probably not as advanced as I made it sound.

I remember we used integration by parts and I remember learning some

of that stuff now that irspow figured out, where you rotate a function

around an axis to get surface area in 3-d, or to get volume.

But I still must relearn now to understand it.]]>

In my oppinion computer science is something every mathematician should take. It allows you to truely harness its power!

Ah the power of math!

Uh oh... I feel it coming again..... ugh.... hold on... keep it together man... keep it to-....oh no!

wigout();

OH MY FREAKIN GOSH! MATH ROXXORS T3H B1G ONEZ!1111111

]]>was taught in college in the late 80's due to lack of usage and never a great

understanding, mostly did it through just memorizing steps. Now I plan on

actually learning something, but I have no due date, as I'm not using the

information for anything except my own pleasure. Also, in my first year at

college I made a mistake by not starting with Calc I, I started with Calc III, since

I already had Calculus in high school (grade 12). I think I should have started

from scratch to see if you learn theory that you missed the first time around.]]>

The author of my mathbook, like many great mathematiciains he was a genius, and he's dead. :-/

Well actually Saxon wasn't a 100% mathematician, I think he was an engineer and computer scientist, and a very good teacher. He wrote mathbooks for grades 1 though 12, including algebra 1, 2, trig and calculus. He also wrote books on other subjects like phonics. (yuck!) Saxon's philosophy was that math is not difficult, math is differant, and that people often call things that are unfamiliar to them, difficult, and things that are familiar easy. Time and practice is required for things that are differant to become things that are familiar and thus, easy. This philosophy is the basis for saxons teaching. Where some mathbooks would try to teach all the aspects of a topic in one lesson, saxon presents each topic little by little. Begining with the simplelest problems of that type and saving the more advanced problems for a later lesson. Each short lesson has 30 practice problems to do before moving on. This ensures you become familiar with each concept making comprehension of the more advanced problems easy since you have a solid foundation on the previous topics. "An Incremental Development" is written under the title of each saxon book. I taught myself algebra 1, 2 trigonometry with these books and am now working on calculus. Very good books, I highly reccomend them to anyone.

But I was mostly thinking out loud. It really needs to be theroughly tested and perfected first.

Oh, and John E. Franklin, the book is called Calculus - with Trigonometry and Analytical Geometry by John H. Saxon Jr and Frank Y.H. Wang. Its supposedly an "elementary calculus book" >:-( so I don't know how much you'll get out of it if you've been through college. Its for AP math students who've completed trig before 12th grade. But I don't think its what you might call Calculus 1/2. I guess just level one calculus. Not calculus 2 or advanced calculus or Calculus For Higher Life Forms. (which I WILL write someday! :-D)

Anyways, go to www.saxonpublishers.com and have a look around. The book I ordered is listed on their site as Saxon Calculus 2nd Edition: Student Edition. Its a revised version of the first book with an additional 30 lessons, (for a grizzly 158 lessons ) I've taught myself algebra 1, 2 and Trig, and I'm now nearly halfway through calculus. I think that speaks for itself. Hang on a second I'll dig up my review for the saxon books.

]]>With a bit of formatting that could become a page on the mathsisfun.com website

]]>Maclaurin series. I am trying to learn new stuff, so I really

appreciate it. Who is the author of the Calculus book you are

studying from? Is it worth getting, do you think?]]>

`:sum`

If you want to be flash, you could also do it with the math tag, but you'd need to do the rest of the equation with it as well, or it would look silly, and the math tag is quite complicated.

After producing

and , I finally managed to get it right: ]]>Still the polynomial seems kind of strange:

(insert big sideways M here) (-1)^(n+1) ((x-1)^n) / n

Note I replaced (x) with (x -1)

I suppose it straightens out if you use enough terms.

]]>If we were to write ln (x + 1) we could do it.

f(x) = ln (x + 1) f(0) = 0

f'(x) = 1/(x + 1) f'(0) = 1

f''(x) = -1 / (x + 1)^2 f''(0) = -1

f''(x) = 2 / (x + 1)^3 f'''(0) = 2

f'''(x) = -6 / (x + 1) 4 f''''(0) -6

Ok I think thats enough to see the pattern is 0!, -1!, +2!, -3!

Now lets write the maclaurin polynomial:

0x^0/0! + (0!x^1)/1! - (1!x^2)/2! + (2! x^3)/3!

we can eliminate the first term since its zero. We notice that in each term, the factorial above is one less then the factorial below, this is fortunate. Because we can always rewrite (n-1)!/n! as 1/n.

So we have -x^1/1 + (x^2)/2 - (x^3)/3 + (x^4)/4

This can be writtin in summation as:

(insert big sideways M here) (-1)^n (x^n) / n from n = 1 to infinity

Sadly, this only gives us a formula for ln (x + 1)

But I suppose we could simply insert (x - 1) in our expression for x to get:

(insert big sideways M here) (-1)^n ((x-1)^n) / n from n = 1 to infinity

Hmmm... I checked this series and it doesn't sem to be working. Either I messed up somewhere or maybe the maclaurin polynomail of ln x cannot be found the traditional way.

I'll try it again on paper.

]]>The maclaurin polynomial of f(x)

f(0)/0! + f'(0)(x^1)/1! + f''(0)( x^2 ) /2! + f'''(0)( x^3 )/3! ...

note you begin by multiplying the first term by f(0), you take the first derivative, evaluate it at zero, and multiply that by the second term, you take the third derivative, evaluate it at zero, and multiply it by the third term, and so on.

Supposedly, this creates a sort of replica of the function, but in the form of the polynomial. If the polynomial does not terminate, the accuracy of the function depends on how many terms you choose to use.

If you have a polynomial like 3x^2 + 2x + 6

f(x) = 3x^2 + 2x +6 so f(0) = 6

f'(x) = 6x + 2 so f'(0) = 2

f''(x) = 6 so f''(0) = 6

The third derivative and anything beyond that will be 0 so we can stop here.

Now lets evaluate the maclaurin polynomial for the function:

f(0)/0! + f'(0)(x^1)/1! + f''(0)( x^2 ) /2! + f'''(0)( x^3 )/3! ...

= 6/0! + 2x/1! + (6x^2)/2!

= 6 + 2x + 3x^2

We found the maclaurin polynomail of 3x^2 +2x +6 and got 3x^2 + 2x + 6! From this we see that the maclaurin polynomail of a polynomial, is the polynomial itself!

Now take e^x

f (x) = e^x f(0) = 1

f'(x) = e^x f''(0) = 1

f''(x) = e^x f''(0) = 1

Obviously the first second third and nth derivative of e^x will always be e^x, evaluated at zero, they are one, so if write the maclaurin polynomail:

f(0)/0! + f'(0)(x^1)/1! + f''(0)( x^2 ) /2! + f'''(0)( x^3 )/3! ...

We have:

x^0/0! + x^1 + (x^2)/2! + (x^3)/3! + (x^4)/4! ....

and so on! Note the first term could have just been written as 1, I just wrote it that way to show the pattern is consistant.

So now we have a polynomial formula for e^x! :-D

You can also use this to find the formula's for sine and cosine, and other trigonometric functions.

I'm not sure how to find the formula for a function such as y = ln x because ln 0 does not exist. But maybe I'll learn more later.

The problems in my book often give me a functon and ask me to find the maclaurin polynomial and write the answer in summation notation. Writing them in summation notation can be a tricky buisiness at times, becuase the signes alternate, some terms disappear when their coefficiants are zero, and sometimes the factorials cancel eachother out. Its tricky but they're kind of fun.

]]>