- #1
sponsoredwalk
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Hi, I have three questions about the application of quadratic approximation, what it is & when to use it. It ties in with a question about linear approximation also, I'll give an example first of what I'm talking about, just for you to evaluate if I'm wrong in the way I see the whole process, I just worry & want to get the logic right you know ![Red Face :redface: :redface:](https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/redface.gif)
So, the formula for linear approximation is [tex]f(x) = f'(x_0)(x - x_0) + f(x_0)[/tex]. I see a relationship to the slope formula, [tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex] (via algebra you obtain the linear approximation, & the slope is an infinitesimal which is logical) and this explains to me why the linear approximation will approximate a function about a certain point and not be so accurate as you go further and further from [tex]x_0[/tex]
Also, from my understanding of this concept, it is a method to approximate a function about a point, say you want to approximate [tex]\sqrt{9.1}[/tex] you follow the above framework, setting x = 9 (a known function close by).
[tex]f(x) = \sqrt{x}[/tex]
[tex]f'(x) = \frac{1}{2\sqrt{x}}[/tex]
[tex]f(x) = f'(x_0)(x - x_0) + f(x_0)[/tex]
[tex]f(x) = \frac{1}{2\sqrt{9}}(x - 9) + \sqrt{9}[/tex]
[tex]f(x) = \frac{1}{6}}(x - 9) + 3[/tex]
[tex]f(x) = \frac{1}{6}}x - \frac{9}{6}} + \frac{18}{6}}[/tex]
[tex]f(x) = \frac{1}{6}}x + \frac{9}{6}} [/tex]
1.The calculator matches the answer here very closely an all is good. How would quadratic approximation fit in here? Does quadratic approximation have any relevance to this equation at all? Am I applying it where it doesn't have any meaning. I only heard about it yesterday & my intuition tells me that it's just a means of getting a more exact answer, is this correct?
2.Where does the derivation for the quadratic approximation come from? The way I understand the linear approx is coming from the slope formula & point-slope formula but I see no way a quadratic fits in.
3. What is the deal with the formula [tex](1 + x)^{r} = 1 + rx[/tex]? Using this type of formulation a teacher in an mit lecture finds a linear approximation & then he gets quadratic factors in the answer and then he tells the class to throw them away. http://www.youtube.com/watch?v=BSAA0akmPEU&feature=SeriesPlayList&p=590CCC2BC5AF3BC1 (27th minute). What I wonder is why use this when the normal linear approximation works fine, why memorize unnecessary baggage?
![Red Face :redface: :redface:](https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/redface.gif)
So, the formula for linear approximation is [tex]f(x) = f'(x_0)(x - x_0) + f(x_0)[/tex]. I see a relationship to the slope formula, [tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex] (via algebra you obtain the linear approximation, & the slope is an infinitesimal which is logical) and this explains to me why the linear approximation will approximate a function about a certain point and not be so accurate as you go further and further from [tex]x_0[/tex]
Also, from my understanding of this concept, it is a method to approximate a function about a point, say you want to approximate [tex]\sqrt{9.1}[/tex] you follow the above framework, setting x = 9 (a known function close by).
[tex]f(x) = \sqrt{x}[/tex]
[tex]f'(x) = \frac{1}{2\sqrt{x}}[/tex]
[tex]f(x) = f'(x_0)(x - x_0) + f(x_0)[/tex]
[tex]f(x) = \frac{1}{2\sqrt{9}}(x - 9) + \sqrt{9}[/tex]
[tex]f(x) = \frac{1}{6}}(x - 9) + 3[/tex]
[tex]f(x) = \frac{1}{6}}x - \frac{9}{6}} + \frac{18}{6}}[/tex]
[tex]f(x) = \frac{1}{6}}x + \frac{9}{6}} [/tex]
1.The calculator matches the answer here very closely an all is good. How would quadratic approximation fit in here? Does quadratic approximation have any relevance to this equation at all? Am I applying it where it doesn't have any meaning. I only heard about it yesterday & my intuition tells me that it's just a means of getting a more exact answer, is this correct?
2.Where does the derivation for the quadratic approximation come from? The way I understand the linear approx is coming from the slope formula & point-slope formula but I see no way a quadratic fits in.
3. What is the deal with the formula [tex](1 + x)^{r} = 1 + rx[/tex]? Using this type of formulation a teacher in an mit lecture finds a linear approximation & then he gets quadratic factors in the answer and then he tells the class to throw them away. http://www.youtube.com/watch?v=BSAA0akmPEU&feature=SeriesPlayList&p=590CCC2BC5AF3BC1 (27th minute). What I wonder is why use this when the normal linear approximation works fine, why memorize unnecessary baggage?