Dr.Z.'s Email Message to Calc2 Students, Nov. 27, 2012 about memorizing Maclaurin Series of Fundamental Functions


Dear Students,

If you hate to memorize things, then all you need to
memorize is the GENERAL definition of a Maclaurin series of
a function f(x):

f(x)=Sum(f^(n)(0)/n!*x^n, n=0..infinity)

or, more explicitly:

f(x)=f(0)+ f'(0)*x+ (f''(0)/2!)*x^2 + (f'''(0)/3!)*x^3+ (f''''(0)/4!)*x^4+ ...

You can always derive the Maclaurin expansion FROM SCRATCH.

But, since e^x, sin(x), cos(x), ln(1+x), 1(1-x) and (1+x)^a (general a) occur SO OFTEN
it is a VERY GOOD IDEA to memorize them. If you do enough problems,
it would come naturally.

e^x=1+x+x^2/2+x^3/6+ ...  +x^n/n!+ ... (all the powers and the coeff. of x^n is  1/n!, ALWAYS positive)

(valid for all x, i.e. interval of convergence is (-infty,infty) )

sin(x)=x-x^3/3!+x^5/5!-x^7/7!+ ... (all the ODD powers (it makes sense, since sin(-x)=-sin(x), i.e. it is an ODD function; also the coeff. are ALTERNATING)

(valid for all x, i.e. interval of convergence is (-infty,infty), but it is stupid to use for x larger than Pi/2, you can always use
trig. identities and plug-in something between 0 and Pi/2(

cos(x)=1-x^2/2!+x^4/4!-x^6/6!+ ... (all the EVEN powers (it makes sense, since cos(-x)=cos(x), i.e. it is an EVEN function; also the coeff. are ALTERNATING)

(valid for all x, i.e. interval of convergence is (-infty,infty),  ditto)

1/(1-x)=1+x+x^2+x^3+ ... (all the POWERS, the coefficients are all 1, all positive)

(only valid for -1<x<1, i.e. the interval of convergence (-1,1))

ln(1+x)=x-x^2/2+x^3/3-x^4/4 (all the POWERS starting at x^1, the coefficient of x^n is (-1)^n/n, ALTERNATING)

(only valid for -1<x<=1, i.e. the interval of convergence (-1,1])

(1+x)^a= 1+a*x+ a*(a-1)/2*x^2+ a*(a-1)*(a-2)/3!*x^3+ ...

(only valid for -1<x<1, i.e. the interval of convergence (-1,1))

Finally, it is up to you whether you want to memorize

arctan(x)=x-x^3/3+x^5/5-x^7/7+...

(only valid for -1<x<=1, i.e. the interval of convergence (-1,1] )
This is historically very important, since that's how Leibniz found the formula for Pi

when you plug-in $x=1$, you get


arctan(1)=1-1/3+1/5-1/7+1/9-1/11+ ...

hence

Pi/4=1-1/3+1/5-1/7+1/9-1/11+ ...

and hence

Pi=4*( 1-1/3+1/5-1/7+1/9-1/11+ ...)

P.S. These functions are the "atoms" of calculus, and show up everywhere in science and engineering.
It is a GOOD idea to memorize them (and as I said, if you do enough problems, you will memorize
them naturally, without effort), not just for doing well in this class, but for your future
classes in science and engineering, and for some of you, in your professional career 
after you graduate.


  Best wishes

            Dr. Z.