sin 2πα cos αx-sin πα cos α + sin αx cos πα-cos 2πα sin αx da α sin α (2π- x)-sin α (π-x) da | α II
sin 2πα cos αx-sin πα cos α + sin αx cos πα-cos 2πα sin αx da α sin α (2π- x)-sin α (π-x) da | α II
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I don't understand where sina(2pi-x) and -sina(pi-x) came from. Can you please explain it to me? Thank you
![12:12 AM Thu May 13
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Math / Bundle: Differential Equations with Boundary-Value Problems, Loo... | In Problems 1-6 find the Fourier integral repr...
i In Problems 1-6 find the Fourier integral representation of the given function. 2. f ( x ...
= 4
cos axdx
27
= 4[-
sin ax
It
sin 2πα-sin πα
A (a) :
= 4
And the value of B (a) is,
00
В (а) —
f (x) sin axdx
J.
2n
00
0 · sin axdx +
(4) · sin axdx +
0 . sin axdx
2л
2n
4
sin axdx
:4[
- cos axr
||
cos πα-cos 2πα
B (a) = 4
Thus, the value of f (x) is,
00
sin 2πα- sin πα
coS TA –
cos 2na
f(x)
cos ax + 4
- sin ax| da
00
sin 2ra cos ax
S1n πα COS αx+ S1n αx cOS πα-cos 2πα sin αχ
da
4
00
sin a (2n – x) – sin a (n – x)
da
Hence, the Fourier integral representation of given function is
sin a (2л — х) - sin a (л — х)
00
4
f(x)
da.
E Chapter 14.3, Problem 1E
Chapter 14.3, Problem 3E >
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Transcribed Image Text:12:12 AM Thu May 13
18% I
AA
bartleby.com
M G
Goog G
Google
1-6 find the...
A Classwork fo...
G akshata mud... mer Mem... l inje...
= bartleby
Q Search for textbooks, step-by-step explanatio...
Ask an Expert
Math / Bundle: Differential Equations with Boundary-Value Problems, Loo... | In Problems 1-6 find the Fourier integral repr...
i In Problems 1-6 find the Fourier integral representation of the given function. 2. f ( x ...
= 4
cos axdx
27
= 4[-
sin ax
It
sin 2πα-sin πα
A (a) :
= 4
And the value of B (a) is,
00
В (а) —
f (x) sin axdx
J.
2n
00
0 · sin axdx +
(4) · sin axdx +
0 . sin axdx
2л
2n
4
sin axdx
:4[
- cos axr
||
cos πα-cos 2πα
B (a) = 4
Thus, the value of f (x) is,
00
sin 2πα- sin πα
coS TA –
cos 2na
f(x)
cos ax + 4
- sin ax| da
00
sin 2ra cos ax
S1n πα COS αx+ S1n αx cOS πα-cos 2πα sin αχ
da
4
00
sin a (2n – x) – sin a (n – x)
da
Hence, the Fourier integral representation of given function is
sin a (2л — х) - sin a (л — х)
00
4
f(x)
da.
E Chapter 14.3, Problem 1E
Chapter 14.3, Problem 3E >
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