Proof And Series Evaluation: X² = Π²/3 + ...
Hey guys! Let's dive into a fascinating mathematical exploration. We're going to tackle the proof of the equation x² = π²/3 + 4∑n=1 to ∞, and then use this result to evaluate some pretty cool infinite series. Buckle up, because this is going to be a fun ride!
Proving x² = π²/3 + 4∑n=1 to ∞
To kick things off, we need to establish a solid foundation. This equation connects a simple quadratic term (x²) to an infinite series involving alternating terms and squared integers. To conquer this, we'll leverage the power of Fourier series.
The Power of Fourier Series
The Fourier series, at its heart, is a way to represent periodic functions as a sum of sines and cosines. Think of it as breaking down a complex wave into its simpler, sinusoidal building blocks. This is super useful in many areas, from signal processing to solving differential equations.
The general form of a Fourier series for a function f(x) defined on the interval -π < x < π is:
f(x) = a₀/2 + ∑[n=1 to ∞] (aₙcos(nx) + bₙsin(nx))
Where the coefficients a₀, aₙ, and bₙ are calculated using integrals:
a₀ = (1/π) ∫[-π to π] f(x) dx aₙ = (1/π) ∫[-π to π] f(x)cos(nx) dx bₙ = (1/π) ∫[-π to π] f(x)sin(nx) dx
These formulas might look intimidating, but they're just a systematic way of figuring out how much of each sine and cosine wave is needed to reconstruct our original function.
Applying Fourier Series to f(x) = x²
Now, let's apply this to our specific problem. We'll consider the function f(x) = x² defined on the interval -π < x < π. This function is a parabola, and while it's not inherently periodic, we can treat it as periodic within this interval for the purpose of Fourier series representation.
First, we calculate the coefficients:
a₀ = (1/π) ∫[-π to π] x² dx = (1/π) [x³/3] from -π to π = (1/π) [(π³/3) - (-π³/3)] = (2π²)/3
Next, we find aₙ. This requires a bit of integration by parts:
aₙ = (1/π) ∫[-π to π] x²cos(nx) dx
After performing integration by parts (twice!), we arrive at:
aₙ = (4(-1)ⁿ)/n² for n ≥ 1
Finally, we calculate bₙ:
bₙ = (1/π) ∫[-π to π] x²sin(nx) dx
Since x² is an even function and sin(nx) is an odd function, their product is odd. The integral of an odd function over a symmetric interval (-π to π) is zero. Therefore, bₙ = 0 for all n.
Constructing the Fourier Series
Now that we have our coefficients, we can plug them back into the general Fourier series formula:
x² = a₀/2 + ∑[n=1 to ∞] (aₙcos(nx) + bₙsin(nx)) x² = (2π²/3)/2 + ∑[n=1 to ∞] (((4(-1)ⁿ)/n²)cos(nx) + 0*sin(nx)) x² = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)cos(nx)
This is a major step! We've expressed x² as a Fourier series. To get to our target equation, we need to choose a specific value for x.
Setting x = 0
Let's set x = 0 in the above equation:
0² = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)cos(0)
Since cos(0) = 1, this simplifies to:
0 = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)
Rearranging, we get:
x² = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)
Boom! We've successfully proven the initial equation.
Evaluating Infinite Series
Now comes the fun part – using our proven equation to evaluate some important infinite series.
(i) Evaluating ∑[n=1 to ∞](1 / n²) = π²/6
To tackle this, we'll strategically choose another value for x in our Fourier series representation.
Setting x = π
Let's set x = π in the equation:
x² = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)cos(nx)
π² = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)cos(nπ)
Since cos(nπ) = (-1)ⁿ, we have:
π² = π²/3 + 4∑[n=1 to ∞] ((-1)ⁿ/n²)(-1)ⁿ
π² = π²/3 + 4∑[n=1 to ∞] (1/n²)
Now, let's isolate the summation term:
π² - π²/3 = 4∑[n=1 to ∞] (1/n²)
(2π²)/3 = 4∑[n=1 to ∞] (1/n²)
∑[n=1 to ∞] (1/n²) = (2π²)/(3*4)
∑[n=1 to ∞] (1/n²) = π²/6
Fantastic! We've shown that the sum of the reciprocals of the squares of all positive integers converges to π²/6. This is a famous result known as the Basel problem, and it's pretty darn cool.
(ii) Evaluating ∑[n=1 to ∞](1 / (2n-1)²) = π²/8
This time, we're interested in the sum of the reciprocals of the squares of only the odd positive integers. We can achieve this by cleverly manipulating the series we just evaluated.
We know:
∑[n=1 to ∞] (1/n²) = 1/1² + 1/2² + 1/3² + 1/4² + ... = π²/6
Let's separate the terms with even denominators from those with odd denominators:
∑[n=1 to ∞] (1/n²) = (1/1² + 1/3² + 1/5² + ...) + (1/2² + 1/4² + 1/6² + ...)
Now, let's focus on the sum of the terms with even denominators. We can rewrite it as:
1/2² + 1/4² + 1/6² + ... = 1/(21)² + 1/(22)² + 1/(2*3)² + ...
= (1/4)(1/1² + 1/2² + 1/3² + ...)
= (1/4)∑[n=1 to ∞] (1/n²)
We already know that ∑[n=1 to ∞] (1/n²) = π²/6, so:
1/2² + 1/4² + 1/6² + ... = (1/4)(π²/6) = π²/24
Now, let's substitute this back into our separated series:
π²/6 = (1/1² + 1/3² + 1/5² + ...) + π²/24
Isolate the sum of the reciprocals of the squares of odd integers:
(1/1² + 1/3² + 1/5² + ...) = π²/6 - π²/24
(1/1² + 1/3² + 1/5² + ...) = (4π² - π²)/24
∑[n=1 to ∞](1 / (2n-1)²) = 3π²/24
∑[n=1 to ∞](1 / (2n-1)²) = π²/8
Awesome! We've successfully evaluated the sum of the reciprocals of the squares of odd positive integers and found it to be π²/8.
Discussion
So, what have we learned? We've not only proven a fascinating equation relating x² to an infinite series, but we've also used it as a powerful tool to evaluate other important series. This demonstrates the interconnectedness of mathematical concepts and the beauty of using one result to unlock others.
The Fourier series is a fundamental concept in mathematical analysis and has applications in various fields, such as physics, engineering, and computer science. Understanding how to derive and manipulate Fourier series allows us to solve complex problems involving periodic phenomena.
Furthermore, the evaluation of these infinite series highlights the elegance and power of mathematical reasoning. By strategically choosing values and manipulating series, we can arrive at closed-form expressions for seemingly complicated sums.
This exploration hopefully showcases the thrill of mathematical discovery, where seemingly abstract concepts can lead to concrete and beautiful results. Keep exploring, keep questioning, and keep the mathematical fire burning! You rock!