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Surface Area (Using Formulas) Math Videos for Kids - Grades 6-8
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WHAT ARE FORMULAS FOR SURFACE AREA?

You have already learned how to find surface area by unfolding 3D shapes into 2D shapes and adding up the areas of all their parts. In this lesson, you learn how to use variables to label the sides in a shape and find surface area more efficiently by using formulas.

To better understand surface area (using formulas)…

WHAT ARE FORMULAS FOR SURFACE AREA?. You have already learned how to find surface area by unfolding 3D shapes into 2D shapes and adding up the areas of all their parts. In this lesson, you learn how to use variables to label the sides in a shape and find surface area more efficiently by using formulas. To better understand surface area (using formulas)…

## LET’S BREAK IT DOWN!

### Surface Area of a Cube

A cube is a rectangular prism that has 6 identical faces, and each face is a square. Since all sides have the same length, and all the faces have the same area, we can find the area of one face and multiply by 6 to find the surface area! If we let a be the length of the sides, then we can use the formula 6a2 to find the surface area of any cube. Now you try: What is the area of a cube with side length 4 cm?

Surface Area of a Cube A cube is a rectangular prism that has 6 identical faces, and each face is a square. Since all sides have the same length, and all the faces have the same area, we can find the area of one face and multiply by 6 to find the surface area! If we let a be the length of the sides, then we can use the formula 6a2 to find the surface area of any cube. Now you try: What is the area of a cube with side length 4 cm?

### Surface Area of a Rectangular Prism

A rectangular prism is made up of 6 faces. In a rectangular prism, we have a length, a width, and a height, and we will call them l,w, and h. Since opposite faces have the same dimensions, we only need to find the areas of the 3 different faces, multiply each area by 2, and add the totals together to find the surface area. Then the surface area of a rectangular prism is 2l×w+2w×h+2(l×h). Now you try: Find the surface area of a rectangular prism that has length 6 m, width 3 m, and height 2 m.

Surface Area of a Rectangular Prism A rectangular prism is made up of 6 faces. In a rectangular prism, we have a length, a width, and a height, and we will call them l,w, and h. Since opposite faces have the same dimensions, we only need to find the areas of the 3 different faces, multiply each area by 2, and add the totals together to find the surface area. Then the surface area of a rectangular prism is 2l×w+2w×h+2(l×h). Now you try: Find the surface area of a rectangular prism that has length 6 m, width 3 m, and height 2 m.

### Surface Area of a Triangular Prism

A triangular prism has a total of 5 faces: 2 triangular faces that are opposite each other and are the same, and 3 rectangular faces. We can find the area of one of the triangular faces, multiply it by 2, and then add the area of each of the rectangular faces. Then the formula for the surface area of a triangular prism is 2 x [ggfrac]b×h/2[/ggfrac] 2×b × h2+l×a+l×b+l×c which we can also simplify to b×h+l×a+l×b+l×c. It is important to remember that the width might be different for each of the 3 rectangles. Now you try: Find the surface area of a triangular prism where the triangular faces are triangles with side lengths 3 in., 4 in., and 5 in. The triangle base is 3 in. and its height is 4 in., and the rectangular faces have length 2 in.

Surface Area of a Triangular Prism A triangular prism has a total of 5 faces: 2 triangular faces that are opposite each other and are the same, and 3 rectangular faces. We can find the area of one of the triangular faces, multiply it by 2, and then add the area of each of the rectangular faces. Then the formula for the surface area of a triangular prism is 2 x [ggfrac]b×h/2[/ggfrac] 2×b × h2+l×a+l×b+l×c which we can also simplify to b×h+l×a+l×b+l×c. It is important to remember that the width might be different for each of the 3 rectangles. Now you try: Find the surface area of a triangular prism where the triangular faces are triangles with side lengths 3 in., 4 in., and 5 in. The triangle base is 3 in. and its height is 4 in., and the rectangular faces have length 2 in.

### Surface Area of a Cylinder

A cylinder is like a tube with a circular cap on each end. If we unravel it into a 2D shape, we get one long rectangle and two identical circles. Notice that as we unravel the cylinder, the circumference of the circle turns out to be the length of the rectangle! The other dimension of the rectangle is the height of the cylinder. Then the surface area of a cylinder is 2 times the area of a circle, plus the area of the rectangle. Now you try: Find the surface area of a cylinder that has radius 3 mm and height 6 mm. Use 3.14 for pi and round to the nearest tenth.

Surface Area of a Cylinder A cylinder is like a tube with a circular cap on each end. If we unravel it into a 2D shape, we get one long rectangle and two identical circles. Notice that as we unravel the cylinder, the circumference of the circle turns out to be the length of the rectangle! The other dimension of the rectangle is the height of the cylinder. Then the surface area of a cylinder is 2 times the area of a circle, plus the area of the rectangle. Now you try: Find the surface area of a cylinder that has radius 3 mm and height 6 mm. Use 3.14 for pi and round to the nearest tenth.

### Surface Area of a Cone

Cones are shaped like ice cream cones with a circular lid that covers the opening at the top. An unrolled cone looks like a part of a circle, and the lid is a circle. The unrolled cone can be decomposed into a rectangle that has side lengths πr and s, where s is the slant height of the cone. To find the surface area, we can find the area of the rectangle and add it to the area of the circular lid. Then the formula for surface area of a cone is πr2 + πrs. Now you try: What is the surface area of a cone with radius 4 feet and slant height 7 feet?

Surface Area of a Cone Cones are shaped like ice cream cones with a circular lid that covers the opening at the top. An unrolled cone looks like a part of a circle, and the lid is a circle. The unrolled cone can be decomposed into a rectangle that has side lengths πr and s, where s is the slant height of the cone. To find the surface area, we can find the area of the rectangle and add it to the area of the circular lid. Then the formula for surface area of a cone is πr2 + πrs. Now you try: What is the surface area of a cone with radius 4 feet and slant height 7 feet?

## SURFACE AREA (USING FORMULAS) VOCABULARY

Rectangular Prism
A shape made of 6 rectangular faces. The opposite faces always have the same dimensions.
Cube
A rectangular prism whose sides all have the same length so that all 6 faces are squares.
Face
Each flat part of a shape that we can calculate an area for.
Triangular Prism
A shape that has 2 parallel triangular faces connected by 3 rectangular faces.
Cylinder
A shape that looks like a tube with a circular lid on each end.
Cone
A shape that has a circular base and a slanted edge that extends from around the base to meet at a point.
In a 3D cone or pyramid, the slant height is the length from the edge of base to the point at the top. It is not the same as the actual height of the object.


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### Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

### Nim

Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.

### Binary Squares

If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?

##### Stage: 3, 4 and 5

We are accustomed to writing numbers in base ten, using the symbols for 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. For example, 75 means 7 tens and five units. However numbers can be written in any number base.

If we use base 8 instead of base ten, then 75 is written as 113 which denotes one sixty-four (82 ), one eight (81 ) and 3 units (instead of hundreds, tens and units).

Base 2 is particularly useful as it only requires two symbols, for zero and one, and it is the way numbers are represented in computers.

Just as, in base ten, the columns represent powers of 10 and have 'place value' 1, 10, 102 , 103 etc. (reading from right to left), so in base 2, the columns represent powers of 2. Hence the number 1001011 denotes (reading from right to left):

1 unit (20 ), 1 two (21 ), no fours (22 ), 1 eight (23 ), no sixteens (24 ), no thirty-twos (25 ), 1 sixty- four (26 ).

The number 1001011 in base 2 is the same as the number 75 in base ten.

As another example, we use the symbols 0, 1, 2, 3 and 4 to represent numbers in base 5. The columns in base 5 have 'place value' 1, 5, 25, 125, 625 etc reading from right to left. The number 75 in base ten is the same as the number 300 in base five, that is 3 twenty-fives, no fives and no units.

The number 4102 in base 5 denotes 2 units, no fives, 1 twenty-five and 4 one hundred and twenty-fives making altogether 527 in base ten.

Writing the number 75 in base six we get 203, which represents 2 thirty-sixes, no sixes and 3 units.

We have seen that 75 (base10), 1001011 (base 2), 300 (base 5), 113 (base 8), and 203 (base 6) all represent the same number.

Similarly, we can write 75 in any base we choose and we can write all numbers in any base.

To write numbers between 0 and 1, we use negative powers of the base. For example, in base 2 we use halves, quarters, eighths, sixteenths etc instead of the tenths, hundredths, thousandths etc. which we use in base ten.

So if we write 11.11 in base 2 this denotes $2^1 + 2^0 + 2^{-1} + 2^{-2}$. The equivalent in base 10 is $2 + 1 + {1\over 2} + {1\over 4}$, that is 3.75 in base 10.

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# The Programming Language Quiz, Mark II - Robbers

Cops challenge

## The Robbers' Challenge

1. Find a vulnerable answer. That is, an answer which hasn't been cracked yet and which isn't safe yet either.

2. Crack it by figuring out its language. That is, find any language in which the given program is a valid program (subject to the rules outlined in the Cops' Challenge above). It doesn't matter if this is the language the cop intended.

Once you have found a working language, post an answer to the Robbers' thread and comment below the Cop's answer. If you don't have enough reputation to comment, say this in your answer and someone will comment for you.

The output format must be the same as the cop's intended solution in order to constitute a valid crack.

Every user only gets one guess per answer. This includes telling other people the answer, or hints towards your guesses. You must not crack your own answer (obviously...).

The user who cracked the largest number of answers wins the robbers' challenge. Ties are broken by the sum of bytes of cracked answers (more is better).

Good luck Robbers!

• Important: Please be nice to TIO and don't just (write a script or manually) test all of the languages there. TIO is very busy already. Feb 7 '18 at 11:37
• I feel like the whole Cops/Robbers thing is backwards. Shouldn't the Robbers be hiding their programs and Cops be trying to bust them?
– Jo King
Feb 8 '18 at 6:29
• @JoKing ... just PPCG terminology. There is no reason to be incompatible, and there is absolutely no reason to change all the past cops-and-robbers ---questions--- challenges. Feb 8 '18 at 10:19

## Whitespace, posted by Jo King

def v(n=[]): #[

#!"⠖⠔⠄⠑⠃⡆⠊⡬⠀⠞⠈
#;;;
print(n*chr(33))

for		n in range(100):
#
#"
#"<<;?)*aanlll>1#<-#.:_:*aa@#.#!9fo"
v(n)
>d$/")!;\ xXxxxXXXxXXxX>({.< xx%c++=t=+~\ D.+L0~-tt/ ..R;MU ]+@x+++++++++[->++++++++++<]> [->+[->+>.<<]> [-<+>] <.< ] ~-<:<<<<:<||~[:?~-]|  Output in unary spaces separated by a single tab. Try it online! • Well... that was a lot faster than I has hoping for :( – Jo King Feb 14 '18 at 13:11 # oOo CODE, posted by MickyT If UR lEEt NUfF 2 W0rK 0Ut W0T 'd4 L4NgUAgE Is It Will 5h0W y4 ZEr0 T0 NiN37Y nIN3 pLUs 0NE, bU7 Us A AsCIi Ch4rACt0r 4 tH3 Numb3r Rath3R ThAN Th3 numB3r itS3Lf  Try it online! # Cascade, posted by emanresUA v>@/< 1%_\| /]8\< d+^}| _/{-/ ^*d|1 9#h\/ k-Q.| e3d+/ 4!5|5  Try it online! My first instincts were right lol. # K (oK), 10 bytes, cracks emanresu A's answer {1+x}'!"d"  Try it online! # dc, posted by moonheart08 0[1+pd100>a]salax  Try it online! # Deadfish~, posted by Mr. Xcoder {{iow}}  Try it online! # Java, posted by wvdz \u0070\u0075\u0062\u006c\u0069\u0063\u0020\u0063\u006c\u0061\u0073\u0073\u0020\u0058\u0020\u007b\u0070\u0075\u0062\u006c\u0069\u0063\u0020\u0073\u0074\u0061\u0074\u0069\u0063\u0020\u0076\u006f\u0069\u0064\u0020\u006d\u0061\u0069\u006e\u0028\u0053\u0074\u0072\u0069\u006e\u0067\u005b\u005d\u0061\u0029\u007b\u0066\u006f\u0072\u0028\u0069\u006e\u0074\u0020\u0069\u003d\u0031\u003b\u0069\u003c\u003d\u0031\u0030\u0030\u003b\u0069\u002b\u002b\u0029\u007b\u0053\u0079\u0073\u0074\u0065\u006d\u002e\u006f\u0075\u0074\u002e\u0070\u0072\u0069\u006e\u0074\u006c\u006e\u0028\u0069\u0029\u003b\u007d\u007d\u007d  • dammit, was just working on writing this same answer Feb 6 '18 at 23:06 # Labyrinth, posted by Martin Ender [##!] [\X_] @-99@  Try it online! # foo, posted by DLosc @10+10*+10(10@1>+$i<$c10-1)+@0  Try it online! # ELVM-IR, posted by MD XF  .text main: mov D, SP add D, -1 store BP, D mov SP, D mov BP, SP .file 1 "-" .loc 1 2 0 mov A, 1 mov B, BP add B, 2 store A, B .L0: mov B, BP add B, 2 load A, B mov D, SP add D, -1 store A, D mov SP, D mov A, 101 mov B, A load A, SP add SP, 1 lt A, B jeq .L3, A, 0 jmp .L4 .L3: jmp .L2 .L4: mov B, BP add B, 2 load A, B mov D, SP add D, -1 store A, D mov SP, D putc A add SP, 1 .L1: mov B, BP add B, 2 load A, B mov D, SP add D, -1 store A, D mov SP, D add A, 1 mov B, BP add B, 2 store A, B load A, SP add SP, 1 jmp .L0 .L2: exit  Try it online! # foo, posted by NoOneIsHere (100+1$c)

Try it online!

# Cubically, posted by MD XF

Try it online!

## Forth (gforth), posted by jmarkmurphy

: | 0 do i 1 + 0 do 42 emit loop cr loop ; 100 |

Try it online!

("v"|% le*)
..(""+,1e2)

Try it online!

• Less than a minute to post the crack!! Feb 7 '18 at 20:49

# Perl 5, posted by msh210

$,=$
";*a&*z;*aa&*az;*ba&*bz;*ca&*cv;s;;
$.++;e##cv] ;;; print++$
_..          $=* 5/3 ;;; 6/4 ;;; 7/5 ;;; 8/6 ==*cv  Try it online! # Triangular, posted by MickyT ,i#%)A;A^A@&NN*L!o|_\Om;;-?!>/  Try it online! Hah, should've recognized one of my languages at first sight. Formatted, it looks like this:  , i # % )  A ; A ^ A @ & N N * L ! o | _ \ O m ; ; - ? ! > / ÿ ÿ ÿ ÿ ÿ  I was trying to write a Triangular program for the cops' thread but l (less than) was broken, making it rather hard to end when the number hits 100. # Python 3, posted by MEE #import<stdio> 1//1;f=__import__; 2//1;String=str 0//1;i = 1; while(i<101):#{ f("sys").stdout.write(String(i)+" "); i+=1; #}  Try it online! # MATL, posted by Stewie Griffin Flatten[{1,100}] (* From x: 1 To 100: huhu% *)  Try it online! # 05AB1E, posted by Magic Octopus Urn TnL,q,a,b,c,d,I,t=input();v=c;u=b*c*TnL exec"if v>=30:v=c;u+=d\nv=v*v/25+6*v+140-u+I;u+=a*(b*v-u);print v\n"*t  Try it online! # MY, posted by Zacharý AȦi←  Try it online! This is the representation in MY's encoding. • This should've been easy, I created the language after all :p Feb 8 '18 at 22:00 # pb, posted by Erik the Outgolfer w[T!100]{t[T+1]b[T]>}  There's a Python interpreter here. # Brain-Flak, 664 bytes, posted by Christopher for(each num in #1/#100) ( do #pointOfNum(num) Push (#pointer@#num) if(result(pointer@(num/pointOfNum(run2)+#diff))) while(each num of pointer@num) do (ITS GO TIME!) DO THE MATH ) def fun add{ while add #(point@100) }) def { do@#point() }) open{ref}add(42)) do {SAY IT TO THE NAND}) { push why1998@point }) if(output!ready (push@#out {buy}) onGler<@spoin> pushet) <Divmod@point@uf2X3> discov def n{ <div@23> point@(4524#f34) (@#do(@3h) {point*723 numdif} )#huCX5DBP^h~0_GG1<h32X542P[18F18h42X%AAP[h!.X%OOSM a@<euws [give@point (for@point13)]) div<pointmod> divide((42/9 { over@point27 }) [pointto(27)] giv@12)} div<mod> def run{ pointOfnum fun@#(23) why1998 }  Try it online! # Pari/GP, posted by MD XF #for (o=1,100,print(o))  Try it online! # Implicit, posted by MD XF int f(int x_1) { int i, foóßar = 1; int *arr = 0x23d3f44e >> 0xa; for (i = ']'; 100 % foóßar++, *(arr+i/1234) += ']'; 0-101) putchar(&arr[i]); return -8; } int main(void) { f(42); }  Try it online! # Prelude, posted by Erik the Outgolfer 91+ v+ vv++ (1-v!) ^ ^^+ 0 1+  Try it online! # ><>, posted by Wurlitzer 0>1+:aa*1+=?;:\ ^ o"|"n/  Try it online! # GolfScript, posted by Lynn 2.upto(100){|i| p i}; printf RUBY_VERSION,$$> 0; p$_

Try it online!

# Brain-Flak (BrainHack), posted by Mr. Xcoder

#begin(
#import(Integer)
#import(String)
#import(STDOUT)
#define(Name:Main)(In:None)(Out:String,Int)(Mixed:TRUE)(Flags:None)(Requires:"String.format")(Coding:"UTF-8"))
#begin Main(
Integer.ranges.from$100.forEach{STDOUT.write(String.format{$0 }.with*$0=Self)(dropWhere{!$0}[ignoreIf:(0)])}
Except{OutOfBounds}
)
STDOUT.stream.read.forEach{with*$0=Self(Integer.ranges.from$100.filter(by:{1...101~=$0})[ignoreIf:(0)])}(if!Error){exit$0;}

Try it online!

There was a lot of balanced brackets and such made me suspicious. In Brain-Flak it errors but when using BrainHack it works.

• Dammit, I tried regular BF but I didn't realize they had prominent differences. :| Feb 9 '18 at 20:55
• @totallyhuman just the comments Feb 9 '18 at 23:01

# Triangular, posted by MD XF

1%P
*>\
A+.A
@~~.@A)][(An!!?_
*/-S+1%U<

Try it online!

# JavaScript, posted by iPhoenix

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text:
# norm

Quaternion norm

## Description

example

N = norm(quat) returns the norm of the quaternion, quat.

Given a quaternion of the form $Q=a+b\text{i}+c\text{j}+d\text{k}$, the norm of the quaternion is defined as $\text{norm}\left(Q\right)=\sqrt{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}}$.

## Examples

collapse all

Create a scalar quaternion and calculate its norm.

quat = quaternion(1,2,3,4);
norm(quat)
ans = 5.4772

The quaternion norm is defined as the square root of the sum of the quaternion parts squared. Calculate the quaternion norm explicitly to verify the result of the norm function.

[a,b,c,d] = parts(quat);
sqrt(a^2+b^2+c^2+d^2)
ans = 5.4772

## Input Arguments

collapse all

Quaternion for which to calculate the norm, specified as a scalar, vector, matrix, or multidimensional array of quaternions.

Data Types: quaternion

## Output Arguments

collapse all

Quaternion norm. If the input quat is an array, the output is returned as an array the same size as quat. Elements of the array are real numbers with the same data type as the underlying data type of the quaternion, quat.

Data Types: single | double

## Extended Capabilities

### Topics

Introduced in R2018b


text:
# Which set of ordered pairs could be generated by an exponential function? A. (-1,-1/2), (0, 0),(1,1/2), (2, 1) B. (–1, –1), (0, 0), (1, 1), (2, 8) C. (-1,1/2), (0, 1), (1, 2), (2, 4) D. (–1, 1), (0, 0), (1, 1), (2, 4)

Which set of ordered pairs could be generated by an exponential function?
A. $\left(-1,-\frac{1}{2}\right)$, (0, 0),$\left(1,\frac{1}{2}\right)$, (2, 1)
B. (–1, –1), (0, 0), (1, 1), (2, 8)
C. $\left(-1,\frac{1}{2}\right)$, (0, 1), (1, 2), (2, 4)
D. (–1, 1), (0, 0), (1, 1), (2, 4)
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An exponential function has the property that the y-values are multiplied by some common factor a when x incresses by 1.
We note that ench set of ordered pairs contains -1, 0, 1, and 2 as x-values, which are increments of 1.
Let us next determine the ratio of each pair of consecutive y-values for each option.
$\frac{0}{-\frac{1}{2}}=0$
$\left(\frac{1}{2}\right)0=$ Does not: exist
$\frac{1}{\frac{1}{2}}=2$
$\frac{0}{-1}=0$
$\frac{1}{0}$= Does not exist
$\frac{8}{1}=8$
$\frac{1}{\frac{1}{2}}=2$
$\frac{2}{1}=2$
$\frac{4}{2}=2$
$\frac{0}{1}=0$
$\frac{1}{0}$=Does no exist
$\frac{4}{1}=4$
We note that the three ratios are only constant for answer option C and thus answer option C could be generated by an exponential function.
C. $\left(-1,\frac{1}{2}\right)$,(0,1),(1,2),(2,4)


text:
• anonymous
Is the following quotation set up properly? Why or why not? Dr. Leonard McCredie stated, "'Literature's purpose within our society is an important one'" (32). No, the citation is incorrectly done. No, the double quotation marks are unnecessary. Yes, we can understand the quotation. Yes, literature is important to society.
English

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text:
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15

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text:
1. ## proofs

given that S=( 0,1), show that the max S doesnt exist.

to show that, my notes states that:

let M ∈ S and M= max S.

take M* = (M+1)/2

then we need to show that M* ∈ S and that M < M*...

mayb i know if there is a fixed method that i can adopt whener i have to solve proving questions?

2. Originally Posted by alexandrabel90
given that S=( 0,1), show that the max S doesnt exist.
to show that, my notes states that:
let M ∈ S and M= max S.
take M* = (M+1)/2
then we need to show that M* ∈ S and that M < M*...
This is always true.
If $\displaystyle a<b$ then $\displaystyle a<\frac{a+b}{2}<b$.

3. Well suppose there was an $\displaystyle x \in S$ with $\displaystyle y<x$ for all $\displaystyle y \in S$. Clearly we would need to have $\displaystyle x<1$, because all elements of $\displaystyle S$ are $\displaystyle <1$. Thus the open interval $\displaystyle (x,1)$ is not empty, and all of its elements are in $\displaystyle S$ and all of them are greater than $\displaystyle x$ - this clearly contradicts the definition of $\displaystyle x$.


text:
ASYMPTOTIC EQUIVALENCE FOR LINEAR DIFFERENTIAL SYSTEMS

Title & Authors
ASYMPTOTIC EQUIVALENCE FOR LINEAR DIFFERENTIAL SYSTEMS
Choi, Sung-Kyu; Koo, Nam-Jip; Lee, Keon-Hee;

Abstract
We investigate the asymptotic equivalence for linear differential systems by means of the notions of $\small{t_{\infty}}$-similarity and strong stability.
Keywords
linear differential system;adjoint system;strong stability;$\small{t_{\infty}}$-similarity;linear asymptotic equilibrium;asymptotic equivalence;
Language
English
Cited by
1.
ASYMPTOTIC EQUIVALENCE BETWEEN TWO LINEAR DYNAMIC SYSTEMS ON TIME SCALES,;;

대한수학회보, 2014. vol.51. 4, pp.1075-1085
1.
ASYMPTOTIC EQUIVALENCE BETWEEN TWO LINEAR DYNAMIC SYSTEMS ON TIME SCALES, Bulletin of the Korean Mathematical Society, 2014, 51, 4, 1075
References
1.
G. Ascoli, Osservazioni sopra alcune questioni di stabilita. I, Atti Accad. Naz. Lincei. Rend. Cl. Sci. Fis. Mat. Nat. (8) 9 (1950), 129-134.

2.
L. Cesari, Asymptotic Behavior and Stability Problems in Ordinary Differential Equations, Academic Press Inc., Publishers, New York; Springer-Verlag, Berlin-Gottingen-Heidelberg, 1963.

3.
S. K. Choi and N. J. Koo, Variationally stable difference systems by $n_{\infty}-similarity$, J. Math. Anal. Appl. 249 (2000), no. 2, 553-568.

4.
S. K. Choi and N. J. Koo, Asymptotic equivalence between two linear Volterra difference systems, Comput. Math. Appl. 47 (2004), no. 2-3, 461-471.

5.
S. K. Choi, N. J. Koo, and S. Dontha, Asymptotic property in variation for nonlinear differential systems, Appl. Math. Lett. 18 (2005), no. 1, 117-126.

6.
S. K. Choi, N. J. Koo, and Y. H. Goo, Asymptotic property of nonlinear Volterra difference systems, Nonlinear Anal. 51 (2002), no. 2, Ser. A: Theory Methods, 321-337.

7.
S. K. Choi, N. J. Koo, and D. M. Im, Asymptotic equivalence between linear differential systems, Bull. Korean Math. Soc. 42 (2005), no. 4, 691-701.

8.
S. K. Choi, N. J. Koo, and H. S. Ryu, h-stability of differential systems via $t_{\infty}-similarity$, Bull. Korean Math. Soc. 34 (1997), no. 3, 371-383.

9.
R. Conti, Sulla t-similitudine tra matrici e la stabilita dei sistemi differenziali lineari, Atti Accad. Naz. Lincei. Rend. Cl. Sci. Fis. Mat. Nat. (8) 19 (1955), 247-250.

10.
R. Conti, Linear differential equations asymptotically equivalent to x = 0, Riv. Mat. Univ. Parma (4) 5 (1979), part 2, 847-853.

11.
W. A. Coppel, Stability and Asymptotic Behavior of Differential Equations, D. C. Heath and Company, Boston, 1965.

12.
G. A. Hewer, Stability properties of the equation of first variation by $t_{\infty}-similarity$, J. Math. Anal. Appl. 41 (1973), 336-344.

13.
L. A. Lusternik and W. S. Sobolev, Functional Analysis, Science, Moscov, 1965.

14.
G. Sansone and R. Conti, Non-linear Differential Equatioins, The Macmillan Company, New York, 1964.

15.
W. F. Trench, On $t_{\infty}$ quasisimilarity of linear systems, Ann. Mat. Pura Appl. (4) 142 (1985), 293-302.

16.
W. F. Trench, Linear asymptotic equilibrium and uniform, exponential, and strict stability of linear difference systems, Advances in difference equations, II. Comput. Math. Appl. 36 (1998), no. 10-12, 261-267.


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Counting sheep using stones

Primitive people used to count their sheep using stones, such that each stone would represent a sheep.

So for example, if we have 10 sheep, we can represent this by 10 stones, i.e.:

This might be simple enough, but usually, for even a fairly small flock of sheep, there may be hundreds. Using this method makes it almost impossible to represent, then!

Because of the clumsiness of this system, some improvements were made to the system.

Egyptian number system

Egyptians used certain symbols to represent 1, 10, 100, 1000 (and also beyond).

These include:

• representing the number “1”
• representing the number “10”
• representing the number “100”
• representing the number “1000”

So for example, the number  is equal to $1000+1000+100+100+100+10+1=2311$

Notice the order in which the Egyptian numbers are written make no difference.

That is, we can rewrite the number above (2311) as , back-to-forth, which would still be the same number! In fact, you can write the symbols in any order and it would make no difference! This is unlike “2311” which will be different if written “1132”.

The Egyptian number system however, was still relatively clumsy, so was introduced the Roman system.

Roman number system

The Roman system, rather than using strange symbols, uses letters from the English alphabet to represent various numbers, including:

• I representing the number “1”
• V representing the number “5”
• X representing the number “10”
• L representing the number “50”
• C representing the number “100”
• D representing the number “500”
• M representing the number “100”

Note that starting with the Roman number system, order is now important.

In particular, if a smaller symbol precedes a larger one, you need to subtract it from the larger one.

For example, instead of writing the letter “4” as IIII (4 I’s), we write it as IV, which means $5-1=4$

Applying what we know, the number 19 would be broken down into its units, 10+9. 10 is X, and 9 is IX, or together, 19 is XIX.

Another example, 149, which is made up of 100+40+9. Now we know 100 is C. 40 is 50-10, which is XL. Also, 9 is 10-1, or IX. Putting these together, is CXLIX.

To solve Roman numerals, you must look at each unit/tens/hundreds/thousands separately.

Going backwards, if we are given CDIX, we know C is 100, D is 500, I is 1 and X is 10. Note C is less than D, and I is less than X, meaning we must the first listed from the second listed. So CD together is 500-100=400, and IX together is 10-1=9. Together, these are $400+9=409$

Hindu/Arabic number system

The number system we use is called the Hindu/Arabic number, which is separated into the units/tens/hundreds/thousands/etc columns.

For example, 12345 is 12,345, or “Twelve thousand, three hundred and forty five”.

Like the Roman system (which order is important due to the use of subtraction), in the Hindu/Arabic system, the position (place value) is important.

Therefore, the Arabic number system uses the zero (0).

The effect is that the number “5” is different from “50”, which is again different from “500”.


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# Easiest way to show positive semi-definite equivalence

For an $$x \in \mathbb{R}^n$$, and $$n$$-by-$$n$$ identity matrix $$I_n$$, we are given that $$\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix} \succeq 0.$$

What is the easiest way to show that $$\begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix} \succeq 0$$ holds?

• Symmetric permutation? – user251257 Mar 22 at 18:37
• seems so. is there such a theory which concludes? – independentvariable Mar 22 at 18:40
• Schur complement? – user251257 Mar 22 at 19:02
• They dont reduce to the same condition, do they? – independentvariable Mar 22 at 19:22

This is due to the identity

$$\underbrace{\begin{pmatrix} 0_{1,n} & 1 \\ I_n & 0_{n,1} \end{pmatrix}}_{J^T} \underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A} \underbrace{\begin{pmatrix} 0_{n,1} & I_n \\ 1 & 0_{1,n} \end{pmatrix}}_{J} = \underbrace{\begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix}}_{B}$$

(notation $$0_{m,n}$$ is for a zero block with $$m$$ lines and $$n$$ columns).

Indeed, $$J$$ being a permutation matrix, it is an orthogonal matrix, with $$J^T=J^{-1}$$. We can conclude that $$A$$ and $$B$$ are similar, thus have the same spectrum (Similar matrices have the same eigenvalues with the same geometric multiplicity) with positive eigenvalues, thus are both semi-definite positive.

Besides, $$A$$ being symmetric, one can conclude from $$J^TAJ=B$$ that $$B$$ is symmetric as well.

Appendix : There is a pending question : is there a criteria on $$x$$ for positive semi-definiteness of $$A$$. ? The answer is yes :

$$A$$ is semi-definite positive iff $$\|x\| \leq 1$$.

This is due, as we are going to see it, to an analysis of the rather particular spectrum of $$A$$. Let us obtain it explicitly.

First of all, let us establish that $$A$$ (which is a $$(n+1) \times (n+1)$$ matrix) has eigenvalue $$1$$ with order of multiplicity at least $$n-1$$.

Consider hyperplane $$x^{\perp}$$ of $$\mathbb{R}^n$$ defined as the set of vectors $$y$$ that are orthogonal to $$x$$. Let $$(y_1,y_2,\cdots y_{n-1})$$ be a basis of $$x^{\perp}$$ ; then,

$$\underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k}=1\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k} \ \ \ \text{for} \ \ k=1,2, \cdots (n-1),$$

proving that $$V_k$$ is an eigenvector associated with eigenvalue $$1$$.

Due to the fact that trace$$(A)=n+1$$, the two remaining eigenvalues are of the form $$\alpha$$ and $$\beta:=2-\alpha$$. We can assume, WLOG that $$\alpha \leq 1 \leq \beta$$.

Besides, using the so-called Schur determinant identity (Eigenvalues of a Block Matrix from Schur Determinant Identity) for the computation of the determinant of a $$2 \times 2$$ block matrix, we obtain :

$$\det(A)=1-x^Tx$$

As the determinant is also the product of eigenvalues, we get the following identity :

$$\det(A)=1-\|x\|^2=\alpha(2-\alpha)\tag{1}$$

Thus, one can compute explicitly the two remaining eigenvalues by solving quadratic equation (1), with the following explicit solutions (if we assume that $$\alpha$$ is the smallest eigenvalue)

$$\alpha=1 - \|x\| \ \ \ \implies \ \ \ \beta:=2-\alpha=1 + \|x\|\tag{2}$$

As the criteria for a symmetric matrix do be semi-definite positive is that must have all eigenvalues $$\geq 0$$, this criteria becomes $$\alpha \geq 0$$, i.e., $$\|x\| \leq 0$$. $$\square$$

Remark : eigenvalues $$\alpha$$ and $$\beta$$ can be associated with eigenvectors $$\begin{pmatrix} x\\ -\|x\| \end{pmatrix}$$ and $$\begin{pmatrix} x\\ \|x\| \end{pmatrix}$$ resp.

Let us take an example in the case $$n=4$$ ; let $$m=1/n$$ ; consider matrix :

$$A:=\left(\begin{array}{rrrr|r} 1 & & & & m \\ & 1 & & & m \\ & & 1 & & m\\ & & & 1 & m\\ \hline m & m & m & m & 1 \\ \end{array}\right)$$

One can check, using (2), that the spectrum of $$A$$ is

$$(\tfrac12, 1 , 1, 1, 1, \tfrac32).$$

Just now, I "googled" with keywords "bordered identity matrix" : I found in (Eigenvalues of a certain bordered identity matrix) a somewhat similar computation that I did in the Appendix.

• Thank you for your answer! My main purpose was showing $||x||_2 \leq 1$ holds iff the matrices I gave are Psd. I proved one by Schur complement theory on PSD matrics, but for the equivalence I just followed the definition of PSD and by contradiction showed that if one holds, the other one should hold etc.. – independentvariable Mar 23 at 19:28
• But yours seem the better way, not the 'dirty' $a^T X a \geq 0$ for all $a$ approach... I don't like it, it seems too manual – independentvariable Mar 23 at 19:35


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## Saturday, 6 February 2016

### Videos online

I posted videos of the proof of the Inverse Function Theorem, Implicit Function Theorem  and higher dimensional derivatives in the left hand margin. Please note that there is an annoying typo in the last line of the proof of the Inverse Function Theorem: in the first posted version of chapter 1, at the end of the proof of the derivative, F has the variable y as argument but it should of course be x (identically copied from the line above). As you see in the video, one finally inserts $$x=G(y)$$ to get the result.


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# Adding PDF bookmarks with novaPDF

This tutorial shows how to add automatically bookmarks to a PDF file using novaPDF, even if in the original document there are no bookmarks. This way the resulting PDF will have shortcuts to the most important sections in the document.

novaPDF Pro has the ability to add bookmarks in a PDF file based on formatting detection rules. This means that you can configure novaPDF to convert a word/sentence/paragraph that has a specific format to a PDF bookmark by using bookmark definitions.  Example: you have a 3 page document and on each page a title that has a font size of 22. You enable in novaPDF bookmark detection and add a definition to recognize text that has the font size of 22 and convert it to a bookmark.
When you print your document to novaPDF, it will look through your entire text for words/sentences/paragraphs that respect the formatting rule (font size=22) and convert those to bookmarks in the PDF file. Of course this is a simple example, as novaPDF can detect formatted text based on the font type used and/or font size, font style (italic/bold) or even font color, and convert that to bookmarks that can have different style or color options.
The PDF bookmarking feature is very useful when needing to structure the resulting PDF file. For instance, when you have a document with multiple pages, you will need it to be better structured so that the recipient will be able to see a list of bookmarks on the left side of the document and easily access different key sections in your document.

### What are PDF bookmarks?

PDF bookmarks appear on the left side of the pdf reader window (in Adobe Reader pressing F4 will toggle the visibility of the Bookmark panel) given that the created PDF contains this functionality.
The usability of the PDF bookmarks comes from the fact that it can be used as a shortcut menu, allowing users to easily go to other sections in the PDF document. Consider a 300 page document and you'll see how this presents an advantage as users don't have to scroll looking for a specific section.
PDF bookmarks can be considered a clickable table of contents list, the difference being that a table of contents is more difficult to use because it's present at the top of the document, while bookmarks are available on the left side of the document (outside of it) and you can access them no matter on what page of the document you are (no need to scroll all the way to the top to see the table of contents). Using PDF bookmarks you can give the user an easy way to access all important sections and sub-sections of the PDF when they need to (Bookmarks panel can be switched on and off), as well as an easy way to print a specific page (right-clicking a boomark will show the option to print the pages where it links to - all pages until next bookmark will be printed).
Here's an example of how bookmarks look in a PDF opened with Adobe Reader:

### How to enable bookmark detection and add bookmark definitions

Since novaPDF supports different profiles, configuring novaPDF to add bookmarks when creating PDF files can be done through its Profile Manager window. Note that once the feature is enabled for a profile and bookmark levels are defined, those will be added when creating a PDF file with novaPDF from any document that supports printing (so you don't have to go in manually each time you create a PDF to enable bookmarks). To enable bookmarks and add definitions:
• Open Printer Manager (go to Start and type "printer manager" in Windows 10/8 or go to Start->novaPDF->Printer Manager in Windows 7 or previous) then click on Profiles->Manage Profiles. If you have an administrator password set, you'll be required to enter it to make changes.
• Select the profile that you want to enable bookmarks for (or if you don't have one profile defined enter a name for it in the General tab) and click on the Bookmarks tab.
• Check the Detect bookmarks option to enable bookmark detection.
You'll see the other settings in that page become available under Bookmarks. Under What Bookmarks to detect select one of the existing presets for bookmark detection. You can select Microsoft Office heading styles to have the same type of heading definitions from Microsoft Office detected and converted to bookmarks (useful when you convert a Word document for instance to a PDF).
Before you start adding bookmark definitions you need to understand the bookmark levels (called Headings) included as predefined bookmark presets. There are 9 bookmark levels that can be added (Heading 1 through Heading 9), each with one or more possible definitions. The first entry in the Bookmarks list (Heading 1) will be shown on the resulting PDF file on the first level, the second entry in the Bookmarks list (Heading 2) will be shown in the resulting PDF file on the second level (as a child of the first level) and so on. Below you can see how Heading 1, Heading 2 and Heading 3 bookmark levels look like in the resulting PDF file (each level can be formatted with a different style or color, the font size though will remain the same):
In order to add a bookmark and define its detection rules and display options:
• In the Bookmarks tab after you check the Detect bookmark option you you can configure the detection and display options. Options that you can set when adding bookmark definitions:
• Definitions list - this list shows the Name, Description and Heading Level of the currently selected bookmark detection presets. A checkbox is also displayed in front of each definition Name, that you can use to enable/disable that particular definition (checked means Enabled). To the right you also have buttons to Add a new definition, Copy an existing one or Remove them.
• Detection - this section lets you choose what detection rules to use for the text in the original document for the selected bookmark definition item. For instance, if you select the Arial font type and set a Size of 20, then if novaPDF encounters in the original document a word or sentence that has these particular formatting rules it will make that a clickable bookmark in the resulting PDF file. Detection options you can set are based on: the Font type (Arial, Verdana, Times New Roman, ...), font Size, font Style (Bold, Italic) and Color. You may note that near the Size box there's an additional +- field - this is needed because some programs even if they contain fonts of a specific size they will send to novaPDF smaller or larger sized fonts with a margin of up to 2 points - by adjusting this novaPDF will be able to accurately detect that specific font size (e.g. if you set a Size of 20 and a +- margin of 1, novaPDF will look for text with the size between 19.01 and 20.99). As you change the detection options, you can see in the Detection preview window how the text that will be converted to bookmarks looks in the original document.
• Display- here you can set the font style and color for the Bookmark items. This will only change the font style and color of the clickable text that will be available in the PDF via the Bookmarks panel (it will not modify the actual text from the PDF, it is used only for the Bookmarks panel). As an example, if you are defining a Heading 1 bookmark level and in the Display you check Bold and Italic, in the PDF file these clickable bookmark links will be bolded and italicized (only for the Heading 1 level).
1. After you have added and configured the desired definitions, click Save in the Profile Manager window (top-left corner) save these changes for the current profile.
2. If you want to test how the PDF bookmarking works, open a document that supports printing (a Word document for instance) and that contains text formatted as set in the Detection options, go to File ->Print, select novaPDF from the Printers list and click on Print. Note: You cannot add PDF bookmarks to an existing PDF file with novaPDF if the PDF file had fonts embedded.

### Setting additional options for the PDF bookmarking feature

On the Bookmarks tab in novaPDF's Profile Manager window you also have some additional options: Allow multiline bookmarks, Match bookmarks regardless of level, Use Levels and Open to level.
Description and examples for these options are given below.
Allow multiline bookmarks
1. Option checked - if in the original document (you're converting to PDF) there is text that spans on two consecutive lines and this text matches a bookmark definition added in novaPDF, in the PDF file this will become a single clickable bookmark.
2. Option unchecked - if in the original document (you're converting to PDF) there is text that spans on two consecutive lines and this text matches a bookmark definition added in novaPDF, in the PDF file each of the two lines will be a clickable bookmark.
Match bookmarks regardless of level
• Option checked - this option has an effect only if there are at least 3 headings defined (bookmark levels). Let's say that novaPDF's Bookmarks window you have 3 definitions added: Heading 1, Heading 2 and Heading 3. If in the original document (you're converting to PDF) there is a portion of text that matches the definition of Heading 1, and after that (not "right after" that) a portion of text that matches the definition of Heading 3, in the PDF file there will be a Heading 1 bookmark, underneath it a blank bookmark ("dummy node"), and after that a Heading 3 bookmark.
• Option unchecked - let's assume the same example as above (you have 3 bookmark definitions Heading 1, Heading 2 and Heading 3). If in the original document (you're converting to PDF) there is a portion of text that matches the definition of Heading 1, and after that (not "right after" that) a portion of text that matches the definition of Heading 3, in the PDF it will be shown as bookmark only the text portion that matched the Heading 1 bookmark definition (the text that matched the Heading 3 bookmark will not be shown because there's no Heading 2 bookmark).
Use levels - this allows you to specify which bookmark levels will be shown in the PDF. By default, as you add bookmark levels (Headings) this will increment automatically. For instance, if you have three bookmarks levels (Heading 1, Heading 2 and Heading 3) and the value in Use levels is 2, only Heading 1 and Heading 2 will appear in the PDF (if the original document has text that matches these definitions). For the same example, if Use levels was set to 3 all three Headings would have appeared in the PDF.
Open to level - this lets you specify until which level should the PDF reader expand the bookmarks in the PDF file. By default this is set to 1 (meaning only the Heading 1 bookmark level will be expanded) but it can be set up to the value of Use levels. For example, if you have 3 bookmark levels defined and Open to level is set to 3, all 3 bookmark levels will be expanded in the PDF file.


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# How many moles of helium are present in a 5 L container at STP?

Jun 7, 2016

The molar volume of an ideal gas at STP is $22.4 \cdot L$.
${n}_{\text{He}}$ $=$ $\text{Volume"/"Molar volume}$ $=$ $\frac{5.0 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}$ $=$ ??mol