How do you find a power series representation for # f(x)=(x / (x^(2)-3x+2) )# and what is the radius of convergence?

Answer 1

#S(x) = sum_{n=0}^oo(1-1/2^n)x^n#.
This serie is convergent to #f(x)# for #abs(x) < 1#

#f(x) = x/(x^2 - 3 x + 2) = x/((x-1)(x-2)) = A/(x-1)+B/(x-2)#
#(A(x-2)+B(x-1))/((x-1)(x-2))=((A+B)x-(2A+B))/((x-1)(x-2))#
Solving for #A,B#

#{ (A+B=1), (2A+B=0) :}#

we obtain #A=-1,B=2# So
#f(x) = -1/(x-1)+2/(x-2)#
Now considering that #(x^{n+1}-1)/(x-1)=1+x+x^2+cdots+x^n#
if #abs(x) < 1# we know
#lim_{n->oo}(x^{n+1}-1)/(x-1) = -1/(x-1) = sum_{n=0}^oo x^n#
Using those results and supposing that #abs(x) < 1# we propose
#S(x) = -1/(x-1)+2/(x-2) = sum_{n=0}^oo x^n-sum_{n=0}^oo (x/2)^n#
so #S(x) = sum_{n=0}^oo(1-1/2^n)x^n#.
This serie is convergent to #f(x)# for #abs(x) < 1#
Supposing you need a series representation for #f(x)# in the surroundings of #x=3# you can proceed as follows.
Making #y = x-3# then #f(y) = -1/(y+2)+2/(y+1)#
so with #abs(y) < 1#
#f(y)=lim_{n->oo}{-(1-(-y/2)^n)/(2(1-(-y/2)))+2(1-(-y)^n)/(1-(-y))}#

or

#f(y) = -1/2 sum_{n=0}^oo(-y/2)^n+2sum_{n=0}^n(-y)^n#

Finally

#S(x) = sum_{n=0}^oo (-1)^n(2-1/2^{n+1})(x-3)^n#
This series converges to #f(x)# for #2 < x < 4#
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Answer 2

To find the power series representation for ( f(x) = \frac{x}{x^2 - 3x + 2} ), we first factor the denominator to ( (x - 1)(x - 2) ). Then, we express ( f(x) ) as a partial fraction:

[ f(x) = \frac{x}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} ]

Solving for ( A ) and ( B ), we get ( A = -2 ) and ( B = 3 ). Now, we can rewrite ( f(x) ) as:

[ f(x) = \frac{-2}{x - 1} + \frac{3}{x - 2} ]

Next, we express each term as a geometric series:

[ \frac{-2}{x - 1} = -2 \sum_{n=0}^{\infty} (x - 1)^n ] [ \frac{3}{x - 2} = 3 \sum_{n=0}^{\infty} (x - 2)^n ]

Combining these, we have:

[ f(x) = -2 \sum_{n=0}^{\infty} (x - 1)^n + 3 \sum_{n=0}^{\infty} (x - 2)^n ]

Finally, we obtain the power series representation for ( f(x) ):

[ f(x) = -2 \sum_{n=0}^{\infty} (x - 1)^n + 3 \sum_{n=0}^{\infty} (x - 2)^n ]

The radius of convergence of this power series is the distance from the center of the series to the nearest singularity. In this case, the singularities occur at ( x = 1 ) and ( x = 2 ). Therefore, the radius of convergence is the distance from the center, which is ( x = 1.5 ), to the nearest singularity, which is ( x = 1 ) or ( x = 2 ). So, the radius of convergence is ( 0.5 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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