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anil86
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Find nth Derivative - Quick Tutorial
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- Derivative Tutorial
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SUMMARY
The discussion focuses on proving the recurrence relation for the nth derivative rather than calculating it directly. The key equation presented is $$(1-x^2)y_{n+1} - 2(\gamma + nx) y_n -n(n-1)y_{n-1} = 0.$$ The initial condition is established with $$y_1 = \frac{2\gamma y}{1-x^2}.$$ The method of proof by induction is recommended, starting from the base case derived from differentiating the established equations.
PREREQUISITES- Understanding of recurrence relations in calculus
- Familiarity with differentiation techniques
- Knowledge of proof by induction methodology
- Basic grasp of functions and their derivatives
- Study the method of proof by induction in mathematical analysis
- Explore advanced differentiation techniques in calculus
- Review recurrence relations and their applications in mathematical proofs
- Investigate the properties of functions related to the equation $$y_1 = \frac{2\gamma y}{1-x^2}$$
Mathematicians, calculus students, and educators looking to deepen their understanding of derivatives and recurrence relations in mathematical proofs.
- #2
Ackbach
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Could you please re-scan your attachment? The current one is so out-of-focus as to be nearly useless.
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anil86
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Opalg
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The assignment does not ask you to find the $n$th derivative, but to prove that $$(1-x^2)y_{n+1} - 2(\gamma + nx) y_n -n(n-1)y_{n-1} = 0.$$
You have shown that $$y_1 = \frac{\gamma y}{1+x} + \frac{\gamma y}{1-x} = \frac{2\gamma y}{1-x^2}.$$ Write that as $$(1-x^2)y_1 - 2\gamma y = 0.$$ Differentiate, to get $$(1-x^2)y_2 - 2xy_1 - 2\gamma y_1 = 0.$$ Now use that as the base case for a proof by induction.
You have shown that $$y_1 = \frac{\gamma y}{1+x} + \frac{\gamma y}{1-x} = \frac{2\gamma y}{1-x^2}.$$ Write that as $$(1-x^2)y_1 - 2\gamma y = 0.$$ Differentiate, to get $$(1-x^2)y_2 - 2xy_1 - 2\gamma y_1 = 0.$$ Now use that as the base case for a proof by induction.
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