$\endgroup$ – André Nicolas Apr 30 '11 at 18:58 $\begingroup$ @shino: Or else if you are doing everything correctly, and exponential is a poor fit, look for a better fit from one of the Weibull distributions. The probability density function is $$f(x) = me^{-mx}$$. distribution is a discrete distribution closely related to the binomial distribution and so will be considered later. First consider λ = 1 λ = 1. Here e is the mathematical constant e that is approximately 2.718281828. Distributions with CV < 1 (such as an Erlang distribution ) are considered low-variance, while those with CV > 1 (such as a hyper-exponential distribution ) are considered high-variance [ … The ‘moment generating function’ of an exponential random variable X for any time interval t<λ, is defined by; M X (t) = λ/λ-t 0.5. c. 0.25. d. 2.0. e. the means of the two distributions can never be equal. Scientific calculators have the key “e … Therefore, X ~ Exp(0.25). Sample means from an exponential distribution do not have exponential distribution. The exponential distribution is one of the widely used continuous distributions. To describe the time between successive occurrences when all occurrences follow an exponential. μ = σ. Study notes and guides for Six Sigma certification tests. In fact, the mean and standard deviation are both equal to A. f ( x) = e-x/A /A, where x is nonnegative. Exponential Distribution Modelling Of Wet-day Rainfall Totals Assume An Exponential Distribution Can Be Used To Model Precipitation Totals On Wet Days. (Taken from ASQ sample Black Belt exam.). The distribution notation is X ~ Exp(m). Required fields are marked *. Exponential distribution is the time between events in a Poisson process. Exponential distribution is the time between events in a Poisson process. Construct a histogram of the dat The mean or expected value of an exponentially distributed random variable X with rate parameter λ is given by Simply, it is an inverse of Poisson. Log in or Sign up in seconds with the buttons below! Exponential Distribution Variance. http://www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf, Your email address will not be published. The amount of time (may be in months) a car battery lasts. Learn how your comment data is processed. μ = σ. 100% of candidates who complete my study guide report passing their exam! Therefore, $$X \sim Exp(0.25)$$. This section requires you to be logged in. That is, for an Exponential distribution, the mean and the standard deviation are equal, and equal to the reciprocal of the rate parameter. Questions, comments, issues, concerns? E ( X k) = ∫ 0 ∞ x k e − x d x = k! The mean of exponential distribution is 1/lambda and the standard deviation is also 1/lambda. This section requires you to be a Pass Your Six Sigma Exam member. (. it describes the inter-arrival times in a Poisson process.It is the continuous counterpart to the geometric distribution, and it too is memoryless.. This statistics video tutorial explains how to solve continuous probability exponential distribution problems. It can be shown for the exponential distribution that the mean is equal to the standard deviation; i.e., μ = σ = 1/λ Moreover, the exponential distribution is the only continuous distribution that is I have seen this question on one of the websites (I guess ASQ, not sure). 16. Therefore, X ~ Exp(0.25). IASSC Lean Six Sigma Green Belt Study Guide, Villanova Six Sigma Green Belt Study Guide, IASSC Lean Six Sigma Black Belt Study Guide, Villanova Six Sigma Black Belt Study Guide, Where e is base natural logarithm = 2.71828. I’ll investigate the … If it is a negative value, the function is zero only. $$\mu = \sigma$$ The distribution notation is $$X \sim Exp(m)$$. Your instructor will record the amounts in dollars and cents. If the number of occurrences follows a Poisson distribution, the lapse of time between these events is distributed exponentially. Login to your account OR Enroll in Pass Your Six Sigma Exam. It is a number that is used often in mathematics. The exponential distribution can be used to determine the probability that it will take a given number of trials to arrive at the first success in a Poisson distribution; i.e. This site uses Akismet to reduce spam.

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# How can I write this power series as a power series representation?

How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?

• This is S(1+x) where $S=1+2x^2+3x^4+...$ for a start. – Paul Dec 19 '14 at 23:18
• @paul: What about x? – Mathy Person Dec 19 '14 at 23:21
• @Paul Saying $(1+x)S(x)$ might be more clear. – Tim Raczkowski Dec 19 '14 at 23:25

Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots)$$ $$=(1+x)(1+2y+3y^2+4y^3 +\ldots)$$ $$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$ $$= (1+x)\left( \frac{y}{1-y}\right)'$$

Edit: $$= (1+x) \frac{1}{(1-y)^2}$$ $$= \frac{1+x}{(1-x^2)^2}$$ $$= \frac{1}{(1-x)(1+x^2)}.$$

• I believe $(y+y^2+y^3+y^4+....)$ = $\frac{1}{1-y}$, if I am not mistaken. Then is it: $\frac{1+x}{1-y}$? – Mathy Person Dec 19 '14 at 23:28
• Oh wait, I see that you had $\frac{y}{1-y}$ instead. Is it: $\frac{(1+x)(y)}{1-y}$? – Mathy Person Dec 19 '14 at 23:30
• And $\frac{(1+x)(y)}{1-y}$, and $y=x^2$, then $\frac{(1+x)(x^2)}{1-x^2}$? – Mathy Person Dec 19 '14 at 23:30
• Simplifying would result in: $\frac{x^2}{1-x}$? – Mathy Person Dec 19 '14 at 23:31
• Just take derivative in y of $y/(1-y)$, then replace $y$ with $x^2$. – ir7 Dec 19 '14 at 23:33

I would go about this by first splitting the series up:

$$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$

Letting $s=1+2x^2+3x^4$ we can do a few tricks:

$$s-x^2s=\begin{array}{c} 1&+2x^2&+3x^4+... \\ &-x^2&-2x^4-...\end{array}$$ $$=1+x^2+x^4+...$$

Which converges to $\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving this is not hard, and can be done by a technique like the above). This gives

$$s -x^2s=\frac{1}{1-x^2}\Leftrightarrow s=\frac{1}{(1-x^2)^2}=$$

Thus the original series converges to:

$$(1+x)s=\frac{(1+x)}{(1-x^2)^2}$$

For $-1 < x - 1$.

• Hmm..I got $\frac{x^2}{1-x}$ (see what I posted in reply to ir7's comment). Did I make a mistake somewhere? – Mathy Person Dec 19 '14 at 23:33
• Whoops my bad, I forgot a step – SBareS Dec 19 '14 at 23:36
• Apparently I am to tired for maths right now... – SBareS Dec 20 '14 at 0:05

Unless I'm mistaken, it is $$\sum_{n=1}^\infty nx^{2n-2} + \sum_{n=1}^\infty nx^{2n-1}$$ If you can compute one of the two terms, e.g. $\sum_{n=1}^\infty nx^{2n-2} = \sum_{n=0}^\infty (n+1)x^{2n} = \sum_{n=0}^\infty (n+1)(x^{2})^n$ (see e.g. this, then you'll also get the other term (by multiplying it by $x$), and thus the sum.

One way is to look at $$1+2x^2+3x^4+4x^6+5x^8+\dots$$ as $$1+2t+3t^2+4t^3+5t^4+\dots$$ where $t=x^2$. The last series is the derivative of $$1+t+t^2+t^3+t^4+t^5+\dots=\frac1{1-t}$$ Therefore, $$1+2t+3t^2+4t^3+5t^4+\dots=\frac1{(1-t)^2}$$ and $$1+2x^2+3x^4+4x^6+5x^8+\dots=\frac1{(1-x^2)^2}$$ Now, just multiply by $1+x$: \begin{align} 1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+\dots &=\frac{1+x}{(1-x^2)^2}\\ &=\frac1{(1-x)(1-x^2)} \end{align}


# How to expand binomials?

I'm working on a few proofs and am missing how this algebra works....

So, how does one expand $(k+1)^3\,$? Can I use FOIL? What does it expand to?

And how to expand $(k+1)^5\,$?

Thanks!

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look up Pascal's Triangle, and Binomial Theorem. Also no, you can't use FOIL, at very least you must use distributive property. – Joseph Skelton Jan 22 '13 at 20:59

Check out the entry Binomial Theorem in Wikipedia.

Putting $y = 1$, that will give you the tools you need to expand $(k+1)^3,\; (k+1)^5, \;$ and $\,(k + 1)^n\,$ for any non-negative integer $n$.

FOIL works fine for $(k + 1)^2 = k^2 + 2k + 1$

One can go a step further by distributing $(k+1)$ over $(k^2 + 2k + 1)$ to get $$(k^3 + 3k^2 + 3k + 1) = (k + 1)^3.$$

But for large exponents, it's handy to know the pattern of coefficients that correspond to different powers of $k$ in the expansion of $(k+1)^n$: Pascal's triangle shows this handy relationship.

I'll include an animation and image of "Pascal's Triangle" which displays the coefficients of expansions of a binomial $(k + 1)$ (these coefficients are referred to as: binomial coefficients): up to and including fourth and fifth degree binomials, respectively:

$\quad\quad\quad\quad$ $\quad\quad\quad\quad$

$$\text{Each number in the triangle is the sum of the two directly above it.}$$

To see how this "plays out" in the expansion of $(x + 1)^n,\;0 \le n \le 6$:

$$(x + 1)^0 = \color{blue}{\bf{1}}$$ $$(x + 1)^1 = \color{blue}{\bf{1}}\cdot x +\color{blue}{\bf{1}}$$ $$(x + 1)^2 = \color{blue}{\bf{1}}\cdot x^2 + \color{blue}{\bf{2}}x + \color{blue}{\bf{1}}$$ $$(x+1)^3 = \color{blue}{\bf{1}}\cdot x^3 + \color{blue}{\bf{3}}x^2 + \color{blue}{\bf{3}}x + \color{blue}{\bf{1}}$$ $$(x+1)^4 = \color{blue}{\bf{1}}\cdot x^4 + \color{blue}{\bf{4}} x^3+ \color{blue}{\bf{6}}x^2 + \color{blue}{\bf{4}}x +\color{blue}{\bf{1}}$$ $$(x+1)^5 = \color{blue}{\bf{1}}\cdot x^5 + \color{blue}{\bf{5}}x^4 + \color{blue}{\bf{10}} x^3 + \color{blue}{\bf{10}} x^2 + \color{blue}{\bf{5}}x + \color{blue}{\bf{1}}$$ $$(x + 1)^6 = \color{blue}{\bf{1}}\cdot x^6 + \color{blue}{\bf{6}}x^5 +\color{blue}{\bf{15}}x^4 + \color{blue}{\bf{20}}x^3 +\color{blue}{\bf{15}}x^2 + \color{blue}{\bf{6}}x + \color{blue}{\bf{1}}$$ $${\bf{\vdots}}$$

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Image source: Wikipedia, found at link to "Pascal's Triangle". – amWhy Jan 22 '13 at 22:17
Nice animated GIF! The answer is good, too :-) (+1) – robjohn Jan 22 '13 at 22:53
Thanks, @robjohn! It says more concisely, in animation, what would take a bit of explaining to write! – amWhy Jan 22 '13 at 22:56
@amWhy: It is unbelievable!! Great!! I cannot say anything. Nothing such this can describe the problem soooo well. +10! – Babak S. Jan 23 '13 at 15:50
This is incredible! Perfect explanation. Thank you! – user56763 Jan 23 '13 at 16:44

Before you jump to the binomial theorem (still the best way to go, in general, for expressions of the form $(a+b)^n$), let's start at the beginning. You undoubtedly know that $$(k+1)^3=(k+1)(k+1)(k+1)$$ We'll start by expanding $(k+1)(k+1)$. You can use FOIL here so we have $$(k+1)(k+1)=k^2+2k+1$$ We're two-thirds of the way to the answer. Now we have $$(k+1)^3=(k+1)(k^2+2k+1)$$ and by the distributive property, namely that $(a+b)c=ac+bc$, we have \begin{align} (k+1)^3=(k+1)(k^2+2k+1)&=(k)(k^2+2k+1)+(1)(k^2+2k+1)\\ &=(k^3+2k^2+k)+(k^2+2k+1)\\ &=k^3+3k^2+3k+1 \end{align} This will work for any positive integer exponent but, as Michael notes, you wouldn't want to do this for $(a+b)^n$.

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Thanks, makes sense. – user56763 Jan 23 '13 at 16:45
@ user56763: From what Rick Decker posted you can see that $(k+1)^3 = (k+1)(k+1)(k+1)$ can be expanded in the following way. Write down all possible 3-fold products of the form $ABC,$ where $A$ is a choice of one of the terms in the first $(k+1)$ factor, $B$ is a choice of one of the terms in the second $(k+1)$ factor, and $C$ is a choice of one of the terms in the third $(k+1)$ factor. Then add the 8 products. Thus, among the things you'll add will be $(k)(k)(1)$ and $(1)(k)(k)$ and $(k)(1)(k).$ Note that there are 3 of these with exactly two $k$'s and exactly one $1,$ so you get $3k^2$ ... – Dave L. Renfro Jan 28 '13 at 19:30

You can use "FOIL" twice. You should get $$k^3+3k^2+3k+1.$$ More generally, $$(a+b)^3 = a^3 + 3a^2b+3ab^2+ b^3.$$

Please don't vacilate between lower-case $k$ and capital $K$ in mathematical notation. Pick one and stick to it. Mathematical notation is case sensitive. Sometimes one uses lower-case $k$ and capital $K$ for two different things in the same problem, and you need to be clear about which is which.

But it would take a while to use FOIL to get $$(a+b)^9 = a^9+9a^8b+36a^7b^2+84a^6b^3+126a^5b^4+126a^4b^5+84a^3b^6+36a^2b^7+9ab^8+b^9.$$ That's one reason to be aware of the binomial theorem, which explains the pattern.

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What you're looking for is the binomial theorem, where y = 1.

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This is very scant. Something more would be good: some explanation or at least a link to an article. Of course, an answer should be more than simply a link. Look at the other answers for examples. – robjohn Jan 22 '13 at 22:56
Yeah, I was going to, but was having trouble typing out the math for a more exhaustive answer. – MITjanitor Jan 23 '13 at 2:10


# This is not a necessary part of the course i'm taking, but an interesting problem.

## Homework Statement

Let a; b; c $\in$ (1,∞) and m; n $\in$ (0,∞). Prove that

\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}

## The Attempt at a Solution

I do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.

Last edited:

Do you mean $$\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}$$

Or something else?

And what does "a; b; c (1;1) and m; n (0;1)" mean?

To your first post yes and thank you. Also, Let a; b; c $\in$ (1,∞) and m; n $\in$ (0,∞).

Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Let a; b; c $\in$ (1,∞) and m; n $\in$ (0,∞). Prove that

$\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}$

## The Attempt at a Solution

I do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.

See how you might use the "Change of base" formula:
$\displaystyle \log_T(P)=\frac{\log_R(P)}{\log_R(T)}=\frac{\log_{10}(P)}{\log_{10}(T)}=\frac{\ln(P)}{\ln(T)}$

I think you could use this property $$\log_b x = \frac {\log_k x} {\log_k b}$$ Using that, for example, $$\log_{b^mc^n} a = \frac {\log_a a}{\log_a b^mc^n} = \frac {1}{\log_a b^mc^n}$$ You could deal with the other logs similarly, so you would get some rational expression involving sums and products of $m\log_a b$ with permutations of a, b, c, m and n. Another consequence of the formula above is that $$\log_b a = \frac {\log_a a} {\log_a b} = \frac 1 {\log_a b}$$ so you should be able to express everything in terms of $A = \log_a b$, $B = \log_a c$ and $C = \log_b c$ and m and n. After some algebra (probably, quite some algebra) you should be able to simplify that into something manageable and prove the inequality.

Well, that certainly clears a lot up for me, but I would very much appreciate some examples of where to go from there. (As you can tell, I'm clearly not versed well enough in mathematics to attempt such a problem). Thank you for any more input and helping me get a better grasp on mathematics.

SammyS
Staff Emeritus
Homework Helper
Gold Member

M4th,

Where did you get this problem from? What is the level of the mathematics that might be used to solve this problem?

BTW: Notice what you get if a = b = c .

I am in a pre-calculus course but my teacher chose me to be part of a separate "group" of students who solves problems outside of the classroom. I would just like to be able to simply understand what is going on in some of these problems.

Last edited:

Using the properties of the logarithm as we already discussed, you can show that $$\\ \log_{b^mc^n} a + \log_{c^ma^n} b + \log_{a^mb^n} c = \frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B}$$ where $$1 \le A \le B$$ Then you will need to prove $$\frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B} \ge \frac 3 {m + n}$$

This is a great forum, I already have learned a bit more about logarithms just watching how you've broken this problem up into its subsequent pieces. I thank you for such rapid responses and hospitality. Im looking forward to learning more about mathematics from you all and would greatly appreciate some more information on this and many more problems to come in my quest to broaden my grasp on this subject. I would also greatly appreciate a worded summary as to how you get to these results. I understand alot of it, but am still missing some peices and any more details would be way more than I expected to get but would also be extremely helpful.

Thank you again

This forum is great because it helps you understand the problem and learn how to solve it. Learn, but not just copy a complete solution. That would be against the rules. The rules require that you make an attempt at solving the problem. Here that would be trying to get the expression to the form given above. You have been given enough information to do so. Now you should try and tell us whether that worked and when not, what exactly went wrong.

Yes my point exactly. My goal is to be able to arrive at, and transform equations with ease into more workable ones as you have above and really have a good grasp at how to do so with any other similar problems.

Now one problem in my understanding is I grasp how logba=logaa/logab and thus I see where

\\log_{b^mc^n} a = \frac {\log_a a}{\log_a b^mc^n} = \frac {1}{\log_a b^mc^n}

but how is that transformed into the last mathematical statement you made

explain how you can transform logb^mc^na = 1/logabmcn into the statement below

Using the properties of the logarithm as we already discussed, you can show that $$\\ \log_{b^mc^n} a + \log_{c^ma^n} b + \log_{a^mb^n} c = \frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B}$$ where $$1 \le A \le B$$ Then you will need to prove $$\frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B} \ge \frac 3 {m + n}$$

explain how you can transform logb^mc^na = 1/logabmcn into the statement below

Basic logarithm properties:
$$\log xy = \log x + \log y \\ \log x^y = y \log x$$

Apply them to $\log_a b^mc^n$ and see where that gets you.

Of course, now it's all starting to come together. Thanks again for your help.


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They wrote about it on the chalkboard using set notation: P = {Kyesha, Angie and Eduardo} When Angie's mother came to pick her up, she looked at the chalkboard and asked: What does that mean? If you … ALT Codes for Math Symbols: Set Membership & Empty Sets Read More » Admin Igcse Mathematics Revision Notes, O Level Mathematics Revision Notes 2 Comments 12,074 Views. In, sets theory, you will learn about sets and it’s properties. When picking a symbol, best to trust the symbol's unicode name for its meaning, not appearance. Occasionally we will introduce a new symbol to cater for an unusual requirement of a client. Created: Jan 19, 2018 | Updated: Feb 6, 2020. Symbols Used in this Book; Glossary; While a comprehensive list of notation is included in the appendix, that is meant mostly as a reference tool to refresh the reader of what notation means. Researchers and mathematicians have developed a language and system of notation around set theory. To fully embrace the world of professional Venn diagrams, you should have a basic understanding of the branch of mathematical logic called ‘set theory’ and its associated symbols and notation. The table below contains one example set… 8 February 2019 OSU CSE 1. Solution: Let P be the set of all members in the math Problem 1: Mrs. Glosser asked Kyesha, Angie and Eduardo to join the new math club. A variant solution, also based on mathtools, with the cooperation of xparse allows for a syntax that's closer to mathematical writing: you just have to type something like\set{x\in E;P(x)} for the set-builder notation, or \set{x_i} for sets defined as lists. Google Classroom Facebook Twitter. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. and symbols. Some notations for sets are: {1, 2, 3} = set of integers greater than 0 … Because rarely used symbol may look very different on another computer. Probability and statistics symbols table and definitions - expectation, variance, standard deviation, distribution, probability function, conditional probability, covariance, correlation After school they signed up and became members. CCSS.Math: HSS.CP.A.1. Quiz & Worksheet Goals A set is a collection of objects, things or symbols which are clearly identified.The individual objects in the set are called the elements or members of the set. Demo. In this notation, the vertical bar ("|") means "such that", and the description can be interpreted as "F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive". IGCSE 9-1 Exam Question Practice (Sets + Set Notation) 4.9 34 customer reviews. Relative complement or difference between sets. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic. That is OK, it is just the "Empty Set". 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# Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer?

I was having a discussion with a friend about the probability, and we came up with very different methods to solve it that lead to the same answer. The problem is pretty simple: you have two teams A and B playing a best of 7 series where wins are independent and the probability that team B wins is .35. What's the probability of team B winning th series? Friend says, "this is just a binomal random variable $X$ with $n=7$ and $p=.35$ and we are looking for $P(X\geq 4)$, which according to my TI-83 here is .1998".

I was rather convinced that this must be wrong. If we think of the series as a sequence of A's and B's, then we are looking for the probability of obtaining a sequence of length $n=4,5,6,7$ with 4 B's and the last element is a B. I was sure that his method will include the probability of obtaining, say, {B,A,B,B,B,A,A}, which we are not interested in and in fact couldn't even ever occur. So I figure that what we really want for a sequence of length $n$ is the probability of obtaining a sequence of length $n-1$ with exactly 3 B's, and then tacking a B onto the end. So for a sequence of length $n$, the probability should be $\binom{n-1}{3}.35^4 .65 ^{n-4}$, and then the answer should be $\sum_{n=4}^{7}\binom{n-1}{3}.35^4 .65 ^{n-4}$. I was confident that I was right and he was wrong, but then I plugged that into Wolfram Alpha and got... .1998.

What's going on here? Is it a coincidence?

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Why do you say that {B,A,B,B,B,A,A} could not ever occur? This looks like a valid sequence of wins to me... –  Code-Guru Jul 18 '12 at 19:22
Once team B wins 4 games the series is over. –  crf Jul 18 '12 at 19:24
See David Spencer's answer below. –  Code-Guru Jul 18 '12 at 19:32
To make more money, the league has decided that the series will go $7$ games, but the usual rules for determining the winner (first to win $4$) apply. Then Team B wins the modified series iff it wins the real series. –  André Nicolas Jul 18 '12 at 20:54

This is not a coincidence, you are looking at the same problem in two different ways. The situation {B,A,B,B,B,A,A} is irrelevant in your friend's argument, because you are not counting the number of different possible sequences of events.

Rather, you are attempting to estimate the probability of at least $n$ successes in $k$ trials, which is exactly what the binomial distribution does.

In your second approach, you are essentially estimating the number of possible legal sequences terminating in a victory condition, and then computing the total fraction out of all possible sequences.

Both approaches are equivalent. One is a little easier to implement ;)

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He is correct. Although this does include series that would be over after fewer than $7$ games "in real life", they would still all result in team B winning, no matter what happens after team B gets $4$ wins.

In your example, the probability that one of

{B,A,B,B,B,A,A}

{B,A,B,B,B,A,B}

{B,A,B,B,B,B,A}

{B,A,B,B,B,B,B}

happening without terminating series after four wins is equivalent to the probability of {B,A,B,B,B} happening across the first five games because later events do not affect anything.

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How to find the argument of $\cos 2 - i\sin 2$

How would you find the argument of the following number. $$\cos{2}-i\sin{2}$$

I'm aware that complex numbers in the form $r(\cos{\theta}+i\sin{\theta})$ have an argument of $\theta$, but what do you do with the $-$ sign?

Regardless of method, you will demonstrate some knowledge of Euler's identity and/or trigonometric identities.

Method I: Arctangent of slope

The argument is the arctangent of the ratio of the imaginary component to the real component, accounting for quadrant. If you forget to account for quadrant, this gives arguments in the interval $(-\pi/2, \pi/2)$, "half" of which are wrong. (Recall that arctangent is not a single-valued inverse. This is obvious when one remembers that tangent is periodic.)

Since $\cos 2 < 0$ and $-\sin 2 < 0$, this is in quadrant III.
\begin{align*} \arctan \frac{-\sin 2}{\cos 2} &= \arctan(-\tan 2)) \\ &= - \arctan \tan 2 &&\text{$\arctan$ is an odd function}\\ &= - \arctan \tan (2 - \pi) &&\text{$\tan$ has period $\pi$}\\ &= -(2 - \pi +\pi k), k \in \mathbb{Z} &&-\pi/2 < 2 - \pi < \pi/2 \text{.} \end{align*} Choosing $k=1$, this lands in quadrant III, giving the argument $-2$. If we had just mechanically evaluated, say with a calculator, \begin{align*} \arctan \frac{-\sin 2}{\cos 2} &= \arctan(2.18504\dots) \\ &= 1.14159\dots{} + \pi k, k \in \mathbb{Z} \text{.} \end{align*} The subset of these in quadrant III have $k$ odd. Picking $k = -1$, we get $-2$.

Of all the $k$ that give a result in the correct quadrant, which do you pick? You pick the one that conforms to your conventions for the range of value of an argument, if you have such a convention. If you do not, pick the one that makes your subsequent steps easier (or don't pick and leave the result as an equivalence class $\mod 2 \pi$).

Method II: Conjugation

If your only problem is the wrong sign of imaginary component, use conjugation. \begin{align*} \arg(\cos 2 - \mathrm{i} \sin 2) &= \arg(\overline{\cos 2 + \mathrm{i} \sin 2}) \\ &= \arg(\overline{\mathrm{e}^{\mathrm{i} 2}}) \\ &= -\arg(\mathrm{e}^{\mathrm{i} 2}) \\ &= -2 \text{.} \end{align*}

Method III: Even-odd identities

Sine is odd. Cosine is even. This is expressed in the even-odd identities. So $\cos(-2) = \cos(2)$ and $\sin(-2) = -\sin(2)$. Consequently, \begin{align*} \cos 2 - \mathrm{i} \sin 2 &= \cos -2 + \mathrm{i} \sin -2 \\ &= \mathrm{e}^{\mathrm{i}(-2)} \text{,} \end{align*} having argument $-2$.

Brahadeesh's and MrYouMath's answers use this method, without identifying what was done. Michael Rozenberg's answer combines this with the periodicity (by $2\pi$) identity, again without identifying what was done.

• Thank you very much. This has really cleared things up for me! – Ewan Miller Sep 9 '17 at 20:53

You can write this complex number as $\cos(-2) + i \sin(-2)$. Now you can see that the argument is $-2$.

$z=r(\sin\theta+i\sin\theta)$, where $r\geq0$ and $\theta=\arg{z}\in[0,2\pi)$.

$$\cos2-i\sin2=\cos(2\pi-2)+i\sin(2\pi-2).$$

Thus, $\arg(\cos2-i\sin2)=2\pi-2$.

• Thank you. Am I right in thinking then that there is a general rule that r(cosA-isinA)=r(cos(2π-A)+isin(2π-A))? – Ewan Miller Sep 9 '17 at 16:05
• @Ewan Miller It depends on definition. Sometimes $\arg{z}\in[0,2\pi)$ and sometimes $\arg{z}\in[-\pi,\pi)$. You need to ask your teacher. I like the first versa. – Michael Rozenberg Sep 9 '17 at 16:13
• In the case where $\arg{z}\in[-\pi,\pi)$, would it be r(cosA-isinA)=r(cos(-A)+isini(-A))? – Ewan Miller Sep 9 '17 at 16:17
• Also, I haven't got a teacher as I have just left school and am teaching this to myself in my gap year – Ewan Miller Sep 9 '17 at 16:19
• @Ewan Miller It's not so easy. If $A\in(-\pi,\pi)$ then you are right:$r(\cos(-A)+i\sin(-A))$ , but for $A=-\pi$ we get $r(\cos{A}+i\sin{A})$. All this depend on the level of your teacher. Maybe he don't see here problems. – Michael Rozenberg Sep 9 '17 at 16:23


# Thread: Counting, Permutation, and Combinations

1. ## Counting, Permutation, and Combinations

Guys I have several questions that I would like to ask about:

1. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, if the bride and the groom are among these 10 people, if:
a. the bride must be in the picture
b. exactly one of the bride and the groom is in the picture

2. How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other?

3. A professor writes 40 discrete mathematics true/false questions. Of these statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible.

1a. because the bride is already included among 6 people, therefore there are 5 more people that we can choose and arrange from, so it's 5!. I am not sure if this is right
1b. I don't quite understand the question

2. P(10,10) x P(11,6)

3. C(40,17)

I am not quite sure of all these answers, but suggestions/help to my questions are highly appreciated

2. a. the bride must be in the picture
Let us divide this problem into cases.
Code:
BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB
Where "X" stands for any other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.

In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.

b. exactly one of the bride and the groom is in the picture
Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.

3. Originally Posted by ThePerfectHacker
Let us divide this problem into cases.
Code:
BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB
Where "X" stands for any other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.

In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.

Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.
as for case 1:

the bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.

and for case 2:

the bride is not in picture but the groom is has the same possibilities as case 1 right?

then we just add case 1 and case 2?

4. Originally Posted by TheRekz
as for case 1:

the bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.

and for case 2:

the bride is not in picture but the groom is has the same possibilities as case 1 right?

then we just add case 1 and case 2?
I agree with you.

5. what if now the question is both the bride and the groom must be in the picture:

(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080

is this right?

6. Hello, TheRekz!

1. In how many ways can a photographer at a wedding
arrange 6 people in a row from a group of 10 people,
if the bride and the groom are among these 10 people, and:
a. the bride must be in the picture
b. exactly one of the bride and the groom is in the picture

a) The bride is in the picture.
Select five more from the other nine people: .$\displaystyle C(9,5)$ ways.
These six people can be arranged in $\displaystyle 6!$ ways.

Therefore, there are: .$\displaystyle C(9,5) \times 6!$ ways.

b) Either bride or the groom is in the picture (but not both).
There are 2 choices for the newlywed to be in the picture.
The other 5 people are chosen from the remaining 8 people (not the other newlywed).
. . There are: .$\displaystyle C(8,5)$ ways.
The six people can be arranged in $\displaystyle 6!$ ways.

Therefore, there are: .$\displaystyle 2 \times C(8,5) \times 6!$ ways.

2. How many ways are there for 10 women and six men to stand in a line
so that no two men stand next to each other?

Arrange the tenwomen in a row. .There are $\displaystyle P(10,10) = 10!$ possible arrangements.

Leave spaces between the women: .$\displaystyle \_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\ _\;W\;\_\;W\;\_\;W\;\_$

Arrange the six men in any six of the eveleven spaces.
. . There are: .$\displaystyle P(11,6)$ ways.

Therefore, there are: .$\displaystyle P(10,10) \times P(11,6)$ ways.

3. A professor writes 40 discrete mathematics true/false questions.
Of these statements in these questions, 17 are true.
If the questions can be positioned in any order,
how many different answer keys are possible?

Your answer is correct . . . $\displaystyle C(40,17)$

7. Originally Posted by TheRekz
what if now the question is both the bride and the groom must be in the picture:

(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080

is this right?
If both the bride and groom are in the picture then:

fix the bride in the first position:

BGXXXX = 1*1*8*7*6*5
BXGXXX = 1*8*1*7*6*5
.
.
.
BXXXXG = 1*8*7*6*5*1

Then there are 5*(1*1*8*7*6*5) ways to arrange them with the bride in the first position.

Now repeat with the bride in the second, third,....,sixth position and you get the same result each time. Therefore, if you add these all up you get (5*(1*1*8*7*6*5))*6 = 50,400

,

,

,

# A professor writes 40 discrete mathematics true/false

Click on a term to search for related topics.


# Integration by Trigonometric Substitution vs Table of Integral Solution

I'm not sure how to phrase this question, but I find myself confused over the correctness of a particular solution in the table of integral, specifically: $$\int\frac{dx}{\sqrt{x^2-a^2}}=\ln|\sqrt{x^2-a^2}+x|+C$$

I find that if you try to solve it with trigonometric substitution, you get a different answer:

So, for the example, since the form $$\sqrt{x^2-a^2}$$ is present, we use the substitution $$x=asec \theta$$, and $$dx=asec\theta tan \theta d\theta$$. Doing so:

$$\int\frac{asec\theta tan \theta d\theta}{\sqrt{(asec \theta)^2-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{\sqrt{a^2sec^2 \theta-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{a\sqrt{sec^2 \theta-1}}$$ Using the trigonometric identity $$tan^2 \theta=sec^2 \theta -1$$ $$\int\frac{asec\theta tan \theta d\theta}{atan \theta}$$

Cancelling like terms, we get $$\int sec \theta$$, which finally integrates to $$ln|sec \theta + tan \theta|+C$$

Undoing the substitution, using this triangle:

https://i.stack.imgur.com/6tkR7.png

With $$sec \theta = \frac{x}{a}$$ and $$tan \theta = \frac{\sqrt{x^2-a^2}}{a}$$, the final answer then is:

$$ln\biggl|\frac{x}{a} + \frac{\sqrt{x^2-a^2}}{a}\biggl|+C$$

Which is evidently different from the one from the table of integrals. Is my solution wrong or something?

• looks the same to me, since $-\ln |a|$ can be absorbed into $C$? – Calvin Khor Feb 7 at 13:48
• Oh right, that can be done. Thanks for pointing that out – Xyzar Feb 7 at 13:52

$$\ln \left|\dfrac{x+\sqrt{x^2-a^2}}{a}\right| +c= \ln |x+\sqrt{x^2-a^2}| - \ln|a| +c = \ln(x+\sqrt{x^2-a^2}) +k$$
$$k = c-\ln |a| =$$ constant


# Different Ways of Integrating $3\sin x\cos x$

I am asking this question for my son who is in (equivalent) twelfth grade and I failed to answer his query.

When he tries to integrate $3\sin x\cos x$, he finds that this can be done in at least following three ways. And these three ways do not produce equivalent results.

ONE

Let us assume, $\sin x = z$.

This gives, \begin{align*} \cos x &= \frac{dz}{dx}\\ \cos x dx &= dz \end{align*}

So, we can write,

\begin{align*} \int 3\sin x\cos x dx &=3 \int zdz\\ &=3 \frac{z^2}{2}\\ &=\frac{3}{2} \sin^2 x\\ &=\frac{3}{4}\times 2\sin^2 x\\ &=\frac{3}{4} (1 -\cos 2x)\\ \end{align*}

TWO

Let us assume, $\cos x = z$.

This gives, \begin{align*} -\sin x &= \frac{dz}{dx}\\ \sin x dx &= -dz \end{align*}

So, we can write,

\begin{align*} \int 3\sin x\cos x dx &=-3 \int zdz\\ &=-3 \frac{z^2}{2}\\ &=-\frac{3}{2} \cos^2 x\\ &=-\frac{3}{4}\times 2\cos^2 x\\ &=-\frac{3}{4} (1 +\cos 2x)\\ \end{align*}

THREE

\begin{align*} \int 3\sin x\cos x dx &=\frac{3}{2}\int 2\sin x\cos x dx\\ &=\frac{3}{2}\int \sin 2x dx\\ &=-\frac{3}{2}\times\frac{1}{2} \cos 2x\\ &=-\frac{3}{4} \cos 2x\\ \end{align*}

The results found in above three methods are not the same.

If we try a simple approach of evaluating the integration results at, $x = \frac{\pi}{6}$, we get as follows.

From the first one,

$\frac{3}{4} (1 -\cos 2x) = \frac{3}{4} (1 -\cos \frac{2\pi}{6}) = \frac{3}{4} (1 -\cos \frac{\pi}{3}) = \frac{3}{4} (1 - \frac{1}{2}) = \frac{3}{4}\times\frac{1}{2} = \frac{3}{8}$

From the second one,

$-\frac{3}{4} (1 +\cos 2x) = -\frac{3}{4} (1 +\cos \frac{2\pi}{6}) = -\frac{3}{4} (1 +\cos \frac{\pi}{3}) = -\frac{3}{4} (1 + \frac{1}{2}) = -\frac{3}{4}\times\frac{3}{2} = -\frac{9}{8}$

From the third one,

$-\frac{3}{4} \cos 2x=-\frac{3}{4} \cos \frac{2\pi}{6} = -\frac{3}{4} \cos \frac{\pi}{3} = -\frac{3}{4} \times \frac{1}{2} = -\frac{3}{8}$

Clearly, we are getting some nonequivalent results. We have failed to find the mistakes or explanations behind this. Your help will be appreciated.

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You can locate an several primitives of a function, they differ in a constant, but that does not mean that the results are wrong. –  Hiperion Sep 16 '13 at 5:50
Indefinite integration can be 'misleading' in this sense. –  copper.hat Sep 16 '13 at 5:52
A great question to ask students to see if they can come up with Zev's answer (or equivalent). –  nbubis Sep 16 '13 at 6:01
–  Jonas Meyer Sep 16 '13 at 6:07
When I was doing A-levels, I came across a similar example, but the two "different" answers to the integral involved respectively $\tan^2(x)$ and $\sec^2(x)$. Of course, these two expressions differ by a constant, but the explanation was non-obvious to me (and my dad) then, and I had to get my teacher to explain it. –  Hammerite Sep 16 '13 at 9:29

You're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that $$\frac{d}{dx}\left(-\frac{3}{4}\cos(2x)+C\right)=3\sin(x)\cos(x),$$ and that is precisely the relationship captured by the statement that

$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$

-
Indeed this is a great demonstration of that's so important. –  David H Sep 16 '13 at 5:50
So, if is asked to solve the problem in the examination, which approach he should follow? His textbook contains only one of the answers. And as you can see, some graders may like only the conventional answers. No disrespect intended. –  Masroor Sep 16 '13 at 6:01
@MMA - All the answers are equally "wrong" since non of them contain a constant. When you let the constant C "swallow" the other constant term, all three give the same result. –  nbubis Sep 16 '13 at 6:02
@nbubis But his text book contains only one of the answers, (I can not get the book now), and given the prevailing circumstances, it will be difficult to modify the system. I can put down the answer which is the book in six/seven hours when the book comes back from school. –  Masroor Sep 16 '13 at 6:07
@MMA I sympathize with your situation, especially if your particular school has you feeling that cynical towards the system. But I have no way of predicting what which technique a grader would consider most appropriate, because there isn't one (and if the grader doesn't know it then your son knows more calculus than his teacher). He should learn all three methods. Solving integrals is like choosing a golf-club for a given shot. The best one is the one that gets the job done. –  David H Sep 16 '13 at 6:23

All three answers are correct provided you add a constant to each one of those.

Because from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value. You can't evaluate the value of an indefinite integral without including constant.

And I am sure that in the examination, your son won't be asked to evaluate the value of an integral without providing limits of integration or providing its value at some other point.

For instance, in question it may be mentioned that evaluate the value of expression at x=π/6 , given its value at x=0 is 1. So in this case, all three answers will give the correct value i.e. 11/8

-


# [SOLVED]chain rule and e^(u)

#### DeusAbscondus

##### Active member
Hi folks,
I don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.

Here is the nub of my difficulty:

1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$

This I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.
But the following I do not understand; could someone help me towards understanding?

2. $\text{If }g(x)=e^{x^{4}} \text{then }g'(x)=4e^{x^4}x^3$

Am I on the right track to observe that, if 2. is correct, then g(x) is a composite function, hence subject to the chain rule?

And if so, is the following a generally valid way to work through this and all such problems:

$\text{If }g(x)=e^{x^{4}} \text{find }g'(x)$

$\text{Now}g'(x)=u'v' \text{by Chain Rule}$

$\text{So, let }u=x^4 \text{and }v=e^u$

$\text{Then }u'=4x^3 \text{and }v'=e^u \text{ (by some rule which currently exceeds my understanding)}$

$\text{Therefore }g'(x)=u'v'=4x^3*e^u=\text{(via substitution) }4x^3*e^{x^{4}}$

$\text{Which, simplified }=4e^{x^{4}}x^3$

Finally, I have a similar hesitation/scruple/sense of vertigo when it comes to dealing with another unusual derivative, that of:

$ln(x)$

If anyone can see why, having the read the foregoing, I would feel unsure of myself around this animal, could they possibly add some notes to help me tame it?

Regs,
DeusAbs

Last edited:

#### Jameson

Staff member
Hi DeusAbscondus,

1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that $$\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$.

In this case that becomes $$\displaystyle \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\frac{e^x \cdot e^h-e^x}{h}=\frac{e^x(e^h-1)}{h}$$.

Ok, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.

So the last bit becomes $$\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right)$$.

From here you need to compute that limit and you can do it a few ways. For a less rigorous approach that is more intuitive as well, I suggest graphing $$\displaystyle \frac{e^h-1}{h}$$ on a graphing calculator and you'll see that as h approaches zero, then the y value approaches 1.

So at the end we get that $$\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right) = e^x \cdot (1) = e^x$$, therefore $$\displaystyle \frac{d}{dx} e^x = e^x$$

Please let me know if any of that is unclear. Again, it's not a rigorous proof by any means, more of a "walk through".

2) You are completely correct that $$\displaystyle \frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot f'(x)$$. As you said this is an application of the chain rule.

#### Sudharaka

##### Well-known member
MHB Math Helper
$\text{Then }u'=4x^3 \text{and }v'=e^u \text{ (by some rule which currently exceeds my understanding)}$
Hi DeusAbscondus,

The only rule that is used here is the chain rule.

\begin{eqnarray}

g(x)&=&e^{x^{4}}\\

g'(x)&=&\frac{d}{d(x^4)}e^{x^{4}}\frac{d}{dx}x^4\\

&=&e^{x^{4}}(4x^3)\\

\therefore g'(x)&=&4x^3e^{x^{4}}

\end{eqnarray}

Kind Regards,
Sudharaka.

#### Fantini

##### "Read Euler, read Euler." - Laplace
MHB Math Helper
Perhaps to avoid particularizing and help you think of the exponential as any other function, think of $e^{x^4}$ as $h(x) = e^x$ and $g(x) = x^4$, then $(h \circ g)(x) = h(g(x)) = e^{x^4}$ and $(h \circ g)' (x) = h'(g(x)) \cdot g'(x) = e^{x^4} \cdot (4x^3) = 4 e^{x^4} x^3.$

#### chisigma

##### Well-known member
Hi DeusAbscondus,

1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that $$\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$.

In this case that becomes $$\displaystyle \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\frac{e^x \cdot e^h-e^x}{h}=\frac{e^x(e^h-1)}{h}$$.

Ok, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.

So the last bit becomes $$\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right)$$.

From here you need to compute that limit and you can do it a few ways...
One possible 'rigorous' way to demonstrate that...

$\displaystyle \lim_{h \rightarrow 0} \frac{e^{h}-1}{h}=1$ (1)

... is the following...

a) You start from the definition of exponential...

$\displaystyle e^{h}= \lim_{n \rightarrow \infty} (1+\frac{h}{n})^{n}$ (2)

b) You demonstrate the identity...

$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{h}{n})^{n}= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac {h^{k}}{k!} = 1 + h + \frac {h^{2}}{2}+... \frac{h^{n}}{n!}+...$ (3)

c) from (3) You derive...

$\displaystyle \frac{e^{h}-1}{h}= 1 + \frac{h}{2} +... + \frac{h^{n-1}}{n!}+...$ (4)

... and the (1) follows immediately...

Kind regards

$\chi$ $\sigma$

#### DeusAbscondus

##### Active member
thanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \circ g)$ and some other stuff besides.

Just a quick follow up:
can one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?

and, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?

thx again gentelmen,

DeusAbs

#### Jameson

Staff member
It's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.

Anyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.

Let me know what you think!

#### Sudharaka

##### Well-known member
MHB Math Helper
thanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \circ g)$ and some other stuff besides.

Just a quick follow up:
can one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?

and, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?

thx again gentelmen,

DeusAbs
Hi DeusAbscondus,

A good place filled with video lessons is Khan Academy. Here is a link to the Epsilon Delta limit definition introduction video.

Kind Regards,
Sudharaka.

#### CaptainBlack

##### Well-known member
Hi folks,
I don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.

Here is the nub of my difficulty:

1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$

This I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.
But the following I do not understand; could someone help me towards understanding?
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:

$\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n$

but it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..

So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).

CB

#### DeusAbscondus

##### Active member
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:

$\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n$

but it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..

So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).

CB
I just had a look at graphs of these functions: given that they are mirror images of each other graphically, I would have thought their inter-definability (if i can put it thus) bordered the trivial.

I'd also have thought my post here (indeed, the combined oeuvre of my posts to date ) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.
Prescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?

I always enjoy your sparse, ascetical take on things, Cap'n.

So, thanks for the input,
Deus Abs

#### DeusAbscondus

##### Active member
It's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.

Anyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.

Let me know what you think!
Thanks Jameson; just spent some hours there! I suffer from a dearth of really first-rate material and my course notes are a nest of errors and poorly edited material, so I appreciate such a rich fund of examples, proofs, worked examples and tips.

I'll be sure to make a fuller response once I've had more time to check it.

DeusAbs

#### chisigma

##### Well-known member
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:

$\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n$

but it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..

So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).

CB
More that 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...

$\displaystyle e^{z}= \lim_{n \rightarrow \infty} (1+\frac{z}{n})^{n}$

$\displaystyle \ln z= \lim_{n \rightarrow \infty} n\ (z^{\frac{1}{n}}-1)$ (1)

... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
I'd also have thought my post here (indeed, the combined oeuvre of my posts to date ) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.
Prescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?
You must have at least some idea of how the exponential function is defined, or at the very least how "e" is defined, otherwise you are starting in the middle.

(I would recommend you get a copy of Morris Kline's book "Calculus; An intuitive and physical approach", the only flaw of which is its use of US Customary Units)

CB

#### Opalg

##### MHB Oldtimer
Staff member
More than 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...

$\displaystyle e^{z}= \lim_{n \rightarrow \infty} \Bigl(1+\frac{z}{n}\Bigr)^{n}$

$\displaystyle \ln z= \lim_{n \rightarrow \infty} n\ (z^{\frac{1}{n}}-1)$ (1)

... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...
In my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\mathbb{R}\to(0,\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)

#### chisigma

##### Well-known member
In my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\mathbb{R}\to(0,\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)
On the basis of the [easily enough demonstrable...] identity...

$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{z}{n})^{n} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{z^{k}}{k!}$ (1)

... the two definitions must be considered fully equivalent. On the purely conceptual basis however the original Euler's definition requires only the concept of limit and is 'more elementary' than any other definition and that is an excellent reason for defining it as 'the best'... at least for me..

Kind regards

$\chi$ $\sigma$

#### QuestForInsight

##### Member
1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$
This is very easy to get using the power series definition from Opalg's post.

\begin{aligned} & (e^x)' = \bigg(\sum_{k \ge 0}\frac{x^k}{k!}\bigg)' = \sum_{k \ge 0}\frac{kx^{k-1}}{k!} = \sum_{k \ge 1}\frac{kx^{k-1}}{k!} = \sum_{k+1 \ge 1}\frac{(k+1)x^{k}}{(k+1)!} = \sum_{k \ge 0}\frac{x^{k}}{k!}.\end{aligned}

#### chisigma

##### Well-known member
This is very easy to get using the power series definition from Opalg's post.

\begin{aligned} & (e^x)' = \bigg(\sum_{k \ge 0}\frac{x^k}{k!}\bigg)' = \sum_{k \ge 0}\frac{kx^{k-1}}{k!} = \sum_{k \ge 1}\frac{kx^{k-1}}{k!} = \sum_{k+1 \ge 1}\frac{(k+1)x^{k}}{(k+1)!} = \sum_{k \ge 0}\frac{x^{k}}{k!}.\end{aligned}
That is true of course but [conceptually...] the preliminary demonstration of the convergence theorem of the series of derivatives is required. The computation of derivative of $e^{x}$ using the standard derivative definition require only to demonstrate that...

$\displaystyle \lim_{h \rightarrow 0} \frac {e^{h}-1}{h}=1$ (1)

... and that requires elementary instruments...

Kind regards

$\chi$ $\sigma$


# 77 to use the alternating series test consider a n f

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Chapter 11 / Exercise 77
Single Variable Calculus
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77. To use the alternating series test, consider a n = f ( n ) , where f ( x ) = arctan( a/x ) . We need to show that f ( x ) is decreasing. Since f 0 ( x ) = 1 1 + ( a/x ) 2 - a x 2 , we have f 0 ( x ) < 0 for a > 0 , so f ( x ) is decreasing for all x . Thus a n +1 < a n for all n , and as lim n →∞ arctan( a/n ) = 0 for all a , by the alternating series test, X n =1 ( - 1) n arctan( a/n ) converges. 78. The n th partial sum of the series is given by S n = 1 - 1 2 + 1 3 - · · · + ( - 1) n - 1 n , so the absolute value of the first term omitted is 1 / ( n + 1) . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than 1 / ( n +1) . Thus, we want to choose n large enough so that 1 / ( n +1) 0 . 01 . Solving this inequality for n yields n 99 , so we take 99 or more terms in our partial sum. 79. The n th partial sum of the series is given by S n = 1 - 2 3 + 4 9 - · · · + ( - 1) ( n - 1) 2 3 ( n - 1) ,
##### We have textbook solutions for you!
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Chapter 11 / Exercise 77
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750 Chapter Nine /SOLUTIONS so the absolute value of the first term omitted is (2 / 3) n . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than (2 / 3) n . Thus, we want to choose n large enough so that (2 / 3) n 0 . 01 . Solving this inequality for n yields n 11 . 358 , so taking 12 or more terms in our partial sum is guaranteed to be within 0 . 01 of the sum of the series. Note: Since this is a geometric series, we know the exact sum to be 1 / (1+2 / 3) = 0 . 6 . The partial sum S 12 is 0 . 595 , which is indeed within 0 . 01 of the sum of the series. Note, however, that S 11 = 0 . 6069 , which is also within 0 . 01 of the exact sum of the series. Theorem 9.9 gives us a value of n for which S n is guaranteed to be within a small tolerance of the sum of an alternating series, but not necessarily the smallest such value. 80. The n th partial sum of the series is given by S n = 1 2 - 1 24 + 1 720 - · · · + ( - 1) n - 1 (2 n )! , so the absolute value of the first term omitted is 1 / (2 n + 2)! . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than 1 / (2 n +2)! . Thus, we want to choose n large enough so that 1 / (2 n +2)! 0 . 01 . Substituting n = 2 into the expression 1 / (2 n + 2)! yields 1 / 720 which is less than 0.01. We therefore take 2 or more terms in our partial sum. 81. Since 0 c n 2 - n for all n , and since 2 - n is a convergent geometric series, c n converges by the Comparison Test. Similarly, since 2 n a n , and since 2 n is a divergent geometric series, a n diverges by the Comparison Test. We do not have enough information to determine whether or not b n and d n converge. 82. (a) The sum a n · b n = 1 /n 5 , which converges, as a p -series with p = 5 , or by the integral test: Z 1 1 x 5 dx = lim b →∞ x - 4 ( - 4) b 1 = lim b →∞ b - 4 ( - 4) + 1 4 = 1 4 . Since this improper integral converges, a n · b n also converges. (b) This is an alternating series that satisfies the conditions of the alternating series test: the terms are decreasing and have limit 0 , so ( - 1) n / n converges. (c) We have a n b n = 1 /n , so a n b n is the harmonic series, which diverges.


# Are these two optimization problems equivalent to each other?

Let $\mathbf{x}=[x_1,\ldots,x_K]^T$. For a fixed vector $\mathbf{a}$, I have the following optimization problem : \begin{array}{rl} \min \limits_{\mathbf{x}} & | \mathbf{a}^T \mathbf{x} | \\ \mbox{s.t.} & x_k\ge 0, \forall k \\ & \sum_k x_k=1 \end{array} The second optimization problem is: \begin{array}{rl} \min \limits_{\mathbf{x}} & ( \mathbf{a}^T \mathbf{x})^2\\ \mbox{s.t.} & x_k\ge 0, \forall k \\ & \sum_k x_k=1 \end{array}

My question: are these two problems equivalent to each other? if yes, how to solve them ? and which one is easier to solve ?

• yes they are equivalent. have you tried the Lagrange multiplier method? – user251257 Aug 23 '15 at 16:07
• The objective function of the problem 1 is $\ell^1$-norm and problem 2 is $\ell^2$-norm, so although equivalent, the objective function of the second problem is continuously differentiable, therefore the second problem is much easier to solve. – guille_NP Aug 23 '15 at 21:22
• @guille_NP: The second problem is indeed much easier to solve, but definitely not for the reason you provide. Your statements about norms are simply wrong (for example, the second objective is not a norm...). Just say $t \mapsto t^2$ is increasing on $\mathbb{R}_+$ (so that the problems are equivalent) and differentiable (so that the second problem is easier). – dohmatob Aug 24 '15 at 2:35
• @mat: For solving the second problem, checkout this thread which treats a slightly more general problem math.stackexchange.com/q/1309671/168758 – dohmatob Aug 24 '15 at 3:07

Which is easier? Neither. Both of these models can be solved analytically, and in the exact same way. First, let's knock out the easy cases:

1. If any $a_i=0$ for some $i$, then the optimal value of either model is clearly $0$, as demonstrated by selecting setting $x$ to be the $i$th unit vector (the vector with $1$ at position $i$ and zeros everywhere else).

2. If $a_i>0$ and $a_j<0$ for some pair $(i,j)$, then again the optimal value is zero. Just set $$x_i=-a_j/(a_i-a_j), \quad x_j = a_i/(a_i-a_j)$$ and the other elements of $x$ to zero.

3. The cases that remain are where $a$ is entirely positive or entirely negative. But if $a$ is negative, then substituting $a$ for $-a$ will not change the optimal value or the values of $x$ that achieve this value, thanks to the presence of the absolute value or square.

4. So we now have one case remaining: when $a$ is a positive vector. But in this case, $a^Tx$ is positive over the simplex, allowing us to drop the absolute value! \begin{array}{ll} \text{minimize} & a^T x \\ \text{subject to} & \mathbf{1}^T x = 1 \\ & x \geq 0 \end{array} We can solve this by inspection: the optimal value is $\min_i a_i$ for the first model, and $\min_i a_i^2$ for the second. If the minimizing $a_i$ is unique, then the unique solution is the $i$th unit vector. But if there are multiple elements of $a$ with the same magnitude, any convex combination of those corresponding unit vectors is a solution.

Putting it all together, the solution is $\min_i |a_i|$ or $\min_i a_i^2$ if $a\succeq 0$ or $a\preceq 0$, and $0$ otherwise. There really was no need for case 1, but it was easy to see.

• I am interested in the case where the elements of $\mathbf{a}$ are $\ge 0$. I suppose that adding a constant $b$ to the objective function, i.e. $\mathbf{a}^T\mathbf{x}+b$, will not affect the solution ($i$th unit vector) but it will modify the optimal value to $\min_i a_i +b$. Am I right? – tam Aug 24 '15 at 9:37
• Yes, that is right. – Michael Grant Aug 24 '15 at 10:56
• Please, I have an additional question (I don't know if I have to post it in a seperate question): I want to maximize $|\mathbf{a}^T\mathbf{x}-b|$, with $\mathbf{1}^T\mathbf{x}=1$, $\mathbf{x} >0$ and $\mathbf{a}^T\mathbf{x} >b$. How to solve this kind of problems. (note that $b>0$) – tam Aug 29 '15 at 10:42
• I think that this problem can be re-written as: maximize $\mathbf{a}^T \mathbf{x}-b$, with $\mathbf{1}^T\mathbf{x}=1$ and $\mathbf{x}>0$. So the solution is similar to that in 4), with the difference of putting $\max_i a_i-b$ instead of $\min_i |a_i-b|$. Is it right ? – tam Aug 29 '15 at 16:03

Let's go for an explicit analytic construction of the set of all the solutions to your problem.

Basic notation: $e_K := \text{ column vector of }K\text{ }1'$s. $\langle x, y \rangle$ denotes the inner product between two vectors $x$ and $y$. For example $\langle e_K, x\rangle$ simply amounts to summing the components of $x$.

Recall the following definitions, to be used without further explanation.

Preimage: Given abstract sets $X$, $Y$, $Z \subseteq Y$., the preimage of $Z$ under $f$ is defined by $f^{-1}Z := \{x \in X | f(x) \in Z\}$. This has nothing to do with function inversion.

Convex hull: Given a subset $C$ of $\mathbb{R}^K$, its convex hull is defined by $\textit{conv }C :=$ smallest convex subset of $\mathbb{R}^K$ containing $C$.

Indicator function: Given a subset $C$ of $\mathbb{R}^K$ its indicator function $i_C:\mathbb{R}^K \rightarrow (-\infty, +\infty]$ is defined by $i_C(x) = 0$ if $x \in C$; otherwise $i_C(x) = +\infty$.

Simplex: The $K$-dimensional simplex is defined by $\Delta_K := \{x \in \mathbb{R}^K|x\ge 0, \langle e_K, x\rangle = 1\}$. Equivalently, $\Delta_K = \{\text{rows of the }K \times K\text{ identity matrix }I_K\}$. Each row of $I_K$ is a vertex of $\Delta_K$.

Faces of a simplex: Given a subset of indices $I \subseteq \{1,2,...,K\}$, the face of $\Delta_K$ spanned by $I$, denoted $F_K(I)$, is defined to be the convex hull of those rows of $I_k$ indexed by $I$. Trivially, $\Delta_K$ is a face of itself. Geometrically, $F_K(I)$ is a $\#I$-dimensional simplex whose vertices are those vertices of $\Delta_K$ which labelled by $I$.

Let $f:\mathbb{R}^K \rightarrow (-\infty,+\infty]$ be an extended real-value function.

Subdifferential: $\partial f(x) := \{g \in \mathbb{R}^K| f(z) \ge f(s) + \langle g, z - x\rangle, \forall z \in \mathbb{R}^K\}$.

Fenchel-Legengre transform: $f^*(x) := \underset{z \in \mathbb{R}^K}{\text{max }}\langle x, z\rangle - f(z)$.

The optimal value of your objective under the given constraints can be conveniently written as \begin{eqnarray} v = \underset{x \in \mathbb{R}^K}{\text{min }}g(x) + f(Dx), \end{eqnarray} where $D := a^T \in \mathbb{R}^{1 \times K}$, $g := i_{\Delta_K}$ and $f:s \mapsto \frac{1}{2}s^2$. One recognizes the above problem as the primal form of the saddle-point problem \begin{eqnarray*} \underset{x \in \mathbb{R}^K}{\text{min }}\underset{s \in \mathbb{R}}{\text{max }}\langle s, Dx\rangle + g(x) - f^*(s) \end{eqnarray*} Given a dual solution $\hat{s} \in \mathbb{R}$, the entire set of primal solutions $\hat{X}$ is given by (one can check that sufficient conditions that warrant the strong Fenchel duality Theorem) \begin{eqnarray*} \hat{X} = \partial g^*(-D^T\hat{s}) \cap D^{-1}\partial f^*(\hat{s}). \end{eqnarray*} For your problem, one easily computes, $g^*(y) = \underset{x \in \Delta_K}{\text{max }}\langle x, y\rangle$, and so by the Bertsekas-Danskin theorem (see Proposition A.22 of Bertsekas' PhD thesis) for subdifferentials, we have \begin{eqnarray*} \begin{split} \partial g^*(-D^T\hat{s}) &= \partial g^*(-\hat{s}a) = ... \text{( some computations masked )} \\ &= F_K\left(\{1 \le i \le K | \hat{s}a_i\text{ is a minimal component of }\hat{s}a\}\right). \end{split} \end{eqnarray*} Also $\partial f^*(\hat{s}) = \{\hat{s}\}$, and so $D^{-1}\partial f^*(\hat{s}) = \{x \in \mathbb{R}^K|\langle a, x\rangle = \hat{s}\}$. Thus the set of all primal solutions is \begin{eqnarray*} \hat{X} = F_K\left(\{1 \le i \le K | \hat{s}a_i\text{ is a minimal component of }\hat{s}a\}\right) \cap \{x \in \mathbb{R}^K|\langle a, x\rangle = \hat{s}\}. \end{eqnarray*} It remains now to find a dual solution. Define $\alpha := \underset{1 \le i \le K}{\text{min }}a_i \le \beta := \underset{1 \le i \le K}{\text{max }}a_i$. One computes \begin{eqnarray*} \begin{split} v &= \underset{x \in \Delta_K}{\text{min }}\underset{s \in \mathbb{R}}{\text{max }}s\langle a, x\rangle -\frac{1}{2}s^2 = \underset{s, \lambda \in \mathbb{R}}{\text{max }}\underset{x \ge 0}{\text{min }}\langle sa + \lambda e_K, x\rangle - \frac{1}{2}s^2 - \lambda\\ &=\underset{s, \lambda \in \mathbb{R}}{\text{max }}-\frac{1}{2}s^2 - \lambda\text{ subject to }sa + \lambda e_K \ge 0\\ &=-\underset{s, \lambda \in \mathbb{R}}{\text{min }}\frac{1}{2}s^2 + \lambda\text{ subject to }\lambda \ge -\underset{1 \le i \le K}{\text{min }}sa_i\\ &= -\underset{s \in \mathbb{R}}{\text{min }}\frac{1}{2}s^2 -\underset{1 \le i \le K}{\text{min }}sa_i = -\frac{1}{2}\underset{s \in \mathbb{R}}{\text{min }}\begin{cases}(s - \alpha)^2 - \alpha^2, &\mbox{ if }s \ge 0,\\(s - \beta)^2 - \beta^2, &\mbox{ otherwise}\end{cases} \end{split} \end{eqnarray*} Thus \begin{eqnarray*} \hat{s} = \begin{cases}\beta, &\mbox{ if }\beta < 0,\\0, &\mbox{ if }\alpha \le 0 \le \beta,\\\alpha, &\mbox{ if }\alpha > 0.\end{cases} \end{eqnarray*} Therefore the set of all solutions to the original / primal problem is \begin{eqnarray*} \hat{X} = \begin{cases}F_K\left(\{1 \le i \le K | a_i = \beta\}\right), &\mbox{ if }\beta < 0,\\ \Delta_K \cap \{x \in \mathbb{R}^K| \langle a, x\rangle = 0\}, &\mbox{ if } \alpha \le 0 \le \beta,\\ F_K\left(\{1 \le i \le K | a_i = \alpha\}\right), &\mbox{ if }\alpha > 0,\end{cases} \end{eqnarray*}

Now that you have the hammer, find the nails...

• Case (1): $\beta < 0$. Choose any $k \in \{1,2,...,K\}$ such that $a_k = \beta$, and take $\hat{x} = k$th row of $I_K$.

• Case (2a): $a_k = 0$ for some $k$. Take $\hat{x} = k$th row of $I_K$.

• Case (2b): $\alpha < 0 < \beta$, and $a_k \ne 0 \forall k$. Choose $k_1, k_2 \in \{1,2,...,k\}$ such that $a_{k_1} < 0 < a_{k_2}$, and take $\hat{x}_k = \frac{1}{{a_{k_2} - a_{k_1}}}\begin{cases}a_{k_2}, &\mbox{ if }k = k_1,\\-a_{k_1},&\mbox{ if }k = k_2,\\0, &\mbox{ otherwise.}\end{cases}$

• Case (3): $\alpha > 0$. Choose any $k \in \{1,2,...,K\}$ such that $a_k = \alpha$, and take $\hat{x} = k$th row of $I_K$.

• Dude. If I were a professor I'd want you in my research group cranking out papers. This is impressive, and you have my vote. But don't you think this is like bringing a fire truck to put out a campfire? – Michael Grant Aug 24 '15 at 14:41
• @MichaelGrant: Your solution is actually more elegant in many respects. Mine is indeed probably an overkill. Mindful of this, I tried to be particularly pedagocial here :) On such toy problems ($D$ is a vector, $g^*$ is a polyhedral function, etc.), it turns out that one can explicitly construct the entire solution set using Fenchel's machinery. Unfortunately, this is not always the case. :) – dohmatob Aug 25 '15 at 6:33

Your problem can be equivalently modeled to LP with an extra variable, say $\epsilon$ and two more simple linear constraints, $| \mathbf{a}^T \mathbf{x} |\leq \epsilon$. Now your problem can be rewritten as $$\begin{array}{rl} \min \limits_{\mathbf{x}} & \epsilon \\ \mbox{s.t.} & \mathbf{a}^T \mathbf{x} \leq \epsilon \\&- \mathbf{a}^T \mathbf{x} \leq \epsilon \\ & x_k\ge 0, \forall k \\ & \sum_k x_k=1 \end{array}$$

Now you can solve this problem using simplex method or any other method you like!


# For which parameter values the function is continuous and differentiable

#### Yankel

##### Active member
For which values of a,b and c, the next function is continuous and differentiable at x=2 ?

$$\left\{\begin{matrix} 3x-1 & x\leq 2\\ ax^{2}+bx+c & x>2 \end{matrix}\right.$$

1. b=2-c
2. b=6+2c+2a
3. 7+c-2a
4. b=3-a-(3/4)c

I know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.
I have calculated the limit when f goes to 2 from the right side, and I got:

4a+2b+c

and so for continuously I need:

4a+2b+c = 5

but here I got stuck...

#### Bacterius

##### Well-known member
MHB Math Helper
The only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).

Recall that for a function to be continuous at $a$, you need:

$$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$

So in your case, you need to show that the following is true:

$$\lim_{x \to 2^{-}} 3x - 1 = \lim_{x \to 2^{+}} ax^2 + bx + c ~ ~ ~ \Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$

Which is what you got, so this is correct. But, you also need differentiability.

Recall that for a function $f(x)$ to be differentiable at $a$, you need:

$$\lim_{x \to a^{-}} f'(x) = \lim_{x \to a^{+}} f'(x)$$

So, differentiate both pieces of the function, and show that the following is true:

$$\lim_{x \to 2^{-}} 3 = \lim_{x \to 2^{+}} 2ax + b ~ ~ ~ \Longleftrightarrow ~ ~ ~ 3 = 4a + b$$

You are now left with two equations in three unknowns:

$$4a + 2b + c = 5$$

$$4a + b = 3$$

Subtract the second from the first, to obtain:

$$b + c = 2 ~ ~ ~ \Longleftrightarrow b = 2 - c$$

And substitute this back into the second:

$$4a + 2 - c = 3 ~ ~ ~ \Longleftrightarrow ~ ~ ~ a = \frac{1}{4} \left ( c + 1 \right )$$

This gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:

$$a = \frac{3}{4} ~ ~ ~ \text{and} ~ ~ ~ b = 0$$

That is one solution, out of infinitely many. This is also the only class of solutions.

Last edited:

#### chisigma

##### Well-known member
For which values of a,b and c, the next function is continuous and differentiable at x=2 ?

$$\left\{\begin{matrix} 3x-1 & x\leq 2\\ ax^{2}+bx+c & x>2 \end{matrix}\right.$$

1. b=2-c
2. b=6+2c+2a
3. 7+c-2a
4. b=3-a-(3/4)c

I know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.
I have calculated the limit when f goes to 2 from the right side, and I got:

4a+2b+c

and so for continuously I need:

4a+2b+c = 5

but here I got stuck...
You have to impose first that $\displaystyle \lim_{x \rightarrow 2 +} f(x) = \lim_{x \rightarrow 2 -} f(x)$ and after that $\displaystyle \lim_{x \rightarrow 2 +} f ^{\ '} (x) = \lim_{x \rightarrow 2 -} f^{\ '}(x)$...

Kind regards

$\chi$ $\sigma$

#### HallsofIvy

##### Well-known member
MHB Math Helper
The only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).

Recall that for a function to be continuous at $a$, you need:

$$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$

So in your case, you need to show that the following is true:

$$\lim_{x \to 2^{-}} 3x - 1 = \lim_{x \to 2^{+}} ax^2 + bx + c ~ ~ ~ \Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$

Which is what you got, so this is correct. But, you also need differentiability.

Recall that for a function $f(x)$ to be differentiable at $a$, you need:

$$\lim_{x \to a^{-}} f'(x) = \lim_{x \to a^{+}} f'(x)$$
This is true, but not an obvious statement. It looks like the definition of "continuous" but derivatives are NOT necessarily contuinuous. What is true is that any derivative, even if not continuous, satisfies the "intermediate value theorem".

So, differentiate both pieces of the function, and show that the following is true:

$$\lim_{x \to 2^{-}} 3 = \lim_{x \to 2^{+}} 2ax + b ~ ~ ~ \Longleftrightarrow ~ ~ ~ 3 = 4a + b$$

You are now left with two equations in three unknowns:

$$4a + 2b + c = 5$$

$$4a + b = 3$$

Subtract the second from the first, to obtain:

$$b + c = 2 ~ ~ ~ \Longleftrightarrow b = 2 - c$$

And substitute this back into the second:

$$4a + 2 - c = 3 ~ ~ ~ \Longleftrightarrow ~ ~ ~ a = \frac{1}{4} \left ( c + 1 \right )$$

This gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:

$$a = \frac{3}{4} ~ ~ ~ \text{and} ~ ~ ~ b = 0$$

That is one solution, out of infinitely many. This is also the only class of solutions.


How to compute $\sum_{k =1}^{100}(-1)^k$

Today I tried to compute $$\sum_{k =1}^{100}(-1)^k$$

Is there a way to find the result more quickly ?

Below if my attempt to find the result.

Especially without considering the case of odd and even numbers like I did ?

Let's consider the following sum:

$$\sum_{k=1}^{n}(-1)^k = (-1)^1 + (-1)^2 + (-1)^3 + ... +(-1)^n$$

If $n$ is even then $\frac n2$ terms have an even exponent and $\frac n2$ terms have an odd exponent.

$$(-1)^p = -1 \tag{when p is odd}$$ $$(-1)^p = 1 \tag{when p is even}$$

Then when $n$ is even

\begin{align} \sum_{k=1}^{n}(-1)^k &= \frac n2 \times(-1) + \frac n2 \times 1 \\ & = \frac n2 - \frac n2 \\ & = 0 \\ \end{align}

100 is an even number, so

$$\sum_{k =1}^{100}(-1)^k = 0$$

• Call the sum $S$, so that $S=(-1)+(1)+\dotsb+(-1)+(1)$. Notice that $-S=(1)+(-1)+\dotsb+(1)+(-1)$—that is, $-S$ is just $S$ reversed. So, $S=-S$, which means that $S=0$. – Akiva Weinberger Sep 23 '14 at 20:15
• If you say that $n/2$ terms have an even exponent, then you are already assuming that $n$ is even. – Dietrich Burde Sep 23 '14 at 20:15
• The quick way is to cancel $1$s with $-1$s. You have equally many of both, and if you didn't, you'd have just one term left after canceling. But of course it's also a finite geometric series, and there's a standard formula for that. – Michael Hardy Sep 23 '14 at 20:57

yes the series has a direct summation formula: $$S_k=\sum_{i=1}^k (-1)^i={1 \over 2}\Big((-1)^k-1\Big)$$ you can prove this easily using induction. \begin{aligned} \mbox{k=1: }\ \ \ &true\\ \mbox{k }\to\mbox{ k+1: }\ \ \ &S_{k+1}=S_{k}+(-1)^{k+1}&={1 \over 2}\Big((-1)^k-1\Big)+(-1)^{k+1} =\\ & &={1 \over 2}\Big((-1)^k+2(-1)^{k+1}-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(1+2(-1)^{1})-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(1-2)-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(-1)-1\Big) =\\ & &={1 \over 2}\Big((-1)^{k+1}-1\Big) \ \ \ \square \end{aligned}

Note that $(-1)^{2k-1}=-1$ and $(-1)^{2k}=1$ for all $k\geq1$. Hence the sum upto any even power should equal $0$.

The easiest way to find the result is to look at the first few partial sums

$$-1, 0, -1, \ldots$$

and identify the pattern. Everything beyond that is just seeking to give a more rigorous justification of it.

• Sure. I think that's what I am looking for. A rigorous justification but explained quickly. – alexandrekow Sep 23 '14 at 20:37

Since $(-1)^{2n}=1$ and $(-1)^{2n-1}=-1$, we have $$\sum_{k=1}^n (-1)^k = -1+1-1+1-\cdots \pm 1=\left\{\begin{array}{ll}-1 & \text{if}\,n\,\text{is odd}\\ 0 & \text{if}\,n\,\text{is even}\end{array}\right.$$

$$\sum_{k=1}^{n}(-1)^k=\sum_{k=1}^{n}(-1)(-1)^{k-1}=-\sum_{k=1}^{n}(-1)^{k-1}=$$ $$=-\sum_{k=0}^{n-1}(-1)^{k}=-\frac{1-(-1)^{n}}{1-(-1)}=\frac{(-1)^{n}-1}{2}$$ for $n=100$ $$\sum_{k=1}^{100}(-1)^k=\frac{(-1)^{100}-1}{2}=\frac{1-1}{2}=0$$

• How do you do to transform the third part of the first line into the fourth one? – alexandrekow Sep 23 '14 at 20:39
• This is of course the hard way. – Michael Hardy Sep 23 '14 at 20:57
• you can see my edit! – Adi Dani Sep 23 '14 at 20:59


Question

# Assertion :For $$n\ge 4$$. Let $$\displaystyle{ a }_{ n }=\sum _{ j=1 }^{ n }{ \sum _{ k=j }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } } \begin{pmatrix} k \\ j \end{pmatrix}$$ and $${ b }_{ n }={ a }_{ n }+{ 2 }^{ n+1 }$$;$${ b }_{ n+1 }=5{ b }_{ n }-6{ b }_{ n-1 }$$ for all $$n\ge 2$$ Reason: $${ a }_{ n }=5{ a }_{ n }-6{ a }_{ n }$$ for all $$n\ge 2$$

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct

Solution

## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion$$\displaystyle \begin{pmatrix} n \\ k \end{pmatrix}\begin{pmatrix} k \\ j \end{pmatrix}=\frac { n! }{ k!\left( n-k \right) ! } .\frac { k! }{ j!\left( k-j \right) ! }$$$$\displaystyle=\frac { n! }{ j!\left( n-j \right) ! } \frac { \left( n-j \right) ! }{ \left( n-k \right) !\left( k-j \right) ! } =\begin{pmatrix} n \\ k \end{pmatrix}\begin{pmatrix} n\quad - & j \\ n\quad - & k \end{pmatrix}$$Thus $$\displaystyle { a }_{ n }=\sum _{ j=1 }^{ n }{ \begin{pmatrix} h \\ k \end{pmatrix} } \sum _{ j=1 }^{ n }{ \begin{pmatrix} n\quad - & j \\ n\quad - & k \end{pmatrix} }$$$$\displaystyle=\sum _{ j=1 }^{ n }{ \begin{pmatrix} n \\ j \end{pmatrix} } { 2 }^{ n-j }={ \left( 2+1 \right) }^{ n }-{ 2 }^{ n }={ 3 }^{ n }-{ 2 }^{ n }$$$$\Rightarrow { b }_{ 1 }={ 3 }^{ n }{ -2 }^{ n }+2\left( { 2 }^{ n } \right) ={ 3 }^{ n }+{ 2 }^{ n }$$Now $${ a }_{ n+1 }={ 3 }^{ n+1 }-{ 2 }^{ n+1 }=\left( 3+2 \right) \left( { 3 }^{ n }-{ 2 }^{ n } \right) -2\left( { 3 }^{ n } \right) +3\left( { 2 }^{ n } \right)$$$$=5\left( { 3 }^{ n }-{ 2 }^{ n } \right) -6\left( { 3 }^{ n-1 }-{ 2 }^{ n+1 } \right) ={ 5a }_{ n }-{ 6a }_{ n-1 }$$And $${ 5b }_{ n }-{ 6b }_{ n-1 }=5\left( { a }_{ n }+{ 2 }^{ n+1 } \right) -6\left( { a }_{ n-1 }+{ 2 }^{ n } \right)$$$$={ 5a }_{ n }-6{ b }_{ n-1 }+{ 2 }^{ n }\left( 10-6 \right)$$$$\Rightarrow { a }_{ n+1 }+{ 2 }^{ n+2 }={ b }_{ n+1 }$$Mathematics

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