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# Quadratic approximation formula, part 1

How to create a quadratic function that approximates an arbitrary two-variable function. Created by Grant Sanderson.

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• Is this the analogy for multivariable functions to what taylor and maclaurin series are for single variable functions?
• What was the need to extend the linear approximation and add other 3 terms: ax^2+bxy+y^2 ? or even if it was for the quadratic approximation, why would we need linear terms then?
• You do still want the first deriviative to be the same, not just the second ones.
• The notations confuse me a little. To be clear, when he writes fx and fy, it's f'x and f'y right? With the ' for denoting a derivative? And f"xx and f"yy instead of fxx and fyy?
• At around , aren't we messing up the differentiation? What if the partial derivative of f with respect to x, or y contains x's and y's - shouldn't we be using the product rule to differentiate? Something seems fishy!
(1 vote)
• The partial derivative of f with respect to x probably does have an x and a y in it. However, in the linear approximation, we plug x0 into every x and y0 into every y. This leaves us with a constant that goes to zero when differentiated because, well, it's constant :). The reason why we are left with a partial derivative at x0 and y0 is because that constant was being multiplied by a variable.
• Does anyone know if this is related to quadratic forms from linear algebra? They seem to have the same expressions.
(1 vote)
• At to , rather than using (x-x0)^2, (x-x0)(y-y0), and (y-y0)^2 to make sure that the quadratic approximation will be equal to the function at (x0,y0), couldn't you also use (x^2 - x0^2), (x0y0 - xy), and (y^2-y0^2)? It is more compact.

for clarification: x0 = x-naught, y0 = y naught
(1 vote)
• No.
(x - x_0)^2 = x^2 - 2x.x_0 + x_0^2
This is not the same as:
(x^2 - x_0^2).
For the 'offset' descibed by nlauseng to work, you have to replace all the 'x' terms with '(x - x_0)' before any operation is conducted on them.
(1 vote)