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

[JanClaesen]

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!

 

[CFDFEAGURU]

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

 

[tiny-tim]

Hi JanClaesen!

(have a pi: π and a square-root: √ )

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 )

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

And what is f?

 

[JanClaesen]

Hello , 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.

 

[Elucidus]

Rephrase:

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

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

\frac{3}{(sin(2x))^{5/2}}.​

--Elucidus

 

[tiny-tim]

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

Hi JanClaesen! Hi Elucidus!

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.

 

[JanClaesen]

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?

 

[tiny-tim]

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 …

f'' + f = 3f^α

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

ok, since f = 1/√(sin(2x)), what is f'' + f?

 

[JanClaesen]

Because I only thought of that while I was writing the newest post
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!

 

[srijithju]

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

 

[JanClaesen]

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