Taking the sequence as f(x) = {1, 2, 4, 7}, the DCT is C(u) = {7, −4.461, 1, −0.317}.
(The "4.7" in the question is read as "4, 7", giving a 4-point sequence.)
1-D DCT formula for \(N = 4\):
\[C(u) = \alpha(u)\sum_{x=0}^{3} f(x)\cos\frac{(2x+1)u\pi}{8}\]
with \(\alpha(0) = \sqrt{1/N} = 0.5\) and \(\alpha(u) = \sqrt{2/N} = 0.7071\) for \(u = 1, 2, 3\).
Useful cosines: \(\cos\frac{\pi}{8} = 0.9239\), \(\cos\frac{3\pi}{8} = 0.3827\), \(\cos\frac{\pi}{4} = 0.7071\).
u = 0
\[C(0) = 0.5(1 + 2 + 4 + 7) = 7\]
u = 1 (cosines 0.9239, 0.3827, −0.3827, −0.9239):
\[\Sigma = 0.9239 + 0.7654 - 1.5307 - 6.4672\]
\[\Sigma = -6.3086,\quad C(1) = 0.7071\times(-6.3086) = -4.461\]
u = 2 (cosines 0.7071, −0.7071, −0.7071, 0.7071):
\[\Sigma = 0.7071(1 - 2 - 4 + 7) = 1.4142\]
\[C(2) = 0.7071\times1.4142 = 1\]
u = 3 (cosines 0.3827, −0.9239, 0.9239, −0.3827):
\[\Sigma = 0.3827 - 1.8478 + 3.6956 - 2.6789\]
\[\Sigma = -0.4484,\quad C(3) = 0.7071\times(-0.4484) = -0.317\]
Result: \(C = \{7,\ -4.461,\ 1,\ -0.317\}\).
Check (energy is preserved): \(1^2 + 2^2 + 4^2 + 7^2 = 70\) and \(7^2 + 4.461^2 + 1^2 + 0.317^2 \approx 49 + 19.90 + 1 + 0.10 = 70\). Most of the energy is in the first two coefficients, which shows the energy-compaction property of the DCT.