I'm having a doubt on solving the below question.
Question: drive google(dot)com/file/d/1OOL_WecZgLk3fcd65UxVzec4T9Y7ztuB/view?usp=sharing
Apparently we can write t as, t = (4/pi)* ( (-1)^(n+1) ) * [sigma n=1-infinity { (sin((n*pi*t)/2) } ]
Hw can I find the Fourier series of f(t) = t^2?
Can I take bn = (4/pi)*((-1)^(n+1)) and use Parsvel's identity to find the Fourier series of f(t) = t^2, which is the square of t?
Or else, if we're to use the property that t^2 = integration of 0.5*t, how can we do it?
What's the most ideal way to solve this problem?
Thank you in advance for your help!
Well, I must admit, I haven't seen such a specific question in a while! But fear not, I'll do my best to clown my way through it!
To find the Fourier series of f(t) = t^2, you're on the right track! You can indeed use Parseval's Identity to determine the coefficients. However, the expression you mentioned for bn seems a little off. It should actually be bn = (2/T) * ∫[T/2,-T/2] (t^2 * sin(nωt)) dt, where T is the period and ω = 2π/T. Just a minor adjustment there!
Now, finding the Fourier series using the integration property f(t) = integration of 0.5*t is an intriguing idea. However, it might make things a tad complicated. You see, integrating over t in this case would give you a function with a cubic term. And trust me, working with cubes can get quite dicey. So, I'd suggest sticking with the traditional method I mentioned earlier.
I hope this helps you shine some light on your doubt! If you have any more questions or need further clarification, feel free to ask. I'm here to tickle your funny bone and assist you with your doubts!
To find the Fourier series of f(t) = t^2, we can use the property of linearity of the Fourier series. However, it is important to note that the Fourier series represents periodic functions, and f(t) = t^2 is not periodic. Therefore, we need to find a periodic extension of f(t) before computing its Fourier series.
One way to make f(t) periodic is to consider its odd extension. In other words, we can extend f(t) to be an odd function in the interval [-L, L], where L is a sufficiently large number. Since f(t) = t^2 is even around the origin, its odd extension will be symmetric about the origin.
To find the Fourier series coefficients using the odd extension, we can evaluate the following integrals:
an = (2/L) ∫[-L, L] f(t) * cos((nπt)/L) dt
bn = (2/L) ∫[-L, L] f(t) * sin((nπt)/L) dt
In this case, f(t) = t^2, and we can choose any suitable value for L.
Using Parseval's identity, which relates the energy of a function to its Fourier series coefficients, is another approach to finding the Fourier series.
Parseval's identity states that for a periodic function g(t) with period T and its Fourier series representation g(t) = ∑[n=-∞ to +∞] (a_n * cos((nπt)/T) + b_n * sin((nπt)/T)), the energy of g(t) over one period is given by:
(1/T) ∫[0, T] |g(t)|^2 dt = (1/2) (|a0|^2 + ∑[n=1 to +∞] (|an|^2 + |bn|^2))
By integrating |g(t)|^2 over a period and setting it equal to (1/2) (|a0|^2 + ∑[n=1 to +∞] (|an|^2 + |bn|^2)), we can solve for the Fourier series coefficients.
Now, to compute the Fourier series of f(t) = t^2 using the property f(t) = 0.5 * t, you can express f(t) as a sum of infinite sine functions using the identity t = 2[(sin(t/2))^2]. Then, you have to find the Fourier series of each term separately and sum them up to obtain the Fourier series of f(t) = t^2.
The most ideal way to solve this problem would be to use the approach of finding the odd extension of f(t) and then evaluating the integrals for the Fourier series coefficients as mentioned earlier. This method accounts for the periodicity of the functions involved and provides a direct way to obtain the coefficients.
Hope this explanation helps! Let me know if you have further questions.
To find the Fourier series of the function f(t) = t^2, we can use a combination of approaches. Let's take a step-by-step approach to solve this problem.
Step 1: Express f(t) as an odd periodic function
The function f(t) = t^2 is an even function, so we need to express it as an odd periodic function. We can do this by extending f(t) to its periodic extension in the interval -π to π, with f(-π) = f(π).
Step 2: Express f(t) in terms of the sine function
Since f(t) is an odd function, we can express it as a sum of sine functions. We can use the formula for the Fourier series of an odd periodic function:
f(t) = a0 + Σ(n=1 to ∞) [an * sin(nωt)]
Step 3: Calculate the Fourier coefficients
To find the Fourier coefficients, we need to calculate the coefficients an. We can use the formula:
an = (2/π) ∫[-π, π] f(t) * sin(nωt) dt
where ω represents the fundamental frequency, which is 2π/T with T being the period (in this case, 2π).
Since f(t) = t^2, we can substitute this into the Fourier coefficient formula and evaluate the integral.
Step 4: Simplify the result
Once we have the Fourier coefficients, we can simplify the result by combining terms and using mathematical properties.
The integral in Step 3 can be evaluated using integration techniques, such as integration by parts or trigonometric identities, depending on the complexity of the integrand.
Using the property t^2 = (1/2) * t * t, as mentioned in your question, can also be a valid approach. In this case, you can rewrite f(t) = t^2 as f(t) = (1/2) * t * t and find the Fourier series of (1/2) * t. Then, you can square the resulting series to find the Fourier series of f(t) = t^2.
Both approaches should yield the same result, so choose the method that you are more comfortable with and that suits your problem best.
Remember to be careful with the algebraic manipulations and calculations along the way. It's recommended to double-check your work and simplify the final result as much as possible.
I hope this explanation helps you solve the problem. If you have any further questions, please let me know!