Taylor series of two variable ?

AI Thread Summary
To calculate the required number of terms for a given decimal accuracy in a two-variable Taylor series, one can adapt the error term used in the one-variable case. The error term for two variables involves a summation of partial derivatives of the function, evaluated at a point 'e' between the center 'c' and the variable points. To achieve a specific accuracy, such as three decimal places, the error term must satisfy an inequality similar to the one-variable case. Resources or tutorials on multivariable calculus can provide further guidance on applying this concept. Understanding the two-variable Taylor series and its error estimation is crucial for accurate calculations.
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Homework Statement




I want to know that how to calculate the required number of terms to obtain a given decimal accuracy in two variable Taylor series .
In one variable case i know there is an error term R(n)=[ f(e)^(n+1)* (x-c)^(n+1)] / (n+1)! where 'e' is between x and c (c is the center value ) so if we want a 3 decimal place accuracy we can have the inequality below

R(n)<=5*10^(-4) then we can obtain particular ' n ' to have that accuracy

But i don't know how to use this in two variable case .please give some explanation or a good tutorial where i can learn it .
 
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The corresponding formula for two variables is
R(n)= \sum_{i=0}^{n+1} \frac{\frac{\partial^{n+1} f(e)}{\partial x^{i}\partial y^{n+1-i}}(x-c)^{n+1}}{(n+1)!}
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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