How do you find a power series representation for #(x-2)^n/(n^2) # and what is the radius of convergence?

Answer 1

#x in (1,3)# and #sum_{i=1}^{infty}(x-2)^i/i^2 = "PolyLog"(2,x-2)#

Suppose that our quest is for #sum_{i=1}^{infty}(x-2)^i/i^2#.
In this case a variable change #y = (x-2)# give
#sigma(y)=sum_{i=1}^{infty}y^i/i^2#.
In the next steps we will try to build such a power series.
We begin with #sigma_1(y)= sum_{i=0}^{infty}y^n#.
This power series is convergent for #abs(y) < 1# and in this interval is equivalent to

#sigma_1^a(y)= 1/(1-y)#

now integrating #sigma_1(y)# we get

#f(y) = int sigma_1(y)dy = sum_{i=1}^{infty}y^i/i = y sum_{i=1}^{infty}y^{i-1}/i = y sigma_2(y)#

#sigma_2(y)=sum_{i=1}^{infty}y^{i-1}/i# is equivalent to

#sigma_2^a(y)=(int sigma_1^a(y)dy)/y = -(log_e(1-y))/y#

The last step is covered by integrating #sigma_2(y)#.

#int sigma_2(y)dy = sum_{i=1}^{infty}y^i/i^2 = sigma(y)#

or equivalently

#sigma^a(y) = int sigma_2^a(y)dy = int -(log_e(1-y))/y dy= "PolyLog"(2,y)#

The description of function PolyLog can be found in

https://tutor.hix.ai

The convergence interval in #x# is a consequence of #y# interval and is #x in (1,3)#.

The attached figure shows the agreement between #sigma(y)# and #sigma^a(y)#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2
To find the power series representation for \(\frac{(x-2)^n}{n^2}\), we can start with a known power series representation and perform algebraic manipulations to match the given function. Since \((x-2)^n\) resembles the binomial theorem, we'll use it as a starting point. The binomial theorem states: \((1 + t)^n = \sum_{k=0}^{\infty} \binom{n}{k} t^k\) We can substitute \(t = x - 2\) into this formula: \((1 + (x - 2))^n = \sum_{k=0}^{\infty} \binom{n}{k} (x - 2)^k\) So, we have: \(\frac{(x - 2)^n}{n^2} = \frac{1}{n^2} \cdot \sum_{k=0}^{\infty} \binom{n}{k} (x - 2)^k\) Now, let's take \(\frac{1}{n^2}\) outside the summation: \(\frac{(x - 2)^n}{n^2} = \sum_{k=0}^{\infty} \frac{\binom{n}{k}}{n^2} (x - 2)^k\) Using the fact that \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), we can rewrite it as: \(\frac{(x - 2)^n}{n^2} = \sum_{k=0}^{\infty} \frac{n!}{k!(n-k)!n^2} (x - 2)^k\) Now, let's consider \(n^2 = n \cdot n\), we can rewrite it as: \(\frac{(x - 2)^n}{n^2} = \sum_{k=0}^{\infty} \frac{n!}{k!(n-k)!n \cdot n} (x - 2)^k\) This simplifies to: \(\frac{(x - 2)^n}{n^2} = \sum_{k=0}^{\infty} \frac{1}{k!n} \cdot \frac{n!}{(n-k)!} (x - 2)^k\) Further simplification gives: \(\frac{(x - 2)^n}{n^2} = \sum_{k=0}^{\infty} \frac{1}{k!n} \cdot \frac{n(n-1)(n-2)...(n-k+1)}{(n-k)!} (x - 2)^k\) Finally, we observe that \(\frac{n(n-1)(n-2)...(n-k+1)}{(n-k)!} = n(n-1)(n-2)...(n-k+1)\), so the series becomes: \(\frac{(x - 2)^n}{n^2} = \sum_{k=0}^{\infty} \frac{1}{k!} \cdot \frac{n(n-1)(n-2)...(n-k+1)}{n^k} (x - 2)^k\) Now, the radius of convergence \(R\) can be found using the ratio test. The ratio test states that if \(\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = L\), then the series converges if \(L < 1\) and diverges if \(L > 1\). Here, \(a_k\) represents the \(k\)-th term of the series. Let's apply the ratio test to our series: \[ L = \lim_{k \to \infty} \left| \frac{\frac{1}{(k+1)!} \cdot \frac{n(n-1)(n-2)...(n-k)}{n^{k+1}} (x - 2)^{k+1}}{\frac{1}{k!} \cdot \frac{n(n-1)(n-2)...(n-k+1)}{n^k} (x - 2)^k} \right| \] Simplify this expression and then take the limit as \(k\) approaches infinity to find \(L\). After finding \(L\), set it less than 1 and solve for \(x\), then the distance between \(x\) and 2 will give you the radius of convergence \(R\).
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7