Start by representing the Taylor series as a power series
Sometimes we’ll be asked because that the radius and interval that convergence of a Taylor series. In stimulate to discover these things, we’ll first have to uncover a power series representation because that the Taylor series.
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Once we have the Taylor series represented together a power series, we’ll recognize ???a_n??? and also ???a_n+1??? and also plug them into the limit formula from the proportion test in order to say where the series is convergent.
Step-by-step solution for detect the radius and interval the convergence
Example
Using the graph below, find the third-degree Taylor series about ???a=3??? because that ???f(x)=\ln(2x)???. Then find the power series representation of the Taylor series, and the radius and interval that convergence.
Taylor series
Since we currently have the graph done, the worth in the far right obelisk becomes the coefficient on each term in the Taylor polynomial, in the form
???\fracf^(n)(a)n!(x-a)^n???
With the whole chart filled in, us can develop each hatchet of the Taylor polynomial.
Putting all of the state together, we acquire the third-degree Taylor polynomial.
???\ln6+\frac13(x-3)-\frac118(x-3)^2+\frac181(x-3)^3???
Power collection representation
We desire to discover a power collection representation because that the Taylor series above. The very first thing we have the right to see is the the exponent of each ???(x-3)??? is same to the ???n??? value of the term, which means that
???(x-3)^n???
will be component of the power series representation. The fountain coefficient in front of the ???(x-3)??? terms deserve to be represented by
???\frac1n3^n???
Finally, we need to resolve the negative sign in front of the???n=2??? term. If we multiply our terms by
???(-1)^n+1???
the ???n=2??? term will certainly be an unfavorable and the ???n=1??? and ???n=3??? terms will be positive. Remember, nobody of this generalizations use to our ???n=0??? term, therefore we’ll leave this term outside of the power series representation.
???\ln6+\sum^\infty_n=1\frac(-1)^n+1(x-3)^nn3^n???
Notice the amount starts at ???n=1???, due to the fact that the ???n=0??? term is not included in the sum.
Radius and also interval of convergence
To discover the radius the convergence, we’ll recognize ???a_n??? and ???a_n+1??? making use of the power collection representation we simply found.
???a_n=\frac(-1)^n+1(x-3)^nn3^n???
???a_n+1=\frac(-1)^n+2(x-3)^n+13^n+1(n+1)???
We deserve to plug ???a_n??? and also ???a_n+1??? right into the border formula native the ratio test.
???L=\lim_n\to\infty\left|\fraca_n+1a_n\right|???
???L=\lim_n\to\infty\left|\frac\frac(-1)^n+2(x-3)^n+1(n+1)3^n+1\frac(-1)^n+1(x-3)^nn3^n\right|???
???L=\lim_n\to\infty\left|\frac(-1)^n+2(x-3)^n+1(n+1)3^n+1\cdot\fracn3^n(-1)^n+1(x-3)^n\right|???
???L=\lim_n\to\infty\left|\frac(-1)^n+2(-1)^n+1\cdot\frac(x-3)^n+1(x-3)^n\cdot\fracnn+1\cdot\frac3^n3^n+1\right|???
???L=\lim_n\to\infty\left|(-1)^n+2-(n+1)\cdot(x-3)^n+1-n\cdot\fracnn+1\cdot3^n-(n+1)\right|???
???L=\lim_n\to\infty\left|(-1)^n+2-n-1\cdot(x-3)^n+1-n\cdot3^n-n-1\cdot\fracnn+1\right|???
???L=\lim_n\to\infty\left|(-1)^1\cdot(x-3)^1\cdot3^-1\cdot\fracnn+1\right|???
???L=\lim_n\to\infty\left|-\frac13(x-3)\fracnn+1\right|???
Since we’re managing absolute value, the ???-1??? deserve to be removed.
???L=\lim_n\to\infty\left|\fracn(x-3)3(n+1)\right|???
The border only impacts ???n???, for this reason we have the right to remove the ???(x+3)???.
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???L=|x-3|\lim_n\to\infty\left|\fracn3(n+1)\right|???
???L=|x-3|\lim_n\to\infty\left|\fracn3n+3\right|???
Since we’ll get the indeterminate type ???\infty/\infty??? if we try to advice the limit, we’ll divide the numerator and denominator by the highest-degree change in order to mitigate the fraction.
???L=|x-3|\lim_n\to\infty\left|\fracn3n+3\left(\frac\frac1n\frac1n\right)\right|???
???L=|x-3|\lim_n\to\infty\left|\frac\fracnn\frac3nn+\frac3n\right|???