Problem Solving - Analysis

Problem Solving - Analysis

M. Velrajan

Understanding the problem, Mathematical thinking, Recalling all the results and examples related to the problem are essential to solve any problem in Mathematics.

This way of problem solving is illustrated with 3 problems related to Analysis.

1. Let p be a real polynomial of the real variable x of the form 

p(x) = xn + an-1xn-1 + . . . + a1x -1. Suppose p has no root in the open unit disc and p(-1) = 0. Then

(a) p(1) = 0     (b) lim x → ∞ p(x) =  ∞   (c) p(2) > 0        (d) p(3) = 0

First look into what is given and then what has to be solved/found.

Then recall whatever we know and that follow from the given, that are  useful to proceed to solve.

Here we are given p is a real monic polynomial of the real variable x with constant term - 1 and p has no root in the open unit disc.

We have to find whether 1, 3 are roots of p, p(2) > 0 and lim x → ∞ p(x) = ∞

Think what follows from the given.

Yes,

Product of all the roots of p(x) = (-1)n(-1)/1 = (-1)n + 1 and 

the modulus of each root of p(x) is ≥ 1.

Connect them.  

Product of modulus of all the roots of p(x) = 1 and

hence no root of p(x) has modulus > 1 

i.e. no root of p(x) lies outside of the open unit disc. 

So, the modulus of each root of p(x) is 1.

Hence only 1 and -1 are the possible real roots of p(x) and (d) is false.

Now we see whether 1 is a root.

But we are not able to proceed.

So, go by contrapositively.

Suppose 1 is not a root of p(x). 

Then - 1 is the only real root of p(x).

A real polynomial may have complex roots. 

Since complex roots of p(x) occur in conjugate pairs, 

the number of complex roots of p(x) is even. 

In the factorisation of p(x), root -1 contributes 1 to the constant term, even if it is a repeated root, the pairs of complex roots and their conjugate contribute their modulus, which is 1 for all pairs.

So, the constant term of p(x) is 1, a contradiction. (or)

Suppose z1, 1/z1, . . . , zm, 1/zm be the complex roots of p(x), 2m < n.

(since modulus of each roots of p(x) is 1, conjugate of zi is 1/zi)

Then 

p(x) = (x + 1)k(x - z1)(x - 1/z1)...(x - zm)(x - 1/zm), for some k ≥ 1.

The constant term of p(x) = (-1)2m = 1, a contradiction, 

since the constant term of p(x) = - 1.

Hence 1 is a root of p(x) and (a) is true.

Note that p(x) = (x + 1)r(x - 1)s(x - z1)(x - 1/z1)...(x - zm)(x - 1/zm),

since the constant term of p(x) = - 1, s is an odd number and

1 is a root of p(x) an odd number of times.

For (b), recall

lim x → ∞ (anxn + an-1xn-1 + . . . + a1x + a0) =  lim x → ∞ anxn. 

So, lim x → ∞  p(x) = lim x → ∞  xn = ∞ (b) is true and

hence the graph of p(x) should be only in the first quadrant for all x > a.

Since only 1 and -1 are the real roots of p(x), a = 1 and p(x) > 0 for all x > 1.

(otherwise, if p(x) < 0 for some x > 1 then since p is continuous, 

the graph of p(x) have to cross x axis from the fourth quadrant to first quadrant 

at some x = c > 1 i.e. p(c) = 0 for some c > 1)

So, (c) is true and Ans. a, b, c.   

Note that lim x → - ∞  (anxn + an-1xn-1 + . . . + a1x + a0) = lim x → - ∞ anxn.

So, lim x → - ∞  p(x) = lim x → - ∞ xn = (-1)n ∞ = ∞ or - ∞ according as n is even or odd.

Oh! We got more than we wanted.

Of course it will be a nice exercise to see how the graph of p(x) looks 

if n is even and if n is odd.

2. Let S be the set of (𝛂, 𝛃) ∈ ℝ2 such that x𝛂y𝛃 / √(x2 + y2) → 0 as (x, y)  → (0, 0).

Then S is contained in 

1. {(𝛂, 𝛃) : 𝛂 > 0, 𝛃 > 0}                   2. {(𝛂, 𝛃) : 𝛂 > 2, 𝛃 > 2}

3. {(𝛂, 𝛃) : 𝛂 + 𝛃 > 1}                       4. {(𝛂, 𝛃) : 𝛂 + 4𝛃 > 1}

Here given x𝛂y𝛃 / √(x2 + y2) → 0 as (x, y) → (0, 0), for all (𝛂, 𝛃) ∈ S 

and we have to determine S is contained in which of the sets given.

If f(x, y) is a real function of two real variables 

and f(x, y)  → l as (x, y) → (0, 0)

then for any ε > 0 there exists δ > 0 such that 

f(x,y) ⊆ (l - ε, l + ε) for all (x, y) ∈ B((0, 0), δ).

Hence f(x,y) ⊆ (l - ε, l + ε) for all (x, y) ∈ B((0, 0), δ) and 

any continuous curve that pass through (0, 0), 

in particular y = mx.

i.e.  if f(x, y)  → l as (x, y) → (0, 0) then 

f(x, y) → l as x → 0 along the line y = mx, for all m or 

along any continuous curves that pass through (0, 0). — (A)

So, if  f(x, y) → a value that depend on m as x → 0 along 

the line y = mx or depends on the continuous curves 

that pass through (0, 0) then we conclude that 

lim f(x, y) does not exist as (x, y) → (0, 0) — (B).

Also note that the converse of (A) is not true.

Usually (B) is used to solve many problems.

For illustrations view the earlier post   

https://velrajanm.blogspot.com/2024/03/real-analysis-functions-of-several.html

(𝛂, 𝛃) ∈ S ⇒  x𝛂y𝛃 / √(x2 + y2) → 0 as (x, y) → (0, 0)  

⇒  x𝛂y𝛃 / √(x2 + y2) → 0 as x → 0 along the line y = mx, for all m (by (A))

⇒  m𝛃x𝛂 +𝛃 - 1 / (1 + m2) → 0 as x → 0, for all m 

                   (since on y = mx,  

x𝛂y𝛃 / √(x2 + y2) = x𝛂m𝛃x𝛃 / √(x2 + m2x2)

=  m𝛃x𝛂 +𝛃 - 1 / (1 + m2) )

⇒ 𝛂 + 𝛃 > 1

S is contained in {(𝛂, 𝛃) : 𝛂 + 𝛃 > 1}.

Note that if 𝛂 + 𝛃 > 1 then m𝛃x𝛂 +𝛃 - 1 / (1 + m2) → 0 as x → 0, for all m. 

But as the converse of (A) is not true, (𝛂, 𝛃) need not be in S. 

Hence S may be a proper subset of {(𝛂, 𝛃) : 𝛂 + 𝛃 > 1}. 

Ans. (3).

3.

 

V = {(𝛂, 𝛂 + 𝛃, 𝛂 + 𝛃) / 𝛂, 𝛃 ∈ ℝ} 

    = {(a, b, b) / a, b ∈ ℝ}, the plane y = z, 

which is closed in ℝ3.

Any plane is closed in ℝ3.

ℝ3 - V = {(a, b, c) / b ≠ c} 

  = points in the space which are not in the plane y = z, is open in ℝ3, 

having two components, y < z and y > z. 

Any point with y < z and any point with y > z can not be joined 

by a polygonal path in ℝ3 - V, hence ℝ3 - V is not connected.

Complement of any plane in the space is union of 

two disjoint open half spaces, which are on the 

two opposite sides of the plane, hence it is not connected. 

A point in one of the half spaces can not be joined with 

a point in the other half space by a polygonal path that 

lies in the complement of the plane.

But any two points in the same half space can be joined 

by a polygonal path in that half space. 

Hence each of the two open half spaces determined 

by any plane is connected and hence they are 

the connected components of the complement of the plane.

But (0, 0, 0) lies in the plane y = z. 

And any point with y < z and any point with y > z can be 

joined by a polygonal path through (0, 0, 0). 

Hence any two points in (ℝ3 - V) ∪ {(0, 0, 0)} can be joined 

by a polygonal path.

So, (ℝ3 - V) ∪ {(0, 0, 0)} is connected and 1 is false.

Note that {tu1 + (1 - t)u3 : 0 ≤ t ≤ 1} is the line segment joining u1 and u3.

Since u1 and u3 are in the same component y < z of ℝ3 - V, 

the line segment joining u1 and u3 lie in the same component, 

hence (ℝ3 - V) ∪ {tu1 + (1 - t)u3 : 0 ≤ t ≤ 1}= ℝ3 - V and it is not connected.

2 is false.

Since u1 and u2 are in the different components y < z and y > z of ℝ3 - V, 

the line segment joining u1 and u2 connects the two components of ℝ3 - V.

Any point with y < z and any point with y > z can be joined 

by a polygonal path through the line segment joining u1 and u2. 

Hence (ℝ3 - V) ∪ {tu1 + (1 - t)u2 : 0 ≤ t ≤ 1} is connected. 3 is true.

{(t, 2t, 2t) : t ∈ ℝ} is the line 2x = y = z on the plane y = z, 

hence any point with y < z and any point with y > z can be joined by 

a polygonal path through (0, 0, 0) or any point on the line 2x = y = z. 

Hence (ℝ3 - V) ∪ {(t, 2t, 2t) : t ∈ ℝ} is connected and 4 is true.   

Ans. 3, 4.

So, develop mathematical thinking and learn properly.

Enjoy Mathematics.