summesion from n=0 to n = infinite of the expresion
cos(ntehtha)/(3 to the power n) = 1 find cos Thetha

We are tasked with solving the infinite sum:

\[
\sum_{n=0}^{\infty} \frac{\cos(n\theta)}{3^n} = 1
\]

This is a geometric series involving the cosine function. Let's analyze the sum step by step.

### Step 1: General Formula for an Infinite Geometric Series
The sum of an infinite geometric series with the first term \( a_0 \) and common ratio \( r \) is given by:

\[
S = \frac{a_0}{1 - r}
\]

This formula holds if \( |r| < 1 \). In our case, each term involves \( \cos(n\theta) \) and the factor \( \frac{1}{3^n} \), so we need to treat the sum more carefully.

### Step 2: Apply Geometric Series to the Given Problem
We have the series:

\[
S = \sum_{n=0}^{\infty} \frac{\cos(n\theta)}{3^n}
\]

This can be treated as a geometric series with the first term \( \frac{\cos(0 \cdot \theta)}{3^0} = 1 \), and the common ratio involving the cosine term and \( \frac{1}{3} \).

We can recognize this sum as a standard result involving the sum of cosines in the context of infinite series. The formula for such a sum is:

\[
\sum_{n=0}^{\infty} \frac{\cos(n\theta)}{r^n} = \frac{1 - r\cos(\theta)}{1 - 2r\cos(\theta) + r^2}
\]

Here, \( r = \frac{1}{3} \). So the sum becomes:

\[
\sum_{n=0}^{\infty} \frac{\cos(n\theta)}{3^n} = \frac{1 - \frac{1}{3}\cos(\theta)}{1 - 2\frac{1}{3}\cos(\theta) + \left(\frac{1}{3}\right)^2}
\]

### Step 3: Simplify the Expression
Simplifying the denominator:

\[
S = \frac{1 - \frac{1}{3}\cos(\theta)}{1 - \frac{2}{3}\cos(\theta) + \frac{1}{9}}
\]

We are given that this sum equals 1:

\[
\frac{1 - \frac{1}{3}\cos(\theta)}{1 - \frac{2}{3}\cos(\theta) + \frac{1}{9}} = 1
\]

### Step 4: Solve for \( \cos(\theta) \)
Multiply both sides by the denominator:

\[
1 - \frac{1}{3}\cos(\theta) = 1 - \frac{2}{3}\cos(\theta) + \frac{1}{9}
\]

Simplify the equation:

\[
1 - \frac{1}{3}\cos(\theta) = \frac{10}{9} - \frac{2}{3}\cos(\theta)
\]

Multiply the entire equation by 9 to eliminate the fractions:

\[
9 - 3\cos(\theta) = 10 - 6\cos(\theta)
\]

Simplify:

\[
-3\cos(\theta) + 6\cos(\theta) = 10 - 9
\]

\[
3\cos(\theta) = 1
\]

Thus:

\[
\cos(\theta) = \frac{1}{3}
\]

### Final Answer:
\[
\cos(\theta) = \frac{1}{3}
\]

Do you need any details or clarifications on this?

Here are 5 related questions:
1. How do we derive the sum of cosines in a geometric series?
2. What are the conditions for convergence of an infinite series?
3. How can we solve similar trigonometric series with different coefficients?
4. What happens if the sum of the series were different from 1?
5. How would this problem change if \( \cos(\theta) \) were replaced by another trigonometric function?

**Tip:** In problems involving infinite series, ensure that the common ratio satisfies \( |r| < 1 \) for convergence.

summation from n=0 to n = infinite of the expression cos(n theta)/(3 to the power n) = 1, find cos(theta)

Explore how to solve an infinite series involving cosines. This problem uses advanced concepts like geometric series and trigonometric identities to determine cos(θ) in the equation Σ(cos(nθ)/3^n) = 1. Suitable for university-level students studying advanced calculus or mathematical analysis.

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