# Solving the infinite series $1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$

I have the following question:

Evaluate the infinite series: $$S=1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$$ (a) $\displaystyle\frac1e$ (b) $\displaystyle\frac{-1}e$ (c) $\displaystyle\frac{2}e$ (d) $\displaystyle\frac{-2}e$

Now in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:

1. How did they get this "trick"?
2. If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?
• The trick seems to be typical when you want to see $An(n-1)(n-2)+Bn(n-1)+Cn+D$. Just identify. Sep 5, 2015 at 14:54

You want to evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)^3}{n!}$.

Now the idea (at the latest if you see some $e$ as a possible answer) should be to relate this to the well-known formula $$\sum_{n=0}^{\infty} \frac{x^n}{n!}=e^x$$ In this particular case (given the $(-1)^n$ included) it is useful to plug in $x=-1$ and obtain $$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}=e^{-1}=\frac{1}{e}$$ But you still have a $(n+1)^3$ in the numerator where you wish to have $1$.

So you might want to split this into several parts such that you get one (or several) related sums.

As an example of this technique, if you want to know $\sum_{n=0}^{\infty} \frac{n+1}{n!}$ you might try: $$\sum_{n=0}^{\infty} \frac{n+1}{n!}=\sum_{n=0}^{\infty} \frac{n}{n!}+\sum_{n=0}^{\infty} \frac{1}{n!}$$ Now, the second sum is just $e$ and the first one is (after cancelling) $$\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1}{n!}=e$$ and so the total sum is $2e$. Can you now understand the "trick" and tackle your problem by yourself?

Concerning the second question:

If one happens to know this "trick", it is not hard to get the result by just a few computations so it would be my preferential choice.

But if you sit in the exam and have absolutely no idea of how to solve this algebraically (and especially if you don't have to give a proof of your result!) it might be a good idea to estimate the sum by computing a few terms. At least the sign of the result ($\pm$) should be approachable...

• Yes thanks a lot, Now I understood the trick. But can you please answer the 2nd part of the question? Sep 5, 2015 at 14:59
• @Kartik by learning the trick now.Its a very common question :-)!! I felt the same as you the first time...!
– user220382
Sep 5, 2015 at 15:29
• I changed "nominator" into "numerator Sep 5, 2015 at 15:37

In general, if $p_i(x)$ is a sequence of polynomials with $\deg p_i=i$, then it is relatively easy to prove that every polynomial can be written as a linear combination of a finite number of $p_i$. When the $p_i$ are monic, you can do this easily. Let $P(x)=a_kx^k + ... a_0$. Then $P(x)-a_kp_k(x)$ is a polynomial of lower degree. Find it's first coefficient of that, and proceed inductively.

The standard basis for polynomials is $e_i(x)=x^i$.

So in your case; $p_i(n)=x(x-1)\cdots(x-(i-1))$ with $p_0(x)=1$. We chose this particular basis because $$\frac{p_i(n)}{n!} =\begin{cases}\frac{1}{(n-i)!} & n\geq i\\0&0\leq n<i\end{cases}$$ So:$$\sum_{n=0}^\infty \frac{p_i(n)z^n}{n!} = z^ie^{z}$$

In your case, $z=-1$.

With these polynomials, we get: \begin{align}(n+1)^3 &= n^3+3n^2+3n+1\\ (n+1)^3-p_3(n) &= n^3 +3n^2+3n+1 - (n^3-3n^2+2n)\\& = 6n^2+n+1\\ (n+1)^3-p_3(n)-6p_2(n) &= 6n^2+n+1 - 6n(n-1)\\&= 7n+1\\ \end{align}

So you get: $$(n+1)^3=p_3(n) + 6p_2(n) + 7p_1(n) +p_0(n)$$

The key is that the leading coefficient of the cubic $(n+1)^3$ is $1$, so we subtract $1\cdot p_3(n)$. Then the leading coefficient of the remaining quadratic is $6$, so we subtract $6p_2(n)$. The leading coefficient of the remaining linear polynomial is $7$, so we subtract $7p_1(x)$. You are left with a constant.

The relationship between the standard basis of the polynomials and the basis $p_i(x)=x(x-1)\cdots(x-(i-1))$ is actually fairly interesting and comes up again and again. The $p_i(x)$ are often called "falling factorials."

My way to solve it:

If the numerators were all $1$, we would have the well-known series for $e^{-1}$, obtained from the entire series

$$e^{-x}=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}.$$

Now we want to make numerators $(k+1)^3$ appear. Multiply by $x$ and derive, to get

$$(xe^{-x})'=\left(\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k!}\right)'=\sum_{k=0}^\infty(-1)^k\frac{(k+1)x^k}{k!}.$$

Repeating this three times,

$$(x(x(xe^{-x})')')'=\sum_{k=0}^\infty(-1)^k\frac{(k+1)^3x^k}{k!}.$$

Then evaluate at $x=1$.

$$e^{-x}\to (1-x)e^{-x}\to (1-3x+x^2)e^{-x}\to (1-7x+6x^2-x^3)e^{-x}\to -\frac1e.$$

For such problems, you can often start from a known Taylor series (in particular $e^x$ and $1/(1-x)$)and perform the following operations

• multiply or divide by a power of $x$ (shifts the exponents),
• derive on $x$ (multiplies by the exponents),
• integrate on $x$ (divides by the exponents),

to modify the coefficients.

Since $(n+1)^3 = \color{red}{1}\cdot n(n-1)(n-2)+\color{red}{6}\cdot n(n-1) +\color{red}{7}\cdot n+\color{red}{1}$ we have:

$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{(-1)^n (n+1)^3}{n!} &=& \frac{13}{2}+\sum_{n\geq 3}\frac{(-1)^n (n+1)^3}{n!}\\&=&\frac{13}{2}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-3)!}+\color{red}{6}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-2)!}+\color{red}{7}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-1)!}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\frac{13}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!}+6\cdot\sum_{n\geq 1}\frac{(-1)^n}{n!}-7\cdot\sum_{n\geq 2}\frac{(-1)^n}{n!}+\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\sum_{n\geq 0}(-1+6-7+1)\frac{(-1)^n}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n}{n!}=\color{red}{-\frac{1}{e}}.\end{eqnarray*}$$

We simply get the initial decomposition by induction. Obviously $(n+1)^3-n(n-1)(n-2)$ is a quadratic polynomial in $n$, whose leading term is $6n^2$. So we substract $6n(n-1)$ and we are left with a linear polynomial and so on. Another chance is to tackle our problem this way: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n+1)^3}{n!}&=&\sum_{n\geq 1}\frac{(-1)^n n(n+1)^2}{n!}+\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}\\&=&\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}-\sum_{n\geq 0}\frac{(-1)^n(n+2)^2}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n(2n+3)}{n!}\\&=&-3\sum_{n\geq 0}\frac{(-1)^n}{n!}+2\sum_{n\geq 0}\frac{(-1)^n}{n!}=\frac{2-3}{e}.\end{eqnarray*}$$

Looking at the given identity: \begin{align} (n+1)^3 &= [n(n-1)(n-2)+6n(n-1)+7n+1] \\ &= n(n^2 - 3n + 2) + 6 n^2 + n + 1 \\ &= n^3 + 3 n^2 + 3 n + 1 = (n+1)^3 \end{align}

Now, consider differentiation of the exponential series: \begin{align} D \, e^{t} &= \sum_{n=0}^{\infty} \frac{1}{n!} \, \frac{d}{dt} \, t^{n} = \sum_{n=0}^{\infty} \frac{n \, t^{n-1}}{n!} \end{align} Further differentiation can be applied.

Now, consider the series in question: \begin{align} S &= 1 - \frac{2^{3}}{1!} + \frac{3^{3}}{2!} - \cdots \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3}}{n!} \\ &= \left. \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3} \, t^{n}}{n!} \, \right|_{t=1} \\ &= \left. t^{3} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1)(n-2) \, t^{n-3}}{n!} + 6 t^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1) \, t^{n-2}}{n!} + 7 t \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n \, t^{n-1}}{n!} + e^{-t} \, \right|_{t=1} \\ &= \left. t^3 \, D^{3} e^{-t} + 6 \, t^{2} \, D^{2} e^{-t} + 7 \, t \, D e^{-t} + e^{-t} \right|_{t=1} \\ &= \left. (-t^3 + 6 \, t^2 - 7 \, t + 1) \, e^{-t} \right|_{t=1} \\ &= - \frac{1}{e} \end{align}

• Thanks for the answer but this explanation is already given in the book which I am solving. I just wanted the explanation of the "Trick" in the first identity, Sep 5, 2015 at 15:16


## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 1

#### Answer

As $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.

#### Work Step by Step

It means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


# Should I use a binomial cdf or a normal cdf when flipping coins?

A coin needs to be tested for fairness. 30 heads come up after 50 flips. Assuming the coin is fair, what is the probability that you would get at least 30 heads in 50 flips?

The right way to do this problem, according to my teacher, is to do

normalcdf(min = .6, max = ∞, p = .5, σ = sqrt(.5 * .5 / 50) = 0.0786

However, I took a binomial cumulative distribution function like this

1 - binomcdf(n = 50, p = .5, x = 29) = 0.1013

I believe the criteria for a binomial distribution are satisfied: the individual events are independent, there are only two possible outcomes (heads vs. tails), the probability is constant for the question (0.5), and the number of trials is fixed at 50. Yet obviously, the two methods give different answers, and a simulation supports my answer (at least the few times I ran it; obviously, I can't guarantee that you'd get the same results).

Is my teacher wrong in assuming that a Normal distribution curve would also be a valid way to do this problem (at no point is it said that the distribution is Normal, but n*p and n*(1-p) are both greater than 10), or have I misunderstood something about binomial distributions?

• A person with experience in using Normal approximations to the Binomial would proceed a little differently: they would apply the (usual) continuity correction, as in 1 - pnorm((30-0.5)/50, mean=0.5, sd=sqrt(0.5*(1-0.5)/50)) (this is an R expression), whose value is 0.1015, in quite close agreement with the Binomial cdf. – whuber Jan 18 '12 at 21:23

Here is an illustration of the answers of whuber and onestop.

In red the binomial distribution $\mathcal Bin(50,0.5)$, in black the density of the normal approximation $\mathcal N(25, 12.5)$, and in blue the surface corresponding to $\mathbb P(Y > 29.5)$ for $Y \sim \mathcal N(25, 12.5)$.

The height of a red bar corresponding to $\mathbb P(X=k)$ for $X\sim\mathcal Bin(50,0.5)$ is well approximated by $\mathbb P\left( k -{1\over 2} < Y < k + {1\over 2}\right)$. To get a good approximation of $\mathbb P(X \ge 30)$, you need to use $\mathbb P(Y>29.5)$.

(edit) This is $$\mathbb P(Y>29.5) \simeq 0.1015459,$$ (obtained in R by 1-pnorm(29.5,25,sqrt(12.5))) whereas $$\mathbb P(X \ge 30) \simeq 0.1013194:$$ the approximation is correct.

This is called continuity correction. It allows you to compute even "point probabilities" like $\mathbb P(X=22)$ : \begin{align*} \mathbb P(X=22) &= {50 \choose 22} 0.5^{22} \cdot 0.5^{28} \simeq 0.07882567, \\ \mathbb P(21.5 < Y < 22.5) & \simeq 0.2397501 - 0.1610994 \simeq 0.07865066.\end{align*}

The normal distribution gives a closer approximation to the binomial if you use a continuity correction. Using this for your example, I get 0.1015. As this is homework, I'll leave it to you to fill in the details.

Consider this. In the discrete binomial distribution you have actual probabilities for individual numbers. In the continuous normal that isn't the case, you need a range of values. So... if you were going to approximate the probability of an individual value, let's say X, from the binomial with the normal how would you do that? Look at a probability histogram of the binomial distribution with the normal curve laid over it. You would need to actually select from X ± 0.5 to capture something similar to what the binomial probability of X is with the normal approximation.

Now extend that to when you're selecting a tail of the distribution. When you use the binomial method you're selecting your entire value's probability (30 in your case) plus everything higher. Therefore, when you do the continuous you have to make sure you capture that and select 0.5 less as well, so the cutoff on the continuous distribution is 29.5.

• Actually, the question exhibits a thoughtful understanding of the problem and does not appear to be looking for an answer to a routine homework question. Although it's tagged homework, consider making an exception here. In particular, a good discussion of using the Normal distribution to approximate discrete distributions (such as Binomials and Poissons with large N's) would be appropriate and most welcome here. – whuber Jan 18 '12 at 23:37


# Prove that $(l^\infty,\|.\|_\infty)$ is a Banach space.

$(l^\infty,\|.\|_\infty)$ is a Banach space. In the proof, $\mathbb{F}$ is either the field of complex numbers or the field of real numbers.

Proof: Let $(x^n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $l^\infty$, where $x^n=(x_k^n)_{k\in\mathbb{N}}$. Consider the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ of $k$th coordinates of the sequence $(x^n)_{n\in\mathbb{N}}$.

Let $\epsilon>0$. Since $(x^n)_{n\in\mathbb{N}}$ is Cauchy, there exists $n_0\in\mathbb{N}$ such that $$\forall m,n>n_0,\|x^m-x^m\|_\infty<\epsilon.$$ Therefore for each $m,n>n_0$: $$|x^m_k-x^n_k|<\epsilon.$$

So the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy. Therefore it converges to some $y_k\in\mathbb{F}\$ ($\mathbb{F}$ is complete).

Let $y=(y_k)_{k\in\mathbb{N}}$. Since $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy it is bounded. Choose $M>0$ such that for each $n\in\mathbb{N}$, $|x^n_k|<M$. But since $$y_k=\lim_{n\to\infty}x_k^n,$$ we have $|y_k|\leq M$ for each $k$. Therefore, $y\in l^\infty$.

Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.

Therefore for each $m>n_0,\ \|x^m-y\|_\infty<\epsilon$. Hence $(x^n)_{n\in\mathbb{N}}$ converges in $l^\infty$ and the proof is complete.

Could someone please tell me if the above proof is alright? Thanks.

• Looks alright to me. The readability was fine even before it was over-pedantically edited. – uniquesolution Aug 23 '17 at 7:35
• The general procedure seems right, however near the end $\|x^m - x^m\|_\infty$ is something you should fix. – mlk Aug 23 '17 at 7:36
• @uniquesolution Then you are just more skilled than I am in reading wall-of text style questions. But I hope even you will agree that the readability is much better now. – 5xum Aug 23 '17 at 7:37
• When I first looked at it it was properly Latex formatted. If it was text-style and you converted to Latex, then thank you. – uniquesolution Aug 23 '17 at 7:39
• @mlk fixed it, and thanks everyone. – Janitha357 Aug 23 '17 at 7:40

Your proof is very rigorous and very detailed all the way up to the point where you say

Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.

Now I know the inequality stands, but as you were very thorough with all your other inequalities, I think it would be nice if you wrote a little more justification for this one as well - it is not entirely obvious how the right inequality follows from the left one.

Other than that, the proof is very well written and easy to follow.

Rather a comment, but my reputation does not allow me to comment yet. You write

choose $M>0$ that for each $n\in\mathbb{N}, |x^n_k|<M$

I am curios whether you have to write $M_k$, since hypothetically $M$ can depend on the choice of the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$. But then we have difficulties with proving that $y\in l^\infty$.

EDIT: the proof is more subtle than I initially thought. First we have to settle the issue 5xum mentioned. We start with $$\forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n,p>n_0:|x^n_k-x^{n+p}_k|<\epsilon.$$ This is inequality in $\mathbb{F}$, so we can let $p\to\infty$. Thus we obtain $$\forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n>n_0:|x^n_k-y_k|<\epsilon.$$ That means $\|x^n-y\|_\infty<\epsilon$, thus $(x^n)_{n\in\mathbb{N}}$ is converging to $y$.

We still have to prove that $y\in l^\infty$. But $$\|y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$$

• Could you explain the last bit please? – Janitha357 Aug 23 '17 at 9:53
• I am using triangle inequality and addition of zero to prove that $y\in l^\infty$: $\|y\|_\infty \le \|x^n - x^n+y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$ – EugenR Aug 23 '17 at 11:10


# An interesting problem of polynomials

In the polynomial $$(x-1)(x^2-2)(x^3-3) \ldots (x^{11}-11)$$ what is the coefficient of $$x^{60}$$?

I've been trying to solve this question since a long time but I couldn't. I don't know whether opening the brackets would help because that is really a mess. I have run out of ideas. Would someone please help me to solve this question?

• Think of how many ways the terms can form a term $x^{60}$ ... It could be $x^1 \cdot x^{59}$, or $x^2 \cdot x^{58}$, or even $x^3 x^8 x^{49}$ ... Can you think of a systematic way to create $x^{60}$ like this? Nov 5 '19 at 10:38
• Old question, see here. Also on this site there are several such question. Please have a look, say here. Next time, please search before posting. Nov 5 '19 at 10:39
• @DietrichBurde it's not easy to search for mathematical expressions (unless you know to use approach 0 or something like it), and there isn't a lot of text to go off of for these problems. I don't think that there's much reason to assume that the asker didn't at least try searching. Nov 5 '19 at 10:44
• @Omnomnomnom I agree, that it is not always immediate to see these duplicates, but honestly my experience on this site is that several users are not even aware that it is possible to find the same question with several beautiful answers. This is true in particular for these "standard questions". One of the goals for students also is to learn how to search. Nov 5 '19 at 10:47
• @Aryan The previous edit was better actually. Nov 5 '19 at 10:48

The highest exponent possible is $$1+2+ \cdots + 11 = 66$$.

Now, to create the exponent $$60$$, you can only leave out the factors containing $$(1,2,3),(2,4),(1,5)$$ and $$6$$. Hence,

• $$1+2+3 \Rightarrow$$ gives coefficient $$(-1)(-2)(-3) = -6$$
• $$2+4 \Rightarrow$$ gives coefficient $$(-2)(-4) = 8$$
• $$1+5 \Rightarrow$$ gives coefficient $$5$$
• $$6 \Rightarrow$$ gives coefficient $$-6$$

Summing up gives $$13-12 = \boxed{1}$$.

Hint : $$1+2+3 +...+11= \frac {11×12}{2} =66$$ so we must find how we can construct number $$6=6+0=5+1=4+2=3+3=1+2+3$$ and note that $$3+3$$ impossible.

Expanding the brackets is definitely not the way to go, but thinking about what would happen if you did is helpful. Every term in the expanded polynomial will come from multiplying one term from each bracket (e.g. the highest degree term will come from multiplying $$\,x \cdot x^2 \cdot x^3 \cdot x^4 \dots \cdot x^{11}\,$$), and since each bracket is a binomial that doesn't leave many options.

The first thing to notice is that the aforementioned highest-degree term will be $$\,x^{66};\,$$ thus, to get a term with $$\,x^{60},\,$$ we will need to use MOST BUT NOT ALL of the $$\,x$$'s in the factors. We need to reduce the maximum degree by $$\,6\,$$; one obvious way to do that is to not use the $$\,x^6\,$$ in the sixth factor, so instead we use the $$\,-6\,$$ from that bracket. However, there are other ways, e.g. not using the $$\,x^2\,$$ or the $$\,x^4\,$$ from the second and fourth factors, respectively. Hopefully by now you're getting the sense that what we need to do is come up with all the different ways to make $$\,x^6\,$$ with any of the powers available to us, and then for each way find the corresponding coefficient.

There are not that many ways of making $$\,x^6;\,$$ note that the most we can use is three powers of $$\,x\,$$, as the minimum power would then be $$\,x\cdot x^2\cdot x^3=x^6.\,$$ The list of ways is: $$\,x^6, x\cdot x^5, x^2 \cdot x^4,\,$$ and $$\,x\cdot x^2\cdot x^3.\,$$

Now we just need to find the coefficient of each term that EXCLUDES one of the above combinations. As an example, let's consider the term that excludes $$\,x^2\,$$ and $$\,x^4.\,$$ Written out in detail, this term would be

$$(x)(-2)(x^3)(-4)(x^5)(x^6)(x^7)(x^8)(x^9)(x^{10})(x^{11}) = 8x^{60}$$

Similarly, it should be easy to see that the coefficients for the terms excluding the other combinations are:

• For $$\,x^6,\,$$, the coefficient will just be $$\,-6\,$$;
• For $$\,x\cdot x^5,\,$$ the coefficient will be $$\,(-1)(-5)=5\,$$;
• For $$\,\,x\cdot x^2\cdot x^3,\,$$ the coefficient will be $$\,(-1)(-2)(-3)=-6.\,$$

Thus the four terms that will have $$\,x^{60}\,$$ are $$\,-6x^{60}, 5x^{60}, 8x^{60}\,$$ and $$\,-6x^{60}.\,$$ The sum of these will be just $$\,x^{60},\,$$ so the answer to the question is $$\,\boxed{1}.\,$$

• Thanks a lot A.J. Nov 6 '19 at 9:12


# Sum in a sudoku with sum total of 15 and 45

I have created a sudoku puzzle with the following restrictions:

• Each row and column sum to $$45$$.
• Each row and column in the nine $$3$$ by $$3$$ sub-grids sum to $$15$$.

Is such a sudoku unique?

• "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. – Hugh Feb 11 at 5:04
• "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45? – Kryesec Feb 11 at 5:33
• @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15. – Hugh Feb 11 at 5:47

No.

Of course, there's rotation and reflection, but even beyond that you can swap any two rows within its group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.

• I wonder if there's any kinda of transformation that would work here that doesn't work on sudoku in general. – Rawling Feb 11 at 12:00

I was beaten to a very similar answer, but you can create such a square based on the magic square below:

abc
def
ghi

Turn it into this:

abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde

Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).

One such arrangement of the 3x3 grid could be:

1 5 9
8 3 4
6 7 2

This would then be repeated along to the right, but rotating the order of the rows to create:

1 5 9 8 3 4 6 7 2
8 3 4 6 7 2 1 5 9
6 7 2 1 5 9 8 3 4

The same could be done to repeat downwards but rotating the columns this time:

1 5 9 8 3 4 6 7 2
8 3 4 6 7 2 1 5 9
6 7 2 1 5 9 8 3 4
5 9 1 3 4 8 7 2 6
3 4 8 7 2 6 5 9 1
7 2 6 5 9 1 3 4 8
9 1 5 4 8 3 2 6 7
4 8 3 2 6 7 9 1 5
2 6 7 9 1 5 4 8 3

However:

This creates one such possibility of your given solution, as each of the 3x3 sections could be placed in the top left and the rotations occur from there.

The answer is no. A Sudoku satisfying the puzzle requirement (i.e. each 3x3 sub-grid is a semi-magic square) is not unique even if we do not make a distinction between Sudokus differing just by rotations or reflections. And the simplest reason, as pointed out by @DR Xorile, is that there are natural Sudoku transformations (namely certain rows and/or columns swaps) which are not rotations and reflections but which preserve the integrity of the Sudoku and the semi-magic property of the sub-grids. This brings up a question alluded to by @Rawling: what if we extend a rotations and reflections group of transformations with row and column permutations suggested by @DR Xorile (let's call this group RRRCP for brevity) and not make distinctions between Sudokus if they can be transformed into each other via any RRRCP transformations. Will it make a Sudoku built out of 3x3 sub-grid of semi-magic squares "unique"? At first I thought it might. On a smaller scale, a 3x3 semi-magic square is "unique" in a sense that all 72 semi-magic squares can be generated by applying transformations from a similar extended group (rotations, reflections, columns/rows permutations) to just one semi-magic square. But it turns out that for a full Sudoku the answer is still no - a Sudoku built out of 3x3 sub-grid of semi-magic squares is not "unique" even modulo RRRCP transformations. Here is why.

For the sake of precision here is the description of an RRRCP group. Let's call each set of three columns (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a "macro" column and each set of three rows (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a "macro" row. "Macro" columns and "macro" rows constitute a 3x3 "macro" grid. A RRRCP group consists of compositions of the following Sudoku transformations:

• rotations and reflections;

• permutations of the columns within a "macro" column (ex. columns 4, 5, 6)

• permutations of the rows within a "macro" row (ex. rows 7, 8, 9)

• permutations of whole "macro" columns within a 3x3 "macro" grid (ex swap columns 1 2 3 with columns 7, 8, 9)

• permutations of whole "macro" rows within a 3x3 "macro" grid (ex swap rows 1 2 3 with rows 4, 5, 6).

The following two Sudokus S1 and S2 are built out of semi-magic squares but cannot be transformed into each other by RRRCP transformations.

2 7 6 1 9 5 3 8 4
9 5 1 8 4 3 7 6 2
4 3 8 6 2 7 5 1 9

3 8 4 2 7 6 1 9 5
7 6 2 9 5 1 8 4 3
5 1 9 4 3 8 6 2 7

1 9 5 3 8 4 2 7 6
8 4 3 7 6 2 9 5 1
6 2 7 5 1 9 4 3 8

-------------------------------

2 7 6 5 1 9 8 4 3
9 5 1 3 8 4 6 2 7
4 3 8 7 6 2 1 9 5

8 4 3 2 7 6 5 1 9
6 2 7 9 5 1 3 8 4
1 9 5 4 3 8 7 6 2

5 1 9 8 4 3 2 7 6
3 8 4 6 2 7 9 5 1
7 6 2 1 9 5 4 3 8

S1 and S2 cannot be transformed into each other by rotations or reflections because they have an identical central 3x3 sub-grid.
S1 and S2 cannot be transformed into each other by permutations within a "macro" row/column because they have an identical "macro" diagonal.
S1 and S2 cannot be transformed into each other by permutations of "macro" rows/columns because they are built out of different semi-magic squares.


# What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$

What is the sum of number of digits of the numbers $$2^{2001}$$ and $$5^{2001}$$? (Singapore 1970)

I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $$2^{2001}\mod{10}\equiv 4$$ and $$2^{2001}\pmod{10}\equiv 5$$. Could you please explain to me how to solve this question? This question is multiple choice with options $$1999, 2003, 4002, 6003, 2002$$

• @samerivertwice yes base 10 Mar 15 at 17:15
• Not hard to do with a computer program, but perhaps that's not allowed. Mar 15 at 17:18
• @Sil I am certain that that is what the question asks Mar 15 at 17:36
• @Sil I'll ask the author for an explanation, as soon as I get a response, I'll post it here Mar 15 at 18:05
• @sil I worked it out given the multi choice 😉 Mar 15 at 18:08

It's $$2002$$. It's asking for the sum of the number of digits of $$2^{2001}$$ and $$5^{2001}$$ in base $$10$$, so just take the log base $$10$$ of each, take the ceiling function and hey presto:

$$603+1399=2002$$

• remark: without taking the ceiling function, $$2001\log_{10}(2)+2001\log_{10}(5) = 2001(\log_{10}(2)+\log_{10}(5)) = 2001\log_{10}(10) = 2001$$ so no calculator is needed to know the answer is in $[2001,2003)$ Mar 15 at 18:14
• "the sum of the number of digits" Well, that's just nasty..... Mar 15 at 18:16
• @hgmath nice. I figured there was something like that although I just used the calculator ;) Mar 15 at 18:17
• @samerivertwice your solution is brilliant, thank you very much Mar 15 at 18:18
• @MichaelBlane for the logarithm I just used a calculator. If needed to do without a calculator this can be done, using an infinite series for the log function but the real magic is in hgmath's comment - if you want to learn something, that comment is the thing to think about. Mar 15 at 18:20

It looks like you don't need logarithms or any calculator to solve this problem. Let's start.

First, observe that the following inequalities hold:

$$10^m<\underbrace {2^{2001}}_{m+1 ~ \text{digits}}<10^{m+1}$$

$$10^n<\underbrace{5^{2001}}_{n+1 ~ \text{digits}}<10^{n+1}$$

You get,

$$10^{m+n}<10^{2001}<10^{m+n+2}$$

$$2001=m+n+1$$

$$m+n=2000$$

Finally, the sum of digits of $$2^{2001}$$ and $$5^{2001}$$ is equal :

\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=2000+2 \\\qquad \qquad \qquad \thinspace=2002.}}}\end{align}

The answer is $$\overbrace{\lfloor2001\log_{10}(2)\rfloor+1}^\text{digits in 2^{2001}}+\overbrace{\lfloor2001\log_{10}(5)\rfloor+1}^\text{digits in 5^{2001}}$$ However, we also have, using Iverson Brackets, $$\lfloor x\rfloor+\lfloor y\rfloor=\lfloor x+y\rfloor-[\{x\}+\{y\}\ge1]$$ So we need to know $$\{2001\log_{10}(2)\}+\{2001\log_{10}(5)\}$$, but since $$2001\log_{10}(2)+2001\log_{10}(5)=2001$$, we know that the sum of their fractional parts is exactly $$0$$ or exactly $$1$$. Since the fractional parts are both positive, we must have exactly $$1$$.

Therefore, \begin{align} \lfloor2001\log_{10}(2)\rfloor+1+\lfloor2001\log_{10}(5)\rfloor+1 &=\lfloor2001\log_{10}(2)+2001\log_{10}(5)\rfloor+1\\ &=2002 \end{align}

• I see that I answered the question after it was edited to be correct. It was probably that edit which brought this question to the top of the front page just as I was looking.
– robjohn
Mar 15 at 22:09

Generalization of the problem:

• What is the sum of number of digits of the numbers $$2^N$$ and $$5^N$$?

$$10^m<\underbrace {2^{N}}_{m+1 ~ \text{digits}}<10^{m+1}$$

$$10^n<\underbrace{5^{N}}_{n+1 ~ \text{digits}}<10^{n+1}$$

$$10^{m+n}<10^{N}<10^{m+n+2}$$

$$N= m+n+1$$

$$m+n=N-1$$

The sum of digits of the numbers $$2^{N}$$ and $$5^{N}$$ will be equal :

\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=N-1+2 \\\qquad \qquad \qquad \thinspace=N+1.}}}\end{align}

• Short answer: $$N+1$$ digits.


# help factoring

#### slowle4rner

##### New member
I'm having trouble factoring the following binomial... can someone try to point me in the right direction please?

4y^3+4

It has been 7 years since I took algebra and I am trying to get my review done. This one just does not make sense to me right now...

Thanks!

#### MarkFL

Staff member
First factor the 4 out, to get:

$$\displaystyle 4y^3+4=4\left(y^3+1 \right)=4\left(y^3+1^3 \right)$$

Now apply the sum of cubes formula:

$$\displaystyle a^3+b^3=(a+b)\left(a^2-ab+b^2 \right)$$

What do you get?

#### slowle4rner

##### New member
So...A^3+b^3 = (a+b)(a^2-ab+b^2)

4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)

That's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply (4y+4)(y^2-y+1) or multiply 4(y+1)(y^2-y+1) and distribute the 4.

Would I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...

Thanks for the help. I wasn't seeing it. Gratitude.

#### SuperSonic4

##### Well-known member
MHB Math Helper
So...$$\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$$

$$\displaystyle 4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)$$

That's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply $$\displaystyle (4y+4)(y^2-y+1)$$ or multiply $$\displaystyle 4(y+1)(y^2-y+1)$$ and distribute the 4.
The 4 is a bit of a red herring because, as you said, it's not a perfect cube but you can leave that alone out the front. You could say $$\displaystyle (4y+4)(y^2-y+1)$$ (as well as $$\displaystyle (y+1)(4y^2-4y+4)$$) but it's good form to leave it fully factored which means not distributing the four. If you ever do exams they usually want it fully factored.

Would I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...

Thanks for the help. I wasn't seeing it. Gratitude.
Yes, that is the correct answer.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Even if you were not able to factor that "4" out, it just a number! If it had been, say, $$x^3+ 4$$ you could write it as $$x^3+(\sqrt[3]{4})^3$$ and use that same fornula.


# Common logarithm question

I'm studying logarithms and am doing an exercise where you're supposed to evaluate the solutions of common logarithms without using a calculator. I'm very stuck on this one particular question. I know the answer because I used my calculator, but I'd like to know how to solve it without one. The question is $$\log\left(\frac{10}{\sqrt[\large3]{10}}\right)$$

How do I solve this without a calculator? (Please provide a step-by-step solution, this has really confused me.)

• Can you write $10/\root3\of {10}$ as $10^x$ for some $x$? If so, what would $\log 10^x$ be? – David Mitra Jul 29 '14 at 10:39
• @DavidMitra I got as far as $10(10^{-\frac13})$ but don't know where to go from there – imulsion Jul 29 '14 at 10:40

Hint

$$\frac{10}{\sqrt[3]{10}} = \frac{10}{10^{1/3}} = 10^{1-1/3} = 10^{2/3}$$

Now, what would the logarithm (assuming base 10) of that final expression be?

• Thanks for your answer, but I don't understand why $\frac{10}{10^{1/3}} = 10^{1-1/3}$. Probably me being a moron, but I don't understand how you got there. – imulsion Jul 29 '14 at 10:46
• @imulsion The following rule is very good to know: $\frac{x^a}{x^b} = x^{a-b}$. Let $x=10$, $a=1$ and $b=1/3$. (Remember that $10 = 10^1$.) – naslundx Jul 29 '14 at 10:48
• I can't believe that I forgot that :facepalm:. Thanks very much for your help though, I will accept your answer when I can – imulsion Jul 29 '14 at 10:49
• @imulsion No need to feel stupid, we are all here to learn. :) – naslundx Jul 29 '14 at 10:50

Remember that the logarithm of a quotient is the difference of logarithms: $$log\left(\frac{10}{\sqrt[3]{10}}\right)=log(10)-log(\sqrt[3]{10})=1-log(10^{1/3})=1-\frac {1}{3}\cdot log(10)=1-\frac {1}{3}=\frac {2}{3}$$

To get a good understanding of logarithms it is good to realize that the following two questions are equivalent:

1) What is the logarithm of $a$ on base of $g$? I.e. $\log$$_{g}\left(a\right)=?$.

2) To what power must $g$ be raised to get $a$ as outcome? I.e. $g^{?}=a$.

Here $a>0$, $g>0$ and $g\neq1$.

So $10^{\frac{2}{3}}=\frac{10}{\sqrt[3]{10}}$ is the same information as $\log_{10}\frac{10}{\sqrt[3]{10}}=\frac{2}{3}$


# Is the action of $G$ on $H_1(T^n, \mathbb{Z}) = \mathbb{Z}^n$ faithful?

Let $G$ be a finite group of diffeomorphisms of the torus $T^n$ fixing some point $p$, i.e. $p$ is fixed by every element of $G$. I have two questions.

1. Is the action of $G$ on $H_1(T^n, \mathbb{Z}) = \mathbb{Z}^n$ faithful?
2. What if $G$ is not assumed to fix a point $p$?
• 2. If $G$ acts as a finite group of translations, then the action on $H_1$ is trivial (while if $G$ is nontrivial it fixes no point).
– YCor
Feb 26, 2016 at 9:14
• This question is not really research level and should be asked at math stack exchange. Nevertheless, the techniques used to answer such a question may not be well-known. Nov 25, 2016 at 15:50
• @user99720 but my comment is really immediate, there are already nice answers on the less trivial part of the question.
– YCor
Nov 26, 2016 at 22:10

From Edmonds' notes on Transformation groups: http://www.indiana.edu/~jfdavis/seminar/transformationgroupsb.pdf

Problem 9. [p.28] Show that if a [finite] group $G$ acts on the torus $T^n$ with a fixed point x and induces the trivial action on $\pi_1(T^n, x)$, then the action is trivial.

So the answer to question 1 is yes. Question 2 is answered by @YCor's comment and also discussed in the above notes.

This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $m$ must move a point of $M$ at least a distance $d$.

Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\tilde{f}$ of $f$ to the universal cover $\mathbb{R}^n$ which fixes a point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). The homotopy of $f$ to the identity lifts to a homotopy of $\tilde{f}$ to the identity, so $\tilde{f}$ also has the property that it moves points at most $D$ and has order $m$. Let $d$ be the constant from Newman's theorem for the manifold $\mathbb{R}^n$ with Euclidean metric and $m$, then rescale the metric by $d/D$, we see that $\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity.

Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.

We can expand the answer from user83633, following hints in the notes by Edmonds, as follows.

Let $G$ be a finite group acting faithfully on an $n$-dimensional torus $T$ and fixing a point $p$. We can write $T$ as $V/L$ for some vector space $V$ and lattice $L<V$. As $T$ is homogeneous, we can assume that this identifies $p$ with the zero element. Let $\pi\colon V\to T$ be the obvious projection, which is a universal covering. Covering theory tells us that there is a unique way to let $G$ act on $V$ such that $0$ is fixed and $\pi$ is equivariant. This action preserves $\pi^{-1}\{0\}=L$, and we can identify $L$ with $H_1(T)$ equivariantly. We want to show that the action of $G$ on $L$ is faithful. If not, we can choose a subgroup $C\leq G$ of prime order $p$ that acts trivially on $L$. We will show that this leads to a contradiction.

First note that the free action map $L\times V\to V$ is $G$-equivariant, and $C$ acts trivially on $L$, so we get an induced free action of $L$ on $V^C$. The map $\pi$ gives an injection $V^C/L\to T^C$; we claim that this is a homeomorphism. As everything is compact and hausdorff, we need only check that it is surjective. Suppose that $x\in T^C$, and choose $v\in V$ with $\pi(v)=x$. If $g$ is a generator of $C$, we must then have $g.v=a+v$ for some $a\in L$. As $g$ acts trivially on $L$ this gives $g^k.v=ka+v$ for all $k$, so in particular $v=g^p.v=pa+v$ so $pa=0$ so $a=0$ so $g.v=v$ as required.

For the rest of the argument we will use various cohomology groups; these are always taken with coefficients $\mathbb{Z}/p$.

Next, recall that $H^*(BC)$ is the tensor product of a polynomial algebra on a class $x$ of degree $2$ with an exterior algebra on a class $a$ of degree $1$. For any $C$-space $X$ we have a map $EC\times_CX\to EC/C=BC$, which makes the Borel cohomology group $H_C^*(X)=H^*(EC\times_CX)$ into an algebra over $H^*(BC)$. We write $\widehat{H}^*_C(X)$ for the ring obtained by inverting $x$ in $H_C^*(X)$. If $Y$ is a $C$-subspace of $X$ then we have relative groups $\widehat{H}_C^*(X,Y)$ defined in a similar way. If $C$ acts freely on $X\setminus Y$ then these relative groups are trivial, as one can prove by cellular induction starting from the case where $X=C\times B^d$ and $Y=C\times S^{d-1}$. On the other hand, if $C$ acts trivially on $X$ then $\widehat{H}^*_C(X)=H^*(X)\otimes R^*$, where $$R^*=\widehat{H}^*_C(\text{point})=\mathbb{Z}/p[x,x^{-1}]\otimes E[a].$$ Because $C$ has prime order, it must act freely on $X\setminus X^C$, so $\widehat{H}^*_C(X,X^C)=0$, so $$\widehat{H}^*_C(X) \simeq \widehat{H}^*_C(X^C) \simeq H^*(X^C)\otimes R^*.$$ This is essentially what is called Smith theory.

Now take $X$ to be our vector space $V$, so $V$ is contractible, and the usual spectral sequence $H^*(G;H^*(X))\Longrightarrow H^*_G(X)$ gives $H^*_G(V)=H^*(BG)$ and therefore $\widehat{H}^*_G(V)=R^*$. By comparison with Smith theory, we get $H^*(V^C)=\mathbb{Z}/p$. As $L$ acts freely on $V^C$ with $V^C/L=T^C$ we get a spectral sequence $$H^*(L;H^*(V^C)) \Longrightarrow H^*(T^C)$$ This admits a map from the similar spectral sequence $$H^*(L;H^*(V)) \Longrightarrow H^*(T).$$ As $H^*(V^C)=H^*(V)=\mathbb{Z}/p$, the map is an isomorphism on the initial page, and there is no room for any differentials anyway, so the map $H^*(T)\to H^*(T^C)$ is an isomorphism. In particular, the map $H^n(T)\to H^n(T^C)$ is an isomorphism.

On the other hand, if $X$ is any proper subset of $T$ then the restriction $H^n(T)\to H^n(X)$ factors through the group $H^n(T\setminus \{x\})=0$ for some $x\in T$. Thus, we must have $T^C=T$, which means that $C$ acts trivially on $T$. This contradicts our assumption that the action of $G$ is faithful.

For a simpler and more geometric proof: It suffices to show the only element which acts as the identity is the identity. Suppose $g$ is our element which acts as the identity on $H_1$ and thus on $\pi_1$. Lift $g$ to act on $\mathbb{R}^2$. Define a homotopy $g_t(p) = (1 - t) p + t g(p)$ between $g$ and the identity. The homotopy descends to $\mathbb{T}^2$. By surface theory we can upgrade this to an isotopy between $g$ and the identity on $\mathbb{T}^2$. Then, and I'm not sure how to prove this (and in some sense this seems to be the crux of the issue) one can show that maps isotopic to the identity have infinite order.


# Mistake in solving $-\int \frac{1}{x\sqrt{x^2-1}}$

I have this function $$f:(-\infty ,-1)\rightarrow \mathbb{R}, f(x)=\frac{1}{x\sqrt{x^{2}-1}}$$ and I need to find the primitives of $$f(x)$$.So because $$x<-1$$ I need to calculate $$-\int \frac{1}{x\sqrt{x^2-1}}\,dx=-\arctan(\sqrt{x^2-1})+C$$ but in my book the correct answer is $$\arcsin\left(\frac{1}{x}\right)+C$$

Where is my mistake?I solved the integral using $$u=\sqrt{x^{2}-1}$$

It is known that $$\arctan(x)+\arctan(1/x)=C$$ where $$C$$ is a constant. So your solution can be written as $$\arctan\left(\frac1{\sqrt{x^2-1}}\right):=u$$Then using some trig identities, $$\frac1{\sqrt{x^2-1}}=\tan u\\\sec^2 u=1+\tan^2 u=\frac{x^2}{x^2-1}\\\sin^2 u=1-\cos^2 u=1-\left(\frac{x^2-1}{x^2}\right)=\frac1{x^2}\\u=\arcsin\left(\frac1x\right)$$

Hint:

You might want to plot both $$\arcsin(\frac{1}{x})$$ and $$-\arctan(\sqrt{x^2 - 1})$$, from, say, $$-3$$ to $$-1$$. You could do this using Desmos (https://www.desmos.com/). You might even insert an extra minus sign here or there...

$$\arcsin(\frac{1}{x}) +C = \operatorname{arccot}(\sqrt{x^2 - 1}) + C$$

$$=\frac{\pi}{2} - \arctan(\sqrt{x^2 - 1})+ C$$

You can form another Constant $$C_1 = \pi/2 + C$$.

So you solved correctly, the book answer and your answer are equivalent, you just had to manipulate it a little to get there. The difference is just of a constant which doesn't matter when calculating indefinite integrals.

• Yes, I understand what you're saying but my book gives both results as anwers and I just need to pick one answer.For example, at exam how to know which is good ? – DaniVaja May 1 '19 at 18:48
• @Vizag: I think that the relationship you've asserted in the first line is only correct for $x > 0$; you might want to check that by plotting both sides when $x < 0$. – John Hughes May 1 '19 at 18:50
• @DaniVaja Both are fine, since they are both correct. So you shouldn't worry too much. You could use these answers to see how to show that both forms are equivalent (up to constants). – John Doe May 1 '19 at 18:50
• @DaniVaja: Both the answers look correct to me. The expressions are equivalent up to constants. Are you sure it's not a multiple answers correct kind of question? – Vizag May 1 '19 at 18:53
• Here is the original exercise: imgur.com/hhQUzoi (i don't know the code for posting images).The book says that one answer from 5 is correct and in this exercise I have both answers, which are both good like you said.But for example in an exam, how to know which answer to pick ?In this case they chose arcsin(1/x) + c as correct answer and not -arctan... – DaniVaja May 1 '19 at 18:56


# [SOLVED]Laplace Transform

#### nacho

##### Active member
How do you find the laplace transform of this without expanding it?

$L(t^2+1)^2$

Last edited:

#### chisigma

##### Well-known member
How do you find the laplace transform of this without expanding it?

$L(t^2+1)^2$
You can use the fact that for n positive integer is...

$\displaystyle \int_{0}^{\infty} t^{n}\ e^{- s\ t}\ dt = \frac{n!}{s^{n+1}}\ (1)$

... expanding $\displaystyle (1 + t^{2})^{2}$ in powers of t...

Kind regards

$\chi$ $\sigma$

I didn't realize immediately that it was requested 'without expanding' ... very sorry!...

#### chisigma

##### Well-known member
How do you find the laplace transform of this without expanding it?

$L(t^2+1)^2$
The possible solution is the use of a 'forgotten formula' ['forgotten' in the sense that it is neglected from most of the Complex Analysis 'Holybooks'...] according to which if You have two L-transformable functions $f_{1}(t)$ and $f_{2} (t)$ with L-transforms $F_{1} (s)$ and $F_{2} (s)$ and abscissas of convergence $\sigma_{1}$ and $\sigma_{2}$, then the L-transform of the product is given by the integral...

$\displaystyle \mathcal {L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F_{1} (z)\ F_{2}(s - z)\ dz\ (1)$

... where $\displaystyle \sigma_{1} < \gamma < \sigma - \sigma_{2},\ \sigma> \sigma_{1} + \sigma_{2}$. In Your case is $\displaystyle f_{1}(t) = f_{2} (t) = 1 + t^{2} \implies F_{1}(s) = F_{2}(s) = \frac{1}{s} + \frac{2}{s^{3}},\ \sigma_{1}=\sigma_{2}=0$. May be it exists a more comfortable way but till now I didnn't succed to find it!...

Kind regards

$\chi$ $\sigma$

#### Chris L T521

##### Well-known member
Staff member
How do you find the laplace transform of this without expanding it?

$L(t^2+1)^2$
To compute this without expanding requires a little trick (especially if you're allowed to do this without using the integral definition of $\mathcal{L}\{f(t)\}$).

Let $f(t) = (t^2+1)^2$. We note that $f(0)=1$.

Thus, $f^{\prime}(t) = 4t(t^2+1)$ and hence $f^{\prime}(0) = 0$.

Finally, we see that $f^{\prime\prime}(t) = 4(t^2+1) + 8t^2 = 12t^2+4$.

Why did we compute all these derivatives? Well, we know that $\mathcal{L}\{f^{\prime\prime}(t)\} = s^2F(s) - sf(0) - f^{\prime}(0)$, which in our case would be $\mathcal{L}\{f^{\prime\prime}(t)\}=s^2F(s)-s$.

Therefore,

$\mathcal{L}\{12t^2+4\} = \mathcal{L}\{f^{\prime\prime}(t)\} = s^2F(s)-s$

We now compute the appropriate Laplace transforms and solve for $F(s)$:

$s^2F(s) - s = \mathcal{L}\{12t^2+4\} = 12\cdot\frac{2!}{s^3}+4\cdot\frac{1}{s}=\frac{24}{s^3}+\frac{4}{s} \implies F(s) = \frac{24}{s^5} + \frac{4}{s^3}+\frac{1}{s}$

Therefore, $\mathcal{L}\{(t^2+1)^2\} = \dfrac{24}{s^5}+\dfrac{4}{s^3}+\dfrac{1}{s}$.

I hope this makes sense!

#### Jester

##### Well-known member
MHB Math Helper
Question for the OP. Why would you want to compute this transform without expanding? I mean expanding would be the easiest approach.

BTW - Nice work Chris!

Following Chris's lead, you could keep going
\begin{align}
f(t) &= (t^2+1)^2,\;\;\; &f(0)&=1\\
f'(t) &= 4t(t^2+1)\;\;\; &f'(0)&=0\\
f''(t) &=4(t^2+1) + 8t^2,\;\;\;&f''(0)&=4\\
f'''(t) &= 8t + 16t,\;\;\;&f'''(0)&=0\\
f^{(4)}(t)&=24,\;\;\;&f^{(4)}(0)&=24\\
f^{(5)}(t)& = 0,
\end{align}

Then

$\mathcal{L}\{f^{(5)}(t)\} = s^5F(s) - f(0) s^4 - f'(0) s^3 - f''(0) s^2 - f'''(0) s - f^{(4)}(0) = 0.$

Substitute $f$ and its derivative at zero and solve for $F$ (same answer)

Last edited:

#### nacho

##### Active member
Thanks for the responses.

@Jester, really no other reason than for the sake of it. I've just started to learn laplace transforms and am finding a little out of the box thinking helps a lot.

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# How do I prove that two empty sets are equivalent?

I need to prove that two empty sets have the same cardinality, I know that proving equivalence is done by defining an onto and one-to-one function from one group to the other, but how can I define a function from an empty set to an empty set?

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Do you mean define a function from an empty set to an empty set? – Git Gud Feb 6 '13 at 22:36
yes of course, sorry. – Georgey Feb 6 '13 at 22:37
@amWhy That is correct. I will will delete my comment. – Git Gud Feb 6 '13 at 23:49

Let $A=B=\varnothing$. A function from $A$ to $B$ is a subset $f$ of $A\times B$ with the property that if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$. The only subset of $A\times B=\varnothing\times\varnothing=\varnothing$ is $\varnothing$. Does it have the function property? Yes: it contains no ordered pairs at all, so the hypothesis

if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$

is never satisfied, and the property

if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$

never actually requires anything to be true $-$ it holds vacuously.

Thus, $\varnothing$ is a function from $\varnothing$ to $\varnothing$. Is it a bijection? Is it true that for each $a\in A=\varnothing$ there is a unique $b\in B=\varnothing$ such that $\langle a,b\rangle\in f$? Again the answer is yes, vacuously: there’s no $a\in A=\varnothing$ at all, so the requirement is never actually invoked.

However, all of that work is completely unnecessary, because it’s always true that if $A=B$, then $|A|=|B|$. And if $A$ and $B$ are both empty, then they have exactly the same members, so they are the same set, a fact that I implicitly used above when I wrote $A=B=\varnothing$.

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Nice answer, Brian. +1 – Rick Decker Feb 7 '13 at 2:00
Thanks, @Rick. ${}$ – Brian M. Scott Feb 7 '13 at 2:03

Hint: Recall that there are no elements in an empty set. Show that if $A$ and $B$ are empty sets then for every $x\in A$ we have $x\in B$, and vice versa. Therefore $A\subseteq B$ and $B\subseteq A$ and so $A=B$.

If two sets are equal then certainly they have the same cardinality.

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Hint:

Use the definition of the (an) empty set - a set with no elements - to show, it is vacuously true that for all $x$ in the domain, $\;x \in \emptyset_1 \implies x \in \emptyset_2,\;$ so $$\emptyset_1 \subseteq \emptyset_2$$ and do similarly to show that it is vacuously true that $$\emptyset_2 \subseteq \emptyset_1$$

Hence, $\emptyset_1 = \emptyset_2$.

Hence, $|\emptyset_1| = |\emptyset_2|$

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The extension and specification axioms implies that there is only one set with no elements.

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# Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors

I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$ into partial fractions.

Now, I thought it could be solved into the following

$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(x-3)^2}$$

but this is apparently incorrect.

According to the text, the decomposition is

$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$

I discussed this with my friend that the fraction first decomposes into

$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{Bx +C}{(x-3)^2}$$

but I can't see how he derived this.

I don't understand how he is correct.

• Anything of your textbook's form can easily be expressed in your friend's form and vice versa. – Lord Shark the Unknown Sep 1 '17 at 20:08

Both decompositions used by your friend and the text are analogous since $$\frac{Bx+C}{(x-3)^2}=\frac{Bx}{(x-3)^2}+\frac{C}{(x-3)^2}=\frac{B(x-3)+3B}{(x-3)^2}+\frac{C}{(x-3)^2}$$ $$=\frac{B}{x-3}+\frac{3B+C}{(x-3)^2}=\frac{B'}{x-3}+\frac{C'}{(x-3)^2}$$

When you decompose, the degree of the numerator will be less than the degree of the denominator. How much less is yet to be determined. Assume that it is one degree less, and then if you get a zero coefficient, so be it.

If you had something like...$\frac {P(x)}{(x^2 +x + 1)(x-1)}$ your first step would be $\frac {Ax + B}{x^2+ x+ 1} + \frac {C}{(x-1)}$

With each numerator of degree 1 less than each denominator.

And the you might want to break down the first term to make it easier to integrate.

e.g. $\frac {A(x+\frac12)}{x^2 + x + 1} + \frac {B}{(x+\frac 12)^2 + \frac 34}$

With this problem.

$(x-3)^2$ makes for a degree 2 denominator, and you need a degree 1 numerator.

So $\frac {A}{x+1} + \frac {Bx+C}{(x-3)^2}$ would be a good place to start.

but $\frac {B}{(x-3)} + \frac {C}{(x-3)^2}$ is easier to integrate, so you might want to skip the intermediate step.


# Let $f(x)$ be continous in $x_0$ and $f(x_0)\ne 0$. Prove that $\exists C>0$ and neighbouhood of $x_0$ such that $|f(x)| \ge C$

Let a function $$f(x)$$ be continuous in some point $$x_0$$ and $$f(x_0) \ne 0$$. Prove that there exists a number $$C > 0$$ and neighbourhood of $$x_0$$ such that forall $$x$$ in that neighbourhood the following inequality holds: $$|f(x)| \ge C$$

The problem statement says that $$f(x_0) \ne 0$$, so it is either greater than $$0$$ or less than $$0$$. Let's consider the case for $$f(x_0) > 0$$. In such case by continuity of $$f(x)$$ we know: $$\lim_{x\to x_0} f(x) = f(x_0) > 0$$

Or in other words: $$\forall \epsilon > 0\ \exists \delta_\epsilon > 0\ \forall x: |x-x_0| < \delta_\epsilon \implies |f(x) - f(x_0)| < \epsilon$$

Since $$f(x_0) \ne 0$$ we may let $$\epsilon = {f(x_0)\over 2}$$. In such case: $$|f(x) - f(x_0)| < \epsilon \\ |f(x) - f(x_0)| < { f(x_0)\over 2 } \\ -{ f(x_0)\over 2 } < f(x) - f(x_0) < { f(x_0)\over 2 }\\ 0<{f(x_0)\over 2} < f(x) < {3f(x_0)\over 2}$$

So we may now choose any $$C \in \left(0; {f(x_0)\over 2}\right)$$, which would imply $$f(x) \ge C$$.

Similar reasoning is applied to the case when $$f(x_0) < 0$$.

I'm not sure my reasoning above is valid, so I would like to kindly ask for verification. Or for a correct proof if the above makes no sense. Thank you!

• Yes, this absolutely fine ;) The neighbourhood would be $(-f(x_0)/2,f(x_0)/2)$ when $f(x_0)>0$. – weirdo May 24 at 16:24
• This looks correct. You have $f(x) \geq C$ for all $|x-x_0|\leq\delta_\epsilon$. The reason this works is because of continuity, i.e. the statement is for all $\epsilon >0$..., the the particular choice of $\epsilon = f(x_0)/2$ works. – Dayton May 24 at 16:24

Yep, the above proof is valid. But for a couple suggestions in terms of proof writing, in the last line you said that $$f(x_0) \geq C$$, but this is only true for all $$x$$ such that $$|x-x_0|<\delta$$. Just make sure you include that in your last line so that people don't think you are saying that you are proving it for every $$x$$.
Have you been introduced to the neighbourhood notation? It is basically $$B_r(x_0) := \{|x-x_0| < r \quad| \quad \forall x \text{ in your domain.}\}$$ This way you can say $$|f(x) - f(x_0)|<\epsilon$$ as $$B_\epsilon (x_0)$$


# Show convergence and find the limit of the sequence given by $a_1=1$ and $a_{n+1}=\frac{1}{3+a_n}$

I've been trying to solve this exam question on an exam in real analysis. Thus, only such methods may be used. The problem is as follows.

Show that the sequence $a_n$ defined by $a_1=1$ and $a_{n+1}=(3+a_n)^{-1}$ for $n=1,2,\dots$ converges and determine the limit.

I'm not really able to find a closed form for the sequence. I guess that computing that and then showing that it's a Cauchy sequence works for showing convergence.

I just have to find that closed form first, which I'm unable to do. If there's a general strategy I'd love to see how that works as I've been unable to find any.

• Hint: monotonicity. Maybe monotonicity for subsequences. – xbh Aug 19 '18 at 19:56

Recursively $a_n>0$, $f(x)={1\over{x+3}}$, $f'(x)=-{1\over{(x+3)^2}}$, implies that $|a_{n+2}-a_{n+1}|=|f'(c_n)||a_{n+1}-a_n|\leq {1\over 3^2}|a_{n+1}-a_n|$, $c_n$ is an element of $(a_n,a_{n+1})$ or $(a_{n+1},a_n)$. You deduce that the sequence is a Cauchy sequence. (recursively you can show that $|a_{n+1}-a_n|\leq {1\over 3^{2(n-2)}}|a_2-a_1|$.

so it converges, the limit verifies $f(l)=l$ which implies that $l^2+3l-1=0$ ans is the positive root of this quadratic equation.

There need not be such a closed form for the sequence. If a limit $l$ exists it must be that$$l=\dfrac{1}{l+3}$$which yields to $$l=-1.5\pm\sqrt{3.25}$$where $l=-1.5-\sqrt{3.25}$ is not possible. To show that the sequence converges to $\sqrt{3.25}-1.5$ lets define $$e_n=a_n-(\sqrt{3.25}-1.5)$$therefore $$e_{n+1}=a_{n+1}-(\sqrt{3.25}-1.5)=\dfrac{1}{a_n+3}-(\sqrt{3.25}-1.5)=\dfrac{1-(e_n+1.5+\sqrt{3.25})(\sqrt{3.25}-1.5)}{a_n+3}=\dfrac{-e_n(\sqrt{3.25}-1.5)}{a_n+3}$$therefore $$|e_{n+1}|<\dfrac{|e_n|(\sqrt{3.25}-1.5)}{3}$$which shows that $e_n\to 0$ and $$a_n\to \sqrt{3.25}-1.5$$

• +1. I had thought of the same approach. Do you have something in mind to show that the convergence happens without having to assume that the limit exists first? – Vizag Aug 19 '18 at 20:18
• Well! Consider $$a_{n+2}=\dfrac{1}{3+a_{n+1}}=\dfrac{a_n+3}{3a_n+10}<\dfrac{4}{10}a_n$$ therefore we can conclude what we want – Mostafa Ayaz Aug 19 '18 at 20:23

Monotonicity method.

Rewrite the recursive relation as $$a_{n+2} = \frac 1 {3+ \dfrac 1 {3 + a_n}} = \frac {3+a_n} {3(3+a_n) + 1} = \frac {a_n + 3} {3 a_n + 10},$$ or $$a_{n+2} = \frac 13 - \frac 1{9a_n + 30} \leqslant \frac 13.$$ Since $$a_{n+4}-a_{n+2} = \frac 1 {9a_n + 30} - \frac 1{9a_{n+2} +30} = \frac {9(a_{n+2}-a_n )} {(9a_n +30) (9a_{n+2}+30)},$$ and clearly $a_n \geqslant 0$, we conclude that subsequences $a_{2k}, a_{2k+1}$ are monotonic [not necessarily the same type]. Since $a_1 = 1, a_2 = 1/4, a_3 = 4/13, a_4=13/34, a_1 > a_3, a_2 < a_4$, we know that $a_{2k} < a_{2k+2} \leqslant 1/3, 1 \geqslant a_{2k+1} > a_{2k+3}$. Hence $a_{2k}, a_{2k+1}$ both converge [since both are monotonic and bounded]. By the recursive relation, they converge to the same limit $L > 0$.

Let $n \to \infty$ in the first formula, $$L = \frac {L + 3}{3L + 10},$$ or simply, $$L(L+3) =1.$$

Here is a somewhat systematic approach. The crux of the method is to separate the numerator and denominator. Write $a_n = p_n / q_n$. Can we work out $p_{n+1}, q_{n+1}$? We have $$x_{n+1} = \frac{1}{3 + p_n/q_n} = \frac{q_n}{3q_n + p_n}$$ So $p_{n+1} = q_n$ and $q_{n+1} = 3q_n + p_n = 3q_n + q_{n-1}$. Now we can see that the important value is the denominator. We are lucky because our second equation is a second order recurring relation for $q_n$, so we can solve for the general form using the following theorem:

Let $(x_n)_{n \ge 0}$ be a sequence such that there exists $(a,b)\in \mathbb{C}^2$ $x_{n+2} = ax_{n+1} + bx_n$. Then the sequence $(u_n)_{n \ge 0}$, $u_n = r^n$ is a solution iff $r$ satisfies the characteristic equation: $$r^2 - ar - b = 0$$ Furthermore if the characteristic equation admits two distinct roots then any solution is of the form $$\alpha r_1^n + \beta r_2 ^n$$

We don't mention the case of repeated roots. In our case the characteristic equation is $$r^2 - 3r - 1 = 0$$

The two roots are $j_1 = \frac{3 + \sqrt{13}}{2}, j_2 = \frac{3-\sqrt{13}}{2} =$. The best part is that we don't need to solve for $\alpha$ and $\beta$ (though you do need to check $\alpha \ne 0$, which is easily done); notice that $|j_2| < 1$. Now we can write $$x_n = \frac{p_n}{q_n} = \frac{q_{n-1}}{q_n} = \frac{\alpha j_1^{n-1} + \beta j_2^{n-1}}{\alpha j_1^{n} + \beta j_2^{n}}$$ Since $|j_2|<1$, as $n\to \infty$, this goes to $1/j_1$. You can check that $1/j_1$ satisfies $L = 1/(3 + L)$


Q&A

# difference between quotient rule and product rule

+2
−1

Product rule :

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

Quotient rule :

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

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

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

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

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

Why does this post require moderator attention?
Why should this post be closed?

+1
−0

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

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

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

Why does this post require moderator attention?

+1
−0

is it OK to use Product rule instead of Quotient rule in University and Real Life?

Sure. For whatever reason, I long had a hard time remembering the quotient rule, and instead used the process you describe.

Why does this post require moderator attention?

+1
−0

If you work out deriving the quotient rule yourself using the exact trick you're highlighting, you can see that the quotient rule is nothing more than the product rule and the chain rule used together. If you rearrange a quotient to use the product rule, then you'll very likely be using the chain rule shortly thereafter on the $(\cdots)^{-1}$ part, and you will inevitably arrive at exactly the same result as if you had used the quotient rule but with more steps. The only potential difference is whether your result looks like $\frac{a}{b} + \frac{c}{b^2}$ or $\frac{ab + c}{b^2}$—but of course, as I'm sure you can see, that's no difference at all.

Why does this post require moderator attention?


# Invertible Matrix and Linearly Independent Vectors Proof

Trying to do this one:

Suppose $A$ is an invertible $n$ x $n$ matrix and the vectors $v_1$, $v_2$, ..., $v_n$ are linearly independent. Show that the vectors $Av_1$, $Av_2$, ..., $Av_n$ are linearly independent.

I know that A's column vectors are linearly independent since A is invertible. I also know there is no relation amongst the $v_i$ because they're linearly independent.

My idea is to write the product of A and a given $v_i$ in terms of the columns of A. Not sure if this is right, any guidance much appreciated!

Thanks, Mariogs

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Maybe you can do a proof by contradiction? If they are lin dependent, then the zero vector can be written as a lin combination (with nonzero scalars) of your $Av's$ That means that a particular $Av$ vector can be written as a lin combination of the remaining $Av's$. Now what does that mean? –  imranfat Feb 27 '14 at 16:33

Let $w_i=Av_i$ for $i=1,\dots,n$. To show linear independence, we must show that $$c_1w_1+\cdots+c_nw_n=0 \Longrightarrow c_i=0,\quad i=1,\dots n.$$ Since $A$ is invertible, we can left-multiply $c_1w_1+\dots=0$ by $A^{-1}$ to get...

Can you take it from there? Remember that the vectors $v_1,\dots v_n$ are linearly independent.

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Ah this makes sense to me, thanks! –  bclayman Feb 27 '14 at 16:56
To finish it: We multiply on the left by $A^-1$ and get $c_1 v_1$ + ... + $c1 v_n$ = 0. Since $v_1, ..., v_n$ are linearly independent, we must have $c_i$ = 0 for all i. Yes? –  bclayman Feb 27 '14 at 17:00
That's right, yes. –  user12477 Mar 3 '14 at 12:05

Notice that since $A$ is invertible then $$Ax=0\iff x=0$$ Now let $a_1,a_2,\ldots,a_n\in \Bbb R$ such that $$\sum_{k=1}^na_k Av_k=0\iff A\left(\sum_{k=1}^na_k v_k\right)=0\iff \sum_{k=1}^na_k v_k=0\Rightarrow a_i=0\;\forall i$$ since $v_1$, $v_2$, ..., $v_n$ are linearly independent. Conclude.

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Whoa! You've been busy with answers! –  amWhy Feb 28 '14 at 13:34

Why not use the definition directly? Justify/explain the following:

$$\text{For scalars}\;\;c_1,...,c_n\;:\;\;0=\sum_{k=1}^nc_kAv_k=\sum_{k=1}^nA(c_kv_k)=A\left(\sum_{k=1}^nc_kv_k\right)\iff$$

$$\iff\sum_{k=1}^nc_kv_k=0\iff c_1=c_2=\ldots=c_n=0$$

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Miscellaneous questions: Part I: tutorial practice for preRMO and RMO

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.

Problem 2:

Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.
B: I see four white caps.
C: I see one black and three white caps.
D: I see four black caps.

Problem 5:

Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times.

Problem 6:

Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.

Problem 7:

There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.

Problem 8:

In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?

Problem 9:

Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

Try these !!

Regards,
Nalin Pithwa

Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance

$\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}$

$\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}$

$\bf{Section 1}$:

$\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$

The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\it{finite}$.$\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}$

On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\it{infinite}$ sets.

Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.

To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.

$\bf{Section 2}$:

$\bf{Countable \hspace{0.1in} Sets}$.

The simplest infinite set is the set $\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots$. By an $\bf{uncountable}$ set we mean, of course, an infinite set which is not countable.

We now give some examples of countable sets:

$\bf{Example 1}$:

The set $\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\mathscr{Z}$ and $\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is,

$n \leftrightarrow (2n+1)$, if $n \geq 0$, and $n \in \mathscr{Z}$
$n \leftrightarrow 2|n|$, if $n<0$, and $n \in \mathscr{Z}$

$\bf{Example 2}$:

The set of all positive even numbers is countable, as shown by the obvious correspondence $n \leftrightarrow 2n$.

$\bf{Example 3}$:

The set 2,4,8,$\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \leftrightarrow 2^{n}$.

$\bf{Example 4}: The set$latex \mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\alpha$ can be written as a fraction $\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the “height” of the rational number $\alpha$. For example, $\frac{0}{1}=0$ is the only rational number of height zero, $\frac{-1}{1}$, $\frac{1}{1}$ are the only rational numbers of height 2, $\frac{-2}{1}$, $\frac{-1}{2}$, $\frac{1}{2}$, $\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\mathscr{Z^{+}}$ of all positive integers. $\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}$ $\bf{Theorem1}$. $\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}$. $\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \leftrightarrow a_{n_{i}}$). $\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}$ $\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\bigcup A_{2})$, $\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \ldots$ in the form of an infinite table $\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to “left”, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\begin{array}{cccccccc} a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\ a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\ \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\ \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \ldots$ and the set $\mathscr{Z^{+}}$ of all positive integers. $\bf{QED.}$ $\bf{Theorem3}$ $\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}$ $\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset $A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\}$ of the set $M$. $\bf{QED.}$ $\bf{Remark}$ Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\bf{Section3}$ $\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\bf{Definition}$ Two sets M and N are said to be $\bf{equivalent}$ (written $M \sim N$) if there is a one-to-one correspondence between the elements of M and the elements of N. The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set $\mathscr{Z^{+}}$ of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent. $\bf{Example1}$ The sets of points in any two closed intervals$[a,b]$and$[c,d]\$ are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect.

$\bf{Example2}$

The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence $z \leftrightarrow \alpha$ can be established by using stereographic projection. The origin is the North Pole of the sphere.

$\bf{Example3}$

The set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula $y=\frac{1}{\pi}\arctan{x}+\frac{1}{2}$ establishes a one-to-one correspondence between these two sets. $\bf{QED}$.

The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval $(0,1)$ as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:

$\bf{Theorem4}$

$\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}$

$\bf{Proof}$

According to Theorem 3, every infinite set M contains a countable subset. Let this subset be $A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \}$ and partition A into two countable subsets $A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \}$ and $A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}$.

Obviously, we can establish a one-to-one correspondence between the countable subsets A and $A_{1}$ (merely let $a_{n} \leftrightarrow a_{2n-1}$). This correspondence can be extended to a one-to-one correspondence between the sets $A \bigcup (M-A)=M$ and $A_{1} \bigcup (M-A)=M-A_{2}$ by simply assigning x itself to each element $x \in M-A$. But $M-A_{2}$ is a proper subset of M. $\bf{QED}$.

More later, to be continued,

Regards,
Nalin Pithwa

A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

$P_{1}+P_{2}=P_{3}+P_{4}$

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

$P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}$

$P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}$

$P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}$

$P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}$

You will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$

Hence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.

Regards,
Nalin Pithwa.

Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff $f^{-1}(f(A))=A$ for all $A \subset X$. Given that $f: X \longrightarrow Y$.

Problem 2:

Prove that a function if is onto iff $f(f^{-1}(C))=C$ for all $C \subset Y$. Given that $f: X \longrightarrow Y$.

Problem 3:

(a) How many functions are there from a non-empty set S into $\phi$\?

(b) How many functions are there from $\phi$ into an arbitrary set $S$?

(c) Show that the notation $\{ X_{i} \}_{i \in I}$ implicitly involves the notion of a function.

Problem 4:

Let $f: X \longrightarrow Y$ be a function, let $A \subset X$, $B \subset X$, $C \subset Y$ and $D \subset Y$. Prove that

i) $f(A \bigcap B) \subset f(A) \bigcap f(B)$

ii) $f^{-1}(f(A)) \supset A$

iii) $f(f^{-1}(C)) \subset C$

Problem 5:

Let I be a non-empty set and for each $i \in I$, let $X_{i}$ be a set. Prove that

(a) for any set B, we have $B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})$

(b) if each $X_{i}$ is a subset of a given set S, then $(\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}$ where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) $A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)$

(ii) $(A-B) \times C=(A \times C)-(B \times C)$

🙂 🙂 🙂

Nalin Pithwa

Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if $A \bigcup B=A$ and $A \bigcap B=A$, then $A=B$.

Problem 2:

Show that in general $(A-B)\bigcup B \neq A$.

Problem 3:

Let $A = \{ 2,4, \ldots, 2n, \ldots\}$ and $B= \{ 3,6,\ldots, 3n, \ldots\}$. Find $A \bigcap B$ and $A - B$.

Problem 4:

Prove that (a) $(A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)$

Prove that (b) $A \Delta B = (A \bigcup B)-(A \bigcap B)$

Problem 5:

Prove that $\bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})$

Problem 6:

Let $A_{n}$ be the set of all positive integers divisible by $n$. Find the sets (i) $\bigcup_{n=2}^{\infty}A_{n}$ (ii) $\bigcap_{n=2}^{\infty}A_{n}$.

Problem 7:

Find (i) $\bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}]$ (ii) $\bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$

Problem 8:

Let $A_{\alpha}$ be the set of points lying on the curve $y=\frac{1}{x^{\alpha}}$ where $(0. What is $\bigcap_{\alpha \geq 1}A_{\alpha}$?

Problem 9:

Let $y=f(x) = $ for all real x, where $$ is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval $\frac{1}{4} \leq y \leq \frac{3}{4}$? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let $\mathscr{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a \in M$, and let $aRb$ if and only if $(a,b) \in \mathscr{R}$. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

• Reflexive and symmetric, but not transitive.
• Reflexive, but neither symmetric nor transitive.
• Symmetric, but neither reflexive nor transitive.
• Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

Set Theory, Relations, Functions: Preliminaries: part VIIIA

(We continue from part VII of the same blog article series with same reference text).

Theorem 4:

A set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M.

Proof of Theorem 4:

Every partition of M determines a binary relation on M, where $aRb$ means that “a belongs to the same class as b.” It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M.

Conversely, let R be an equivalence relation on M, and let $K_{a}$ be the set of all elements $x \in M$ such that $xRa$ (clearly, $a \in K_{a}$, since R is reflexive). Then, two classes $K_{a}$ and $K_{b}$ are either identical or disjoint. In fact, suppose that an element c belongs to both $K_{a}$ and $K_{b}$, so that $cRa$ and $cRb$. But by symmetry of R, being an equivalence relation, we can infer that $aRc$ also and, further by transitivity, we say that $aRb$. If now, $x \in K_{a}$ then we have $xRa$ and hence, $xRb$ (since we already have $aRb$ and using transitivity).

Similarly, we can prove that $x \in K_{b}$ implies that $x \in K_{a}$.

Therefore, $K_{a}=K_{b}$ if $K_{a}$ and $K_{b}$ have an element in common. Therefore, the distinct sets $K_{a}$ form a partition of M into classes.

QED.

Remark:

Because of theorem 4, one often talks about the decomposition of a set M into equivalence classes.

There is an intimate connection between mappings and partitions into classes, as illustrated by the following examples:

Example 1:

Let f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image $b=f(a) \in B$. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point $(x,y)$ into the point $(x,0)$. Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f.

Example 2:

Given any partition of a set A into classes, let B be the set of these classes and associate each element $a \in A$ with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line $[0, \infty)$ each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line $[0,\infty)$.

Example 3:

Suppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of “winding” the real line onto a circle of unit circumference. (Note: The largest integer $\leq x$ is called the integral part of x, denoted by [x], and the quantity $x -[x]$ is called the fractional part of x).

In the next blog article, let us consider a tutorial problem set based on last two blogs of this series.

🙂 🙂 🙂

Nalin Pithwa

A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that $\tan {A}$ and $\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of

$\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}$

Solution:

Let $\alpha=\tan{A}$ and $\beta=\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$

By Viete’s relations between roots and coefficients:

$\alpha+\beta=\tan{A}+\tan{B}=-p$ and $\alpha \beta = \tan{A}\tan{B}=q$ but we also know that $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}$

Now, let us call $E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)}$ which in turn is same as

$\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)$

We have already determined $\tan{(A+B)}$ in terms of p and q above.

Now, again note that $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ which in turn gives us that $\tan^{2}{\theta}+1=\sec^{2}{\theta}$ so we get:

$\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that

$\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$

Hence, the given expression E becomes:

$(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q)$, which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.


# How would I find a point on the y-axis equidistant from two other points?

The points are$$(5,-5) and (1,1)$$

I tried doing this visually and came up with (0,-5). This wasn't correct once I applied the distance formula to check the distance between that point and the two others.

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Let the point be $(0,y)$. Then $5^2+(y+5)^2=1^2+(y-1)^2$. Expand and solve. (There are other ways.) – André Nicolas Aug 26 '14 at 11:49
As for what you did wrong, I'm not sure, but $(-5,0)$ is not on the $y$-axis... – fixedp Aug 26 '14 at 11:51
that was a typo – Cherry_Developer Aug 26 '14 at 11:52

A point on the $y-$axis is of the form $(0,y)$.

The distance between $(0,y)$ and $(5,-5)$ is: $$\sqrt{(5-0)^2+(-5-y)^2}=\sqrt{25+(5+y)^2}$$

The distance between $(0,y)$ and $(1,1)$ is: $$\sqrt{(1-0)^2+(1-y)^2}=\sqrt{1+(1-y)^2}$$

The two distances are equal, so also their squares:

$$25+(5+y)^2=1+(1-y)^2$$

Now you have to solve for $y$.

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It's probably more convenient to equate the squares of the distances, rather than the distances themselves. – fixedp Aug 26 '14 at 11:50
I have doubts but I think it's (0,-4) – Cherry_Developer Aug 26 '14 at 11:59
It is correct!! – Mary Star Aug 26 '14 at 12:04

You want the point where the perpendicular bisector of the two points cuts the $y$-axis.

The slope of the line between $(5,-5)$ and $(1,1)$ is $-\frac{3}{2}$, so the slope of the normal is $\frac23$. Hence the normal through the mid-point $(3,-2)$ cuts the $y$-axis at $(0,-4)$.

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+1 because on seeing the question I thought to myself, "well personally I'd just solve the equation, but what would that teach you about the set of points equidistant between A and B?" :-) – Steve Jessop Aug 26 '14 at 13:19
@Steve: Yes! I was disturbed by all those quadratic terms in the other answers, which always cancelled each other out. – TonyK Aug 26 '14 at 13:22
That's an insight in itself, mind you. They cancel out so that the set of equidistant points can be a straight line, as opposed to something curved if there were quadratic terms left in there! – Steve Jessop Aug 26 '14 at 13:23
This also illustrates why it's always possible to find such a point, unless the given points lie on a horizontal line (which would make the perpendicular bisector parallel to the $y$-axis. – cjm Aug 26 '14 at 14:31