How do you find a power series representation for \[\dfrac{10x}{14+x}\] and what is the radius of convergence?
Answer
581.4k+ views
Hint: From the given question we have been asked to find a power series and the radius of convergence for a given expression. For this question we will bring the ten out and bring the given expression into the form of geometric power series which is \[\Rightarrow \dfrac{1}{1-x}=1+x+{{x}^{2}}+......\infty \] and we use the condition for this power series and find the required radius of convergence. So, we proceed with our solution as follows.
Complete step-by-step solution:
We are given in the question that,
\[\Rightarrow \dfrac{10x}{14+x}\]
We will bring the ten out side of the bracket and keep the rest of the term inside the bracket and simplify the remaining term. So, we get it as follows.
\[\Rightarrow 10\left( \dfrac{x}{14+x} \right)\]
We can rewrite it as follows.
\[\Rightarrow 10\left( 1-\dfrac{14}{14+x} \right)\]
Here we divide the fractional term inside bracket in both numerator and denominator with the integer \[14\]. So, we get the expression reduced as follows.
\[\Rightarrow 10-10\left( \dfrac{1}{1+\dfrac{x}{14}} \right)\]
Now comparing with the geometric power series \[\Rightarrow \dfrac{1}{1-x}=1+x+{{x}^{2}}+......\infty \] and writing \[-\dfrac{x}{14}\] for \[x\], the series would become as follows.
So, now we will use the substitution method and substitute the value of \[-\dfrac{x}{14}\] in the place of \[x\] in the geometric power series mentioned above. So, we get,
\[\Rightarrow 10-10\left( 1-\dfrac{x}{14}+{{\left( \dfrac{x}{14} \right)}^{2}}-{{\left( \dfrac{x}{14} \right)}^{3}}+......\infty \right)\]
\[\Rightarrow 10\left( \dfrac{x}{14}-{{\left( \dfrac{x}{14} \right)}^{2}}+{{\left( \dfrac{x}{14} \right)}^{3}}-......\infty \right)\]
The implied condition for a convergent geometric series represented by \[\dfrac{1}{1-x}\] is \[\Rightarrow -1 < x < 1\], hence in the present case it would be \[\Rightarrow -1 < -\dfrac{x}{14} < 1\] or \[\Rightarrow -14 < -x < 14\]
Which is \[\Rightarrow 14>x>-14\].
Thus, the radius of convergence is \[14\].
Note: Students must be very careful in doing the calculations. Students should have good knowledge in the concept of convergent geometric series and its properties. We must know the formulae given below to solve these kinds of problems.
The implied condition, \[\Rightarrow -1 < x < 1\].
The geometric power series \[\Rightarrow \dfrac{1}{1-x}=1+x+{{x}^{2}}+......\infty \]
Complete step-by-step solution:
We are given in the question that,
\[\Rightarrow \dfrac{10x}{14+x}\]
We will bring the ten out side of the bracket and keep the rest of the term inside the bracket and simplify the remaining term. So, we get it as follows.
\[\Rightarrow 10\left( \dfrac{x}{14+x} \right)\]
We can rewrite it as follows.
\[\Rightarrow 10\left( 1-\dfrac{14}{14+x} \right)\]
Here we divide the fractional term inside bracket in both numerator and denominator with the integer \[14\]. So, we get the expression reduced as follows.
\[\Rightarrow 10-10\left( \dfrac{1}{1+\dfrac{x}{14}} \right)\]
Now comparing with the geometric power series \[\Rightarrow \dfrac{1}{1-x}=1+x+{{x}^{2}}+......\infty \] and writing \[-\dfrac{x}{14}\] for \[x\], the series would become as follows.
So, now we will use the substitution method and substitute the value of \[-\dfrac{x}{14}\] in the place of \[x\] in the geometric power series mentioned above. So, we get,
\[\Rightarrow 10-10\left( 1-\dfrac{x}{14}+{{\left( \dfrac{x}{14} \right)}^{2}}-{{\left( \dfrac{x}{14} \right)}^{3}}+......\infty \right)\]
\[\Rightarrow 10\left( \dfrac{x}{14}-{{\left( \dfrac{x}{14} \right)}^{2}}+{{\left( \dfrac{x}{14} \right)}^{3}}-......\infty \right)\]
The implied condition for a convergent geometric series represented by \[\dfrac{1}{1-x}\] is \[\Rightarrow -1 < x < 1\], hence in the present case it would be \[\Rightarrow -1 < -\dfrac{x}{14} < 1\] or \[\Rightarrow -14 < -x < 14\]
Which is \[\Rightarrow 14>x>-14\].
Thus, the radius of convergence is \[14\].
Note: Students must be very careful in doing the calculations. Students should have good knowledge in the concept of convergent geometric series and its properties. We must know the formulae given below to solve these kinds of problems.
The implied condition, \[\Rightarrow -1 < x < 1\].
The geometric power series \[\Rightarrow \dfrac{1}{1-x}=1+x+{{x}^{2}}+......\infty \]
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