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Taylor expansion in physics 
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#1
Apr112, 03:01 AM

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i am very confuse how my profs always use taylor expansion in physics which somehow doesn't follow the general equation of
f(x) = f(a) + f'(a)(xa) + 1/2! f''(a)(xa)^{2} and so on... like for example, what is the taylor expansion of x  kx where k is small it was given as something like x  kx f'(x) + (1/2) k^{2} x^{2} f''(x) + ... is this taylor expansion? but there is no 'about which point, i.e, a=? ' i don't even understand how the first term x is gotten. f(a) = x?? please help thank you! 


#2
Apr112, 04:36 AM

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hi quietrain!
the (x)s should be a, and the x at the start should be f(a) (and k = x  a) 


#3
Apr112, 05:03 AM

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Hi quietrain! :)
Suppose you expand f(akx). You would get: f(akx) = f(a)  kx f'(a) + (1/2) k^{2} x^{2} f''(a) + ... Now replace "a" with "x"... Apparently you've got a typo in the first term. EDIT: Oops, ^{TM} got here first! 


#4
Apr112, 05:34 AM

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Taylor expansion in physics



#5
Apr212, 04:39 AM

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g(0) = f(x), g'(k) = x*f'(xkx) > g'(0) = x*f'(x), g''(k) = x^2*f''(xkx) > g''(0) = x^2*f''(x), etc. RGV 


#6
Apr212, 05:29 AM

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hi everyone thanks for helping
but the taylor expansion was for xkx or perhaps only kx was expanded? but anyway it was given as x  kx f'(x) + ... because i only know this f(x) = f(a) + f'(a)(xa) + 1/2! f''(a)(xa)^{2} as the general formula? in any case, does it mean if i want to expand say 1 + kx then it would be f(1+kx) = f(1) + kx f'(1) + ... so f(1) is 1 + k(1) = 1+k? then f'(1) is k? then where do i put my 1 since i don't have x also does it mean f'' onwards are all 0? but in particular, how did you get g'(k) and g'(k)? is it through the chain rule differentiation? thanks! 


#7
Apr212, 05:54 AM

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f(x) = f(a) + f'(a)(xa) + 1/2! f''(a)(xa)^{2} + ... but let's rewrite it with y instead of x to eliminate the ambiguity between the x's. So: f(y) = f(a) + f'(a)(ya) + 1/2! f''(a)(ya)^{2} + ... You want to expand f(xkx). To do this, first define y=xkx, and define a=x. Then replace all occurrences of y by (xkx), and replace all occurrences of a by x. What do you get? Edit: Btw, an alternative formula for Taylor expansion is: f(x+h)=f(x) + h f'(x) + 1/2! h^{2} f''(x) + ... 


#8
Apr212, 06:27 AM

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#9
Apr212, 07:30 AM

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does it mean i can define a as kx or k also? but i thought a is the point that we evaluate the taylor expansion on? am i right to say if i taylor expand on a point say x =1 for a curve graph. then as my orders of taylor expansion become greater, the approximation is better too? 


#10
Apr212, 07:33 AM

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so if i had a curve, would taylor expansion ( approximation? ) make more sense here? that means the first term of the expansion gives me a straight line, the 2nd makes it more curve, the 3rd makes it even more like the original curve function? does it then mean i cannot taylor expand things like 1+kx? i have to expand only curves? 


#11
Apr212, 07:34 AM

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You can replace them by anything you want, as long as you do so consistently and do not mix letters up. (Note that "x" has 2 different meanings in your problem statement. That's why I introduced "y"  to eliminate one of the two.) In your original formula f(x)=f(a)+f'(a)(xa)+... the expansion is around "a". In the formula f(xkx)=f(x)+f'(x)((xkx)x)+...= f(x)  kx f'(x) + ... the expansion is around "x". 


#12
Apr212, 08:20 AM

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So using your formula to expand it, you get: f(x)=1+kx f'(x)=k f''(x)=0 f'''(x)=0 So: f(x)=f(a)+f'(a)(xa)+1/2! f''(a)(xa)^{2}+... f(x)=f(0)+f'(0)(x0)+1/2! f''(0)(x0)^{2}+... f(x)=1+k(x0)+1/2! 0.(x0)^{2}+ 0 + ... f(x)=1+kx Hey! But you already had that! :) 


#13
Apr212, 02:47 PM

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f(x)  kx f'(x) + (1/2) k^{2} x^{2} f''(x) + ... [with f(x) as the first term, not just x]. You did say the expression was "something like...", which says to me that you were not sure. RGV 


#14
Apr312, 08:21 PM

P: 651

alright thanks everyone!



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