# What do we actually prove using induction theorem?

Here is the picture of the page of the book, I am reading:

$$P_k: \qquad 1+3+5+\dots+(2k-1)=k^2$$ Now we want to show that this assumption implies that $P_{k+1}$ is also a true statement: $$P_{k+1}: \qquad 1+3+5+\dots+(2k-1)+(2k+1)=(k+1)^2.$$ Since we have assumed that $P_k$ is true, we can perform operations on this equation. Note that the left side of $P_{k+1}$ is the left side of $P_k$ plus $(2k+1)$. So we start by adding $(2k+1)$ to both sides of $P_k$: \begin{align*} 1+3+\dots+(2k-1) &= k^2 &P_k\\ 1+3+\dots+(2k-1)+(2k+1) &= k^2+(2k+1) &\text{Add $(2k+1)$ to both sides.} \end{align*} Factoring the right side of this equation, we have $$1+3+\dots+(2k-1)+(2k+1) =(k+1)^2 \qquad P_{k+1}$$ But this last equation is $P_{k+1}$. Thus, we have started with $P_k$, the statement we assumed true, and performed valid operations to produce $P_{k+1}$, the statement we want to be true. In other words, we have shown that if $P_k$ is true, then $P_{k+1}$ is also true. Since both conditions in Theorem 1 are satisfied, $P_n$ is true for all positive integers $n$.

In the top line there is written that we have to show that 'the assumption $P_k$ is true implies $P_{k+1}$ is true'.

And what I think is that: as long as we know that the state of a proposition being true for any positive integer $k$ after Base number implies that proposition is true for integer $k+1$, we have to show that $P_k$ is true. However I don't have an idea yet how to show the truth of $P_k$.

So, my first question is that who is right? My book or me? And if I am right then how can I show the truth of $P_k$?

2) This may be taken as the second question but it is also bit annoying.

In the second paragraph it is written, "Since we have assumed that $P_k$ is true, we can perform operations on it".

Why is it necessary for an equation to be true for performing operations on it? Well, this is not much annoying as compared to the first question because whole the induction theory depends upon that.

• Who is right about what? You haven't made a claim that contradicts the book, Feb 8 '16 at 13:32
• I have tried to retype the text from your picture. It would be rather difficult task, considering the low quality of the image. Luckily Google and Google Books returns some places with the same text. Could you perhaps add to your post a more precise reference than "the book I am reading"? Feb 8 '16 at 16:27
• You prove the $P_1$ is true. We don't know that $P_n$ is true for all $n$. Only that it is true for $n = 1$. Then we prove that IF it is true for a specific $P_k$ then it is true for $P_{k+1}$. Once we prove these two things: 1) $P_1$ is true. and 2) $P_k \implies P_{k+1}$. Then we can conclude and we have proven 3) $P_n$ is true for all n. Your comment. If we prove it is true for all $P_k$ then it is true for $P_{k+1}$ is irrelevant as we are not assuming and certainly haven't proven it is true for all $P_k$. We have assumed it for ONE $P_k$. Feb 8 '16 at 17:25
• @MarkJoshi Induction works on many structures. Induction of the form "start at zero and add one" also works on finite fields, that is arithmetic $\pmod p$. It is not correct at all to say that natural numbers are precisely the set for which induction works. Feb 9 '16 at 6:02
• Once there is enough evidence to make you believe something is true, it can suddenly become much easier to understand why it is true. Given the multitude of experts here (and the authoer himself) telling you the book is right I recommend, as strange as it sounds, to reread the chapter on induction trying to believe it (and everyone on this site) is right and understand why that is, rather than trying to argue why it is wrong. Feb 9 '16 at 13:25

Induction is, intuitively, an outline of an infinite proof.

You first prove $P(0)$, the base case.

Then you prove $P(1)$ follows from $P(0)$.

Then you prove that $P(2)$ follows from $P(1)$.

Et cetera.

In general, if you know that $P(k+1)$ follows from $P(k)$, and you know that $P(0)$ is true, then you know how to prove $P(n)$ for any natural number. For example, $P(4)$ follow because $P(0)\implies P(1)\implies P(2)\implies P(3)\implies P(4)$.

So you don't need to know how to show $P(k)$ to show that $P(k)$ implies $P(k+1)$. Once you've shown this implication, and $P(0)$, we know how to write a non-inductive finite proof of $P(10000)$ or $P(10^{99})$.

Your book is right. You have to show that $$P_k \implies P_{k+1}$$

Because if you prove that and since you said it is true for base number, the proposition is automatically proved true for every $n \in S$.

Let us see how!

To prove "$P_k \implies P_{k+1}$" means you are proving that a particular proposition has a quality of being true for $k+1$ because of being true for $k$.

Improvement (Edit): Just as you do in case of domino effect. Suppose it is reported that the cornered domino has been pushed over. And then you find out that dominoes are arranged such that whenever $k^{th}$ domino falls, $(k+1)^{th}$ falls. You would at once say that all dominoes must fall.

Remember that these two conditions of induction theorem simultaneously imply that a certain proposition is true.

i) It is true for base number

ii) It is true for an integer $k$ implies it is true for integer $k+1$ after it.

For the proposition in your picture, if you would take base number $2$ then after doing all that (done in your book), you would prove that proposition is true for integer $2$ and integers after it. But since we usually ask to prove that conjecture for every $n \in \mathbb N$, we take base number $1$ not $2$ because the smallest number of set $\mathbb N$ is $1$.

• It may help to think that you are supposed to prove the arrow in-between the two propositions. Feb 9 '16 at 4:00

(Wanted to post as a comment, but it is too long :/)

When we prove something by induction we prove that our claim is correct for a base case (for example, n=1). Afterwards we assume (not proving, only assuming) that our claim stands for some arbitrary value k and than, based on the assumption we prove it holds for k+1.

Note that based on the proof of base case and the proof that if $P_k$ is true we show that $P_{k+1}$ is true. Example: you prove that some claim is correct for $n=1$, then you assume it is correct for some n=k. Now you prove for n=k+1.

We proved for n=1 and using the assumption that the claim stands for n=k we proved for n=k+1. In this case, we know that the claim stands for n=1, thus we can now say it stands for n=2. Afterwards we can say that it stands for n=3 and so on, hence it stands for any n.

That is how we use induction - you don't prove that $P_k$ is true, you assume it it true - that is a big difference.

If my explanation is not coherent enough, you might want to google "Induction" and read a little more.

Good luck.

• "...that our claim stands for some arbitrary $k$ and then, based on the assumption we prove that claim holds for $k+1$" . Okay.. if we prove the claim for $k+1$ . At what place in the picture, I can see such prove? Feb 8 '16 at 14:12
• All that shown is how they are assuming that $P_k$ is true and how they use it to prove that $P_{k+1}$ is true. Feb 8 '16 at 14:23
• After reading this "$P_k$ holds implies $P_{k+1}$ holds", I guess that we should show truth of $P_k$ and not of $P_{k+1}$ because it is an implication and once we prove $P_k$, then according to our claim ($P_k \implies P_{k+1}$), $P_{k+1}$ would get proved automatically. Feb 8 '16 at 14:30
• @Man_Of_Wisdom, the induction assumption is that $P_k$ is true, we don't need to prove it. We show that because it is true (take it as a fact), than $P_{k+1}$ is also true. We know that "$P_k$ holds", we want to show that $P_{k+1}$ holds. Feb 8 '16 at 14:43
• You show the truth of $P_1$. Than you show that if it's true for $P_k$ which may or may not be the same as $P_1$ then it must be true for $P_{k+1}$. From that we know that $P_1 \implies P_2$. And therefore $P_2 \implies P_3$. Now we can hire an undergraduate to sit in an office to continue "$...\implies P_{2,345}$ and $P_{2,345} \implies P_{2,346}$ and $P_{2346} \implies ...$". Or we can conclude: 1) ($P_1$ is true). and 2) ($P_k \implies P_{k+1}): \implies$ 3) $P_n$ is true for all $n$. Feb 8 '16 at 17:31

## What do we actually prove using induction theorem?

Probably an example is needed here. Let's suppose you want to prove the little Gauss. Suppose we want to prove this Theorem for all $$n\in\mathbb{N}$$:

$$1+2+3+ \cdots+ n = \sum_{k=1}^n k = \frac{n(n+1)}{2}$$

The statement $$A(n)$$ is thus given by: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ With induction this is really simple. Step one lets suppose $$n=1$$: $$1 = \sum_{k=1}^1 k = \frac{1(1+1)}{2} = \frac{n(n+1)}{2}$$ Done the statement is true for $$n=1$$. The good part about induction is we can select the number for which we want to prove it so we could also have shown with $$n=0$$. So dont try to prove it for $$n$$ beeing any number but prove it for one number you selected, almost everytime $$n=0$$ or $$n=1$$.

Now we have to prove that if the statement is true for $$n$$ it is also true for $$n+1$$. Meaning if $$A(n)$$ is true, $$A(n+1)$$ is also true:

$$\underbrace{1+ 2+ \cdots + n}_{\sum_{k=1}^n k} + n+1 = \sum_{k=1}^n k + n+1 \overset{A(n) \text{ is true}}{=} \frac{n(n+1)}{2} + n+1 = \frac{n^2+n+2n+2}{2} = \frac{n^2+3n+2}{2} = \frac{(n+1)(n+2)}{2}$$

So now we know the statement is true for $$n+1$$ or $$A(n+1)$$ is true.

### Prove without induction

Proving this statement without induction is not that simple. You can do it like this. Suppose $$n\in \mathbb{N}$$, then we can look at the sum and rearange that:

$$n$$ is even

$$1+2+\cdots + n = (1 + n) + (2+n-1) + \cdots + \left(\frac{n}{2} + \frac{n}{2} +1\right)=\frac{n}{2}(n+1) = \frac{n(n+1)}{2}$$

and for $$n$$ is odd

$$1+2+\cdots + n = (1+n) + (2+n-1) + \cdots + \left(\frac{n-1}{2} + \frac{n+1}{2} + 1\right) + \frac{n+1}{2} = (n+1)\frac{n-1}{2} + \frac{n+1}{2} = \frac{n(n+1)}{2}$$

### Conclusion

So let's conclude, we can use induction here but we can go without it too. The first though is much simpler because it uses the induction and if you get your head around it, it's easier to verify that it is true. The other one is using induction internally for the rearanging and you have to think alot if this is true or if I made a mistake.

## What does the operation part mean?

The funny thing is that induction can be generalized in a lot of things. If we look at the set $$\mathbb{N}$$ than we can define it as follows $$1$$ is in $$\mathbb{N}$$ and if $$n\in\mathbb{N}$$ than $$n+1$$ is also in $$\mathbb{N}$$, meaning lets call that $$(\star)$$:

1. $$1 \in \mathbb{N}$$
2. $$\forall n\in\mathbb{N} \Rightarrow n+1\in\mathbb{N}$$

If you look at that you will find this is looking a lot like induction. So basically you can say if you have the statement $$A(n)$$ and you want to show that it is true for all $$n\in\mathbb{N}$$ you do the following, you determine the set $$\mathfrak{A}$$ for which the statement is true. This set is given by: $$\mathfrak{A} = \{n\in\mathbb{N} | A(n) \text{ is true}\}$$ And now if we know $$1\in\mathfrak{A}$$ and for all $$n\in\mathfrak{A}$$ it follows $$n+1\in\mathfrak{A}$$ then it follows from $$(\star)$$: $$\mathfrak{A} = \mathbb{N}$$ so the statement is true for all $$n\in\mathbb{N}$$

## More generalized induction shemes

The example where this is generalized is the "principle of the good set" (this is the german name translated directly), it's used in measure theory and works with sets that are way bigger then $$\mathbb{N}$$ in fact with sets that have the cardinality of $$\mathbb{R}$$. And it works as follows. A $$\sigma$$-Algebra $$\mathcal{A}$$ of a set $$\Omega$$ fullfills the following axioms:

1. $$\Omega \in \mathcal{A}$$
2. $$A\in \mathcal{A} \Rightarrow \Omega\setminus A\in\mathcal{A}$$
3. $$A_1,A_2,\cdots \in \mathcal{A} \Rightarrow \bigcup_{k=1}^\infty A_k \in \mathcal{A}$$

Now we have a function $$\sigma$$ that generates $$\sigma$$-Algebras for a given subset $$\mathcal{G}\subset \mathscr{P}(\Omega)$$ just using the axioms 1-3 just more often than $$\mathbb{R}$$ has elements. Like $$\mathbb{N}$$ is generated by $$\{1\}$$, if we take $$1$$ and then we add $$1$$ onto it put the result $$2$$ in our set. Then we take $$2$$ and add $$1$$ onto it and put it in our set until we cannot add any more element because we have all of them already. Back to the example, $$\mathcal{G}$$ is called a generator if $$\sigma(\mathcal{G}) = \mathcal{A}$$, this also means $$\Omega \in \mathcal{G}$$. Now if we want to show a statement $$B(A)$$ is true for all $$A\in\mathcal{A}$$ we just show:

1. $$B(G)$$ is true for all $$G\in \mathcal{G}$$
2. $$B(A)$$ is true then it is also true for $$B(\Omega\setminus A)$$
3. $$B(A_1),\ B(A_2),\ \cdots$$ is true than its also true for $$B\left(\bigcup_{k=1}^\infty A_k\right)$$

So here 1. shows that we know the statement is true for at least a generating set of $$\mathcal{A}$$, like in induction we know the statement is true for $$\{1\}$$ which generates $$\mathbb{N}$$ using $$n+1$$. And then we show it for the operations 2. and 3. like we would in induction where we show it for $$n+1$$.

## Conclusion

I hope I didn't confuse you too much. But I am certain that after some time you will accept inductions and learn how to use them. And if you once return here or have to explain it to somebody else, after using it a lot of times. You will fully understand it. People often find induction counter-intuitiv, I have some ideas why but I think that would really go to far. So my hint, accept it first, use it, try to explain it und you will have understood this and similar shemes.

Intuitively you can view this as a chain reaction, because once you show that $\color{green}{P_0}$ is true, then by applying $\color{blue}{P_k \Rightarrow P_{k+1}}$ repeatedly you can reach any number $n \geq 0$ and see that $P_n$ is indeed true.

$$\color{green}{P_0} \color{blue}{\Rightarrow} P_1 \color{blue}{\Rightarrow} \dots \color{blue}{\Rightarrow} P_n \color{blue}{\Rightarrow} \dots .$$

At no point are you directly proving that $P_1$, $P_2$, $\dots$ are true, instead they follow from base case and induction step.

• What is $P_k \implies P_{k+1}$? I think my question actually asks about this. Feb 8 '16 at 14:17
• It is the implication, you are proving $P_{k+1}$ under assumption of $P_{k}$, i.e. you are assuming that $P_{k}$ is true, and you show that then $P_{k+1}$ is true. Once you prove that, you can apply this for case $k=0$ (because you know that $P_0$ is true). By that you immediately get that $P_1$ is true. And you can repeat that...
– Sil
Feb 8 '16 at 14:20
• You must combine it with the base case, in which you proved that $P_0$ is true. Both base case AND induction step then imply that $P_1$ is true. But now you have $P_1$ is true and you can use it same way with induction step, to arrive at $P_2$ being true, etc... Reaching any $n$ if you repeat the process.
– Sil
Feb 8 '16 at 14:25
• And that is exactly what you do, in case of $P_0$. You have $P_0$ and $P_0 \implies P_1$, therefore $P_1$. The trick is that you ONLY need to show this for $P_0$, everything else follows from the $P_k \implies P_{k+1}$.
– Sil
Feb 8 '16 at 14:48
• You prove $P_k$ is true when $k = 1$. You do not need to assume that $k \ne 1$ or that $P_k$ is true for any value of $k$ other than 1. But you don't want to assume that $k$ has any characteristic that pertains only to 1. You want to prove that it is true for some $k$ of which you don't know the value of, then it must also be true of $k + 1$. You know it is true for some $k$ because it is true for $k = 1$. It might also be true for $k = 7,921$ but you don't know that. You can't say I know it is true for $P_1$ so I will prove it for $P_2$ because you want something more general. Feb 8 '16 at 17:39

The part about assuming $P_k$ is an application of basic rules of propositional logic.

Suppose you have a proposition of the form $A \implies B$. One way you might be able to prove this proposition is by using the rule for Introduction of an Implication:

1. Assume that $A$ is true.
2. Using $A$ as a statement already known to be true, prove $B$.
3. "Discharge" the assumption that $A$ is true.
4. Conclude that $A \implies B$ is true.

Of course this works only if you are able to come up with a proof of $B$ in the context of Step 2. But while you are in that context, you already know $A$ is true, and you do not have to prove it.

Once you have proven $A \implies B$, if you later find out that $A$ actually is true (not just assumed to be true in the limited context of Step 2, above), you can apply a rule called Modus Ponens:

1. Given that $A \implies B$ is known to be true.
2. Given that $A$ is known to be true.
3. Conclude that $B$ is true.

None of these rules of logic has used anything from the method of induction, but the method of induction uses those rules of logic.

We prove $P_0$.

We prove $P_k \implies P_{k+1}$ for any possible value of $k$, using the rule for Introduction of an Implication.

Now we know $P_0$, and since $P_k \implies P_{k+1}$ for any $k$, including the case $k=0$, we have $P_0 \implies P_1$. Apply Modus Ponens:

1. Given that $P_0 \implies P_1$ is true.
2. Given that $P_0$ is true.
3. Conclude that $P_1$ is true.

Now we know $P_1$, and $P_k \implies P_{k+1}$ is still true in the case $k=1$, so we can apply Modus Ponens again:

1. Given that $P_1 \implies P_2$ is true.
2. Given that $P_1$ is true.
3. Conclude that $P_2$ is true.

We can repeat this process as many times as needed to show $P_3$, then $P_4$, then $P_5$.

The principle of induction merely summarizes this long, repetitive, boring process so that we can relatively quickly show $P_n$ for any given non-negative integer $n$. And since we can always do this no matter what non-negative value we choose for $n$, we conclude $P_n$ is true for all $n$.

By the way, I would write at least one of the sentences in the book a little differently. Instead of "perform operations on this equation" I would write "use algebraic manipulation of $P_k$ to prove that other equations are true". That is,

Since we have assumed that $P_k$ is true, we can use algebraic manipulation of $P_k$ to prove that other equations are true.

I hope this clears up one of the points of confusion.

• According to point no. 2 we have to prove that $P_{k+1}$ is true based on the assumption that $P_k$ is true. The author shows us that $$P_{k}+T_{k+1}=P_{k+1}$$. Is it the way to implement point no.2? And, if it iss then how? I mean in what way proving $P_{k}+T_{k+1}=P_{k+1}$ is equivalent to proving $P_{k+1}$ under the assumption?? Feb 9 '16 at 10:41
• What does $P_k + T_{p+1}$ mean? $P_k$ is an equation. Does it make sense to write $(5^2 = 25) + 3$? I would never write math that way. But do you agree that if $a$ and $b$ are numbers and $a=b$, then $a$ and $b$ are the same number, so $a + 3$ is the same as $b +3$, that is, $a+3=b+3$? That's what we mean by "add ___ to both sides of the equation." And yes, when you start with a true equation, then (the left side plus something) is equal to (the right side plus the same something), which is how the author derived $P_{k+1}$. Feb 9 '16 at 12:52
• DavidK first of all, adding something to an equation always means it is added to its both sides. In other words, it can never mean to add something to only side of equation because two sides make up an equation. So $P_k+T_{k+1}$ is not wrong anyway. It is not necessary for something to be right. When it is only "not wrong", we can adopt it. Feb 9 '16 at 18:41
• Secondly, you said that since we assumed $P_k$ is true, we add $T_{k+1}$ to its both sides and then if it becomes $P_{k+1}$, it is proved that $P_{k+1}$ is true. And to support it, you said that if something is added to a true equation such as $3^{2}=9$ then the new equation for example $3^{2}-1=9-1$ is necessarily true (because it is the property of equation). In case of $3^{2}=9$ we are certain that the equation is true but in case of $P_k$ we are at the sake of assumption. These two things are different. I am sure I am missing something but please explain more accurately. Feb 9 '16 at 18:54
• On the second point, that's the whole point of Introduction of an Implication. To prove $P\implies Q$, assume $P$ is true and show that $Q$ is true under this assumption. Of course now all you have actually shown is that $Q$ is true if $P$ is true; but that is what $P\implies Q$ means. Feb 9 '16 at 19:09

There is another way of thinking of this.

Throw out the standard induction theorem, and replace it with this one:

Any nonempty set of natural numbers has a smallest element.

This seems obviously true, right? This is called the well-ordering principle, and it is equivalent to induction. Here's how you do induction with the well-ordering principle (WOP for short):

1. Prove that $P_0$ holds (or $P_1$, if you don't consider zero to be a "natural number").
2. Take this set: $$S = \{x | x \in \mathbb{N} \wedge \neg P_x \}$$ That's the set of natural numbers for which our proposition does not hold.
3. Assume $S$ is non-empty. Then, by WOP, it has a least element, which we can call $k+1$. In particular, we know that $\neg P_{k+1}$ (we know that the proposition does not hold for $k+1$, since it's an element of $S$).
4. By step (1), we know $k+1 > 0$ (or $k+1 > 1$, if you started with $P_1$). So $k$ is a natural number. If we had not proven step (1), $k$ could be equal to negative one (or zero), which would invalidate the rest of the proof.
5. Prove that $\neg P_{k}$; that is, prove that the proposition is not true for $k$, based on our knowledge that it is not true for $k+1$.
6. Since $k < k + 1$ and $k \in S$, we have a contradiction ($k + 1$ isn't the smallest element of the set). That means our assumption in step (3) must be false. The set of counterexamples is empty, meaning the proposition must hold for every number.

Note that these statements are equivalent: $$P_k \implies P_{k+1} \\ \neg P_{k+1} \implies \neg P_k$$ If you can prove one, you can prove the other quite easily. Standard induction asks you to prove the first, while WOP-based induction requires you to prove the second.

In the end, it's just a matter of notation. But you might find one form of induction more intuitive than the other. If you are going to use this method, you should note that we usually talk about $k$ and $k - 1$ instead of $k+1$ and $k$; I used the latter because it corresponds more obviously to standard induction.

"as long as we know that the state of a proposition being true for any positive integer k after Base number implies that proposition is true for integer k+1, we have to show that Pk is true. "

1) No, you don't. If I wanted to show that IF Godzilla were 10 times as large in all dimensions as a T. Rex THEN Godzilla would be weigh 1,000 times as much, I do NOT need to prove that Godzilla is 10 times as large in all dimensions as a T. Rex. I CAN'T prove Godzilla is 10 times as large in all dimensions as a T. Rex because Godzilla isnt 10 times as large in all dimensions as a T. Rex.

and

2) You already have shown $P_k$ is true for some k. You have shown it is true for $k = 1$. Reread that statement. It does not say "we will assume this is true for all $P_k$ and show that means it is true for all $P_{k+1}$". It says "we will show that if it is true for (one) $P_k$ than that means it is true for $P_{k+1}$".

Thus we have proven an infinite loop: If it is true for k, then it is true for k + 1: Relabel k+1 as k and it is true for k + 1 (which is the old k + 2): Relabel and repeat indefinitely and we have proven it is true for all $n \ge k$. Now we just need to demonstrate that it is true for a single base case k. Oh, wait. It's true for k = 1. Therefore it is true for $n \ge 1$.

As for you second question....

"Why is it necessary for an equation to be true for performing operations on it?"

It isn't. You can perform operations on untrue equations. You just can't assume any of the results are true.

Let me write down: "2 + 2 = 5". Let me perform operations on them:

$7^{2+2} - 37*5 = 7^5 - 37*{2 + 2}$

$2401 - 185 = 16,807 - 74 -74$

$2,216=16,659$

I can do that. But I can't get any meaningful results.

====

In a very abstract and philosophical way a proof by induction is a tiny bit like a proof by contradiction.

In both you assume a statement without proof and you reach a conclusion that would be true if your assumption were.

In a proof by induction you might prove that following: "If any even number is divisible by two then they all are divisible by two."

And in a proof by contradiction you might prove the following: "If any odd number is divisible by two then they all are divisible by two."

What's the difference in these statements? They are both shown in the exact same way ("if $2|n$ then $2|n + 2$ and $2|n - 2$"). They are both equally valid. And furthermore, they are both TRUE.

But in the proof by induction you verify a single confirmation: "2 is divisible by two. Therefore all even numbers are."

While in a proof by contradiction you contradict with a contradiction: "3 is not divisible by two. Therefore no odd numbers are."

Think about Peano system (see footnotes) for defining natural numbers. You take an initial number, which you call 1, and then you say:

• Element 1 exists in the set.
• Any successor of an element n in the set, say $suc(n)$, exists in the set.

With these axioms, you construct the whole set of natural numbers. You don't need to prove such elements exist in a deductive was as you think, but you must construct the proof and you have the building blocks for that (while in the deductive method, the path is the inverse).

So, the analogous of Peano system applies for the induction:

• Prove that $P(1)$ exists, where 1 is just the base case you want to start the proof with. This is the equivalent of the Peano system for the value 1.
• The constructive step: Prove that $P(n+1)$ is true when $P(n)$ was true. In this case, you move forward in the demonstration (induction actually means move towards the end) to construct the successive cases in the proof.

So you initially prove $P(1)$. If you prove $P(n) => P(n+1)$ then you can start traversing infinitely:

• $P(2) because P(1)$
• $P(3) because P(2)$
• ...

You never prove the $n$ case, because $P(n)$ is in hypotetic case which derives ultimately from the case $P(1)$, and formerly being $P(n)$, it was $P(n+1)$ in former... iterations.

So, the way to think why you should assume $P(n)$ is true, is because:

• In an initial case, $P(1)$ has been proven true by you. If you didn't, then you should start for it.
• Since the construction is upward, instead of downward, you should not think that you have to prove the n-case, but that you have alreadr proven it.

Footnotes:

• You are not forced to start from 1 and upwards. You can take any number you want, and so you will restrict the proof on it and further.
• Other induction processes can traverse entire integer set instead of just natural numbers. Such proof system involves prooving $n+1$ and $n-1$.


# Compute $\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$

How to evaluate $$\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ ?$$

where $$\displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}$$ , $$|x|\leq1$$

I came across this integral while I was working on $$\displaystyle \displaystyle\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\$$ and here is how I established a relation between these two integrals:

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\underbrace{\int_1^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx}_{x\mapsto 1/x}$$

$$=\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\int_0^1 \frac{\operatorname{Li}_3(1/x)}{1+x^2}\ dx$$ $$\left\{\color{red}{\text{add the integral to both sides}}\right\}$$

$$2\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx+\int_0^1 \frac{\operatorname{Li}_3(x)-\operatorname{Li}_3(1/x)}{1+x^2}\ dx$$

$$\{\color{red}{\text{use}\ \operatorname{Li}_3(x)-\operatorname{Li}_3(1/x)=2\zeta(2)\ln x-\frac16\ln^3x+i\frac{\pi}2\ln^2x}\}$$

$$=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx+2\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}\ dx}_{-G}-\frac16\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}+i\frac{\pi}2\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}\ dx}_{2\beta(3)}$$

$$=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-2\zeta(2)G+\beta(4)+i\pi \beta(3)$$

Then

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\frac12\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\zeta(2)G+\frac12\beta(4)+i\frac{\pi}2 \beta(3)\tag{1}$$

where $$\displaystyle\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\$$ is the the Dirichlet beta function.

So any idea how to evaluate any of these two integrals?

Thanks.

• What are you asking? – clathratus Jun 13 '19 at 2:17
• Note the formula you got from the triglogarithmic identity involved you being cavalier about branch cuts; you are missing a factor of $\frac{16 i \pi^4}{512}$ – Brevan Ellefsen Jun 13 '19 at 4:04
• @clathratus I am not sure if you are being sarcastic or not but I'll take as not. If you want to know what I'm asking just ignore the body and read the title please. – Ali Shather Jun 13 '19 at 4:28
• @BrevanEllefsen wolfram gives the closed form. – Ali Shather Jun 13 '19 at 4:30
• @BrevanEllefsen I dont think I'm missing any. To make sure, just take x=1/2 and compare the two sides. – Ali Shather Jun 13 '19 at 4:45

Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.

$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $$n=2$$ we get

$$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2 u}{1-xu}\ du$$

we can write

$$\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx=\frac12\int_0^1\ln^2u\left(\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx\right)\ du$$ $$=\frac12\int_0^1\ln^2u\left(-\frac12\left(\frac{\pi u}{1+u^2}+\frac{2\ln(-u)}{1+u^2}\right)\right)\ du,\quad \color{red}{\ln(-u)=\ln u+i\pi}$$

$$=-\frac{\pi}{4}\underbrace{\int_0^1\frac{u\ln^2u}{1+u^2}\ du}_{\frac3{16}\zeta(3)}-\frac12\underbrace{\int_0^1\frac{\ln^3u}{1+u^2}\ du}_{-6\beta(4)}-i\frac{\pi}2\underbrace{\int_0^1\frac{\ln^2u}{1+u^2}\ du}_{2\beta(3)}$$

Then

$$\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx=-\frac{3\pi}{64}\zeta(3)+3\beta(4)-i\pi\beta(3)\tag{2}$$

Bonus:

By combining $$(1)$$ in the question body and $$(2)$$, the imaginary part $$i\pi\beta(3)$$ nicely cancels out and we get

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=2\beta(4)-\zeta(2)G-\frac{3\pi}{128}\zeta(3)$$

where $$\beta(4)$$ $$=\frac{1}{768}\psi^{(3)}(1/4)-\frac{\pi^4}{96}$$

• The formula \begin{align}\Re\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx&=\frac12\int_0^1\ln^2u\left(\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx\right)\ du\end{align} is weird. In the right side it's $\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$ not its real part. There is no indication for the values of x involved in the formula about polylogarithm in the beginning of your post. Anyway your closed-form matches numeric evaluation. – FDP Nov 22 '19 at 8:45
• its edited now. thanks for pointing this out. – Ali Shather Nov 22 '19 at 8:58
• The formula for polylogarithm in the beginning of your post is valid for $|x|<1$ only. If you take $x=2$ the denominator is $1-2u$ and $1-2u=0$ when $u=1/2\in [0;1]$ – FDP Nov 22 '19 at 15:42
• Yes .. actually i got the right formula yesterday and I'll fix the problem today. – Ali Shather Nov 22 '19 at 16:18
• When you consider the integral $$\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx$$ as a CauchyPrincipalValue you don't get the imagniary part. It arises from the $\epsilon$ contour around that singularity. – Diger Nov 22 '19 at 22:05

For a different solution, use the first result from A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by Cornel Ioan Valean.

Essentially, the main new results in the presentation are:

Let $$a\le1$$ be a real number. The following equalities hold: $$\begin{equation*} i) \ \int_0^1 \frac{\log (x)\operatorname{Li}_2(x) }{1-a x} \textrm{d}x=\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}; \end{equation*}$$ $$\begin{equation*} ii) \ \int_0^1 \frac{\log^2(x)\operatorname{Li}_3(x) }{1-a x} \textrm{d}x=20\frac{\operatorname{Li}_6(a)}{a}-12 \zeta(2)\frac{\operatorname{Li}_4(a)}{ a}+\frac{(\operatorname{Li}_3(a))^2}{a}. \end{equation*}$$ For a fast proof, see the paper above (series expansion combined with the Cauchy product of squared Polylogarithms)

The use of these new results with integrals allows you to obtain your result elegantly, but also other results that are (very) difficult to obtain by other means, including results from the book, (Almost) Impossible Integrals, Sums, and Series.

BONUS: Using these results you may also establish that (or the versions with integration by parts applied).

$$i) \ \int_0^1 \frac{\arctan(x) \operatorname{Li}_2(x)}{x}\textrm{d}x$$ $$=\frac{1}{384}\left(720\zeta(4)+105\pi\zeta(3)+384\zeta(2)G-\psi^{(3)}\left(\frac{1}{4}\right)\right),$$ $$ii)\ \int_0^1 \frac{\arctan(x) \operatorname{Li}_2(-x)}{x}\textrm{d}x$$ $$=\frac{1}{768}\left(\psi^{(3)}\left(\frac{1}{4}\right)-384\zeta(2)G-126\pi\zeta(3)-720\zeta(4)\right).$$

EXPLANATIONS (OP's request): The following way in large steps shows the amazing possible creativity in such calculations.

We'll want to focus on the integral, $$\displaystyle \int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x)}{x}\textrm{d}x$$ which is a translated form of the main integral.

Now, based on $$i)$$ where we plug in $$a=i$$ and then consider the real part, we obtain an integral which by a simple integration by parts reveals that

$$\int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x)}{x}\textrm{d}x=\int_0^1 \frac{\arctan(x)\log(1-x) \log(x)}{x}\textrm{d}x+\frac{17}{48}\pi^2 G+\frac{\pi^4}{32}-\frac{1}{256}\psi^{(3)}\left(\frac{1}{4}\right).$$

Looks like we need to evaluate one more integral and we're done. Well, if you read the book (Almost) Impossible Integrals, Sums, and Series (did you?), particularly the solutions in the sections 3.24 & 3.25 you probably observed the powerful trick of splitting the nonnegative real line at $$x=1$$ with the hope of getting the same integral in the other side but with an opposite sign. Therefore, with such a careful approach (since we need to avoid the divergence issues), we obtain immediately that $$\int_0^1 \frac{\arctan(x)\log(1-x) \log(x)}{x}\textrm{d}x$$ $$=\frac{1}{2} \underbrace{\int_0^1 \frac{\arctan(x)\log^2(x)}{x}\textrm{d}x}_{\text{Trivial}}+\frac{\pi}{4}\underbrace{\int_0^1 \frac{\log(x)\log(1-x)}{x}\textrm{d}x}_{\text{Trivial}}$$ $$-\frac{1}{2}\Re\left \{\int_0^{\infty}\frac{\arctan(1/x) \log(1-x)\log(x)}{x}\textrm{d}x\right \},$$

and the last integral works simply nice with Cornel's strategy described in the second part of this post (it involves the use of Cauchy Principal Value) https://math.stackexchange.com/q/3488566.

• Nice (+1) but it's not clear how to get the integral in the question from using the first result. – Ali Shather Dec 30 '19 at 5:39
• The explanation is really amazing and helpful. thank you. – Ali Shather Dec 30 '19 at 20:07


# Example of a continuous function $f:\mathbb{R} \to [0,1]$ such that f is not uniformly continuous.

I am trying to find an example of a continuous function $f:\mathbb{R} \to [0,1]$ such that f is not uniformly continuous. I have been playing around with the $sin$ function and I have noticed that $|sin(x^2)|$ is a continuous function from $\mathbb{R}$ to $[0,1]$. This function I am thinking is not uniformly continuous since it behaves a lot like $\sin(x^2)$.

What I am most interested is in how you come up with a function that meets the criteria. For example, when I first saw this question, I immediately thought of the $\sin$ function. Are there any other functions that meet this requirement and are easier to work with in case that I want to prove that it is not uniformly continuous. Is this how you come up with functions, that is, by looking at the rate of oscillations? It seems to me that the only functions that would work are variations of the $sin$ function. I would appreciate it if you could give advice on how to approach this problem.

• you cannot just say "it oscillates a lot" to prove a function is not uniformly continuous, start with the definition
– Nick
Nov 17 '16 at 6:02
• you are right. Right now I am just constructing graphs that I think are not uniformly continuous and trying to come up with possible functions that meet the requirements. I am just trying to see how people approach the problem rather than proving it. Nov 17 '16 at 6:05
• You can use the mean value theorem to prove that $\frac 1 2 + \frac 1 2 \sin(x^2)$ isn't uniformly continuous since the derivative becomes arbitrarily large as $x \to \infty$. Nov 17 '16 at 6:15

The key to finding such a function, is to look for a function with "arbitrarily large derivative" (this doesn't work for all such functions, as the comment below points out) . This is to say, we want some problems with infinity, to be created on the end points.

For example, let us consider $f(x) = \frac{1}{2} + \frac{1}{2}\sin x^2$, which has a derivative of $2x \cos x^2$, and also has range in $[0,1]$ (the $\frac{1}{2}$ is just some adjustment to ensure that the range is in $[0,1]$). This derivative blows up as $x$ increases. Hence, we consider this as a candidate.

Fix a natural number $N$. We will find a pair of points $x,y$ such that $|f(x)-f(y)| \geq N|x-y|$.

Now, all we need is to use the mean value theorem: We know that $f(x)-f(y) = (x-y) f'(z)$ where $z$ is some point between $x$ and $y$. Now, suppose we chose $x$ and $y$ in a region where $f'(z)$ could only be a large positive quantity, greater than the given $N$. Then of course, $|f(x)-f(y)| >N|(x-y)|$.

I leave you to formalize the details, but it is clear, that $f$ cannot be uniformly continuous, despite being differentiable.

• "The key to finding such a function, is to look for a function with "arbitrarily large derivative". " That doesn't always work. Consider $\sin(e^x)/(1+x^2).$ Also the derivative of your function doesn't "blow up" as I understand the term. Rather it oscillates unboundedly.
– zhw.
Nov 17 '16 at 7:14
• @zhw I see. The point is, my function oscillates unboundedly, as you have said. I mistakenly wrote it as blowing up, but the meaning should be clear. But I understood the example of $\sin (e^x)/ (1+x^2)$, so thank you for pointing it out. It is a function with arbitrarily large derivative, but reducing oscillations, hence could be uniformly continuous. Nov 17 '16 at 7:25

"Right now I am just constructing graphs that I think are not uniformly continuous and trying to come up with possible functions that meet the requirements. I am just trying to see how people approach the problem rather than proving it."

This is a good strategy, and it's often how I approach analysis problems as well. Having an intuition that a result is "correct" (even if it's due to hand-waving reasoning) is good to have before starting a rigorous proof.

Uniform continuity is sort of like a stronger, "global" version of regular continuity. To wit, if you start with a given $\varepsilon > 0$, then you can find a $\delta > 0$ such that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$, regardless of where you are in the domain. On the other hand, we talk about regular continuity "at a given point", and the $\delta$ depends on that chosen point.

Your idea with $f(x) = |\sin(x^2)|$ is perfect. Indeed, if we suppose that this is uniformly continuous, then we can arrive at a contradiction for essentially the reason you cite: it oscillates faster and faster as we move away from the origin.

To take advantage of this problem, choose $\varepsilon$ less than the amplitude of the wave; e.g. $\varepsilon = 1/2$. If we suppose there exists a $\delta > 0$ such that $|x-y| < \delta$, then $||\sin(x^2)| - |\sin(y^2)|| < 1/2$, we can make this fail by going out far enough away from the origin so that our function goes through an entire oscillation in a $\delta$-interval. Recall that a sine wave reaches a "crest" or a "trough" at every $(2k + 1/2) \pi$ radians and every $(2k + 3/2) \pi$ radians. Solving $x^2 = (2k + 1/2) \pi$ gives the values at which $\sin(x^2)$ reaches a crest, and a similar equation for the troughs. Taking an absolute value results in all of these becoming crests (with new troughs in between each at integer multiples of $\pi$). Now just go out sufficiently far so that the distance between these crests is less than the chosen $\delta$. I'll leave the details to you to crunch out (one can show that the distance between the crests limits to zero).

• Apparently, thge OP added the absolute value to achieve the desired codomain $[0,1]$ Nov 17 '16 at 6:36
• Oh!! I lost sight of the forest for the trees. Thanks for pointing that out @HagenvonEitzen. Nov 17 '16 at 6:38
• Thank you Kaj Hansen your explanation is very helpful! Nov 17 '16 at 7:00
• Glad I could help @AnP. ! Nov 17 '16 at 7:18


# If I say $\{g_1,\dots,g_n\}$ freely generates a subgroup of $G$, does that mean the elements $g_i$ are all distinct?

Let $G$ be a group and let $g_1,\dots,g_n$ elements of $G$. If I say that $\{g_1,\dots,g_n\}$ freely generates a free subgroup of $G$, say $H$, do I mean that the rank of $H=n$ or should I consider the case when two elements are equal (for example $g_1=g_2$) and so the rank of $H<n$ ? thank you

• If the elements freely generate $\;H\;$ it can't be $\;g_i=g_j\;,\;\;i\neq j$, nor $\;g_i=1\;$ for some $\;i\;$ – DonAntonio Feb 28 '16 at 1:31
• @Joanpemo I agree that it can't be $g_i=1$ but why it can't be $g_i=g_j$ for $i\not=j$? thank you – Richard Feb 28 '16 at 1:34
• Else you would have the ralation $g_ig_j^{-1}$. – Daniel Robert-Nicoud Feb 28 '16 at 1:38
• @Joanpemo It depends on the context, but the way the question is phrased, i.e., without context, the statement is not strictly false. There is a perfectly nice free group which is free on the set $\{x,x,y\}$. It is the free group on $\{x,y\}$, since $\{x,x,y\}=\{x,y\}$. – Zach Blumenstein Feb 28 '16 at 1:44
• @ZachBlumenstein Thank you. Yet, as sets, $\;\{x,x,y\}=\{x,y\}\;$ so no problem, indeed. I though understand something different when I read "a set of free generators", and as far as I know this implies $\;g_i\neq g_j\;$ for $\;i\neq j\;$, otherwise the universal property of a free group in the number of free generators given isn't fulfilled. – DonAntonio Feb 28 '16 at 2:42

Strictly speaking, what you wrote means that the set $\{g_1,\dots,g_n\}$ freely generates $H$, so the rank of $H$ is the number of (distinct) elements of that set, which may or may not be $n$. However, people frequently abuse language and say what you wrote when they really mean that additionally the $g_i$ should all be distinct. In context, there is usually not too much risk of confusion, and as a practical matter, if you see someone write this, you should judge for yourself which interpretation would make more sense in context. I don't really know of a succinct and completely standard way to express the latter meaning; I might write it as "the elements $(g_1,\dots,g_n)$ freely generate a subgroup", writing it as a tuple to stress that you are actually saying that the map $\{1,\dots,n\}\to H$ sending $i$ to $g_i$ satisfies the universal property to make $H$ a free group on the set $\{1,\dots,n\}$. What's really going on is that a collection of (potential) "free generators of a group $H$" should not be thought of as just a subset of $H$ but as a set together with a map to $H$ (and if the group really is freely generated by a map, the map must be injective).
(This terminological difficulty is not restricted to this context. For instance, a similar issue quite frequently comes up when talking about linear independence: if $v\in V$ is a nonzero element of a vector space, then the set $\{v,v\}$ is linearly independent (since it actually has only one element), but the tuple $(v,v)$ is not.)
• I don't exactly agree. When we say that a (finite or infinite) set $X$ is a set of free generators of a group $F$, we don't need to specify any function, because the function is just inclusion. There is no mistake there. But I do agree that the notation $\{g_1,\ldots,g_n\}$ taken to mean that $g_i$ are distinct is bad, because it literally means what that $X=\{g_1,\ldots,g_n\}$ is a free generating set of $F$, while it probably really means that $g_1, \ldots, g_n$ generate $F$ freely, which is not the same thing if $\lvert X\rvert\neq n$. – tomasz Feb 28 '16 at 8:34
• Sure, there is nothing wrong with saying that a subset freely generates a group. But it is more natural to formulate it in terms of maps from a set, and the failure to do so is the cause of the terminological awkwardness when saying "$(g_1,\dots,g_n)$ freely generates" a group, and the frequency with which people incorrectly write "$\{g_1,\dots,g_n\}$ freely generates" when that's not really what they mean. – Eric Wofsey Feb 28 '16 at 8:41


# Math Help - probability of success

1. ## probability of success

I am trying to answer this problem for an exam.Can anyone please help me.I wonder if I should be using the formula prob=nCr (p^r) q^(n-r).

The probability for students from a certain university to pass Mathematics and Economics are 3/7 and 5/7 respectively. If one student fails both subjects and there are four students who passed both subjects find the number of students who took the test.

2. Hello, hven191!

This doesn't require any fancy formulas . . .

The probability for students at a certain university
to pass Mathematics and Economics are 3/7 and 5/7 respectively.

One student failed both subjects and four students passed both subjects.

Find the number of students who took the test.
Make a Venn diagram of the students who passed.
Code:
      *---------------------------------------*
|                                       |
|   *-----------------------*           |
|   | Math                  |           |
|   | only  *---------------+-------*   |
|   |   M   |     Both      |       |   |
|   |       |       4       |       |   |
|   |       |               |       |   |
|   *-------+---------------*  Eco  |   |
|           |             E   only  |   |
|           *-----------------------*   |
|  Neither 1                            |
*---------------------------------------*

Let $M$ = number of students that passed Math only.
Let $E$ = number of students that passed Economics only.

We know there are 4 that passed both, and 1 that failed both.

The total number of students is: . $N \:=\:M + E + 5$

The number of students that passed Math is: . $M + 4$
The probability that a student passed Math is $\tfrac{3}{7}$
We have: . $\frac{M+4}{M+E+5} \:=\:\frac{3}{7} \quad\Rightarrow\quad 4M - 3E \:=\:-13\;\;{\color{blue}[1]}$

The number of students that passed Economics is $E + 4$
The probability that a student passes Economics is $\tfrac{5}{7}$
We have: . $\frac{E+4}{M+E+5} \:=\:\frac{5}{7} \quad\Rightarrow\quad 5M - 2E \:=\:3\;\;{\color{blue}[2]}$

$\begin{array}{cccccc}\text{Multiply {\color{blue}[1]} by -2:} & \text{-}8M + 6E &=& 26 & {\color{blue}[3]}\\
\text{Multiply {\color{blue}[2]} by 3:} & 15M - 6E &=& 9 & {\color{blue}[4]}\end{array}$

$\text{Add {\color{blue}[3]} and {\color{blue}[4]}: }\;7M \:=\:35 \quad\Rightarrow\quad\boxed{ M \:=\:5}$

Substitute into [1]: . $4(5) - 3E \:=\:-13 \quad\Rightarrow\quad\boxed{ E \:=\:11}$

Therefore, the number of students is: . $N \;=\;M + E + 5 \;=\;5 + 11 + 5 \;=\;{\color{red}21}$

3. ## Thanks

Thanks a lot Soroban! The solution you gave was very simple and easy to understand.The illustration was a big help. Although the answer does not match the answer in the book I'm using I'm still convinced with your answer. The answer in the book which is 28 may just be a typo-graphical error. Thanks!


# Find the infinite sum $\sum_{n=1}^{\infty}\frac{1}{2^n-1}$

How to evaluate this infinite sum? $$\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

-
Why the expression appears so small? How can I enlarge that? : ( –  Ryan Dec 28 '12 at 18:54
use \displaystyle –  sxd Dec 28 '12 at 18:55
If I'm not wrong, this one appears in one volume of Ramanujan's notebook. (Chris) –  Chris's sis Dec 28 '12 at 18:59
$1.606695152415291$ –  Henry Dec 28 '12 at 19:12
Not an answer, but it is easy to convert the formula to: $$\sum_{m=1}^\infty \frac{\tau(m)}{2^m}$$ where $\tau(m)$ is the number of distinct divisors of $m$. –  Thomas Andrews Dec 28 '12 at 19:33
show 6 more comments

I think you wanna see this:

Ramanujan’s Notebooks Part I

Click me and try Entry $14$ (ii) / pag 146 where you set $x=\ln2$

Chris.

-
I want to write out that particular equation just because it's so . . . out there: $$\sum_{k \ge 1}\frac{1}{e^{kx}-1}=\frac{\gamma}{x}-\frac{\log x}{x}+\frac{1}{4}-\sum_{1 \le k \le n}\frac{B^2_{2k}x^{2k-1}}{(2k)(2k)!}+R_n,$$ where $\gamma$ is Euler's constant, $B_{2k}$ are the conventionally defined Bernoulli numbers, $x>0$, $n\ge 1$, and $R_n$ is a constant bounded by the inequality $$|R_n|\le \frac{|B_{2n}B_{2n+2}|x^{2n}}{(2n)!}\left(\frac{x^2}{4\pi^2}+\frac{\pi^2}{6} \right).$$ For our case, let $x=\ln 2$ as Chris's sister stated. –  000 Dec 29 '12 at 18:00
@Limitless: thank you for the details provided! (+1) –  Chris's sis Dec 29 '12 at 18:03

Yes. I found it. It is called the Erdős-Borwein Constant.

$$E=\sum_{n\in Z^+}\frac{1}{2^n-1}$$

According to the page, Erdős showed that it is irrational.

-
I almost see in each step in the link a possible nice question to ask. :D (+1) –  Chris's sis Dec 28 '12 at 19:24
Is it known whether the Erdos-Borwein Constant is transcendental or not? –  Rustyn Dec 28 '12 at 19:27
I have no idea. I guess no. –  Amr Dec 28 '12 at 19:35
@Rustyn: Please use markdown formatting for non-mathematical things like italics and bold in normal text. I've edited your comment. –  Zev Chonoles Dec 28 '12 at 19:42
@ZevChonoles Ok, I will in the future. –  Rustyn Dec 28 '12 at 19:44
show 1 more comment

$$\displaystyle \sum _{k=1}^n \frac{1}{\left(\frac{1}{q}\right)^k-\frac{1}{r}}=\frac{r}{\log (q)} \left(\psi _q^{(0)}\left(1-\frac{\log (r)}{\log (q)}\right)-\psi _q^{(0)}\left(n+1-\frac{\log (r)}{\log (q)}\right)\right)$$

In trying to get Mathematica to solve the series, I eventually found the preceding form which assumes $0<q<1$. If we take $q=1/2$, $r=1$ and let n approach infinity, we get the same solution that Amr references. The partial sum solution utilizes the function, $\psi _q^{(n)}(z)$.

$$\displaystyle \lim_{n\to \infty } \, \frac{1}{\log (1/2)}\left(\psi _{\frac{1}{2}}^{(0)}(1)-\psi _{\frac{1}{2}}^{(0)}(n+1)\right)=1+\frac{\psi _{\frac{1}{2}}^{(0)}(1)}{\log (1/2)}=E$$

-

(This is not meant as an answer but it is too long for the comment.box)

Since you mentioned interest in variations of the problem: here is a text, in which L. Euler discussed that sum:

"Consideratio quarumdam serierum quae singularibus proprietatibus sunt praeditae" (“Consideration of some series which are distinguished by special properties”)
L. Euler Eneström-index E190.

You can find it online.

A further discussion of this by Prof. Ed Sandifer, where he sheds light on a very interesting discussion about a "false series for the logarithm" at which that constant pops up (and which actually had pointed me originally to L.Euler's article):

A false logarithm series (Discussion of E190)
Ed. Sandifer in: "How Euler did it" Dec 2007
http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2050%20false%20log%20series.pdf

I've fiddled then with it myself a bit further, maybe you find that amateurish explorations interesting too. The constant is part of a consideration on page 5.

-


# Is $\Bbb Z_3 [x] / \langle x^2 - 1 \rangle \cong \Bbb Z_3 \times \Bbb Z_3$?

I have paired off the zero divisors in these two rings, and there are exactly 4 pairs of distinct zero divisors in both rings:

In $\Bbb Z_3[x]/\langle x^2 - 1\rangle$ the zero divisor pairs are $$x+1, x+2 \qquad x+1, 2x+1, \qquad 2x+1, 2x+2 \qquad x+2, 2x+2$$

Similarly, in $\Bbb Z_3 \times \Bbb Z_3$ the zero divisor pairs are $$(0, 1), (1, 0) \qquad (0,1), (2,0) \qquad (0, 2), (1, 0) \qquad (0, 2), (2, 0)$$ The fact that the zero divisors match up indicates to me that these rings should indeed be isomorphic, as zero divisors are one of the structures which have failed checks in the past on rings such as $\Bbb Z_2 [x] / \langle x^2 \rangle \ncong \Bbb Z_2 \times \Bbb Z_2$.

My Question

Is the only homomorphism (and thus isomorphism) between these two rings an enumerative one, where I explicitly map each of the nine elements of $\Bbb Z_3[x]/ \langle x^2 - 1\rangle$ to an element in $\Bbb Z_3 \times \Bbb Z_3$? I can do that if necessary, but it seems an inelegant and brute force method. However, since these rings are so small, it may be the best way to go about it. Thanks!

$Z_3[x]/(x^2-1)\cong Z_3[x]/(x-1)\times Z_3[x]/(x+1)\cong Z_3\times Z_3$. The map is induced by $ax+b\rightarrow (a+b, -a+b)$

• Thank you all for the answers! They give me much more intuition about the result than a simple enumerative homomorphism. – Moderat Nov 9 '12 at 0:28
• Actually I am finding it hard to see that this map is onto. Can you elaborate? – Moderat Nov 10 '12 at 21:03
• @jmi4 $x-1\rightarrow (0,1)$, $-x-1\rightarrow (1,0)$ – i. m. soloveichik Nov 11 '12 at 0:02

In $R:=(\mathbb Z/3\mathbb Z)[x]$, $x-1$ and $x+1$ generate two distinct maximal ideals $\mathfrak m_1, \mathfrak m_2$. We have $(x^2-1)R=\mathfrak m_1 \mathfrak m_2$. By Chinese Remainder Theorem, we have a canonical isomorphism $$R/(x^2-1)R \to R/\mathfrak m_1\times R/\mathfrak m_1.$$ Now $R/\mathfrak m_i$ is isomorphic to $\mathbb Z/3\mathbb Z$. So your rings are isomorphic.

There are exatly two isomorphisms. One is given as above, the other one is obtained by composition this isomorphism with the permutation in the second ring. (Hints: the only automorphism of $\mathbb Z/3\mathbb Z$ is the identity).

1. A homomorphism from a quotient $A/\theta\to B$ may be given by its lift $A\to B$ such that its kernel factors through $\theta$.
2. A homomorphism from a polynomial ring $R[x_1,..,x_i,..]$ to an $R$-algebra is unquely determined by the image of the indeterminants $x_i$.
3. $1$ must be mapped to the unit, that is $(1,1)$, and by 2., the image of $x$ can be freely chosen, and it determines all other images.

Since the ideals $(x-1)$ and $(x+1)$ both contain $(x^2-1)$ (because $x-1,x+1\mid x^2-1$ in $\mathbf{F}_3[x]$), there are natural surjective homomorphisms $\pi_1:\mathbf{F}_3[x]/(x^2-1)\rightarrow\mathbf{F}_3[x]/(x-1)$ and $\pi_2:\mathbf{F}_3[x]/(x^2-1)\rightarrow\mathbf{F}_3[x]/(x+1)$. Explicitly $\pi_1(f+(x^2-1))=f+(x-1)$ and $\pi_2(f+(x^2-1))=f+(x+1)$. This gives a homomorphism

$\pi:\mathbf{F}_3[x]/(x^2-1)\rightarrow\mathbf{F}_3[x]/(x-1)\times\mathbf{F}_3[x]/(x+1)$.

Note that $f+(x^2-1)$ is in the kernel of this map if and only if both $x-1$ and $x+1$ divide $f$, and since the polynomials $x-1,x+1$ are relatively prime, their product $x^2-1$ divides $f$, so $f\in (x^2-1)$, and thus $f+(x^2-1)=0$ in $\mathbf{F}_3[x]/(x^2-1)$. Thus $\pi$ is injective. There are a couple ways to see that $\pi$ is surjective, and which one you prefer depends I guess on what you know. For example, since both the source and the target of $\pi$ are $\mathbf{F}_3$-vector spaces of dimension $2$, and the map is $\mathbf{F}_3$-linear, it has to be an isomorphism. Or, since the ideals $(x-1)$ and $(x+1)$ together generate the unit ideal $\mathbf{F}_3[x]$, since the ideal they generate contains $(x+1)-(x-1)=1+1=2$, a unit in $\mathbf{F}_3$, you can invoke the Chinese remainder theorem (which will actually tell you injectivity of the above map as well as surjectivity). Or you can try to directly find, for given $f_1,f_2\in\mathbf{F}_3[x]$, a polynomial $f$ such that $f-f_1\in(x-1)$ and $f-f_2\in(x+1)$. Then $f+(x^2-1)$ will map to $(f_1+(x-1),f_2+(x+1))$.

In any case, the map is an isomorphism.


Spiral of Archimedes area and sketch in polar coordinates

This is an exercise from Apostol's Calculus, Volume 1. It asks us to sketch the graph in polar coordinates and find the area of the radial set for the function:

$$f(\theta) = \theta$$

On the interva $0 \leq \theta \leq 2 \pi$. I think to find the area we should just integrate $\theta \ d\theta$ from 0 to $2\pi$ like any other function? Is that right? Also I'm not sure how to think about sketching a function in polar coordinates.

The problem is the book gives the answer as $4\pi^3/3$ which is not what I get if I just integrate the function.

• In the 2nd edition of this book (of which I have a copy), this is exercise 5 on page 111. You should review the previous three or four pages, especially the section about "Polar coordinates" and the section about "The integral for area in polar coordinates" (or whatever those sections are called in the edition you're using). If those pages do not answer your question, perhaps you can identify something in those pages that did not make sense to you, and add that to your question. – David K Nov 25 '15 at 17:15
• By the way, I think the [calculus] tag was fine for this question. – David K Nov 25 '15 at 17:15
• Did you find the theorem yet that says, in part, "Let $R$ denote the radial set of a nonnegative function $f$ ... the area of $R$ is ..."? It gives you the formula. In my edition of the book, this is on the page just before the exercise. – David K Nov 25 '15 at 18:55

First, to sketch such a graph, you want to consider the distance from the origin as the angle from the $x$-axis changes. Just like when sketching the graph of a function in rectangular coordinates it is good to evaluate at particular values of $x$ and see the height of the function, when sketching a curve in polar coordinates, evaluate the function are a few values of the angle and find the radius, i.e., the distance from the origin at the angle. So, doing that we obtain the following graph:

Then, to calculate the area of the radial set, you must integrate $\frac{1}{2} r^2$, where the radius is the value of the function. So we have, \begin{align*} \text{Area} &= \frac{1}{2}\int_0^{2 \pi} \theta^2 \, d\theta \\ &= \left. \frac{1}{2} \cdot \frac{\theta^3}{3} \right|_0^{2 \pi} \\ &= \frac{(2 \pi)^3}{6}\\ &= \frac{4 \pi^3}{3}. \end{align*}

Sketching this in polar coordinates is pretty straightforward. Draw your axes and you know that the radial value is equal to the angle, so your curve would start at the origin when $\theta$ is zero, it would be $\frac{\pi}{2}$ at 90 degrees, etc.

As for calculating the area, you are correct that you integrate it, but I'm not sure what it means physically because the curve does not close on itself since $f(0) \neq f(2\pi)$ (perhaps this is the insight they are trying to get from you).

• The "radial set" is defined in the book. Graphically, it is formed by the line segments from the origin to each point on the curve. All the points those segments pass through are in the area to be measured. As long $f(\theta)$ is never negative, it's clear what region's area we're supposed to measure. – David K Nov 25 '15 at 17:32

commenting on Daves's answer "but I'm not sure what it means physically because the curve does not close on itself since f(0)≠f(2π)f(0)≠f(2π)"

The area calculated is the area bounded by the curve and the terminal vector. ie the curve is subtended by the sweeping vector, which in this case is at 2pi- the positive x axis. .


# How can I find the work on a block when it is pulled up in a curve?

#### Chemist116

The problem is as follows:

The figure from below shows a force acting on block as it is pulled upwards in a curve from point $A$ to $B$. It is known that the block is pulled by a force which its modulus is $100\,N$. Find the work (in Joules) of such force between the points indicated. Consider that the angle given is with respect of the vertical with the floor.

The alternatives given in my book are:

$\begin{array}{ll} 1.&143\,J\\ 2.&312\,J\\ 3.&222\,J\\ 4.&98\,J\\ 5.&111\,J\\ \end{array}$

I attempted to decompose the force given as such:

$F\cos 37^{\circ} \times d = W$

But the result doesn't seem to yield an adequate result:

$W= 100\times \cos 37^{\circ} \times 2.1= 100 \times \frac{4}{5}\times 2.1=168$

Assuming the gravity does negative work?

$W= - 100 \times \sin 37^{\circ}= - 100 \times \frac{3}{5}\times 1.2= -72$

Anyways the sum doesn't yield the result which supposedly is option $5$. Can somebody help me here? :help:

#### DarnItJimImAnEngineer

Treat $\vec{F}$ as two separate forces. Horizontal force $F \sin 37Â°$ acts over a horizontal distance of 2.1 m. The vertical force acts over a vertical distance of 1.2 m. Try finding the work from each and adding them.

Last edited by a moderator:
2 people

#### Chemist116

Treat $\vec{F}$ as two separate forces. Horizontal force $F \sin 37Â°$ acts over a horizontal distance of 2.1 m. The vertical force acts over a vertical distance of 1.2 m. Try finding the work from each and adding them.
I attempted to do such thing.

Let's see:

$F\sin 37^{\circ}\cdot 2.1 = 100 \cdot \frac{3}{5} \cdot 2.1 = 126 J$

$F\cos 37^{\circ}\cdot 1.2 = 100 \cdot \frac{4}{5} \cdot 1.2 = 96 J$

The summing both would give me:

$W_{1}+W_{2}=126+96=222\,J$

which supposedly is the third option. I'm not doing any sort of vector sum here as work is scalar (thanks to romsek for that remark). Mind to confirm whether what I'm doing is what you intended to say?

I'm sorry if I mentioned that the answer is $111\,J$. I checked with the answers sheet and it lists it to be $222\,J$. I feel so dumb now, as when I compared with what I did in my notes I got an error in the summation. Learning, yikes!

But I'm still confused, by comparing it to a similar problem which I posted before this, why shouldn't I subtract the "angle" which could result by joining $AB$? Is it because it's a curve? Can you help me with this part please? :help:

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# Matrix of $T$ over $P_2$

I'm having a bit of trouble following the logic my professor used to construct the matrix of a given transformation $T$ in class, and was wondering if anyone could share any further intuition or insight.

Given $P_2$, the set of polynomials of degree $\leq2$, define a linear transformation $T:P_2\rightarrow \mathbb{R}^3$ such that $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$.

The matrix of $T$ with respect to the basis $\mathcal{B}=\left\{1,x,x^2\right\}\\$

is given by $T(1)=\begin{pmatrix} 1\\1\\1\end{pmatrix}$, $T(x)=\begin{pmatrix} 0\\1\\2\end{pmatrix}$, and $T(x^2)=\begin{pmatrix} 0\\1\\4\end{pmatrix}$,

i.e. $\,A=\begin{pmatrix}1&0&0\\1&1&1\\1&2&4\end{pmatrix}$.

I get the whole business of the coefficients on this particular transformation being the "trivial relation" such that $c_1=c_m=0$, and that $ImT=\mathbb{R}^{3}$ such that $T$ is an isomorphism.

What I'm struggling with, quite frankly, is how he got the coefficients for $A$. The coefficients for $p(x)=1$ in particular are a bit counter-intuitive. Is the idea that, if we define $p(x)=1$, then no matter what, $p(0)=p(1)=p(2)=1$?

The notation for expressing the various matrices associated with an isomorphism is also a bit nebulous for me. My textbook defines a $\mathcal{B}$-coordinate transformation as

$T^{-1}\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=c_1f_1+\cdots + c_nf_n$.

What would the inverse of the transformation $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$ be?

What would the matrix associated with the inverse be? Or would it be a set of three different polynomials (linear equations)?

For those wondering, we're using Bretscher's Linear Algebra with Applications, 5th edition. Thanks!

• $A = [[T(1)]_B|[T(x)]_B|[T(x^2)]_B]$ where $B=\{ e_1,e_2,e_3 \}$ is just the standard basis for $\mathbb{R}^3$. More generally, you just find the coordinate vectors of the image of each vector in the domain basis and glue them together. There are good reasons for doing this, but you ought to have seen them in lecture I think... – James S. Cook Mar 22 '15 at 4:42
• My silly notation $[v]_B = v$ in this case since the coordinate system in the codomain is just the plain-old-standard Cartesian coordinate system. – James S. Cook Mar 22 '15 at 4:43

To give you a much more complicated example. Let us consider $T(f(x)) = f'(x)$ where we consider $T:P_2 \rightarrow P_2$ and to be perverse let's use $\beta = \{ 1,x,x^2 \}$ as the domain basis, but $\gamma = \{ x^2,x,1 \}$ as the codomain basis. To understand $T$ I like to consider $f(x)=a+bx+cx^2$ and see what happens: $$T(f(x)) = b+2cx$$ Then the coordinate vector of the image is easy enough to see: $$[T(f(x))]_{\gamma}= [b+2cx]_{\gamma} = [0(x^2)+2c(x)+b(1)]_{\gamma} = [0,2c,b]^T.$$ Then, the matrix $[T]_{\beta,\gamma}$ is the matrix which when multiplied on $$[f(x)]_{\beta} = [a+bx+cx^2]_{\beta} = [a,b,c]^T$$ yields $[T(f(x))]_{\gamma}=[0,2c,b]^T$. A moments reflection yields: $$[T]_{\beta, \gamma} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 0\end{array}\right].$$ Alternatively, just use: $$[T]_{\beta, \gamma} = [ [T(1)]_{\gamma} | [T(x)]_{\gamma}| [T(x^2)]_{\gamma} ] = [[0]_{\gamma}|[1]_{\gamma}|[2x]_{\gamma}] = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 0\end{array}\right].$$ Some books also use $[T]_{\beta}^{\gamma}$ to reflect the differing roles the domain and codomain bases play especially in regard to coordinate change. Perhaps your text does that.
• The text doesn't explicitly use the terms "domain" and "codomain", but I think I understand. Follow-up question: if we define some other basis $\mathcal{C}=\left\{x^2,xy,y^2\right\}$ and $T(x)=\begin{pmatrix}f(1,0)\\f(0,1)\\f(1,1)\end{pmatrix}$, does the computation still move along similar lines? see here for more. – Benjamin Loya Mar 22 '15 at 5:05


# Integration of a rational function

stripes

## Homework Statement

find
$$\int\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)}dx$$

None

## The Attempt at a Solution

$$\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{Cx + D}{x^{2}+1}$$

$$x^{2}-2x-1 = A(x-1)(x^{2}+1) + B(x^{2}+1) + (Cx + D)(x-1)^{2}$$

$$x^{2}-2x-1 = Ax^{3}-Ax^{2}+Ax-A+Bx^{2}+B+Cx^{3}-2Cx^{2}+Cx+Dx^{2}-2Dx+D$$

$$x^{2}-2x-1 = x^{3}(A+B+C) + x^{2}(-A+B-2C+D) + x(A+C-2D) - A + B +D$$

so A+B+C = 0, -A+B-2C+D = 1, A+C-2D=-2 and -A+B+D=-1

solving for coefficients, we get
A = 5/3
B = -2/3
C = -1
D = 4/3

so $$\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \frac{5/3}{x-1} - \frac{2/3}{(x-1)^{2}} - \frac{x-(4/3)}{x^{2}+1}$$

but apparently, the last statement is not correct. Did I break down the rational function incorrectly?

Tedjn
Check A+B+C=0.

Homework Helper
i think that's ok? but -A+B-2C+D = (-5 -2 +6-4)/3 = -5/3?

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stripes
so did I get all my coefficients messed up? Arghhh...

The Chaz
Count Iblis

Cheating to the third power

1) Obviously the Prof. doesn't want students to use Wolfram alpha.

2) The "show steps" feature adds insult to injury. Just knowing the correct answer is going to give away nontrivial information to the student.

3) And Wolfram alpha is lying to you when it shows the steps. Not that there is anything wrong in these steps, but these are not the steps Wolfram alpha uses itself when it computed the answer. If you ask Wolfram alpha to show the steps, what it does is crank up the high school algorithm that gives the same answer, as it is most likely that this will yield the desired steps (I mean, who else other than students having difficulties with their assignments would want to see this?).

Wolfram alpha uses series expansion methods to compute the answer, which is way more efficient than the usual high school method. The general rule of thumb is that methods that involve solving equations for variables are less efficient compared to methods that don't involve solving equations.

Mentor
so $$\frac{x^{2}-2x-1}{(x-1)^{2}(x^{2}+1)} = \frac{5/3}{x-1} - \frac{2/3}{(x-1)^{2}} - \frac{x-(4/3)}{x^{2}+1}$$

but apparently, the last statement is not correct. Did I break down the rational function incorrectly?
Finding the constants is tedious and error-prone, but checking them is relatively simple, and you should do this. If you get a common denominator for what you have on the right, you should end up with what you have on the left. If you do, then you have the right constants.

stripes
I ended up finding the right constants, they were 1, -1, 1, and -1 I think. Did the question again and got it right!

And as for the Wolfram Alpha thing...I use it when I am completely stumped. It gives me hints and guides me in the right direction. And this "high school method" you speak of makes me sound like a high-schooler, which I am most definitely not lolll.

Staff Emeritus
Homework Helper
One technique I like to use when solving for the coefficients is the Heaviside coverup method. When you get to this point

$$x^{2}-2x-1 = A(x-1)(x^{2}+1) + B(x^{2}+1) + (Cx + D)(x-1)^{2}$$

Substitute a value for x to eliminate some terms. In this case, x=1 works. This leaves you with $-2 = 2B$, and you can immediately see that B=-1. Ideally, you can use different values to solve to isolate different coefficients quickly. Alas, that's not the case in this problem. If you plug in the value you found for B, you get

$$2x^{2}-2x = A(x-1)(x^{2}+1) + (Cx + D)(x-1)^{2}$$

so you only have three equations and three unknowns to solve.

Count Iblis
I see! Well, this is how I would do this problem. I would start by writing:

x^2 - 2 x - 1 = x^2 - 2 x + 1 - 2 = (x-1)^2 - 2

So, we have:

(x^2 - 2 x - 1)/[(x-1)^2 (x^2 + 1)] =

1/(x^2+1) - 2/[(x-1)^2 (x^2 + 1)]

Expanding the last term about the point x = 1 and keeping only the singular terms yields:

- 2/[(x-1)^2 (x^2 + 1)] = -2/(x-1)^2 [expansion of 1/(x^2+1) around x = 1]

Put x = 1 + t:

1/(x^2+1) = 1/[(1+t)^2 + 1] = 1/(2 + 2 t + t^2) =

1/2 1/[1+t+t^2/2] = 1/2 [1-t +t^2/2 + ...]

So, we have the expansion:

- 2/[(x-1)^2 (x^2 + 1)] =

-1/(x-1)^2 + 1/(x-1) + nonsingular terms

If we now do the same at the singularities of 1/(x^2+1) and keep the singular terms, then the sum of all the singular terms from all the expansions, S(x), is the partial fraction expansion. This is because the difference between the rational function R(x) and S(x) will be a function that has no singularities, therefore it is a polynomial. But since both R(x) and S(x) tend to zero at infinity, we have that
R(x) = S(x).

So, we could now proceed to find the singular terms in the expansion around the singularities of 1/(x^2+1) at x = ±i. However, we can skip that as the partial fraction expansion is already determined by the above terms. This is because the large x behavior of

- 2/[(x-1)^2 (x^2 + 1)]

is -2/x^4, while the part of the partial fracton expansion we have so far is

-1/(x-1)^2 + 1/(x-1)

So, clearly the partial fraction expansion due to the singular terms coming from 1/(x^2+1) must cancel the asymptotic 1/x and 1/x^2 behavior of the above two terms. We know that the the remaining terms of the partial fraction expansion will be of the form:

(A x + B)/(x^2 +1)

Expanding around x = infinity gives:

(A x + B) 1/x^2 1/(1+1/x^2) =

A/x + B/x^2 + O(1/x^3)

Expanding the two terms we got so far about infinity gives:

-1/(x-1)^2 + 1/(x-1) =

-1/x^2 + 1/x 1/(1 - 1/x) + O(1/x^3) =

-1/x^2 + 1/x + 1/x^2 + O(1/x^3) =

1/x + O(1/x^3)

So, A = -1 and B = 0 and we have:

(x^2 - 2 x - 1)/[(x-1)^2 (x^2 + 1)] =

(1-x)/(x^2+1) -1/(x-1)^2 + 1/(x-1)

stripes
count iblis, B ≠ 0, B = -1

Count Iblis
count iblis, B ≠ 0, B = -1

Your B is my 1/(x-1)^2 coefficient.

Staff Emeritus
Homework Helper
I see! Well, this is how I would do this problem.
I wonder what the grader would think if stripes were to turn in his homework using that solution.

Count Iblis
I wonder what the grader would think if stripes were to turn in his homework using that solution.

It's not that different from the Heaviside coverup method you mentioned. If we have a rational function R(x) which has a linear factor in the denominator 1/(x-a)^n, then we can take that factor out and write:

R(x) = P(x) 1/(x-a)^n

So, we see that P(a) is the coefficient of 1/(x-a)^n (to do this "by the book", you would write out the equations as you did and then put x = a). To find the coeficient of 1/(x-a)^(n-1), using the Heaviside coverup method, one would subtract p(a)/(x-a)^n from R(x), simplify the rational function; you know that the numerator will contain a factor x-a that cancels against the denominator (you can use synthetic division to divide the numerator by x - a), so we can write:

R2(x) = R(x) - p(a)/(x-a)^n = P2(x)/(x-a)^(n-1)

And then we can repeat this by subtracting P2(a)/(x-a)^(n-1) from R2(x), simplify the rational function, divide numerator and denominator by x-a to obtain:

R3(x) = R2(x) - p2(a)/(x-a)^(n-1) = P3(x)/(x-a)^(n-2)

etc. etc.

But this whole process is equivalent to finding the first part of the Laurent expansion of R(x) around x = a, so you could just as well do that using any method that is most convenient, not necessarily using the above iterative process...

Count Iblis
So, in the way you wrote it down the next step would be to factor the left hand side of the last equation

2x^2 - 2x = 2x(x-1)

and cancel (x-1) on both sides:

2x = A(x^2+1) + (Cx+D) (x-1)


# Definition of 'average inclination'?

1. Mar 27, 2014

### dilloncyh

It's not a homework problem, just a problem that suddenly popped out of my mind.

1. The problem statement, all variables and given/known data

So, my question is : how to calculate, or how to define 'average inclination'? Suppose I am given an equation y=f(x) that resembles the shape of a section of a hill, and I want to calculate the average inclination (something I always see when I watch cycling or alpine skiing race), how do I do that? Let's use f(x) = x^2 as an example for the following discussion.

2. Relevant equations

I know the average of a function is defined as:
So I suppose the correct equation should be similar, with f(x) the function that gives the shape of the hill I am calculating?

3. The attempt at a solution

Here comes the problem: Do I use dx for ds? And for (b-a) in the image, I need to replace it with the actual length of the function from x=a to x=b, right?
Since I don't really know I want to calculate (to find the slope at each point of the function and add them together, and then divide it by the total length of the slope?), my question may seem very silly, but please give me some idea.

thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 27, 2014

### Simon Bridge

The average slope is usually just the rise-over-run for two positions on the slope.
$$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ ... that would be what they did on the ski races.

You want to be a bit more detailed than that:

If $y(x)$ is the height of the slope at position $x$,
Then $$\bar y = \frac{1}{b-a}\int_a^b y(x)\;\text{d}x$$ would be the average height of the hill in a<x<b.

The slope (gradient) function would be g(x)=dy/dx ... tells you the gradient at position x.

3. Mar 27, 2014

### dilloncyh

I'm still a bit confused.

average slope = detla y / delta x seems very legit, but take y=x^2 and y=x as example.

If I want to calculate the average slope between x=0 and x=1, then by calculating the difference in height and horizontal displacement of the two end points, then both should give the same result (slope=1), but the length of the curve of y=x^2 is obviously longer than y=x (which is just sq root of 2), so by calculating the the sine of the 'triangle' if I straighten the curve section of y=x^2, I will get difference result.

4. Mar 27, 2014

### haruspex

It's a matter of how you choose to define the average. The usual would be to define it as average over horizontal distance, but as you note you could instead chose to define it as average over path distance.

5. Mar 28, 2014

### Simon Bridge

I'm with haruspex - there is no reason that two different averaging methods should produce the same value.
They are just producing a different kind of average.


## Expansions of $e^x$

A very basic question was asked recently: What is a better approximation to $e^x$, the usual Taylor approximation, or a similar approximation involving $1/e^{-x}$?

More precisely, given an integer $m$, which is a better approximation to $e^x$:

$$f_1(x) = \sum_{k=0}^m \frac{x^k}{k!}$$

or

$$f_2(x) = \frac1{\displaystyle \sum_{k=0}^m \frac{(-1)^k x^k}{k!}}$$

The answer is amazingly simple: if $m$ is even, then $f_2$ is more accurate, while if $m$ is odd, $f_1$ is more accurate. The proof below is, in my opinion, really nifty and worth showing here. Keep in mind that this result follows from the fact that we are dealing with a remarkable function with very special properties. Other functions will not exhibit such easily calculable results as shown below. (Anyone familiar with Pade approximates can attest to how sticky rational approximations to functions can get.)

Keep in mind that, by “accuracy,” I refer to truncation error rather than roundoff error. Truncation error is that which results from substituting a Taylor series with a polynomial of finite order. The accuracy, i.e., truncation error, of any Taylor expansion of a given order is given by the next order of the expansion. For example, consider the first order approximation to $1/e^{-x}$, $f_2$, and its second-order error:

$$\frac1{1-x} = 1+x+x^2+O(x^3)$$

This approximation has twice the error as the first order approximation to $e^x$, $1+x$, whose second order term is $x^2/2!$.

Now consider the second order approximations with third order errors:

\begin{align}\frac1{1-x+x^2/2} &= 1+\left (x-\frac{x^2}{2} \right ) + \left (x-\frac{x^2}{2} \right )^2+ \left (x-\frac{x^2}{2} \right )^3 + \cdots \\&= 1+x+\frac{x^2}{2} + O(x^4)\end{align}

Note that, for the $f_2$ approximation, the third order error vanishes! This is obviously not the case for $f_1$. Thus, for the second-order expansion, $f_2$ has a smaller truncation error than $f_1$. However, for the next order approximation, the expansion of $f_2$ is

$$1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12} + O(x^5)$$

whereas the fourth order term in the direct, Taylor expansion of the exponential is $x^4/24$. Thus, as with the first order approximation, $f_2$ has twice the truncation error as $f_1$.

There is a systematic way to prove that the accuracy alternates with order oddness or evenness by considering the error in the expansion of $e^x \cdot e^{-x}=1$. The proof is actually not all that hard. Consider the finite approximations to the equation $e^x \cdot e^{-x}=1$:

$$\left (\sum_{k=0}^m \frac{x^k}{k!} \right ) \left (\sum_{k=0}^m (-1)^k \frac{x^k}{k!} \right )$$

It may be easily shown that the coefficients of $x$, $x^2$, …, $x^m$ are zero. We now consider the first error term, i.e., the coefficient of $x^{m+1}$:

$$\frac{x^1}{1!} \frac{(-1)^m x^m}{m!} + \frac{x^2}{2!} \frac{(-1)^{m-1} x^{m-1}}{(m-1)!} +\cdots + \frac{x^m}{m!} \frac{(-1)^1 x^1}{1!} = \sum_{k=1}^m \frac{(-1)^{m-k+1}}{k! (m-k+1)!} x^{m+1}$$

Rewrite this sum, sans the $x^{m+1}$ term, as

$$\sum_{k=0}^{m-1} \frac{(-1)^{m-k}}{(k+1)! (m-k)!} = (-1)^m \sum_{k=0}^{m-1} \frac{(-1)^k}{k! (m-k)!} \frac1{k+1}$$

We can evaluate this sum by realizing that

$$(1-x)^m = \sum_{k=0}^m (-1)^k \binom{m}{k} x^k$$

so that

$$\sum_{k=0}^m (-1)^k \binom{m}{k} \frac1{k+1} = \int_0^1 dx \, (1-x)^m = \frac1{m+1}$$

Therefore, by subtracting off the last term in the sum, we find that

$$(-1)^m \sum_{k=0}^{m-1} \frac{(-1)^k}{k! (m-k)!} \frac1{k+1} = – \frac{1-(-1)^m}{(m+1)!}$$

Therefore, we now may say that

$$\frac1{\displaystyle \sum_{k=0}^m (-1)^k \frac{x^k}{k!} } = \sum_{k=0}^m \frac{x^k}{k!} + \frac{1-(-1)^m}{(m+1)!} x^{m+1} + O(x^{m+2})$$

For even values of $m$, the next term error is zero, which is smaller than that for the direct Taylor series, which has error $x^{m+1}/(m+1)!$. On the other hand, for odd values, the error is double that of the direct Taylor series. This agrees with the above examples.

#### One Comment

• This was indeed very surprising and beautiful (and at the same time not too difficult to understand).


# Probability: Determining Which Phone Plan Is Better

A consumer is trying to decide between two long-distance calling plans. The first one charges a flat rate of $10$ cents per minute, whereas the second charges a flat rate of $99$ cents for calls up to 20 minutes in duration and then $10$ cents for each additional minute exceeding 20 (assume that calls lasting a non-integer number of minutes are charged proportionately to a whole-minute’s charge). Suppose the consumer’s distribution of call duration is exponential with parameter.

• Which plan is better if expected call duration is 10 minutes? 15 minutes? [Hint: Let $h_1(x)$ denote the cost for the first plan when call duration is x minutes and let $h_2(x)$ be the cost function for the second plan. Give expressions for these two cost functions, and then determine the expected cost for each plan.]

For the first function of cost, I got $h_1(x) = 0.10X$; and for the second one, I got the piece-wise function $h_2(x) = .99$ for $0 \le X \le 20$, and $h_2(x) = .99$ for $X > 20$

Are these correct? Also, how do I calculate the expected cost for each plan? I used the formula given in this link http://en.wikipedia.org/wiki/Expected_value#Functional_non-invariance; however, it didn't work.

EDIT: $E[h_1(x)] = lim_{a \rightarrow \infty} \int^a_0 .10x \lambda e^{-\lambda x}dx$

After doing integration by parts, and simplifying, I get to the step $lim_{a \rightarrow \infty}[-.10ae^{-\lambda a} + \frac{.1}{\lambda}(e^{-\lambda a})]$. I know that the second term goes to zero, and, according to my teacher, the first term goes to zero as well, because $e^{-\lambda a}$ goes to zero more quickly than $-.10a$ goes to negative infinity, although I don't really understand why. Note, the answer is not zero

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If the second plan charges $99$ cents plus an additional $10$ cents per minute after $20$ minutes, I don't think $h_2(x) = 0.99$ for $X > 20$ is correct. –  TMM Mar 12 '13 at 11:32
@TMM Should it be $h_2(x) = .99 + 0.10X$ for $X > 20$? –  Mack Mar 12 '13 at 11:55
@Eli No It is $0.99 + .1(X-20)$ –  Gautam Shenoy Mar 12 '13 at 11:59
@GautamShenoy Are you sure? Can you explain why that is the answer? –  Mack Mar 12 '13 at 12:12
@Eli: What I meant was: He says it is 10 cents for every minute exceeding 20. So in place of X, shouldn't it be X-20 ? It is not the answer. I am just correcting the expression you gave. –  Gautam Shenoy Mar 13 '13 at 9:15

All costs are in dollars.

Your $h_1(X) = 0.1X$ is correct. The second term is $$h_2(X) = 0.99 + 0.1(X-20)1_{X > 20}$$

Now if you compute the expected cost, for $\frac{1}{\lambda} = 10$, it is

$\mathbb{E}h_1(X) = \frac{0.1}{\lambda}= 1$

But for $$\mathbb{E}h_2(X) = 0.99 + 0.1 \mathbb{E}(X-20)1_{X>20}$$ $$= .99 - 2\mathbb{P}(X>20) + .1\mathbb{E}(X1_{X>20})$$ $$= .99 - 2e^{-\lambda 20} + .1\mathbb{E}(X1_{X>20})$$ Thus we just need to compute $\mathbb{E}(X1_{X>20})$. You could do it directly, but I will provide an alternate way.

$$E[X1_{X>20}] = \int_0^\infty x1_{x>20}f_x(x)dx$$ $$= \int_{x=0}^\infty \int_{u=0}^\infty 1_{u<x}1_{x>20}f_x(x)dudx$$ $$= \int_{u=0}^\infty \int_{x=0}^\infty 1_{u<x}1_{x>20}f_x(x)dudx$$ (Fubini theorem) $$= \int_{u=0}^\infty \int_{x=u}^\infty 1_{x>20}f_x(x)dxdu$$ $$= \int_{u=0}^\infty \int_{x=\max(u,20)}^\infty f_x(x)dxdu$$ $$= \int_{u=0}^\infty P(X > \max(u,20))du$$

Now split the integral as $$= \int_{u=20}^\infty P(X > u)du + \int_{u=0}^{20}P(X > 20)du$$ $$= \int_{20}^\infty e^{-\lambda u}du + \int_{u=0}^{20}e^{-\lambda 20}du$$

I could work out the rest but can you proceed from here? Also in reply to your question: why is $\lim_{x \to \infty} xe^{-x} = 0$, do you know L-Hospital's rule?

-


# How to prove this series converges?

What test do you use to prove that

$$\sum_{n=2}^\infty \frac{\ln(n)}{n^{3/2}}$$

converges?

I tried the limit comparison test using $\frac{1}{n^{3/2}}$ as the comparison, but it did not converge.

-

Hint: You can use integral test.

Added: You need to consider the integral

$$\int_{2}^{\infty} \frac{\ln(x)}{x^{3/2}}dx ,$$

which can be integrated using integration by parts.

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@MhenniBenghorbal, there is a typo, the lower bound should be 2, but it doesn't seem like it's going to matter that much for this question. –  Nameless Apr 30 '14 at 2:47
@Nameless: It is corrected. Thanks for the comment. –  Mhenni Benghorbal Apr 30 '14 at 2:48
Using the integral test I got the answer to be = 25.55275762. –  khap93 Apr 30 '14 at 2:49
@khap93, you are not supposed to directly calculate the sum...well I guess that shows you the sum does converge. –  Nameless Apr 30 '14 at 2:49
@MhenniBenghorbal, no problem. I'll upvote you. –  Nameless Apr 30 '14 at 2:50

Apply Cauchy Condensation Test.

Let $a_n = \dfrac{\ln(n)}{n^{3/2}}$, then $a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}}$

Therefore, $$a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}} = \frac{n\ln 2 }{2^{3n/2}} \leq \ln2\frac{n}{2^n}$$

Thus $$\sum_{n=2}^{\infty}\dfrac{\ln 2^n }{(2^n)^{3/2}} \leq \sum_{n=2}^{\infty} \ln2\frac{n}{2^n}.$$

The RHS is a convergent series, by the Ratio Test, as you should verify.

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Neat footwork there :-) –  Carl Witthoft Apr 30 '14 at 13:17

You can use that $\ln n$ goes to infinitely much more slowly that any (positive) power of $n$, that is: for every $\alpha>0$, $\ln n < n^\alpha$ for $n$ large enough. You can apply this to $\alpha=1/4$ says (any $\alpha$ strictly between $0$ and $1/2$ will do, so let's say $1/4$), so $\ln n < n^{1/4}$ for $n$ large enough, hence $\ln n / n^{3/2} < 1/n^{5/4}$ for $n$ large enough, and since the series $\sum 1/n^{5/4}$ converges (as does $\sum 1/n^\beta$ for any $\beta>1$), you're done.

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+1 the most direct and conceptually simple solution. –  jwg Apr 30 '14 at 10:19


+0

0
330
8

Let $b$ and $c$ be constants such that the quadratic $-2x^2 +bx +c$ has roots $3+\sqrt{5}$ and $3-\sqrt{5}$. Find the vertex of the graph of the equation $y=-2x^2 + bx + c$.

Guest Feb 19, 2018
#1
+12560
+2

There MIGHT (probably) be an easier way, but here is ONE way:

(x+(-3-sqrt5))(x+(-3+sqrt5))   is given in the question for the roots

if you multiply these two term out you will get

x^2-6x+4    now multiply by -2

-2x^2 +12x-8 =0     (a parabola)

to find vertex take derivative and set equal to zero

-4x+12 = 0     x = 3    substitue this into the equation

to get      y=10     for the vertex    3,10

ElectricPavlov  Feb 19, 2018
edited by ElectricPavlov  Feb 19, 2018
#3
0

Should it be (x+(3-sqrt 5))(x+(3+sqrt5))?

Guest Feb 21, 2018
#4
0

Shouldnt it be (-3,10)?

Guest Feb 21, 2018
#5
+12560
+2

The roots are given   as   3+-sqrt5

so one part would be

x+(- 3+sqrt5) =0

x-3+sqrt5  = 0     solve for x   (add  (3 and subtract sqrt5) from both sides )

x= 3 -sqrt5

and the other would be  x+( -3-sqrt5)  =0

x-3-sqrt5    solve for x  (add 3 and sqrt 5 to both sides

x = 3+sqrt5

These are the roots given in the question....

ElectricPavlov  Feb 21, 2018
#6
+12560
+2

Here is a graph

ElectricPavlov  Feb 21, 2018
#7
+12560
+2

y=  a (x-h)^2  +k        and

y= -2 (x-3)^2 +10

Now can you see that a =-2    h=3   k = 10  ???

ElectricPavlov  Feb 21, 2018
#2
+12560
+2

OK.....another way...

AFTER you get to

-2x^2 +12x -8                arrange into vertex form  y=a(x-h)^2 +k    the vertex is h,k

-2{ x^2 -6x +4}

-2{(x-3)^2 +4 -9}

-2 {(x-3)^2 -5}

-2 (x-3)^2 +10             so  h,k is the vertex   =   3,10

ElectricPavlov  Feb 19, 2018
#8
+92805
+1

I am quite sure EP will be correct but I will take a look:

Let b and c be constants such that the quadratic $$-2x^2 +bx +c$$

has roots $$3+\sqrt{5}$$  and $$3-\sqrt{5}$$.

Find the vertex of the graph of the equation

$$y=-2x^2 + bx + c$$

The roots of a quadratic   $$ax^2+bx+c$$  is the answers to   $$ax^2+bx+c=0$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=3\pm\sqrt5$$

This means that  $$x=\frac{-b}{2a}$$ = 3      must be exactly halfway between the two roots.!!

So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3

The y value will be    $$y=-2*3^2+bx+c = -18+3b+c$$

This is fine but you need to find the vleu of b and c

From the equation above I can see that

$$\frac{-b}{2a}=3 \qquad and \qquad \pm\frac{\sqrt{b^2-4ac}}{2a}=\pm\sqrt5\\ a=-2\qquad so\\ \frac{-b}{-4}=3 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{-4}=\pm\sqrt5\\ b=12 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{4}=\pm\sqrt5\\ b=12 \qquad and \qquad \frac{b^2+8c}{16}=5\\ b=12 \qquad and \qquad b^2+8c=80\\ b=12 \qquad and \qquad 144+8c=80\\ b=12 \qquad and \qquad 18+c=10\\ b=12 \qquad and \qquad c=-8\\$$

so

$$y= -18+3b+c\\ y= -18+3*12+-8\\ y=-18+36-8\\ y=10$$

So the vertex is   (3,10)

Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!

Thanks EP :)

Melody  Feb 21, 2018


# Convergence of $a_{n+1}=\frac{1}{1+a_n}$

We define

$$a_{n+1}=\frac{1}{1+a_n}, a_0=c> 0$$

I stumbled across that sequence and Mathematica gives me that it converges against $\frac{1}{2} \left(\sqrt{5}-1\right)$ which doesn't even depend on $a_0$. Normally I show that sequences converge by seeing that they are monotonic and then the limit can be easily found by setting $a_n=a_{n+1}$, but this one seems to be alternating. Also by checking OEIS I noticed that $a_n=1/g(n)$ where $g(n)$ gives the $n+1$'th Golden Rectangle Number.

I would be glad if someone can show me how to show that such sequences converge and how to find the limit, also maybe this is a well known sequence, as it has a quite simple form, too bad that google is very bad for looking for sequences and OEIS doesn't mention anything.

• These numbers are the convergents of the continued fraction of the golden ratio. If you assume that it converges you can calculate the limit by replacing the $a_k$s by $a$. You can prove convergence by proving that the odd/even subsequences are monotonous. May 12 '11 at 18:33
• Right, but 3123 is asking why $\frac{\sqrt{5}-1}{2}$ is a(n attractive) fixed point of the function $f(x)=\frac1{a+x}$ for "any" starting value... May 12 '11 at 18:37
• The reason why $a_n$ converges to a fixed value irrespective of $a_0 = c$ is actually not really surprising. For instance, consider $a_0 = \frac{c}{10}$ where $c \in [0,10]$ $a_{n+1} = \frac{9+a_n}{10}$ Note that $\displaystyle \lim_{n\rightarrow \infty} a_n = 1$ for any starting value $c \in [0,10]$
– user17762
May 12 '11 at 18:50
• May 13 '11 at 11:18
• math.stackexchange.com/questions/235578/… Apr 20 '13 at 13:19

## 3 Answers

The function $1/(1+x)$ is strictly decreasing and thus order-reversing on the positive reals. If you apply it twice it conserves order.

Therefore, the odd and even subsequences are monotonic. They are also bounded as all your values except the first are bounded by 1.

Now you can just replace $a_n$ and $a_{n+1}$ by $a$ to find the limit.

• Thank you for your answer, this was very useful too. May 12 '11 at 18:45

As for why the first term does not matter, consider the $6$th term,

$$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+c}}}}}$$

as $c$ varies between $0$ and $\infty$ the value of the $6$th term varies only by $0.1$. As $c$ varies between $0$ and $\infty$ the $100$th term varies only by very small $\varepsilon$. In the infinite limit this variation is zero.

This also explains convergence. Since convergence occurs call the number $\varphi$, it satisfies the fixed point equation

$\varphi = \frac{1}{1+\varphi}$

which, multiplying both sides by the denominator, is a quadratic equation hence the square root.

• Thank you for the answer, writing it as a continued fraction makes things much more obvious May 12 '11 at 18:44

For $x\geq 0$, $f(x)=1/(1+x)$ is a monotonically decreasing function, bounded between 0 and 1. The fixed point (if it exists) of the iterative map defined by your function, is the solution of $x=\frac{1}{1+x}$. There is only one solution for $x\geq 0$, and that is $\tilde{x}=(\sqrt{5}-1)/2$.

If we look at the evolution of $a_n=f(f(f(...(a_0))))$, we get

$$a_n=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a_0}}}}}$$

and as $n\to\infty$, what $a_0$ was becomes irrelevant. In other words, no matter the value of $a_0$, it converges to a fixed point, which has to be the only fixed point, $\tilde{x}$. Hence $\tilde{x}$ is an attractive fixed point and also the limit of the sequence.

The Mathematica code that produced the above figure is below (it's modified from an answer to a question on SO)

Clear[fixedPoint]
fixedPoint[a0_, nSteps_] :=
Module[{f, a, n},
f = RecurrenceTable[{a[n] == 1/(1 + a[n - 1]), a[1] == a0},
a, {n, 1, nSteps}];
Plot[{1/(1 + x), x}, {x, 0, 1},
PlotStyle -> {Darker@Blue, {Dashed, Black}},
Epilog -> {Darker@Green,
Line[Riffle[Partition[f, 2, 1], {#, #} & /@ Rest[f]]]}]]