The other day I was looking at series and noticed that cos(i) would actually be a real number, and the series for cos(ix) would be a part of e^x . Is there a function which has the required series for the missing terms (i.e. x + x³ /3! + ...) to make up e^x , besides [sin(ix)]/i?
Also - is there a way to derive the series of cos^-1 and sin^-1 (if such series exist) from the series for cos and sin?

I hope the above makes sense - I haven't actually studied imaginary numbers yet and am just treating them as ordinary numbers.

Thanks,
Olof.

Your observations are excellent. A few points:

cos(i) is a real number. However I very much doubt it is rational - it is actually equal to (e + 1/e)/2 (we'll see why a bit later). To prove that it is irrational I would have thought you could do a similar trick to the one in the proof that e is irrational. I'll try this a bit later. Do you know the proof that e is irrational by the way? I could write it out, or if you have STEP Maths III, 1997, Q7 proves it (this is how I know the proof!).

Alternatively you can use the fact that e is transcendental. This means it is not the root of any polynomial equation with integer coefficients. I must confess I can't prove this, but it is known that e (and also pi) are transcendental. Then note that if (e + e^-1 )/2 is rational we can write

(e + e^-1 )/2 = a/b

for integers a,b which gives:

b e² + b = 2ae

be² - 2ae + b = 0

so e is a root of a polynomial equation which then contradicts the fact that e is transcendental. Therefore (e + e^-1 )/2 is irrational.

Secondly you are right that cos(ix) has part of the series for e^x , and you are also right that sin(ix)/i has the "missing bit". If you put these two together you get:

cos(ix) + sin(ix)/i = e^x

Now there is a far more usual way of writing this - let x = it:

cos(-t) + sin(-t)/i = e^it

So:

e^it = cos(t) + i sin(t)

This is an extremely useful (and well known) formula, called Euler's theorem. One particularly interesting consequence is that if you set t = pi:

e^ipi = -1.

Richard Feynman was especially entralled by this result.

If you take Euler's theorem and replace t by -t you get:

e^-it = cos t - i sin t

And if you add this to e^it = cos t + i sin t you get:

cos t = [e^it + e^-it ]/2

(you can now see why cos(i) = (e + 1/e)/2

If you subtract one equation from the other:

sin t = [e^it - e^-it ]/2i

These two formulae can be very useful. One application (though by no means the most important) is you can now work out things like:

sin x + sin 2x + sin 3x + ... + sin nx

Simply rewrite the sum in terms of exponentials then arrange it to get two geometric progressions, which you know how to sum.

Also you ask whether x + x³ /3! + x⁵ /5! + ... has another name. The answer is yes - it is called sinh(x) (pronounced as shine x). This is a hyperbolic function - two others are cosh x (pronounced as spelt) and tanh x (I'm not really sure how this is pronounced!). They are highly analogous to the normal trigonometric functions. They satisfy:

cosh(x) = cos(ix)
sinh(x) = -isin(ix)
tanh(x) = sinh(x) / cosh(x)

and are reasonably useful. You can write them very simply in terms of exponentials:

cosh(x) = (e^x + e^-x )/2
sinh(x) = (e^x - e^-x )/2
tanh(x) = (e^2x - 1)/(e^2x + 1)

Also they obey very similar identities to sin,cos and tan. For instance:

sinh(2x) = 2sinh(x)cosh(x)

This of course is no co-incidence! You can show it is equivalent to sin(2x) = 2sin(x)cos(x) by making a very simple substitution.

Hope this helps slightly,

Michael

By the way, you might like to have a think about other ways of showing that

e^ix = cos x + i sin x

Here is one way.

Let y = cos x + i sin x

Now y'' + y = 0

where y'' = second derivative of y wrt x.

You can rewrite this as:

y'' + iy' - iy' + y = 0

which then becomes:

(y' + iy)' - i(y' + iy) = 0

So z = y' + iy satisfies:

z' - iz = 0

Now you can solve this for z in terms of exponentials (applying initial conditions). Then solve for y (again with initial conditions). You should find y = e^ix , thus proving e^ix = cos x + i sin x.

The first way I came across how to find Euler's theorem, was when we were set an integration question - integrate 1/(x² +1). Not knowing about the derivative of arctan, I tried writing it as:

1/(x² +1) = 1/[(x+i)(x-i)]

Then split the right hand expression into partial fractions (using normal rules of algebra) and integrate! If you equate the result to arctan x you can derive the result:

arctan x = i/2 ln((i+x)/(i-x))

from which you can get Euler's formula after a lot of tedious manipulation.

It's funny that you should mention STEP III '97 - I just got ahold of this paper yesterday! I'll have a look at Question 7 tonight, thanks for telling me about it.

e^pi=-1 is certainly amazing. After all what has the base ofnatural growth got to dowith the ratio between the circumferenceand diameter of a circle, and the square root of -1!? I've heard people even use this to argue the existence of God! (Let's not get side-tracked by that though!) It generally tends to surprise non-mathematicians more than mathematicians.

This is perhaps because, if you think about it for a while, you realise that e^pi=-1 is actually little more than a tautology, in thick disguise. But to realise why you need to know a bit aboutthe geometry of complex numbers. Have you heard of the Argand diagram?

Next... z ' -i z=0 is a first order differential equation. Another way of writing is is:

dz/dx=i z

Have you learnt how to solve equations of this type yet? Here is one way of getting z in terms of x:

1/z dz/dx=i

Notice that by the chain rule:

d(lnz)/dx = 1/z dz/dx

So our equation:

d(lnz)/dx=i

In other wordsthe gradient of lnz is i everywhere. This means that lnz is a straight line:

lnz = i x+C

where C is just some constant.

z=e^{i x+C}=e^{i x}e^C=A e^{i x}

where A is another constant (=e^C).

We really want to get rid of that constant A, as it is unknown at the moment. So what do we know? Well we know that when x=0, z=2i. How do we know this? Well y=cosx+isinx=1 (when x=0) and y ' =-sinx+icosx=i (when x=0. So z=i+i=2i when x=0 (this is called an initial condition).

Also we know z=A e^{i x}, so substituting x=0:

2i = A e⁰, so A=2i

z=2i e^{i x}

Therefore the differential equation for y is:

y ' +i y=2i e^{i x}

I'll show you how to solve this for y if you like, but you'll probably work it out for yourself first! There are actually two approaches, and I'll give a hint for each one:

Approach 1: Multiply the equation by e^{i x}. See if you can do anything with the result.

Approach 2: Start by finding any old solution to the equation. Then use this to find all solutions.

Yours,

Michael

Well done for getting STEP 1997, Q7 so quickly. I remember struggling with the third part of Q7 for a long time for some reason, when I first tried the paper about 1-2 years ago. Euler was the first to prove that e is irrational in the 1730s I think. I'm not sure whether that was his method or not (but I'd guess so).

For y' + iy = 2ie^ix then if you multiply by e^ix :

e^ix y' + e^ix iy = 2ie^2ix

This can now be written:

(e^ix y)' = 2ie^2ix

(you can check the left hand side using the product rule)

Now you can integrate:

e^ix y = e^2ix + C

Because 2ie^2ix integrates to e^2ix . To check this you can differentiate e^2ix by the chain rule. Is this bit OK? If so we then have:

y = e^ix + Ce^-ix

where C is an arbitrary constant. So in fact we can't yet conclude y = e^ix ; that is only one possibility. To show that C = 0 we need more information.

Remembering the definition of y as cos x + i sin x, you can apply the condition y = 1 when x = 0:

1 = e⁰ + C

So C = 0, and we get:

y = e^ix

But y = cos x + i sin x so:

e^ix = cos x + i sin x as we were hoping!

Now there is another method for solving:

y' + iy = 2ie^ix

You have already spotted one solution is y = e^ix . (If you hadn't spotted this you could "guess" a solution of the form Ae^ix and see if it works for any A.) Once you've got one possible solution, the problem becomes easier.

Suppose there are two different functions y₁ (x) and y₂ (x) that both satisfied the given equation.

Well

y₁ ' + iy₁ = 2ie^2ix
y₂ ' + iy₂ = 2ie^2ix

Subtracting:

(y₁ - y₂ )' + i(y₁ - y₂ ) = 0

So q = y₁ - y₂ satisfies:

q' + iq = 0

So:

1/q q' = -i

(ln q)' = -i
ln q = -ix + C
q = e^{-ix + C}
q = Ae^-ix

And this means that any two solutions to the equation satisfy:

y₁ - y₂ = Ae^-ix

It therefore follows that if y₂ (x) = e^ix then for all y₁ satisfying the equation:

y₁ = e^ix + Ae^-ix

which is the same solution as before. Once again you can show A = 0 for this particular problem - sometimes you can't determine A, and have to leave it as a final result.

In this case you wouldn't need to bother with any of this, because you had already found y = e^ix and this happens to be the solution with the correct value at x = 0. For simple 1st order differential equations (i.e. ones not involving y'', y''', y'''') it is always true that if you find a function that satisfies the equation, and it has the correct value at x = 0 (or indeed anywhere else) then this function gives the correct values everywhere.

By the way, I've just noticed you asked about the series for cos^-1 x and sin^-1 x. These can't really be derived from the series from sin and cos (or at least I can't!) The problem is:

sin x = x - x³ /3! + x⁵ /5! - ...

Now if you replace x by sin^-1 x:

x = sin^-1 x - (sin^-1 x)³ /3! + ...

Unfortunately this gives us a polynomial equation of infinite order to solve for sin^-1 x, which is not easy. It might just be possible though if we assume sin^-1 x to have a series solution, then plug in to get a recurrence relationship for the coefficients. I can't quite make this work at the moment, but it is certainly a good idea.

An easier way to get sin^-1 x (and you can apply an identical method for cos^-1 x) is to look at its derivative.

If y = sin^-1 x

Then:

x = sin y

Differentiate wrt y:

dx/dy = cos y

So dy/dx = 1/cos y

Now we try to get this in terms of x:

dy/dx = 1/sqrt(1-sin² y) = 1/sqrt(1-x² )

Strictly you should check to make sure you take the positive root, but it should be clear why this is so, given that sin^-1 x returns principle values.

So if y = sin^-1 x, dy/dx = 1/sqrt(1-x² )

Now try writing out the denominator of the RHS using the binomial expansion. Then integrate to find y (remember you get an arbitrary constant of integration, which you need to calculate) and then that's the series expansion for sin^-1 x!

Hope this helps,

Michael

Another way to find p is via tan^-1 x. The most obvious one is by setting x=1 in the Maclaurin series of tan^-1 x, which is

x-x³/3+x⁵/5-¼

and multiply by 4. However, this series converges very slowly, and one can show that you will need about 1 million terms to get to 6 d.p accuracy. On the other hand, if you play around with tan(a±p/4)=tanb or various other equations, you should be able to find a method that will give p in a reasonable amount of time.

Kerwin

Another way of seeing that x⁴ = 1 has a root of -i as well as 1,-1,i is the following theorem:

If x is a solution to a polynomial equation with real coefficients then x* is also a solution. x* is the complex conjugate of x, defined as follows:

(a + ib)* = a - ib

where a,b are real.

This theorem is pretty easy to prove if you know the binomial expansion. Here the theorem tells us that as i is a solution to x⁴ = 1, i* is also a solution. And of course i* = (0 + 1i)* = (0 - 1i) = -i. This can be a useful check if you're solving polynomials in the complex domain.

Out of interest, when you programmed your calculator with the series for sin^-1 x and then tested it out, which value of x did you use? The obvious one to try I think is x = 1/2 as this returns pi/6. If this is what you did then I don't really see why it didn't converge... It should converge for all -1 < x < 1.

There are loads of formulae for calculating pi. One other is:

^{pi 2} /6 = 1/1² + 1/2² + ...

Actually that one is extremely important for other reasons. To see where it comes from, note that sin x has roots at n pi. Therefore it is conceivable that the sin function is:

sin x = ...(1 - x/(2pi))(1 - x/pi)x(1 + x/pi)(1 + x/(2pi))...

Then you can combine factors pairwise to get (by difference of two squares):

sin x = x(1-x² /pi² )(1-x² /(4pi² ))...

Then write sin x using its usual Maclaurin expansion and equate the coefficients of x³ on each side of the equation, and you should get:

1/1² + 1/2² + ... = pi² /6

This method also allows you to work out 1/1⁴ + 1/2⁴ + ... (simply equate coefficients of x⁵ ) and in fact all the series with even powers, but the result for odd powers is unknown.

This series does converge more quickly than the arctan one Kerwin gave, but this is nothing compared to one derived in the 1960s. In this algorithm you have a simple programming loop (of about 6 lines) which gives an approximation for p each loop. The amazing thing is: the number of correct digits of p it gives doubles with each loop!! I haven't got the program on me at the moment unfortunately, but you can find it in the Penguin Dictionary of Curious and Interesting Numbers (a very useful reference book).

I can't remember any of the rest of the 1997 paper (I don't think I did all the questions, but the prove e is irrational one was too interesting to miss out!). If you have any specific difficulties with the paper please post them!

Your teacher was obviously in a very bad mood to make you calculate that derivative. Out of interest, how did you differentiate 4^{sin x} ?

Yours,

Michael

By the way, I should have mentioned earlier that the standard way of deriving Euler's formula (the one you'll find in every such textbook without fail) also uses series expansions:

e^ix = 1 + ix + (ix)² /2! + (ix)³ /3! + (ix)⁴ /4! + ...
= (1 - x² /2! + x⁴ /4! - ...) + i(x - x³ /3! + x⁵ /5! - ...) = cos x + i sin x

I used to dislike this method, but then I distrusted series expansions till quite recently. And anyway it gives no indication of why the result is true. The best way of seeing why it is true is to think geometrically. So I'll give a bit of introductory information about the Argand diagram.

Each complex number is represented as a point on a plane. So the number x + iy is displayed as the point with co-ordinates (x,y).

The modulus of a complex number is defined as the length from the origin to the number in the Argand diagram. So modulus of a + ib = sqrt(a² + b² )

The argument of a complex number is the angle between the x axis and the number. So we can write all complex numbers in modulus-argument form:

z = r(cos x + i sin x)

where r is the modulus and x is the argument.

Next... can you see what is going to happen when you add two complex numbers? It is very, very similar to vector addition!

Also we need to know about multiplication. See if you can show that when you multiply two complex numbers the result has modulus equal to the product of the moduli of the first two complex numbers, while the argument of the product is the sum of the arguments of the original complex numbers. Hint: use the modulus - argument form of the numbers given above.

I think that not many people would have seen your way of deriving Euler's formula, because substituting a complex quantity into a cosine (which is to do with real angles) is not at all obvious!

Yours,

Michael

I see what you mean about -i satisfying x⁴ = 1 as well. Complex numbers open up so many new interesting possibilities.

Is the series for tan^-1 x really that slow to converge to pi /4? Is there a way to prove how quickly a series converges for different values?

I found what the problem was with my program: there was a bracket missing on one of the lines! It only caused a slight error, but it was enough to cause the series to diverge after 10 or so terms. I might have a go at programming it onto my PC, so that I can get more decimal places (my calculator only does it up to 9 I think), and compare the efficiencies of all the different methods I can find for generating pi .

The Penguin Dictionary of Curious and Interesting Numbers is a very good book. They have it at my school's library - I'll get it out on monday and have a look for the loop you mention - it sounds amazing! The only problem with the book is that you find so many things you want to investigate further. I had no idea how to get to the series for pi ² /6, which I first found in this book, so thanks once again for showing me. One thing - I'm having a little trouble equating the coefficients for x⁴ and reducing it all to 1/1⁴ + 1/2⁴ .... What is it that I'm missing?



To differentiate y = 4^{sin x} I said:

y = e^u ,
u = ln4 xsin x,

Then
du/dx = ln4 xcos x,
dy/du = e^u ,

therefore

dy/dx = ln 4 x cos x x4^{sin x} . Hope that's right.

It wasn't actually the teacher who set the question: she told us students to set each other challenges, and my 'so-called' friend gave me that :).

Thanks for the information on the Argand diagram - I'll be sure to have a go at working with complex numbers in the ways you mentioned (when I'm not this tired).

Regards,

Olof.

[Part of this post was actually from Kerwin Hui: the correction he pointed out has been inserted! - The Editor]

Regarding 1/1⁴+1/2⁴+¼,

Actually this is a bit tricky to write out so perhaps a better way is as follows:

sinx=x(1-x²/p²)(1-x²/4p²)¼

or we can write this:

sinx=¼(1-x/(2p))(1-xp)x(1+x/p)(1+x/(2p))¼

Take logs and differentiate:

cosx/sinx=¼+1/(x+2p)+1/(x+p)+1/x+1/(x-p)+¼

We have to be a bit careful here about convergence. To make sense, we have to evaluate this infinite sum from the middle outwards. So we can think of the RHS as the limit of the sequence:

1/x

1/(x+p)+1/x+1/(x-p)

1/(x+2p)+1/(x+p)+1/x+1/(x-p)+1/(x-2p)

...

Differentiate the expression again:

cosec² x=¼+1/(x+p)²+1/x²+1/(x-p)²+¼

cosec2=

¥ å r=-¥

(x+rp)-2

Differentiating once gives

-2cosec2 xcotx=-2

¥ å r=-infty

(x+rp)-3

Simplify and differentiate again gives

-2cosec2 xcot2 x-cosec4 x=-3

¥ å r=-¥

(x+rp)-4

Substitute x=p/2 and remember that cot(p/2)=0, we have

1=3

¥ å r=-¥

(p/2)+rp-4=96p4

å all positive odd r

r-4

so we have

å all positive odd r

r-4=p4/96

Hence

1/14+1/24+¼ = z(4)=

¥ å r=1

r-4=p/90

Can you see why this method fails for 1/1³+1/2³+1/3³+¼ (and indeed all odd powers)? Whether this number has an analytical form is unknown. In the 1970s it was proved to be irrational but that is about all we know!

Yours,

Michael

Just to clarify:

z(n) is the Zeta function, defined as:

z(n)=1/1ⁿ+1/2ⁿ+1/3ⁿ+¼

This function is connected to possibly the most important unsolved problem in mathematics today - the Riemann hypothesis. This states that the only complex solutions to the equation:

z(x)=0

are of the form x=1/2+t i where t is real.

So if you're finding your A-Levels a bit easy and have a bit of spare time, why not give this a crack?

Ah, so that's what the Riemann hypothesis is all about. Looks like fun.
What are cosec and cot? From my differentiation (which is probably wrong), cosec x would equal 1/sin x; is that right? I'm not even going to hazard a guess for cot x (ok go on then, 1/tan x?).
Do they have special derivative connections, or do you have to go through using product rule and chain rule about 10 times over when doing all the above?

/Olof

Olof,

Indeed, cosec x = 1/sin x, and cot x = 1/tan x = cos x/sin x.

To find the derivative of cot x, we differentiate the quotient to get

(-sin² x - cos² x)/sin² x, which is, of course, -1/sin² x, otherwise known as -cosec² x.

Similarly, we can differentiate cosec x (I'll leave it as an exercise for you).

Note that in some books, cosec is abbreviated further as csc.

Kerwin

Thanks Kerwin.

y = cosec x

dy/dx = -cot x/sin x?

Does 1/cos x also have another name?


Michael-
I've had a look at what you mentioned about complex arithmetic and the Argand diagram, and I'm fine with that.
I'm still working on showing Euler's formula geometrically using this.


Regards,
Olof.

Olof,

That's correct. The derivative of cosec x is -cot x cosec x. 1/cos x is written as sec x, where sec is the abbreviation of secant.

Going back to the question of convergence, there is no easy way of estimating how quickly a series converges. However, for alternating series that converge, the truncation error is less than or equal to the first neglected term providing the remaining terms are each less than the first neglected term in magnitude. Hence, as a crude approximation, the number of terms needed to get N decimal place accuracy out of

tan^-1 (1)=1-3^-1 +5^-1 -7^-1 +...

is when the first neglected term is less than (0.5)10^-N , which means that 10^N terms is necessary.

Kerwin

To remember what cot, sec andcosec are the reciprocal of,just look at their third letter. We similarly have hyperbolic equivalents:

coth,sech,cosech

Anyway, I've just realisedI made a bit of a mess of thegeometry - I used the letter x both as the argument, and as the x co-ordinate. I think you knew what I meant. Just to re-iterate:

A complex number x+i y is represented by a point (x,y) in the Argand diagram. And it can also be specified by the length from the origin (the modulus r), and the angle between the x-axis and the line joining the origin to the complex number (this is called the argumentand I'll call it q this time). The angle is measured anti-clockwise, by the way.

So the modulus-argument form of our complex number z=x+i y is:

z=r(cosq+isinq)

where

r=

_____ Öx2+y2

and tanq = y/x.

And we know that complex numbers add like vectors and when you multiply two complex numbers you multiply the moduli and add the arguments.

To see how to explain Euler's formula with the Argand diagram, remember what y=e^{m x} means. It means the rate of change of the function y is equal to m times the value of the function y.

Now if let m=i, we have a function, the rate of change of which wrt x is equal to i y. And we recall that in the Argand diagram i y is perpendicular to y and of the same magnitude.

Now imagine you are walking in the Argand diagram, and your position after a time t is given by the complex number y=e^{i t}. You know the rate of change of the function (your position) with respect to time is given by i y. In othe words, your velocity is i y and is perpendicular ot the line joining you and the origin. So what kind of curve are we moving in?

Would you be moving in a circle (radius 1, centre the origin) on the Argand diagram? I'm not sure if I went about it the right way : I plotted the complex numbers z, where z = e^it .

Sorry for taking so long to get back to you; I have a Physics exam on Monday so I haven't been on the PC much.

Regards,
Olof.

Sorry, I missed your message! Anyway, your answer is correct but unfortunately you've had to assume the result we're trying to prove!

What's happening is that we're moving in such a way that our velocity is perpendicular to the line joining us and the origin (see earlier). In other words our distance from the origin isn't changing . (If our distance from the origin was changing this would require our velocity had an inward or outward component. And we know this isn't true.) Therefore we are clearly walking in a circle, centred on the origin.

But what is the radius of the circle? Well our position is z = e^it . So at t = 0 we are at z = e^i*0 = 1. We start off at a distance one from the origin, and as we're moving in a circle we stay a distance 1 from the origin.

Therefore e^ix is a circle (radius 1, centre origin) in the Argand diagram. The only question is which values of x correspond to which points on the unit circle. If you're with me so far, I'll try and explain this next bit...

Good luck in the physics!

Michael

Of course! It's so obvious now!
I missed the key bit about e^mt meaning that the rate of change of the function y is equal to m times the value of the function y, which you mentioned in your previous post. That cleared it all up.

I kept doing the same thing (assuming the thing you're trying to prove) when I had a go a proving the fundamental theorem of arithmetic. Now that was annoying. Got there in the end though, fortunately.

I had the Physics exam this morning (Materials and Waves). I think it went OK, but everyone thought it was an easy exam, so the examiners may not be very kind in their marking. It'll be a while before we find out what we get anyway, so I'm not going to worry about it.

Thanks again,

Olof