Sequences and Series of Functions MCQ Quiz in বাংলা - Objective Question with Answer for Sequences and Series of Functions - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

We have for any positive integer n, f_n ∶ [0, 1] → ℝ defined by

f_n(x) = \(\rm\frac{x}{nx+1}\), ∀ x ∈ [0, 1]

∴ f(x) = \(\rm\displaystyle\lim_{n \rightarrow−\infty}\) f_n(x)

= \(\rm\displaystyle\lim_{n \rightarrow \infty} \frac{x}{nx+1}\)

= 0, ∀x ∈ [0, 1]

Hence, sequence {f_n(x)} converges to 0.

(1) Now, M_n = \(\rm\displaystyle\sup_{x \in \mid 0, 1]}\) |f_n(x) − f(x)|

= \(\rm\displaystyle\sup_{x \in[0, 1]}\left|\frac{x}{n x+1}\right|\)

Let t = \(\rm\frac{x}{nx+1}\)

⇒ \(\rm\frac{dt}{dx}=\frac{(nx+1).1−xn}{(n x+1)^2}=\frac{1}{(nx+1)^2}\) > 0,   ∀x ∈ [0, 1]

Let y = \(\rm\frac{1}{(nx+1)^2}\)

∴ \(\rm\frac{dy}{dx}=\frac{(nx+1)^2 \cdot 0−2(nx+1).n}{(nx+1)^4}\)

= \(\rm −\frac{2n(nx+1)}{(nx+1)^4}\)

Putting \(\rm\frac{dy}{dx}\) = 0 ⇒ nx + 1 = 0 ⇒ x = \(\rm−\frac{1}{n}\)

Now,

\(\rm\frac{d^2 y}{d x^2}=\frac{(n x+1)^4\left(−2 n^2\right)−(2 n(n x+1)) 4(n x+1)^3 \cdot n}{(n x+1)^8}\)

= \(\rm\frac{(nx+1)^3\left\{−2 n^3 x−2 n^2−8 n^3 x−8 n^2\right\}}{(n x+1)^8}\)

= \(\rm\frac{−10 n^3 x−10 n^2}{(n x+1)^5}\)

∴ \(\rm\left(\frac{d^2 y}{d x^2}\right)_{x=−\frac{1}{n}} = \frac{−10 n^3\left(−\frac{1}{n}\right)−10 n^2}{\left(n\left(−\frac{1}{n}\right)+1\right)^5}\) = ∞

⇒  M_n = ∞

∴ \(\rm\displaystyle\lim_{n \rightarrow \infty}\) M_n = ∞

∴ f is an increasing function in [0, 1].

Therefore, M_n = \(\rm\frac{1}{n.1+1}=\frac{1}{n}\)

Now, \(\rm\displaystyle\lim_{n \rightarrow \infty} M_n=\lim_{n \rightarrow \infty} \frac{1}{n}\) = 0

Hence, by M_n-test sequence {f_n(x)} is uniformly convergent in [0, 1].

Hence, option (1) is correct.

(2) Now, \(\rm f_n^{\prime}(x)=\frac{1}{(nx+1)^2}\)

and M_n = \(\rm\displaystyle\sup_{x \in[0, 1]}\left|f_n^{\prime}(x)−f^{\prime}(x)\right|\)

= \(\rm\displaystyle\sup_{x \in[0, 1]}\left|\frac{1}{(n x+1)^2}−0\right| \left[\because f^{\prime}(x)=0\right]\)

= \(\rm\displaystyle\sup_{x \in[0, 1]}\left|\frac{1}{(n x+1)^2}\right|\)

Hence, sequence \(\rm\left\{f_n^{\prime}\right\}\) of derivatives of {f_n} does not converges uniformly on [0, 1].

Hence, option (2) is incorrect.

(3) Since, each f_n(x) is Riemann integrable. Hence, 

\(\displaystyle\int_0^1\) f_n(x) dx = \(\displaystyle\int_0^1\) f(x) dx

= \(\displaystyle\int_0^1\) 0 dx

= 0

∴ \(\rm\displaystyle\lim_{n \rightarrow \infty} \int_0^1\) f_n(x) dx = \(\rm\displaystyle\lim_{n \rightarrow \infty}\) 0 = 0, ∀ x ∈ [0, 1]

Hence, the sequence \(\rm\left\{\displaystyle\int_0^1 f(x) d x\right\}\) is convergent.

Hence, option (3) is correct.

(4) Since, each \(\rm f_n^{\prime}\)(x) is Riemann integrable. Hence,

\(\rm\displaystyle\int_0^1 f_n^{\prime}\)(x) dx = \(\rm\displaystyle\int_0^1\) f'(x) dx

= \(\displaystyle\int_0^1\) 0 dx

= 0

∴ \(\rm\displaystyle\lim_{n \rightarrow \infty} \int_0^1 f_n^{\prime}\) (x) dx = \(\rm\displaystyle\lim_{n \rightarrow \infty}\) 0 = 0, ∀ x ∈ [0, 1]

Hence, the sequence \(\rm\left\{\int_0^1 f_n^{\prime}(x)dx\right\}\) is convergent.

Hence, option (4) is correct.

Concept:

Pointwise Convergence:

Let \(\{f_n(x)\}\) be a sequence of functions defined on a domain  D . The sequence \(\{f_n(x)\}\) is said to

converge pointwise to a function  \(f(x)\) on  D  if, for every point \(x \in D\)  and for every \(\epsilon >0\), there exists an integer  N

 such that for all  \(n \geq N\), the following condition holds

\(|f_n(x) - f(x)| < \epsilon.\)
 

In other words, for each fixed point \(x \in D\), the values of the functions \(f_n(x) \) get arbitrarily close to \(f(x)\) as \(n \to \infty \).

Uniform Convergence:

Let \(\{f_n(x)\}\) be a sequence of functions defined on a domain D. The sequence \(\{f_n(x)\}\) is said to converge uniformly to a function

 f(x) on  D if for every \(\epsilon >0\), there exists an integer N  such that for all \(n \geq N\) and for all \(x \in D\), the following condition holds

\(|f_n(x) - f(x)| < \epsilon.\)
 

Explanation:

\(f(x) = \frac{1}{1 - x} for x \in [0, 1)\)

For \(n \geq 1\) , let \(p_n(x) = 1 + x + x^2 + \dots + x^n \). This is the sum of the first n+1 terms of a geometric series.

We know that the infinite geometric series \(1 + x + x^2 + \dots = \frac{1}{1 - x} for |x| < 1\) , which means \(p_n(x)\) is an approximation of f(x) as \(n \to \infty\)  .

Option 1: The function \(f(x) = \frac{1}{1 - x}\) blows up as \(x \to 1\) , meaning that as we get closer to 1, the function

value grows without bound. This implies that f(x) is not uniformly continuous because the difference in

function values for close inputs near 1 can be arbitrarily large.

Option 1) is true.

Option 2: The sequence \(p_n(x) = 1 + x + \dots + x^n\) is a partial sum of the geometric series, and we know that

the infinite sum converges to \(f(x) = \frac{1}{1 - x}\). Therefore, \(p_n(x)\) converges to f(x) pointwise on [0, 1) .

Option 2 is true.

Option 3: Uniform convergence requires that the convergence happens uniformly across the entire interval.

Near x = 1 , the function f(x) grows very large, and the sequence \(p_n(x)\) does not converge uniformly

because the convergence slows down significantly near x = 1 .

Option 3 is false.

Option 4: This is true because for any c < 1 , the function \(f(x) = \frac{1}{1 - x}\) is continuous and well-behaved

on the interval [0, c] , and the sequence \(p_n(x)\)  will converge uniformly on compact subintervals away from 1.

Option 4 is true.

The correct options are Option 1), Option 2), and Option 4).

Concept:

Pointwise convergent: A sequence of functions f₁, f₂, … , f_n, … : E → ℝ (E is a subset of ℝ) is said to converge pointwise on E to function f: E → ℝ if and only if

f(x) = \(\lim_{n\to \infty}f_n(x)\) for all x ∈ E 

Uniform convergent: Given a sequence of functions f_n: E → ℝ, we say fn converges uniformly to f if and only if \(\lim_{n\to \infty}\left(\sup_{x∈ E}|f_n(x)-f(x)|\right)\) = 0

Explanation:

\(f_n(x)=x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right) .\)

(1): For x = 0

f(x) = \(\lim_{n\to \infty}f_n(x)\) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) = 0

For x = 1                                                  

f(x) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) = \(\lim_{n\to \infty}\)\(1^n \log \left(\frac{4+{1^{\frac{1}{4}}}}{5}\right)\) = log 1 = 0

For 0 < x < 1,

f(x) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) = 0

So, (fn) converges pointwise on [0, 1].

(1) is correct.

(4): f_n(x) = \(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) is a sequence of continuous functions converges pointwise to a continuous function f(x) = 0 on a compact set [0, 1]

xⁿ⁺¹ ≤ xⁿ ∀ x ∈ [0, 1]

\(x^{n+1} \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) ≤ \(x^n \log \left(\frac{4+{x^{\frac{1}{4}}}}{5}\right)\) ∀ x ∈ [0, 1]

f_n+1(x) ≤ f_n(x) ∀ x ∈ [0, 1]

So, {fn(x)} is a increasing sequence

hence (fn) converges uniformly in [0, 1]

(4) is correct

Concept:

Pointwise convergent: A sequence of functions f₁, f₂, … , f_n, … : E → ℝ (E is a subset of ℝ) is said to converge pointwise on E to function f: E → ℝ if and only if

f(x) = \(\lim_{n\to \infty}f_n(x)\) for all x ∈ E 

Uniform convergent: Given a sequence of functions f_n: E → ℝ, we say fn converges uniformly to f if and only if \(\lim_{n\to \infty}\left(\sup_{x∈ E}|f_n(x)-f(x)|\right)\) = 0

Explanation:

\(f_n(x)=x^n \log \left(\frac{1+\sqrt{x}}{2}\right)\)

(1): For x = 0

f(x) = \(\lim_{n\to \infty}f_n(x)\) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{1+\sqrt{x}}{2}\right)\) = 0

For x = 1                                                  

f(x) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{1+\sqrt{x}}{2}\right)\) = \(\lim_{n\to \infty}\)\(1^n \log \left(\frac{1+\sqrt{1}}{2}\right)\) = log 1 = 0

For 0 < x < 1,

f(x) = \(\lim_{n\to \infty}\)\(x^n \log \left(\frac{1+\sqrt{x}}{2}\right)\) = 0

So, (fn) converges pointwise on [0, 1].

(1) is correct.

(4): f_n(x) = \(x^n \log \left(\frac{1+\sqrt{x}}{2}\right)\) is a sequence of continuous functions converges pointwise to a continuous function f(x) = 0 on a compact set [0, 1]

xⁿ⁺¹ ≤ xⁿ ∀ x ∈ [0, 1]

\(x^{n+1} \log \left(\frac{1+\sqrt{x}}{2}\right)\) ≤ \(x^n \log \left(\frac{1+\sqrt{x}}{2}\right)\) ∀ x ∈ [0, 1]

f_n+1(x) ≤ f_n(x) ∀ x ∈ [0, 1]

So, {fn(x)} is a increasing sequence

hence (fn) converges uniformly in [0, 1]

(4) is correct

Explanation:

\(f(x)=\sqrt{2+x}\), x ≥ -2 and the iteration

xn+1 = f(xn); n ≥ 0 for x0 = 0

Then we get

xn+1 = \(\sqrt{2+x_n}\)

Putting n = 0, 1, 2, ... we get

x1 = \(\sqrt{2+x_0}\) = \(\sqrt{2}\)

x2 = \(\sqrt{2+x_1}\) = \(\sqrt{2+\sqrt2}\)

x3 = \(\sqrt{2+x_2}\) = \(\sqrt{2+\sqrt{2+\sqrt2}}\)

Continuing this process we get the iteration

\(\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}\)

(1) is correct

Let $\lim x_n = l$ from the recurrence $l = \sqrt{2 + l}$squaring gives $l^2 - l - 2 = 0$, solving  $l = -1, 2$, but since $x_0 = 0$ only $l = 2$ is valid

(3) is correct

Explanation:

If the sequence of continuous function {fn} converges to a continuous function f(x) on a compact set A and {fn} ≥ {fn+1} for all n belongs to \(\mathbb N\), then {fn} tends to f(x) uniformly on A.

(3) is correct

Explanation:

Option (1):

Let f_n(x)  = \(x\over n\), which is a sequence of real-valued functions on \(\mathbb{R}\) and it converges uniformly to 0 on each closed and bounded subset of \(\mathbb{R}\) .

Now sup\(|\frac x n|→\infty \) as x → \(\infty\). So sequence {fn} not converges to f uniformly on \(\mathbb{R}\).

(1) is false.

Option 2:

Since for any point in \(\mathbb{R}\) we will get an interval and {fn} converges to a continuous function f uniformly on each closed and bounded subset of \(\mathbb{R}\). So the sequence {fn} converges to f pointwise on \(\mathbb{R}\)

Option (2) is correct.

Option (3):

Let f_n(x) = \(x\over n\), which is a sequence of real-valued functions on \(\mathbb{R}\) and it converges uniformly to 0 on each closed and bounded subset of \(\mathbb{R}\) . but f_n is not bounded for sufficiently large n.

(3) is false.

Option (4)

\(f(x)=\begin{cases}\frac1n, x\neq0\\0, x=0\end{cases}\) then f_n converges to a continuous function f uniformly on each closed and bounded subset of \(\mathbb{R}\) but f_n is not continuous at x = 0.

(4) is false.

Option (2) is correct.

Concept:

If f_n is a sequence of continuous functions that converges uniformly to a function f , then f is continuous. 

Uniform convergence means that for every ϵ > 0, there exists an N such that for all n ≥ N and for all x in the domain, |f_n(x) - f(x)| < ϵ.

Explanation:

1. Uniform Convergence of Continuous Functions:

When a sequence of functions {f_n} converges uniformly to a function f , it means the rate of convergence of f_n  to f does not depend on x .

This uniform convergence preserves the continuity of the limit function f .

2. Continuity of the Limit Function:

Since each f_n  is continuous and the convergence to f is uniform, f will also be continuous. This result is a well-known theorem in real analysis.

Given that f is the uniform limit of a sequence of continuous functions {f_n} , f is continuous.

Therefore, the correct answer is Continuous.

Concept:

Absolute and Uniform Convergence of Function Series:

• Given: \( f_n(x) = \dfrac{x}{(1 - x)^n} \), for all integers \( n \geq 0 \) and \( x \in [-1, 0] \).
• We are asked to examine the convergence behavior of \( \sum_{n=0}^\infty f_n(x) \) on \( [-1, 0] \) — both in terms of absolute convergence and uniform convergence.
• Absolute convergence: A function series \( \sum f_n(x) \) is absolutely convergent if \( \sum |f_n(x)| \) converges for all \( x \in [-1, 0] \).
• Uniform convergence: A function series converges uniformly if \( \sup_{x \in [-1, 0]} |S_N(x) - S(x)| \to 0 \) as \( N \to \infty \).

Calculation:

Given,

\( f_n(x) = \dfrac{x}{(1 - x)^n} \)

Step 1: Absolute Convergence

For each fixed \( x \in [-1, 0] \), we examine \( \sum_{n=0}^\infty |f_n(x)| = \sum_{n=0}^\infty \left| \frac{x}{(1 - x)^n} \right| \)

Note:

• For \( x = 0 \), we have \( f_n(0) = 0 \), so series trivially converges.
• For \( x \in [-1, 0) \), \( 1 - x \geq 1 \Rightarrow (1 - x)^n \geq 1 \), and \( x < 0 \Rightarrow |f_n(x)| = \frac{-x}{(1 - x)^n} \).
• The function \( |f_n(x)| \) decays exponentially fast, hence \( \sum |f_n(x)| \) converges for each \( x \in [-1, 0] \).

⇒ So the series is absolutely convergent on \( [-1, 0] \).

Step 2: Uniform Convergence

• Uniform convergence of a sequence \( f_n(x) \to 0 \) means: \( \sup_{x \in [-1, 0]} |f_n(x)| \to 0 \) as \( n \to \infty \).

We analyze: \( \sup_{x \in [-1, 0]} |f_n(x)| = \sup_{x \in [-1, 0]} \left| \frac{x}{(1 - x)^n} \right| \).

Let us define: \( g_n(x) = \left| \frac{x}{(1 - x)^n} \right| \).

Since \( x \le 0 \), we write: \( g_n(x) = \frac{-x}{(1 - x)^n} \).

To check uniform convergence, examine: \( \sup_{x \in [-1, 0]} g_n(x) \).

Let’s maximize \( g_n(x) \) over \( [-1, 0] \).

Set: \( g_n(x) = \frac{-x}{(1 - x)^n} \), and take derivative with respect to \( x \) to find maximum.

Alternatively, test values:

At \( x = -1 \Rightarrow f_n(-1) = -\frac{1}{2^n} \)

At \( x = -\frac{1}{n} \), we get: \( |f_n(-\frac{1}{n})| = \frac{1/n}{(1 + \frac{1}{n})^n} = \frac{1}{n} \cdot \left( \frac{n}{n+1} \right)^n \approx \frac{1}{n e} \)

So \( \sup_{x \in [-1, 0]} |f_n(x)| \ge \frac{1}{n e} \not\to 0 \)

This implies: \( \sup_{x \in [-1, 0]} |f_n(x)| \not\to 0 \) as \( n \to \infty \)

∴ Since the supremum of \( |f_n(x)| \) does not tend to zero, the sequence \( f_n(x) \) is not uniformly convergent  on \( [-1, 0] \).

Concept:

A sequence of function f_n(x) is said to be uniformly convergent to its limit point f(x) in a domain D if, for all ϵ > 0, there exists m ∈ \(\mathbb N\) depend upon ϵ but independent of x ∈ D such that |f_n(x) - f(x)| < ϵ for all n ≥ m

Explanation:

Cc(ℝ) = { f: ℝ → ℝ | f is continuous and there exists a compact set K such that f(x) = 0 for all x ∈ Kc}.

g(x) = \(\rm e^{x^2}\) for all x ∈ ℝ.

Let f_n(x) = \(\begin{cases}e^{-x^2}-e^{-n^2}&x∈ [-n, n]\\0& x∈ (-∞, -n]∪[n,∞)\end{cases}\)

Now, 

for x ∈ [-n, n]

|f_n(x) - g(x)| = |\(e^{-x^2}-e^{-n^2}-e^{-x^2}\)| = \(e^{-n^2}\) < ϵ for all n ≥ n₀

for x ∈ (-∞, -n] ∪ [n, ∞)

|fn(x) - g(x)| = |\(0-e^{-x^2}\)| =\(e^{-x^2}\) ≤  \(e^{-n^2}\) < ϵ for all n > n0