# Moderate Averages Solved QuestionAptitude Discussion

 Q. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?
 ✔ A. 57 ✖ B. 56.8 ✖ C. 58.2 ✖ D. 52.2

Solution:
Option(A) is correct

Let the average weight of the 59 students be $A$.

Therefore, the total weight of the 59 of them will be 59 $A$.

The questions states that when the weight of this student who left is added, the total weight of the class

= $59A + 45$

When this student is also included, the average weight decreases by 0.2 kgs.

$59A + \dfrac{45}{60} = A - 0.2$

$\Rightarrow 59A + 45 = 60A - 12$

$\Rightarrow 45 + 12 = 60A - 59A$

$\Rightarrow A = \textbf{57}$

Edit: For an alternative solution, check comment by Durga.

## (9) Comment(s)

Siddharth Sharma
()

60*0.2+45=57 thats the shortest method to do it.

Shivakumar
()

divided by 60 for 59A+45 not only for 45

Vijay
()

simply,

60*(x-.20)-59x=45;

=>x=45+12=57.

Sanajit Ghosh
()

@DURGA's ans is right

Zoha Amjad
()

answer is 56.8

0.2 kgs increase means

$0.2*59=11.8$

$11.8+45 = 56.8$

Shanu
()

The ans should be 56.8

it can be solved as

$60A-((A-0.2)*59)=45$

so $A(\text{avarage}) = 56.8$

Kiruthi
()

ans should be 56.8 because in question they have given increased by 200g but while solving they took it as decreased by 200g so 57

Durga
()

Hi ,

The answer is 57 only.

Let say the average of 60 students is $x$.

Total weight $=60*x$

If one man left average should increase 200gr

i.e $\dfrac{200}{1000} =0.2 \text{ kg}$

So, $60 \times x-45=59(x+0.2)$

$60x-59x=59*0.2+45$

$\Rightarrow x=56.8 \text{ kg}$

But here we need to calculate the average of 59 students

i.e $56.8+0.2= \textbf{57 kg}$

Elmira
()

Thank you for creating a very informative website, much appreciate it.

Please review the solution once again, when multiplying the sides by 60, 59A was left without effect, therefore I think the response is wrong. Oh maybe I am missing something here.

Please explain.

Elmira


# Find the matrix representation of a linear map

Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix} 4 & 6\\ 6 & 3 \end{pmatrix}$$ with respect to the basis $B=\bigg\{ \begin{pmatrix} 2 \\ 2 \end{pmatrix} , \begin{pmatrix} 3\\ 2 \end{pmatrix}\bigg\}$. Find the matrix representation of $f$ with respect to the canonical basis.

MY ATTEMPT: Let $b_1=\begin{pmatrix} 2 \\ 2 \end{pmatrix}$, $b_2=\begin{pmatrix} 3\\ 2 \end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis. Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix} \Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$

$f(b_2)=\begin{pmatrix} 6\\ 3 \end{pmatrix} \Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$

Solving this system I obtain that $f(e_1)=2e_1 -3e_2=\begin{pmatrix} 2\\ -3 \end{pmatrix}$ and $f(e_2)= 6e_2=\begin{pmatrix} 0\\ 6 \end{pmatrix}$. So the matrix rappresentation is $M_{E}^{E}=\begin{pmatrix} 2 & 0\\ -3 & 6 \end{pmatrix}$.

BUT my theacher says that the solution is $M_{E}^{E}=\begin{pmatrix} 13 & 10\\ 0 & 0 \end{pmatrix}$. Where I am wrong?

$$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$

This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.

• How can I do that? – fcoulomb Jun 1 '18 at 18:47
• @Besh00 write $\color{red}{=4b_1+6b_2}$ instead of the red part, and then write both $b_1$ and $b_2$ in terms of $e_1$, $e_2$. – Arnaud Mortier Jun 1 '18 at 18:50
• Why $f(b_1)=4b_1+6b_2$? – fcoulomb Jun 1 '18 at 21:10
• You wrote yourself "Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix}$". These coordinates are with respect to the basis $B$ since they are taken from a matrix wrt $B,B$. – Arnaud Mortier Jun 1 '18 at 21:14


# How many $4$-digit numbers that do not have $5$ and have $7$ in the hundreds position?

I encountered this problem:

How many $4$-digit numbers are there that do not have $5$ and have $7$ in the hundreds position? Digits cannot be repeated.

My thinking: first, let's count numbers that do not have $0$. It's going to be $7\cdot 6 \cdot 5$ arrangements. Now, let's count numbers that have $0$ in the tens position. We have two positions open so there are $7 \cdot 6$ arrangements. We get the same number of arrangements for numbers that end with $0$. The total number of arrangements is $7\cdot 6 \cdot 5+ 2\cdot 7 \cdot 6=7\cdot 7 \cdot 6=294$.

However, my friend argues that there are $8$ choices for the last digit, $7$ choices for the tens digits and $6$ choices for the first digit so the total number should be $8\cdot 7 \cdot 6=336$.

I am asking who is right not to prove my friend wrong but to find the truth. Thanks for listening!

294 is correct.

First digit (most significant), can not be 7,0,5, so 7 choices; 3rd digit, cannot be 7,5, and the one chosen for 1st, so 7 choices; 4th digit, 6 choices;

$7\times7\times6=294$

• Welcome to the site. Great first answer! – B. Mehta Apr 14 '18 at 23:17

You are right and your friend is wrong. If you choose the units digit and then choose the tens digit (and are forced to use $7$ for the hundreds digit) then the number of choices for the thousands digit not always $6$: it is either $5$ or $6$, depending on if you've used $0$ yet or not.

For example, there are $5$ numbers satisfying the condition which have the form $x789$ (since $x$ can be any of $1,2,3,4,6$) but $6$ numbers satisfying the condition which have the form $x780$ (since $x$ can be any of $1,2,3,4,6,9$).

All this tells us is that the number of solutions is between $8\cdot7\cdot5 = 280$ and $8\cdot7\cdot6 = 336$, and for a more precise answer we do need to look at whether $0$ gets used.

I think the key disagreement here is that your friend is thinking that $0$ is allowed to be the first digit (say, for a 4 digit passcode), while you are not. If $0$ can be the first digit, your friend is correct, but if not (as other posters explain), you are correct.

You don't have to choice the positions in order so don't worry about $0$s.

Do the Thousandth place first. It can not be $0$ and it can not be $7$ or $5$. So there are $7$ options. Then do either the tens or the ones, it doesn't matter which. It can be $0$ but it can't be $7$ or $5$ and it can't be what the thousandth place was. That is $7$ options as well. And the final position can't be $7$ or $5$ or either of the previous two positions. so that is $6$ options.

So there are $7*7*6 = 294$.

You were correct but you didn't have to work so hard.

Your freind is wrong in that $0$ is not an option for the first digit.


# Telescoping Series Partial Fractions

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How to get the partial fractions of higher degree numerators. An in nite series can be represented as such: P 1 n=1 a n. The Sigma notation for summation of series. 3 The Integral Test: The section includes the integral test and a discussion on the remainder estimate for the integral test. Review for Test 3 Math 1552, Integral Calculus The series is telescoping, so it converges and we can ﬁnd its sum using partial fractions: ". Say we have something we want to sum up, let's call it a k. Be sure to review the Telescoping Series page before continuing forward. Partial Fractions. For the telescoping series X1 k=3 2 (2k 1)(2k + 1); nd a formula for the nth term of the sequence of partial sums fS ng. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. (1) and ask whether the sum is convergent. ~~~~~ AP Cal BC Students: Telescoping Series and Partial Fractions Tests. A telescoping series (or telescoping sum) is one that "expands" in such a way that most of its terms cancel away. ) (Now evaluate the limit. 2: p 2] EXAMPLES Does the series converge or diverge? 1. Definitions: Let{ } 1 n n a. Retrieved from " https: The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series terms. Alternating Series Test cannot be applied  26. In nite Series SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. TELESCOPING SERIES Now let us investigate the telescoping series. ) = 1 - 0 = 1. So: s1=a1 s2=a1+a2 s3=a1+a2+a3 s4=a1+a2+a3+a4. This is a challenging sub-section of algebra that requires the solver to look for patterns in a series of fractions and use lots of logical thinking. Improper integrals of type 2. This time there will be a few more terms that do not cancel. Lecture 17: Series (II) Telescoping Series A telescoping series is a special type of series for which many terms cancel in the nth partial sums. (The first Maplet may take a little longer to open because it needs to start Java. Back to Thomas O'Sullivan's Homepage. Creating the telescoping effect frequently involves a partial fraction decomposition. Telescoping series. 3 Telescoping Series. You - ProProfs Discuss. We will now look at some more examples of evaluating telescoping series. If you are looking for more in partial fractions, do check in: Partial fractions of lower degree numerators. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Partial fractions Fractions in which the denominator has a quadratic term Sometimes we come across fractions in which the denominator has a quadratic term which. Be able to determine the nth partial sum of any geometric series. Partial Fractions Stewart Chapter 7. Test for convergence of a series using an appropriate test: divergence, integral, comparison and limit comparison, ratio, root, or alternating series. High School AP Calculus BC Curriculum. For telescoping series, intermediate terms 'drop. For example, using partial fractions and cancelling a bunch of terms, we find that; An infinite series that arises from Parseval’s theorem in Fourier analysis. Most telescopic series problems involve using the partial fraction decomposition before expanding it and seeing terms cancel out, so make sure you know that very well before tackling these questions. Calculus 2 Syllabus. Note that the fundamental concepts of functions, graphs, and limits, which are studied at the beginning of courses in differential calculus, are often first introduced in earlier classes (most notably intermediate algebra and precalculus). Also note that it is possible to tell that this last series. Inﬁnite series 1: Geometric and telescoping series Main ideas. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. 26, 29, 39, 40 21 – 42 Solve applications involving series. Exercise 6. 3) Cancel duplicates. If you do not own Maple, click Use MapleNET 12 at the top-right corner of this page. Note: is a telescoping series. Thus, as $$k$$ increases, the partial sum $$S_k$$ increases (the series is a sum of positive terms), but is always smaller than \(2\text{. Find the sum of the series • Â k=0 4 k2 +3k+2 if it exists. That is, s k → -∞. Hence, Therefore, by the definition of convergence for infinite series, the above telescopic series converges and is equal to 1. Find the sum of the series • Â k=1 1 k 1 k+2 if it exists. With geometric series, we carried out the entire evaluation process by finding a formu for the sequence of partial sums and evaluating the limit of the sequence. Infinite Series Chapter 1: Sequences and series Section 4: Telescoping series Page 2 We can use the method of partial fractions to rewrite the general term as: 4 1 1 2 2 2 2 ab n n n n n n §· ¨¸ ©¹ For your practice, check that this decomposition is correct! Now we can write any partial sum of this series as: 11 4 1 1 2 22 kk nn n n n n. This video is a great one on learning about evaluating fractions. Telescoping: Transform by partial fractions Procedure for Determining Convergence No Series Diverges nth-Term Test Yes or maybe Yes No Yes Nonnegative terms and/or absolute convergence No Yes No Can the Integral Test, the Ratio Test, or the Root Test be applied? Does the Integral Test apply? Yes Does the Ratio Test apply? No Yes Is Is Does the. (a) X1 n=3 1 2n 1 1 2n 4 As I stated in my email follow-up to this problem, this is not actually an easy series to nd a. Now pop in the first term (a 1) and the common ratio (r). The telescoping Series is a method for examining the convervence of infinite series of the form: This method, combined with partial fraction decomposition, is frequently effective. Be able to determine the nth partial sum of any geometric series. (a) X1 n=1 cos 1 n cos 1 n+ 1. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as: #3. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. According to Ferraro [4], Mengoli did so by making frequent use of the following. — Try Before you Buy To start a Maplet, click on its name. if integral of series reaches ∞ or DNE then series diverges. Telescoping Series and Partial Fractions. In this case, the series is convergent and the sum isS lim n→ sn b1 −lim n→ bn 1. See if you can ﬁgure it out. Welcome to Maplets for Calculus. 4 Comparison Tests. Can the series be compared favorably to one of the special types?. EXAMPLE 8:. 50 46 – 53 Evaluate telescoping series. telescoping series to ﬁnd out. This is best demonstrated with an example. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. Now let us investigate the telescoping series. (a)Sequences and Series, partial sums. Example: Determine whether the given series converge. Use comparison test if numerator fluctuates between two constants. From Wikipedia, the free encyclopedia.


# Introduction

Although we have functions that can compute our $\gcd$easily it's important enough that we need to give and study an algorithm for it: the euclidean algorithm.

It's extended version will help us calculate modular inverses which we will define a bit later.

# Euclidean Algorithm

Important Remark

If $a = b \cdot q + r$and $d = \gcd(a, b)$ then $d | r$. Therefore $\gcd(a, b) = \gcd(b, r)$

## Algorithm

We write the following:

$a = q_0 \cdot b + r_0 \\ b = q_1 \cdot r_0 + r_1 \\ r_0 = q_2 \cdot r_1 + r_2 \\ \vdots \\ r_{n-2} = r_ {n-1} \cdot q_{n - 1} + r_n \\ r_n = 0$

Or iteratively $r_{k-2} = q_k \cdot r_{k-1} + r_k$ until we find a $0$. Then we stop

Now here's the trick:

$\gcd(a, b) = gcd(b, r_0) = gcd(r_0, r_1) = \dots = \gcd(r_{n-2}, r_{n-1}) = r_{n-1} = d$

If $d = \gcd(a, b)$ then $d$ divides $r_0, r_1, ... r_{n-1}$

Pause and ponder. Make you you understand why that works.

Example:

Calculate $\gcd(24, 15)$

$24 = 1 \cdot 15 + 9 \\ 15 = 1 \cdot 9 + 6 \\ 9 = 1 \cdot 6 + 3 \\ 6 = 2 \cdot 3 + 0 \Rightarrow 3 = \gcd(24, 15)$

Code

def my_gcd(a, b):    # If a < b swap them    if a < b:         a, b = b, a    # If we encounter 0 return a    if b == 0:         return a    else:        r = a % b        return my_gcd(b, r)print(my_gcd(24, 15))# 3

Exercises:

1. Pick 2 numbers and calculate their $\gcd$by hand.

2. Implement the algorithm in Python / Sage and play with it. Do not copy paste the code

# Extended Euclidean Algorithm

This section needs to be expanded a bit.

Bezout's identity

Let $d = \gcd(a, b)$. Then there exists $u, v$ such that $au + bv = d$

The extended euclidean algorithm aims to find $d = \gcd(a, b), \text{ and }u, v$given $a, b$

# In sage we have the xgcd functiona = 24b = 15g, u, v = xgcd(a, b)print(g, u, v)# 3 2 -3 print(u * a + v * b)# 3 -> because 24 * 2 - 15 * 3 = 48 - 45 = 3


# 3.1 Functions and function notation  (Page 4/21)

 Page 4 / 21
 $n$ 1 2 3 4 5 $Q$ 8 6 7 6 8

[link] displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.

 Age in years, (input) 5 5 6 7 8 9 10 Height in inches, (output) 40 42 44 47 50 52 54

Given a table of input and output values, determine whether the table represents a function.

1. Identify the input and output values.
2. Check to see if each input value is paired with only one output value. If so, the table represents a function.

## Identifying tables that represent functions

Input Output
2 1
5 3
8 6
Input Output
–3 5
0 1
4 5
Input Output
1 0
5 2
5 4

[link] and [link] define functions. In both, each input value corresponds to exactly one output value. [link] does not define a function because the input value of 5 corresponds to two different output values.

When a table represents a function, corresponding input and output values can also be specified using function notation.

The function represented by [link] can be represented by writing

Similarly, the statements

represent the function in [link] .

[link] cannot be expressed in a similar way because it does not represent a function.

Input Output
1 10
2 100
3 1000

yes

## Finding input and output values of a function

When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.

When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.

## Evaluation of functions in algebraic forms

When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function $\text{\hspace{0.17em}}f\left(x\right)=5-3{x}^{2}\text{\hspace{0.17em}}$ can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.

Given the formula for a function, evaluate.

1. Replace the input variable in the formula with the value provided.
2. Calculate the result.

## Evaluating functions at specific values

Evaluate $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}+3x-4\text{\hspace{0.17em}}$ at

1. $2$
2. $a$
3. $a+h$
4. $\frac{f\left(a+h\right)-f\left(a\right)}{h}$

Replace the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the function with each specified value.

1. Because the input value is a number, 2, we can use simple algebra to simplify.
$\begin{array}{ccc}\hfill f\left(2\right)& =& {2}^{2}+3\left(2\right)-4\\ & =& 4+6-4\hfill \\ & =& 6\hfill \end{array}$
2. In this case, the input value is a letter so we cannot simplify the answer any further.
$f\left(a\right)={a}^{2}+3a-4$
3. With an input value of $\text{\hspace{0.17em}}a+h,\text{\hspace{0.17em}}$ we must use the distributive property.
$\begin{array}{ccc}\hfill f\left(a+h\right)& =& {\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ & =& {a}^{2}+2ah+{h}^{2}+3a+3h-4\hfill \end{array}$
4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that
$f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h-4$

and we know that

$f\left(a\right)={a}^{2}+3a-4$

Now we combine the results and simplify.

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5


# Fibonacci identity: $f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$

Consider this Fibonacci equation: $$f_{n+1}^2 - f_nf_{n+2}$$

The problem asked to write a program with given n, output the the result of this equation. I could use the formula $$f_n = \frac{(1+\sqrt{5})^n - ( 1 - \sqrt{5} )^n}{2^n\sqrt{5}}$$

However, from mathworld, I found this formula Cassini's identity $$f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$$

So, I decided to play around with the equation above, and I have:

$$\text{Let } x = n + 1$$ $$\text{then the equation above becomes } f_x^2 - f_{x-1}f_{x+1}$$ $$\Rightarrow -( f_{x-1}f_{x+1} - f_x^2 ) = -1(-1)^x = (-1)^{x+1} = (-1)^{n+1+1} = (-1)^{n+2}$$

So this equation either is 1 or -1. Am I in the right track?

Thanks,
Chan

-
Yes. $\mbox{}$ –  Mariano Suárez-Alvarez Feb 8 '11 at 5:44
@Mariano Suárez-Alvarez: Thanks ^_^! Happy! –  Chan Feb 8 '11 at 5:47
In your first expression, should the subscript on the square be "n+1" instead of "n-1"? If so, then you are on the right track. (Also, you'll want parentheses around your "-1"s.) –  Blue Feb 8 '11 at 5:47
Note: You should write $(-1)^{n+2}$, and not $-1^{n+2}$; exponentiation takes precedence, so the former is either $-1$ or $1$ depending on the parity of $n$, but the latter is $-1^{n+2} = -(1^{n+2}) = -1$ always. –  Arturo Magidin Feb 8 '11 at 5:50
@Day Late Don: nice catch. It was my typo. I was practicing using Latex:(. –  Chan Feb 8 '11 at 5:51

We have the following (easily proved by induction):

$\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \\ \end{pmatrix}$

Equating the determinants of the matrices gives us the identity immediately.

-
NOTE $\$ This is precisely the same answer I gave a couple months ago in this duplicate thread –  Bill Dubuque Feb 28 '11 at 4:06

Let us try to find gcd of $F_n$ and $F_{n+1}$ using Extended Euclidean algorithm. I will write the steps algorithm in a table; this table method was also explained in some Bill Dubuque's posts.

$\begin{array}{|l||c|c|} \hline F_{n+1} & 1 & 0 \\\hline F_{n} & 0 & 1 \\\hline F_{n-1} & 1 & -1 \\\hline F_{n-2} & -1 & 2 \\\hline F_{n-3} & 2 & -3 \\\hline \vdots & \vdots & \vdots \\\hline F_{n-k} & (-1)^{k+1}F_k & (-1)^kF_{k+1} \\\hline \end{array}$

After a few steps we can guess the $k$-th line, which gives us the following formula:
$F_{n-k}=(-1)^{k+1}F_kF_{n+1}+(-1)^kF_{k+1}F_n=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)$.

For $k=n-1$ we get Cassini's identity $F_1=(-1)^n(F_{n-1}F_{n+1}-F_n^2)$.

So the only thing we have to do is to verify the above formula, which can be done easily by induction on $k$.

Inductive step: We know that:
$F_{n-k}=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)=-(-1)^{k}(F_kF_{n+1}-F_{k+1}F_n)$
$F_{n-(k-1)}=(-1)^{k}(F_{k-1}F_{n+1}-F_{k}F_n)$

Since $F_{n-(k+1)}=F_{n-(k-1)}-F_{n-k}$, we get
$F_{n-(k+1)}=(-1)^{k}[(F_{k-1}+F_k)F_{n+1}-(F_k+F_{k+1})F_n]=(-1)^{k+2}(F_{k+1}F_{n+1}-F_{k+2}F_n)$
which completes the inductive step.

-
I tried to google a little, and I found almost the same derivation of Cassini's identity in the text Yiu: Recreational Mathematics. –  Martin Sleziak Mar 16 '12 at 11:09


# Math Help - series

1. ## series

Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman

Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))

2. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman

Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
1)

$\displaystyle\huge\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}$

$=\displaystyle\huge\frac{n+1}{(n+1)n!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$

Therefore

$\displaystyle\huge\frac{1}{2!}+\frac{2}{3!}+......$

$=1-\displaystyle\huge\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+.......-\frac{1}{(n+1)!}$

$=1-\displaystyle\huge\frac{1}{(n+1)!}$

3. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman

Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
2)

$\displaystyle\huge\frac{1}{(n+2)n!}=\frac{(n+1)}{( n+2)(n+1)n!}=\frac{n+1}{(n+2)!}$

$\displaystyle\huge\frac{n+2-1}{(n+2)!}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}$

$\displaystyle\huge\frac{n+2}{(n+2)(n+1)!}-\frac{1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$

Use this to form the telescope.

4. Thanks very much for the kind help.
May I know how and what is the idea which you employ in your solution making it so neat and simply. Is the f(r) unique and is there a general method to find f(r) ?.
Lastly is it possible to find a g(r) such that the nth term= g(r)-g(r-1) allowing us to find the sum?

5. If you are told to evaluate an infinite sum you should either change it into a known form of a function or you should make it a telescoping sum. Now to make it a telescoping sum you need to look for the following things:

- an obvious way to apply partial fractions decomposition
- add and subtract something in the numerator so that part of it looks like something that we can cancel on the bottom.

That is usually a good start. There is a lot of intuition involved in some examples and less in others.

6. Thanks very much for the reply.
Can I know how to find g(r) such that the nth term= g(r)-g(r-1) thus allowing us to find the sum of the above 3 problem.
How to apply telescope method in question 3?
I cannot wrap around my head to get started in problem3
Thanks

7. 3)

The sum can be written as...

$\displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \sum_{k=1}^{n} \frac{1}{k+5}$ (1)

In order to solve the second term of (1) we have to introduce the function digamma, defined as...

$\displaystyle \psi(x)= \sum_{j=1}^{\infty} \frac{x}{j\ (j+x)} - \gamma$ (2)

... where $\gamma$ is the 'Euler's constant'. From (2) You can derive the following basic identity...

$\displaystyle \psi(1+x) - \psi (x) = \frac{1}{1+x}$ (3)

Now setting in (3) $x=k+4$ You obtain...

$\displaystyle \psi(k+5) - \psi (k+4) = \frac{1}{k+5}$ (4)

... so that is...

$\displaystyle \sum_{k=1}^{n} \frac{1}{k+5} = \sum_{k=1}^{n} \{ \psi(k+5) - \psi (k+4) \} = \psi (n+5) - \psi(5)$ (5)

... and finally You can write...

$\displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \psi(n+5) + 5\ \psi(5)$ (6)

Digamma function - Wikipedia, the free encyclopedia

Kind regards

$\chi$ $\sigma$

8. That's pretty cool, chisigma, but I think there is an even simpler answer.

$\displaystyle\sum_{k=1}^n \frac{k}{k+5} = \sum_{k=1}^n \frac{k+5-5}{k+5} = \sum_{k=1}^n \left(\frac{k+5}{k+5}-\frac{5}{k+5}\right)$

$\displaystyle = \left(\sum_{k=1}^n 1\right)-5\left(\sum_{k=1}^n \frac{1}{k+5}\right) = n - 5\sum_{k=6}^{n+5} \frac{1}{k+5} = n - 5\left(H_n - \frac{1}{1}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right) = \frac{137}{12}+n-5H_n$

where [LaTeX ERROR: Convert failed] is the n-th Harmonic number. There is no simpler way to write this sum.


What is the probability that the max of $n$ numbers from some distribution is a greater than another number from the same distribution?

Independently choose $n$ numbers $X_1, X_2, \ldots, X_n$ from some distribution. And Independently choose another number $X_{n+1}$ from the same distribution. Let $P(\max(X_1, X_2, \ldots,,X_n) > X_{n+1}))$ means the probability that the maximum of $X_1, X_2, \ldots, X_n$ is bigger than $X_{n+1}$. What is $P(\max(X_1, X_2, \ldots,X_n) > X_{n+1}))$? I think that $P(\max(X_1, X_2, \ldots,X_n) > X_{n+1})) = \frac{n}{n+1}$. Here is my proof.

The case that any one of $X_1, X_2, \ldots X_n, X_{n+1}$ is the maximum is symmetric. So $P(X_i\text{ is the maximum)} = \frac{1}{n+1}, (i = 1,\ldots,n+1)$.

$P(\max(X_1, X_2, \ldots,X_n) > X_{n+1})) = P(X_i \text{ is the maximum}, i =1,\ldots,n)= \sum_{i=1}^{n}P(X_i \text{ is the maximum}) = n \cdot \frac{1}{n+1} = \frac{n}{n+1}$

But I think that my proof is not rigorous. https://math.stackexchange.com/a/20841/16625 gives a rigorous proof for the $n=1$ case. Can anyone give a rigorous proof for the general case?

• You should get $\frac{n}{n+1}$. $\frac{1}{n+1}$ is the probability that $\max(X_i)<X_{n+1}$. – Thomas Andrews Jan 20 '16 at 3:53
• This only works for continuous random variables, of course. For instance, if the $X_i$ are die rolls, you will get a very different result. – Thomas Andrews Jan 20 '16 at 3:58
• @ThomasAndrews Yes. For discrete random variables, the case that random variables are equal can not be ignored. – Jingguo Yao Jan 21 '16 at 7:21

Probably easier to compute the probability that the max is less. Specifically, $P(\max(X_i)<x)=P(X<x)^n$. So if $p$ is the CDF for $X$ then the probability is:

$$\int p(x)^n\,dp(x) = \frac{1}{n+1}p(x)^{n+1}\Bigg{|}_{-\infty}^{\infty}=\frac{1}{n+1}$$

This is sort of obvious - the probability that the last one is the largest is the same as the probability that any other is the largest. (If the pdf is discontinuous, it is possible, with non-zero probability, for two to be equal, of course. So we assume continuous.)

So the probability that you were seeking is $\frac{n}{n+1}$.

Here is a slightly more formal argument. Suppose $X_1, X_2, \ldots,$ are independent and identically distributed random variables. Then the maximum order statistic $X_{(n)}$ of a sample of size $n$ has CDF $$F_{X_{(n)}}(x) = \Pr[X_{(n)} \le x] = \prod_{i=1}^n \Pr[X_i \le x] = F_X(x)^n.$$ Suppose we are now interested in the random variable $Y = X_{(n)} - X_{n+1}$, and in particular, $$\Pr[Y > 0] = 1 - \Pr[Y \le 0] = 1- \int_{x = -\infty}^\infty F_{X_{(n)}}(x) f_X(x) \, dx = 1 - \int_{x = -\infty}^\infty F_X(x)^n f_X(x) \, dx.$$ With the substitution $$u = F_X(x), \quad du = f_X(x) \, dx, \quad F_X(-\infty) = 0, \quad F_X(\infty) = 1,$$ we arrive at $$\Pr[Y > 0] = 1 - \int_{u=0}^1 u^n \, du = 1 - \frac{1}{1+n} = \frac{n}{n+1},$$ as claimed.

Assuming that the $P(X_i=X_j)=0$ for $i\ne j$, $$P(\max(X_1,X_2,\dots,X_n)\gt X_{n+1})$$ is the probability that $X_{n+1}$ is not the greatest of the $n+1$ samples. The probability that it is the greatest of $n+1$ samples is $$P(\max(X_1,X_2,\dots,X_n)\le X_{n+1})=\frac{n!}{(n+1)!}=\frac1{n+1}$$ because there are $n!$ arrangements where $X_{n+1}$ is greatest and $(n+1)!$ arrangements altogether.

Therefore, $$P(\max(X_1,X_2,\dots,X_n)\gt X_{n+1})=1-\frac1{n+1}=\frac n{n+1}$$

Your argument is correct if the $n+1$ observations are independent, an assumption that you did not state.


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# Daily Challenge #230 - Beeramid

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Let's pretend your company just hired your friend from college and paid you a referral bonus. Awesome! To celebrate, you're taking your team out to the terrible dive bar next door and using the referral bonus to buy, and build, the largest three-dimensional beer can pyramid you can. And then probably drink those beers, because let's pretend it's Friday too.

A beer can pyramid will square the number of cans in each level - 1 can in the top level, 4 in the second, 9 in the next, 16, 25...

Complete the beeramid function to return the number of complete levels of a beer can pyramid you can make, given the parameters of:

2) the price of a beer can

For example:

beeramid(1500, 2); // should === 12
beeramid(5000, 3); // should === 16

Tests:
beeramid(9, 2)
beeramid(21, 1.5)
beeramid(-1, 4)

This challenge comes from kylehill on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

## Discussion (6)

Here's my C solution,
Notice that there is no need for any iteration because (taking number of cans = N and number of levels = L)

$N = 1 + 2^2 + 3^2 + ... + L^2 = L(L+1)(2L+1)/6 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L^3/3 < N < (L+1)^3/3 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L < (3N)^{1/3} < L+1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L = \lfloor (3N)^{1/3} \rfloor \quad or \quad L = \lfloor (3N)^{1/3} \rfloor - 1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$

int beeramid(double bonus, double price) {
if (bonus < price) return 0;
int cans = floor(bonus / price);
int levels = floor(cbrt(3.0*cans));
return levels*(levels + 1)*(2*levels + 1) > 6*cans ? levels - 1 : levels;
}

Vidit Sarkar

Here is a Python Solution,

def beeramid(bonus, price):
canCount = int(bonus/price)
level = 0

while((level+1)**2 <= canCount):
level += 1
canCount -= level*level

return level

Test Cases,

print(beeramid(1500, 2)) # output -> 12
print(beeramid(5000, 3)) # output -> 16
print(beeramid(9, 2)) # output -> 1
print(beeramid(21, 1.5)) # output -> 3
print(beeramid(-1, 4)) # output -> 0

Michael Kohl • Edited

Ruby, using a lazy, infinite enumerator:

ROWS = (1..).lazy.map { |n| n * n }

def beeramid(bonus, price)
cans = bonus / price
ROWS.take_while { |n| (cans -= n) >= 0 }.count
end

Valts Liepiņš • Edited

Haskell solution using lazy, infinite list of cans:

import Data.List (findIndex)

beeramid :: Double -> Double -> Int
beeramid m p =
case findIndex (> cans) levels of
Just level -> level
Nothing    -> 0
where
cans = m / p
levels = scanl1 (+) . fmap (^2) \$ [1..]

Paula Gearon • Edited

Clojure. Not very terse, but it gets a solution in constant time

(defn faul [n]
(let [n2 (* n n)
n3 (* n2 n)]
(/ (+ n3 n3 n2 n2 n2 n) 6)))

(defn layers [n]
(let [est (int (Math/pow (* 3 n) (/ 1 3)))]
(first (filter #(>= n (faul %)) (range est 0 -1)))))

(defn beeramid [bonus price] (or (layers (/ bonus price)) 0))

Testing:

=> (beeramid 1500 2)
12
=> (beeramid 5000 3)
16
=> (beeramid 9 2)
1
=> (beeramid 21 1.5)
3
=> (beeramid -1 4)
0

Paula Gearon

To follow up on my solution, I borrowed from the Wikipedia article on Square Pyramidal Numbers. This provides a formula for the sum of squares as a special case of Faulhaber's formula:

$P_n = \sum_{k=1}^n k^2 = \frac {2n^3 + 3n^2 + n} 6$
So if we have n layers, then that will have a total of Pn cups.
As n becomes large then: 2n3 ≫ 3n2 + n
$P_n = \frac {2n^3} 6 + \Big( \frac {3n^2 + n} 6 \Big) \newline P_n > \frac {n^3} 3 \newline$
Since the layers is n and the cups in the pyramid is Pn then this gives a bound of:
$layers < \sqrt[3]{3 \cdot cups}$
Consequently, my code creates an estimate of the number of layers using this formula, and counts down from there applying the Faulhaber formula to check if the result is less than or equal to the provided number of cups.

Using this method, a small numbers of cups will require 2 guesses, but as the number of cups gets higher (and n3n2) then it is more likely to get it on the first guess. I told it to keep testing while the number of layers until 1, but it never actually gets beyond 2 attempts.


# Math Help - definite integral

1. ## definite integral

a little confused on this one

"solve the definite integral"

$\int^2_0 2x^2e^{x^3+5}$

2. Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"

$\int^2_0 2x^2e^{x^3+5}$
Well, again, use substitution $z := x^3+5$, which gives that $dz=3x^2\,dx$ and hence you have

$\int 2x^2e^{x^3+5}\,dx=\tfrac{2}{3}\cdot\int e^{x^3+5}\cdot 3x^2\, dx=\tfrac{2}{3}\cdot\int z\,dz=\ldots$

3. Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
What exactly are you confused about? Have you made an attempt at it?

Also the integral should be $\int^2_0 2x^2e^{x^3+5}dx$

I could tell you how to do it but i think i it's only fair that an attempt is made first

Hint: use substitution

4. ah perfect! i couldn't figure out how i could change that $2x^2$ to a $3x^2$ . thanks!

5. actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

$\int^2_0 2x^2e^{x^3+5}$

u= $x^3 + 5$
h(u) = $3x^2$

$\frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$

when x = 0, u = 5
when x = 2, u = 13

$\frac{2}{3} \int^{13}_5 e^u du$
$

\frac{2}{3} e^{13} - \frac{2}{3} e^5$

i'm getting 294,843.32 which makes me think i'm doing something wrong.

6. Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

$\int^2_0 2x^2e^{x^3+5}$

u= $x^3 + 5$
h(u) = $3x^2$

$\frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$

when x = 0, u = 5
when x = 2, u = 13

$\frac{2}{3} \int^{13}_5 e^u du$
$

\frac{2}{3} e^{13} - \frac{2}{3} e^5$

i'm getting 294,843.32 which makes me think i'm doing something wrong.
No, that result is ok. Of course, $e^{x^3+5}$ is a fast growing function.

7. Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

$\int^2_0 2x^2e^{x^3+5}$

u= $x^3 + 5$
h(u) = $3x^2$

$\frac{2}{3} \int^2_0 3x^{2} e^{x^3+5}$

when x = 0, u = 5
when x = 2, u = 13

$\frac{2}{3} \int^{13}_5 e^u du$
$

\frac{2}{3} e^{13} - \frac{2}{3} e^5$

i'm getting 294,843.32 which makes me think i'm doing something wrong.
Apart from [tex]3x^{2} it lookes fine to me.

8. Originally Posted by JGaraffa
a little confused on this one

"solve the definite integral"

$\int^2_0 2x^2e^{x^3+5}$
Again observe that $3x^2$ is the derivative of $x^3+5$, so consider:

$\frac{d}{dx}e^{x^3+5}$

CB


# Evaluating $\int_0^2(\tan^{-1}(\pi x)-\tan^{-1} x)\,\mathrm{d}x$

Hint given: Write the integrand as an integral.

I'm supposed to do this as double integration.

My attempt:

$$\int_0^2 [\tan^{-1}y]^{\pi x}_{x}$$

$$= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}$$

$$= \int_2^{2\pi} \int_{\frac{y}{\pi}}^2 \frac { \mathrm{d}x \mathrm{d}y} {y^2+1}$$

$$= \int_2^{2\pi} \frac { [x]^2_{\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$

$$= \int_2^{2\pi} \frac { 2- {\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$

Carrying out this integration, I got, $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]$$

But I'm supposed to get $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]+ [\frac {\pi-1}{2 \pi}] \ln 5$$

Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.

• At third line it seems to me the integral must be $$\int_0^{2\pi}\int_{\frac{y}{\pi}}^{y}\frac{\operatorname d\!x \operatorname d\!y}{y^2+1}$$ – Ángel Mario Gallegos Nov 4 '14 at 6:13
• but that is taking my answer even further away from what I require. – Diya Nov 4 '14 at 6:50
• I got it! You were right! The integral is divided into two regions. One is the one I used, and the other is the one you mentioned. Thank you so much! :) – Diya Nov 4 '14 at 6:57

Use the change of variables $y = xt.$

$$I = \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \, \mathrm{d}x} {y^2+1}\\=\int_0^2 \int_1^{\pi } \frac { x} {x^2t^2+1}\mathrm{d}t \, \mathrm{d}x\\=\int_1^{\pi} \int_0^{2 } \frac { x} {x^2t^2+1}\mathrm{d}x \, \mathrm{d}t\\=\int_1^{\pi} \frac{\ln(1+4t^2)}{2t^2} \mathrm{d}t$$

Now use integration by parts.

$$I = -\left.\frac{\ln(1+4t^2)}{2t}\right|_1^{\pi}+4\int_1^{\pi}\frac1{1+4t^2} \, dt$$

A more straight forward approach uses integration by parts.

Define: \begin{align} & I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2} \end{align}

using partial fraction this reads: \begin{align} I(c)=\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\log(1+y^2)|_{ac}^{bc} \end{align}

taking $I(\pi)-I(0)$ with $a=0$ and $b=2$ we are done

For the sake of an alternative approach, recall the formula for the integral of an inverse function $$\int f^{-1}(x)\mathrm{d}x=xf^{-1}(x)-F\circ f^{-1}(x)$$ Where $$F'(x)=f(x)$$. Plugging in $$f(x)=\tan(x)$$, $$I=\int\arctan(x)\mathrm{d}x=x\arctan(x)-\int_0^{\arctan(x)}\tan(t)\mathrm{d}t$$ Then recall that $$(-\log|\cos(x)|)'=\tan(x)$$ So we have $$I=x\arctan(x)+\log|\cos(\arctan(x))|$$ then using trig, $$I=x\arctan(x)-\frac12\log(x^2+1)$$ So $$I_1=\int_0^2\arctan(x)\mathrm{d}x=2\arctan2-\frac12\log5$$ And \begin{align} I_2=&\int_0^2\arctan(\pi x)\mathrm{d}x\\ =&\frac1\pi\int_0^{2\pi}\arctan(x)\mathrm{d}x\\ =&\frac1\pi\bigg(2\pi\arctan2\pi-\frac12\log(4\pi^2+1)\bigg)\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)\\ \end{align} So \begin{align} \int_0^2(\arctan\pi x-\arctan x)\mathrm{d}x=&I_2-I_1\\ =&2\arctan2\pi-\frac1{2\pi}\log(4\pi^2+1)-2\arctan2+\frac12\log5\\ =&2\arctan2\pi-\frac1{2}\log\sqrt[\pi]{4\pi^2+1}-2\arctan2+\frac12\log5\\ =&2(\arctan2\pi-\arctan2)+\frac12\log\frac{5}{\sqrt[\pi]{4\pi^2+1}}\\ \end{align}


# The number of different triangles and each side has a different length

I'm trying to solve the following problem:

What is the number of different triangles we can form from numbers $$4,5,6,7,8,9$$ (lengths of sides), where every side has a different length (for an example $$4,5,6$$ or $$4,5,7$$)..

My solution is the following:

The number of all possible permutations is $$\frac{6!}{(6-3)!}$$. We subtract the numbers which don't make a triangle (which is $$6*2$$) and then divide by $$2$$, because $$(4,5,6)$$ makes the same triangle is $$6,5,4$$. My answer is $$54$$.

However, the correct answer should be $$53$$. Can anyone tell me where I did a mistake?

Thanks

• $6 \choose 3$ is only $20$ and there is one choice $(4,5,9)$ that doesn't make a triangle, so I get $19$ Your expression (ignoring the factorial in the number to choose) gives $8$. Oct 20, 2019 at 16:42
• I think you'll have to show more details for someone to point out your mistake. What do you mean by $\binom{6}{(6-3)!}$? It look to me like ${6\choose6}=1$. Did you mean to say ${6\choose3}?$ Also, why do you say there are $12$ where the triangle inequality doesn't hold. How did you get that? Oct 20, 2019 at 16:43
• Sorry, I made a mistake when writing my answer here. Please wait a minute and I will fix it Oct 20, 2019 at 16:44
• If you write out all the triples $(4,5,6), (4,5,7), \ldots, (7,8,9)$, you'll see that there are only $20$ in the list, one of which is degenerate. How can there be $53$ or $54$ triangles? We could consider "right-handed" and "left-handed" triangles to be different, effectively doubling the number of triangles, but even then, we would have at most $40$, still falling short of the given answer. Have you written out the problem exactly as it was given to you? Oct 20, 2019 at 16:56
• You should have gotten six crossed off, as you describe. 2 for 4, 2 for 5 and 2 for 9 (not 6) makes six. Why do you double that? Then you should divide by $6$ because 456 comes six ways-456,465,546,564,654,645 not two Oct 20, 2019 at 17:04

There are $$\frac {6!}{3!}$$ ordered choices of three numbers, which is $$120$$. Six of those do not make a triangle, all the permutations of $$(4,5,9)$$, which leaves $$114$$. Each unordered triangle gives $$3!$$ permutations, so we divide by $$6$$ and get $$19$$. I don't know where a number in the $$50$$s comes from.
I think it is easier to just choose unordered combinations to start with, which is $${6 \choose 3}=20$$ and subtract the one that doesn't make a triangle. That also gets $$19$$.


# In $\mathbb{R}^n$, is the dot product the only inner product?

Just what the title says. I'm reading, from various resources, that the inner product is a generalization of the dot product. However, the only example of inner product I can find, is still the dot product.

Does another "inner product" exist on the $$\mathbb{R}^n$$ space?

• This might be useful. – StackTD Dec 14 '18 at 11:07
• @StackTD: thanks!!! Exactly what I was looking for – blue_note Dec 14 '18 at 11:15
• – StackTD Dec 14 '18 at 11:37
• @StackTD: thanks again – blue_note Dec 14 '18 at 11:38
• As some other answers point out, there are infinitely many inner products (i.e., symmetric, positive-definite bilinear forms) on $\Bbb R^n$. But for any of them one can choose a basis of $\Bbb R^n$ with respect to which the bilinear form is the standard one: $({\bf x}, {\bf y}) = x_1 y_1 + \cdots + x_n y_n$. So up to isomorphism there is only one inner product on $\Bbb R^n$. (This is a special case of Sylvester's Law of Inertia.) – Travis Willse Dec 25 '18 at 0:51

Yes and are all in the form $$\langle x, y\rangle=x^T\cdot A\cdot y$$ where $$x^T$$ is the transpose vector of $$x$$ and $$A$$ is a $$n\times n$$ symmetric definite positive matrix.

In fact let $$\langle x, y\rangle$$ a generic inner product on $$\mathbb R^n$$ then for every $$y$$ the function $$f_y:x\rightarrow \langle x, y\rangle$$ is linear from $$\mathbb R^n$$ to $$\mathbb R$$ then exists a vector $$\alpha(y)\in\mathbb R^n$$ such that $$\langle x, y\rangle = \alpha(y)^T\cdot x$$

Observe that $$\langle x, ay+by'\rangle = a\langle x, y\rangle + b\langle x, y'\rangle\Rightarrow \alpha(ay+by')=a\alpha(y)+b\alpha(y')$$ then $$\alpha$$ is a linear operator from $$\mathbb R^n$$ in itself then exists an $$n\times n$$ matrix $$A$$ such that $$\alpha(y)=A\cdot y$$ so $$\alpha(y)^T\cdot x=y^T\cdot A^T\cdot x$$

Now remember that $$\langle x, y\rangle=\langle y, x\rangle$$ then you can easly prove that $$A^T=A$$ and $$A$$ must be symmetric.

Why now $$A$$ must be definite positive? Because $$\langle x, x\rangle\geq 0$$ and holds equality if and only if $$x=0$$. Applying it to the initial formula we obtain the definition of a definite positive matrix.

• We need $A$ to be positive definite also. – littleO Dec 14 '18 at 11:26
• Oh yes, I forgotten it! – Jihlbert Dec 14 '18 at 11:28
• Note that in some contexts, one only requires that an inner product be nondegenerate (and not necessarily positive definite). – Travis Willse Dec 25 '18 at 0:52
• If it's semidefinite positive it becomes a semi-inner product and not a inner product. – Jihlbert Dec 28 '18 at 11:30

The dot product on $$\mathcal{R}^n$$ is defined as follows:

$$(a,b) = a^i b^j (e_i,e_j) = a^i b^j \delta_{ij} = a^i b^i ,$$

where $$a,b \in \mathcal{R}^n$$ and $$e_i,e_j$$ standard basis vectors. I used Einstein summation convention here.

In general we can express $$a,b$$ in a different basis i.e. $$a=\tilde{a}^i \tilde{e}_i$$ and $$b = \tilde{b}^i \tilde{e}_i$$ so now not the standard basis but an arbitrary basis of $$\mathcal{R}^n$$ assuming still $$(,)$$ is positive-definite. This then gives:

$$(a,b) = \tilde{a}^i \tilde{b}^j (\tilde{e}_i,\tilde{e}_j) = \tilde{a}^i \tilde{b}^j A_{ij} \equiv a^T A \ b.$$

Note that $$A$$ now is not the identity matrix like in the standard inner product.

The tensor metric allows for the vector norm to remain constant under change of basis vectors, and is an example of inner product (page 15). In the simple setting of basis vectors constant in direction and magnitude from point to point in $$\mathbb R^2,$$ here is an example:

Changing basis vectors from Euclidean orthonormal basis $$\small\left\{\vec e_1=\begin{bmatrix}1\\0 \end{bmatrix},\vec e_2=\begin{bmatrix}0\\1 \end{bmatrix}\right\}$$ to

$$\left\{\vec u_1=\color{red}2\vec e_1 + \color{red}1\vec e_2,\quad \vec u_2=\color{blue}{-\frac 1 2 }\vec e_2+\color{blue}{\frac 1 4} \vec e_2\right\}\tag 1$$ would result in a different norm of the vector $$\vec v=\begin{bmatrix} 1\\1\end{bmatrix}_\text{e basis},$$ i.e $$\Vert \vec v \Vert^2=v_1^2 + v_2^2 = 2,$$ when expressed with respect to the new basis vectors.

In this regard, since vector components transform contravariantly, the change to the new coordinate system would be given by the backward transformation matrix:

$$B=\begin{bmatrix} \frac 1 4 & \frac 1 2\\-1&2\end{bmatrix}$$

i.e. the inverse of the forward transformation for the basis vectors as defined in $$(1),$$ which in matrix form corresponds to

$$F=\begin{bmatrix} \color{red} 2 & \color{blue}{-\frac 1 2}\\\color{red}1& \color{blue}{\frac 1 4}\end{bmatrix}.$$

Hence, the same vector $$\vec v$$ expressed in the new basis vectors is

$$\vec v_{\text{u basis}}=\begin{bmatrix} \frac 1 4 & \frac 1 2\\-1&2\end{bmatrix}\begin{bmatrix} 1\\1\end{bmatrix}=\begin{bmatrix} \frac 3 4\\1\end{bmatrix}.$$

And the norm would entail an inner product with the new metric tensor. In the orthonormal Euclidean basis this is simply the identity matrix. Now the matrix tensor is a $$(0,2)-\text{tensor},$$ and transforms covariantly:

$$g_{\text{ in u basis}}=(F^\top F) I= \begin{bmatrix} 2 & 1\\-\frac 1 2& \frac 1 4\end{bmatrix}\begin{bmatrix} 2 & -\frac 1 2\\1& \frac 1 4\end{bmatrix}\begin{bmatrix} 1 & 0\\0& 1\end{bmatrix}=\begin{bmatrix} 5 & -\frac 3 4\\- \frac 3 4& \frac{5}{16}\end{bmatrix}$$

The actual multiplication of basis vectors to obtain the metric tensor is explained in this presentation by @eigenchris. This metric tensor indeed renders the right norm of $$\vec v:$$

$$\begin{bmatrix}\frac 3 4 & 1\end{bmatrix}\begin{bmatrix} 5 & -\frac 3 4\\- \frac 3 4& \frac{5}{16}\end{bmatrix}\begin{bmatrix}\frac 3 4 \\ 1\end{bmatrix}=2$$

following the operations here.


# Proof of the statement “The product of 4 consecutive integers can be expressed in the form 8k for some integer k”

I am slowly diving into simple number theory and learning how to craft direct proofs. I needed to proof the statement "The product of 4 consecutive integers can be expressed in the form 8k for some integer k".

My approach was to use the quotient-remainder theorem, as it was surely intended in the textbook. The proof looks like follows.

Proof:

Suppose n is any particular but arbitrarily chose integer. We must show that n(n+1)(n+2)(n+3) is divisible by 8. By the quotient-remainder theorem, n can be written in one of the forms 4q or (4q+1) or (4q+2) or (4q+3) for some integer q. We divide into cases accordingly:

Case 1 (n = 4q for some integer q):

\begin{align} n(n+1)(n+2)(n+3) & = 4q(4q+1)(4q+1)(4q+2)(4q+3)\\ & = 8[q(32q^3+48q^2+18q+3)] \end{align}

Let $m=q(32q^3+48q^2+18q+3)$. Then m is an integer because sums and products of integers are integers. By substitution, $n(n+1)(n+2)(n+3) = 8m$ where m is an integer. Hence n(n+1)(n+2)(n+3) is divisible by 8.

Case 2 (n = (4q+1) for some integer q): \begin{align} n(n+1)(n+2)(n+3) & = (4q+1)(4q+2)(4q+3)(4q+4)(4q+5)\\ & = 8[(4q+1)(2q+1)(4q+3)(q+1)] \end{align}

Let $m=q((4q+1)(2q+1)(4q+3)(q+1))$. Then m is an integer because sums and products of integers are integers {...} {See case 1, its basically the same reasoning here...}

Case 3 (n = (4q+2) for some integer q): {...}

Case 4 (n = (4q+3) for some integer q): {...}

Conclusion:

I each of the above cases, n(n+1)(n+2)(n+3) was to be shown to be a multiple of 8. By the quotient-remainder theorem, one of these cases must occur, hence n(n+1)(n+2)(n+3) can be written in the form 8k for some integer k. q.e.d. [End Proof]

But honestly, I consider this proof somehow clumsy and too inconvenient [I am a bloody amateur, maybe I am wrong]. Are there any better/shorter ways to prove the above statement?

• Any four consecutive numbers will have one of each of the possible four remainders mod $4$. This means that one will be divisible by $4$ and another will be divisible by $2$, which means that the product is divisible by $8$. – Tobias Kildetoft Jul 12 '13 at 12:34
• Wow, this is straightforward! – Nikolai Tschacher Jul 18 '13 at 8:41

In each four consecutive integers there is exactly one pair of even integers, and exactly one of them is $\;2\pmod 4\;$ and the other one is $\,0\pmod 4\;$ , so the product of these two even (consecutive even) integers is already divisible by $\,8\,\ldots\ldots$

• Very good, sir. – Don Larynx Jan 9 '15 at 20:53

Note that if you have four consecutive integers, then one must be divisible by 4 and another must be divisible by 2: they must be equivalent to $0,1,2,\text{ and }3\pmod{4}$.

One way is by proving that the binomial coefficients $$\dbinom n r=\cfrac {n!}{r!(n-r)!}$$ are integers, whoch can be done in various ways including using the binomial recurrence from Pascal's Triangle. Then$$\binom n 4=\frac {n(n-1)(n-2)(n-3)}{4!}$$ which proves that the product of four consecutive integers is divisible by $24$.

Another way is by induction.$$(n+1)n(n-1)(n-2)-n(n-1)(n-2)(n-3)=4n(n-1)(n-2)$$

So you can prove the result using that the product of three consecutive integers is even. Or prove the result for $24$, by showing that the product of three consecutive integers is divisible by $6$ (reducing the number of factors again by taking the difference of successive terms). And remembering to prove the base case - but this can be arranged to have zero as a factor, so divisible by the integer we require.

$$(n+1)n(n-1)-n(n-1)(n-2)=3n(n-1)$$is divisible by 3 and $$(n+1)n-n(n-1)=2n$$ is divisible by 2.

• What is the reasoning of binomial coefficient proof? We have $4! \cdot {}^nC_4 = {}^n P_4$ because every combination of 4 objects in 4 places has 4! permutation. But how does this totally unrelated fact, without invoking any properties of numbers, prove a property of numbers i.e. the product of any four consecutive numbers is a multiple of 24? – user103816 Nov 7 '15 at 16:09
• @user103816 You can easily show that Pascal's Triangle encapsulates the formula $\binom nr+\binom n{r+1}=\binom {n+1}{r+1}$ and this shows that all the binomial coefficients are integers. Then suppose you have the consecutive integers $n-3, n-2, n-1, n$, you will find, as I have put, that $\binom n4$ is the product of those integers divided by $4!$ and is an integer - so the product of four successive integers is divisible by $24$. Negative numbers are not a problem, and if one of the numbers is zero, so is the product. – Mark Bennet Nov 7 '15 at 16:42

Your formal argument looks fine to me, but you're right: it does seem clumsy and inconvenient. This may be because you are making exactly the same argument in each case: Writing the four numbers, finding one factor of four and another factor of two, and pulling those out to get a factor of $8$.

This suggests a simpler argument. In any four consecutive integers, by the quotient-remainder theorem, there must be one multiple of four. There must also be two multiples of two. One of the multiples of two will also be a multiple of four; the other will not. Therefore, we may factor a four and another two from the product of the four consecutive integers and we see that the product is a multiple of eight.


# Find a pattern in a recursive definition

A set M $\in$ $\mathbb{Z}^2$ is defined by

(Rule 1) $(0, 1)$ $\in$ M

(Rule 2) If $(x, y)$ $\in$ M, then $(x + 1, y + 2x + 3)$ $\in$ M

(a) (2 points) Starting with $(0, 1)$, write out the six pairs with the smallest first coordinates.

(b) (2 points) State the (simple) relationship that holds between the first and second coordinates of all pairs in M.

Ok for a I got: $(0,1)$,$(1,4)$,$(2,9)$,$(3,16)$,$(4,25)$,$(5,36)$,$(6,49)$

However for b, I cant see the relationship. Can anyone give me any hints?

EDIT:

Ok I found the relation: $y=(x+1)^2$. However, I must now prove this claim using structural induction. I can get up to the inductive hypothesis, but after that I'm not really sure what to do.

Base case: $(0,1), 1=(0+1)^2 = 1$

Inductive Hypothesis: For any element $x,y$ if $x\in M$ then $y=(x+1)^2$. We must show that $y+1=(x+1)^2 + 1$

I'm not really sure that my IH is correct either.

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It appears you are using the new x in the calculation of y. I believe you should use the old x. So the second pair should be (1,4). All will become clearer. –  Ross Millikan Nov 8 '10 at 18:21
Ahh ok yes you are right..now I see the relation, thanks! –  maq Nov 8 '10 at 18:26
@Ross please see edits –  maq Nov 8 '10 at 18:41

Your inductive step is to show that if $(x,(x+1)^2) \in M$, then $(x+1,(x+2)^2) \in M$. So apply Rule 2 to $(x,(x+1)^2)$

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Wait x, x^2+1 is an element of M? I thought it was x, (x+1)^2 is an element of M –  maq Nov 8 '10 at 18:56
Right you are. Corrected. –  Ross Millikan Nov 8 '10 at 19:04
Thanks, I was able to get the rest.. –  maq Nov 8 '10 at 19:06

Your inductive hypothesis is wrong. You should be showing that if $(x,(x+1)^2)\in M$, then $(x+1, ((x+1)+1)^2) \in M$. From there you should be able to see how to get the result (just use Rule 2).

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HINT $\rm\ \ f\ (x,\:y)\ = \ (x+1,\ y+2x+3)\ \Rightarrow\ f\ (x,\ (x+1)^2)\ =\ (x+1,\ (x+2)^2)$

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# State the domain

#### shamieh

##### Active member
state the domain

$$\displaystyle y = x\sqrt{1 - x^2}$$

so

$$\displaystyle 1 - x^2 >= 0$$
$$\displaystyle -x^2 >= -1$$
$$\displaystyle x^2 <= 1$$

$$\displaystyle d = {x <= (+)(-) 1}$$ or should I say $$\displaystyle d = {-1, 1}$$

which one is correct?

#### MarkFL

Staff member
I have moved this topic here to our Pre-Calculus subforum as questions on function domains does not involve the calculus. I suspect that this question is part of a calculus question though, or may even come from a calculus textbook, but when choosing the subforum in which to post, the nature of the question itself, rather than from where it originates should be considered firstly.

You are correct in stating:

$$\displaystyle 1-x^2\ge0$$

What I would suggest doing next is factoring the left side:

$$\displaystyle (1+x)(1-x)\ge0$$

Now, determine the critical values, plot them on a number line and test the 3 resulting intervals.

A quicker method would be to plot the expression (the original radicand), and observe where it is non-negative.

What do you find?

#### soroban

##### Well-known member
Hello, shamieh!

State the domain: .$$y \:=\: x\sqrt{1 - x^2}$$

So: .$$1 - x^2 \:\ge\:0$$
. . . . . $$-x^2 \:\ge\:-1$$
. . . . . . .$$x^2\:\le\:1$$

$$\begin{array}{cc}\text{I would write:} & |x| \:\le\:1 \\ & \text{or} \\ & \text{-}1 \:\le\:x\:\le\:1 \\ & \text{or} \\ & [\text{-}1,\,1]\end{array}$$

#### HallsofIvy

##### Well-known member
MHB Math Helper
state the domain

$$\displaystyle y = x\sqrt{1 - x^2}$$

so

$$\displaystyle 1 - x^2 >= 0$$
$$\displaystyle -x^2 >= -1$$
$$\displaystyle x^2 <= 1$$

$$\displaystyle d = {x <= (+)(-) 1}$$ or should I say $$\displaystyle d = {-1, 1}$$

which one is correct?
NOT $$\displaystyle \{-1, 1\}$$ because that means the numbers -1 and 1 only, not the numbers between. You could say $$\displaystyle \{x| -1\le x\le 1\}$$ or [-1, 1] as MarkFL said.

("math" uses the "{" and "}" and as separators itself. To get "{" or "}" in the actual message, use "\{" and "\}".)

#### Deveno

##### Well-known member
MHB Math Scholar
A naive approach:

We know that there's "something special" about the points $x = -1, x = 1$. So let's do this:

We'll split the real number line into 5 parts:

Part 1: all numbers less than -1
Part 2: -1
Part 3: all numbers between -1 and 1
Part 4: 1
Part 5: all numbers greater than 1.

Now let's pick numbers in each part, to see what happens:

The 5 real numbers I will use are:

Part 1: -4
Part 2: -1 (no choice, here)
Part 3: 1/2
Part 4: 1 (again, no other choice)
Part 5: 6

Now we will look at $f(x_0)$ for $x_0$ being each one of these 5 numbers:

$f(-4) = 4\sqrt{1 - (-4)^2} = 4\sqrt{-15} = \text{bad}$ (undefined)
$f(-1) = (-1)\sqrt{1 - (-1)^2} = (-1)\sqrt{1 - 1} = (-1)\sqrt{0} = (-1)(0) = 0$ (OK!!!!)
$f(\frac{1}{2}) = \frac{1}{2}\sqrt{1 - (\frac{1}{2})^2} = \frac{1}{2}\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{4}$ (OK!!!)
$f(1) = (1)\sqrt{1 - 1^2} = (1)(0) = 0$ (OK again!!!!)
$f(6) = 6\sqrt{1 - 6^2} = 6\sqrt{-35} = ???$ (not so good).

So it looks as if what we want is parts 2,3, and 4, and NOT parts 1 and 5. This is:

$\{-1\} \cup (-1,1) \cup \{1\} = [-1,1]$

If you prefer, you can write this as:

$\text{dom}(f) = \{x \in \Bbb R: |x| \leq 1\}$

the absolute value of $x$, written $|x|$ is just another way of saying:

$\sqrt{x^2}$, if one understand square roots as always being non-negative. So if:

$x^2 \leq 1$, then

$\sqrt{x^2} \leq \sqrt{1} = 1$

so:

$|x| \leq 1$.