Show that ( 2 l + 1 ) x 2 − 1 P l ′ ( x ) = l ( l + 1 ) P l + 1 ( x ) − P l − 1 ( x ) . Hint : Integrate (5.8e) and (7.2) and combine the results. Thus show that P 1 + 1 ( x ) = P l − 1 ( x ) at maximum and minimum points of P l ( x ) and at ± 1.
Show that ( 2 l + 1 ) x 2 − 1 P l ′ ( x ) = l ( l + 1 ) P l + 1 ( x ) − P l − 1 ( x ) . Hint : Integrate (5.8e) and (7.2) and combine the results. Thus show that P 1 + 1 ( x ) = P l − 1 ( x ) at maximum and minimum points of P l ( x ) and at ± 1.
Show that
(
2
l
+
1
)
x
2
−
1
P
l
′
(
x
)
=
l
(
l
+
1
)
P
l
+
1
(
x
)
−
P
l
−
1
(
x
)
.
Hint: Integrate (5.8e) and (7.2) and combine the results. Thus show that
P
1
+
1
(
x
)
=
P
l
−
1
(
x
)
at maximum and minimum points of
P
l
(
x
)
and at
±
1.
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.
Finite Mathematics & Its Applications (12th Edition)
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.