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On the basic interval the graph is the pair of straight segments joining $(-\pi ,0)$, $(0,1)$ and $(\pi ,0)$. Its periodic extension is an even triangular wave.

The Fourier coefficients satisfy $b_n=0$ by evenness, andFor the second formula, integration by parts gives ${\int }_0^{\pi }x\cos (nx)dx=[(-1{)}^n-1]/n^2$. Only odd indices survive, giving this rescaled Fourier cosine series of a triangular wave:The pointwise convergence of a piecewise smooth Fourier series applies throughout the interval, including the continuous joined endpoints. The coefficients also imply uniform convergence by the Weierstrass M-test. At $x=0$, the value is one, so

Evenness removes every sine coefficient. In the period-two Fourier cosine series convention $f=a_0/2+{\sum }_{n\ge 1}{a}_n\cos (n\pi x)$, the constant coefficient is $a_0=2{\int }_0^1(1-x)dx=1$. Integration by parts givesThus the Fourier cosine series of a triangular wave isThe periodic function is continuous and piecewise smooth, so the Dirichlet-Jordan convergence theorem gives its value at every point. The summable absolute coefficient bound also gives uniform convergence by the Weierstrass M-test.

Evaluate the Fourier cosine series of a triangular wave at its continuous peak $x=0$, where $f(0)=1$. Every cosine is one, soThe convergence established above justifies substituting this point into the series.