*rocks back and forth*

The Taylor Series is the general case. It's terms are

(1/n!) d

^{n}f(a)/dx

^{n} (x - a)

^{n}The MacLaurin Series is the case where we take a = 0, so the sum becomes...

(1/n!) d

^{n}f(0)/dx

^{n} x

^{n}... and then we sum up to n = infinity.

So, basically, reaching back to the idea that the derivative is a slope:

(df(a)/dx) (x-a) is a good approximation of the value of f at x = a... it turns out that it is only good up to the second derivative...

[(df(a)/dx) (x-a)] + [(1/2) d

^{2}f(0)/dx

^{2} x

^{2}]

... you can think of needing to multiply the second derivative by x

^{2} as a matter of units (or a matter of order): the derivative of nth order has units of [f/x

^{n}]. So, to be able to add them all up (in terms of units) we need to multiply by x

^{n}. Basically, we are approximating f so we need to have the same units/order/degree/highest power in x be the same in each term.

So, to actually take a MacLaurin series, all we have to do is evaluate the derivative at 0 for successive derivatives up to whatever degree polynomial we want to find, or you might want to find the general form for the function. Such as,

cos x = SUM( 1/(2n)! (-1)

^{n} x

^{2n} )

so for this, you just need to recognize that every odd derivative will be sin(0) which allows us to replace all the n's with 2n's. Each of the even derivatives will alternate between -cos x and cos x with x=0 so it's just (-1)

^{n}. Tuh-duh... so, I was going to keep going until I derived Euler's equation from the MacLaurin series of sin x and cos x... but it's 1am... so more on this later maybe

(Posted on: March 06, 2009, 02:00:13 AM)

*is back to continue said derivation now*

So, I need two more things to get to Euler's equation: namely, the Taylor series for the exponential function and the sine function.

Let's do sin(x) now!

sin(0) = 0

(sin x)' = cos x => cos(0) = 1

(cos x)' = -sin x => -sin(0) = 0

-(sin x)' = -cos x => -cos(0) = -1

So, once again, we alternate signs ( (-1)

^{n}) but in this case, we pick off the ODD powers instead. So,

sin x = SUM( 1/(2n+1)! (-1)

^{n} x

^{2n+1} )

ok, now, for something completely different: exp(x)

I am going to assume that you know [exp(x)]' = exp(x)

This happy little fact is the crux of so much higher mathematics AND in truth, trigonometry! I know, it's amazing, isn't it?

Anyway, this means that the Taylor series for exp(x) is really simple

exp(x) = SUM( (1/n!) x

^{n})

It turns out (by the chain rule) that I can pick up an alternating sign in my exp(x) sum by taking the Taylor Series of exp(-x) instead.

Now, if sine picks off the odd numbers... and cosine picks off the even...

cos x = [exp(x) + exp(-x)]/2

sin x = [exp(x) - exp(-x)]/2

BUT THIS IS WRONG!!!!!!!!! Seriously, don't believe the two lines above... they are the hyperbolic versions of the sine and cosine. What's missing?

I didn't account for the fact that the cos and sine sums also alternate sign!

The solution is trivial: insert the factor i = sqrt(-1).

Since sin (x) picked off the odd powers, the exp(ix) derivatives now give us also an (i)

^{2n+1}... which equals (-1)

^{n}*i... which is right within a factor of i, and cos(x) does not have this issue because it picks off the even powers so it in turn only gets the even powers of i which are all real (they're -1 and 1).

That was a fairly handwavy argument, but typing out the expressions like this is fairly awkward. I hope you can do it out yourself... if not, maybe I'll do it out in an equation editor and attach it at some later time.

If you follow/believe the above, you get that

sin(x) = [exp(ix) - exp(-ix))]/2i

and

cos(x) = [exp(ix) + exp(-ix)]/2

So, what does cos(x) + isin(x)?

(Posted on: March 24, 2009, 02:56:12 PM)

*has a story to tell and realizes he needs to start linking up pictures, etc.*

Today in the car, I was telling my dad: "Spanish really missed an opportunity to have inclusive and exclusive 'we' forms: the we-form they have is nosotros,' which is [etymologically at least: nos utros] 'we others.' With a nostodos, they could have had an inclusive form to contrast with it." ...pause in conversation... "They really missed a golden opportunity to improve on something Latin didn't do right."

To which he responded: "But, Brandon, you know how languages work. They don't improve: they

*decline*."

I grunting and groaning admitted: "Yes, all right, and declensions are often the first thing to decline."

(Posted on: August 19, 2009, 09:41:33 AM)

/~Picture Time~\

Handel's keyboard, on which he first performed Handel's

*Messiah*...

I think I took the same picture of Gothic arches extending back and away in every Gothic structure I visited if possible. Both of these pictures are from St. Patrick's in Dublin.

I swear the thought process here was: "Ooh! Swords!

" ...which now that I think of it... is a strange thought for a belenophobe to be having... oh, well, I'll ponder that later. (This picture is from Enniskillen Castle which has some sort of war museum in it... and pointy objects, cannons, a silver display, etc., etc.)

Galway Bay as seen from the first B&B the tour stayed at in Ballyshannon. I took a walk into town from the B&B. I still think this was the best B&B by far... that probably has something to do with the time spent looking at really old Gaelic schoolbooks.

Muckross House was constructed during An tOcras Mor (The Great Hunger, which is what they call it now since Britain had more than enough food to feed the starving dying masses in Ireland... This is why I prefer Grainne Mhaol to Elizabeth I now... that and Good Queen Bess called Gaelic barbaric

). Anyway, it was constructed to be a sort of Irish home away from home for Queen Victoria, who came and stayed for a couple of days, hated it, and left.

(Posted on: August 30, 2009, 02:02:20 PM)

*skipping to some good stuff* (not that there isn't more good stuff in the Ireland album

)

York Minster

Tintern Abbey