An exercise with fft

6.26.4 An exercise with fft

Given temperatures T at time t, in degrees Celcius:

t	0	3	6	9	12	15	19	21
T	11	10	17	24	32	26	23	19

What was the temperature at 13h45 ?

Here N=8=2*m. The interpolation polynomial is

p(t)=

p_−m(exp(−2i

π mt

)+ exp(2i

π mt

))+

m−1

∑

k=−m+1

p_k exp(2i

π kt

)

and

p_k=

N−1

∑

k=j

T_k exp(2i

π k

)

Input:

q:=1/8*fft([11,10,17,24,32,26,23,19])

Output:

⎡
⎣

20.25,−4.48115530061+1.72227182413 i,0.375+0.875i,

−0.768844699385+0.222271824132, i,0.5,

−0.768844699385−0.222271824132, i,

0.375−0.875i, −4.48115530061−1.72227182413i

⎤
⎦

hence:

p₀=20.25
p₁=−4.48115530061+1.72227182413i=p₋₁,
p₂=0.375+0.875i=p₋₂,
p₃=−0.768844699385+0.222271824132i=p₋₃,
p₄=0.5

Indeed

q=[q₀,… q_N−1]=[p₀,..p

−1

−

,..,p₋₁]=

F_N([y₀,..y_N−1])=

fft(y)

Input:

pp:=[q[4],q[5],q[6],q[7],q[0],q[1],q[2],q[3]]

Here, p_k=pp[k+4] for k=−4…3. It remains to compute the value of the interpolation polynomial at point t0=13.75=55/4.
Input:

t0(j):=exp(2*i*pi*(13+3/4)/24*j)

T0:=1/2*pp[0]*(t0(4)+t0(-4))+sum(pp[j+4]*t0(j),j,-3,3)

evalf(re(T0))

Output:

29.4863181684

The temperature is predicted to be equal to 29.49 degrees Celsius.

Remark.
Using the Lagrange interpolation polynomial (the polynomial is not periodic):
Input:

l1:=[0,3,6,9,12,15,18,21]

l2:=[11,10,17,24,32,26,23,19]

subst(lagrange(l1,l2,13+3/4),x=13+3/4)

evalf(ans())

Output:

30.1144061688