Explanation Given: Let x be fixed and let f ( z ) = ∑ n = 0 ∞ P n ( x ) z n Approach: The Legendre polynomials satisfy the recurrence formula given by, ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) … ( 1 ) Calculation: Multiply each term of formula ( 1 ) by z n as, ( n + 1 ) P n + 1 ( x ) z n = ( 2 n + 1 ) x P n ( x ) z n − n P n − 1 ( x ) z n Take the summation from n = 1 to ∞ as, ∑ n = 1 ∞ ( n + 1 ) P n + 1 ( x ) z n = ∑ n = 1 ∞ ( 2 n + 1 ) x P n ( x ) z n − ∑ n = 1 ∞ n P n − 1 ( x ) z n Solve the above expression as, ∑ n = 1 ∞ ( n + 1 ) P n + 1 ( x ) z n = ∑ n = 1 ∞ ( 2 n + 1 ) x P n ( x ) z n − ∑ n = 1 ∞ n P n − 1 ( x ) z n = 2 ∑ n = 1 ∞ n x P n ( x ) z n + ∑ n = 1 ∞ x P n ( x ) z n − ∑ n = 1 ∞ ( n + 1 ) P n ( x ) z n + 1 = 2 x ∑ n = 1 ∞ n P n ( x ) z n + x ∑ n = 1 ∞ P n ( x ) z n − ∑ n = 1 ∞ n P n ( x ) z n + 1 − ∑ n = 1 ∞ P n ( x ) z n + 1 = 2 x ∑ n = 1 ∞ n P n ( x ) z n + x ∑ n = 1 ∞ P n ( x ) z n − z ∑ n = 1 ∞ n P n ( x ) z n − z ∑ n = 1 ∞ P n ( x To determine (b) To obtain: The solution to the differential equation in part (a) as, f ( z ) = ( 1 − 2 x z + z 2 ) − 1 / 2

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 8.8, Problem 35E

To determine

(a)

To show:

The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as,

dfdz=x−z1−2xz+z2f

To determine

(b)

To obtain:

The solution to the differential equation in part (a) as,

f(z)=(1−2xz+z2)−1/2

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.8, Problem 34E

Chapter 8.8, Problem 36E

Students have asked these similar questions

9. (a) Use pseudocode to describe an algo- rithm for determining the value of a game tree when both players follow a minmax strategy. (b) Suppose that T₁ and T2 are spanning trees of a simple graph G. Moreover, suppose that ₁ is an edge in T₁ that is not in T2. Show that there is an edge 2 in T2 that is not in T₁ such that T₁ remains a spanning tree if ₁ is removed from it and 2 is added to it, and T2 remains a spanning tree if 2 is removed from it and e₁ is added to it. (c) Show that a degree-constrained spanning tree of a simple graph in which each vertex has degree not exceeding 2 2 consists of a single Hamiltonian path in the graph.

Chatgpt give wrong answer No chatgpt pls will upvote

@when ever one Point sets in x are closed a collection of functions which separates Points from closed set will separates Point. 18 (prod) is product topological space then VaeA (xx, Tx) is homeomorphic to sul space of the Product space (Txa, prod). KeA © The Bin Projection map B: Tx XP is continuous and open but heed hot to be closed. A collection (SEA) of continuos function oha topolgical Space X se partes Points from closed sets inx iff the set (v) for KEA and Vopen set in Xx from a base for top on x.

Chapter 8 Solutions

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

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.

(a) To show: The recurrence formula ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) when multiplied by z n gives the differential equation as, d f d z = x − z 1 − 2 x z + z 2 f

Solution Summary: The author explains that the recurrence formula (n+1)P_n-1 (x ) when multiplied by

Videos

(a)

To show:

The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as,

dfdz=x−z1−2xz+z2f

(b)

To obtain:

The solution to the differential equation in part (a) as,

f(z)=(1−2xz+z2)−1/2

Want to see the full answer?

Chapter 8 Solutions

(a) To show: The recurrence formula ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) when multiplied by z n gives the differential equation as, d f d z = x − z 1 − 2 x z + z 2 f

Solution Summary: The author explains that the recurrence formula (n+1)P_n-1 (x ) when multiplied by

Videos

(a) To show: The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as, dfdz=x−z1−2xz+z2f

(b) To obtain: The solution to the differential equation in part (a) as, f(z)=(1−2xz+z2)−1/2

Want to see the full answer?

Chapter 8 Solutions

(a)

To show:

The recurrence formula (n+1)Pn+1(x)=(2n+1)xPn(x)−nPn−1(x) when multiplied by zn gives the differential equation as,

dfdz=x−z1−2xz+z2f

(b)

To obtain:

The solution to the differential equation in part (a) as,

f(z)=(1−2xz+z2)−1/2