To derive: The Legendre polynomial of degree 6 and 7. The Legendre polynomial of degree 6 is P 6 ( x ) = 1 16 ( 231 x 6 − 315 x 4 + 105 x 2 − 5 ) _ and the Legendre polynomial of degree 7 is P 7 ( x ) = 1 16 ( 429 x 7 − 693 x 5 + 315 x 3 − 35 x ) _ . Result used: “The equation ( 11 ) is, P n ( x ) = ∑ m = 0 M ( − 1 ) m ( 2 n − 2 m ) ! 2 n m ! ( n − m ) ! ( n − 2 m ) ! x n − 2 m where M = n 2 or n − 1 2 whichever is an integer.” Calculation: Substitute n = 6 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 6 ( x ) = { ( − 1 ) 0 ( 2 ( 6 ) − 2 ( 0 ) ) ! 2 6 ( 0 ) ! ( 6 − ( 0 ) ) ! ( 6 − 2 ( 0 ) ) ! x ( 6 ) − 2 ( 0 ) + ( − 1 ) 1 ( 2 ( 6 ) − 2 ( 1 ) ) ! 2 6 ( 1 ) ! ( 6 − ( 1 ) ) ! ( 6 − 2 ( 1 ) ) ! x ( 6 ) − 2 ( 1 ) + ( − 1 ) 2 ( 2 ( 6 ) − 2 ( 2 ) ) ! 2 6 ( 2 ) ! ( 6 − ( 2 ) ) ! ( 6 − 2 ( 2 ) ) ! x 6 − 2 ( 2 ) + ( − 1 ) 3 ( 2 ( 6 ) − 2 ( 3 ) ) ! 2 6 ( 3 ) ! ( 6 − ( 3 ) ) ! ( 6 − 2 ( 3 ) ) ! x 6 − 2 ( 3 ) } = { 12 ! 64 ( 6 ) ! ( 6 ) ! x 6 − 10 ! 64 ( 1 ) ! ( 5 ) ! ( 4 ) ! x 4 + 8 ! 64 ( 2 ) ! ( 4 ) ! ( 2 ) ! x 2 − 6 ! 64 ( 3 ) ! ( 3 ) ! ( 0 ) ! x 0 } = 231 16 x 6 − 315 16 x 4 + 105 16 x 2 − 5 16 = 1 16 ( 231 x 6 − 315 x 4 + 105 x 2 − 5 ) Thus, the Legendre polynomial of degree 6 is P 6 ( x ) = 1 16 ( 231 x 6 − 315 x 4 + 105 x 2 − 5 ) _ . Substitute n = 7 in equation (11) to obtain the Legendre polynomial of degree zero as shown below. P 7 ( x ) = { ( − 1 ) 0 ( 2 ( 7 ) − 2 ( 0 ) ) ! 2 7 ( 0 ) ! ( 7 − ( 0 ) ) ! ( 7 − 2 ( 0 ) ) ! x 7 − 2 ( 0 ) + ( − 1 ) 1 ( 2 ( 7 ) − 2 ( 1 ) ) ! 2 7 ( 1 ) ! ( 7 − ( 1 ) ) ! ( 7 − 2 ( 1 ) ) ! x ( 7 ) − 2 ( 1 ) + ( − 1 ) 2 ( 2 ( 7 ) − 2 ( 2 ) ) ! 2 7 ( 2 ) ! ( 7 − ( 2 ) ) ! ( 7 − 2 ( 2 ) ) ! x 7 − 2 ( 2 ) + ( − 1 ) 3 ( 2 ( 7 ) − 2 ( 3 ) ) ! 2 7 ( 3 ) ! ( 7 − ( 3 ) ) ! ( 7 − 2 ( 3 ) ) ! x 7 − 2 ( 3 ) } = { 14 ! 128 ( 7 ) ! ( 7 ) ! x 7 − 12 ! 128 ( 1 ) ! ( 6 ) ! ( 5 ) ! x 5 + 10 ! 128 ( 2 ) ! ( 5 ) ! ( 3 ) ! x 3 − 8 ! 128 ( 3 ) ! ( 4 ) ! ( 1 ) ! x 1 } = 429 16 x 7 − 693 16 x 5 + 315 16 x 3 − 35 16 x = 1 16 ( 429 x 7 − 693 x 5 + 315 x 3 − 35 x ) Thus, the Legendre polynomial of degree 7 is P 7 ( x ) = 1 16 ( 429 x 7 − 693 x 5 + 315 x 3 − 35 x ) _ .

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

5.12 A steel wire 300 feet long and 1½-½ inch in diameter weighs 0.042 #/ft. If the wire is suspended vertically from its upper end, calculate (a) the maximum tensile stress due to its own weight and (b) the maximum weight W that can be safely supported assuming a safety factor of three and an ultimate tensile stress of 65 ksi. a. The maximum tensile stress due to its own weight b. Maximum weight

5.6 The turnbuckles in the diagram shown are tightened until the compression block DB exerts a force of 10,000# on the beam at B. Member DB is a hollow shaft with an inner diameter of 1.0 inch and outer diameter of 2 inches. Rods AD and CD each have cross-sectional areas of 1.0 in.². Pin C has a diameter of 0.75 inch. Determine: a. The axial stress in BD- b. The axial stress in CD- c. The shearing stress in pin C. B UB CD PIN VIEWED FROM BELOW

5.15 The ends of the laminated-wood roof arch shown are tied together with a horizontal steel rod 90 feet, "10 inches" long, which must withstand a total load of 60 k. Two turn- buckles are used. All threaded rod ends are upset. a. Determine the required diameter D of the rod if the maximum allowable stress is 20 ksi. b. If the unstressed length of the rod is 90 feet, "10 inches" and there are four threads per inch on the upset ends, how many turns of one turnbuckle will bring it back to its unstressed length after it has elongated under full allowable tensile stress? E = 29 × 103 ksi. a. Required diameter D b. Number of turns LAMINATED WOOD ROOF ARCH TIE ROD TURNBUCKLE DETAIL: UPSET END