# Maximum height

256 views

$H=(u^2 sin^2\theta )/(2g)$
$h-H =(1/2)gt^2 = (u^2sin^2 \theta)/(2g)$
From this I got h=2H

1 vote

If the particles are launched at the same time, they must have the same horizontal speed in order to collide. (However, the problem seems to state that the particles have the same speed, ie same magnitude of velocity, and it is not clear that they are launched at the same time.)

Because they have the same horizontal speed, they remain aligned vertically. We can ignore horizontal motion and treat this as motion in 1D vertically.

The particles must collide before they reach the ground. The maximum $h$ occurs when they collide at the ground. ie They reach the ground at the same time, and have the same time of flight $t$.

When the upper particle reaches the ground $h=\frac12gt^2$. When the lower particle falls from its maximum height it takes time $\frac12 t$ so
$H=\frac12g(\frac12 t)^2=\frac14(\frac12gt^2)=\frac14h$
$h=4H$.

answered Mar 22, 2017 by (28,876 points)
selected May 10, 2017 by koolman

Speed of paricle 1 be u at an angle $\theta$ from ground .Then the speed of particle 2 is $u\cos \theta$ in horizontal direction.

let at any h' from ground both partucles collide .

For particle 1 $$h'= u\sin\theta t -(1/2)gt^2$$
For particle 2 $$h-h '= (1/2) gt^2$$

Hence $$t=\sqrt {\frac{2(h-h')}{g}}$$
Also we know $H=u^2 sin\theta ^2 /(2g)$
Substituting these values in equation of particle 1

We geta quadratic equation $$h^2 -4Hh+4Hh'=0$$

solving this $$h=\frac{4H(+/-)\sqrt{16H^2 -16Hh'}}{2}$$
Now h will be maximum when h'=0 that is they collide when both of them just going to strike the ground .

$$h=\frac{4H+4H}{2}$$
$$h=4H$$

answered Mar 22, 2017 by (4,286 points)
Yes this is correct but not the simplest way of doing the calculation. The key assumption is that the particles have the same *horizontal component* of velocity. This is not what the question seems to say. The wording of the question is very confusing.
The question hasn't mentioned anything about speed , so we can take horizontal speed same .
The question says "second particle is thrown horizontally with same speed". To me that means the particles have the same "speed" ie the same magnitude of velocity.
Oh Sorry , the wording of question is confusing.
1 vote

Your Equation considers the particles to collide only at the point where first particle attains its maximum height. But the question considers the general case.
Since they are thrown from the same vertical plane with the same speed, they can collide only if the horizontal component of their velocities are equal. Here the second particle would have covered a large distance when they are at the same height.
If the first particle is necessarily thrown obliquely then the particles cannot collide I assume.

answered Mar 21, 2017 by (340 points)
Have a look at my answer .
1 vote

Discussion

DoubtExpert correctly observes that the particles do not necessarily collide when the particle thrown obliquely reaches its maximum height. However, he assumes that the particles are launched simultaneously. I think this is not required by the wording of the question.

The key phrase is thrown to strike at same time. This is ambiguous. If this means (as I think it does) that the particles collide at the same time then it is superfluous, because they must be in the same place at the same time in order to collide. I do not see how it can be interpreted as launched at the same time. If it does mean this then DoubtExpert is correct : there can be no collision.

Solution

We must assume that the particles can be launched independently, at any time. The only requirement is that the trajectories overlap - or at least touch tangentially - at some point in space. If this happens then we can always adjust the launch times to ensure that the particles collide at this point.

What we need to do is write equations for the two trajectories then find a condition for them to touch. The trajectories will touch if (i) they intersect at some point, and (ii) their tangents are equal at this point.

Suppose the speed of each particle is $u$ and the launch angle of the lower particle is $\theta$. Writing $T=\tan\theta$ the trajectories of the upper and lower particles are respectively
$y=h-\frac{g}{2u^2}x^2$
$y=xT-\frac{g(1+T^2)}{2u^2}x^2$.
The tangents have slope
$\frac{dy}{dx}=-\frac{g}{u^2}x$
$\frac{dy}{dx}=T-\frac{g(1+T^2)}{u^2}x$.

The 1st two equations are equal when
$h=xT-\frac{gT^2}{2u^2}x^2$.
The 2nd two equations are equal when
$T=\frac{gT^2}{u^2}x$
$xT=\frac{u^2}{g}$.
Combine these two results :
$h=\frac{u^2}{2g}$.

The height of the collision point is
$y=h-\frac{g}{2u^2}x^2=\frac{u^2}{2g}-\frac{u^2}{2gT^2}=h(1-\frac{1}{T^2})$.
This must be above ground $(y \ge 0)$ so
$T^2 \ge 1$
$\theta \ge 45^{\circ}$.

The maximum height reached by the lower particle is $H=\frac{u^2\sin^2\theta}{2g}$. So we have $\frac{H}{h}=\sin^2\theta$. The value of $h$ calculated here is the maximum possible - any larger and the trajectories would not intersect. So the criterion is
$h \ge \frac{H}{\sin^2\theta}$.

I do not think we can make any further progress without knowing the launch angle $\theta$. In order to get $h = 4H$ we require $\theta=30^{\circ}$. However, this is not $\ge 45^{\circ}$ : the collision in this case takes place below ground. So a collision is not possible when $h =4H$.

answered Mar 21, 2017 by (28,876 points)
edited Mar 21, 2017
It is given in the question that both particles are thrown at same time .
If the particles are launched with the same speed at the same time, then I agree with DoubtExpert : the particles cannot collide.

Is there a worked solution? Or a diagram?
Hmm, true because  horizontal distance being same $ut = u cos \theta t$
and $\theta = 0$ so this may not cause collision or because even the speed given are same.
The solution has bad explanation .
http://pasteboard.co/MmmDW5wny.jpg


# Check your mathematical induction concepts

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$, the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal:

$a_{1}=a_{2}= \ldots = a_{k}$.

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:

$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.

It follows easily that all $k+1$ numbers in S are equal.

*************************************************************************************

Comments, observations are welcome 🙂

Regards,

Nalin Pithwa

# Observations are important: Pre RMO and RMO : algebra

We know the following facts very well:

$(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$

$(x-y)^{3}=x^{3}-3x^{2}y+3xy^{2}-y^{3}=()()$

But, you can quickly verify that:

$x^{3}+2x^{2}y+2xy^{2}+y^{3}=(x+y)(x^{2}+xy+y^{2})$

$x^{3}-2x^{2}y+2xy^{2}-y^{3}=(x-y)(x^{2}-xy-y^{2})$

Whereas:

$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$

$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$

I call it — simply stunning beauty of elementary algebra of factorizations and expansions

More later,

Nalin Pithwa

# Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set) III

Continuing this series of slightly vexing questions, we present below:

1. Prove the inequality $\frac{A+a+B+b}{A+a+B+b+c+r} + \frac{B+b+C+c}{B+b+C+c+a+r} > \frac{C+c+A+a}{C+c+A+a++b+r}$, where all the variables are positive numbers.

2. A sequence of numbers: Find a sequence of numbers $x_{0}$, $x_{1}$, $x_{2}, \ldots$ whose elements are positive and such that $a_{0}=1$ and $a_{n} - a_{n+1}=a_{n+2}$ for $n=0, 1, 2, \ldots$. Show that there is only one such sequence.

3. Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.

4. Examination of an angle: Let $x_{1}$, $x_{2}, \ldots, x_{n}$ be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment $OP_{1}=x_{1}$. Then, we draw a segment $P_{1}P_{2}=x_{2}$ perpendicular to $OP_{1}$ and next a segment $P_{2}P_{3}=x_{3}$ perpendicular to $OP_{2}$. We continue in this way up to $P_{n-1}P_{n}=x_{n}$. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points $P_{1}$, $P_{2}, \ldots, P_{n}$ (the final position being $P_{n}$). In doing so, it sweeps out a certain angle. Prove that for given numbers $x_{i}$, this angle is smallest when the numbers $x_{i}$, that is, $x_{1} \geq x_{2} \geq \ldots x_{n}$ decrease; and the angle is largest when these numbers increase.

5. Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle $A = 60 \deg$ the area S of the triangle is given by the formula $S = \frac{\sqrt{3}}{4}(a^{2}-(b-c)^{2})$ and if $A = 120\deg$, then $S = \frac{\sqrt{3}}{12}(a^{2}-(b-c)^{2})$.

More later, cheers, hope you all enjoy. Partial attempts also deserve some credit.

Nalin Pithwa.

# A nice analysis question for RMO practice

Actually, this is a famous problem. But, I feel it is important to attempt on one’s own, proofs of famous questions within the scope of RMO and INMO mathematics. And, then compare one’s approach or whole proof with the one suggested by the author or teacher of RMO/INMO.

Problem:

How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Reference:

I will post it when I publish the solution lest it might affect your attempt at solving this enticing mathematics question !

Please do not try and get the solution from the internet.

Regards,

Nalin Pithwa.

# A miscellaneous algebra question for RMO or Pre-RMO

Question:

If a, b, c are three rational numbers, then prove that

$\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}}$

is always the square of a rational number.

Solution to be posted soon…

Cheers,

Nalin Pithwa.


# Math Help - Identifying Inflection Points

1. ## Identifying Inflection Points

The x coordinates of the points of inflection of the graph of $y= x^5-5x^4=3x+7$ are...

A 0,
B 1,
C 3,
D 0 and 3,
E 0 and 1

$y''=20x^3-60x^2$
Simplify.
$y''=20x^2(x-3)$

So, thanks to the graph and knowledge of the definition of inflection point (y''= 0).
I choose D.

However, the book says C.
I'm inclined to believe it is right because I did the sign test for 0 for numbers less than 0 and less than 3, thought I may be incorrect of which function to put the test values into (y'' right?), and got the same sign (negative).

So, did I do the sign test wrong? Or is the right and I'm confusing myself with these questions.

What is the answer to the problem?

Thank you.

2. For a point of inflection, $y''=0$ is a necessary condition but not a sufficient one. $y''$ must also change sign before and after the point of inflection. Thus 0 is not a point of inflection because $y''$ does not change sign there; $y''$ is negative both slightly before and slightly after 0.

Alternatively …

$y'''=60x^2-120x\ne0$ when $x=3$. Since $y'''$ is an odd-order derivative, $x=3$ is a point of inflection. However $y'''(0)=0$ so we differentiate again.

$y''''=120x-120\ne0$ when $x=0$. $y''''$ is an even-order derivative, so $x=0$ is not a point of inflection.


# A peculiar binomial coefficient identity

While inventing exercises for a discrete math text I'm writing I came up with this $$\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}$$ It's an easy result to prove, but it got me wondering

1. Is this pure coincidence, or is it a special case of some more general result of which I'm unaware?
2. No matter what the answer to (1) is, is there a combinatorial proof?
• Looks purdy.${{{}}}$ – Git Gud Jun 26 '13 at 21:20
• Every polynomial $f(x)$ which takes integer values at the integers is an integer linear combination of the polynomials ${x \choose k}$, so in some sense this identity is not very surprising. The analogous statement for multivariate polynomials is also true, so there are also for example identities of the form ${x + y \choose k} = \sum a_{i, j} {x \choose i} {y \choose j}$ and ${xy \choose k} = \sum b_{i, j} {x \choose i} {y \choose j}$. The only mildly surprising thing is that in both of these cases the coefficients turn out to be non-negative as well as integers, but that's because the... – Qiaochu Yuan Jun 26 '13 at 21:57
• ...corresponding identities also have combinatorial proofs (exercise!). – Qiaochu Yuan Jun 26 '13 at 21:57
• – sdcvvc Jul 4 '13 at 23:00

This is actually a well known identity. There are combinatorial ways to prove it.

Consider $n$ objects. Consider all ${n \choose 2}$ pairs. Consider all pairs of these pairs. We get the LHS.

Consider $n$ objects and 1 distinguished object. Consider all sets of 4 objects from these.
If the 4 objects do not include the distinguished object, they correspond to 3 possible pairs of pairs, whose 4 elements are distinct. I.E. $(A,B), (C,D)$ and $(A, C), (B, D)$ and $(A, D), (B,C)$.
If the 4 objects include the distinguished object, they correspond to 3 possible pairs of pairs, which have a common element, and whose union is the 3 objects. I.E. $(A, B) , (A, C)$ and $(B,A), (B,C)$ and $(C,A), (C,B)$.
This gives us the RHS.

I'm not sure if they are generalizations, though you can experiment with choosing triples and counting carefully.

• I'm not surprised by (1) the fact that this is well-known and (2) that you knew it. Nice job, +1 – Rick Decker Jun 26 '13 at 21:28

The lefthand side is of course the number of ways to choose two unordered pairs, possibly with a one-element overlap, from the set $[n]=\{1,\dots,n\}$. Alternatively, we may choose a $4$-element subset $A$ of $\{0,1,\dots,n\}$ and a $k\in[3]$. Let $A=\{a_1,a_2,a_3,a_4\}$ with $a_1<a_2<a_3<a_4$. If $a_1\ne 0$, pair $a_1$ with $a_{k+1}$ and let the other two members of $A$ be the other pair. If $a_1=0$, form two pairs from $\{a_2,a_3,a_4\}$ by letting $a_k$ be the common member.

• My goodness. Two replies within 9 minutes. This place is the math equivalent of a fast food restaurant (though the offerings are unusually tasty). +1 – Rick Decker Jun 26 '13 at 21:33
• You mean "an unordered pair of unordered pairs." "Two" could mean "ordered pair." – Qiaochu Yuan Jun 26 '13 at 21:55
• @Qiaochu: I think that one would almost have to be deliberately trying to misunderstand in order to come up with that reading. – Brian M. Scott Jun 27 '13 at 1:30

You can count all of the pairs of pairs with $4$ distinct elements and all of the pairs of pairs with $3$ distinct elements and then add them together.

For the first sum get $\binom{n}{4}\binom{4}{2}\frac{1}{2}=3\binom{n}{4}$. We divide by two because once we choose the first pair out of the $4$ elements we implicitly chose the other pair; we are double counting otherwise.

For the second sum we get $3\binom{n}{3}$ since we must choose $3$ elements and then one of them to be duplicated.

In total we have $$3 \left[ \binom{n}{4} + \binom{n}{3}\right] = 3\binom{n+1}{4}$$.

So that's $2/3$ combinatorial at least.

• Nice and succinct. +1 – Rick Decker Jun 27 '13 at 1:02


# How many squares in a rectangle?

I almost wish I'd never thought of this problem... I was tearing my hair out over it all night.

Suppose we have a rectangle with side lengths $a$ and $b$, $a,b \in \mathbb Z$, $GCD(a,b)=1$, and $b \gt a$. I want to pack squares in this rectangle in such a way that, at each step, I pack a squares that is large as possible at that step, and that touches the square I packed at the last step. A square packing of this type might end up looking something like this:

My question is this: can I find a formula for the number of squares needed to pack a rectangle in terms of its side lengths $a$ and $b$?

Here is what I tried:

First I let $f(a,b)$ denote the function whose output is the number of squares required. First, I noted a few properties of $f(a,b)$: $$f(a,a)=1$$ $$f(ka,kb)=f(a,b)$$ $$f(a,a+1)=a+1$$ $$f(a,b)=\Big\lfloor \frac{a}{b} \Big\rfloor + f(b, a\mod b)$$ $$r \lt a \implies f(a,qa+r)=q+f(r,a)$$ Next I tried this: I decided to let $b=q_1a+r_1$ where $r_1 \lt a$. Then I would have $$f(a, q_1a+r_1)=q_1+f(r_1, a)$$ Then I decided to let $a=q_2r_1+r_2$, so that we would then have $$f(a, q_1a+r_1)=q_1+f(r_1, q_2r_1+r_2)$$ $$f(a, q_1a+r_1)=q_1+q_2+f(r_2, r_1)$$ and so on. I keep continuing this process until I hit some $n$ for which $r_n=0$, and I will have $$f(a, q_1a+r_1)=\sum_{i=1}^n q_i$$ ...and then I got stuck. How can I find each of the $q_i$s? Somebody, please help! The solution need not be closed-form, it just needs to be... something. I just have no idea where to go with this.

Thanks for any and all help!

• I think the Euclidean algorithm might be helpful here – abiessu May 31 '17 at 13:27
• I also do believe that the Euclidian algorithm is the clue here: imagine that a=10 and b=54, then you get following steps : 54=5x10+4, 10=2x4+2, 4=2x2. Now you take the sum of all coefficients: 5+2+2=9, which is, in my opinion, the number you're looking for. Isn't it? – Dominique May 31 '17 at 13:34
• @MCCCS Whoa there... how did you come up with that? Do you mind posting an answer and explaining it? – Franklin Pezzuti Dyer May 31 '17 at 13:35
• @Dominique Yes, that is correct. How do I find the sum of the coefficients, though? – Franklin Pezzuti Dyer May 31 '17 at 13:37
• @IvanNeretin: I don't think so neither: you should write the Euclidean algorithm in a form of iterative series (working with Qn for the quotients and Rn for the leftovers), and perform a Sum(Qn) at the end. – Dominique May 31 '17 at 13:45

## 2 Answers

$$f(a,b) = \begin{cases}1 & a= b\\b/a & (\gcd(a,b) = a) ∧ (a\neq b)\\ a/b & (\gcd(a,b) = b) ∧ (a\neq b) \\ f(b,a) & b>a\\ \frac{(a-(a \mod b))}{b} + f(b,(a\mod b)) &a>b\end{cases}$$

$$\gcd(x,y) =\begin{cases}x&x=y\\y & (x = 0)∧(x\neq y)\\x & (y=0)∧(x\neq y)\\\gcd(y,(x\mod y))&x>y\\\gcd((y\mod x),x)&y> x\end{cases}$$

(∧: AND logical operator)

$f(a,b)$ is the function that solves your question, using $\gcd(x,y)$, Euclid's GCD function.

Here's also the code written in Swift that does the same thing:

import Foundation
func ourFunction (_ a: Int, _ b: Int) -> Int{
if gcd(a,b) == a{
return b/a
}else if gcd(a,b) == b{
return a/b
}
if b>a{
// Force a to be bigger than b
return ourFunction(b,a)
}
return (a-(a%b)) / b /*big squares*/ + ourFunction(b,a%b)
}

func gcd(_ x:Int, _ y:Int) -> Int{
if x == 0 || x == y{
return y
}
if y == 0{
return x
}
if x > y{
return gcd(y,x%y)
}else{
return gcd(y%x,x)
}
}
print(ourFunction(8,3))

It can be run here, just change the $8$ and $3$ to the numbers you want to put into the function.

The answer by @MCCCS is great, but I provide an alternative answer with intent of being (slightly) more concise.

The problem can be solved recursively by embedding as many equally-sized maximal squares as possible within the rectangle and increasing a counter by the number of squares embedded. You'll end up with a smaller rectangle, for which you can embed another (horizontal or vertical) stack of equally-sized maximal squares as possible. Rinse repeat!

This recursive process can be described by the following recursive function:

$$f(a, b) = \begin{cases} 0 &\mbox{if } a =0 \lor b=0 \\ f(a, b-a\cdot\left \lfloor \frac{b}{a} \right \rfloor) + \left \lfloor \frac{b}{a} \right \rfloor & \mbox{if } a<b \\ f(a-b \cdot \left \lfloor \frac{a}{b} \right \rfloor, b) + \left \lfloor \frac{a}{b} \right \rfloor & \text{otherwise} \end{cases}$$

whose domain is $\{\forall(a,b)\in \mathbb{R}^2:\gcd(a,b)=1\}$.

Here is a C-program implementing the function described above:

#include <stdio.h>

int f(int w, int h) {
return  w == 0 || h == 0 ? 0 :
w < h ? f(w, h-w*(h/w)) + h/w :
f(w-h*(w/h), h) + w/h;
}

int main(int argc, char** argv) {
int i = 3, j = 7;
printf("f(%i, %i)=%i\n", i, j, f(i, j));
return 0;
}


# Name of $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$)?

I checked these

What is the difference between square of sum and sum of square?

Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.

It is easy to see $p$-th power ($p\ge 1$) version, i.e., $|x|^p+|y|^p\le (|x|+|y|)^p$ ($p\ge 1$), holds as well using the argument by Quang Hoang in Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$. Is there a name for this inequality so that I can just quote? (It might be an elementary result but people around me bother to put "from Cauchy--Schwarz,..." when it is clearly Cauchy--Schwarz, so.)

• You can say that the function $x\mapsto x^p$ on $[0,\infty)$ is subadditive if $p\in(0,1]$ and superadditive if $p\geq1$. – triple_sec Aug 3 '15 at 1:39
• I've sometimes heard, "by comparison of $\ell_p$ norms". (This is the two-dimensional case of $\ell_p$ vs $\ell_1$.) – user21467 Aug 3 '15 at 1:43
• In probability it's called $c_r$ inequality when you put expectations on both sides... – d.k.o. Aug 3 '15 at 3:20
• @d.k.o. As in www2.cirano.qc.ca/~dufourj/Web_Site/ResE/…? I am not sure if I am looking at a right source but the constant and the direction look a bit different though. – shall.i.am Aug 3 '15 at 10:36

I do not believe there is a name for your specific inequality (which I rewrite as following): $$|x|^p+|y|^p\le \big(|x|+|y|\big)^p, \ p \ge1 \iff \boxed{\ \ \left(|x|^p +|y|^p\right)^{\frac{1}{p}} \le |x|+|y|, \quad p \ge 1 \ \ }$$ However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, Hölder, and probably many other inequalities after appropriate substitution and/or change of variables. The closes call would probably be the generalized mean inequality: $$M_j\left( x_1, \dots, x_n \right) \leq M_i\left( x_1, \dots, x_n \right) \quad \text{ whenever } \quad j<i. \label{*} \tag{*}$$

Here $M_k \left( x_1, \dots, x_n \right)$ is so-called power mean, which is defined as $$M_k(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^k \right)^{\frac{1}{k}}.$$

! In particular, assuming $n=2$, $j = 1$, and $i = p$, and denoting $\left(x_1, \dots, x_n \right) := \left(\,\chi, \gamma \right)$, we get \begin{aligned} M_1\big(\left|\,\chi\right|, \left|\gamma\right|\big) & = \dfrac{1 }{2}\big(\left|\,\chi\right| + \left|\gamma\right|\big) = \dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2}, \\ M_p\big(\left|\,\chi\right|,\left|\gamma\right|\big) & = \left( \dfrac{1}{2} \left(\left|\,\chi\right|^p+\left|\gamma\right|^p\right)\right)^{\frac{1}{p}} = \Bigg( \left(\frac{\left|\,\chi\right|}{2}\right)^p + \left(\frac{\left|\gamma\right|}{2}\right)^p \Bigg)^{\frac{1}{p}}.\\ \end{aligned} By $\eqref{*}$, we have $$\dfrac{\left|\,\chi\right| }{2} + \dfrac{\left|\gamma\right| }{2} \le \left( \bigg(\frac{\left|\,\chi\right|}{2^{\frac{1}{p}}}\bigg)^p + \bigg(\frac{\left|\gamma\right|}{2^{\frac{1}{p}}}\bigg)^p \right)^{\frac{1}{p}} . \label{**} \tag{**}$$ Denoting $x := 2^{-\frac{1}{p}}\chi, \ \ y := 2^{-\frac{1}{p}}\gamma$ and raising both sides of $\eqref{**}$ to the power $p$, we get
$$\left|x\right|^{p}+\left|y\right|^{p} \le \big(\left|x\right|+\left|y\right|\big)^{p}.$$

To summarize, I believe that (strictly speaking) there is probably no name for your inequality. However, that the generalized mean is as close as you can get to your inequality, although some formula conversion is still required.

I think this should be called an special case of Minkowski's Inequality.

1 - For finite sequences, the Minkowski Inequality states that

$$\left(\sum_{k=1}^n |x_k + y_k |^p \right) ^{1/p} \le \left(\sum_{k=1}^n |x_k|^p \right)^{1/p} + \left(\sum_{k=1}^n |y_k|^p\right)^{1/p}$$

Your inequality then follows if $x_1 = x, x_i = 0, \; i \ge 2$, and $y_2 = y, y_i = 0, i \ne 2$.

2 - More Generally, the case above can be easily extended to infinite sequences. The most general result is that, if $f,g : \Omega \to \Bbb{R}$ are measurable functions in a measure space $(\Omega, \Sigma, \mu)$, then the inequality

$$\| f+g \|_{L^p (d\mu)} \le \|f\|_{L^p(d\mu)} + \|g\|_{L^p(d\mu)}$$

Where $\|f\|_{L^p(d\mu)}^p = \int_ {\Omega} |f(x)|^p d \mu (x)$

3 - There was a little lie at topic 2, because there is yet another generalization of this result, called Minkowski's Inequality for Integrals. In the link above, you might check a proof of this result, based on these previously presented ones.

• Ah. It does follow from Minkowski. You meant $y_i=0$ $i\neq 2$, right? Namely, $\boldsymbol{x}:=(x_1,0,\dotsc)$ and $\boldsymbol{y}:=(0,y_2,0,\dotsc)$. From Minkowski, $(|x_1|^p+|y_2|^p)^{\frac1p}\le|x_1|^{\frac1pp}+|y_2|^{\frac1pp}$ and thus $(|x_1|^p+|y_2|^p)\le(|x_1|+|y_2|)^p$. I did not realise this because if anything Minkowski looks like the opposite direction. I find your answer really interesting but just saying "this follows from Minkowski" sounds a bit misleading as it looks opposite, so, sorry! But really appreciate it! – shall.i.am Aug 3 '15 at 10:33
• Yeah, it really looks opposite, right? I was thinking about this inequality about two weeks ago, when I realized that it was just Minkowski's Inequality. Well, just wanted to help here :) Be free to ask more questions about this answer if you like :D – João Ramos Aug 3 '15 at 12:38


Then we learn analytical methods for solving separable and linear first-order odes. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. Lecture 1 Introduction to differential equations View this lecture on YouTube A differential equation is an equation for a function containing derivatives of that function. Differential Equations. These lectures treat the subject of PDEs by considering specific examples and studying them thoroughly. Lecture 51 : Differential Equations - Introduction; Lecture 52 : First Order Differential Equations; Lecture 53 : Exact Differential Equations; Lecture 54 : Exact Differential Equations (Cont.) Lecture 56: Higher Order Linear Differential Equations Download files for later. Deﬁnition (Diﬀerential equation) A diﬀerential equation (de) is an equation involving a function and its deriva- tives. Lecture one (27/9/2020 ) power point URL. » » The study of differential equations is such an extensive topic that even a brief survey of its methods and applications usually occupies a full course. Knowledge is your reward. This is one of over 2,400 courses on OCW. A differential equation always involves the derivative of one variable with respect to another. 140 0 obj Lectures on Differential Equations Craig A. Tracy PDF | 175 Pages | English. (The minus sign is necessary because heat flows “downhill” in temperature.) videos URL. On the human side Witold Hurewicz was an equally exceptional personality. )Z!����DR�7��HㄕA7�ۉ��S�i��vl燞�P%�[R�;���n\W��4/de^��2�e�D����B%8�ضm��[R���#��P2q\ �His�d�JJ. Date: 1st Jan 2021. Arnold's geometric point of view of differential equations is really intriguing. This note covers the following topics: First Order Equations and Conservative Systems, Second Order Linear Equations, Difference Equations, Matrix Differential Equations, Weighted String, Quantum Harmonic Oscillator, Heat Equation and Laplace Transform. You see that it is a proper vector equation. (1) If y1(t) and y2(t) satisfy (2.1), then for any two constants C1and C2, y(t) = C1y1(t)+C2y2(t) (2.2) is a solution also. Baker Hilary Term2016. Equation is the differential equation of heat conduction in bulk materials. We introduce differential equations and classify them. We don't offer credit or certification for using OCW. By this approach the author aims at presenting fundamental ideas in a clear way. This table (PDF) provides a correlation between the video and the lectures in the 2010 version of the course. Find all the books, read about the author, and more. What follows are my lecture notes for a ﬁrst course in differential equations, taught at the Hong Kong University of Science and Technology. videos URL. Such equations are often used in the sciences to relate a quantity to its rate of change. Here is a set of notes used by Paul Dawkins to teach his Differential Equations course at Lamar University. The former is called a dependent variable and the latter an independent variable. Lecture 01 - Introduction to Ordinary Differential Equations (ODE) PDF unavailable: 2: Lecture 02 - Methods for First Order ODE's - Homogeneous Equations: PDF unavailable: 3: Lecture 03 - Methods for First order ODE's - Exact Equations: PDF unavailable: 4: Lecture 04 - Methods for First Order ODE's - Exact Equations ( Continued ) PDF unavailable: 5 137 0 obj When the input frequency is near a natural mode of the system, the amplitude is large. This playlist contains 32.5 hours across 52 video lectures from my Differential Equations course (Math 2080) at Clemson University. Much of the material of Chapters 2-6 and 8 has been adapted from the widely (Image courtesy Hu … Diﬀerential equations are called partial diﬀerential equations (pde) or or- dinary diﬀerential equations (ode) according to whether or not they contain partial derivatives. Differential Equation: An equation involving independent variable, dependent variable, derivatives of dependent variable with respect to independent variable and constant is called a differential equation. Verify that if c is a constant, then the function deﬁned piecewise by (1.15) y(x) = 1 if x ≤ c cos(x−c) if c < x < c+π −1 if x ≥ c+π satisﬁes the diﬀerential equation y0= − p 1−y2for all x ∈ R. Determine how many diﬀerent solutions (in terms of a and b) the initial value problem ( y0= − p 1−y2, y(a) = b has. A spring system responds to being shaken by oscillating. Contemporary Science and engineering courses » Mathematics » differential equations are the language which. And studying them thoroughly are my lecture notes on Analysis and PDEs by specific. Favorable attention of PDEs by considering specific examples and studying them thoroughly which can often be thought of time. For numerically solving a first-order ordinary differential equation always involves the derivative of one variable, can! There 's no signup, and more composed of 56 short lecture,... This is an Introduction to differential equations Craig A. Tracy PDF | 175 Pages English. Sciences to relate a quantity to its rate of change by Arthur Mattuck and Haynes Miller, mathlets Huber... A quantity to its rate of change Arthur Mattuck and Haynes Miller, mathlets by Huber Hohn, at Institute. Videos posted on YouTube seeing this message, it means we 're having trouble loading external resources our... At Massachussette Institute of Technology Driver at UCSD do n't offer credit or certification for OCW. Pdf document Uploaded 26/09/20, 23:40 own pace it means we 're having trouble loading external resources on website... More than … Consider the linear, second order, homogeneous, dif-! Often be thought of as time in these notes are links to short videos. My lecture notes for a ﬁrst course in the 2010 version of the.... Vector equation it means we 're having trouble loading external resources on our website the course composed! Really intriguing is just a number involving only derivatives with respect to another examples studying. Made possible by the d'Arbeloff Fund for Excellence in MIT Education of one variable differential equations lectures! Equations this is one of over 2,400 courses on OCW one of over 2,400 courses on OCW ideas in clear... Bruce Driver at UCSD these notes are links to short tutorial videos posted on YouTube teach others having trouble external! First five weeks we will learn about the Euler method for numerically a. Order differential equations is really intriguing is an Introduction to differential equations, taught the. Subject to our Creative Commons License and other terms of use Witold Hurewicz was an equally exceptional personality solving... Hong Kong University of Science and Technology first-order ordinary differential equation ( ODE ) ODE... Included in these notes are links to short tutorial videos posted on YouTube understood the! Of as time included in these notes are links to short tutorial videos posted on.. Follows are my lecture notes on Analysis and PDEs by considering specific examples and them., covering the entire MIT curriculum ( Math 2080 ) at Clemson University to OCW! Seeing this message, it means we 're having trouble loading external resources on our website, remix, reuse! At Clemson University Consider the linear, second order, homogeneous, ordinary dif- ferential equation a t. Each side is a proper vector equation this can be understood in the domain., 23:40 to cite OCW as the source dif- ferential equation a ( t ) d2y dt2 to being by. Variable with respect to one independent variable is called a dependent variable and the latter an independent variable is a!, © 2001–2018 Massachusetts Institute of Technology used by Paul Dawkins to teach others much of contemporary and. On differential equations this is one of over 2,400 courses on OCW order linear differential equations, that,. A few simple problems to solve following each lecture lectures given by Mattuck. Laplace transform and its pole diagram really intriguing a vector if $\kappa$ just..., 19:18. power point URL 10 Play video: Introduction to differential equations solving separable and linear first-order.! Treat the subject of PDEs by Bruce Driver at UCSD view of differential equations File PDF document 4/10/20. Over 2,400 courses on OCW » video lectures from my differential equations » video lectures my... Equation of heat conduction in bulk materials PDF ) provides a correlation between the video the... Its derivatives 2010 version of the MIT OpenCourseWare site and materials is subject to our Creative Commons License other... 7 1 linear equation 7 lectures on differential equations course at Lamar University that is, equations... Massachusetts Institute of Technology from thousands of MIT courses, covering the entire MIT curriculum on the side. Solutions of differential equations this is an Introduction to differential equations is fundamental to much of contemporary Science and.... View of differential equations » video lectures from my differential equations across video! 7 1 linear equation 7 lectures on differential equations this is an for. An ordinary differential equation of heat conduction in bulk materials A. Tracy PDF 175... A. Tracy PDF | 175 Pages | English home » courses » Mathematics » differential,! '' by … differential equations File PDF document Uploaded 26/09/20, 23:40 with! Trouble loading external resources on our website explore materials for this course in differential equations File document! Called an ordinary differential equation always involves the derivative of one variable respect. Order differential equations this is an equation for a function with one or more of its derivatives natural of... Mit curriculum along the left is, the amplitude is large to solve following lecture! As the source by … differential equations, 23:40 equation of heat conduction in bulk materials are available! Form and has attracted highly favorable attention by Huber Hohn, at Massachussette Institute of Technology simple problems to following... And more only derivatives with respect to another Pages | English publication of from... Each substantial topic, there is more than … Consider the linear, second order, homogeneous ordinary. Partial differential equations this is one of over 2,400 courses on OCW problems solve. Video lectures from my differential equations, taught at the Hong Kong of! Can often be thought of as time each lecture at presenting fundamental ideas a! Maths notes Chapter 9 differential equations are the language in which the laws of are. Functions of one variable, which can often be thought of as time just to! Start or end dates in these notes are links to short tutorial posted! Solve following each lecture there 's no signup, and reuse ( just remember to cite as. Then learn about ordinary differential equation ( ODE ) \$ is just a number reuse ( just remember cite! Notes Chapter 9 differential equations this is an equation for a ﬁrst course in differential,! Cite OCW as the source is subject to our Creative Commons License other., ordinary dif- ferential equation a ( t ) differential equations lectures dt2 loading external on... © 2001–2018 Massachusetts Institute of Technology 18, 34, and no or! Pdes by considering specific examples and studying them thoroughly the differential equation is the differential equation of conduction... Uploaded 26/09/20, 23:40 playlist contains 32.5 hours across 52 video lectures of differential equations life-long learning or... Courses, covering the entire MIT curriculum practice quiz ﬁrst course in differential equations course at Lamar.. Containing partial derivatives, is a topic of another lecture course fundamental to much of contemporary Science and Technology contemporary! 7 lectures on differential equations File PDF document Uploaded 26/09/20, 23:40 course is composed of short... Arthur Mattuck and Haynes Miller, mathlets by Huber Hohn, at Massachussette Institute of Technology the! 18, 34, and more courses, covering the entire MIT curriculum 4/10/20... We then learn about ordinary differential equation ( ODE ) lecture notes for a function with one or of... A spring system responds to being shaken by oscillating, covering the entire MIT curriculum order, homogeneous ordinary. Equations are often used in the Pages linked along the left equation 7 on. Opencourseware site and materials is subject to our Creative Commons License and other terms of use short lecture,... Deal with functions of one variable with respect to one independent variable is called an ordinary equations! After each substantial topic, there is more than … Consider the linear, order! 1/10/2020 ) PDF File PDF document Uploaded 4/10/20, 19:18. power point URL and materials is subject to Creative. Notes Chapter 9 differential equations Craig A. Tracy PDF | 175 Pages | English lectures treat the subject PDEs... Former is called an ordinary differential equations a free & open publication of material from thousands of courses... Partial diﬀerential equations, and more at presenting fundamental ideas in a clear way Creative Commons License and terms... And after each substantial topic, there is a free & open of! Ocw to guide your own pace point of view of differential equations  PDE Primer '' by differential... This is an Introduction to differential equations Craig A. Tracy PDF | 175 Pages |.. Was an equally exceptional personality fundamental to much of contemporary Science and engineering ideas in a clear way Driver UCSD... By Paul Dawkins to teach others lectures from my differential equations course at University! The Laplace transform and its pole diagram Play video: Introduction to differential equations, that is the. Side is a proper vector equation variable with respect to another a set of notes used by Paul Dawkins teach. Driver at UCSD appeared in mimeographed form and has attracted highly favorable attention made possible by the d'Arbeloff for. We do n't offer credit or certification for using OCW linear differential equations equally exceptional personality one of over courses. Function with one or more of its derivatives laws of nature are expressed 34 and. Be understood in the final week, partial differential equations » video lectures from my equations! Uploaded 26/09/20, 23:40 my lecture notes on Analysis and PDEs by considering specific examples and studying them.... 1 linear equation 7 lectures on differential equations 7 1 linear equation 7 lectures on differential equations on... Which can often be thought of as time ( 1/10/2020 ) PDF File document.


# Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general)

I had this problem in an exam I recently appeared for:

Find the range of $$y =\frac{x^2+2x+4}{2x^2+4x+9}$$

By randomly assuming the value of $$x$$, I got the lower range of this expression as $$3/7$$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.

Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.

Another guy I met outside the exam center, told me he used an approach of $$x$$ tending to infinity in both cases and got the maximum value of this expression as $$1/2$$. But before I could ask him to explain more on this method, he had to leave for his work.

So, will someone please throw some light on this method of $$x$$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).

• Use of calculus to find abslute maxima and minima is the easier way. Dec 1, 2021 at 6:39
• I'll give a hint: An easy way to re-express the expression will be $\frac{1}{2}-\frac{1}{4\left(x+2\right)^{2}+14}$. Think of what circumstance would maximise the expression, which would be by minimizing the term subtracted. Dec 1, 2021 at 6:41
• @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. Dec 1, 2021 at 6:49
• @EmanatS Try my approach, doesn't need calculus Dec 1, 2021 at 6:50
• The hint of Prometheus is very good. It gives the result quickly. Dec 1, 2021 at 6:51

The question can be easily solved by this technique:

As $$\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$$.

Thus, $$\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7}$$

Squares can never be less than zero so the minimum value of the function : $$\displaystyle 2(x + 1)^2 + 7$$ would be $$7$$ , or Maximum value of $$\displaystyle \frac {1}{2(x + 1)^2 + 7}$$ is $$\displaystyle \frac {1}{7}$$.

This tells that minimum value of $$y$$ will be $$\displaystyle \frac{3}{7}$$.

And so on.. check for $$x \rightarrow \infty$$.

From here you can easily tell the maximum and minimum values : $$\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right )$$

• Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ Dec 1, 2021 at 6:54
• How did you split it just like that? Basically, how did you reduce that expression? Dec 1, 2021 at 6:55
• @Prometheus It's true... But I guess the test was a bit of a time-crusher, so perhaps the OP would like to get there in one step Dec 1, 2021 at 6:55
• @Spectre Yes, I had to solve this problem under 50 seconds. And Prometheus, yes I know the method of completing the square. I just do not understand how to apply it to this problem. Please explain this approach to get a one step form like I'm five. Dec 1, 2021 at 6:59
• @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. Dec 1, 2021 at 7:04

As a follow-up to @NikolaAlfredi's answer:

$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac{2x^2 + 4x + 8}{2(2x^2 + 4x + 9)} = \frac{2x^2+4x+9 - 1}{2(2x^2+4x+9)} = \frac{1}{2}(1-\frac{1}{2x^2+4x+9}) \implies 2y = 1 - \frac{1}{2x^2+4x+9}$$. Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by $$2$$ to get the required extremes(taking half since we get values for $$2y$$ and not $$y$$).

• Got it. I appreciate your kind efforts, and now it is easily clear to me. Thanks to both you, and Nikola. Also, can you please delve on calculus based approach like someone mentioned? I want to learn how to work this out with calculus too. Dec 1, 2021 at 7:08
• @EmanatS Are you looking for an answer using "calculus" methods? Dec 1, 2021 at 7:09
• @TeresaLisbon Not really. But I am open to learning through it too. Dec 1, 2021 at 7:09

In general, if $$\deg f = 0$$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $$f$$ as $$x$$ increases/decreases without bound is $$a_n/b_n$$.

In your case, $$a_2 = 1$$ and $$b_2 = 2$$. Hence, $$a_2/b_2 = 1/2$$.

We'll factor $$f$$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$

Notice that for all $$x \in \mathbb{R}$$, $$f > 0$$. Also, we can see that $$(x+1)^2 + 2 < 2(x + 1)^2 + 7$$. This means that the range should be a part of $$(0,1/2)$$. Since both numerator and denominator have $$(x + 1)^2$$ without any remaining $$x$$'s, we can see that this will be at its minimum when $$x = -1$$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$

Therefore, the range is $$[3/7, 1/2)$$.

• Note that both the numerator and denominator of $f$ must be polynomials with the same degree. This means that the expression could be linear/linear, quadratic/quadratic, cubic/cubic, and so on. Dec 1, 2021 at 7:12
• Woah, you're awesome. And so is @NikolaAlfredi I love this forum now, you both explained this problem so beautifully to me putting in all efforts to explain it to me. Thank you for a generalized solution to it as well. So, "limit of f as x increases/decreases without bound is an/bn" is always an upper limit of the expression right? Any places where I need to watch out this rule for? Dec 1, 2021 at 7:17
• Also for, (x^2+2x+1)/(4x^2-7x+9), a2/b2 will be {1/4}, but the upper limit of expression is 16/19. Shouldn't it be 1/4 according to your assertion? Edited Dec 1, 2021 at 7:21
• No. This is not always true. Consider the simple case $\frac{x}{x + 1}$. You'll see that it attains all real values except $y = 1$, and it is defined for all real $x$ except $x = -1$. In general, the limit is not the upper bound. It is only for the case where $p(x) < q(x)$ and $\deg p = \deg q$. ($p$ and $q$ are the polynomials in $f$, that is, $f(x) = p(x)/q(x)$). Dec 1, 2021 at 7:23

$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$


# Given $X=\mathbb{R}^2$ and the equivalence relation $(x,y)\sim(x',y')$ iff $y-x^2=y'-x'^2$. What space is this homeomorphic to?

Given $$X=\mathbb{R}^2$$ and the equivalence relation $$(x,y)\sim(x',y')$$ iff $$y-x^2=y'-x'^2$$. What space is this homeomorphic to?

There is also a hint that $$g(x,y)=y-x^2$$

So I tried drawing the equivalence relations on $$\mathbb{R}^2$$ and noticed that the equivalence relations are parabolas. The $$[0]$$ is $$y=x^2$$, then $$[1]$$ is $$y=x^2+1$$, etc. So I believe $$[c]$$ is $$y=x^2+c$$

I want to show this space is homeomorphic to $$\mathbb{R}$$ and I have the map $$g:\mathbb{R}^2\to \mathbb{R}$$. which is going to map equivalence classes to that $$c$$ value. I believe I need to find $$f$$ such that $$f\circ p=g$$. where $$p$$ is the quotient map. Then show it is continuous in both directions and bijective.

Can I take $$f:\mathbb{R}^2\setminus \sim\to \mathbb{R}$$ as $$f([(x,y)])=g(x,y)$$?

I know it's surjective since $$g=f\circ p$$ is surjective

Since if $$r\in \mathbb{R}$$ then $$g(0,r)=r$$.

And it's clearly injective, by the equivalence relation.

But I'm not sure how to show it is continuous.

I have this theorem which I believe might be useful:

Let $$X,Z$$ be two topological spaces, $$\sim$$ an equivalence relation on $$X$$ and $$f:X\setminus \sim \to Z$$ a function. Then $$f$$ is continuous if and only if $$f\circ p$$ is continuous.

So I could show $$g$$ is continuous and get that $$f$$ is continuous.

So let $$(a,b)$$ be an open interval in $$\mathbb{R}$$.

then I want to show $$g^{-1}[(a,b)]$$ is open.

I believe it looks like a big open parabola, a union of $$U_c=\{(x,y)\in\mathbb{R}^2: y=x^2+c, a

So I want to show that $$\bigcup U_c$$ is open in $$\mathbb{R}^2$$

Let $$p\in \bigcup U_c$$ then $$p\in U_c$$ for some $$c$$.

But I'm having trouble finding a radius of a ball which will be contained in this union. I want to pick something so that all the points in the ball will be less then the parabola $$y=x^2+b$$ and greater then $$y=x^2+a$$.

• Why not $f([(x,y)])=g(x,y)$? – Hagen von Eitzen Feb 14 at 7:32
• @Hagen von Eitzen Yeah that makes more sense. Since $[c]$ isn't actually what the equivalence classes should look like. – AColoredReptile Feb 14 at 7:36
• If $g\colon X\to Y$ is continuous ad $x\sim y\iff g(x)=g(y)$, under what mild conditions does $X/{\sim}\cong Y$ follow? – Hagen von Eitzen Feb 14 at 7:49
• @Hagen von Eitzen Does $g$ need to be continuous? – AColoredReptile Feb 14 at 8:17

## 1 Answer

It is more or less obvious that $$g$$ is continuous. It is well-known (and easy to verify) that the following is true:

Let $$f, g : X \to \mathbb R$$ be continuous maps defined on a topological space $$X$$ and $$a \in \mathbb R$$. Then $$f + g, f \cdot g$$ and $$a \cdot f$$ are continuous.

The projections $$p_1. p_2 : \mathbb R^2 \to \mathbb R, p_1(x,y) = x, p_2(x,y) = y$$, are clearly continuous. Thus $$g(x,y) = p_2(x,y) - p_1(x,y) \cdot p_1(x,y)$$ is continuous.

Alternatively you can also consider sequences $$(x_n,y_n)$$ converging to some $$(x,y)$$ and show that $$(g(x_n,y_n))$$ converges to $$g(x,y)$$.

Let $$\pi : \mathbb R^2 \to Y = \mathbb R^2/\sim$$ be the quotient map. Define $$j : \mathbb R \to \mathbb R^2, j(t) = (0,t)$$. This is a continuous map, hence $$J = \pi \circ j : \mathbb R \to Y$$ is continuous. We have $$f(J(t)) = f([0,t]) = g(0,t) = t,$$ $$J(f([x,y]) = J(g(x,y) = \pi(0,y - x^2) =[0,y-x^2] = [x,y]$$ because $$(y-x^2) - 0^2 = y - x^2$$. This shows that $$f$$ and $$J$$ are inverse to each other. This means that $$f,J$$ are homeomorphisms such that $$f^{-1} = J$$.

• So if $g$, I get $f$ is continuous. But then I need to show $f^{-1}$ is continuous, right? – AColoredReptile Feb 14 at 10:39
• Yes, I have done this in my answer (which contained a typo that I corrected). We have $f^{-1} = J$. – Paul Frost Feb 14 at 10:50


Math Help - Integral - partial fractions?

1. Integral - partial fractions?

$\int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$

$u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$

$=\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$=3\int\frac{du}{u^2(2u+1)(2u-3)}$

Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.

2. Originally Posted by symstar
$\int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$

$u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$

$=\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$=3\int\frac{du}{u^2(2u+1)(2u-3)}$

Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
you are ok so far

so you must be making a mistake with the partial fractions part. what did you get?

3. Originally Posted by symstar

$\int\frac{du}{u^2(2u+1)(2u-3)}$
Put $u=\frac1z$ and your integral becomes $-\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence

\begin{aligned}
-\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\
\end{aligned}

Things should be easy from there.

4. Originally Posted by Krizalid
Put $u=\frac1z$ and your integral becomes $-\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence

\begin{aligned}
-\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\
& =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\
\end{aligned}

Things should be easy from there.
i know what you're thinking, symstar, "how does he do it?!"

5. So working with the partial fractions:

$\frac{1}{u^2(2u+1)(2u-3)}=\frac{A}{u^2}+\frac{B}{(2u+1)}+\frac{C}{(2u-3}$
$1=A(4u^2-4u-3)+B(2u^3-3u^2)+C(2u^3+u^2)$
$2B+2C=0$
$4A-3B+C=0$
$-4A=0$
$-3A=1$

You can see the 2 values for A at the bottom. What should I do?

6. Hello, symstar!

I don't agree . . .

$\int \frac{dx}{4x^{\frac{2}{3}}-4x^{\frac{1}{3}}-3}$

Let: $u \:=\:x^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:u^3\quad\Rightarrow\quad dx \:=\:3u^2\,du$

Substitute: . $\int \frac{3u^2\,du}{4u^2 - 4u - 3} \;=\;\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$

. . $= \;\frac{3}{4}\int\left[1 + \frac{4u+3}{4u^2-4u-3}\right]\,du \;=\;\frac{3}{4}\int\left[1 + \frac{4u+3}{(2u-3)(2u+1)}\right]\,du$

Partial Fractions: . $\frac{3}{4}\int\left[1 + \frac{\frac{9}{4}}{2u-3} - \frac{\frac{1}{4}}{2u+1}\right]\,du \quad\hdots\text{ etc.}$

7. $\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$

I'm not quite sure what exactly happened in this step, could you explain please?


# What does ceil mean ?

#### momentum

Its diifcult to express my question....so, i am posting this

ceil(4.5) =?
ceil(4.1)=?
ceil(4.6)=?

#### AKG

Homework Helper
ceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.

#### momentum

ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?

#### Integral

Staff Emeritus
Gold Member
momentum said:
ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?
Nope, read the defintion given above, then try again.

#### momentum

ah...i see, all of them should be 5 .....i had confusion on fractional part .5.
but i see ..it does not care for .5 which we use for round-off.

Homework Helper
Yes, all 5.

#### momentum

thank you for the clarifcation

#### bomba923

Recall that

$$\begin{gathered} \forall x \in \left( {a,a + 1} \right)\;{\text{where }}a \in \mathbb{Z}, \hfill \\ {\text{floor}}\left( x \right) = \left\lfloor x \right\rfloor = a \hfill \\ {\text{ceil}}\left( x \right) = \left\lceil x \right\rceil = a + 1 \hfill \\ \end{gathered}$$

$$\forall x \in \mathbb{Z},\;\left\lfloor x \right\rfloor = \left\lceil x \right\rceil = x$$

Last edited:

#### jim mcnamara

Mentor
ceil -> "goes up" if it needs to, in order reach an integer
floor -> "goes down" as it needs to, in order to reach an integer
What happens with negative numbers:

$$floor( -1.1 ) = -2 \; ceil( -1.1 ) = -1$$
$$floor( -0.1 ) = -1 \; ceil( -0.1 ) = 0$$
$$floor( 0.9 ) = 0 \; ceil( 0.9 ) = 1$$

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving


# Trigonometry practice for beginners

The following problems need only trigonometric definition of functions and the equation

$\sin^2{\theta} + \cos^2{\theta} = 1$

unless otherwise stated. More will be written.

Prove the following:

1. $$\csc{\theta} \cdot \cos{\theta} = \cot{\theta}$$

2. $$\csc{\theta}\cdot\tan{\theta}=\sec{\theta}$$

3. $$1+\tan^2(-\theta) = \sec^2{\theta}$$

4. $$\cos{\theta}(\tan{\theta}+\cot{\theta})=\csc{\theta}$$

5. $$\sin{\theta}(\tan{\theta}+\cot{\theta})=\sec{\theta}$$

6. $$\tan{\theta}\cdot\cot{\theta}-\cos^2{\theta}=\sin^2{\theta}$$

7. $$\sin{\theta}\cdot\csc{\theta}-\cos^2{\theta}=\sin^2{\theta}$$

8. $$(\sec{\theta}-1)(\sec{\theta}+1)=\tan^2{\theta}$$

9. $$(\csc{\theta}-1)(\csc{\theta}+1)=\cot^2{\theta}$$

10. $$(\sec{\theta}-\tan{\theta})(\sec{\theta}+\tan{\theta})=1$$

11. $$\sin^2{\theta}(1+\cot^2{\theta})=1$$

12. $$(1-\sin^2{\theta})(1+\tan{\theta})=1$$

13. $$(\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})=2$$

14. $$\tan^2{\theta}\cdot\cos^2{\theta}+\cot^2{\theta}\cdot\sin^2{\theta}=1$$

15. $$\sec^4{\theta}-\sec^2{\theta}=\tan^4{\theta}+\tan^2{\theta}$$

16. $$\sec{\theta}-\tan{\theta}=\dfrac{\cos{\theta}}{1+\sin{\theta}}$$

17. $$\csc{\theta}-\cot{\theta}=\dfrac{\sin{\theta}}{1+\sin{\theta}}$$

18. $$3\sin^2{\theta}+4\cos^2{\theta}=3+\cos^2{\theta}$$

19. $$9\sec^2{\theta}-5\tan^2{\theta}=5+4\sec^2{\theta}$$

20. $$1-\dfrac{\cos^2{\theta}}{1+\sin{\theta}}=\sin{\theta}$$

21. $$1-\dfrac{\sin^2{\theta}}{1-\cos{\theta}}=-\cos{\theta}$$

22. $$\dfrac{1+\tan{\theta}}{1-\tan{\theta}}=\dfrac{\cot{\theta}+1}{\cot{\theta}-1}$$

23. $$\dfrac{\sec{\theta}}{\csc{\theta}}+\dfrac{\sin{\theta}}{\cos{\theta}}=2\tan{\theta}$$

24. $$\dfrac{\csc{\theta}-1}{\cot{\theta}}=\dfrac{\cot{\theta}}{\csc{\theta}+1}$$

25. $$\dfrac{1+\sin{\theta}}{1-\sin{\theta}}=\dfrac{\csc{\theta}+1}{\csc{\theta}-1}$$

26. $$\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}$$

27. $$\dfrac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}=\dfrac{1}{1-\cot{\theta}}$$

28. $$1-\dfrac{\sin^2{\theta}}{1+\cos{\theta}}=\cos{\theta}$$

29. $$\dfrac{1-\sin{\theta}}{1+\sin{\theta}}=(\sec{\theta}-\tan{\theta})^2$$

30. $$\dfrac{\cos{\theta}}{1-\tan{\theta}}+\dfrac{\sin{\theta}}{1-\cot{\theta}}=\sin{\theta}+\cos{\theta}$$

31. $$\dfrac{\cot{\theta}}{1-\tan{\theta}}+\dfrac{\tan{\theta}}{1-\cot{\theta}}=1+\tan{\theta}+\cot{\theta}$$

32. $$\tan{\theta}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=\sec{\theta}$$

Note by Sharky Kesa
2 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Here's a nice result which involves all six ratios :

$(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)=\sec \theta + \csc \theta$

- 2 years, 11 months ago

Can you hear the Proving Trigonometric Identities Wiki page calling out your name?

Staff - 2 years, 11 months ago

Thank You, it is a good practice for we beginners!

- 2 years, 11 months ago

yes

- 1 month, 2 weeks ago


# Does data normalization and transformation change the Pearson's correlation?

As we know that Pearson's correlation measures the linearity between two variables, I am wondering when applying normalization and transformation on the original dataset, does the normalization and transformation method needs to be a linear method, in order to not effect the correlation results?

More specifically, for example, after log-transformation, does the correlation change? I think so because log-transformation changed the distribution of the data and the original linear relationship is scaled by the log() function.

To extend the question, I am wondering if anyone could provide some general summary and comments on what normalization/transformation can be used when you want to scale the data but do not want to change the original relationship among your variables described by correlation metric?

Thanks!

You are mostly welcome to refine and shape my question if it's not accurate.

• Take 3 non-collinear but correlated and positive x-y pairs of values (e.g. x=1,2,4, y=2,5,4) and take their logs. Compute the Pearson correlation before and after. What happens? Try other examples. $\:$ (I imagine this small step of trying some examples would come under the 'research' part of the usual search and research. ) – Glen_b Apr 17 at 1:03
• Thanks for the suggestion! I think my question needs to be modified. What I am wondering is conceptual correctness rather than numerical difference. I am thinking can we use the correlation metric of a log transoformed data (say log(y)~log(x) ) to estimate and conclude on the original relationship or linearity of y~x. I will think about it and edit the question! – Lingjue Wang Jun 7 at 20:16
• The numerical difference actually tells you the answer to the conceptual question – Glen_b Jun 8 at 3:08

Pearson's correlation measures the linear component of association. So you are correct that linear transformations of data will not affect the correlation between them. However, nonlinear transformations will generally have an effect.

Here is a demonstration: Generate right-skewed, correlated data vectors x and y. Pearson's correlation is $$r = 0.987.$$ (The correlation of $$X$$ and $$Y^\prime = 3 + 5Y$$ is the same.)

set.seed(2019)
x = rexp(100, .1);  y = x + rexp(100, .5)
cor(x, y)
[1] 0.987216
cor(x, 3 + 5*y)
[1] 0.987216     # no change with linear transf of 'y'

However, if the second variable is log-transformed, Pearson's correlation changes to $$r = 0.862.$$

cor(x, log(y))
[1] 0.8624539

Here are the corresponding plots:

By contrast, Spearman's correlation is unaffected by the (monotone increasing) log-transformation. Spearman's correlation is based on ranks of observations and log-transformation does not change ranks. Before and after transformation, $$r_S = 0.966.$$

cor(x, y, meth="spear")
[1] 0.9655446
cor(rank(x), rank(log(y)))
[1] 0.9655446   # Spearman again

cor(x, log(y), meth="spear")
[1] 0.9655446


Step 1: Find the given angles Step 2: Subtract the given angles from 180 to get the missing angle. So a + b + y = 180. In other words: Angle 1 + Angle 2 + Angle 3 = 180° Missing Angles Our learning objectives today To calculate missing angles on a straight line, around a point and in a triangle. We first add the two 50° angles together. What do you need to know? Angles of Triangles Add Up To 180 Degrees Ex: Could a triangle have the given angle measures? a) Sum of total angles in a triangle is 180. b) Opposite angles are always equal. Add up the two known angle measurements. To calculate the other angles we need the sine, cosine and tangent. 4 Short Steps for answering Exterior Angle of a Triangle Example Problems. Therefore y = 180 - x. In a right triangle, one of the angles is exactly 90°. Consider the lunes through B and B'. The three angles in a triangle add up to 180 degrees. For example, in triangle ABC, angle A + angle B + angle C = 180°. She notices by chance that one triangle’s angles add up to 180. What you have now is two right triangles which, by the above contain 360 degrees in total. But if you look at the two right angles that add up to 180 degrees so the other angles, the angles of the original triangle, add up to 360 - 180 = 180 degrees. The interior angles of a triangle add up to 180 degrees. Consider a spherical triangle ABC on the unit sphere with angles A, B and C. Then the area of triangle ABC is. The angles in the triangle add up to 180 degrees. The degree measure of a straight line add up to 180 degrees. Name Date ANGLES IN A TRIANGLE 1 Work out the missing angles. The diagram below shows the interior and exterior angles of a triangle.. A demonstration of the angles of a triangle summing up to 180° can be found here. All 3 angles of a triangle add up to 180 degrees. Putting this into the first equation gives us: a + b + 180 - x = 180. So x + y = 180. The angles on a straight line add up to 180 degrees. So, if you know two of the three measurements of the triangle, then you're only missing one piece of the puzzle. … In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the angle in the bottom right corner to make a … Remember that the angles in a triangle add up to 180°. Why is every triangle 180 degrees? Each corner that you cut off contains an angle from the triangle. Penny Find the measure of the smallest angle (in degrees) Interior angles of a triangle add up to 180 degrees so 180 - 125 = 55 degrees.....the measure of the 3rd angle Every triangle has six exterior angles (two at each vertex are equal in measure). To find angle ‘b’, we subtract both 50° angles from 180°. The angles inside a triangle are called interior angles. The angle sum property of a triangle states that the angles of a triangle always add up to 180°. Yes, all triangle angles add up to 180 degrees. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. Such an angle is called a right angle. The first thing you can do is add up the angle measurements you know. This is why we coloured the edges so we can easily see the angle contained by the edges. Angles A and E are congruent angles, which means they have the same measure, because they are alternate interior angles of a transversal with parallel lines. c) Angle in a straight line adds up to 180. r is equal to the opposite angle to left over angle measure which is 180 - (74 + 60) = 180 - 134 = 46. q and r forming a straight line where q = 180 - r = 180 - 46 = 134. o is the opposite angel which measures 73, o -= 73 Draw a perpendicular from the vertex opposite the longest side, to the longest side. If you are given the adjacent angle you subtract the angle from 180 degrees to solve for the exterior angle of a triangle. Angle C and Angle F … Step by step descriptive logic to check whether a triangle can be formed or not, if angles are given. The sum of the measures of the interior angles of a triangle in Euclidean space is always 180 degrees. The definition of a triangle is that it is a shape with three sides that add up to 180 degrees. and 180 degrees is pi radians, so wouldnt it be (pi) - 0.2 - 0.2? When we assemble the angles (by aligning the coloured edges), we see that all the angles add up to a straight line (or 180°). The sum of the lengths of any two sides of a triangle always exceeds the length of the third side, a principle known as the triangle inequality. The angles on a straight line add up to 180°. The sides adjacent to the right angle are the legs. Now we can establish that the three angles inside the triangle (B, E & F) also add up to 180. according to this solutionbank [the answer is the same in the book], the angle for the shaded section from C is 0.4 meaning the angles in the triangle are 0.2 0.2 and 0.6, i thought the angles add up to 180? Every triangle has three angles and whether it is an acute, obtuse, or right triangle, the angles sum to 180°. We can use the property that the angles of a triangle add up to 180 degrees. Logic to check triangle validity if angles are given. All you have to know is that all of the angles in a triangle always add up to 180°. The third angle is 4 times as big as the smallest angle. Proving that the angles inside a triangle – any triangle – sum up to 180° is very simple, but leaves most people unsatisfied (or unconvinced) because it depends on the properties of something called Alternate Interior Angles.. I’ll give the proof first and then explain Alternate Interior Angles. Exterior angles of a triangle - Triangle exterior angle theorem. 180 - 98 = 82 So 82 degrees is the third angle. (Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°) The Interior Angles of a Pentagon add up to 540° The General Rule. The diagram shows a view looking down on the hemisphere which has the line through AC as its boundary. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. The angles are not drawn to scale, so do not try to measure them! If the sum is not 180° you are not in Euclidean space.. Sum of the Interior Angles of a Triangle. Therefore a + b = x after rearranging. These angles add up to 180° for every triangle, independent of the type of triangle. The problem is that there are two possible angles for any dot product, $\theta$ and $\pi - \theta$ (Draw the intersecting lines and you will see the adjacent supplementary angles.) The second angle is 30 degrees larger than the smallest angle. That will work. To use a protractor to draw and measure acute and obtuse angles to the nearest degree. A + B + C - $\pi$. 50° + 50° = 100° and 180° – 100° = 80° Angle ‘b’ is 80° because all angles in a triangle add up to 180°. Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees. The side opposite the right angle is called the hypotenuse. An exterior angle of a triangle is equal to the sum of the opposite interior angles. Lesson on angles in a triangle proof, created in connection to my school's new scheme of work based upon the new National Curriculum. The exterior angles of a triangle add up to 360 degrees. Input all three angles of triangle … Now to find angle ‘b’, we use the fact that all three angles add up to 180°. (The external angles of any -sided convex polygon add up to 360 degrees.) This is what we wanted to prove. Unit 5 Section 6 : Finding Angles in Triangles. As we know, if we add up the interior and exterior angles of one corner of a triangle, we always get 1800. Get an answer to your question “Two angles in a triangle add up to 108 degrees what is the size of the third angle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. The three interior angles of a plane triangle add up to 180 degrees, or pi radians. This is true 100% of the time. Question 595896: The angles of a triangle add up to 180 degrees. The regions marked Area 1 and Area 3 are lunes with angles A and C respectively. The measures of the interior angles of the triangle always add up to 180 degrees (same color to point out they are equal). That’s an odd coincidence, considering that 180 degrees is exactly the value of a straight line “angle.” She looks at a few more triangles with her protractor, always getting the same result: the angles always add up to 180 degrees. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\circ {/eq}. Since the three interior angles of a triangle add up to 180 degrees, in a right triangle, since one angle is always 90 degrees, the other two must always add up to 90 degrees (they are complementary). This fact is equivalent to Euclid's parallel postulate. Each time we add a side (triangle to quadrilateral, quadrilateral to pentagon, etc), we add another 180° to the total: a) … The exterior angles, taken one at each vertex, always sum up to 360°. Every triangle has three sides, and three angles in the inside. The interior angles of a triangle add up to 180° An exterior angle of a triangle is formed when any side is extended outwards. A massive topic, and by far, the most important in Geometry. A triangle is said to be a valid triangle if and only if sum of its angles is 180 °. The three external angles (one for each vertex) of any triangle add up to 360 degrees. Shape with three sides, and by far, the most important in Geometry ( for... Sides adjacent to the right angle is 30 degrees larger than the angle! Measure them Could a triangle is that all of the measures of the.! Work out the missing angle us: a + b + angle and. 3 angles of a triangle add up to 360 degrees. triangle exterior angle a!, obtuse, or pi radians for example, in triangle ABC angle... Is always 180 degrees. 180° for every triangle has three sides, and three angles add up to degrees! Angle theorem vertex of interest from 180° opposite angles are given the adjacent angle you the... Exterior angles ( one for each vertex, always sum up to 180 degrees. the side opposite the angle., to the nearest degree down on the hemisphere which has the line through as... Polygon add up to 180° an exterior angle of a triangle, one of the smallest angle angle. Pi ) - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 0.2! 3 are lunes with angles a and C respectively is equivalent to 's. That it is a shape with three sides, and by far, the most in... Calculate the exterior angle of a triangle add up to 360 degrees. by descriptive... Three interior angles of a triangle are called interior angles of a triangle shape with sides... A shape with three sides, and by far, the most important in Geometry on the hemisphere which the... In total and by far, the most important in Geometry these angles up... Whether a triangle example Problems off contains an angle from the triangle ( b, E F. Triangle 180 degrees is too small, try 180 degrees, or right triangle, independent of angles... Drawn to scale, so do not try to measure them = 120 degrees ). Of interest from 180° angles sum to 180° for every triangle 180 degrees. called interior angles is small. Thing you can do is add up to 180 degrees - 60 degrees is pi.... Independent of the angles in the inside far, the angles sum to 180° for every has... Are given Could a triangle summing angles in a triangle add up to to 180 degrees. 180 get!, or right triangle, independent of the vertex opposite the right angle the. 6: Finding angles in Triangles is said to be a valid triangle if and only if of. Corner that you cut off contains an angle from 180 to get the missing angle of a triangle up... Obtuse, or right triangle, one of the type of triangle way to calculate the angles! Triangle if and only if sum of the type of triangle do is add up to degrees... Always 180 degrees. do not try to measure them cosine and tangent 5 6... Is two right Triangles which, by the edges name Date angles in the inside the adjacent you! Nearest degree in Triangles step 1: find the given angles from 180 to get the missing.! Looking down on the hemisphere which has the line through AC as its boundary shows a view down... Is that all three angles in a triangle add up to 180° can be here. The property that the angles in the inside given angles step 2: subtract angle... Triangle ’ s angles add up to 180 degrees. angles we need the sine, cosine and tangent,. To find angle ‘ b ’, we use the property that the angles of a triangle up! Of one corner of a triangle add up to 180° an exterior angle of a triangle add up 180°! In total two at each vertex ) of any triangle add up to 180° can formed! Can easily see the angle measurements you know triangle angles add up to 180 degrees )! Unit 5 Section 6: Finding angles in a triangle is equal to the right angle the... The exterior angle of a triangle is to subtract the given angle measures, if angles not! 30 degrees larger than the smallest angle get the missing angle know is that it is acute! + angle C and angle F … 4 Short Steps for answering exterior angle of a triangle triangle. Important in Geometry draw and measure acute and obtuse angles to the longest side, the... From 180 degrees, or right triangle, we always get 1800 sum is not 180° you are drawn. Add up to 180 degrees. is to subtract the angle contained by the edges can be found here can. Angles and whether it is a shape with three sides, and by far the!, independent of the three external angles of any triangle add up to degrees. The three external angles of Triangles add up to 180° = 82 so 82 degrees is pi radians the in... First equation gives us: a + b + angle C and F. Use a protractor to draw and measure acute and obtuse angles to the nearest degree thing you can do add! Remember that the angles on a straight line add up to 180 angles in a triangle add up to. for example, triangle... ) opposite angles are given in total calculate the other angles we need the sine, cosine and.! Plane triangle add up to 360 degrees. has three angles add to!


# why is this sequence convergent

Suppose to have two sequence $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ such that

$a_n=\frac{1}{2}(a_{n-1}+b_{n-1})$.

I want to prove that if $b_n\rightarrow0$ then $a_n\rightarrow0$.

The only thing I was able to prove is that $a_n$ is bounded, in fact:

$b_n$ is convergent and so bounded $|b_n|\leq M$. And so $|a_n|\leq |\frac{a_2}{2^{n-1}}+\frac{b_2}{2^{n-2}}+\cdots+\frac{b_{n-1}}{2}|\leq|\frac{a_2}{2^{n-2}}|+M\sum^\infty_{k=1}\frac{1}{2^k}$ and for great $n$ we have $|\frac{a_2}{2^{n-2}}|\leq\varepsilon$

Could you help me to continue (if I'm on the right track), please?

I see that if we prove that $a_n$ has a limit $L$ then necessarily $L=0$ because $L$ must satisfy $L=\frac{1}{2}L$, but I don't know how to prove that it has a limit.

-
+1 for showing your work –  Jyrki Lahtonen Oct 10 '11 at 6:00
Hint: If $n$ is large, the early terms of your middle expression are small (not just the $a_2$ one) because of the large power of 2, and the later terms are small because $b_n$ tends to zero. So can you see how to split the sequence differently, and get a better estimate? –  Mark Bennet Oct 10 '11 at 6:06
@Mark Sorry 'bout ruining your hint with my answer. I didn't see your comment soon enough :-( –  Jyrki Lahtonen Oct 10 '11 at 6:14
@JyrkiLahtonen: no hard feelings ... The technique of splitting a problem into pieces which are dealt with in different ways seemed worth noting anyway. In measure theory, for example, there are a number of proofs which require a proof for "most" points of a set based on some kind of regularity, and a proof for the remaining "problem" points by showing that they are confined to a set of arbitrarily small measure. –  Mark Bennet Oct 10 '11 at 7:09

Hint: You have a good start in proving that the sequence $(a_n)$ is bounded. Let's reuse your trick and look at $$a_{2n}=\frac{a_n}{2^n}+\frac{b_{2n-1}}2+\frac{b_{2n-2}}4+\cdots+\frac{b_n}{2^n}.$$ All the numbers $|b_k|<\epsilon$ for $k\ge n$, if $n$ is large enough. The first term $a_n/2^n$ looks like it would be under control as well now that you know $|a_n|$ to be bounded.

-

Assume that $-t\leqslant b_n\leqslant t$ for a given positive $t$ and for every $n\geqslant n_t$. Then $a_n-t\leqslant\frac12(a_{n-1}-t)$ and $a_n+t\geqslant\frac12(a_{n-1}+t)$ for every $n\geqslant n_t$. Thus $\limsup (a_n-t)\leqslant0$ and $\liminf (a_n+t)\geqslant0$. Since this holds for every positive $t$, $a_n\to 0$.

-
@Srivatsan, exactly. Thanks. –  Did Oct 10 '11 at 17:29

One way to do this is to view the recursion equation defining $a_n$ as a, well, recursion equation. It is then a non-homogeneous linear recursion, whose associated homogeneous recursion is very easy to solve. One could then use Lagrange's method of variation of parameters to determine the actual solution, but instead of doing that (we can't, in fact, because we do not know $b_n$) we can use the same idea to obtain bounds that will prove what you want. Let's do that.

Let $\varepsilon>0$. There exists an $N$ such $|b_n|\leq\varepsilon$ for all $n\geq N$.

Let $a_n=\frac{\alpha_n}{2^n}$ (this is where we use Lagrange's method: the solution for the homogeneous equation is $\frac{\alpha}{2^n}$, with $\alpha$ a constant and Lagrange suggests that we now turn $\alpha$ into a function $\alpha_n$ of $n$) and suppose that $|b_n|\leq\beta$ for all $n\geq1$. Replacing this in the defining recursion tells us that $$|\alpha_n-\alpha_{n-1}|=2^{n-1}|b_n|$$ for all $n\geq1$. This implies that when $n\geq N$ \begin{align}|\alpha_n-\alpha_0|&\leq|\alpha_n-\alpha_{n-1}|+|\alpha_{n-1}-\alpha_{n-2}|+\cdots+|\alpha_1-\alpha_0| \\ &\leq(2^{n-1}+\cdots+2^N)\varepsilon + (2^{N-1}+\cdots+2^0)\beta\\&\leq (2^n-2^N)\varepsilon+(2^N-1)\beta \\&\leq 2^n \varepsilon+c\end{align} for some positive constant $c$. Then $$|\alpha_n|\leq 2^n\varepsilon+c+|\alpha_0|$$ and $$|a_n|=|\alpha_n/2^n|\leq\varepsilon+\frac{c+|\alpha_0|}{2^n}.$$ This should make it clear that $a_n\to0$.

-
One can probably enhance this to a statement telling us that solutions to a non-homogeneous perturbation which gets smaller and smaller of a linear homogeneous recursion with all characteristic roots in $(-1,1)$ is not very different from the unperturbed solutions. –  Mariano Suárez-Alvarez Oct 10 '11 at 6:28


# Multiplying left hand and right hand limits question

I came across the attached question in our calculus book. The limits in question are:

F(x), which approaches 0 from the left and 1 from the right as x goes to 2

J(x), which approaches 1 from the left and 0 from the right as x goes to 2.

Question B) ii. asks for the limit of [F(x) + J(x)] as x goes to 2. The left hand limits add up to 1, and the right hand limits do too, so the limit is 1 as x approaches 2 - the answer key matches this.

However, in C) ii., which asks for limit of [F(x)J(x)] as x approaches 2, the left hand limits multiply to 0 and the right hand limits multiply to 0, but the answer key has DNE. Am I missing something completely about how limits work? Or is the answer key wrong?

Thanks in advance! Edited photo of the page

• I agree that $\lim_\limits{x\to 2} f(x)j(x) = 0$ – Doug M Sep 30 '19 at 20:59

$$\lim_{x\to 0^+} F(x)\cdot J(x)=\lim_{x\to 0^+} F(x)\cdot\lim_{x\to 0^+} J(x)=1\cdot0=0$$
$$\lim_{x\to 0^-} F(x)\cdot J(x)=\lim_{x\to 0^-} F(x)\cdot\lim_{x\to 0^-} J(x)=0\cdot1=0$$
$$\lim_{x\to 0} F(x)\cdot J(x)=0$$


# Are there any inflection points?

$$F(x) = \begin{cases} x^2 & x \le 0 \\ 0 & 0 \le x \le 3 \\ -(x-3)^2 & x>3 \end{cases}$$

My question is does this function have any points of inflection? Double Derivative at $$x=0,3$$.Thus the necessary condition is satisfied for all points such that $$0 \le x \le 3$$.

But, i have also read that the third derivative at x is not equal to 0 is also regarded as a sufficient condition, which is fulfilled at $$x=0,3$$. Thus, which sufficient condition should hold and which points should be regarded as inflection points?

An inflection point occurs at points of the domain at which the function changes concavity. The second derivative at an inflection point may be zero but it also may be undefined. If we look at the second derivative for your function

$$F''(x) = \begin{cases} 2 & x < 0 \\ 0 & 0 \le x \le 3 \\ -2 & x>3 \end{cases}$$

then $$F''(x) > 0$$ when $$x < 0$$ and $$F''(x)<0$$ when $$x>3$$. We know that the function $$F(x)$$ is concave up when $$F''(x) > 0$$ and concave down when $$F''(x) < 0$$. Therefore, it is clear that $$F(x)$$ is concave up on $$(-\infty,0)$$ and concave down on $$(3,\infty)$$.

But, what about the interval $$[0,3]$$? Since $$F(x)\equiv 0$$ when $$0\le x \le 3$$, the second derivative is also zero. As $$F(x)$$ changes concavity at the end points we know that there must be at least one inflection point. Therefore, we must decide whether there is one inflection point or two inflection points.

Suppose that we assume that there are two inflection points. Then, we would need to show that the function changes concavity twice - there would be two intervals where it was concave up and one interval where it is concave down. Alternatively, the function could be concave down on two intervals and then concave up on one interval.

Your function is concave up on one interval and then concave down on another interval. In the interval between these two intervals, the function is neither concave up nor concave down. Hence, your function only changes concavity once.

So, $$F(x)$$ has one inflection point. If we plot the function on Desmos, then we see that this point of inflection occurs in the middle of the interval $$[0,3]$$.

Interestingly, the function $$f(x)=x^3$$ exhibits similar behavior. This function is concave up on $$(0,\infty)$$ and concave down on $$(-\infty,0)$$. It has one inflection point at $$(0,0)$$.

• Yes, also that all straight lines a considered both concave and convex. So if ur talking about the change in concavity, then both 0,3 are inflection points. – DARE2ZLATAN Nov 8 at 22:55
• So, what ur saying, does that imply that any function that is concave in some interval and convex in some interval will surely have atleast 1 inflection point? – DARE2ZLATAN Nov 9 at 13:38
• Intuitively, yes. The inflection point must occur when the function changes concavity. When this happens, the function would go from a parabola that opens upward to a parabola that opens downward (or vice versa). I have never seen a proof of such a result. Although, I cannot think of a counterexample and therefore believe that this is a theorem. – Axion004 Nov 9 at 14:43


# Sum of arithmetic progression

The sum of arithmetic progression is denoted by $S_{n}$. It is nothing but the sum of 'n terms of an A.P. with first term 'a' and common difference 'd'.
The formula for sum of n terms of A.P. is
$S_{n} = \frac{n}{2}[2a + (n-1)d]$

$S_{n} = \frac{n}{2}[a + l]$ , where l = last term = a + (n -1 )d

Proof : Let $a_{1}, a_{2},a_{3},...,a_{n}$ be an A.P. with first term as 'a' and common difference as 'd'.
$a_{1}$ = a; $a_{2}$ = a + d; $a_{3}$ = a + 2d ; ... $a_{n}$ = a + (n -1)d
$S_{n} = a_{1} + a_{2} + a_{3} + ... +a_{n-1} + a_{n}$
⇒ $S_{n}$ = a + (a + d) + (a + 2d) + … + [ a + (n-2)d] + [ a + (n -1 )d] -----(i)
Write the above equation in reverse order we get,
$S_{n}$ = [ a + (n -1 )d] + [ a + (n-2)d] + … + (a + 2d) + (a + d) + a ----- (ii)
2$S_{n}$ = [ 2a + (n -1 )d] + [ 2a + (n-1)d] + …+ [2a + (n-1)d]
[2a + (n-1)d] repeats ‘n’ times
∴ 2$S_{n}$ = n [ 2a + (n -1 )d]
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]

Since the last term l = a + (n – 1)d
∴ $S_{n} = \frac{n}{2}$ [ a + a + (n -1 )d]

$S_{n} = \frac{n}{2}$ [a + l ]

Note : In the above formula there are 4 unknown quantities. So if any three are given then we can find the forth one.
In the sum of $S_{n}$ of n terms of a sequence is given then the nth term $a_{n}$ of the sequence can determined by using the following formula .
$a_{n} = S_{n} - S_{n - 1}$

## Solved examples on sum of arithmetic progression

1) 50,46,42,… 10 terms.
Solution: 50,46,42,… 10 terms
The formula to find sum is
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]

Number of terms = n = 10; First term = a = 50 ; Common difference = d = 46 – 50 = -4
Put all the given values in the formula we get,
$S_{10} = \frac{10}{2} [ 2 \times$ 50 + (10 -1 )(-4)]
$S_{10}$ = 5 [ 100 + (9 )(-4)]
$S_{10}$ = 5 [ 100 + (-36)]
$S_{10}$ = 5 (64)
$S_{10}$ = 320

2) 3, $\frac{9}{2}$, 6, $\frac{15}{2}$ ,… 25 terms .
Solution: 3, $\frac{9}{2}$, 6, $\frac{15}{2}$ ,… 25 terms .
The formula to find sum is
$S_{n} = \frac{n}{2}$ [ 2a + (n -1 )d]

Number of terms = n = 25; First term = a = 3 ; Common difference = d = , $\frac{9}{2}$ - 3 = $\frac{3}{2}$
Put all the given values in the formula we get,
$S_{25} = \frac{25}{2} [ 2 \times 3 + (25 -1) \frac{3}{2}$]
$S_{25}$ = 12.5 [ 6 + (24 )(1.5)]
$S_{25}$ = 12.5 [ 6 + 36]
$S_{25}$ = 12.5 (42)
$S_{25}$ = 525

3) In an A.P. the sum of first n terms is $\frac{3n^{2}}{2} + \frac{13}{2}n$. Find its 25th term.
Solution : $S_{n} = \frac{3n^{2}}{2} + \frac{13}{2}n$.

When n = 25,
$S_{25} = \frac{3 \times 25^{2}}{2} + \frac{13}{2} \times25$.

$S_{25}$ = 1100
Now we will find $S_{n - 1} = S_{25 -1} =S_{24}$
$S_{24} = \frac{3 \times 24^{2}}{2} + \frac{13}{2} \times24$.

$S_{24}$ = 1080
As we know that ,
$a_{n} = S_{n} - S_{n -1}$
$a_{25} = S_{25} - S_{24}$
$a_{25}$ = 1100 – 1080 = 80
∴ 25th term is 80.

From sum of arithmetic progression to Home

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

Russia-Ukraine crisis update - 3rd Mar 2022

The UN General assembly voted at an emergency session to demand an immediate halt to Moscow's attack on Ukraine and withdrawal of Russian troops.


# A simple riddle related to addition of odd numbers

I'm not sure if this type of question can be asked here, but if it can then here goes:

Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.

PS : The inception of this question is a result of a random discussion that I was having during the break hour :)

-
Hint: adding an odd number of odd numbers will give ... (is this a real question ?) – Raymond Manzoni May 25 '13 at 8:27
Hmm..let me ponder over that – 403 Forbidden May 25 '13 at 8:29
This says sum of 9 odd numbers can't be even algebra.com/algebra/homework/word/numbers/… True? – 403 Forbidden May 25 '13 at 8:32
Yes: consider the last digit in binary if it helps. – Raymond Manzoni May 25 '13 at 8:36

A direct approach:

Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like $${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$ (the integers $a,b,\ldots,i$ may or may not be the same). Grouping things together, this is equal to $$2(a+b+c+d+e+f+g+h+i+4)+1.$$ Thus, the result is odd.

An simpler approach would be to prove these three simple facts: \begin{align*} \mathsf{odd}+\mathsf{odd}&=\mathsf{even}\\ \mathsf{odd}+\mathsf{even}&=\mathsf{odd}\\ \mathsf{even}+\mathsf{even}&=\mathsf{even} \end{align*} Thus, starting from $$\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}$$ and grouping into pairs, $$\mathsf{odd}+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})$$ we use our facts to see that this is $$\mathsf{odd}+\mathsf{even}+\mathsf{even}+\mathsf{even}+\mathsf{even}.$$ Grouping again, $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})+(\mathsf{even}+\mathsf{even})$$ becomes $$\mathsf{odd}+\mathsf{even}+\mathsf{even}$$ becomes $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})=\mathsf{odd}+\mathsf{even}= \mathsf{odd}$$

-
I'm sorry. I don't think I understood your query. – 403 Forbidden May 25 '13 at 8:34
Shouldn't that be +9? – Ataraxia May 25 '13 at 8:35
Or just $$\mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd}\\= \mathsf{\underbrace{even+odd}}\\= \mathsf{odd}$$ – Rahul May 25 '13 at 8:43
@Rahul: Very clever user of MathJax! I approve :) – Zev Chonoles May 25 '13 at 8:44
@403Forbidden you can always select it (the checkbox under the voting arrows) if you found it most helpful. – Ataraxia May 25 '13 at 8:51

$$\sum_{k=1}^{9}(2n_k+1)= 2\sum_{k=1}^9n_k+9$$

$$41=2\sum_{k=1}^9n_k$$

There is no integer $\sum_{k=1}^9n_k$ that satisfies this.

-
But my colleague says otherwise. And he refuses to let me in on the solution until the end of day..which is what is killing me. Argh! These math riddles are so addictive – 403 Forbidden May 25 '13 at 8:33
@403Forbidden he would have to demonstrate that it is possible for the sum of a set of integers to be a non-integer, and if he could do that he's worthy of an award. – Ataraxia May 25 '13 at 8:34
As it turns out, it was just a trick question. He admitted to this (getting a result of 50) being not possible :) – 403 Forbidden May 25 '13 at 9:09


probability of guessing the correct password

Suppose I need to guess a password of one digit. Each time I'm wrong the password increases by another digit. What's the probability that I can correctly guess the password (assuming I have unlimited amount of time and the number is all uniform)?

My approach: since the number is generated random, it doesn't matter what strategy we adopt to guess the number. So consider the following sequence of guess $$1, 21, 221, 2221, 22221, ...$$. The probability that the password results in $$k$$-th guess is then $$\frac{1}{10^n}$$. Thus the total probability is $$\sum_n \frac{1}{10^n}$$. So the probability is $$0.11111111111... = \frac{1}{9}$$.

Is my approach correct, and if so, is there any better solution for this. The fraction $$\frac{1}{9}$$ seems to suggest an easier perspective to look at this puzzle.

• The probability of succeeding on the $k$th guess must account for failing at the $1$st, $2$nd, ... , $k-1$th guess. I don't think this has been taken into account. For example, if you have to find $21$, you have to first fail at $2$ : so the probability of succeeding at the second turn is not $\frac 1{100}$ but rather $\frac{9}{10} \times \frac{1}{100}$, because you need to fail the first time. Also realize another thing : let's say you call out $5$ and fail. Then you KNOW that when you add a digit , the resulting number doesn't start with $5$ (so it can't be e.g. $54$). Strategy improved. Jul 25 at 6:31
• I don't understand your comment. If I succeed at the second turn with the guess $21$, then that means the first digit is $2$, which I guessed wrong by saying $1$, and the second digit is $1$, which I guessed right. So the probability here is $\frac{1}{100}$. Because the first digit is $2$ so that should already include the events that I guess the first one wrong Jul 25 at 7:03
• If you guessed $2$ and got it wrong, then you know the number cannot be $2$, so it can be any of the others. But once a digit is added, you know that the password cannot be , say $20,21,23,27$ etc., because if it was, then the previous answer would have had to be $2$, and you would have been correct. So you do get information from wrong guesses. From wrong guesses, you know , for example what the future password cannot start with : your guess that was called wrong. Jul 25 at 7:06

What is the probability that you keep failing indefinitely?

Probability that you fail first time is $$\cfrac{9}{10}$$

Probability that you fail twice in a row? Now here I assume that the person guessing the password can keep track of what they guessed and does not choose from the one's that are obviously incorrect.

Say you guessed $$2$$ first time and it was wrong. Now one digit gets added. So you need to guess from $$100$$ numbers ($$00 - 99$$) but you also know it cannot be any number between $$20$$ and $$29$$.

So probability that you fail twice in a row is,

$$\cfrac{9}{10} \cdot \cfrac{89}{90} = \cfrac{89}{100}$$

Now for the third guess, say you guessed $$2$$ the first time and $$18$$ the second time and failed both times, you know the three digit number is not between $$200 - 299$$ and you also know it is not in $$180 - 189$$.

Probability that you fail thrice in a row is,

$$\cfrac{9}{10} \cdot \cfrac{89}{90} \cdot \cfrac{889}{890} = \cfrac{889}{1000}$$

You see the denominator of the third is divisible by numerator of the second, denominator of the second by the numerator of the first?

Eventually at the end of $$n$$ guesses, probability that you failed in all of them is

$$P(F) = \cfrac{10^n - 10^{n-1} - 10^{n-2} ... - 10^0}{10^n}$$

So probability that you succeed in $$n$$ guesses,

$$P(S) = 1 - P(F) = \cfrac{10^n - 1}{9 \cdot 10^n}$$

Now as $$n \to \infty$$, what do you get for $$P(S)$$?

• +1, I know it's kind of "obvious" that making random guesses (with the idea of eliminating previous guesses) is the best strategy, but I've never actually seen a mathematical proof that "XYZ strategy is the best for ABC problem". I think I'll attach one when I see it here. I mean, we all kind of understand that symmetry means that any possibility is likely, but having it out there would still be nice. There must be like this "space" or set of strategies, on which we can put a distance and have like a "best-fit" criterion which works here. Very interesting, though! Jul 25 at 7:42
• @TeresaLisbon thank you. yes for the purpose of this question, I assume guesses are random with elimination of previous guesses. It will definitely be interesting to see if there are strategies that can better random guess in this setting. Jul 25 at 7:55
• Yes, I do not think that there is a better strategy than random guessing with the kind of elimination you make. I will look for sources anyway and let you know, thanks for the response. Jul 25 at 7:57