# Why does $\frac{1}{x} < 4$ have two answers?

Solving $\frac{1}{x} < 4$ gives me $x > \frac{1}{4}$. The book however states the answer is: $x < 0$ or $x > \frac{1}{4}$.

My questions are:

Why does this inequality has two answers (preferably the intuition behind it)?

When using Wolfram Alpha it gives me two answers, but when using $1 < 4x$ it only gives me one answer. Aren't the two forms equivalent?

• $x$ can be negative...^^ – Probabilitytheoryapprentice Aug 27 '16 at 17:01
• There is just one answer here, not two as you seem to think. It is not the case that $x>1/4$ is one answer and $X<0$ is another answer. Neither inequality by itself would be a correct answer. – Andreas Blass Aug 27 '16 at 17:14
• "when using 1<4x it only gives me 1 answer. Aren't the two forms equivalent?" Nope. They are not. If $1/x < 1/4$ it's possible that $x < 0$. For $1 < 4x$ it is not possible. – fleablood Aug 28 '16 at 0:14
• Draw a graph of the function $x \mapsto 1/x$ should give some intuition why there are two "zones" for the solution. – quid Aug 28 '16 at 0:26
• Just try, for instance, $x=-1$. Is it true in this case that $1/x < 4$? Is it true that $1 < 4x$? – TonyK Aug 28 '16 at 12:04

You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$ If $x>0$ and $x>1/4$, then $x>1/4$.

Now suppose $x<0$. Then $$\frac{1}{x}<4\iff4x<1\iff x<1/4.$$ If $x<0$ and $x<1/4$, then $x<0$. So the solution set is $(-\infty,0)\cup(1/4, \infty).$

• Wow!This answer got so many upvotes! ;-) – tatan Mar 24 '18 at 17:28

Here is the solution $$\frac { 1 }{ x } <4$$$$\frac { 1-4x }{ x } <0$$$$\frac { x\left( 1-4x \right) }{ { x }^{ 2 } } <0$$$$x\left( 1-4x \right) <0$$$$x\left( 4x-1 \right) >0$$

so $$x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 4 } ,+\infty \right)$$

• nice approach to avoid the case division (+1) – b00n heT Aug 27 '16 at 17:06
• As soon as I read that there were two answers, I started thinking "there has to be a quadratic hidden here." Thanks for revealing it! – Ross Presser Aug 28 '16 at 0:50
• Well this is misleading. But the same means you could get solution of $x-1>0$ to be $(-\infty,0)\cup(1,+\infty)$. Your final presentation does need some justification (checking validity of each interval). – Ruslan Aug 28 '16 at 15:05
• @Ruslan - Not quite. The solution multiplies by $x^2$ between lines 3 and 4, and we know $x^2 > 0$ for all real $x \neq 0$. So multiplying by $x^2$ is preserves the direction of the inequality for all $x \neq 0$. With $x - 1 > 0$ however, you'd multiply by $x$ which would require splitting the equality into two cases. However, I do concede that this maybe should have been emphasized in the answer. (Still, it's a very nice answer) – David E Aug 29 '16 at 18:39
• @RossPresser The hidden quadratic can also be seen by inspection of the graph. 1/x is a hyperbola, therefore is a quadratic form. The obvious 2nd degree terms are just hidden by rotation :) – QuantumMechanic Aug 30 '16 at 17:12

I am following the suggestion given by @quid in the comments because I like pictures:

The orange/red line is the $x$-axis. The yellow line is the line $y=4$. The two blue curves are the graph of $y=1/x$. The solution to the inequality is the set of $x$ values for which the blue curve is below the yellow line. As @quid predicted, this picture gives some intuition for why there are two "zones" in the solution.

Note: originally I had $y=\frac{1}{4}$ which was incorrect, so I changed the answer to reflect the fact that it should be $y=4$ and re-plotted the graph.

• I agree a picture helps, but feel this would be more helpful if you (a) indicated the solution set explicitly, (b) displayed a smaller $x$-range, (c) used the same scale on both axes and (d) made the markings legible – PJTraill Aug 31 '16 at 17:24
• @PJTraill I don't feel that the first criticism is relevant to the actual question the OP asked, which was not for the exact value of the solution, but the intuition behind its two-part structure. The second and third criticism strike me as pedantic and unhelpful. The fourth criticism is just wrong -- click on the picture, and the full-size version opens in a new tab, where all of the markings are clearly legible. – Chill2Macht Aug 31 '16 at 18:46
• It is (of course) up to you, though I thought of them as suggestions rather than criticisms; I hope you were not offended by them. I meant “readily legible as they appear in the answer”, as I prefer to see all relevant information at once. P.S. I do like the clean look of your diagram. – PJTraill Sep 1 '16 at 19:11

The accepted answer is good, but I feel like you're really asking: why is there only one piece to the question, but two pieces to the answer?

This is actually a great question. Sometimes it happens that one piece turns into two (or more), like when you try to solve $x^2 = 9$ (which has "one piece") and get the two-piece solution $x = 3, -3$. Here, to figure out why one piece becomes two, you have to think about how the equation $y = x^2$ works.

So in our case we should think about how the equation $y = 1/x$ works. And when you think about it, you realize that you didn't really start with one piece. No matter what you plug in for $x$, the value of $1/x$ can never be zero. And that means when you write $1/x < 4$, really this gives you the TWO pieces

$$0 < 1/x < 4$$

and

$$1/x < 0$$

Basically, everything smaller than 4 but with zero removed. And that's why you end up with two pieces at the end -- because that's actually how many you started with!

• Yeah that was exactly what I was confused about! – Augusto Dias Noronha Aug 28 '16 at 11:53
• Another way to look at it is that 1/x is a hyperbola and therefore is really a quadratic form, so you'd expect two "pieces" to the solution. – QuantumMechanic Aug 30 '16 at 17:10
• Though you did well to nail what the questioner was confused about, I think you underplay the fact that turning one piece into one piece is only a reasonable expectation of a continuous function while $x \mapsto 1/x$ is not continuous (or even defined) at $0$, though it is continuous on the pieces you start with and and bijective on its domain. – PJTraill Aug 31 '16 at 17:38
• Just an idea: you are asked when an expression in x is smaller than 4, and this question has two answers: 1) the value of the expression is smaller than 4 and 2) the expression is negative, hence the two answers. – Dominique Sep 1 '16 at 11:06
• $y = 5 - {(8x - 1)}^{2}$ is a continuous function but it still yields two solutions to $y < 4$. – joeytwiddle Sep 7 '16 at 6:12

Just draw a graph of $1/x$ and you'll see 'why'.

Here is an image by WolframAlpha, with appropriate parts enhanced:

Whenever you're writing $x \gt \frac{1}{4}$ you're assuming $x \gt 0$.

But for $x \lt 0$ you have $\frac{1}{x}\lt0\lt4$

English doesn't have good words for this, so exactly what's going on can be a bit tricky to describe if you don't already understand the meaning.

It is true that every $x$ satisfying $x > \frac14$ does in fact satisfy $\frac 1x < 4$.

However, the converse fails: there are some $x$ that satisfy $\frac 1x < 4$ that do not satisfy $x > \frac 14$.

So, you have found a simple description of some of the $x$ that satisfy $\frac 1x < 4$ — but (presumably) you were being asked to describe all of the $x$ that satisfy $\frac 1x < 4$.

And there are indeed more of them: every $x$ satisfying $x < 0$ satisfies $\frac 1x < 4$.

Now, what is true is the following: if $\frac 1x < 4$, then it follows that at least one of the two statements "$x > \frac 14$" and "$x < 0$" is true. Thus, this gives a complete description of the solutions to $\frac 1x < 4$.

Put differently, for every $x$, the following bullet points are either both true or both false:

• $x$ satisfies $\frac 1x < 4$
• $x$ satisfies one of the statements "$x > \frac 14$", "$x < 0$".

Regarding your solution method, you forgot that multiplying by negative numbers reverses the sign of an inequality, and multiplying by zero turns any inequality into an equality. Here, we know that $x$ can't be zero, but it still could be either positive or negative, so you don't know the effect that multiplying by $x$ will have on the inequality.

The typical way to fix this problem is to break the problem into two parts: one part where you solve the case with the assumption $x<0$, and one part where you solve the case with the assumption $x>0$, and then you put the results together.

we have $$\frac{1}{x}<4$$ is true if $$x<0$$ and if $x>0$ we get $$x>\frac{1}{4}$$

• What? $\frac{1}{x} < 4$ if $x < 0$? – nbro Aug 31 '16 at 17:19

No, $$\frac 1x<4\text{ and }1<4x$$ are not equivalent.

You could think so just multiplying by $x$. But a rule says that an inequality is preserved when you multiply by a positive number and inverted with a negative one. So the right thing is

$$\begin{cases}x>0\to1<4x,\\x<0\to1>4x.\end{cases}$$

we have $\frac{1}{x}>4$ then $\frac{1}{x}-4>0$

that is , $$\frac{1-4x}{x}>0$$ but the domain of definition is $x\neq 0$

first of all you need to find the zeros and then study the its signs $$1-4x=0$$ then $$x=1/4$$

\begin{align} & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1/4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\ & \underline{\left. 1-4x\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \left. \frac{1-4x}{x}\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}

as we observe from the above table that inequality is positive only when $0<x<\frac{1}{4}$

I would like to harp on the meaning of your "two answers". When a question asking for a number satisfying some conditions leads to a unique number having that property we are fine. When there are two numbers having the required property then we can say two answers.

Here we have uncountably many real numbers $x$ such that $\frac1x < 4$. So even the region $x>4$ is infinitely many answers. One possible interpretation that could justify "two" could be number of connected components of the solution space of the question. This happens many times. The set $GL(n,\mathbf{R})$ of non-singular real $n\times n$ matrices has two connected components.

$$\frac{1}{x} < 4,$$ $$\frac{1}{x} - 4 < 0,$$ $$\frac{1 - 4x}{x} < 0,$$ $1 - 4x < 0$ or $x > 0$. Because if $1 - 4x$ is negative then $x$ must be positive. So we must write the solution in that way.

Here is an important aspect which should be always considered. If someone asks me:

Problem: Find the solution of \begin{align*} \frac{1}{x}<4 \end{align*} I would not answer the problem, but instead ask: What is the domain of $x$?

Please note the problem is not fully specified if the domain of $x$, the range of validity, is not given. This is crucial to determine the set of solutions.

Some examples:

Find the solution of


# How to show $\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$

$$\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$$

Can someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this. WolframAlpha says it is true but I don't know what the gamma function is.

$$\sum_{k=n}^\infty{\frac{1}{k!}} = \frac{1}{n!} + \sum_{k = n+1}^\infty \frac{1}{k!}$$ So then I need to show that$$\sum_{k=n+1}^\infty{\frac{1}{k!}} \leq \frac{1}{(n+1)!} ~~\Big[\leq \frac{1}{n!}\Big]$$

Is it possible to do this by induction? I don't really know how to approach this now.

• Just in case: the inequality doesn't hold for $n = 0$. For $n \geqslant 1$, note that $(n+k)! \geqslant n!\cdot (n+1)^k$. – Daniel Fischer Dec 3 '15 at 10:44
• $$\sum_{k=n}^{\infty}{\frac{1}{k!}}\le\frac{2}{n!}\Longleftrightarrow$$ $$e-\frac{e\Gamma(n,1)}{\Gamma(n)}\le\frac{2}{n!}\space\text{for}\space n>-1$$ – Jan Dec 3 '15 at 10:51

It suffices to show that $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \frac{1}{n!}$$

For all $k \ge n+1$, we have $$k! \ge 2^{k-n} \cdot n!$$

Therefore, we have $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \sum_{k=n+1}^\infty \frac{1}{n! \cdot 2^{k-n}} = \frac{1}{n!} \sum_{i=1}^\infty \frac{1}{2^i} = \frac{1}{n!}$$

• Does the inequality $k! \geq 2^{k-n} \cdot n!$ have a name? I think I need to prove that in order to use it. – elfeck Dec 3 '15 at 11:06
• Not really. For a proof note that $\frac{k!}{n!} = k(k-1) \cdots (n+1) \ge 2 \cdot 2 \cdots 2 = 2^{n-k}$. Of course, this doesn't hold when $n=0$, so you need to exclude that. – Gyumin Roh Dec 3 '15 at 11:06
• Okay thanks. Since in my case I substitute $n$ for $n+1$ anyway $n=0$ does not cause issues. Thanks for your help! – elfeck Dec 3 '15 at 11:10

Another way:

$$\sum_{k=n}^{\infty} \frac{1}{k!} = \frac{1}{n!} \bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots \bigg) < \frac{1}{n!} \bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots \bigg)\\ = \frac{1}{n!} \cdot \frac{n+1}{n} = \frac{1}{n!} \cdot \bigg(1 + \frac{1}{n} \bigg)$$ Now compare this expression to what you have on the RHS: certain terms cancel out. What do you get?

• Small typo, that $k$ should start from $n$, not $n+1$. – Gyumin Roh Dec 3 '15 at 11:18
• yeah, that's how I solved it anyway – Alex Dec 3 '15 at 11:44

$$\dfrac{1}{(k+1)!} = \dfrac{1}{(k-1)!}(\dfrac{1}{k} - \dfrac{1}{k+1}) \leq \dfrac{1}{(n-1)!}(\dfrac{1}{k}-\dfrac{1}{k+1})$$ when $k\geq n$

So

$$\sum_{k=n}^\infty\dfrac{1}{(k+1)!} \leq \dfrac{1}{(n-1)!}\sum_{k=n}^\infty (\dfrac{1}{k}-\dfrac{1}{k+1}) = \dfrac{1}{(n-1)!} *\frac{1}{n} = \dfrac{1}{n!}$$


# Math Help - Definite integral ln(1+t)dt

1. ## Definite integral ln(1+t)dt

I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..

$\int_{0}^{5} ln(1+t)dt$

My first way of thinking led me to a dead end. I used integration by parts with $U=ln(1+t)$ and $U=1/(1+t)$ and $V=1dt$ and $V=t$

This led me to the following

$= [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt$

and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.

I think my line of thinking is wrong.. is my assumption that $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?

2. Originally Posted by mkelly09
I need some help with a particular question that i can't seem to figure out no matter how much time i've spent on it this evening..

$\int_{0}^{5} ln(1+t)dt$

My first way of thinking led me to a dead end. I used integration by parts with $U=ln(1+t)$ and $U=1/(1+t)$ and $V=1dt$ and $V=t$

This led me to the following

$= [tln(1+t)]_{0}^{5} - \int_{0}^{5} t/(1+t)dt$

and this is where i'm stuck.. can't seem to figure out the integration for the second integral here.

I think my line of thinking is wrong.. is my assumption that $U=ln(1+t)$ and $U=1/(1+t)$ incorrect?
You're correct...but you need to think a little "outside of the box" for this integral.

$\int\frac{t}{1+t}\,dt$

Let $u=t+1\implies t=u-1$

Thus, $\,du=\,dt$

Therefore, the integral becomes $\int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du$

Can you take it from here?

--Chris

3. Originally Posted by Chris L T521
You're correct...but you need to think a little "outside of the box" for this integral.

$\int\frac{t}{1+t}\,dt$

Let $u=t+1\implies t=u-1$

Thus, $\,du=\,dt$

Therefore, the integral becomes $\int\frac{t\,du}{u}=\int\frac{u-1}{u}\,du=\int\left(1-\frac{1}{u}\right)\,du$

Can you take it from here?

--Chris
Here's another way I just thought of: add "zero" to the numerator

$\int\frac{t}{t+1}\,dt=\int\frac{t+1-1}{t+1}\,dt=\int\left[\frac{t+1}{t+1}-\frac{1}{t+1}\right]\,dt=\int\left(1-\frac{1}{t+1}\right)\,dt$

--Chris

4. using integration by parts on that guy directly is overkill. (that is, if you know the integral of ln(x) by heart--which you should).

$\int_0^5 \ln (1 + t)~dt$

Let $u = t + 1$, this substitution yields

$\int_1^6 \ln u~du$

which we know is $u \ln u - u \bigg|_1^6$ ...

see post #7 here for how to integrate ln(x) using integration by parts. note that you could also figure out the integral by contemplating the derivative of xln(x)

5. Thanks for your replies, they certainly solved my problem. In fact, now i know of a few different ways of doing it when before i couldn't figure out one.

I think i will be committing to memory the antiderivative of ln(x)


# What is the reasoning behind why the ratio test works?

The ratio test says that if we have

$$\sum_{n=1}^{\infty}a_n$$

such that $\lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$, then if:

1) $L < 1$, then $\sum_{n=1}^{\infty}a_n$ is absolutely convergent,

2) $L > 1$, then $\sum_{n=1}^{\infty}a_n$ is divergent, and

3) $L = 1$, then the ratio test gives no information.

I want to understand the mathematics behind the ratio test. I want to know the reasons behind why and how this works, rather than just memorising a formula.

I have attempted to reason about this theorem myself. It can be seen that we're taking the ratio between the second term in the series, $a_{n+1}$, and the first term in the series, $a_n$. This is exactly how one finds the common ratio in a geometric sequence ($r = \dfrac{a_{n+1}}{a_n} )$.

We then take the limit of this ratio, which I assume is to find the relative rate of change between $a_{n+1}$ and $a_n$. In which case, It is more effective to take the absolute value of the limit, $\lim_{n \to \infty} \begin{vmatrix}{ \dfrac{a_{n+1}}{a_n} }\end{vmatrix} = L$. This seems analagous to how the limit comparison test theorem works. Therefore, I presume that if $L < 1$, this implies that $a_n$ is has a greater rate of change than $a_{n+1}$, which implies that each successive term in the series is getting smaller, and as we go to infinity, each successive term is converging towards $0$. As such, the series should converge to some value. Analogously, I presume that if $L > 1$, this implies that $a_{n+1}$ has a greater rate of change than $a_n$, which implies that each successive term in the series is getting larger, and as we go to infinity, the terms diverge towards infinity.

Is this reasoning correct? If anything is incorrect, please clarify why and what the correct reasoning is.

Thank you.

• "each succesive term is converging towards zero. As such, the series should converge." This is a famously fallacious statement. The harmonic series tends termwise to zero, but the series diverges. – Matthew Leingang Nov 11 '16 at 1:13
• @MatthewLeingang I forgot about that. You're absolutely correct. In which case, what is the correct way to reason about that section? – The Pointer Nov 11 '16 at 1:14
• @GCab Which is not less than $1$ . . . – Noah Schweber Nov 11 '16 at 1:19
• @GCab That's correct, the ratio test doesn't apply to the harmonic series. But the statement I quoted is still invalid. – Matthew Leingang Nov 11 '16 at 1:20
• next time read wikipedia's entry en.wikipedia.org/wiki/Ratio_test#Proof – reuns Nov 11 '16 at 1:35

You have some ideas correct, but remember that $a_n \rightarrow 0$ does not imply that $\sum a_n$ converges. So we need something more than that. However, for the case when $L>1$, your reasoning is correct.

For the case $L<1$, we know that there exists some $r < 1$ and $N \in \mathbb{N}$ such that $$\frac{a_{n+1}}{a_n} \leq r < 1,$$ whenever $n \geq N$.

Therefore,

$$a_{N+1} \leq ra_N,$$ $$a_{N+2} \leq ra_{N+1} \leq r^2a_N$$ $$a_{N+3} \leq ra_{N+2} \leq r^3a_N,$$ and in general $$a_{N+n} \leq r^na_N.$$

Therefore, at least eventually, we can compare the series from above with a convergent geometric series $\sum a_N r^n$, which implies the convergence of the original series $\sum a_n$.

• I'm not sure if your notation has the same meaning as mine or if you've defined it differently. For instance, how can $\dfrac{ a_{n+1} }{a_n} < r$? I've defined $r = \dfrac{ a_{n+1} }{a_n}$. – The Pointer Nov 11 '16 at 1:54
• My notation is different. If you set $r$ equal to the ratio, then it is changing with $n$. But when $L < 1$, such a fixed $r$ exists as I employed it in my answer to your question. – user333870 Nov 11 '16 at 2:00
• Can you please elaborate on what $r$ and $N$ are? – The Pointer Nov 11 '16 at 2:03
• Since $lim \frac{a_{n+1}}{a_n} = L < 1$, take $\epsilon = \frac{1-L}{2}$, then by the definition of the limit, there exists $N\in \mathbb{N}$ such that $n \geq N \Longrightarrow -\epsilon < \frac{a_{n+1}}{a_n} - L < \epsilon$, so you can take for example $r = \epsilon + L = \frac{1+L}{2}$ – user333870 Nov 11 '16 at 2:18
• I'm sorry but this still doesn't tell me what $N$ and $r$ are supposed to be. I cannot see the relevance between your answer and my original question. – The Pointer Nov 11 '16 at 5:28


Rationalize the denominator and simplify

So the problem I have says rationalize the denominator and simplify. $$\frac{ \sqrt{15}}{\sqrt{10}-3}$$

My answer I got was $\frac{5 \sqrt6}{7}$.

Am I doing this wrong or is this the wrong answer I was told it was incorrect?

-
So it is pretty easy to check that your answer is wrong, just use a calculator. You will get $23.87$ for the question and $1.75$ for your answer. – David Aug 27 '14 at 0:13

It seems that you tried multiplying by $\frac{\sqrt{10}}{\sqrt{10}}$. Instead, you should try multiplying by the conjugate and take advantage of difference of squares: $$\frac{\sqrt{15}}{\sqrt{10} - 3} = \frac{\sqrt{15}}{\sqrt{10} - 3} \cdot \frac{\sqrt{10} + 3}{\sqrt{10} + 3} = \frac{\sqrt{150} + 3\sqrt{15}}{(\sqrt{10})^2 - 3^2} = 5\sqrt{6} + 3\sqrt{15}$$
I think wht happened was that you correctly multiplied the denominator by $\sqrt{10}+3$, but incorrectly multiplied the numerator by $\sqrt{10}$. The numerator should also have been multiplied by $\sqrt{10}+3$
$$\frac{\sqrt{15}}{\sqrt{10}-3}=\frac{\sqrt{15} \cdot (\sqrt{10}+3)}{(\sqrt{10}-3) \cdot (\sqrt{10}+3)}=\frac{\sqrt{15} \cdot (\sqrt{10}+3)}{10-9}=\sqrt{15} \cdot (\sqrt{10}+3)$$


# CharacteristicFunction

CharacteristicFunction[dist,t]

gives the characteristic function for the distribution dist as a function of the variable t.

CharacteristicFunction[dist,{t1,t2,}]

gives the characteristic function for the multivariate distribution dist as a function of the variables t1, t2, .

# Examples

open allclose all

## Basic Examples(4)

Characteristic function (cf) for the normal distribution:

Characteristic function for the binomial distribution:

Characteristic function for the bivariate normal distribution:

Characteristic function for the multinomial distribution:

## Scope(8)

Characteristic function for a specific continuous distribution:

Characteristic function for a specific discrete distribution:

Characteristic function at a particular value:

Characteristic function evaluated numerically:

Obtain a result at any precision:

Compute the characteristic function for a formula distribution:

Find the characteristic function for a parameter mixture distribution:

Characteristic function for the slice distribution of a random process:

## Applications(7)

Compute the raw moments for a Poisson distribution:

First 5 raw moments using derivatives of the characteristic function at the origin:

Use Moment directly:

Compute mixed raw moments for a multivariate distribution:

Use Moment to obtain raw moments directly:

Find raw moments of a Student distribution from its characteristic function:

Compute to extract moments by taking limits from the right:

Evaluate the limits from the left:

Only the first four moments are defined, as confirmed by using Moment directly:

Use inverse Fourier transform to compute the PDF corresponding to a characteristic function:

Illustrate the central limit theorem on the example of symmetric LaplaceDistribution:

Find the characteristic function of the rescaled random variate:

Compute the large limit of the cf of the sum of such i.i.d. random variates:

Compare with the characteristic function of a standard normal variate:

Use smooth characteristic function to construct the upper bound for the distribution density of ErlangDistribution:

Plot the upper bounds and the original density:

Verify that the sum where are independent identically distributed variates tends in distribution to for large :

Use a combinatorial equality for product :

Evaluate the sum:

Take the limit and compare it to the characteristic function of the UniformDistribution:

## Properties & Relations(5)

CharacteristicFunction is the Expectation of for real :

The characteristic function is related to all other generating functions when they exist:

The cf of a continuous distribution is equivalent to FourierTransform of its PDF:

The cf of a discrete distribution is equivalent to FourierSequenceTransform of its PDF:

The PDF is the inverse Fourier transform of the cf for continuous distributions:

The PDF is the inverse Fourier sequence transform of the cf for discrete distributions:

## Possible Issues(1)

Symbolic closed forms do not exist for some distributions:

## Neat Examples(1)

Visualize real and imaginary parts of CharacteristicFunction for random instances of BinomialDistribution:

Wolfram Research (2007), CharacteristicFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CharacteristicFunction.html (updated 2010).

#### Text

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#### CMS

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#### APA

Wolfram Language. (2007). CharacteristicFunction. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/CharacteristicFunction.html

#### BibTeX

@misc{reference.wolfram_2022_characteristicfunction, author="Wolfram Research", title="{CharacteristicFunction}", year="2010", howpublished="\url{https://reference.wolfram.com/language/ref/CharacteristicFunction.html}", note=[Accessed: 11-August-2022 ]}

#### BibLaTeX

@online{reference.wolfram_2022_characteristicfunction, organization={Wolfram Research}, title={CharacteristicFunction}, year={2010}, url={https://reference.wolfram.com/language/ref/CharacteristicFunction.html}, note=[Accessed: 11-August-2022 ]}


# equation custom horizontal alignment & numbering each row

As I know, "align" has the fixed horizontal alignment right-left-right-left-..., and we cannot change this alignment. So if we want to align equations with custom horizontal alignment, e.g., right-center-center-left, we may use "equation" or "gather", "array", "arraycolsep" (for spacing "=" like "align"), and "displaystyle" (for proper handling of "frac" or "lim").

For example,

\newcommand{\argmax}{\operatornamewithlimits{arg\,max}}
...
\begin{gather}
\arraycolsep=1.4pt\def\arraystretch{2.2}
\begin{array}{rccl}
p_{\mathrm{MLE}}(x) & = & \displaystyle \max_{m} &P(X = x | \theta = m) \\
m_{\mathrm{MLE}}(x) & = & \displaystyle \argmax_{m} &P(X = x | \theta = m)
\end{array}
\end{gather}

produces below.

However, this has the only one numbering label. It cannot split numbers for each row. It may be one BAD choice to use "align" and adjust spacing MANUALLY by using "\,", "\phantom{}", "\quad", or "\qquad".

For this case, how can we label different numbers for each row?

• You probably already know about the option \notag that removes the numbering of one row in align environment. I think there must be something similar, but instead it forces a tag for each entry in multi-rows of math environments. What you ask is indeed interesting and I am waiting to find the answer – Al-Motasem Aldaoudeyeh Jun 30 '17 at 2:54
• Off-topic: Don't use | to denote "given that" items. Instead, write \mid. – Mico Jun 30 '17 at 8:04
• A newenvironment here could help... But I am waiting for an answer in a problem I found trying to manipulate the rows of my newenvironment... I will come back for an answer... But have to study a little bit about newenvironments – koleygr Jun 30 '17 at 8:08

Since you're using the amsmath package, I would use that package's \DeclareMathOperator* directive to create two new "operators": \argmax and \midmax, where the latter displays the word "max" centered in a box of width equal to "arg max". I would also use a split environment instead of an array environment, an equation environment instead of a gather environment, and \mid instead of \.

If you need to number each row separately, use an align environment instead of the nested equation/ split environments.

\documentclass{article}
\usepackage{amsmath}
%% Create two new math opertors: \argmax and \midmax
\DeclareMathOperator*{\argmax}{arg\,max}
\newlength\mylen
\settowidth\mylen{arg\,max}
\DeclareMathOperator*{\midmax}{\parbox{\mylen}{\centering\upshape max}} % center-set "max"
\begin{document}

%% Single equation number for both rows:
$$\begin{split} p_{\mathrm{MLE}}(x) &= \midmax_{m} P(X = x \mid \theta = m) \\ m_{\mathrm{MLE}}(x) &= \argmax_{m} P(X = x \mid \theta = m) \end{split}$$

% Separate equation numbers, one per row:
\begin{align}
p_{\mathrm{MLE}}(x) &= \midmax_{m} P(X = x \mid \theta = m) \\
m_{\mathrm{MLE}}(x) &= \argmax_{m} P(X = x \mid \theta = m)
\end{align}
\end{document}


# Volume of a split log

When solving the following problem, I could not understand why my reasoning came up with an answer that's different than the one on the solution's manual.

Question: Consider $(x,y,z)$ such that $x^2+y^2<1, x>0, 0 \le z \le 5$. This describes one half of a cylinder (a split log). Chop out a wedge out of the log along $z=2x$. Find the volume of the wedge.

My reasoning:

Wedge Volume $=$ (half the volume of the cylinder of height 2) $-$ (volume of solid of revolution found revolving the curve $x=z/2$, from 0 to 2, over the z axis).

Half Cylinder Volume: $(\pi r^2h)/2 = \pi/2 \int_0^2 (1)^2 dz = \pi$

(Volume of solid of revolution found revolving the curve x=z/2, from 0 to 2, over the z axis): $(\pi r^2h)/2 = \pi/2 \int_0^2 (z/2)^2 dz = \pi/3$

Wedge Volume: $\pi - \pi/3 = 2\pi/3$

However, the solution's manual answer is $4/3$ and it displays a different reasoning:

The slice perpendicular to the $xz$-plane are right triangles with base of length $x$ and height $z=2x$. Therefore the area of a slice is $x^2$. The volume is:

$$\int_{-1}^1 x^2 dy = \int_{-1}^1 (1-y^2) dy = 4/3.$$

While I understand the solution's manual reasoning and even find it simpler than mine, I still cannot understand why the two approaches results in different answers. What am I missing?

Your reasoning in your first method goes wrong in that chopping a wedge isn't the same thing as taking the solid of revolution of a curve. The solid of revolution of $z = 2x$ is a cone, so if you take a cone out of the cylinder, you're not left with a wedge but a somewhat weird-looking anti-cone shape.
If we were to analyze the cross-section of the wedge that we want in cylindrical coordinates, we'd see that it grows as the angle approaches $0$, and shrinks as it approaches either $\pi/2$ or $-\pi/2$. Whereas with your solid of revolution, the cross-section stays constant at all angles.


# Determinant of $N \times\ N$ matrix

For $n \geq 2$, compute the determinant of the following matrix: $$B = \begin{bmatrix} -X & 1 & 0 & \cdots & 0 & 0 \\ 0 & -X & 1 & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & \cdots & 0 & -X & 1 \\ a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X) \end{bmatrix}$$

Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix:

$\det \begin{bmatrix} -X & 0 \\ 0 & (a_1-X) \end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X$

by expansion along the first row:

$\det \begin{bmatrix} -X & 1 & 0 \\ 0 & -X & 0 \\ 0 & 0 & (a_2-X) \end{bmatrix} = (-X) \times\det \begin{bmatrix} -X & 0 \\ 0 & a_2-X \end{bmatrix} - 1 \det\begin{bmatrix} 0 & 0 \\ 0 & a_2-X \end{bmatrix}$

$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2$

So it looks like:

$\det \begin{bmatrix} -X & 1 & 0 & \cdots & 0 & 0 \\ 0 & -X & 1 & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & \cdots & 0 & -X & 1 \\ a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X) \end{bmatrix} = X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$

Does this look right? Is "prove by induction" valid to use here?

• Your $2\times 2$ matrix must be $\begin{bmatrix} -X & 1 \\ a_0 & (a_1-X) \end{bmatrix}$. You have also done the same mistake for the next case. I do not think you can use induction here, but you can get an intuitive idea about the general expression. Apr 8, 2016 at 4:08
• That's essentially the companion matrix. The induction step that works is expansion by minors along the first column. Apr 8, 2016 at 4:29

Let $v= \begin{pmatrix} 1 \\ x \\ x^{2} \\ \cdot \\ \cdot \\ \cdot \\ x^{n-2} \\x^{n-1} \end{pmatrix}$.
Then $Bv = (x-X)v \iff a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} -x^{n} = p(x) =0$. Thus, all the roots $\alpha$ of the monic polynomial $p(x)$ of degree $n$ noted here give us our eigenvectors $v_{\alpha}$, which are linearly independent since they are columns of a Vandermonde matrix. The associated eigenvalues are $\lambda = \alpha - X$.
• Thanks! Since the characteristic polynomial consists the trace and the determinant. So the detminant of this Vandermonde matrix is $a_0$. Since B is similar to the matrix, so det(B) = $a_0$. Am I right? Apr 8, 2016 at 11:42
• You are not right. You've already conjectured correctly what the determinant of $B$ should be: a polynomial in $X$. For example in your case for $n=2$, you found out that $\det B = X^{2} - a_{1} X - a_{0}$. In my answer I found linearly independent eigenvectors with associated eigenvalues. Determinant is the product of eigenvalues. Let $n = 2$ and $\alpha_{1}, \alpha_{2}$ are the roots of $p(x) = x^{2} - a_{1}x - a_{0}$, then by my answer above $\det B = (\alpha_{1} - X)(\alpha_{2} -X) = X^{2} - (\alpha_{1} +\alpha_{2})X + \alpha_{1} \alpha_{2}$ - roots and coefficients are related (how?) Apr 8, 2016 at 13:14
• Note that this proof requires a bit of elementary algebraic geometry in order to work. Per se, the roots of the polynomial $p$ may fail to be distinct, in which case you do not get $n$ linearly independent eigenvectors. So you need to first WLOG assume that the resultant of the polynomial $p$ and its derivative $p'$ is $\neq 0$ (this is true on a Zariski-dense subset of the space of all polynomials, so you can WLOG assume it). Apr 11, 2019 at 18:55
To add a final touch somewhat as a synthesis of the (thorough) answers of @user5713492 and @akech: the global result is that the companion matrix of polynomial p(X) is diagonalized with a Vandermonde matrix V(r_1,r_2,\cdots r_n) where the $r_k$ are the roots of p(X)" see https://en.wikipedia.org/wiki/Vandermonde_matrix


# Find the length x such that the two distances in the triangle are the same

I have been working on the following problem

## Statement

Assume you have a right angle triangle $$\Delta ABC$$ with cateti $$a$$, $$b$$ and hypotenuse $$c = \sqrt{a^2 + b^2}$$. Find or construct a point $$D$$ on the hypothenuse such that the distance $$|CD| = |DE|$$, where $$E$$ is positioned on $$AB$$ in such a way that $$DE\parallel BC$$ ($$DE$$ is parallel to $$BC$$).

## Background

My background for wanting such a distance is that I want to create a semicircle from C onto the line $$AB$$. This can be made clearer in the image below

To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem

## Solution

Using similar triangles one arrives at the three equations

\begin{align*} \frac{\color{blue}{\text{blue}}}{a - x} & = \frac{b}{a} \\ \frac{\color{red}{\text{red}}}{x} & = \frac{c}{a} \\ \color{red}{\text{red}} & = \color{blue}{\text{blue}} \end{align*}

Where one easily can solve for $$\color{blue}{\text{blue}}$$, $$\color{red}{\text{red}}$$, $$x$$.

## Question

I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $$D$$ in a simpler matter?

• You still have not described what $F$ is either from the statement or from the graph. – Hw Chu Apr 17 at 16:42
• Right when I said $F$ i meant $E$. I will fix it in the problem statement =) – N3buchadnezzar Apr 17 at 16:47
• cateti is Italian for legs – J. W. Tanner Apr 17 at 16:56
• Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function! – David K Apr 17 at 20:25

Let the angle bisector of $$\angle ACB$$ intersect side $$\overline{AB}$$ at point $$E$$.

Let the measure of $$\angle ACB$$ be $$\alpha$$. Then $$m \angle ACE = \dfrac{\alpha}{2}$$.

Let the line perpendicular to side $$\overline{AB}$$ at point $$E$$ intersect side $$\overline{AC}$$ at point $$D$$.

Since $$\overline{ED}$$ is parallel to $$\overline{BC}$$, then $$\angle ADE \cong \angle ACB$$.

By the exterior angle theorem, $$m\angle DEC = \dfrac{\alpha}{2}$$.

Hence $$\triangle EDC$$ is isoceles.

So $$CD = DE$$.

(Added later). Assuming that the sides have lengths of $$x$$ and $$y$$, and that $$r = \sqrt{x^2+y^2}$$, the lengths of the segments are displayed below.

• This is exactly what I was looking for =) – N3buchadnezzar Apr 17 at 18:24

The curve that traces out the points that are equidistant to $$C$$ and the line extension of $$AB$$ is a parabola with focus $$C$$ and directrix $$AB$$.

Choosing a convenient coordinate system (parallel to $$AB$$ with $$B$$ as the origin), I get the equation $$y = \frac{x^2}{2b} + \frac{b}{2}$$, which you want to intersect with the line $$y = \frac{b}{a}x + b$$.

Solving, I get that the point $$D$$ has $$(x,y)$$ coordinates of $$x = \frac{b(b - c)}{a}$$ and $$y = \frac{bc(c - b)}{a^2}$$.

Let $$CD=DE=y$$ then we get $$\frac{b}{c}=\frac{y}{c-y}$$ so $$y=\frac{bc}{b+c}$$

• @mathmandan There was an edit after my comment. – Michael Biro Apr 17 at 21:17

Not sure if this is less barbaric but using simple trig: $$DE=(a-x)\tan A$$, $$DC=\frac{x}{\cos A}$$ so the equation to solve is $$(a-x)\frac{b}{a}=\frac{x\sqrt{a^2+b^2}}{a}$$ or $$x=\frac{ab}{\sqrt{a^2+b^2}+b}$$

Just another idea to construct point $$E$$: since $$\triangle{DCE}$$ is isosceles, it's easy to find $$\angle{ACE}=(90°-A)/2$$


## SectionInverse Functions

Recall that in Brief Intro to Composite and Inverse Functions we gave the following definition of an inverse function:

###### Inverse Functions

Suppose the inverse of $f$ is a function, denoted by $f^{-1}\text{.}$ Then

\begin{equation*} f^{-1}(y) = x \text{ if and only if }f(x) = y. \end{equation*}

We also stated the following property about inverse functions:

###### Functions and Inverse Functions

Suppose $f^{-1}$ is the inverse function for $f\text{.}$ Then

\begin{equation*} f^{-1}\left(f(x)\right) = x\text{ and }f\left(f^{-1}( y)\right) = y \end{equation*}

as long as $x$ is in the domain of $f\text{,}$ and $y$ is in the domain of $f^{-1}\text{.}$

In this section we will explore the invertability of a function. In other words, by the end of this section we will be able to test if a function is invertable.

### SubsectionGraph of the Inverse Function

In Example161, we used a graph of $h$ to read values of $h^{-1}\text{.}$ But we can also plot the graph of $h^{-1}$ itself. Because $C$ is the input variable for $h^{-1}\text{,}$ we plot $C$ on the horizontal axis and $F$ on the vertical axis. To find some points on the graph of $h^{-1}\text{,}$ we interchange the coordinates of points on the graph of $h\text{.}$ The graph of $h^{-1}$ is shown in Figure192.

 $C=h(F)$ $F$ $C$ $14$ $-10$ $32$ $0$ $50$ $10$ $68$ $20$
 $F=h^{-1}(C)$ $C$ $F$ $-10$ $14$ $0$ $32$ $10$ $50$ $20$ $68$

###### Example193

The Park Service introduced a flock of $12$ endangered pheasant into a wildlife preserve. After $t$ years, the population of the flock was given by

\begin{equation*} P = f (t) = 12 + 2t^3. \end{equation*}
1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $t = f^{-1}(P)\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.
Solution
1. The graph of $f$ is shown in Figure194, with $t$ on the horizontal axis and $P$ on the vertical axis.

2. We solve $P = 12 + 2t^3$ for $t$ in terms of $P\text{.}$ \begin{align*} 2t^3 \amp = P - 12\amp\amp\text{Substract 12 from both sides.}\\ t^3 \amp = \frac{P - 12}{2}\amp\amp\text{Divide both sides by 2.}\\ t \amp = \sqrt[3]{\frac{P - 12}{2}}\amp\amp\text{Take cube roots.} \end{align*} The inverse function is $t = f^{-1}(P) =\sqrt[3]{\dfrac{P - 12}{2}}\text{.}$ It tells us the number of years it takes for the pheasant population to grow to size $P\text{.}$
3. The graph of $f^{-1}$ is shown in Figure195, with $P$ on the horizontal axis and $t$ on the vertical axis.

The formula $T = f(L) = 2\pi \sqrt{\dfrac{L}{32}}$ gives the period in seconds, $T\text{,}$ of a pendulum as a function of its length in feet, $L\text{.}$

1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $L = f^{-1}(T )\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.

### SubsectionWhen Is the Inverse a Function?

We can always find the inverse of a function $y=f(x)$ simply by solving for $x$ thus interchanging the role of the input and output variables. In the preceding examples, this process created a new function. However, this process does not always lead to be a function.

For example, to find the inverse of $y = f (x) = x^2\text{,}$ we solve for $x$ to get $x = \pm\sqrt{y}\text{.}$ When we regard $y$ as the input and $x$ as the output, the relationship does not describe a function. This can be seen as plugging in $y=4 \text{,}$ for example, gives two outputs $x=2$ and $x=-2\text{.}$ The graphs of $f$ and its inverse are shown in Figure197. (Note that for the graph of the inverse, we plot $y$ on the horizontal axis and $x$ on the vertical axis.) Because the graph of the inverse does not pass the vertical line test, it is not a function.

For many applications, it is important to know whether or not the inverse of $f$ is a function. This can be determined from the graph of $f\text{.}$ When we interchange the roles of the input and output variables, horizontal lines of the form $y = k$ become vertical lines.

Thus, if the graph of the inverse is going to pass the vertical line test, the graph of the original function must pass the horizontal line test, namely, that no horizontal line should intersect the graph in more than one point. Notice that the graph of $f(x) = x^2$ does not pass the horizontal line test, so we would not expect its inverse to be a function.

###### Horizontal Line Test

If no horizontal line intersects the graph of a function more than once, then its inverse is also a function.

We have been talking about how to tell if the inverse of a function is also a function, but in practice this is not the language typicaly used. Usually we ask this same question in the form "Is the function invertible?" The following definition explains this relationship:

###### Invertible Function

If $y=f(x)$ is a function such that its inverse, $x=f^{-1}(y)\text{,}$ is also a function then we say that $f(x)$ is an invertible function.

###### Example200

Which of the functions in Figure201 are invertible?

Solution

In each case, apply the horizontal line test to determine whether the function is invertible. Because no horizontal line intersects their graphs more than once, the functions pictured in Figures201(a) and (c) are invertible. The functions in Figures201(b) and (d) are not invertibe.

### SubsectionMathematical Properties of the Inverse Function

The inverse function $f^{-1}$ undoes the effect of the function $f\text{.}$ In Example152, the function $f(t) = 6 + 2t$ multiplies the input by $2$ and then adds $6$ to the result. The inverse function $f^{-1}(H) = \dfrac{H -6}{2}$ undoes those operations in reverse order: It subtracts $6$ from the input and then divides the result by $2\text{.}$

If we apply the function $f$ to a given input value and then apply the function $f^{-1}$ to the output from $f\text{,}$ the end result will be the original input value. For example, if we choose $t = 5$ as an input value, we find that \begin{align*} f(\alert{5})\amp= 6 + 2(\alert{5}) = \blert{16}\amp\amp\text{ Multiply by 2, then add 6.}\\ \text{and } f^{-1}(\blert{16}) \amp = \frac{\blert{16} - 6}{2} = \alert{5}.\amp\amp\text{Subtract 6, then divide by 2.} \end{align*}

We return to the original input value, $5\text{,}$ as illustrated in Figure202.

Example203 illustrates the fact that if $f^{-1}$ is the inverse function for $f\text{,}$ then $f$ is also the inverse function for $f^{-1}\text{.}$

###### Example203

Consider the function $f(x) = x^3 + 2$ and its inverse, $f^{-1}(y) = \sqrt[3]{y - 2}\text{.}$

1. Show that the inverse function undoes the effect of $f$ on $x = 2\text{.}$
2. Show that $f$ undoes the effect of the inverse function on $y = -25\text{.}$
Solution
1. First evaluate the function $f$ for $x = 2\text{:}$
Then evaluate the inverse function $f^{-1}$ at $y = 10\text{:}$
\begin{equation*} f^{-1}(\blert{10}) = \sqrt[3]{\blert{10} - 2} = \sqrt[3]{8}= \alert{2}. \end{equation*}
We started and ended with $2\text{.}$
2. First evaluate the function $f^{-1}$ for $y = -25\text{:}$
\begin{equation*} f^{-1}(\alert{-25}) = \sqrt[3]{-25 - 2} = \blert{-3}. \end{equation*}
Then evaluate the function $f$ for $x = -3\text{:}$
\begin{equation*} f (\blert{-3}) = (\blert{-3})^3 + 2 = \alert{-25}. \end{equation*}
We started and ended with $-25\text{.}$
###### Example204
1. Find a formula for the inverse of the function $f(x)=\dfrac{2}{x - 1}\text{.}$
2. Show that $f^{-1}$ undoes the effect of $f$ on $x = 3\text{.}$
3. Show that $f$ undoes the effect of $f^{-1}$ on $y = -2\text{.}$
Solution
1. To find the inverse, we solve for $x\text{:}$

\begin{equation*} \begin{aligned} y \amp= \frac{2}{x-1} \\ y(x-1) \amp= 2 \\ x-1 \amp= \frac{2}{y} \\ x \amp =\frac{2}{y}+1 \end{aligned} \end{equation*}

Therefore $f^{-1}(y)=\frac{2}{y}+1\text{.}$

2. First evaluate the function $f$ for $x=3\text{:}$

Then evaluate the inverse function $f^{-1}$ at $y=1\text{:}$

We started and ended with $3\text{.}$

3. First evaluate the inverse function $f^{-1}$ for $y=-2\text{:}$

Then evaluate the original function $f$ at $x=0\text{:}$

We started and ended with $-2\text{.}$

### SubsectionSymmetry

So far we have been careful to keep track of the input and output variables when we work with inverse functions. This is important when we are dealing with applications; the names of the variables are usually chosen because they have a meaning in the context of the application, and it would be confusing to change them.

However, we can also study inverse functions purely as mathematical objects. There is a relationship between the graph of a function and the graph of its inverse that is easier to see if we plot them both on the same set of axes.

A graph does not change if we change the names of the variables, so we can let $x$ represent the input for both functions, and let $y$ represent the output. Consider the function $C = h(F)$ from Example161, and its inverse function, $F = h^{-1}(C)\text{.}$ The formulas for these functions are \begin{align*} C \amp = h(F) = \frac{5}{9}(F - 32)\\ F \amp = h^{-1}(C) = 32 + \frac{9}{5}C. \end{align*} But their graphs are the same if we write them as \begin{align*} y \amp = h(x) =\frac{5}{9}(x - 32)\\ y \amp= h^{-1}(x) = 32 + \frac{9}{5}x. \end{align*} The graphs are shown in Figure205.

Now, for every point $(a, b)$ on the graph of $f\text{,}$ the point $(b, a)$ is on the graph of the inverse function. Observe in Figure205 that the points $(a, b)$ and $(b, a)$ are always located symmetrically across the line $y=x\text{.}$ The graphs are symmetric about the line $y=x$, which means that if we were to place a mirror along the line $y=x\text{,}$ each graph would be the reflection of the other.

###### Example206

Graph the function $f (x) = 2\sqrt{x + 4}$ on the domain $[-4, 12]\text{.}$ Graph its inverse function $f^{-1}$ on the same grid.

Solution

The graph of $f$ has the same shape as the graph of $y = \sqrt{x}\text{,}$ shifted $4$ units to the left and stretched vertically by a factor of $2\text{.}$ Figure207a shows the graph of $f\text{,}$ along with a table of values. By interchanging the rows of the table, we obtain points on the graph of the inverse function, shown in Figure207b.

If we use $x$ as the input variable for both functions, and $y$ as the output, we can graph $f$ and $f^{-1}$ on the same grid, as shown in Figure208. The two graphs are symmetric about the line $y = x\text{.}$

Graph the function $f (x) = x^3 + 2$ and its inverse $f^{-1}(x) = \sqrt[3]{x - 2}$ on the same set of axes, along with the line $y = x\text{.}$


# Evaluate $\int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}$.

I need to solve the following integral: $$I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.$$ Wolfram Alpha gives the answer as $\dfrac{\pi}{2 \sqrt{2}}$.

I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?

• The integrand has a closed-form antiderivative. Nov 13 '15 at 2:55
• You don't need to use complex analysis. Nov 13 '15 at 3:03
• if you want to use contour integration, take a keyhole around with a slit at the line $(1,\infty)$ Nov 13 '15 at 10:30

Let $u=\sqrt{x-1}$, $du=\frac{1}{2\sqrt{x-1}}$, then

$$I=2\int_0^\infty\frac{u^2}{(u^2+2)^2}du$$

We can apply partial fractions here.

$$=2\int_0^\infty\left(\frac{1}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$

$$=2\int_0^\infty\frac{1}{u^2+2}du-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$

The first integrand is almost $\tan^{-1}$, so we can factor the 2 and apply the substitution $s=\frac{u}{\sqrt{2}}$.

$$I=\sqrt 2\int_0^\infty\frac{1}{s^2+1}ds-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$

$$=\frac{\pi}{\sqrt 2}-4\int_0^\infty\frac{1}{(u^2+2)^2}du$$

Now to tackle the second integral we can use a trig sub.

Let $u=\sqrt{2}\tan (t)$, $du=\sqrt 2 \sec^2(t)dt$.

$$I=\frac{\pi}{\sqrt 2}-4\sqrt 2 \int_0^{\pi/2}\frac{\sec ^2 t}{4\sec^4t}dt$$

$$=\frac{\pi}{\sqrt 2}-\sqrt 2\int _0^{\pi/2}\cos^2(t) dt$$

$$=\frac{\pi}{\sqrt 2}-\sqrt 2 \int_0^{\pi/2}\left(\frac{1}{2}\cos(2t)+\frac{1}{2} \right)dt$$

If we substitute $v=2t$ and split the integral up we get

$$I=\frac{\pi}{\sqrt 2}-\frac{1}{2\sqrt 2} \int_0^\pi \cos(v)dv-\frac{1}{\sqrt 2}\int_0^{\pi/2}1dt$$

The first integral clearly goes to 0 and the second integral becomes $\frac{\pi}{2\sqrt 2}$.

Therefore

$$I=\frac{\pi}{2\sqrt 2}$$

• There's a much easier way to find the $u$ integral. Using IBP gives $$\int u\cdot\frac{2u}{(u^2+2)^2}du = -u\cdot\frac{1}{u^2+2} + \int \frac{1}{u^2+2}\,du = -\frac{u}{u^2+2} + \arctan u$$ Nov 18 '15 at 5:51

Using a keyhole countour with the origin of the keyhole at $z=1$ and the small circle enclosing the value $z=1$ and using $$f(z) = \frac{\exp(1/2\log(z-1))}{(1+z)^2}$$ with the branch cut of the logarithm on the positive real axis and the argument from $0$ to $2\pi$ we get for the integral $$I=\int_1^\infty \frac{\sqrt{x-1}}{(1+x)^2} dx$$ that $$I(1-\exp(\pi i)) = 2\pi i\mathrm{Res}_{z=-1} f(z)$$ or $$I = \pi i\mathrm{Res}_{z=-1} f(z).$$

Now the logarithmic term is certainly analytic in a neighborhood of $z=-1$ and we have $$\mathrm{Res}_{z=-1} f(z) = \left.\left(\exp(1/2\log(z-1))\right)'\right|_{z=-1} = \left. \left(\exp(1/2\log(z-1))\right) \frac{1}{2}\frac{1}{z-1}\right|_{z=-1} \\= \exp(1/2(\log 2 + \pi i)) \times -\frac{1}{2}\frac{1}{2} = -\frac{1}{4} \sqrt{2} i.$$

This yields $$I = -\frac{1}{4} \sqrt{2} i \times\pi i=\frac{\sqrt{2}\pi}{4}.$$

Remark. The estimates for the circular components are done using ML same as at this MSE link. We get for the large circle parameterized by $z=R e^{it}$ $$\lim_{R\rightarrow \infty} 2\pi R \frac{\sqrt{R+1}}{(R-1)^2} = 0.$$

The small circle is a parameterized with $z=1+\epsilon e^{it}$ and we get $$\lim_{\epsilon\rightarrow 0} 2\pi\epsilon \frac{\sqrt{\epsilon}}{4} = 0.$$

$$I=\int_{1}^{\infty}\frac{\sqrt{x-1}}{(x+1)^2}dx$$ Use substitution $\frac{1}{x+1}=t$ which implies $\frac{dx}{(x+1)^2}=-dt$ So

$$I=-\int_{0.5}^{0}\sqrt{\frac{1}{t}-2}\:dt=\int_{0}^{0.5}\frac{\sqrt{1-2t}}{\sqrt{t}}dt$$

Again use substitution $\sqrt{t}=y$ which implies $\frac{dt}{\sqrt{t}}=2dy$

$$I=2\int_{0}^{\sqrt{0.5}}\sqrt{1-2y^2}dy=2\sqrt{2}\int_{0}^{\sqrt{0.5}}\sqrt{(\sqrt{0.5})^2-t^2}$$

Use standard integral $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})$$

we get $$I=\frac{\pi}{2\sqrt{2}}$$

Hint Substitution $x=t^2+1$...

This is the general antiderivative. Just take the limits and you're good. $$\int{\frac{\sqrt{x-1}}{(x+1)^2}dx}$$ Substitute $u = \sqrt{x-1}$ $$= \int{\frac{u}{(u^2+2)^2}du}$$ $$= 2\int\left(\frac{u}{u^2+2}-\frac{2}{(u^2+2)^2}\right)du$$ $$= \int\frac{u}{\frac{u^2}{2}+1}du-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}$ $$= \sqrt{2}\int\frac{1}{s^2+1}ds-4\int\frac{2}{(u^2+2)^2}du$$ $$= \sqrt{2}\arctan(s)-4\int\frac{2}{(u^2+2)^2}du$$ Substitute $u = \sqrt{2}\tan (p)\quad$ and $\quad du = \sqrt{2}\sec^2(p) dp\quad$ and $\quad(u^2 + 2)^2 = (2\tan^2(p)+2)^2 = 4\sec^4(p)\quad$ and $\quad p=\arctan\frac{u}{\sqrt{2}}$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\cos^2(p)du$$ $$= \sqrt{2}\arctan(s)-\sqrt{2}\int\left(\frac{1}{2} \cos(2p) + \frac{1}{2}\right)du$$ $$= \sqrt{2}\arctan(s)-\frac{1}{\sqrt{2}}\int\cos(2p) - \frac{1}{\sqrt{2}}\int du$$ $$= \sqrt{2}\arctan(s)-\frac{p}{\sqrt{2}} - \frac{\sin(p)\cos(p)}{\sqrt{2}}$$ $$= \sqrt{2}\arctan(s)-\frac{\arctan\frac{u}{\sqrt{2}}}{\sqrt{2}} - \frac{\sin(\arctan\frac{u}{\sqrt{2}})\cos(\arctan\frac{u}{\sqrt{2}})}{\sqrt{2}}$$ Note that $\cos(\arctan(z)) = \frac{1}{\sqrt{z^2 + 1}}$ and $\sin(\arctan(z)) = \frac{z}{\sqrt{z^2+1}}$ $$= \frac{2\sqrt{2}(u^2+2)\arctan(s)+\sqrt{2}(u^2+2)\arctan\frac{u}{\sqrt{2}}+2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(u^2+2)\arctan \frac{u}{\sqrt{2}}-2u}{2(u^2+2)}$$ $$=\frac{\sqrt{2}(x+1)\arctan \frac{\sqrt{x-1}}{\sqrt{2}}-2\sqrt{x-1}}{2(x+1)}$$ $$=\frac{\arctan \frac{\sqrt{x-1}}{\sqrt{2}}}{\sqrt{2}}-\frac{\sqrt{x-1}}{x+1}$$

I think I have a different answer.

Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then

\begin{align} \int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx &= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\ &= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du \end{align} Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing. \begin{align} \phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx} &= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\ &= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\ &= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\ &= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\ &= \frac{\pi}{2\sqrt{2}}. \end{align}

Original image of work by hand.

• Honestly, with 1500 points I expect you know the customs here, and I see many answers where you used LaTeX. You could make an effort and write your answer correctly. Nov 14 '15 at 0:48


# [SOLVED]a linear equation in 3 variables

#### karush

##### Well-known member
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$

#### chisigma

##### Well-known member
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$
You have the linear system...

$\displaystyle a + b + c = d$

$\displaystyle 2 b = d$

$\displaystyle a = d\ (1)$

... the solution of which is $a=d,\ b= \frac{d}{2},\ c=- \frac{d}{2}$, so that You can write...

$\displaystyle a x + \frac{d}{2} y - \frac{d}{2} z = d \implies x + \frac{y}{2} - \frac{z}{2} = 1\ (2)$

Kind regards

$\chi$ $\sigma$

#### HallsofIvy

##### Well-known member
MHB Math Helper
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$
I would have done this just a little differently from chisigma. First, note that there is no "unique" solution. for any ax+ by+ cz= d describing the line, you could multiply through by any number and get a new equation for the same line. You could. for example, divide both sides by d to get (a/d)x+ (b/d)y+ (c/d)z= 1, then let a'= a/d, b'= b/d, and c'= c/d so that the equation is a'x+ b'y+ c'z= 1, with only three unknown values.

I suspect that the answer you are asked for is the one that has all integer coefficients with no common factor.

Now, from ax+ by+ cz= d, (1, 1, 1) on the line means we must have a+ b+ c= d. (0, 2, 0) on the line means 0a+ 2b+ 0c= 2b= d. (1, 0, 0) on the line means 1a+ 0b+ 0c= a= d. Since both 2b and a are equal to d, a= 2b and we can replace both a and d with 2b in the first equation: 2b+ b+ c= 2b so that b+ c= 0 or b= -c.

Now we can write everything in terms of b: a= 2b, c= -b, d= 2b, and so the equation is
2bx+ by- bz= 2b. Here, b can be any (non-zero) number and still give an equation describing the line but dividing both sides of the equation by b gives
2x+ y- z= 2, the simplest form for the equation as all numbers are integers and they do not all have a common factor so cannot be reduced.

#### Prove It

##### Well-known member
MHB Math Helper
write a linear equation in 3 variables that is satisfied by all 3 of the given ordered triples

$(1,1,1), (0,2,0), (1,0,0)$

the examples in book are all on the $ax+by+cz=d$ equation but with just ordered triples there is no $d$ or can it be found from them... otherwise I would solve this by simultaneously.

the answer to this is $2x+y-z=2$
There will be a unique plane which satisfies these three points simultaneously. The plane will have the same coefficients as its normal vector, and the normal vector will also be normal to any vectors in the plane. So if we can get two vectors that lie in the plane, taking their cross product will give the normal vector. Call two of these vectors \displaystyle \begin{align*} \mathbf{a} \end{align*} and \displaystyle \begin{align*} \mathbf{b} \end{align*}. We could have

\displaystyle \begin{align*} \mathbf{a} &= ( 1 - 0, 1 - 2, 1 - 0) \\ &= ( 1 , -1, 1 ) \\ \\ \mathbf{b} &= (1 - 0, 0 - 2, 0 - 0 ) \\ &= ( 1 , -2, 0) \end{align*}

\displaystyle \begin{align*} \mathbf{n} &= \mathbf{a} \times \mathbf{b} \\ &= \left| \begin{matrix} \mathbf{i} & \phantom{-}\mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -2 & 0 \end{matrix} \right| \\ &= \mathbf{i} \, \left| \begin{matrix} -1 & 1 \\ -2 & 0 \end{matrix} \right| - \mathbf{j} \, \left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| + \mathbf{k} \, \left| \begin{matrix} 1 & -1 \\ 1 & -2 \end{matrix} \right| \\ &= \mathbf{i} \, \left[ -1 \cdot 0 - 1 \cdot (-2) \right] - \mathbf{j} \, \left( 1 \cdot 0 - 1 \cdot 1 \right) + \mathbf{k} \, \left[ 1 \cdot (-2) - (-1) \cdot 1 \right] \\ &= 2\,\mathbf{i} + \mathbf{j} - \mathbf{k} \end{align*}

The plane will have the same coefficients as the normal vector, so that gives \displaystyle \begin{align*} 2x + y - z = d \end{align*}. Since we have three points that lie on the plane, any of them can be substituted to find \displaystyle \begin{align*} d \end{align*}. If we substitute \displaystyle \begin{align*} (1, 0, 0) \end{align*} that gives

\displaystyle \begin{align*} 2 \cdot 1 + 0 - 0 &= d \\ 2 &= d \end{align*}

and thus the plane is \displaystyle \begin{align*} 2x + y - z = 2 \end{align*}.


I'm having an issue plotting torus links in Mathematica. What I'm trying to generate is a (2,8) torus link like the one in the picture in Wikipedia:

using equations such as those outlined here.

However, I appear to only get half of the respective torus link.

a = 1; d = 4; p = 2; q = 8;
ParametricPlot3D[{(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*
Cos[p*t], a*Cos[q*t]}, {t, 0, 2*Pi}, PlotStyle -> Orange,
PlotRange -> All] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]

(a and d are parameters that control the width of the tube and the distance of the curve from the origin respectively, outlined here.)

Can anyone explain why I am only seeing half of the link?

• You're only plotting one set of parametric equations and you can see in the image that you don't expect there to be any intersections so the smooth parametric equations clearly can't generate the structure you want. On the other hand, it ought to be clear that the others may be generated via a simple rotation in the x-y plane. – b3m2a1 Aug 2 '18 at 8:09

The second torus is created by a rotation of $$\pi/4$$ (around {0, 0, 1}):

R = RotationMatrix[Pi/4, {0, 0, 1}];
torus = {(a*Sin[q*t] + d)*Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]};
ParametricPlot3D[{torus, R.torus} // Evaluate, {t, 0, 2*Pi}, PlotStyle -> {Orange, Blue}, PlotRange -> All]

Here's I think a generalization of what Ulrich gave to an arbitrary $(p, q)$ torus (if I read Wikipedia right):

plotPQTorus[{p_, q_}, a : _?NumericQ : 1, d : _?NumericQ : 4,
ops : OptionsPattern[]] :=
Block[{t},
ParametricPlot3D[
Evaluate@
Table[
RotationMatrix[
i*2 \[Pi]/q, {0, 0, 1}].{(a*Sin[q*t] + d)*
Sin[p*t], (a*Sin[q*t] + d)*Cos[p*t], a*Cos[q*t]},
{i, 0, If[Divisible[q, p], p - 1, 0]}
],
{t, 0, 2*Pi},
PlotRange -> All,
ops
] /. Line[pts_, rest___] :> Tube[pts, 0.2, rest]
]

Here are a few plots:

Table[plotPQTorus[{p, Fibonacci[q]}, Boxed -> False,
Axes -> None], {p, 1, 4}, {q, 2, 6, 2}] // Grid

Note that relatively prime things are single connected loops (as Wikipedia suggests they should be)

The (1,4) torus knot is known to KnotData[], so we can use that to build the (2,8) knot:

knot14 = First[KnotData[{"TorusKnot", {1, 4}}, "ImageData"]];


# Logic problem

##### Active member
Consider the following sequence of statements:
$$S_1: \text{at least 1 of the statements }S_1-S_n \text{ is false}\\ S_2: \text{at least 2 of the statements }S_1-S_n \text{ are false}\\ \vdots \\ S_n: \text{at least } n \text{ of the statements }S_1-S_n \text{ are false}$$
Where $n$ is some integer.

Question: for which $n$ are these statements self-consistent? In those cases: what is the truth value of each statement?

I got this off of a blog I tend to frequent. I will wait before posting the solution this time.

EDIT:
Changed the question; I had written the statements wrong

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Suppose $k$ out of $n$ statements are true.
Then $S_1$ up to $S_k$ have to be true and the rest has to be false.
This appears to be consistent for any $n$ and any $0\le k \le n$.

##### Active member
Suppose $k$ out of $n$ statements are true.
Then $S_1$ up to $S_k$ have to be true and the rest has to be false.
This appears to be consistent for any $n$ and any $0\le k \le n$.
Sorry about that, you were absolutely right about the question as phrased.

However, this new version should prove to be a bit more interesting. This is what I had meant; I had accidentally written "true" instead of "false".

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If $S_n$ is true, then $n$ statements are false, including $S_n$.
Therefore $S_n$ is false.

We now know that at least $1$ statement is false.
Therefore $S_1$ is true.
For $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.

For $n \ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.
Since $S_1$ is true, this implies that $S_{n-1}$ is false.
Therefore $S_{n-1}$ is false.

So at least $2$ statements are false.
Therefore $S_2$ is true.
For $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.

Etcetera.

In other words, we get a consistent consistent solution if and only if $n$ is even.
In that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\qquad \blacksquare$

##### Active member
If $S_n$ is true, then $n$ statements are false, including $S_n$.
Therefore $S_n$ is false.

We now know that at least $1$ statement is false.
Therefore $S_1$ is true.
For $n=1$ this is a contradiction, and for $n=2$ this is a consistent solution.

For $n \ge 3$ we can say, that if $S_{n-1}$ were true, then $n-1$ statements are false.
Since $S_1$ is true, this implies that $S_{n-1}$ is false.
Therefore $S_{n-1}$ is false.

So at least $2$ statements are false.
Therefore $S_2$ is true.
For $n=3$ this is a contradiction, and for $n=4$ this is a consistent solution.

Etcetera.

In other words, we get a consistent consistent solution if and only if $n$ is even.
In that case $S_1$ up to $S_{n/2}$ are true and $S_{n/2+1}$ up to $S_{n}$ are false. $\qquad \blacksquare$
Couldn't have phrased it better myself.

The source, for anybody interested:
The Parity Paradox – Futility Closet

I highly recommend the website as a time-wasting tool.


# Thread: Is it a field? set of real numbers in the form...

1. ## Is it a field? set of real numbers in the form...

Hi,

problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.

(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?

attempt:

(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?

(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..

Thanks

2. Originally Posted by Mollier
Hi,

problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.

(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?

attempt:

(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?

(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..

Thanks
just go through each property for a field and make sure that they all hold. you have to verify these explicitly.

and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?

3. Originally Posted by Mollier
Hi,

problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.

(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?

attempt:

(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?

(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..

Thanks
Firstly, if every real number was of the form $\alpha+\beta \sqrt{2}$ with $\alpha, \beta \in \mathbb{Q}$ then we can easily define a bijection $\mathbb{R} \rightarrow \mathbb{Q} \times \mathbb{Q}$, $\alpha + \beta \sqrt{2} \mapsto (\alpha, \beta)$. As there exists a bijection between $\mathbb{Q}$ and $\mathbb{Q} \times \mathbb{Q}$ we have that $|\mathbb{R}| = |\mathbb{Q}|$, a contradiction.

Secondly, in both of your problems you know that "most" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.

What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?

What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)

4. Originally Posted by Jhevon
and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?
Thanks

Originally Posted by Swlabr
The one to concentrate on will be multiplicative inverses.

What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?

What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)
It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).

Thanks guys!

5. Originally Posted by Mollier
Thanks

It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).

Thanks guys!
You want to try and write $\frac{1}{\alpha + \beta \sqrt{2}}$ as $\alpha^{\prime} + \beta^{\prime}\sqrt{2}$. I mean, $\sqrt{2} = \frac{2}{\sqrt{2}}$...numbers are strange things...just because something isn't obviously in a certain form does not mean you cannot write it in that form.

I shall give you a tantalising hint: It is a field.

6. $
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$

$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$

$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$

Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?

Thanks mate!

7. Originally Posted by Mollier
$
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$

$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$

$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$

Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?

Thanks mate!
Yeah, that's it! Although to be thorough you would need to make sure there is no division by 0 going on.

8. So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..

As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?

Thank you very much.

9. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..

As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?

Thank you very much.
For part (b) you need to find a counter-example - $\alpha^{\prime}$ and $\beta^{\prime}$ may look like non-integers, but this may not always be the case.

So, to complete the problem simply take an integer value of $\alpha$ and an integer value of $\beta$ such that $\alpha^{\prime} \notin \mathbb{Z}$ or $\beta^{\prime} \notin \mathbb{Z}$. For simplicity, let one of them be zero.

10. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.

11. $\alpha^2-2\beta^2\neq 0$ because of the irrationality of $\sqrt{2}$, it can be easily proved in number theory.

12. Originally Posted by Swlabr
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$

13. Originally Posted by Mollier
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$
Precisely!

Sorry to be pedantic about it, but in maths the proof is the king...

14. Please don't apologize. I have a BS in mechanical engineering from a horrible school, so I have now decided to actually learn some math. You being "pedantic" is exactly what I need! Thanks!


# Math Help - rational numbers

1. ## rational numbers

Find two rational numbers with denominators 11 and 13, respectively, and a sum of 7/143.

I got x/11 + y/13=7/143.

Two unknowns and one equation. Kind of stuck.

2. Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.

Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?

3. Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.

Laurent.

4. Originally Posted by Matt Westwood
Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.

Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm

5. Originally Posted by rmpatel5
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm

6. Originally Posted by Laurent
Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.

Laurent.
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?

7. Originally Posted by rmpatel5
gcd(11,13)=1 and there is a thm that says ax+by=1.
There is a theorem that says that there exists x and y such that 11x+13y=1. Now, you should find an example of such a couple (x,y). Either you guess it, or you make use of the Euclidean algorithm as you may have done in class.

8. Originally Posted by rmpatel5
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?
Ok, so solve for x and y in 13x+11y=1, using the Euclidean algorithm (http://en.wikipedia.org/wiki/Euclidean_algorithm). Actually, do the Euclidian algorithm over 13 and 11, and observe...

Then, multiply by 7 :
13*(7x)+11*(7y)=7

Uh.

9. Use the Euclidean Algorithm to get the gcd of 11 and 13 (yes, you know this is 1, but the calculations you did in the algorithm help you find out what values of a and b give you 11a + 13b = 1.

You now want two numbers x and y such that 11x + 13y = 7.

Well, you just got 11a + 13b = 1, so multiply everything by 7:

$11 \times 7a + 13 \times 7b = 7$

So 7a and 7b are the numbers you want for x and y. Job done.

10. You do not need to use Euclidean algorithm.
It is quite obvious that $13(6) + 11(-7) = 1$.
Thus, $13(42) + 11(-49) = 7$.

This means the solutions to $13x+11y=7$ are: $x=42 - 11t \mbox{ and }y=-49 + 42t$.

11. 6 and -5 work as well, so are there many solutions to this problem. I got them by doing the EA


Topic: Discrete-space Fourier transform computation

## Question

Compute the discrete-space Fourier transform of the following signal:

$f[m,n]= 2^{-(n+m)} u[n] u[m]$

First off notice that this equation can easily be separated into two functions $g[m]=2^{-m}u[m]$ and $h[n]=2^{-n}u[n]$ where $f[m,n]=g[m]h[n]$. Then since both equations are the same except for a change of variable where m=n or vice verse we can show the DTFT of either g[m] or h[n] and it should suffice for the other. If they were not the same we would have to evaluate both of them.

\begin{align} G[u]&=\sum_{m=-\infty}^{\infty}2^{-m}u[m]e^{-j u m} \\ &=\sum_{m=0}^{\infty}( \frac{1}{2e^{ju}} )^{m} \\ &=\frac{1}{1-\frac{1}{2e^{ju}}} \\ &=\frac{-2e^{ju}}{1-2e^{ju}} \\ \end{align}

Thus

$F[u,v]= \frac{4e^{j(u+v)}}{(1-2e^{ju})(1-2e^{jv})}$

\begin{align} F [u,v] &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f[m,n]e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} 2^{-(n+m)} u[n] u[m] e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} 2^{-m} u[m] e^{-j(mu)} \sum_{n=-\infty}^{\infty} 2^{-n} u[n] e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} 2^{-m} e^{-j(mu)} \sum_{n=0}^{\infty} 2^{-n} e^{-j(nv)}\\ &= \sum_{m= 0}^{\infty} (2e^{ju})^{-m} \sum_{n=0}^{\infty} (2e^{jv})^{-n}\\ &= \frac{1}{1-\frac{1}{2e^{ju}}}\cdot\frac{1}{1-\frac{1}{2e^{jv}}}\\ &= \frac{1}{1-\frac{1}{2}e^{-ju}}\cdot\frac{1}{1-\frac{1}{2}e^{-jv}}\\ &= \frac{1}{(1-\frac{1}{2}e^{-ju})(1-\frac{1}{2}e^{-jv})} \end{align}

--Xiao1 23:03, 19 November 2011 (UTC)

Instructor's comments: Solutions 1 and solutions 2 are the two main ways to answer the question. The second one is a bit longer to write, in my opinion, but it does not require knowing the separation property of the continuous-space Fourier transform. Note however that, in the exam, if you do not have the separation property in the table, you will need to prove it in order to get full credit. -pm