Dominated Convergence Theorem

1. Formula
   Observation

   Board shows $f_n \to f$, $x \in [a,b]$.

2. Audio
   Observation

   The speaker says there is a class of functions $f_n$ that tend to some particular function $f$ on an interval $a,b$.

$f_n$

Sequence or class of functions converging to $f$ on the stated interval.

Functions indexed by $n$, evaluated at $x \in [a,b]$.

1. Formula
   Observation

   Board shows $f_n \to f$, $x \in [a,b]$.

2. Audio
   Observation

   The speaker names the limiting function as $f$.

f

Limit function approached by $f_n$.

Function on $[a,b]$.

1. Formula
   Observation

   Board shows $x \in [a,b]$ in both hypothesis lines.

2. Audio
   Observation

   The speaker repeats “on an interval $a,b$” and “for all $n$ and $x$ on the interval $a,b$.”

$x \in [a,b]$

Point variable $x$ restricted to the closed interval from $a$ to $b$.

Real interval $[a,b]$.

1. Formula
   Observation

   Board shows $\exists g$ s.t. $|g| \ge |f_n|$, $x \in [a,b]$, $\forall n$, $\int_a^b g dx < \infty$.

2. Audio
   Observation

   The speaker says there exists a function $g$ such that the absolute value of $g$ is greater than or equal to the absolute value of $f_n$ for all $n$ and $x$ on $[a,b]$, and that $\int_a^b g dx$ does not diverge.

1. The board writes $|g|$ rather than the more common $g$ as a nonnegative dominating function; the video does not separately discuss measurability or sign conventions.

g

Dominating function used in the DCT hypotheses.

Function on $[a,b]$ with finite integral as written.

1. Formula
   Observation

   Board shows $\int_a^b g dx < \infty$ and $\int_a^b \sum f_n dx = \sum \int_a^b f_n dx$.

2. Audio
   Observation

   The speaker reads the integral from $a$ to $b$ of $g$ and then the integral from $a$ to $b$ of the sum of $f_n$.

$\int_a^b$

Definite integral over the interval $[a,b]$.

Applied to $g$, $f_n$, and $\sum f_n$ in the displayed formulas.

1. Formula
   Observation

   Board conclusion line shows $\int_a^b \sum f_n dx = \sum \int_a^b f_n dx$.

2. Audio
   Observation

   The speaker states that the integral of the sum equals the sum of the integrals.

1. The summation index is not explicitly written on the conclusion line; context indicates a sum over $n$.

2. The board uses $\sum f_n$ rather than a fully explicit $\sum_{n=1}^{\infty} f_n(x)$.

$\sum f_n$

Sum over the sequence of functions $f_n$, integrated term by term in the DCT conclusion.

Series of functions on $[a,b]$.

1. Formula
   Observation

   Board lower-left area shows $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$.

2. Audio
   Observation

   The speaker says they will evaluate the integral of cosine using power series.

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$

Maclaurin power series for cosine.

Real variable $x$; in the example the integration interval is $[0,\pi/2]$.

1. Audio
   Observation

   The speaker says “evaluate the integral of cosine of $x$ from zero to pi over two using power series.”

2. Formula
   Observation

   After the cut, the board shows $\int_0^{\pi/2} \cos x\,dx$ and later $\sin x\big|_0^{\pi/2}$.

[0,$\pi/2$]

Closed integration interval used in the worked example.

Real interval from $0$ to $\pi/2$.

1. Formula
   Observation

   Power-series denominator is written $(2n)!$.

(2n)!

Factorial of $2n$, appearing in the cosine Maclaurin series.

Defined for integer $n \ge 0$.

1. Formula
   Observation

   The board shows the power-series identity $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$.

$\cos x$

Cosine function represented by its Maclaurin power series in the worked integral.

Real variable $x$; used on the interval $[0,\pi/2]$ in the example.

1. Formula
   Observation

   The summation index is written as $n=0$ to $\infty$ throughout the series expressions.

$\sum_{n=0}^{\infty}$

Infinite sum over nonnegative integers indexing the terms of the power series.

$n\in\{0,1,2,\dots\}$.

1. Audio
   Observation

   The speaker says “two n factorial times two n plus one” can be pulled together into one factorial.

(2n)!

Factorial of $2n$, appearing as the coefficient denominator in the cosine series.

Defined for integer $n\ge 0$.

1. Audio
   Observation

   Speaker introduces “the Dominated Convergence Theorem, or DCT.”

2. Formula
   Observation

   Board title reads “Dominated Convergence Theorem.”

The video presents DCT as a theorem about interchanging an integral with an infinite sum under domination hypotheses. The speaker explicitly says the full measure-theoretic proof is not covered and instead gives the essential statement “as it pertains to Calculus BC.”

1. Presented in a calculus-oriented form rather than a full measure-theory proof.

1. Formula
   Observation

   Board line: $f_n \to f$, $x \in [a,b]$.

2. Audio
   Observation

   Speaker says there is a class of functions $f_n$ that tend to some particular function $f$ on an interval $a,b$.

1. The type of convergence is not specified beyond the notation $f_n \to f$; the video does not state pointwise, uniform, or almost-everywhere convergence explicitly.

The first stated condition is that a sequence of functions $f_n$ converges to a function $f$ on the interval $[a,b]$. This is shown on the board and read aloud as the setup for the theorem.

1. $f_n \to f$ on $x \in [a,b]$.

1. Dominated Convergence Theorem (DCT) as a calculus-level statement

1. Formula
   Observation

   Board line: $\exists g$ s.t. $|g| \ge |f_n|$, $x \in [a,b]$, $\forall n$.

2. Audio
   Observation

   Speaker says there exists a function $g$ such that the absolute value of $g$ is greater than or equal to the absolute value of $f_n$ for all $n$ and $x$ on the interval $a,b$.

1. The board uses $|g| \ge |f_n|$ rather than the more common $|f_n| \le g$ with $g \ge 0$; the video does not discuss this notational choice.

The second condition is the existence of a function $g$ that dominates the absolute values of all $f_n$ on the interval, uniformly in $n$. In the video’s wording, $|g|$ is greater than or equal to $|f_n|$ for every $n$ and every $x \in [a,b]$.

1. There exists $g$ such that $|g(x)| \ge |f_n(x)|$ for all $n$ and all $x \in [a,b]$.

1. Dominated Convergence Theorem (DCT) as a calculus-level statement

1. Formula
   Observation

   Board line includes $\int_a^b g dx < \infty$.

2. Audio
   Observation

   Speaker says the integral from $a$ to $b$ of $g$ does not diverge and has a finite integral.

1. The board writes the condition with $g$ rather than $|g|$; the video does not separately justify whether integrability of $g$ or $|g|$ is intended.

The final hypothesis shown is that the integral of the dominating function over $[a,b]$ is finite. The speaker paraphrases this as saying the integral “does not diverge.”

1. $\int_a^b g(x)\,dx < \infty$ as written on the board.

1. Second DCT hypothesis: domination by $g$

1. Formula
   Observation

   Board conclusion: $\int_a^b \sum f_n dx = \sum \int_a^b f_n dx$.

2. Audio
   Observation

   Speaker states that the integral from $a$ to $b$ of the sum of $f_n$ equals the sum of the integrals from $a$ to $b$ of $f_n$.

1. The summation index is implicit on the board; context indicates a sum over $n$.

2. The video does not explicitly connect the left side $\sum f_n$ to the earlier limit function $f$.

Under the stated hypotheses, the theorem allows one to exchange the definite integral over $[a,b]$ with the infinite summation, yielding term-by-term integration.

1. Assumes the previously stated convergence, domination, and finite-integral conditions.

1. First DCT hypothesis: convergence of $f_n$ to $f$ on $[a,b]$
2. Second DCT hypothesis: domination by $g$
3. Third DCT hypothesis: finite integral of the dominating function

1. Formula
   Observation

   Board shows $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$.

2. Audio
   Observation

   Speaker says they will evaluate the integral of cosine using power series.

The example begins from the standard Maclaurin series for $\cos x$, written explicitly on the board as an alternating even-power series with factorial denominators.

1. Used as the starting identity for the worked integral example.

1. Audio
   Observation

   The speaker says, “now we can make use of the DCT.”

The clip uses the Dominated Convergence Theorem as the justification for interchanging an infinite sum and a definite integral. In this example, the theorem is invoked after verifying a simple uniform domination condition for the series terms on the integration interval.

1. Used to justify swapping summation and integration.

2. The domination check is stated for the interval $[0,\pi/2]$.

1. Audio
   Observation

   The speaker refers to the term “negative one to the n over two n factorial times x to the two n.”

The example starts from the Maclaurin series for $\cos x$, writing cosine as an infinite power series in even powers of $x$ with alternating signs and factorial denominators.

1. Standard power-series representation of $\cos x$.

2. In the worked example it is used on $[0,\pi/2]$.

1. Audio
   Observation

   He then says, “So g of x equals one actually fulfills our conditions.”

1. The board does not show a separate written inequality line for the domination step; it is spoken while pointing at the series term.

2. The precise absolute-value notation is not visible, though the intended domination condition is clear from context.

To apply DCT, the presenter checks that each term of the cosine series is bounded in magnitude by $1$ on the interval $[0,\pi/2]$. This makes the constant function $g(x)=1$ an admissible dominating function for the termwise integration argument.

1. Applies on the closed interval $[0,\pi/2]$.

2. Holds for every $n\ge 0$.

3. Used specifically to satisfy the domination hypothesis needed before swapping sum and integral.

1. Dominated Convergence Theorem (DCT)
2. Power series for cosine

1. Audio
   Observation

   The speaker says DCT allows us to “swap sum and integral.”

After the domination condition is verified, the integral of the cosine series is rewritten as the sum of the integrals of its individual terms. This is the central move enabled by DCT in the clip.

1. Requires the domination condition checked just before this step.

2. Applied on $[0,\pi/2]$.

1. Dominated Convergence Theorem (DCT)
2. Domination by the constant function 1
3. Power series for cosine

1. Audio
   Observation

   The speaker says the factor can be removed from the integral “as that’s just a constant with regards to dx.”

Once the sum and integral have been swapped, the coefficient $\frac{(-1)^n}{(2n)!}$ no longer depends on the integration variable $x$, so it is taken outside the integral, simplifying the remaining task to integrating $x^{2n}$.

1. Valid because $\frac{(-1)^n}{(2n)!}$ is constant with respect to $x$ for each fixed $n$.

2. Performed after term-by-term integration has been justified.

1. Interchanging sum and integral

1. Audio
   Observation

   The speaker says, “we can just integrate x to the two n in general,” producing $x$ to the $2n+1$ over $2n+1$ evaluated at the endpoints.

Each remaining integral is a basic monomial integral. The power rule raises the exponent by one and divides by the new exponent, giving an explicit antiderivative for every term in the series.

1. Used for each fixed integer $n\ge 0$.

2. Evaluated here on $[0,\pi/2]$.

1. Factoring the $n$-dependent coefficient out of the integral

1. Formula
   Observation

   Board lists the hypotheses and conclusion in sequence.

2. Audio
   Observation

   Speaker verbally states each condition and then the result.

1. The video does not specify the exact mode of convergence in $f_n \to f$.

2. The board’s use of $|g| \ge |f_n|$ and $\int_a^b g dx < \infty$ differs from some standard textbook formulations; the clip does not explain this choice.

If $f_n \to f$ on $[a,b]$, if there exists $g$ such that $|g| \ge |f_n|$ for all $n$ and all $x \in [a,b]$, and if $\int_a^b g\,dx < \infty$, then $\int_a^b \sum f_n\,dx = \sum \int_a^b f_n\,dx$.

1. $f_n \to f$ for $x \in [a,b]$

2. There exists $g$ such that $|g| \ge |f_n|$ for all $n$ and all $x \in [a,b]$

3. $\int_a^b g\,dx < \infty$

For all $n$ and all $x \in [a,b]$ in the domination condition; the conclusion is an equality of an integral of a sum and a sum of integrals.

1. Audio
   Observation

   Speaker says these conditions are important because if they are not met, the nice result doesn’t necessarily hold.

2. Formula
   Observation

   Boxed counterexample text: $e^{2x}$ “dominates” $e^x$ on $[0,\infty)$ but $\int_0^{\infty} e^{2x} dx$ diverges.

1. The example is presented as a warning illustration rather than a fully formal counterexample to the exact theorem statement on the board.

If the DCT hypotheses are not satisfied, the equality between the integral of the sum and the sum of the integrals need not hold.

1. At least one DCT hypothesis fails.

2. The video illustrates failure using a dominating function whose integral diverges.

Existential warning: there are cases where the conclusion fails when hypotheses are absent.

1. Formula
   Observation

   The board displays the transition from $\int_0^{\pi/2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\,dx$ to $\sum_{n=0}^{\infty}\int_0^{\pi/2}\frac{(-1)^n}{(2n)!}x^{2n}\,dx$.

1. The full formal statement of DCT is not written out; only its application to this example is shown.

Because the series terms are dominated on $[0,\pi/2]$, the Dominated Convergence Theorem permits interchanging the infinite sum and the definite integral: $\int_0^{\pi/2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\,dx = \sum_{n=0}^{\infty}\int_0^{\pi/2}\frac{(-1)^n}{(2n)!}x^{2n}\,dx$.

1. The integrand is represented by the cosine power series.

2. Each term is bounded in magnitude by an integrable dominating function on $[0,\pi/2]$.

3. Here the dominating function is taken to be $g(x)=1$.

For all $n\ge 0$ and all $x\in[0,\pi/2]$ in the domination check; the conclusion is an equality of a definite integral and an infinite sum of definite integrals.

1. Audio
   Observation

   The speaker states that the answer is one.

Using the displayed power-series and DCT argument, $\int_0^{\pi/2}\cos x\,dx = 1$.

1. Start from $\cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$.

2. Apply the domination check and termwise integration shown on the board.

3. Recognize the resulting series as the sine series.

A definite numerical equality for the specific integral over $[0,\pi/2]$.

1. Audio
   Observation

   Speaker says they will evaluate $\int \cos x$ from $0$ to $\pi/2$ using power series, then integrate both sides from $0$ to $\pi/2$ with respect to $dx$.

2. Formula
   Observation

   Before the cut, the board shows $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$.

3. Formula
   Observation

   After the cut, the board shows the integrated series, the antiderivative step, and the final boxed answer $1$.

1. The actual spoken explanation of the post-cut algebra is not present in the provided audio; only the visual board content after the cut is available.

2. The clip does not audibly verify the DCT hypotheses for this specific cosine-series example.

1. Expression

   $$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$$
   Explanation

   Start from the Maclaurin series for cosine.

   Justification

   Standard power-series identity shown on the board.

   Shown in the video
2. Expression

   $$\int_0^{\pi/2} \cos x\,dx = \int_0^{\pi/2} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}\,dx$$
   Explanation

   Integrate both sides of the cosine series over $[0,\pi/2]$.

   Justification

   Stated verbally by the speaker and shown on the board after the cut.

   Shown in the video
3. Expression

   $$= \sum_{n=0}^{\infty} \int_0^{\pi/2} \frac{(-1)^n}{(2n)!} x^{2n}\,dx$$
   Explanation

   Interchange the integral and the infinite sum.

   Justification

   This is the term-by-term integration step motivated by the DCT discussion, though the clip does not audibly check the hypotheses for this example.

   Shown in the video
4. Expression

   $$= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \cdot \frac{x^{2n+1}}{2n+1} \Big|_0^{\pi/2}$$
   Explanation

   Integrate each monomial $x^{2n}$ to obtain $x^{2n+1}/(2n+1)$ and keep the constant factor $(-1)^n/(2n)!$.

   Justification

   Power rule for integration, visible on the board after the cut.

   Shown in the video
5. Expression

   $$= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{2}\right)^{2n+1}$$
   Explanation

   Simplify the coefficient using $(2n+1)(2n)!=(2n+1)!$ and evaluate at the upper limit; the lower limit contributes $0$.

   Justification

   Algebraic simplification and Fundamental Theorem of Calculus, shown on the board after the cut.

   Shown in the video
6. Expression

   $$= \sin\left(\frac{\pi}{2}\right)$$
   Explanation

   Recognize the resulting series as the Maclaurin series for sine evaluated at $\pi/2$.

   Justification

   Standard identification of the sine power series; visible on the board after the cut.

   Shown in the video
7. Expression

   $$= 1$$
   Explanation

   Conclude the value of the definite integral.

   Justification

   Direct evaluation of $\sin(\pi/2)$, boxed on the board after the cut.

   Shown in the video

The worked example yields $\int_0^{\pi/2} \cos x\,dx = 1$ by integrating the cosine power series term by term and recognizing the resulting sine series.

1. Audio
   Observation

   The speaker narrates each major step in order: domination check, swap sum and integral, pull out constants, integrate $x^{2n}$, combine factorials, recognize sine, evaluate endpoints.

1. Some intermediate algebra is compressed visually; for example, the domination inequality is spoken rather than separately written out on the board.

1. Expression

   $$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$$
   Explanation

   Begin with the Maclaurin series for cosine.

   Justification

   Standard power-series expansion shown on the first board line.

   Shown in the video
2. Expression

   $$\left|\frac{(-1)^n}{(2n)!}x^{2n}\right| \le 1 \quad \text{for } x\in[0,\pi/2],\ n\ge 0$$
   Explanation

   Check that each series term is bounded by $1$ on the integration interval.

   Justification

   Spoken domination condition used to meet the hypothesis needed for DCT.

   Shown in the video
3. Expression

   $$\int_0^{\pi/2}\cos x\,dx = \int_0^{\pi/2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\,dx$$
   Explanation

   Substitute the cosine series into the definite integral.

   Justification

   Direct replacement of $\cos x$ by its power series.

   Shown in the video
4. Expression

   $$= \sum_{n=0}^{\infty}\int_0^{\pi/2}\frac{(-1)^n}{(2n)!}x^{2n}\,dx$$
   Explanation

   Interchange the sum and the integral.

   Justification

   Allowed by the Dominated Convergence Theorem after the domination check.

   Shown in the video
5. Expression

   $$= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\int_0^{\pi/2}x^{2n}\,dx$$
   Explanation

   Move the $n$-dependent coefficient outside each integral.

   Justification

   Linearity of integration with respect to constants in the variable $x$.

   Shown in the video
6. Expression

   $$= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\left.\frac{x^{2n+1}}{2n+1}\right|_0^{\pi/2}$$
   Explanation

   Integrate the monomial $x^{2n}$.

   Justification

   Power rule for integration.

   Shown in the video
7. Expression

   $$= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\Big|_0^{\pi/2}$$
   Explanation

   Combine $(2n)!$ and $2n+1$ into $(2n+1)!$.

   Justification

   Factorial identity $(2n)!(2n+1)=(2n+1)!$.

   Shown in the video
8. Expression

   $$= \sin x\Big|_0^{\pi/2}$$
   Explanation

   Recognize the resulting series as the Maclaurin series for $\sin x$.

   Justification

   Standard sine power series identification stated by the speaker.

   Shown in the video
9. Expression

   $$= \sin\left(\frac{\pi}{2}\right)-\sin(0)=1-0=1$$
   Explanation

   Evaluate at the endpoints to obtain the final numerical answer.

   Justification

   Fundamental theorem of calculus endpoint evaluation plus standard sine values.

   Shown in the video

The worked derivation proves, within the displayed method, that $\int_0^{\pi/2}\cos x\,dx=1$.

1. Formula
   Observation

   Boxed board text: $e^{2x}$ “dominates” $e^x$ on $[0,\infty)$ but $\int_0^{\infty} e^{2x} dx$ diverges.

2. Audio
   Observation

   Speaker says that if the conditions are not met, the nice result doesn’t necessarily hold, and gives this exponential example.

1. The example is used illustratively; the clip does not spell out a full formal contradiction to a specific series equality.

Show why the DCT-type hypotheses matter by considering a case where a larger function dominates a smaller one but its integral diverges.

1. On $[0,\infty)$, $e^{2x}$ is said to dominate $e^x$.

2. $\int_0^{\infty} e^{2x}\,dx$ diverges.

Explain that without suitable domination plus finite-integral conditions, the desired interchange result may fail.

1. Expression

   $$e^{2x} \text{ dominates } e^x \text{ on } [0,\infty)$$
   Explanation

   The larger exponential bounds the smaller one on the half-line.

   Justification

   Stated on the board and in the audio.

   Shown in the video
2. Expression

   $$\int_0^{\infty} e^{2x}\,dx \text{ diverges}$$
   Explanation

   Despite domination, the proposed controlling function has an infinite integral.

   Justification

   Stated on the board and in the audio.

   Shown in the video
3. Expression

   $$\text{Therefore the “nice result” need not hold.}$$
   Explanation

   The example is used to motivate why the finite-integral hypothesis is essential.

   Justification

   Speaker’s explanatory conclusion.

   Shown in the video

The example illustrates that domination alone is insufficient if the dominating function’s integral diverges.

No separate numerical verification is given in the clip beyond the stated divergence.