12.72 & 41.03 & 30.08 & 1 \
1.48 & 2.2 & 3.56 & 6.69 & 22.18 & 62.88 & 1 \
cdots & cdots &cdots & cdots &cdots & cdots & 1 \
endarray
right);;
b_7times m = left(
beginarrayc
47 \
17 \
16 \
10 \
7 \
2 \
1 \
cdots
endarray
right)
X_7times 1 = left(
beginarrayc
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
x_6 \
x_7 \
endarray
right)
$$



we can adjust a hyper-plane with coefficients $X$ such that the error



$$
E(X) = ||Acdot X-b||^2
$$



is minimum



This can be obtained by solving



$$
X^* = left((A^topcdot A)^-1cdot Aright)cdot b
$$



NOTE



There are many ways to obtain this kind of data adjusting.






share|cite|improve this answer





















  • Can you please provide more details with a full example. It is still hard for me to code.
    – Ilya Gazman
    Jul 28 at 16:54












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Calling



$$
A_mtimes 7 = left(
beginarrayccccccc
28.78 & 15.38 & 6.58 & 1.94 & 0.29 & 0.02 & 1 \
22.05 & 23.26 & 20.2 & 12.94 & 4.06 & 0.48 & 1 \
20.89 & 23.11 & 20.83 & 14.04 & 4.56 & 0.57 & 1 \
13.88 & 18. & 22.72 & 22.99 & 10.54 & 1.89 & 1 \
9.92 & 13.72 & 19.24 & 28.67 & 17.35 & 4.07 & 1 \
2.99 & 4.33 & 6.86 & 12.72 & 41.03 & 30.08 & 1 \
1.48 & 2.2 & 3.56 & 6.69 & 22.18 & 62.88 & 1 \
cdots & cdots &cdots & cdots &cdots & cdots & 1 \
endarray
right);;
b_7times m = left(
beginarrayc
47 \
17 \
16 \
10 \
7 \
2 \
1 \
cdots
endarray
right)
X_7times 1 = left(
beginarrayc
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
x_6 \
x_7 \
endarray
right)
$$



we can adjust a hyper-plane with coefficients $X$ such that the error



$$
E(X) = ||Acdot X-b||^2
$$



is minimum



This can be obtained by solving



$$
X^* = left((A^topcdot A)^-1cdot Aright)cdot b
$$



NOTE



There are many ways to obtain this kind of data adjusting.






share|cite|improve this answer













Calling



$$
A_mtimes 7 = left(
beginarrayccccccc
28.78 & 15.38 & 6.58 & 1.94 & 0.29 & 0.02 & 1 \
22.05 & 23.26 & 20.2 & 12.94 & 4.06 & 0.48 & 1 \
20.89 & 23.11 & 20.83 & 14.04 & 4.56 & 0.57 & 1 \
13.88 & 18. & 22.72 & 22.99 & 10.54 & 1.89 & 1 \
9.92 & 13.72 & 19.24 & 28.67 & 17.35 & 4.07 & 1 \
2.99 & 4.33 & 6.86 & 12.72 & 41.03 & 30.08 & 1 \
1.48 & 2.2 & 3.56 & 6.69 & 22.18 & 62.88 & 1 \
cdots & cdots &cdots & cdots &cdots & cdots & 1 \
endarray
right);;
b_7times m = left(
beginarrayc
47 \
17 \
16 \
10 \
7 \
2 \
1 \
cdots
endarray
right)
X_7times 1 = left(
beginarrayc
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
x_6 \
x_7 \
endarray
right)
$$



we can adjust a hyper-plane with coefficients $X$ such that the error



$$
E(X) = ||Acdot X-b||^2
$$



is minimum



This can be obtained by solving



$$
X^* = left((A^topcdot A)^-1cdot Aright)cdot b
$$



NOTE



There are many ways to obtain this kind of data adjusting.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 14:18









Cesareo

5,6412412




5,6412412











  • Can you please provide more details with a full example. It is still hard for me to code.
    – Ilya Gazman
    Jul 28 at 16:54
















  • Can you please provide more details with a full example. It is still hard for me to code.
    – Ilya Gazman
    Jul 28 at 16:54















Can you please provide more details with a full example. It is still hard for me to code.
– Ilya Gazman
Jul 28 at 16:54




Can you please provide more details with a full example. It is still hard for me to code.
– Ilya Gazman
Jul 28 at 16:54












 

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