How do you find a power series representation for #ln(1x^2) # and what is the radius of convergence?
The radius of convergence is equal to
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To find a power series representation for ( \ln(1x^2) ), you can use the known Maclaurin series expansion for ( \ln(1+x) ).

Maclaurin series for ( \ln(1+x) ) is: [ \ln(1+x) = x  \frac{x^2}{2} + \frac{x^3}{3}  \frac{x^4}{4} + \dots ] for ( x < 1 ).

Replace ( x ) with ( x^2 ): [ \ln(1x^2) = x^2  \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3}  \frac{(x^2)^4}{4} + \dots ]

Simplify each term: [ \ln(1x^2) = x^2  \frac{x^4}{2}  \frac{x^6}{3}  \frac{x^8}{4} + \dots ]
This is the power series representation for ( \ln(1x^2) ).
The radius of convergence ( R ) for this series can be found using the ratio test:
[ R = \lim_{n \to \infty} \left \frac{a_{n}}{a_{n+1}} \right ]
For this series, the general term ( a_n ) is ( \frac{(1)^n x^{2n}}{n} ).
[ \lim_{n \to \infty} \left \frac{\frac{(1)^n x^{2n}}{n}}{\frac{(1)^{n+1} x^{2(n+1)}}{n+1}} \right ]
[ = \lim_{n \to \infty} \left \frac{n+1}{n} x^2 \right ]
[ = x^2 ]
Since this limit should be less than 1 for convergence: [ x^2 < 1 ]
Thus, the radius of convergence ( R ) is 1.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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