Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

f(x+iy)=sum_(k=1)^(oo)(1)/(k^((x+iy)))

f(x+iy)=k=1(1k(x+iy))f(x+iy)=\sum_{k=1}^{\infty}\left(\frac{1}{k^{(x+iy)}}\right)

Full solution

Q. f(x+iy)=k=1(1k(x+iy))f(x+iy)=\sum_{k=1}^{\infty}\left(\frac{1}{k^{(x+iy)}}\right)
  1. Complex Exponents Properties: The given function f(x+iy)f(x+iy) is a complex series where each term is of the form 1kx+iy\frac{1}{k^{x+iy}}. To evaluate this series, we need to understand the properties of complex exponents.
  2. Euler's Formula Application: The term k(x+iy)k^{(x+iy)} can be rewritten using Euler's formula, which states that eiθ=cos(θ)+isin(θ)e^{i\theta} = \cos(\theta) + i\sin(\theta). Therefore, k(x+iy)=kxkiy=kx(cos(ylog(k))+isin(ylog(k)))k^{(x+iy)} = k^x \cdot k^{iy} = k^x \cdot (\cos(y\log(k)) + i\sin(y\log(k))).
  3. General Term Expression: Now we can express the term (1)/(k(x+iy))(1)/(k^{(x+iy)}) as (1)/(kx(cos(ylog(k))+isin(ylog(k))))(1)/(k^x * (\cos(y\log(k)) + i\sin(y\log(k)))). This is the general term of the series f(x+iy)f(x+iy).
  4. Dirichlet Series Convergence: The series f(x+iy)=k=11kx+iyf(x+iy) = \sum_{k=1}^{\infty}\frac{1}{k^{x+iy}} is known as the Dirichlet series, which converges if the real part of the exponent, xx, is greater than 11. We need to ensure that x > 1 for the series to converge.
  5. Convergence Conditions: If x > 1, the series converges absolutely and uniformly on compact subsets of the half-plane \text{Re}(s) > 1, where s=x+iys = x + iy. This is a condition for the convergence of the Dirichlet series.
  6. Series Evaluation Approach: Since the problem does not specify the value of xx, we cannot directly evaluate the series without knowing whether x > 1. If x1x \leq 1, the series may diverge or require a different approach to evaluate.
  7. Special Cases Consideration: Assuming x > 1, the series converges, but evaluating it to a closed-form expression is generally not possible except for special cases like the Riemann zeta function, where y=0y = 0 and xx is a positive integer greater than 11.
  8. Final Evaluation Conclusion: Without additional information on the value of xx and yy, we cannot provide a simplified exact answer for the series f(x+iy)f(x+iy). The series is defined and converges for x > 1, but its evaluation is not straightforward.

More problems from Evaluate definite integrals using the chain rule