text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
by separability of h(·, ·), measurability of a(·)
and continuity of b(·), they can be restricted to a countable (dense) set in the complement
of C. Then for any θ /∈ C,
1
n
ha(θ, Xi) ≥ (1 − ε)b(θ) ≥ γ(θ0) + ε.
(3.3.12)
�n
i=1
On the other hand, for n large enough
�n
inf θ
1
n
ha(θ, Xi) ≤ 1
n
i=1
�n
i=1
ha... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
fine it as ψ. It can
also be assumed that θ0 ∈ C by adjoining it if necessary, and the proof below will show
that θ0 had to be in C.
Let U be an open neighborhood of θ0. It follows from (A-2(cid:6)) that γ(·) is lower semi-
continuous. Thus its infimum on the compact set C \ U is attained: let θk be a sequence
in C \... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
n
�n
i=1 ha(θ0, Xi) ≤ γ(θ0) + ε. It follows that
�n ha(θ, Xi) : θ ∈ C \ U } ≥ 1
�n
i=1
n
1
inf{ n
i=1
inf{
−1
and n
(3.3.14)
≥ inf{ n
1
�n ha(θ, Xi) : θ ∈ U } + ε,
i=1
�n
i=1
ha (θ0, Xi) + ε
n
inf{ha(φ, Xi) : φ ∈ Uj } ≥ γ(θ0) + 2ε
so Pr{Tn ∈ U for all n large enough} = 1. This completes the proof.
�
Next... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
+ Q,
by the Radon-Nikodym theorem in measure theory.
Suppose then that P has a density f = dP/dµ and Q has a density g = dQ/dµ with
respect to µ. Then the likelihood ratio of Q to P is defined as RQ/P (x) = g(x)/f (x),
or as +∞ if g(x) > f (x) = 0, or as 0 if g(x) = f (x) = 0. Then the likelihood ratio is
well-defin... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
By derivatives, it’s easy to check that log x ≤ x − 1 for all x ≥ 0, with log x = x − 1
if and only if x = 1. Thus
�
�
I(P, Q) = − log(RQ/P )dP ≥
1 − RQ/P dP ≥ 0,
with equality if and only if RQ/P = 1 a.s. for P , and then Q = P .
�
Although I(P, Q) is sometimes called a metric or distance, it is not symmetric ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
.11), then the Tn are consistent.
Proof. If (A-1) through (A-5) hold then (A-2(cid:6)) and (3.3.11) hold by Lemmas 3.3.9 and
3.3.10, and then Theorem 3.3.13 applies. So just (A-3) and (A-4) need to be proved. Let
a(x) := − log f (θ0, x). We have 0 < f (θ0, x) < ∞ a.s. for P , and so −∞ < log f (θ0, x) <
∞. Thus h(θ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
R, a point m is a median of P iff both P ((−∞, x]) ≥ 1/2
and P ([x, +∞)) ≥ 1/2. Thus if P is a continuous distribution without atoms, m is a
median if and only if P ((−∞, m]) = 1/2. If P is any law on R having a unique median
θ0 and h(θ, x) := |x − θ|, show that conditions (A-1) through (A-5) hold for some a(·)
and ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
Kullback (1983) gives an update.
REFERENCES
Cram´er, Harald (1945). Mathematical Methods of Statistics. Almqvist & Wicksells, Upp-
sala, Sweden; Princeton University Press, 1946; 10th printing 1963.
Dudley, R. M. (1998). Consistency of M -estimators and one-sided bracketing. In High
Dimensional Probability, Progres... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
. A. (1951). On information and sufficiency. Ann. Math.
Statist. 22, 79-86.
Wald, A. (1949). Note on the consistency of the maximum likelihood estimate. Ann.
Math. Statist. 20, 595-601.
8 | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/3f81a744094897380591c3d2b21201e8_m_estimates.pdf |
6.S096 Lecture 2 – Subtleties of C
Data structures and Floating-point arithmetic
Andre Kessler
Andre Kessler
6.S096 Lecture 2 – Subtleties of C
1 / 16
Outline
1
Memory Model
2
Data structures
3
Floating Point
4
Wrap-up
Andre Kessler
6.S096 Lecture 2 – Subtleties of C
2 / 16
... | https://ocw.mit.edu/courses/6-s096-effective-programming-in-c-and-c-january-iap-2014/3f8f2bc5d4ee804b3bce651a210a940f_MIT6_S096IAP14_Lecture2.pdf |
allocated
int array[10];
int array2[] = { 1, 2, 3, 4, 5 };
char str[] = "Static string";
Dynamically allocated
#include <stdlib.h>
int *array = malloc( 10 * sizeof( int ) );
// do stuff
array[5] = 5;
//when done
free( array );
Andre Kessler
6.S096 Lecture 2 – Subtleties of C
5 / 16
... | https://ocw.mit.edu/courses/6-s096-effective-programming-in-c-and-c-january-iap-2014/3f8f2bc5d4ee804b3bce651a210a940f_MIT6_S096IAP14_Lecture2.pdf |
.
Andre Kessler
6.S096 Lecture 2 – Subtleties of C
7 / 16
Memory Model
Array Indexing: Syntactic sugar
C doesn’t know what an array is, really.
T array[] and T *array = malloc(...) are both pointers to
contiguous blocks o... | https://ocw.mit.edu/courses/6-s096-effective-programming-in-c-and-c-january-iap-2014/3f8f2bc5d4ee804b3bce651a210a940f_MIT6_S096IAP14_Lecture2.pdf |
Data structures
Structs and Typedef
If C only knows about memory, how do we get it to understand a data
structure?
typedef struct IntPair_s {
int first;
int second;
} IntPair;
// in code:
IntPair pair;
pair.first = 1;
pair.second = 2;
IntPair *pairPtr = &pair;
// use pairPtr->first and pairPtr->second
// ... | https://ocw.mit.edu/courses/6-s096-effective-programming-in-c-and-c-january-iap-2014/3f8f2bc5d4ee804b3bce651a210a940f_MIT6_S096IAP14_Lecture2.pdf |
0 0 0 0 0 0 0
= 0.15625
31
23
0
double (64 bits)
Sign
Exponent
(11-bit)
63
52
Fraction (52-bit)
0
Images by MIT OpenCourseWare.
Andre Kessler
6.S096 Lecture 2 – Subtleties of C
12 / 16
Floating Point
Subtleties
Rounding and precision
Denormals
Long d... | https://ocw.mit.edu/courses/6-s096-effective-programming-in-c-and-c-january-iap-2014/3f8f2bc5d4ee804b3bce651a210a940f_MIT6_S096IAP14_Lecture2.pdf |
6.252 NONLINEAR PROGRAMMING
LECTURE 8
OPTIMIZATION OVER A CONVEX SET;
OPTIMALITY CONDITIONS
Problem: minx∈X(cid:160)f (x),(cid:160)where:
(a) X(cid:160)⊂ (cid:2)n(cid:160)is nonempty, convex, and closed.
(b) f(cid:160)is continuously differentiable over X.
• Local and global minima. If f(cid:160) is convex local
mi... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/3fab988e50bec05b31655d6cccd8af53_6252_slides08.pdf |
160)shown.
PROOF
Proof: (a) Suppose that ∇f (x(cid:160)∗)(cid:3)(x(cid:160)− x(cid:160)∗) <(cid:160)0 for
some x(cid:160)∈ X. By the Mean Value Theorem, for
every (cid:1) > (cid:160)0 there exists an s(cid:160)∈ [0,(cid:160)1] such that
(cid:1)
∗ +(cid:1)(x−x
x
(cid:1)
∗ )+(cid:1)∇f x
∗ +s(cid:1)(x−x
∗ ) = f... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/3fab988e50bec05b31655d6cccd8af53_6252_slides08.pdf |
160)− x(cid:160)∗)
for every x(cid:160)∈ X. If the condition ∇f (x(cid:160)∗)(cid:3)(x−x(cid:160)∗) ≥
0 holds for all x(cid:160)∈ X, we obtain f (x) ≥ f (x(cid:160)∗), so
x ∗ minimizes f(cid:160)over X. Q.E.D.
OPTIMIZATION SUBJECT TO BOUNDS
• Let X = {x | x ≥ 0}. Then the necessary
1, . . . , x∗
condition for x∗ =... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/3fab988e50bec05b31655d6cccd8af53_6252_slides08.pdf |
)
∂xi
∗
i ) ≥ 0,
(xi−x
∀ xi ≥ 0 with
n(cid:3)
i=1
xi = r.
i > 0 and let j be any other index.
m for
i , and xm = x∗
j + x∗
• Fix i with x∗
Use x with xi = 0, xj = x∗
all m (cid:7)= i, j:
(cid:7)
∂f (x∗)
∂xj
− ∂f (x∗)
∂xi
(cid:8)
x∗
i
≥ 0,
i > 0 =⇒ ∂f (x∗)
x∗
∂xi
≤ ∂f (x∗)
∂xj
,
∀ j.
OPTIMAL ROUTING
• Given a data net,... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/3fab988e50bec05b31655d6cccd8af53_6252_slides08.pdf |
i,j) xp
be the travel time of link (i, j).
• User-optimization principle: Traffic equilibrium
is established when each user of the network chooses,
among all available paths, a path requiring mini-
mum travel time, i.e., for all w ∈ W and paths
p ∈ Pw,
∗
∗
p > 0 =⇒ tp(x
x
∗
) ≤ tp(cid:1) (x
),
∀ p
(cid:2) ∈ Pw, ∀ w ∈ W
... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/3fab988e50bec05b31655d6cccd8af53_6252_slides08.pdf |
Introduction to C++
Massachusetts Institute of Technology
January 26, 2011
6.096
Lecture 10 Notes: Advanced Topics II
1 Stuff You May Want to Use in Your Project
1.1 File handling
File handling in C++ works almost identically to terminal input/output. To use files, you
write #include <fstream> at the top of your so... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
using the close() method when you’re done using them. This is automat
ically done for you in the object’s destructor, but you often want to close the file ASAP,
without waiting for the destructor.
You can specify a second argument to the constructor or the open method to specify what
“mode” you want to access the fi... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
S = 0 , DIAMONDS = 1 , HEARTS = 2 , SPADES = 3;
2 void print_suit ( const int suit ) {
3
4
5
6 }
const char * names [] = { " Clubs " , " Diamonds " ,
" Hearts " , " Spades " };
return names [ suit ];
The problem with this is that suit could be integer, not just one of the set of values we
know it should be re... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
.
• Every other item defaults to the previous item plus 1.
Just like any other type, an enum type such as suit t can be used for any arguments,
variables, return types, etc.
1.4 Structuring Your Project
Many object-oriented programs like those you’re writing for your projects share an overall
structure you will l... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
the reference variable
is used.
The syntax for setting a reference variable to become an alias for another variable is just like
regular assignment:
1 int & x = y ; // x and y are now two names for the same variable
Similarly, when we want to pass arguments to a function using references, we just call the
function... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
avoids copying over the entire object. You could also then
use your method to do something like:
1 vector < Card > & cardList
4
= deck . getList () ; // getList declared to return a reference
2
3 cardList . pop_back () ;
The second line here modifies the original list in deck, because cardList was declared as a ... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
when we start talking about const objects.
Clearly, no fields on a const object should be modifiable, but what methods should be
available? It turns out that the compiler can’t always tell for itself which methods are safe
to call on const objects, so it assumes by default that none are. To signal that a method is
sa... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
12
13
14
15
16 }
}
// ... Some other code ...
* arrPtr = new int [ divide (5 , x ) ];
} catch ( int error ) {
try {
// cerr is like cout but for error messages
cerr << " Caught error : " << error ;
The code in main is executing a function (divide) that might throw an exception. In
anticipation, the potenti... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
" new failed to allocate memory " ;
}
catch ( runtime_exception & error ) {
// cerr is like cout but for error messages
cerr << " Caught error : " << error . what () ;
In such a case, the exception’s type is checked against each of the catch blocks’ argument
types in the order specified. If line 2 causes an except... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
if we want to
make the fields of the USCurrency type from the previous lecture private, we can still have
our stream insertion operator (the output operator, <<) overloaded:
1 class USCurrency {
2
friend ostream & operator < <( ostream &o , const USCurrency & c )
;
int dollars , cents ;
3
4 public :
5
USCurrenc... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
’ve already seen C-style casts
– e.g. 1/(double)4. Such casts are not recommended in C++; C++ provides a number of
more robust means of casting that allow you to specify more precisely what you want.
All C++-style casts are of the form cast type<type>(value), where type is the type you’re
casting to. The possible c... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
to an object so you can
force use of the const version of a member function.
7 That’s All!
This is the end of the 6.096 syllabus, but there are lots of really neat things you can do with
C++ that we haven’t even touched on. Just to give you a taste:
• Unions – group multiple data types together; unlike classes/str... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/3fb8a922fc4283863e29eeed55d7e664_MIT6_096IAP11_lec10.pdf |
18.175: Lecture 10
Zero-one laws and maximal inequalities
Scott Sheffield
MIT
18.175 Lecture 10
1 Outline
Recollections
Kolmogorov zero-one law and three-series theorem
18.175 Lecture 10
2 Outline
Recollections
Kolmogorov zero-one law and three-series theorem
18.175 Lecture 10
3 Recall Borel-Cantelli lemm... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/3fed030f143799a396c58ad269a4b19b_MIT18_175S14_Lecture10.pdf |
at stuff arbitrarily far into the
future. Intuitively, membership in tail event doesn’t change
when finitely many Xn are changed.
Event that Xn converge to a limit is example of a tail event.
Other examples?
Theorem: If X1, X2, . . . are independent and A ∈ T then
P(A) ∈ {0, 1}.
18.175 Lecture 10
8 Kolmogorov ... | https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/3fed030f143799a396c58ad269a4b19b_MIT18_175S14_Lecture10.pdf |
18.01 Calculus
Jason Starr
Fall 2005
Math 18.01 Lecture Summaries
Homework. These are the problems from the assigned Problem Set which can be completed using
the material from that date’s lecture.
8 Velocity and derivatives
9 Limits
Practice Problems. Practice problems are not to be written up or turned in. These ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Lecture 26. Nov. 18 Partial fraction decomposition
Lecture 27. Nov. 22
4 Related rates problems
6 Newton’s method
13 Antidifferentiation
14 Riemann integrals
18 The Fundamental Theorem of Calculus
20 Properties of the Riemann integral
21 Separable ordinary differential equations
25 Numerical integration
28 Applications ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
v(t0 ) = lim vave = lim
→
→
0
0
Δt
Δt
s(t0 + Δt) − s(t0)
.
Δt
This is a derivative, v(t) equals s�(t) = ds/dt. The derivative of velocity is acceleration,
a(t0) = v�(t0 ) = lim
→
0
Δt
v(t0 + Δt) − v(t0)
.
Δt
Example. For s(t) = −5t2 + 20t, first computed velocity at t = 1 is,
v(1) = lim 10 − 5Δt = 10.
Δt
0... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Δx 0
f (x0 + Δx) − f (x0)
Δx
.
3. Examples in science and math.
(i) Economics. Marginal cost is the derivative of cost with respect to some other variable, for
instance, the quantity purchased.
(ii) Thermodynamics. The ideal gas law relating pressure p, volume V , and temperature T of a
gas is,
Under isothermal c... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2 =
A.
This is very reasonable. In some sense, this explains the classical formula for the volume of
a cone.
Lecture 2. September 9, 2005
Homework. Problem Set 1 Part I: (f)–(h); Part II: Problems 3.
Practice Problems. Course Reader: 1C2, 1C3, 1C4, 1D3, 1D5.
1. Tangent lines to graphs. For y = f (x), the equatio... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
3 = 0.
4
18.01 Calculus
Jason Starr
Fall 2005
Factoring (x0 − 3)(x0 + 1), the solutions are x0 equals −1 and x0 equals 3. The corresponding
tangent lines are,
and
For general (x, y), the solutions are,
y = −2x − 1,
y = 6x − 9.
x0 = x ± �
x2 − y.
f (x)
2. Limits. Precise definition is on p. 791 of Appendix ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
limits, use a laser that illuminates only to the left of x0. For righthand limits,
use a laser that illuminates only to the right of x0.
3. Continuity. A function f (x) is continuous at x0 if f (x0) is defined, limx
f (x) is
f (x) equals f (x0). Also, f (x) is continuous on an interval if it is contin
defined, and l... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
k + 1)
,
k(k − 1) · · ·
3 · 2 ·
1
is the number of ways to choose a subset of k elements from a collection of n elements. A funda
mental fact about binomial coefficients is the following,
� � � �
n
k
n + 1
k
n
k − 1
�
.
�
=
+
This is known as Pascal’s formula. This link is to a webpage produced by MathWorld, p... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
n
a n−k bk + · · · + bn
�
� �
n
k
�
.
Multiplying each term in the second factor first by a and then by b gives,
� �
n an+1−k bk +
�
� k
n an+1−kbk +
k−1
�
�
n an−k bk+1
� �
k+1
n an−k bk+1
k
an+1 + nanb +
+ anb +
. . . +
. . . +
+
+
n
. . . + ab
. . . + nab
n + bn+1
Summing in columns gives,
a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
every integer.
3. The derivative of x . Let f (x) = xn where n is a positive integer. For every a and every h,
the binomial theorem gives,
n
f (a + h) = (a + h)n = a + na n−1h + · · · +
n
a n−k hk + · · · + hn .
� �
n
k
Thus, f (a + h) − f (a) equals,
(a + h)n − a = na
n−1h + · · ·
n
� �
n
k
+
a n−k hk + ·... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(f (x)g(x))� = f �(x)g(x) + f (x)g�(x).
The crucial observation in proving this is rewriting the increment of f (x)g(x) from a to a + h as,
f (a+h)g(a+h)−f (a)g(a) = f (a+h)[g(a+h)−g(a)]+f (a+h)g(a)−f (a)g(a) = f (a+h)[g(a+h)−g(a)]+[f (a+h)−
f (a)]g(a).
5. The quotient rule. Let f (x) and g(x) be differentiable funct... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
another induction proof that for every positive integer n, d(xn)/dx equals
nxn−1 . For n = 1, we proved this by hand. Let n be some specific positive integer, and make the
induction hypothesis that d(xn)/dx equals nxn−1 . The goal is to deduce the formula for n + 1,
d(xn+1)
dx
= (n + 1)x n .
By the Leibniz rule,
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
un . From above, (u2)� equals 2uu�. By a similar computation, (u3)� equals
3u2u�. This suggests a pattern,
d(un)
dx
= nu
n−1 du
.
dx
8
18.01 Calculus
Jason Starr
Fall 2005
This can be proved by induction on n. For n = 1, 2 and 3, it was checked. Let n be a particular
integer (for instance, 7011920947293305... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
fraction m/n and let u(x) be x . Then u
equals xm . Thus,
a
n
d(un)
dx
=
d(xm)
,
dx
which equals mxm−1 . By the above, d(un)/dx equals nun−1(du/dx). Thus,
n−1
nu
du
dx
= mx
m−1
.
Solving for du/dx,
du mxm−1
nun−1
dx
=
=
mxm−1
n(xm/n)n−1
.
One of the basic rules of exponents is that (a
equals nxm/n(... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(g ◦ f )(x) = g(f (x)).
This is often easier to remember in the form,
(g ◦ f )�(x) = g�(f (x)) · f �(x).
du
dx
=
du dy
·
.
dy
dx
This also suggests the proof,
(g ◦ f )�(x0) = lim
→
Δx 0 Δx Δx 0 Δy
= lim
→
Δu Δy
·
Δx
,
Δu
where y0 equals f (x0), u0 equals g(y0) = g(f (x0)), Δy equals f (x0 + Δx) − f (x... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
.
5. Implicit differentiation. This method has already been used many times. Given a function
y(x) satisfying some equation involving both x and y, formally differentiate each side of the equation
with respect to x and then try to solve for y�.
10
18.01 Calculus
Jason Starr
Fall 2005
Lecture 5. September 16, 200... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
=
1
x2
·
1
(2 − 1/x)2
=
1
.
(2x − 1)2
2. Rules for exponentials and logarithms. Let a be a positive real number. The basic rules
of exponentials are as follows.
Rule 1. If ab equals B and a equals C, then ab+c equals B · C, i.e.,
c
b+c
c
a = a a .
·
b
Rule 2. If ab equals B and Bd equals D, then abd equa... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
equals,
L(a) = lim
→
h 0
h
a − 1
.
h
ax0+h − ax0
h
.
lim
→
h 0
By Rule 1, ax0+h equals a
x0 ah
. Thus the limit factors as,
− ax0
ax0 ah
h
lim
→
h 0
= a x0 lim a h − 1h.
h 0
→
Therefore, for every x, the derivative of ax is,
d(ax)
dx
= L(a)a x .
What is L(a)? To figure this out, consider how L(a) chang... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a0)x, it is geometrically clear that L(a0) is positive (though we have not
proved that L(a0) is even defined). Thus the graph of L(a) looks qualitatively like the graph of
loga0 (a). In particular, for a less than 1, L(a) is negative. The value L(1) equals 0. And L(a)
approaches +∞ and a increases. Therefore, there m... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ln(a)a u = 1.
dx
13
18.01 Calculus
Solving gives,
Jason Starr
Fall 2005
d loga(x)
dx
=
1
1
ln(a) au
=
1/(ln(a)x) .
In particular, for a = e, this gives,
d ln(x)
dx
=
1/x .
What is the derivative of ln(x) at x = 1? On the one hand, since the derivative of ln(x) equals 1/x,
the derivative at x = 1 is ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
of n factors, say f1(x)·
fn(x), the derivative
of y can be computed by the product rule. However, it seems to be a fact that multiplication is
more errorprone than addition. Thus introduce,
f2(x)·· · ··
u = ln(y) = ln(f1(x)) + ln(f2(x)) + · · · + ln(fn(x)).
14
18.01 Calculus
The derivative of u is,
du
dx
=
d ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
√
x)� = (
x) = (1/2x−1/2)/(1+
√
x).
So, finally,
y� = yu� =
(1 + x3)(1 +
√
x3/7
x) �
3x2
(1 + x3)
+
1
x(1 +
√
√
2
x)
−
�
.
3
7x
Lecture 6. September 20, 2005
Homework. Problem Set 2 Part I: (f)–(j); Part II: Problems 1, 3 and 4.
Practice Problems. Course Reader: 1J1, 1J2, 1J3, 1J4
1. Trigonometric functi... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
functions are sin(θ), cos(θ), tan(θ), sec(θ), csc(θ) and cot(θ). Full descriptions of these
are in §9.1 of the textbook by Simmons. The same information is contained in the webpage on
Trigonometry produced by MathWorld, part of Wolfram Research.
2. Trigonometric identities. For today, the most important identities a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
) − 1
θ
=
0.
cos(θ) − 1 cos(θ) + 1
cos(θ) + 1
θ
=
cos2(θ) − 1
.
θ · (cos(θ) + 1)
By Identity (v), cos2(θ) − 1 equals − sin2(θ), so the term equals,
− sin2(θ)
θ · (cos(θ) + 1)
=
−
sin(θ)
θ
1
cos(θ) + 1
sin(θ).
16
18.01 Calculus
Jason Starr
Fall 2005
As θ → 0, this limit tends to,
By a similar c... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
cos(x).
An entirely similar computation gives,
cos(a + h) − cos(a)
h
= cos(a)
cos(h) − 1
h
− sin(a)
sin(h)
,
h
which leads to,
�
cos
a
( ) = cos( ) lim
h 0→
a
cos(h) − 1
h
− sin(a
) lim
h→0
sin(h)
h
= cos(a) × 0 − sin(a) × 1.
Thus the derivative of cos(x) equals,
d cos(x)
dx
=
− sin(x).
17
18.01 Ca... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x = a is the linear function,
f (a) + f �(a)(x − a).
In a precise sense, this is the best approximation of f (x) by a linear function near x = a. For x
close to a, the value of f (x) is close to the value of the linearization. The notation for this is,
f (x) ≈ f (a) + f �(a)(x − a) for x ≈ a.
Example. The lineariz... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
x
ln(1 + x)
≈
1 − x
+ x2/2 − x3/3 + . . .
3. Combining basic approximations. The basic approximations can be combined to get new
linear approximations.
(i) The linear approximation of f (x) for x ≈ a can be converted to a linear approximation at 0 by
setting g(u) = f (a + u). In symbols,
f (a
) +
f �(a)(x −... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
=
df
dx
(x) +
dg
dx
(x).
(iv) The linear approximation of f (x)g(x) for x ≈ a is the product of the linear approximations,
disregarding all quadratic terms,
f (x)g(x) ≈ (f (a) + f �(a)(x − a))(g(a) + g�(a)(x − a)),
which simplifies to,
f (x)g(x) ≈ f (a)g(a
) + (
This is equivalent to Leibniz’s rule,
f �(a)g(a
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x)/g(x)) =
1
g(x)
(x)g(x) − f (x)
(x)).
dg
dx
df
(
dx
20
18.01 Calculus
Jason Starr
Fall 2005
(vi) The linear approximation of g(f (x)) for x ≈ a is obtained from the linear approximation of
g(u) for u ≈ f (a) by substituting in for u the linear approximation of f (x) for x ≈ a and ignoring
quadratic te... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
approximation is not good enough. One
example is the linear approximation of cos(x) as 1 for x ≈ 0. The linear approximation gives no
idea whether cos(x) is greater than 1, less than 1, concave up, concave down, etc. This is remedied
by the quadratic approximation,
f (x) ≈ f (a
) +
f �(a)(x − a
) +
1
2 f ��(a)(x... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
an input signal x and returns an output signal y. If
the input signal is always in the range −1/2 ≤ x ≤ 1/2 and if the output signal is,
y = f (x) =
1
,
1 + x + x2 + x3
what precision of the input signal x is required to get a precision of ±10−3 for the output signal?
If the ideal input signal is x = a, and if t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
b. In symbols, f is increasing, respectively
decreasing, if
f (a) < f (b) whenever a < b, resp. f (a) > f (b) whenever a < b.
22
18.01 Calculus
Jason Starr
Fall 2005
If f (a) is less than or equal to f (b), resp. greater than or equal to f (b), whenever a is less than
b, then f (x) is nondecreasing, resp. non... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x) ≤ f (a) for all x near a, then x is a local maximum. If f (x) ≥ f (a)
for all x near a, then x is a local minimum. Because of the First Derivative Test, if f �(a) > 0 and
f is defined to the right of a, the graph of f rises to the right of a. Thus a is not a local maximum.
Similarly, if f �(a) < 0 and f is defined ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
0, rexp.
f �(a) ≤ 0.
(v) The function f is defined to the left and right of a, and f �(a) equals 0. In other words, a is a
critical point of f .
Example. For the function y = x3 + x − x − 1, the critical points are x = −1 and x = 1/3. By
examining where y is increasing and decreasing, x = −1 is a local maximum and ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
secant line segment to f (x) crosses below the graph of f , resp. above the graph
of f . In symbols, f is concave up, resp. concave down, if
(f (c) − f (a))/(c − a) ≤ (f (b) − f (a))/(b − a) whenever a < c < b,
resp. (f (c) − f (a))/(c − a) ≥ (f (b) − f (a))/(b − a) whenever a < c < b.
For a differentiable function ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
second derivative is y�� = 6x + 2. Since y��(−1) = −4 is
negative, the critical point x = −1 is a local maximum. Since y��(1/3) = 4 is positive, x = 1/3 is a
local minimum.
2
5. Inflection points. If f is differentiable, but for every neighborhood of a, f is neither concave
up nor concave down on the entire neighborh... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
bounded, in some sense. There are two main examples. The function f has a vertical asymptote
x = a if at least 1 of the following holds,
lim f (x) = +∞, lim f (x) = −∞, lim f (x) = +∞, lim f (x) = −∞.
x a−
→
x a−
→
x a+
→
x a+
→
In each case, the graph of y = f (x) becomes unbounded, and becomes arbitrarily clos... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Two long walls meet at right angles making a corner. Using a length of 10 meters of
fence to form the other 2 sides of a rectangle, what is the largest area that can be enclosed in this
corner?
Step 1. Identify parameters. A parameter is a constant or variable. The constant in this
problem is 10 meters. Two variabl... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the graph of A(l) is a segment of a parabola opening down. The vertex
of the parabola is contained in the segment. Thus the vertex is the maximum.
Step 5. Compute the derivative. In this case,
A�(l) = −2l + 10.
Step 6. Find all critical points, endpoints, discontinuity points, etc. In most cases, it
suffices to find ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
meters per second and runs v2 meters per second, at what distance x from the closest
point on shore should she aim to minimize her time to the target? Mathematically, the swimmer
is at point (0, b1) and wants to reach point (a, −b2), where the shore is the xaxis. At what point
(x, 0) should she aim?
The constants ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x 2 + b2
1)1/2 +
1
v1
1
v2
((a − x)2 + b2
2)1/2
.
The derivative of T with respect to x is,
(x 2 + b2
1)−1/2(2x) +
((a − x)2 + b2
2)−1/2(−2(a − x))
�
.
dT
dx
=
This simplifies to,
�
1
1
v1 2
�
�
1
1
v2 2
dT
dx
=
x
v1d1
−
a − x
.
v2d2
Observe that x/d1 equals sin(θ1) and (a − x)/d2 equals s... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
I: (g) and (h).
Practice Problems. Course Reader: 2E4, 2E8, 2E9.
1. Related rates. A situation that arises often in practice is that two quantities, say x and y,
depend on a third independent variable, say t. The quantities x and y are related through some
constraint. Using the constraint, if the rateofchange ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(iii) Label all dependent variables. In the example, x and P are dependent variables.
(iv) Draw a diagram and carefully label it.
(v) Write the given rateofchange and the unknown rateofchange. In the example, dx/dt(T ) is
given as 5cm/s, and dP /dt is unknown.
(vi) Using the diagram and any other information, fi... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(T ). The constraint is
the Pythagorean theorem,
2
2
2
r = x + y .
28
18.01 Calculus
Implicit differentiation with respect to t yields,
dx
2r = 2x + 0 = 2x
dt
dr
dt
dx
.
dt
At time t = T , x(T ) equals a, because the angle θ is π/4. Thus r(T ) equals
gives,
√
2(
2a)50 = 2(a)
(T ).
dx
dt
Jason Starr
F... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
b) on the circle. Because the tangent line to the
circle at (a, b) is perpendicular to the radius through (a, b), the triangle with vertices (0, 0), (a, b)
and the point (x, 0) is a right triangle. The angle of the triangle at (x, 0) is θ/2. Since the radius
has length 1 and the hypotenuse has length x, the constrai... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
point (x, y) on the unit circle in the first
quadrant. What is the maximal area enclosed by such a trapezoid?
2
The parameters are x and y. The height of the trapezoid is y. The area of a trapezoid is the
product of the height with the average of the parallel sides. Thus,
A = y
(2 + 2x)
2
= (x + 1)y.
This is the ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x = 0, x = 1 or at the critical point
x = 1/2. Plugging in gives,
A(0) = 1, A(1/2) = 3
√
3/4, A(1) = 0.
This gives the answer,
A achieves its maximum 3
√
3/
4 for the point (x, y) = (1
/2,
√
3/
2).
30
18.01 Calculus
Jason Starr
Fall 2005
Two other methods were given in lecture. The fastest among the three is ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
dF
dx
(x) = f (x),
which means the same thing as the differential notation. It may look like the first and second
equation are obtained by dividing and multiplying by the quantity dx. It is crucial to remember
that dF/dx is not a fraction, although the notation suggests otherwise.
In differential notation, some deriv... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2 + 1) =
1/2)dx .
+ 1)
2. Antidifferentiation. Recall, the basic problem of differentiation is the following.
Problem (Differentiation). Given a function F (x), find the function f (x) satisfying dF = f (x).
dx
The bais problem of antidifferentiation is the inverse problem.
Problem (Antidifferentiation). Given a function ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
simple antiderivative exists, and some tools to compute
the antiderivative.
3. Antidifferentiation. Guessandcheck. The main technique for antidifferentiation is edu
cated guessing.
Example. Find an antiderivative of f (x) = x2 + 2x + 1. Since the derivative of xn is nxn−1, it is
reasonable to guess there is an ant... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
less effective.
By roughly the same logic in the last example, we might guess an antiderivative has the form
Ax3 sin(x2). The derivative is 3Ax2 sin(x2) + 2Ax4 cos(x2). The first term is good, but the second
term is bad. We can try to correct our guess by adding a term, Ax3 sin(x2) − 2/5Ax5 cos(x2), whose
derivative ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
dx, try to find du = u�(x)dx as a factor.
(iv) Try to write f (x)dx as g(u)du. If you cannot do this, the method does not apply with the
given choice of u.
(v) Find an antiderivative G(u) = g(u)du for the simpler integrand g(u) (if this is possible).
�
(vi) Backsubstitute u = u(x) to get an antiderivative F (x) = ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
. The area of the large region is the sum
of the areas of the small regions. But the ancient Greeks also knew the area of a circle, which
cannot be dissected exactly into finitely many polygonal regions. Their method was exhaustion:
finding polygonal regions approximately equal to the original region, and computing th... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
For a region partly above the xaxis and partly below the
xaxis, the signed area is the sum of the signed area of the region above the xaxis and the signed
area of the region below the xaxis.
2. Partitions. A partition of an interval [a, b] is a finite decomposition of the interval as a union
of nonoverlapping s... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
1/2n, but has one interval, [1/2, 1], of size 1/2. The number
1/2 may not seem large, but as n increases, it is quite large compared to 1/2n.
Because of such examples, a better measure of fineness is mesh size: The mesh size of a partition
is the maximal length of any subinterval in the partition,
mesh = max Δxk |k ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Fall 2005
The sum above is a Riemann sum. It is an approximation of the signed area of the curvilinear
region.
There are many choices of partition. And for each partition, there are many choices for the numbers
x∗
k. However, there are some special choices. On the kth interval, the smallest value f (x) takes on
i... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
The mesh size measures the fineness
of the partition, thus also the fit of the union of vertical strips to the curvilinear region. If the two
limits,
lim Amin,
mesh→0
lim Amax,
mesh→0
are defined and equal, it is said the Riemann integral exists, and the common limit is called the
Riemann integral,
� b
a
f (x)d... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
),
and,
k=1
Amax =
n
�
k=1
yk,maxΔxk =
n
� kL L
n n
k=1
=
n
L2 �
n2
k=1
k.
To evaluate these sums, use the wellknown formula,
This also gives,
k=1
n
�
k =
n(n + 1)
.
2
n
�
(k − 1) =
n−1
n−1
� �
l =
l =
k=1
l=0
l=1
(n − 1)n
2
,
by making the substitution l = k − 1. Substituting the formula g... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
5
5. Rules for Riemann integrals. There are several rules for Riemann integrals, summarized
below.
� b
(f
(x) + g(x))dx =
a
�
b
a (r ·
=
f (x))dx
�
c
f (x)dx + b f (x)dx =
� b
a
� b
a
� b
f (x)dx + a g(x)dx,
� b
·
r
a f (x)dx,
�
f (x)dx.
c
a
Lecture 15. October 18, 2005
Homework. Problem Set 4 Part ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
/n
b
.
n
Amax =
n
�
k=1
yk,maxΔxk =
n
�
k=1
ekb/n
b
.
n
To evaluate each of the sums, make the substitution c = eb/n . Then the lower sum is,
Amin =
The sum is a geometric sum,
n
b �
n
k=1
c k−1 =
n−1
b �
n
l=0
l
c .
Plugging this in gives,
(1 + c + c 2 + · · · + c n−2 + c n−1) =
n
c − 1
.
c ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, this gives,
he − 1
lim
h 0 h
→
= 1.
Inverting gives,
h
lim
h→0 eh − 1
=
�
h
e − 1 �−1
lim
h 0 h
→
= (1)−1 = 1.
Also, because ex is continuous,
Putting this together gives,
lim e h = e 0 = 1.
h 0→
lim Amin = (e b − 1) lim
n→∞
h→0 eh
h
− 1
= (e b − 1)(1) = e b − 1.
Similarly,
lim Amax = (e b − 1)(... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
subinterval is,
Δxk = xk − xk−1 = q k − q k−1 = q k−1(q − 1).
Observe this increases as k increases. So this is not the partition of [1, b] into n equal subintervals.
The mesh size is,
mesh = max(Δx1, . . . , Δxn) = Δxn = (q − 1)b(n−1)/n ≤ q − 1.
As n tends to infinity, the mesh size tends to,
lim mesh = lim q − 1... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
k=1
Amax = (q − 1)q r
n
�
q(k−1)(r+1)
.
To evaluate the sum, make the substitution c = qr+1 . Then the sum is,
k=1
n
�
c k−1 = 1 + c + c + · · · + c n−2 + c
2
n−1
.
k=1
This geometric sum equals,
n
c − 1
c − 1
Thus the upper and lower sums simplify to,
=
qn(r+1) − 1
qr+1 − 1
.
Amin = (q − 1)(q n(r+1) ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
5
Also, since x is continuous,
r
Substituting this in gives,
lim q r = 1r = 1.
q→1
lim Amin = (br+1 − 1)
n→∞
lim Amax = (br+1 − 1)
n→∞
lim q r
q→1
�
qr+1 − 1 −1
�
lim
q→1 q − 1
� �
�
=
br+1
− 1
,
r + 1
qr+1 − 1 −1
�
lim
q→1 q − 1
=
br+1
− 1
,
r + 1
Since the limit of Amin and the limit of ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
is defined on all of [a, x]. If f (x) is continuous, the
Fundamental Theorem of Calculus asserts F (x) is differentiable and,
dF
dx
(x) =
x
�
d
dx a
f (t)dt = f (x).
The proof of the second part is very easy. Consider the increment in F from x to x + Δx,
F (x + Δx) − F (x) =
� x+Δx
�
f (t)dt −
x
f (t)dt =
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
�x.
Thus the Riemann sum is squeezed between,
k=1
yminΔx ≤
n
�
y
k=1
∗Δ ≤x
k
k
y
maxΔx.
Because the Riemann integral is a limit of Riemann sums, it is also squeezed,
yminΔx ≤
� x+Δx
x
f (t)dt ≤ ymaxΔx.
Substituting in F (x + Δx) − F (x) and dividing each term by Δx gives,
ymin ≤
F (x + Δx) − F (x)
Δx
≤ ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x) is a known antiderivative of f (x), then,
To see this, observe that,
� b
a
f (t)dt = G(b) − G(a).
�
x
f (t)dt,
F (x) =
a
is also an antiderivative of f (t) by the Fundamental Theorem of Calculus. Thus, since the general
antiderivative is G(x) + C, there is a constant C such that F (x) = G(x) + C. But als... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
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