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MacLaurin exercice
- Thread starter nestor95
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HallsofIvy
Elite Member
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- Jan 27, 2012
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Okay, first do you know what a "MacLaurin series" is?
A "MacLaurin series" for a given function, f(x), is an infinite sum of constants times powers of x, \(\displaystyle \sum_{n=0}^\infty c_mx^n\), where the constants are constructed in a particular way. If you are asked to find the MacLaurin series of \(\displaystyle f(x)= 1/(b- ax)\) then I have to assume that you have been taught that "particular way" of constructing those coefficients!
That way is \(\displaystyle c_n= \frac{1}{n!}\frac{d^nf}{dx^n}(0)\). That is, the coefficient of \(\displaystyle x^n\) is one over n factorial times the nth derivative of the function evaluated at x= 0 (the "zeroth" derivative being the value of the function itself).
Here, \(\displaystyle f(x)= \frac{1}{a+ bx}= (a+ bx)^{-1}\). The value of the function at x= 0 is \(\displaystyle c_0= f(0)= a^{-1}= \frac{1}{a}\). The first derivative is \(\displaystyle f'(x)= -(a+ bx)^{-2}(b)= \frac{-b}{(a+ bx)^2}\) and its value at x= 0 is \(\displaystyle c_1= \frac{-b}{a^2}\). The second derivative is \(\displaystyle 2(a+ bx)^{-3}b^2= \frac{2b^2}{(a+ bx)^3}\). Its value at x= 0 is \(\displaystyle \frac{2b^2}{a^3}\) and, since 2!= 2, the coefficient of \(\displaystyle x^2\) is \(\displaystyle \frac{b^2}{a^3}\).
The first three terms of the MacLaurin Series for \(\displaystyle \frac{1}{a+ bx}\) are \(\displaystyle \frac{1}{a}- \frac{b}{a^2}x+ \frac{b^2}{a^3}x^2\).
To get the entire MacLaurin series you will want to calculate enough terms explictiely so that you can "guess" the general formula for all n (so far we have \(\displaystyle c_0= \frac{1}{a}\), \(\displaystyle c_1= -\frac{b}{a^2}\), and \(\displaystyle c_2= \frac{b^2}{a^3}\) so that shouldn't be too hard!)
A "MacLaurin series" for a given function, f(x), is an infinite sum of constants times powers of x, \(\displaystyle \sum_{n=0}^\infty c_mx^n\), where the constants are constructed in a particular way. If you are asked to find the MacLaurin series of \(\displaystyle f(x)= 1/(b- ax)\) then I have to assume that you have been taught that "particular way" of constructing those coefficients!
That way is \(\displaystyle c_n= \frac{1}{n!}\frac{d^nf}{dx^n}(0)\). That is, the coefficient of \(\displaystyle x^n\) is one over n factorial times the nth derivative of the function evaluated at x= 0 (the "zeroth" derivative being the value of the function itself).
Here, \(\displaystyle f(x)= \frac{1}{a+ bx}= (a+ bx)^{-1}\). The value of the function at x= 0 is \(\displaystyle c_0= f(0)= a^{-1}= \frac{1}{a}\). The first derivative is \(\displaystyle f'(x)= -(a+ bx)^{-2}(b)= \frac{-b}{(a+ bx)^2}\) and its value at x= 0 is \(\displaystyle c_1= \frac{-b}{a^2}\). The second derivative is \(\displaystyle 2(a+ bx)^{-3}b^2= \frac{2b^2}{(a+ bx)^3}\). Its value at x= 0 is \(\displaystyle \frac{2b^2}{a^3}\) and, since 2!= 2, the coefficient of \(\displaystyle x^2\) is \(\displaystyle \frac{b^2}{a^3}\).
The first three terms of the MacLaurin Series for \(\displaystyle \frac{1}{a+ bx}\) are \(\displaystyle \frac{1}{a}- \frac{b}{a^2}x+ \frac{b^2}{a^3}x^2\).
To get the entire MacLaurin series you will want to calculate enough terms explictiely so that you can "guess" the general formula for all n (so far we have \(\displaystyle c_0= \frac{1}{a}\), \(\displaystyle c_1= -\frac{b}{a^2}\), and \(\displaystyle c_2= \frac{b^2}{a^3}\) so that shouldn't be too hard!)
HallsofIvy
Elite Member
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- Jan 27, 2012
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- 7,763
That doesn't match the values I got. I got that the first three coefficients are \(\displaystyle c_0= \frac{1}{a}\), \(\displaystyle c_2= -\frac{b}{a^2}\), and \(\displaystyle c_3= \frac{b^2}{a^3}\) while your formula gives \(\displaystyle c_0= \frac{a^0}{b^{-1}}= b\), \(\displaystyle c_2= \frac{a^2}{b^0}= a^2\), and \(\displaystyle c_3= \frac{a^3}{b^2}\).