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Use this list of Basic Taylor Series to find the Taylor Series for f(x) = 2x 1+x2 based at 0. Give your answer using summation notation and give the largest open interval on which the series converges. (If you need to enter ∞ , use the ∞ button in CalcPad or type "infinity" in all lower-case.) The Taylor series for f(x)= 2x 1+x2 is: ∞ k=0 The Taylor series converges to f(x) for |x|< . Use the Taylor series you just found for f(x)= 2x 1+x2 to find the Taylor series for ln(1+x2) based at 0. Give your answer using summation notation and give the largest open interval on which the series converges. The Taylor series for ln(1+x2) is: ∞ k=0 The Taylor series converges to ln(1+x2) for |x|< .

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Recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]

Then for [tex]|-x^2|<1[/tex], or [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^nx^{2n}[/tex]

Multiply this series by [tex]2x[/tex] to get the Taylor series for [tex]f(x)[/tex]:

[tex]f(x)=\dfrac{2x}{1+x^2}=\displaystyle2\sum_{n=0}^\infty(-1)^nx^{2n+1}[/tex]

Notice that

[tex]\dfrac{\mathrm d(\ln(1+x^2))}{\mathrm dx}=\dfrac{2x}{1+x^2}[/tex]

so to find the Taylor series for [tex]g(x)=\ln(1+x^2)[/tex], we integrate the Taylor series for [tex]f(x)[/tex]:

[tex]g(x)=\displaystyle\int f(x)\,\mathrm dx=C+2\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{2n+2}[/tex]

Since [tex]g(0)=\ln(1+0^2)=\ln1=0[/tex], it follows that [tex]C=0[/tex] and

[tex]g(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{n+1}[/tex]

which converges for [tex]|x|<1[/tex] as well.

Following are the calculation to the Taylor Series:

Geometric series:

[tex]\to \Sigma_{\infty}^{n=0} \ a+ar+ar^2+.....+[/tex] which converges to [tex]\frac{a}{1-r} \ \ for\ \ |r| < 1[/tex].

Remembering that:

[tex]\to \frac{2x}{1+x^2}=\frac{2x}{1-(-x^2)}[/tex]  Taking  [tex]a=2x \ \ \ \ \ \ \ \ \ r=-x^2\\\\[/tex]

Using the Taylor series:

[tex]\to \frac{2x}{1+x^2}= \Sigma_{\infity}^{n=0} \ 2x \times (-x^2)^{n} = \Sigma_{\infity}^{n=0} 2x \times (-1)^{n} \times (x^{2n})[/tex]

[tex]\to \frac{2x}{1+x^2}= \Sigma_{\infty}^{n=0} (-1)^{n} \times 2 \times x^{2n+1} = 2x -2x^3+2x^5+............+[/tex]

In the given scenario we will converge the [tex]|x^2| < 1 |x| < 1[/tex].  Now, realize:

[tex]\to \frac{2x}{1+x^2} \ dx = \In (1+x^2) \\\\[/tex]

Integrating the series for [tex]\frac{2x}{1+x^2}[/tex]  :

[tex]\to \In( 1+x^2)=\int \Sigma_{\infty}^{n=0} \ (-1)^{n} \times 2 \times x^{2n+1}\ dx\\\\[/tex]

                  [tex]=\Sigma_{\infity}^{n=0} \frac{(-1)^{n} \times 2 \times x^{2n+2}}{ 2n+2}\\\\=\Sigma_{\infity}^{n=0} \frac{(-1)^{n} \times x^{2n+2}}{ n+1}\\\\= x^2 -\frac{x^4}{2}+\frac{x^6}{3}+............+[/tex]

Since integrating a number has no effect on its radius of converge, this series similarly converges for [tex]\bold{|x| < 1}[/tex].

Learn more about Basic Taylor Series:

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