Suppose that f ( x ) and its derivative f ′ ( x ) are both expanded in Fourier series on ( − π , π ) . Call the coefficients in the f ( x ) series a n and b n and the coefficients in the f ′ ( x ) series a n ′ and b n ′ . Write the integral for a n [equation (5.9)] and integrate it by parts to get an integral of f ′ ( x ) sin n x . Recognize this integral in terms of b n ′ [equation (5.10) for f ′ ( x ) and so show that b n ′ = − n a n . (In the integration by parts, the integrated term is zero because f ( π ) = f ( − π ) since f is continuous sketch several periods.). Find a similar relation for a n ′ and b n . Now show that this is the result you get by differentiating the f ( x ) series term by term. Thus you have shown that the Fourier series for f ′ ( x ) is correctly given by differentiating the f ( x ) series term by term (assuming that f ′ ( x ) is expandable in a Fourier series).
Suppose that f ( x ) and its derivative f ′ ( x ) are both expanded in Fourier series on ( − π , π ) . Call the coefficients in the f ( x ) series a n and b n and the coefficients in the f ′ ( x ) series a n ′ and b n ′ . Write the integral for a n [equation (5.9)] and integrate it by parts to get an integral of f ′ ( x ) sin n x . Recognize this integral in terms of b n ′ [equation (5.10) for f ′ ( x ) and so show that b n ′ = − n a n . (In the integration by parts, the integrated term is zero because f ( π ) = f ( − π ) since f is continuous sketch several periods.). Find a similar relation for a n ′ and b n . Now show that this is the result you get by differentiating the f ( x ) series term by term. Thus you have shown that the Fourier series for f ′ ( x ) is correctly given by differentiating the f ( x ) series term by term (assuming that f ′ ( x ) is expandable in a Fourier series).
Suppose that
f
(
x
)
and its derivative
f
′
(
x
)
are both expanded in Fourier series on
(
−
π
,
π
)
.
Call the coefficients in the
f
(
x
)
series
a
n
and
b
n
and the coefficients in the
f
′
(
x
)
series
a
n
′
and
b
n
′
.
Write the integral for
a
n
[equation (5.9)] and integrate it by parts to get an integral of
f
′
(
x
)
sin
n
x
.
Recognize this integral in terms of
b
n
′
[equation (5.10) for
f
′
(
x
)
and so show that
b
n
′
=
−
n
a
n
.
(In the integration by parts, the integrated term is zero because
f
(
π
)
=
f
(
−
π
)
since
f
is continuous sketch several periods.). Find a similar relation for
a
n
′
and
b
n
.
Now show that this is the result you get by differentiating the
f
(
x
)
series term by term. Thus you have shown that the Fourier series for
f
′
(
x
)
is correctly given by differentiating the
f
(
x
)
series term by term (assuming that
f
′
(
x
)
is expandable in a Fourier series).
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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