Showing property for the derivative partial_x T of a trigonometric polynomial

Arendrogfkl 2022-11-21 Answered
Showing property for the derivative x T of a trigonometric polynomial
Let be
T = n = 0 N T ^ ( n ) e i n x
a trigonometric polynomial of grade N without negative frequencies.
I wanna show that
x T = i N ( F N T T )
Where F N T meas the convolution of the Fejer Kernel and T.
might be easy..but I just can't work out the right conversion for this property..
SO
x T = n = 0 N T ^ ( n ) i n e i n x
= n = 0 N ( 1 2 π π π T ( y ) e i n y d y ) e i n x i n
= 1 2 π π π T ( y ) ( n = 0 N e i n ( x y ) ) i n
from there on I get carried away in the wrong direction. is the derivative right?
Also..the Fejer Kernel can be expressed as the mean arithmetic value of the dirichtlet kernel so:
F N = 1 n + 1 k = 0 n D k ( x )
Where D k = n = k k e i n x is the Dirichtlet Kernel
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Answers (1)

yen1291kp6
Answered 2022-11-22 Author has 12 answers
Step 1
1). The last equality you wrote for the derivative is weird, since you have the in outside a term that involves summing in n, so that's a mistake.
2). The definition of F N should involved 'N' rather than 'n'.
3). Either FN should be defined to be 1 N k = 0 N 1 D k or the result you wish to prove should be x T = i ( N + 1 ) ( F N T T ).
I will keep your definition of F N (with point (2) taken into account of course) and prove ∂xT=−i(N+1)(FN∗T−T). We show the fourier transform of both sides is the same -- this suffices.
Step 2
Note [ i ( N + 1 ) ( F N T T ) ] ^ ( m ) = i ( N + 1 ) [ F N ^ ( m ) T ^ ( m ) T ^ ( m ) ] (the latter you can see from your first equality for the derivative). So, immediately, if m < 0 or if m N + 1, both sides are 0 and we're good. So suppose 0 m N. Then, F N ^ ( m ) = 1 N + 1 k = 0 N + 1 D k ^ ( m ) = 1 N + 1 k = 0 N + 1 1 m k = 1 N + 1 [ N + 1 m ], so we get [ i ( N + 1 ) ( F N T T ) ] ^ ( m ) = i ( N + 1 ) T ^ ( m ) m N + 1 = i m T ^ ( m ), as desired.
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