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Are you familiar with how to obtain the Fourier series of an arbitrary periodic function? Because if you are, I'm confused at your confusion.

This is simply the Fourier series expansion of the triangle wave with fundamental frequency $f_0=1/T$. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (the time-average) is zero, and that the function is odd, so it has no cosine terms. So, let's attempt to expand it as the sum of sines or in other words as $$ x(t) = \sum_{k=0}^{\infty} b_k \sin\left(\frac{2 \pi k t}{T}\right) $$ where $x(t)$ is the triangle wave defined on the interval $-T/2$ to $T/2$ as $$ x(t) = \left\{\begin{array}{cc} \frac{4 t + 2 T}{T} & -\frac{T}{2} < t < -\frac{T}{4} \\ \frac{4 t}{T} & -\frac{T}{4} < t < \frac{T}{4} \\ \frac{4 t - 2 T}{T} & \frac{T}{4} < t < \frac{T}{2} & \end{array}\right. $$ and elsewhere by the fact that $x(t) = x(t+T)$.

To calculate our Fourier series coefficients, we must compute the integral $$ b_k = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

Because the function we are integrating is a product of two odd functions, it is even, so we can reduce this to $$ b_k = \frac{4}{T} \int_{0}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

We notice that $x(t)$ exhibits even symmetry about the point $\frac{T}{4}$ while even frequency sine waves of the form $\sin(4\pi k t)$ exhibit odd symmetry about the same point. This is quarter wave symmetry and tells us that the even harmonics will all be zero. So, we only need to solve this for odd values of $k$. To that end, we replace $k$ with $2 k + 1$.

We can once again exploit the even symmetry of the odd harmonics to further reduce the range of our integral to $$ b_{2 k + 1} = \frac{8}{T} \int_{0}^{\frac{T}{4}} \frac{4 t}{T} \sin\left(\frac{2 \pi (2 k + 1) t}{T}\right) dt,$$ and, skipping the calculus, we arrive at $$ b_{2 k + 1} = \frac{8}{\pi^2 (2 k + 1)^2} \cos(\pi k). $$

This reduces to the formula given on Wikipedia. If you don't see it, let me know and I'll be explicit.

Are you familiar with how to obtain the Fourier series of an arbitrary periodic function? Because if you are, I'm confused at your confusion.

This is simply the Fourier series expansion of the triangle wave with fundamental frequency $f_0=1/T$. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (the time-average) is zero, and that the function is odd, so it has no cosine terms. So, let's attempt to expand it as the sum of sines or in other words as $$ x(t) = \sum_{k=0}^{\infty} b_k \sin\left(\frac{2 \pi k t}{T}\right) $$ where $x(t)$ is the triangle wave defined on the interval $-T/2$ to $T/2$ as $$ x(t) = \left\{\begin{array}{cc} \frac{4 t + 2 T}{T} & -\frac{T}{2} < t < -\frac{T}{4} \\ \frac{4 t}{T} & -\frac{T}{4} < t < \frac{T}{4} \\ \frac{4 t - 2 T}{T} & \frac{T}{4} < t < \frac{T}{2} & \end{array}\right. $$ and elsewhere by the fact that $x(t) = x(t+T)$.

To calculate our Fourier series coefficients, we must compute the integral $$ b_k = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

Because the function we are integrating is a product of two odd functions, it is even, so we can reduce this to $$ b_k = \frac{4}{T} \int_{0}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

We notice that $x(t)$ exhibits even symmetry about the point $\frac{T}{4}$ while even frequency sine waves of the form $\sin(4\pi k t)$ exhibit odd symmetry about the same point. This is quarter wave symmetry and tells us that the even harmonics will all be zero. So, we only need to solve this for odd values of $k$. To that end, we replace $k$ with $2 k + 1$.

We can once again exploit the even symmetry of the odd harmonics to further reduce the range of our integral to $$ b_{2 k + 1} = \frac{8}{T} \int_{0}^{\frac{T}{4}} \frac{4 t}{T} \sin\left(\frac{2 \pi (2 k + 1) t}{T}\right) dt,$$ and, skipping the calculus (for brevity), we arrive at $$ b_{2 k + 1} = \frac{8}{\pi^2 (2 k + 1)^2} \cos(\pi k). $$

This reduces to the formula given on Wikipedia. If you don't see it, let me know and I'll be explicit.

Are you familiar with how to obtain the Fourier series of an arbitrary periodic function? Because if you are, I'm confused at your confusion.

This is simply the Fourier series expansion of the triangle wave with fundamental frequency $f_0=1/T$. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (the time-average) is zero, and that the function is odd, so it has no cosine terms. So, let's attempt to expand it as the sum of sines or in other words as $$ x(t) = \sum_{k=0}^{\infty} b_k \sin\left(\frac{2 \pi k t}{T}\right) $$ where $x(t)$ is the triangle wave defined on the interval $-T/2$ to $T/2$ as $$ x(t) = \left\{\begin{array}{rl} \frac{4 t + 2 T}{T} & -\frac{T}{2} < t < -\frac{T}{4} \\ \frac{4 t}{T} & -\frac{T}{4} < t < \frac{T}{4} \\ \frac{4 t - 2 T}{T} & \frac{T}{4} < t < \frac{T}{2} & \end{array}\right. $$ and elsewhere by the fact that $x(t) = x(t+T)$.

To calculate our Fourier series coefficients, we must solve the equation $$ b_k = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

Because the function we are integrating is a product of two odd functions, it is even, so we can reduce this to $$ b_k = \frac{4}{T} \int_{0}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

We notice that there is even symmetry about the point $\frac{T}{4}$ while even frequency sine waves of the form $\sin(4\pi k t)$ exhibit odd symmetry about the same point. This is quarter wave symmetry and tells us that the even harmonics will all be zero. So, we only need to solve this for odd values of $k$. To that end, we replace $k$ with $2 k + 1$.

We can once again exploit the even symmetry of the odd harmonics to further reduce the range of our integral to $$ b_{2 k + 1} = \frac{8}{T} \int_{0}^{\frac{T}{4}} \frac{4 t}{T} \sin\left(\frac{2 \pi (2 k + 1) t}{T}\right) dt,$$ and, skipping the calculus, we arrive at $$ b_{2 k + 1} = \frac{8}{\pi^2 (2 k + 1)^2} \cos(\pi k). $$

This reduces to the formula given on Wikipedia. If you don't see it, let me know and I'll be explicit.

Are you familiar with how to obtain the Fourier series of an arbitrary periodic function? Because if you are, I'm confused at your confusion.

This is simply the Fourier series expansion of the triangle wave with fundamental frequency $f_0=1/T$. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (the time-average) is zero, and that the function is odd, so it has no cosine terms. So, let's attempt to expand it as the sum of sines or in other words as $$ x(t) = \sum_{k=0}^{\infty} b_k \sin\left(\frac{2 \pi k t}{T}\right) $$ where $x(t)$ is the triangle wave defined on the interval $-T/2$ to $T/2$ as $$ x(t) = \left\{\begin{array}{cc} \frac{4 t + 2 T}{T} & -\frac{T}{2} < t < -\frac{T}{4} \\ \frac{4 t}{T} & -\frac{T}{4} < t < \frac{T}{4} \\ \frac{4 t - 2 T}{T} & \frac{T}{4} < t < \frac{T}{2} & \end{array}\right. $$ and elsewhere by the fact that $x(t) = x(t+T)$.

To calculate our Fourier series coefficients, we must compute the integral $$ b_k = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

Because the function we are integrating is a product of two odd functions, it is even, so we can reduce this to $$ b_k = \frac{4}{T} \int_{0}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

We notice that $x(t)$ exhibits even symmetry about the point $\frac{T}{4}$ while even frequency sine waves of the form $\sin(4\pi k t)$ exhibit odd symmetry about the same point. This is quarter wave symmetry and tells us that the even harmonics will all be zero. So, we only need to solve this for odd values of $k$. To that end, we replace $k$ with $2 k + 1$.

We can once again exploit the even symmetry of the odd harmonics to further reduce the range of our integral to $$ b_{2 k + 1} = \frac{8}{T} \int_{0}^{\frac{T}{4}} \frac{4 t}{T} \sin\left(\frac{2 \pi (2 k + 1) t}{T}\right) dt,$$ and, skipping the calculus, we arrive at $$ b_{2 k + 1} = \frac{8}{\pi^2 (2 k + 1)^2} \cos(\pi k). $$

This reduces to the formula given on Wikipedia. If you don't see it, let me know and I'll be explicit.

Are you familiar with how to obtain the Fourier Series of an arbitrary periodic function? Because if you are, I'm confused at your confusion. This is simply the Fourier series expansion of the triangle wave with fundamental frequency $f_0=1/T$. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (the time-average) is zero, and that the function is odd, so it has no cosine terms. So, let's attempt to expand it as the sum of sines or in other words as $$ x(t) = \sum_{k=0}^{\infty} b_k \sin\left(\frac{2 \pi k t}{T}\right) $$ where $x(t)$ is the triangle wave defined on the interval $-T/2$ to $T/2$ as $$ x(t) = \left\{\begin{array}{rl} \frac{4 t + 2 T}{T} & -\frac{T}{2} < t < -\frac{T}{4} \\ \frac{4 t}{T} & -\frac{T}{4} < t < \frac{T}{4} \\ \frac{4 t - 2 T}{T} & \frac{T}{4} < t < \frac{T}{2} & \end{array}\right. $$ and elsewhere by the fact that $x(t) = x(t+T)$.

To calculate our Fourier series coefficients, we must solve the equation $$ b_k = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

Because the function we are integrating is a product of two odd functions, it is even, so we can reduce this to $$ b_k = \frac{4}{T} \int_{0}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

We can also notice that there is even symmetry about the point $\frac{T}{4}$. This is quarter wave symmetry and tells us that the even harmonics will all be zero (since they will exhibit odd symmetry around this point). So, we only need to solve this for odd values of $k$.

We can once again exploit the even symmetry of the odd harmonics to further reduce the range of our integral to $$ b_{2 k + 1} = \frac{8}{T} \int_{0}^{\frac{T}{4}} \frac{4 t}{T} \sin\left(\frac{2 \pi (2 k + 1) t}{T}\right) dt,$$ and, skipping the calculus, we arrive at $$ b_{2 k + 1} = \frac{8}{\pi^2 (2 k + 1)^2} \cos(\pi k). $$

This reduces to the formula given on Wikipedia. If you don't see it, let me know and I'll be explicit.

Are you familiar with how to obtain the Fourier series of an arbitrary periodic function? Because if you are, I'm confused at your confusion. 

This is simply the Fourier series expansion of the triangle wave with fundamental frequency $f_0=1/T$. This can be found by taking the inner product of the function with the basis functions. We note that the DC value (the time-average) is zero, and that the function is odd, so it has no cosine terms. So, let's attempt to expand it as the sum of sines or in other words as $$ x(t) = \sum_{k=0}^{\infty} b_k \sin\left(\frac{2 \pi k t}{T}\right) $$ where $x(t)$ is the triangle wave defined on the interval $-T/2$ to $T/2$ as $$ x(t) = \left\{\begin{array}{rl} \frac{4 t + 2 T}{T} & -\frac{T}{2} < t < -\frac{T}{4} \\ \frac{4 t}{T} & -\frac{T}{4} < t < \frac{T}{4} \\ \frac{4 t - 2 T}{T} & \frac{T}{4} < t < \frac{T}{2} & \end{array}\right. $$ and elsewhere by the fact that $x(t) = x(t+T)$.

To calculate our Fourier series coefficients, we must solve the equation $$ b_k = \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

Because the function we are integrating is a product of two odd functions, it is even, so we can reduce this to $$ b_k = \frac{4}{T} \int_{0}^{\frac{T}{2}} x(t) \sin\left(\frac{2 \pi k t}{T}\right) dt.$$

We notice that there is even symmetry about the point $\frac{T}{4}$ while even frequency sine waves of the form $\sin(4\pi k t)$ exhibit odd symmetry about the same point. This is quarter wave symmetry and tells us that the even harmonics will all be zero. So, we only need to solve this for odd values of $k$. To that end, we replace $k$ with $2 k + 1$.

We can once again exploit the even symmetry of the odd harmonics to further reduce the range of our integral to $$ b_{2 k + 1} = \frac{8}{T} \int_{0}^{\frac{T}{4}} \frac{4 t}{T} \sin\left(\frac{2 \pi (2 k + 1) t}{T}\right) dt,$$ and, skipping the calculus, we arrive at $$ b_{2 k + 1} = \frac{8}{\pi^2 (2 k + 1)^2} \cos(\pi k). $$

This reduces to the formula given on Wikipedia. If you don't see it, let me know and I'll be explicit.