Use the power seriesTo determine a power series centered at 0, for the function.Identify the interval of convergence. f ( x ) = ln ⁡ ( x 2 + 1 )

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Use the power seriesTo determine a power series centered at 0, for the function.Identify the interval of convergence.  f  (  x  )  =  ln  ⁡  (      x          2        +  1  )

Jaycee Figueroa · Accepted Answer

g  (  x  )  =      1          1      +      x        =  1  −  x  +      x          2        −      x          3        +      x          4      We must recognize a connection between 1/(1+x) and the naturallog. well if we integrate what we have now we&#039;d end up with   ln  ⁡  (  1  +  x  ), so that&#039;s a start, but we want to end up with   ln  ⁡  (      x          2        +  1  ). So we&#039;d need   2  x      /    [  1  +      x          2        ].   So we have  g  (      x          2        )  =      1          1      +              x                  2                      =  1  −      x          2        +      x          4        −      x          6        +      x          8      Multiplying everything by x  x  g  (      x          2        )  =      x          1      +              x                  2                      =  x  −      x          3        +      x          5        −      x          7        +      x          9      And by 2  2  x  g  (      x          2        )  =            2      x              1      +              x                  2                      =  2  x  −  2      x          3        +  2      x          5        −  2      x          7        +  2      x          9      Integrating letting   u  =  1  +      x          2        d  u  =  2  x  d  x, we find  ln  ⁡  (  1  +      x          2        )  =      x          2        −            x              4              2    +            x              6              3    −            x              8              4    +            x              10              5  Hence  ln  ⁡  (  1  +      x          2        )  =      ∑          n      =      1              ∞        (  −  1      )          n      +      1                  x              2        n              n        lim          n      →      ∞            |              x              2        n        +        1                    n      +      1        ∗      n          x              2        n                  |    &amp;lt;  1      |    x      |        lim          n      →      ∞            |        n          n      +      1            |    &amp;lt;  1|x|&amp;lt;1Now we must check the endpoints individually. For x =1, we have a alternating harmonic series, which isconvergent.   Same for x = -1. So the intervalis [-1,1]

Greyson Landry · Accepted Answer

1  −  x  +      x    2    −      x    3    +      x    4    −      x    5    +      x    6    −      x    7    +      x    8    −      x    9    +      x          10        −      x          11        +      x          12        −      x          13        +      x          14        −      x          15        +  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  f  (  x  )  =  ln  ⁡  (      x    2    +  1  )  =      x    2    −      x          4              /            2        +      x          6              /            3        −      x          8              /            4        +      x          10              /            5        −      x          12              /            6        +  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .

Use the power series To determine a power series centered at 0, for the function.Identify the interval of convergence. f(x)=ln(x^2+1)

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