Download Series P series – converges if P >1 diverges if P ≤ 1 Integral only

Series
P series –
1
p
x
converges if P >1
diverges if P ≤ 1
Integral
only convergent if
0 < ∫ f(x) < ∞
Comparison An ≤ Bn
if ∑ An diverges then ∑ Bn diverges
if ∑ Bn converges then ∑ An converges
Limit Comparison Lim n → ∞ of
An
Bn
= L will converge only if 0 < L < ∞

proves they have the same behavior
An 1
Ratio test
Lim n → ∞ of |
|=L
0 ≤ L < 1 convergent
An
L > 1 divergent
Root test
Lim n → ∞ of n√ (| An| ) = L 0 ≤ L < 1 convergent
L > 1 divergent
Alternating series
1) Lim n → ∞ of An = 0
2) f '(x) < 0 makes sure its decreasing
An 1
3) ∑ |
| = L 0 ≤ L < 1 proves absolute convergence
An
Sequences- Use limits first then Lh's rule and if Lim n → ∞ An = L
If -∞ < L < ∞ it converges
a
Geometric Series n=1 to ∞ ∑ ar(n-1) = 1− r
If |r| < 1 it is convergent
Trigonometry Formulas
Sin² θ = 1 - Cos² θ
Cos² θ = 1 - Sin² θ
Tan² θ = Sec² θ – 1
Double Angle Formulas
Sin(2θ) = 2SinθCosθ
Cos(2θ) = Cos² θ – Sin² θ
Cos(2θ) = 2Cos² θ – 1
Cos(2θ) = 1 – 2Sin² θ
2Tanθ
Tan(2θ) = 1− Tan² θ
Half Angle Formulas
[1 – cos 2θ
Sin² θ =
2
[1 cos 2θ
Cos² θ =
2
[1 – cos 2θ
Tan² θ = [1 cos 2θ
]
]
]
]
Limits of Inverse trigonometry functions
∏
−∏
Sin ̄ ¹(x) = 2
to 2
Cos ̄ ¹(x) = 0 to ∏
∏
−∏
Tan ̄ ¹(x) = 2
to 2
Derivatives(multiply by same, then subtract one)
[e˟]' = (x)' e˟
[a˟]' = a˟ ln(a)
[a ̾ ̽ ]' = a ̾ ̽ ln(a) s'(x)
[x² ]' = 2x
Log and ln derivatives
[log ₐ(x)]' = 1/ ln(a) * 1/x
[u ' x ]
1
[log ₐu(x)]' = [log a ] x [u x ]
[u ' x ]
[ln(u(x))]' = [u x ]
1
[ln(x)]' = x
ex [ln(x)]' =
Basic Trigonometry Derivatives
[Cos(x)]' = - Sin(x)
[Sin(x)]' = Cos(x)
[Tan(x)]' = Sec²(x)
[Sec(x)]' = Sec(x) Tan(x)
Inverse trigonometry Derivatives
1
[Sin ̄ ¹(x)]' = √ 1− x²
−1
[Cos ̄ ¹(x)]' = √ 1− x²
1
[Tan ̄ ¹(x)]' = 1 x²
−1
[Csc ̄ ¹(x)]' = x √ x²− 1
1
[Sec ̄ ¹(x)]' = x √ x²− 1
1
[Cot ̄ ¹(x)]' = 1 x²
x'
x
Integrals(add one, divide by the same)
x³
3
∫u dv = u * v - ∫ v du
∫Sin(x) = -Cos(x)
∫Cos(x) = Sin(x)
1
∫ x dx = ln|x| + c
∫x ² =
a̽
̽ dx = ln a + c
∫ln(x) = xln(x) – x + c
∫a
Inverse trigonometry Integrals
1
∫ √ 1− x²
∫
∫
∫
∫
∫
= Sin ̄ ¹(x)
−1
√ 1− x² = Cos ̄ ¹(x)
1
1 x² = Tan ̄ ¹(x)
−1
x √ x²− 1 = Csc ̄ ¹(x)
1
x √ x²− 1 = Sec ̄ ¹(x)
1
= Cot ̄ ¹(x)
1 x²
Trigonometric Substitutions
a² x² ; use x = aTanθ
a² – x² ; use x = aSinθ
x² a² ; use x = aSecθ
Arc length
b
2
a∫ √( 1 + [f '(x)] ) dx
Surface Area
b
2
a∫ f(x) √(1+[f ' (x) ] ) dx
Integral for length
b
[ y ' ]2 [ x' ]2
a∫
Power Series expansion
f x
= ∑ an (x-a)n
fn a
n!
Geometric series formula
1
= 1 x x 2 x 3. .... x n = ∑ ∞n= 0 x n
1− x
1
= ∑ ∞n= 1 n − x n− 1
2
1 x
an =
Power series
f x
∞
=
∑ n= 0 a n
x− c n = a 0 a1 x− c
1
a 2 x− c
2
a3 x – c
3
th
where a n represents the coefficient of the n term, c is a constant, and x varies around c (for this
reason one sometimes speaks of the series as being centered at c). This series usually arises as the
Taylor series of some known function; the Taylor series article contains many examples.
In many situations c is equal to zero, for instance when considering a Maclaurin series. In such cases,
the power series takes the simpler form
Maclaurin series
∞
2
3
f x = ∑
a X n = a 0 a1 X a 2 X a 3 X
n= 0 n
X 2n 1
n= 0
2n 1 !
∞
x 2n
cos x = ∑ n= 0 − 1 n
2n !
∞
xn 1
ln 1 x = ∑ n= 0 − 1 n
n 1
n
∞
X
e x = ∑ n= 0
n!
∞
x 2n
cos x− 1 = ∑ n= 0
−1
2n !
sin x = ∑
∞
−1
example f(x) =
find
f
n
n
1
x
n
a = 9, find the Taylor series centered at ( a )
(9)
−1
f x =
X 2
−3
−1
f1 x =
X 2
2
− 1 −3
f 2 x =
X
2
2
−5
2
− 1 −3 − 5
f x =
X
2
2
2
f n x =
3
derivatives
[
1
]' =
x n
1
× n'
x n
−7
2