For radius of convergence and interval of convergence, find the limit of the sequence in terms of x by applying the ratio test or another test. For the series to converge, the limit of the sequence at a given x must be 0. So, you can imagine a plot of the graph where, at each x, your series has converged.
The value of the function at any given x is determined by a very large polynomial evaluated at that x value. Now, for some values of x, your sequence may not approach zero so your
For each given x value, you're essentially taking the evaluation of a very large polynomial and entering it as y.
For substitution, we have to let t=sqrt(x) and then solve for dx in terms of t. For instance, integral of cos(sqrt(x))dx would be t=sqrt(x), then dt = 1/2sqrt(x)dx, so dx=2tdt. Then we plug in the ts for xs and then evaluate. Then REMEMBER TO SUB BACK IN
1/(1-r) = sum from n=0 to infinity of r^n
Do some partial fractions practice. First try long division. Have all of the cases written down.
All linear (of the form (ax+b)) is A/(ax+b) + B/(cx+D) etc…
stuff/(x^2*(x-3)^3) would be A/x + B/x^2 + C/(x-1) + D/(x-1)^2 + E/(x-1)^3
stuff/(x^2+1)^2 then (Bx+C)/(x^2+1)+(Dx+E)=(x^2+1)^2
if lim n->infinity of abs(a_n+1/a_n) = L < 1, then the series is absolutely convergent. i.e. if the limit of the absolute value of the limit of the next term divided by the previous is less than one.
if greater than 1 or infinity, diverges.
If = 1, then inconclusive (such as the harmonic series).
if lim n-> infinity of nth root of abs(a_n) = L < 1, then the series is absolutely convergent
if greater than 1 or infinity, then divergent
if =1, inconclusive.
Have the arc length function
Differential equations…a bit
Have notes on completing the square
Really go over the last .5 chapters.