Need help with some antenna maths

[PčéJäy]

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j

 

[budgie]

On Thu, 23 Oct 2003 18:36:58 +1000, "PčéJäy" <wiseguyatiinetnetau>
wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

geez, it's so long since I did that sort of trig that I'd get a sore
head. But surely you know which way the maxrad is?

 

[PčéJäy]

"budgie" <me@privacy.net> wrote in message
news:3f979c4e.33095293@news.individual.net...

On Thu, 23 Oct 2003 18:36:58 +1000, "PčéJäy" <wiseguyatiinetnetau
wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

geez, it's so long since I did that sort of trig that I'd get a sore
head. But surely you know which way the maxrad is?

Yes, it's one of those flower shaped patterns, but unfortunately I need to
be able to prove that it's at exactly 42 and 137 degrees. I've never had to
deal with embedded trig functions yet!

 

[The real Andy]

"budgie" <me@privacy.net> wrote in message
news:3f979c4e.33095293@news.individual.net...

On Thu, 23 Oct 2003 18:36:58 +1000, "PčéJäy" <wiseguyatiinetnetau
wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

geez, it's so long since I did that sort of trig that I'd get a sore
head. But surely you know which way the maxrad is?

Gee, I thought I was real stupid after seeing that calculus test, but
surprisingly, I do remember this stuff. I would give the answer too, but I
am afraid I would be doing someone elses homework. I suggest heading to the
local tech bookstore and investigating schaum!! Gee he musta' been smart...

 

[Roger Lascelles]

If you like, you can use a spreadsheet. Fill a column with angles from 0 to
90 degrees in one degree steps. Copy the formula into the cells all the way
down the next row.

Now insert a graph into the spreadsheet which plots your two columns X,Y.
You can even make it a polar plot to get a nice radiation pattern.

This isn't "real" maths, but it tells you what you want to know.

Roger


"PčéJäy" <wiseguyatiinetnetau> wrote in message
news:3f979462$1_1@news.sydney.pipenetworks.com...

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j

 

[Phil Allison]

"PčéJäy"

Could some kind soul point me in the right direction to solve or simplify
the following:


S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x.

** Seems you have your function in its simplest form - now you just
evaluate it.

To find maxima one uses differentiation - that is not solving.




............. Phil

 

[Vermin]

On Fri, 24 Oct 2003 19:57:47 +1000, "Phil Allison"
<philallison@optusnet.com.au> wrote:

"PčéJäy"


Could some kind soul point me in the right direction to solve or simplify
the following:


S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x.


** Seems you have your function in its simplest form - now you just
evaluate it.

To find maxima one uses differentiation - that is not solving.




............ Phil

I managed to stumble through the EW quiz you posted Phil, but this has
me stymied. It would take me an afternoon to work it out even if I
could be bothered dragging out my old maths text books.

But... just from looking at it, isn't it going to be maximum as x
approaches 0?

As sin(0)=0 so as x and therefore the denominator in the above
equation approaches 0 the expression approaches infinity.

 

[David Segall]

"PčéJäy" <wiseguyatiinetnetau> wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j

When I found myself in a similar situation I used GnuCalc to transform

my equation despite my total inability to do so myself. GnuCalc is
free and open source but plugs into Emacs so is probably not suitable
if you don't know Emacs and/or you don't use Linux or Unix. I had
already tried Mathematica which had failed to transform the equation.

You will find a list of computer algebra systems at
http://sal.kachinatech.com/A/1/. It distinguishes between commercial,
shareware and free programs but I'm sure that some of the commercial
programs will provide a trial version that you can use. The URL above
contains a link to http://www.symbolicnet.org/systems/demos.html which
may save you the trouble of installing the software by using the web
to solve your equation.

 

[Franc Zabkar]

On Thu, 23 Oct 2003 19:22:46 +1000, "PčéJäy" <wiseguyatiinetnetau> put
finger to keyboard and composed:

"budgie" <me@privacy.net> wrote in message
news:3f979c4e.33095293@news.individual.net...
On Thu, 23 Oct 2003 18:36:58 +1000, "PčéJäy" <wiseguyatiinetnetau
wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

geez, it's so long since I did that sort of trig that I'd get a sore
head. But surely you know which way the maxrad is?

Yes, it's one of those flower shaped patterns, but unfortunately I need to
be able to prove that it's at exactly 42 and 137 degrees. I've never had to
deal with embedded trig functions yet!

Normally one could determine the maxima and minima of a function S(x)
by solving the equation S'(x)=0. However, in this particular case the
derivative appears to be more complicated than the original function.

dS/dx = a . [3pi . sin(3pi cosx) / (2 . sinx)
- (1 + cos(3pi cosx)) . cosx / (sinx)^3]

= a . 1/(sinx)^3 . [3pi/2 . sin(3pi cosx) . (sinx)^2
- (1 + cos(3pi cosx)) . cosx]

I'd approach this problem empirically by closing in on the solution
using successive approximation. Obviously the answer lies somewhere in
the range from x = 0 to x = arc cos(1/3) = 70.53 deg, ie the angles at
which the big lobe intersects the origin. You could plug x values into
S(x) and look for a maximum, or you could plug the same values into
S'(x) and look for the angle where the gradient switches from positive
to negative. The latter method converges on the solution (42.5643)
very rapidly.

BTW, the max value of the small lobe is "a" and occurs at 90 deg. And
of course, as S(x) is symmetrical, a solution for x will be repeated
at pi-x, pi+x, and -x radians.


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[Franc Zabkar]

On Sat, 25 Oct 2003 07:46:32 GMT, Vermin <Vermin@nowhere.com> put
finger to keyboard and composed:

On Fri, 24 Oct 2003 19:57:47 +1000, "Phil Allison"
philallison@optusnet.com.au> wrote:


"PčéJäy"


Could some kind soul point me in the right direction to solve or simplify
the following:


S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x.


** Seems you have your function in its simplest form - now you just
evaluate it.

To find maxima one uses differentiation - that is not solving.




............ Phil


I managed to stumble through the EW quiz you posted Phil, but this has
me stymied. It would take me an afternoon to work it out even if I
could be bothered dragging out my old maths text books.

But... just from looking at it, isn't it going to be maximum as x
approaches 0?

No. S(x) -> 0 as x -> 0.

As sin(0)=0 so as x and therefore the denominator in the above
equation approaches 0 the expression approaches infinity.

No. You are ignoring the behaviour of the numerator, which also
approaches 0.

ie cos(3pi/2 . cos(0)) = cos(3pi/2 . 1) = 0

In this case one needs to examine the lim of S(x) as x -> 0. To do
this we need the following info:

sinx -> x as x -> 0
cosx -> 1 - 1/2 . x^2 as x -> 0
cos(90-x) = sinx

Therefore

lim S(x) = a . ( cos(3pi/2 - 3pi.x^2/4) / x )^2
x -> 0

= a . (sin(3pi.x^2/4) / x)^2
= a. (3pi.x^2/4/x)^2

= lim a. (3pi . x / 4)^2
x -> 0

= 0


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[PčéJäy]

"David Segall" <DavidSegall@nowhere.net> wrote in message
news:ft0lpv4n3pbprpnkupvf032e1mau3k6k84@4ax.com...

"PčéJäy" <wiseguyatiinetnetau> wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j

When I found myself in a similar situation I used GnuCalc to transform
my equation despite my total inability to do so myself. GnuCalc is
free and open source but plugs into Emacs so is probably not suitable
if you don't know Emacs and/or you don't use Linux or Unix. I had
already tried Mathematica which had failed to transform the equation.

You will find a list of computer algebra systems at
http://sal.kachinatech.com/A/1/. It distinguishes between commercial,
shareware and free programs but I'm sure that some of the commercial
programs will provide a trial version that you can use. The URL above
contains a link to http://www.symbolicnet.org/systems/demos.html which
may save you the trouble of installing the software by using the web
to solve your equation.

Yes, I've got Linux here. I'll give it a go. Thanks!

 

[PčéJäy]

"Franc Zabkar" <fzabkar@optussnet.com.au> wrote in message
news:r0hlpv0dp7cim6dvfg3a4q4qpjuhqf3r69@4ax.com...

Normally one could determine the maxima and minima of a function S(x)
by solving the equation S'(x)=0. However, in this particular case the
derivative appears to be more complicated than the original function.

dS/dx = a . [3pi . sin(3pi cosx) / (2 . sinx)
- (1 + cos(3pi cosx)) . cosx / (sinx)^3]

= a . 1/(sinx)^3 . [3pi/2 . sin(3pi cosx) . (sinx)^2
- (1 + cos(3pi cosx)) . cosx]

I'd approach this problem empirically by closing in on the solution
using successive approximation. Obviously the answer lies somewhere in
the range from x = 0 to x = arc cos(1/3) = 70.53 deg, ie the angles at
which the big lobe intersects the origin. You could plug x values into
S(x) and look for a maximum, or you could plug the same values into
S'(x) and look for the angle where the gradient switches from positive
to negative. The latter method converges on the solution (42.5643)
very rapidly.

BTW, the max value of the small lobe is "a" and occurs at 90 deg. And
of course, as S(x) is symmetrical, a solution for x will be repeated
at pi-x, pi+x, and -x radians.

Hmmm... Yes, I got the first derivative but couldn't solve it. Our Emag
tutor said we should be able to solve it and started to get upset when we
all had no idea. He's going to give us the answer on Thursday and apparently
he's done it numerically so I'll let you know how he did it.

 

[Franc Zabkar]

On Sun, 26 Oct 2003 14:56:22 +1100, "PčéJäy" <wiseguyatiinetnetau> put
finger to keyboard and composed:

"Franc Zabkar" <fzabkar@optussnet.com.au> wrote in message
news:r0hlpv0dp7cim6dvfg3a4q4qpjuhqf3r69@4ax.com...


Normally one could determine the maxima and minima of a function S(x)
by solving the equation S'(x)=0. However, in this particular case the
derivative appears to be more complicated than the original function.

dS/dx = a . [3pi . sin(3pi cosx) / (2 . sinx)
- (1 + cos(3pi cosx)) . cosx / (sinx)^3]

= a . 1/(sinx)^3 . [3pi/2 . sin(3pi cosx) . (sinx)^2
- (1 + cos(3pi cosx)) . cosx]

I'd approach this problem empirically by closing in on the solution
using successive approximation. Obviously the answer lies somewhere in
the range from x = 0 to x = arc cos(1/3) = 70.53 deg, ie the angles at
which the big lobe intersects the origin. You could plug x values into
S(x) and look for a maximum, or you could plug the same values into
S'(x) and look for the angle where the gradient switches from positive
to negative. The latter method converges on the solution (42.5643)
very rapidly.

BTW, the max value of the small lobe is "a" and occurs at 90 deg. And
of course, as S(x) is symmetrical, a solution for x will be repeated
at pi-x, pi+x, and -x radians.


Hmmm... Yes, I got the first derivative but couldn't solve it. Our Emag
tutor said we should be able to solve it and started to get upset when we
all had no idea. He's going to give us the answer on Thursday and apparently
he's done it numerically so I'll let you know how he did it.

I've taken the liberty of posting your question to alt.math and
sci.math.


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[Vermin]

On Sun, 26 Oct 2003 07:09:03 +1100, Franc Zabkar
<fzabkar@optussnet.com.au> wrote:

On Sat, 25 Oct 2003 07:46:32 GMT, Vermin <Vermin@nowhere.com> put
finger to keyboard and composed:

-snip-

But... just from looking at it, isn't it going to be maximum as x
approaches 0?

No. S(x) -> 0 as x -> 0.

As sin(0)=0 so as x and therefore the denominator in the above
equation approaches 0 the expression approaches infinity.

No. You are ignoring the behaviour of the numerator, which also
approaches 0.

ie cos(3pi/2 . cos(0)) = cos(3pi/2 . 1) = 0

Yep I can see that now.

In this case one needs to examine the lim of S(x) as x -> 0. To do
this we need the following info:

sinx -> x as x -> 0
cosx -> 1 - 1/2 . x^2 as x -> 0
^^^^^^^^

Where did the second term (above) come from? I thought it was just
cosx -> 1 as x ->0.

cos(90-x) = sinx

Therefore

lim S(x) = a . ( cos(3pi/2 - 3pi.x^2/4) / x )^2
x -> 0

= a . (sin(3pi.x^2/4) / x)^2
= a. (3pi.x^2/4/x)^2

= lim a. (3pi . x / 4)^2
x -> 0

= 0


- Franc Zabkar

 

[Franc Zabkar]

On Sun, 26 Oct 2003 09:28:28 GMT, Vermin <Vermin@nowhere.com> put
finger to keyboard and composed:

On Sun, 26 Oct 2003 07:09:03 +1100, Franc Zabkar
fzabkar@optussnet.com.au> wrote:

On Sat, 25 Oct 2003 07:46:32 GMT, Vermin <Vermin@nowhere.com> put
finger to keyboard and composed:

-snip-

But... just from looking at it, isn't it going to be maximum as x
approaches 0?

No. S(x) -> 0 as x -> 0.

As sin(0)=0 so as x and therefore the denominator in the above
equation approaches 0 the expression approaches infinity.

No. You are ignoring the behaviour of the numerator, which also
approaches 0.

ie cos(3pi/2 . cos(0)) = cos(3pi/2 . 1) = 0


Yep I can see that now.

In this case one needs to examine the lim of S(x) as x -> 0. To do
this we need the following info:

sinx -> x as x -> 0
cosx -> 1 - 1/2 . x^2 as x -> 0
^^^^^^^^
Where did the second term (above) come from? I thought it was just
cosx -> 1 as x ->0.

If you follow that line of reasoning, then sinx -> 0 as x -> 0, which
gets you nowhere. You need to show that the numerator approaches zero
faster than the denominator.

To show that cosx -> 1 - 1/2 . x^2 as x -> 0, we know that

(sinx)^2 + (cosx)^2 = 1

So, as x -> 0

(cosx)^2 = 1 - (sinx)^2
= 1 - x^2

Therefore, for small values of x

cosx = 1 - 1/2 . x^2

To confirm that this the case,

cosx . cosx = (1 - 1/2 . x^2) . (1 - 1/2 . x^2)
= 1 - x^2 + 1/4 x^4
= 1 - x^2
= 1 - sinx . sinx

cos(90-x) = sinx

Therefore

lim S(x) = a . ( cos(3pi/2 - 3pi.x^2/4) / x )^2
x -> 0

= a . (sin(3pi.x^2/4) / x)^2
= a. (3pi.x^2/4/x)^2

= lim a. (3pi . x / 4)^2
x -> 0

= 0


- Franc Zabkar

- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[Vermin]

On Mon, 27 Oct 2003 07:13:30 +1100, Franc Zabkar
<fzabkar@optussnet.com.au> wrote:

In this case one needs to examine the lim of S(x) as x -> 0. To do
this we need the following info:

sinx -> x as x -> 0
cosx -> 1 - 1/2 . x^2 as x -> 0
^^^^^^^^
Where did the second term (above) come from? I thought it was just
cosx -> 1 as x ->0.

If you follow that line of reasoning, then sinx -> 0 as x -> 0, which
gets you nowhere. You need to show that the numerator approaches zero
faster than the denominator.

To show that cosx -> 1 - 1/2 . x^2 as x -> 0, we know that

(sinx)^2 + (cosx)^2 = 1

So, as x -> 0

(cosx)^2 = 1 - (sinx)^2
= 1 - x^2

Therefore, for small values of x

cosx = 1 - 1/2 . x^2

To confirm that this the case,

cosx . cosx = (1 - 1/2 . x^2) . (1 - 1/2 . x^2)
= 1 - x^2 + 1/4 x^4
= 1 - x^2
= 1 - sinx . sinx

cos(90-x) = sinx

Therefore

lim S(x) = a . ( cos(3pi/2 - 3pi.x^2/4) / x )^2
x -> 0

= a . (sin(3pi.x^2/4) / x)^2
= a. (3pi.x^2/4/x)^2

= lim a. (3pi . x / 4)^2
x -> 0

= 0

Thanks for taking the time to explain that Frank, much appreciated. I
haven't done that stuff for years, makes me wonder what else I've
forgotten.

V.

 

[Dana]

Hi David. You say that Mathematica was giving you a hard time. I have no
idea how to solve this manually, but here is my attempt w/ Mathematica...
I believe this is the equation:

equ = (Cos[((3*Pi)/2)*Cos[x]]/Sin[x])^2

It returns the following...
Cos[(3/2)*Pi*Cos[x]]^2*Csc[x]^2

I wasn't sure if this was correct, so I did a test. Since there are
multiple peeks with the smaller lobes, I limited the search between 0 and 1.
(A plot helps)

sol = NMaximize[equ, {x, 0, 1}]
{1.9572148603660808, {x -> 0.7428876577624516}}

This does nothing more than extract the "x" value

x /. sol[[2]]
0.7428876577624516

This converts it to degrees:
(x /. sol[[2]])*(180/Pi)
42.564327206391006

This is close to the given solution, but it is not exactly 42 degrees.
Here's the other one...

sol = NMaximize[equ, {x, 2, 3}]
{1.9572148603660806, {x -> 2.398704995180533}}

(x /. sol[[2]])*(180/Pi)
137.43567252079302

The other solution given was 137.

Here is its attempt at a derivative:
v2 = D[equ, x]
-2*Cos[(3/2)*Pi*Cos[x]]^2*Cot[x]*Csc[x]^2 +
3*Pi*Cos[(3/2)*Pi*Cos[x]]*Csc[x]*Sin[(3/2)*Pi*Cos[x]]

This was its "best" solution. This does not seem like a simple equation to
solve for x when the equation =0. By converting this to Exponential form,
and then simplifying it, I got a slightly shorter answer for the derivative:

v3 = FullSimplify[TrigToExp[v2]]
(1/2)*Csc[x]*(-4*Cos[(3/2)*Pi*Cos[x]]^2*Cot[x]*Csc[x] +
3*Pi*Sin[3*Pi*Cos[x]])

A plot shows these two are the same.
Here, I search for a solution out to 50 digits to see if it is exactly 42
degrees.

(x /. FindRoot[v3 == 0, {x, 0.1, 1}, WorkingPrecision -> 50,
MaxIterations -> 200])*(180/Pi)
42.564327442147583437573369699132730650952776941648402067792

And the other...
(x /. FindRoot[v3 == 0, {x, 2, 3}, WorkingPrecision -> 50, MaxIterations ->
200])*(180/Pi)
137.435672557852416562426630300867269349047442770547199228163

Again, I have no idea how to manually solve for the Derivative, or to solve
for x when the equation equals 0.
--

Dana

= = = = = = = = = = = = = = = = =


"David Segall" <DavidSegall@nowhere.net> wrote in message
news:ft0lpv4n3pbprpnkupvf032e1mau3k6k84@4ax.com...

"PčéJäy" <wiseguyatiinetnetau> wrote:

Hi all,

Could some kind soul point me in the right direction to solve or simplify
the following:

S = a ( cos(3pi/2 cos(x)) / sinx )^2

Need to solve for x. We never learnt this in maths and none of my books
seem
to cover it either. It's in relation to finding the direction of maximum
radiation from a dipole antenna.

Thanks!
P.j

When I found myself in a similar situation I used GnuCalc to transform
my equation despite my total inability to do so myself. GnuCalc is
free and open source but plugs into Emacs so is probably not suitable
if you don't know Emacs and/or you don't use Linux or Unix. I had
already tried Mathematica which had failed to transform the equation.

You will find a list of computer algebra systems at
http://sal.kachinatech.com/A/1/. It distinguishes between commercial,
shareware and free programs but I'm sure that some of the commercial
programs will provide a trial version that you can use. The URL above
contains a link to http://www.symbolicnet.org/systems/demos.html which
may save you the trouble of installing the software by using the web
to solve your equation.

 

[David Segall]

"Dana" <ng_only@hotmail.com> wrote:

Hi David. You say that Mathematica was giving you a hard time. I have no
idea how to solve this manually, but here is my attempt w/ Mathematica...
I believe this is the equation:

equ = (Cos[((3*Pi)/2)*Cos[x]]/Sin[x])^2

It returns the following...
Cos[(3/2)*Pi*Cos[x]]^2*Csc[x]^2

I wasn't sure if this was correct, so I did a test. Since there are
multiple peeks with the smaller lobes, I limited the search between 0 and 1.
(A plot helps)

sol = NMaximize[equ, {x, 0, 1}]
{1.9572148603660808, {x -> 0.7428876577624516}}

This does nothing more than extract the "x" value

x /. sol[[2]]
0.7428876577624516

This converts it to degrees:
(x /. sol[[2]])*(180/Pi)
42.564327206391006

This is close to the given solution, but it is not exactly 42 degrees.
Here's the other one...

sol = NMaximize[equ, {x, 2, 3}]
{1.9572148603660806, {x -> 2.398704995180533}}

(x /. sol[[2]])*(180/Pi)
137.43567252079302

The other solution given was 137.

Here is its attempt at a derivative:
v2 = D[equ, x]
-2*Cos[(3/2)*Pi*Cos[x]]^2*Cot[x]*Csc[x]^2 +
3*Pi*Cos[(3/2)*Pi*Cos[x]]*Csc[x]*Sin[(3/2)*Pi*Cos[x]]

This was its "best" solution. This does not seem like a simple equation to
solve for x when the equation =0. By converting this to Exponential form,
and then simplifying it, I got a slightly shorter answer for the derivative:

v3 = FullSimplify[TrigToExp[v2]]
(1/2)*Csc[x]*(-4*Cos[(3/2)*Pi*Cos[x]]^2*Cot[x]*Csc[x] +
3*Pi*Sin[3*Pi*Cos[x]])

A plot shows these two are the same.
Here, I search for a solution out to 50 digits to see if it is exactly 42
degrees.

(x /. FindRoot[v3 == 0, {x, 0.1, 1}, WorkingPrecision -> 50,
MaxIterations -> 200])*(180/Pi)
42.564327442147583437573369699132730650952776941648402067792

And the other...
(x /. FindRoot[v3 == 0, {x, 2, 3}, WorkingPrecision -> 50, MaxIterations -
200])*(180/Pi)
137.435672557852416562426630300867269349047442770547199228163

Again, I have no idea how to manually solve for the Derivative, or to solve
for x when the equation equals 0.
Thanks Dana. My experience using Mathematica with a completely

different equation was not so successful. I left it running overnight
but it neither transformed my equation nor told me that it could not
do so. GnuCalc solved it in a minute or two. I have just failed to
find the equation in my files otherwise I would have posted it for you
to try with a newer version of Mathematica. It was much simpler than
this equation and did not involve any trigonometric functions.

Perhaps the OP can use your results. Mathematica does not seem to
offer a trial version.

 

[Dana]

... GnuCalc solved it in a minute or two. .... It was much simpler than
this equation and did not involve any trigonometric functions.

Thanks for the feedback with what GnuCalc came up with! That sounds very
interesting. I'll keep playing with it then. Right now, I can't seem to
get it any shorter. Sounds like the professor is expecting some type of
shorter code in order to solve it. I sure don't see it yet though. :&gt

--
Dana
= = = = = = = = = = = = = = = = =

<snip>

Thanks Dana. My experience using Mathematica with a completely
different equation was not so successful. I left it running overnight
but it neither transformed my equation nor told me that it could not
do so. GnuCalc solved it in a minute or two. I have just failed to
find the equation in my files otherwise I would have posted it for you
to try with a newer version of Mathematica. It was much simpler than
this equation and did not involve any trigonometric functions.

Perhaps the OP can use your results. Mathematica does not seem to
offer a trial version.

 

[Franc Zabkar]

On Mon, 27 Oct 2003 10:13:56 -0500, "Dana" <ng_only@hotmail.com> put
finger to keyboard and composed:

Hi David. You say that Mathematica was giving you a hard time. I have no
idea how to solve this manually, but here is my attempt w/ Mathematica...
I believe this is the equation:

equ = (Cos[((3*Pi)/2)*Cos[x]]/Sin[x])^2

It returns the following...
Cos[(3/2)*Pi*Cos[x]]^2*Csc[x]^2

.... which is just (Cos[((3*Pi)/2)*Cos[x]])^2 * (1/Sin[x])^2

This is not a "simplification", just a minor rewrite.

I wasn't sure if this was correct, so I did a test.

No need for a test as cosec(x) = 1/sin(x) by definition.

The only "simplified" expression I could come up with is the
following. It gets rid of the squared terms, but doesn't help answer
the question.

S(x) = a . (1 + cos(3pi cos(x)) / (1 - cos(2x))

Since there are
multiple peeks with the smaller lobes, I limited the search between 0 and 1.
(A plot helps)

sol = NMaximize[equ, {x, 0, 1}]
{1.9572148603660808, {x -> 0.7428876577624516}}

This does nothing more than extract the "x" value

x /. sol[[2]]
0.7428876577624516

This converts it to degrees:
(x /. sol[[2]])*(180/Pi)
42.564327206391006

- Franc Zabkar
--
Please remove one 's' from my address when replying by email.