# Does every number not ending with zero have a multiple without zero digits at all?

Does every number not ending with zero have a multiple without zero digits at all?

We consider numbers in their usual base 10 representation. we don't consider numbers ending with $0$, because all multiples of such numbers end with $0$.

It can be proven that all powers of $2$ have such a multiple: Indeed, for every number of the form $2^n$, we can find a multiple $l$ of it whose last $n$ digits are nonzero (This can be done by repeatedly multiplying a number by $10^k+1$, as explained here), and then forget all other digits; these last $n$ digits constitute a number that is still divisible by $2^n$, since it is obtained from $l$ by subtracting a multiple of $10^n$ which is in particular a multiple of $2^n$.

The same argument shows that every power of $5$ has a multiple without zeros, but this does not seem to generalize to other numbers.

My intuition is actually that the claim is not true, since the density of numbers without any zero digits approaches $0$ for big numbers. This is because the probability of a random number with at most $n$ digits to contain only nonzero digits is $(\frac{9}{10})^n$. Thus big numbers "nearly always" contain a zero. This hints on the existence of a counterexample, but does not prove its existence.

• I'm not sure what the answer is, but I think your heuristic argument misses something big: Sure, the proportion of $n$ digit numbers with only non-zero digits is $(9/10)^n$, but the number of multiples of a number $k$ with $n$ digits is roughly $\frac{10^n}k$ - so multiply those together and we "expect" there to be $\frac{9^n}k$ multiples of $k$ with $n$ digits and no zeros. (This is certainly true of numbers coprime to 10) – Milo Brandt Dec 21 '16 at 20:51
• User @lulu seems to have deleted his correct observation that this is also true for all numbers coprime to $10$, by Fermat's little theorem. – Emolga Dec 21 '16 at 21:06
• I deleted the post because someone pointed out this question which generalizes the straightforward argument I gave. – lulu Dec 21 '16 at 21:08
• Just a note: in fact you can get stronger results for multiples of $2^n.$ For any odd digit $a$ and even digit $b$ you can find a multiple of $2^n$ using only $a$ and $b$ – cats Dec 21 '16 at 21:12
• Found it - see "104 Number Theory Problems from the training of the USA IMO Team" Introductory Problems #52 – cats Dec 21 '16 at 21:18

If $n$ is relatively prime to $10$ then the answer is easy, see for example this post.

If, $n=2^k m$ with $m$ relatively prime to $10$ or $n=5^km$ with $m$ relatively prime to $10$, then prove first the following by induction:

Claim 1 $2^k$ has a $k$ digits multiple containing only the digits $1$ and $2$.

Claim 2 $5^k$ has a $k$ digits multiple containing only the digits $1,2,3,4$ and $5$.

Both are easy induction exercises.

Once you prove this, proceed as follows (same idea as in the posted link):

$n=2^k m$ or $n=5^km$ with $gcd(m,10)=1$.. Then, we know that $2^k$ respectively $5^k$ has a multiple of the form $\overline{a_1...a_k}$

Look at the following numbers $$\overline{a_1...a_k} , \overline{a_1...a_ka_1...a_k} , \overline{a_1...a_ka_1...a_ka_1...a_k} , ...$$

In this infinite list, there exists two numbers with the same reminder when divided by $m$. Their difference is a multiple of $m$.

Therefore $$m| \overline{a_1...a_ka_1...a_ka_1...a_k00...0}=\overline{a_1...a_ka_1...a_ka_1...a_k}\cdot 10^l$$ Since $m$ is relatively prime to $10$ it follows that $$m|\overline{a_1...a_ka_1...a_ka_1...a_k}$$ Since $2^k$ respectively $5^k$ also divide $\overline{a_1...a_ka_1...a_ka_1...a_k}$, and they are relatively prime to $m$, it follows that $$n|\overline{a_1...a_ka_1...a_ka_1...a_k}$$

P.S. We proved that $n$ has a multiple which can be written only with the digits $1,2,3,4,5$. These digits can be replaced by any $5$ digits which cover all classes $\pmod{2}$ and $\pmod{5}$.

• Could you please give a proof of the claims? – J. Abraham Feb 25 '17 at 19:56
• @J.Abraham $P(k) \Rightarrow P(k+1)$. Let $A$ be the $k$ digit number which is divisible by $a^k$ where $a=2$ or $5$. Then $$A=a^k \cdot m$$ and $$10^{k}=a^k \cdot l$$ where $l$ is invertible modulo $a$. Then the equation $$x l +m \equiv 0 \pmod{a}$$ has an unique solution in the listed digits. Then $$x \cdot 10^k+A$$ is the desired $k+1$ digit number. – N. S. Feb 26 '17 at 23:07


1. ## congruences 2

Slove the system of congruences

3x = 2 mod 4
4x = 1 mod 5
6x = 3 mod 9

is it no solution since 6 and 9 do not have an inverse since the gcd of 6 and 9 is not one thus u cant write it in the form x=a mod p

2. I have been trying to find the inverse and solve it that way but i dont know if that is right?

3. If $ac \equiv bc \ (\text{mod } m)$ and $(c,m) = d$, then $a \equiv b \left(\text{mod } \frac{m}{d}\right)$

Keeping in this in mind, I'm going to simplify the congruences first:
$\begin{array}{rcll} 3x & \equiv & 2 & (\text{mod } 4) \\ 3x & \equiv & 6 & (\text{mod } 4) \\ x & \equiv & 2 & (\text{mod } 4) \end{array}$........ $\begin{array}{rcll} 4x & \equiv & 1 & (\text{mod } 5) \\ 4x & \equiv & -4 & (\text{mod } 5) \\ x & \equiv & -1 & (\text{mod } 5) \\ x & \equiv & 4 & (\text{mod } 5) \end{array}$........ $\begin{array}{rcll} 6x & \equiv & 3 & (\text{mod } 9) \\ 6x & \equiv & 12 & (\text{mod }9) \\ x & \equiv & 2 & (\text{mod } 3) \end{array}$

So we're left with the system:
$\begin{array}{rcll} x & \equiv & 2 & (\text{mod } 4) \\ x & \equiv & 4 & (\text{mod } 5) \\ x & \equiv & 2 & (\text{mod } 3) \end{array}$

And by the Chinese Remainder theorem, since $(4, 5, 3) = 1$, there is a unique solution modulo $4\cdot 5 \cdot 3 = 60$. Should be pretty simply to solve this one.

4. Originally Posted by o_O
If $ac \equiv bc \ (\text{mod } m)$ and $(c,m) = d$, then $a \equiv b \left(\text{mod } \frac{m}{d}\right)$

Keeping in this in mind, I'm going to simplify the congruences first:
$\begin{array}{rcll} 3x & \equiv & 2 & (\text{mod } 4) \\ 3x & \equiv & 6 & (\text{mod } 4) \\ x & \equiv & 2 & (\text{mod } 4) \end{array}$........ $\begin{array}{rcll} 4x & \equiv & 1 & (\text{mod } 5) \\ 4x & \equiv & -4 & (\text{mod } 5) \\ x & \equiv & -1 & (\text{mod } 5) \\ x & \equiv & 4 & (\text{mod } 5) \end{array}$........ $\begin{array}{rcll} 6x & \equiv & 3 & (\text{mod } 9) \\ 6x & \equiv & 12 & (\text{mod }9) \\ x & \equiv & 2 & (\text{mod } 3) \end{array}$

So we're left with the system:
$\begin{array}{rcll} x & \equiv & 2 & (\text{mod } 4) \\ x & \equiv & 4 & (\text{mod } 5) \\ x & \equiv & 2 & (\text{mod } 3) \end{array}$

And by the Chinese Remainder theorem, since $(4, 5, 3) = 1$, there is a unique solution modulo $4\cdot 5 \cdot 3 = 60$. Should be pretty simply to solve this one.
i get the right answer after applying the CRT to the reduced congruences, but i still dont know how u reduced the congruences or simplified them. Could you maybe explain it a little bit. Could u find the inverse modulo of each congruence and and reduce it to the form x=a mod m

5. Hello,

Divide the formula of o_O into 2 formulae:

1) If $ac\equiv bc\pmod m$ and $(c, m)=1$ then $a\equiv b\pmod m$.
2) If $ac\equiv bc\pmod {mc}$ then $a\equiv b\pmod m$.

1) corresponds to the case where you can find an inverse.
2) is a technique to decrease the modulus.
You add the modulus as many times as you want so that the coefficient of x and the right-hand-side have a common divisor.

Bye.

6. What I did was play around with the congruences a little. Remember, they sort of act like equality signs as well. Let's take the middle one I did:

$
\begin{array}{rcll} 4x & \equiv & 1 & (\text{mod } 5) \\ 4x & \equiv & -4 & (\text{mod } 5) \qquad \text{1. Subtracted 5} \\ x & \equiv & -1 & (\text{mod } 5) \qquad \text{2. Divided both sides by 4} \\ x & \equiv & 4 & (\text{mod } 5) \qquad \text{3. Added 5 to get a least residue} \end{array}
$

Basically, I wanted to get rid of the coefficient in front of the x and what I wanted is to ultimately divide it out:
1. If something is congruent to 4x mod 5, then adding or subtracting multiple of 5's will still be congruent to 4x. Remember: $a \equiv b \ (\text{mod m}) \ \iff \ a = b + km$

2. This refers to the very first statement I had in my first post

3. Again, adding and subtracting multiples of the modulus will not change the congruence as we did in step 1. The main purpose in doing so was to get a least residue (i.e. the "remainder")

Edit: Oh beaten!


## Anita505 2 years ago Suppose that we have a white urn containing four white balls and one red ball and have a red urn containing one white ball and four red balls. An experiment consists of selecting at random a ball from the white urn and then (without replacing the first ball) Selecting at random a ball from the urn having the colour of the first ball. Find the probability that the second ball is red. The probability that the second ball is red is ______.

1. kropot72

There are two situations to consider: (1) If a white ball is selected on the first draw then the second draw will be from the white urn. Probability of white on the first draw is 4/5 and the probability of red on the second draw is 1/4. P(red second ball) = 4/5 * 1/4 = 4/20 ...............(1) (2) If a red ball is selected on the first draw then the second draw will be from the red urn. Probability of red on the first draw is 1/5 and the probability of red on the second draw is 4/5. P(red second ball) = 1/5 * 4/5 = 4/25 ...............(2) The probability that the second ball is red is the sum of the fractions (1) and (2).

2. Anita505

so the sum of (4/20)+(4/25) So in this case the answer would be 0.36 correct?

3. kropot72

Correct, or alternatively 9/25.

4. Anita505

thank you for showing me the process

5. kropot72

You're welcome :)

6. Anita505

An urn contains 3 one-dollar bills, 1 five- dollar bill and 1 ten dollar bill. A player draws bills one at a time without replacement from the urn until a ten dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine: A) The probability of winning $15 B)The probability of winning all bills in the urn C) The probability of the game stopping at the second draw. 7. Anita505 can you help me with this and show me the process? 8. satellite73 as far as i can see there is only one way to win$15: first draw the five, then draw the ten

9. satellite73

since there are 5 bills, and one is a five, the probability of drawing the five first is $$\frac{1}{5}$$ then there are 4 bills of which one is a ten, the probability of drawing a ten second given that the first bill was a five is $$\frac{1}{4}$$ the probability of both things occurring is $\frac{1}{5}\times \frac{1}{4}$

10. Anita505

so for part a) it would be 1/20? the answer

11. satellite73

yes

12. Anita505

okay thank you but i need help with part b and c

13. satellite73

for b) it means you pick the ten dollar bill last right?

14. satellite73

the probability you pick the ten dollar bill last is the same as the probability you pick the ten dollar bill first, namely $$\frac{1}{5}$$

15. Anita505

so b then in this case is 1/5

16. satellite73

and for the last one, that means you pick something other then the ten dollar bill, and then you pick the ten dollar bill that is also $$\frac{1}{5}$$ via $\frac{4}{5}\times \frac{1}{4}=\frac{1}{5}$

17. satellite73

more simply put, the probability you pick the ten first, second, third, fourth or fifth are all the same, namely $$\frac{1}{5}$$

18. Anita505

Okay thank you for your assistance i have one last question to ask,

19. satellite73

do me a favor and post in a new thread this is hard so scroll down to

20. Anita505

A grade 11 art class is offering students two choices for a project: a pottery project and a mixed media project. Of the 46 students in the class, 23 have selected to do the pottery project and 33 have selected to do the mixed media project (notice some students have decided to do both) It two students are selected at random from the class to show their finished project(s), what is the probability that at least one pottery project and at least one mixed media project will be shown? Probability (given to three decimal places)


1. ## L'Hopital's Rule question

I have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:

Find the limit of [ln(x)]/[ln(2x)] as x goes to infinity.

I tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?

2. Originally Posted by pianopiano
I have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:

Find the limit of [ln(x)]/[ln(2x)] as x goes to infinity.

I tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?
Note that $\frac{\ln x}{\ln(2x)}=\frac{\ln x}{\ln 2+\ln x}$. Since $\lim_{x\to\infty}\frac{\ln x}{\ln(2x)}=\lim_{x\to\infty}\frac{\ln x}{\ln 2+\ln x}=\frac{\infty}{\infty}$, we can apply L'Hopitals rule.

So $\lim_{x\to\infty}\frac{\ln x}{\ln(2x)}=\lim_{x\to\infty}\frac{\ln x}{\ln 2+\ln x}=\lim_{x\to\infty}\frac{\displaystyle\frac{1}{x} }{\displaystyle\frac{1}{x}}=\lim_{x\to\infty}\frac {x}{x}=1$

Does this make sense?

3. Thanks, it does make sense!

However, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?

4. Originally Posted by pianopiano
Thanks, it does make sense!

However, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?
You could...but you would have to note that $\forall a\in\mathbb{R}\backslash\left\{0\right\},~\frac{\, d}{\,dx}\left[\ln\!\left(ax\right)\right]=\frac{1}{x}$

5. $\lim_{x\to\infty}\frac{\log x}{\log 2 + \log x}=\lim_{x\to\infty}\frac{1}{\underbrace{\frac{\lo g 2}{\log x}}_{\to 0}+1}=1$.

Or you must use L'H?

6. Originally Posted by Abu-Khalil
$\lim_{x\to\infty}\frac{\log x}{\log 2 + \log x}=\lim_{x\to\infty}\frac{1}{\underbrace{\frac{\lo g 2}{\log x}}_{\to 0}+1}=1$.

Or you must use L'H?

I was going to point that out.
There's no need for L'Hopital's Rule at all, just divide by the common term, log x.


# Using the Mean Value Theorem

The (classical) Mean Value Theorem states that if $f$ is a continuous function on $[a,b]$ and is differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ where $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.

Here is a "real-world" application of the theorem. Suppose two racers, Barry and Harry, start a race at time $t=0$ and end the race in a tie, both crossing the finish line at, say, time $t = 10$. Let $s(0)$ be the position of the starting line, and let $s(10)$ be the position of the finish line.

The position of each racer is continuous on [0,10] and (presumably) differentiable on (0,10). So the MVT tells us that at some time $t_0 \in (0,10)$, Barry's velocity equaled $s'(t_0)=v(t_0)=\dfrac{s(10)-s(0)}{10}$, and similarly there is a time $t_1 \in (0,10)$ where Harry's velocity equaled $s'(t_1)=v(t_1)=\dfrac{s(10)-s(0)}{10}$.

Question: Does the MVT also guarantee that $t_0=t_1$? That is, is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race? If so, why?

• I think the MVT doesn't say that the point $c$ is unique, hence you could find other times say $t_2$ ro $t_3$ where that holds. But I don't know if this helps – Euler_Salter Nov 6 '16 at 1:49
• Let $x(t),y(t)$ be $x(t)=t$, $y(t)=\sqrt{t}$. Clearly $x(0)=y(0)=0$ and $x(1)=y(1)=1$, but $x'(t)=1=\frac{x(1)-x(0)}{1}=1$ for all $t$, while $y'(t)=1$ only when $t=\sqrt{1/2}$, so they need not be equal – user160738 Nov 6 '16 at 1:53

## 3 Answers

You're being ambiguous. If you mean to ask if their velocities will be the same at some time in the interval $[0,10]$, yes. The result follows by applying the mean value theorem to the function $x(t) = s_{b}(t) - s_{h}(t)$ on the interval $[0, 10]$. $s_b$ and $s_h$ are the position functions of Barry and Harry respectively.

However, if you apply the mean value theorem to the functions independently, the "mean value" times may not be identical. Consider the following example, where we suppose the velocities are continuous. If Barry runs very fast right at the start, almost up to the finish line, over the course of $1$ second and then slows down like a caterpillar for the other $9$ seconds, the mean value time for Barry will be somewhere around $t = 1$ seconds while he is slowing down.

If Harry does the opposite, that is, he's extremely slow for the first $9$ seconds and then lightning fast to catch up to Barry at the final second, then his "mean value time" will be somewhere around $t=9$ as he speeds up.

They will share the same velocity at some time ($g(t) = v_{b}(t) - v_{h}(t)$ is continuous, initially positive and negative at the end), but this shared velocity will not be the "mean value" velocity $\frac{s(10) - s(0)}{10}$.

• Thank you for answering my question. I don't see how I was being ambiguous (my question was very clearly stated, was it not?), but thank you nonetheless. – Mr Toad Nov 6 '16 at 5:43
• Your question was ambiguous because you ask two different things in the last paragraph of the question, and in fact they happen to have different answers. As MathematicsStudent1122 wrote, the answer to "does the MVT also guarantee that $t_0 = t_1$?" is no, but the answer to "is it necessarily the case that Barry and Harry have the same velocity at the same exact time at some point during the race?" is yes. – David Z Nov 6 '16 at 7:40
• @David Z, thanks for the clarification. – Mr Toad Nov 6 '16 at 14:53

You can use Cauchy's mean value theorem. It says that if $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $c\in (a,b)$ such that $$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).$$ If we use $f$ to denote the position of Barry and $g$ to denote the position of Harry then we have $f(0)=g(0)=0$ and $f(T)=g(T)=L=$length. ($T$ denotes the total time.) So, there exists a point $c\in (0,T)$ such that $$Lg'(c)=(f(T)-f(0))g'(c)=(g(T)-g(0))f'(c)=Lf'(c).$$ That is, there exists $c$ such that $f'(c)=g'(c).$ So they must have had the same velocity at some point.

• +1 for the perfect answer. It does not appear at first that a single value of $c$ will work for $f$ and $g$ but Cauchy's MVT ensures that this is possible. – Paramanand Singh Nov 6 '16 at 7:09

Assume runner 1 runs at speed $2t$ feet per second, so that in ten seconds, runner 1 travels $\int_0^{10}2tdt=100$ feet. Assume runner $2$ runs at speed $\frac{3t^2}{1000}$ so that in $10$ seconds he also travels $100$ feet. However, $2t=\frac{3t^2}{1000}$ has roots $0$ and $\frac{2000}{3}>10$, so the runners never simultaneously travel the same speed during the race (besides the start, speed $0$). (There is nothing special about this example, I just started with a simple speed function $2t$ and tried to fudge another function to make it work. Im sure there are many more counterexamples)

• I think the example doesn't hold for runner 2. His initial speed isn't zero. – Cehhΐro Nov 6 '16 at 1:55
• @O.VonSeckendorff thanks for the catch. It is fixed. – TomGrubb Nov 6 '16 at 2:00


# Find the volume of the region defined by $|x|+|y|+|z|<4$

I want to find the volume of the region defined by the following inequality:

$\quad \quad |x|+|y|+|z|<4$

In addition to calculating the volume of this region, I would also like to be able to obtain its representation as a plot.

I have started by plotting the vertices, i.e., the points (4, 0, 0), (-4, 0, 0), (0, 4, 0), (0, -4, 0), (0, 0, 4), and (0, 0, -4), but I am not sure how to continue from there.

• Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. – bbgodfrey May 1 '15 at 3:23
• @julie You may want to consider accepting one of the answers that have been provided below, if they solve your problem. – MarcoB May 3 '15 at 17:45

volume = Integrate[1,  Element[{x, y, z}, ImplicitRegion[Abs[x] + Abs[y] + Abs[z] < 4, {x, y, z}]]]

(*256/3*)

ContourPlot3D[Abs[x] + Abs[y] + Abs[z] == 4, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}]

## A small addition from Mathematica v. 10

In his answer @Ivan has already nailed the most important part of the question, i.e. the formal representation of your region as an ImplicitRegion object:

region = ImplicitRegion[Abs[x] + Abs[y] + Abs[z] < 4, {x, y, z}]

I just wanted to add that, if you are on Mathematica 10 or newer, you can also use the RegionMeasure or Volume functions to calculate the volume of a region:

RegionMeasure[region]
Volume[region]

(* 256/3 *)

## How to arrive at the shape of this region

In response to @julie's question in the comment, here is how one can imagine constructing the 3D region.

Consider the corresponding equality in 2D, disregard the absolute values for now, and let's just concentration on the first quadrant ($x>=0$,$y>=0$). The resulting $x + y = 4$ is just the equation of a line, i.e. $y = 4 - x$. The region in the first quadrant in which the inequality $x+y<4$ holds is the region between that line and the axes:

RegionPlot[
x + y < 4 && x >= 0 && y >= 0,
{x, -5, 5}, {y, -5, 5},
Axes -> True, Frame -> False
]

Taking the absolute value of $x$ is geometrically equivalent to mirroring that triangle with respect to the $y$ axis as shown below. Since we are taking its absolute value, we will also remove the $x>=0$ restriction.

RegionPlot[
Abs[x] + y < 4 && y >= 0,
{x, -5, 5}, {y, -5, 5},
Axes -> True, Frame -> False
]

Taking the absolute value of $y$ is similarly equivalent to mirroring this figure with respect to the $x$ axis. The result is the square region below:

RegionPlot[
Abs[x] + Abs[y] < 4,
{x, -5, 5}, {y, -5, 5},
Axes -> True, Frame -> False
]

In the three-dimensional case, you can work your way through in a similar way: $x+y+z=4$ is the equation of a plane that defines a trangular wedge in the first quadrant, and so forth.

As for the numerical value, you could consider that your figure is made up of two square pyramids. The volume of a square pyramid is $V=a^2h/3$ where $a$ is the length of the base edge, and $h$ is the height. In your case, you can calculate the base edge as the hypotenuse of the smallest 2D triangle we introduced above ($a^2=4^2+4^2=32$), and the height of the pyramid is 4. The volume of this pyramid is 128/3, and your figure is made of two such pyramids, so $2\times128/3 = 256/3$.

• thank you! but how was the final answer calculated? how did you know what all the sides were and where those points went? – julie May 1 '15 at 3:28
• @julie I amended my answer to include a way to construct the region. Let me know if that answers your question appropriately. – MarcoB May 1 '15 at 4:27

An alternate approach to calculating the volume

Integrate[
Boole[Abs[x] + Abs[y] + Abs[z] < 4],
{x, -Infinity, Infinity},
{y, -Infinity, Infinity},
{z, -Infinity, Infinity}]

256/3

As a cone (with a square base), the volume is one-third the base times the height: 1/3 * (4 Sqrt[2])^2 * 8.

Another plot, based on the OP's coordinates:

ConvexHullMesh@Join[4 IdentityMatrix[3], -4 IdentityMatrix[3]]


So the x value is 0. You will get a point now. Back Function Institute Mathematics Contents Index Home. A parabola is a curve where any point is at an equal distance from: a fixed point (the focus), and ; a fixed straight line (the directrix) Get a piece of paper, draw a straight line on it, then make a big dot for the focus (not on the line!). Hence the equation of the parabola in vertex form may be written as $$y = a(x - 2)^2 + 3$$ We now use the y intercept at $$(0,- 1)$$ to find coefficient $$a$$. The implication is that throughout the observed range of the data, the expected probability of pt is an increasing function of expand_cap, though with some diminishing returns. When the function has been re-written in the form y = r(x + s)^2 + t, the minimum value is achieved when x = -s, and the value of y will be equal to t. x-intercepts in greater depth. A parabola can have either 2,1 or zero real x intercepts. Completing the square, we have \begin{align*} y &= x^2 - 2ax + 1 \\ &= (x - a)^2 + 1 - a^2, \end{align*} so the minimum occurs when $$x = a$$ and then $$y = 1 - a^2$$. What do you notice? You can take x= -1 and get the value for y. Example 2 Graph of parabola given vertex and a point Find the equation of the parabola whose graph is shown below. Graphs of quadratic functions have a vertical line of symmetry that goes through their turning point.This means that the turning point is located exactly half way between the x-axis intercepts (if there are any!).. The easiest way to find the equation of a parabola is by using your knowledge of a special point, called the vertex, which is located on the parabola itself. Only vertical parabolas can have minimum or maximum values, because horizontal parabolas have no limit on how high or how low they can go. Now play around with some measurements until you have another dot that is exactly the same distance from the focus and the straight line. Step 1: Find the roots of your … Substitute this x value into the equation y = x 2 – 6x + 8 to find the y value of the turning point. Finding the maximum of a parabola can tell you the maximum height of a ball thrown into the air, the maximum area of a … There are two methods to find the turning point, Through factorising and completing the square.. Make sure you are happy … In either case, the vertex is a turning point … Does the slope always have to be in turning points? If the function is smooth, then the turning point must be a stationary point, however not all stationary points are turning points, for example has a stationary point at x=0, but the derivative doesn't change sign as there is a point of inflexion at x=0. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In the first two examples there is no need for finding extra points as they have five points and have zeros of the parabola. Any point, ( x 0 , y 0 ) on the parabola satisfies the definition of parabola, so there are two distances to calculate: Distance between the point on the parabola to the focus Distance between the point on the parabola to the directrix To find the equation of the parabola, equate these two expressions and solve for y 0 . The x-intercepts are the points or the point at which the parabola intersects the x-axis. If the parabola only has 1 x-intercept (see middle of picture below), then the parabola is said to be tangent to the x-axis. Let’s work it through with the example y = x 2 + x + 6. A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). This is a second order polynomial, because of the x² term. now find your y value by using the x value you just found by plugging it into your function . Example 1 . Yes, the turning point can be (far) outside the range of the data. Use this formula to find the x value where the graph turns. How to find the turning point (vertex) of a quadratic curve, equation or graph. STEP 1 Solve the equation of the derived function (derivative) equal to zero ie. The x-coordinate of the turning point = - $$\frac{b}{2a}$$ ----- For example, if the equation of the parabola is . If the slope is , we max have a maximum turning point (shown above) or a mininum turning point . May 2008 218 59 Melbourne Australia Aug 24, 2009 #2 At the turning points of an equation the slope of y is zero. (Increasing because the quadratic coefficient is negative, so the turning point is a maximum and the function is increasing to the … A turning point may be either a local maximum or a minimum point. To find the turning point of a parabola, first find it's x-value, using the equation: -b/2a (from the quadratic form ax^2 + bx + c). Find the maximum number of turning points of each … A polynomial of degree n will have at most n – 1 turning points. In other words the differential of the equation must be zero. How do I find the coordinates of a turning point? This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. The equation is y=4xsquare-4x+4. When the equation of the parabola is in this form: y = ax 2 + bx + c . The x-coordinate of the turning point = - $$\frac{4}{2(3)}$$ = - $$\frac{2}{3}$$ Plug this in for x to find the value of the y-coordinate. Does slope always imply we have a turning point? K. Kiwi_Dave. or the slope just becomes for a moment though you have no turning point. To graph a parabola, visit the parabola grapher (choose the "Implicit" option). Given that the turning point of this parabola is (-2,-4) and 1 of the roots is (1,0), please find the equation of this parabola. … Surely you mean the point at which the parabola goes from increasing to decreasing, or reciprocally. In this case, b = 0, since there is no b term, and a is 1 (the number before the x squared) : -b/2a = -0/2. This means: To find turning points, look for roots of the derivation. To find the turning point of a quadratic equation we need to remember a couple of things: The parabola ( the curve) is symmetrical; If we know the x value we can work out the y value! You therefore differentiate f(x) and equate it to zero as shown below. The coordinate of the turning point is (-s, t). Solution to Example 2 The graph has a vertex at $$(2,3)$$. Example 7: Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function . If we look at the function . Free functions turning points calculator - find functions turning points step-by-step This website uses cookies to ensure you get the best experience. If, on the other hand, you suppose that "a" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y … If y=ax^2+bx+c is a cartesian equation of a random parabola of the real plane, we know that in its turning point, the derivative is null. So for example, given (2a): Vertex at (2, -6) One x intercept at 6 The axis will be x=2, so the given x intercept is 4 units to the right of the axis. Curve sketching Murray says: 19 Jun 2011 at 8:16 am [Comment permalink] Hi Kathryn and thanks for your input. … $0=a(x+2)^2-4$ but i do not know where to put … Such a point is called saddle point. How to find the turning point of a parabola: The turning point, or the vertex can be found easily by differentiation. Remember that the axis of symmetry is the straight line that passes through the turning point (vertex) of the parabola. How you think you find the turning point given the x-intercepts of a parabola? So, our starting or reference parabola formula looks like this: y = x 2. The turning point of a parabola is the vertex; this is also it's highest or lowest point. Reactions: … So the turning point is at $(a, 1 - a^2).$ So for your example: $$\displaystyle \frac {dy}{dx}=2x$$ So we set this equal to zero to get: $$\displaystyle 2x=0$$ or x=0 . The vertex is at point (x,y) First find x by using the formula -b/2a <--- a = 2, b= -5 and c= 1 (because it is quadratic) So -(-5)/2(2) = 5/4 <--- your x value at the vertex or turning point is 5/4. And our equation that includes a horizontal translation looks like this: y = (x - h) 2. If the coefficient of the x 2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “ U ”-shape. Write down the nature of the turning point and the equation of the axis of symmetry. And the lowest point on a positive quadratic is of course the vertex. When the parabola opens down, the vertex is the highest point on the graph — called the maximum, or max. A second approach is to find the turning point of the parabola. Turning Points and Intercepts of a Parabola Function. Published in: Education. The S.K.A. Horizontal translation for the parabola is changed by the value of a variable, h, that is subtracted from x before the squaring operation. It’s hard to see immediately how this curve will look just by looking at the function. how to i find the turning point of that parabola? So remember these key facts, the first thing we need to do is to work out the x value of the turning point. If the parabola is upright - as these examples are - then it will be laterally symmetrical about its axis, which is the vertical line through the vertex. is the set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. GeoGebra can be used very easily to find the equation of a parabola: given three points, A, B, C input the command FitPoly[{A, B, C}, 2]. Worked examples. In … Did You Know That...? 17 Comments 2 Likes ... – 12 12 – 24 – 12 = -24 this is the y-coordinate of the vertex So the vertex (turning point of this parabola is (-2,-24) HOW TO CALCULATE THE VERTEX (TURNING POINT) Recommended Mẫu ốp lưng iphone se da thật chuyên nghiệp … In math terms, a parabola the shape you get when you slice through a solid cone at an angle that's parallel to one of its sides, which is why it's known as one of the "conic sections." Here is a typical quadratic equation that describes a parabola. The turning point is when the rate of change is zero. To find the axis of symmetry, use this formula: x = -b/2a. solve dy/dx = 0 This will find the x-coordinate of the turning point; STEP 2 To find the y-coordinate substitute the x-coordinate into the equation of the graph ie. … Turning Points of Quadratic Graphs. I started off by substituting the given numbers into the turning point form. The graph below has a turning point (3, -2). y = 3x 2 + 4x + 1 . A parabola The set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. Parabola, Horizontal Translation. TURNING POINT The formula to find the x value of the turning point of the parabola is x = –b/2a. This can help us sketch complicated functions by find turning points, points of inflection or local min or maxes. In example 3 we need to find extra points. There is also a spreadsheet, which can be used as easily as Excel. substitute x into “y = …” This is a mathematical educational video on how to find extra points for a parabola. A turning point can be found by re-writting the equation into completed square form. No. The Parabola. In the case of a vertical parabola (opening up or down), the axis is the same as the x coordinate of the vertex, which is the x-value of the point where the axis of symmetry crosses the parabola. Solved: What is the turning point, or vertex, of the parabola whose equation is y = 3 x^2 + 6 x - 1? The Vertex of a Parabola The vertex of a parabola is the point where the parabola crosses its axis of symmetry. A parabola can have 2 x-intercepts, 1 x-intercept or zero real x intercepts. Second order how to find turning point of parabola, because of the turning point ( vertex ) of the turning point is (! Below has a vertex at \ ( ( 2,3 ) \ ) extra... To graph a parabola uses cookies to ensure you get the best.! 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Question

# Let $$a_{n}$$ denote the number of all n-digit positive integers formed by the digits $$0,1$$ or both such that no consecutive digits in them are $$0$$. Let $$b_{n}=$$ the number of such $$n$$-digit integers ending with digit $$1$$ and $$c_{n}=$$ the number of such $$n$$-digit integers ending with digit $$0$$. Which of the following is correct?

A
a17=a16+a15
B
c17c16+c15
C
b17b16+c16
D
a17=c17+b16

Solution

## The correct option is A $$a_{17}=a_{16}+a_{15}$$In such a number either last digit is $$'0'$$ or $$'1'$$When you consider $$'a_1'$$, only one number is possible i.e. $$1$$When you consider $$'a_2'$$, 2 such numbers are possible i.e. $$10$$, $$11$$When you consider $$'a_3'$$, 3 such numbers are possible i.e. $$101$$, $$111$$, $$110$$When you consider $$'a_4'$$, 5 such numbers are possible i.e. $$1010$$, $$1011$$, $$1110$$, $$1101$$, $$1111$$By observing this we get a relationship which is $$a_n$$ $$=$$ $$a_{n-1}$$ $$+$$ $$a_{n-2}$$So, $$a_{17}$$ $$=$$ $$a_{16}$$ $$+$$ $$a_{15}$$(Alternate Method)Using Recursion formula$$a_n=a_{n-1}+a_{n-2}$$Similarly, $$b_n=b_{n-1}+b_{n-2}$$ and $$c_n=c_{n-1}+c_{n-2}$$    $$\forall\ n\geq 3$$and $$a_n=b_n+c_n$$  $$\forall\ n\geq 1$$So, $$a_1=1, a_2=2, a_3=3, a_4=5, a_5=8$$ .......$$b_1=1, b_2=1, b_3=2, b_4=3, b_5=5, b_6=8$$ .......$$c_1=0, c_2=1, c_3=1, c_4=2, c_5=3, c_6=5$$ .......Using this we get $$b_{n-1}=c_n$$$$\therefore a_{17}=a_{16}+a_{15}$$Mathematics

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# Prove that the set $C = \{(x,y) \in \mathbb{R}^2:\max \{ |x|,|y|\}\leq 1 \}$ is convex.

Prove that $C = \{(x,y) \in \mathbb{R}^2:\max \{ |x|,|y|\}\leq 1 \}$ is a convex set.

I am using the following definition for a convex set

Let $D \subseteq \mathbb{R}^n$. The set $D$ is said to be convex if, given $\bar{x},\bar{y}\in D$, we have that $$(1-t)\bar{x} \ +t\bar{y} \in D$$ for all $t\in [0,1].$

Let $\bar{x}=(x_1,y_1)$ and $\bar{y}=(x_2,y_2)$ be arbitrary points such that $\bar{x},\bar{y} \in C$, and let $t\in [0,1].$ Now, we want to see that $(1-t)\bar{x} \ +t\bar{y} \in C.$ So then \begin{align} (1-t)\bar{x} \ +t\bar{y} &=(1-t)(x_1,y_1)+t(x_2,y_2)\\ &= ((1-t)x_1,(1-t)y_1)+ (tx_2,ty_2)\\ &= ((1-t)x_1+tx_2,(1-t)y_1+ty_2) \end{align}

And here is where I get stuck. I want to prove that $((1-t)x_1+tx_2,(1-t)y_1+ty_2)\in C$, which I believe is the same as proving that $\max\{|(1-t)x_1+tx_2|,|(1-t)y_1+ty_2| \}\leq 1$, but I don't know how to proceed.

Any help would be appreciated!

• Use triangle inequality. – Youem Apr 13 '18 at 3:57

So $|(1-t)x_{1}+tx_{2}|\leq(1-t)|x_{1}|+t|x_{2}|\leq(1-t)+t=1$ and $|(1-t)y_{1}+ty_{2}|\leq(1-t)|y_{1}|+t|y_{2}|\leq(1-t)+t=1$ because $|x_{1}|,|x_{2},|y_{1}|,|y_{2}|\leq 1$, this shows that $(1-t)\overline{x}+t\overline{y}\in C$.
$|(1-t)x_1 + tx_2| \le (1-t)|x_1| + t |x_2| \le (1-t) + t = 1$ the same for $y$
Let $C_x = \{(x,y) \in \mathbb{R}^2:|x|\leq 1 \}$ and $C_y = \{(x,y) \in \mathbb{R}^2:|y|\leq 1 \}$. Then,
• $C_x$ and $C_y$ are obviously convex
• $C = C_x \cap C_y$ is convex, too


# 3-colored triangulations of the sphere $S^2$, and Sperner's Lemma

I noticed something about colored triangulations of the topological sphere $$S^2$$ and have a question about this.

Observation. If you triangulate the sphere $$S^2$$ and color the vertices with three colors: then the number of 3-colored triangles is always even (or zero). In particular, there is no coloring with exactly one 3-colored triangle.

For a proof, view $$S^2$$ as two triangulated disks with matching coloring of the boundaries that are glued together. As their boundaries have the same number of color changes, we know from Sperner’s Lemma that their triangulations have the same number (mod 2) of 3-colored triangles. So the total number of 3-colored triangles is even or zero.

As an interesting corollary, we get the characterization: A triangulated sphere has zero 3-colored triangles iff all cycles of the triangulation have an even number of color changes.

I looked at the torus, the Klein bottle, and the projective plane, and I find that the observation is also true for them.

Edit: Just for contrast, adding an example below of a "soap bubble" surface, where the two soap bubbles share a common disk. This surface allows for triangulations with even and odd numbers of 3-colored triangles (but like the other surfaces I looked at, cannot have just one).

Question. I wonder whether this also follows from more general theorems about triangulations of surfaces, or about maximal planar graphs? I have consulted algebraic topology and graph theory texts, but could not find any results in that direction. Would you have a suggestion where else to look, or maybe a reference for that?

• This looks related to Sperner's lemma: en.wikipedia.org/wiki/Sperner%27s_lemma Jun 24, 2020 at 22:51
• @JanKyncl you are definitely right. It is like Sperner’s Lemma “without boundary”. Or the way I looked at it, 2 x Sperner’s Lemma and then glued together. Just in case you are interested, see this mathoverflow.net/q/362025/156936 Jun 25, 2020 at 3:47
• "Uneven" numbers are commonly called odd. Jun 28, 2020 at 19:57
• @VictorProtsak thanks a lot for your comment! I corrected the wording in the text now. Jun 29, 2020 at 11:32

A counting proof shows that this observation is unrelated to the global topology.

Every edge is monochromatic or dichromatic. How many dichromatic edges are there? If each triangle tells you its number of dichromatic edges (either 0, 2, or 3), you can add these up and divide by two to get the total number of dichromatic edges (since every edge contributes to two triangles). So the number of trichromatic triangles must be even.

This proof works for $$k$$-dimensional manifolds when $$k$$ is even, since the number of $$k$$-colored $$(k-1)$$-simplices bounding any $$k$$-simplex must be 0, 2, or $$k+1$$.

Your corollary similarly transfers to higher even dimensions, at least for orientable manifolds, replacing "cycles of edges" with "hypersurfaces of $$(k-1)$$-simplices", and "even number of color changes" with "even number of $$k$$-colored $$(k-1)$$-simplices".

• I'm not sure which direction you want to go in with this, but I think it falls pretty squarely in combinatorics and graph theory. What you (not incorrectly) call "triangulations of $S^2$" are often referred to as maximal planar graphs. My proof is very similar to the handshaking lemma. People have also considered coloring and planarity on tori and other manifolds.
– Matt
Jun 24, 2020 at 19:32
• thanks a lot for this helpful comment and great links. This is excellent Jun 24, 2020 at 20:20

Just to close the loop on this: the double-counting argument in the answer of user Matt allows for a nice visual proof of the (2-dim.) Lemma of Sperner. Just want to capture it here, as it connects nicely with the triangulation of the sphere / the maximal planar graph in my OP question.

Start with a triangulated polygon in the plane, and label each vertex with one of 3 colors. The example just shows the boundary of such a triangulated, 3-colored polygon. Claim (Sperner’s Lemma): If the boundary has an odd number of color changes, then a 3-colored triangle exists in the polygon triangulation. In fact, more generally, an odd number of such 3-colored triangles exists.

Proof: Go to 3-dimensional space, and build a “tent” over the polygon like in the diagram: add a colored vertex, and add the edges between this additional vertex and the boundary vertices of the polygon. This way, we have effectively created a triangulation of the topological sphere $$S^2$$.

If the boundary of the polygon has an odd number of color changes, this gives an odd number of 3-colored triangles in the “tent” over the polygon. But from the double-counting argument in user Matt’s answer, we know an even number of 3-colored Sperner triangles must exist. Hence the polygon at the bottom must have an odd number of 3-colored triangles (at least one) in its triangulation, which completes the proof.


# Dimension of the sum of two vector subspaces

$\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2).$

I want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3).

Can we generalize this notion to $\mathbb{F^{n}}$?

Suppose we have an additional case where $U_1$ and $U_2$ are planes in $\mathbb{R^3}$, but $U_1 \subseteq U_2$. In this instance, $dim(U_1 + U_2) < 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.

Is this proper intuition?

• The intuition is correct. Sep 21, 2013 at 17:17

In the latter case, they are actually the same plane, so their sum is again the same plane (as they are closed under addition).

Here is another (analogous) way to think about it. Let's start with a basis $B_0$ for $U_1\cap U_2.$ We can extend $B_0$ to a basis $B_1$ for $U_1$ and a basis $B_2$ for $U_2$. Then $B_1\cup B_2$ is a basis for $U_1+U_2,$ and $B_1\cap B_2=B_0,$ so \begin{align}\dim(U_1+U_2) &= |B_1\cup B_2|\\ &= |B_1|+|B_2|-|B_1\cap B_2|\\ &= |B_1|+|B_2|-|B_0|\\ &= \dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2).\end{align} This generalizes nicely to $\Bbb F^n$, and allows us to avoid geometric arguments that may be less sensible for an arbitrary field $\Bbb F$.

Also, it will never be the case that the intersection of two planes in space is precisely $\{0\}.$ If there were two such planes $U_1$ and $U_2,$ then we would have $$\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2)=2+2-0=4>3=\dim(\Bbb R^3),$$ which is not possible, since $U_1+U_2$ is a subspace of $\Bbb R^3$.

• Thanks for the answer. Could it be the case that a pair of two dimensional planes in $\mathbb{R^4}$ intersect at a point? Sep 21, 2013 at 17:27
• Yes, indeed. Consider $$U_1=\{(w,x,y,z)\in\Bbb R^4:y=z=0\}$$ and $$U_2=\{(w,x,y,z)\in\Bbb R^4:w=x=0\}.$$ The upshot in $\Bbb R^3$ is that there "isn't enough room" for two planar subspaces to avoid each other that well. Sep 21, 2013 at 17:38
• Why can we conclude that $B_{1} \cap B_{2} = B_{0}$? It seems perfectly reasonable that the extensions share vectors outside of $B_{0}$. Sep 14, 2016 at 0:40
• @Jonathan: If there is some $\vec v\in B_1\cap B_2,$ then $\vec v\in U_1\cap U_2.$ What then can we conclude by definition of $B_0$? Sep 14, 2016 at 0:46
• Ok, then $v$ must be a linear combination of $B_{0}$. But, this forces both $B_{1}$ and $B_{2}$ to be linearly dependent, right? Sep 14, 2016 at 0:53

Cameron Buie's answer does answer the original question sufficiently. However, I'm adding the formal proof of the theorem in context (taken from "Linear Algebra Done Right" by Sheldon Axler).

If $$U_1$$ and $$U_2$$ are subspaces of a finite dimensional vector space then: $$\dim(U_1+U_2)=\dim U_1 + \dim U_2 - \dim(U_1 \cap U_2)$$

Let $$u_1,...,u_m$$ be a basis of $$U_1\cap U_2$$; thus $$\dim (U_1\cap U_2)=m$$. Because $$u_1,...,u_m$$ is a basis in $$U_1\cap U_2$$, it is linearly independent in $$U_1$$. Hence this list can be extended to a basis $$u_1,...,u_m,v_1,...,v_j$$ of $$U_1$$. Thus, $$\dim U_1 = m+j$$. Also extend $$u_1,..,u_m$$ to a basis $$u_1,...,u_m, w_1,...,w_k$$ of $$U_2$$. $$\dim U_2 = m +k$$.

We will show that $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is a basis of $$U_1+U_2$$. This will complete the proof because then we will have $$\dim(U_1+U_2)=m+j+k=\dim U_1 + \dim U_2 - \dim (U_1\cap U_2)$$

We just need to show that the list $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is linearly independent. To prove this, suppose:

$$a_1u_1+...+au_m+b_1v_1+...+b_jv_j+c_1w_1+...+c_kw_k=0$$

where all $$a,b,c$$'s are scalars. We need to show that all the $$a,b$$ and $$c$$'s are $$0$$.

The equation can be rewritten as

$$c_1w_1+...+c_kw_k=-a_1u_1 - ... -a_mu_m -b_1v_1 - ... - b_j v_j$$

Which shows that $$c_1w_1+...+c_kw_k\in U_1$$. But actually all $$w$$'s are in $$U_2$$. So the LHS must be an element of $$U_1\cap U_2$$.

$$c_1w_1+...+c_kw_k=d_1u_1+...+d_mu_m$$ for some choice of scalars $$d_1,d_2,...,d_m$$. But $$u_1,...,u_m,w_1,...,w_k$$ is linearly independent. So our last equation implies that all the $$c$$'s equal $$0$$.

Thus our original equation involving $$a,b,c$$ becomes

$$a_1u_1+...+a_mu_m+b_1v_1+...+b_jv_j=0$$

But we already knew that the list $$u_1,...,u_m,v_1,...,v_j$$ is linearly independent. This equation implies that all the $$a$$'s and $$b$$'s are $$0$$. We now know that all $$a,b$$ and $$c$$'s are $$0$$, hence proving our original claim.


# Taylor Approximation

For $f(x)=e^x$, find a Taylor approximation that is in error by at most $10^-7$ on [-1,1]. Using this approximation, write a function program to evaluate $e^x$. Compare it to the standard value of $e^x$ obtained from the MATLAB function exp(x); calculate the difference between your approximation and exp(x) at 21 evenly spaced points in [-1,1].

So I am stuck on this problem. I know for the first part, I needed to find n smallest such that $e/(n+1)!$ $\leq$ $10^-7$, so I found n=10 to satisfy that part.

So do I need to find a polynomial of degree 10 using a nested loop that evaluates $e^x$? What is my next movie in this problem?

• Yes, a loop to evaluate $1+x+x^2/2+...$ up to 10. – lemon Oct 7 '14 at 18:49
• Thank you! and then I just compare my loop with the MATLAB function given and see where they are different? – user141745 Oct 7 '14 at 18:52
• Very relevant wiki example. – BeaumontTaz Oct 7 '14 at 18:59
• @user141745 Yes - the purpose of comparing is to make sure that your routine is accurate to within $10^{-7}$. – lemon Oct 7 '14 at 19:02

I posted the link in the comments, but I'll flush out the example on the wiki here with some added commentary.

We have $f(x)=e^x$ and we know by Taylor'r theorem that

$$f(x)=P_k(x)+R_k(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^k}{k!}+R_k(x)$$

where $R_k(x)$ is our error term and $P_k(x)$ is the Taylor polynomial. We want to find the $k$ such that $R_k(x)<10^{-7}$ for $x\in[-1,1]$.

We know that the remainder term $R_k(x)$ can be expressed as

$$R_k(x)=\frac{e^\xi}{\left(k+1\right)!}x^{k+1}$$

for some $\xi\in[-1,1]$. We need to find the value of $\xi$ though. And since we don't know $e^x$ (otherwise we'd just use that function) we again approximate it using a lower order Taylor expansion

$$e^x=P_1(x)+R_1(x)=1+x+\frac{e^\xi}{2}x^2$$

Since $e^x$ is an increasing function, we know that $e^\xi<e^x$ when $-1<\xi<x$. So we can say

$$e^x=1+x+\frac{e^\xi}{2}x^2<1+x+\frac{e^x}{2}x^2$$

Which we can solve for $e^x$ and get

$$e^x<2\frac{x+1}{2-x^2}$$

Some basic calculus gives us that the maximum of this expression for $x\in[-1,1]$ is at $x=1$ and $e^x<4$.

Therefore

$$R_k(x)=\frac{e^\xi}{\left(k+1\right)!}x^{k+1}<\frac{4x^{k+1}}{\left(k+1\right)!}$$

And since $x^{k+1}$ is a maximum on $[-1,1]$ at $x=1$ and we want $R_k(x)$ to be less than $10^{-7}$ we have

$$R_k(x)<\frac{4}{\left(k+1\right)!}<10^{-7}$$

So basically, we just need to find $k$ such that $(k+1)!>4\cdot10^7$. And $11!=39916800=3.99\cdot10^7$ which is reallllly close. But, really we need $(k+1)!=12!$ so $k=11$.

Now... this is a very modest upper bound. If we really think about it (and this is what you did), the upper bound on the interval $[-1,1]$ is $e$. So if we did this with $e^x<e$ on $x\in[-1,1]$ instead, then we'd end up with

$$\frac{e}{\left(k+1\right)!}<10^{-7}\implies (k+1)!>2.718\cdot10^7\implies k=10$$

Same as you got. However, this relies on the fact that we know $e$ already.

If we don't know $e$ we can use a higher order approximation than $2$. So we find that

$$e^x<\frac{6+6x+3x^2}{6-x^3}$$

Which means it's either $e^x<\frac{15}{5}=3$ if it's at $x=1$ or it's at the maximum of the interval which involves find the roots of a fourth order polynomial $3x^4+12x^3+18x^2+36x+36$ whose critical points turn out not to be on our bound. So we can use this higher order approximation the same way we did before and get that

$$\frac{3}{\left(k+1\right)!}<10^{-7}\implies (k+1)!>3\cdot10^7\implies k=10$$

From there, you can either hard code the approximation

$$e^x\approx \text{myFunc}(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^9}{9!}+\frac{x^{10}}{10!}$$

or do a sum like this in a for loop

$$e^x\approx \text{myFunc}(x)=\sum_{i=0}^{10}\frac{x^i}{i!}$$

It appears as though Arkamis gave you some MATLAB code that'll do this for you, though.

Just for fun, here's a matlab script that computes the answer in one line -- in MATLAB we almost never need to loop!

The function takes three parameters: n, the maximum order of the polynomial to try; m, the number of data points in the interval we wish to evaluate; and tol, the tolerance we wish to achieve.

function Order = TaylorExp(n,m,tol)
Order = min(find(max(abs(tril(kron(ones(n,1),1./factorial(0:(n-1))))*power(kron(ones(n,1),linspace(-1,1,m)),kron(ones(1,m),(0:(n-1))')) - kron(ones(n,1),exp(linspace(-1,1,m)))),[],2) < tol)) - 1;


# Line equation of a tangent line of $f(x) = x\cos(3x)$

I'm new here so maybe I'll need some help with formatting with MathJax, as well.

So question asks for tangent line of $f(x) = x\cos(3x), x= \pi$

So:

$$f(x) = x.\cos(3x)$$ $$f'(x) = -x.\sin(3x)+\cos(3x).3$$

$$f(\pi) = -\pi$$ $$f'(\pi) = -3$$

$$y = -3x - 2\pi$$

The answer of the book: $$y = -x$$

Considering: $$y = ax + b$$ $$y = f(x)$$ $$a = f'(x)$$ What am I missing?

• you miscalculated the derivative. – mesel May 17 '14 at 17:51

You made a couple of mistakes in using the product rule.

$$f(x) = x\cos(3x) \implies f'(x) = x \cdot \frac{d}{dx}(\cos(3x)) + \dfrac{d}{dx}(x)\cdot \cos(3x)$$

$$= x\cdot (-\sin (3x))\cdot 3 + \cos 3x$$

$$= -3x\sin(3x)+ \cos (3x)$$

Now substitute $x = \pi$ into $f'(x)$ to obtain slope at that point: $-3\pi\cdot \sin(3\pi) + \cos (3\pi) = -1$

So the slope of the desired line needs to be $-1$.

Now, given $x_0 = \pi$, $y_0 = f(x_0) = f(\pi) = -\pi$.

So we have the point on the tangent line (the point of tangency): $(\pi, -\pi)$. That gives you the line \begin{align} y - y_0 = -1(x - x_0) &\iff y - (-\pi) = -1(x - \pi)\\ \\ & \iff y + \pi = -x + \pi \\ \\ & \iff y = -x\end{align}

• Your answer showed me something that I was repeating question after question and haven't given attention. Derivative was wrong and how to find line equation was misunderstood. Thank you a lot. – Fabiano Araujo May 17 '14 at 18:15
• You're welcome, Fabiano! – amWhy May 17 '14 at 18:15

HINT:

$$f'(x)=\frac{d(x\cos3x)}{dx}=\frac{dx}{dx}\cdot\cos3x+x\cdot\frac{d(\cos3x)}{dx}=\cos3x-3x\sin3x$$


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# Two important formulas for mixture problems

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Two important formulas for mixture problems  [#permalink]

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Updated on: 19 Aug 2009, 06:52
6
9
I came across two methods which I found very handy in solving some types of mixture problems. So I thought of sharing it with the gc community.
here they are

Rule of Alligation
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.

Mean Price: The cost price of a unit quantity of the mixture is called mean price.

Rule of Alligation

If two quantities are mixed, then

$$\frac{Quantity of cheaper}{Quantity of dearer}=\frac{(C.P.) of dearer - (Mean Price)}{(Mean Price) - (C.P.)}$$

Taking a simple example

In what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?

C.P of cheaper=9.30
C.P of dearer=10.80
C.P of mean or mean price = 10.0

so putting the values in the formula
$$\frac{Qc}{Qd}=\frac{Cd-Cm}{Cm-Cc}$$

=$$\frac{10.80-10}{10-9.3}$$

=8:7 Ans

Originally posted by joebloggs on 19 Aug 2009, 06:10.
Last edited by joebloggs on 19 Aug 2009, 06:52, edited 2 times in total.
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19 Aug 2009, 06:25
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11
Suppose a container contains x units of liquid from which y units are taken out and replaced with water.

After n operations, the quantity of pure liquid = $$[x(1-\frac{y}{x})^n]$$

Example: A container contains 40 liters of milk. From this container 4 litres of milk was taken and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Amount of milk left after 3 operations = $$[40(1-\frac{4}{40})^3]liters = (40 *\frac{9}{10}^3)$$ = 29.16 liters
##### General Discussion
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Re: Two important formulas for mixture problems  [#permalink]

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19 Aug 2009, 06:37
3
you also might wanna see a short method for mixture problems used by Economist
check out this problem

http://gmatclub.com/forum/how-much-of-the-mixture-is-to-be-removed-and-replaced-82423.html

Hope this post helps

Soon I'll be posting some mixture problems which I couldn't solve in one go ...

see ya ....
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Re: Two important formulas for mixture problems  [#permalink]

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19 Aug 2009, 12:40
nice post .. thanks .. +1
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Re: Two important formulas for mixture problems  [#permalink]

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19 Aug 2009, 12:53
Keep up the good work !
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14 Aug 2019, 02:59
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Two important formulas for mixture problems  [#permalink]

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28 Aug 2019, 00:40
Quote:
In what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?

C.P of cheaper=9.30
C.P of dearer=10.80
C.P of mean or mean price = 10.0

For those who love a more visual approach, consider the following, which may be even simpler.

Just subtract cheaper and dearer from the mean and find the difference.

---9.3----10.8--
-----\----/------
-------10-------
------/---\------
---0.8---0.7---
Two important formulas for mixture problems   [#permalink] 28 Aug 2019, 00:40
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## anonymous one year ago Find circle which is tangent to x-axis and path through points (1,-2) and (3,-4). I worked it through analytic geometry and it gives two possible circles!!! and I can not figure how 3 conditions for a circle gives 2 answers or how to draw it geometrically?

1. anonymous

@ganeshie8

2. welshfella

general equation is (x - a)^2 + ( y - b)^2 = r^2 you can create 2 equations in a , b and r by substituting the 2 given points

3. anonymous

Circle is clearly defined by 3 conditions. How can I draw 2 circles which path through two points and tangent to a line!!

4. welshfella

The circle touches the x -axis so that another point on (x , 0)

5. welshfella

- good question - I'm trying to figure that

6. anonymous

Can you give a picture? that is puzzling me

7. welshfella

|dw:1432910060214:dw|

8. welshfella

lo1 - not a great diagram

9. anonymous

I'm sorry I meant two possible cirlces

10. welshfella

|dw:1432910234711:dw|

11. welshfella

thats worse than the first

12. anonymous

:) Yeah that is amazing.

13. ganeshie8

two circles can have the same common tangent, yes ?

14. welshfella

yes

15. ganeshie8

may be forget about x-axis and look at below diagram |dw:1432910495765:dw|

16. ganeshie8

both circles meet below 3 conditions : 1, 2) pass through two points (intersection) 3) having that blue line as tangent

17. welshfella

ah - yes the common tangenytis the x-axis

18. anonymous

I got it now thanks

19. ganeshie8

your question about "why 3 conditions are not giving me an unique circle" is very interesting, im still thinking of a better explanation

20. anonymous

I don't know. In such cases I just change that thinking. I think the tangent line condition isn't tough enough ((unless the tangent point is given))

21. anonymous

anyway that for your helping :)

22. ganeshie8

hmm the tangent point is not so random, we only have two choices so its still a bit mysterious

23. ganeshie8

|dw:1432911213933:dw|

24. anonymous

But the result is the same tangent line condition isn't unique as you have said.

25. ganeshie8

geometrically how do you know there doesn't exist a third circle that meets the given conditions ?

26. ganeshie8

|dw:1432911454842:dw|

27. ganeshie8

how do i convince more than two circles are never possible given 1) two points 2) tangent line

28. anonymous

the center isn't on locus of chord made by these 2 points

29. anonymous

I mean when you bisect it and make a perpendicular which is the locus of the center

30. ganeshie8

perpendicular bisector of the chord passes through the center but again how do you know the 3 centers are not collinear ?

31. ganeshie8

|dw:1432911763458:dw|

32. welshfella

the line joining the points of contact to the centers of the circles are perpendicular to the tangent - can that be used to answer the question?

33. ganeshie8

BINGO!!!

34. ganeshie8

|dw:1432912228871:dw|

35. ganeshie8

hmm idk

36. welshfella

yes - its a puzzle

37. welshfella

|dw:1432912469311:dw|

38. welshfella

why can't there be an other larger circle like the above?

39. anonymous

for any triangle it has only one unique circle paths through its vertices |dw:1432912656462:dw| so if we moved the point along the locus the lengths won't be equal and if we shifted it up or down, the the distance to 2pionts won't be equal.

40. ganeshie8

brilliant!

41. welshfella

i'm afraid i gotta go and leave this interesting discussion ..

42. myininaya

I know you wanted geometric stuff or whatever but I think I have found two answers with a combination of algebra and calculus.

43. myininaya

It is pretty long that way. :p

44. myininaya

$\text{ our circle has points: } (1,-2);(3,-4);(a,0) \\ \text{ we have horizontal tangent at } (a,0) \\ (a-h)^2+(0-k)^2=r^2 \\ (1-h)^2+(k+2)^2=r^2 \\ (3-h)^2+(k+4)^2=r^2 \\ y'=\frac{h-x}{y-k} \\ y'|_{x=a}=0=\frac{h-a}{0-k} \implies h=a \\ (a-a)^2+k^2=r^2 \text{ so } k=-r \text{ or } k=r \\ \text{ so going with the } k=r \text{ thing we have } \\ (1-h)^2+(r+2)^2=r^2 \\ (3-h)^2+(r+4)^2=r^2 \\ (1-h)^2+4r+4=0 \\ (3-h)^2+8r+16=0 \\ 2(1-h)^2+8=(3-h)^2+16 \\ h^2+2h-15=0 \\ (h+5)(h-3)=0 \\ h=-5 \text{ or } h=3$ so pluggin some things back in we can find r then k $h=-5 \\ a=-5 \\ (1+5)^2+4r+4=0 \\ r=10 \\ k=10,-10$ $h=3 \\ a=3 \\ (1-3)^2+4r+4=0 \\ r=2 \\ k=2 ,-2$ some of these solutions need to be checked you will see only 2 of the 4 will work and like i said I know this isn't the approach you wanted but I thought it would be nice to try another approach

45. welshfella

looks good to me


# Diagonal Crossing

#### mathmaniac

##### Active member
Challenge:
How many boxes are crossed by a diagonal of an 800 X 200 rectangle?

Solution with proof required...

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Challenge:
How many boxes are crossed by a diagonal of an 800 X 200 rectangle?

Solution with proof required...

Hello Mathmaniac
I don't understand the question. What us meant by 'boxes are crossed'.

Also, if this is a challenge problem then shouldn't this go in the 'Challenge and Puzzles' forum?

#### MarkFL

Staff member
Yes, I sent the OP a VM asking if this is a challenge or if it is a problem for which he needs help shortly after it was posted, but have not gotten a response yet. Once the matter is settled, I will move it if need be, then remove this post so that the topic is not cluttered.

#### mathmaniac

##### Active member
I don't understand the question. What us meant by 'boxes are crossed'.
Here is an example:

The boxes inside the rectangle are meant to be squares....

#### MarkFL

Staff member
Its a challenge,Mark...
I have thus moved the topic to the Challenge Questions and Puzzles sub-forum. I know you used the word "Challenge" but wanted to make sure it fit the criteria, i.e., you have the correct solution ready to post in the event no one solves it.

Last edited:

Yes!!!

#### Bacterius

##### Well-known member
MHB Math Helper
[JUSTIFY]The rectangle is made of $800 \times 200$ "boxes". The diagonal thus has gradient $\pm \frac{200}{800} = \pm \frac{1}{4}$, where a "box" has unit dimensions. Note the diagonal starts at the top left corner of the top-left-most box. This is important. So, after 4 units of width travelled, the diagonal will intersect the top left corner of another box:[/JUSTIFY]

[JUSTIFY]And this section of the diagonal intersects four boxes. Since the rectangle is 800 boxes wide, this section of the diagonal will repeat $\frac{800}{4} = 200$ times, and so the diagonal intersects $4 \times 200 = 800$ boxes.[/JUSTIFY]

[HR][/HR]

[JUSTIFY]A more interesting problem is to consider a rectangle of dimensions $p \times q$ where $p$ and $q$ are distinct primes. Then the diagonal never intersects the top-left corner of a box within the rectangle, and a different approach is called for:[/JUSTIFY]

Last edited:

#### mathmaniac

##### Active member
Thats right,Bacterius...
And can you show your approach for distinct primes?

#### Bacterius

##### Well-known member
MHB Math Helper
Thats right,Bacterius...
And can you show your approach for distinct primes?
[JUSTIFY]It's simple enough, in fact. Because we know that the diagonal will, in this case, never intersect a corner, we can count the number of boxes crossed by simply counting the number of times the diagonal intersects both the vertical sides and the horizontal sides of the boxes. The diagonal will intersect $p - 1$ horizontal sides, and $q - 1$ vertical sides, simply by virtue of being a line crossing the rectangle from top-left to bottom-right (or top-right to bottom-left).

But not so fast - the diagonal always starts inside the rectangle, and so automatically intersects the top-left box (or whichever corner of the rectangle you start your diagonal from), which we haven't yet considered, giving a total of $(p - 1) + (q - 1) + 1 = p + q - 1$ boxes intersected. Checking with the diagram above gives $11 = 7 + 5 - 1$ boxes intersected, as expected.

Not very rigorous, just a proof sketch showing the general approach.[/JUSTIFY]

Last edited:

#### mathmaniac

##### Active member
Here we have a winner!!!


# Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$

How can I find the domain of:

$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$

I think the hard part will be to find:

$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$

So far I have: not sure how to preceed: $$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$ For $\sqrt{g(x)}$ to be valid, $g(x) \ge 0$

For $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \ne 0$

Thus, $x \ne \sqrt 2, 2, \sqrt 3$

$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \ge 0$$

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A friendly note: You have posting pictures in which you presumably had written the Math. I'd suggest you learn $\TeX$. Use the community to learn the commands. You can see the basic codes by clicking on them as seeing Math as $\TeX$ commands here. It will pay you in the long run and you'll become effective without having to write those crappy pics. Will you? (I like the way write your questions showing us what you have done! +1 for that!) –  user21436 Feb 23 '12 at 12:09
@KannappanSampath, I know TeX but it takes longer to type them :) perhaps I need more practice. I also have put the more important parts of the question in TeX, the not as important stuff, I thought providing an image will suffice –  Jiew Meng Feb 23 '12 at 12:12
I see you already know $\TeX$ from the first part of your question. Why don't you do the same thing for the Math in the pic as well? –  user21436 Feb 23 '12 at 12:13
Is the numerator $(x^2-1)(x^2-3)(x^2-5)$ or $(x^2-1)(x^3-1)(x^5-1)$? –  01000100 Feb 23 '12 at 12:13
Its the 1st, $(x^2-1)(x^2-3)(x^2-5)$ –  Jiew Meng Feb 23 '12 at 12:14

The first thing you should note is that the expression $$\Phi=\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}$$ is undefined at any zero of the denominator. So, the points $x=\pm \sqrt 2$, $\pm 2$, and $\pm\sqrt 6$ are not in the domain of $\sqrt\Phi$.

Now, to find the domain of $\sqrt\Phi$, we need to find when $\Phi$ is nonnegative.

Towards solving $\Phi\ge0$, the fundamental observation to make is that the only points "across which" $\Phi$ can change signs are at the zeros of its numerator or at the zeroes of its denominator.

The zeroes of the numerator are: $$\pm 1, \pm \sqrt 3, \pm \sqrt 5$$ and the zeroes of the denominator are $$\pm \sqrt 2, \pm 2 \pm\sqrt 6.$$

As already mentioned, the only points across which $\Phi$ can change sign are at one of the zeroes above. Let's also note the expression $\Phi$ is "even": if $\Phi(x)\ge 0$, then $\Phi(-x)\ge 0$. So, let's find the $x$ values on the nonnegative $x$-axis that satisfy $\Phi(x)\ge0$. Then by "reflection" we'll obtain the points on the negative $x$-axis where $\Phi(x)\ge0$.

So, draw a number line with the zeroes listed above:

The zeroes subdivide the nonnegative $x$-axis into intervals, the endpoints of which, except the endpoint 0, are the zeroes of either the numerator or the denominator of $\Phi$. In each subinterval, if you pick a point $x_0$ (not an endpoint, save for 0), evaluate $\Phi(x_0)$, and note its sign, then across that entire subinterval, the expression $\Phi$ will have that sign.

For example in $[0,1)$, picking $x=1/2$ it is easy to see that the sign of $\Phi(1/2)$ is

$$\frac{((1/2)^2-1)((1/2)^2-3)((1/2)^2-5)}{((1/2)^2-2)((1/2)^2-4)((1/2)^2-6)} ={ (-)(-)(-)\over(-)(-)(-)} \ge 0$$ So on all of $[0,1)$, the expression $\Phi$ is positive. (Note, you only need to find the sign, there is no need to do the actual arithmetic.)

The complete "sign chart" is shown below:

And we see that $\Phi>0$ on $$[0, 1)\cup(\sqrt2,\sqrt3)\cup(2,\sqrt5)\cup(\sqrt6,\infty).$$ By symmetry, $\Phi>0$ on $$( -\infty,-\sqrt6)\cup(-\sqrt5,2)\cup (-\sqrt3,-\sqrt2)\cup (-1,0].$$

So, the domain of $\sqrt\Phi$ is the union of the two sets above, together with the zeroes of the numerator of $\Phi$ that aren't zeros of the denominator of $\Phi$: $x=\pm1$, $x=\pm \sqrt3$, and $x=\pm \sqrt5$.

Below is a plot of $y=\Phi$ and $y=\sqrt\Phi$. Note that the zeroes of the denominator of $\Phi$ are vertical asymptotes of the graph of $y=\Phi$ and the zeroes of the numerator of $\Phi$ are the zeroes of $\Phi$.

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1. $\frac{f(x)}{g(x)} \ge 0$ is not so different from $f(x)\times g(x) \ge 0$ (except for zeros of $g$.)
2. With $(x^2-1) = (x-1)(x+1)$ etc, your problem reduces to the form of $(x-a)(x-b)(x-c)(x-d)...(x-z) \ge 0$

Edit: oops I only read the hand-written part! Anyways thanks to the monotonicity of $x^5-1$ etc, you can still use similar argument.

Edit 2: Plot it in google.

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+1 for the link. I didn't know Google could plot. –  user23211 Feb 23 '12 at 13:37

$(x^2-1)$ is negative iff $x\in(-1,1)$. Similarly, $(x^2-2)$ is negative for $x\in(-\sqrt{2},\sqrt{2})$, $(x^2-3)$ for $x\in(-\sqrt{3},\sqrt{3}), \dots,(x^2-6)$ for $x\in(-\sqrt{6},\sqrt{6})$.

Consider the open intervals $(-\infty,-\sqrt{6}), (-\sqrt{6},-\sqrt{5}),(-\sqrt{5},-2),\dots$ separately. Don't forget to look at how $f(x)$ behaves on the boundaries. E.g $x=\pm\sqrt{n}$ for $n=1,2,3,4,5,6$.

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Ok, now I get the domain as $-\sqrt{5} \lt x \lt -2$, $-\sqrt{3} \lt x \lt -\sqrt{2}$, $-1 \lt x \lt 1$ ... but how do I find out without graph about what happens $x \lt -\sqrt{6}$ and $x \gt \sqrt{6}$ –  Jiew Meng Feb 23 '12 at 14:23
When x>$\sqrt{6}$ all the factors in the numerator and denominator are positive, so the whole expression is positive. The situation is the same for $x<−\sqrt{6}$ –  01000100 Feb 23 '12 at 14:49


Let $f:X\to Y$ be a mapping. We will prove that $f^{-1}(Y-f(X-W))\subseteq W$, with equality when $f$ is injective. Note that $f$ does not have to be closed, open, or even continuous for this to be true. It can be any mapping.

Let $W\subseteq X$. The mapping of $W$ in $Y$ is $f(W)$. As for $f(X-W)$, it may overlap with $f(W)$, we the mapping be not be injective. Hence, $Y-f(X-W)\subseteq f(W)$.

>Taking $f^{-1}$ on both sides, we get $f^{-1}(Y-f(X-W))\subseteq W$.

How can we take the inverse on both sides and determine this fact? Is the reasoning valid? Yes. All the points in $X$ that map to $Y-f(X-W)$ also map to $W$. However, there may be some points in $f^{-1}(W)$ that do not map to $Y-f(X-W)$.

Are there other analogous points about mappings in general? In $Y$, select two sets $A$ and $B$ such that $A\subseteq B$. Then $f^{-1}(A)\subseteq f^{-1}(B)$


Q&A

# Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern?

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I was playing around with the reciprocals of some positive integers and found these interesting patterns:

$$\frac{1}{19} = 0.\overline{052631578947368421}$$

Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows:

\begin{align} 1& \\ 21& \\ 421& \\ 8421& \\ {\color{red}{1}}68421 \\ {\color{red}{3}}{\color{green}{3}}68421 \\ \vdots\quad & \end{align} Notice that here I think of the red $\color{red}{1}$ of $16$ and $\color{red}{3}$ of $32$ as being "carried over", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the left. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.

How can I go about making this idea precise?

Some more data for this pattern: \begin{align} \vdots\quad & \\ {\color{red}{3}}{\color{green}{3}}68421 \\ {\color{red}{6}}{\color{green}{7}}368421 \\ {\color{red}{1}}{\color{green}{34}}7368421 \\ {\color{red}{2}}{\color{green}{69}}47368421 \\ {\color{red}{5}}{\color{green}{38}}947368421 \\ {\color{red}{10}}{\color{green}{77}}8947368421 \\ \vdots\quad & \end{align}

In this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.

I have also observed this to be the case for the decimal expansion of $1/29$: $$\frac{1}{29} = 0.\overline{0344827586206896551724137931}$$ In this case, we concatenate the powers of $3$ in a similar fashion.

Is there any significance to the fact that $19 = {\color{red}{2}}*10 - 1$ and $29 = {\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?

I also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up: $$\frac{1}{39} = 0.\overline{025641}$$ and concatenating the powers of $4$ indeed gives $\dotsc 025641$.

The last case I checked was $1/49$, and it also obeys this pattern: $$\frac{1}{49} = 0.\overline{020408163265306122448979591836734693877551}$$ and concatenating the powers of $5$ gives us this very repeating set of digits.

Of course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the forward direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that $$\frac{1}{49} = \frac{2}{100} + \frac{2^2}{100^2} + \frac{2^3}{100^3} + \dotsb$$ which is true by the formula for the sum of an infinite geometric series. So, "forward concatenating" patterns can be explained in this manner, but I'm stumped regarding how to explain the "backward concatenating" ones.

If anyone can throw some light on this, it would be greatly appreciated!

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First of all, there is indeed a pattern.

To figure out why, there's an easier example than what you've found: $$0.\bar{1} = \frac{1}{9}.$$

How does that work? $1$ is the only power of $1$: $1^n = 1$.

Now, how do we prove this? if $x = 0.\bar{1}$, then $10x = 1.\bar{1}$ and so, $9x=1$.

Hidden in that proof lies the answer for the other cases - you have some repeating number and in order to show what fraction that repeating number is equal to, here $1/\left(10n-1\right)$, you multiply it by $10n$ and subtract the number (again, I'll just call it $x$).

To take your example of $1/19$, we have that the fraction is $x = 0.\overline{052631578947368421}$, so we multiply by $20$ to get $20x = 1.\overline{052631578947368421}$, subtract $x$ and $19x = 1$. Crucially, the number left on the RHS after subtracting is exactly $1$, as everything after the decimal point cancels. It must do this, because the fraction is some integer divided by some other integer. That is, if $x=n/m$, then $mx=n$.

As we're in a decimal system, multiplying by $10$ is shifting one digit to the left. So, in doing this multiplication by $10n$, you're shifting the digit to the left, then multiplying by $n$, which is exactly the procedure outlined in the question.

Now, each particular digit (after this multiplication procedure) must cancel with the digit to the left so at this point, they must be equal and we have the exact behaviour outlined in your question.

To explain this generally and in more detail (taking $x<1$ wlog.) I'll use a fraction $$x = \frac{m}{10n-1}.$$

When written in decimal format, this is $x = 0.x_1x_2\ldots x_k\ldots$, with $x_{k+r}=x_k$ for some $r$, where each $x_k$ is a single digit (i.e. an integer in the range $0-9$ as we're in a decimal basis).

Now, we know that $\left(10n-1\right)x = m$, or $10nx - x = m$. However, m is an integer, so all the digits after the decimal point (the 'fractional part', in other words) are $0$.

This means that the digits in the fractional part of $10nx - x$ must also be $0$.

So, we look at the digits $x_r$ and $x_1\ldots x_{r-1}$.

The fractional part of $10nx -$ the fractional part of $x$ equals $0$, or the fractional part of $10nx$ equals the fractional part of $x$. For this to be the case, all the digits must be equal.

So, we take the digit $x_r$. To calculate $x_{r-1}$, we use the properties of the fractional parts being equal and multiplication by 10 as shifting the digit to the left to give that the fractional part of $10n\left(0.x_1\ldots x_r\right) - 0.x_1\ldots x_r = 0.0\ldots x_r$ (where the $0$s after the point are repeated $r-1$ times).

This is exactly what's written in the question, just in a different way: $nx_r \mod 10 = x_{r-1}$. Or, the digit to the left of $x_{r-1}$ is $nx_r$, with change. Accounting for this change gives that $x_{r-j} = n^jx_r$, which is where the powers of $n$ come from, the change being what needs to be added to digits further to the left.

To explain what I mean by phrases such as "Accounting for this change", let's take the initial stream of digits again and write them out more explicitly, where the number in brackets to the right of each line represents the number of times the 0 is repeated after the decimal point:

Now, we apply this procedure of multiplying by $10n$. It is a requirement that, because the fractional part of $x$ must equal the fractional part of $10nx$, $10nx-m=x$, that is

\begin{align*}10nx - m &= 0.0\ldots x_k \qquad(k-1)\\ &+ 0.0\ldots x_{k-1} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2} \qquad(k-3) + \ldots\end{align*}

Taking this line by line (where the term in brackets on the LHS has $j-1$ $0$s after the decimal point and the superscripts denote that this series arises from the $j^{th}$ term in the similar expansion of $x$) gives that

We can add this to the next term in the expansion of $x$ to get that (unfortunately at this point, notation gets very confusing but the first term on the left hand side has $j-1$ $0$s and the second, $j-2$) \begin{align*}10n\left(0.0\ldots x_j + 0.0\ldots x_{j-1}\right) &= 0.0\ldots x_{k-1}^{(j)} \qquad(k-2)\\ &+ 0.0\ldots x_{k-2}^{(j)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-2}^{(j-1)} \qquad(k-3)\\ &+ 0.0\ldots x_{k-3}^{(j)} \qquad(k-4)\\ &+ 0.0\ldots x_{k-3}^{(j-1)} \qquad(k-4) + \ldots\end{align*}

Now, having done this, the $k^{th}$ term is exactly $$\sum_{j>k}0.0\ldots x_k^{(j)} \qquad(k-1),$$

which is the sum over $10n\times$ all the digits to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.

However, you can recursively apply this to get that the $k^{th}$ digit is the sum over $\left(10n\right)^p\times$ all the digits, $p$ positions to the right of the $k^{th}$ digit, where you're only summing the resulting $k^{th}$ digit.

This is precisely what's done in the question, although written in a different way and shows that the behaviour arises because the terms are required to be sums of powers of $10n$ of previous terms because this is a fraction with a denominator of $10n-1$.

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I'm confused how this answers the question of the pattern of digits. You only use the fact that x = 1/19 in your answer and don't explain what the significance of the powers of n is. (Or maybe my reading comprehension skills are poor...) Quintec about 1 month ago

Thank you for your answer! I hope you won't mind if I say that I somehow feel this answer doesn't go all the way towards explaining why this "backwards concatenation" process appears. Specifically, I do understand that if $x = 0.\overline{x_1 x_2 x_3 \dotso x_r}$, then $x$ is a rational number equal to some $n/m$ that can be found as you outlined. However, why does this method of specifically concatenating the powers of $n$ work when considering $1/(10n - 1)$? Abheri Ragam about 1 month ago

Secondly, though I say this "works", how do I adequately make sense of that? After all, the string that I get by my method of concatenation strictly speaking gives an "infinite number" such as $\dotsc 052631578947368421$. So perhaps, my claim is something like $$\frac{1}{10n-1} = \lim_{k \to \infty} \frac{(10n)^0 + (10n)^1 + (10n)^2 + \dotsb + (10n)^k}{10^k}\ ?$$ I'm not sure... But I do think there's a bit more to be said from where you leave off in your post :) Abheri Ragam about 1 month ago

I've updated my answer (although honestly, I feel that the first bit is about 100 times clearer). I've included the bit about the powers of $n$ and what you're referring to as 'backwards concatenation' is what I'm referring to as 'multiplying a digit by $n$ to get the digit to the left'. It works because the '-1' is the bit that causes the fractional part to cancel, hence be equal Mithrandir24601 about 1 month ago

But, $x_{r-j}$ is not equal to $n^j x_r$! (I presume you meant modulo $10$?) For instance, for the decimal expansion of $1/19$, $x_{r-5} = 3$ but $2^5 x_r = 32 \neq 3 \pmod{10}$. In fact, they are not equal precisely because of the carrying-over happening in the concatenation process. I'm sorry, but I'm still not satisfied that your answer fully explains this phenomenon. Abheri Ragam about 1 month ago

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I was pretty curious about this, so I posted a question on Math.SE. Most of this answer will draw from the idea of J. W. Tanner's answer there.

There is actually a simple reason why this pattern holds.

Lets start by examining how the digits are generated: each digit is a power of two (with carrying). We can withhold decimal shifting for a second and simply look at a simplified expression with the same digits, only generated to the left of the decimal point:

\begin{aligned}\sum_{k=0}^{n-1} 2^k\cdot 10^k&=1+20+400+16000+320000+\cdots+2^{n-1}\cdot10^{n-1}\\ &=\cdots7368421\end{aligned}

You can see how the pattern generated is the same. Let's look at it another way, using the geometric series formula.

\begin{aligned}\sum_{k=0}^{n-1} 2^k\cdot10^k&=\frac{1(1-20^n)}{1-20}\\ &=\frac{20^n-1}{19}\end{aligned}

At this point, it's pretty clear why the digits generated become the repeated decimal of $1/19$. You can see the denominator of $19$ right there!

If you're not satisfied with that observation, we can further note the following:

\begin{aligned}\frac{20^n-1}{19}&=\frac{20^n}{19}-\frac{1}{19}\\ &=\frac{2^n}{19}\cdot10^n-\frac{1}{19}\end{aligned}

Let's specifically focus on $2^n/19$. Since we are interested in repeated groups of digits, let's let $n$ be a multiple of the totient of $19$, which is $18$. The reason we chose the totient is because of a useful property of modular arithmetic: $a^b\equiv 1\mod c$ if $a$ and $c$ are coprime and $b$ is a multiple of the totient of $c$ (Euler's theorem).

Clearly, $2$ is coprime to $19$, so letting $n$ be a multiple of $18$ gives us a remainder of 1 when dividing $2^n$ by $19$. Therefore, the digits after the decimal point of $2^n/19$ are the same as the digits of $1/19$.

Going back to the whole expression, we then multiply by $10^n$, bringing $n$ of those repeated digits over to the left of the decimal point. Everything to the right is canceled by the subtraction of $1/19$ (the digits must cancel, since the expression is an integer).

The result is the final number has its last $n$ digits be the first $n$ digits of $1/19$ when $n$ is a multiple of $18$, which is what we wanted to prove. (It's not difficult to generalize this for any $n$, but that's beyond this answer).

A similar result can be produced for any $10c-1$, where the digits will repeat every totient of $10c-1$ (though it may repeat before that) and the repeating portion will be the same as the digits of $1/(10c-1)$. Since it's basically the same process, I won't repeat it here.

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#### 1 comment

Thank you! I've been considering what you say in your answer, and I'm now satisfied with the explanation :) Abheri Ragam about 1 month ago

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I think the simplest way to see how it works is as follows:

Given $x=\frac{1}{10n-1}$, it is easy to check that $\frac{x+1}{10}=nx$.

Now $\frac{x+1}{10}$ means shifting the digits to the right, with an $1$ added as the first decimal. For example, in the case of $x=1/19$, we have $$\frac{x+1}{10} = 0.1\overline{052631578947368421}.$$ But now we see that the additional digit matches the one at the end of the period, so we just can shift the period one to the left. At the same time on the left we use the equation derived above: $$2\cdot\frac{1}{19} = 0.\overline{105263157894736842}.$$ Thus multiplying by $2$ (or, in the general case, by $n$) is equivalent to shifting the digits one position to the right (and adding a $1$ at the first decimal, which gives the carry from the multiplication). Thus since the last digit of the original period was $1$, the digits to the left of it are successive powers of $2$ (with carry), or successive powers of $n$ in the general case.

So the only remaining question is why the period always ends in $1$. Well, since multiplying the number with $19$ (or, in the general case, with $10n-1 = 10(n-1)+9$) has to give $1=0.\overline{999999999999999999}$, the last digit of the period, when multiplied with the last digit of the denominator, that is $9$, must give $10k+9$ for some integer $k$. But that is clearly only possible if that last digit is $1$.

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