text stringlengths 30 4k | source stringlengths 60 201 |
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lying both sides by 2z gives,
1
2
(ey − e−y ) = x.
1
2
(z − z−1) = x.
2 z − 1 = 2xz ⇔ z 2 − 2xz − 1 = 0.
Completing the square gives,
Taking square roots gives,
(z − x)2 = x 2 + 1.
z = x ±
√
x2 + 1.
92
18.01 Calculus
Jason Starr
Fall 2005
Since z equals ey , z is positive. Thus, the correct square root... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
bolic functions
x2 − 1) equals cosh−1(x) + C.
allow us to compute more antiderivatives than before, e.g., dx/(
Essentially this comes down to making a direct substitution of an inverse function, e.g., u =
cosh−1(x). However, this is logically equivalent to making an inverse substition, x = cosh(u).
When the integra... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
gives,
� √
a2 − x2dx = (1/
2)(
a2 sin−1(x/a
) +
x
√
a2 − x2
) +
C.
4. Three different kinds of integrals, three kinds of inverse substitution. The type of
antiderivative where inverse substitution is most successful has the form,
�
√
√
F (x,
G(x,
Ax2 + Bx + C)
Ax2 + Bx + C)
dx,
where A, B and C are constants,... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
II(u,
GII(u,
u2 − a2)
u2 − a2)
du,
√
√
FIII(u,
GIII(u,
a2 + u2)
a2 + u2)
du.
For each of these types, there are 3 possible inverse substitutions: trigonometric, hyperbolic and
rational. A flow chart of the 9 possible outcomes will be posted on the course webpage. Here are
a couple of examples. In each example,... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
h2(t)
�
a2sech2(t)
asech2(t)dt.
Simplifying gives,
�
2
a
tanh2(t)sech(t)dt = a
2
�
sinh2(t)
cosh3(t)
dt.
Jason Starr
Fall 2005
This can be simplified a bit by multiplying numerator and denominator by cosh(t) and then ex
pressing in terms of sinh(t) as much as possible,
�
�
2 a
sinh2(t)
cosh4(t)
cosh... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ifies to,
�
8a 2
t2
(1 + t2)3
dt = 8a 2
�
1
(1 + t2)2
dt − 8a 2
�
1
(1 + t2)3
dt.
Notice, these two integrals are the same type that occurred with inverse hyperbolic substitution.
But they came up more quickly: rational inverse substitution is more efficient than inverse hyper
bolic substitution for this p... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(1 − u2)2
du.
This can be computed using partial fractions (not yet discussed).
Alternatively, the hyperbolic inverse substitution is,
x = a cosh(t), dx = a sinh(t)dt.
The new antiderivative is,
�
�
a2 cosh2(t)
a2 cosh2(t) − a
a sinh(t)dt.
2
Since cosh2(t) − 1 equals sinh2(t), simplifying gives,
�
2 a
cosh2... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a(1 − t2)
dt.
a(1 − t2)
2t2
This simplifies to,
This evaluates to,
�2
a
−
4
(1 + t2)2
t3
dt =
2
a
−
4
�
1
t3
+ + tdt.
2
t
−
2a
4
�
−1
2t2
+ 2 ln(t) +
�
t
2
+ C.
This is clearly the easiest of the 3 methods for computing the antiderivative, for this problem.
However, there still remains the formid... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Complete the square of the expression Ax2 + Bx + C, inside the
radical. In the example,
x − 2ax + 2a 2 = (x − a)2 + a .
2
2
Step 2. Make a linear change of coordinates. Make a linear change of coordinates to simplify
the quadratic term to one of the 3 types: a2 − x , x − a , or x2 + a . In the example, this means ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
18.01 Calculus
Jason Starr
Fall 2005
This simplifies to,
�
2 a
(tan 2(θ) + 2 tan(θ) + 1) sec(θ)dθ.
This can be written as a sum of 3 terms,
�
2
a
2
tan (θ) sec(θ)dθ + 2a 2
�
sec(θ) tan(θ)dθ + a 2
�
sec(θ)dθ.
Step 4. Compute the new antiderivative. If this were only as simple as it sounds, how much
easier... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
terms,
�
2
a
2
tan (θ) sec(θ)dθ + 2a 2
�
sec(θ) tan(θ)dθ + a 2
�
sec(θ)dθ.
The last antiderivative was actually Problem 3(b) from Part II of Problem Set 4. It turns out to
be,
�
2 a
sec(θ)dθ = a 2 ln(u +
√
u2 + a2) + C = a2 ln(x − a +
�
√
x2 − 2ax
a2
+ 2 ) +
C.
The middle antiderivative is simply the d... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(θ),
The new antiderivative is,
z = sin(θ), dz = cos(θ)dθ.
�
2z
(1 − z2)2
dz.
How do we compute this antiderivative? That is the topic of partial fractions.
Remark: In lecture the solution was done a bit differently. This led to a slightly different an
tiderivative,
�
1
(1 − z2)2
dz.
Notice the difference of ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
.03. Both polynomials and partial fractions are (relatively) easy
to antidifferentiate. The antiderivative of a polynomial is,
�
q(x)dx =
an
+1)
(n
xn+1 + an−1
n xn + + a1
· · ·
2 x2 + a0x + C.
101
18.01 Calculus
Jason Starr
Fall 2005
The antiderivative of the first kind of partial fraction is,
�
A
(
x −... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
rational expressions: division and factoring Many rational expressions that
come up are not of the simple kinds above. The goal is to express an arbitrary rational expression
as the sum of a polynomial and partial fractions. The first step is polynomial division. Given a
fraction F (x)/G(x), apply polynomial division... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
are nonnegative integers and m1, . . . , mk , n1, . . . , nl are positive integers. Also,
a1, . . . , ak , α1, . . . , αl, and β1, . . . , βl are real numbers. The last l factors were not discussed in
lecture until the end of lecture. Although they are important, they do not often come up in this
course.
The Fundam... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)m1
· · · · · (x − ak )mk
r(x)
· ((x − α1)2 + b2
1)n1
· · · · · ((x − αl)2 + b2
l )nl
,
where q(x) is a polynomial, the degree of r(x) is less than the degree of H(x), and r(x) has no
common factor with H(x). This can be further simplified using partial fraction decomposition. It
is a fact that every rational ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
real constants. This sum of partial fractions is called the
partial fraction decomposition of r(x)/H(x). The difficulty is precisely to find the constants Ci,j ,
Di,j , and Ei,j .
103
18.01 Calculus
Jason Starr
Fall 2005
One approach, which always works but is quite inefficient, is simply to multiply all terms by th... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
=
⇔
1
.
2
Thus B = −A = −1/2. So the partial fraction decomposition is,
1
1 − x2
=
1
2
1
x+1 + −1
2
1
x−1 .
The Heaviside coverup method. The Heaviside coverup method is a method for deter
5.
mining many of the coefficients Ci,j . For each highest power of a linear factor occuring in H(x),
say (x − ai... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
− 1)2
+
C1,1
z + 1
+
C2,1
.
z − 1
Using the Heaviside coverup method,
C1,2 =
2
�
z
�
�
�
(z − 1)2
z=−1
=
(−1)2
(−2)2
=
1
.
4
Also,
C2,2 =
�
�
2z
�
�
(z + 1)2
z=+1
=
(+1)2
(+2)2
=
1
.
4
Thus the partial fraction decomposition is,
z2
(1 − z2)2
= 1
4
1
+1)
2 + 1
4
(z
1
(z−1)2 +... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
+
z + 1
−C1,1 .
z − 1
Finally, plug in z = 0 to get,
0 =
1
1
4 (+1)2
+
1
1
4 (−1)2 +1
C1,1 +
+
−
C1,1
−
1
= + 2C1,1.
1
2
Solving gives C1,1 = 1/4. Finally this gives the full partial fraction decomposition,
−
z2
(1 − z2)2
= (1/
4)(1
/(z
2
+ 1) + 1
/(z − 1)2 − 1/(z
+ 1) + 1
/(z −
1)).
105
18.01 C... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
is,
d(uv) = udv + vdu.
udv = d(uv) − vdu.
This gives a very useful antidifferentiation formula,
�
�
udv = uv −
vdu.
This formula is integration by parts.
Example. Compute the antiderivative of,
�
x cos(x)dx.
Set u to be x and dv to be cos(x)dx. Then u, v, du and dv are,
u = x
du = dx
dv = cos(x)dx,
v = s... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= x
du = dx/x,
Using integration by parts,
�
�
udv = uv −
�
ln(x)dx = x ln(x) −
dx.
vdu,
�
The new integral is easy to evaluate. Altogether this gives,
�
ln(x)dx =
x ln(x) − x + C.
Notice this example does not follow the general rule. The integral v = x is strictly more complicated
than dv = dx. However... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
of integration by parts. It sometimes happens that integration by parts gives the
induction step to solve infinitely many integrals. In this case, the formula given by integration by
parts is called a reduction formula.
Example. Use integration by parts to give a reduction formula for,
�
[ln(x)]ndx.
Now there is m... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
v = uv −
vdu,
�
tn e tdt = tnet − n �
tn−1etdt.
Notice how similar this answer was to the answer of the previous example. The connection comes
from the inverse substitution,
so that,
x = e , dx = e tdt,
t
�
[ln(x)]ndx =
�
tn e tdt.
3. Advanced reduction formulas. Sometimes a reduction formula can only be ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
for. However, bringing like terms to one side of the equation gives,
�
�
�
[sin(x)]ndx + (n − 1)
[sin(x)]n = −[sin(x)]n−1 cos(x) + (n − 1)
[sin(x)]n−2dx.
Cleaning this up a bit gives the reduction formula,
�
[sin(x)]ndx = −[sin(x)]n−1 cos(x)/n
+ (
n − 1)/n �
[sin(x)]n−2dx.
Lecture 28. December 1, 2005
Homewo... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
limit failed to give any
useful information. The reason is that the general formula,
lim[g(x) + h(x)] = lim g(x) − lim h(x),
x→a
x→a
x→a
only holds if all three limits are defined, which they are not in our case.
Of course F (x) is simply the constant function with value b. Therefore,
lim F (x) = lim b =
→
x 0
→... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
) is continuous on [a, b]. Since f (x) and g(x)
are differentiable on (a, b), also F (x) is differentiable on (a, b). Moreover,
F (a) = F (b) = 0.
Thus, by the Mean Value Theorem, there exists a value c strictly between a and b such that
F �(c) = 0. By a straightforward computation,
F �(c) = (f (b) − f (a))g�(c) − (... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x) − g(a)) by f �(c)/g�(c). Then x
approaches a as c approaches a.
111
18.01 Calculus
Jason Starr
Fall 2005
3. L’Hospital’s rule. The most important case of the proposition is L’Hospital’s rule. This is
exactly the case when f (a) = g(a) = 0. In this case, a naive computation would give,
f (x)
lim
x a+ g(x)
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x→2 2x − 1
=
12 · 4
2 · 2 − 1
=
48
3
=
16.
= lim sin(x)2x = lim cos(x)2 =
1/2.
x 0
→
x 0
→
4. L’Hospital’s rule for other indeterminate forms. L’Hospital’s rule can be used to compute
limits that naively lead to indeterminate forms other than 0/0. For instance, if
then the naive computation gives,
lim f (x... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
→a+ g�(x)
� �
·
�
f (x) −2
lim
x a+ g(x)
→
.
limx→a+ f (x)/g(x
) = lim
x→a+ f �(x)/g�(x),
if both limits are defined and nonzero. In fact, a better result is true (with a more subtle proof): if
the second limit is defined, then the first limit is defined and the 2 are equal (whether or not they
are zero).
Example. ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
most often is that for a piecewise continuous function f (x) on a bounded interval
[a, b], the Riemann integral,
� b
f (x)dx,
exists (and equals a finite number). What if the interval is unbounded, e.g., [a, ∞)? Quite simply,
the Riemann integral is not defined. This isn’t a problem with our methods for computing in... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
new definition. It is not a theorem about Riemann integrals.
Example. Let p > 1 be a real number. Then for every t > 1, the integral,
�
t 1
xp
1
dx,
exists and equals,
�
�
�
1
�
�
(p − 1)xp−1
t
1
−
=
1
p − 1
−
1
(p − 1)tp−1
.
Since p is greater than 1, the limit,
exists and equals 0. Therefore,
exists... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
with the Riemann
integral. Let [a, b] be a bounded interval. Let f (x) be a function that is bounded on [t, b] for every
a < t < b, but which is unbounded on [a, b]. According to the definition of the Riemann integral,
is not defined. However, it may happen that for every a < t < b, the integral,
� b
a
f (x)dx,
is... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the Riemann integral,
is defined equals,
Since 0 < p < 1, the limit,
exists and equals 0. Therefore,
� 1 1
xp
t
dx,
1 − t1−p
.
1 − p
lim t1−p,
0→
t
� 1 1
xp
lim
t→
t
0
dx,
exists and equals 1/(1 − p). Therefore the improper integral,
exists and its value is,
� 1 1
xp
0+
dx,
� 1 1
xp
0+
dx =
1/(1... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
F (x) exists,
→
and if F (x) is monotone decreasing and bounded above, then limx a+ F (x) exists.
→
→
→
These give the following tests for convergence of an improper integral.
Squeezing lemma. If f (x) ≤ g(x) ≤ h(x) on the interval [a, ∞), and if the improper integrals,
�
∞
a
f (x)dx and
�
∞
a
h(x)dx,
exist and... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, n = 1, 2, . . .
The most common notation for a sequence is (an)n≥1.
117
18.01 Calculus
Jason Starr
Fall 2005
A sequence (an)n≥1 converges to a limit L if the sequence becomes arbitrarily close to L, and stays
arbitrarily close to L. More precisely, the sequence converges to L if for every positive number �,
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
0, 1, 2, . . . converges to 0 if |r| < 1 and
diverges if |r| > 1. There are 2 remaining cases. If r = −1, then an = (−1)n diverges. If
r = 1, then an = 1 converges to 1.
2. Tests for convergence/divergence. One useful test for convergence is the Squeezing Lemma.
The squeezing lemma. Let (an)n≥1, (bn)n≥1 and (cn)n≥1... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
. A sequence (an) is bounded below if there exists a real number l such that for every index
n, an ≥ l. The number l is a lower bound for the sequence.
118
18.01 Calculus
Jason Starr
Fall 2005
Monotone Convergence Test. A nondecreasing sequence converges if and only if it is bounded
above. In this case, the li... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
verges absolutely. Although it is not obvious, if the
series converges absolutely, then the series converges (this is a basic theorem from course 18.100).
If a series converges but does not converge absolutely, sometimes it is said the series converges
conditionally.
k ak
Examples. The harmonic sequence is the seq... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the
alternating harmonic series to make the partial sum bigger than, say, 106 . Now add only the first
odd term −1/2. This has a negligible effect. Now add a large number of the remaining even terms
to make the partial sum bigger than 107 . Now add one more odd term, −1/3. Continuing in this
way, eventually every ter... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
n
�
|an|
= bn − bn−1 = (bn − L) − (bn−1 − L)
|
|
|
| ≤ |
bn − L + bn−1 − L < �/2 + �/2 = �.
|
|
|
Thus the sequence (an) converges to 0. Contrapositively, if the sequence (an) does not converge
n an diverges. This is the most basic test for divergence of a series. For
to 0, then the series
example, it immediately... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
|
converges absolutely.
n=1 an
| ≤ |
bn|.
�
n=1
n
an = r , n ≥ 0,
120
18.01 Calculus
Jason Starr
Fall 2005
(by convention, if r = 0, the first term a0 is defined to be 1). By high school algebra, if r = 1, the
partial sums are
bn = 1 + r + · · · + r n =
1 − rn+1
1 − r
Observe this sequence depends on n o... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a given sequence (an)n≥∞ to a
The ratio test.
geometric sequence (rn)n≥1. If the following limit,
�
�
an+1
�
lim
�
n→∞ an
can be compared to a sequence (Crn)n≥1 for some
exists, call it r. Then the sequence (an)n≥1
an converges absolutely if the sequence
choice of C. This leads to the ratio test : The series ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
get useful
information about a series. Let (an)n≥1 be a sequence. Let f (x) ≥ 0 be a function on [1, ∞) such
that for every integer n, f (x) ≥ an
for all n ≤ x ≤ n + 1. If the improper integral,
�
converges, then the series
function on [1, ∞) such that for every integer n, g(x) ≤ an
integral,
∞
n=1
an converges ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, n + 1]. By the Fundamental Theorem of Calculus, the partial sums of the
sequence (cn) are,
n
�
k=1
n
�
1
1 x
ck =
dx = ln(n).
As n tends to ∞, the natural logarithms ln(n) also tend to ∞ (although very slowly – ln(n) does
not get bigger than a fixed real number R until n gets bigger than the much larger number... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
1). Therefore, by the comparison test, the series,
∞ 1
�
,
ns
n=1
122
18.01 Calculus
Jason Starr
Fall 2005
converges absolutely to a value bounded by 1/(s − 1). The value of this limit is called the Riemann
zeta function at s, denoted
ζ(s) :=
∞
� 1
.
ns
n=1
This function is of fundamental importance i... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
. Given a power series, for which real numbers x does the corresponding series absolutely
converge?
Examples. 1. Consider the power series,
0 + 11 x 1 + 22 x 2 + 33 x
3 + · · ·
=
∞
�
n n
n x .
Of course this converges to 0 for x = 0. But for any x other than 0, the sequence nnxn = (nx)n
diverges. Therefore the... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
cn(x − a)n be a power series about x = a. Exactly one of the following hold.
�
n=0
(i) For every x different from a, the series does not converge absolutely.
(ii) There exists a real number R such that the series converges absolutely if x − a < R and
|
|
does not converge absolutely if x − a > R.
|
|
(iii) For every... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
function f (x) is differentiable and the derivative has a
power series converging absolutely with the same radius R,
f �(x) =
∞
�
cnn(x − a)n−1 =
∞
�
(m + 1)cm+1(x − a)m .
n=0
m=0
We can iterate the theorem, i.e., f �(x) is differentiable and f ��(x) has a power series converging
absolutely with radius R. Iterat... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
) + c2(a − a)2
+ · · ·
= c0 + 0 + 0 + · · · = c0.
Thus the first coefficient of the power series is simply,
Moreover, from the power series for the kth derivative,
c0 = f (a).
f (k)(a) = k!ck + (k + 1)!/1!ck+1(a − a) + (k + 2)!/2!ck+2(a − a)2
+ · · ·
= k!ck + 0 + 0 + · · · = k!ck .
Solving for ck , the kth coefficien... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x). This reduces the original
question to 2 new questions. Does the Taylor series have a positive radius of convergence? If so,
does the analytic function defined in this way equal the original function f (x)?
125
18.01 Calculus
Jason Starr
Fall 2005
The radius of convergence question is precisely the radius of ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(1 − x)−1 .
f �(x) = −(1 − x)−2(−1) = (1 − x)−2 .
f ��(x) = (−2)(1 − x)−3(−1) = (1 − x)−3 .
This begins to suggest a pattern: The kth derivative of f (x) will be,
f (k)(x) = bk (1 − x)−k−1 ,
for some real number bk . Having made this guess, it is easy to verify by induction. By computation,
the result is true for... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= n(n−1)(n−2)·· · ··
3·2·
1.
This number has come up before in this class. It is the nth factorial number,
This gives the final formula for the nth derivative of f (x),
cn = n!.
f (n)(x) =
n!(1 − x)−n−1 .
Step 2. Substitute x = a into the derivatives. Compared to the work of finding the derivatives,
this is very si... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
centered at x = 0 not containing
x = 1 is the interval (−1, 1). This interval has radius R = 1.
4. More examples. What is the Taylor series expansion for (1 − x)−1 about a point x = a
different from x = 1? The fortunate fact is that Step 1 allows to compute the derivatives f (n)(a)
127
18.01 Calculus
Jason Starr
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
only if this limit has absolute
value less than 1,
(1 − a)−1(x − a) ≤ 1.
|
|
Rearranging, the series converges if and only if,
Thus the radius of convergence is,
|
x − a
| ≤ |
|
.
1 − a
R =
|1 − a|.
This is perfectly reasonable. The function (1 − x)−1 has a vertical asymptote at x = 1. Therefore,
the power seri... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
It has certain advantages to our original
definition of ex . Importantly, it is easy for a computer to determine ex to very high precision using
this formula.
x
Example 3. Having computed the Taylor series expansion for ex about x = 0, the next question
is to compute the Taylor series expansion for ex about x = a. A... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
f �(x) =
f ��(x)
=
f (3)(x)
=
f (n+4)(x)
=
sin(x),
cos(x),
− sin(x),
− cos(x),
f (n)(x)
Together, these give all the derivatives of f (x). Write n = 4l, 4l + 1, 4l + 2 or 4l + 3 for some
nonnegative integer l. Then the rules above give,
f (n)(x) =
⎧
⎪⎪⎨
⎪⎪⎩
n = 4l,
sin(x)
cos(x) n = 4l + 1,
− sin(x)... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x
2m+3
/(2m + 3)!]/[(−1)m x 2m+1/(2m + 1)!] = −x /(4m 2 + 8m + 3).
2
This sequence converges to 0. Therefore, by the ratio test, the power series converges absolutely to
sin(x) for every choice of x,
sin(x) = �∞
m=0(−1)m/(2m + 1)!x2m+1 .
130
18.01 Calculus
Jason Starr
Fall 2005
There is an exactly similar form... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
addition formulas,
sin(x) = sin(a + (x − a)) = cos(a) sin(x − a) + sin(a) cos(x − a),
cos(x) = cos(a + (x − a)) = cos(a) cos(x − a) − sin(a) sin(x − a).
This gives the Taylor series expansions,
sin(x) = �∞
m=0
(−1)m cos(a)
(2m+1)! (x − a)2m+1 + �∞
m=0
(−1)m sin(a)
(2m)!
(x − a)2m ,
cos(x) = �∞
m=0
(−1)m c... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
to write down a power series expansion
for f (x). First of all, the Taylor series for e−t2 about t = 0 is obtained by substituting x = −t2 in
the Taylor series for ex about x = 0,
e−t2 =
∞
�
(−1)nt2n/n!.
n=0
Because this series converges absolutely, the integral of the series is the series of the termbyterm
in... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)(a)
n!
n=0
(x − a)n + RN,a(x).
The precise version of the questions above is, what bounds exist for RN,a(x)?
To understand the answer, consider the simplest case where N = 0. Then the remainder term is
simply,
R0,a(x) = f (x) − f (a).
By the Mean Value Theorem, for every x there exists a real number c (dependin... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x
if we want to compute e for x in the interval (−1, 1), then M equals e. To make the remainder
term less than 10−10, it suffices to take N = 12.
3. Review problems. Each of the following problems was discussed in lecture. Here are the
problems and answers, without the discussion.
Problem 1. Let a and b be positive r... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the solid obtained by revolving about the xaxis the region
in the first quadrant bounded by the curve y = x2 and the curve x = y .
2
The volume of this solid is,
Volume =
3π/10.
Problem 4. Using a trigonometric substitution and a trigonometric identity, compute the an
tiderivative,
√
�
2
1 − x
2x
dx.
The antid... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Lecture 16
8.324 Relativistic Quantum Field Theory II
Fall 2010
8.324 Relativistic Quantum Field Theory II
MIT OpenCourseWare Lecture Notes
Hong Liu, Fall 2010
Lecture 16
Firstly, we summarize the results of the vertex correction from the previous lecture:
Γµ
1 (k1, k2) ≡
q
k2 + l
µ
k1 + l
l
k2
k1
= γµA(... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
)
(3)
The calculation then proceeds by imposing an infrared cut-off λ on the photon momenta. The divergences in the
λ −→ 0 limit cancel among virtual and real soft photon emissions, and we can safely take the λ −→ 0 limit in the
end.
3.4: VACUUM POLARIZATION
We will now evaluate the one-loop correction to the phot... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
with p ≡ q + xk, D ≡ x(1 − x)k2 + m2 − iϵ, and
4N µν
[
]
= tr γµ(ik/ + i/q + m)γν (i/q + m)
[
γµ(i/p + i(1 − x)k/ + m)γν (i
= tr
[
= −tr
]
]
p/ − ixk/ + m)
γµpγ/ ν p/
+ m 2tr [γµγν ] + x(1 − x)tr [γµ /
+terms linear in p+terms with an odd number of γmatrices.
kγν /
k]
We note that the trace of a term with an odd ... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
(8)
,
where we have set A ≡ m2 − x(1 − x)k2 . We note that the first and third terms in the numerator are logarithmically
divergent, and the second term is quadratically divergent. We now apply a Wick rotation p0
iddpE
2
and p
→ E
ipd , dd
p →
→
2
pE . We recall that
ˆ
ddpE
(2π)d (pE
(p2 )a
E
2 + D)b
=
Γ(... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
− 1)k2P µν
T
− i(Z3 − 1)k2P µν
T .
2
Lecture 16
8.324 Relativistic Quantum Field Theory II
Fall 2010
Thirdly, we use dimensional regularization, setting d = 4 − ϵ, e → eµ 2 ,
(
ϵ
iΠµν (k) =
dx x(1 − x)
− γ + log
−8ie2k2P µν
16π2
T (k)
ˆ 1
0
[
2
ϵ
)]
µν
− i(Z3 − 1)k2P ,
T
4πµ2
D
and so
with
Π... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
charged particle, we must add their respective contributions, and the smallest m
contributes most.
The internal propagator is given by
2Dµν (q)
e
→
=
ηµν
−ie2
q2 − iϵ 1 − Π(q2)
−ie2ηµν
q2 − iϵ
(1 + Π(q 2) + . . .).
We see that for q2 ≫ m2 , a large spacelike momentum, from (17) we have
Π(q 2) ≈
=
2
e
2
... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
associated with the lightest intermediate particle.
S(1, 2 −→ 1′ , 2′) = (−ie)2 u¯1′ (p′
1)γµu1(p1)Dµν (q)¯u2′ (p′
2)γν u2(p2),
where
Dµν (q) =
−i
P T
µν (q)
q2 − iϵ 1 − Π(q2)
+ DL
µν (q),
(20)
qν
, q = p′
1
µν = ηµν − qµ
q2
and P T
p2)u2(p2) = 0. We now
want to consider corrections to the Coulomb pot... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
⃗q2
=
d3⃗r e−i⃗q.⃗rV0(⃗r),
(21)
(22)
(23)
where V0(⃗r) =
1e
e
2
|⃗r
|
π
4
e2 Π(⃗q2), where, from (17)
is the Coulomb potential. The one-loop correction is given by e1
q⃗
2
Π(⃗q 2) =
2
e
2π2
ˆ 1
0
(
dx x(1 − x) log
1 +
⃗2q
m2
x(1 − x)
)
,
and the term provides a correction to the Coulomb potential
δV ... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
Lecture 16
8.324 Relativistic Quantum Field Theory II
Fall 2010
Figure 1: Clockwise contour for the integral in dz in (26). The semi-circular arc is taken to infinity and gives a
vanishing contribution.
If we now let z ≡ qr, a ≡ √
mr
x(1−x)
≥ 2mr, the result reduces to
The integral in dz, I
giving
=
´
∞ dz e ... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
δV (r) =
e1e2 e2 e−2mr
4π 16π 2 (mr) 2
3
3
+ . . . .
(27)
Z(mr) increases with decreasing r.
off at mr = e ≪ 1, we find
and thus
where
As r −→ 0, I(r) −→ ∞. If we consider putting a short-distance cut
Z(ϵ) =
e2
6π2
log
1
ϵ
+ . . . ,
V (r) =
=
=
(1 + Z(mr))
V0 + δV (r)
e1e2
4πr
˜e1(r)˜e2(r)
,
4πr ... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/4011862a0e45c9328c9eae315bab4517_MIT8_324F10_Lecture16.pdf |
Turbulent Flow and Transport
6
6.1
Introduction to Turbulent Boundary Layers
The nature of flow in turbulent boundary layers. Inner and outer regions, eddy
diffusivity distributions, intermittency, etc.
6.2
Integral form of the mean flow boundary layer equations.
6.3
6.4
6.5
Reasons for why the turbulent bou... | https://ocw.mit.edu/courses/2-27-turbulent-flow-and-transport-spring-2002/401aed17397ef5afafa60b237670a4ca_Boundary_layers.pdf |
): 87−103.
•For drag reduction by riblets, see for example: Walsh & Lindemann, AIAA−84−0347;
Gallagher & Thomas, AIAA−84−2185; Bechert et al, AIAA−85−0546. | https://ocw.mit.edu/courses/2-27-turbulent-flow-and-transport-spring-2002/401aed17397ef5afafa60b237670a4ca_Boundary_layers.pdf |
Welcome to ...
2.717J/MAS.857J
Optical Engineering
MIT 2.717J
wk1-b p-1
This class is about
• Statistical Optics
– models of random optical fields, their propagation and statistical
–
properties (i.e. coherence)
imaging methods based on statistical properties of light: coherence
imaging, coherence tomography
... | https://ocw.mit.edu/courses/2-717j-optical-engineering-spring-2002/4020129e21b47c48dbf70c986d0f9508_wk1_b.pdf |
• You have taken a class elsewhere that covered Geometrical Optics,
Diffraction, and Fourier Optics
• Some background in probability & statistics is helpful but not
necessary
MIT 2.717J
wk1-b p-6
Syllabus (summary)
• Review of Fourier Optics, probability & statistics 4 weeks
• Light statistics and theory of coher... | https://ocw.mit.edu/courses/2-717j-optical-engineering-spring-2002/4020129e21b47c48dbf70c986d0f9508_wk1_b.pdf |
– Wednesdays 2-3pm (examples and discussion)
– presentations only: Wednesdays 7pm-??, pizza served
MIT 2.717J
wk1-b p-9
The 4F system
1f
1f
2f
2f
( , )
yx
g 1
object plane
MIT 2.717J
wk1-b p-10
′′
′′
x
y
G 1
,
λf 1
λf 1
Fourier plane
g
1
−
f 1 y ′
f
x
′ −
,
f 1
f 2
Image plane
2
... | https://ocw.mit.edu/courses/2-717j-optical-engineering-spring-2002/4020129e21b47c48dbf70c986d0f9508_wk1_b.pdf |
f
θx
)
(
v u
,
G
1
( , )
x
y
g 1
object plane
MIT 2.717J
wk1-b p-12
G 1
′′
x
λf 1
′′
y
,
λf 1
×
′′
r
circ
R
(
g
1
)
∗ h −
f 1
f
2
′
−
,
x
f
′
1 y
f
2
Fourier plane: aperture-limited
Image plane: blurred
(i.e. low-pass filtered)
The 4F system with ... | https://ocw.mit.edu/courses/2-717j-optical-engineering-spring-2002/4020129e21b47c48dbf70c986d0f9508_wk1_b.pdf |
herent transfer function
(FT of intensity in ⇒ FT of intensity out)
(
x h
, )
y
H ( v u ) = FT{
,
( ,
x h
)}
y
2
~
( ,
) =
( , y
)
x h
y
x h
H ( v u ) = FT{
)}
~
~
( ,
x h
y
,
= H ( v u )⊗ H ( v u )
,
,
~
H ( v u )
,
:
Modulation
~
H ( v u )
,
:
Optical
Transfer
Function
(MTF)
Transfer
Fun... | https://ocw.mit.edu/courses/2-717j-optical-engineering-spring-2002/4020129e21b47c48dbf70c986d0f9508_wk1_b.pdf |
delay to the impulse response
(
h aberrated
) = h
y
)e
y
diffractio
n
aberration
x
,
x
,
i ϕ
(
(
,
x
)
y
Effect of aberrations
on the MTF
limited
( )~
u H
1
unaberrated
diffraction
(
limited
)
c2u−
MIT 2.717J
wk1-b p-18
aberrated
u
c2u | https://ocw.mit.edu/courses/2-717j-optical-engineering-spring-2002/4020129e21b47c48dbf70c986d0f9508_wk1_b.pdf |
9
Fourier Transform
Properties
The Fourier transform is a major cornerstone in the analysis and representa-
tion of signals and linear, time-invariant systems, and its elegance and impor-
tance cannot be overemphasized. Much of its usefulness stems directly from
the properties of the Fourier transform, which we discuss... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
domain of a linear contraction (expan-
sion). In other words, linear scaling in time is reflected in an inverse scaling in
frequency. As we discuss and demonstrate in the lecture, we are all likely to
be somewhat familiar with this property from the shift in frequencies that oc-
curs when we slow down or speed up a tap... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
procedure for implementing or
calculating the Fourier transform of a signal, then exactly the same procedure
with only minor modification can be used to implement the inverse Fourier
transform. This is in fact very heavily exploited in discrete-time signal analy-
sis and processing, where explicit computation of the Fo... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
cessing systems are the convolution and modulation properties. According to
the convolution property, the Fourier transform maps convolution to multi-
plication; that is, the Fourier transform of the convolution of two time func-
tions is the product of their corresponding Fourier transforms. For the analy-
sis of line... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
and the foundation for many modern signal processing systems
using digital and other discrete-time technologies.
We will spend several lectures exploring further the ideas of filtering,
modulation, and sampling. Before doing so, however, we will first develop in
Lectures 10 and 11 the ideas of Fourier series and the Fo... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
+->.
X(W)
X(w) = Re IX(w) + j
Im
(j)[
= IX(eo)ej x("
PROPERTIES OF THE FOURIER TRANSFORM
TRANSPARENCY
9.2
Symmetry properties
of the Fourier
transform.
Symmetry:
X(t)
X(j)
x(t) real
=> X(-w) = X*()
Re X(o)
= Re X(-o)
IX(co)I
=
IX(-w)I
Im X(o)
=
-Im X(-4)
'4X(o)
even
odd
Fourier Transform Properties
9-5
Example 4.7... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
duality property.
Example 4.11:
2sin wt
2 2rt
Example 4.10:
X(W)
w
w
X((o)
2
co
TRANSPARENCY
9.7
Parseval's relation for
the Fourier transform
and the Fourier series.
Parseval's relation:
+00
f
00
Ix(t)| 2dt
if
0O
I'(t)I12dt
IX(c) |2 do
+00)
f-00
1
27r
+00
lak12
k=- 00
Signals and Systems
9-8
Time shifting:
TRANSPA... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
x(t)
H(w)
H(w) X(W)
e*Jcot H(w o )
H(w) = frequency response
FILTERING
Ideal lowpass filter:
-I
Co Co
Differentiator:
y(t) =t
dx(t)
=> H(co) =
jco
IH(M)I
Oppo
Signals and Systems
9-10
DEMONSTRATION
9.2
Lowpass filtering of an
image.
DEMONSTRATION
9.3
Effect of
differentiating an
image.
Fourier Transform Properties
... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/403577752d3007ed4ea30e8c61cf90d7_MITRES_6_007S11_lec09.pdf |
6.851: Advanced Data Structures
Spring 2012
Lecture 3 — February 23, 2012
Prof. Erik Demaine
1 Overview
In the last lecture we saw the concepts of persistence and retroactivity as well as several data
structures implementing these ideas.
In this lecture we are looking at data structures to solve the geometric pr... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
case, when we query a
point (x, y) we are interested in the first line segment lying above it. Equivalently, if we imagine
shooting a vertical ray from the query point, we want to return the first map segment that it
intersects (see Figure 2).
We can use vertical ray shooting to solve the planar point location proble... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
dimensional problem will allow us to solve the d-dimensional problem.
In the case of vertical ray shooting, in one dimension we can solve the problem with a balanced
binary search tree. More specifically, we are doing successor queries. In the last lecture we saw how
to make a partially persistent balanced binary s... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
. See [4] and [10].
Several variants of the vertical ray shooting problem are still open. Examples include:
• OPEN: Can we do O(log n) dynamic vertical ray shooting in a general planar graph?
• OPEN: Can we do O(log n) static ray shooting when the rays do not have to be vertical?
Note that the three dimensional ver... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
add or delete any more
segments, it would have to involve two segments that are currently adjacent in the tree order. Then
for each segment, we track when it would cross its successor in the tree and each internal node
tracks the earliest crossing in its subtree. It is not difficult to maintain this extra data in O(lo... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
same efficiency. Once exception
is the last question - for a given input, the answer may involve returning every point in the set,
which would take O(n) time. We’re therefore looking for a solution something like O(log n + k),
where k is the size of the output.
3.1 Range Trees
Let’s start with the 1-dimensional case... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
(refer to Figure 5). As we search, we’ll move down the
tree and branch at a number of points. (As before, finding pred(a) and succ(b) takes O(log n)
time.) Once the paths to pred(a) and succ(b) diverge, any time we turn left at a node while
Figure 5: Branching path to pred(a) and succ(b)
searching for pred(a), we kn... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
1 and b1.
We now want to sort the points by y coordinate. We can’t use a range tree over all the points,
because we’re only interested in the points which we know are between a1 and b1, rather than every
point in the set. We can restrict our attention to these points by creating, for each x subtree, to
a correspond... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
y coordinate. We’re
searching through O(log n) y subtrees, so the query runs in O(log2 n) time.
6
We can extend this idea to d dimensions, storing a y subtree for each x subtree, a z subtree for each
y subtree, and so on. This results in a query time of O(logd n). Each dimension adds a factor of
O(log n) in spac... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
Figure 8: Storing arrays instead of range trees
We can do the same thing as before, taking O(log n) time to search through each array by y; but
we can also do better: we can search through just one array in y, the array corresponding to the
root of the tree, which contains every element, sorted by y.
We want to mai... | https://ocw.mit.edu/courses/6-851-advanced-data-structures-spring-2012/405bf317ea6dffc96e77badf1833f56f_MIT6_851S12_L3.pdf |
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