
How do you find a power series representation for \[\dfrac{10x}{14+x}\] and what is the radius of convergence?
Answer
535.5k+ 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 \[\Rightarrow \dfrac{1}{1-x}=1+x+{{x}^{2}}+......\infty \]
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

