7. Suppose that the equal_listsfunction (page 49 of Sebesta) is called with the

Question

7. Suppose that the equal_listsfunction (page 49 of Sebesta) is called with the lists ((A(B))(C)) and ((A(B))(C)) as the arguments. How many calls of equal_lists will be performed altogether, including the original call and all recursive calls?

Page (49) LISP Example function The following code defines a LISP predicate function that takes two lists as arguments and returns True if the two lists are equal, and NIL (false) otherwise (DEFUN equal sts (lis1 lis2) (COND ((ATOM lis1) (EQ lis1 lis2)) (ATOM lis2) NIL) (equal lists (CAR lis1) (CAR lis2)) (equal lists (CDR lis1) (CDR lis2))) NIL)

Explanation / Answer

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Using this thread, and also the indisputable fact that if f1f1 and f2f2 ar 2 integrable functions, F(fg)=F(f)F(g)F(fg)=F(f)F(g), we have
F(ddx(fg))(x)=ixF((fg))(x)=ixF(f)(x)F(g)(x),
F(ddx(fg))(x)=ixF((fg))(x)=ixF(f)(x)F(g)(x),
and
F((ddxf)g)(x)=(F(ddxf))(F(g)(x))=ixF(f)(x)F(g)(x).
F((ddxf)g)(x)=(F(ddxf))(F(g)(x))=ixF(f)(x)F(g)(x).
We conclude by individuality of Fourier remodel.

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answered Jul thirty one '12 at 16:57

Davide Giraudo
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How you'll be able to take Fourier remodel whereas we do not comprehend it has Fourier remodel or not? In alternative words, we do not understand ddx(fg)ddx(fg) is in L1L1? For the second Fourier remodel, it's correct since we all know that fgfg is in L1L1. – rfvahid Jul thirty one '12 at 17:09
     
Indeed, it deserves additional details. i feel associate degree approximation argument will work (approximate in L1L1 ff and gg by C1C1 functions with compact support). – Davide Giraudo Jul thirty one '12 at 17:20
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Note that, if fL1(R)fL1(R) then it's Fourier translatable. Since,

f(x)eixw|f(x)|<
|f(x)eixw||f(x)|<
.

To prove that the convolution of 2 L1(R)L1(R) performs is once more associate degree L1(R)L1(R) function, let

h(x)=f(t)g(xt)dt
h(x)=f(t)g(xt)dt
|h(x)|dx|f(t)||g(xt)|dtdx=|f(t)||g(xt)|dxdt=|f(t)|||g||1dt=||f||1||g||1hL1(R).
|h(x)|dx|f(t)||g(xt)|dtdx=|f(t)||g(xt)|dxdt=|f(t)|||g||1dt=||f||1||g||1hL1(R).
The modification of the order of integration is even by Fubini's theorem. So, you'll be able to use the Fourier technique as in Davide's answer.

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edited August one '12 at 16:30
answered Jul thirty one '12 at 18:46

Mhenni Benghorbal
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Definition:
h(x)=fg(x)=f(xt)g(t)dt
h(x)=fg(x)=f(xt)g(t)dt
Let's calculate derivative:

dhdx=limdx>0(f(x+dxt)g(t)dtf(xt)g(t)dt)dx=limdx>0((f(x+dxt)f(xt))dxg(t)dt
dhdx=limdx>0(f(x+dxt)g(t)dtf(xt)g(t)dt)dx=limdx>0((f(x+dxt)f(xt))dxg(t)dt
If assume that for
(f(x+dxt)f(xt))dx
(f(x+dxt)f(xt))dx
exist some integrable perform q(t), that is edge on this expression, perhaps except potential on set with live zero then by Lebesgue dominated convergence theorem we will push the limit within integral.

dhdx=fg(x)=f(xt)g(t)dt=fg

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