Fourier cosine series coefficients

Question

If I have $λ_{-k-1}=-λ_k$ for $k\in\mathbb N$ and$$u(x,t)=\sum_{k=-∞}^∞a_k{\rm e}^{-λ_k^2t}\cos(λ_kx),$$then when I write$$u(x,t)=\sum_{k=0}^∞(a_k+a_{-k-1}){\rm e}^{-λ_k^2t}\cos(λ_kx)=\sum_{k=0}^∞b_k{\rm }e^{-λ_k^2t}\cos(λ_kx),$$ do I find $b_k$ as $b_k=2\int_0^1u(x,0)\cos(\lambda_kx)\,\mathrm dx$ or as $b_k=\int_0^1u(x,0)\cos(\lambda_kx)\,\mathrm dx$? (I have $\lambda_k=\frac{\pi}{2}+\pi k$)

Answer 1

$\def\d{\mathrm{d}}\newcommand{\ip}[2]{\left\langle#1,#2\right\rangle}$Note that $\{\cos(λ_k x) \mid k \in \mathbb{N}\}$ is a system of orthogonal functions on $[0, 1]$, thus for any $k \in \mathbb{N}$,\begin{gather*} \ip{u(x, 0)}{\cos(λ_k x)} = \ip{\sum_{j = 0}^∞ b_j \cos(λ_j x)}{\cos(λ_k x)} = \sum_{j = 0}^∞ b_j \ip{\cos(λ_j x)}{\cos(λ_k x)}\\= b_k \ip{\cos(λ_k x)}{\cos(λ_k x)} = b_k \int_0^1 (\cos(λ_k x))^2 \,\d x = \frac{b_k}{2}, \end{gather*} and$$ b_k = 2\ip{u(x, 0)}{\cos(λ_k x)} = 2 \int_0^1 u(x, 0) \cos(λ_k x) \,\d x. $$