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anonymous
 5 years ago
Can anybody help me with linear approximation...
anonymous
 5 years ago
Can anybody help me with linear approximation...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Our calculus book uses the formula \[f(x)\approx f(Xo) + f \prime(Xo)(XXo)\] I kinda know what to do on basic problems, but I don't know how to associate this formula in some of the more complicated problems. Can anybody help me understand it more by giving an example function and explain to me how to associate the formula? Thanks in advance!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What the above is saying is that, for a point x close to x_0, your *actual* function, f(x), can be approximated by the tangent line at that point. So, take a parabola. Then \[f(x)=x^2\]and let's say we want a *linear* function that will approximate f(x) for xvalues *close* to x_0=2. The equation of the tangent is given by \[yy_1=m(xx_1) \rightarrow yf(x_0)=f'(x_0)(xx_0)\]then\[y2^2=2(2)(x2) \rightarrow y = 4x4\]For *small* deviations of x from 2, \[y \approx f(x)\]so\[f(x) \approx f'(x_0)(xx_0)+f(x_0)\]or\[f(x) = x^2 \approx 4x4\]around x = 2. You can test this by comparing the actual function with the linear approximation at a small deviation, say, x=2.1. Then\[f(2.1)=(2.1)^2=4.41\]exactly, while\[f(2.1) \approx 4(2.1)4=4.40\] in the linear approximation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can see why if you plot a parabola and draw a tangent at the point (2,4). You'll see geometrically that the distance between the tangent and the parabola is small when you move only small distances away from x, but as you move out either way, the difference between the tangent value at x and the actual function value at x becomes markedly different. You now might ask, "What is the point of this?" Well, there are a few reasons. One, not all functions are as easy as y=x^2 to calculate everything for. For example, we use linear approximations at different parts of a system in chaos theory to extract information on a system that may be incredibly difficult to compute. Also, more complicated objects can be built up from linear approximations, and sometimes, a linear approximation for a variable over a small interval may be very useful when performing other forms of calculus, particularly in physics, chemistry, engineering and economics. We would usually use linear approximations where the functions are complicated too, when we want to obtain a qualitative understanding of how the function or system is behaving. Hope this helps :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for taking the time to answer my question! You did help me understand the idea behind the concept. But I was trying to understand on how to solve more complicated problems by using the formula.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For example, Find the local linear approximation of \[f(x) = \tan^{1} x\] at Xo= 1 and my solution: \[f(1) = \tan^{1} (1) = \pi /4\]\[f \prime (Xo) = 1/1+x ^{2} = 1/2\] \[(XXo) = \Delta x\] now, following the formula provided: we get \[\pi/4 +(1/2) \Delta x\] I was wondering if you guys know a more complicated function that can be solve using the formula. anyways Thanks again!
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