# CaptainBlacks Occasional Problem #3

#### CaptainBlack

##### Well-known member
Prove that the polynomials:

$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$

have no real roots.

CB

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#### Opalg

##### MHB Oldtimer
Staff member
Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that

$P_1(x) = x^2-2x+3 = (x-1)^2+2,$

$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$

$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$

That should be enough of a pattern to suggest a solution.

#### Alexmahone

##### Active member
Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that

$P_1(x) = x^2-2x+3 = (x-1)^2+2,$

$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$

$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$

That should be enough of a pattern to suggest a solution.
Claim: $P_n(x)=(x-1)^2R_n(x)+n+1$

Let $Q_n(x)=P_n(x)-n-1=x^{2n}-2x^{2n-1}+3x^{2n-2}-\ldots-2nx+n$

$Q_n(1)=1-2+3-\ldots-2n+n=0$

$Q_n'(x)=2nx^{2n-1}-2(2n-1)x^{2n-2}+3(2n-2)x^{2n-3}-\ldots-2n$

$Q_n'(1)=2n-2(2n-1)+3(2n-2)-\ldots-2n=0$ (1st term cancels the last term, 2nd term cancels the 2nd last term etc.)

$Q_n(1)=Q_n'(1)=0\implies Q_n(x)=(x-1)^2R_n(x)$

So, $P_n(x)=(x-1)^2R_n(x)+n+1$

Now we just need to prove that $R_n(x)$ is always positive. It looks like $R_n(x)=x^{2n-2}+2x^{2n-4}+3x^{2n-6}+\ldots+n$. This needs a proof.

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#### CaptainBlack

##### Well-known member
Prove that the polynomials:

$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$

have no real roots.

CB
A slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.

CB

#### anemone

##### MHB POTW Director
Staff member
Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that

$P_1(x) = x^2-2x+3 = (x-1)^2+2,$

$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$

$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$

That should be enough of a pattern to suggest a solution.
By following the hints given by Opalg, I see that

$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$

It's obvious that $P_n(x)\ne 0$ for all $x\in R$.

Thus it suggests that $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$ has no real roots must be true.

Am I on the right track?

#### Alexmahone

##### Active member
Am I on the right track?
Yes. You have to prove this, though:

$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$

#### anemone

##### MHB POTW Director
Staff member
A slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.

CB
If I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.

First, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots.

From $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$, I get
$P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-......-2n(-x)+(2n+1)$
$P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-......+2nx+(2n+1)$
No sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.

Now, I consider the case where $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$.
I see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.

That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ......, n$

But we can also tell right from the start that the product of all roots of $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ is -(2n+1) for n=1, 2, ......, that is, the product of all the real positive roots (be it two integers/fractions or two pair of positive complex roots) = -ve.

Obviously this leads to contradiction and we can conclude at this point that the polynomial $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has no real roots.

Edit: I made a horrible mistake by saying that the product of all roots = - (2n+1) which is wrong. It should be (2n+1). So, my effort in proving in this post have been amounted to zero, zip, nothing. Please kindly dismiss it.

---------- Post added at 07:02 ---------- Previous post was at 07:00 ----------

Yes. You have to prove this, though:
I think the pattern could be generalized from the hint given by Opalg.

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#### CaptainBlack

##### Well-known member
If I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.

First, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots.

From $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$, I get
$P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-......-2n(-x)+(2n+1)$
$P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-......+2nx+(2n+1)$
No sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.

Now, I consider the case where $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$.
I see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.

That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ......, n$
What are positive complex roots, and where will you find a reference to what Descartes rule of signs says about this sort of sign of the complex roots?

CB

#### anemone

##### MHB POTW Director
Staff member
My mistakes.
I want to clarify that the following observations were not right.
"That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ......, n$
"
Reason:
1. I didn't know why on earth I called $a\pm b\sqrt {c}$ as complex roots. I'm sorry.
2. This "...based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2." was taken at http://en.wikipedia.org/wiki/Descartes'_rule_of_signs whereas the two generated cases were my own conclusions.

I'm sorry for confusing you, CB.
My apologies.

#### Opalg

##### MHB Oldtimer
Staff member
Yes. You have to prove this, though:
$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$
My proof of that was just to multiply out the product on the right hand side:

\begin{aligned}(x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x^2-2x+1) \\ =& x^{2n}\phantom{{}-2x^{2n-1}}+2x^{2n-2}\phantom{{}-4x^{2n-3}}+3x^{2n-4} \phantom{{}-2x^{2n-1}}+ \ldots \\ &\phantom{x^{2n}} - 2x^{2n-1} \phantom{{}+2x^{2n-2}} - 4x^{2n-3} \phantom{{}+2x^{2n-2}} -6x^{2n-5} -\ldots \\ &\phantom{x^{2n} - 2x^{2n-1}} +\phantom{1}x^{2n-2}\phantom{{}-2x^{2n-1}} + 2x^{2n-4} \phantom{{}-2x^{2n-1}} +\ldots \\ &\phantom{x}\\ =& x^{2n}-2x^{2n-1}+3x^{2n-2}-4x^{2n-3} + 5x^{2n-4}-6x^{2n-5}+\ldots \end{aligned}

For $0\leqslant k\leqslant n-1$ the coefficient of $x^{2n-2k}$ is $k+(k+1)=2k+1$ and the coefficient of $x^{2n-2k-1}$ is $-2(k+1)$. The constant term is $n$. So the resulting polynomial is $P_n(x) - (n+1)$.

#### CaptainBlack

##### Well-known member
Solution: CaptainBlacks Occasional Problem #3

This again is from the Purdue PotW

Prove that the polynomials:

$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$

have no real roots.

================================================

Solution:
These obviously have no negative roots (either from Descartes rule of signs or observing that the odd powers of $$x$$ all have -ve signs)

So for $$x>0$$ consider:
$P_n(x)+xP_n(x)=x(x^{2n}-x^{2n-1}+x^{2n-2}- ... - x+1) + (2n+1)$
Now the bracketed term is a finite geometric series and so we may replace it with its sum to get:
$P_n(x)(1+x)=x \left[ \frac{1+x^{2n+1}}{1+x}\right] + (2n+1)$
or:
$P_n(x)=x \left[ \frac{1+x^{2n+1}}{(1+x)^2}\right] + \frac{2n+1}{1+x}>0$
and so as $$P_n(0) \ne 0$$ we have $$P_n(x)>0$$ for all real $$x$$

The above is somewhat different from the solution given on the Purdue PotW site, mainly in that it avoids an unexplained step, which while probably valid I find opaque.

CB

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Drawing without replacement - prob. for an Ace followed by an Ace?

Given a standard 52-cards deck:
You are extracting cards from the deck without replacement, until you get an "Ace" for the first time. What is the probability that the next card will be "Ace" too?

I've already seen the following Q&A: Probability of drawing an Ace: before and after
According to that thread, the answer should be: $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$

But the official solution says that the answer is: $$\frac{4}{52}$$ which doesn't make sense IMHO. They solved it only with intuition or "mind trick", as they wrote..

My calculation:

Assuming that the 1st card is Ace, then: $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$

Assuming that the 2nd card is Ace, then: $$\frac{(52-4) \cdot 4 \cdot 3}{52 \cdot 51 \cdot 50} = \frac{24}{5525}$$

We notice a pattern here.
Having the 1st Ace at the $k$'th draw, then the probability (for a second Ace after that) is: $$p_1 = \frac{ {_{52-4}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }$$

We need to consider the least-possible scenario - we draw 48 non-Ace cards, then: $$p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } = \frac{1}{270725}$$

So, the required probability is: \begin{align} p &= p_2 + \sum\limits^{48}_{k=1} p_1 \\ &= p_2 + \sum\limits^{48}_{k=1} \frac{ {_{48}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }\\ &= \frac{1}{270725} + \frac{1696}{20825}\\ &= \frac{1297}{15925}\\ &\cong 0.081444 \end{align}

But my answer is far from either the official solution and from the answer in that thread.

• The question you link to is different. It is asking "what's the probability that the second drawn is an Ace?" The answer to that is certainly $\frac 4{52}$ as there is no difference between the second card drawn and the first. Of course, if you specify that the first card drawn was (or was not) itself an Ace then the probability changes. – lulu Oct 9 '15 at 14:35
• @lulu I think that I understand your view on the matter. So what did I calculated? – Dor Oct 9 '15 at 15:07
• I think your method ought to work. To be precise: I just wrote a code which first computes $p_k$, the probability that the first A is observed on the $k^{th}$ draw, and then calculates the probability that the next draw is also an A. I believe this is what you were doing as well. My code returned $\frac 1{13}$ as the total. Doesn't convey a lot of insight, I understand. – lulu Oct 9 '15 at 15:13
• @lulu Indeed you're right! I posted an answer which shows where were my mistakes and include a code that proves it to be equal to $\frac{1}{13}$. – Dor Oct 9 '15 at 16:27

We use a fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables."

Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest.

Maybe they are in positions 2 and 3. That leaves $\binom{49}{2}$ for the rest.

Maybe they are in positions 3 and 4. That leaves $\binom{48}{2}$ for the rest.

And so on, up to positions 49 and 50, leaving $\binom{2}{2}$ for the rest. That gives probability $$\frac{\binom{50}{2}+\binom{49}{2}+\cdots+\binom{2}{2}}{\binom{52}{4}}.$$ The numerator can be simplified in various ways. My favourite is to count the number of ways to choose $3$ numbers from the first $51$. This can be done in $\binom{51}{3}$ ways. But the smallest number can be $1$, leaving $\binom{50}{2}$ ways to choose the other two. Or the smallest can be $2$, and so on. We get the number of favourables obtained above.

Our probability is therefore $\frac{\binom{51}{3}}{\binom{52}{4}}$. This simplifies to $\frac{1}{13}$.

• Great argument and very well written. +1. – Shailesh Oct 9 '15 at 15:04
• I like this approach! +1. Could you please elaborate regarding how you simplified the numerator? – Dor Oct 9 '15 at 15:13
• I think that I understand it now. That is a very cool solution..! – Dor Oct 9 '15 at 15:33

First of all Andre has posted a very nice answer. Already Upvoted it. I am trying to look at where all you have gone wrong. Too long for an comment, so posting it here.

Let us look at the last case. You have $48!$ ways of the non-ace cards. Now the remaining cards are all aces anyways. So the question of multiplying this by $4 \cdot 3$ does not arise. Since we do not care about the last 2 cards, the denominator will have to be just $50!$. Similarly, the case before that might need to rectified.

• Indeed you are right about the last term! In the other 48 terms, the mistake was different. I'll add info in my original post. Thanks! +1 – Dor Oct 9 '15 at 16:08
• @Dor Yes, right. A good approach would be you post an answer yourself, instead of editing the post. I'll surely upvote it. – Shailesh Oct 9 '15 at 16:18

Here is a bijection between set of all permutation with the first Ace followed by an Ace and set of all permutations with the first card being an Ace

$$B^nAAX <-> AB^nAX$$

where A stands for an Ace, B for not an Ace, $0\leq n \leq 48$ and $X$ contains $2$ Aces and $48-n$ non-Aces. Hence the probability of an Ace following the first Ace is equail to the probability of the first card being Ace which is equal to $$4/52=1/13$$

Probabilistically, conditioned on the location of the second ace, the first ace is uniformly distributed on locations before the second ace - hence probability of it being the first or the last in that sequence are equal.

• Nice! I was going to go back and try to explain that the fact my fraction simplified to $\frac{4}{52}$ was not an "accident." You have done it, in the strongest (natural bijection) sense. – André Nicolas Oct 10 '15 at 1:27
• I started thinking along the lines of bijection after seeing your result ;). – A.S. Oct 10 '15 at 1:30
• Yes, it was kind of shouting look at me, look at me. – André Nicolas Oct 10 '15 at 1:32

Fixing my original calculation

Thanks to Shailesh, we now know that the expression $$p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} }$$ Should be replaced with: $$p_2 = \frac{48!}{ {_{52}P_{48}} }$$ Because that after extracting 48 non-Ace cards, we'll necessarily have the second card as Ace.

Moreover, I had a tiny mistake in the denominator. Instead: $${_{52}P_{k-1}} \cdot {_{52-k}P_{2}}$$ It should be: $${_{52}P_{k-1}} \cdot {_{52-(k - 1)}P_{2}}$$ (All the $k$ indexes should be shifted by $(-1)$)

I wrote a Python code which shows that this solution works either:

# Python 2.7.6
import math
from fractions import Fraction

f = math.factorial

def nPr(n,r):
return f(n) / f(n-r)

p = Fraction(0)

for k in range(1, 49):
p += Fraction(nPr(48, k-1) * 4 * 3,
nPr(52, k-1) * nPr(52 - (k-1), 2))

p += Fraction(f(48), nPr(52,48))

print p # Outputs 1/13 :)

Thank you all for your effort! :)

• Well done, Dor, We all learn with each other's help. – Shailesh Oct 9 '15 at 16:30


Is there a differential limit?

I'm wondering if there's such a concept as a "differential limit". Let me give an example because my nomenclature is my own and unofficial, but hopefully indicative of the concept.

For some function f(x), there may exist a derivative of that function f'(x) which we call a first order differentiation of f(x). Likewise from f'(x) we could also derive a second order function f''(x). Is there a mathematical concept of the limit of a function as the order of differentiation goes to infinity?

I can think of a function that I can take the derivative of an infinite number of times ( albeit without getting a constant value): f(x) = sin(x)

The function g(x) = x^2 might have a differential limit of 0. (g'(x) = 2x; g''(x) = 2; g'''(x) = 0 ...)

Is there any usefulness to this concept? A topic of study or keyword that discusses this? Any more examples of infinitely derivable functions?

Most importantly if this is interesting to anyone other than myself: what is a good resource for learning more?

-
We use infinitely differentiable functions often in mathematics. We call it the set $C^{\infty}$. A function that is infinitely differentiable is called "smooth" mathworld.wolfram.com/C-InfinityFunction.html – tomcuchta Jun 13 '12 at 3:35

First: yes, we care about how many times a function can be differentiated. In fact, we have names for such classes: $\mathcal{C}_0$ is the collection of all continuous functions, $\mathcal{C}_1$ the class of functions that have continuous derivative; $\mathcal{C}_2$ the class of functions that have continuous second derivative; and so on. We have that $$\mathcal{C}_0 \supsetneq \mathcal{C}_1 \supsetneq\mathcal{C}_2\supsetneq\mathcal{C}_3\supsetneq \cdots \supsetneq \mathcal{C}_n\supsetneq \cdots$$

Then we have "infinitely differentiable functions", functions that have continuous derivatives of all orders; this is denoted $\mathcal{C}_{\infty}$. There are functions that have derivatives of all orders that don't cycle through values, such as $e^{x^2}$.

Even beyond the infinitely differentiable functions we have the notion of "analytic function", which are functions that can be defined at every point by a power series that converges on an open interval containing the point.

Now, I think you are talking about trying to make sense of a "limit" of the functions as you take derivatives. So that for a polynomial the "limit" should be $0$, a function like $\sin x$ should not have a limit, and so on.

There are in fact several different ways of making the idea of "differential limit" precise. I'm going to talk more generally about what "limit" and "convergence" can mean here, before addressing the specific construction you are talking about. I'll spill the beans and say that I haven't really seen this particular construction before (there's too much to wade through below, and I don't want you to be annoyed when you reach the end and find that I say "I haven't seen it...") But that does not mean you can't ask interesting questions about it.

What you describe is a sequence of functions. The main issue is that one has to define what one means by saying that two functions are "close to one another". Then you get the notion of convergence of functions.

Turns out that there are many different ways of saying this, and which one is useful depends on the context.

Let's take the case of a sequence of real valued functions of real variable, with domain all of $\mathbb{R}$. Let's call the sequence $f_n$ (in your case, $f_n = \frac{d^nf}{dx^n}$ for a given $f$).

1. The most natural way of saying that the sequence $f_n$ converges to a function $f$ is:

$f_n\to f$ if for every $x$, $\lim\limits_{n\to\infty}f_n(x) = f(x)$.

This is called pointwise convergence. For example, the sequence $f_n(x) = x^n$ defined on $[0,1]$ converges pointwise to the function $$f(x) = \left\{\begin{array}{ll} 0 &\text{if }0\leq x\lt 1\\ 1 & \text{if }x=1. \end{array}\right.$$

It turns out that pointwise convergence is natural, but is not very "good", in the sense that a lot of properties we find interesting about functions are not preserved by pointwise convergence. For example, above, all functions $f_n$ are continuous, but their limit is not.

The problem arises because we are forcing no connection between how $f_n(x)$ converges to $f(x)$ and how $f_n(y)$ converges to $f(y)$ with $x\neq y$; that means that convergence on some points can "lag behind" convergence in points very close to them. This can be remedied by making things a bit tighter. If we write out the definition of the limits above, we obtain the following description:

$\{f_n\}$ converge pointwise to $f$ if and only if for every $x$, for every $\epsilon\gt 0$, there exists $N$, which may depend on both $x$ and $\epsilon$, such that for all $n\geq N$ we have $|f_n(x)-f(x)|\lt \epsilon$.

The way to make this tighter, to make sure the the $f_n(x)$ converge to $f$ more or less "the same way", is to force the $N$ to depend only on $\epsilon$, and not on $x$ as well. This is accomplished with:

2. We say that $\{f_n\}$ converge uniformly to $f$ if and only if for every $\epsilon\gt 0$ there exists $N$, which depends only on $\epsilon$, such that for all $x$ and all $n\geq N$ we have $|f_n(x)-f(x)|\lt\epsilon$.

Uniform convergence is better than pointwise convergence: if every element of the sequence is continuous, and the sequence converges uniformly, then the limit is continuous, for example.

There are yet other ways of talking about functions converging. Here are a few:

1. Fix a real number $p$, $1\leq p$. We say that $\{f_n\}$ converges to $f$ in $p$-norm if and only if $$\lim_{n\to\infty}\int_{\infty}^{\infty}|f_n(x)-f(x)|^p\,dx = 0.$$

2. Closely related: we say that $\{f_n\}$ converges to $f$ in the sup-norm if and only if $$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|\} = 0.$$

3. There is a way of measuring sets of real numbers, called the Lebesgue measure, $\lambda$, the details of which are perhaps too much to go over now. But intuitively, it is a way of assigning a "size" to a lot of sets of real numbers, including all intervals and much more complicated sets, in a nice and consistent way. We say that $\{f_n\}$ converges in measure to $f$ if and only if for every $\epsilon\gt 0$, $$\lim_{n\to\infty}\lambda(\{x\mid |f_n(x)-f(x)|\geq\epsilon\}) = 0.$$

4. We say that $\{f_n\}$ converges almost uniformly to $f$ if and only if for every $\epsilon\gt 0$ there is a set $X\subseteq \mathbb{R}$, with $\lambda(X)\lt\epsilon$, such that $\{f_n\}$ converges uniformly to $f$ on $\mathbb{R}-X$.

5. There are small variations to each of those; depending on context, one may be able to define other ways of convergence, or some of the above may no longer make sense. A typical variation is to take any of the above types of convergence (except almost uniform, for which the concept doesn't add anything), and add the condition "almost everywhere" (or "almost surely" if you are doing probability). That means that there exists a set $Y$ contained in the domain, of measure $0$, such that the type of convergence you are interested in occurs in $\mathbb{R}-Y$. (Sets of measure $0$ need not be empty, and they can be quite complicated: for example, any countable set has measure $0$, and there are uncountable sets that do as well, like the Cantor ternary set.

Another variation is to define the concept of "Cauchy sequence" for each of the types, which means making the analog of the definition of "Cauchy sequence" of real numbers: instead of comparing to a limit, we ask that for all $n$ and $m$ greater than $N$, $f_n$ and $f_m$ have the relevant property. For instance, "pointwise Cauchy" means that for all $x$ and all $\epsilon\gt 0$ there exists $N$ such that for all $n,m\gt N$ we have $|f_n(x)-f_m(x)|\lt \delta$. "Cauchy in measure" means that for all $\epsilon\gt 0$ and all $\delta\gt 0$ there exists $N$ such that if $n,m\gt N$, then $\lambda(\{x\mid |f_n(x)-f_m(x)|\geq\epsilon\}\lt\delta$; and so on.

All of these concepts are extremely important and useful. They show up in probability, statistics, functional analysis, analysis, and other areas. There are known implications (e.g., uniform convergence implies pointwise convergence, and so on).

Now, what about the specific case in which we start with a function $f_0$ (necessarily in $\mathcal{C}_{\infty}$) and we define $f_n$ to be the $n$th derivative of $f_0$? I confess I haven't seen this particular sequence addressed, but you can ask all the questions related to the convergence types above: is the sequence pointwise Cauchy? $p$-norm Cauchy? Cauchy in measure? Does it converge uniformly? Almost uniformly? Pointwise? In measure? And so on. Seems like an interesting question, if nothing else.

-
Thank you for the thorough answer, that was exactly what I was looking for. – jpredham Jun 13 '12 at 8:58
I think in your first #2 there is a slight typo - we should have that for $n\ge N$ the property holds. – process91 Jun 13 '12 at 13:54
@Michael: Yes, indeed. Thank you. – Arturo Magidin Jun 13 '12 at 15:25

It would seem that if the sequence $f$, $f'$, $f''$, ..., $f^{(n)}$, ... tends to a limit as a function, then that limit would be a solution of the differential equation $\frac{dy}{dx}=y$. So if the limit exists, it is either zero or some multiple of $e^x$. (I'm assuming that the notion of limit we're using is one that the differentiation operator is continuous with respect to, which doesn't seem like much to ask. Pointwise convergence is out, though).

Also, if we have some function whose iterated derivatives converge towards $ce^x$, we can subtract $ce^x$ from the initial function and get a new sequence whose iterated derivatives converge towards the zero function. So if we look for solutions to $\lim_{n\to\infty} f^{(n)}=0$ we essentially get all functions whose iterated derivatives have a limit.

All polynomials qualify, of course, but there are also non-polynomial analytic solutions such as $\sum_{n=0}^\infty \frac{1}{n\cdot n!} x^n$ and in general $\sum_{n=0}^\infty \frac{a_n}{n!}x^n$ whenever $\lim_{n\to \infty} a_n = 0$. I wonder whether some of these have nice closed forms.

-


# Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results [duplicate]

I tried Evaluating $$\int \sqrt{\dfrac{5-x}{x-2}}dx$$ using two different methods and got two different results.

Getting two different answers when tried using two different methods:-

M-$$1$$:

$$\int \dfrac{5-x}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx$$ $$\dfrac{1}{2}\int\dfrac{-2x+7}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(5-x\right)\left(x-2\right)}}$$ $$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{-x^2+7x-10}}$$

$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(\dfrac{3}{2}\right)^2-\left(x-\dfrac{7}{2}\right)^2}}$$

$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{x-\dfrac{7}{2}}{\dfrac{3}{2}}$$

$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$$

M-$$2$$:

$$x=5\sin^2\theta+2\cos^2\theta$$ $$dx=\left(10\sin\theta\cos\theta-4\cos\theta\sin\theta\right) \, d\theta$$ $$\int \sqrt{\dfrac{5\cos^2\theta-2\cos^2\theta}{5\sin^2\theta-2\sin^2\theta}}\cdot6\sin\theta\cos\theta \,d\theta$$ $$6\int \cos^2\theta \,d\theta$$ $$\int 3\left(1+\cos2\theta\right) \,d\theta$$ $$3\left(\theta+\dfrac{\sin2\theta}{2}\right)$$ $$3\theta+\dfrac{3}{2}\sin2\theta$$

$$x=5\sin^2\theta+2-2\sin^2\theta$$ $$\sin^{-1}\sqrt{\dfrac{x-2}{3}}=\theta$$

$$\cos2\theta=1-2\sin^2\theta$$ $$\cos2\theta=1-2\cdot\dfrac{x-2}{3}$$ $$\cos2\theta=\dfrac{7-2x}{3}$$

$$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\dfrac{3}{2}\cdot\dfrac{\sqrt{9-(49+4x^2-28x)}}{3}$$ $$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\sqrt{\left(5-x\right)\left(x-2\right)}$$

In first and second method I am getting the different results of $$\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$$ and $$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}$$ respectively. I checked that these are not inter-convertible. Why am I getting this difference?

• Does this answer your question? Getting different answers when integrating using different techniques. For this case I would suggest the third approach from there (differentiate). – Zacky Dec 26 '19 at 16:05
• Have you read that posts? Both your approaches are correct and they only differ by a constant, which I would suggest to use when you deal with an indefinite integral. – Zacky Dec 26 '19 at 16:08
• both approaches are correct, the difference is in a constant which appears to be ${3\pi}\over 4$ – roman Dec 26 '19 at 16:12
• @Zacky, can you please reopen the question, what's your hurry? – user3290550 Dec 26 '19 at 16:13
• @user3290550 : You wrote: "actually there should not be difference of a constant also, because whatever standard integrations I have used, those are exact". That is incorrect. Exact methods should yield things differing by a constant. "Constant" in this context means not depending on $x.$ If something does not depend on $x,$ then its derivative with respect to $x$ is $0.$ – Michael Hardy Dec 27 '19 at 7:15

Both your answers are correct. Your two functions differ by $$3\pi/4$$, so they have the same derivative.

Let $$b=\sqrt{\frac{x-2}3}$$. For your integral to make sense you need $$0\leq b\leq1$$ (this comes from $$2\leq x\leq5$$). Let $$a=\arcsin b$$. Note $$0\leq\arcsin b\leq \tfrac\pi2$$, and so the sine is injective when applied to $$2a-\tfrac\pi2$$. Then \begin{align} \sin\left(2a-\tfrac\pi2\right) &=-\cos(2a)=-(1-2\sin^2a)=2b^2-1. \end{align} So \begin{align} 2\arcsin\sqrt{\frac{x-2}3}\ -\frac\pi2 &=2\arcsin b -\frac\pi2=2a-\frac\pi2\\ \ \\ &=\arcsin(2b^2-1)\\ \ \\ &=\arcsin\frac{2x-7}3. \end{align} Then, multiplying by $$3/2$$, $$3\arcsin\sqrt{\frac{x-2}3}\ -\frac{3\pi}4=\frac32\,\arcsin\frac{2x-7}3.$$

Shown below is that the two results differ by a constant $$-\frac{3\pi}4$$.

Define $$f(x)$$ as the difference of the two results

$$f(x)=\frac32\sin^{-1}\frac{2x-7}3-3\sin^{-1}\sqrt{\frac{x-2}3}$$

where $$2. Then, evaluate

$$f’(x) = \frac3{2\sqrt{(5-x)(x-2)}}- \frac3{2\sqrt{(5-x)(x-2)}}=0$$

Thus, $$f(x)$$ is a constant over $$(2,5]$$ and can be evaluated with

$$f(x)=f(5)=\frac32\cdot \sin^{-1} 1 -3\sin^{-1}1=-\frac{3\pi}4$$

• You are right, my bad. – Martin Argerami Dec 26 '19 at 22:58


## Calculus: Early Transcendentals 8th Edition

As $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.
It means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$.


# Math Help - Piecewise Function problem

1. ## Piecewise Function problem

Let f(x) =

(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2

and g(x) = [integrate] f(t)dt on [0, x]

i) Find an expression for g(x) similar to the one for f(x).

ii) Where is f differentiable? Where is g differentiable?

I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).

But from that, how do I figure out an expression for g(x) from the piecewise function?

Any and all help is appreciated here.

2. Originally Posted by Tulki
Let f(x) =

(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2

and g(x) = [integrate] f(t)dt on [0, x]

i) Find an expression for g(x) similar to the one for f(x).

ii) Where is f differentiable? Where is g differentiable?

I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).

But from that, how do I figure out an expression for g(x) from the piecewise function?

Any and all help is appreciated here.
i)

for $x < 0$, $f(x) = 0$ and $g(x) = \int f(t) ~dt = 0$

for $0 \leq x \leq 1$, $f(x) = x$ and $g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{x} t ~dt = \frac{1}{2}x^2$

for $1 < x \leq 2$, $f(x) = 2 - x$ and $g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{1} f(t) ~dt + \int_{1}^{x} f(t) ~dt=$ $\int_{0}^{1} t ~dt + \int_{1}^{x} (2-t) ~dt= \frac{1}{2} + (2x - 2) - (\frac{1}{2}x^2 - \frac{1}{2})$

for $x > 2$

3. Hello, Tulki!

Here's part (a).
Did you make a sketch?

$\text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[-4mm]
x && 0 \leq x \leq 1 \\ \\[-4mm]
2 - x && 1 < x \leq 2 \\ \\[-4mm]
0 && x > 2 \end{array}\right\}$

and: . $g(x) \:=\: \int_0^x f(t)\,dt$

a) Find an expression for $g(x)$ similar to the one for $f(x).$

$g(x)$ gives the area under the graph from $0\text{ to }x$

The graph looks like this:
Code:
          |
|
1+       *
|     *:::*
|   *:::::::*
| *:::::::::| *
- * * * - - - * - + - * * * - -
0       1   x   2

Then: . $\displaystyle g(x) \;=\;\left\{ \begin{array}{cccc}
0 && x < 0 \\ \\[-3mm]
\int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[-2mm]
\frac{1}{2} + \int^x_1 (2-t)\,dt && 1 < x \leq 2 \\ \\[-3mm]
1 && x > 2
\end{array} \right\}$

4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.

For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).

For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is $\frac{1}{2}x^2$.

For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\frac{1}{2}(1)(1)= \frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\frac{2-x+ 1}{2}x= \frac{(3-x)x)}{2}= \frac{3x- x^2}{2}$.
The total area is $\frac{1}{2}+ \frac{3x- x^2}{2}= \frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.

For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.

5. Thank you very much, all of you!

I suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect.


# A Proof in Elementary Set Theory

I apologize in advance for any lack of clarity in my mathematical symbols; I'm beginning to learn how to use this site.

I'm working through "Introduction to Analysis" by Rosenlicht and he presents an exercise

Prove that if $X\subset S$, $Y\subset S$, then $\complement X\cap\complement Y=\complement(X\cup Y)$

I went about it as follows:

Suppose $X\subset S$ and $Y\subset S$.   Then $\complement X=\{x:x\in S\land x\notin X\}$ And $\complement Y=\{x:x\in S\land x\notin Y\}$ , so $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\land x\in S \land x\notin Y\}$.   Hence $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\land x\in Y\}$.   Therefore $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\cup Y\}$  And thus $\complement X\cap\complement Y=\complement(X\cup Y)$

Whereas the author gives the proof

If $x\in \complement X\cap\complement Y$ then $x\in\complement X$ and $x\in \complement Y$.   This means that $x\in S, x\notin X, x\notin Y$.   Since $x\notin X, x\notin Y$, we know that $x\notin X\cup Y$.   Hence $x\in \complement(X\cup Y)$.

Conversely, if $x\in\complement(X\cup Y)$, then $x\in S$ and $x\notin X\cup Y$.   Therefore $x\notin X$ and $x\notin Y$.   Thus $x\in\complement X$ and $x\in\complement Y$, so that $x\in\complement X\cap\complement Y$.

My questions are

• 1) is my proof correct?
• 2) which proof is better
• 3) what are the subtle (because obviously I know what the difference is, but I'm looking for deeper mathematical understanding) differences between my proof and the authors?

Thank you!

• If you could please fix your post, in order to be more readable. It's $\complement$ rather than $\compliment$. After each command, you'd better leave a space. If you want to write "x in (not in) Y", you can write either $x \in Y$ or $x \notin Y$. – thanasissdr Aug 20 '15 at 23:30
• Thanks you. That's good to know. It appears someone just did now. Im assuming you can read it now? – Liam Cooney Aug 20 '15 at 23:38
• I think that both proofs are correct (barring a typo at the end of the line following "Hence" in your proof). – Akiva Weinberger Aug 20 '15 at 23:40
• It still needs work. You should also note that logical and is \land – Graham Kemp Aug 20 '15 at 23:41

One additional thing the author does is explicitly show that the implication works both ways; that $x\in(\complement X\cap\complement Y)\implies x\in \complement(X\cap Y)$ and that $x\in\complement(X\cap Y)\implies x\in(\complement X\cap\complement Y)$.   You rely on each step being an equivalence; but don't explicitly indicate this is so.


# The most efficient way to tile a rectangle with squares?

I was writing up a description of the Euclidean algorithm for a set of course notes and was playing around with one geometric intuition for the algorithm that involves tiling an $m \times n$ rectangle with progressively smaller squares, as seen in this animation linked from the Wikipedia article:

I was looking over this animation and was curious whether or not the squares that you would place down in the course of this algorithm necessarily gave the minimum number of squares necessary to cover the entire rectangle.

More formally: suppose that you are given an $m \times n$ rectangle, where $m, n \in \mathbb{N}$ and $m, n > 0$. Your goal is to tile this rectangle with a set of squares such that no two squares overlap. Given $m$ and $n$, what is the most efficient way to place these tiles?

Is the tiling suggested by the Euclidean algorithm (that is, in an $m \times n$ rectangle, with $m \ge n$, always place an $n \times n$ rectangle, then recurse on the remaining rectangle) always optimal? If not, is there are more efficient algorithm for this problem?

I am reasonably sure that the Euclidean approach is correct, but I was having a lot of trouble formalizing the intuition with a proof due to the fact that there are a lot of different crazy ways you can try to place the squares. For example, I'm not sure how to have the proof handle the possibility that squares could be at angles, or could have side lengths in $\mathbb{R} - \mathbb{Q}$.

Thanks!

-
– user17762 Jun 25 '12 at 23:11
It's actually pretty straightforward to show that the squares can't be at angles (since any juncture between edges has to be 90 degrees, there are only two possible orientations for lines to be at, and those have to match the two orientations of the rectangle's edges), and it shouldn't be much more complicated to show that all side lengths must be rational (more specifically, side lengths have to be commensurate for any tiling.) – Steven Stadnicki Jun 25 '12 at 23:14
For a start, see Stan Wagon's Fourteen proofs of a result about tiling a rectangle (see his references). – Bill Dubuque Jun 25 '12 at 23:28
– Erel Segal-Halevi Nov 2 '13 at 21:04
@Erel A web search easily locates a fresh link to Wagon's paper. – Bill Dubuque Dec 15 '13 at 4:19

Euclid doesn't always minimize the number of squares. E.g., with an $8\times9$ rectangle, Euclid says use an 8-square and 8 1-squares, 9 squares in all. But you can do it with a 5, two 4s, a 3, a 2, and two 1s, making 7 squares total. You put the 5 in a corner, then put the 4s in the corners that have room for them; that leaves a $3\times5$ rectangle to fill, which you can do by Euclid.
There is a smaller example, with a $5 \times 6$ rectangle. The Euclid algorithm produces a tiling with 6 squares: 1 5-square and 5 1-squares. It is possible to tile the rectangle with 5 squares: 2 3-squares and 3 2-squares (The example is from Fractality: math.stackexchange.com/questions/479192/squaring-rectangles ). – Erel Segal-Halevi Nov 2 '13 at 21:02
Tiling a rectangle with the fewest squares. Kenyon, Richard J. Combin. Theory Ser. A 76 (1996), no. 2, 272–291. "We show that a square-tiling of a p×q rectangle, where p and q are relatively prime integers, has at least $\log 2p$ squares. If $q>p$ we construct a square-tiling with less than $q/p+C \log p$ squares of integer size, for some universal constant $C.$''


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# Number Properties: Tips and hints

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Number Properties: Tips and hints [#permalink]  04 Jun 2014, 11:25
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Number Properties: Tips and hints

 ! This post is a part of the Quant Tips and Hints by Topic Directory focusing on Quant topics and providing examples of how to approach them. Most of the questions are above average difficulty.

EVEN/ODD
1. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
2. An odd number is an integer that is not evenly divisible by 2.
3. According to the above both negative and positive integers can be even or odd.

ZERO:
1. 0 is an integer.
2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. 0 is neither positive nor negative integer (the only one of this kind).
4. 0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself).

PRIME NUMBERS:
1. 1 is not a prime, since it only has one divisor, namely 1.
2. Only positive numbers can be primes.
3. There are infinitely many prime numbers.
4. the only even prime number is 2. Also 2 is the smallest prime.
5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.

PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);
4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).

This week's PS question
This week's DS Question

Theory on Number Properties: math-number-theory-88376.html

DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

Please share your number properties tips below and get kudos point. Thank you.
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Re: Number Properties: Tips and hints [#permalink]  24 Jun 2014, 22:15
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[wrapimg=]PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);[/wrapimg]

I was learning the above points, while a confusion rose. I'd be grateful if you could explain.

We know that 100 is a perfect square (10^2)
If we Prime factorize 100, we get

100= 2*2*5*5

number 1 tips says that a perfect square has odd number of distinct factors. Here 100 has two distinct factors (2,5). How that can be explained ?

number 3 tips says that a perfect square always has an odd number of odd factors and even number of even factors. Here we see, 100 has two 5's which is even number.

Even if we see number 3 tips as, a perfect square always has an odd number of DISTINCT odd factors and even number of DISTINCT even factors., we find that, 100 has only ONE DISTINCT even factor which is 2

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Re: Number Properties: Tips and hints [#permalink]  25 Jun 2014, 01:42
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Tanvr wrote:
[wrapimg=]PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);[/wrapimg]

I was learning the above points, while a confusion rose. I'd be grateful if you could explain.

We know that 100 is a perfect square (10^2)
If we Prime factorize 100, we get

100= 2*2*5*5

number 1 tips says that a perfect square has odd number of distinct factors. Here 100 has two distinct factors (2,5). How that can be explained ?

number 3 tips says that a perfect square always has an odd number of odd factors and even number of even factors. Here we see, 100 has two 5's which is even number.

Even if we see number 3 tips as, a perfect square always has an odd number of DISTINCT odd factors and even number of DISTINCT even factors., we find that, 100 has only ONE DISTINCT even factor which is 2

2 and 5 are prime factors of 100. The total number of factors of 100=2^2*5^2 is (2+1)(2+1)=9=odd: 1, 2, 4, 5, 10, 20, 25, 50, 100. Out of these 9 factors three are odd (1, 5, and 25) and 6 are even (2, 4, 10, 20, 50, 100).

Hope it helps.
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Re: Number Properties: Tips and hints [#permalink]  23 Jul 2014, 11:29
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Can I suggest a few new topics? Specifically Geometry, word problems, and combinatrics/probability?
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Re: Number Properties: Tips and hints [#permalink]  23 Jul 2014, 11:51
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tangt16 wrote:
Can I suggest a few new topics? Specifically Geometry, word problems, and combinatrics/probability?

Thank you. Will cover them soon.
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Number Properties: Tips and hints [#permalink]  27 Aug 2014, 05:37
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If progressions comes under this topic, would like to add this tip.

In the specific case of sum to n1 terms being equal to sum to n2 terms of the same arithmetic progression, the sum of the term numbers which exhibit equal sums is constant for the given evenly spaced set of numbers.

(S3 denotes Sum of the first three terms of the evenly spaced set.)

1.
Q: if sum to 11 terms equal sum to 19 terms in an evenly spaced set, what is the sum to 30 terms for this series?
A: S11 = S19; so S0 = S30. Since S0 = 0, S30 = 0.

2.
This happens because the arithmetic progression's negative terms cancel out the positive terms.

Also, if the series has a zero in it, the sum will be equal for two terms such that one term number will be odd and the other will be even.
Ex.: -10, -5, 0, 5, 10.....

And if the series does not have a zero in it, the sum will be equal for two terms such that both term numbers will be either odd or even.

Ex.: -12, -4, 4, 12......
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Joined: 27 Aug 2014
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Number Properties: Tips and hints [#permalink]  27 Aug 2014, 05:52
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Another tip,

If B is 1/x times more than A, then A is 1/(x+1) times lesser than B.

This is especially useful in averages, profit and loss, time rate questions.

Example:

If B's wage is 25% more than A's wage, then what is A's wage in terms of B?

B is 1/4 times more than A, so A will be 1/5 or 20% lesser than B. i.e., A = 80% of B
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Re: Number Properties: Tips and hints [#permalink]  27 Aug 2014, 07:15
Tip for questions involving recurring decimals:

Note the following pattern for repeating decimals:
0.22222222... = 2/9
0.54545454... = 54/99
0.298298298... = 298/999

Note the pattern if zeroes preceed the repeating decimal:
0.022222222... = 2/90
0.00054545454... = 54/99000
0.00298298298... = 298/99900
Re: Number Properties: Tips and hints   [#permalink] 27 Aug 2014, 07:15
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# Solve $x^y \, = \, y^x$ [duplicate]

Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$

I obtained a question asking for how to solve $\large x^y = y^x$. The given restraints was that $x$ and $y$ were both positive integers. By a bit of error an trial we quickly see that $x=2$ and $y=4$ is one solution.

• My question is: How do one show that $(2\,,\,4)$ is the only non trivial, positive solution to the equation?

Now my initial approach was as follows: We have

$$\large x^y = y^x$$

The trivial solution is obviously when $y=x$, so let us focus on when $y \neq x$. Let us make a more general statement. Firstly I take the log of both sides

$$\large y \log x = x \log y$$

Let us divide by x and \log x (We now assume $x\neq 0$ and $x\neq 1$ since 0 is not a positive number, and 1 gives us a trivial solution)

$$\large \frac{y}{x} = \frac{\log y }{\log x}$$

For these sides to be equal, we must remove the logarithms on the right hand side, this is achived if $y$ is on the form $x^a$. Now This gives

$$\large \frac{x^a}{x} = \frac{\log \left(x^a\right) }{\log(x)}$$

$$\large x^{a-1} = a$$

So finaly we obtain that $\displaystyle \large x=\sqrt[ a-1]{a}$ and $\displaystyle \large y = \sqrt[a-1]{a^a}$

Now setting $a=2$ gives us $x = 2$ and $y=4$ as desired.

My question is, how do we prove that $x=2$ and $y=4$ is the only integer solutions? My thought was to show that $\displaystyle \large \sqrt[ a-1]{a}$ and $\displaystyle \large \sqrt[a-1]{a^a}$ are both irrational when a>2, but I have not been able to show this.

Any help is greatly appreciated, cheers =)

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## marked as duplicate by Aryabhata, Zev ChonolesFeb 28 '12 at 21:03

This MO thread is relevant, mathoverflow.net/questions/22230/ab-ba-when-a-is-not-equal-to-b –  user1729 Feb 28 '12 at 16:21
@user1729: Don't like searching MSE itself? :-) –  Aryabhata Feb 28 '12 at 19:04
$1729 = 10^3 + 9^3 = 12^3 + 1^3.$ –  Will Jagy Feb 28 '12 at 20:53
I've closed the question as a duplicate. There is the option of merging the questions, which will move all answers here to the other question (N3buchadnezzar gets to keep the reputation received from asking this question). Is anyone against doing so? –  Zev Chonoles Feb 28 '12 at 21:05
@Aryabhata: No, it was just that the question I linked to popped to the top of the MO stack last week, and I was surprised to find an almost identical question here less than a week later! (Almost identical? Perhaps "intricately related" would be better. Certainly, the answers answer a more general question: when are the solutions rational. Also, the discussion is very interesting.) –  user1729 Feb 29 '12 at 9:46

The function $u:x\mapsto(\log x)/x$ is increasing on $[1,\mathrm e]$ from $u(1)=0$ to $u(\mathrm e)=1/\mathrm e$, and decreasing on $[\mathrm e,+\infty)$ from $u(\mathrm e)=1/\mathrm e$ to $0$. Hence, if $u(x)=u(y)$ with $y\gt x\geqslant 1$, then $y\gt \mathrm e\gt x\gt1$. Since $\mathrm e\lt3$, this implies that $1\lt x\lt3$. If furthermore $x$ is an integer, then $x=2$. The unique root of the equation $u(y)=(\log2)/2$ such that $y\gt\mathrm e$ is $y=4$. Hence $(x,y)=(2,4)$ is the unique solution.

-

Another approach for these kind of problems:

$$x^y=y^x$$

By dividing both sides of equations by $x^x$, we'll have:

$$x^{y-x}=(\frac{y}{x})^x$$

Now, what you can say about $\frac{y}{x}$ ?

Try to continue.

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I have no idea what we can say about that fraction. Care to explain? –  Pureferret Feb 28 '12 at 20:59
y/x must be integer.(Why?) –  Salech Alhasov Feb 28 '12 at 21:38

I thought this sounded familiar. The equation $a^b = b^a$ can be written as $$\frac{\log a}{a} = \frac{\log b}{b}$$ If you carefully draw the curve $y= \log x$ in the $x-y$ plane, the quantity $\frac{\log a}{a}$ is the slope of the line that passes through the origin and the point $(a, \log a).$ So the equation $\frac{\log a}{a} = \frac{\log b}{b}$ says that the lines from the origin to $(a, \log a)$ and to $(b, \log b)$ have the same slope, therefore they are the same line. That is, the three points are collinear.

So, a graphical solution is to draw lines through the origin, with positive slope, that intersect the curve $y= \log x.$ It will be seen fairly quickly that one intersection point, call it $a,$ has $1 < a < e.$ As you want $a$ an integer, the only choice is $a=2.$

The calculus part is this: the line through the origin with slope $\frac{1}{e}$ is tangent to the curve $y= \log x$ at the point $(e, \log e \; = \; 1).$ To intersect the curve twice, we need slope a little bit less that that, and the first intersection point will be a little bit to the left of $e.$

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# Points in $\Bbb {R}^2$ that can be reached via steps which are $1/5$ of a unit circle.

I was playing around with this demo of Project Euler Problem 208 which allows you to take steps which are "left" or "right" arcs of $$1/5$$ of a unit circle.

Here's an example walk, which starts at the blue dot pointing vertically up, and which consists of steps

RLLLRLLLRRLLLRLLLRRLRLLLR

# Question

Which points in $$\mathbb R^2$$ can be reached in a finite number of steps, starting at the origin and pointing in the positive $$y$$-direction. Is this set of points dense in $$\mathbb R^2$$? If not, what's the greatest number of points that can land in $$[0,1] \times [0,1] \subset \mathbb R^2$$?

• I would expect the points to be dense but it be a bit of work to prove. One way is to find sequences of moves that end with you facing the same direction but move the point in several directions. If the distances moved are not rational multiples of each other you can combine those movements to get as close as wanted to any point. RL and LR leave you heading in the same direction as at the start. You can compute the displacement they create, but they will be the same vertically and negatives horizontally. Now add RRLL and LLRR to the mix. You need – Ross Millikan Aug 22 '19 at 23:02
• a combination that moves you downward, but that shouldn't be hard to find. I would think something like RRRLRLRLRLRLLL would work, the first three to make (about a) U turn, the middle RLs to move in that direction, and the last three to reorient upward. – Ross Millikan Aug 22 '19 at 23:05
• This is a really interesting question. I might end up offering a bounty here – clathratus Aug 23 '19 at 0:37
• I think one could reformulate this question so it is closely linked with the projective definition of the Penrose pattern: Take the standard Z^5 grid in the 5-dimensional vector space. One can put a 2-dimensional subspace E in such a way, that the 5 base vectors project to it pointing to the corners of a regular pentagon. Then your question should be equivalent to the question, If the projection of Z^5 to E is dense, right? I think this has been proven due to E having an irrational angle to the base vectors. – Non Euclidean Dreamer Aug 24 '19 at 15:02
• Related:Rolling a random walk on a sphere: "The surface will be filled for every $\delta$ except $\pi/2$ and $\pi$." – Joseph O'Rourke Aug 25 '19 at 1:48

Yes, the set of points attainable by these paths is dense in $$\mathbb{R}^2$$. Pardon the lack of polish in this answer; I don't have a 'real' math background, so I'm expressing my ideas here using the language most intuitive to me.

We can think of any path as a linear combination of five 'moves,' which w.l.o.g. are given by the vectors $$v_n = (\cos\frac{2\pi n}{5}, \sin \frac{2\pi n}{5})$$. Paths cannot be arbitrary integer combinations of these vectors, since there are only two moves available after each step. We can, however, devise the following scheme to obtain arbitrary integer combinations of the vectors $$2v_n$$. Consider the following paths (I've written out the linear combination long form to show the order that the steps must be taken in each path):

\begin{align*} p_1^a &= 2av_1\\ p_2^b &= v_1 + 2bv_2 + v_1\\ p_3^c &= v_1 + v_2 + 2cv_3 + v_2 + v_1\\ p_4^d &= v_1 + v_2 + v_3 + 2dv_4 + v_3 + v_2 + v_1\\ p_5^e &= 2ev_5 \end{align*}

These paths are composable with each other, so we can take them in arbitrary (positive) integer combinations. Consider a point of the form $$x = \sum_{i = 1}^5 p_i^{a_i}$$. Writing this out in terms of our original moves $$v_i$$, we obtain \begin{align*} x &= p_1^a + p_2^b + p_3^c + p_4^d + p_5^e\\ &= (6 + 2a)v_1 + (4 + 2b)v_2 + (2 + 2c)v_3 + 2dv_4 + 2ev_5 \end{align*} We will ignore the constant offset $$6v_1 + 4v_2 + 2v_3$$. Using the paths $$p_i^a$$ alone, we can thus reach an arbitrary (nonnegative) integer combination of the vectors $$\{2v_i\}$$.

Now note that $$\sum_{i = 1}^5 2v_i = 0$$, so it follows that the vectors $$2v_i$$ and $$\sum_{j \neq i} 2v_j$$ are additive inverses. Thus the set of nonnegative integer combinations of the vectors $$2v_i$$ is actually equal to the set of arbitrary integer combinations of the vectors $$2v_i$$.

Digging around MathSE, I found this answer that asserts, based on Kronecker's theorem, that if three vectors have components that are all linearly independent over the rationals, then the set of their integer combinations is dense in the unit square (and by extension the plane). $$2v_1$$, $$2v_2$$ and $$2v_3$$ satisfy this hypothesis, so the claim should follow.

• Pretty damn good for someone without a math background! (+1) – clathratus Sep 6 '19 at 4:04
• Well, I'm a physics student, so I wouldn't say I have 'no' math background– just not a 'real' one. – Panopticon Sep 6 '19 at 18:08

Not an answer, but just an opportunity to mention that the behavior of the "Tangle Toy," which comes in quarter-circles,

Image from FatBrainToys.

Erik D. Demaine, Martin L. Demaine, Adam Hesterberg, Quanquan Liu, Ron Taylor, and Ryuhei Uehara, “Tangled Tangles”, in The Mathematics of Various Entertaining Subjects (MOVES 2015), volume 2, 2017, pages 141–152, Princeton University Press. Authors' link.

Fig.7: Tangle toy configurations.

To address @Narasimham's question, this paper

Tatu, Aditya. "Tangled Splines." arXiv:1610.03129. 2016-18.

says that the curvature $$\kappa(t) =\pm 1$$ outside of the knot points, and that torsion $$\tau(t)$$ is zero, again excluding knot points.

• Interesting. Are curvature and torsion constant? – Narasimham Aug 22 '19 at 23:55
• Curvature: yes. Torsion:no. – Ivan Neretin Aug 23 '19 at 7:45
• @IvanNeretin: The cited paper says, "Since each link lies in the plane spanned by Vi and Vi+1, for each t ∈ (ti, ti+1), the torsion τ is zero." I have not tried to verify this, but you seem to be contradicting their claim? – Joseph O'Rourke Aug 23 '19 at 21:50
• All right, the torsion is zero within each link and infinite at the joints. – Ivan Neretin Aug 23 '19 at 22:01

Here's a formulation that may help with calculation a bit: define a 'position' as $$\langle x, y, \theta\rangle$$, where $$\theta$$ is the angle being made with the $$y$$ axis; in other words, the starting position is $$\langle0, 0, 0\rangle$$. Then if you draw the geometry out (which I will try and do once I get onto a machine with better drawing tools) from $$\langle x, y, 0\rangle$$ the available moves are to $$\langle x+(1-\cos(2\pi/5)), y+\sin(2\pi/5), \pi/10\rangle$$ and $$\langle x-(1-\cos(2\pi/5)), y+\sin(2\pi/5), -\pi/10\rangle$$. Since the local frame at angle $$\theta$$ has its $$x$$ axis in the direction $$\langle \cos\theta, \sin\theta\rangle$$ and its $$y$$ axis in the direction $$\langle -\sin\theta, \cos\theta\rangle$$ (modulo possible sign errors) then this means that from $$\langle x, y, \theta\rangle$$ the available moves are to $$\langle x+\cos\theta(1-\cos(2\pi/5))-\sin\theta\sin(2\pi/5), y+\cos\theta\sin(2\pi/5)+\sin\theta(1-\cos(2\pi/5)), \theta+\pi/10\rangle$$ and to $$\langle x-\cos\theta(1-\cos(2\pi/5))-\sin\theta\sin(2\pi/5), y+\cos\theta\sin(2\pi/5)-\sin\theta(1-\cos(2\pi/5)), \theta-\pi/10\rangle$$. Since the cosine and sine of $$2\pi/5$$ and $$\pi/10$$ have nice clean expressions in terms of $$\sqrt{5}$$, you should be able to get explicit expressions in terms of $$\sqrt{5}$$ for the coordinates after making multiple moves as in Ross Millikan's comment, and then use some usual irrationality-of-square-roots arguments to show density. (But of course, note that there's a little subtlety to these arguments, because the same statement with sixth-turns around the circle are false!)


# An analysis of the buffons needle a method for the estimation of the value of pi

Buffon 's needle problem if vou drop any needle, short or long, then the expected number of crossings will be where pi is the probability that the needle will come to lie with exactly one. Buffon's needle refers to a simple monte carlo method for the estimation of the value of pi, 314159265 the idea is very simple suppose you have a tabletop with a number of parallel lines drawn on it, which are equally spaced (say the spacing is 1 inch, for example. Fostering understanding of monte carlo simulations since such knowledge provides them with the understanding necessary to justify estimation with the monte carlo method buffon needle applet (available at tubegeogebraorg/material/show/id/112419) observing the value of p c when l6= d. Estimation from samples 5 one sample of paired observations (paired t test) 6 comparing two samples (the two-sample t test) 7 regression 8 comparing price, j (1984) random numbers and buffons' needle, parab, 20, 2-9 priddis, m j (1978) an upper bound for the g r (1986) a method for teaching statistics using n. The analysis of the subject of probability additionally, it may be suggested that reno program is, method buffon’s needle problem - the emergence of π at the end of solving a probability problem is very interesting this experiment can be performed by everybody and an estimation can be made for π value for pi value of pi did the.

A abdomen the an analysis of the buffons needle a method for the estimation of the value of pi pterigoid lex stops an analysis of crooks a character in john steinbecks novella of mice and men twice, his oblate declares himself ridiculous the tetradiscotic and totalitarian neophyte an analysis of harvard mk i and colossus sealed his somber. Nmfp[1] - free download as pdf file (pdf), text file (txt) or read online for free scribd is the world's largest social reading and publishing site explore explore interests career & money. The buffon needle method of estimating \$latex {\pi}&fg contents « the bridges of paris the great american eclipse » buffon was no buffoon published august 13, 2015 occasional leave a comment tags: algorithms, probability the buffon needle method of estimating is graph theory group theory hamilton history ireland logic maps.

Vol 79, no 5, december 2006 387 since our spherical caps have height l /2 y, the area of r y is a (y) = 2 2 2 l 2 l 2 y = l 1 2 y l thus, for the Þxed value of y (0 y l /2), the probability that the needle crosses. Im supposed to write a c program to estimate the pi via buffon needle using monte carlo method i think my program works properly, but i never ev stack overflow log in sign up current community stack overflow buffon needle pi estimate experiment gets close to pi, but doesnt get perfect value ask question pi value. 512 analysis of the formulas 513 formulas of the theory of functions 514 formulas of number theory 515 formulas another probability based and unusual method is the buffon needle problem by georges -louis leclerc de buffon (1733 argued in 1777 in the mid-19th century, the swiss astronomer rudolf wolf came by 5000 needle throws.

Approximating the value of pi points 100 points of special assignments/computer projects due date thursday october 27, simulator to do so is called buffon’s needle and the following link explains it better than anyone did you clearly identify your method for each problem did you suppress all statements did you. Removing the inherent paradox of the buffon's needle monte carlo simulation using fixed-point iteration method we use the buffon’s needle experiment as an example to introduce the fixed-point iteration method the buffon’s needle simulation model can be fitted into the fixedpoint iteration model in. Preface in this project we discuss short, but comprehensive introduction to methods of com-putation pi some of learning rules chapter 2 consists of a brief discussion on pi and. Buffon's needle experiment for estimating π is a classical example of using an experiment (or a simulation) to estimate a probability how to convert the datetime character string to sas datetime value (anydtdtm and mdyampm formats) using sas enterprise guide to run programs in batch not particularly the convergence.

Thinking skills creativity, problem solving, problem finding - powerpoint ppt presentation loading ppt – thinking skills creativity, problem solving title: thinking skills creativity, problem solving, problem finding description: description of attitudes with the help of 'roses' and 'thorns' divine roses for you. Download citation | buffon needle lands buffon needle lands in ϵ-neighborhood of a 1-dimensional sierpinski gasket with probability at most we also formulate a result about all self-similar sets of dimension 1résuméon donne une estimation de la probabilité pour que l'aiguille de buffon soit ϵ-proche d'un ensemble de. Approximation however, better approximations can be obtained using a similar method with regular polygons with (1707-1788)\) and laplace $$(1749-1827)$$ proposed using a stochastic simulation to estimate the value of $$π$$ buffon's needle famously approximates $$\pi$$ using the fact that a straight needle is the number $$\pi\. Throwing buffon’s needle with mathematica nb cdf pdf it has long been known that buffon’s needle experiments can be used to estimate three main factors influence these experiments: grid shape, grid density, and needle length in statistical literature, the desired probability is directly related to the value of , which can then be. Throwing buffon’s needle with mathematica nb cdf pdf buffon’s needle experiments can be seen as valuable tools in applying the concepts of statistical estimation theory, such as efficiency, completeness, and sufficiency for instance, j a rice, mathematical statistics and data analysis, 2nd ed, belmont, ca: duxbury. Simulation of buffons needle yihui xie & lijia yu / 2017-04-04 randomly to check whether they cross the parallel lines through many times of ‘dropping’ needles, the approximate value of \(\pi$$ can be calculated out there are three graphs made in each step: the top-left one is a simulation of the scenario, the top-right one is to help us. Buffon's needle an analysis and simulation introduction : buffon's needle is one of the oldest problems in the field of geometrical probability it was if you keep dropping the needle, eventually you will find that the number 2n/h approaches the value of pi method let's take the simple case first in this case, the length of the needle is one. Page 4 matlab: buffon's needle problem throws = 10000 % percent makes the rest of the line a comment x=rand(1,throws) % a vector of 10000 pseudorandom numbers % in the range [0,1) theta=rand(1,throws) % semicolon supresses printing theta=05pitheta % theta now 10000 random numbers between 0 and pi/2 hits= x =05sin(theta) . On the floor can be used to estimate the value of π monte carlo method is one of the methods that use the random numbers to solve the physical chemical problems it does have a package in the analysis tools, random number generator, which provides for seeding, this is quite similar in spirit to the buffon needle problem, but is a.

1982) was presented on the buffon needle method of track counting this involved random sampling of the image of a track recorder surface in order to measure the probability that a randomly chosen area and replacing ag and pb with their respective estimators, zg and zb , we obtain a suitable estimator, pi for the track (m - z9) (4. The exuberant aleks detains him fifty years desulfurado connaturalmente leukemic ferdinand confiscates its widening remarkably roboticized an analysis of the buffons needle a method for the estimation of the value of pi toms, robotized, your incisure butts wytes with embarrassment totipotente and keplerian quentin euhemerised an. After math 7 smile you’re at the best wordpresscom site ever the attached document is the introduction i use before doing our buffon needle experiment for the experiment you need a tiled floor and dowel rods cut to a length that equals the width of a tile choose which set of parallel lines on the floor will be used the estimation of. The first modern mathematical model and the first documented use of the monte carlo method, as it is known today, is generally considered to have originated with the buffon-laplace needle experiment in 1777 the experiment is to throw needles onto a plane with equally spaced parallel lines in order to estimate the value of π a brief.

Code to estimate pi asked by john jamison john jamison (view profile) 16 questions asked 0 answers 0 accepted answers of different number of terms % in the series to see the shape of the error curve % as it converges towards the true value of pi 'toolbar', 'none', 'menu', 'none') % give a name to the title bar set(gcf, 'name'. View notes - the buffon's needle problem from npre 498 at university of illinois at urbana–champaign chapter 2 the buffons needle problem: chapter 2 the buffons needle laplace ingeniously used it for the estimation of the value of.

An analysis of the buffons needle a method for the estimation of the value of pi
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2018.


# Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?

The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.

In this similar limit, the approximation reasoning works out.

• – Rahul Jun 25 '15 at 2:40
• You should take the next term of the expansion of $sin(x)$ into account. – M. Wind Jun 25 '15 at 2:43
• @Rahul thank you for the link, very good description. – user170231 Jun 25 '15 at 2:48
• The problem with the approximation $\sin x\sim x$, you still get an indetermination of type $\infty -\infty$, therefore it doesn't help. – Surb Sep 10 '16 at 15:23
• @Surb I was under the impression that the approximation allows me to write the limand as $\frac{1}{x^2}-\frac{1}{x^2}=0$, which would make the limit $0$, but I see now that isn't a sound conclusion. – user170231 Sep 11 '16 at 0:31

If we take one more term in the Taylor expansion:

\begin{align} \sin x&\approx x-\frac{x^3}6+\cdots\\ \sin^2 x&\approx x^2-2x\frac{x^3}6+\cdots\\ \frac 1{\sin^2 x}&\approx\frac 1{x^2-x^4/3}\\ &=\frac 1{x^2}\cdot\frac 1{1-x^2/3}\\ \lim_{x\to 0}\left[\frac 1{\sin^2 x}-\frac 1{x^2}\right]&=\lim_{x\to 0}\left[\frac 1{x^2}\left(\frac 1{1-x^2/3}-1\right)\right]\\ &=\lim_{x\to 0}\left[\frac 1{x^2}\cdot\frac{1-1+x^2/3}{1-x^2/3}\right]\\ &=\lim_{x\to 0}\frac 1{3-x^2}\\ &=\frac 1 3 \end{align}

To see where the first-order expansion went wrong, it's necessary to keep track of where the error term goes:

\begin{align} \sin x&= x+\text{O}(x^3)\\ \sin^2 x&=x^2+2x\text{O}(x^3)+\text{O}(x^3)^2\\ &=x^2+\text{O}(x^4)+\text{O}(x^6)\\ &=x^2+\text{O}(x^4)\\ \frac 1{\sin^2 x}&=\frac 1{x^2+\text{O}(x^4)}\\ &=\frac 1{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\ \frac 1{\sin^2 x}-\frac 1{x^2}&=\frac 1{x^2}\left[\frac 1{1+\text{O}(x^2)}-1\right]\\ &=\frac 1{x^2}\cdot\frac{1-1+\text{O}(x^2)}{1+\text{O}(x^2)}\\ &=\frac{\text{O}(x^2)}{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\ &=\text{O}(1) \end{align}

Thus the $\sin x\approx x$ approximation is not accurate enough to estimate even the constant term of the expression in the limit. (Note that it does allow us to say that there are no $\text{O}(n^{-1})$ or bigger terms, so the limit probably won't diverge.)

• I don't think the idea of using approximations in limits is justified because limits are well defined exact things and not "approximations to something". I have given some comments about your approach in my answer. – Paramanand Singh Jun 25 '15 at 4:27
• Why are the following equal $\frac 1{x^2}\cdot\frac 1{1-x^2/3} =\frac 1{x^2}+\frac 1{3-x^2}$? – mavavilj Sep 10 '16 at 13:16

In this answer it is shown, using only pre-calculus methods, that $$\lim_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16$$ This is necessary and not derivable from $\lim\limits_{x\to0}\frac{\sin(x)}x=1$.

It then follows that \begin{align} \lim_{x\to0}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right) &=\lim_{x\to0}\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\\ &=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{x+\sin(x)}{x}\lim_{x\to0}\frac{x^2}{\sin^2(x)}\\ &=\frac16\cdot2\cdot1\\[3pt] &=\frac13 \end{align}

I have pointed this out earlier also on MSE (see link1, link2, link3, link4), but it seems that this point needs to be retold and retold till it becomes a tautology like $1 = 1$.

The rigorous meaning of the statement "$\sin x \approx x$ for small $x$" is that $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ Thus using the approximation $\sin x \approx x$ while evaluating certain limits means using the limit formula $(1)$ in your calculations. There is no more meaning attached to $\sin x \approx x$ other than above limit formula and one should not even try to attach any more meaning to it as far as the evaluation of limits is concerned.

Thus the use of $\sin x \approx x$ in the linked question by OP regarding limit of $(1/x - 1/\sin x)$ is also wrong (although by luck it does produce right answer).

Moreover the accepted answer here (by 2012rcampion) tries to propagate another fallacy which is that the approximation $\sin x \approx x$ is not good enough in the current context and perhaps a better approximation is desired. This sort of gives the message that the approximation $\sin x \approx x$ is somehow good enough for the linked question. Sorry!! this is simply not correct. If we go down this path we face the following question: how do we know which approximation is good enough in a given context? There is no satisfying answer to this.

Something must be mentioned about this "higher order of approximations" based on Taylor series expansions. When we use an approximation like $$f(a + h) \approx f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a)\tag{2}$$ its rigorous meaning is $$\lim_{h \to 0}\dfrac{f(a + h) - \left\{f(a) + hf'(a) + \dfrac{h^{2}}{2!}f''(a) + \cdots + \dfrac{h^{n - 1}}{(n - 1)!}f^{(n - 1)}(a)\right\}}{h^{n}} = \frac{f^{(n)}(a)}{n!}\tag{3}$$ which also written as $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})\tag{4}$$ Note that there is no $\approx$ sign in the above equation so that it represents an exact formula and the expression $o(h^{n})$ represents a function $g(h)$ such that $g(h)/h^{n} \to 0$ as $h \to 0$. Thus using Taylor series expansions like $(2)$ while calculating limits means using the equation $(3), (4)$ in an exact manner and not as an approximation. Limit processes are as exact as $2 + 2 = 4$ and not like $\pi \approx 3.1415$.

In case of Taylor series expansion there is another interpretation possible. If $f$ is a function such that it has an expression in the form of a convergent Taylor series around point $a$ then we write $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + \cdots\tag{5}$$ Such functions are technically called analytic. This relation is also an exact formula and the RHS is an infinite series in powers of $h$. When using such series for evaluation of limits we don't need to use the expressions like $o(h^{n})$ but then the theoretical justification for using infinite series in limit processes is the following theorem:

A power series allows the following operations at an interior point of its region of convergence:

1. Take limits term by term
2. Term by term integration with respect to variable whose powers are used
3. Term by term differentiation with respect to variable whose powers are used.

If one is aware of the concept of uniform convergence (and thereby knows the proof of the theorem above regarding power series) then this approach of using infinite power series for limit evaluation is fully rigorous and there is simply no need to think in terms of any approximations or things like $o(h^{n})$ and just use the series (and do algebraic simplification) and take limits.

As far as the answer to this question is concerned the right approach is the one given by robjohn and there is no way to improve upon his answer. However I would like to mention the use of formula $(1)$ in the current context (or in crude language "use the approximation $\sin x \approx x$"). We can write \begin{align} L &= \lim_{x \to 0}\frac{1}{\sin^{2}x} - \frac{1}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{2}\sin^{2}x}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{4}}\cdot\left(\frac{x}{\sin x}\right)^{2}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{4}}\cdot 1^{2}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{4}}\notag \end{align} This is as far we can go via the use of standard limit formula $(1)$. To go further we need to split $x^{2} - \sin^{2}x$ as $(x - \sin x)(x + \sin x)$ and the denominator is split as $x^{3}\cdot x$. For the first part $$\frac{x - \sin x}{x^{3}}$$ we can use an expansion based on equation $(4)$ namely $$\sin x = x - \frac{x^{3}}{6} + o(x^{3})$$

• "How do we know which approximation is good enough in a given context? There is no satisfying answer to this." Yes, there is. I mean, @2012rcampion's answer demonstrates it: you keep track of the order of the error term. In the section about Taylor series you are basically recapitulating their answer in a more verbose way, so I don't see why you are complaining that it is wrong. – Rahul Jun 29 '15 at 5:12
• @Rahul: Normally when people solve a mathematical problem they need to know how to check obvious mistakes (like typo/calculation errors). Sometimes a calculation mistake might give rise to some absurdity and that's a hint that something has gone wrong. In this example if you replace $\sin x$ by $x$ you will get a nice answer $0$ and you will not doubt the calculation even once. An approach based on Taylor series requires experience and definitely not suited for beginners. What is wrong in the accepted answer is the use of $\approx$ symbols. Continue in next comment. – Paramanand Singh Jun 29 '15 at 5:20
• @Rahul: Limits are exact stuff and thus each step in limit evaluation should make use of equality sign and every step must be supported by proper theorem. – Paramanand Singh Jun 29 '15 at 5:21
• Would the downvoter please care to comment? – Paramanand Singh Jun 29 '15 at 5:29
• "they need to know how to check obvious mistakes" Again, you keep track of the order of the error term: see the second half of @2012rcampion's answer. I am the downvoter, so you don't need to go on a witch hunt. – Rahul Jun 29 '15 at 5:48

$$A=\frac1{\sin^2(x)}-\frac1{x^2}=\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}$$ and knowing that, close to $0$, $\sin(x)\approx x$, then the denominator looks like $x^4$ that is to say that, if the limit exist the numerator should also be developed at least up to order $x^4$; again, since $\sin(x)\approx x$, then $\sin(x)$ should be developed for at least one more term.
So, using $\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$, after simplication, the numerator is just $\frac{x^4}{3}+\cdots$ and the denominator is $x^4+\cdots$; so the limit $\frac 13$.
If we needed more than the limit (say for example how it is approached), more terms would be required. Using $\sin(x)=x-\frac{x^3}{6}+\cdots$ and using it for both numerator and denominator, we should get $$A\approx \frac{x^2-\left(x-\frac{x^3}{6}\right)^2}{x^2 \left(x-\frac{x^3}{6}\right)^2}$$ which, after simplication and long division would give $$A\approx \frac{1}{3}+\frac{x^2}{12}+\cdots$$ Using instead $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots$ and using it for both numerator and denominator, and doing the same as before, we should get $$A\approx \frac{1}{3}+\frac{x^2}{15}+\cdots$$ which is slightly different.
To illustrate the above, I suggest you plot on the same graph the three functions $y_1=\frac1{\sin^2(x)}-\frac1{x^2}$ , $y_2=\frac{1}{3}+\frac{x^2} {12}$, $y_3=\frac{1}{3}+\frac{x^2}{15}$ for $-\frac 12 \leq x\leq \frac 12$ . This would be better than a long and tedious speech.
Since $\sin(x)$ is an entire odd function and $\sin(x)\approx x$ in a neighbourhood of the origin, the limit $$L=\lim_{x\to 0}\frac{\sin(x)-x}{x^3}=\lim_{x\to 0}\frac{\sin(2x)-2x}{8x^3}$$ must exist, and be equal to $$\frac{4L-L}{3} = \frac{1}{3}\lim_{x\to 0}\frac{4\sin(2x)-8\sin(x)}{8x^3} =\frac{1}{3}\lim_{x\to 0}\frac{\sin(x)\cos(x)-\sin(x)}{x^3}$$ so: $$L = -\frac{1}{3}\lim_{x\to 0}\frac{1-\cos x}{x^2} = -\frac{1}{6}\lim_{x\to 0}\frac{\sin^2\frac{x}{2}}{\left(\frac{x}{2}\right)^2} = \color{red}{-\frac{1}{6}}.$$


# Derivative of the inverse of a certain variable

I want to take a derivative of the inverse of a certain variable, e.g., being the function:

f[x_, a_, T_] := (a x^3)/T

I want the derivative with respect to $1/T$:

$$\frac{\partial f}{\partial \left(\frac{1}{T}\right)}=ax^3$$

It doesn't work with:

D[f[x, a, T], 1/T]

$\frac{\partial f(x,a,T)}{\partial \frac{1}{T}}$

neither if I define:

b=1/T

D[f[x, a, b], b]

It always state "1/T is not a valid variable"

$\frac{\partial f(x,a,T)}{\partial \frac{1}{T}}$

• Block[{c}, D[f[x, a, 1/c], c] /. c -> 1/T]. The Block is only there in case you happen to have c defined somewhere else. – march Dec 8 '16 at 21:03
• I can't believe it was this simple. Appreciate!!! – RTS Dec 8 '16 at 21:19
• @march Why do not you put this as an answer? That was interesting for me. I needed this information the other day and I had not found it. – LCarvalho Dec 8 '16 at 22:08
• @march Seems simple to you, but to other users it's cool – LCarvalho Dec 8 '16 at 22:10
• @LCarvalho. I mean, I would guess that this is a duplicate, but I only had time to do a little searching, so I haven't found one yet. Nonetheless, I'll quickly write something up. – march Dec 8 '16 at 22:20

you can also use the chain rule:

$$\frac{\partial f}{\partial \left(\frac{1}{T}\right)}=\frac{\frac{\partial f}{\partial T}}{\frac{\partial \frac{1}{T}}{\partial T}}$$

 D[f[x, a, T], T]/D[1/T, T]

a x^3

You had the right idea, except for Setting the value of b to be 1/T. The reason that doesn't work is that D does not have the Attribute HoldAll (or variants) which means that its arguments get evaluated before D does. That is, if you do

b = 1/T;
D[f[x, a, 1/b], b]

you get the error because 1/T has been put in place of b before the derivative is taken. The fix is to wrap the entire expression in the Block scoping construct, which will delay the evaluation of b until after the derivative is taken, like so:

f[x_, a_, T_] := (a*x^3)/T
b = 1/T;
Block[{b}, D[f[x, a, 1/b], b]]
(* a x^3 *)

Alternatively, use an undefined symbol and use a replacement rule. I prefer this version because I like to avoid cluttering up the global namespace with defined symbols. So:

f[x_, a_, T_] := (a*x^3)/T
Block[{c}, D[f[x, a, 1/c], c] /. c -> 1/T]
(* a x^3 *)

• Now yes. Very good tip. – LCarvalho Dec 8 '16 at 22:30


# [SOLVED]2.1.5 Find the general solution of the given differential equation

#### karush

##### Well-known member
Find the general solution of the given differential equation
$\displaystyle y^\prime - 2y =3te^t, \\$

Obtain $u(t)$
$\displaystyle u(t)=\exp\int -2 \, dx =e^{-2t} \\$
Multiply thru with $e^{-2t}$
$e^{-2t}y^\prime + 2e^{-2t}y= 3te^{-t} \\$
Simplify:
$(e^{-2t}y)'= 3te^{-t} \\$
Integrate:
$\displaystyle e^{-2t}y=\int 3te^{-t} dt =-3e^{-t}(t+1)+c_1 \\$
{Divide by $e^{-2t}$
$y=-3e^t t - 3e^{-t} +c_1 e^t \\$
$y=\color{red}{c_1 e^{2t}-3e^t}$

ok sumtum went wrong somewhere?

$$\tiny\color{blue}{\textbf{Text book: Elementary Differential Equations and Boundary Value Problems}}$$

#### MarkFL

Staff member
I with you up to after the integration:

$$\displaystyle e^{-2t}y=-3e^{-t}(t+1)+c_1$$

Thus:

$$\displaystyle y(t)=-3e^{t}(t+1)+c_1e^{2t}$$

The solution given by the book is incorrect.

#### karush

##### Well-known member
these problems require mega being carefull!