- #1
anil86
- 10
- 0
Find nth Derivative - Quick Tutorial
- Context: MHB
- Thread starter anil86
- Start date
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- Tags
- Derivative Tutorial
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Discussion Overview
The discussion centers around finding the nth derivative of a function, with a focus on a specific assignment that requires proving a recurrence relation involving derivatives rather than directly calculating the nth derivative. The context includes mathematical reasoning and proof techniques.
Discussion Character
- Homework-related
- Mathematical reasoning
Main Points Raised
- One participant requests a clearer attachment to facilitate understanding of the nth derivative problem.
- Another participant points out that the assignment is not about finding the nth derivative but proving a specific recurrence relation involving derivatives.
- The same participant provides a derivation of the first derivative and suggests using it as a base case for a proof by induction.
Areas of Agreement / Disagreement
Participants do not seem to reach a consensus on the primary task; while one participant emphasizes the need for a proof, others focus on the derivative calculation. The discussion remains unresolved regarding the best approach to the assignment.
- #2
Ackbach
Gold Member
MHB
- 4,148
- 94
Could you please re-scan your attachment? The current one is so out-of-focus as to be nearly useless.
- #3
anil86
- 10
- 0
Attachments
- #4
Opalg
Gold Member
MHB
- 2,778
- 13
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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