Progressions & series
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Terminology
A sequence is an ordered, enumerated collection of objects. Elements can be
repeated. Sequences can be finite or infinite.
A progression is a sequence such that there is a formula to derive each element
in terms of its index (position in the sequence).
A series is the sum of a progression.
Summation notation
i — the index of summation, the ith term in a sequence. Use j to distinguish
nested or separate loops.
ai – represents each term of the sum
m – the lower bound of summation, first value of i
n – the upper bound of summation, inclusive final value of i
n
∑ ai
i=m
⚠️
The variables used seem to not be standardized. A lot of things will say
there are “n” elements, but if m isn’t 1, this isn’t the case. Watch out: “n”
also gets used as the index all the time too. Sometimes, a0 is used as the
first term, others a1 .
Arithmetic
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An arithmetic progression has a common difference between successive terms,
starting from an initial value.
ai = a1 + d(i − 1)
The sum of a finite arithmetic progression is called an arithmetic series.
n
∑=
i=1
n(a1 + an )
2
Geometric
A geometric progression has a common ratio (i.e. multiple) between successive
terms: ar0 + ar1 + ar2 + ... + arn−1
Given coefficient a (a/k/a scale factor or initial term) and common ratio r, ai =
ari−1
There’s some inconsistent indexing involved here. If you’re zero-indexing the
elements, then it’s ar^n. The consistent thing is that the first term is just the
coefficient a, with r to the zero power.
Note that no term in a geometric progression can be zero.
The sums of the first n terms of the progression are called partial sums: Sn =
n
a 1−r
1−r
Proof
sn+1 = sn + arn and sn+1 = a + r ⋅ sn
This latter equation bumps all the terms “up by one spot” and then reapplies
the 0 power term.
Setting those two equal to each other and solving for s_n gives the formula.
Unlike arithmetic series (which is the sum of a finite arithmetic progression), a
geometric series is the sum of an infinite geometric progression:
∞
S = ∑ ari−1
i=1
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The series diverges (i.e. is infinite) if |r| ≥ 1, and it converges to a finite value if |r|
< 1.
a
The sum of an infinite geometric progression is the series: 1−r
(the limit of r^n
tends to 0 if r < 1 and n→∞).
Other progressions
A progression does not need to have a common difference or ratio, but can be a
function of n, for example ∑ n1 .
A series in most cases is implied to be the sum of an infinite sequence. It is the
limit as n→∞ of the partial sum of n terms:
n
lim ∑ ai
n→∞
i=1
Like improper integrals, a series can converge or diverge. A simple divergence
test is whether limi→∞ ai =
0.
Mathematical induction
1. Guess a formula for the partial sums of a progression.
2. Verify the formula works for the base case, here: n=1.
3. Inductive step: Assume it works for sn and verify that the formula works for
sn+1 in terms of sn
4. (Optional) Take the limit as n→∞ to find the series.
Comparisons
Direct comparison simply means that if 0 < ai < bi for all i ≤ n, then they can use
each other to determine convergence or divergence. If b converges, so does a. If
a diverges, so does b.
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Limit comparison is the same as with improper integrals. If f (i) ∼ g(i) (”is
(i)
similar”, that is, limi→∞ fg(i)
= c where c is non-zero and finite), then their infinite
series either both diverge or both converge. This is used to simplify expressions,
1
e.g. i51−i2 ∼ i5/2
.
(i)
Another form of limit comparison: If limi→∞ fg(i)
= 0, then f(i) decays much faster
than g(i). Thus, if g converges, so does f. If f diverges, so does g.
Ratio test: A series converges if limi→∞ ∣ aai+1i ∣ < 1 and diverges if > 1. (No
conclusion if = 1.)
Root test: A series converges if limi→∞ ∣ai ∣ < 1 and diverges if > 1. (No
conclusion if = 1.) This one seems useful for cases with variable exponents.
i
Integral comparison
A partial sum can be thought of as a Riemann sum estimate (∆x=1) of a definite
integral.
b
b
∫ f (x) ⋅ dx ≈ ∑ f (i) ⋅ Δx
a
i=a
This also works for b=∞ (improper integrals and infinite series) but only if f(x) is
positive, continuous, and decreasing over [0,∞).
As for how close this estimate gets, we can “trap” the integral between an upper
and lower bound.
Example: For a positive f(x) decreasing over (a,b), we know the following:
The left/upper Riemann sum will be greater than the integral. It can be
expressed ∑b−1
i=a f (i).
Remember, i increments by integer, so the width of the slices ∆x is 1.
The upper limit on the sum is b-1 because this is the left Riemann.
The right/lower Riemann sum will be slightly less than the integral:
b
∑i=a+1 f (i)
The lower limit is a+1 because this is the right Riemann sum.
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Thus,
b
b−1
b
∑ f (i) < ∫ f (x) ⋅ dx < ∑ f (i)
a
i=a+1
i=a
When b=∞, b-1 is still ∞. Also, you can make both partial sums use the same
expression by summing from a to b and then just subtracting f(a) from the
right/lower sum.
∞
∑ f (i) − f (a) < ∫
i=a
∞
∞
a
f (x) ⋅ dx < ∑ f (i)
i=a
This is predominately applied to estimate the series by solving the integral, not the
other way around, since the series are harder. Reversing the formula, you get:
∫ < ∑ < ∫ +f (a)
Importantly, if the integral converges or diverges, so does the series.
Conditional & absolute convergence
Absolute convergence is when ∑∞ ∣ai ∣ converges. Converging series with all
positive terms imply absolute convergence.
If a series converges absolutely, it converges regularly too.
Conditional convergence is when a series converges but is not absolutely
convergent.
You can use the root test and ratio test (above) to determine absolute
convergence by just putting the a_i terms in absolute value brackets.
i
An alternating series is of the form ∑∞
i=m (−1) ⋅ ci . An alternating series can
diverge, conditionally converge, or absolutely converge. For example, the
i
progression where ai = (−1)
i conditionally converges. You should first check for
absolute convergence by eliminating the alternating part.
Otherwise, an alternating series conditionally converges if limi→∞ ci = 0.
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Power series
⚠️
Now using “n” as the index.
A power series is a single-variable function represented as an infinite series of
terms of x.
∞
f (x) = a0 + a1 x + a2 x + a3 x + .... = ∑ an (x − c)n
2
3
n=0
where c is the “center” of the series.
A power series converges to a finite number (i.e. the sum of infinite terms has a
finite output, more officially that the limit exists for the first m terms of the series
as m→∞) within a certain domain of x, called the interval of convergence. The
interval is always at least c and may contain more values. This interval is always
centered on c and extends for the radius of convergence R in either direction: (c–
R, c+R).
Outside this interval, the series diverges (is infinite). Determining the behavior at
endpoints of the interval requires specific tests.
If the series converges at some value x0 , then it converges everywhere within a
radius of x0 − c. If it diverges at x0 , it diverges everywhere outside a radius of
x0 − c .
Note that the interval of convergence is not the same as the usual domain of the
resulting function, e.g. ∑x^n converges to 1/(1-x) when |x|<1 only, while that
function’s domain is all real except 1.
You can find the radius of convergence by conducting the ratio test or root
test and determining for which values of x the ratio test is true. It is possible
that it is true for all values of x, or none.
n
Alternatively, the reciprocal of the ratio test ∣ aan+1
∣ = R.
Polynomials can be represented as a power series where the coefficient for all
“unused” powers of x is 0. Fractional (root) and negative powers cannot be
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represented in this format.
Operations
Convergent power series are regular functions, basically just polynomials only
infinite. They can be added, subtracted, multiplied, divided, composited,
differentiated, and integrated.
Add/subtract: Simply add together the summation notation. Converges for at
least the smaller of their radii of convergence.
Multiply: Find the coefficients for each power of x by .
Divide: If h(x) = f(x)/g(x), then set up f(x) = g(x)*h(x) with variables a_n as the
coefficients in h(x). Conduct as a system of linear equations so solve for each
a_n.
n−1
Differentiate: f ′ (x) = ∑∞
, note that the lower bound of n is
n=1 nan (x − c)
now 1.
Integrate: Remember that antidifferentiating requires a constant of integration
c, that may or may not be able to be identified.
When you add two power series, they converge over the smaller of their radii.
Just like how, when differentiating and integrating polynomials, you can operate
on each term individually: the power series lets you operate on just the term
formula. Derivatives and antiderivatives of power series always have the same
interval of convergence as the original.
Taylor series
Linear and quadratic approximations are simply partial sums of a power series that
can completely represent the original function.
Approximations can be of an order as high as however many times you can
differentiate the original function, ultimately perfectly replicating it. This is Taylor’s
formula:
∞
f (n) (x0 )
(x − x0 )n
f (x) = ∑
n!
n=0
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where f^(n) is the order of the derivative. (By convention, 0! = 1.)
The idea here is that, by definition, this approximation will have the same nth
derivative as the original function, for all n.
If possible depending on what f(x) is, try writing f n (x0 ) as a formula in terms of n.
Because the Taylor series is infinite, the center of the approximation doesn’t seem
to matter as much. Thus, for simplicity, putting 0 as x_0 is helpful when calculating
the derivatives. (This special case is called a Maclaurin series.)
A function f is analytic at x = c if it has a Taylor series expansion about x = c that
has a radius of convergence r > 0.
Examples:
a
n
= ∑∞
n=0 ax , the geometric series
1−x
n
∞
ex = ∑n=0 xn! when approximated at 0 because the derivative of e^x is
always itself, and e^0 = 1.
3
5
sin x = x − x3! + x5! − ..., alternating-sign odd, resulting from the periodicity
of sin’s derivatives at x=0.
2
4
cos x = 1 − x2! + x4! − ..., alternating-sign even
3
5
sinh x = x + x3! + x5! + ..., odd no alternating
2
4
cosh x = 1 + x2! + x4! + ..., even no alternating
The Taylor series of a polynomial has a radius of convergence of ∞ because the
polynomial is finite for all x.
Manipulation
To find the Taylor series for a function, see if you can u-substitute to get it in the
1
1
form of a known Taylor series, and just plug it back in, e.g. 1−x
3 = 1−u =
∑ un = ∑ x3n . This doesn’t require the whole du/dx conversion.
Other direction: If you have a Taylor series that does not match a known function,
try to manipulate the summation so that it does. Remember:
∑ an + bn = ∑ an + ∑ bn
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You can factor out non-variable multiples of x, e.g. ∑ xn+2 = x2 ∑ xn . n is
the variable for the summation, not x.
Try differentiating or antidifferentiating to get you to a series you do recognize,
make the conversion, then reverse the steps.
If n starts at some non-zero value like 2, and the summand is a product of a
term of n, you can move the lower bound down, because that term would be a
zero.
Polynomials & remainders
The Taylor polynomial of degree n of f(x) Pn is the function you get when you
truncate the Taylor series to degree n: that is, the partial sum.
The difference between a function and its Taylor polynomial (nth-degree
approximation centered at x_0) is a function called the Taylor remainder.
f (n+1) (c)
R(x) = f (x) − Pn (x) =
(x − x0 )n+1 , c ∈ [x0 , x]
(n + 1)!
This is typically used to find an upper bound for the error, or to determine a
polynomial degree n required to restrict error to some range, by finding the
maximum absolute value of f (n+1) over the interval in question.
Deriving Taylor remainder theorem:
The objective is to bound R(x) over an interval, ∣R(x)∣ ≤ K .
Repeatedly take the derivative of the definition of R(x). Recall that, at x_0,
the values of f and P are equal. That also goes for all derivatives of those
up to n. (This is the definition of the approximation.) Thus, the derivatives
up to degree n of R(x) are all zero.
However, because P (n+1) (x) always equals zero (P’s highest power is n),
R(n+1) = f (n+1) , not zero.
The strategy is to find an upper bound for this n+1’th derivative of R and f,
and integrate it all the way back up to the original remainder function,
which is facilitated because we know that the derivatives of R for all
degrees [0,n] are zero.
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M ≥ ∣R(n+1) (x)∣. Integrate both sides: ∫ M dx ≥ ∫ ∣R(n+1) (x)∣dx. The
integral of an absolute value will always be ≥ the absolute value of the
integral, so the inequality holds if we move the absolute value brackets
outside. Thus: M x + c ≥ Rn (x), which we know = 0 at x=x_0, so c ≥
−M x0 . To minimize c, we set it equal to this value, so M*(x-x_0) ≥ R^n(x).
Repeat this integration step all the way back up to degree 0, and we get:
M ⋅(x−x0 )n+1
≥ ∣R(x)∣.
K = (n+1)!
Remember that M is the maximum absolute value of the n+1’th derivative
of R(x) over the interval in question, which is the same as that derivative of
f(x).
So there is some x-value c between x_0 and x, f (n+1) (c), c ∈ [x0 , x], so
that you get the exact remainder. Thus, the range in values of f^(n+1) over
that interval can give the bounds of the remainder.
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