How to Solve for the Nth Derivative of a Trigonometric Function?

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Discussion Overview

The discussion revolves around finding the nth derivative of the function sin(2x)^(-1/2) within the interval (0, π/2). Participants explore a differential equation involving the second derivative and the function itself, seeking a relationship to determine the value of n.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation D2(f) + f = 3*D(a) and seeks a quick method to solve for n, expressing uncertainty about the nth derivative.
  • Another participant questions the meaning of "a'd derivative" and clarifies the notation used for derivatives.
  • A participant suggests that the nth derivative refers to 1/√(sin(2x)) and asks for clarification on what f represents.
  • One participant reformulates the problem, stating that if y = 1/√(sin(2x)), then the equation y'' + y = 3y^{(n)} needs to be solved for n, noting that their calculations for the first 10 derivatives do not satisfy the equation.
  • Another participant expresses skepticism about the existence of a solution, suggesting that the characteristic equation derived from the differential equation cannot yield 1/√(sin(2x)) as a solution.
  • Some participants discuss the possibility of misunderstanding the exercise, with one suggesting that the answer should be 5 and speculating about the role of α as a power in the equation.
  • Clarifications are made regarding the notation and the nature of the problem, with participants reflecting on their earlier posts and misunderstandings.

Areas of Agreement / Disagreement

Participants express differing views on the solvability of the problem and the interpretation of the notation used. There is no consensus on the correct approach or solution, and some participants acknowledge potential misunderstandings.

Contextual Notes

There are unresolved questions regarding the assumptions made in the problem, particularly concerning the notation and the nature of the derivatives involved. The discussion reflects varying levels of understanding and interpretation of the mathematical expressions.

Who May Find This Useful

This discussion may be of interest to those studying differential equations, trigonometric functions, or mathematical notation, particularly in the context of higher-order derivatives.

JanClaesen
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For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!
 
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Hello,

For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!

What are you trying to do here? Are you trying to find the derivative of

sin(2x)^(-0.5) ?

What is a'd derivative?

Thanks
Matt
 
Hi JanClaesen! :smile:

(have a pi: π and a square-root: √ :wink:)
JanClaesen said:
For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!

(we say the nth derivative, not the n'd derivative :wink:)

Do you mean "what is the nth derivative of 1/√(sin2x)?"

And what is f? :confused:
 


Hello :smile:, I'm sorry if I was a little unclear:

So there is a certain relation for the function sin(2x)^(-1/2): the second derivative of this function plus the function itself gives the (n'th derivative)*3, the question is to determine n.
 


Rephrase:

If [itex]y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}[/itex], solve for n. (Note [itex]y^{(n)} = \frac{d^n y}{{dx}^n}[/itex].)

I have hammered out the first 10 derivatives of y but none seem to be equal to (y'' + y)/3. The exponent of the denominator factor is of the order 1/2 + n, while the numerator is a trigonometric polynomial in cosine of order n. I do not forsee this collapsing into something nice. This exercise seems fishy.

N.B.: The expression on the left of the differential equation is

[tex]\frac{3}{(sin(2x))^{5/2}}.[/tex]​

--Elucidus
 
Elucidus said:
If [itex]y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}[/itex], solve for n. (Note [itex]y^{(n)} = \frac{d^n y}{{dx}^n}[/itex].)

Hi JanClaesen! Hi Elucidus! :smile:

I don't think there can be a solution …

just try to solve 3y(n) - y'' - y = 0 the usual way …

that gives you a characteristic equation, of which 1/√sin can't be an answer. :redface:
 


I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg
It's exercice 17a, the answer should be 5, perhaps alpha is a power?
 
Last edited by a moderator:
JanClaesen said:
I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg
It's exercice 17a, the answer should be 5, perhaps alpha is a power?

D'oh!

Yes of course α is a power … Question 17a says so :rolleyes: …

f'' + f = 3fα

Why did you write D2(f) + f = 3*D(a) in your first post??

ok, since f = 1/√(sin(2x)), what is f'' + f?
 
Last edited by a moderator:


Because I only thought of that while I was writing the newest post :smile:
I'm sorry, but I never saw a power-variable being named alpha before, that's why I thought it was perhaps a way to note the n'th derivative, which seemed to me like a quite hard thing to solve.
Hope I didn't waste too much of your time!
 
  • #10


Just curious ... what language is that question written in ... and are you from the Netherlands ?
 
  • #11


Hah nice, you're right, it's Dutch :wink:, but I'm not from the Netherlands but from its little brother, Belgium. Most people here are American I guess?
 

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