How to find minimum turning points

[weetabixharry]

I have a function:

f(x) = Asin²(x) + Bcos²(x) + Csin(2x)

and I want to find the minimum turning point(s). To start with, I calculated:

f'(x) = (A - B)sin(2x) + 2Ccos(2x)

Therefore, turning points occur when f'(x)=0, or:

tan(2x) = 2C / (B - A)

To find the minima, I then want to look at the 2nd derivative:

f''(x) = 2(A - B)cos(2x) - 4Csin(2x)

So, as far as I understand, minima will occur when both f'(x)=0 and f''(x)>0. So the second inequality can be expressed as:

2(A - B)cos(2x) > 4Csin(2x)

Now I'm having trouble knowing how to proceed. Since sin and cos can be positive or negative depending on x, I can't see a neat way of really understanding where the minima will be. Could anyone help?

 

[pasmith]

I have a function:

f(x) = Asin²(x) + Bcos²(x) + Csin(2x)

and I want to find the minimum turning point(s). To start with, I calculated:

f'(x) = (A - B)sin(2x) + 2Ccos(2x)

Therefore, turning points occur when f'(x)=0, or:

tan(2x) = 2C / (B - A)

To find the minima, I then want to look at the 2nd derivative:

f''(x) = 2(A - B)cos(2x) - 4Csin(2x)

So, as far as I understand, minima will occur when both f'(x)=0 and f''(x)>0. So the second inequality can be expressed as:

2(A - B)cos(2x) > 4Csin(2x)

Now I'm having trouble knowing how to proceed. Since sin and cos can be positive or negative depending on x, I can't see a neat way of really understanding where the minima will be. Could anyone help?

Best to put $$f(x) = \frac12(A + B) + \frac12(B - A)\cos(2x) + C \sin(2x) = \frac12(A + B) + R\sin(2x + \alpha)$$
using $$R\sin(2x + \alpha) = R\sin(\alpha)\cos(2x) + R\cos(\alpha)\sin(2x).$$
The minima then occur when $\sin(2x + \alpha) = -1$.

 

[weetabixharry]

Okay that's really clever... it's taken me a while to figure out how that works.

So, if I have $0 \leq x < 2\pi$, I think taking the arcsin would give:$$\alpha+2x=\frac{3\pi}{2}$$Then, given how we've defined $\alpha$, I feel like we can say:$$\tan\alpha = \frac{(B-A)}{2C}$$Putting those things together, I get:$$x=\frac{3\pi}{4}-\frac{1}{2}\tan^{-1}\left\{\frac{(B-A)}{2C}\right\}$$Hmm... but given that tan(x) repeats itself twice on the interval I'm interested in $(0 \leq x < 2\pi)$ ... I feel like I've just reintroduced ambiguity.

I'm a bit out of my depth, so I've crunched some numbers with some computer simulations... and this x seems to sometimes give the minimum, sometimes the maximum. I feel like I need to somehow constrain (B - A)/2C ... but at the moment I'm somewhere on a slippery slope to proving 1 = 0...

 

[weetabixharry]

I've looked more closely at my problem and have determined three further constraints:$$A\geq0\\B\geq0\\C\sin(2x)\geq0$$Imposing these constraints seems to provide a unique solution in my computer simulations... but I'm not really certain why. Can anyone offer any insight?

 

[pasmith]

Okay that's really clever... it's taken me a while to figure out how that works.

So, if I have $0 \leq x < 2\pi$, I think taking the arcsin would give:$$\alpha+2x=\frac{3\pi}{2}$$Then, given how we've defined $\alpha$, I feel like we can say:$$\tan\alpha = \frac{(B-A)}{2C}$$Putting those things together, I get:$$x=\frac{3\pi}{4}-\frac{1}{2}\tan^{-1}\left\{\frac{(B-A)}{2C}\right\}$$Hmm... but given that tan(x) repeats itself twice on the interval I'm interested in $(0 \leq x < 2\pi)$ ... I feel like I've just reintroduced ambiguity.

That is to be expected, because $f(x)$ has a period of $\pi$, not $2\pi$. Thus there are two minima in $0 \leq x \leq 2\pi$.