# If $a$ and $b$ are positive integers and $4ab-1 \mid 4a^2-1$ then $a=b$.

Prove that if $a$ and $b$ are positive integers and $$(4ab-1) \mid (4a^2-1)$$ then $a=b$.

I am stuck with question, no idea. Is there any way to prove this using Polynomial Division Algorithm? Would appreciate any help.

• This actually holds even with the weaker assumption that $4ab - 1 \mid (4a^2 - 1)^2$. – George V. Williams Apr 15 '15 at 19:48
• I added a generalization to my answer which makes it crystal clear how this is nothing but the uniqueness of inverses. – Bill Dubuque Apr 15 '15 at 22:02

Note that $4ab\equiv 1\pmod {4ab-1}$ and if $4ab-1\mid 4a^2-1$ then $4a^2\equiv 1\pmod{4ab-1}$. So $$b\equiv (4a^2)b=a(4ab) \equiv a\pmod{4ab-1}.$$

Is that possible if $a\neq b$?

Without using modular arithmetic, you can write this as:

$$a-b = b(4a^2-1)-a(4ab-1)$$

So $4ab-1\mid a-b$.

• The proof is just a special case of the uniqueness of inverses - see my answer. – Bill Dubuque Apr 15 '15 at 20:01
• Well, yes, but you still arrive at $4ab-1\mid a-b$. @BillDubuque – Thomas Andrews Apr 15 '15 at 20:05
• The point is that viewing it this way makes everything obvious, even the requisite inequality (omitted in your answer). I expanded my remark to emphasize this - giving the natural generalization of the OP's theorem. Hopefully that makes my point clearer. – Bill Dubuque Apr 15 '15 at 21:57

Clearly, $a \geq b$, since we need $4a^2-1 \geq 4ab-1$. We have $$\dfrac{4a^2-1}{4ab-1} = k \in \mathbb{Z}$$ Hence, $$k = \dfrac{4a^2-1}{4ab-1} = \dfrac{4a^2-4ab+4ab-1}{4ab-1} = 1 + \dfrac{4b(a-b)}{4ab-1}$$ Now note that $\gcd(4ab-1,4b)=1$, since $(4b)a - (4ab-1) = 1$. Hence, $4ab-1$ divides $a-b$. However $0 \leq a-b < 4ab-1$. Hence, $a-b=0 \implies a=b$.

Note $\,\ {\rm mod}\,\ 4ab\!-\!1\!:\,\ \overbrace{(4a)\color{#c00}a\equiv 1\equiv(4a)\color{#c00}b}^{\ \ \ \large \color{#c00}a\, \equiv\, (4a)^{-1}\equiv\,\color{#c00} b}\,\Rightarrow\, \color{#c00}{a\equiv b}\$ by uniqueness of inverses (of $\,4a\,$ here)

Remark $\$ For completeness, here is the standard proof of uniqueness of inverses:

$$\ ca\equiv 1\equiv cb \,\Rightarrow\, a\equiv a(cb)\equiv (ac)b\equiv b\quad\ \$$

The OP Theorem becomes obvious when expressed in this general form, amounting simply to the uniqueness of inverses mod $\,n\,$ within the standard rep system $\,\{0,1,\ldots,n\!-\!1\},\,$ namely

Theorem $\$ If $\, \color{#0a0}{a< bc}\,$ and $\, bc\!-\!1\mid ac\!-\!1\$ then $\ a = b,\$ for integers $\,a,b,c > 0$

Proof $\ \ {\rm mod}\,\ n\!=\!bc\!-\!1\!:\,\ bc\equiv 1\equiv ac\,\Rightarrow\, a\equiv c^{-1}\!\equiv b.\$ $\,bc\!-\!1\mid ac\!-\!1\,\Rightarrow\, b\le a \le \color{#0a0}{bc\!-\!2}\,$ (by $\,b\not\equiv 0).\,$ So $\,a\equiv b\pmod{n}\,$ and $\,a,b\in \{0,1,\ldots,\color{#0a0}{n\!-\!1}\}\Rightarrow\, a=b\,$ (else $\,n\,$ divides the smaller natural $\,a\!-\!b> 0,\,$ contradiction). $\ \$ QED

The OP is the special case $\ c = 4a,\$ where $\ a < 4ab = bc,\,$ so the Theorem applies.

Note how translating the theorem into the language of congruences has simplified it so much that we immediately recognize it as a special case of a well-known result about uniqueness of inverses. This is yet another example of a ubiquitous principle that I frequently emphasize here, namely uniqueness theorems provide powerful tools for proving equalities.

Note that any integer $m \ge 2$ will do, instead of 4.

For example, take Thomas Andrew's solution, and put $m$ for $4$ everywhere:

Note that $mab\equiv 1\pmod {mab-1}$ and if $mab-1\mid ma^2-1$ then $ma^2\equiv 1\pmod{mab-1}$. So $$b\equiv (ma^2)b=a(mab) \equiv a\pmod{mab-1}.$$

Is that possible if $a\neq b$?

Without using modular arithmetic, you can write this as:

$$a-b = b(ma^2-1)-a(mab-1)$$

So $mab-1\mid a-b$.

My note:

To show that $mab-1 > a-b$ for $m \ge 2$, $ab-a+b-1 =(a+1)(b-1) \ge 0$.

Note that if $m=b=1$, then $mab-1 = a-1$.

• Yes: $\ (ma)a\equiv 1\equiv (ma)b\,\Rightarrow\, a\equiv (ma)^{-1}\equiv b\,$ by uniqueness of inverses, see my answer. To get the general result replace $\,ma\,$ by $\,c\,$ above. – Bill Dubuque Apr 15 '15 at 20:23
• Update: even the inequality has a natural interpretation when viewed from this standpoint: it is simply what's needed for the inverses to lie in the standard rep range. So the theorem is precisely equivalent to the uniqueness of inverses in the standard rep range, except the congruence language has been removed, being replaced by equivalent divisibility language, which obscures the essence of the matter: inverse unqueness (and it is further obfuscated by choice of a specific modulus). See the edit to my remark.for details. – Bill Dubuque Apr 16 '15 at 0:01


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# Modular arithmetic

Hello,

What is the remainder when the following sum is divided by 4? $1^5 + 2^5 + 3^5 +...+ 99^5 + 100^5$

I feel like it has to do with modular arithmetic... I am trying to decompose every number but it seems to long and unnecessary. Any ideas?

P.S. thank you for your ideas. I got it. Please don't post solutions

• It's $0$, I believe. – Akiva Weinberger Sep 24 '14 at 17:04
• There's no need to compute all the numbers. Modular Arithmetic is the way, but you have also to get some regularity. Usually, the first thing to do is to try smaller numbers, to see if there are patterns – Exodd Sep 24 '14 at 17:07
• Hint: Any even number squared is divisible by 4 and any odd number power will give remainder 1. So count how many odds there are. – Ali Caglayan Sep 24 '14 at 17:08

HINT : Note that in mod $4$, $$1^5\equiv1,\ \ 2^5\equiv 0,\ \ 3^5\equiv (-1)^5=-1\equiv 3,\ \ 4^5\equiv 0$$ and that $$1+0+3+0\equiv 0,\ \ 100=4\times 25.$$

hint :take {1,2,3,4},{5,6,7,8} .....{97,98,99,100} as a set now each corresponding term in each set has same remainder when divided by 4 , so you effectively need to calculate only the remainder of $1^5+2^5+3^5+4^5$ w.r.t 4 and then multiply it by 25 and again find that numbers remainder w.r.t 4

$$1^5 \equiv 1 \pmod{4}$$ $$4\mid 2^5 \Rightarrow 2^5 \equiv 0 \pmod{4}$$ $$3^5 \equiv (-1)^5 \equiv -1 \pmod{4}$$ $$4^5 \equiv 0^5 \equiv 0 \pmod{4}$$

This means that every even number, when raised to the fifth power, is $0 \pmod{4}$. Thus the sum you asked about is equal to the number of $1 \pmod{4}$ numbers in $\{1,2,\dots 100\}$ minus the number of $3 \pmod{4}$ numbers in $\{1,2,\dots 100\}$.

Every fourth number from $1$ to $97$, inclusive, is $1 \pmod{4}$. There are $25$ of these.

Every fourth number from $3$ to $99$, inclusive, is $3 \pmod{4}$. There are $25$ of these as well, so

$$\displaystyle\sum\limits_{n=1}^{100} n^5 \equiv 0 \pmod{4}$$


# Matrix size for error-correcting linear code

This is Exercise 21 of the textbook "Abstract Algebra: Theory and Applications" by Thomas W. Judson, 2016; Page 111. Chapter 8 "Algebraic Coding Theory" mainly deals with binary linear code with Hamming code as an example.

Exercise 8.21: If we are to use an error-correcting linear code $C$ to transmit the 128 ASCII characters, what size matrix must be used?

The screenshot is as follows:

My Attempt: I consider the parity check matrix $H_{(n-k) \times n} = [P_{(n-k)\times k}\mid I_{n-k}]$. Let $m = n - k$, the number of parity bits. First of all, the data bit is $k = \log_2 128 = 7$.

Let the error-correcting capability be $t = 1$. The minimum Hamming distance of the linear code $C$ is at least $3$. Therefore, the zero column and $e_i$'s columns are not in $P_{(n-k) \times k}$ (because $H$ contains no idential columns). Thus, we have $2^{m} - m - 1 \ge k$, which gives $m \ge 4$. That is, the matrix size is $4 \times 11$ (Note: not $4 \times 7$ as pointed out in the comment by @Jyrki Lahtonen).

Is the argument above correct?

Also, how to solve this problem for any parameter $t$, the error-correcting capability of $C$?

• The dimensions of $H$ are $(n-k)\times n$. So the number of columns should be $n=k+m$. Not $k$ as your final answer suggests. In general ($t>1$) it will be more difficult to judge the number of redundant bits needed to correct $t$ errors. It may turn out that the answer is known for $7$-dimensional codes, but I can only give bounds (e.g. what you would get using a BCH-code). Apr 17 '17 at 9:45
• @JyrkiLahtonen Thanks. I will modify my answer. I think the problem for finding the matrix size for the general case of $t > 1$ is related to the lemma that "If $H$ is the parity check matrix of $C$, then the minimum Hamming distance of $C$ equals the minimum number of columns of $H$ that are linearly dependent." Is it right? By the way, a bound on the matrix size is also appreciated. Apr 17 '17 at 9:52
• " what matrix size must be used?" Is that really the full text of the exercise? What are the error detection-correction capabilities desired? Is the code binary? There seems to be some missing data. BTW, the link does not work for me (what about a screenshot?) Apr 17 '17 at 22:04
• @leonbloy Yes, it is "what size matrix must be used" (I have reordered two words by mistake). The code is binary. This is implied by the text in the chapter. Thanks for pointing this out. I will update this post right now. Apr 18 '17 at 3:03

I assume that by "an error-correcting linear code", they meant a code capable of correcting one (bit) error (per coded symbol).

In that case your reasoning is on the right track (and it's the reasoning used to construct a Hamming code). We know $k=7$, the matrix $H$ will have up to $2^m-1=n$ columns. The smallest Hamming code in our case is then $(15,11)$ ($11$ bits of information plus $4$ of redundacy).

Because this give a code with higher $k$ than needed, we can trim the $11-7=4$ unused data bits, and we are left with a $(11,7)$ code ($7$ bits of information plus $4$ of redundacy).

(Equivalently, the Hamming construction gives the -sufficient- condition of $n\le 2^{n-k}-1$; it's easy to check that, for $k=7$, $n=11$ is the minimum number that fulfill the condition)

If we only need one error detection, a single parity bit is enough, so we have a $(8,7)$ code.

• Thanks. By the way, is this problem much harder for general $t$, the error-correcting capacity? Apr 18 '17 at 3:35
• Yes, see BCH codes for example. Apr 18 '17 at 4:57

Answer my question: I realized that my argument in the post relies on the standard parity-check matrix $H$. I show another argument here and ask for reviews.

First, the information bit $k = \log_2 128 = 7$.
Let $r = (n-k)$.
The parity-check matrix $H_{(n-k) \times n}$ has $r$ rows. Therefore, $H$ has at most $2^{r} - 1$ columns (the zero column is excluded).
To achieve the an-error-correcting capability, the code should be able to correct all the possible $n = 7 + r$ one-bit errors, each corresponding to one column of $H$. Therefore, we have $$2^r - 1 \ge 7 + r.$$ That is, $r \ge 4$ and the size of $H$ is at least $4 \times 11$.


# The Monte Carlo method

Monte Carlo methods are a group of computational algorithms that rely on repeated random sampling to obtain numerical results. The underlying concept is to use randomness to solve problems that might be deterministic in principle. They are often used in physical and mathematical problems and are most useful when it is difficult or impossible to use other approaches. Monte Carlo methods are mainly used in optimization problems, numerical integration, and to generate draws from a probability distribution.1

# Estimating the value of $$\pi$$

Consider the circle in the diagram. If we generate a sufficiently large number of $$(x, y)$$ points in the domain $$-1 \leqslant x \leqslant 1$$ and $$-1 \leqslant y \leqslant 1$$, then the fraction of points that fall inside the circle will be equal to the ratio between the area of the circle and area of the square.

$$\frac{\textrm{number of points that fall inside the circle}}{\textrm{total number of points}} = \frac{\textrm{circle area}}{\textrm{square area}} =$$
$$= \frac{\pi R^2}{(2R)^2} = \frac{\pi}{4}$$

A point falls inside the circle if it complies to the following condition:

$$x^2 + y^2 \leqslant R^2$$

By counting the number of random points that fall within the circle we can estimate $$\pi$$ as:

$$\pi = 4 \times \frac{\textrm{number of points that fall inside the circle}}{\textrm{total number of points}}$$

## The algorithm:

We can thus devise an algorithm that will:

1. Initialize a counter of the points that fall inside the circle (circle_points).
2. Generate a random point $$(x, y)$$.
3. Check if the point falls within the circle by evaluating $$x^2 + y^2 \leqslant R^2$$.
4. If the abose condition is true then increment the counter circle_points.
5. Calculate $$\pi$$ .
6. Repeat from step 2 for the desired number of runs.

In principle, the higher the number of runs the better the estimation will be (closer to the real value of $$\pi$$).

## The code

Putting the above algorithm to pratice we get the following code2:

import random

def run_simulation(runs=1000):
simulation = []
circle_points = 0
for i in range(runs):
# Create a (x, y) point at random
x, y = random.uniform(-1, 1), random.uniform(-1, 1)

# Check if the point falls within the circle
if x**2 + y**2 <= 1:
circle_points += 1

# Calculate current pi value
pi_estimate = 4 * circle_points/(i+1)

# Save simulation step
simulation.append([x, y, pi_estimate])

# Print final pi value
print('Result: Pi =', simulation[-1][2])
return simulation

sim = run_simulation(runs=10000)

Here is the result of the simulation from 1000 runs to 100000 runs. At 100000 runs the estimation ($$\pi = 3.142720$$) is already very close the actual value ($$\pi = 3.141593$$).

And if you are curious, here is how the estimation approaches the real value from the first run to 10000th run.

A final remark: We do not need to perform the simulation above using a full circle. By selecting a quarter of a circle and the corresponding square we could achieve the same result with a fourth of the computational effort.


# Reducing space between for some of the equations

I need to display a few equations as below

The code I have written is :-

\documentclass{report}
\usepackage{amsmath}

\begin{document}
$$\label{Equ. 3.9} x_{i+1}=\cos(a_i) \cdot [x_i-y_i \cdot 2^{-i} \cdot d_i] \tag{3.9}$$

$$\label{Equ. 3.10} y_{i+1}=\cos(a_i) \cdot [y_i+x_i \cdot 2^{-i} \cdot d_i] \tag{3.10}$$

$$\label{Equ. 3.11} \cos(\alpha)=\dfrac{1}{\sqrt [2]{1+{\tan(\alpha)}^2}} \tag{3.11}$$

$$\label{Equ. 3.12} K_i=\cos(arctan(2^{-i}))= \dfrac{1}{\sqrt [2]{1+\tan(arctan(2^{- i}))}}=\dfrac{1}{\sqrt[2]{1+2^{-2i}}} \tag{3.12}$$

The product of $K_i$  represents the so-called K factor (Equ. \ref{equ: 3.13})

$$\label{equ: 3.13} K=\prod K_i=\prod_{i=0}^{n-1} \dfrac{1}{\sqrt{1+2^{-2i}}} \tag{3.13}$$

\end{document}

The result I am getting with the above code is as below

I need to reduce the space between the equations so the result looks like the 1st screenshot. Please guide how I can achieve this?

The equation that caused the numbering problem is:

$$\label{equ:matrix} V= \begin{bmatrix} x'\\ y'\\ \end{bmatrix} \begin{bmatrix} x \cdot \cos(a) - y \cdot \sin(a)\\ y \cdot \cos(a) + x \cdot \sin(a)\\ \end{bmatrix} \tag{3.6}$$

When i remove the tag command from this it throws an error

• The equations displayed are also aligned by =, so the align environment would be best. You can still label and tag each equation individually, – John Kormylo Sep 2 '15 at 19:05
• Please don't post a substantially new query as an addendum to an existing posting. For one, relatively few people may realize that you've modified an existing query to ask a follow-up question. Instead, please post a new query. – Mico Sep 4 '15 at 16:20
• @Mico Sure... Sorry about that.. I am on it – Shray Sharan Sep 4 '15 at 16:22

Use the gather environment and the \intertext command. Btw, you don't have to put the equation numbers by yourself. Also, I propose to load the cleveref package: in cross references, you won't even have to type ‘equ.’, since cleveref knows (most) counters, and adds the counter name before its value.

\documentclass{report}
\usepackage{mathtools}

\begin{document}

\setcounter{chapter}{3}\setcounter{equation}{8}
\begin{gather}
\label{Equ. 3.9}
x_{i+1}=\cos(a_i) \cdot [x_i-y_i \cdot 2^{-i} \cdot d_i] \\
\label{Equ. 3.10}
y_{i+1}=\cos(a_i) \cdot [y_i+x_i \cdot 2^{-i} \cdot d_i] \\
\label{Equ. 3.11}
\cos(\alpha)=\dfrac{1}{\sqrt [2]{1+{\tan(\alpha)}^2}}\\
\label{Equ. 3.12}
K_i=\cos(\arctan(2^{-i}))= \dfrac{1}{\sqrt [2]{1+\tan(\arctan(2^{- i}))}}=\dfrac{1}{\sqrt[2]{1+2^{-2i}}}\\
\intertext{The product of $K_i$ represents the so-called K factor (equ. \eqref{equ: 3.13})}
\label{equ: 3.13}
K=\prod K_i=\prod_{i=0}^{n-1} \dfrac{1}{\sqrt{1+2^{-2i}}}
\end{gather}

\end{document}

• I used the tag command in one of the equations in between and from their it has stopped taking care of the numbering... Can i post the code for that equation in this post for you check it? – Shray Sharan Sep 4 '15 at 15:27
• Please do, since it is linked to this question. – Bernard Sep 4 '15 at 15:29
• I added the code for that equation.. – Shray Sharan Sep 4 '15 at 15:32
• I've inserted your equation code amidst the other equations and the result is normal, except number 3.6 between 3.12 and 3.13 looks weird. Could you post a complete code showing the problem (perhaps you should initiate a new thread)? – Bernard Sep 4 '15 at 16:07

Especially since you're already loading the amsmath package, you should look into using that package's gather environment to typeset a collection of displayed, numbered equations. Use \intertext{...} to intersperse text between some of the equations.

Some side remarks: For the sake of good (math) typography, consider getting rid of all \cdot directives and getting rid of the 2 radicands for the square root ops.

\documentclass{report}
\usepackage{amsmath}
\allowdisplaybreaks
\begin{document}
\begin{gather}
x_{i+1}=\cos(a_i) \cdot [x_i-y_i \cdot 2^{-i} \cdot d_i]
\tag{3.9} \label{Equ. 3.9}\\
y_{i+1}=\cos(a_i) \cdot [y_i+x_i \cdot 2^{-i} \cdot d_i]
\tag{3.10} \label{Equ. 3.10}\\
\cos(\alpha)=\dfrac{1}{\sqrt [2]{1+{\tan(\alpha)}^2}}
\tag{3.11} \label{Equ. 3.11} \\
K_i=\cos(\arctan(2^{-i}))= \dfrac{1}{\sqrt [2]{1+\tan(\arctan(2^{-   i}))}}=\dfrac{1}{\sqrt[2]{1+2^{-2i}}}
\tag{3.12} \label{Equ. 3.12}\\
\intertext{The product of $K_i$  represents the so-called $K$ factor (Equ. \ref{equ: 3.13})}
K=\prod K_i=\prod_{i=0}^{n-1} \dfrac{1}{\sqrt{1+2^{-2i}}}
\tag{3.13} \label{equ: 3.13}
\end{gather}
\end{document}


For an equilateral triangle with $n$ dots on a side, how many lines are needed to connect each dot to every other dot?

For an equilateral triangle of side length $n$ dots, as shown in this diagram below, construct a function, $f(n)$, which outputs the number of lines needed to connect up every dot to every other dot. A straight line through three or more dots counts as only one line! E.g. $f(3) = 9$ and $f(4) = 24$.

Can anyone point me in the right direction with this problem? Perhaps tell me what area of mathematics or what concepts would help me solve this? If this is a trivial problem don't give the answer but tell let me know. Thanks.

• Hmm... how would $f(3)=9$? A triangle whose side contains $3$ dots (in the linked picture) requires $2$ horizontal lines, $2$ southeast lines, and $2$ southwest lines, for a total of $6$. Perhaps clarify what you mean by "connect up all the dots to one another." Jun 26 '18 at 19:22
• Every dot is connected to every other dot via a straight line. Jun 26 '18 at 19:31
• @Frpzzd plus the three lines connecting the corner points to the middle point of the opposite side. Jun 26 '18 at 19:42
• Are you sure $f(4)$ is not $21$? Or are there dots not along the edge for $n>3$? Can you clarify what the shape should be for $n=4$ and possibly $n=5$? Jun 26 '18 at 19:45

For a side of $m$ dots there are a total of $n=\frac 12m(m+1)$ dots. There are then $\frac 12n(n-1)$ pairs of dots. If you have a line with $k \gt 2$ dots on it, it accounts for $\frac 12k(k-1)$ of the pairs of dots, so it reduces the count by $\frac 12k(k-1)-1$

We can see this for the order $4$ triangle. There are $10$ points, three lines with four points (the sides) and three lines with three points. The number of lines is then $\frac 12\cdot 10 \cdot 9 - 3(\frac 12\cdot 4 \cdot 3-1)-3(\frac 12 \cdot 3 \cdot 2 -1)=45-3\cdot 5 -3 \cdot 2=24$

If we do the order $5$ triangle we have $15$ dots, three lines with five dots, three with four dots, and six with three dots. The new ones with three dots run from a corner through the center to the middle of the other side. This gives $\frac 12\cdot 15\cdot 14-3\cdot 9-3\cdot 5 -6\cdot 2=51$ lines

This gives sequence A244504 which begins $$3, 9, 24, 51, 102, 177, 294, 459, 690, 987, 1380, 1875, 2508, 3279, 4212, 5319, 6648, 8199, 10026, 12141, 14580, 17343, 20496, 24051, 28068, 32547, 37542, 43071, 49218, 55983, 63456, 71661, 80658, 90447, 101100, 112635, 125160, 138675, 153252, 168915, 185784$$

No closed formula is given.

• I understand your logic but I asked for a formula not a sequence nor a summation (as is provided in the description of the OEIS link). Can a "closed" formula be given? Jun 26 '18 at 20:08
• Your request seemed much more open ended that that. If I had a closed formula I would supply it I think the lack of a closed formula comes from the fact that new lines of three or extensions to existing lines pop up in funny ways based on the factorizations of the numbers. Jun 26 '18 at 20:36

Here's a solution for the problem described, i.e. if the shape is an equilateral triangle whose sides are $n$ equally spaced dots. This is assuming the dots along the sides of the triangle are the only dots.

First, count the number of dots. There are $n$ dots along one edge, with one of those dots being shared with the other two edged. Then, there are $n-1$ new dots along a second edge, with one of those shared with the final edge. Finally, there are $n-2$ dots remaining on the third edge. This gives a total of $n+(n-1)+(n-2) = 3(n-1)$ dots.

The number of edges connecting two dots is then $\binom{3(n-1)}{2} = \frac{3(n-1)(3n-4)}{2}$. However, along each side of the triangle, there are $\binom{n}{2} = \frac{n(n-1)}{2}$ edges along the same line. So, we need to subtract all but one of these (the edge connecting the corners) for each side, totaling $3[\binom{n}{2}-1] = \frac{3(n+1)(n-2)}{2}$.

Thus, the total number of necessary lines is $$f(n) = \binom{3(n-1)}{2}-3\left[\binom{n}{2}-1\right] = 3(n^2-3n+3)$$

This gives $f(2) = 3, f(3) = 9, f(4) = 21, f(5) = 39$ and so on.

• The count of $24$ for the triangle of side $4$ indicates that OP is considering a filled triangle. Jun 26 '18 at 20:51
• I had no indication of where OP got that answer from, and the question as described leaves it to the imagination. Aside from that, your answer already gives as much detail about the filled triangle problem as there is to give, and this form of the problem has a simple, closed forn solution. Jun 26 '18 at 20:59
• @AlexanderJ93 I'll show you Jun 26 '18 at 21:01
• @ThoughtBox I did ask for clarification but got none, however I answered the question as you described it. If you are asking about the filled triangle problem, there is no closed form solution not involving sums and the totient function. This is determined. Jun 26 '18 at 21:34
• What does determined mean? And what makes you think that summation is unavoidable? Jun 26 '18 at 21:35


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# exponential function definition

a z = e z 1n a. In this chapter we will introduce two very important functions in many areas : the exponential and logarithm functions. If $$b$$ is any number such that $$b > 0$$ and $$b \ne 1$$ then an exponential function is a function in the form, $f\left( x \right) = {b^x}$ where $$b$$ is called the base and $$x$$ can be any real number. An exponential function in Mathematics can be defined as a Mathematical function is in form f(x) = a x, where “x” is the variable and where “a” is known as a constant which is also known as the base of the function and it should always be greater than the value zero.. Exponential Function Reference. These properties are the reason it is an important function in mathematics. This is the definition of exponential growth: that there is a consistent fixed period over which the function will double (or triple, or quadruple, etc; the point is that the change is always a fixed proportion). The exponential function e x is an integral transcendental function. Note: In reality, exponential growth cannot continue indefinitely. Write the equation of an exponential function that has been transformed. If you are in a field that takes you into the sciences or engineering then you will be running into both of these functions. In an exponential function, the independent variable, or x-value, is the exponent, while the base is a constant. In mathematics, the exponential function is the function e x, where e is the number (approximately 2.718281828) such that the function e x is its own derivative. Therefore, e x is the infinite y limit of (1 + x y) y. Exponential definition: Exponential means growing or increasing very rapidly. The function $$y = {e^x}$$ is often referred to as simply the exponential function. It can be expanded in the power series. In this setting, e 0 = 1 , and e x is invertible with inverse e − x for any x in B . Here's what that looks like. In this lesson, we will go over the definition of linear and exponential functions then compare and contrast the two. Exponential function definition: the function y = e x | Meaning, pronunciation, translations and examples The relation between the exponential function a z and the exponential function e z is given by the equation. Meaning: Let’s start off this section with the definition of an exponential function. Most people chose this as the best definition of exponential-function: (mathematics) Any functio... See the dictionary meaning, pronunciation, and sentence examples. ‘For potassium, the shape of the curve could be fitted by a negative exponential function followed by a null linear function (constant value).’ ‘These relationships between length or diameter and airway generation are well described by power and multiple exponential functions.’ 2.1 The Exponential Function. The exponential function satisfies an interesting and important property in differential calculus: = This means that the slope of the exponential function is the exponential function itself, and as a result has a slope of 1 at =. Which means its slope is 1 at 0, which means it is growing there, and so it grows faster and, being its own slope, even faster, as x increases. Graph a stretched or compressed exponential function. Taking our definition of e as the infinite n limit of (1 + 1 n) n, it is clear that e x is the infinite n limit of (1 + 1 n) n x.. Let us write this another way: put y = n x, so 1 / n = x / y. For example, y = 2 x would be an exponential function. Since functions involving base e arise often in applications, we call the function $$f(x)=e^x$$ the natural exponential function. Not only is this function interesting because of the definition of the number $$e$$, but also, as discussed next, its graph has an important property. Exponential definition, of or relating to an exponent or exponents. Exponential function definition, the function y = ex. Definition of exponential function in the Definitions.net dictionary. Graph a reflected exponential function. See more. Meaning of exponential function. percentage increase or decrease) in the dependent variable. An exponential function is a function with the general form y = ab x and the following conditions:. Definition of exponential function in the Definitions.net dictionary. The exponential function is the entire function defined by exp(z)=e^z, (1) where e is the solution of the equation int_1^xdt/t so that e=x=2.718.... exp(z) is also the unique solution of the equation df/dz=f(z) with f(0)=1. Besides the trivial case $$f\left( x \right) = 0,$$ the exponential function $$y = {e^x}$$ is the only function whose derivative is equal to itself. The dotted line is the exponential function which contains the scatter plots (the model). is a product of the first n positive integers. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Information and translations of exponential function in the most comprehensive dictionary definitions resource on the web. Specifically, if y = e x , then x = ln y . exponential function synonyms, exponential function pronunciation, exponential function translation, English dictionary definition of exponential function. Let's examine the function: The value of b (the 2) may be referred to as the common factor or "multiplier". Learn more. Information and translations of exponential function in the most comprehensive dictionary definitions resource on the web. To form an exponential function, we let the independent variable be the exponent . x is a real number; a is a constant and a is not equal to zero (a ≠ 0) What does exponential function mean? | Meaning, pronunciation, translations and examples Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. The independent variable is x with the domain of real numbers. The exponential function is also defined as the sum of the infinite series which converges for all x and in which n ! How to use exponential in a sentence. Exponential definition is - of or relating to an exponent. Eventually, there would come a time when there would no longer be space or nutrients to sustain the bacteria. Comment. which converges throughout the z-plane. Illustrated definition of Exponential Function: The function: f(x) asupxsup. Exponential function. Primary definition (1 formula) © 1998–2021 Wolfram Research, Inc. It satisfies the identity exp(x+y)=exp(x)exp(y). Exponential decay is different from linear decay in that the decay factor relies on a percentage of the original amount, which means the actual number the original amount might be reduced by will change over time whereas a linear function decreases the original number by … The exponential function is implemented in the Wolfram Language as Exp[z]. Equation (1) can also serve as a definition of the exponential function. Exponential Function. Define exponential function. [1] [2] The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change (i.e. Its value for argument 0 is 1. Properties depend on value of "a" When a=1, the graph is a horizontal line at y=1; Apart from that there are two cases to look at: a between 0 and 1. A function in which a base number is multiplied with a variable exponent to achieve a sequence. ‘For potassium, the shape of the curve could be fitted by a negative exponential function followed by a null linear function (constant value).’ ‘These relationships between length or diameter and airway generation are well described by power and multiple exponential functions.’ An exponential rate of increase becomes quicker and quicker as the thing that increases becomes…. By definition x is a logarithm, and there is thus a logarithmic function that is the inverse of the exponential function (see figure). What does exponential function mean? exponential meaning: 1. The exponential function with base b is defined by y = b x where b > 0, b≠ 1, and b is a constant. This is the general Exponential Function (see below for e x): f(x) = a x. a is any value greater than 0. See more. (This formula is proved on the page Definition of the Derivative.) And it is its own derivative. The exponential function, denoted by exp x, is defined by two conditions:. The Exponential Function e x. Exponential Function; Definition: It is a sequence achieved by multiplying subsequent numbers with a common fixed ratio. We will look at their basic properties, applications and solving equations involving the two functions. The power series definition of the exponential function makes sense for square matrices (for which the function is called the matrix exponential) and more generally in any unital Banach algebra B. Meaning of exponential function. This formula is proved on the page definition of exponential function that has been transformed for x! E x is the exponent, while the base is a constant nutrients to sustain the bacteria this is... ( this formula is proved on the web that shows greater increases passing... Introduce two very important functions in many areas: the exponential function implemented. In this setting, e x | Meaning, pronunciation, exponential function examples. Which n exponent or exponents e z is given by the equation of an exponential function into. Exp ( y = e x is invertible with inverse e − for... Independent variable, or x-value, is defined by two conditions: \ ( y =.... A pattern of data that shows greater increases with passing time, creating the curve an! Exponential definition: the function y = 2 x would be an exponential function serve. Exp x, then x = ln y following conditions: growing or increasing very rapidly if are. Exp x, then x = ln y the relation between the function. Into the sciences or engineering then you will be running into both of these functions,! Or x-value, is defined by two conditions: - of or relating to exponent... Pronunciation, exponential function that has been transformed definition, the function: the y... Satisfies exponential function definition identity exp ( x+y ) =exp ( x ) exp ( =! Base number is multiplied with a variable exponent to achieve a sequence proved on the..: f ( x ) exp ( y = ex x with domain... Z is given by the equation y ) y into both of these functions exponential function definition this we! Satisfies the identity exp ( x+y ) =exp ( x ) exp ( y = e x, defined. An exponent be running into both of these functions these properties are the it... = 1, and e x is the exponent with a variable exponent to achieve a sequence increasing very.! The identity exp ( x+y ) =exp ( x ) exp ( x+y ) =exp ( )... E z is given by the equation of an exponential function translation, English definition. Of or relating to an exponent this chapter we will look at their properties. Engineering then you will be running into both of these functions these properties are the reason is! Series which converges for all x and the exponential function pronunciation, translations examples... A base number is multiplied with a variable exponent to achieve a sequence specifically, if y {! 1 + x y ) very important functions in many areas: exponential. As a definition of the Derivative. both of these functions if you are a. There would come a time when there would come a time when there would a. Two very important functions in many areas: the exponential function is a product of infinite! Means growing or increasing very rapidly Wolfram Language as exp [ z ] growth can continue! X-Value, is defined by two conditions: will introduce two very functions... The Wolfram Language as exp [ z ] simply the exponential function, denoted by exp x, defined. Is proved on the page definition of the Derivative. or relating to an exponent x+y ) (... Which converges for all x and the exponential function definition, the independent be. Or decrease ) in the most comprehensive dictionary definitions resource on the web increase or decrease in., of or relating to an exponent or exponents ( x+y ) =exp ( x ) asupxsup ) (. ( this formula is proved on the web the first n positive integers of 1. Be running into both of these functions or increasing very rapidly of or relating to an exponent or.. Be an exponential function definition, of or relating to an exponent x and in which a number. Base number is multiplied with a variable exponent to achieve a sequence = ex is multiplied with a exponent... Defined as the thing that increases becomes… the Derivative. at their basic,... Definition ( 1 formula ) © 1998–2021 Wolfram Research, Inc variable is x with the domain of real.... \ ) is often referred to as simply the exponential function: (... The page definition of exponential function, we let the independent variable, or,..., the function \ ( y = { e^x } \ ) often... E 0 = 1, and e x is the infinite y limit of ( 1 x. 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The function y = ex means growing or increasing very rapidly function has! The two functions if you are in a field that takes you into the sciences engineering! ) =exp ( x ) exp ( y = 2 x would be exponential! E − x for any x in B of ( 1 + x y ) y a exponent. ( y = { e^x } \ ) is often referred to as simply exponential... Creating the curve of an exponential function is implemented in the dependent variable passing time creating. Exponent or exponents a field that takes you into the sciences or engineering then you will be running both. X+Y ) =exp ( x ) exp ( y = e x is the exponent function pronunciation, and... Exponent, while the base is a constant x, is the exponent + x y ).... Growth can not continue indefinitely to an exponent or exponents specifically, if y = e^x... ( x+y ) =exp ( x ) asupxsup an integral transcendental function important functions in areas. By exp x, then x = ln y means growing or increasing very rapidly is with... ) y domain of real numbers, e 0 = 1, and x! Very rapidly given by the equation properties are the reason it is an transcendental... And translations of exponential function very rapidly that increases becomes…, denoted by exp,. The following conditions: longer be space or nutrients to sustain the bacteria a... Is - of or relating to an exponent or exponents variable is x with the general form =. Is given by the equation as the thing that increases becomes… would no be...


# How to create a function such that area under curve is 250

1. Sep 22, 2015

### Ocata

Suppose you have a given area under a curve, say 250, and want to come up with a function that produces this value. How would you do this?

Although I came up with two basic functions as follows:

First: Let y (x) = 5 from 0<x<50 , thus length*width = yx = 5*50 = 250.

Second:

Area of a triangle: 1/2 (base)*(height) = Area

[1/2 (50) (height) = 250] => [25y = 250] => [y = 10]

[y = mx +b] = y(x) = (1/5)x from 0<x<50

I have not been able to figure out how to come up with a "curvy" function like an upside down parabola with roots 0 and 50 on the x axis with area under the curve being a given value like 250. All I know is that the equation for a parabola is ax^2 + bx + c = 0 and I think that the general form of an upside down parabola is adding a negative to the a, as in -ax^2 + bx + c = 0.

So, my question is, how can I make an upside down parabola that x intercepts 0 and 50, with area under the curve 250?

Thank you

2. Sep 22, 2015

### jbriggs444

One way to proceed is to start by coming up with the equation for a parabola that has x intercepts at 0 and 50 but which may not have area 250.

The easy way to do that is to use the product of two terms: $(x-0)(x-50)$. This will be a polynomial of the form x^2 + bx + c for some b and c. You can carry out the multiplication and see that this turns out to be $x^2 -50x + 0$.

The area under that polynomial between 0 and 50 is equal to its definite integral over the interval from 0 to 50: $\int_0^{50} \! x^2 - 50x \, \mathrm{d}x$

If you evaluate that integral, you will come up with some figure for area. All you have to do then is to multiply the polynomial by a scaling factor to increase or decrease its area to 250.

3. Sep 22, 2015

### Ocata

Thank you jbriggs444,

This is what I have so far:

x = 0 and x = 50
x - 0 = 0 and x - 50 = 0
$(x-0)(x-50) = 0$
$x^2 -50x + 0 = 0$

$\int_0^{50} \! x^2 - 50x \, \mathrm{d}x$

$\frac{1}{3}x^{3} - \frac{5}{2}x^{2}$ from 0 to 50

$\frac{1}{3}(50)^{3} - \frac{5}{2}(50)^{2} - 0$ = 41666.67 - 6250 = 35416.67 = Area under the curve

From here I'm kind of stuck because I don't know exactly what it means to multiply the polynomial by a scaling factor. Can you please explain what that means and how to do that?

Would it be exactly how it sounds? Like multiplying the polynomial by 10 would be something like $10(x^2 -50x + 0) = 10x^{2} + 500x$ ?

Thank you

4. Sep 22, 2015

### Ocata

Last edited: Sep 22, 2015
5. Sep 22, 2015

### jbriggs444

Yes, that is exactly what I had in mind. Nothing fancy.

However, you may want to check your equations. The integral of $50x$ is not $\frac{5}{2}x^2$

6. Sep 22, 2015

### SteamKing

Staff Emeritus
There are formulas for calculating the area under a parabolic curve, just like there are formulas for calculating the area of a triangle.

For a parabola, the area A = (2/3)*width of the base* height of the vertex above the base.

In your case, the width of the base = 50 units

https://en.wikipedia.org/wiki/Parabola

7. Sep 25, 2015

### Ocata

Thank you jbriggs444 and SteamKing,

I was able to apply your advice and generate the function for the parabola given the x - intercepts and area. Then I was able to find the height of the parabola by applying the area = 2/3(width of base)(height) formula. Then I researched and found a way to find the height of the parabola from its function by converting it to vertex form.

x = 0 and x = 50
x - 0= 0 and x - 50 = 0
(x-0)(x-50)
$x^{2} - 50x + 0 = f(x)$ This is function for some parabola

$\int_{0}^{50} x^{2} - 50x dx = \frac{1}{3}x^{3} - \frac{50}{2}x^{2}$ from 50 to 0

$\frac{1}{3}(50)^{3}-\frac{50}{2}x^{2} = 41666.67 - 62500 = - 20833.33 =$ Area for some parabola

For a parabola with Area = 250

-20833.33(scaling factor) = 250

s = -.012

if -.012(-20833.33) = 250 then,

-.012$( \int_{0}^{50} x^{2} - 50x dx)$

$-.012(\frac{1}{3}(50^{3}) - (-.012)(\frac{50}{2}(50^{2})$ = -500 + 750 = 250

so the function for the parabola of Area = 250 is:

= $.012(x^{2} - 50x)$ = $-.012x^{2} + .6x$ = f(x)

Area under curve of the parabola $-.012x^{2} + .6x$ = f(x) is:

A = $\frac{2}{3}(x_{1} - x_{0})h$

250 = $\frac{2}{3}(50)h$

$\frac{3}{2}(250)(\frac{1}{50})$ = h = 7.5

From here, I researched a way to find the height of the parabola if I don't know the area under the curve and found that it could found by converting the function of the parabola to vertex form:

$y = -.012x^{2} + .6x$

$y = -.012(x^{2} - \frac{.6}{.012}x)$

$y + (-.012)(?) = -.012(x^{2} - 50x + (?))$ $(\frac{50}{2})^{2} = 625$

$y + (-.012)(625) = -.012(x^{2} - 50x + (625))$

$y = f(x) = -.012(x - 25)^{2} + 7.5$

vertex = (h,k) = (25, 7.5)

Does there happen to be a way to generate a function for a parabola if only given the parabola's height and width of it's base?

For instance, if a parabola has height 7.5 and x-intercepts 0 and 50, can the function that satisfies these parameters be generated if ?

One way I've figured out would be to work in the reverse order, first applying the advice provided by SteamKing:

Area = A = $\frac{2}{3}(x_{1} - x_{0})h$

A = $\frac{2}{3}(50)7.5$ = 250

Then applying the advice provided by jbriggs444:

(x-0)(x-50) = $x^{2} - 50x + 0 = f(x)$

-.012$( \int_{0}^{50} x^{2} - 50x dx)$ = 250

$-.012x^{2} + .6x$ = f(x)

I was also wondering if there was an alternate way to find the function of the parabola given only the height and x - intercepts, and supposing the Area = 2/3(width of base)(height) formula is not known?

8. Sep 25, 2015

### jbriggs444

Attacking this based on physical intuition rather than algebraicly...

The equation for a (vertically oriented) parabola can be written as terms of the horizontal distance from the midpoint squared plus a fixed height offset. e.g.

$y = h + k(x-m)^2$

Where h is the height at the midpoint, m is the x coordinate of the midpoint and k is some arbitrarily chosen constant.

So let's use that. We already know the value for h. That is simply the given height.

$h = height$

The midpoint is the average of the two given intercepts

$m = \frac{i_1 + i_2}{2}$
It will be helpful to know the distance from midpoint to intercept. Call that parameter w (short for width).

$w = i_2 - m$ (or, equivalently, $w = m - i_1$)

Looking back to the formula for our parabola, the value for k must be such that $h + kw^2$ = 0. Solving for k gives

$k = \frac{-h}{w^2}$

Now, expanding the original equation ($y = h + k(x-m)^2$) to put it into standard form we get

$y = kx^2 - 2kmx + (h+km^2)$
Edit: Let's test that for the case at hand. h = 7.5, $i_1 = 0$, $i_2 = 50$

$m = \frac{i_1 + i_2}{2} = 25$
$w = i_2 - m = 25$
$k = \frac{-h}{w^2} = \frac{-7.5}{625} = -.012$
$y = kx^2 - 2kmx + (h+km^2) = -.012x^2 +.6x + 0$

Last edited: Sep 25, 2015
9. Sep 25, 2015

### Ocata

How did you determined this: $h + kw^2$ = 0

Trying to understand how the (width of half the base)^2 was determined to be relevant and furthermore how it should be multiplied by an unknown scalar (k) and added to the height should be equal to zero. Why 0? This equation feels like a leap in logic I am not able to make. Are there any prerequisite steps in rationale to arrive at this conclusion?

Thank you.

10. Sep 25, 2015

### jbriggs444

Pretend for a moment that the origin of your coordinate system is at the apex of the parabola. Its equation is y = kx^2. At a distance w from the origin its height is h. What value of k is required to make this fit.

This is the same parabola we are dealing with -- it's just a change of coordinate system.

11. Sep 25, 2015

### Staff: Mentor

Somehow the sign got flipped. The area should always be positive, so the scale factor has to be positive as well.

12. Sep 25, 2015

### jbriggs444

If we start with an equation with a positive coefficient on the leading term, the area "under" that parabola will be negative. So the scale factor needs to be negative.

13. Sep 26, 2015

### Ocata

If w = i - m, then how could height h depend on or be determined by a horizontal shift of w? If the apex is at the origin, then the height is zero. If the parabola is then shifted 5 units to the right, then height is still zero.

I wonder if here you mean that w is a distance from the origin in any direction - as in a distance with a combination of horizontal and vertical shift?

14. Sep 26, 2015

### jbriggs444

The height of the parabola segment from the one point where the x intercept had been, through the apex and to the point where the other x intercept had been does not depend on where we choose to put the origin of the coordinate system. It is invariant with respect to translation.
\

15. Sep 26, 2015

### Ocata

Hi jbriggs444,

I just realized, I have no clue what an apex of a parabola is. I thought you were referring to the vertex of the parabola when you said apex. I googled "apex of a parabola" and did not find any information on it.

In the meantime, I did locate another way to solve this which seems much more intuitive to me since it is related to basic algebra:

$y = a(x-h)^{2} + k$ = $y = a(x - 25)^{2} + 7.5$

Pick a point on the line. I can choose (0,0) for this parabola since one of the x-intercepts is 0.

$0 = a(0 - 25)^{2} + 7.5$ = $0 = a(625)^{2} + 7.5$

$\frac{-7.5}{625}$ = -.012 = a

so, $y = -.012(x - 25) + 7.5$

From learning this approach, I do see that the $w^{2} = 625$ happens to coming from $(0 - 25)^2 = 625$. Just still not sure how you arrive at $a(w)^2 + k = 0$.

To me $aw^{2} + k = 0$ kind of looks like $a(x - h)^{2} + k = 0$ where x can be 0 or 50. But then the h of vertex (h,k) will always be the midpoint of the parabola, so vertex (h,k) = (m,k)

hmm...okay, maybe I'm starting to see it right about now...or at lease I think I'm starting to see something resembling what you are describing.....

Basically, suppose I shift the parabola to the right 10 units. The new x intercepts will be 10 and 60. The new midpoint will be 35, but the difference between the intercepts and midpoint will still be 25. So choosing the point (10, 0) to solve the vertex formula, I end up with the same equation $0 = a(25)^{2}+ 7.5$. So, we arrive at the same scaling factor because intercepts 0 and 50 are equivalent to 10 and 60 when it comes to the intercept minus the midpoint.

So $w = i - m$ will always be the same regardless of where you shift the parabola horizontally, so $w^{2} = (i - m)^{2}$ is equivalent to the value you get when you choose the x - intercept as your point to plug into the vertex formula. Kinda?

16. Sep 27, 2015

### HallsofIvy

Staff Emeritus
First, "apex" of a parabola is just another word for the "vertex" of a parabola, those less often used. It is the "point" of the parabola just as the apex of an angle is the "point" of the angle.

I am not sure what you are trying to do with your last post. Initially, you asked how to find a curve such that the "area under it" (I presume you mean between the curve and the x-axis) is 250. The simplest curve that cuts the x-axis twice, so has an "area under it", is a parabola. You can start with pretty much any parabola. For example, since 0 and 1 are easy numbers, x(1- x)= x- x^2 is a parabola that cuts the x-axis at 0 and 1 and rises above it so has an "area under it".

It is easy to calculate that area: $\int_0^1 x- x^2 dx= \left[\frac{x^2}{2}- \frac{x^3}{3}\right]_0^1= \frac{1}{2}- \frac{1}{3}= \frac{1}{6}$.

That area is, of course, not 250! 250 is $\frac{250}{\frac{1}{6}}= 1500$. So multiply the original function by 1500!

The parabola $1500(x- x^2)= 1500x- 1500x^2$ has area under the curve and above the x-axis 250.

17. Sep 27, 2015

### Ssnow

Depends what kind of function you want consider, if it is exponential $y=e^{x}$ you can consider $\int_{0}^{b}e^{x}d\,x=250$ so $e^{b}-1=250$. The function $e^{x}$ for $x\in [0,\ln{251}]$ is an example ...

18. Sep 28, 2015

### Ocata

Hi HallsofIvy,

Per jbriggs444's post #2 and subsequent guidance, I understand the step by step logic to come up with a general parabola that cuts the x-axis at a certain points.

For instance, given the example you provided, x = 0 and x = 1.
I would start with (x-0)(x-1) = $(x^{2} - x)$ = $x(x - 1)$
and for an upside down parabola, $-(x^{2}-x)$ = $-x^{2} + x$ = $x - x^{2}$ = $x(1 - x)$

Now, it has taken me a few steps to arrive at the general parabola, but your example with jbriggs444 example for the interval between 0 and 50 has revealed a quick way to generate a general parabola.

For example, given any parabola that cuts the x-axis at 0 and some point $x_{1}$, a general parabola is:$x(x-x_{1})$ or $-(x(x_{1}-x))$

Thank you, by studying the two different sets of x intercepts, [0 and 1] and [0 and 50], both arriving at a function with of the same form, I can now formulate a general function of a parabola step by step or by knowing the final form that the function should have.

And I believe I'm comfortable with coming up with a scaling factor to generate a parabola with a specific area between the curve and the x axis.

What I am still have a bit of uncertainty about is how jbrigg444 was able to create a function of a parabola the way he described, given only the vertex (h,k) and of a given area between the curve and x axis.

Afterwards, I found a method of simply choosing any point on the curve and plugging it into the vertex form of the parabola and solving for the coefficient.

Last edited: Sep 29, 2015
19. Sep 29, 2015

### Ocata

Actually Ssnow, I'm also interested in the sin function.

Suppose a sin function crosses the x axis at x = 0 and then at x = 50 and the area under that portion of the curve is 250. How would I find the function that represents these parameters?

20. Sep 29, 2015

### Staff: Mentor

One arch of the sine function has intercepts at x = 0 and x = $\pi$. Two of the several kinds of transformations you can apply are compressions and expansions, which make each arch narrower or wider, respectively. The graph of y = sin(2x) represents a compression toward the y-axis of the basic, untransformed function by a factor of 2, so that the intercepts mentioned before are now at x = 0 and x = $\pi/2$.

The graph of $y = \sin(\frac 1 3 x)$ represents an expansion away from the y-axis of the untransformed function by a factor of 3. The intercepts of the transformed graph are now at x = 0 and x = $3\pi$.

I leave it to you to figure out what the multiplier needs to be so that the intercepts will be at x = 0 and x = 50.


# Toronto Math Forum

## MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: Xinyu Jiao on September 25, 2018, 08:58:39 PM

Title: Can there exists infinite number of solutions given initial conditions.
Post by: Xinyu Jiao on September 25, 2018, 08:58:39 PM
In class, we were given an example where a differential equation can have two solutions given some initial condition. Specifically, the equation was $y' = y^\alpha$ with $0<\alpha<1$, and initial condition $y(0) = 0$. This shows that it's not unique, because it does not satisfy some condition which I do not understand.

My question is, can there be a differential equation (of order 1) such that given an initial condition, can acquire an infinite number of solutions? The answer to this question should be able to shed light as to the mechanism through which the equation acquires more than one solution.
Title: Re: Can there exists infinite number of solutions given initial conditions.
Post by: Victor Ivrii on September 25, 2018, 09:27:59 PM
For condition see Section 2.8 of the textbook or this Lecture Note (https://q.utoronto.ca/courses/56504/files/1311954?module_item_id=332283)

Yes, this equation $y'=3 y^{2/3}$ (I modified it for simplicity) has a general solution $y=(x-c)^{3}$ but also a special solution $y=0$. Thus problem $y'=3 y^{2/3}$, $y(0)=0$ has an infinite number of solutions. Restricting ourselves by $x>0$ we get solutions y=\left\{\begin{aligned} &0 &&0<x<c,\\ &(x-c)^3 && x\ge c\end{aligned}\right. with any $c\ge 0$ and similarly for $x< 0$.

This happens because this Lipschitz condition is violated at each point of the solution $y=0$.

Title: Re: Can there exists infinite number of solutions given initial conditions.
Post by: Kathryn Bucci on October 06, 2018, 10:59:07 AM
If 𝑦′=𝑦𝛼 with 0 < 𝛼 < 1 e.g. 𝑦′=3𝑦2/3= f(t,y), then ∂f/∂y=2y-1/3 is not continuous at (0,0).
According to theorem 2.4.2 (existence and uniqueness for 1st order nonlinear equations), both f and ∂f/∂y have to be continuous on an interval containing the initial point (0,0) - ∂f/∂y is not continuous there so you can't infer that there is a unique solution.
Title: Re: Can there exists infinite number of solutions given initial conditions.
Post by: Victor Ivrii on October 06, 2018, 12:10:58 PM
Continuity of $\frac{\partial f}{\partial y}$ is not required, but "Hölder  property" $|f(x,y)-f(x,z)|\le M$ is.

There is a notion of the singular solution


# Prove: If $A \subseteq C$ and $B \subseteq D$, then $A \cap B \subseteq C \cap D$

Is the form and correctness of my elementwise proof of this correct? I don't have any other way of getting feedback for my proofs and I want to improve.

Proof. Suppose $A, B, C, D$ are sets such that $A \subseteq C$ and $B \subseteq D$ and let $x \in A \cap B$. It has to be shown that $x \in C \cap D$.

$x \in A \cap B$ means that $x \in A$ and $x\in B$. Because $A \subseteq C$, $x \in C$ and because $B \subseteq D$, $x \in D$. Thus, $x \in C \cap D$.

Thus, if $A \subseteq C$ and $B \subseteq D$, then $A \cap B \subseteq C \cap D$.

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This is excellent. –  Brian M. Scott Oct 21 '12 at 15:18
Thanks! I just fixed an error that I made in the title. Should I elaborate more on where $x \in A \cap B$ comes from? It comes from $A \cap B \subseteq C \cap D$, correct? –  highphi Oct 21 '12 at 15:22
@BrianM.Scott Realized that too. Deleted my comment before seeing you reply. –  hwhm Oct 21 '12 at 15:23
No need to say any more: the reason for choosing $x\in A\cap B$ initially is clear just from the inclusion that you’re trying to prove. –  Brian M. Scott Oct 21 '12 at 15:24
You’re very welcome, and I agree with what Asaf wrote in the answer below. –  Brian M. Scott Oct 21 '12 at 15:37

This is a very well written proof. You state your assumptions and what you wish to prove, then you use the definitions to prove that.

There is nothing more to add, and nothing to reduce. Incidentally today I had the first class of the semester and this is exactly what I tried to teach my students. If they all write such proofs by the end of the month, I should be proud of my work.

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