Mathematicalmethods Lecture Slides Week1 LimitsAndDerivatives

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    AMATH 460: Mathematical Methods for

    Quantitative Finance

    1. Limits and Derivatives

    Kjell KonisActing Assistant Professor, Applied Mathematics

    University of Washington

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    Outline

    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    Outline

    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    Course Information

    Instructors:

    Lecturer: Kjell Konis

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    Recommended Texts

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 5 / 68

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    Recommended Texts

    Calculus, Early TranscendentalsJames Stewart

    Any Edition

    (Or equivalent big, fat calculus textbook)Kjell Konis (Copyright 2013) 1. Limits and Derivatives 6 / 68

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    Quarter Overview

    Topics

    1 Limits and Derivatives

    2 Integration

    3/4 Multivariable Calculus

    5/6 Vectors, Matrices, Linear Algebra

    7 Lagranges Method, Taylor Series

    8 Numerical Methods

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    Outline

    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    Simple and Compound Interest Rules

    Simple Interest

    Money accumulates interest proportional to the total time of the

    investment.V = (1+rn) A0

    Compound Interest

    Interest is paid regularly.V = [. . . (1+r)[(1+r) A0]] = [. . . (1+r)A1] = (1+r)

    n A0

    Compounding at Various Intervals

    Interest rate ris yearly but interest is paid more often (e.g., monthly).

    V = [1+ (r/m)]k A0

    A0 =Principal, r=rate, n=# years, m=periods/year, k=# periodsKjell Konis (Copyright 2013) 1. Limits and Derivatives 9 / 68

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    Present and Future Value

    Suppose the interest rate r=4%.

    In one years time, the future value (FV) of$100 will be $104.Conversely, the present value (PV) of a $104 payment in one yearstime is $100.

    SinceFV= (1+r) PV,

    an expression for the present value is

    PV= 1

    1+rFV= d1FV

    where d1 is the 1-year discount factor.More generally,

    dk= 1

    [1+ (r/m)]k

    the present value of a payment Akreceived after kperiods is dkAk.

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    Annuities

    An annuity is a contract that pays money regularly over a period oftime.

    Question: suppose we would like to buy an annuity that pays $100 atthe end of the year for each of the next 10 years. At an interest rateof 4%, how much should we expect to pay?

    Solution: appropriately discount each of the 10 future payments and

    take the sum.The present value of the kth payment is

    PVk=dkAk= 100

    (1+0.04)k

    The value of the annuity is then

    V =PV1+PV2+ +PV10=10k=1

    100

    (1+0.04)k

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    Annuities (continued)

    Recall: V =10k=1

    100

    (1+0.04)k

    (1+0.04) V = 100+9

    k=1

    100

    (1+0.04)k

    V = 9k=1

    100

    (1+0.04)k +

    100

    (1+0.04)10

    0.04V = 100 100

    (1+0.04)10

    = V = 10.04

    100 100

    (1+0.04)10

    = 811.09

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    Perpetual Annuities

    A perpetual annuity or perpetuity pays a fixed amount periodicallyforever.

    Actually exist: instruments exist in the UK called consols.

    (1+0.04) V = 100+

    k=1

    100

    (1+0.04)k

    V =k=1

    100

    (1+0.04)k

    0.04V = 100

    = V = 1000.04

    =2500

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    Summary: Present Value

    Discount factor dk for a payment received after k periods, annual

    interest rate r

    , interest paidm

    times per year.

    dk= 1

    [1+ (r/m)]k

    Value Vof an annuity that pays an amount Aannually, annualinterest rate r, n total payments.

    V = A

    r

    1 1

    (1+r)n

    Value Vof a perpetual annuity that pays an amount A annually,annual interest rate r.

    V = A

    r

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    Outline

    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    A Diverging Example

    CalculateV =1

    1+1

    1+1 . . .

    Rewrite as

    V ?=

    k=1

    (+1 1) ?=k=1

    0= 0

    Also could rewrite as

    V ?=

    k=1

    1 k=1

    1 ?=0

    But

    k=11 ?

    =1+1+1+

    k=41= 1+1+1+

    j=11

    V ?=

    k=1

    1 k=1

    1 ?=1+1+1+

    j=1

    1 k=1

    1 ?=3

    Can make V any integer value.Kjell Konis (Copyright 2013) 1. Limits and Derivatives 16 / 68

    N

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    Notation

    R the set of real numbers

    Z the set of integers

    in: indicator of set membership for all

    there exists goes tog : R R a real-valued function with a real argument

    x floor: the largest integer less than or equal toxx ceiling: the smallest integer greater than or equal tox

    not (e.g. not inwould be x Z)

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    D fi i i f li i

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    Definition of limit

    Let g : R R.The limit ofg(x) as x

    x0 exists and is finite and equal to l if and

    only if for any >0 there exists >0 such that|g(x) l|< for allx(x0 , x0+).Alternatively,

    limxx0

    g(x) =l iff

    >0,

    >0 s.t.

    |g(x)

    l

    |< ,

    |x

    x0

    |<

    Similarly,

    limxx0

    g(x) = iff C>0, >0 s.t. g(x)> C,

    |x x0|< limxx0

    g(x) = iff C0 s.t. g(x)

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    Modification for x

    How to make sense of

    |x

    |< ?

    limx

    g(x) =l iff >0,b s.t. |g(x) l|< , x>b

    Example, recall pricing formulas:

    Annuity: Vn = A

    r

    1 1

    (1+r)n

    Perpetuity: V = A

    r

    Question: is limn

    Vn equal to V?

    Strategy: given >0, find N such that|Vn V|< forn>N.

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    E l

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    Example

    Given >0, find N such that|Vn V|< forn>N

    |Vn V| =Ar

    1 1

    (1+r)n

    A

    r

    =

    A

    r A

    r(1+r)n A

    r

    = A

    r(1+r)n

    Next, solve for N such that

    A

    r(1+r)n

    2

    forn

    >N

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    E l ( ti d)

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    Example (continued)

    A

    r(1+r)n =

    2

    (1+r)n = 2A

    r

    n log(1+r) = log2A

    r

    n = log(2A) log(r)

    log(1+r)

    Choose N=

    log(2A) log(r)

    log(1+r)

    then for n>N

    |Vn V| Ar(1+r)n

    2

    <

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    Summary: Limits

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    Summary: Limits

    Given >0, can find N such that|Vn V|< forn>N, thuslimn

    Vn =V

    Definition of a limit

    limxx0

    g(x) =l iff >0, >0 s.t. |g(x) l|< , |x x0|<

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    Outline

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    Outline

    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    Evaluating Limits

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    Evaluating Limits

    Observation: working with the definition = not fun!

    Suppose that c is a real constant and the limits

    limxa

    f(x) and limxa

    g(x)

    exist. Then

    (i) limxa

    [f(x) +g(x)] = limxa

    f(x) + limxa

    g(x)

    (ii) limxa

    cf(x) =c limxa

    f(x)

    (iii) limxa

    [f(x)g(x)] = limxa

    f(x) limxa

    g(x)

    (iv) limxa

    f(x)

    g(x)=

    limxa

    f(x)

    limxa

    g(x) if lim

    xag(x)=0

    (v) limxa

    n

    f(x) = nlimxa f(x)Kjell Konis (Copyright 2013) 1. Limits and Derivatives 24 / 68

    Evaluating Limits

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    Evaluating Limits

    IfP(x) and Q(x) are polynomials and c>1 is a real constant then

    (a) limx

    P(x)

    cx = lim

    xP(x) cx =0

    (b) limx

    log |Q(x)|P(x)

    =0

    Examples

    (a) limx

    x2ex =0

    (b) limx

    log(x3)

    x =0

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    Evaluating Limits

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    Evaluating Limits

    Notation: x0 means x0 with x>0; k! =k (k 1) 2 1

    Let c>0 be a positive constant, then

    (a) limx

    c 1x =1

    (b) limx

    x 1x =1

    (c) limx0

    xx =1

    Ifk is a positive integer and ifc>0 is a positive constant, then

    (a) limk

    k 1k =1

    (b) limk

    c 1k =1

    (c) limk

    ck

    k! =0

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    Example

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    Example

    limx

    1x2 4x+1 x

    = limx

    1x2 4x+1 x x

    2

    4x+1+xx2 4x+1+x

    = limx

    x2 4x+1+x

    x2

    4x+1

    x2

    = limx

    x2 4x+1+x1 4x

    1x1x

    = limx1

    4x

    + 1x2

    +1

    1x 4

    =

    limx

    1 4

    x + 1

    x2

    +1

    limx 1x 4

    =1+1

    4

    =12

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    Example

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    Example

    limx1

    x2 4x+1 x+2

    = limx

    1x2 4x+1 (x 2)

    x2 4x+1+ (x 2)x2 4x+1+ (x 2)

    = limx

    x2 4x+1+x 2x2 4x+1 (x 2)2

    = limx

    x2 4x+1+x 2

    x

    2

    4x+1 [x2

    4x+4]= lim

    x

    x2 4x+1+x 2

    3

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    Outline

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    Outline

    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    Continuity

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    Continuity

    Most of the time: limxa

    f(x) =f(a)

    x

    y

    f(x)

    a

    f(a)

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    Continuity Definitions

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    Continuity Definitions

    A function f is continuous at a if

    limxa f(x) =f(a)

    A function is right-continuous at a if

    limxa f(x) =f(a)

    A function is left-continuous at a if

    limxa f(x) =f(a)

    A function is continuous on an interval (c, d) if it is continuous atevery number a

    (c, d).

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    (Dis)continuity

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    ( ) y

    x

    yf(x)

    a

    Is f(x) continuous at a?

    Is f(x) right-continuous at a?

    Is f(x) left-continuous at a?

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    Continuity Theorems

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    y

    Let f and gbe continuous at aand cbe a real-valued constant. Then

    1. f +g 2. f g 3. cf4. fg 5. fg

    (g(a)=0)are continuous at a.

    Any polynomial is continuous on R.

    The following functions are continuous on their domains:

    polynomials trigonometric functionsrational functions inverse trigonometric functionsroot functions exponential functions logarithmic functions

    Ifg is continuous at a, and f is continuous at g(a), thenf

    g(x) =fg(x) is continuous at a.

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    Example

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    p

    Where is the following function continuous?

    f(x) = log(x) +tan1(x)x2 1

    tan1(x) means arctan(x), not 1tan(x) = cos(x)

    sin(x) =cot(x)

    log(x) is continuous for x>0 and arctan(x) forx Rthus log(x) +arctan(x) is continuous for x(0, )

    x2

    1 is a polynomial =

    continuous everywhere

    f(x) is continuous on (0, ) except where x2 1= 0= f(x) is continuous on the intervals (0, 1) and (1, )

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    Horizontal Asymptotes

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    y p

    What happens to f(x) = x2 1x2 +1

    as xbecomes large?

    x

    y

    The line y=L is called a horizontal asymptote if

    limx

    f(x) =L or limx

    f(x) =L

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    Outline

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    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes

    6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives9 Bond Duration

    10 lHopitals Rule

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    Tangent Line

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    Goal: find the slope of the line tangent to y=f(x) at a point a.

    a

    f(a)

    x

    f(x)

    A line y=l(x) is tangent to the curve y=f(x) at a point aif thereis >0 such that

    f(x)> l(x) on (a , a) and(a, a+) orf(x)< l(x) on (a , a) and(a, a+)

    and f(a) =l(a).Kjell Konis (Copyright 2013) 1. Limits and Derivatives 37 / 68

    Strategy

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    x

    f(x)

    x + h

    f

    (x + h

    )rise

    run

    x

    f(x)

    Use the slope of a line connecting a nearby point as an estimate ofthe slope of the tangent line.

    slope= rise

    run =

    f(x+h) f(x)h

    Get successively better estimates by letting h0Kjell Konis (Copyright 2013) 1. Limits and Derivatives 38 / 68

    Definition of Derivative

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    A function f : R R is differentiable at a point x R if

    limh0

    f(x+h) f(x)h

    exists.

    When this limit exists, the derivative off(x), denoted by f(x), isdefined as

    f(x) = limh0

    f(x+h) f(x)h

    A function f(x) is differentiable on an open interval (a, b) if it isdifferentiable at every point x(a, b).

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    Example

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    Compute the derivative off(x) =3x2 +5x+1

    f(x) = limh0

    f(x+h) f(x)h

    = limh0

    3(x+h)2 +5(x+h) +1 [3x2 +5x+1]h

    = limh0

    3x2 +6xh+3h2 +5x+5h+1 3x2 5x 1h

    = limh0

    6xh+3h2 +5h

    h

    = limh0

    6x+3h+5

    = 6x+5

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    Section Summary: Differentiation

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    Summary

    Definition: f(x) = limh0

    f(x+h) f(x)h

    Derivative of a constant: iff(x) =c then f(x) =0.

    Linearity:

    a f(x) +b g(x)

    =a f

    (x) +b g

    (x)

    Power Rule: (xc) =c xc1 forc=0.

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    Outline

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    1 Course Organization

    2 Present Value

    3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives9 Bond Duration

    10 lHopitals Rule

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    Product Rule

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    Suppose f(x) and g(x) are differentiable functions.

    Let p(x) =f(x) g(x)

    Then p(x) is differentiable, and

    p

    (x) =

    f(x) g(x)

    =f

    (x) g(x) +f(x) g

    (x)

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    Examples

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    Recall: p(x) =

    f(x) g(x)

    =f(x) g(x) +f(x) g(x)

    p(x) =x ex

    f(x) = x, g(x) =ex

    p(x) = 1 ex +x ex

    = (x+1)ex

    p(t) =

    t(1 t)f(t) =

    t=t

    12 , g(t) = (1 t)

    p(x) = 12

    t 12 (1 t) +t12 (1)

    = (1 t)

    2

    t t= 1 3t

    2

    t

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    Chain Rule

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    Suppose f(x) and g(x) are differentiable functions.

    The composite function (g f)(x) =g(f(x)) is differentiable, and

    (g f)(x) =g(f(x))f(x)

    Alternative notation: let u=f(x) and g=g(u) =g(f(x)), then

    dg

    dx =

    dg

    du

    du

    dx

    (Leibniz notation)

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    Examples

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    Recall:

    (g f)(x) =g(f(x))f(x)

    (g f)(x) = x2 +1= (x2 +1) 12

    (g f)(x) = 12

    (x2 +1)12 2x

    = xx2 +1

    (g f)(x) = (x3 1)100

    (g f)(x) = 100(x3 1)99 3x2= 300x2 (x3 1)99

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    Substitution Method

    d d d

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    Recall Leibniz notation: dg

    dx =

    dg

    du

    du

    dx

    Compute d

    dxsin(x2

    ); let u=x2

    , g(u) =sin(u) and du

    dx =2x

    d

    dx sin(u) =

    d

    dusin(u) du

    dx

    = cos(u)

    2x

    = 2xcos(x2)

    Compute dd

    esin(); let u=sin(), g(u) =eu and dud

    =cos()

    ddx

    esin() = ddu

    eu dud

    = eu cos()

    = cos()esin()

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    Derivative of the Inverse Function

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    The inverse function off(x) is the function such that f1(f(x)) =x

    Let f : [a, b][c, d] be a differentiable functionLet f1 : [c, d][a, b] be the inverse function off(x)f1(x) is differentiable for x[c, d] where f(f1(x))=0 and

    f1(x)

    = 1

    f(f1(x))

    Example

    log(ex) =x =

    f(x) =ex , f(x) =ex , and f1(x) =log(x)

    log(x)

    =

    f1(x)

    = 1

    elog(x) =

    1

    x

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    Summary: Product Rule and Chain Rule

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    Product Rule f(x) g(x)

    =f(x) g(x) +f(x) g(x)

    Chain Rule (g f)(x) =g(f(x))f(x)

    dg

    dx =

    dg

    du

    du

    dx

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    Outline

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    1 Course Organization

    2 Present Value3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives9 Bond Duration

    10 lHopitals Rule

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    Utility Functions

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    A utility function is a real-valued function U(x) defined on the realnumbers.

    Used to compare wealth levels: ifU(x)>U(y) then U(x) ispreferred.

    Exponential U(x) =eax

    (a>0)

    Logarithmic U(x) =log(x)

    Power U(x) =bxb

    b

    (0, 1]

    Quadratic U(x) =x bx2(b>0)

    x

    U(x)

    Exponential

    Logarithmic

    Power

    Quadratic

    What is the derivative telling us?

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    Critical Points

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    A critical point of a function f(x) is a number a in the domain off(x) where either f(a) =0 orf(a) does not exist.

    Iffhas a local min or max at a, then a is a critical point.

    xa

    U(x)

    U'(x)

    xa

    f(x)

    f'(x)

    local max iff(x) decreasing at a, local min if increasing

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    Higher Derivatives

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    Iff(x) is a differentiable function, f(x) is also a function.

    Iff(x) is also differentiable, its derivative is denoted by

    f(x) =

    f(x)

    f(x) is called the second derivative off(x)

    In Leibniz notation:

    d

    dx

    d

    dxf(x)

    =

    d2

    dx2f(x) =

    d2f

    dx2

    Alternative notation: f(x) =D2f(x)

    No reason to stop at 2:

    f(n)(x) = dn

    dxnf(x) =

    dnf

    dxn =Dnf(x)

    providedf(n1)(x) differentiable

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    First and Second Order Conditions

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    xa

    U(x)

    U'(x)

    U''(x)

    xa

    f(x)

    f'(x)

    f''(x)

    f(x) =0 and f(x)< 0 = local maximumf(x) =0 and f(x)> 0 = local minimum

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 55 / 68

    Absolute Minimum and Maximum Values

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    Want to invest a fixed sum in 2 assets to maximize expected return.

    Find the absolute maximum value off(x) on the interval [0, 1]More generally, on a closed interval [a, b]

    1 Evaluate f(x) at the criticalpoints in (a, b)

    2 Evaluate f(x) at aand b

    3 The global max is the max ofthe values in steps 1 & 2.

    4 The global min is the min of thevalues in steps 1 & 2.

    ExpectedReturn

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 56 / 68

    Outline

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    1 Course Organization

    2 Present Value3 Limits

    4 Evaluating Limits

    5 Continuity and Asymptotes6 Differentiation

    7 Product Rule and Chain Rule

    8

    Higher Derivatives9 Bond Duration

    10 lHopitals Rule

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 57 / 68

    Bond Pricing Formula

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    Price

    0 5% 10% 15% 20%0

    100

    200

    300

    400

    The price Pof a bond is

    P = F

    [1+]n +

    nk=1

    C

    1+k = F

    [1+]n +C

    1 1

    [1+]n

    where: C= coupon payment n= # coupon periods remainingF = face value = yield to maturity

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    Duration and Sensitivity

    cn n

    c

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    Let PVk= ck

    [1+]k, P=

    k=1

    PVk=k=1

    ck

    [1+]k

    The duration of a bond is a weighted average of times that paymentsare made.

    D=n

    k=1

    PVk

    P k or D P=

    nk=1

    k PVk

    Sensitivity: how is price affected by a change in yield?

    PVk = ck

    [1+]k =ck[1+]

    k

    d PVk

    d = k ck[1+]k1 = k ck

    [1+]k+1 = k

    1+PVk

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    Duration and Sensitivityn

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    P =k=1

    PVk

    d

    dP =

    d

    d

    nk=1

    PVk=n

    k=1

    d PVk

    d

    =

    n

    k=1

    k

    1+PV

    k

    = 11+

    nk=1

    k PVk

    = 11+

    D P DMP

    DMis called the modified duration

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 60 / 68

    Duration

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    Price Sensitivity Formula

    dPd

    =DMP

    Since1

    P

    dP

    d =DMDMmeasures the relative change in price as changes.

    Duration measures interest rate sensitivity.

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    Linear Approximation

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    Price

    0 5% 10% 15% 20%0

    100

    200

    300

    400

    The tangent line can be used to approximate the price.

    The approximation can be improved by adding a quadratic termbased on the convexity of the price-yield curve.More on this later

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    Summary: Bond Duration

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    Duration

    D=n

    k=1

    PVk

    P k or D P=

    nk=1

    k PVk

    Modified Duration

    DM= 1

    1+D

    Price Sensitivity Formula

    dP

    d =DMP

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 63 / 68

    Outline

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    1 Course Organization

    2 Present Value3 Limits

    4 Evaluating Limits

    5

    Continuity and Asymptotes6 Differentiation

    7 Product Rule and Chain Rule

    8 Higher Derivatives

    9 Bond Duration

    10 lHopitals Rule

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    lHopitals Rule

    i ( )

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    Problem: how to evaluate limx0

    sin(x)

    x ?

    4 3 2 2 3 4

    1

    limx0

    sin(x)x

    =limx0

    sin(x)

    limx0

    x = 0

    0

    Not allowed since limx0

    denominator=0

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 65 / 68

    lHopitals Rule

    Let x0 be a real number (including ) and let f (x ) and g (x ) be

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    Let x0 be a real number (including) and let f(x) and g(x) bedifferentiable functions.

    (i) Suppose limxx0f(x) =0 and limxx0g(x) =0. If limxx0f (x)g(x) exists

    and there is an interval (a, b) containing x0 such that g(x)=0 for all

    x(a, b)\ 0, then limxx0 f(x)g(x) exists and

    limxx0

    f(x)

    g(x)

    = limxx0

    f(x)

    g

    (x)

    (ii) Suppose limxx0f(x) = and limxx0g(x) =. If limxx0 f (x)g(x)

    exists and there is an interval (a, b) containing x0 such that g(x)=0

    for all x(a, b)\ 0, then limxx0 f(x)g(x) exists and

    limxx0

    f(x)

    g(x)= lim

    xx0

    f(x)

    g(x)

    When x0 =, intervals are of the form (, b) and (a, ).Kjell Konis (Copyright 2013) 1. Limits and Derivatives 66 / 68

    Examples

    lisin(x)

    i li i ( ) 0 d li 0

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    limx0

    ( )

    x since lim

    x0sin(x) =0 and lim

    x0x=0

    limx0

    sin(x)x

    = limx0

    ddxsin(x)

    ddx

    x= lim

    x0

    cos x1

    =1

    limx

    x2 1x

    2

    +1

    since limx

    x2

    1=

    and lim

    x

    x2 +1=

    limx

    x2 1x2 +1

    = limx

    ddx

    x2 1ddx

    x2 +1= lim

    x

    2x

    2x =?

    Since limx

    2x=

    have to use lHopitals rule again:

    limx

    x2 1x2 +1

    = limx

    ddx

    x2 1ddx

    x2 +1= lim

    x

    2x

    2x = lim

    x

    ddx

    2xddx

    2x= lim

    x

    2

    2=1

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 67 / 68

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    http://computational-finance.uw.edu

    Kjell Konis (Copyright 2013) 1. Limits and Derivatives 68 / 68

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