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Jet001 [13]
4 years ago
15

Find the radius of convergence and the internval of convergence:

Mathematics
1 answer:
o-na [289]4 years ago
5 0
You can use the root test here. The series will converge if

L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}

You have

L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}=|4-x|\lim_{n\to\infty}\frac1{\sqrt[n]{4^n+9^n}}

Notice that

\dfrac1{\sqrt[n]{4^n+9^n}}=\dfrac1{\sqrt[n]{9^n}\sqrt[n]{1+\left(\frac49\right)^n}}=\dfrac1{9\sqrt[n]{1+\left(\frac49\right)^n}}

so as n\to\infty, you have \left(\dfrac49\right)^n\to0, which means you end up with

L=\dfrac{|4x-1|}9

This is the interval of convergence. The radius of convergence can be determined by finding the half-length of the interval, or by solving the inequality in terms of |x-c| so that R is the ROC. You get

|4x-1|
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We are told to translate; (x, y) to (x -8, y). This means we have to add - 8 to each value of x in P(-5,1), Q(-4,6), and R(-2,3).

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For the dilation centered at the origin k =2, simply multiply the value of k, which is 2 into the translations.

\begin{gathered} At\text{ K = 2,  }P^i(-13,\text{ 1)  }\rightarrow\text{   }2(-13,\text{ 1)  }\rightarrow\text{  }P^{ii}(-26,\text{ 2)} \\ Q^i(-12,\text{ 6)  }\rightarrow\text{  }2(-12,\text{ 6)  }\rightarrow\text{   Q}^{ii}(-24,\text{ 12)} \\ R^i(-10,\text{ 3)   }\rightarrow\text{   }2(-10,\text{ 3)  }\rightarrow\text{   R}^{ii}(-20,\text{ 6)} \end{gathered}

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1 year ago
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