Recent content by IMDerek

Is there any nice trick for finding the Taylor polynomial of a composition of 2 functions, both of which can be expressed as taylor polynomials themselves? For example, finding the taylor polynomial for e^{\cos x}. Thanks.

Perhaps you should try using \sin (2x) = 2 \sin (x) \cos (x) edit: sorry, didn't see the new problem

Along x=y, the limit becomes \lim _{x \to 0} \frac {2x^3}{2x^4} = \lim _{x \to 0} \frac 1x

Try comparing sin n to n

Yes it's correct

Let's suppose it converges, then t_(n+1) and t_n will have a infinitesimally small difference as n approaches infinity, and you can consider them the same number. (aka the limit of this sequence)

u = 2x du = 2 dx \frac 12 \int \sec ^3 u \tan u du Now let v = \sec ^2 u What is dv?

Maybe using the power series for arctan (x) would be useful here. arctan x = x - x^3/3 + x^5/5 - ... Part b is just the definition of derivative for arctan (x) at x=1. d/dx (arctan x) = (1+x^2)^-1 EDIT: part a is also the definition of derivative for arctan(x) at x=0. It should be more clear...

The integral is equal to: \int \frac {\cos x + 1}{-\sin ^2 x} dx I suggest you use some trigonometric identities to change it into \int - \cot x \csc x - \csc ^2 x dx

The answer is right. Just plug in .01 or .001 into the limit using the calculator and compare it to the exact value.

let u = sin x. then du = cos x dx. Integration is just experience. You just need to practice.

arctan (1/x) = arccot x

What is arctan x as x approaches negative infinity? It is not 1/2. For b, consider that if the right and left sides of the limit don't agree, then there is no limit.

Simplify 1+2+3+...k.

a. x^2-x^4 approaches negative infinity as x approaches positive infinity. b. Look at both sides of the limit.