I am new to ANN. I am trying out several 'simple' algorithms to see what ANN can (or cannot) be used for and how. I played around with Conv2d once and had it recognize images successfully. Now I am looking into trend line analyses. I have succeeded in training a network where it solved for linear equations. Now I am trying to see if it can be trained to solve for $y$ in the formula $y = b + x^2$.

No matter what parameters I change, or the number of dense layers, I get high values for loss and validation loss, and the predictions are incorrect.

Is it possible to solve this equation, and with what network? If it is not possible, why not? I am not looking to solve a practical problem, rather build up understanding and intuition about ANNs.

See the code I tried with below

#region Imports
from __future__ import absolute_import, division, print_function, unicode_literals
import math 
import numpy as np 
import tensorflow as tf
from tensorflow.keras import models, optimizers
from tensorflow.keras.models import Sequential
from tensorflow.keras.layers import Activation, Dropout, Flatten, Dense, Lambda
import tensorflow.keras.backend as K

#region Constants
learningRate = 0.01
epochs: int = 1000
batch_size = None
trainingValidationFactor = 0.75
nrOfSamples = 100
activation = None

#region Function definitions
def CreateNetwork(inputDimension):
  model = Sequential()
  model.add(Dense(2, input_dim=2, activation=activation))
  model.add(Dense(64, use_bias=True, activation=activation))
  model.add(Dense(32, use_bias=True, activation=activation))
  adam = optimizers.Adam(learning_rate=learningRate)
  # sgd = optimizers.SGD(lr=learningRate, decay=1e-6, momentum=0.9, nesterov=True)
  # adamax = optimizers.Adamax(learning_rate=learningRate)
  model.compile(loss='mse', optimizer=adam)
  return model

def SplitDataForValidation(factor, data, labels):
  upperBoundary = int(len(data) * factor)

  trainingData = data[:upperBoundary]
  trainingLabels = labels[:upperBoundary]

  validationData = data[upperBoundary:]
  validationLabels = labels[upperBoundary:]
  return ((trainingData, trainingLabels), (validationData, validationLabels))

def Train(network, training, validation):
  trainingData, trainingLabels = training
  history = network.fit(

  return history

def subtractMean(data):
  mean = np.mean(data)
  data -= mean
  return mean

def rescale(data):
  max = np.amax(data)
  factor = 1 / max
  data *= factor
  return factor

def Normalize(data, labels):
  dataScaleFactor = rescale(data)
  dataMean = subtractMean(data)

  labels *= dataScaleFactor
  labelsMean = np.mean(labels)
  labels -= labelsMean

def Randomize(data, labels):
  rng_state = np.random.get_state()

def CreateTestData(nrOfSamples):
  data = np.zeros(shape=(nrOfSamples,2))
  labels = np.zeros(nrOfSamples)

  for i in range(nrOfSamples):
    for j in range(2):
      randomInt = np.random.randint(1, 5)
      data[i, j] = (randomInt * i) + 10
    labels[i] = data[i, 0] + math.pow(data[i, 1], 2)
  Randomize(data, labels)
  return (data, labels)

allData, allLabels = CreateTestData(nrOfSamples)
Normalize(allData, allLabels)
training, validation = SplitDataForValidation(trainingValidationFactor, allData, allLabels)

inputDimension = np.size(allData, 1)
network = CreateNetwork(inputDimension)

history = Train(network, training, validation)

prediction = network.predict([
  [2, 2], # Should be 2 + 2 * 2 = 6
  [4, 7], # Should be 4 + 7 * 7 = 53
  [23, 56], # Should be 23 + 56 * 56 = 3159
  [128,256] # Should be 128 + 256 * 256 = 65664

2 Answers 2


$f(x) = x^2 + b$ is a polynomial (more precisely, a parabola) so it is continuous, thus, a neural network (with at least one hidden layer) should be able to approximate that function (given the universal approximation theorem).

After a very quick look at your code, I noticed you aren't using an activation function for your dense layers (i.e. your activation function is None). Try to use e.g. ReLU.

  • 1
    $\begingroup$ Thank you @nbro for your answer. I will test more, with different activation functions, including ReLU. +1 for pointing out the universal approximation theorem, of which I was not aware yet. That is really helpful to know. $\endgroup$ Commented Feb 20, 2020 at 6:48
  1. if you know in advance that your data is polynomial and you want to fit the best coefficient, then this is an easier (classical machine learning) problem . you would just assume a maximum degree e.g $y_{pred} = a_0+a_1.x+a_2x^2+...+a_n.x^n $
    you can just initialize $a_i$s and consider MSE loss $(y_{groundtruth}-y_{pred})^2$.
    then derivative with respect to $a_i$s is $2(y_{groundtruth}-y_{pred}).(- d(y_{pred})/d(a_i)) = 2(y_{groundtruth}-y_{pred}).(- x^i)$
    so having access to the gradinet, you can easily do gradient descent which is trivial.

  2. So I assume this (the general form of the data generator) is not known. in this case as you did, we assume a general form and let neural network learn it. it is important to note that you should limit the support (range) of your data by normalization as you did by normalization because it is impossible to find an approximation which holds true every where. but in a small range of [0,1] it is much easier to approximate the desired function.

in general it is possible to approximate this function but on a limited range e.g on [0,1], unless we have the correct general form (in this case polynomial) and then just fit the correct parameters using gradient descent or other methods.

I hope these two cases shed some lights on your question


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