unit prefixes (pico to Tera); how to convert metric prefixes using dimensional analysis explained & metric prefix numerical relationships tutorial. Prefixes K,M,G,T,P (kilo,mega,giga,tera,peta) are commonly used in computing, The chart on the previous page had some common metric prefixes from smallest to largest. Bulk agricultural products, such as grain, beer and wine, are often measured in hectolitres (each 100 litres in size). Start studying Metric Prefix Chart. These SI prefixes or metric prefixes are in widespread use in all areas of life. The LaTeX typesetting system features an SIunitx package in which the units of measurement are spelled out, for example, \SI{3}{\tera\hertz} formats as "3 THz". for this purpose. [Note 1] When both are unavailable, the visually similar lowercase Latin letter u is commonly used instead. Converts the metric prefix in one unit to the others. there are 1000 metres in a kilometre (km). Power of ten Prefix Prefix Abbrev. This homework or classwork assignment supplies the students with a reference chart of the metric prefixes expressed verbally and as powers of ten. If they have prefixes, all but one of the prefixes must be expanded to their numeric multiplier, except when combining values with identical units. There are gram calories and kilogram calories. They are also occasionally used with currency units (e.g., gigadollar), mainly by people who are familiar with the prefixes from scientific usage. When typing your answer, use scientific notation. Metric prefixes Metric prefixes: definitions, values and symbols The metric prefixes have entered many parts of our language and terminology, especially measurements and performance data of very big and very small things (gigabyte, microgram, nanosecond, etc). Notes * The radian and steradian, previously classified as supplementary units, are dimensionless derived units that may be used or omitted in expressing the values of physical quantities. one thousandth of a metre (mm). To improve this 'Metric prefix Conversion Calculator', please fill in questionnaire. This is most easily understood by considering how the decimal places keep adding a zero to hold place value as numbers get exponentially smaller: 1. Jan 21, 2014 - This won’t give all the metric prefixes to you, but then, you won’t generally need them all. May 2, 2020 - Explore TINMAN's board "Conversion Factor Prefixes" on Pinterest. The metric system provides a logical way to organize numbers and mathematical thinking. Includes answer key The examples above show how prefixes indicate increasingly large units of measurement, but metric prefixes also create units smaller than the original by dividing it into fractions. Likewise, milli may be added may be added to metre to form the word millimetre, i.e. Here you can make instant conversion from this unit to all other compatible units. Long time periods are then expressed by using metric prefixes with the annum, such as megaannum or gigaannum. The metric prefixes have entered many parts of our language and terminology, especially measurements and performance data of very big and very small things (gigabyte, microgram, nanosecond, etc). The metric system charts in this ScienceStruck post will help kids understand converted values quite easily. Except for the early prefixes of kilo-, hecto-, and deca-, the symbols for the multiplicative prefixes are uppercase letters, and those for the fractional prefixes are lowercase letters. 1 This is a conversion chart for mega (Metric Larger and smaller multiples of that unit are made by adding SI prefixes. 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When is a Sudoku like table solvable

Given a $n\times n$ table is it possible to fill each cell with one of the numbers $1,2,3,\cdots,n$ such that in each column,each row and each diagonal (i.e Denoting $(x,y)$ as number of column and row $(2,1)$ and $(1,2)$ form the first diagonal) every number appears exactly once? For which $n$ can we fill the table?

Context: I've been given this problem on a contest few months ago but just for $n=4,5$ which I solved easily since $n=4$ is impossible and for $n=5$ we have \begin{array}{|c|c|c|c|c|} \hline 1&2&3&4&5\\ \hline 3&4&5&1&2\\ \hline 5&1&2&3&4\\ \hline 2&3&4&5&1\\ \hline 4&5&1&2&3\\ \hline\end{array} But I was interested in a more general statement I think I've also proved that for $n=6$ it's impossible by trying to fill the table manually. My guess is that for even $n$ it's not solvable and for odd $n$ it's solvable but I have no idea how to approach it except to fill it manually.

EDIT: For prime $n$ we can fill each cell $(i,j)$ with $i+2j\pmod{n}$ except when $i+2j\equiv0\pmod{n}$ then we write $n$ instead for example such filling with $n=7$ (the $n=5$ example is the same filling if you look at $(j,i)$ instead of $(i,j)$) \begin{array}{|c|c|c|c|c|c|c|} \hline 3&5&7&2&4&6&1\\ \hline 4&6&1&3&5&7&2\\ \hline 5&7&2&4&6&1&3\\ \hline 6&1&3&5&7&2&4\\ \hline 7&2&4&6&1&3&5\\\hline 1&3&5&7&2&4&6\\\hline2&4&6&1&3&5&7\\\hline\end{array}

PROOF OF THE EDIT: For the same row if cells $(i_1,j)$ and $(i_2,j)$ have the same value we have that $$i_1+2j\equiv i_2+2j\pmod{n}$$ implies $i_1\equiv i_2$ which is possible only if $i_1=i_2$. Same logic applies to the column for cells $(i,j_1),(i,j_2)$ we get $$i+2j_1\equiv i+2j_2\pmod{n}$$ when $n$ is prime it implies $j_1=j_2$ if $(i_1,j_1),(i_2,j_2)$ are on a diagonal we have $$|i_1-i_2|=|j_1-j_2|$$ now assuming they have the same value $$i_1+2j_1\equiv i_2+2j_2\pmod{n}$$ then $i_1-i_2\equiv 2(j_2-j_1)\pmod{n}$ which implies $1\equiv \pm 2\pmod{n}$ which is absurd.

• What does the sentence "$(2,1)$ and $(1,2)$ form the first diagonal" mean? – 5xum Jun 12 '17 at 12:51
• @5xum I mean position second cell in first column or first cell in second column, like in matrix. – kingW3 Jun 12 '17 at 12:55
• This seems to by highly related to the 8 queens problem, if you think of all ones as queens, they may not be in the same row, column or diagonal. The same holds for all twos and so on. – mlk Jun 12 '17 at 13:00
• I played with this once. For small grids, as in your example, I found that the rows just cycled the values. However, at 12x12, I found some solutions that did not do this. – badjohn Jun 12 '17 at 13:01
• I found an old Java program that I wrote for this puzzle. It just uses brute force and ignorance. Assume that row 1 is 1, 2, 3, etc. Sizes 1, 2, 3, 4, 6, 8 have no solution. 5, 7, have a unique solution. This might suggest that there are no solutions for the even cases but there are multiple solutions for the 12 case. I am running the 9, 10, and 11 cases but, as you may expect, the program gets slower for these larger values. – badjohn Jun 12 '17 at 15:12

This is not an answer but hopefully a contribution to an answer.

Double diagonal Latin squares or just diagonal Latin squares (the terminology seems to vary) are Latin squares where both main diagonals (sometimes called the main and the anti-main) also have the property that all $N$ symbols occur exactly once. I realize that your requirement is that all "minor" diagonals also don't have repeating symbols, but it should be clear that a necessary condition for this, is that the square must be a diagonal Latin square.

In this paper there is a proof on page $4$ which shows that, if there are numbers A and B from the range $[0, N-1]$ which satisfy the properties:

• A is relatively prime to N
• B is relatively prime to N
• (A + B) is relatively prime to N
• (A - B) is relatively prime to N

then you can generate a diagonal Latin square with the following rule:

Cell$(i,j) = (A * i + B * j) \mod N$

This is like the rule you found but without the strict requirement that $N$ is prime. A corollary to the above theorem is that if $N$ is an odd number not divisible by $3$, there is a diagonal Latin square of order $N$. So I tried the formula with the first odd non-prime fulfilling the corollary's requirement $(N=25)$ and got the following:

It seems to me this is a square of the type you are looking for, and with $N$ odd, but not a prime.

Edit

We can also show that with an even $N$, no diagonal Latin square can be generated using the method above. If $N$ is even, both $A$ and $B$ must be odd. But then both $(A+B)$ and $(A-B)$ must be even and can therefore not be relatively prime to $N$.

Edit 2

I made a program to generate diagonal Latin squares based on the formula above and then to check if all diagonals were without repeats. I ran the program for all odd $N$ between $3$ and $1001$ and the result is that all squares, where $N$ is not divisible by $3$, fulfilled the requirements! I therefore conjecture that the corollary above is not only true for diagonal Latin squares but also for "kingW3" squares.

Edit 3

Ladies and gentlemen, I have found a very nice document which answers many of our questions. In fact, if we use the definition of "diagonal" assumed by @Ewan Delanoy (called "broken diagonals" in the document), it basically solves the OP:

1. It proves the conjecture I made above
2. It proves that if the definition of "diagonal" is "broken diagonals", no solutions exist for even $N$
3. It gives an outline of a proof (leaving the details as homework!) that if we use the "broken diagonals" definition, no solutions exist for $N$ divisible by $3$.

Enjoy!

• Thanks for the update. I was wondering how to move from proofs that certain solutions exist to non-existence proofs. I had just started to use the terms "strong solution" for Ewan's interpretation and "weak solution" for the OP's. – badjohn Jun 16 '17 at 16:45

This is only a partial answer but it is too large to be a comment. I looked at this problem a long time ago and I wrote a Java program which tried to crack it by brute force and ignorance.

I only looked at cases in which the first row is the selected symbols in order. Any other solution is "isomorphic" to one of this form.

When the size is above 9, I use A, B, C, etc in a hexadecimal style.

For size 1, there is a trivial solution

$$\begin{array} {|c|} \hline 1\\ \hline \end{array}$$

for sizes 2, 3, and 4, there is no solution.

Size 5 has a solution as posted by kingW3 in his original post. There is a second which is in a sense a reflection. Each row is offset by 3 to the right which can be viewed as 2 to the left hence the reflection comment.

Size 6 has no solution.

Size 7 has 4 solutions here is one which is similar to kingW3's solution for size 5. Each row is offset 2 to the right. The others are offset by 3, 4, and 5.

$$\begin{array} {|c|c|c|c|c|c|c|} \hline 1&2&3&4&5&6&7\\ \hline 3&4&5&6&7&1&2\\ \hline 5&6&7&1&2&3&4\\ \hline 7&1&2&3&4&5&6\\ \hline 2&3&4&5&6&1&2\\ \hline 4&5&6&7&1&2&3\\ \hline 6&7&1&2&3&4&5\\ \hline \end{array}$$

Note the simple cyclic pattern shared by size 5.

Size 8 has no solution.

This hints at a pattern: even unsolvable, odd solvable with a cyclic pattern.

The pattern does not continue. At size 9, that cyclic style does not work. My cracking program just completed for size 9; there is no solution.

I won't run the cracking program on size 10. It would probably not finish before I die.

The cyclic patterns work for size 11. So, as kingW3 has found, a prime size helps which makes sense once you know.

However, this does not cover all solutions. Here is one for size 12. I have a memory, but no records, of others.

$$\begin{array} {|c|c|c|c|c|c|c|} \hline 1&2&3&4&5&6&7&8&9&A&B&C\\ \hline 5&6&C&1&B&2&3&A&7&4&8&9\\ \hline 9&A&4&8&6&1&B&5&C&2&7&3\\ \hline B&7&9&3&A&C&8&1&4&6&5&2\\ \hline 8&C&5&2&7&4&9&6&A&1&3&B\\ \hline 3&B&7&6&C&A&5&4&2&8&9&1\\ \hline A&4&1&9&8&3&2&B&5&C&6&7\\ \hline C&8&2&5&1&B&6&9&3&7&A&4\\ \hline 4&1&6&A&3&8&C&7&B&9&2&5\\ \hline 6&3&B&C&9&7&4&2&8&5&1&A\\ \hline 2&9&8&7&4&5&A&3&1&B&C&6\\ \hline 7&5&A&B&2&9&1&C&6&3&4&8\\ \hline \end{array}$$

I have a solution for size 25. It is the cyclic style. I think that will work if the size is coprime to 2 and 3 but I have not proved that yet.

The 12 case remains interesting as it is not of the cyclic style.

Update

On further thought, I now believe for size $n$ and a shift per line of $s$, a cyclic solution exists provided that all of $s - 1$, $s$, and $s + 1$ have order $n$ in $\mathbb{Z}_n$ (as an additive group). In other words, they are all non-zero and either $1$ or coprime to $n$. This can be simplified to Jens's rule of odd and not divisible by $3$.

The $12\times12$ example above remains the only exceptional case that we know.

• Your $12\times 12$ example is wrong : for example, the $(11,1)$ and $(9,11)$ entries are both equal to $2$, yet they are on the same diagonal. – Ewan Delanoy Jun 15 '17 at 11:47
• @EwanDelanoy I see what you mean but we seem to have different notions of the diagonal. I guess that you are imagining that the two sides are implicitly connected and a diagonal can travel off one side and continue on the other. My diagonals were not doing that. This comment from the OP, "(2,1) and (1,2) form the first diagonal" suggested my interpretation. – badjohn Jun 15 '17 at 11:57
• The OP has been quiet for a while. We need him to say whether my example qualifies. Nonetheless, I will consider this interpretation as well. – badjohn Jun 15 '17 at 12:06
• I too see what you mean. I think the OP should clarify this point. Personally if the sides are not implicitly connected I find the problem has less symmetry and beauty – Ewan Delanoy Jun 15 '17 at 12:08
• @EwanDelanoy I like your interpretation and I will be trying it. I guess that you are connecting the top and bottom as well so, topologically, you are working with a torus. – badjohn Jun 15 '17 at 12:37


# Easy Time, Speed & Distance Solved QuestionAptitude Discussion

 Q. A race course is 400 m long. $A$ and $B$ run a race and $A$ wins by 5m. $B$ and $C$ run over the same course and $B$ win by 4m. $C$ and $D$ run over it and $D$ wins by 16m. If $A$ and $D$ run over it, then who would win and by how much?
 ✔ A. $D$ by 7.2 m ✖ B. $A$ by 7.2 m ✖ C. $A$ by 8.4 m ✖ D. $D$ by 8.4 m

Solution:
Option(A) is correct

If $A$ covers 400m, $B$ covers 395 m

If $B$ covers 400m, $C$ covers 396 m

If $D$ covers 400m, $C$ covers 384 m

Now if $B$ covers 395 m, then $C$ will cover $\dfrac{396}{400}\times 395=391.05$m

If $C$ covers 391.05 m, then $D$ will cover $\dfrac{400}{384}\times 391.05 = 407.24$

If $A$ and $D$ run over 400 m, then $D$ win by 7.2 m (approx.)

Edit: For an alternative solution, check comment by Vejayanantham TR.

## (5) Comment(s)

Vaibhav Varish
()

Sa-Speed of a in meter/minute

eq1---

400/Sb-400/Sa=5

eq2---

400/Sc-400/Sb=4

eq3--

400/Sc-400/Sd=16

400/Sc-400/Sa=9;

eq3-eq2---

400/Sa-400/Sd=7

that' s why D wins by 7 min why 7.2 (Whats wrong in my solution)..pls help

Vejayanantham TR
()

A - 5m

B - 4m

D -16 m

We need to find from A -> D

A->B->C = 5+4

D -> 16m

16-9 = 7 => D wins by 7m (approx)

Manohar Tangi
()

Nayak Sowrabh
()

Since the number 16 is too small in comparison to 400 u get an approximate value to be 7.

If it was a 100meter race then u would have got the answer to be -> D wins the race by 8.5m.

This is because 16 makes a lot of difference to the number 100.

so we cannot use this methodology in all those cases.

Yogesh
()

can we apply this method on these type of question


Can the area enclosed by two curves be infinite?

This is a question from my test: Find the area enclosed by the graph of $y=x^3-6x^2+11x-6$ and $y=0$.

This is actually very simple but the way I look at it, I can see two different answers to this question, depending on the answer to my title.

The first is where I assume area can be infinite and use a boundary from $-\infty$ to $\infty$. This is what I ended up writing as my answer in the test ($x=1;2;3$; are points of intersection): $$-\lim_{a\to-\infty}\int_{a}^{1}y \ dx + \int_{1}^{2}y \ dx - \int_{2}^{3}y \ dx+\lim_{b\to\infty}\int_{1}^{b}y \ dx = \infty$$

The other one is what my teacher told me the answer to the test question is, which is just: $$\int_{1}^{2}y \ dx - \int_{2}^{3}y \ dx = \frac{1}{2}$$ Now my question is, can the area enclosed actually be infinite? I personally think an infinite area should be possible, which is why we normally use boundaries in integrals to limit the area from becoming infinite.

From Wikipedia, it says something like $\int_{a}^{b}f(x) \ dx = \infty$ means that $f(x)$ doesn't bound a finite area between $a$ and $b$, while $\int_{-\infty}^{\infty}f(x) \ dx = \infty$ means the area under $f(x)$ is infinite.

So what do you think about this? I'd really appreciate your opinions or even facts on this matter. Can an area enclosed be infinite? And thus, would an answer of $\infty$ be a legitimate or false answer to the test question? (I don't plan on complaining for marks, this is just for my own curiosity and self-learning).

Sorry for the long question, and thanks in advance!

• Perhaps the question would have been better stated as "find the area of the region bounded by" the two graphs. A bounded region in the plane is a region which can be enclosed within a circle. – John Wayland Bales Nov 28 '16 at 4:25
• @JohnWaylandBales Since you got here first, feel free to expand that into an answer. – Mark S. Nov 29 '16 at 1:37
• OK, I have submitted an answer. – John Wayland Bales Nov 29 '16 at 2:46

It is a reasonable question. Although usually if asked to find the area of a region or regions bounded by two graphs what is meant by "bounded" is that the regions all lie within the interior of some circle.

This is analogous to a bounded set on the number line being contained in some interval $[a,b]$. It is completely circumscribed.

However it is possible for to graphs to enclose a finite, yet unbounded region.

There are many examples, but one is as follows.

Find the area of the region "bounded" by the graphs of $y=0$ and $y=\dfrac{x}{x^4+1}$

Here is the graph of the region.

This region is not bounded in the sense stated above. It cannot be contained in the interior of a circle. Yet it has a finite area.

$$\int_{-\infty}^\infty\dfrac{|x|}{x^4+1}\,dx=\int_{0}^\infty\dfrac{2x}{x^4+1}\,dx\\$$

Make the substitution $u=x^2$, $du=2x\,dx$ and this becomes

\begin{eqnarray} \int_{0}^\infty\dfrac{1}{u^2+1}\,du&=&\frac{1}{2}\arctan(u){\Large\vert}_{0}^\infty\\ &=&\left(\dfrac{\pi}{2}-0\right)\\ &=&\frac{\pi}{2} \end{eqnarray}

Therefore it is acceptable to say that, in a sense, an unbounded region is "bounded" by two graphs so long as the area enclosed is finite.

• How about when the area enclosed is infinite? I think an unbounded region with a finite area would be considered as a convergent integral, so we can calculate the finite area. But what about an unbounded region which is divergent? When asked to calculate the area, would the area be infinite, or should we only consider the finite area regions? – Gyakenji Nov 29 '16 at 2:57
• If the area approaches infinity then we can say that the region has infinite area. In your original problem, however, it was a question of what the teacher meant by saying the region "bounded" by the function. I do not know the policy of your teacher, but when I was teaching I was always happy to further explain the meaning of a question if the student thought the question was unclear. – John Wayland Bales Nov 29 '16 at 3:02
• Actually I notice that your professor said "enclosed" not bounded. That is actually not a mathematical term so you could certainly ask what he meant by "enclosed." – John Wayland Bales Nov 29 '16 at 3:07
• My professor doesn't speak English as a first language so perhaps it was just a mistake, but on the lectures, he is usually asking for the area between two curves (so the area bounded by the two functions?). I will ask him further about it. Sorry if I'm repeating, but assuming he meant the area bounded by the two functions, then would it be mathematically correct to say the area is infinite, given the integral will be divergent and area will approach infinite? Or would the area bounded by the functions be the finite areas only? Thanks so much in advance. – Gyakenji Nov 29 '16 at 3:17
• The term "enclosed" is widely used in textbooks when talking about regions bounded by two curves, but I have never seen a textbook give an explicit definition of the term. There is a common mathematical concept of a "closed curve" which is a curve in the plane which begins and ends at the same point. So one could define an enclosed region as a region whose boundary consisted of one or more closed curves. With that definition, even the example I gave in my answer is not enclosed. – John Wayland Bales Nov 29 '16 at 3:18


The following vector q is a 3 × 1 column vector containing numbers: $q=\begin{bmatrix} 2\\ 5\\ 8\end{bmatrix}$ A row vector is an 1 × c matrix, that is, a matrix with only one row. A column vector is an r × 1 matrix, that is, a matrix with only one column. 2. Column and row vectors a matrix with one column, i.e., size n×1, is called a (column) vector a matrix with one row, i.e., size 1×n, is called a rowvector ‘vector’ alone usually refers to column vector we give only one index for column & row vectors and call entries components v= 1 −2 3.3 0.3 w= −2.1 −3 0 4. Two matrices of the same order whose corresponding entries are equal are considered equal. A column matrix is a matrix that has only one column. Rectangular Matrix A matrix of order m x n, such that m ≠ n, is called rectangular matrix. To understand what this number means, take each column of the matrix and draw it as a vector. One column matrix. A matrix with only one row is called a.....matrix, and a matrix with only one column is called a.....matrix. Suppose that A has more columns than rows. The matrix derived from a system of linear equations is called the..... matrix of the system. The entries are sometimes A matrix is said to be a column matrix if it has only one column. 5. A matrix with only one row is called a _____ matrix, and a matrix with only one column is called a _____ matrix. The entries of a vector are denoted with just one subscript (since the other is 1), as in a3. A matrix with only one column, i.e., with size n × 1, is called a column vector or just a vector. 3. Column Matrix A matrix having only one column and any number of rows is called column matrix. 3) Square Matrix. For example, four vectors in R 3 are automatically linearly dependent. In general, B = [b ij] m × 1 is a column matrix of order m × 1. one column (called a column vector). Converting Systems of Linear Equations to Matrices. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). INCLUDES THE SOLUTIONS. Sometimes the size is speciﬁed by calling it an n-vector. A column matrix has only one column but any number of rows. Below, a is a column vector while b is a row vector. Each equation in the system becomes a row. For example, $$A =\begin{bmatrix} 0\\ √3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 × 1. The variables are dropped and the coefficients are placed into a matrix. Each variable in the system becomes a column. Determinants. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. row column. Fill in the blanks. A matrix with only one row or one column is called a vector. a = 7 2 3 , b = (− 2 7 4) A scalar is a matrix with only one row and one column. I have the matrix as follows. Example: D is a column matrix of order 2 × 1 A zero matrix or a null matrix is a matrix that has all its elements zero. augmented. The determinant takes a square matrix and calculates a simple number, a scalar. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. Example: C is a column matrix of order 1 × 1 A column matrix of order 2 ×1 is also called a vector matrix. Then A cannot have a pivot in every column (it has at most one pivot per row), so its columns are automatically linearly dependent. a = 1 3 2 5 3 2 4 8 5 9 I want to sort the second column in the a matrix. A vector is almost often denoted by a single lowercase letter in boldface type. A matrix, that has many rows, but only one column, is called a column vector. Horizontal Matrix A matrix in which the number of rows is less than the number of columns, is called a horizontal matrix. 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# difference between quotient rule and product rule

Product rule :

$$\frac{d}{dx} \big(f(x)\cdot g(x)\big)=f'(x)\cdot g(x)+f(x)\cdot g' (x)$$

Quotient rule :

$$\frac{d}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$$

Suppose, the following is given in question. $$y=\frac{2x^3+4x^2+2}{3x^2+2x^3}$$

Simply, this is looking like Quotient rule. But, if I follow arrange the equation following way

$$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$

Then, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?

• Your just deriving the quotient rule on the fly, rather than assuming it is true and then applying it directly; there is nothing wrong with doing the former. May 8 at 13:39
• First thing a mathematician would do is write $$y=1+\frac{x^2+2}{2x^3+3x^2}$$ before taking the derivative. May 8 at 16:14
• What do you mean "I think both answers are correct."? If you do the math properly, you'll get the same answer with both methods. May 9 at 0:10
• Real physicists and mathematicians stick the whole thing into their computer algebra system and don't worry about the precise algorithms it follows, unless they have a particular reason to risk silly errors by doing that kind of calculation by hand. Symbolic differentiation is a solved problem; one doesn't earn any "purity points" in the real world by doing it the hard way. May 9 at 0:33
• @JosephSible-ReinstateMonica Yes! I got same answer :)
– user876873
May 9 at 4:30

Note that $$((g(x))^{-1})'=-g'(x)(g(x))^{-2}$$. Then, applying the product rule: $$\left(\frac{f(x)}{g(x)}\right)'=\left(f(x)\cdot \frac{1}{g(x)}\right)'=\frac{f'(x)}{g(x)}+\frac{-g'(x)f(x)}{(g(x))^2}=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$ which is the quotient rule

• That's very helpful. But, it was my main question How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?
– user876873
May 8 at 13:44
• @Istiak Please don't keep making that bold. Sure, it's OK, because any valid proof is OK; there are no "rules" beyond that. To be honest, I hardly ever see mathematicians or physicists explicitly using the quotient rule. It seems to be more for teaching than anything else.
– J.G.
May 8 at 13:45
• I'd always use the quotient rule on a quotient because it is usually much simpler to work with $g'$ than $(\frac1g)'$ May 8 at 13:45
• @J.G. Actually, I had read in a meta post that is saying,you should bold when it may need to be attracted or, something just like this. I don't remember. That's why I was just making that bold
– user876873
May 8 at 13:48
• @Istiak I see, cool. Thanks, though, for your edit that means you now only bold it in the question.
– J.G.
May 8 at 13:52

How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?

is that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time.

I would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $$(3x^2+2x^3)^{-1}$$ then the product rule would be more appropriate.

the quotient rule is not a separate and non-compatible rule. It is just the product rule inserting $$g^{-1}$$ instead of $$g$$.

let's see...

to make it clear... I show you, prove it for you, how quotient rule is just compatible with product law.

$$\left(\frac f h\right)'=\left(f\cdot \frac 1 h\right)'=f' \cdot \frac 1 h +f\cdot \left (\frac 1 h\right)'$$

Before continuing I want to recall that, Since $$h\cdot (1/h)=1$$, by product rule, we ould get $$0=1'=(h\cdot \frac 1 h )'=h' \cdot \frac 1 h + h\cdot (1/ h)'$$. Therefore we have $$h\cdot ( 1/ h )'=-h'\cdot ( 1/ h)$$.

So, $$\left(1 /h\right)'=-h'/h^2$$.

and on another hand, we know $$\frac 1 h$$, could be written as $$\frac h{h^2}$$, for a reason.

so,

$$\left(\frac f h\right)'=f' \cdot \frac 1 h +f\cdot \left (\frac 1 h\right)'=f'\cdot \frac h {h^2}+f\cdot \frac{-h'}{h^2}=\frac{f'\cdot h-f\cdot h'}{h^2}$$

that was the proof for the quotient rule.

• $g^{-1}$ is a poor choice of notation here, as it usually means the inverse of $g$, and not the reciprocal. May 9 at 0:19


# Math Help - Problem finding area

1. ## Problem finding area

i am having a problem finding the area of the three triangles outside the Pitagoras drawing
you can see what i am meaning in this picture
http://i51.tinypic.com/28gxmyd.jpg
so i need the area of all the triangles in the middle of the squares

2. ## Re: Problem finding area

Hello, leart369!

http://i51.tinypic.com/28gxmyd.jpg
i need the area of all the triangles.

Label the inner right triangle like this:

Code:
      A
*
|   *     c
b |       *
|           *
* - - - - - - - * B
C         a

Note that for any triangle, its area is: . $A \;=\;\tfrac{1}{2}ab\sin C$
. . (One-half the product of two sides and the sine of the included angle)

The right triangle at the lower-left has sides $a$ and $b$.
. . Its area is: . $\boxed{A_1 \:=\:\tfrac{1}{2}ab}$

The triangle at the right has sides $a$ and $c$, and included angle $(180^o - B).$
. . Its area is: . $A_2 \:=\:\tfrac{1}{2}ac\sin(180^o-B) \:=\:\tfrac{1}{2}ac\sin B$
In the above diagram, we see that: . $\sin B \,=\,\tfrac{b}{c}$
. . Hence: . $A_2 \;=\;\tfrac{1}{2}ac\left(\tfrac{b}{c}\right) \quad\Rightarrow\quad \boxed{A_2 \:=\:\tfrac{1}{2}ab}$

The third triangle has sides $b$ and $c$, and included angle $(180^o - A).$
. . Its area is: . $A_3 \:=\:\tfrac{1}{2}bc\sin(180^o-A) \:=\:\tfrac{1}{2}bc\sin A$
In the above diagram, we see that: . $\sin A \,=\,\tfrac{a}{c}$
. . Hence: . $A_3 \;=\;\tfrac{1}{2}bc\left(\tfrac{a}{c}\right) \quad\Rightarrow\quad \boxed{A_3 \:=\:\tfrac{1}{2}ab}$

Fascinating! . . . All four triangles have the same area.

3. ## Re: Problem finding area

does exist any way to find the area off all that drawing?

4. ## Re: Problem finding area

Originally Posted by leart369
does exist any way to find the area off all that drawing?
1. I assume that you mean the area of the hexagon, containing 3 squares, 2 right triangles and 2 obtuse triangles. If so:

2. As Soroban has shown you have:

$\text{complete area} = a^2+b^2+c^2+4 \cdot \frac12 \cdot a \cdot b = c^2 +(a+b)^2$

3. Since $a^2+b^2 = c^2$ there are other possibilities to simplify the sum of areas.

5. ## Re: Problem finding area

thank you very much that was what i was looking for


# Pascal’s Triangle

In mathematics, Pascal’s triangle is a triangular array of the binomial coefficients. It can also be viewed as: each number in Pascal’s triangle is the sum of the two numbers directly above it as shown:

          1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Problem statement

Given an integer n as input, print first n lines of the Pascal’s triangle.

#### Approach

Let's consider Pascal's triangle as a matrix pascal[i][j] in the following way:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

We could solve this problem using Dynamic Programming.

Optimal substructure

From the above matrix we could see that:

pascal[i][j] is the sum of previous element in the row i-1 and element just above the current element in the row i-1. Consider out of bound indices as 0
ie pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j]

#### Code Implementation

//
//  main.cpp
//  Pascal Triangle
//
//  Created by Himanshu on 20/09/21.
//

#include <iostream>
using namespace std;

void printPascalTriangle (int n) {
int pascal[n+1][n+1];

//Base case
pascal[1][1] = 1;

for (int i=0; i<=n; i++) {
for (int j=0; j<=n; j++) {
pascal[i][j] = 0;
}
}

for (int i=1; i<=n; i++) {
for (int j=1; j<=i; j++) {
// first and last binomial coefficients are
// always 1
if (i == 1 || j == i) {
pascal[i][j] = 1;
} else {
pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j];
}
}
}

for (int i=1; i<=n; i++) {
for (int j=1; j<=i; j++) {
cout<<pascal[i][j]<<" ";
}
cout<<endl;
}

}

int main() {
int n = 7;
printPascalTriangle (n);

return 0;
}

Output

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

Time Complexity: O(n^2)
Auxiliary Space: O(n^2)

#### Approach with O(1) auxiliary space

By using the definition of binomial coefficient: we know that

C(n, i) = (n!)/((n-i)! * i!)
similarly,
C(n, i-1) = (n!)/((n-i+1)! * (i-1)!)

Now, C(n, i) could also be written as:
(n!)*(n-i+1)/(n-i+1)!*(i)*(i-1!)
which is nothing but

(n!)*(n-i+1)/((n-i+1)!*(i-1)!*(i)) ie,
C(n, i) =  C(n, i-1)*(n-(i-1))/(i)

Using the above equation, we just need the previous binomial coefficient C(n, i-1) to calculate C(n, i). Hence we need not save all the binomial coefficients to print current coefficient.

#### Code Implementation

//
//  main.cpp
//  Pascal Triangle 2
//
//  Created by Himanshu on 20/09/21.
//

#include <iostream>
using namespace std;

void printPascalTriangle (int n) {

for (int i=1; i<=n; i++) {
int C = 1;
for (int j=1; j<=i; j++) {

cout<<C<<" ";
// We are using (i-j) instead of (i-(j-1)) because
// calculation is for next or (j+1)th element
C = C * (i-j)/ j;
}
cout<<endl;
}

}

int main() {
int n = 7;
printPascalTriangle (n);
return 0;
}

Output

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

Time Complexity: O(n^2)
Auxiliary Space: O(1)

Here’s a working example: Pascal’s Triangle

Practice Problem
Pascal’s Triangle [LeetCode]


# Approximating $\log_{10}7$

The Moderators have been talking about methods to approximate logarithms given only a little bit of information; namely, $\log_{10}2\approx0.301$ and $\log_{10}3\approx0.477.$ I'm going to go through a couple methods to provide estimates.

It should be noted that the true value of $\log_{10}7$ is $0.845098...$

Let's start off on a large scale. Obviously, $\log_{10}6<\log_{10}7<\log_{10}8.$ To find $\log_{10}6$ and $\log_{10}8,$ we use the given information. $\log_{10}6=\log_{10}2+\log_{10}3\approx0.301+0.477=0.778.$ Also, $\log_{10}8=3\log_{10}2\approx3\times0.301=0.903.$ So $0.778<\log_{10}7<0.903.$ Not a very good estimate. Even if we take the average of these, we get $0.8405,$ which is only accurate to two decimal places.

Let's be a little more creative in our thinking. The set of numbers $\{48,49,50\}$ includes numbers whose logarithms can be expressed with our given information and a power of $7.$ Once again, $\log_{10}48<\log_{10}49\log_{10}50\Rightarrow$ $\log_{10}48<2\log_{10}7<\log_{10}50.$ The only difference here is that $50$ has a power of $5$ in its factorization, but $\log_{10}5$ can easily be eliminated by noticing that $\log_{10}50=\log_{10}\frac{100}{2}=\log_{10}100-\log_{10}2=2-\log_{10}2.$

$\log_{10}48=4\log_{10}2+\log_{10}3\approx1.681.$ Also, $\log_{10}50=2-\log_{10}2\approx1.699.$ Dividing by $2,$ we find that $0.8405<\log_{10}7<0.8495.$ This is better. Not to mention, taking the average of these values yields $\log_{10}7\approx0.8450,$ accurate to $4$ decimal places.

What method would you use to calculate $\log_{10}7?$ Please share what you think!

Note by Trevor B.
5 years, 1 month ago

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Sort by:

\begin{aligned} 2400 & \approx & 2401 \\ \log 2400 & \approx & \log 2401 \\ \log (2^3 \times 3 \times 10^2) & \approx & \log 7^4 \\ \log 2^3 + \log 3 + \log 10^2 & \approx & 4 \log 7 \\ 3 \log 2 + \log 3 + 2 & \approx & 4 \log 7 \\ \log7 & \approx & 0.8450 \\ \end{aligned}

- 5 years, 1 month ago

So, that begs the question of
1. How do I make such an amazing observation that $2400 \approx 2401$?
2. How do I know what the next approximation should be?

Staff - 5 years, 1 month ago

1. I just found numbers that are close to powers of $7$ that are can be factored in terms of $2,3,5,7$

2. I don't have a solid answer, but I'm thinking we should find larger pairs of numbers such that their percentage difference is as small as the previous, like: $6^5 \approx 7777 = \frac {7}{9}(10^4 - 1) \approx \frac {7}{9} \times 10^4$ or $6^7 \approx 280000 = 10^4 \times 2^2 \times 7$, but both methods yields $0.844$ as the approximation, so I'm kinda stumped right now.

In the meantime, lemme peruse your other comment. =)

- 5 years, 1 month ago

There's a 0.04% error in approximating 2401 as 2400, but works out quite well :) +1

- 5 years, 1 month ago

@Calvin Lin and @Daniel Liu were the moderators discussing this in our messageboard.

- 5 years, 1 month ago

Cool!! So you guys have a message board of your own?

- 5 years, 1 month ago

The general idea is that if you can bound $a < 7 ^ n < b$, then you should also look at how $ab$ compares to $7^{2n}$. This is akin to the Root approximation - bisection method.

For example, from above, we have $6 < 7 < 8$, which tells us we should compare $48$ with $49$, and work with $48 < 49 < 8^2$.

Explicitly, we can use the following series of inequalities:
Step 1: $6 < 7 < 8$. Comparing $6 \times 8$ with $7^2$, we have $48 < 7^2$. Hence, this leads to:
Step 2: $48 < 7^2 < 8^2 = 64$. Comparing $48 \times 64$ with $7^4$, we have $7^4 < 3072$. Hence, this leads to:
Step 3: $2304 = 48^2 < 7^4 < 3072$. Comparing $2304 \times 3072$ with $7 ^ 8$, we have $7^ 8 < 7077888$. Hence this leads to:
Step 4: $5308416 = 2304^2 < 7^8 < 7077888$.

We can continue this process indefinitely to bound $\log 7$ as tightly as we want to. At step 1, we have $0.778 < \log 7 < 0.903$, and at step 4 we have $0. 840 < \log 7 < 0.856$.

Question: Why did I say that this is "akin to the Root approximation - bisection method"?
(see comments below for the explanation)

Note: As it turns out, we could make the observation that 50 a much better bound. With $48 < 49 < 50$, and so we should compare $48 \times 50$ with $49^2$.

Staff - 5 years, 1 month ago

Because you want to find values $a, b$ such that the inequality is satisfied $a < 7^n < b$

Taking log to both sides $\log a < n \log 7 < \log b$

Bisection method implies we must minimize the difference between $\frac { \log a + \log b }{2}$ and $n \log 7$.

Set both of them to be equal implies we want to compare $ab$ to $7^{2n}$

- 5 years, 1 month ago

Yes. We are performing the bisection method on $\log a < n \log 7 < \log b$. The midpoint of this interval is $\frac{ \log a + \log b } { 2 } = \log \sqrt{ ab }$. We then compare this to $n \log 7$, which is in essence comparing $ab$ to $7^{2n}$.

In this way, we don't have to think about what the next number is, or figure out a fancy combination to bound it tightly. We just have to do the bisection method many many times (or automate it with a program), to get the interval down to as small as we want it to.

Staff - 5 years, 1 month ago

"It is independent of the number of sig figs / decimal places that we started off with."

I was under the impression, you can only find up to the number of SF given. But this just blew my mind. It's like saying:

"Hey, my calculator is kinda bad, it only let shows that $\log 2 = 0.3$ and $\log 13 = 1.1$ to one decimal place, can you help me find the value of $\log 11$ to a gazillion decimal places?"

Thanks, I learned something new today!!

- 5 years, 1 month ago

Indeed. See my conversation with Kenny Lau below. It is a common misconception (likely from Physics error analysis) that your calculations cannot be more accurate than the degree of significance that you started out with.

For example, even if your calculator can only multiply whole numbers, and you want to approximate $\sqrt{2}$, you can do much better than $1 < \sqrt{2} < 2$. For example, by showing that $1414^2 < 2000000 < 1415^2$, we can conclude that $1.414 < \sqrt{2} < 1.415$. We can get $\sqrt{2}$ to any degree of accuracy (with sufficient patience / computing power).

Staff - 5 years, 1 month ago

543543

- 5 years, 1 month ago

Well, any method based on the two given information can only be correct to 3 significant figures (or decimal places), because the information itself is correct to only 3 significant figures.

- 5 years, 1 month ago

Not necessarily. For example, if we know $\log 7 ^ {10}$ to 3 decimal places, then we can get $\log 7$ to 4 decimal places.

Staff - 5 years, 1 month ago

It would not be accurate to 3 decimal places. Log 7^10 would only be accurate to 3 significant figures which is 2 decimal places.

- 5 years, 1 month ago

I've added a way to show how we can get an arbitrarily tight bound for $\log 7$. It is independent of the number of sig figs / decimal places that we started off with.

Staff - 5 years, 1 month ago


# A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond ?

A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond?

To me this seems like we can set up a fraction-like equation:

$$\frac{60 \ \text{seconds}}{1} = \frac{x \ \text{seconds}}{1/4}$$ then $x = 15$ seconds. But the answer is $58$ seconds which really makes no sense to me. Any suggestions are greatly appreciated.

• Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen Sep 18 '17 at 17:31

I think it's easiest to work backwards: if the area doubles every second and the pond is totally covered at time $t=60$, then it must be half covered at $t=59$, and therefore one quarter covered at $t=58$.

Alternately, let $f(t)$ be the fraction of the pond's area covered at time $t\leq 60$. Then $f(t)=f(0)2^t$ since the area doubles every second, and since $f(60)=1$ we get $f(0)=2^{-60}$. Therefore $f(t)=2^{-60}2^t=2^{t-60}$. Then setting $2^{t-60}=\frac{1}{4}$ and solving for $t$ yields $t=58$.

• shouldnt $f(0)=2^{t-60}$. I think you may be missing a t – Derek Sep 16 '17 at 3:03
• @Derek: No, $f(0)=2^{-60}$. $f(0)$ is a number rather than a function of $t$. – carmichael561 Sep 16 '17 at 3:06
• @Derek $f(0)=2^{t-60} \text{ at } t=0$ so then $f(0)=2^{0-60}=2^{-60}$ – Wolfie Sep 18 '17 at 6:40

Forget formulas for this one!

If going forward 1 second the area gets doubled, then going back 1 second the area gets halved.

So, 1 second before the pond was filled the pond must have been half filled, and 1 second before that it must have been quarter filled.

• Yes. Use formulas if they help you to get the answer. If you can get the answer easily without formulas, great! – Wildcard Sep 15 '17 at 1:25
• And 5s before that, we can hear economists say : "See, an exponential growth is perfectly possible in a finite world!". – Eric Duminil Sep 16 '17 at 8:21
• If there was one complaint about my pre-college math education it's that there was such an emphasis on formulas (which was great when it came to calculus, for sure) that there was so little critical thinking. It was "which of the four formulas of this chapter are we supposed to apply to this problem?" instead of thinking about the problem first. – corsiKa Sep 16 '17 at 15:18

This is exponential rather than linear. If $A$ is the initially covered area, then after one second the covered area will be $2A$, after two second $2\cdot 2A=4A$, after three seconds $2\cdot 4A=8A$. And so on: after $t$ seconds the covered area will be $2^tA$.

After $60$ seconds it will be $2^{60}A$, by assumption this is the whole pond. A quarter of this is $$\frac{2^{60}A}{4}=2^{58}A$$

Of course Carmichael’s answer is slicker.

• While other answers just solve the problem, your also points (although briefly) at the flaw in OP's reasoning. Thanks for that. – pajonk Sep 15 '17 at 5:51

Your 'fraction-like equation' has nothing to do with the problem, because the lily pad doubles every second – its growth is exponential, not linear.

$$Area(t) = 2\cdot Area(t-1)$$ where $t$ is a number of seconds since 'some moment', hence $$Area(t) = \color{red}{2^t}\cdot Area(0)$$ where zero is an arbitrary 'some moment'.

That implies $$\frac{Area(60)}{Area(t)} = \frac{2^{60}}{2^t} = 2^{60-t}$$

Then if they ask at what $t$ is $$Area(t)=1/4\cdot Area(60)$$ you have $$2^{60-t} = \frac{Area(60)}{1/4\cdot Area(60)} = 4=2^2$$ so $$60-t = 2$$ and finally $$t=60-2 = 58$$ – the pond is quater-filled at $58$ seconds.

Take a look at the table below: \begin{array}{|c|c|}\hline \text{Area } (A) & 1 & \dfrac12 & \dfrac14 & \dfrac18 & \cdots\\ \hline \text{Time } (t) & 60 & 59 & 58 & 57 & \cdots\\ \hline \end{array} As you might see, you could easily deduce the relationship between $A$ and $t$, i.e., $A(t)=\dfrac{1}{2^{60-t}}$.

Chronological reasoning might help.

• At the beginning or $t=0$, its area is $A(0)$.
• After 1 second elapses or $t=1$, its area becomes $A(1)=2\times A(0)$.
• The next 1 second or $t=2$, its area becomes $A(2)=2 \times A(1) = 2\times2\times A(0)=2^2 A(0)$.

• At $t=60$ the area is $A(60)=2^{60} A(0)$. The pond is fully filled.

• The quarter of the fully filled pond is $A(60)/4 = 2^{60}A(0)/4=2^{58}A(0)$. This area is equal to $A(58)$, so $t=58$.

We want a function relating the current time to the size of the lily pad; let's call it $f$. From the problem, we know that

$f(60\ \mathrm{seconds})=1\ \mathrm{pond}$
$f(x\ \mathrm{seconds})=1/4\ \mathrm{pond}$

so we can indeed write an equation like the one you want by solving for $1\ \mathrm{pond}$ in each equation; then:

$\frac{f(60\ \mathrm{seconds})}1=\frac{f(x\ \mathrm{seconds})}{1/4}$

But note the mediating $f$ that you are missing in your equation! If we assumed $f(x)=x$, then we would get your equation; but the problem also tells us that the lily pad doubles in size every second, that is, that:

$f((x+1)\ \mathrm{seconds}) = 2f(x\ \mathrm{seconds})$

If we choose $f(x)=x$, then this equality is not validated, since $(x+1)\ \mathrm{seconds}=2x\ \mathrm{seconds}$ is not validated.

Luckily we can make progress even without assuming $f(x)=x$. Simplifying the corrected version of your equation, we have:

$f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$

Now we can apply the other equation given in the problem twice:

$f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$
$\phantom{f(60\ \mathrm{seconds})}=2(2f(x\ \mathrm{seconds}))$
$\phantom{f(60\ \mathrm{seconds})}=2f((x+1)\ \mathrm{seconds})$
$\phantom{f(60\ \mathrm{seconds})}=f((x+2)\ \mathrm{seconds})$

Then we can conclude that $60=x+2$ would be sufficient to validate this equation, so $x=58$ is one possible solution.

The area of a lily pod can be described by a function:

$$f:[0,60]\to[0,1]\\ f(x)=2^{x-60}$$

Now we have to compute the argument $x$ for which the value of the function $f(x)$ is equal to $\frac{1}{4}$:

$$\frac{1}{4}=2^{x-60}\\ \log_2 \frac{1}{4}=\log_2 2^{x-60}\\ -2 = x-60\\ x=58$$

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