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φιλ = Σ m=3,5,7.... (2a ) sin (mm) π? ᾖ 2mat το (1 - m2))

φιλ = Σ m=3,5,7.... (2a ) sin (mm) π? ᾖ 2mat το (1 - m2))

Advanced Engineering Mathematics
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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Could you explain how we expanded the sum ? Why m got out of the sum as a constant?

= 2 a = = 91¹ = 2 √² ( a 4maa π² ħ² 4maa π² ħ² 24 ) sin ( Σ a π² ħ² 2ma² m=3,5,7.... ħ² (sin (³x) + a ma =√√() (sin (²x) - :) Therefore, the first three terms are МП 2 (1 - m²)) -1 -1 ·) (193° + √2/545° +1=4947° + ....) 43 1-25 4 sin (5x) + — sin (7x) + ....) -1 -48 -24 Pm0 sin (2x) + π² ħ² sin (²5x) + ...) a ma ) (sin ( ³7x) — — sin ( 5x) + ½-sin (77x) + ...). -
Transcribed Image Text:= 2 a = = 91¹ = 2 √² ( a 4maa π² ħ² 4maa π² ħ² 24 ) sin ( Σ a π² ħ² 2ma² m=3,5,7.... ħ² (sin (³x) + a ma =√√() (sin (²x) - :) Therefore, the first three terms are МП 2 (1 - m²)) -1 -1 ·) (193° + √2/545° +1=4947° + ....) 43 1-25 4 sin (5x) + — sin (7x) + ....) -1 -48 -24 Pm0 sin (2x) + π² ħ² sin (²5x) + ...) a ma ) (sin ( ³7x) — — sin ( 5x) + ½-sin (77x) + ...). -
Expert Solution
Step 1

It is given that:

                                                  φ11=∑m=3,5,7....2αasinmπ2π2h22ma21-m2φm0

We have to expand the given sum.

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