Solving Equations with e and ln x
1 10x2 1 1 y = 10x2 + 1 10x2 1 It’s a good idea to check our work by plugging y = 10x 2 +1 10x2 1 back into the original equation 3 2lny = ln(y + 1) + x Once again, we apply the inverse function ex to both sides We could use the identity e2lny = (elny)2 or we could handle the coe cient of 2 as shown below 2lny = ln(y + 1) + x lny2 = ln(y
Algebraic Properties of ln(
I = 1 2 ln(x 2 + 1) ln(x) Example 2 Express as a single logarithm: lnx + 3ln(x + 1) 1 2 ln(x + 1): I We can use our four rules in reverse to write this as a single
Limits involving ln(
x1 lnx = 1; lim x0 lnx = 1 : I We saw the last day that ln2 > 1=2 I Using the rules of logarithms, we see that ln2m = mln2 > m=2, for any integer m I Because lnx is an increasing function, we can make ln x as big as we choose, by choosing x large enough, and thus we have lim x1 lnx = 1: I Similarly ln 1 2n = nln2 < n=2 and as x approaches
AP Calculus BC - College Board
1 3 x < for x
Jiwen He 11 Geometric Series and Variations
Power Series Expansion of ln(1+x) Note: d dx ln(1+x) = 1 1+x = X∞ k=0 (−1)kxk for x < 1 Integration: ln(1+x) = X∞ k=0 (−1)k k +1 xk+1(+C = 0) = X∞ k=1 (−1)k k xk = x− 1 2 x2 + 1 3 x3 − 1 4 x4 +··· The interval of convergence is (−1,1] At x = 1, ln2 = X∞ k=1 (−1)k k = 1− 1 2 + 1 3 − 1 4 +··· Power Series
59 Representations of Functions as a Power Series
x n+1 n+ 1 = X 0 X ( 1) x +2 n+ 1 3 f(x) = ln(x2 + 1) This series is similiar to ln(x+ 1), except we replaced x with x2 So all we need to do is replace x with x2 in our power series representation for ln(x+ 1) from part (1) f(x) = ln(x2 + 1) = X 0 ( 1)n (x2) n+1 n+ 1 = X 0 ( 1)n x2 +2 n+ 1 4 f(x) = ln(1 x) Again, recall that we know ln(1
Techniques of Integration
cos2 x = 1−sin2 x sec2 x = 1+tan2 x tan2 x = sec2 x −1 If your function contains 1−x2, as in the example above, try x = sinu; if it contains 1+x2 try x = tanu; and if it contains x2 − 1, try x = secu Sometimes you will need to try something a bit different to handle constants other than one EXAMPLE10 2 2 Evaluate Z p 4− 9x2 dx We
ln(1+ x
x∈ R Example 1 Relative Rates of Growth and Decay Prove the following limit (1) lim x→∞ lnx x =0 Proof We may assume that x>1throughout the proof Now we claim that (2) 0
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Lecture 2711.7 Power Series 11.8 Differentiation and
Integration of Power Series
Jiwen He
1 Power Series
1.1 Geometric Series and Variations
Geometric Series
Geometric Series:?∞
k=0xk k=0x k= 1 +x+x2+x3+···? ?11-x,if|x|<1, diverges, if|x| ≥1.Power Series
Define a functionfon the interval (-1,1)
f(x) =∞? k=0x k= 1 +x+x2+x3+···=11-xfor|x|<1As the Limit
fcan be viewed as the limit of a sequence of polynomials: f(x) = limn→∞pn(x), wherepn(x) = 1 +x+x2+x3+···+xn.Variations on the Geometric Series (I)
Closed formsfor many power series can be found by relating the series to the geometric seriesExamples1.f(x) =∞? k=0(-1)kxk= 1-x+x2-x3+··· k=0(-x)k=11-(-x)=11 +x,for|x|<1. f(x) =∞? k=02 kxk+2=x2+ 2x3+ 4x4+ 8x5+··· =x2∞? k=0(2x)k=x21-2xfor|2x|<1.1Variations on the Geometric Series (II)
Closed formsfor many power series can be found by relating the series to the geometric seriesExamples2.f(x) =∞? k=0(-1)kx2k= 1-x2+x4-x6+··· k=0(-x2)k=11-(-x2)=11 +x2,for|x|<1. f(x) =∞? k=0x 2k+13 k=x+13 x3+19 x5+127 x7+··· =x∞? k=0? x23 k =x1-(x2/3)=3x3-x2for|x2/3|<1.1.2 Radius of Convergence
Radius of Convergence
There are exactly three possibilities for a power series: ?akxk.Radius of Convergence: Ratio Test (I) The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.2Example3.f(x) =∞?
k=1k2xk=x+ 4x2+ 9x3+···
Ratio Test :
????a k+1a k? ???=????(k+ 1)2xk+1k 2xk? (k+ 1)2k2|x| → |x|ask→ ∞
Thus the series converges absolutely when|x|<1 and diverges when|x|>1.Radius of Convergence: Ratio Test (II)
The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.Example4.f(x) =∞? k=1(-1)kk!xk= 1-x+12 x2-16 x3+···=e-xRatio Test :
????a k+1a k? ???=????xk+1/(k+ 1)!x k/k!? k!(k+ 1)!? ???xk+1x k? ???=1k+ 1|x| →0<1 for allxThus the series converges absolutely for allx.
Radius of Convergence: Ratio Test (III)
The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.Example5.f(x) =∞? k=1? k+ 1k k2 x k= 2x+ (3/2)4x2+ (4/3)9x3+···Ratio Test :(|ak|)1k
?k+ 1k k2 |x|k? 1k =?k+ 1k k |x| 1 +1k k |x| →e|x|<1 if|x|<1/e Thus the series converges absolutely when|x|<1/eand diverges when|x|> 1/e.3Interval of Convergence
For a series with radius of convergencer, the interval of convergence can be[-r,r], (-r,r], [-r,r), or (-r,r).Example6.In general,the behavior of a power series at-rand atris not
predictable. For example, the series ?xk,?(-1)kk xk,?1k xk,?1k 2xk all have radius of convergence 1, but the first series converges only on (-1,1), the second converges on (-1,1], but the third converges on [-1,1), the fourth on [-1,1].