I'm trying to get the condition number of a multivariate function $f(a,b,c)$ to see if it is stable. I am reading the information here.

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I know how to do it for a $1$-dimensional function. But for a multivariate one, I am not sure how. Also how do I even get the Jacobian because there is only 1 function here. Wouldn't that mean the Jacobian is a $1 \times 3$ matrix?

  • $\begingroup$ Provided that "there is only 1 function here" means that the function $f$ is a scalar function, then simply compute the Jacobian as if $f$ was a 1-component vector function (that is, you get the gradient of f). $\endgroup$ Apr 3, 2014 at 2:09
  • $\begingroup$ Doesnt that give you a column vector. Doesn't jaccobian need to be a square matrix? $\endgroup$
    – omega
    Apr 3, 2014 at 2:12
  • $\begingroup$ Why should it be square? $\endgroup$ Apr 3, 2014 at 2:14
  • $\begingroup$ Actually since its really like [f(a,b,c)] which is a 1 by 1 vector. The jaccobian will be a row vector (1 by 3). Is that right? $\endgroup$
    – omega
    Apr 3, 2014 at 2:18
  • $\begingroup$ Yes that's right. $\endgroup$ Apr 3, 2014 at 2:27

1 Answer 1


If $f(x)=f(x_1,\ldots,x_n)$ is a scalar differentiable function, then the condition number of $f$ at $x^*=(x_1^*,\ldots,x_n^*)$ is given by $$ \frac{\|J(x^*)\|}{|f(x^*)| / \|x^*\|} = \frac{\|x^*\| \|J(x^*)\|}{|f(x^*)|} $$ Assuming you choose $L_2$ norm: $$ \frac{\sqrt{ \sum_{i=1}^n (f'_{x_i})^2(x_1^*,\ldots,x_n^*) }}{|f(x_1^*,\ldots,x_n^*)| / \sqrt{\sum_{i=1}^n (x_i^*)^2}} = \frac{ \sqrt{\sum_{i=1}^n (x_i^*)^2} \; \sqrt{\sum_{i=1}^n (f'_{x_i})^2(x_1^*,\ldots,x_n^*)}}{|f(x_1^*,\ldots,x_n^*)|} $$

  • $\begingroup$ Is there a textbook covering such topics? $\endgroup$
    – vkonton
    Jan 19, 2017 at 17:00

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