HSC 2013 MX2 Marathon (archive) (9 Viewers)

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

barbernator said:
In how many ways can 1 person be put into 1 room.

I will use my rote learned formula:

$^1 C _ 1 = \frac{1!}{(1-1)! \cdot 1!} = \frac{1}{1 \cdot 1} = 1$

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

Might want to be a bit careful about the concept of unique solutions.

y=k e^x satisfies the ODE y'=y, for all real k.

Also, the differential equation y''+y=0 has the general solution y=Asin(x)+Bcos(x), for all real A and B.

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Might want to be a bit careful about the concept of unique solutions.

y=k e^x satisfies the ODE y'=y, for all real k.

Also, the differential equation y''+y=0 has the general solution y=Asin(x)+Bcos(x), for all real A and B.

Heh, well that failed then. Am I still allowed to assume a function is sin x or cos x if the power series satisfies the definition y'' + y = 0 ?
If not, how can we derive the taylor series for sin and cos using only the derivative definition?

And for ke^x, am I allowed to:

E(x) = k (infinite series)

Then let k=1?

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Heh, well that failed then. Am I still allowed to assume a function is sin x or cos x if the power series satisfies the definition y'' + y = 0 ?
If not, how can we derive the taylor series for sin and cos using only the derivative definition?

And for ke^x, am I allowed to:

E(x) = k (infinite series)

Then let k=1?

If the power series satisfies the definition y''+y=0 (although there are some serious issues with infinite applications of the derivative if you are differentiating a power series), then it isn't necessarily sin(x) or cos(x). In fact, it could be ANYTHING in the form Asin(x)+Bcos(x). Of course, we could use Auxiliary Angles to make it a single trig function, but then we have Asin(x+alpha) for example, not sin(x) as you said.

Also, for ke^x, E(x) is already predefined to be the taylor series for e^x, so we cannot possibly have E(x) = k (taylor series), otherwise it just becomes k*E(x).

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
If the power series satisfies the definition y''+y=0 (although there are some serious issues with infinite applications of the derivative if you are differentiating a power series), then it isn't necessarily sin(x) or cos(x). In fact, it could be ANYTHING in the form Asin(x)+Bcos(x). Of course, we could use Auxiliary Angles to make it a single trig function, but then we have Asin(x+alpha) for example, not sin(x) as you said.

Also, for ke^x, E(x) is already predefined to be the taylor series for e^x, so we cannot possibly have E(x) = k (taylor series), otherwise it just becomes k*E(x).

Oh alright I see.

Ah well, I can't seem to find a rigorous way to introduce Euler's formula into 4U

GoldyOrNugget · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Show that $ \ \ y=e^x \ \ $is the only function which satisfies$ \ \ \ y'=y \ \ \ \ \ \fbox{1}$

Oooh I love this proof! It's so clean and elegant.

$Let $y=f(x)$ be a function that satisfies $y=y'$, so $f(x) = f'(x)$, i.e. $f(x) - f'(x) = 0$. \\ Define $g(x) = \frac{f(x)}{e^x}$. By the quotient rule , $g'(x) = \frac{e^xf'(x) - f(x)e^x}{e^{2x}} = \frac{f'(x) - f(x)}{e^x} = 0$, so $g(x)$ is constant -- let $k = g(x)$. \\ Substituting this into our definition of $g(x)$, $k = \frac{f(x)}{e^x}$ thus $f(x) = ke^x$ as required.$

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Oh alright I see.

Ah well, I can't seem to find a rigorous way to introduce Euler's formula into 4U

Very hard to do so. The closest thing you can do is this...

$\\ $Let $ y= \cos \theta + i \sin \theta $ and we notice that $ \frac{y'}{y} = i $ so integrating both sides with respect to theta gives $ \ln(y) = i \theta $ and therefore by making both sides the exponents of $ e $ we have $ e^{i \theta} = \cos \theta + i \sin \theta$

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Oooh I love this proof! It's so clean and elegant.

$Let $y=f(x)$ be a function that satisfies $y=y'$, so $f(x) = f'(x)$, i.e. $f(x) - f'(x) = 0$. \\ Define $g(x) = \frac{f(x)}{e^x}$. By the quotient rule , $g'(x) = \frac{e^xf'(x) - f(x)e^x}{e^{2x}} = \frac{f'(x) - f(x)}{e^x} = 0$, so $g(x)$ is constant -- let $k = g(x)$. \\ Substituting this into our definition of $g(x)$, $k = \frac{f(x)}{e^x}$ thus $f(x) = ke^x$ as required.$

Nice proof, I would of just done:

$\frac{dy}{dx} = y \\ \frac{dx}{dy}= \frac{1}{y} \\ x = \ln y + C \\ x-C = \ln y \ \ \ y=e^{x-C} = e^x \div e^C = ke^x \ \ \ \ \ $as C is constant$$

EDIT: Happy to have 1000th post in the Maths section =)

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Oooh I love this proof! It's so clean and elegant.

$Let $y=f(x)$ be a function that satisfies $y=y'$, so $f(x) = f'(x)$, i.e. $f(x) - f'(x) = 0$. \\ Define $g(x) = \frac{f(x)}{e^x}$. By the quotient rule , $g'(x) = \frac{e^xf'(x) - f(x)e^x}{e^{2x}} = \frac{f'(x) - f(x)}{e^x} = 0$, so $g(x)$ is constant -- let $k = g(x)$. \\ Substituting this into our definition of $g(x)$, $k = \frac{f(x)}{e^x}$ thus $f(x) = ke^x$ as required.$

A useful technique for separable differential equations like this is to do something like this:

$\\ y'=y $ so $ \frac{dy}{dx} = y $ and by re-arranging, we have $ \frac{dy}{y} = dx $ and integrating both sides with respect to their variables gives $ \ln (y) = x + C $ and therefore $ y = e^{x+c} = e^{x} \times e^C. $ We now let $ e^C = k $ since it is just a constant, so therefore $ y = k e^x$

EDIT: Too slow =p

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Very hard to do so. The closest thing you can do is this...

$\\ $Let $ y= \cos \theta + i \sin \theta $ and we notice that $ \frac{y'}{y} = i $ so integrating both sides with respect to theta gives $ \ln(y) = i \theta $ and therefore by making both sides the exponents of $ e $ we have $ e^{i \theta} = \cos \theta + i \sin \theta$

Ooh, one of the better proofs I have seen (other than the original, which was more a deduction)

EDIT: Can we assume that for the logarithm though? I read somewhere that someone once derived $\ln(\cos x + i\sin x ) = ix$
but due to the periodicity of trig functions the complex logarithm can have many values or something like that.

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Very hard to do so. The closest thing you can do is this...

$\\ $Let $ y= \cos \theta + i \sin \theta $ and we notice that $ \frac{y'}{y} = i $ so integrating both sides with respect to theta gives $ \ln(y) = i \theta $ and therefore by making both sides the exponents of $ e $ we have $ e^{i \theta} = \cos \theta + i \sin \theta$

This is not legit though, as I have explained on these forums before.

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
This is not legit though, as I have explained on these forums before.

Yep still not legitimate, but it's the closest thing I can think of to being anywhere near it.

I don't think there exists a full proof of Euler's formula using techniques purely within the MX2 course (or not that I know of).

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
This is not legit though, as I have explained on these forums before.

Originally Posted by bleakarcher
Hey seanie, just a question. Would you consider this a satisfactory proof of Euler's formul? I remember reading somewhere it's like a poor man's proof of it lol.

Let f(x)=cos(x)+isin(x)
Consider f'(x)/f(x)=[-sin(x)+icos(x)]/[cos(x)+isin(x)]=i
Integrating both sides, log(e)[f(x)]=ix+C
When x=0, f(x) => C=0 => f(x)=cos(x)+isin(x)=e^(ix)
K, going to be really careful and really critical here so you see the sort of issues that can happen when you 'prove' things without care.

First line is fine, we have defined a function from the reals to the complex numbers, this isn't really done in mx2 but it is perfectly legit.

The second is fine, it uses the fact that the derivative of functions from R to C behave like the regular derivative, in that (f+g)'=f'+g' and (cf)'=cf' for functions f,g constant c. This should be proven by first principles before taking it as fact, but the proof is pretty much identical to the real valued analogue.

The third line is my first and main issue. Firstly, you are integrating a complex function. This, like differentiation is legit...it turns out we can integrate the real and imaginary parts separately and add them, although this requires proof.
More importantly, you are integrating the LHS using a pattern that is valid for real functions, but how do we know this is the case for complex functions? In fact this line does not even make sense, because we have not defined what log(z) is for complex z! (And f(x) is certainly complex for all real x). In this line we are basically crossing our fingers and hoping that there is a function called log defined on the complex plane (except possibly at 1) which has derivative 1/z everywhere, then we are pretending this function exists!

The next step involves raising e to the power of both sides. But each side is complex? What does it mean to raise a number to a complex power?

We are also crossing our fingers and pretending that our magical function log on the complex plane from before cancels out when e is raised to the power of it (in the sense of our magical way of raising numbers to complex powers.) Because we are using the same letters for e and log as we did for our familiar e and log from the real line to the real line, they must obey all the same properties right?

On the final line we have arrived at the conclusion...but wait a minute, what does the RHS even mean? We have gone from not having complex powers defined, to knowing what a certain number to a complex power is?! That's the most immediate way to tell that this argument is balls, it doesn't 'prove' anything.

Hope this helps . Feel free to ask for clarification on anything.

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Yep still not legitimate, but it's the closest thing I can think of to being anywhere near it.

I don't think there exists a full proof of Euler's formula using techniques purely within the MX2 course (or not that I know of).

Well of course not, because it is proving a statement about an object that is not even defined in mx2 (the complex exponential). It is much more convenient to define the trig and exponential functions via power series, whence Euler's formula is a triviality.

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Yep still not legitimate, but it's the closest thing I can think of to being anywhere near it.

I don't think there exists a full proof of Euler's formula using techniques purely within the MX2 course (or not that I know of).

Could we prove the Taylor Series formula from the generic power series representation:

$f(x) = \sum_{k=0}^{\infty} c_k x^k$

And then evaluating the constants in terms of the derivatives of f. This does of course assume that power series exists and that it converges, but if we can prove to be legit using ext 2 methods, we could theoretically take Euler's path to the proof as a nifty 15 mark question?

(apologies if it sounds like non-rigorous garbage, I'm just curious)

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Could we prove the Taylor Series formula from the generic power series representation:

$f(x) = \sum_{k=0}^{\infty} c_k x^k$

And then evaluating the constants in terms of the derivatives of f. This does of course assume that power series exists and that it converges, but if we can prove to be legit using ext 2 methods, we could theoretically take Euler's path to the proof as a nifty 15 mark question?

(apologies if it sounds like non-rigorous garbage, I'm just curious)

Well what do you mean by "proving the taylor series formula"? Also power series don't have to converge.

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Well what do you mean by "proving the taylor series formula"? Also power series don't have to converge.

This thing

seanieg89 · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
This thing

I know what a taylor series is, I was asking what you are trying to prove about this infinite sum?

Sy123 · Jan 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
I know what a taylor series is, I was asking what you are trying to prove about this infinite sum?

Oh, the taylor series representations for sin x and cos x and e^x, which is what I was implying when I talked about 'Euler's path'

But yes there are too many rigour issues according to the earlier post, that complex exponentiation cannot be accounted for and there is no way around it.

Carrotsticks · Jan 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Could we prove the Taylor Series formula from the generic power series representation:

$f(x) = \sum_{k=0}^{\infty} c_k x^k$

And then evaluating the constants in terms of the derivatives of f. This does of course assume that power series exists and that it converges, but if we can prove to be legit using ext 2 methods, we could theoretically take Euler's path to the proof as a nifty 15 mark question?

(apologies if it sounds like non-rigorous garbage, I'm just curious)

If you do manage to properly prove the Taylor Series formula (for all R and C), then Euler's formula should follow immediately with no issues because the radius of convergence for sin(x) and cos(x) are infinity.

However, the issue is HOW you are going to properly prove the formula using methods within Extension 2?

The Taylor Series for e^x has nearly been proven in the HSC before in the 2005 HSC Q6(a) and touched in 2007 HSC Q7(a).

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