[Zhen Lin]

(Original post by Mark13)
Doesn't the problem remain of showing that the power series that defines sin(x) is the same as the quotient of the opposite and hypotenuse sides of a right-angled triangle (for example)?

Yes, but then that's secondary to differentiating it. The point is, there is a function called sin we define by a power series, and we can differentiate it term-by-term to get another function we call cos. We can prove later that they are the same as the functions we define geometrically.

(Original post by DeanK22)
However, we do define the trigonmetric functions as power series so there is no need for the above as we can just differentitate the power series term by term.

Naturally, it needs to be proven that a power series does indeed have a derivative equal to what you get by differentiating term-by-term.

[Oh I Really Don't Care]

(Original post by Mark13)
Doesn't the problem remain of showing that the power series that defines sin(x) is the same as the quotient of the opposite and hypotenuse sides of a right-angled triangle (for example)?

Sureely that is just coincidence? For example if you didn't know what a triangle was or a circle was yet you know the power series for this function;

exp(ix) you could seperate it up into two terms;

exp(ix) = Re(exp(ix)) + Im(exp(ix)) and define the function Re(exp(ix)) to be sin(x) and Im(exp(ix)) to be (sin(x))' which you could just as well define as cos(x). So no mater which way you look at it whereever you define it from you probably will just have to accept that there is a deep connection between the two definitions [geometrical definition and power series definition].

[Mark13]

(Original post by DeanK22)
Sureely that is just coincidence? For example if you didn't know what a triangle was or a circle was yet you know the power series for this function

But we haven't shown that they are the same function. If we define sin(x) as , then we can easily show that its derivative is , and define this function as cos(x). But we haven't shown that the original power series is the same as the function which takes an angle in a right-angled triangle and divides the opposite side by the hypotenuse. Probably haven't phrased that that well, sorry.

[SimonM]

(Original post by DeanK22)
Sureely that is just coincidence? For example if you didn't know what a triangle was or a circle was yet you know the power series for this function;

Errmm.. I'm not really following you? There are lots of equivalent definitions of the trigonometric functions. The first one we're given is as the ratio of sides on a given triangle. That is the best motivated definition so it's not really "coincidence" we define our power series to be related to those useful functions

[Mark13]

(Original post by Zhen Lin)
Yes, but then that's secondary to differentiating it. The point is, there is a function called sin we define by a power series, and we can differentiate it term-by-term to get another function we call cos. We can prove later that they are the same as the functions we define geometrically.

How would you go about proving they are the same function, out of interest?

[Zhen Lin]

(Original post by Mark13)
But we haven't shown that the original power series is the same as the function which takes an angle in a right-angled triangle and divides the opposite side by the hypotenuse. Probably haven't phrased that that well, sorry.

True. However this is actually more subtle than it looks. The point of analysis is to make explicit and rigorous what intuition tells us. So, in order to do this, one must first define what an angle is and how to measure it -- and it has to be defined in the same language as analysis, so you would have to do it in coordinate geometry.

[Oh I Really Don't Care]

[n radicals];

prove

[Unbounded]

(Original post by DeanK22)
[n radicals];

prove

Spoiler:

Show

Drops out quickly with induction, I think. I'll have a look for a different route

Side note: for a nice 'cos' sign, use \cos in LateX [making sure to leave a gap after it - i.e. \cos x, not \cosx].

[Zhen Lin]

Actually, \cos\theta gives the same result as \cos \theta.

[Unbounded]

(Original post by Zhen Lin)
Actually, \cos\theta gives the same result as \cos \theta.

Hmm perhaps it's just when writing \cosx

Spoiler:

Show

LateX test

Unparseable or potentially dangerous latex formula. Error 6: Image was not produced or one of its dimensions is too small.

\cosx

Edit: Right you are.

[My Alt]

\cos{x} also does the same job. \ denotes an "action" and {} denotes what you are doing that action to (well kind of anyway)

[miml]

Not as formal as some would like, but it'll do for me.

Spoiler:

Show

Starting from

[Oh I Really Don't Care]

(Original post by miml)
Not as formal as some would like, but it'll do for me.

Spoiler:

Show

Starting from

meh pretty clear - tbh this isn't an exam or anything

Spoiler:

Show

for a more rigorous approach let the statement be P(n). P(1) is trivially true. Assume that P(k) is true. Adding two to both sides and rewriting in terms of cos^2(x) and square rooting shows that P(k) implies P(k+1)

[Unbounded]

(Original post by My Alt)
\cos{x} also does the same job. \ denotes an "action" and {} denotes what you are doing that action to (well kind of anyway)

Cheers

[Oh I Really Don't Care]

Which integers (a,b,c) will be acceptable below;

[My Alt]

Is that not merely an extension of Fermat's last Theorum? There are no solutions.

[Daniel Freedman]

(Original post by My Alt)
Is that not merely an extension of Fermat's last Theorum? There are no solutions.

The theorem states that no three positive integers a, b and c can satisfy for any integer value of n greater than 2.

[Oh I Really Don't Care]

(Original post by My Alt)
Is that not merely an extension of Fermat's last Theorum? There are no solutions.

Well you cannot really do that. For instance;

will have a solution is a=b=c=2. FLT requires that in the equation x^n + y^n = z^n, x,y,z are integers. third rooting a prime is going to be irrational which is not an intneger).

Spoiler:

Show

in this problem you need to use the principle of well ordering

[Unbounded]

(Original post by DeanK22)
Well you cannot really do that. For instance;

will have a solution is a=b=c=2. FLT requires that in the equation x^n + y^n = z^n, x,y,z are integers. third rooting a prime is going to be irrational which is not an intneger).

Spoiler:

Show

in this problem you need to use the principle of well ordering

Spoiler:

Show

Does it turn out there are no solutions, except (0, 0, 0)?

Spoiler:

Show

If so:



with

Assume there is some hypothetical, least value of c, by the well-ordering principle, called c'





So we have generated a smaller number, h, which satisfies c in the original equation, which is a contradiction. ?

Haven't got it down awfully rigourously, but I think that works.

[Oh I Really Don't Care]

(Original post by GHOSH-5)
aewtea

The only solution is a=b=c=0