Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that we have a multivariate function $f(a,b,c,a_1,b_1)$, where $a,b,c,a_1,b_1$ are all real numbers.

$f = \frac{{c({a^2} + {b^2}) + {{({a^2} + {b^2})}^{1/2}}}}{{c{{({a^2} + {b^2})}^2}}}{a_1}{b_1}$

I generate a vector using the multivariate function $f(a,b,c,a_1,b_1)$. This is a set of evaluations of $f$ that I can plot as a 1D curve.

Taking the numerical partial derivative (i.e. Numerical differentiation), of the vector with respect to $a_1$ yields an approximation to:

$\frac{{\partial f}}{{\partial {a_1}}} = \frac{{c({a^2} + {b^2}) + {{({a^2} + {b^2})}^{1/2}}}}{{c{{({a^2} + {b^2})}^2}}}{b_1}$

However, suppose that all $a_1$ is unknown, but I do know $ca_1$, which is the product of $c$ with $a_1$. I also know $cb_1$, but not $a,b,c$ or $b_1$.

Under what conditions can I approximate the denominator of the numerical derivative using only $ca_1$? What are the error bounds associated with taking:

$\frac{{\partial f}}{{\partial (c{a_1})}}$

So rather than using the change in $a_1$, I would use some other type of change such as the change in $ca_1$.

I also know the ratio $a_1/b_1$ for all $a_1$ and $b_1$.

I also know $c{({a^2} + {b^2})^{1/2}}$ for all $c,a,b$

Can anything be done to approximate $a_1$, even using numerical or statistical methods? What additional information might be required?

share|improve this question
1  
It is difficult to see what you are trying to accomplish here. –  copper.hat Oct 11 '12 at 16:12
    
I would like to take the numerical derivative with respect to $a_1$, but I don't know $a_1$, and I would like to approximate it in some way. However, I do know $ca_1$, and well as $a_1/b_1$ and $c(a^2 + b^2)^(1/2)$. Is there any numerical or statistical procedure that can be used here, or perhaps the problem can be re-formulated in some way? –  Nicholas Kinar Oct 11 '12 at 16:19
    
I am still at a loss here. How can you take the numerical derivative if you cannot evaluate at specific points? And why are you computing a numerical derivative when you have a formula for $f$? –  copper.hat Oct 11 '12 at 16:29
    
I am sorry if I am not being very explicit; thanks for your help, copper.hat, and please continue to ask questions if further clarification is required. What I have is a vector as a set of evaluations of $f$. Now suppose that I have only the vector, but suppose that I do not know $a_1,b_1,a,b,c$. Also suppose that I know $a_1/b_1$ and $c{({a^2} + {b^2})^{1/2}}$ for each element of the vector. I want to approximate the numerical derivative of the vector with respect to $a_1$, but suppose that I do not know $a_1$ for each element of the vector. But I have a vector of $ca_1$. –  Nicholas Kinar Oct 11 '12 at 16:43
    
I do not understand what you mean by ' a vector as a set of evaluations of $f$'. Do you mean you only have the values of $f$ but not the points at which it was evaluated? I don't understand what information you have available. –  copper.hat Oct 11 '12 at 16:53
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.