# Finding Fourier coefficients of function $f$.

Let's define function $$u(x,t)$$ where $$x \in [0, 1], t \ge 0$$.
We know that $$u$$ satisfies the equation $$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2},$$ with boundary condition $$u(x, 0) = f(x)$$.

Let $$f \in C^2[0, 1]$$. Moreover $$f$$ has period $$1$$.

I know that $$f(x) = u(x, 0) = \sum \limits_{n=0}^{\infty} \big( C_n \sin(2 \pi n x) + D_n \cos(2 \pi n x) \big) \tag{1}.$$

Of course $$(1)$$ says that $$C_n$$ and $$D_n$$ are Fourier coefficients of function $$f$$.

Thus $$C_n = \int \limits_{0}^{1} f(y) \sin(2 \pi n y) \, dy, \\ D_n = \int \limits_{0}^{1} f(y) \cos(2 \pi n y) \, dy.$$

That leads to $$u(x, t) = \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \big( C_n \sin(2 \pi n x) + D_n \cos(2 \pi n x) \big) \\ = \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \bigg( \int \limits_{0}^{1} f(y) \sin(2 \pi n y) \, dy \, \sin(2 \pi n x) + \int \limits_{0}^{1} f(y) \cos(2 \pi n y) \, dy \, \cos(2 \pi n x) \bigg) \\ = \int \limits_{0}^{1} f(y) \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \bigg(\sin(2 \pi n x) \sin(2 \pi n y) + \cos(2 \pi n x)\cos(2 \pi n y) \bigg) \, dy \\ = \int \limits_{0}^{1} f(y) \sum \limits_{n=0}^{\infty} e^{-4 \pi^2 n^2 t} \cos\big(2 \pi n (x-y)\big) \, dy \\ = \int \limits_{0}^{1} f(y) \bigg(1 + \sum \limits_{n=1}^{\infty} e^{-4 \pi^2 n^2 t} \cos\big(2 \pi n (x-y)\big) \bigg) \, dy.$$

However I know that the the kernel should be equal $$1 + 2\sum \limits_{n=1}^{\infty} e^{-4 \pi^2 n^2 t} \cos\big(2 \pi n (x-y)\big).$$

Where am I wrong?

The Fourier coefficients of a function $$f$$ having period $$p$$ are

$$C_n =\frac2p\int_P f(t)\cos(\tfrac{2\pi n t}{p})\,dt \qquad \text{and} \qquad D_n =\frac2p\int_P f(t)\sin(\tfrac{2\pi n t}{p})\,dt,$$

where $$P$$ denotes some interval of measure $$p$$. In your case, the integrals are missing the coefficient of $$\frac21=2$$.

Reference: See the formula for coefficients here, and a proof on the site here.

• Thanks. Can you provide a proof of give some references, please? – Hendrra Jun 9 at 12:35
• @Hendrra I updated my answer to include reference. – Luke Collins Jun 9 at 12:50
• Thank you very much! I truly appreciate your help! :) – Hendrra Jun 9 at 13:35
• @Hendrra No problem :) – Luke Collins Jun 9 at 13:39