Solve nth Derivative of cos ax with Respect to a

[vin-math]

The nth derivative...

I don't know how to proof the nth derivative of cos ax with respect to a is a^ncos (ax+ na/2). Can anyone help me?

 

[arildno]

Well, it isn't. Where's your pi?

 

[vin-math]

ooops, i type wrongly. it should be "proof the nth derivative of cos (pi)x with respect to a is (pi)^ncos [(pi)x+ n(pi)/2)]"
actually i don't know hot to type the notation of pi here, so i replace the pi with a

 

[d_leet]

ooops, i type wrongly. it should be "proof the nth derivative of cos (pi)x with respect to a is (pi)^ncos [(pi)x+ n(pi)/2)]"
actually i don't know hot to type the notation of pi here, so i replace the pi with a

I think you could prove this by induction, show it's true for n=1, assume it's true for n=k then show that it being true for n=k implies it being true for n=k+1.

 

[vin-math]

I think you could prove this by induction, show it's true for n=1, assume it's true for n=k then show that it being true for n=k implies it being true for n=k+1.

But can i just calculate the final answer (the nth derivative), apart from using the mathematical induction?

 

[HallsofIvy]

You might want to keep in mind that
cos(x+ \frac{n\pi}{2})
is equal to
(-1)^{\frac{n}{2}}cos(x)
if n is even,
(-1)^{\frac{n-1}{2}}sin(x)
if n is odd.

 

[d_leet]

But can i just calculate the final answer (the nth derivative), apart from using the mathematical induction?

What do you mean? The way you stated the problem was that you didn't know how to prove it, and it seems that the best way to prove this would be to use mathematical induction and HallsofIvy's suggestion.

 

[vin-math]

You might want to keep in mind that
cos(x+ \frac{n\pi}{2})
is equal to
(-1)^{\frac{n}{2}}cos(x)
if n is even,
(-1)^{\frac{n-1}{2}}sin(x)
if n is odd.

thx!
I think i now hv some idea in solve the Q.

 

[vin-math]

What do you mean? The way you stated the problem was that you didn't know how to prove it, and it seems that the best way to prove this would be to use mathematical induction and HallsofIvy's suggestion.

I mean that like proofing identities, we will proof the L.H.S equal to the R.H.S. I want to do the same thing in this Q.

 

[d_leet]

I mean that like proofing identities, we will proof the L.H.S equal to the R.H.S. I want to do the same thing in this Q.

And how do you expect to do this? I don't really think that you can do something like that in this situation, and so I will again recommend that you try to prove this using mathematical induction, it really doesn't seem like it would be that hard or long of a proof given the suggestion that HallsofIvy gave you.

 

[lurflurf]

I don't know how to proof the nth derivative of cos ax with respect to a is a^ncos (ax+ na/2). Can anyone help me?

I pressume the result you intend is something like
(0)
( \frac{\d}{\d x} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})
use the following basic results
(1) An obvious trigonometry identity
<br /> \frac{\Delta}{\Delta x} \cos(a x+b)=a \cos(a x+b+a \frac{h}{2}+\frac{\pi}{2})\frac{\sin(h/2)}{h/2}
(2) Continuity of cosine
\lim_{h-&gt;0}(\cos(x+h)-\cos(x))=0
(3) sin(x)~x x small
\lim_{h-&gt;0}\frac{\sin(h)}{h}=1
(4) apply (1) n times
<br /> (\frac{\Delta}{\Delta x})^n \cos(a x+b)=a \cos(a x+b+a n \frac{h}{2}+n \frac{\pi}{2})(\frac{\sin(h/2)}{h/2})^n
(5) passage to the limit (h->0) of (4) using (2) and (3) give the desired result
( \frac{\d}{\d x} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})

 

[HallsofIvy]

By the way, you keep saying "the derivative with respect to a". If that's what you really mean, then the answer you give is NOT correct. But everyone has been assuming that you meant "with respect to x" anyway.

 

[vin-math]

By the way, you keep saying "the derivative with respect to a". If that's what you really mean, then the answer you give is NOT correct. But everyone has been assuming that you meant "with respect to x" anyway.

Yes, u are right

 

[lurflurf]

I don't know how to proof the nth derivative of cos ax with respect to a is a^ncos (ax+ na/2). Can anyone help me?

I pressume the result you intend is something like
(0)
( \frac{d}{dx} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})
use the following basic results
(1) An obvious trigonometry identity
<br /> \frac{\Delta}{\Delta x} \cos(a x+b)=a \cos(a x+b+a \frac{h}{2}+\frac{\pi}{2})\frac{\sin(a h/2)}{a h/2}
(2) Continuity of cosine
\lim_{h-&gt;0}(\cos(x+h)-\cos(x))=0
(3) sin(x)~x x small
\lim_{h-&gt;0}\frac{\sin(h)}{h}=1
(4) apply (1) n times
<br /> (\frac{\Delta}{\Delta x} )^n \cos(a x+b)=a^n \cos(a x+b+a n \frac{h}{2}+n \frac{\pi}{2})(\frac{\sin(a h/2)}{a h/2})^n
(5) passage to the limit (h->0) of (4) using (2) and (3) give the desired result
( \frac{d}{dx} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})