## Tuesday, April 14, 2009

### Dear TRS: Reality sets in

That's right folks, it's time for another exciting edition of ask the RS. Incidentally, all of these are real questions, sent in by real people. Scary, I know. Here we go:

Dear Sir:

Can you think of an English rhyme to aid baalei tshuva in understanding the talmudic and practical concept of "der oilam's ah goilem?"
sincerely,

Shliach to the poets-in-residence
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Dear Shliach to the poets-in-residence,

Are you trying to suggest that the masses are asses? And not even the asses of Rebbi Pinchas Ben Yair?
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Dear Sir:
Is the prohibition of music in effect on hallel days that occur during sfira? If so does yom ha'atzmaut (known to certain chagas kreizin as yoim huhhatzomos) count?

What about stuff that is technically produced on instruments but is not music by any sane definition? (Insert singer/group you dislike)
sincerely,

Metal Fan
--
Dear Metal Fan,

The answer to your first question is that if you're willing to suspend rational thought and hold that you can't listen to whatever it is you're not listening to then the same would apply on hallel days as well, obviously excepting the 5th of Iyar, when it's a mitzvah to play Hatikvah on loudspeakers from a Mitzvah Tank in Williamsburg.

Any music which is not music by any sane definition shouldn't be listened to anyway, so what's the question? You mean if you're of a snaggish disposition and have masochistic tendencies?
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Dear Sir:

What is the fundamental question of integral calculus and what does it have to do with differential calculus? And oh yeah, what is the point of that little carrot shaped button on my blue calculator?

Sincerely,

Mathematically Mystified
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Dear Mathematically Mystified,

I.
Let f be a continuous real-valued function defined on a closed interval a,">b. If F is the function defined for x in a,">b by
$F\left(x\right) = \int_a^x f\left(t\right)\, dt$
then
$F\text{'}\left(x\right) = f\left(x\right)\,$
for every x in a,">b.

II.
Let F be a continuous real-valued function defined on a closed interval a,">b. If f is the function such that
$f\left(x\right) = F\text{'}\left(x\right)\,$ for all x in a,">b
then
$\int_a^b f\left(x\right) dx = F\left(b\right) - F\left(a\right)$.

### Corollary

Let F be a real-valued continuous function defined on a closed interval a,">b. If f is the function defined by
$f\left(x\right) = F\text{'}\left(x\right)\,$ for all x in a,">b
then
$F\left(x\right) = \int_a^x f\left(t\right) dt + F\left(a\right)$
and
$f\left(x\right) = \frac\left\{d\right\}\left\{dx\right\} \int_a^x f\left(x\right) dx$.

## Proof

### Part I

It is given that
$F\left(x\right) = \int_\left\{a\right\}^\left\{x\right\} f\left(t\right) dt$
Let there be two numbers x1 and x1 + Δx in a,">b. So we have
$F\left(x_1\right) = \int_\left\{a\right\}^\left\{x_1\right\} f\left(t\right) dt$
and
$F\left(x_1 + \Delta x\right) = \int_\left\{a\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt$.
Subtracting the two equations gives
$F\left(x_1 + \Delta x\right) - F\left(x_1\right) = \int_\left\{a\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt - \int_\left\{a\right\}^\left\{x_1\right\} f\left(t\right) dt \qquad \left(1\right)$.
It can be shown that
$\int_\left\{a\right\}^\left\{x_1\right\} f\left(t\right) dt + \int_\left\{x_1\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt = \int_\left\{a\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt$.
(The sum of the areas of two adjacent regions is equal to the area of both regions combined.)
Manipulating this equation gives
$\int_\left\{a\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt - \int_\left\{a\right\}^\left\{x_1\right\} f\left(t\right) dt = \int_\left\{x_1\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt$.
Substituting the above into (1) results in
$F\left(x_1 + \Delta x\right) - F\left(x_1\right) = \int_\left\{x_1\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt \qquad \left(2\right)$.
According to the mean value theorem for integration, there exists a c in x1,">x1 + Δx such that
$\int_\left\{x_1\right\}^\left\{x_1 + \Delta x\right\} f\left(t\right) dt = f\left(c\right) \Delta x$.
Substituting the above into (2) we get
$F\left(x_1 + \Delta x\right) - F\left(x_1\right) = f\left(c\right) \Delta x \,$.
Dividing both sides by Δx gives
$\frac\left\{F\left(x_1 + \Delta x\right) - F\left(x_1\right)\right\}\left\{\Delta x\right\} = f\left(c\right)$.
Notice that the expression on the left side of the equation is Newton's difference quotient for F at x1.
Take the limit as Δx → 0 on both sides of the equation.
$\lim_\left\{\Delta x \to 0\right\} \frac\left\{F\left(x_1 + \Delta x\right) - F\left(x_1\right)\right\}\left\{\Delta x\right\} = \lim_\left\{\Delta x \to 0\right\} f\left(c\right)$
The expression on the left side of the equation is the definition of the derivative of F at x1.
$F\text{'}\left(x_1\right) = \lim_\left\{\Delta x \to 0\right\} f\left(c\right) \qquad \left(3\right)$.
To find the other limit, we will use the squeeze theorem. The number c is in the interval x1,">x1 + Δx, so x1cx1 + Δx. Also, $\lim_\left\{\Delta x \to 0\right\} x_1 = x_1$ and $\lim_\left\{\Delta x \to 0\right\} x_1 + \Delta x = x_1$. Therefore, according to the squeeze theorem,
$\lim_\left\{\Delta x \to 0\right\} c = x_1$.
Substituting into (3), we get
$F\text{'}\left(x_1\right) = \lim_\left\{c \to x_1\right\} f\left(c\right)$.
The function f is continuous at c, so the limit can be taken inside the function. Therefore, we get
$F\text{'}\left(x_1\right) = f\left(x_1\right) \,$.
which completes the proof. (Leithold et al, 1996)

### Part II

This is a limit proof by Riemann Sums. Let f be continuous on the interval a,">b, and let F be an antiderivative of f. Begin with the quantity
$F\left(b\right) - F\left(a\right)$.
Let there be numbers x1 thru xn such that $a = x_0 < x_n =" b. It follows that$
$F\left(b\right) - F\left(a\right) = F\left(x_n\right) - F\left(x_0\right) \,$.
Now, we add each F(xi) along with its additive inverse, so that the resulting quantity is equal:
$\begin\left\{matrix\right\} F\left(b\right) - F\left(a\right) & = & F\left(x_n\right)\,+\,-F\left(x_\left\{n-1\right\}\right)\,+\,F\left(x_\left\{n-1\right\}\right)\,+\,\ldots\,+\,+ F\left(x_1\right)\,-\,F\left(x_0\right) \, \\$
& = & F(x_n)\,-\,F(x_{n-1})\,+\,F(x_{n-1})\,+\,\ldots\,-\,F(x_1)\,+\,F(x_1)\,-\,F(x_0) \, \end{matrix} The above quantity can be written as the following sum:
$F\left(b\right) - F\left(a\right) = \sum_\left\{i=1\right\}^n- F\left(x_\left\{i-1\right\}\right)\qquad \left(1\right)$

Here we employ the Mean Value Theorem. In brief, it is as follows:

Let f be continuous on the closed interval a,">b and differentiable on the open interval (a, b). Then there exists some c in (a, b) such that
$f\text{'}\left(c\right) = \frac\left\{f\left(b\right) - f\left(a\right)\right\}\left\{b - a\right\}$.
It follows that
$f\text{'}\left(c\right)\left(b - a\right) = f\left(b\right) - f\left(a\right) \,$.
The function F is differentiable on the interval a,">b; therefore, it is also differentiable and continuous on each interval xi-1. Therefore, according to the Mean Value Theorem (above),
$F\left(x_i\right) - F\left(x_\left\{i-1\right\}\right) = F\text{'}\left(c_i\right)\left(x_i - x_\left\{i-1\right\}\right) \,$.
Substituting the above into (1), we get
$F\left(b\right) - F\left(a\right) = \sum_\left\{i=1\right\}^n- x_\left\{i-1\right\}\right)$.
As the derivative of the antiderivative is the original function, $F\text{'}\left(c_i\right) = f\left(c_i\right)$. Also, $x_i - x_\left\{i-1\right\}$ can be expressed as $\Delta x$ of partition $i$.
$F\left(b\right) - F\left(a\right) = \sum_\left\{i=1\right\}^nx_i\right)\qquad \left(2\right)$
Notice that we are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the Mean Value Theorem, describes an approximation of the curve section it is drawn over. Also notice that $\Delta x_i$ does not need to be the same for any value of $i$, or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with $n$ rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we will get closer and closer to the actual area of the curve. By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann integral. That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity. So, we take the limit on both sides of (3). This gives us
$\lim_\left\{\| \Delta \| \to 0\right\} F\left(b\right) - F\left(a\right) = \lim_\left\{\| \Delta \| \to 0\right\} \sum_\left\{i=1\right\}^nx_i\right)\,dx$
The expressions F(b) and F(a) are not dependent on ||Δ||, so the limit on the left side remains F(b) - F(a).
$F\left(b\right) - F\left(a\right) = \lim_\left\{\| \Delta \| \to 0\right\} \sum_\left\{i=1\right\}^nx_i\right)$
The expression on the right side of the equation defines an integral over f from a to b. Therefore, we obtain
$F\left(b\right) - F\left(a\right) = \int_\left\{a\right\}^\left\{b\right\} f\left(x\right)\,dx$
which completes the proof.

Happy?

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Dear Sir:

My recently engaged friend was appointed to oversee all dorm-food preparation and thus is at her wits end. How might I go about telling her that raw eggs mixed with mushy bananas doesn't create ice cream, without losing my head?

Sincerely,

A'feared of Wrath
--
Dear A'feared of Wrath

Just give her this simple recipe, and mention that TRS is now accepting donations:

Blend 6 very ripe bananas until smooth. Add 2 egg yolks and 3/4 cup of ground nuts, filberts, or almonds. Blend for one minute. Beat 4 egg whites until stiff. Mix slowly into banana mixture with wooden spoon. Pour into a glass dish and spread remaining 1/4 cup nuts on top. Freeze at least 4 hours. When serving poor 1 teaspoon wine on each scoop or portion.

Serves 12
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Dear Sir:

My brother insists on not cutting his afro and I fear that by the end of sefira all hope, not to mention sight, will be lost. Do you think he's mind if I trim it during the night? Then again, it would make for a lovely sheitel...

Sincerely,

Distressed
--
Dear Distressed,

Do I think he'll mind if you sin against all that is holiness and sacred? Of course, locks for love is a wonderful organization.
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Dear Sir:

What's a Chol Hamoed trip that will be enjoyable for a family aged 1.5 through 55?

Sincerely,

Even more desperate
--
Dear Even more desperate,

What exactly do 1.5 year olds enjoy doing anyway? From what I can tell, they're the same happy whether here or there.
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Dear Sir:

Why is borscht in America made of beets and borscht in Russia made with cabbage?

Sincerely,

Ari
--
Dear Ari,

Perhaps because Americans prefer sour cream nine times out of ten?
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And that's all folks!

yoish said...

thanx for teaching us math!!! (i didn't actually read it, but it looks complicated.)

Modeh B'Miktsas said...

I did read it. It is complicated. There are much simpler ways of writing that which might even make sense. Then again, they don't look half so impressive.

Oh, and the carrot thingy is for exponents.

The Real Shliach said...

yoish: don't worry, I didn't read it either.

Modeh: Carrots are for carrot cake, obviously.

le7 said...

How does that even answer the Calculus question?

Modeh B'Miktsas said...

It doesn't quite. He took a proof of the Fundamental theorem and pasted it in. As far as I can tell it isn't even much of a proof but I didn't actually slog through it. I have better things to do.

The Real Shliach said...

le7: How should I know?

Modeh: Exactly.

le7 said...

Why are there only 6 comments on this post?

(Now 7).

sarabonne said...

Because Pesach rather limits one's intnet time...by te way, that recipe sounds good.

The Real Shliach said...

le7: now 9.

sara: really? shocking!
and yeah, it is a real recipe.

Crawling Axe said...

Next time you want a fake Math theorem talk to a dilettante extraordinaire. I.e., find any Russian Jew. But to an untrained eye looks hilarious.

See, you would tell me it’s a question of taste, but I would answer that your funny posts are objectively funnier than supposedly funny posts of other supposedly funny supposedly frum bloggers.

le7 said...

I definitely second CA.

The Real Shliach said...

Re: math: it's not a fake math theorem! I have no idea what it means, but it's not fake.

re: humor: thanks, CA+le7

Crawling Axe said...

I meant fake math question. Or something like that. All math is real. Even the one proven wrong. It’s just that its source is on such spiritual levels that it cannot cloth itself in this world’s reality.

By the way, you should know that equation. And I should know what brocha to say on raw lemons. Neither of us is doing a good job.

The Real Shliach said...

I should know that equation? Why?

Crawling Axe said...

Nu, nu. You are a shliach of the Rebbe! (Said in the same tone as “I am captain Jack Sparrow!”)

The Real Shliach said...

Which tone would that be?