Can a neural network compute $y = x^2$?












4












$begingroup$


In spirit of the famous Tensorflow Fizz Buzz joke and XOr problem I started to think, if it's possible to design a neural network that implements $y = x^2$ function?



Given some representation of a number (e.g. as a vector in binary form, so that number 5 is represented as [1,0,1,0,0,0,0,...]), the neural network should learn to return its square - 25 in this case.



If I could implement $y=x^2$, I could probably implement $y=x^3$ and generally any polynomial of x, and then with Taylor series I could approximate $y=sin(x)$, which would solve the Fizz Buzz problem - a neural network that can find remainder of the division.



Clearly, just the linear part of NNs won't be able to perform this task, so if we could do the multiplication, it would be happening thanks to activation function.



Can you suggest any ideas or reading on subject?










share|improve this question









New contributor




Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    4












    $begingroup$


    In spirit of the famous Tensorflow Fizz Buzz joke and XOr problem I started to think, if it's possible to design a neural network that implements $y = x^2$ function?



    Given some representation of a number (e.g. as a vector in binary form, so that number 5 is represented as [1,0,1,0,0,0,0,...]), the neural network should learn to return its square - 25 in this case.



    If I could implement $y=x^2$, I could probably implement $y=x^3$ and generally any polynomial of x, and then with Taylor series I could approximate $y=sin(x)$, which would solve the Fizz Buzz problem - a neural network that can find remainder of the division.



    Clearly, just the linear part of NNs won't be able to perform this task, so if we could do the multiplication, it would be happening thanks to activation function.



    Can you suggest any ideas or reading on subject?










    share|improve this question









    New contributor




    Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4


      3



      $begingroup$


      In spirit of the famous Tensorflow Fizz Buzz joke and XOr problem I started to think, if it's possible to design a neural network that implements $y = x^2$ function?



      Given some representation of a number (e.g. as a vector in binary form, so that number 5 is represented as [1,0,1,0,0,0,0,...]), the neural network should learn to return its square - 25 in this case.



      If I could implement $y=x^2$, I could probably implement $y=x^3$ and generally any polynomial of x, and then with Taylor series I could approximate $y=sin(x)$, which would solve the Fizz Buzz problem - a neural network that can find remainder of the division.



      Clearly, just the linear part of NNs won't be able to perform this task, so if we could do the multiplication, it would be happening thanks to activation function.



      Can you suggest any ideas or reading on subject?










      share|improve this question









      New contributor




      Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      In spirit of the famous Tensorflow Fizz Buzz joke and XOr problem I started to think, if it's possible to design a neural network that implements $y = x^2$ function?



      Given some representation of a number (e.g. as a vector in binary form, so that number 5 is represented as [1,0,1,0,0,0,0,...]), the neural network should learn to return its square - 25 in this case.



      If I could implement $y=x^2$, I could probably implement $y=x^3$ and generally any polynomial of x, and then with Taylor series I could approximate $y=sin(x)$, which would solve the Fizz Buzz problem - a neural network that can find remainder of the division.



      Clearly, just the linear part of NNs won't be able to perform this task, so if we could do the multiplication, it would be happening thanks to activation function.



      Can you suggest any ideas or reading on subject?







      machine-learning neural-network






      share|improve this question









      New contributor




      Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 8 hours ago







      Boris Burkov













      New contributor




      Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 12 hours ago









      Boris BurkovBoris Burkov

      1235




      1235




      New contributor




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      New contributor





      Boris Burkov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          Neural networks are also called as the universal function approximation which is based in the universal function approximation theorem. It states that :




          In the mathematical theory of artificial neural networks,
          the universal approximation theorem states that a feed-forward network
          with a single hidden layer containing a finite number of neurons can
          approximate continuous functions on compact subsets of Rn, under mild
          assumptions on the activation function




          Meaning a ANN with a non linear activation function could map the function which relates the input with the output. The function y = x^2 could be easily approximated using regression ANN.



          You can find an excellent lesson here with a notebook example.



          Also, because of such ability ANN could map complex relationships for example between an image and its labels.






          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, this is exactly what I was asking for!
            $endgroup$
            – Boris Burkov
            12 hours ago






          • 2




            $begingroup$
            Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
            $endgroup$
            – Jeffrey
            10 hours ago





















          4












          $begingroup$

          I think the answer of @ShubhamPanchal is a little bit misleading. Yes, it is true that by Cybenko's universal approximation theorem we can approximate $f(x)=x^2$ with a single hidden layer containing a finite number of neurons can approximate continuous functions on compact subsets of $mathbb{R}^n$, under mild assumptions on the activation function.




          But the main problem is that the theorem has a very important
          limitation
          . The function needs to be defined on compact subsets of
          $mathbb{R}^n$
          (compact subset = bounded + closed subset). But why
          is this problematic?
          . When training the function approximator you
          will always have a finite data set. Hence, you will approximate the
          function inside a compact subset of $mathbb{R}^n$. But we can always
          find a point $x$ for which the approximation will probably fail. That
          being said. If you only want to approximate $f(x)=x^2$ on a compact
          subset of $mathbb{R}$ then we can answer your question with yes.
          But if you want to approximate $f(x)=x^2$ for all $xin mathbb{R}$
          then the answer is no (I exclude the trivial case in which you use
          a quadratic activation function).




          Side remark on Taylor approximation: You always have to keep in mind that a Taylor approximation is only a local approximation. If you only want to approximate a function in a predefined region then you should be able to use Taylor series. But approximating $sin(x)$ by the Taylor series evaluated at $x=0$ will give you horrible results for $xto 10000$ if you don't use enough terms in your Taylor expansion.






          share|improve this answer










          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$









          • 1




            $begingroup$
            Nice catch! "compact set".
            $endgroup$
            – Esmailian
            8 hours ago










          • $begingroup$
            Many thanks, mate! Eye-opener!
            $endgroup$
            – Boris Burkov
            8 hours ago










          • $begingroup$
            @Esmailian: Thank you :).
            $endgroup$
            – MachineLearner
            8 hours ago











          Your Answer





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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

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          active

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          active

          oldest

          votes









          7












          $begingroup$

          Neural networks are also called as the universal function approximation which is based in the universal function approximation theorem. It states that :




          In the mathematical theory of artificial neural networks,
          the universal approximation theorem states that a feed-forward network
          with a single hidden layer containing a finite number of neurons can
          approximate continuous functions on compact subsets of Rn, under mild
          assumptions on the activation function




          Meaning a ANN with a non linear activation function could map the function which relates the input with the output. The function y = x^2 could be easily approximated using regression ANN.



          You can find an excellent lesson here with a notebook example.



          Also, because of such ability ANN could map complex relationships for example between an image and its labels.






          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, this is exactly what I was asking for!
            $endgroup$
            – Boris Burkov
            12 hours ago






          • 2




            $begingroup$
            Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
            $endgroup$
            – Jeffrey
            10 hours ago


















          7












          $begingroup$

          Neural networks are also called as the universal function approximation which is based in the universal function approximation theorem. It states that :




          In the mathematical theory of artificial neural networks,
          the universal approximation theorem states that a feed-forward network
          with a single hidden layer containing a finite number of neurons can
          approximate continuous functions on compact subsets of Rn, under mild
          assumptions on the activation function




          Meaning a ANN with a non linear activation function could map the function which relates the input with the output. The function y = x^2 could be easily approximated using regression ANN.



          You can find an excellent lesson here with a notebook example.



          Also, because of such ability ANN could map complex relationships for example between an image and its labels.






          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, this is exactly what I was asking for!
            $endgroup$
            – Boris Burkov
            12 hours ago






          • 2




            $begingroup$
            Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
            $endgroup$
            – Jeffrey
            10 hours ago
















          7












          7








          7





          $begingroup$

          Neural networks are also called as the universal function approximation which is based in the universal function approximation theorem. It states that :




          In the mathematical theory of artificial neural networks,
          the universal approximation theorem states that a feed-forward network
          with a single hidden layer containing a finite number of neurons can
          approximate continuous functions on compact subsets of Rn, under mild
          assumptions on the activation function




          Meaning a ANN with a non linear activation function could map the function which relates the input with the output. The function y = x^2 could be easily approximated using regression ANN.



          You can find an excellent lesson here with a notebook example.



          Also, because of such ability ANN could map complex relationships for example between an image and its labels.






          share|improve this answer









          $endgroup$



          Neural networks are also called as the universal function approximation which is based in the universal function approximation theorem. It states that :




          In the mathematical theory of artificial neural networks,
          the universal approximation theorem states that a feed-forward network
          with a single hidden layer containing a finite number of neurons can
          approximate continuous functions on compact subsets of Rn, under mild
          assumptions on the activation function




          Meaning a ANN with a non linear activation function could map the function which relates the input with the output. The function y = x^2 could be easily approximated using regression ANN.



          You can find an excellent lesson here with a notebook example.



          Also, because of such ability ANN could map complex relationships for example between an image and its labels.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 12 hours ago









          Shubham PanchalShubham Panchal

          34215




          34215








          • 1




            $begingroup$
            Thank you very much, this is exactly what I was asking for!
            $endgroup$
            – Boris Burkov
            12 hours ago






          • 2




            $begingroup$
            Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
            $endgroup$
            – Jeffrey
            10 hours ago
















          • 1




            $begingroup$
            Thank you very much, this is exactly what I was asking for!
            $endgroup$
            – Boris Burkov
            12 hours ago






          • 2




            $begingroup$
            Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
            $endgroup$
            – Jeffrey
            10 hours ago










          1




          1




          $begingroup$
          Thank you very much, this is exactly what I was asking for!
          $endgroup$
          – Boris Burkov
          12 hours ago




          $begingroup$
          Thank you very much, this is exactly what I was asking for!
          $endgroup$
          – Boris Burkov
          12 hours ago




          2




          2




          $begingroup$
          Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
          $endgroup$
          – Jeffrey
          10 hours ago






          $begingroup$
          Although true, it a very bad idea to learn that. I fail to see where any generalization power would arise from. NN shine when there's something to generalize. Like CNN for vision that capture patterns, or RNN that can capture trends.
          $endgroup$
          – Jeffrey
          10 hours ago













          4












          $begingroup$

          I think the answer of @ShubhamPanchal is a little bit misleading. Yes, it is true that by Cybenko's universal approximation theorem we can approximate $f(x)=x^2$ with a single hidden layer containing a finite number of neurons can approximate continuous functions on compact subsets of $mathbb{R}^n$, under mild assumptions on the activation function.




          But the main problem is that the theorem has a very important
          limitation
          . The function needs to be defined on compact subsets of
          $mathbb{R}^n$
          (compact subset = bounded + closed subset). But why
          is this problematic?
          . When training the function approximator you
          will always have a finite data set. Hence, you will approximate the
          function inside a compact subset of $mathbb{R}^n$. But we can always
          find a point $x$ for which the approximation will probably fail. That
          being said. If you only want to approximate $f(x)=x^2$ on a compact
          subset of $mathbb{R}$ then we can answer your question with yes.
          But if you want to approximate $f(x)=x^2$ for all $xin mathbb{R}$
          then the answer is no (I exclude the trivial case in which you use
          a quadratic activation function).




          Side remark on Taylor approximation: You always have to keep in mind that a Taylor approximation is only a local approximation. If you only want to approximate a function in a predefined region then you should be able to use Taylor series. But approximating $sin(x)$ by the Taylor series evaluated at $x=0$ will give you horrible results for $xto 10000$ if you don't use enough terms in your Taylor expansion.






          share|improve this answer










          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$









          • 1




            $begingroup$
            Nice catch! "compact set".
            $endgroup$
            – Esmailian
            8 hours ago










          • $begingroup$
            Many thanks, mate! Eye-opener!
            $endgroup$
            – Boris Burkov
            8 hours ago










          • $begingroup$
            @Esmailian: Thank you :).
            $endgroup$
            – MachineLearner
            8 hours ago
















          4












          $begingroup$

          I think the answer of @ShubhamPanchal is a little bit misleading. Yes, it is true that by Cybenko's universal approximation theorem we can approximate $f(x)=x^2$ with a single hidden layer containing a finite number of neurons can approximate continuous functions on compact subsets of $mathbb{R}^n$, under mild assumptions on the activation function.




          But the main problem is that the theorem has a very important
          limitation
          . The function needs to be defined on compact subsets of
          $mathbb{R}^n$
          (compact subset = bounded + closed subset). But why
          is this problematic?
          . When training the function approximator you
          will always have a finite data set. Hence, you will approximate the
          function inside a compact subset of $mathbb{R}^n$. But we can always
          find a point $x$ for which the approximation will probably fail. That
          being said. If you only want to approximate $f(x)=x^2$ on a compact
          subset of $mathbb{R}$ then we can answer your question with yes.
          But if you want to approximate $f(x)=x^2$ for all $xin mathbb{R}$
          then the answer is no (I exclude the trivial case in which you use
          a quadratic activation function).




          Side remark on Taylor approximation: You always have to keep in mind that a Taylor approximation is only a local approximation. If you only want to approximate a function in a predefined region then you should be able to use Taylor series. But approximating $sin(x)$ by the Taylor series evaluated at $x=0$ will give you horrible results for $xto 10000$ if you don't use enough terms in your Taylor expansion.






          share|improve this answer










          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$









          • 1




            $begingroup$
            Nice catch! "compact set".
            $endgroup$
            – Esmailian
            8 hours ago










          • $begingroup$
            Many thanks, mate! Eye-opener!
            $endgroup$
            – Boris Burkov
            8 hours ago










          • $begingroup$
            @Esmailian: Thank you :).
            $endgroup$
            – MachineLearner
            8 hours ago














          4












          4








          4





          $begingroup$

          I think the answer of @ShubhamPanchal is a little bit misleading. Yes, it is true that by Cybenko's universal approximation theorem we can approximate $f(x)=x^2$ with a single hidden layer containing a finite number of neurons can approximate continuous functions on compact subsets of $mathbb{R}^n$, under mild assumptions on the activation function.




          But the main problem is that the theorem has a very important
          limitation
          . The function needs to be defined on compact subsets of
          $mathbb{R}^n$
          (compact subset = bounded + closed subset). But why
          is this problematic?
          . When training the function approximator you
          will always have a finite data set. Hence, you will approximate the
          function inside a compact subset of $mathbb{R}^n$. But we can always
          find a point $x$ for which the approximation will probably fail. That
          being said. If you only want to approximate $f(x)=x^2$ on a compact
          subset of $mathbb{R}$ then we can answer your question with yes.
          But if you want to approximate $f(x)=x^2$ for all $xin mathbb{R}$
          then the answer is no (I exclude the trivial case in which you use
          a quadratic activation function).




          Side remark on Taylor approximation: You always have to keep in mind that a Taylor approximation is only a local approximation. If you only want to approximate a function in a predefined region then you should be able to use Taylor series. But approximating $sin(x)$ by the Taylor series evaluated at $x=0$ will give you horrible results for $xto 10000$ if you don't use enough terms in your Taylor expansion.






          share|improve this answer










          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          I think the answer of @ShubhamPanchal is a little bit misleading. Yes, it is true that by Cybenko's universal approximation theorem we can approximate $f(x)=x^2$ with a single hidden layer containing a finite number of neurons can approximate continuous functions on compact subsets of $mathbb{R}^n$, under mild assumptions on the activation function.




          But the main problem is that the theorem has a very important
          limitation
          . The function needs to be defined on compact subsets of
          $mathbb{R}^n$
          (compact subset = bounded + closed subset). But why
          is this problematic?
          . When training the function approximator you
          will always have a finite data set. Hence, you will approximate the
          function inside a compact subset of $mathbb{R}^n$. But we can always
          find a point $x$ for which the approximation will probably fail. That
          being said. If you only want to approximate $f(x)=x^2$ on a compact
          subset of $mathbb{R}$ then we can answer your question with yes.
          But if you want to approximate $f(x)=x^2$ for all $xin mathbb{R}$
          then the answer is no (I exclude the trivial case in which you use
          a quadratic activation function).




          Side remark on Taylor approximation: You always have to keep in mind that a Taylor approximation is only a local approximation. If you only want to approximate a function in a predefined region then you should be able to use Taylor series. But approximating $sin(x)$ by the Taylor series evaluated at $x=0$ will give you horrible results for $xto 10000$ if you don't use enough terms in your Taylor expansion.







          share|improve this answer










          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|improve this answer



          share|improve this answer








          edited 8 hours ago





















          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 8 hours ago









          MachineLearnerMachineLearner

          30810




          30810




          New contributor




          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          MachineLearner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.








          • 1




            $begingroup$
            Nice catch! "compact set".
            $endgroup$
            – Esmailian
            8 hours ago










          • $begingroup$
            Many thanks, mate! Eye-opener!
            $endgroup$
            – Boris Burkov
            8 hours ago










          • $begingroup$
            @Esmailian: Thank you :).
            $endgroup$
            – MachineLearner
            8 hours ago














          • 1




            $begingroup$
            Nice catch! "compact set".
            $endgroup$
            – Esmailian
            8 hours ago










          • $begingroup$
            Many thanks, mate! Eye-opener!
            $endgroup$
            – Boris Burkov
            8 hours ago










          • $begingroup$
            @Esmailian: Thank you :).
            $endgroup$
            – MachineLearner
            8 hours ago








          1




          1




          $begingroup$
          Nice catch! "compact set".
          $endgroup$
          – Esmailian
          8 hours ago




          $begingroup$
          Nice catch! "compact set".
          $endgroup$
          – Esmailian
          8 hours ago












          $begingroup$
          Many thanks, mate! Eye-opener!
          $endgroup$
          – Boris Burkov
          8 hours ago




          $begingroup$
          Many thanks, mate! Eye-opener!
          $endgroup$
          – Boris Burkov
          8 hours ago












          $begingroup$
          @Esmailian: Thank you :).
          $endgroup$
          – MachineLearner
          8 hours ago




          $begingroup$
          @Esmailian: Thank you :).
          $endgroup$
          – MachineLearner
          8 hours ago










          Boris Burkov is a new contributor. Be nice, and check out our Code of Conduct.










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