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Ingo Gerth1 styear BSc StudentAerospace EngineeringCalculus—stquarterSummary of formulasJanuary 13, 2009Disclaimer This document was prepared as a summary for myself. However, it might be useful for other students as well, which is why it was published online. No rights can be claimed, the author is...

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Ingo Gerth 1st year BSc Student Aerospace Engineering

Calculus—1st quarter Summary of formulas January 13, 2009 Disclaimer This document was prepared as a summary for myself. However, it might be useful for other

students as well, which is why it was published online. No rights can be claimed, the author is not responsible for any wrong information spread through this document. Also, the author is not affiliated to TU Delft in any other kind than being a student.

1 Prerequisits

Changes the starting value of the function.

1.1 Exponents

Double angle formulas

sin 2α = 2 sin α cos α

xm x x =x = xm−n xn  m n m·n x =x (xy)n = xn y n √ m √ n xm/n = n m = x !n n x x = n y y √ √ √n xy = n x n y m n

m+n

cos2α = cos α2 − sin α2 Other relations

sin (π/2 − α) = cos α

cos(π/2 − α) = sin α Values

1.2 Trigonometric functions

a b sin θ = cosθ = c c a sin θ tan θ = = b cosθ 2 sin + cos2 = 1 sin −θ = − sin θ sin α sin β sin γ = = b a c

θ

rad

sin θ

cosθ

tan θ

0◦

0 π/6 π/4 π/3 π/2

0 1/2 √ 2/2 √ 3/2 1

√1 √3/2 2/2 1/2 0

√0 3/ 3 1 √ 3 —

30◦ 45◦ 60◦ 90◦

1.3 Logarithmic laws

loga x = y ln x = loge x

Functions

ln x = y

ga (x) = a sin (x)

loga (a x ) = x

Changes the amplitude of the function.

⇔ ay = x ⇔ ln e = 1 ⇔ ey = x

alog a x = x eln x = x loga (xy) = loga x + loga y ! x loga = loga x − loga y y  r loga x = r · loga x

gb (x) = sin (bx) Changes the period of the function. gc (x) = sin (x + c)

1

Strategy for solving: Take expression in function, equal it to y, solve it so it cancels out the original function.

1.4 Rationalizing and factoring

Rationalizing square roots

Multiply by conjugate radical √ √  x + 4 − 2  x + 4 + 2    · √ x x+4+2 1 = √ x+4+2

3 Implicit differentiation Implicit differentiation can be applied when it is not easily possible, or not possible at all, to solve for y in a complicated function, such as x3 + y 3 = 6xy. The idea is to differentiate both sides with regard to x, then solve for y ′ .

Factorising quadratic equations

c x2 + |{z} b x + |{z} (r+s)

✔ Example . . .

(r·s)

= (x + r)(x + s)

x2 + y 2 = 25 d 2 d 2 x + y =0 dx dx y is a function of x, must be differentiated! d 2 y = 2y · y ′ |{z} |{z} dx

Example: x2 + 5x − 24 r + s = 5 = (x − 3)(x + 8)

r · s = 24

outer ′

1.5 Limits

lim ex = 0

lim ex = ∞

x→−∞

x→+∞

lim ln x = −∞

x→∞

x→0+

inner

2x + 2yy = 0

Note For expressions like 6xy use multiplication rule,

lim ln x = ∞

too! (6 xy)′ = 6xy ′ + 6y

1.6 Line equations Note If term like

y − y p = m(x − x p )

x−1 + y −1 = 0 ,

y = mx + p

derivative is not (y ′ )−1 but still y ′ . −x−2 − y −2 y ′ = 0

1.7 Quadratic formulas

√ −b ± b2 − 4ac x1,2 = 2a r  p 2 p −q x1,2 = − ± 2 2

4 Linear approximations and differentials 4.1 Linear approximations

2 Inverse trigonometric functions Generally, a function resembles its tangent line at a Because they are not one-to-one (do the horizontal line point. This can be used to approximate values. check/only one x for y), the domain has to be restricted to make them one-to-one. L(x) = f (a) + f ′ (a)(x − a)

2

4.2 Differentials

Differentials express the change of a function over a ✔ Example . . . certain interval. Z4 √ 2x + 1dx

dy = f ′ (x)d x dy represents the amount how much the function rises or falls (change in linearization).

0

= How to find approximate relative errors

Z9

1√ udu 2

1

"

1 2 = · u3/2 2 3

• Express unknown, e.g. area, as a function • Calculate the differential

#9 1

=

26 3

So note that we have not substituted back, but have indeed plugged in the values into the function u = g(x) and calculated the integral, which saved us time!

• Plug in the values, also for the change in x (d x)

5 Substitution rule What can also save us some time is the symmetry of integrals.

5.1 For indefinite integrals ′ If u = g(x) is differentiable, R then du = g (x)dx. Thus, plugging in, the integral f (x)dx turns into Z f (g(x)) g′ (x)dx | {z } | {z } = f (u)

du

Notice that du = g′ (x)dx is a differential that we plug in.

Ra 1. If f is even [ f (−x) = f (x)], then −a f (x)dx = Ra 2 0 f (x)d x. Ra 2. If f is odd [ f (x) = − f (−x)], then −a f (x)d x = 0.

6 Integration by parts

It is permissible to operate with dx and du after integral signs as if they were differentials! You are writing an exam and have suddenly forgotten Always remember that the idea behind the substituthe rule. Don’t worry, you can easily derive it: tion rule is to replace a relatively complicated integral by a simpler one. The strategy is to find a substitution (uv)′ = u′ v + uv′ Z Z Z whose differential occurs in the integral. Also consider ′ ′ to substitute the dx after solving for it. (uv) = (u v) + (uv′ ) Z Z ⇒ udv = uv − vdu

5.2 For definite integrals

A little “trick”, that is possible: Z ln xdx

This rule is more important than you think and very much preferable to solving the indefinite integral. Zb a

Zg (b) f (g(x))g′ (x)dx = f (u)du

u = ln x 1 du = dx x

g (a )

dv = dx

v=x In words: Replace the integrating boundaries by the Now you can easily solve this integral. value of the function of the substitution.

3

Stuf f to remember:

7 Stuck on integrating? Sometimes you can’t find a way to solve an integral. Try out the following strategies. 1. Simplify the integrand (if possible) 2. Look for obvious substitution 3. Look for a not-so-obvious substitution

Zb

f (x)dx = lim

Z∞

f (x)dx) =

t→−∞

−∞

Zb

f (x)dx

t

Za

f (x)dx +

−∞

−∞

Z∞

f (x)dx

a

4. Look for an even less obvious substitution 5. Try to integrate by parts ✔ Example . . .

Doesn’t work at all? That sucks, but don’t give up! Here are some tricks: √ √ tan = sin / cos, e x , u = x, u2 = x

Z5

Always remember that you can still play around with the substitution! Maybe if you square the substitution or take the root, you can substitute it into the integral.

2

dx = lim √ x − 2 t→2+

Z5

dx √ x−2 2 √ √ √ lim+ 2( 3 − t − 2) = 2 3

t→2

⇒ Integral is convergent.

8 Discontinous Integrals Integrals that have a discontinuity in [a, b] are im- Short check R1 proper integrals, for instance −1 1/x2 dx.

If an integral is continous at [a, b) and discontinous at A short check for the con- or divergency of a special b), then integral: Zb Zt Z∞ f (x)dx = lim+ f (x)dx . 1 t→b dx xp a a 1

If f is continous at (a, b] and discontinous at a, then Zb

f (x)dx = lim+ t→b

a

Zb

is convergent if p > 1, and divergent if p ≤ 1,

f (x)dx

Note While 1/x2 is convergent, 1/x is not, since 1/x

t

does not reach the x-axis fast enough.

If a < c < b and f is discontinous at c, then Zb a

f (x)dx =

Zc a

f (x)dx +

Zb

Comparison Theorem

f (x)dx

c

It is impossible to find the exact value of an improper When coming from the right to the left side: a+ (posi- integral, yet you want to find out whether it is improper or not? Use the comparison theorem: tive side). When coming from the left to the right side: a− (negf (x) ≥ g(x) ≥ 0 for x ≥ a ative side). If the corresponding limit exists, the integral is conver- Now, if g(x) is con/divergent, then f (x) is, too! Just gent. If it does not exist (the integral turns out to be an take a comparable, easier function g(x) and evalute it, it will be the same result as f (x). inifite value), it is divergent.

4

9 L’Hospitals rule

and then integrate both sides. Note that the left side will always be an expanded form of the multiplication rule and can be simplfied,

This rule is not on the schedule, but “it can make your life much easier”, so I give it here! If lim f (x) = 0 x→a

or

R

y ′ + P(x)y = Q(x) | · I(x) = e (y · I(x))′ = Q(x) · I(x)

lim f (x) = ±∞ x→a

and the same is true for g(x), then

P(x)dx

which makes it a child’s play to integrate it.

f ′ (x) f (x) lim = lim ′ x→a g(x) x→a g (x)

10.3 Second order differential equations

So this means, that the limit of f (x)/g(x) is the same 10.3.1 Homogenous second order . . . as the fraction of their derivatives. In many cases, this is much easier to solve! ay ′′ + by ′ + cy = 0 ✔ Example . . . This kind of equation is solvable with y = erx → y ′ = rerx → y ′′ = r 2 erx .

ln t t→0 1/t 1/t lim = lim+ −t = 0 t→0 t→0+ −1/t2

lim t ln t = lim + +

t→0

Plugging this in, you get the auxilliary equation, with which you have to solve. ar2 + br + c = 0

10 Differential Equations

Often, a quck way to solve it is to factor it (see section 1.4). Otherwise, use the quadratic formulas to solve it.

10.1 Separable equations

Possible solutions:

A special type of first order differential equations. In such functions, dy/dx can always be factored as a function of x times a function of y .

Roots of aux. eqn.

General solution

r1 , r2 real and distinct r1 = r2 = r r1 , r2 complex: α ± iβ

y = c1 er1 x + c2 er2 x y = c1 erx + c2 xerx y = eαx (c1 cosβx + c2 x sin βx)

h(y )dy = g(x)d x Now simply integrate both sides and solve for y: Z Z h(y )dy = g(x)dx

10.3.2 Nonhomogenous equations

10.2 Linear differential equations

Complementary equation:

To solve such equations, solve the complementary equation, then solve the particular equation, add Note that if the integral yields something like ln y you them. should still solve for y by raising it to the power of e. ay ′′ + by ′ + cy = G(x)

ay ′′ + by ′ + cy = 0

Always have the form

The solution y(x) of a nonhomogenous equation can be written as

dy + P( x)y = Q( x) dx For instance xy ′ + y = 2x, because y ′ + y/x = 2. To solve, multiply by the integrating factor R

I(x) = e

y(x) = y p (x) + y c ( x) where y p is the particular solution and y c the complementary.

P(x)dx

5

✔ Example . . . y ′′ + y ′ − 2y = x2 y c = c1 ex + c2 e−2x

G(x)

y p (x)

const xn enx sin x x cosnx

A Axn + Bxn−1 +. . . + C = y p ( x) Aenx A cos x + B sin x (Ax + B) cosnx + (Cx + D) sin nx

As is the case with the last row, the solutions should be combined in order to find a solution.

Now guess what y p could look like: y p (x) = Ax2 + Bx + C

10.3.3 Initial value problems

Differentiate this equation and substitute it into the original equation:

Those are especially easy, all you have to do is plug in the given values into your solution and solve for the c1 and c2 .

(2A) + (2 Ax + B) − 2( Ax2 + Bx + C) = x2 [. . .]

Just to be sure. . .

− 2A = 1

2A − 2B = 0 2A + B − 2C = 0 1 3 1 ⇒A=− C=− B=− 4 2 2 y (x) = y c (x) + y p (x) = . . .

1 y(|{z} 0 ) = |{z} y

x

11 Complex numbers Complex numbers suck, but you have to know them.

Note The solution y p must never equal the solution y c .

In that case, a x has to be added to the solution. For example: ′′

2x

′

z = |{z} α + βi |{z} real

−2x

√ 2 The i= √ main property 2of i is i =√−1. Therefore: √ √−1 But √ also: (−1) = −1 ⇒ −i = −1 In general: −c = ci

y − 3y + 2y = 2x + e + e x

complex

2x

y c = c1 e + c2 e

y p = ax + b + cxe2x + de2x

The particular solution y p must be “linearly indepen- Fundamental Theorem of Algebra dent” of y c . If you want to show off, this was proofed by Gauss! an xn + an−1 xn−1 + . . . + a1 x + a0 = 0

Attempts for solving nonhomogenous equations

does always have a solution among the complex numbers.

Note If you have an equation like Basic rules

ay ′′ + by ′ + cy = G(x) + H(x)

(a + bi) + (c + di) = (a + c) + (b + d)i

the solution may be acquired by

(a + bi) − (c + di) = (a − c) + (b − d)i

(a + bi)(c + di) = a(c + di) + (bi)(c + di)

ay ′′ + by ′ + cy = G(x) ay ′′ + by ′ + cy = H(x)

Conjugates

then y(x) = y p1 + y p2 .

Complex number: z = a + bi Conjugate: z = a − bi ay ′′ + by ′ + cy = G(x)

z+w = z+w

6

zw = z · w

zn = zn

Division of complex numberx

Roots of complex numbers

To divide complex numbers, multiply them with their Roots of complex numbers suck especially. For the nth conjugate denominator. root, you need n iterations. . . " ! !# z θ + 2kπ θ + 2kπ 1/n z}|{ wk = r cos , + i sin −1 + 3i −1 + 3i 2 − 5i n n · = 2 + 5i 2 2 + 5i − 5i |{z} |{z} where k = 0, 1, . . . , n − 1. z

z

13 + 11i 13 11 + i = 2 2 = 29 29 2 +5

Complex exponentials

ez = ea+bi = ea ebi = ea (cosb + i sin b) Show of f note:

Modulus

Distance from the origin in the complex number diaeiπ + 1 = 0 gram. This formula comprises the very most important mathp |z| = a2 + b2 ematical numbers. Also note: 2  n zz = |z| [r(cosθ + i sin θ)]n = reiθ = r n einθ zw z = Trick for complex numbers with high exponentials: w |w|2 z10 = (z2 )5 Polar form

Expresses a complex number by means of the complex A Recurring Derivatives number diagram. a = r cosθ, b = r sin θ

And hence integrals as well. . .

z = a + bi z = r(cosθ + i sin θ)

d x a = a x ln a dx 1 d (loga x) = x ln a dx 1 d tan x = dx cos2 x 1 d arcsin x = √ dx 1 − x2 d 1 arccos x = − √ dx 1 − x2 1 d arctan x = dx 1 + x2 d x ln | cos x| = tan x dx d 1 ln(1 + x2 ) = ln(1 + x2 ) dx 2

√ Where r = |z| = a2 + b2 and θ is the “argument” of z, θ = arg (z) = arctan(b/a). z1 z2 = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )] z1 r1 = [cos(θ1 − θ2 ) + i sin(θ1 − θ2 )] , z2 , 0 r2 z2 1 1 z−1 = = (cos θ − i sin θ) z r De Moivre’s Theorem

zn = [r(cosθ + i sin θ)]n = r n (cosnθ + i sin nθ)

7...