Friday, January 13, 2017

Unizor - Derivatives - Problem 2

Notes to a video lecture on

Derivatives - Problems 2

Problem 2.1
Artillery officer needs to choose an angle a cannonball should be launched to reach the maximum distance, provided its linear speed at the moment of launching is fixed.
Assume ideal physical conditions (no air resistance, gravity constant is not changed with the height and whatever else can be assumed to simplify the problem).
The maximum distance is reached when a cannonball is launched at an angle of π/4=45o

Problem 2.2
How many zero points does function f(x)=x³−3x+4 have?
Analyze intervals of monotonic behavior of this function and compare the signs of the function on each interval's ends.
This function has only one zero point.

Problem 2.3
Consider pulling an object by a rope on a surface with friction. If the rope is horizontal, the force needed to move it with constant speed should be equal to the force of friction. If a rope is at certain angle to horizon, the vertical component of the force applied to it partially neutralizes the friction.
What angle of a rope to horizon is needed to minimize the force applied to it and still to move forward with constant speed?
Consider ideal conditions and a coefficient of friction equal to k.
The angle or a rope to horizon should be equal to arctan(k).
Assume, the angle of a rope with horizon is φ (obviously, its range is from 0 to π/2), the weight of an object is P and the force applied to a rope is F.
Then vertical component of the pulling force is
Fv = F·sin(φ)
and horizontal component is
Fh = F·cos(φ)
Vertical component Fv reduces the weight and, therefore, reduces the friction. Therefore, the friction equals to
T = (P−Fv)·k
To pull an object with constant speed this friction force must be equal to horizontal component of the force applied to a rope Fh:
T = Fh
The above equation gives a dependency between the force applied to a rope F and angle of a rope to horizon φ. Having this as a function, we can minimize it and find an optimal angle of minimum force.
Fh = (P−Fv)·k
F·cos(φ) = [P−F·sin(φ)]·k
F·cos(φ)+F·sin(φ)·k = P·k
F = P·k/[cos(φ)+sin(φ)·k]
To minimize this function, let's take its derivative by φ and find where it's equal to zero.
dF/dφ = {−P·k/[cos(φ)+sin(φ)·k]²}·

Equation dF/dφ = 0
results in
-sin(φ)+cos(φ)·k = 0
which can be easily solved:
sin(φ) = cos(φ)·k
sin(φ)/cos(φ) = k
tan(φ) = k
φ = arctan(k)
This solution is independent of the weight of an object and means that the greater the friction - the more vertical should be an angle we pull the object to neutralize friction and minimize the force applied to a rope.
If we are talking about practical application of this, when a person pulls something by a rope, for lower friction coefficient we should use longer rope and for higher friction - shorter.

Thursday, January 12, 2017

Unizor - Derivatives - Problem 1

Notes to a video lecture on

Derivatives - Problems 1

Problem 1.1
Consider a sufficiently smooth function f(x).
Is condition f I(x0) = 0
(a) necessary,
(b) sufficient or
(c) necessary and sufficient
for f(x0) to have a local extremum (local maximum or local minimum) at point x=x0?
Prove your answer.
(a) necessary
It is not sufficient because it might be an inflection point, like for f(x)=x3 at x=0.

Problem 1.2
Consider a sufficiently smooth function f(x).
Is condition f II(x0) = 0
(a) necessary,
(b) sufficient or
(c) necessary and sufficient
for f(x0) to have an inflection point at x=x0?
Prove your answer.
(a) necessary
It is not sufficient because it might be a point of local extremum, like for f(x)=x4 at x=0.

Problem 1.3 - Derivative of the inverse function theorem
Consider a sufficiently smooth function y=f(x) with derivative f I(x) not equal to zero.
Prove that its inverse y=g(x) (that is, f(g(x))=x) has a derivative g I(x)such that
g I(x) = 1 / f I(g(x))
Example 1
y = f(x) = xn
y = g(x) = x1/n (inverse)
f I(x) = n·xn−1
g I(x) = (1/n)·x(1/n)−1 =
= (1/n)·x(1−n)/n

1 / f I(g(x)) = 1 / {n·[x1/n]n−1} =
= (1/n)·1/x(n−1)/n =
= (1/n)·x(1−n)/n = g I(x)

Example 2
y = f(x) = ex
y = g(x) = ln(x) (inverse)
f I(x) = ex
g I(x) = 1/x
1 / f I(g(x)) =
= 1 / e ln(x) = 1/x = g I(x)

Problem 1.4
Consider all possible regular square prisms with a given surface area.
Under what condition between the length of a base' side a and altitude h the volume of this prism will be minimum or maximum?

Volume V=a²h
Surface area S=2a²+4ah
Since S is given,
Substitute it into an expression for volume:
V(a) = a²·(S−2a²)/4a =
= (1/4)·a·(S−2a²)

So, we have to find extremum(s) of function
V(a) = (1/4)·a·(S−2a²) =
= (1/4)·(−2a³+S·a)

This is a polynomial function of a, defined on an interval from 0 to a maximum value when the volume is still greater or equal to zero, that is satisfying the condition S−2a² ≥ 0.
Here is how this function looks on a graph (we have chosen S=6 in this case, so we have to consider this function only on an interval [0,√3]):

As seen from the graph, the extremum of our function within the specified domain is a local maximum.
To find its extremum(s), find the stationary points where derivative equals to zero:
dV/dx = (1/4)·(−6a²+S)
To make sure, we are dealing with a local maximum, we can take a second derivative, it's equal to −3a, and it is negative within a domain of our function, which confirms that a stationary point is a local maximum.
Set the first derivative to 0, getting an equation for variable a:
−6a²+S = 0
Its only root within the established domain is a = √S/6.
Now we can find the corresponding value of altitude h in terms of surface area S:
h = (S−2a²)/4a =
= [S−(S/3)]/(4√S/6) = √S/6
As we see, h=a, which means that the maximum volume is reached when our prism is a cube.

Unizor - Derivatives - Exercises 2

Notes to a video lecture on

Derivatives - Exercises 2

Exercise 2.1
Given a function f(x)=x·e -x.
Find an equation of a tangential line that touches this function at pointx=2
y = −e−2·x + 4·e−2

Exercise 2.2
Given a function f(x)=x·e-x.
Find all its maximum, minimum and inflection points.
f I(x) = e−x·(1−x)
f II(x) = e−x·(x−2)
Point x=1 is a point of local maximum.
Point x=2 is a point of inflection.
There is no point of local minimum.

Exercise 2.3
Given a functionf(x)=sin(x)+cos(x).
Find all intervals where it's monotonically increasing and intervals where it's monotonically decreasing.
Intervals of monotonic increasing:
[−3π/4+2πn, π/4+2πn]
Intervals of monotonic decreasing:
[π/4+2πn, 5π/4+2πn]

Exercise 2.4
Given a function f(x)=sin(x).
Find all inflection points of this function.
What are the first derivatives at these points?
x = π·n
First derivatives equal to 1 or −1

Exercise 2.5
Given a function f(x)=x·e-x.
Find an equation of a normal to its graph at point x=2
y = e2·x + 2·(e−2−e2)

Exercise 2.6
Given a sufficiently smooth function f(x).
Find equations of a tangential line and a normal to its graph at pointx=x0
Tangential line:
y = f I(x0)·(x−x0) + f(x0)
y = −[f I(x0)]−1·(x−x0) + f(x0)

Monday, January 9, 2017

Unizor - Derivatives - Exercises 1

Notes to a video lecture on

Derivatives - Exercises 1

Exercise 1.1
Find derivative of a polynomial
P(x) = Σn∈[0,N] An·xn
DxP(x) = Σn∈[1,N] An·n·xn−1

Exercise 1.2
As we know,sin(2x)=2sin(x)cos(x)
Find independently derivatives of two functions:
sin(2x), as a compound function g(f(x)), where f(x)=2x and g(x)=sin(x)
2sin(x)cos(x) as a product of functions.
Compare the results (supposed to be the same).
Use an identity

Exercise 1.3
Hyperbolic sine function is defined as
sinh(x) = (ex−e−x)/2
Hyperbolic cosine function is defined as
cosh(x) = (ex+e−x)/2
Prove that
Dxsinh(x) = cosh(x) and
Dxcosh(x) = sinh(x)
which resembles (except the sign in case of derivative of hyperbolic cosine) the situation with regular sine and cosine.

Exercise 1.4
Find derivative of secant and co-secant using their definitions as reciprocal to cosine and sine:
sec(x) = 1 / cos(x) and
csc(x) = 1 / sin(x)
Dxsec(x) =
= sin(x)/cos²(x) =
= sec(x)tan(x)

Dxcsc(x) =
= −cos(x)/sin²(x) =
= −csc(x)cot(x)

Exercise 1.5
Find derivative of tangent and cotangent functions using their definitions as ratios of sine and cosine:
tan(x) = sin(x) / cos(x) and
cot(x) = cos(x) / sin(x)
There are different variants, all equivalent:
Dxtan(x) =
= 1+tan²(x) =
= 1/cos²(x) =
= sec²(x)

Dxcot(x) =
= −1−cot²(x) =
= −1/sin²(x) =
= −csc²(x)

Exercise 1.6
Find derivative of arc-sine and arc-cosine using their definitions as inverse functions to sine and cosine:
φ=arcsin(x) ⇒
⇒ −π/2 ≤ φ ≤ π/2; sin(φ)=x
φ=arccos(x) ⇒
⇒ 0 ≤ φ ≤ π; cos(φ)=x
Dxarcsin(x) = (1−x²)−1/2
Dxarccos(x) = −(1−x²)−1/2

Unizor - Derivatives - Newton's Method of Finding Zeros of a Function

Notes to a video lecture on

Derivatives - Newton's Method
to Find Zeros of Function

The purpose of this lecture is to address a specific problem of finding an approximate solution of equations f(x)=0 using some methodology suggested by Sir Isaac Newton a few centuries ago.
Obviously, it makes sense to apply this methodology when there is no analytical solution of a given equation, or this analytical solution requires too much efforts.

A word of warning should be given up front. This method might sometimes fail to find an approximate solution. After describing this method it will be obvious that under some circumstances the process of finding a solution is not converging to any number.

Let's start with an impractical but illustrative example of a linear function f(x)=2x−4. Obviously, linear equation f(x)=0 looks like 2x−4=0 and can be immediately solved:

But let's approach it differently using the following method.
The graph of f(x)=2x−4 looks like this (blue line):

To solve the equation 2x−4=0 means to find the X-coordinate of point C, where the graph intersects the X-axis.

Let's choose any point A with X-coordinate x0 and draw a perpendicular to X-axis from this point to an intersection with the graph of our function f(x)=2x−4 at point B with coordinates {x0, f(x0)} (black line).
We notice that segment AC is a cathetus of triangle ΔABC and, therefore, its length equals to
AC = AB/tan(∠ACB)

We also know that tan(∠ACB)is a derivative of a function f(x)=2x−4 at point C, which is the same as a derivative at any other point, like A, since our function is linear. This derivative is equal to f I(x0)=2.

Therefore, knowing this derivative, X-coordinate of point A, which is equal to x0, and the value of our function at point A, which is equal to f(x0), that is the length of segment AB, we can easily find the length of segment AC and then the X-coordinate of point C:
AC = AB / tan(∠ACB) =
f(x0) / f I(x0)

Now, X-coordinate of point C, let's call it x1, equals to X-coordinate of point A minus the length of segment AC:
x1 = x0 − f(x0) / f I(x0)

Point A can be taken arbitrarily. Let's say, x0=4.
Then AB=f(x0)=2·4−4=4.
Since tan(∠ACB)=2,AC=4/2=2.
Finally, knowing AC and coordinate of point A (x0=4), we can determine the coordinate of point C:
x1 = x0 − AC = 4−2=2
This is a solution to an original equation 2x−4=0.

Next step towards Newton's Method is to consider a more complex function and how analogous approach to the one above might lead to a solution.

Our next example is a quadratic polynomial
presented on the next graph (red line).

We have specifically considered such a function that the linear function described above (blue line) is a tangential line to this new function exactly at a point where we chose to start our calculation x0=4.
Incidentally, it's easy to see that the zero point of this quadratic polynomial is x=1. However, we pretend that we don't know this and attempt to approach this value in a process similar to described above for a linear function.

First of all, note that the behavior of parabola is very smooth on a graph. It is smooth not only in terms of differentiability and continuity of a derivative, but also in terms of its derivative being a monotonic function. In particular, as seen from the graph, direction of parabola and direction of its tangential line at x0=4 (blue line) are very close to each other. From the point of tangency x0=4 both parabola and its tangential line go to zero towards the same direction - to the left from point x0.

The main implication of this is that a zero point C of a tangential line with X-coordinate x1=2, as we calculated above, is closer to a zero point of parabola than original point A at X-coordinate x0=4, where we started the process. Therefore, jumping from point A to point C we got closer to a zero point of a parabola.

An obvious continuation of this process is to recursively repeat the step above, starting at point C with X-coordinate x1=2.
We draw a perpendicular to X-axis at point C to intersection with our parabola at point with coordinates {x1, f(x1)} and a tangential line to parabola at point D (green line) that intersects X-axis at point E with X-coordinate x2.

Let's determine x2 from triangle ΔEDC similarly to the above procedure:
CE = CD/tan(∠CED)
CD = f(x1) = (2/9)·2²+(2/9)·2−(4/9) = 8/9
tan(∠CED) = f I(x1) =
= (2/9)·2x1+(2/9) = 10/9

CE = (8/9) / (10/9) = 4/5
x2 = x1 − CE =
= x1 − f(x1) / f I(x1) =
= 2−(4/5) = 6/5

Got closer to a zero point of a parabola that we know is x=1, but pretend we don't know it.

Let's repeat this process once more, starting at x2=6/5.
x3 = x2 − f(x2) / f I(x2) ≅
≅ 1.2 − 0.142222 / 0.755556 ≅
≅ 1.01176

Got very close to x=1!

The recursive process described above is the Newton's method of finding arguments where function equals to zero.
Let's formulate it more rigorously.

Assume, we need to find zeros of some smooth function f(x).

Step 0
Start with some approximation of a value of an argument x0that is relatively close to the one where our function equals to zero.
There is no universal recipe for this approximation, but the better your approximation is - the sooner you will approach the real zero of f(x).
And, to be noted, if your starting value is too far from the real zero, the Newton's method might never converge to real zero of our function.

Step 1
From the chosen value xcalculate the next value of a sequence:
x1 = x0 − f(x0) / f I(x0)
This and subsequent steps are straightforward calculations.

Step 2, 3 etc.
Check two things:
(a) the value of f(xn) - is it close to zero within your chosen precision?
(b) the difference between xand xn−1 - is it smaller than your chosen precision?
If both checks are satisfactory, stop the process. The last xn is your final result.
If your precision requirements are not satisfied yet, continue.
Repeat the previous step according to this recursive formula:
xn+1 = xn − f(xn) / f I(xn)

As we noted, Newton's method is not universal. A lot depends on properties of function, whose zeros we are trying to find, and on a choice of the first approximation x0.
Let's have an example where Newton's method does not converge to a zero point of a function.

An obvious example is when we choose the first approximation point where a tangential line is parallel to the X-axis.
Thus, in case of a parabola f(x)=(2/9)x²+(2/9)x-(4/9), that we considered above, such point would be x=−0.5.

Another obvious example is an attempt to find a zero point if the function has none. Consider function f(x)=1/x with any starting point. The Newton's method will lead you to infinity.

It is very important that the first approximation point should be relatively close to a real zero point of a function. Imagine a function having a zero point and a "hump" near it. If we choose a starting point "behind a hump", the process will never get to a zero point. Here is an example.

This function has a maximum at point x=1. At this point the tangential line is parallel to X-axis. If our first approximation x0 is greater or equal to 1, we will never converge to a function's zero point at x=0.

Tuesday, January 3, 2017

Unizor - Calculus - L'Hospital's Rule

Notes to a video lecture on

L'Hopital's Rule

L'Hopital's Rule is a helpful theorem and a related technique to determine certain indeterminate form limits, like 0/0 or ∞/∞.

It should be noted that this theorem has certain limitations that are not always observed and, therefore, the method of determining the indeterminate limit based on L'Hospital's Rule might not always help.

There are, actually, a set of three theorems related to the L'Hospital's Rule, successively more powerful - basic, intermediate and advanced.

Theorem 1
(basic L'Hospital's Rule)

Given two sufficiently smooth functions, F(x) and G(x), defined on some contiguous interval with a point x=xinside this interval (which means, x0 belongs to this interval together with its immediate neighborhood).

Assume that F(x0)=G(x0)=0, while G(x), to be able to divide by it, is not equal to zero for x ≠ x0. So, as x→0, both functions are infinitesimal variables and, therefore, the limit of their ratio is an indeterminate form 0/0.

Assume further that the derivative of G(x) at point x=x0is not equal to zero:
G I(x0) ≠ 0

limx→x0 F(x)/G(x) = limx→x0 F I(x)/G I(x) = F I(x0)/G I(x0)

In short, a limit of ratio of two infinitesimal functions equals to a ratio of their derivatives at a limit point.


Consider the ratio of two functions depicted on a graph below
F(x)=x²−2x−3 (blue) and
G(x)=−x²+7x−12 (red)
and the limit of their ratio asx→3.
Both functions approach zero asx→3, so their ratio represents an indeterminate form 0/0.
We cannot determine the limit without some clever technique.

In this particular case we can determine the limit using the following method:
F(x) = x²−2x−3 = (x−3)·(x+1)
G(x) = −x²+7x−12 = −(x−3)·(x−4)
F(x)/G(x) = [reducing by (x−3)] = (x+1)/(4−x)
And the limit of this ratio asx→3 equals to
(3+1)/(4−3) = 4

In a more general case this simple method might not work, but we will use these two functions to demonstrate a different method called L'Hospital's Rule.

We have drawn two straight lines tangentially to two given functions exactly at point x=3, where we want to find the limit of their ratio:
dark green line is tangential to F(x) and
orange line is tangential to G(x).
In the neighborhood of point of tangency x=3 function and its tangential line are very close to each other. In fact, they are so close that in the ratio of one function over another we can replace the value of a function with the value of the Y-coordinate on the tangential line for the same argument and, as we get closer to a point of tangency, the difference will be an infinitesimal that we can ignore.
And that is the main reason for L'Hospital's Rule.

Here is a justification for this.

[F(x)−F(3)]/(x−3) → F I(3)
the difference between[F(x)−F(3)]/(x−3) and F I(3) is an infinitesimal variable ε, so
[F(x)−F(3)]/(x−3) = F I(3) + ε
from which follows
F(x)−F(3) = F I(3)·(x−3) + ε·(x−3)
Recall that F(3) = 0.
F(x) = F I(3)·(x−3) + ε·(x−3)
As x→3, the first member of the sum on the right hand side is an infinitesimal of the same order (this dependency is called Big-O) as x−3 because F I(3) is a constant. The second member is, however, an infinitesimal of a higher order (Little-o) because it's a product of two infinitesimal, ε and x−3.
G(x) = G I(3)·(x−3) + δ·(x−3)
where δ is another infinitesimal.

Using all this, the ratio of our two functions after canceling common multiplier x−3 can be written as
F(x)/G(x) = [F I(3) + ε]/[G I(3) + δ]
Since both ε and δ are infinitesimals, the limit of the expression on the right is F I(3)/G I(3).


Since F(x0)=G(x0)=0, we can subtract these from numerator and denominator without changing the value of a ratio.
F(x)/G(x) = [F(x)−F(x0)] / [G(x)−G(x0)]

Now we can divide numerator by x−x0 to obtain an expression that resembles the definition of a derivative for function F(x):
[F(x)−F(x0)]/(x−x0)F I(x0) as x→x0.
Similarly, we can divide denominator by the same x−xto leave the value of a ratio unchanged and to obtain an expression in the denominator that resembles the definition of a derivative for function G(x):
[G(x)−G(x0)]/(x−x0)G I(x0) as x→x0.

Now the original ratio ofF(x)/G(x) is transformed into this:
F(x)/G(x) = [F(x)−F(x0)]/(x−x0) } / { [G(x)−G(x0)]/(x−x0) }

Both numerator and denominator of the last expression have limits as x→x0.
They are, correspondingly, derivatives F I(x0) and G I(x0)with the latter not equal to zero by assumption.
Therefore, the original limit limx→x0 F(x)/G(x) equals to a ratio of limits, which, in turn, equals to a ratio of two derivatives at point x=x0.
End of proof.

What if the derivative of G(x)at point x0 is equal to zero? We cannot use this basic L'Hospital's Rule. Fortunately, we can extend this theorem to a more general case.

Theorem 2
(general L'Hospital's Rule)

Given two sufficiently smooth functions, F(x) and G(x), defined on segment [a,b].
Assume that F(x)→0 andG(x)→0 as x→+a

(which makes the limit of their ratio indeterminate as x→+a).
Assume further that G I(x) ≠ 0 on segment [a,b] and the limit of the ratio of the derivatives of these functions exists as x→+a.

limx→+a F(x)/G(x) = limx→+a F I(x)/G I(x)
In short, a limit of ratio of two infinitesimal functions equals to a limit of ratio of their derivatives.

NOTE: This theorem does not require existence of derivatives at point x=a or the value of a derivative of G(x) not to be equal to zero at point a. Therefore, we can apply this theorem recursively, going from a ratio of functions to a ratio of their derivatives, then to a ratio of their second derivatives etc. until we get rid of indeterminate form of a limit (or give up).


Consider any point x(a,b) and consider our functions on interval [a,x].
We can apply Cauchy Mean Value Theorem (see the previous lecture) that states that there exists point x0[a,x], where the following is true:
F I(x0)/G I(x0) = [F(x)-F(a)]/[G(x)-G(a)] = F(x)/G(x)

Since point x(a,b) can be arbitrarily chosen, let's choose x+a.
Then point x0 would also converge to +a since x0[a,x].
That results in
limx→0F I(x0)/G I(x0) = limx→0F(x)/G(x)
(where x0[a,x]).
Since we are dealing with sufficiently smooth functions, the last equality is equivalent to
limx→0F I(x)/G I(x) = limx→0F(x)/G(x)
End of proof.

Theorem 3
(extended L'Hospital's Rule)

Not only we can use L'Hospital's Rule to resolve indeterminate form 0/0, but also ∞/∞.
We can also extend the limit point of an argument to ±∞.

Let's prove this for a case of ∞/∞.
Assume, F(x)→∞ and G(x)→∞
as x→a.
Assume that F(x)/G(x)→L
as x→a.
We will prove that
F I(x)/G I(x)→L as well.
Recall that
[1/f(x)] I = −[1/f²(x)]·f I(x).
L = limx→aF(x)/G(x) limx→a[1/G(x)]/[1/F(x)] =
(now we have 0/0 form, we can apply L'Hospital's Rule)
limx→a[1/G(x)] I/[1/F(x)] I limx→a [−1/G(x)]²·G I(x) / [−1/F(x)]²·F I(x) =
limx→a [F(x)/G(x)]²·limx→aG I(x)/F I(x) = L²·limx→aG I(x)/F I(x)
So, we have obtained an equality
L = L²·limx→aG I(x)/F I(x)
from which follows that
L = limx→aF I(x)/G I(x)

Thursday, December 22, 2016

Unizor - Derivatives - Limit of Sequence - Big O, Little o

Notes to a video lecture on

Rate of Change: Big-O, Little-o

Consider two infinitely growing sequences
{n} and {n+1}.
Both are infinitely growing, the second one is always (for any order number n) greater than the first, so it is "bigger" in some sense.
But, if you compare the rate of their growth, it is, actually, the same. When one doubles in value, another almost doubles too and the difference gets proportionally smaller and smaller.
The difference between these sequences, as they grow, is always 1, but their values are getting larger and larger, so proportional difference is getting more and more negligible. For order number n=10 the difference of represents 10% of the value, for n=1000 the difference of represents only 0.1% of the value. This allows us to say that both sequence are growing at the same rate.

On the other hand, if you consider these infinitely growing sequences
{n} and {ln(n)},
you will soon see that proportional difference becomes larger and larger as growth much faster than ln(n). For order number n=10 the first sequence has value 10, while the second, approximately, has 2.3 - 77% difference. For n=1000 the first sequence has value 1000, while the second, approximately, has 6.9 - more than 99% difference. And this proportional difference is growing. This allows us to say that the first sequence growth proportionally faster than the second. The rate of growth of the first sequence is greater than of the second.

Similar situation is with infinitesimals.
Consider two infinitesimal sequences
{1/n} and {10/n}.
Both are infinitesimal, though the first one is always smaller. Proportional difference between them is always 9/n, which represents 900% of the value of the first sequence. It does not change and, therefore, we can say that these two infinitesimals are diminishing with the same speed.

On the other hand, if you consider these infinitesimals
{1/n} and {1/ln(n)},
you will see that proportional difference is increasing as 1/n converges to zero much faster than 1/ln(n). Indeed, with order number n=10 the first sequence equals 0.1, while the second, approximately, is 0.4343 - more than 300% difference. For n=100 the first sequence is 0.01, while the second, approximately, is 0.2174 - more than 2000% difference. For n=1000 the first sequence is 0.001, while the second is 0.1448 - more than 14000% difference.

To address the relative rate of growth for infinitely growing sequences or relative rate of diminishing for infinitesimal sequences, there is a special notation called Big-O and Little-o.
Big-O means similar rate of change, while Little-o means "smaller" in terms of rate of change (slower growth for infinitely growing and faster diminishing for infinitesimals).

It should be noted that many sources differentiate between relative rates of change bounded from above (where it's called Big-O), from below (where it's called Ω) and from both sides (where it's called Θ).
For our purposes we will use only Big-O and assume that it refers to a relative rate of change bounded from both upper and lower sides. The definition below clarifies the exact meaning of this.

Thus, considering infinitely growing sequences, we can say that
n+1 = O(n) - to indicate the same rate of change or the same order of growing,
ln(n) = o(n) - to indicate that ln(n) is "smaller" in a sense of the rate of growing of ln(n) to infinity is slower than the rate of growing of n.

For infinitesimal sequences we can say that
10/n = O(1/n) - to indicate that both infinitesimals are diminishing to zero with the same speed, their order of diminishing is the same,
1/n = o(1/ln(n)) - to indicate that 1/n is "smaller" in a sense of the rate of diminishing to zero of 1/n is faster than the rate of diminishing to zero of 1/ln(n).

Let's now define Big-O and Little-o more rigorously.

Consider two sequences
{Xn} and {Yn}.
Definition of Little-o is based on the behavior of their ratio {Xn/Yn} (assuming Yn≠0).
If this ratio is an infinitesimal sequence, we say that
Xn = o(Yn)
Symbolically, the rate of change of Xn is o(Yn) if
ε>0 ∃Nn ≥ N ⇒ |Xn/Yn| ≤ ε

Definition of Big-O for two positive infinitely growing or infinitesimal sequences {Xn} and {Yn} is also based on the ratio {Xn/Yn}.
If, after certain order number, it is bounded from both sides by positive numbers, we say that these two sequences are of the same rate of change (or of the same order).
Symbolically, the rate of change of Xn is O(Yn) if
∃ A > 0B > 0N:
n ≥ N ⇒ A ≤ Xn/Yn ≤ B


1. Infinitely Growing - O()
Xn = 2n²+n−1Yn = n²−1
Xn/Yn = (2n²+n−1)/(n²−1) =
[(2n²−2)+(n+1)]/(n²−1) =
2+[(n+1)/(n²−1)] =
The last expression is, obviously, limited between A=2 and B=3 for n ≥ 2.
Therefore, Xn is O(Yn).

2. Infinitesimals - O()
Xn = 1/nYn = n/(n²−1)
Xn/Yn = (n²−1)/n² =
The last expression is, obviously, limited between A=0.5 and B=1 for n ≥ 2.
Therefore, Xn is O(Yn).

3. Infinitely growing - o()
Xn = ln(n)Yn = n
Xn/Yn = ln(n)/n = ln(n1/n)
Expression n1/n converges to 1.
Here is a proof (with a help of a trick).
Let Zn=n1/n−1.
Obviously, Zn ≥ 0.
Consider an expression
On one hand, it is equal to
(1+n1/n−1)n = n.
On the other hand, let's use Newton's binomial for it:
(1+Zn)n = Σi∈[0,n]CniZni
Replacing the left side with and leaving only one member of the sum on the right - the one with Zn2 - we come up with the following inequality:
n ≥ Cn2Zn2
Since Cn2 = n(n−1)/2,
n ≥ Zn2n(n−1)/2 or
Zn2 ≤ 2/(n−1),
which proves that Zn is an infinitesimal variable and, therefore, n1/n→1.
Therefore, ln(n1/n) is converging to 0.
Therefore, Xn is o(Yn).

4. Infinitesimals - o()
Xn = 1/2nYn = 1/n
Xn/Yn = n/2n;
The expression n/2n is converging to zero.
Here is a proof.
Use Newton's binomial for identity 2n = (1+1)n:
2n = Σi∈[0,n]Cni
Using only one member of the sum on the right, we come up with an inequality
2n ≥ n(n−1)/2
from which follows
n/2n ≤ 2/(n−1).
Therefore, n/2n is an infinitesimal.
Therefore, Xn is o(Yn).