# [SOLVED]How to use Taylor series to represent any well-behaved f(x)

#### DeusAbscondus

##### Active member
Does one assess $x$ at $x=0$ for the entire series? (If so, wouldn't that have the effect of "zeroing" all the co-efficients when one computes?)
only raising the value of $k$ by $1$ at each iteration?
and thereby raising the order of derivative at each iteration?

$$\sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!} x^k= f(0)+\frac{df}{dx}|_0 \ x + \frac{1}{2!}\frac{d^2f}{dx^2}|_0 \ x^2+ \frac{1}{3!}\frac{d^3f}{dx^3}|_0 \ x^3+ ....$$

I have no experience with series or sequences, so, I know I have to remedy this gap in my knowledge.
In the interim, however, I am currently enrolled in a Math course that looks at Calculus by beginning with Taylor series.
I am an adult beginner at Math, having done an introductory crash course in Calculus last year; I wanted to flesh this out: hence my current enrolment.

But I am at a loss to know how to manipulate the notation above and would appreciate a worked solution for some simple $f(x)$ (I won't nominate one, so as to preclude the possibility of my cheating on some set work)
I just need to see this baby in action with a "well-behaved function" of someone else's choosing, with some notes attached if that someone would be so kind.

Thanks,
Deo Abscondo

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#### Bacterius

##### Well-known member
MHB Math Helper
Re: How to use Taylor series to represent any well-behaved $f(x)$

Yes, the idea is that as you add more and more terms to the Taylor series, the series approximation becomes better and better and fits the function more closely for values farther away from $x = 0$.

For instance, let's take the venerable $y = \sin(x)$ function. Let's plot it:

The first term of the Taylor series (i.e. the Taylor series approximation at $k = 0$) is just $f(0) = 0$, hence:

Of course, this approximation sucks. Let's try $k = 1$. Then the Taylor series approximation for $k = 1$ is:
$$\sum_{k = 0}^1 \frac{f^k (0)}{k!} x^k = \sin(0) + x \cos(0) = x$$
And our "first-order approximation" for $\sin(x)$ is the curve $y = x$, as illustrated below:

Still not an awesome approximation, but it works pretty well for $x$ close to zero. In fact this is known as the small-angle approximation which says that $\sin(x) \approx x$ for small $x$.

What about the second order approximation $k = 2$, which is given by:
$$\sum_{k = 0}^2 \frac{f^k (0)}{k!} x^k = \sin(0) + x \cos(0) - \frac{sin(0)}{2!} x^2 = x$$
It turns out that the new term becomes zero. Ok.. fine.. that happens, so what about $k = 3$? Now we see that:
$$\sum_{k = 0}^3 \frac{f^k (0)}{k!} x^k = x - \frac{\cos(0)}{3!} x^3 = x - \frac{x^3}{3!}$$
And let's plot this:

Wow! That's a really good approximation for all $|x| < 1$. And we see a pattern: repeatedly differentiating $\sin(x)$ will end up giving you $\cos(x)$, $- \sin(x)$, $- \cos(x)$, $\sin(x)$ endlessly. Every second term will become zero because of the $\sin(0)$ term, so only odd-numbered terms actually matter. We conclude that:
$$\sin(x) = \sum_{k = 0}^\infty \frac{f^k (0)}{k!} x^k = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
Here's the Taylor series approximation for $k = 7$. Check it out:

And here's the one for $k = 9$:

As you add more and more terms, the approximation tends towards the function for larger and larger values of $x$. In the limit, the Taylor series is equal to the original function.

So in other words, a Taylor series is another representation of a function, as an infinite series (a sum of infinitely many terms). A Taylor series approximation is the Taylor series truncated at a finite number of terms, which has the nice property of approximating the function around $x = 0$, and is often easier to calculate and work with, especially in physics where approximations are often used.

There is in fact a theorem that gives the maximal error of the Taylor series approximation at any point $x$ of the function in terms of $k$. Of course, as $k$ tends to infinity, the error tends to zero. This is Taylor's Theorem.

EDIT: uploaded images to imgur for perennity. W|A does strange things to externally linked images.

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#### DeusAbscondus

##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$

Thanks kindly Bacterius.

But how would this work for a polynomial?
I mean, wouldn't the evaluation $x=0$ lead to "zeroing" all the co-efficients and resulting in non-sense?

If I take the case of $y=x^2+2x$ is this amenable to being described by the same series?
$$\sum_{k=0}^{\infty}\ \frac{f^k (0)}{k!}=0+ 2(0) + \frac{1}{2!}(2\cdot 0) .....$$
This doesn't seem to work! All I get is an infinite string of 0s!

So in other words, a Taylor series is another representation of a function, as an infinite series (a sum of infinitely many terms). A Taylor series approximation is the Taylor series truncated at a finite number of terms, which has the nice property of approximating the function around $x = 0$, and is often easier to calculate and work with, especially in physics where approximations are often used.

There is in fact a theorem that gives the maximal error of the Taylor series approximation at any point $x$ of the function in terms of $k$. Of course, as $k$ tends to infinity, the error tends to zero. This is Taylor's Theorem.

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#### Bacterius

##### Well-known member
MHB Math Helper
Re: How to use Taylor series to represent any well-behaved $f(x)$

Taylor series of polynomials have the interesting property that they are in fact equal to the polynomial itself, and all extra terms are zero. In other words, the Taylor series for $x^2 + 2x$ is $x^2 + 2x + 0 + 0 + \cdots$. To see why:
$$f(x) = x^2 + 2x$$
$$f'(x) = 2x + 2$$
$$f''(x) = 2$$
$$f'''(x) = 0$$
.. and any further differentiation still gives zero

So:
$$\sum_{k = 0}^\infty \frac{f^{(k)} (0)}{k!} x^k = \frac{0^2 + 2 \cdot 0}{0!} x^0 + \frac{2 \cdot 0 + 2}{1!} x^1 + \frac{2}{2!} x^2 + 0 + \cdots = 0 + 2x + x^2 + 0 + \cdots = x^2 + 2x$$
So, yes, it still works, although it would seem to be less useful for polynomials than for other functions (but then, polynomials are easy to compute and are already pretty simple. I don't think "reducing" them to infinite series generally helps).

To calculate Taylor series I recommend you write down the iterated derivative of your function $f(x)$ and then plug in the numbers. Remember, $f^{(k)} (0)$ means "the $k$th derivative of $f$ evaluated at $x = 0$". The $x$ in your Taylor series is a different $x$ and is *not* equal to zero.

Also, there is a generalization of Taylor's series which is centered on arbitrarily values of $x$ instead of $x = 0$. I'll let you work out the general expression, though, it's an interesting exercise.

#### DeusAbscondus

##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$

That has nailed it for me Bacterius!
Mightily obliged to you.

As per usual, the problem has vanished under the gaze of fresh eyes.

(Going now to chew on this strong meat with a cup of medicinal wine "to aid the digestion")

Cheers,
D'abs

Taylor series of polynomials have the interesting property that they are in fact equal to the polynomial itself, and all extra terms are zero. In other words, the Taylor series for $x^2 + 2x$ is $x^2 + 2x + 0 + 0 + \cdots$. To see why:
$$f(x) = x^2 + 2x$$
$$f'(x) = 2x + 2$$
$$f''(x) = 2$$
$$f'''(x) = 0$$
.. a.

#### DeusAbscondus

##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$

The x in your Taylor series is a different x and is *not* equal to zero.
Okay, basically, I'm in the clear .....
but just what is this "different x"?

If it is distinct from the $x$ of my polynomial, by what principle/rule do I distinguish the two when I come to compute the polynomial?

D'abs

#### Bacterius

##### Well-known member
MHB Math Helper
Re: How to use Taylor series to represent any well-behaved $f(x)$

Okay, basically, I'm in the clear .....
but just what is this "different x"?

If it is distinct from the $x$ of my polynomial, by what principle/rule do I distinguish the two when I come to compute the polynomial?

D'abs
The $x$ in the Taylor series is the $x$ at which you are evaluating your Taylor series (or your approximation of it). The derivative is always evaluated at 0 for this version of the Taylor series (sorry I agree it was a bit confusing, there is only one $x$, it's just the derivative is evaluated at a constant).

Essentially you are numerically evaluating derivatives of your original function, and using the resulting values as coefficients for your Taylor series (this is not a particularly useful way to think of it but it may be more intuitive to you to understand what is what)

#### DeusAbscondus

##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$

That makes sense.
Meanwhile, I've been plugging and chugging a few simple polynomial functions through the T. series: hehe! what fun! it works!!! (*broadly grins)

You made my day: not understanding was getting me down (as usual);
and, again as usual, once some fresh light comes, and the aha! moment arrives, and one feels exuberant rather than dispirited.

Have a great weekend Bacteriu!

[QUOTEEssentially you are numerically evaluating derivatives of your original function, and using the resulting values as coefficients for your Taylor series (this is not a particularly useful way to think of it but it may be more intuitive to you to understand what is what)[/QUOTE]


# Minimum number of states in dfa accepting binary number with decimal equivalent divisible by $n$

I was aware of the fact that, if DFA needs to accept binary string with its decimal equivalent divisible by $n$, then it can have minimum $n$ states.

However recently came across following text:

• If $n$ is power of $2$
Let $n=2^m$, so number of minimum states $=m+1$. For $n=8=2^3$, we need $3+1=4$ states.
• Else If $n$ is odd
Number of states $=n$. For $n=5$, we need $5$ states.
• Else if $n$ is even
Keep dividing it with $2$ until you get odd quotient. The result is final odd quotient plus number of divisions done. For $n=20,20/2=10,10/2=5→5+2=7$ states

I was guessing:
Q1. From where all these facts came. (I know this must have come from general pattern that such DFAs follow, but then I have below doubts)
Q2. Are these points indeed correct?
Q3. If yes (for Q2), then what can be the reason for each point?
Q4. Do these three bullet points cover all cases? (I guess yes, since there is even and odd, but I am new to this, and I was guessing if any such more point is left unmentioned in my textbook, especially since I did not find any reference book (by Peter Linz and Hopcroft-Ullman) talking about this topic)

• There are a few things missing here. First, is the number input LSB to MSB or MSB to LSB? (Looks like LSB to MSB.) Second, are you allowing the empty string to represent zero? Dec 23 '17 at 22:41
• The way to prove such results is to twofold: first you prove an upper bound by constructing a DFA accepting the given language, then you prove a lower bound using Myhill–Nerode theory. Dec 23 '17 at 22:42
• For an example, see this: cs.stackexchange.com/questions/85785/…. Dec 23 '17 at 22:53
• Regarding Q4, yes, they cover all cases. I'm certain you can prove this in your own, even if you're new to "this" – this has nothing to do with automata theory. Dec 23 '17 at 22:55
• Can you provide any link discussing these two points: (1) LSB to MSB and MSB to LSB number input (2) Representing zero by empty string and possibly these two points in the context of "divisible by $n$ DFA" . Also I guess first point wont make number of states in min DFA to change. Second point might change it.
– anir
Dec 24 '17 at 7:02

## Case 1: $n$ is a power of 2

If you wanted to check whether a decimal number is divisible by some power of 10, you can just look at the number of trailing zeros. For example, all numbers that are divisible by $100 = 10^2$ end with 2 zeros (this is of course including numbers ending with more than 2 zeros). The same idea can be applied here for binary numbers and powers of 2. Specifically to check for multiples of $n = 2^m$, you can simply check if the string ends in $m$ zeros. For example, if $n = 16 = 2^4$, all multiples of $n$ will end in 0000 (4 zeros). Thus we can create the following DFA for $n = 16$:

This construction can be extended to all multiples of 2, where you have $m+1$ states to ensure that your input ends in at least $m$ zeros (reading a 0 takes you one state forward, reading a 1 takes you back to the start). As suggested in the comments on your question, you can show that you can't do any better than this using a Myhill–Nerode argument.

## Case 2: $n$ is odd

The general idea here is that when a number is divided by $n$, the possible remainders are $0, 1, \ldots, n-2$ or $n-1$. We can give our DFA one state for each of these possible remainders so as we process a string, we're keeping track of the remainder of what we've read so far and transitioning to the appropriate remainder based on what character we read. Then, we accept if we finish in the remainder 0 state.

Design DFA accepting binary strings divisible by a number 'n' walks through a detailed method for constructing such a DFA, and you can find out more about this by searching for "DFA based division". This gives a total of $n$ states, which is optimal by Myhill-Nerode because these remainders are exactly the equivalence classes given by the relation $\equiv_L$ for the language $L = \{\text{binary strings divisible by }n\}$.

## Case 3: $n$ is even (and not a power of 2)

The technique in case 2 also works here but it isn't optimal. Every $n$ in this case can be expressed as $2^km$ where $m$ is odd (this is the division procedure described in your question). Thus, to check whether a binary string is divisible by $n$, we check whether it is divisible by both $2^k$ and $m$.

We know that it takes $k$ states to check divisibility by $2^k$ (case 1) and $m+1$ states to check divisibility by $m$ (case 2). We can take a DFA for divisibility by $m$, unmark the accepting state and make it the start state for a DFA for divisibility by $2^k$, giving our final DFA $k+m$ states.

As an example, here's the DFA for $n = 6$:

The top three states ensure that the number is a multiple of 3, and the final accept state ensures that it is a multiple of 2.

• Revisiting this problem, I dont get how to combine two DFAs in case 3. You said "unmark the accepting state and make it the start state for a DFA for divisibility by $2^k$". But you didnt explain how you came up with transition outgoing (labeled 1) from final state of last DFA. If we are to combine "divisible by $2^k$ DFA" with "divisible by $m$ ($m$ being any odd number)" DFA, "all" outgoing transitions on reading "1" from "all" states of divisible by $2^k$ DFA will go to same second state of divisible by $m$ DFA, (Q1.) right? [to next comment...]
– anir
Dec 3 '19 at 13:03
• [..from earlier comment] For divisible by $2^k$, we need string to end with $k$ zeroes. But, your case 3, divisible by 6 DFA will also let $k=1$ zeroes appear anywhere. (Q2.) How this fits in divisible by 6 string? For example in your divisible by 6 DFA, for 110110, first 11 moves through divisible by 3 DFA, next 01 moves through divisible by 2 DFA, next 1 moves through divisible by 3 DFA and final 0 moves through divisible by 2 DFA. (Q3.) How running through different component DFAs one after other still ensures that the whole string will still be divisible by $2^km$
– anir
Dec 3 '19 at 13:09


# Math Help - Circle questions -- Secants, chords and tangents

1. ## Circle questions -- Secants, chords and tangents

Hello again, just got a few questions pertaining to circles. I'll just post one to start with;

"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).

So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)

Did I do that right?

2. ## Re: Circle questions -- Secants, chords and tangents

Originally Posted by StonerPenguin
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;

"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).

So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)

Did I do that right?
Computation looks right to me. But simply memorizing theorems does not promote understanding.

Let's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.

$Let\ r = length\ of\ radius\ of\ given\ circle,$

$u + r = length\ from\ the\ given\ circle's\ center\ to\ a\ given\ point\ outside\ the\ circle,$

$x = length\ of\ the\ line\ that\ is\ tangent\ to\ the\ given\ circle\ and\ runs\ through\ the\ given\ point.$

By the Pythagorean Theorem, we have:

$(u + r)^2 = x^2 + r^2 \implies u^2 + 2ru + r^2 = x^2 + r^2 \implies x^2 = u^2 + 2ru = u(u + 2r) \implies x = \sqrt{u(u + 2r)}.$

3. ## Re: Circle questions -- Secants, chords and tangents

ISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg
cos 71.35 =d/6724
d km to horizon =2150 km

4. ## Re: Circle questions -- Secants, chords and tangents

Hello, StonerPenguin!

The radius of the earth is approximately 6371 km.
If the international space station (ISS) is orbiting 353 km above the earth,
find the distance from the ISS to the horizon (x).

Code:
o
|\
| \
352 |  \
|   \ x
|    \
* * *   \
*     |     *\
*       |       o
*    6371|     *  *
|   *6371
*         | *       *
*         o         *
*                   *

*                 *
*               *
*           *
* * *
Note the right triangle.

The equation is: . $x^2 + 6371^2 \:=\:6723^2$

5. ## Re: Circle questions -- Secants, chords and tangents

Thank you JeffM, bjhopper and Soroban! It's nice to see different perspectives and the image drawn in code is really cool.
Here's another question I've had trouble with:

"Explain how you know $\overline{AB}$ $\overline{CD}$ given E is the center of the circle. (Include theorem numbers.)"

And here's some pertinent theorems;
Theorem 10.4
If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
Theorem 10.5
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
Theorem 10.6
In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
Obviously from the diagram $\overline{AB}$ $\overline{CD}$ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/

6. ## Re: Circle questions -- Secants, chords and tangents

t is a triangle a is an angle
t AED congruent to BEC isosceles t's same legs and equal altitudes
a DEC = 180- 2* 1 / 2 *AED
a AEB 180-2*1/2 * BEC a AED = aBEC
a AEB = a DEC
AB =DC equal arcs = equal chords


Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$

Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.

I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$.

My reasoning was :

$$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$

So

$$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$

where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$.

Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.

• The LHS behaves like $\int_{k}^{+\infty}\frac{dx}{x^4}=\frac{1}{3k^3}$ while the RHS behaves like $\left(\int_{k}^{+\infty}\frac{dx}{x^2}\right)^3=\frac{1}{k^3}$, so that is not surprising. – Jack D'Aurizio Apr 21 '16 at 15:36

For $k > 1$, $$\sum_{n=k}^\infty \dfrac{1}{n^4} < \int_{k-1}^\infty \dfrac{dx}{x^4} = \dfrac{1}{3(k-1)^3}$$

$$\left(\sum_{n=k}^\infty \dfrac{1}{n^2}\right)^3 > \left(\int_{k}^\infty \dfrac{dx}{x^2}\right)^3 = \dfrac{1}{k^3}$$

$\dfrac{1}{k^3} > \dfrac{1}{3(k-1)^3}$ when $3(k-1)^3 > k^3$, which is true for $k > 3.2612$.

Let's see what we can say about comparing $s_1(k) =\sum_{n=k}^{\infty} \frac1{n^a}$ with $s_2(k) =\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c$ for large enough $k$, where $a > 1$ and $b > 1$ so the sums converge.

Using the integral approximation, $s_1(k) =\sum_{n=k}^{\infty} \frac1{n^a} \approx \int_k^{\infty} \frac{dx}{x^a} =-\frac1{(a-1)x^{a-1}}\big|_k^{\infty} =\frac1{(a-1)k^{a-1}}$.

Therefore $s_2(k) =\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c \approx \left(\frac1{(b-1)k^{b-1}}\right)^c =\frac1{(b-1)^c k^{c(b-1)}}$, so $\dfrac{s_1(k)}{s_2(k)} \approx \dfrac{\frac1{(a-1)k^{a-1}}}{\frac1{(b-1)^c k^{c(b-1)}}} = \dfrac{{(b-1)^c k^{c(b-1)}}}{{(a-1)k^{a-1}}} = \dfrac{{(b-1)^c }}{{(a-1)}}k^{c(b-1)-(a-1)}$.

Therefore if $c(b-1) > a-1$, then $s_1(k) > s_2(k)$ for all large enough $k$; if $c(b-1) < a-1$, then $s_1(k) < s_2(k)$ for all large enough $k$;

If $c(b-1) = a-1$, then $\dfrac{s_1(k)}{s_2(k)} \approx \dfrac{{(b-1)^c }}{{(a-1)}}$, so the result depends on this ratio.

For your case of $a=4$ and $b=2$, the key difference is $c(b-1)-(a-1) =c-3$.

If $c > 3$, then $s_1(k) > s_2(k)$ for large enough $k$; if $c < 3$, then $s_1(k) < s_2(k)$ for large enough $k$.

If $c=3$, which is your case, the ratio is $\dfrac{(b-1)^c }{(a-1)} =\dfrac{1}{3} < 1$, so $s_1(k) \approx \frac13 s_2(k) < s_2(k)$ for large enough $k$, which confirm's Robert Israel's result (good thing too, because any result of mine that differs from a result of his is probably wrong).


# For which values of $\alpha$ is {$z_n$} a bounded sequence?

Where $$\alpha$$ is a real constant, consider the sequence {$$z_n$$} defined by $$z_n=\frac{1}{n^\alpha}$$. For which value of $$\alpha$$ is {$$z_n$$} a bounded sequence?

How do I start with this kind of question? I think that $$\forall\space \alpha\in\Bbb{R}_{\geq0}$$ the sequence is convergent and therefore bounded, but how do I write it out?

• In order for us to tell you how to write things out, it would be helpful if you explained why you believe that the answer is what it is. Why do you think that $\frac 1{n^{\alpha}}$ converges for $\alpha \geq 0$? Are you saying that $\frac 1{n^{\alpha}}$ is not bounded when $\alpha < 0$? If so, then you must say so explicitly in your answer. Also, why do you believe that this is the case? – Ben Grossmann Aug 16 '20 at 10:16
• Because it is clear that for $\alpha\geq0$ the sequence converges to 0. If $\alpha<0$ then the value of $\frac{1}{n^\alpha}$ will become very big unless $\alpha>-\frac{1}{n}$. I might be wrong, but this is what I think. I don't know how to approach this question. – Jess Aug 16 '20 at 10:20
• By giving the answer that you have given, you have not only "approached" the problem correctly, but also have given an almost complete answer. It seems that your only question, then, is how to write this up with sufficient "formality." – Ben Grossmann Aug 16 '20 at 10:25

## 2 Answers

If $$\alpha=0$$, $$(z_n)$$ is constant, hence bounded.

If $$\alpha>0$$, $$(z_n)$$ converges to 0 and is thus bounded.

If $$\alpha<0$$, $$(z_n)$$ diverges to $$+\infty$$ and is thus unbounded.

As I state in the comment, you have the correct answer. The only remaining task is to give a formal explanation of the answer. One way to write an answer is as follows:

First, we note that the function $$f: \Bbb [1,\infty) \to \Bbb R$$ defined by $$f(x) = x^{\beta}$$ satisfies $$\lim_{x \to \infty}f(x) = \begin{cases} 0 & \beta < 0\\ 1 & \beta = 0\\ \infty & \beta > 0. \end{cases}$$ I suspect that you do not need to prove this statement formally: it is likely that there is a statement in the textbook that you can refer to.

With that established, address the problem in $$3$$ cases: in the case that $$\alpha < 0$$, conclude using the above fact that $$\lim_{n \to \infty} z_n = \infty$$, which means that the sequence is not bounded. In the case that $$\alpha = 0$$, conclude that $$z_n \to 0$$, which means that the sequence is convergent and is therefore bounded. Similarly, if $$\alpha > 0$$, conclude that $$z_n \to 0$$, which means that the sequence is convergent and therefore bounded.

Thus, we conclude that the sequence is bounded if and only if $$\alpha \geq 0$$.

• If the sequence is bounded, does that mean that the limit of the sequence exists? Because I just noticed that the next question asks me to find the values of $\alpha$ for which lim$_{n\to\infty}z_n$ exists. I assume the values of $\alpha$ are the same as which $z_n$ is a bounded sequence? – Jess Aug 16 '20 at 10:40
• It is not necessarily the case that a bounded sequence will have a limit. However, we can clearly see by exhausting the possibilities that for this problem, the sequence $z_n$ converges whenever it is bounded. – Ben Grossmann Aug 16 '20 at 10:42


# A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd? [duplicate]

Q. A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd?

My work:

suppose that head comes in first toss so probability of getting head in the first toss $$=\dfrac{1}{2}$$

suppose that first & second tosses show tails & third toss shows head so probability of getting head in the third toss $$=(1-\dfrac12)(1-\dfrac12)\dfrac{1}{2}$$ $$=\dfrac1{2^3}$$

suppose that first 4 tosses show tails & fifth toss shows head so probability of getting head in the fifth toss $$=(1-\dfrac12)^4\dfrac{1}{2}$$ $$=\dfrac1{2^5}$$

suppose that first 6 tosses show tails & seventh toss shows head so probability of getting head in the fifth toss $$=(1-\dfrac12)^6\dfrac{1}{2}$$ $$=\dfrac1{2^7}$$

…………….

and so on

But I am not able to find the final probability of getting head first time so that the number of tosses required is odd. what should do I next to it? please help me.

• You're almost there. Just recognize that you have a geometric series and take its sum. To allow you to check your work, the answer will be $\frac 23$. – Robert Shore Jul 3 at 4:06
• This is an exact duplicate of math.stackexchange.com/q/834344/117057 , which unfortunately doesn't have an accepted answer. – shoover Jul 3 at 22:27

You have a geometric series,

$$\frac12+\frac1{2^3}+\frac1{2^5}+\frac1{2^7}+\ldots=\sum_{n\ge 0}\frac12\cdot\left(\frac14\right)^n=\frac{\frac12}{1-\frac14}=\frac23\;.$$

Alternatively, if $$p$$ is the desired probability, then $$p=\frac12+\frac14p$$: with probability $$\frac12$$ you get a head on the first toss, and with probability $$\frac14$$ you start with two tails and are now in exactly the same position that you were in at the beginning. Solving this for $$p$$ again yields $$p=\frac23$$.

• Your second method is very simple. I have never thought of it. Hats off to you sir – user801303 Jul 3 at 4:44
• @PaulAldrin: You’re welcome. – Brian M. Scott Jul 3 at 4:44
• @TShiong: When you can find a recurrence like that, it usually makes things a lot simpler! – Brian M. Scott Jul 3 at 4:45

Your approach is good and will get you the right answer. Just realize that you're building a geometric series and you want its sum.

I'm drafting an answer to give you an alternative approach. Let $$p$$ be the probability you're looking for. Then your first flip will be heads with probability $$0.5$$. If it's tails, then you will solve your original problem (the first heads occurs on an odd toss) exactly when, from your new starting point, your first heads occurs on an even toss, which happens with probability $$1-p$$.

That means $$p = 0.5 + 0.5(1-p) \Rightarrow 1.5 p = 1 \Rightarrow p = \frac 23$$.

• haven't you used sum of infinite GP? – user805532 Jul 3 at 4:38
• No, I think I've provided an alternative approach that avoids the need, just as does the alternative solution provided by Brian M Scott. – Robert Shore Jul 3 at 8:35

Here's another way to solve the problem by considering pairs of tosses at a time instead of single tosses.

Let $$\sigma$$ be an arbitrary infinite sequence of heads and tails. $$\sigma$$ uses 1-based indexing.

$$\text{e.g.}\;\;\; \sigma = HTHTHTHTHTTTTTHHHH\cdots$$

Imagine grouping the elements of $$\sigma$$ into pairs.

$$\sigma = HT,HT,HT,HT,HT,TT,TT,HH,HH\cdots$$

Let's imagine we have three states, $$S$$, $$E$$, and $$O$$.

• $$S$$ is the start state, we haven't seen a head yet.
• $$E$$ is the state marking that we saw a head at an even index first.
• $$O$$ is the state marking that we saw a head at an odd index first.

$$E$$ and $$O$$ are both absorbing states. Once we enter one of those states, we are never going to leave it.

Our state at the beginning of our process is always $$S$$ because, initially, we haven't observed any tosses at all of our coin.

Next let's consider what happens when we read our first toss-pair from $$\sigma$$.

There are twice as many ways to transition from $$S$$ to $$O$$ than there are to transition from $$S$$ to $$E$$.

     TT
S  ----->   S

TH
S  ----->   E

HT
S  ----->   O

HH
S  ----->   O

As the number of pairs processed approaches infinity, the probability that the current state is $$S$$ approaches zero.

However, the ratio of the probability that that the current state is $$O$$ is always twice the probability that the current state is $$E$$.

Therefore, the limiting probability that the state is $$O$$ is $$2/3$$

You are almost done. Add all the terms

$$\frac12+\frac1{2^3}+\frac1{2^5}+\frac1{2^7}+\ldots$$ above series is an infinite GP with first term $$a=\dfrac{1}{2}$$ and common ratio $$r=\dfrac{1}{4}$$ $$\dfrac{a}{1-r}$$

$$=\frac{\frac{1}{2}}{1-\frac{1}{4}}$$$$=\frac23$$

• does this really have infinite terms which can be added? – user805532 Jul 3 at 4:38
• Of course it has uncountable number of terms. I 've added them in my answer – user801303 Jul 3 at 4:40
• it's of course countable – RiaD Jul 3 at 12:58
• @PaulAldrin none of the terms are infinite in value, but they are infinitely many. Name a large odd number n and I can respond with a term $1/2^n$. – neptun Jul 3 at 13:09
• r should be 1/4, not 1/2 – Mark Pattison Jul 3 at 13:32

Another way to avoid summing a GP.

The probability of the first head happening on toss $$n$$ is $$(1/2)^{n}$$. Let $$A$$ be the event that the first head occurs on an odd toss, and $$E_k$$ be the event that it occurs either on toss $$2k+1$$ or $$2k+2$$. Now $$P(A\mid E_k)=\frac{(1/2)^{2k+1}}{(1/2)^{2k+1}+(1/2)^{2k+2}}=\frac{2}{3},$$ independently of $$k$$. Since exactly one of the $$E_k$$ will occur almost surely, we have $$P(A)=\sum_kP(A\mid E_k)P(E_k)=\sum_k\frac23P(E_k)=\frac23\sum_kP(E_k)=\frac23.$$


# How to average cyclic quantities?

Looking on Internet, I mostly found this definition:

Given quantities on a cyclic domain D, first rescale the domain to [0;2$\pi$[, then, find $z_n$ the point on the unit circle corresponding to the $n$th value, and compute the average by:

$$z_m = \sum_{n=1}^N z_n$$

The average angle is then $\theta_m = \arg z_m$ and to obtain the average value you scale back to your original domain D.

I must say, I have problems with this definition. For simplicity, I will use oriented angles in degree for my examples (i.e. D = [0;360[). With this formula, having angles -90, 90 and 40 will give 40 as mean angle, when I would expect 13.33 as an answer (i.e. (90-90+40)/3).

For my own problems, I usually use:

$$v_m = \mathop{\rm arg\,min}_{v\in D} \sum_{n=1}^{N} d(v_n,v)^2$$

Where $d(x,y)$ is the distance in the cyclic domain, and $\{v_1, v_2, \ldots, v_n\}$ is the set of cyclic data I want to average of.

It has the advantage to work the same way whatever the domain (replace D by a non-cyclic domain and $d$ with the usual euclidean distance, and you find the usual definition of an average). However, it is expensive to compute and I don't know any exact method to do it in general.

So my question is: what is the appropriate way to deal with average of cyclic data? And do you have good pointers that explain the problem and its solutions?

-
en.wikipedia.org/wiki/Circular_mean –  Rasmus Dec 16 '10 at 19:35
I know about this page. However, there is no justification for it. Also, I left a comment in the discussion of this page in the hope of understanding. But for now, I still disagree with this method to calculate the average. –  PierreBdR Dec 17 '10 at 17:38
The choice of distance metric depends crucially on the application. Bearing data, for instance, might be derived from estimates of X and Y with normally distributed errors, and this leads naturally to the circular mean. For other cases this might not be a good choice. –  wnoise Apr 4 '11 at 18:52

Like all averages, the answer depends upon the choice of metric. For a given metric $M$, the average of some angles $a_j \in [-\pi,\pi]$ for $j \in [1,N]$ is that angle $\bar{a}_M$ which minimizes the sum of squared distances $d^2_M(\bar{a}_M,a_j)$. For a weighted mean, one simply includes in the sum the weights $w_j$ (such that $\sum_j w_j = 1$). That is,

$$\bar{a}_M = \mathop{\rm arg\,min}_{x} \sum_{j=1}^{N}\, w_j\, d^2_M(x,a_j)$$

Two common choices of metric are the Frobenius and the Riemann metrics. For the Frobenius metric, a direct formula exists that corresponds to the usual notion of average bearing in circular statistics. See "Means and Averaging in the Group of Rotations", Maher Moakher, SIAM Journal on Matrix Analysis and Applications, Volume 24, Issue 1, 2002, for details. http://lcvmwww.epfl.ch/new/publications/data/articles/63/simaxpaper.pdf

-
Thank you, it is exactly the kind of reference I was looking for! –  PierreBdR May 9 '11 at 13:54

The problem with expecting the mean of 90°, −90°, and 40° to be (90°−90°+40°)/3 = 13.33° is that you would then expect the mean of 10° and 350° to be (10°+350°)/2 = 180°, and not 0° which is the more reasonable answer. It only gets worse when you have more than two angles (What is the mean of 340°, 350°, 360°, 10°, and 20°? What about 340°, 350°, 0°, 10°, and 20°?). Essentially, what you're doing there is equivalent to setting $z_n = e^{i\theta_n}$ and computing $$\bar z = (z_1 z_2 \cdots z_N)^{1/N},$$ and the problem is of course that it's not obvious a priori which of the $N$ possible roots of that equation is the right one, if any.

The "circular mean" definition is not so bad. In fact, it corresponds to the point which minimizes the sum of its squared distances to the points corresponding to the data, $$\bar z = \underset{\lvert z \rvert = 1}{\arg\min} \sum_{n=1}^N \lvert z - z_n \rvert^2.$$ So this is almost the same as the formula you like to use; you only have to define the "distance" between angles as the distance between the corresponding points on the unit circle. That is, $d(\theta, \phi) = \sqrt{2 - 2\cos(\theta-\phi)} = 2 \sin(\lvert\theta-\phi\rvert/2)$. This metric is close to $\lvert\theta-\phi\rvert$ when $\theta$ and $\phi$ are close, and has the advantage of being really easy to find the solution to.

-
Of the $N$ possible roots, one will be the global minimum of his distance functions, so checking all of them is sufficient. (See the reference Rob Johnson gave). –  wnoise Apr 4 '11 at 19:02

The angle is supposed to be the independent variable, not the dependent variable. If your function is cyclic (using degrees), z(-90)=z(270) so it doesn't matter which you use. Then the average value of the function is $$z_m=\sum_{n=1}^Nz(\theta_n)$$ You are right that if you average the angles, it matters which lap of the circle you use (-90 vs 270) because the difference of 360 gets divided by N.

-
This is what I am reading on Internet, but I have issues with this definition (as described in my post) and there is never a good explanation on why we should use this also for cyclic data. Could you expand? –  PierreBdR Dec 16 '10 at 17:54
The definition of a cyclic function is that f(x+T)=f(x) for some period T. The recommendation you saw was to rescale the units of x so that the period is 2pi radians or 360 degrees. I don't think that is important, but was trying to stay with it. As the function is cyclic, it doesn't matter which cycle you take the data from. So if you want the average value over a cycle it would be (if we take T to be 360) $$\frac{1}{360}\int_a^{a+360}f(x)dx$$ You can then approximate this by a sum just by taking equally spaced steps like $$\frac{1}{n}\sum_{i=0}^{n-1}f(\frac{360i}{n})$$ –  Ross Millikan Dec 16 '10 at 22:06
It looks like I am working on a different problem than you. Did you check the Wikipedia page that Rasmus suggested? Is that what you are after? –  Ross Millikan Dec 16 '10 at 22:27

For angles, one can adapt the iterative way of computing means to angles, that is:

given angles v[1] .. v[n]

m[1] = v[1]

m[i] = remainder( m[i-1] + remainder( v[i]-m[i-1], C)/i, C) (i=2..n)

where remainder(x,y) gives the signed remainder on dividing x by y and C is the measure of a circle.

I suspect this gives your vm, but haven't been able to prove it.

-
A quick check indicates it might not work. Applying this formula to the same set of values, simply taking the elements in various order leads to different results. –  PierreBdR Dec 17 '10 at 17:44


How to derive the identity $\sin x/x=\prod_{n=1}^\infty \cos(x/2^n)$ without using telescoping?

I am wondering how to derive the following equality $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n)\tag{1}$$ without using the method of telescoping. I know that there is already a question on deriving this infinite product representation of $$\sin x/x$$, but all the answers in the link telescope the product. Here is the method, for completeness.

First, we can use the trigonometric identity $$\sin x=2\cos (x/2)\sin(x/2)$$ to yield $$\cos(x/2)=\frac{\sin x}{2 \sin(x/2)}$$. More generally, this implies that $$\cos(x/2^{n})=\frac{\sin (x/2^{n-1})}{2 \sin(x/2^{n})}$$

Our infinite product is thus $$\prod_{n=1}^\infty\cos(x/2^n)=\frac{\sin (x)}{2 \sin(x/2)}\cdot \frac{\sin (x/2)}{2 \sin(x/4)} \cdot \frac{\sin (x/4)}{2 \sin(x/8)} \cdots$$ Treating the product as a limit of a finite product $$f_k(x)=\prod_{n=1}^k \cos(x/2^n)$$, we notice that $$f_k(x)=\frac{\sin(x)}{2^k\sin(x/2^k)},$$ with $$\lim_{k\to\infty} f_k(x)=\sin x/x$$. Thus, $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n).$$

Question:

How to show that $$(1)$$ is true without using telescoping?

• Why do you want to avoid telescoping? – Lucas Henrique Jan 2 at 23:33
• Note the Weierstrass factorization theorem:\begin{align*} \frac{\sin x}{x}&=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)\\ \cos\frac{x}{2^n}&=\prod_{m=0}^\infty\left(1-\frac{x^2}{(2m+1)^2 2^{2(n-1)}\pi^2}\right) \end{align*} – Edward H Jan 2 at 23:44
• You could use the fact that $\sum_{n\ge1}\pm 2^{-n}$ (where the signs are chosen independently and uniformly) is uniformly distributed on $[-1,1]$; the desired equality relates two formulas for the characteristic function of such a random variable. – kimchi lover Jan 2 at 23:50
• @kimchilover very nice – Sandeep Silwal Jan 3 at 0:10
• @Lucas Henrique I am curious to know if there are other ways to evaluate this product. – Zachary Jan 3 at 2:13

You can do this by a trick that is essentially the same as looking at this product in frequency domain.

To avoid any analytical difficulty, let's examine finite sums; we have $$\prod_{n=1}^{k}\cos(x/2^n)=\prod_{n=1}^k\left(\frac{e^{ix/2^n}+e^{-ix/2^n}}{2}\right)$$ We can expand the sum on the right as $$\frac{1}{2^k}\sum_{\sigma\in\{-1,1\}^k}\exp\left(ix\cdot \left(\sigma_1\cdot \frac{1}2+\sigma_2\cdot \frac{1}{2^2}+\ldots+\sigma_k\cdot \frac{1}{2^k}\right)\right)$$ where $$\sigma$$ is a string of $$k$$ terms in $$\{-1,1\}$$ representing which side of the sum within the former product was followed.

One can see that for $$n=1$$, the angular frequencies encountered (i.e. the coefficient of $$ix$$) are $$1/2$$ and $$-1/2$$ . For $$n=2$$, the frequencies are $$-3/4,\,-1/4,\,1/4,\,3/4$$. We can prove via induction that the possible values of that coefficient are just the set of numbers of the form $$a/2^k$$ for odd integers $$a$$ between $$-2^k$$ and $$2^k$$. Thus, the partial sum works out to: $$\frac{1}{2^k}\cdot \sum_{\substack{a\text{ odd}\\ -2^k < a < 2^k}}\exp\left(ix \cdot \frac{-a}{2^k}\right)$$ We could bail out at this step and recognize that the sum is actually a geometric series (with ratio $$\exp\left(\frac{ix}{2^{k-1}}\right)$$), which would lead us back to the expression you derived for the partial sums. However, we could also recognize this an average of equally spaced evaluations of the function $$z\mapsto \exp(ix\cdot z)$$ over the interval $$[-1,1]$$ with more evaluations as $$k$$ increases; thus, in the limit, this product becomes an integral giving the average value of $$\exp(ixt)$$ over the interval $$[-1,1]$$: $$\lim_{k\rightarrow\infty}\prod_{n=1}^k\cos(x/2^n) = \frac{1}2\int_{-1}^1\exp(ixt)\,dt$$ Of course, this is just integrating an exponential function, which can be done easily, and works out to $$\frac{\sin(x)}x$$.

Here is a different packaging of the same basic argument presented in Milo Brandt's answer.

Let $$X_k=\sum_{j=1}^k \sigma_j 2^{-j}$$, where the $$\sigma_i$$ are iid $$\pm1$$ random variables. This has uniform distribution on the $$2^k$$ points uniformly spaced $$2^{1-k}$$ apart in the range from $$-1+2^{-k}$$ to $$1-2^{-k}$$, as can be seen from the binary expansions of the integers from $$0$$ to $$2^k$$. One can verify directly that $$X_k$$ converges in distribution to the continuous uniform distribution on $$[-1,1]$$.

The characteristic function of $$X_k$$, namely the function $$\phi_k(t)=E[\exp(itX_k)]$$ is given by $$\prod_{j=1}^k E[\exp(i \sigma_j 2^{-j})] = \prod_{j=1}^k \cos(t2^{-j})$$.

By Lévy's continuity theorem, for each $$t$$, one has $$\lim_{k\to\infty}\varphi_k(t)=\varphi(t)$$, where $$\varphi(t)$$ is the characteristic function of the uniform distribution on $$[-1,1]$$, which is $$\varphi(t)=\frac 12\int_{-1}^1 \exp(itx)\,dx = \frac{\sin(t)}t.$$


# Number of times $2^k$ appears in factorial

For what $n$ does: $2^n | 19!18!...1!$?

I checked how many times $2^1$ appears:

It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$

I checked how many times $2^2 = 4$ appears:

It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$

I checked how many times $2^3 = 8$ appears:

It appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$

I checked how many times $2^{4} = 16$ appears:

It appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$

In all,

$$2^{18} \cdot 2^{32} \cdot 2^{36} \cdot 2^{16} = 2^{102}$$

But that is the wrong answer, its supposed to be $2^{150}$?

• Note that, for example, $6!$ contributes 4 factors of 2 - one from 2, one from 6 and two from 4. You only count 3 of these. – Wojowu Apr 24 '15 at 16:37

A simple trick to compute $k$ such that $2^k|n!$ is to compute $\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$, this is because $n$ has $[n/2]$ numbers divided by $2$, if we pick out these numbers and find out that there're $[n/4]$ numbers divided by $4$.. If we continue this procedure, we see that $$k=1\cdot\left(\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n}{4}\right\rfloor\right)+2\left(\left\lfloor\frac{n}{4}\right\rfloor-\left\lfloor\frac{n}{8}\right\rfloor\right)+\ldots=\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$$. In this case, we have to sum $$0+1+1+3+3+4+4+7+7+8+8+10+10+11+11+15+15+16+16=150.$$ Your fault is that your did not count the contribution of those which is not the power of $2$. For instance, there's $14$ in $14!$..

How many times $2$ divides the product $\prod_{i=1}^{19}i!$ ?

Let's call each term inside a factorial $i$. That way, $i = 1$ occurs in 19 factorials, $i = 2$ occurs in 18 factorials, and $i = 3$ occurs in 17 factorials etc.

$i = 2$ occurs 18 times. $1 \times 18 = 18$
$i = 4$ occurs 16 times. $2 \times 16 = 32$
$i = 6$ occurs 14 times. $1 \times 14 = 14$
$i = 8$ occurs 12 times. $3 \times 12 = 36$
$i = 10$ occurs 10 times. $1 \times 10 = 10$
$i = 12$ occurs 8 times. $2 \times 8 = 16$
$i = 14$ occurs 6 times. $1 \times 6 = 6$
$i = 16$ occurs 4 times. $4 \times 4 = 16$
$i = 18$ occurs 2 times. $1 \times 2 = 2$

$$18 + 32 + 14 + 36 + 10 + 16 + 6 + 16 + 2 = 150$$

It might be more helpful to do this recursively.

Let $T(n) = \prod_{k=1}^n k!$.

We will use the notation: $2^{r} \| m$ to mean that $2^r$ is the largest power of $2$ that divides $m$.

Then we have $2 \| 2! = T(2)$. We also know that $2 \| 3!$, so $2^2 \| T(3) = 3! T(2)$. Continuing:

$$2^3 \| 4!$$ $$2^3 \| 5!$$ $$2^4 \| 6!$$ $$2^4 \| 7!$$ $$2^7 \| 8!$$ $$2^7 \| 9!$$ $$2^8 \| 10!$$ $$2^8 \| 11!$$ $$2^{10} \| 12!$$ $$2^{10} \| 13!$$ $$2^{11} \| 14!$$ $$2^{11} \| 15!$$ $$2^{15} \| 16!$$ $$2^{15} \| 17!$$ $$2^{16} \| 18!$$ $$2^{16} \| 19!$$

If we take the sum of all of those powers, $$2\cdot 1 + 2\cdot 3 + 2\cdot 4 + 2\cdot 7 + 2\cdot8 + 2 \cdot 10 + 2 \cdot 11 + 2 \cdot 15 + 2 \cdot 16$$

$$=2(1+3+4+7+8+10+11+15+16) = 2(75) = 150.$$


1. Aug 6, 2008

### laura_a

1. The problem statement, all variables and given/known data
I am working on a topic called differential equations, and Im stuck on some working out.

2. Relevant equations[/b]

I have to integrate it, and I know I can without factorising, but it is so messy, my professor said to factorise the fraction so it looks a bit like a1/(bu+ c)+ a2/(du+ e)

Well thats what the professor said, I have no idea what it means... I can factorise as far as
= u/(-u^2 +3u +2) - 2/(-u^2 +3u +2)

And I know that -u^2 +3u +2 = (-u + 1)(u-2) +4

But not sure how to put it all together

Here is the working out for the whole question just in case you're interested
\begin{align*} y' &= \frac{y+2x}{y-2x} \\ Let y &= ux \\ \text{Then we have} y' &= \frac{ux + 2x}{ux - 2x} \\ &= \frac{u+2}{u-2} \\ \text{Now } y' &= \frac{dy}{dx} = \frac{d(ux)}{dx} = x \frac{dy}{dx} + u \\ \text{So we can say that} x \frac{du}{dx} + u& = \frac{u+2}{u-2} \\ x \frac{du}{dx} &= \frac{u+2}{u-2} - u \\ x \frac{du}{dx} &= \frac{u+2}{u-2} - \frac{u^2-2u}{u-2} \\ x \frac{du}{dx} &= \frac{u+2- u^2+2u}{u-2} \\ x \frac{du}{dx} &= \frac{-u^2+3u+2}{u-2} \\ \end{align*} \bigskip

Then I have to integrate both sides of this equation...which is what I think I need to factorise in order to make it nice and neat

$$\frac{u-2}{-u^2 + 3u + 2}du &= \frac{1}{x} dx$$

Last edited: Aug 6, 2008
2. Aug 6, 2008

### HallsofIvy

Staff Emeritus
That can't be factored using integer coefficients but you know that if an expression factors as (x-a)(x-b) then a and b must be roots of (x-a)(x-b)= 0. Use the quadratic formula to find the roots of x2- 3x- 2= 0.

3. Aug 6, 2008

### Defennder

Try completing the square of $$-u^2+3u+2$$. Then use a simple algebraic identity to decompose it into partial fractions.

4. Aug 7, 2008

### laura_a

I think i've gotten a little closer, The question I'm trying to do is to solve a d.e. using change of variables, so once I get this part worked out I should be able to get it

So this is how far I've gone
Since $$u^2- 3 u -2= 0$$
has two roots
$$u_1 = (3+ 17^{0.5})/2, u_2= (3- 17^{0.5})/2$$

so

$$u^2- 3 u -2= (u - (3+ 17^{0.5})/2 ) (u- (3- 17^{0.5})/2 )$$

$$\frac{u-2}{-u+3u+2}du = \frac{1}{2}dx$$ - I'll call this equation (2)
So now I've got the LHS as

$$S = -u /((u - (3+ 17^{0.5})/2 ) +2 /(u- (3- 17^{0.5})/2 )$$
I changed the u-2 to -u+2 since I made a similar change on the denominator in order to solve it...

Anyhow, I integrated that and ended up with something really messy invloving logs

$$-2(u + 17^{0.5} ln(u-17^{0.5}-3)+3ln(u-17^{0.5}-3)-17^{0.5}-3+4ln(u+17^{0.5}-3) = ln(x) + C$$

(I integrated both sides of equation (2) above)

Well the prob is now, I'm probably wrong anyway, but the next step is to solve for u, and then I have to sub back in the original change of varaible which was y=xu and end up with an expression for y in terms of x and C... before I go on, can anyone tell me if I should bother working through those logs, or is it wrong? Thanks :)

5. Aug 8, 2008

### arildno

Let us take this as a GENERAL case, shal we?
We are to decompose:
$$\frac{u-u_{0}}{(u-u_{1})(u-u_{2})}=\frac{A}{u-u_{1}}+\frac{B}{u-u_{2}}$$
where the unindexed u is our variable, the indexed u's known numbers, and A and B constants to be determined.
We therefore must have:
$$u-u_{0}=A(u-u_{2})+B(u-u_{1})$$, or by comparing coefficients, we get the system of equations:
$$A+B=1$$
and
$$u_{2}A+u_{1}B=u_{0}$$
whereby we arrive at the solutions:
$$A=\frac{u_{0}-u_{1}}{u_{2}-u_{1}},B=\frac{u_{2}-u_{0}}{u_{2}-u_{1}}$$

This is much simpler than using specific numbers!


# Epsilon-delta definition of limit : set an upper bound to delta?

Am I allowed to set an upper bound to delta in the epsilon-delta definition of limit? Why would it still be equivalent to the original definition ? For example, if the function is defined for all $x$ in $\Bbb R$ :

$f : D(f) = \Bbb R \rightarrow \Bbb R$.

Let $\epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x : 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - l| < \epsilon$

$\Leftarrow\text{?}\Rightarrow$

Let $\epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x : 0 < |x-x_o| < \delta \Rightarrow |f(x) - l| < \epsilon$.

• Lets suppose you found an $\; \delta \;$ then the definition still holds for any $\delta^* \le \delta$. Hence, you can choose your $\; \delta^* \le 3$ – XPenguen Dec 26 '15 at 20:16
• If you find an epsilon which beats a small delta, you've found an epsilon which beats a big one. – Mark Bennet Dec 26 '15 at 20:16

$$\forall\varepsilon > 0\ \exists \delta > 0\ \forall x \Big( 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - \ell| < \varepsilon\Big)$$

I would phrase this differently:

$$\forall\varepsilon > 0\ \exists \delta \in(0,3)\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big)$$

Now the question is whether that is equivalent to this:

$$\forall\varepsilon > 0\ \exists \delta>0\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big)$$

If it is true that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$ then it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, simply because every number in $(0,3)$ is $>0$.

Now the question is whether the converse holds. If it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, does it necessarily follow that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots\,{}$? Here the answer in general is “no”. I.e. there are some things you could put in place of $\text{“}\cdots\cdots\text{''}$ for which the first statement would be true and the second false. However, in the definition of "limit" the thing in place of $\text{“}\cdots\cdots\text{''}$ is “if a certain thing is $<\delta$, then a certain thing follows.” Given a certain $\varepsilon>0$ suppose we know there exists $\delta>0$ such that if a certain thing is less than $\delta$ then a certain conclusion follows. If the given value of $\delta$ is small enough, then the minimum of that value of $\delta$ and $2.9$ is small enough, because everything less than $\min\{\delta,2.9\}$ is less than $\delta$. So when the statement in place of $\text{“}\cdots\cdots\text{''}$ has the form “if a certain thing is $<\delta$, then a certain thing follows.”, then it is true that if $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$ then $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$.

Yes, of course. If the first condition is satisfied, then clearly the second is too. Conversely, if the second condition is satisfied, then one can replace $\delta$ by $\min(\delta, 3)$ and so the first condition is satisfied.

Yes, as you have to exhibit an $\delta$ that works for every $\epsilon$, then you can be as artificial as you want if that works for you. Always in existence proofs the same rule applies, construct all you need and show that it does work.

Yes you can. The idea is that some proposition $P(x,\epsilon)$ is true for all $x$ that are sufficiently close to $x_0$. So if it holds for all $x\in (-d+x_0,d+x_0)$ then it holds for all $x\in (-d'+x_0,d'+x_0)$ whenever $0<d'<d$. Sometimes it is useful to know how large $d$ can be, for a given $\epsilon.$


# Calculus : Study Guide for AP Calculus AB/BC or University Calculus

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### PRODUCT DESCRIPTION

Calculus Study Guide suitable for AP Calculus or First Year College Calculus Students. Main topics covered : Functions Limits and Continuity, Derivatives, Application of the Derivative, Indefinite integration, Definite Integration, Techniques of Integration, Sequences and Series, L’Hopitals Rule, Improper Integrals, Taylor Series, Short Introduction to Differential Equations.

Calculus without Limits is a self-study guide that can be used as a textbook/lecture supplement by students, as a source of homework problems by the instructor or as a source of lecture material. Covers most topics discussed in a first year calculus course (or AP Calculus). The emphasis is on solved problems that show students how to attack the types of problems they will encounter in homework and on exams. Discussion of concepts is informal followed by example problems. Many practice problems are also included for students.

1 - Function Review
What is a function - Graphing - Even and Odd Functions - Increasing and Decreasing Functions - One to One Functions - Powers and Polynomials - Absolute Value Function - Exponential Function - Logarithm - Trigonometric Functions
2- Limits
Definition - Tricks for Doing Limits - Properties of Limits and Rules - Limits at Infinity - One Sided Limits Continuity
3- Derivative
Notation and Definition in terms of limits - Slope of Tangent Line - Power Rule and Derivative of a constant - Chain Rule - Derivatives of Trig Functions - Product and Quotient Rules
4- Applications of the Derivative
Derivative of Inverse Function - Implicit Differentiation - Application: Rates of Change, Velocity and Acceleration - Finding Minima and Maxima First and Second Derivative Test - Inflection Points - The Mean Value Theorem - Related Rates
5- The Integral
Notation - Power Rule of Integration - Integration of Trig Functions - Integration of inverse function - Integrating Exponential Function - The substitution technique - Integrating Hyperbolic Functions - Definite Integration - Sums and Area Under a Curve - Area Between Two Curves -
6- U-Substitution
Solved Problems
7- Integration by Parts
Definition and When to Use - Examples
8- Integration of Rational Functions
Introduction - Examples
9- Integration Using the Method of Partial Fractions
Discussion - Examples
10- Integrating Powers of Trig Functions
Examples
11- Trig Substitution
Three Cases - Basic Technique -Examples
12 - Improper Integrals
Definition - Examples
13 - Sequences and Infinite Series
Sequences - Limits - Infinite Series
14 Ratio, Comparison, and Integral Tests
Examples
15 - Alternating Series
16 - Power Series and Radius of Convergence
17- Taylor Series Expansions
18- L’Hopital’s Rule
When to use and Definition - Examples
19- Introduction to Differential Equations
First Order Differential Equations - Second Order Differential Equations
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# Difference between revisions of "2006 AIME I Problems/Problem 10"

## Problem

Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ $[asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j

## Solutions

### Solution 1

The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\left(1,\frac 12\right)$ and $\left(\frac 32,2\right)$, which can be easily solved to be $6x = 2y + 5$. Thus, $a^2 + b^2 + c^2 = \boxed{065}$.

### Solution 2

Assume that if unit squares are drawn circumscribing the circles, then the line will divide the area of the concave hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the x-axis as $(x, 0)$.

The line divides the region into 2 sections. The left piece is a trapezoid, with its area $\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}$. The right piece is the addition of a trapezoid and a rectangle, and the areas are $\frac{1}{2}((1-x) + (2-x))(3)$ and $2 \cdot 1 = 2$, totaling $\frac{13}{2} - 3x$. Since we want the two regions to be equal, we find that $3x + \frac 32 = \frac {13}2 - 3x$, so $x = \frac{5}{6}$.

We have that $\left(\frac 56, 0\right)$ is a point on the line of slope 3, so $y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5$. Our answer is $2^2 + 5^2 + 6^2 = 65$.

We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.


# How to evaluate $\int_0^\infty e^{itx} e^{-x} dx$?

When calculating the characteristic function of the exponential distribution function, we need to evaluate the complex-integration: \begin{align*} \int_0^\infty e^{itx}e^{-x} dx \end{align*} for any $t \in \mathbb{R}$.

I understand how to evaluate this integral by treating the real part and imaginary part separately, but I am wondering is there any approach that uses the complex analysis theory? I found some answer uses the seemingly unjustified "fundamental theorem of calculus":

$$\int_0^\infty e^{(it - 1)x} dx = \frac{1}{it - 1}\int_0^\infty e^{(it - 1)x} d(it - 1)x = \frac{1}{1 - it}.$$

I think this solution is lacking any theoretic support (maybe I am wrong, please advise if there is any theorem that supports the above calculation). Specifically, can this integral be evaluated using residue calculus (contour integration)?

• Thanks, but for a complex-valued function, how to justify $\int_0^\infty f'(x) dx = f(x)|_0^\infty$? Could you please provide me some references? – Zhanxiong Mar 5 '17 at 4:01
• Integrating a complex-valued function is the same as a real-valued function. In complex analysis we look at $f(z)$ where $z$ is a complex variable, it is different because we integrate over curves in the complex plane, not over real intervals – reuns Mar 5 '17 at 4:07

You're right to demand justification beyond glib use of the FTC. But it is true that $$\int_0^\infty e^{-ax}dx = \frac{1}{a}$$ whenever $\Re(a)>0.$

In complex analysis, we can still affect the change of variables to $z=ax,$ but the resulting integral is along a ray in the complex plane in the direction of $a,$ not the positive real axis (unless $a$ is a positive real). We can write this (using somewhat bad notation) as $$\int_0^\infty e^{-ax}dx = \frac{1}{a} \int_0^{a\infty}e^{-z}dz.$$

(To see formally that the change of variables works, note that we have the parametrization $\gamma(t) = at$ for $0<t<\infty$ for the ray. We can write the integral of $e^{-z}/a$ along that path as $$\int_0^\infty \frac{e^{-\gamma(t)}}{a}\gamma'(t)dt = \int_0^\infty \frac{e^{-at}}{a}adt = \int_0^\infty e^{-at}dt$$).

Now, the difficult part is how we can justify saying $$\int_0^{a\infty}e^{-z}dz = \int_0^\infty e^{-x}dx = 1.$$

In other words we want to be able to rotate the contour back down to the real axis without changing the value of the integral.

To see why we can do this, imagine doing a integral around a large wedge-shaped contour. It goes out along the real axis to $R \gg 1$ and then goes along a circular path to $Re^{i\arg(a)}$ and then back into the origin along the ray $[0,a\infty)$ that our integral is taken over.

By Cauchy's theorem, the integral along this closed path is zero since $e^{-z}$ is analytic. The integral along the circular path goes to zero as $R\to \infty.$ We can see this cause the integrand decays like $e^{-R}.$ More formally the integral is $$\int_0^{\arg a} e^{-Re^{i\theta}}Re^{i\theta}id\theta$$ and we have $$\left|\int_0^{\arg a} e^{-Re^{i\theta}}Re^{i\theta}id\theta\right| \le \arg(a) \max_\theta|ie^{-Re^{i\theta}}Re^{i\theta}| = \arg(a)Re^{-R\cos(\arg(a))} \to 0$$ as $R\to\infty$

Thus as $R\to \infty,$ the integral along the real axis must cancel out the integral along the ray $[0,a\infty)$ in order that the integral along the closed path be zero as Cauchy's theorem demands. We have $$\int_0^{a\infty}e^{-z}dz = \int_0^\infty e^{-x}dx = 1$$ and therefore $$\int_0^\infty e^{-ax}dx = \frac{1}{a} \int_0^{a\infty}e^{-z}dz = \frac{1}{a}.$$

I've intentionally not said explicitly where $\Re(a)>0$ is used in the above argument. Of course, it's essential. See if you can find where it's used.

• Thanks for your answer, I found my own solution is very similar to yours! +1 – Zhanxiong Mar 5 '17 at 5:14

I figured out a rigorous proof by myself.

If $t = 0$, then it is an integration of a real-valued function, and clearly, $\int_0^\infty e^{-x} dx = 1$.

If $t \neq 0$, without losing of generality, assume $t > 0$. Consider the contour below:

In the picture, $n$ is a positive number that will be sent to $\infty$, the top line passes the origin and the point $(-1, t)$. And we set $f(z) = e^z, z \in \mathbb{C}$. By Cauchy's integration theorem, \begin{align} 0 = \int_\Gamma f(z) dz = \int_{\Gamma_1} e^z dz + \int_{\Gamma_2} e^z dz + \int_{\Gamma_3} e^z dz \end{align}

Let's denote the angle between $\Gamma_1$ and the real axis by $\theta_0$.

Clearly, $\int_{\Gamma_3}e^z dz = \int_{-n}^0 e^x dx = 1 - e^{-n}$.

On $\Gamma_1$, $z$ has the representation $z = (it - 1)x, x \in (0, n/|1 - it|)$, thus \begin{align*} \int_{\Gamma_1}e^z dz = \int_0^{n/\sqrt{1 + t^2}}e^{(it - 1)x}(it - 1) dx = (it - 1) \int_0^{n/\sqrt{1 + t^2}} e^{(it - 1)x} dx \end{align*}

To get the desired result, it remains to show $\int_{\Gamma_2} e^z dz \to 0$ as $n \to \infty$. Let $z = ne^{i\theta}$ with $\theta \in (\theta_0, \pi)$. It follows that \begin{align*} & \left|\int_{\Gamma_2} e^z dz\right| = \left|\int_{\theta_0}^\pi e^{ne^{i\theta}}nie^{i\theta} d\theta\right| \\ \leq & \int_{\theta_0}^\pi e^{n\cos\theta}n d\theta \\ \leq & ne^{n\cos{\theta_0}}(\pi - \theta_0) \to 0 \end{align*} as $n \to \infty$. Here we used the fact that $\pi/2 < \theta_0 < \pi$.

The real analysis way : $$f(x) = \frac{e^{(it-1)x}}{it-1}, f'(x) = e^{(it-1)x}, \qquad \int_0^\infty f'(x)dx=f(\infty)-f(0) = \frac{1}{1-it}$$

The complex analysis way, with the (complex) change of variable $z = (it-1)u, dz = (it-1)du$ : $$\int_0^\infty e^{(it-1)u}du = \int_0^{(it-1) \infty } e^{z}\frac{dz}{it-1} = \left.\frac{e^{z}}{it-1}\right|_0^{(it-1)\infty} = \frac{1}{1-it}$$

where $\int_0^{(it-1) \infty }$ is a contour integral


# $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ real?

Question: For $x,y \in \mathbb{C}$, suppose $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ necessarily real?

For instance, if $x+y=xy=3$, then one solution is $x = \frac{3 \pm i \sqrt{3}}{2}$, $y = \frac{3 \mp i \sqrt{3}}{2}$, but $x^3 + y^3 = 0$, which is real.

I've checked this numerically for many values of $w$ that give complex $x$ and $y$ (namely, $w \in (0,4)$.)

• @PatrickStevens Then $x+y$ isn't real. – wythagoras Aug 21 '16 at 10:37
• @PatrickStevens But he required that $x+y=xy\in\Bbb R$.. – BigbearZzz Aug 21 '16 at 10:37
• Sorry all. I wasn't paying attention, clearly. – Patrick Stevens Aug 21 '16 at 10:38
• The answer is yes; the essential idea is that $xy = x+y \in \mathbb{R}$ forces either that $x,y \in \mathbb{R}$, or that $x$ and $y$ are complex conjugates. – Drew N Aug 22 '16 at 2:22

Yes. Since $x + y \in \mathbb{R}$, $y = \overline{x} + r$ for some $r \in \mathbb{R}$. Then $xy = |x|^2 + xr \in \mathbb{R}$, implying that either $r = 0$ or $x \in \mathbb{R}$. Then we do casework:

• If $r = 0$, then $y = \overline{x}$; this leads to

$$x^w + y^w = x^w + \overline{x^w} \in \mathbb{R}.$$

Warning: for this to work, we had to pick the standard branch of the complex logarithm, specifically, the one undefined on the nonpositive real line, whose imaginary part is between $-\pi$ and $\pi$. Once we define $z^w := e^{w \ln z}$, ${(\overline{x})}^w = \overline{x^w}$ is true for this branch of $\ln$ (as $x$ is not nonpositive real), but might not be true for another branch.

This warning does not come into play when $w$ is an integer. But take, for example, $x = 1 + \frac{1 + i}{\sqrt{2}}$, $y = 1 + \frac{1 - i}{\sqrt{2}}$. Then $x + y = xy = 2 + \sqrt{2}$. If we picked a different branch of the complex logarithm, then we could have $x^w + y^w$ not real.

• On the other hand, if $x \in \mathbb{R}$, then $y \in \mathbb{R}$, so $x^w + y^w \in \mathbb{R}$. Since $x + y = xy > 0$, $x,y$ must both be positive, so we have no trouble with a negative base of the exponent.

Note there was nothing special about $w$: we could have reached the stronger conclusion that $x^a + y^a \in \mathbb{R}$ for all $a \in \mathbb{R}$.

• You do need $x,y\geq0$, for your last statement. – wythagoras Aug 21 '16 at 10:54
• @6005: Even with $a>0$, you still need $x,y\not\le 0$. Consider $x=y=-1\in\mathbb R$ and $a=1/2>0$. Then no matter where you do the branch cut, either $x^a=y^a=i$ or $x^a=y^a=-i$. Basically, the equation $\overline x^a = \overline{x^a}$ is nor true for all $x$ if $a$ is not integer. – celtschk Aug 21 '16 at 11:06
• As a second conjecture, I believe the casework depends on the value of $w$. That is, in fact $r = 0$ if and only if $0< w < 4$, and $x \in \mathbb{R}$ if and only if $w \geq 4$ ($w = 0$ is excluded by assumption). – Drew N Aug 21 '16 at 11:28
• @celtschk Thanks for your valuable feedback. I believe I fixed all the problems and provided all the necessary caveats. – 6005 Aug 21 '16 at 11:48
• @DrewN your second conjecture is correct. – 6005 Aug 21 '16 at 11:55

Yes. Write $x=a+bi$, $y=c-di$, then clearly $b=d$ because $x+y$ is real.

So $x=a+bi$, $y=c-bi$. Then $xy=ac-abi+cbi+b^2$, so $a=c$ or $b=0$.

• If $a=c$, then $y = \overline{x}$, i.e. the complex conjugate of $x$.
So $x^w+y^w = x^w+(\overline{x})^w = x^w+\overline{x^w}$, which is real.

• If $b=0$, $x$ and $y$ are real so $x^w+y^w$ is real as $w>0$.

• Thanks for your feedback on my answer earlier. In your answer, note that $(\overline{x})^w = \overline{x^w}$ is only true for a particular branch of complex log, and will never be true for negative real $x$. However, you can guarantee it for all $x$ which aren't negative real by picking the relevant branch of complex log to define your exponential. – 6005 Aug 21 '16 at 11:53

$x+y$ real implies that $\Im(x)=-\Im(y)$. Thus if $x=a+bi$ then $y=c-bi$.

$xy$ real implies that, because $xy=ac+b^2+ib(c-a)$, $b=0\lor c=a$.

If $b=0$ the result is trivial as $x,y\in\mathbb{R}$.

If $c=a$ then $x=\overline{y}$. But then clearly $$x^w+y^w=x^w+\overline{x}^w=\overline{x^w+\overline{x}^w}=\overline{x^w+y^w},$$ where the middle equality holds by symmetry.

But then it must be real, as the only complex numbers satisfying $z=\bar{z}$ are real.

Note: for why/when we are allowed to write $x^w=\overline{\bar{x}^w}$ refer to @6005's answer

• Actually, $x = \overline{(\overline{x})^w}$ is a bit more complicated; it's not sufficient that $w$ is real. We have to pick a definition of complex exponential that allows for the symmetry that you appeal to. But regardless of which definition we take (branch of complex log), $x^w = \overline{(\overline{x})^w}$ will be false for $x$ negative real. Consider $w = \tfrac12$. – 6005 Aug 21 '16 at 11:51
• Yeah. That is indeed true. I'll add in my answer to look at yours for a more detailed explanation on when/why this works. – b00n heT Aug 21 '16 at 12:17


# What is the volume of the $3$-dimensional elliptope?

### My question

Compute the following double integral analytically

$$\int_{-1}^1 \int_{-1}^1 2 \sqrt{x^2 y^2 - x^2 - y^2 + 1} \,\, \mathrm{d} x \mathrm{d} y$$

### Background

The $3$-dimensional elliptope is the spectrahedron defined as follows

$$\mathcal E_3 := \Bigg\{ (x_{12}, x_{13}, x_{23}) \in \mathbb R^3 : \begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{12} & 1 & x_{23}\\ x_{13} & x_{23} & 1\end{bmatrix} \succeq 0 \Bigg\}$$

Using Sylvester's criterion for positive semidefiniteness (i.e., all $2^3-1 = 7$ principal minors are nonnegative), we obtain $1 \geq 0$ (three times), the three quadratic inequalities

$$1 - x_{12}^2 \geq 0 \qquad \qquad \qquad 1 - x_{13}^2 \geq 0 \qquad \qquad \qquad 1 - x_{23}^2 \geq 0$$

and the cubic inequality.

$$\det \begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{12} & 1 & x_{23}\\ x_{13} & x_{23} & 1\end{bmatrix} = 1 + 2 x_{12} x_{13} x_{23} - x_{12}^2 - x_{13}^2 - x_{23}^2 \geq 0$$

Thus, $\mathcal E_3$ is contained in the cube $[-1,1]^3$. Borrowing the pretty figure in Eisenberg-Nagy & Laurent & Varvitsiotis, here is an illustration of $\mathcal E_3$

What is the volume of $\mathcal E_3$?

### Motivation

Why is $\mathcal E_3$ interesting? Why bother? Because $\mathcal E_3$ gives us the set of $3 \times 3$ correlation matrices.

### My work

For convenience,

$$x := x_{12} \qquad\qquad\qquad y := x_{13} \qquad\qquad\qquad z := x_{23}$$

I started with sheer brute force. Using Haskell, I discretized the cube $[-1,1]^3$ and counted the number of points inside the elliptope. I got an estimate of the volume of $\approx 4.92$.

I then focused on the cubic surface of the elliptope

$$\det \begin{bmatrix} 1 & x & y\\ x & 1 & z\\ y & z & 1\end{bmatrix} = 1 + 2 x y z - x^2 - y^2 - z^2 = 0$$

which I rewrote as follows

$$z^2 - (2 x y) z + (x^2 + y^2 - 1) = 0$$

Using the quadratic formula, I obtained

$$z = x y \pm \sqrt{x^2 y^2 - x^2 - y^2 + 1}$$

Integrating using Wolfram Alpha,

$$\int_{-1}^1 \int_{-1}^1 2 \sqrt{x^2 y^2 - x^2 - y^2 + 1} \,\, \mathrm{d} x \mathrm{d} y = \cdots \color{gray}{\text{(magic happens)}} \cdots = \color{blue}{\frac{\pi^2}{2} \approx 4.9348}$$

I still would like to compute the double integral analytically. I converted to cylindrical coordinates, but did not get anywhere.

### Other people's work

This is the same value Johnson & Nævdal obtained in the 1990s:

Thus, the volume is

$$\left(\frac{\pi}{4}\right)^2 2^3 = \frac{\pi^2}{2}$$

However, I do not understand their work. I do not know what Schur parameters are.

Here's the script:

-- discretization step
delta = 2**(-9)

-- discretize the cube [-1,1] x [-1,1] x [-1,1]
grid1D = [-1,-1+delta..1]
grid3D = [ (x,y,z) | x <- grid1D, y <- grid1D, z <- grid1D ]

-- find points inside the 3D elliptope
points = filter (\(x,y,z)->1+2*x*y*z-x**2-y**2-z**2>=0) grid3D

-- find percentage of points inside the elliptope
p = (fromIntegral (length points)) / (1 + (2 / delta))**3

*Main> delta
1.953125e-3
*Main> p
0.6149861105903861
*Main> p*(2**3)
4.919888884723089

Hence, approximately $61\%$ of the grid's points are inside the elliptope, which gives us a volume of approximately $4.92$.

### A new Buffon's needle

A symmetric $3 \times 3$ matrix with

• $1$'s on the main diagonal

• realizations of the random variable whose PDF is uniform over $[-1,1]$ on the entries off the main diagonal

is positive semidefinite (and, thus, a correlation matrix) with probability $\left(\frac{\pi}{4}\right)^2$. Estimating the probability, we estimate $\pi$. Using the estimate given by the Haskell script:

*Main> 4 * sqrt 0.6149861105903861
3.1368420058151125

### References

• This is an astoundingly good question! Nice! – clathratus Apr 16 at 1:08

The integral can be separated:

$$I = 2\int_{-1}^1 \sqrt{1-x^2} dx \cdot \int_{-1}^1 \sqrt{1-y^2} dy = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2$$

This integral is straight-forward using the substitution $$t=\sin\theta$$:

$$\int_{-1}^1 \sqrt{1-t^2} dt = \int_{-\pi/2}^{\pi/2} \sqrt{1-\sin^2\theta} \cos\theta d\theta = \int_{-\pi/2}^{\pi/2} |\cos\theta|\cos\theta d\theta$$

$$=\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta) d\theta = \dfrac{1}{2}\left(\theta + \dfrac{1}{2}\sin 2\theta\right)\Big|_{-\pi/2}^{\pi/2} = \dfrac{\pi}{2}$$

Therefore

$$I = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2 = 2\left(\dfrac{\pi}{2}\right)^2 = \dfrac{\pi^2}{2}$$

• Thanks. Shame on me for not realising that $(1-x^2) (1-y^2) = 1 - x^2 - y^2 + x^2 y^2$. I never expected the double integral to be this easy. – Rodrigo de Azevedo Oct 16 '16 at 19:19
• easy to get lost in the bigger picture, sometimes fresh eyes can help – David Peterson Oct 16 '16 at 19:21

Then integrand factors as $\sqrt{(1-x^2)(1-y^2)}=\sqrt{(1-x^2)}\sqrt{(1-y^2)}$ and every factor can be integrated separately. But you recognize the integral for the area of a half circle of radius $1$, hence

$$I=2\left(\frac\pi2\right)^2.$$


# Is the whole set $\mathbb R$ open?

Is the whole set $\mathbb R$ open or closed? A lot of answers from the following link said that the whole set is closed.

Why is empty set an open set?

However varies notes said that $\mathbb R$ is open, for example:

https://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf

• Open does not mean not closed. – Improve Jun 29 '17 at 2:58
• It is both... some say "clopen" the empty set is the another classic example of a set that is both open and closed. – Doug M Jun 29 '17 at 3:24
• "Is the whole set R open or closed?" both. "Said that the whole set is closed". That is true. "However varies notes said that R is open". That is also true. R is open. And R is closed. – fleablood Jun 29 '17 at 4:00

It's open and closed by definition.

In order to refer to open and closed sets a topology must be made explicit. Given a set $X$ and a collection $\mathcal F$ of subsets of $X$, $\mathcal F$ is a topology on $X$ only if $X, \emptyset \in \mathcal F$. Thus X is open by definition, and as closed sets are defined as sets whose compliments are open we have $X^c=\emptyset$ and we have $X$ is closed.

In the context of a topological space $\mathbb R$ with collection $\mathcal F$, $\mathbb R$ must be open and closed. However if we consider $\mathbb R \times \mathbb R$ with the usual topology, $\mathbb R \simeq \{0\} \times \mathbb R$ is closed but not open. It is important to state context.

Indeed the observation by Michael Hardy does really make sense. Additionally the question is not well posed because it does not state which topology is to be considered. Let us assume that euclidean topology is meant. So the answer by Michael Rozenberg uses axiomatization via open or closed sets.

Using axiomatization via neighborhoods (read balls in this case, due to the euclidean topology of $\mathbb R$)

1. $\mathbb R$ is open because any of its points have at least one neighborhood (in fact all) included in it;

2. $\mathbb R$ is closed because any of its points have every neighborhood having non-empty intersection with $\mathbb R$ (equivalently punctured neighborhood instead of neighborhood).

Equivalently:

1. $\mathbb R$ is open because all its points are interior points of itself

2. $\mathbb R$ is closed because all its points are adherent points of itself (equivalently limit points instead of adherent points)

Using axiomatization via Moore-Smith net convergence (read sequence convergence in this case, due to the euclidean topology of $\mathbb R$),

1. $\mathbb R$ is closed because every point to which at least one net of its points converges belongs to it

2. $\mathbb R$ is open because every point to which every convergent net converges has a non-empty intersection with it. (Or equivalently there are no nets of points of its complement (the empty set) converging to any of its points (in fact there are no nets of points of its complement))

The list can go on and on.

By the definition of a topology on a set, both the empty set and the entire set (which I'm assuming you're taking as $\mathbb R$) are open sets. Since the complement of an open set is a closed set, and $\mathbb R$ is the complement of the empty set, it is also closed.

Notice that "open" and "closed" are not mutually exclusive.

$A$ is open iff $A^c$ is closed.

So $\emptyset$ and $\Omega$ are both open and closed (clopen)

For general topology, it is guaranteed via definition of topology, i.e. $\emptyset, \Omega \in \mathscr T$.

By one commonplace definition of "open set", a set $A\subseteq\mathbb R$ is open if for every point $x\in A$ there is some open interval containing $x$ that is a subset of $A$. That clearly is true if $A=\mathbb R,$ so $\mathbb R$ is open.

By one commonplace definition of "closed set", a set $A\subseteq \mathbb R$ is closed if every limit point of $A$ is a member of $A$. That clearly is true if $A=\mathbb R$, so $\mathbb R$ is closed.

Only two subsets of $\mathbb R$ are both open and closed: $\mathbb R$ and $\varnothing.$


# Integrating arcsech x

1. May 17, 2014

### sooyong94

1. The problem statement, all variables and given/known data
I was asked to prove the integral

$\int_{\frac{4}{5}}^{1} \textrm{arcsech}(x) =2\arctan 2-\frac{\pi}{2}-\frac{4}{5} \ln 2$

2. Relevant equations
Integration by parts

3. The attempt at a solution
Let $u=\textrm{arcsech} (x)$
$\textrm{sech u}=x$
$\textrm{cosh u}=\frac{1}{x}$
Differentiating implicitly,

$\textrm{sinh u} \frac{du}{dx}=\frac{-1}{x^{2}}$
$\frac{du}{dx}=\frac{-1}{x^{2}\textrm{sinh u}}$

Then I simplify it into
$\frac{-1}{x\sqrt{1-x^{2}}}$

Let $\frac{dv}{dx}=1$

$v=x$

Using integration by parts and evaluating the integral, I got
$\frac{\pi}{2}-\sin^{-1} \frac{4}{5} -\frac{4}{5} \ln 2$

Which is numerically correct. But how do I obtain $2\tan^{-1} 2$ as shown in the question above?

2. May 17, 2014

### Curious3141

Draw the standard 3-4-5 right triangle. Observe that $\frac{\pi}{2}-\sin^{-1} \frac{4}{5} = \tan^{-1}\frac{3}{4}$.

You now have to get $\tan^{-1}\frac{3}{4}$ into something with $\tan^{-1}2$ in it.

I found this a little tricky. The best solution I could find was to let:

$\tan^{-1}\frac{3}{4} = x$ so $\tan x = \frac{3}{4}$

Then let x = 2y so that $\tan 2y = \frac{3}{4}$

Solve for y in the form $y = \tan^{-1}z$, where z is something you have to find. Only one value is admissible. Express x as 2y = $2\tan^{-1}z.$

Now observe that $\frac{1 + \tan w}{1 - \tan w} = \tan(w + \frac{\pi}{4})$. Use that to find an alternative form for $\tan^{-1}z$, which will allow you to find x, in the form you need.

There might be a simpler way (indeed, it might start with an alternative solution of the integral), but I can't immediately find one.

Last edited: May 17, 2014
3. May 17, 2014

### AlephZero

To show the answers are the same, you have to show $2 \tan^{-1} 2 + \sin^{-1}\displaystyle\frac 4 5 = \pi$.

From a 3-4-5 triangle, $\sin^{-1}\displaystyle\frac 4 5 = \tan^{-1}\displaystyle\frac 4 3$.

From $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\displaystyle\frac{a+b}{1-ab}$,

$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.

QED.

4. May 22, 2014

### sooyong94

I don't get it, but why
$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.

5. May 22, 2014

### CAF123

I think the equalities should be $$2 \tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi = \pi - \tan^{-1} \left(\frac{4}{3}\right)$$

Using Alephzero's formula with $a=b$ gives $$\tan(\tan^{-1} 2 + \tan^{-1} 2) = -\frac{4}{3} \Rightarrow 2\tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi$$

6. May 22, 2014

### SammyS

Staff Emeritus
Compare the graphs of $\ y=\text{arcsech}(x) \$ and $y=\text{sech}(x)\ .$

Integrate $y=\text{sech}(x)\$ to get an area related to that given by integrating $\ y=\text{arcsech}(x) \ .$

You will have to subtract the area of some rectangle.

7. May 23, 2014

### sooyong94

I plotted two graphs and yet I can't figure it out... :(

8. May 23, 2014

### SammyS

Staff Emeritus
The definite integral you are evaluating represents the area below the y = arcsech(x) graph which is between x = 4/5 and x = 1 . Notice that arcsech(4/5) = ln(2) .

That is the same as the area below the y = sech(x) graph and above y = 4/5, for x ≥ 0. Right?

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# Number TheoryPrimitive root modulo 169

#### tda120

##### New member
How can I find a primitive root modulo 169?
I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide phi(13^2)..?.

#### Opalg

##### MHB Oldtimer
Staff member
How can I find a primitive root modulo 169?
I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide phi(13^2)..?.
Hi tda, and welcome to MHB. You might be interested in this link, which tells you that the answer to your question is either $2$ or $2+13=15$. That still leaves you with the work of testing to see if $2$ works. If it does not, then $15$ does.

#### mathbalarka

##### Well-known member
MHB Math Helper
tda120 said:
How can I find a primitive root modulo 169?
There is no simple method. You can cobble up together some basic theories on primitive roots, find a bit of a rough upperbound (although none is known to be useful for small cases) and some modular exponentiation to get a fast enough algorithm.

As Opalg there indicated, that either 2 or 15 is primitive root modulo 169, can easily be found for this case. Try proving the former and then the later if it doesn't work.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
How can I find a primitive root modulo 169?
I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide phi(13^2)..?.
You don't have to check all the orders that divide $\phi(13^2)$.
It suffices to check each of the orders that are $\phi(13^2)$ divided by one of the distinct primes it contains.

$$\phi(13^2)=2^2\cdot 3 \cdot 13$$
So the orders to verify are:
$$2\cdot 3 \cdot 13,\quad 2^2 \cdot 13, \quad 2^2\cdot 3$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi tda, and welcome to MHB. You might be interested in this link, which tells you that the answer to your question is either $2$ or $2+13=15$. That still leaves you with the work of testing to see if $2$ works. If it does not, then $15$ does.
Nice!

From that link we also get that since 2 is a primitive root mod 13, it follows that the order of 2 mod 169 is either (13-1) or 13(13-1).
So if $2^{13-1} \not\equiv 1 \pmod{169}$ that means that 2 has to be a primitive root mod 169. Or otherwise 15 has to be.

In other words, no need to check any of the other powers.


# Finding the coefficient on the $x$ term of ${\prod_{n = 1}^{20}(x-n)}.$

I am trying to find the coefficient on the $x$ term of $\displaystyle{\prod_{n = 1}^{20}(x-n)}$. The issue is that the binomial theorem can't be applied since our $b$ value is changing from term to term. Is there any simple way to do this problem, perhaps a way to change the expression so that the binomial theorem applies? I've tried to do that, and tried looking for a pattern on similar expressions, but I haven't come up with anything. Any help you might have would be appreciated.

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It will be the 19th elementary symmetric polynomial in your roots. See en.m.wikipedia.org/wiki/Elementary_symmetric_polynomial for symmetric polynomials. – Marc Jun 1 '14 at 17:11
A concept popularly known as sum of roots of a polynomial can be of help. – fermesomme Jun 1 '14 at 17:22

The coefficient is given by $$\sum_{k=1}^{20}\prod_{n\not=k\atop n=1,\dots,20}{(-n)}=-\sum_{k=1}^{20}\frac{20!}{k}=-20! H_{20}$$ where $H_k$ is the $k$-th Harmonic number. Those can be looked up in tables (see A001008 and A002805): $H_{20}=\frac{55835135}{15519504}$ and thus the coefficient is given by $-8752948036761600000$.

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This answer just gets better and better on a moment-by-moment basis! Thanks for adding the value of $H_{20}$! – Robert Lewis Jun 1 '14 at 18:30
Just a question, is there a simple way to calculate the constant of the expanded form? – recursive recursion Jun 1 '14 at 19:47
@Dominik, Also, I'm not sure how you got your first expression. Could you please explain that – recursive recursion Jun 1 '14 at 19:51
You get terms with a single power in $x$ from the product $\prod_{n=1}^{20}(x-n)$ by taking $19$ non-$x$ factors and one $x$. The sum is the sum over the different choices of those non-$x$ and $x$ factors. More formally you could try to prove that $(-1)^{N+1} N! H_{N}$ is the $x$-coefficient of $\prod_{n=1}^N (x-n)$ by induction. – Dominik Jun 1 '14 at 20:14

This is one of Vieta's formulas.

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The coefficient must be less than $-20!$, so this can't be correct? – copper.hat Jun 1 '14 at 17:22
Not to put too fine a point on it, but this answer is incorrect; the coefficient of $x^{19}$ is in fact $-\sum_{n = 1}^{20} n = -(20 \times 21)/2 = -210$; for the coefficient of $x$, see the answer given by Dominik. I do believe it is correct. – Robert Lewis Jun 1 '14 at 17:31
@RobertLewis: Unfortunately no 'Med' time for me this morning... – copper.hat Jun 1 '14 at 17:37
@Robert Lewis. Shoot! You're right! I'll remove everything but the vieta's formula link. – Avi Steiner Jun 1 '14 at 17:40
@Avi Steiner: well done! Glad we got it right! – Robert Lewis Jun 1 '14 at 19:40

One way is to just use Maclaurin's formula: $$f(x) = f(0) + \frac{f'(0)}{1!} x + \ldots$$ In this case: \begin{align} f'(x) &= \sum_{1 \le n \le m} \prod_{\substack{1 \le k \le m\\k \ne n}} (x - k) \\ f'(0) &= (-1)^{m - 1} m! H_m \end{align}

-

The question has been well answered by others. However, I would like to point out that this polynomial has a name---Wilkinson's polynomial---and its own Wikipedia article. It was put forward by Wilkinson as an example of an apparently innocuous polynomial with remarkable numerical-analytic properties.

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This polynomial is a lot more interesting than I thought! Thanks for giving me a name to google. – recursive recursion Jun 1 '14 at 21:29

A generalization to other coefficients (and limits different from $20$) is given in the generating functions of Stirling numbers of the first kind, Pochammer symbols:

$$(x)_n:=\prod_{k=0}^{n-1}(x-k)=\sum_{k=0}^n s(n,k)x^k.$$

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The coefficients are the Stirling numbers of the first kind :

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# Moderate Probability Solved QuestionAptitude Discussion

 Q. Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number?
 ✖ A. 17/24 ✔ B. 3/8 ✖ C. 1/2 ✖ D. 5/8

Solution:
Option(B) is correct

First of all, if we IGNORE the condition about where the objects can be placed, we can arrange the 4 different objects in 4! ways (= 24 ways).

So, we now must determine HOW MANY of those 24 arrangements are such that no objects occupy the location corresponding to its number.

A quick way to do this is to LIST acceptable outcomes.

IMPORTANT: We'll list each arrangement so that the first number represents the object that goes to location #1, the second number represents the object that goes to location #2, and so on.

So, for example, 3421 represents object #3 in location #1, object #4 in location #2, object #2 in location #3, and object #1 in location #4.

Let's be systematic:

Arrangements where object #2 is in location #1

The possible arrangements where NO object is in the correct location are as follows:

2143, 2341, 2413

Total number of arrangements $= 3$

Arrangements where object #3 is in location #1

The possible arrangements where NO object is in the correct location are as follows:

3142, 3412, 3421

Total # of arrangements $= 3$

Arrangements where object #4 is in location #1

The possible arrangements where NO object is in the correct location are as follows:

4123, 4312, 4321

Total # of arrangements $= 3$

Altogether, the number of arrangements where no object is in the correct location,

$= 3 + 3 + 3$

$= 9$

So, $P(\text{no object in correct location}) = \dfrac{9}{24}$

$= \dfrac{3}{8}$

Thus, option (B) is the correct choice.

Edit: Based on Vaibhav's comment, the solution has been updated.

Edit 2: For an alternate solution see Kartik's comment

Edit 3: Sheldon has provided a way to reach the correct answer without even solving the question

Edit 4: Suzen has given an exhaustive solution

Edit 5: For derangements formula, check Himanshu Singh's comment.

Note: For better understanding and different approaches do visit comment section.

## (14) Comment(s)

Gaurav Karnani
()

Total number of cases = 24

let us divide the case into categories where :

1) single number is in right position = 8 cases; two for each $1,2,3,4$

2) Two numbers are in right position = 6 cases $12,13,14,23,24,34$

3) No case for three digits as if three are in correct position fourth one will definitely be, so only one case of $1234$ .

total number of cases =15

Probability = $9/24$

Himanshu Singh
()

Please use this formula for derangements :

$!n = n!\left(1-\dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} ... (-1)^n \times \dfrac{1}{n!} \right)$

Kartik
()

Total number of ways of placing objects into places randomly,

$=4 \times 3 \times 2 \times 1=4!=24$

No. of ways of placing them such that only one of them gets correct place,

\begin{align*} = & 4 \text{ (ways of choosing the correctly} \\ & \text{ placed objects)} \times \\ & \times 2 \text{ (to place next object wrongly)} \\ & \times 1 \text{ (to place next object wrongly)} \\ & \times 1 \text{ (to place next object wrongly)}\\ = & \textbf{8} \end{align*}

Number of ways of placing exactly two objects correctly,

\begin{align*} = & ^4C_2 \text{ (to identify the two correctly placed} \\ & \text{ objects their arrangment are unique}\\ & \text{ for them to be correct)} \times \\ & 1 \text{ (to arrange next object wrongly)} \times \\ & 1 \text{ (to arrange next object wrongly)}\\ = & \textbf{6} \end{align*}

Now,

Number of ways to place exctly 3 objects correct = ways to place all objects correct $= \textbf{1}$

Therefore, number of ways to place all digits wrongly,

$=24-(8+6+1)$

$=24-15$

$=9$

Thus, Required probability,

$=\dfrac{9}{24}$

$=\dfrac{3}{8}$

Raj
()

Since there's only one correct way of placing the objects as per the boxes- can't we work on the lines of using probability of an event happening = 1 - probability of that event not happening?

So therefore, can't the answer be 1 - (1/24) = 23/24?

Brent
()

That's a good idea, but it doesn't apply here.

If event A = NONE of the objects in the correct places, then the complement (event A NOT happening) will consist of any arrangement where some (perhaps all) of the objects are in their correct place(s).

Vaibhav
()

but this is the case for 2 at first position. it can occupy position 3 and 4 as well . Similar case with other digits will be there.

Deepak
()

You are right Vaibhav, updated the solution.

Suzen
()

Here's the full list:

1234

1243

1423

1432

1324

1342

2134

2143 OK

2431

2413 OK

2341 OK

2314

3124

3142 OK (Close to 1000pi, but no correlation really, or is there? Perhaps the nth position is not n for all decimal places. OK, an interesting idea to explore)

3421 OK

3412 OK

3214

3241

4123 OK

4132

4321 OK

4312 OK

4213

4231

Clearly $P = \frac{9}{24} = \frac{3}{8}$

I know it's long winded, but I'm glad it matched!

Tesla
()

Let a particular number (say) number 2 occupies position 1.

Then all possible arrangement are given as:

(2,1,3,4), (2,1,4,3), (2,3,4,1), (2,4,1,4), (2,4,1,3), (2,4,3,1).

Out of these six, three (2,1,3,4), (2,3,1,4), (2,4,3,1) are not acceptable because numbers 3 and 4 occupy the correct positions.

Required probability = 3/6 = 1/2

Sheldon
()

I am not sure what difficulty level this would fall into.

If I see this question in the real exam, I would spend about a minute solving it.

After that, I would guess and move on. because it would be time consuming.

Looking at the question, it is more obvious that at-least 1 number would fall into its corresponding place.

So definitely, the probability that the number wouldn't fall into its numbered place should be less than half.

Looking at options:

A. 17/24

B. 3/8

C. 1/2

D. 5/8

Only option B is less than half. I would pick B and move on.

Brent
()

That's a great approach that allows you to minimize time spent on a question (that you feel is going nowhere) and maximize your guess.

Probability questions are perfect for this, because most people have a gut feeling about how likely something is. As you suggest, it does seem unlikely (probability less than 0.5) that every object would be out of place, so B is the perfect guess.

Steve
()

I agree with sam and abhishek

Abhishek
()

the correct answer will be $\dfrac{9}{24} =\dfrac{3}{8}$

out of

1234 1243 1324 1342 1423 1432
2341 2314 2143 2134 2413 2431
3124 3142 3412 3421 3214 3241
4123 4132 4213 4231 4312 4321

24 combinations

2341 2143 2413
3421 3412 3142
4123 4312 4321

are misplaced(dearranged)

So, the correct answer will be $\dfrac{9}{24}$ i.e. $\dfrac{3}{8}$. :)

Sam
()

I think Abhishek is correct...... answer should be 3/8


Limit involving Series and Greatest Integer Function

If $[$.$]$ denotes the greatest integer function, then find the value of $\lim_{n \to \infty} \frac{[x] + [2x] + [3x] + … + [nx]}{n^2}$

What I did was, I wrote each greatest integer function $[x]$ as $x - \{x\}$, where $\{.\}$ is the fractional part. Hence, you get

$\lim_{n \to \infty} \frac{\frac{n(n+1)}{2}(x-\{x\})}{n^2}$

The limit should then evaluate to $\frac{x-\{x\}}{2}$

But the answer given is $\frac{x}{2}$. What am I missing here?

$$\frac{\lfloor x\rfloor+\ldots+\lfloor nx\rfloor}{n^2}=\frac{x+2x+\ldots nx-\{x\}-\ldots-\{nx\}}{n^2}=$$

$$=\frac{n(n+1)}{2n^2}x-\frac{\{x\}+\ldots+\{nx\}}{n^2}\xrightarrow[n\to\infty]{}\frac12x-0=\frac x2$$

since the second addend above tends to zero:

$$\frac{\{x\}+\ldots+\{nx\}}{n^2}\le\frac n{n^2}=\frac1n\xrightarrow[n\to\infty]{}0$$

• I didn't really get how the second addend (fractional part thing) tends to zero, as n tends to infinity. Isn't it in an inderminate form as it is? Also, what has been done in the last line of the answer? – skb May 28 '18 at 20:20
• @skb Every term $\;\{kx\}\;$ is less than $\;1\;$ , so that whole sum's numerator is less that $\;1+1+\ldots+1=n\;$ ... – DonAntonio May 28 '18 at 20:22
• Have you used the sandwich theorem in the last line? But shouldn't there be another expression less than it for it to work? – skb May 28 '18 at 20:24
• @skb But isn't it obvious that the whole expression is greater than zero or equal to it? You can complete that argument... – DonAntonio May 28 '18 at 20:27
• Got it. Thanks a lot for the help! – skb May 28 '18 at 20:27

Note that by Stolz-Cesaro

$$\lim_{n \to \infty} \frac{\lfloor x\rfloor + \lfloor 2x \rfloor + \lfloor3x\rfloor + … + \lfloor nx]}{n^2}=\lim_{n \to \infty} \frac{\lfloor(n+1)x\rfloor}{(n+1)^2-n^2}=\lim_{n \to \infty} \frac{\lfloor(n+1)x\rfloor}{2n+1}=$$ $$=\lim_{n \to \infty} \frac{(n+1)x}{2n+1}-\frac{\{(n+1)x\}}{2n+1}$$