Fourier Series. This was the super annoying memorizing-the-formulas topic in my class, but it was actually one of the most interesting math concepts ever. So what is the Fourier series? What's the significance of it? For that we first have to understand what a periodic function is. A periodic function is a function where T > 0, and f(x+T) = f(x) for every value of x. The T is the period of f(x). An example of a periodic function is sin(x) and cos(x), which have a period of 2π. Fourier series is essentially a way to expand this periodic function to an infinite series involving a bunch of sines and cosines. So how do you derive the Fourier series of a periodic function? Let p > 0 and f(x) be a periodic function with period 2p, within the bounds of (-p, p). The Fourier series of f(x) is: where the a of n and a of 0 and b of n are Fourier coefficients ALSO, there are two things to keep in mind: assuming x is an integer,
the left graph shows the original f(x) and the right graph shows Fourier series estimation graph. When you graph the certain number of n terms of the fourier series, you will get a close approximation. This series is very similar to Taylor series, except Fourier series also works with discontinuous functions as well. Fourier sine seriesYou can obtain the Fourier sine and cosine series from the general formula. For the Fourier Sine Series, we assume that f(x) is an odd function, which means that f(-x) = -f(x). If thats the case, then the a of 0 and a of n terms become zero because an odd function (f(x)) multiplied by an even function (cos(nπx)) = an odd function. An interval from -p to p over an odd function is 0. an example of an odd function over an interval (-p, p). both the areas cancel each other out, which evaluates the integral to 0. Since a of n and a of 0 are both equal to 0, the function evaluates to the general Fourier series with just the b of n Fourier coefficient. Fourier cosine seriesYou obtain the Fourier cosine series when you assume that f(x) is an even function; which means that f(-x) = f(x). Since f(x) is an even function, we know that the b of n term in the general Fourier series equation is 0 because an even function (f(x)) times an odd function (sin(nπx)) is equal to an odd function. The integral from (from -p to p) of an odd function is always zero since the areas cancel each other. Therefore, with only the a of n an a of 0 terms, the Fourier series becomes the Fourier cosine series (with only cosines).
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The Cantor's Disappearing Table is a famous and super fun series problem regarding a table and whether it'll disappear by cutting only half of the table. Sounds incredibly absurd but it is super interesting to solve! Lets explore this problem in depth: We will start with a table, with the length L. Lets remove 1/4 of the table from the center, leaving two pieces in the sides. The remaining pieces each definitely have a length less than 1/2L. Then, lets remove 1/8 of the table by removing 1/16th from each of the remaining sides. Each of the remaining pieces have a length less than 1/4L. Next, we remove 1/16th of the desk by removing 1/64th from each previous remaining piece. Each of the remaining pieces have a length less than 1/8L. As we continuously remove this portions of the table, will the table eventually disappear? There are two ways to look at this problem. One way is focusing on the amounts removed from the table, and the other way to look at this problem is focusing on the amounts remaining. Amounts RemovedSo if we continuously remove the portions of the table in this pattern, the series works out to be: 1/4L + 1/8L +1/16L + 1/32L + .... = (n=2 to ∞) ∑ (1/2^n)L = 1/2L. This means, that we will actually remove half of the table by following this process. I obtained 1/2L by: S = 1/4L + 1/8L +1/16L + 1/32L + .... S = (1/2)^2 + (1/2)^3 + (1/2)^4 + (1/2)^5... = (1/2)^2 + 1/2( (1/2)^2 + (1/2)^3 + (1/2)^4) = S = (1/2)^2 + 1/2(S) 1/2S = 1/4 S = 1/2 So this leads to the conclusion that only half of the table will ultimately disappear but the entire table won't. Seems really absurd, but it is mathematically proven to be correct in this way. Amounts RemaningSo if we focus on the amounts remaining, we will end up with a different conclusion: that the table does eventually disappear. Lets make a series for the remaining pieces: S = 1/2L + 1/4L +1/8L + 1/16L + .... In order to find out how much is remaining: ; lim(as n approaches ∞) (1/2)^n = (1/2)^∞ = 0. So this means that the remaining lengths will approach 0L. This means that the pieces will ultimately have a length of 0. This proves that the table eventually disappears. We can conclude that there are two different sides to this problem. It just depends on which way you choose to look at it, whether it is the remaining pieces or the removed pieces. This problem was so fun to solve and it was so interesting to prove the two different perspectives!! Good luck as always!
A sequence is a function whose domain is basically a set of positive integers. We denote these list of values with a variable and a subscript of 1, 2, 3,... all the way up to n. This shows a series of a set of numbers. 2 + (-1)^1, 2 + (-1)^2 + 2 + (-1)^3 + 2 + (-1)^4.. 1, 3, 1, 3 Sequences approaching limiting values converge, and sequences approaching a value that does not exist are said to diverge. The definition of a limit of a sequence is: lim (n approaches ∞) a(nth term) = L where L is a limit. If L exists, the sequence converges. If limit L of a sequence does not exist, it diverges. For the above example, because the sequence alternates between 1 and 3, the limit does not exist. Therefore, the above sequence diverges. Squeeze Theorem for Sequences
Monotonic Sequences and Bounded SequencesA sequence is monotonic when its terms are non-decreasing or non-increasing. To determine whether a sequence is monotonic, simply check whether the terms in the series are increasing or alternating. If it is changing between two values, it is not monotonic. A sequence is bounded above when the real number M is greater than the sequence for all numbers. M is the upper bound in this case.A sequence is bounded below when the real number M is less than the sequence for all numbers. M is the lower bound in this case. When a sequence is bounded above and below, it is bounded.
Therefore, from that, an - L < ϵ which means that {an} converges to L. It is super interesting to delve into the depths of series and their applications in the real world too. Explore on! Good luck!
I probably needed to repeat that. I know, though it sounds ridiculous, the sum of all natural numbers leading up to, I suppose infinity is proven to be -1/12 by many of the greatest mathematicians of all time. How about we become a mathematician and prove it once again? Instead of using complicated formulas and other ways of solving this puzzle, we can use a simpler way to prove it. I also wanted to talk about many other ways of solving this and mathematician's efforts to proving this as well. (especially from my favorite and highly regarded mathematician, Srinivasan Ramanujan). I saw this phenomenal fact in a youtube video on proving this (https://www.youtube.com/watch?v=w-I6XTVZXww), and wanted to share it with you all. Let's prove this infinite series in a simple way: Step 1. Write the first sum.First, we are going to write three different sums, labelled as S1, S2, and S3. The last sum, S3 should be the infinite series we are trying to prove (aka: 1 + 2 + 3 + 4 + 5......) The first sum is going to be: S1: 1-1+1-1+1-1+1-1+1....There are two possible answers for this infinite sum: If we end after a subtraction, the answer will be 0. If we end after an addition, the answer if 1. So, to compute the sum of the series, we would take the average of the two values ~ (1+0)/2 = 1/2. Step 2. Write the second sum twice.Write the next sum, which is slightly different, involving both addition and subtraction. We are going to duplicate this sum and add it to each other (2S2) to make the work easier. Also, shift the second S2 to one digit right so that the first number of the second sum is below the second number of the first sum. S2: 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8.....+ S2: 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8.....2S2= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1...When you add both the sums digit by digit, you get the original sum we started with, which is S1. Therefore, we can conclude that 2S2 = S1 = 1/2., which means that S2= 1/4. Step 3. Write the original sum we are solving for and subtract S2 from it.S3: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8.....- [1 - 2 + 3 - 4 + 5 - 6 + 7 - 8.....]0 + 4 + 0 + 8 + 12 + 0 + 16 + 0...Now, we are going to factor this new sum by 4, therefore, we obtain...you got it...the original sum times 4. 4 ( 1 + 2 + 3 + 4 + 5.....) = 4S3S3 - S2 = 4S3 S3 - 1/4 = 4S3 -1/4 = 3S3 -1/12 = S3 And...there is your proof that the sum of the infinite series is -1/12. It is absolutely surprising, but actually proven to be true. I know..weird. But I like it. Good luck,
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