# Oxford Maths Interview 2009

[I was asked some questions – maths questions]

## 22 thoughts on “Oxford Maths Interview 2009”

1. Okay, now I will perform back substitution. The first function that satisfy the above equality is exponential function, thus we may write exp(x+y)=exp(x)exp(y). You are right because exp(0)=1. But can you show me another function?

2. jebavarde says:

I’ve been thinking… to no avail. I’m thinking perhaps some function made of a combination of trigometric functions may exhibit this, since cos(a+a) = 2cos(a)cos(a) – 1, and cos(0) = 1, which seems kind of close.

Considering the ‘opposite’ of this equality:
f(ab) = f(a) + f(b)
I have found two functions which satisfy this.
1. Obviously (it follows from your function) the logarithmic function: log(ab) = log(a) + log(b).

2. The arg function (argument of a complex number)
arg(zw) = arg(z) + arg(w)

I’ll think it over and try to find another function which satisfies the first.

3. jebavarde says:

One function which does satisfy this, but is cheating somewhat, would be f(x) = 1

4. jebavarde says:

Done some reading about, and apparently this is the ‘Cauchy Exponential Function’

5. Really your statement not wrong, because you have introduced the word “when” in the front of f(x+y)=f(x)f(y). With the word “when” as the constrain, so your equality is always satisfied for x=0 and hence f(0)=1. Let we check for cosinus function

cos(x+pi/2)=cos(x)cos(pi/2) , (1)

But we know from expand of the cosinus function that

cos(x+pi/2)=cos(x)cos(pi/2) – sin(x)sin(pi/2) , (2)

Thus, it is clearly that eq.(1) is obtained after cancel out the term of – sin(x)sin(pi/2) in eq.(2) because sin(0)=0. And somebody will agree with you that cos(0)=1 as the proof f(0)=1.

Okay, next we continue with expanding the left side of eq.(1)

cos(x+pi/2)=cos(x)cos(pi/2)

cos(x)cos(pi/2)-sin(x)sin(pi/2)=cos(x)cos(pi/2)

-sin(x)=cos(x)cos(pi/2)

finally,

-sin(x)=0,

We know that there so many values of x that satisfied the sin(x)=0, that are x=0, pi, 2pi, etc but only the even multiple of 2pi that satisfy cos(x)=1. Please take your conclusion to my explanation above.

Next, let consider the following example if the word “when” to be ignored from your statement

sin(x+pi/6)=sin(x)sin(pi/6)

6. Okay my special comment for this statement

%One function which does satisfy this, but is cheating % %somewhat, would be f(x) = 1

Let’s try this following function

f(x)=V*sqrt(1-x)-arctan(sqrt(x/(1-x)))-arctan(sqrt((x+a)/(1-x)))

for any positive of a. You will never the value of x that satisfies f(x)=1 except for V>arctan(sqrt(a)).

Happy with you.

7. Apologize there is small corection, my purpose

Let’s try this following function

f(x)=[V*sqrt(1-x)-arctan(sqrt(x/(1-x)))-arctan(sqrt((x+a)/(1-x)))]/pi

for any positive of a. You will never the value of x that satisfies f(x)=1 except for V>arctan(sqrt(a)).

8. Well, next I will explore more detail the following equality of

tan(x+pi/4)=[tan(x)+tan(pi/4)]/[1-tan(x)tan(pi/4)]

or

tan(x+pi/4)=[tan(x)+1]/[1-tan(x)], (1)

Let’s fill x=pi/4, then we have

tan(pi/2)=[tan(pi/4)+1]/[1-tan(pi/4)]

that gives :

1/0=2/0

Okay we agree to do not say that 1=2, isn’t it.

But let we change the tangent function for the left side with this

tan(t)=sin(2t)/[1+cos(2t)], (2)

hence we obtain the following equality 0/0=2/0, and what do you think withthis? I believe you will agree with me that without L’Hospital Theory, mathematics ever inexactly, isn’t it.

Are you and all visitors here interested to learn more about how to create eq.(2) from arctangent differential equation dy/dt=1+y^2?
The new tangent function only can be create by using SMT (shortened from the stable modulation technique) that recently posted in this address :
http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34&start=20
Please visit to the website maybe useful for you all.

9. My example of f(x) above is the equation of normalized dispersion relation of optical mode guided in a slab waveguide that has been commonly used in designing integrated optics for Photonics application. Please visit to the following address for detail explanation.

The simpliest method to obtain the x value of normalized optical mode propagation constant for every values of V and a that represent normalized thickness of the guide and the different ratio of refrative indices cover to the substrate of the guide respectively, is the secant method that needed two initial guesses x0 and x1. The following script of solving the dispersion relation for searching the value of x that I created using matlab. Maybe useful for you all here, let’s try the following scheme.

function x2=rohedi(func,a,V,x0,x1)
%script used in this routin called frohedi.m

x2=x1-(x1-x0)feval(func,a,V,x1)/(feval(func,a,V,x1)-feval(func,a,V,x0));
while (abs(x2-x1)>=1e-16)
x0=x1;
x1=x2
x2=x1-(x1-x0)feval(func,a,V,x1)/(feval(func,a,V,x1)-feval(func,a,V,x0));
end
%%%%%
function y=frohedi(a,V,x)
f=[V*sqrt(1-x)-atan(sqrt(x/(1-x)))-atan(sqrt((x+a)/(1-x)))]/pi;
y=f-1;
%%%%%

Important to be informed here that the value of x satisfied the dispersion relation :
V*sqrt(1-x)=atan(sqrt(x/(1-x)))+atan(sqrt((x+a)/(1-x)))+pi
is in the range of 0<x>a=1000;V0=atan(sqrt(a))+0.1; x0=0.9; x1=0.91;

If the searching value of x for every value of a is performed iteratively from V>atan(sqrt(a)), there are so many values of x that satisfy f(x)=1. This things to be proved by plotting the x-V curve.

V0=atan(sqrt(a))+0.1;Vf=16;step=0.1;
V=V0:step:Vf;
nV=length(V);
for k=1:nV
x(k)=rohedi(‘frohedi’,a,V(k),x0,x1);
end
plot(V,x);

10. o o o, there is a miss type, the following revisted,

Important to be informed here that the value of x satisfied the dispersion relation :
V*sqrt(1-x)=atan(sqrt(x/(1-x)))+atan(sqrt((x+a)/(1-x)))+pi
is in the range of 0<xa=1000;V0=atan(sqrt(a))+0.1; x0=0.9; x1=0.91;

Sincerely,
Rohedi.

11. apologize, there is still miss display, while the right of x range values is greater than null and lower than 1, or in 0<x<1.
Best regards,
Rohedi.

12. Denaya Lesa says:

Hello rohedi,

can you explain me the difference of arctangent and tangent function?
Thx.

13. jebavarde says:

Hi,
I’m not Rohedi, but I can explain the difference.

arctangent is simply the inverse function to tangent, so that arctan(tan(x)) = x and tan(arctan(x)) = x (assuming a domain of 0 to Pi)

and say one knows that tan(x) = sqrt(3), then one can calculate x by taking the arctan of both sides:
arctan(tan(x)) = x = arctan(sqrt(3)) = PI/3

Not sure if that helps… but yeah, they are inverse functions, just like f(x) = x^2, g(x) = sqrt(x) – f and g are inverse functions.

14. He he he…maybe miss Denaya Lesa regards me is also your member group at oxford math interview 2009 Mr.Jabavarde.

Denaya, if both function are written in infinite series, then

for arctangent function

atan(x)=1-(1/3)x^3+(1/5)x^5-(1/7)x^7+(1/9)x^9+…

while tangent function

tan(x)=x+(2/3!)x^3+(16/5!)x^5+(272/7!)x^7+(7936/9!)x^9
+(353792/11!)x^11+(22368256/13!)x^13+…

here the three dot (…) goes to infinity, and ! is factorial.

You know that if you write y=tan(x), then the two function related each other in form

dy/dx=1+y^2

where ∫dy/(1+y^2) is called arctangent.

Okay Denaya, happy with you.

15. Denaya Lesa says:

@jabavarde,

Sorry I’ve done small blunder by calling you rohedi. Okay, now I ask special question for you.

According to RexHWu@aoL.com the pie number П or is defined as

П=8*atan(1/3) + 4*atan(1/7)

why mathematicians still hunter another defintion of the pie number? Thx for your attention.

16. jebavarde says:

I’m not sure that they are exactly. There are lots of definitions of pi, the simplest by far being:
“Pi is the ratio of a circle’s diameter to its circumference.”

Other definitions include:
# The sum of the infinite series $4/1 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + ...$
# 4 times $\int_0^1 \! \sqrt{1-x^2} \, dx$

For the above, observe that $(x-a)^2 + (y-b)^2 = r^2$ is the general equation of circle centre (a,b) radius r.

Hence the unit circle (centre (0,0) radius 1) has equation $x^2 + y^2 = 1$

The unit circle has area $\pi r^2 = \pi \times 1^2 = \pi$

Now make y the subject of the equation of the unit circle:
$x^2 + y^2 = 1 \Rightarrow y = \sqrt{1-x^2}$ and so integrating this between 0 and 1 yields the area of one quadrant of the circle, hence multiplying this by 4 gives a value for $\pi$

I suppose some mathematicians are always looking for new ways of calculating PI, since it’s a strange number. For info on this I’d look at wikipedia:
http://en.wikipedia.org/wiki/Computing_%CF%80

17. Finding all function satisfying $f(x+y)=f(x)f(y)$ is an interesting problem, unfortunately many of the posts above is unrelated or has little thing to do with the original post. Ok, by putting continuity condition on $f$ , I claim that the functions that satisfying the the condition are exponential functions $a^x$ (with any base $a$).
Let $f(1)=a$. It is easy to see that if $a=0$ then $f(x)=0$ for every $x$. So from now on let $a\neq 0$.
It is obvious that by induction $f(x_1+\cdots +x_n)=f(x_1)\cdots f(x_n)$.
By this property we have $f(n)=f(1)^n=a^n$ for every non negative integer $n$.
Now $1=f(0)=f(1+(-1))=f(1)f(-1)$, which implies that $f(-1)=a^{-1}$. From this it follows that $f(m)=a^m$ also true for negative integer.
Now if $p/q$ is a rational number, then
$a^p=f(p)=f(p/q\cdot q)=f(p/q)^q$. Hence $f(p/q)=a^(p/q)$.
So till now we know that for every rational number $r$ we have $f(r)=a^r$.
Now let $x$ be an irrational number. Let $\{r_n\}$ be a sequence of rationals converging to $r$. By continuity of $f$ and the exponential function we have
$f(x)=\lim_{n\to \infty} f(r_n)=\lim_{n\to \infty}a^{r_n}=a^x$.
The last step we can easily verify that $f(x)=a^x$ satisfies the original condition.
QED.

18. Greetings! I posed your problem to a friend of mine and he came up with this. What do you think?

1. jebavarde says:

Nice :D Can’t say I fully understand it, but have only looked through briefly. I’m not familiar with the style of argument used… I’ll look later – I’ve just returned from a maths exam, so want to relax for a while :P

2. jebavarde says:

Wow, this is now almost 2 years later :D..