Question

d/dx f(x)[f(x)] is the fourier cosine transform of is​

Answer 1

Answer:

lectures we defined the fourier and fourier cosine (for even functions) transforms as follows:

fˆ(ω)=∫∞−∞f(x)e−iωxdx

fˆc(ω)=∫∞0f(x)cos(ωx)dx

It is clear that

fˆ(ω)=2fˆc(ω)

Now we also derived two formulas for taking the fourier transforms and fourier cos transforms of second derivatives:

F{dnf/dxn}=(iω)nfˆ(ω)

Fc{f′′(x)}=−f′(0)−ω2fˆc(ω)

But equating them with taking into account the factor of 2 leads to:

Fc{f′′(x)}=−f′(0)−ω2fˆc(ω)=12F{f′′(x)}=−ω2fˆc(ω)

Which suggest that f′(0)=0, which is not necessarily true, am I assuming something incorrect here?

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Dec 16 '19 at 20:14

analysis1

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f′(0) of a differentiable even function f is? – peterwhy Dec 16 '19 at 20:29

ah right, yeah, thanks – analysis1 Dec 16 '19 at 20:40

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Consider an even function f, i.e. f(−x)=f(x),

f′(x)=df(x)dx=df(−x)dx=df(−x)d(−x)d(−x)dx=−f′(−x)

This shows f′ is odd, and by substituting x=0,

f′(0)=0

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