Re: Smarter Scots

Chicmac wrote:
Chicmac wrote:
Custos Custodum wrote:
On 9 May 2006 04:10:05 -0700, "Chicmac"
<charles.mcgregor@xxxxxxxxxxxx> wrote:

Chicmac wrote:
Chicmac wrote:
Custos Custodum wrote:
On Wed, 3 May 2006 22:11:36 +1200, "Mad Prof" <mad@xxxxxxx> wrote:

"uNkulunkulu" <izulu@xxxxxxxxxx> wrote in message
news:k0Z5g.63407$wl.19314@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
"Mad Prof" <mad@xxxxxxx> wrote in message
news:e39cv8$j2h$1@xxxxxxxxxxxxxxxx
Are Scots any smarter than other nations? We often laugh at the Yanks
who's
knowledge of geography is sadly lacking but if you went on the streets
of
a
major Scottish town and asked a person at random to point to say 'South
Dakota' or Luxemberg on a map would they be able to do it? I doubt it.
Ask
them to solve a simple algebra problem

sin(x)+x^2=0
As this equation ha a trigonometric function (sin(x)) it has to be a
trigonometric problem and not algebraic. At first glance it is obvious to
those with even a smattering of high school maths that the answer can not
be
zero unless x=0


Fraid not - that is the trivial solution - there is also another solution.
You can see this by drawing the graph of sin(x) and x^2 and looking at the
intersections.(one of which is your solution at x=0) But other than a
graphical solution how to find x?

I don't think it can be solved algebraically - you have to resort to
numerical methods such as successive approximation or truncated
series. Alternatively you can learn to use a programmable calculator
and get the result x = -0.8767 to 4 decimal places.
Actually it can very simply, since the question does not specify
radians, (which you have understandably assumed).

So if I choose to use degrees.

Then it can immediately be seen that the solution in that instance must
be <<1.

This allows us to make the small angle approximation, where sin x = x
where x is in radians and x<<1 radian.

there are 57.3 degrees in a radian, therefore if we are working in
degrees with a very small angle then the approximation means that


x/57.3 = x^2

so x=1/57.3 = 0.000345 approx.

Chic
Oops, I forgot the neg. sign.

x/57.3 =-x^2

x= -0.000345
Jings I must have been pissed when I did this.

Having used some mathematical lateral thinking in switching from
radians to degrees in order to obviate the need for a calculator, i.e
trying to be clever, I then go on to demonstrate my intrinsic need of a
calculator in order to do simple arithmetic.

The answer should have been 1/57.3 = 0.01745 degrees.

Also I was talking pish with x as a negative. Not allowed so the
transposition needs to be
x^2 = -sin x

Classic case of concentrating on the clever bit and ignoring the
simple.
Indeed. I think you're barking up the wrong tree (or should that be
log?) with this approach, Chic. Sin(x) = x is a useful approximation
to make when performing calculations using known values, but it is
fraught with dangers when solving equations involving unknowns.
Since |sin(x)| <= 1 for all real x (complex x hurts my brain too
much), regardless of units chosen, we can also state that, for this
problem, |x| <= 1. It doesn't necessarily follow that x << 1. If you
deliberately choose x very small to allow your approximation, you get
x + x^2 = 0. Factoring gives x(1+x) = 0, so x = 0 or -1. Now x = 0 is
the trivial solution. The other solution, x = -1, contains an
indeterminate error because of the error inherent in the original
approximation. Remember that sin(x) = x isn't just an approximation;
it's a limit. i.e. sin(x) --> x as x --> 0. Ignoring arithmetical
errors, the answer you got was basically the x = 0 solution (because
you assumed it to be very nearly so at the outset), offset by the
error introduced by your approximation.

[ObScot: Just in case anyone is wondering what this has to do with
Scottish culture, I give you:
The Maclaurin Series for sin(x):
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ... ]

x = - 0.01745

Shove it into your equation:-

sin(x)+x^2=0

and you get

- 0.0003045 + 0.0003045 = 0

And nope it is not just an error due to approximation.

Gotta go out so can't go into any more detail.

Right, sorry bout that bit busy.

Here is a bit more on this.

Most folk who only take maths up to school level measure angles in
degrees where 1 degree = 1/360th of a circle.

However in higher maths, most of the time, radians are used.
What is a radian?
If you take a piece of string and cut it to be the same length as the
radius of a circle, then place that piece of string around the
circumfrence, then take a line from each end of the piece of string to
the centre of the circle, the angle between those two lines is
approximately 57.3 degrees. This is known as 1 radian.
This is because, as most will remember, the circumference of a circle
is equal to pi (3.1472 etc.) times the diameter, which is obviously 2pi
times the radius.
So 360/2pi = 57.3 degrees.

But why on Earth would folk want to use such a big unit for measuring
angles when the 'normal' much smaller degree is much easier to deal
with? Especially when it doesn't even go into a circle evenly, there
are 6 and bit of them making up one circle.


Well, here is a clue. Sine of an angle is defined as being the length
of the side opposite to the angle divided by the length of the
hypotenuse of a a right angled triangle. (Right angled meaning one
corner of the triangle has an angle of 90 degrees or pi/2 and the
hypotenuse being the longest side of the triangle).
Now imagine a rod of length = 1 lying on the ground. If you lift one
end while the other remains fixed, then the free to move end obviously
describes part of a circle. If you move it up to an angle of 1 radian
(or 57.3 degrees) then the end of the rod will have described part of a
circle equal in length to the radius.
To get the sine of that angle, we woulds need to drop an imaginary line
vertically down to the ground to make our right angle and create our
opposite side. The sine of the angle is then the length of that
imaginary vertical line divided by the length of the radius. In this
special case we have chosen the length of the radius/hypotenuse to be
1, so the sine is simply the length of the imaginary vertical line.
Also because the radius is 1, the length of circumference to the ground
is also the angle in radians. The thing to notice here is that in this
instance the length of this vertical line from the end of the rod to
the ground is quite a bit less than the length of the circumference
from there to the ground. So sin (z) < z where z is in radians.
However if we imagine a much smaller angle, like a 6th of a radian,
then the length of a vertical line dropped from the end of the rod to
the ground, is now very close to the length of the circumference
dropped from there to the ground.
In this case, the sine of the angle, length of the vertical line, is
virtually identical to the length of the circumference from there to
the ground. So sin (z) for small angles is only a very little bit less
than z. As the angle gets closer to zero, then, as we say 'in the
limit' the two become the same.
This is a very useful property because it allows for series expansions
of triginometric functions. THAT is why we use radians in higher
maths.

But it also allows us to use the sin (z) = z (where z is in radians) to
a good approximation for small angles (anything less than 15 degrees or
a quarter of a radian say).

Now the stated problem does not specify degrees or radians, although as
said, normally in higher maths, radians would be assumed. Whether
expressed in degrees or radians, the sine of an angle remains the same.
However, the square of the angle is dependent upon the units used. By
instead choosing to assume it was in degrees, the presence of sin (x)
means that x^2 it would have to be =<1, because sine of anything
cannot be >1 or less than -1. So x must be =<1 degree.

That in turn means we can use the small angle approximation (because we
know it is a small angle). BUT the small angle approximation can only
be used if the angle is expressed in radians, SO the approximation

sin (x/57.3) = x/57.3 is valid where x is in degrees

And it does not matter whether we calculate sin (x) directly using
degrees tables OR sin (x/57.3) using radians tables, we get the same
answer, the same value for the sine of the angle, namely x/57.3.


So for :-
sin(x)+x^2=0

we now have

x/57.3 +X^2 =0
or x=-1/57.3 = - 0.001745 degrees to a very close approximation.

Oh and regarding the Scottish connection, MacLaurin's series was shown
to be a special case of 'Taylor's' series, but the real Scottish
connection is that 'Taylor's' series was originally discovered decades
earlier by the Scotsman James Gregory. Certainly the hardest done by
Scot from a historical perspective, and that is saying something.
Virtually unkown outside esotheric mathematical circles he was the
first to devise methods for calculating gradients and areas under
curves pre-empting both Newton's and Leibnitz' claim to being the
father of calculus (a fact even acknowledged in Newton's extant notes).
His work on series was truly groundbreaking, recognising the
importance of convergence and divising a test for it later attributed
to Leibnitz, as was the first series cacluation for pi. He invented
the reflecting telescoipe later attributed to Newton and the first
diffraction grating, described planetary orbits etc. etc. and died in
mysterious circumstances at a young age.

Aye well, I had to use three dimensional trigonometry while studying crystallography as part of my degree course (I scored 17/35 in a class test and still came second!), and I still don't understand a word of what you've said above.

A very important part of being "smart", IMHO, is being able to explain yourself to others. This is a fact of life that the Tams of this world fail to recognise.

--
Ian O.
http://www.iomorrison.pwp.blueyonder.co.uk
.

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