5 For any m n matrix A, we have A i = eT i A and A j = Ae j. P. Sam Johnson (NITK) Existence of Left/Right/Two-sided Inverses September 19, 2014 3 / 26 In a monoid, if an element has a right inverse… Thus the unique left inverse of A equals the unique right inverse of A from ECE 269 at University of California, San Diego h�b�y��� cca�� ����ِ� q���#�!�A�ѬQ�a���[�50�F��3&9'��0 qp�(R�&�a�s4�p�[���f^'w�P&޶ 7��,���[T�+�J����9�$��4r�:4';m$��#�s�Oj�LÌ�cY{-�XTAڽ�BEOpr�l�T��f1�M�1$��С��6I��Ҏ)w It would therefore seem logicalthat when working with matrices, one could take the matrix equation AX=B and divide bothsides by A to get X=B/A.However, that won't work because ...There is NO matrix division!Ok, you say. Generalized inverse Michael Friendly 2020-10-29. best. %PDF-1.4 given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. 11.1. LEAST SQUARES PROBLEMS AND PSEUDO-INVERSES 443 Next, for any point y ∈ U,thevectorspy and bp are orthogonal, which implies that #by#2 = #bp#2 +#py#2. An inverse that is both a left and right inverse (a two-sided inverse), if it exists, must be unique. U-semigroups endobj Proof: Assume rank(A)=r. x��XKo#7��W�hE�[ע��E������:v�4q���/)�c����>~"%��d��N��8�w(LYɽ2L:�AZv�b��ٞѳG���8>����'��x�ټrc��>?��[��?�'���(%#R��1 .�-7�;6�Sg#>Q��7�##ϥ "�[� ���N)&Q ��M���Yy��?A����4�ϠH�%�f��0a;N�M�,�!{��y�<8(t1ƙ�zi���e��A��(;p*����V�Jڛ,�t~�d��̘H9����/��_a���v�68gq"���D�|a5����P|Jv��l1j��x��&޺N����V"���"����}! Let G G G be a group. left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. �n�����r����6���d}���wF>�G�/��k� K�T�SE���� �&ʬ�Rbl�j��|�Tx��)��Rdy�Y ? 0 Yes. JOURNAL OF ALGEBRA 31, 209-217 (1974) Right (Left) Inverse Semigroups P. S. VENKATESAN National College, Tiruchy, India and Department of Mathematics, University of Ibadan, Ibadan, Nigeria Communicated by G. B. Preston Received September 7, 1970 A semigroup S (with zero) is called a right inverse semigroup if every (nonnull) principal left ideal of S has a unique idempotent … (An example of a function with no inverse on either side is the zero transformation on .) Recall also that this gives a unique inverse. If f contains more than one variable, use the next syntax to specify the independent variable. Viewed 1k times 3. Note that other left In fact, if a function has a left inverse and a right inverse, they are both the same two-sided inverse, so it can be called the inverse. This is generally justified because in most applications (e.g., all examples in this article) associativity holds, which makes this notion a generalization of the left/right inverse relative to an identity. Ask Question Asked 4 years, 10 months ago. Let (G, ⊕) be a gyrogroup. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. eralization of the inverse of a matrix. If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)? By using this website, you agree to our Cookie Policy. Indeed, the existence of a unique identity and a unique inverse, both left and right, is a consequence of the gyrogroup axioms, as the following theorem shows, along with other immediate, important results in gyrogroup theory. Let e e e be the identity. Some easy corollaries: 1. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. Show Instructions. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Actually, trying to prove uniqueness of left inverses leads to dramatic failure! '+o�f P0���'�,�\� y����bf\�; wx.��";MY�}����إ� endstream endobj 54 0 obj <> endobj 55 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/Thumb 26 0 R/TrimBox[79.51181 97.228348 518.881897 763.370056]/Type/Page>> endobj 56 0 obj <>stream Show Instructions. Proof In the proof that a matrix is invertible if and only if it is full-rank, we have shown that the inverse can be constructed column by column, by finding the vectors that solve that is, by writing the vectors of the canonical basis as linear combinations of the columns of . If BA = I then B is a left inverse of A and A is a right inverse of B. share. 6 comments. Note the subtle difference! u (b 1 , b 2 , b 3 , …) = (b 2 , b 3 , …). 3. This may make left-handed people more resilient to strokes or other conditions that damage specific brain regions. Active 2 years, 7 months ago. Recall that$B$is the inverse matrix if it satisfies $AB=BA=I,$ where$I$is the identity matrix. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. Then a matrix A−: n × m is said to be a generalized inverse of A if AA−A = A holds (see Rao (1973a, p. 24). (Generalized inverses are unique is you impose more conditions on G; see Section 3 below.) Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. For any elements a, b, c, x ∈ G we have: 1. Proof: Assume rank(A)=r. inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse) and surjective (since there is a right inverse). We will later show that for square matrices, the existence of any inverse on either side is equivalent to the existence of a unique two-sided inverse. If the function is one-to-one, there will be a unique inverse. Proof. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. Remark When A is invertible, we denote its inverse … Theorem 2.16 First Gyrogroup Properties. There are three optional outputs in addition to the unique elements: However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. (We say B is an inverse of A.) If E has a right inverse, it is not necessarily unique. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Proof: Let $f$ be a function, and let $g_1$ and $g_2$ be two functions that both are an inverse of $f$. u(b_1,b_2,b_3,\ldots) = (b_2,b_3,\ldots). If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. save hide report. inverse. Let (G, ⊕) be a gyrogroup. A.12 Generalized Inverse Deﬁnition A.62 Let A be an m × n-matrix. 1. f is injective if and only if it has a left inverse 2. f is surjective if and only if it has a right inverse 3. f is bijective if and only if it has a two-sided inverse 4. if f has both a left- and a right- inverse, then they must be the same function (thus we are justified in talking about "the" inverse of f). Hence it is bijective. h��[[�۶�+|l\wp��ߝ�N\��&�䁒�]��%"e���{>��HJZi�k�m� �wnt.I�%. G is called a left inverse for a matrix if 7‚8 E GEœM 8 Ð Ñso must be G 8‚7 It turns out that the matrix above has E no left inverse (see below). 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective Theorem. New comments cannot be posted and votes cannot be cast. Yes. ��� See Also. 125 0 obj <>stream Sort by. 53 0 obj <> endobj wqhh��llf�)eK�y�I��bq�(�����Ã.4-�{xe��8������b�c[���ö����TBYb�ʃ4���&�1����o[{cK�sAt�������3�'vp=�$��$�i.��j8@�g�UQ���>��g�lI&�OuL��*���wCu�0 �]l� It's an interesting exercise that if$a$is a left unit that is not a right uni Thus, p is indeed the unique point in U that minimizes the distance from b to any point in U. 100% Upvoted. Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. So to prove the uniqueness, suppose that you have two inverse matrices$B$and$C$and show that in fact$B=C$. 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# Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum\limits_n \frac{1}{n+f(n)}$ is convergent?

Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum_1 ^\infty \frac{1}{n+f(n)}$ is convergent?

I could not solve this. I tried to proceed in following lines:

1) Tried to provide a contradiction:

First let $n \sim m$ iff $\exists k \in \mathbb{Z}$ such that $f^k(n)=m$. This is an equivalence relation. Consider the orbits. For the finite orbits we can compare the series to $\sum_1^\infty \frac{1}{n+n}$. But then I could not figure out how to proceed for infinite orbits.

2) Tried to prove that there is some function:

Let $\{k_n\}$ be a subsequence of $\mathbb{N}$ such that $\sum_0^\infty \frac{1}{k_n}$ converges. Set $f(n)=k_{n}, \forall n \in \mathbb{N}\setminus\{k_n\}$. Then images of each $n$ which are not in the subsequence $k_n$ is defined. Now we have to define images of each $k_n$. Define $f(k_n)=n$ $\forall n \in \mathbb{N}\setminus \{k_n\}.$.

Could not proceed further. I think My second attempt was going in right direction. My plan was use the fact that all elements here are positive and to construct the function $f$ in such a manner that $\forall n\in \mathbb{N}$ either $n$ or $f(n)$ is in $\{k_n\}$.

• Decompose $\mathbb N$ into the disjoint union of $S=\{k^2\mid k\in\mathbb N\}$ and $R=\mathbb N\setminus S$. Decompose $S$ into the disjoint union $\bigcup\limits_{k\in\mathbb N}S_k$, where the size of $S_k$ is $2k$ and every element of $S_k$ is smaller than every element of $S_{k+1}$, for every $k$. Then enumerate $R=\{r_k\mid k\in\mathbb N\}$ with $r_k<r_{k+1}$ for every $k$, reorder $\mathbb N$ as $$S_1\ r_1\ S_2\ r_2\ S_3\ \ldots\ S_k\ r_k\ S_{k+1}\ \ldots$$ and consider the corresponding function $f$. Then $$\sum_{n\in S}\frac1{n+f(n)}<\sum_{n\in S}\frac1{n}=\sum_k\frac1{k^2}$$
– Did
Feb 17 '17 at 12:51
• ... converges and $r_k\geqslant k^2$ for each $k$, hence $$\sum_{n\in R}\frac1{n+f(n)}<\sum_k\frac1{r_k}\leqslant\sum_k\frac1{k^2}$$ converges hence the whole series converges.
– Did
Feb 17 '17 at 12:51
• @Did, I think you want to order $\mathbb{N}$ as $r_1,S_1,r_2,S_2,r_3,S_3,\ldots$ (providing that $0\notin\mathbb{N}$). But well, if $0\in \mathbb{N}$, then we have to modify this solution only slightly. I would give you an upvote if you posted this as a solution. Feb 17 '17 at 13:03
• @Batominovski "I think you want to order N as r1,S1,r2,S2,r3,S3,…" Hmmm, why? What difference does it make?
– Did
Feb 17 '17 at 19:36
• @Did The difference is that $f^{-1}\left(r_k\right)\geq k^2$ may not hold. You have $f^{-1}\left(r_k\right)\geq k^2-1$ instead. (However, nothing really changes.) Feb 18 '17 at 2:26

Originally, I gave this example:

$$f (n)=\begin{cases}k,&\ \text { if } n=3^k , \text { with k  not a power of 2}\\ 2^{n-1} , &\ \text { otherwise }\end{cases}$$ (the idea is to push the small numbers further and further down the road so that when they appear they are compensated by the $n$). Then $$\sum_n\frac1 {n+f (n)}<\sum_{k}\frac 1 {3^k+k}+\sum_n\frac1 {n+2^{n-1}}<\infty.$$

And it is the right idea, but the problem is that such $f$ is not onto. For instance, $2^{26}$ is not in the range of $f$, because when $n=27$, we are using the other branch of $f$ to get $3$.

So we need to tweak the example slightly. Let $$T=\{3^k:\ k\ \text{ is not a power of } 2\}=\{3^3,3^5,3^6,3^7,3^9,\ldots\}$$ and $$S=\mathbb N\setminus T=\{1,\ldots,7,8,10,11,\ldots\}.$$ Write them as an ordered sequence, $T=\{t_1,t_2,\ldots\}$ and $S=\{s_1,s_2,\ldots\}$. Now define $$f(n)=\begin{cases} \log_3 n,&\ \text{ if }\ n=t_k\\ \ \\ 2^{k-1},&\ \text{ if }\ n=s_k \end{cases}$$ One can check explicitly that $$g(m)=\begin{cases} 3^m,&\ \text{ if m is not a power of 2}\ \\ \ \\ s_{k+1},&\ \text{ if }\ m=2^k \end{cases}$$ is an inverse for $f$.

Not a new solution, just writing to clarify for myself how the solution of @Martin Argerami: works.

We want $\sum_{n\in \mathbb{N}} \frac{1}{n + f(n)} < \infty$. Consider a covering $\mathbb{N} = A\cup B$. It's enough to have $\sum_{n\in A} \frac{1}{n + f(n)}$, $\sum_{n \in B} \frac{1}{n + f(n)}< \infty$. So it's enoough to find $A$, $B$ so that $$\sum_{n \in A} \frac{1}{f(n)}= \sum_{n \in f(A)} \frac{1}{n} < \infty \\ \sum_{n \in B} \frac{1}{n} < \infty$$

So it's enough to have $B$ so that $\sum_{n\in B} \frac{1}{n} < \infty$ and $f$ mapping $A = \mathbb{N} \backslash B$ to $B$. There are many possibilities here.

• I would not say that "this is how the other solution works" because the other solution relies heavily on infinite orbits while a simple involution between the powers of two and the non-powers of two works fine. Nov 19 '18 at 10:58


# Is there a formula for this gaussian integral

1. Apr 28, 2014

### skate_nerd

Is there a formula for this gaussian integral:

$$int_{-\infty}^{\infty}{x^4}{e^{-a(x-b)^2}}dx$$

I've tried wikipedia and they only have formulas for the integrand with only x*e^... not x^4e^...
Wolframalpha won't do it either, because I actually have an integral that looks just like that, unknown constants and all.

2. Apr 28, 2014

### Matterwave

Well, we know that:

$$\int_{-\infty}^\infty e^{-bx^2}dx=\sqrt{\frac{\pi}{b}}$$

Taking some derivatives:

$$\frac{d^2}{db^2}\int_{-\infty}^\infty e^{-bx^2}dx=\int_{-\infty}^{\infty}x^4e^{-bx^2}dx=\frac{3}{4}\frac{\sqrt{\pi}}{b^{5/2}}$$

I'm not sure if you can use this for your integral, but it looks possible.

3. Apr 28, 2014

### skate_nerd

Sorry, but you can't :/

4. Apr 28, 2014

### vanhees71

5. Apr 28, 2014

### pwsnafu

You are basically just asking for the moments of the normal distribution. These are well known:
$$\int_{-\infty}^{\infty} \frac{x^p}{\sigma \sqrt{2\pi}} e^{-(x-\mu)^2 / (2 \sigma^2)} dx= \sigma^p (p-1)!$$
when p is even, and zero when p is odd.

So $p=4$, $\sigma = \frac{1}{\sqrt{2a}}$ and $\mu = b$, which gives $\frac{3\sqrt{\pi}}{4 a^{5/2}}$ as Matterwave gave.

6. Apr 28, 2014

### skate_nerd

I'm sorry i just dont think that's correct. Try wolframalpha. I know the formulas for the x, x^2, x^3 cases, but not the x^4. But everything up to x^3 agrees with wolframalpha's output, and this formula you guys are giving doesnt look very similar and does not agree.
$$\int_{-\infty}^{\infty}{x}e^{-a(x-b)^2}dx=b\sqrt{\frac{\pi}{a}}$$
$$\int_{-\infty}^{\infty}{x^2}e^{-a(x-b)^2}dx=(\frac{1}{2a}+b^2)\sqrt{\frac{\pi}{a}}$$
$$\int_{-\infty}^{\infty}{x^3}e^{-a(x-b)^2}dx=(\frac{3b}{2a}+b^3)\sqrt{\frac{\pi}{a}}$$
There's a pattern here but I can't figure it out.

7. Apr 28, 2014

### pwsnafu

I'm an idiot, I looked up the central moments (and even that's wrong because it should be $(p-1)!!$). The non-central moments are

$\mu$
$\mu^2+\sigma^2$
$\mu^3+3\mu\sigma^2$
$\mu^4+6\mu^2\sigma^2+3\sigma^4$,
for p = 1, 2, 3, and 4. Those should agree with what you have.

8. Apr 28, 2014

### skate_nerd

Ahhh thank you thank you thank you! That works 100%! Man I've been trying to get this formula for a week...I really appreciate it.

9. Apr 28, 2014

### eigenperson

The "freshman calculus" way to do this is to integrate by parts:
$$u = x^3 \qquad dv = xe^{-x^2}\,dx$$so that
$$du = 3x^2 \qquad v = -{1 \over 2}e^{-x^2}$$and now the degree of the polynomial has been reduced by 2.

Now, that is for the case where a and b are 0, but the other cases are really the same. You just have to change variables; then you will have a more complicated polynomial in place of $x^4$, but once you know how to do $\int x^{2k}e^{-x^2}$ you can integrate any polynomial times $e^{-x^2}$.

10. Apr 28, 2014

### Matterwave

Another clever way to do this integral is to look at the characteristic function for a Gaussian distribution:

$$C(t)\equiv\left<e^{itx}\right>=\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{itx}e^{-(x-\mu)^2/2\sigma^2}$$

You can do this integral by completing the square in the exponential, or by just looking it up. You will find:

$$C(t)=e^{it\mu-\frac{1}{2}t^2\sigma^2}$$

You can expand this function in powers of t, and match it to the expansion of:

$$\left<e^{itx}\right>\approx 1+\left<itx\right>+\frac{\left<(itx)^2\right>}{2}+...$$

If you match the powers of t on both sides, you can get all the general moments of the Gaussian distribution. You can get all the odd ones too, for which the formula I gave earlier would no longer work. This Characteristic function is closely related to the moment generating function, but I'm more familiar with working with these.


# Trouble with trig integral

#### Kristal

##### New member
I have a few questions and a request for an explanation.

I worked this problem for a quite a while last night. I posted it here.

https://math.stackexchange.com/questions/3547225/help-with-trig-sub-integral/3547229#3547229

The original problem is in the top left. Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.

My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?

My second question is: is this a correct solution? https://imgur.com/2tgEz0O

It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p

Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some advice about how I could improve my math tricks to that point?

#### skeeter

##### Well-known member
MHB Math Helper
Is the algebra in my original work sound?
Looks like you did this (correct me if I'm mistaken) ...

$$\displaystyle -7\int \dfrac{x^2}{\sqrt{4x-x^2}} \,dx = -7\int \dfrac{x^2}{\sqrt{x}\sqrt{4-x}} \,dx = -7\int \dfrac{x^{3/2}}{\sqrt{4-x}} \,dx$$

from here, I see you attempted to use the trig sub $\sqrt{x} = 2\sin{\theta}$.

you made a mistake determining $dx$ ...

$\sqrt{x} = 2\sin{\theta} \implies x = 4\sin^2{\theta} \implies dx = 8\sin{\theta}\cos{\theta} \, d\theta$

#### Kristal

##### New member
I did sub it in as a replacement for $\sqrt{x}$ though.

I posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\theta=\frac{1}{\sqrt{x}}dx$

This is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?

If so, well it would be nice if a teacher told us these things sometimes...

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I did sub it in as a replacement for $\sqrt{x}$ though.

I posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\theta=\frac{1}{\sqrt{x}}dx$

This is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?

If so, well it would be nice if a teacher told us these things sometimes...
Hi Kristal, welcome to MHB!

There are 2 ways that are generally taught to do a substitution.

You have the integral
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx$$
and you want to replace $\sqrt x$ by $2\sin\theta$.

Method 1
Write $x$ as a function of $\theta$ and take the derivative as skeeter suggested:
$$\sqrt x=2\sin\theta \implies x=4\sin^2\theta \implies dx=d(4\sin^2\theta) = 8\sin\theta\cos\theta\,d\theta$$
We can now replace $\sqrt x$ and $dx$ to find:
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{(2\sin\theta)^{3}}{\sqrt{4-(2\sin\theta)^2}}\,d(4\sin^2\theta) = \int \frac{(2\sin\theta)^{3}}{\sqrt{4-(2\sin\theta)^2}}\cdot 8\sin\theta\cos\theta\,d\theta$$

Method 2
Take the derivative left and right, and do what needs to be done to eliminate $x$.
In this case:
$$\sqrt x=2\sin\theta \implies d(\sqrt x)=d(2\sin\theta) \implies \frac 1{2\sqrt x}\,dx= 2\cos\theta\,d\theta \implies \frac{dx}{\sqrt x} = 4\cos\theta\,d\theta$$
Rewrite the integral a bit to match and substitute:
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{x^2}{\sqrt{4-x}}\cdot\frac{dx}{\sqrt x} = \int \frac{(2\sin\theta)^4}{\sqrt{4-(2\sin\theta)^2}}\cdot 4\cos\theta\,d\theta$$

We can see that the result is the same in both cases.

I'm afraid I can't comment on what your teacher is telling you since I don't know what that is.
We can only point out and show what the correct math is.

#### skeeter

##### Well-known member
MHB Math Helper
my calculator’s CAS worked out these antiderivatives ...

#### Attachments

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## Program in Python: USING BISECTION SEARCH TO MAKE THE PROGRAM FASTER

USING BISECTION SEARCH TO MAKE THE PROGRAM FASTER
You'll notice that in problem 2, your monthly payment had to be a multiple of $10. Why did we make it that way? You can try running your code locally so that the payment can be any dollar and cent amount (in other words, the monthly payment is a multiple of$0.01). Does your code still work? It should, but you may notice that your code runs more slowly, especially in cases with very large balances and interest rates. (Note: when your code is running on our servers, there are limits on the amount of computing time each submission is allowed, so your observations from running this experiment on the grading system might be limited to an error message complaining about too much time taken.)
Well then, how can we calculate a more accurate fixed monthly payment than we did in Problem 2 without running into the problem of slow code? We can make this program run faster using a technique introduced in lecture - bisection search!
The following variables contain values as described below:
1. balance - the outstanding balance on the credit card
2. annualInterestRate - annual interest rate as a decimal
To recap the problem: we are searching for the smallest monthly payment such that we can pay off the entire balance within a year. What is a reasonable lower bound for this payment value? $0 is the obvious anwer, but you can do better than that. If there was no interest, the debt can be paid off by monthly payments of one-twelfth of the original balance, so we must pay at least this much every month. One-twelfth of the original balance is a good lower bound. What is a good upper bound? Imagine that instead of paying monthly, we paid off the entire balance at the end of the year. What we ultimately pay must be greater than what we would've paid in monthly installments, because the interest was compounded on the balance we didn't pay off each month. So a good upper bound for the monthly payment would be one-twelfth of the balance, after having its interest compounded monthly for an entire year. In short: Monthly interest rate = (Annual interest rate) / 12 Monthly payment lower bound = Balance / 12 Monthly payment upper bound = (Balance x (1 + Monthly interest rate)12) / 12 Write a program that uses these bounds and bisection search (for more info check out the Wikipedia page on bisection search) to find the smallest monthly payment to the cent (no more multiples of$10) such that we can pay off the debt within a year. Try it out with large inputs, and notice how fast it is (try the same large inputs in your solution to Problem 2 to compare!). Produce the same return value as you did in problem 2.
Note that if you do not use bisection search, your code will not run - your code only has 30 seconds to run on our servers. If you get a message that states "Your submission could not be graded. Please recheck your submission and try again. If the problem persists, please notify the course staff.", check to be sure your code doesn't take too long to run.
The code you paste into the following box should not specify the values for the variables balance orannualInterestRate - our test code will define those values before testing your submission.

Note:
Depending on where, and how frequently, you round during this function, your answers may be off a few cents in either direction. Try rounding as few times as possible in order to increase the accuracy of your result.


# A trivial combinatorics result I found, is my proof correct?

I have just finished highschool and have started learning on my own some combinatorics and how to do proofs, and while messing around with sums and Pascal's triangle I found an interesting yet trivial property that I tried to prove. I'm assuming that it has already been found, but I couldn't find anything mentioning it.
So my questions are: Is my proof correct? Has this property been found already? What can this property be used for?

Let $$f(x,n)= \sum_{i_1=1}^n \sum_{i_2=1}^{i_1} \sum_{i_3=1}^{i_2} \cdots \sum_{i_x=1}^{i_{x-1}}i_x \$$ where $x$ is the number of sigma sums.

My conjecture is that $$f(x,n)={n+x \choose x+1}$$

Proof by induction

Basis step:
\begin{align} f(1,n) &=\sum_{i=1}^ni\\ &=\frac{n(n+1)}{2}\\ &={n+1 \choose 2} \end{align} The basis step works. This first result is already proven for all n.

Inductive step: Assume that $f(x,n)= {n+x \choose x+1}$

Now, \begin{align}f(x+1,n) &=\sum_{m=1}^n \sum_{i_1=1}^{m} \sum_{i_2=1}^{i_1} \cdots \sum_{i_x=1}^{i_{x-1}}i_x\\ &=\sum_{m=1}^n f(x,m)\\ &=\sum_{m=1}^n {m+x \choose x+1}\\ &=\sum_{l=0}^{n-1}{l+1+x \choose x+1}\\ &={n+x+1 \choose x+2}\\ \end{align} Therefore, if $f(x,n)$ holds, then $f(x+1,n)$ holds.
What I am not sure about is whether I have to use induction to prove that this property holds for every integer $n$ as well,

• This is indeed correct! (Even verified it for some test cases just to be extra sure.) – Cameron Williams Jul 29 '16 at 3:59
• In the phrase "where $x$ is the number of sigma sums.", why did you use a capital rather than lower-case "W" in "Where"? That seems to be done ninety-percent of the time both here and in Wikipedia, and it doesn't make sense since it's not the beginning of a new sentence. $\qquad$ – Michael Hardy Jul 29 '16 at 5:06
• Congratulations on working through this and producing a proof. It doesn't matter if the result is already widely known or not. The important thing is that you did it yourself. Being taught mathematics is a pointless occupation. Constructing mathematics yourself is the way to go. Find teachers to guide your discovery so that you don't bang your head against too many walls, and you will become a fine mathematician – Martin Kochanski Jul 29 '16 at 5:14

The proof is correct since you can treat $n$ as fixed. You don't have to induct on $n$ since your base case holds for all $n$. If your base case had been $x=n=0$, then yes, you would have to induct on $n$ as well.
$$f(x, n) = \sum_{i_1=1}^n \sum_{i_2=1}^{i_1} \sum_{i_3=1}^{i_2} \cdots \sum_{i_x=1}^{i_{x-1}} \sum_{i_{x+1}=1}^{i_x}1$$
Now, notice that this counts the number of ways to choose $i_1, i_2, \ldots, i_{x+1}$ such that $1 \le i_{x+1} \le i_x \le \ldots \le i_1 \le n$. But by Stars and Bars (also known as multichoose), this is just $\binom{n+x}{x+1}$ as desired.


# When is the union of infinitely many closed sets, closed?

It is known that, in general, the union of infinitely many closed sets need not be closed. However, in the following case, apparently, the union is closed:

Suppose there is a large closed polygon $C$, inside which there is a square $S$ (green). Consider the set of all closed convex objects that contain $S$ and are contained in $C$. Then, apparently, the union of all these closed objects is closed.

My questions:

• Is the above claim true, and if so, how to prove it?
• In general, what are conditions for infinite set of closed sets to be closed, especially in $\mathbb{R}^2$?
• This doesn't apply in your situation, but a common sufficient condition for the union of a family of closed sets to be closed is that the family be locally finite. This means that for any point, there is a neighbourhood of that point that meets only finitely many of the closed sets. – user49640 Jun 13 '17 at 3:48
• Echoing the comment above, the result that the union of a locally finite family of closed sets is closed, is a result that holds in every topological space. – DanielWainfleet Jun 13 '17 at 6:33

The claim is true. Let $A$ be the set you define and which we wish to prove closed.

A point $x$ belongs to $A$ if and only if the convex hull of $S \cup \{x\}$ is contained in $C$. (In that case, its closure is also contained in $C$.) This convex hull consists in turn of the union $B_x$ of all segments joining $x$ to a point in $S$.

$B_x$ is convex because if $xz_1 z_2$ is a triangle with $z_1, z_2 \in S$, and the points $y_1$ and $y_2$ lie on sides $xz_1$ and $xz_2$, respectively, then for any point $w$ on segment $y_1 y_2$ the ray $xw$ meets side $z_1 z_2$, which is contained in the convex set $S$.

Thus a point $x$ belongs to $A$ if and only if the segment $xz$ is contained in $C$ for every point $z$ in $S$. Therefore $A$ is the intersection, for all $z \in S$, of the set $C_z$ which is the union of all segments with one endpoint at $z$ which are contained in $C$. Hence we need only prove that $C_z$ is closed.

After a translation, we may assume $z = 0$. The set $C_0$ is the intersection of the sets $(1/t)C$ for all $t \in (0,1]$, and all the sets $(1/t)C$ are obviously closed.

• (a) Why must the square $S$ be closed? (b) in the last sentence, why is the set $C_0$ the intersection of all sets $(1/t)C$? – Erel Segal-Halevi Jun 13 '17 at 5:09
• I don't know if $S$ must be a closed square to prove what you want, but I use the fact that it is closed when I show that $B_x$ is closed. Your requirements are that there be a closed convex set containing $x$ and $S$. For your second question, a point $x$ belongs to $C_0$ if and only if $tx \in C$ for all $t \in (0,1]$. – user49640 Jun 13 '17 at 5:40
• If S is not closed, can't you just replace S with its closure and get the same result? – Erel Segal-Halevi Jun 13 '17 at 11:23
• Yes, you're right. Just take the closure of the convex hull of $S \cup \{x\}$. It's contained in $C$ because $C$ is closed. So I didn't need to mention anything about compactness. I'm editing the answer now. – user49640 Jun 13 '17 at 11:39

(0). The closure bar denotes closure in $\mathbb R^2.$

(1). Exercise: If $Y$ is a closed bounded convex subet of $\mathbb R^2$ and $x\in \mathbb R^2$ then the convex hull of $\{x\}\cup Y$ is closed.

(2). Let $A$ be the family of closed convex subsets of $C$ that have $S$ as a subset. Let D be the set of $p\in C$ such that the convex hull of $\{p\}\cup S$ is not a subset of $C.$ We have $(\cup A )\cap D=\phi$ so $$\cup A\subset C \backslash D.$$ Take any $p\in D.$ There exists $q\in S$ such that the line segment joining $p$ to $q$ contains a point $r\not \in C.$ Take an open ball $B(r,d)$, of radius $d>0$, centered at $r$, such that $B(r,d)\cap C=\phi.$ If $s>0$ and $s$ is sufficiently small then for any $p'\in C\cap B(p,s),$ the line segment from $p'$ to $q$ will intersect $B(r,d).$ So $B(p,s)\cap C\subset D.$

So $D$ is open in the space $C.$

Therefore $C$ \ $D$ is closed in the space $C.$ And $C=\overline C$ so we have $$C \backslash D =\overline {C \backslash D}.$$ But for any $x\in C$ \ $D$ the convex hull of $\{x\}\cup S$ is closed, and is a subset of $C$. Therefore $$C\backslash D\subset \cup A.$$

We have now $\cup A\subset C \backslash D \subset \cup A,$ so $$\overline {\cup A}=\overline {C\backslash D}=C\backslash D=\cup A.$$


# What is wrong with my algorithm for finding how many positive integers are divisible by a number d in range [x,y]?

I have been solving basic counting problems from Kenneth Rosen's Discrete Mathematics textbook (6th edition). These come from section 5-1 (the basics of counting), pages 344 - 347.

This question is not specifically about finding an answer to a problem or being given the correct equation, but whether my reasoning is sound. Therefore I would find it hard to argue this is a duplicate of seemingly similar questions like this one or this one.

The problems I have been dealing with come of the form How many positive integers in range [x,y] are divisible by d? All additional questions are based on the composition of the information learned in these, e.g. how many positive integers in range [x,y] are divisible by d or e?

To answer the simple question I wrote this "equation/algorithm," which takes as input an inclusive range of positive integers $[x,y]$ and a positive integer $d$, and returns $n$, the total number of positive integers in range $[x,y]$ which are divisible by $d$.

(1) $n = \left \lfloor{\frac{y}{d}}\right \rfloor - \left \lfloor{\frac{x}{d}}\right \rfloor$

The idea is that in order to count how many positive integers are divisible by $d$ from $[1,m]$, we simply calculate $\left \lfloor{\frac{m}{d}}\right \rfloor$, because every $dth$ positive integer must be divisible by $d$. However, this does not work when given a range $[x,y]$ where $x \not= 1$ or when $x > 1$. So we need to subtract the extra integers we counted, which is $\left \lfloor{\frac{x}{d}}\right \rfloor$, i.e. the number of positive integers divisible by $d$ from $[1,x]$.

For a sanity check, I also wrote a brute force algorithm that does a linear search over every positive integer in the range $[x,y]$ and counts it if $x \text{ mod } d == 0$. It also can list out the integers it picked, in case I am feeling really paranoid.

With (1) I've been getting the correct answers except on this problem/input: How many positive integers between 100 and 999 inclusive are odd? My solution was to calculate how many are even, and subtract this from the total number of positive integers in range $[100,999]$. To find the evens I simply use the algorithm in (1):

$\left \lfloor{\frac{999}{2}}\right \rfloor - \left \lfloor{\frac{100}{2}}\right \rfloor = 499 - 50 = 449$

But this answer is wrong, since there actually $450$ even numbers in range $[100,999]$ by the brute force algorithm. (1) is somehow counting off by 1. My question is, why is (1) failing for this input of $(2, [100,999])$ but so far it's worked on every other input? What do I need to do to fix (1) so it produces the correct answer for this case? Perhaps I'm actually over counting because $x$ should actually be $x - 1$?

(1') $n = \left \lfloor{\frac{y}{d}}\right \rfloor - \left \lfloor{\frac{x - 1}{d}}\right \rfloor$

(1') returns the correct answer for this specific input now, but I am not sure if it will break my other solutions.

After computing the number of positive multiples of $d$ less than or equal to $y,$ you correctly want to subtract the multiples that are not actually in the range $[x,y].$ Those are the multiples that are less than $x.$ When $x$ is divisible by $d,$ then $\left\lfloor \frac xd \right\rfloor$ counts all multiples of $d$ up to and including $x$ itself. So as you surmised, you want to subtract $\left\lfloor \frac {x-1}d \right\rfloor$ instead. This is true for any divisor, not just $d=2.$
Suppose you have $$(k-1)d<x\leq kd<(k+1)d<\ldots<(k+n)d\leq y<(k+n+1)d.$$ Then, you always have $\lfloor{\frac{y}{d}}\rfloor=k+n$ is $k+n$. On the other hand, if $x$ is divisible by $d$, then $\lfloor{\frac{x}{d}}\rfloor=k$, whereas if $x$ isn't divisible by $d$, then $\lfloor{\frac{x}{d}}\rfloor=k-1$. To rectify this, instead use $\lceil{\frac{x}{d}}\rceil$ which always returns $k$. Your solution is then $$n+1=(n+k)-k+1=\color{blue}{\left\lfloor{\frac{y}{d}}\right\rfloor-\left\lceil{\frac{x}{d}}\right\rceil+1}.$$
p.s. $$(k-1)d<x\leq kd\implies(k-1)d\leq x-1<kd\implies\left\lfloor{\frac{x-1}{d}}\right\rfloor=k-1$$ so that $$n+1=(n+k)-(k-1)=\color{red}{\left\lfloor{\frac{y}{d}}\right\rfloor-\left\lfloor{\frac{x-1}{d}}\right\rfloor}.$$ The formulas in blue and red produce the same answer.


# Recurrence relation involving unknown d(n, m) numbers

Problem goes like this: numbers $$d(n, m)$$, where $$n, m$$ are integers and $$0\le m \le n$$ are defined by:$$d(n, 0) = d(n, n) = 1$$ for all $$n\ge 0$$ and$$m\cdot d(n, m) = m\cdot d(n-1, m) + (2n-m)\cdot d(n-1, m-1)$$ for $$0\lt m \lt n$$. Prove that all the $$d(n, m)$$ are integers. Now this statement reminds me of some combinatorial identities, where the $$d(n, m)$$ can be considered, for example, the number of ways to distribute or arrange $$m+n$$ objects in a specific way. Since I don't know what these numbers represent from a combinatorial point of view, I can't prove the identity, but if it turns out that I guess what they represent, proving it should be easy enough. Problem is I struggle to find out what they can relate to. Am I in the right way or is there a completely different approach to solve it? I tried to expand $$d(n, m)$$ using its definition and it looks like binomial coefficients make their appearance.

• Induction is the way to go. – Don Thousand Nov 6 '18 at 19:58
• With 2 variables? Does it mean I should fix one and use induction on the other? – Peanut Nov 6 '18 at 20:00
• That's right. You do two layers of induction – Don Thousand Nov 6 '18 at 20:01
• But how did the proposer derive this formula? I don't think he used induction. – Peanut Nov 6 '18 at 20:04
• I think proving $d(n,m)$ to be an integer is equivalent to showing that $m$ divides $2n \cdot d(n-1,m-1)$. Check if that works – Rakesh Bhatt Dec 24 '18 at 12:11

A good start would be to tabulate some small values. Here is some python code to do that:

>>> def tab(n):
...   r = [[0]*(n+1) for _ in range(n+1)]
...   for i in range(n+1):
...     r[i][i], r[i][0] = 1, 1
...   for i in range(1, n+1):
...     for j in range(1, i):
...       r[i][j] = r[i-1][j]+(2*i-j)*r[i-1][j-1]//j
...   for i in range(n+1):
...     print('\t'.join(map(str, r[i])))
...
>>> tab(6)
1   0   0   0   0   0   0
1   1   0   0   0   0   0
1   4   1   0   0   0   0
1   9   9   1   0   0   0
1   16  36  16  1   0   0
1   25  100 100 25  1   0
1   36  225 400 225 36  1

If you're familiar with them, you'll pretty quickly recognize these as the squares of the Pascal's triangle. Hence, we hypothesize that $$d(n, m) = \binom{n}{m}^2$$.

We can show this holds using induction. The boundary conditions $$d(n, n)$$ and $$d(n, 0)$$ hold trivially since $$1^2 = 1$$.

Now take $$0 < m < n$$, substitute in $$d(n-1, m) = \binom{n-1}{m}^2$$ and $$d(n-1,m-1) = \binom{n-1}{m-1}^2$$ and divide the whole equation by $$m$$. Then:

$$d(n, m) = \binom{n-1}{m}^2 + \frac{2n}{m}\binom{n-1}{m-1}^2 - \binom{n-1}{m-1}^2$$

Recall $$\binom{n}{m} = \frac{n}{m}\binom{n-1}{m-1}$$, so:

$$d(n, m) = \binom{n-1}{m}^2 + 2\binom{n}{m}\binom{n-1}{m-1} - \binom{n-1}{m-1}^2$$

Expand $$\binom{n}{m} = \binom{n-1}{m} + \binom{n-1}{m-1}$$ to get:

$$d(n, m) = \binom{n-1}{m}^2 + 2\binom{n-1}{m}\binom{n-1}{m-1} + 2\binom{n-1}{m-1}\binom{n-1}{m-1} - \binom{n-1}{m-1}^2$$

A mess, but we can cancel the last two terms and factor:

$$d(n, m) = \Big(\binom{n-1}{m} + \binom{n-1}{m-1}\Big)^2 = \binom{n}{m}^2$$

Which is our induction hypothesis. So by induction, $$d(n, m) = \binom{n}{m}^2$$, which certainly is an integer.

• The tricky part is always guess something. Once you have an idea of a suitable function, then it's no big deal – Peanut Dec 24 '18 at 12:45


1. ## Group Combinations Problem

If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

Any thoughts?

2. Originally Posted by mathceleb
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

Any thoughts?
I haven't worked this out rigorously, but I think the probability is $\displaystyle \frac{1}{\binom{149}{2}}$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $\displaystyle \binom{149}{2}$ ways to choose two people out of the remaining 149...

3. I'm not sure. I found this thread from back in the day which is a similar question...

http://www.mathhelpforum.com/math-he...mutations.html

4. Hello, mathceleb!

We have a group of 150 people, broken into groups of 3, and only 3 are blue.
What is the probability that all 3 blue people are put in the same group?

The 150 people are divided into 50 groups of 3 people each.

There are: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.

To have the 3 blue people in one group, there is: . ${3\choose3} = 1$ way.

The other 147 people are divided into 49 groups of 3: . $\dfrac{147!}{(3!)^{49}}$ ways.
. . Hence, there are: . $\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.

The probability that all 3 blue people are in one group is:

. . $\dfrac{\dfrac{147!}{(3!)^{49}}} {\dfrac{150!}{(3!)^{50}}} \;=\;\dfrac{147!}{(3!)^{49}}\cdot\dfrac{(3!)^{50}} {150!} \;=\;\dfrac{6}{150\cdot149\cdot148} \;=\;\dfrac{1}{551,\!300}$

5. Originally Posted by undefined
the probability is $\displaystyle \frac{1}{\binom{149}{2}}$, by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are $\displaystyle \binom{149}{2}$ ways to choose two people out of the remaining 149...
This correct.

Originally Posted by Soroban
[size=3]
The 150 people are divided into 50 groups of 3 people each.
There are: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.
Those are ordered partions.

There are $\dfrac{150!}{(3!)^{50}(50!)}$ unordered partitions.

There are $\dfrac{147!}{(3!)^{49}(49!)}$ of those where the blues are together.

Note that $\dfrac{\dfrac{147!}{(3!)^{49}(49!)}}{ \dfrac{150!}{(3!)^{50}(50!)}}=\dfrac{1}{\binom{149 }{2}}$

6. Hmm since Soroban's answer differs from mine I decided to do a little more research.

Originally Posted by Soroban

There are: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.
I believe this should be $\dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!}$ because the groups of 3 can be permuted.

Originally Posted by Soroban

...Hence, there are: . $\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.
Likewise I believe this should be $\dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}$.

$\dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)\cdot50 !}{49!\cdot\left(\dfrac{150!}{(3!)^{50}}\right)} = 50\left(\dfrac{1}{551,\!300}\right) = \dfrac{1}{11026} = \displaystyle \frac{1}{\binom{149}{2}}$


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# Set K consists of all fractions of the form x/(x+2) where x is a posit

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Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?

A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9

It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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26 Sep 2018, 19:15
1
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?

A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9

It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.

Sounds frustrating! Here is one way. About 30 seconds if just a few set members are listed. If all are listed, maybe 75 seconds.

There's a pattern you should see quickly.

Start by finding the greatest value of $$x$$.
Doing so is smart because we know that $$x$$ must be even, positive, and less than 20. Use that high-end limit.

The greatest that $$x$$ can be is 18. Work backwards from 18.

First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$

Next? Well, the next $$x$$ must be 16 (even, positive, < 20).
Its denominator, $$x+2$$, is 18: $$\frac{16}{18}$$

Next? $$x$$ must be 14. The denominator is 16: $$\frac{14}{16}$$

List just those three: $$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$
18 and 18 cancel
16 and 16 cancel
20 remains. 14 remains.
So the first denominator in the list will remain
And the last numerator in the list will remain.

We know the first denominator. What is the last numerator? The smallest $$x$$.

Smallest integer that $$x$$ could be? Positive, even ... 2?
Yes. So the last member of the set at the end of this list is $$\frac{x}{x+2}=\frac{2}{4}$$

First denominator (20) and last numerator (2) will remain:
$$\frac{2}{20}=\frac{1}{10}$$

Another pattern: numerators and denominators decend in order.

Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2

Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4

Same conclusion as above. All but the first denominator and the last numerator will cancel.

$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=$$

$$\frac{2}{20}=\frac{1}{10}$$

I hope that helps.

Let me know if you have questions.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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02 Oct 2018, 19:15
2
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?

A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9

Let’s begin by writing out this product:

2/4 x 4/6 x 6/8 x 8/10 x 10/12 x 12/14 x 14/16 x 16/18 x 18/20

So we see that everything cancels except the numerator of the first fraction and the denominator of the last fraction.

Thus, the product of all the fractions in set K is 2/20 = 1/10.

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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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02 Oct 2018, 19:26
generis wrote:
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?

A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9

It took more than 2 minutes to solve this question, because I constructed the set before solving for the answer. Is there a way to solve this problem quickly? Please explain your answers.

Sounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)

There's a pattern you should see quickly.

Start by finding the greatest value of $$x$$.
Finding the greatest value of $$x$$ is smart because we have a lot of information:
$$x$$ must be even, positive, and less than 20.

The greatest that $$x$$ can be is 18.
Work backwards from 18.

First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$

Next? Well, the next $$x$$ must be 16 (even, positive, < 20).
Its denominator, $$x+2$$, is 18: $$\frac{16}{18}$$

Next? $$x$$ must be 14. The denominator is 16: $$\frac{14}{16}$$

List just those three.

$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$

All but the first denominator and the last numerator "cancel."**

Pattern: numerators and denominators decend in order.

Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2

Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4

As mentioned, all but one denominator and numerator will "cancel."

$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=$$

$$\frac{2}{20}=\frac{1}{10}$$

I hope that helps.

Let me know if you have questions.

**I wrote only this list of three:
$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$
18 and 18 cancel
16 and 16 cancel
20 remains. 14 remains.
So the first denominator in its list will remain
And the last numerator in its list will remain: $$\frac{2}{20}=\frac{1}{10}$$

Why x cannot be 19?, could you help me please?
Thank you.
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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02 Oct 2018, 19:44
1
jorgetomas9 wrote:
generis wrote:
anupam87 wrote:
Set K consists of all fractions of the form x/(x+2) where x is a positive even integer less than 20. What is the product of all the fractions in Set K ?

A) 1/20
B) 1/10
C) 1/9
D) 1/2
E) 8/9

Sounds frustrating! Here is one way. About 30 seconds (don't write out all the fractions below)

There's a pattern you should see quickly.

Start by finding the greatest value of $$x$$.
Finding the greatest value of $$x$$ is smart because we have a lot of information:
$$x$$ must be even, positive, and less than 20.

The greatest that $$x$$ can be is 18.
Work backwards from 18.

First fraction? $$\frac{x}{x+2}=\frac{18}{20}$$

Next? Well, the next $$x$$ must be 16 (even, positive, < 20).
Its denominator, $$x+2$$, is 18: $$\frac{16}{18}$$

Next? $$x$$ must be 14. The denominator is 16: $$\frac{14}{16}$$

List just those three.

$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}$$

All but the first denominator and the last numerator "cancel."**

Pattern: numerators and denominators decend in order.

Numerators:
18, 16, 14, 12, 10, 8, 6, 4, 2

Denominators:
20, 18, 16, 14, 12, 10, 8, 6, 4

As mentioned, all but one denominator and numerator will "cancel."

$$\frac{18}{20}*\frac{16}{18}*\frac{14}{16}*\frac{12}{14}*\frac{10}{12}*\frac{8}{10}*\frac{6}{8}*\frac{4}{6}*\frac{2}{4}=$$

$$\frac{2}{20}=\frac{1}{10}$$

Why x cannot be 19?, could you help me please?
Thank you.

jorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake.
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Re: Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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03 Oct 2018, 08:39
1
generis wrote:
jorgetomas9 wrote:
generis wrote:
[
jorgetomas9 , x cannot be 19 because the prompt says that x is a positive even integer. See the highlight above. Easy mistake.

Thank you very much I haven't seen that one.
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Joined: 29 Jul 2018
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Set K consists of all fractions of the form x/(x+2) where x is a posit  [#permalink]

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08 Oct 2018, 11:26
1
Here is another, simpler approach to solving the problem in under 1 minute.

You can take several values of x and see the pattern.
For x=2, fraction is 1/2
For x=4, fraction is 2/3
For x=6, fraction is 3/4 etc.

So, the product of the 9 fractions would be 9!/10!*1 = 1/10
Set K consists of all fractions of the form x/(x+2) where x is a posit &nbs [#permalink] 08 Oct 2018, 11:26
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# A stock trader originally bought 300 shares

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A stock trader originally bought 300 shares [#permalink]

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A stock trader originally bought 300 shares of stock from a company at a total cost of m dollars. If each share was sold at 50% above the original cost per share of stock, then interns of m for how many dollars was each share sold?

a) 2m/300
b) m/300
c) m/200
d) m/300 + 50
e) 350/m
[Reveal] Spoiler: OA

Originally posted by sabrina3509 on 09 Jun 2016, 12:02.
Last edited by Vyshak on 09 Jun 2016, 12:13, edited 1 time in total.
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Re: A stock trader originally bought 300 shares [#permalink]

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09 Jun 2016, 12:09
sabrina3509 wrote:
A stock trader originally bought 300 shares of stock from a company at a total cost of m dollars. If each share was sold at 50% above the original cost per share of stock, then interns of m for how many dollars was each share sold?

a) 2m/300
b) m/300
c) m/200
d) m/300 + 50
e) 350/m

Quote:
bought 300 shares of stock from a company at a total cost of m dollars.

Let Cost of 300 shares be $3000 So, Cost of 1 shares be$ 10 =>m/300

Quote:
each share was sold at 50% above the original cost per share of stock

Selling price per share = (100+50)/100 * m/300

Or, Selling price per share = 3/2 * m/300 => m/200

Hence answer will be (C)

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Small leverage words here make a big difference. First, notice that the "total" cost of 300 stocks is equal to $$m$$. Thus, the cost per stock would be $$\frac{m}{300}$$. This fraction matches answer choice (B), and highlights a classic trap of the GMAT: answer choices often include the "right answer to the wrong question." The target of this question is not the original cost per stock, but instead the price at which each stock was sold, after an increase.

The question indicates that each share was sold at 50% above the original cost. This is a percentage increase, adding half of the original value to the value. Thus, if the original cost per stock were $$(\frac{m}{300})$$, the value of the stock after the increase would be $$(\frac{m}{300})(1+\frac{1}{2})$$. Now, don't convert the fraction to decimal values here. Not only are "fractions your friends", but the answer choices clearly keep the math in fractional form.

Now, we need to just simplify the math, looking for common factors in the top and bottom of the equation to make the math even easier:

$$(\frac{m}{300})(1+\frac{1}{2})=(\frac{m}{3*100})(\frac{3}{2})=\frac{m}{200}$$

The answer is (C).
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A stock trader originally bought 300 shares   [#permalink] 21 Dec 2017, 19:16
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# In the figure above, x^2 + y^2 =

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In the figure above, x^2 + y^2 =  [#permalink]

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In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625

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2017-09-27_1016_002.png [ 10.27 KiB | Viewed 879 times ]

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In the figure above, x^2 + y^2 =  [#permalink]

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Updated on: 28 Sep 2017, 07:16
Bunuel wrote:

In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625

Attachment:
2017-09-27_1016_002.png

All four figures are right triangles.

$$x^2 + y^2$$ = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.

Short leg, SL, of inside triangle:
$$(\sqrt{4})^2 + (\sqrt{5})^2 = (SL)^2$$
$$(4 + 5) = 9 = (SL)^2$$
$$SL = 3$$

Long leg, LL, of inside triangle:
$$(\sqrt{6})^2 + (\sqrt{10})^2 = (LL)^2$$
$$(6 + 10) = 16 = (LL)^2$$
$$LL = 4$$

The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5. Let hypotenuse 5 = c:

$$x^2 + y^2 = c^2$$
$$x^2 + y^2 = 5^2$$
$$x^2 + y^2 = 25$$

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Originally posted by generis on 27 Sep 2017, 18:06.
Last edited by generis on 28 Sep 2017, 07:16, edited 1 time in total.
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Re: In the figure above, x^2 + y^2 =  [#permalink]

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28 Sep 2017, 06:04
1
genxer123 wrote:
Bunuel wrote:

In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625

Attachment:
2017-09-27_1016_002.png

All four figures are right triangles.

$$x^2 + y^2$$ = the length of the hypotenuse of the inside triangle. Find side lengths of inside triangle.

Short leg, SL, of inside triangle:
$$(\sqrt{4})^2 + (\sqrt{5})^2 = (SL)^2$$
$$(4 + 5) = 9 = (SL)^2$$
$$SL = 3$$

Long leg, LL, of inside triangle:
$$(\sqrt{6})^2 + (\sqrt{10})^2 = (LL)^2$$
$$(6 + 10) = 16 = (LL)^2$$
$$LL = 4$$

The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.

$$x^2 + y^2 = 5$$

The highlighted part should equal to (hypotenuse)...it should be 25.
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In the figure above, x^2 + y^2 =  [#permalink]

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28 Sep 2017, 07:14
Mo2men wrote:
genxer123 wrote:
Bunuel wrote:
In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625
Attachment:
2017-09-27_1016_002.png

. . .
The inside triangle is a 3-4-5 right triangle - its hypotenuse is 5.

$$x^2 + y^2 = 5$$

The highlighted part should equal to (hypotenuse)...it should be 25.

Mo2men , nice catch. I was thinking of 5 as a fixed value derived from other fixed values, such that 5 = $$c^2$$. I missed a step.

5 = $$(\sqrt{c^2}) = (\sqrt{5^2}) = \sqrt{25}$$.

I'll edit.

Thanks, and kudos.
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Re: In the figure above, x^2 + y^2 =  [#permalink]

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03 Oct 2017, 10:06
Bunuel wrote:

In the figure above, x^2 + y^2 =

(A) 5
(B) 7
(C) 25
(D) 80
(E) 625

We see that x^2 + y^2 is equal to the square of the hypotenuse of a right triangle whose sides are not shown. Let’s first determine the base of this triangle, which we can call b:

(√4)^2 + (√5)^2 = b^2

4 + 5 = b^2

9 = b^2

b = 3

Let’s now determine the height of the triangle, which we can call h:

(√6)^2 + (√10)^2 = h^2

6 + 10 = h^2

16 = h^2

h = 4

Thus, x^2 + y^2 = 3^2 + 4^2 = 25.

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Re: In the figure above, x^2 + y^2 = &nbs [#permalink] 03 Oct 2017, 10:06
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# In the figure above, x^2 + y^2 =

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# Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$

My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?

-
I suppose using the Binet formula is a "mosquito-nuking" solution... but it works. – J. M. Nov 23 '10 at 7:17
@J.M.: "overkill" (FTFY) – Isaac Dec 24 '10 at 22:16
@Isaac, It's an idiom here and MO... – J. M. Dec 24 '10 at 23:58

Fix $m \in \mathbb{N}$. We shall use induction on $n$. For $n=1$, the RHS of the equation becomes $$F_{m-1}F_{1} + F_{m}F_{2} = F_{m-1} + F_{m}$$, which is equal to $F_{m+1}$. When $n=2$, the equation is also true.( I hope you can prove this!).

Now assume, that the result is true for $k=3,4, \cdots , n$. We want to show that the result is true for $k=n+1$. $$\text{For} \ k=n-1 \ \text{we have} \quad F_{m+n-1} = F_{m-1}F_{n-1} + F_{m}f_{n}$$ and $$\text{For} \ k = n \ \text{we have} \quad F_{m+n}=F_{m-1}F_{n} + F_{m}F_{n+1}$$ Adding both the sides you will get $$F_{m+n-1} + F_{m+n} = F_{m+n-1} = F_{m-1}F_{n+1} + F_{m}F_{n+2}$$

Oh, i reversed the notations. This is for proving $$F_{m+n} = F_{m-1}F_{n} + F_{m}F_{n+1}$$

-
@Chandru1 thanks a lot for the help i think i'm slowly starting to understand this stuff. To prove it with $n=2$ i simply broke $F_{n+2}$ down into $F_{n+1} + F_n = F_n + F_{n-1} + F_n$ which is what i obtained when substituting $n=2$. – ghshtalt Nov 23 '10 at 18:45

If pressed, I'd nuke with Binet's formula myself, but here's another approach. By an easy induction, if $$A=\pmatrix{0&1\\\\ 1&1}$$ then $$A^n=\pmatrix{F_{n-1}&F_n\\\\ F_n&F_{n+1}}.$$ Comparing the top right entry in the equation $A^m A^n=A^{m+n}$ gives $$F_{m-1}F_n+F_m F_{n+1}=F_{m+n}.$$

-

HINT $\ \$ If you put the Fibonacci recurrence into matrix form then the result is obvious, viz.

$$M^n\ :=\ \left(\begin{array}{ccc} 1 & 1 \\\ 1 & 0 \end{array}\right)^n\ =\ \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)$$

Now compute $\ M^{n+m}\ =\ M^n \ M^m$

-
ok so what i compute is: $$M_{n+m} = M_n*M_m = \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)*\left(\begin{array}{ccc} F_{m+1} & F_m \\\ F_m & F_{m-1} \end{array}\right) = \left(\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\ F_nF_{m+1} + F_{n-1}F_m & F_{n}F_{m} + F_{n-1}F_{m-1} \end{array}\right)$$ which as you said shows the desired expression in the bottom left(and i assume equivalent) in the top right. is this method something one can usually use with sequences? i guess i just don't know how it works in general... – ghshtalt Nov 23 '10 at 8:37
that is, have i understood it correctly if i were to say that to define a sequence, one creates a $2 X 2$ matrix with the upper left value equal to the next value the bottom left and upper right values equal to the current value, and the bottom right equal to the previous? – ghshtalt Nov 23 '10 at 8:38
@user3711: The matrix $M$ is simply the shift operator viz. $\ (F_{n-1},F_n)\:M^t\ = (F_n,F_{n+1})\:.\$ The same idea works for any linear recurrence. – Bill Dubuque Nov 23 '10 at 17:38

With Fibonacci numbers usually there are multiple ways of proving identities.

One way (which is one of my favourites) to prove your identity is the following:

Consider the following problem:

A person climbs up $\displaystyle n$ steps, by taking either one step, or two steps at a time.

The total number of ways the person can climb up all the $\displaystyle n$ steps is $\displaystyle F_{n+1}$ (Why?)

Now consider climbing $\displaystyle m+n-1$ steps and split into the cases when the person lands on step $\displaystyle n$ and the cases when the person lands on step $\displaystyle n-1$ and takes two steps at that point (and so does not land on step $\displaystyle n$ in those cases). These two cases cover all possibilities, and so we have:

$$\displaystyle F_{m+n} = F_{n+1}F_{m} + F_{n}F_{m-1}$$

-
Right. This is basically the matrix proof translated into the language of walks on graphs, which is then further translated into a tiling problem. – Qiaochu Yuan Nov 23 '10 at 17:46
@Qia: That is one way of looking at it. But I wouldn't say it is basically just a translation... I think it is interesting on its own. – Aryabhata Nov 23 '10 at 18:11
I like this proof for giving a simple combinatorial interpretation of the result - it offers better 'motivation' IMHO than the matrix proof, which might look like magic to someone not familiar with those manipulations. – Steven Stadnicki Nov 23 '10 at 18:46
In any case it was good to try to see it from another angle, although to be honest, i'm still having some trouble really visualizing this, hopefully it hits me later... and yes the matrices seem really efficient but at least for me they still look like magic ;) – ghshtalt Nov 23 '10 at 18:52
@Steven: the two proofs are equivalent. Powers of the Fibonacci matrix count walks on a certain graph, the Fibonacci graph, which can in turn be interpreted in terms of certain tilings. I discuss a generalization to companion matrices in this blog post: qchu.wordpress.com/2009/08/23/… – Qiaochu Yuan Nov 23 '10 at 19:08

There are several good answers already, but I thought I would add the following derivation because it is one of the few uses I know for the sum property of permanents; namely,

If $A$, $B$, and $C$ are matrices with identical entries except that one row (column) of $C$, say the $k^{th}$, is the sum of the $k^{th}$ rows (columns) of $A$ and $B$, then $\text{ per } A + \text{ per } B = \text{per } C$.

Start with the matrices $\begin{bmatrix} F_n & F_{n-1} \\ F_0 & F_1 \end{bmatrix}$ and $\begin{bmatrix} F_n & F_{n-1} \\ F_1 & F_2 \end{bmatrix}$. Since $F_0 = 0$ and $F_1 = F_2 = 1$, they have permanents $F_n$ and $F_n + F_{n-1} = F_{n+1}$, respectively. Applying the Fibonacci recurrence and the permanent sum property, we have $\text{ per } \begin{bmatrix} F_n & F_{n-1} \\ F_2 & F_3 \end{bmatrix} = F_{n+2}$. By continuing to construct new matrices whose second rows are the sums of the second rows of the previous two matrices, this process continues until we have $F_n F_{m+1} + F_{n-1}F_m = \text{ per} \begin{bmatrix} F_n & F_{n-1} \\ F_m & F_{m+1} \end{bmatrix} = F_{n+m}.$

For more on this approach (but with determinants), see this paper I wrote a few years ago: "Fibonacci Identities via the Determinant Sum Property," The College Mathematics Journal, 37 (4): 286-289, 2006.

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+1: Hadn't seen this before. – Aryabhata Nov 23 '10 at 23:49

Try proving this statement: Claim: If $f(n) = f(n-1)+f(n-2)$, then $f(n) = F_n f_1 + F_{n-1} f_0$.

Now "fix" $m$ and think of $F_{n+m}$ as a linear recurrence in $n$ with initial conditions $F_{m+1}$ and $F_m$.

By the way, there is a short and clean proof of Claim, but you should know it uses the fact $F_{-1} = 1$.(Something you can easily verify if you did not already know).

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One simple way is to use the formula for the $n^{th}$ Fibonacci number, viz,

$F_n = \frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$ where $\phi$ is the golden ratio.

$\phi = \frac{1 + \sqrt{5}}{2}$.

Or another equally simple way is to use induction on $n$ and then on $m$ or just using induction on one of these might turn out to suffice.

It will be interesting to see a direct proof...

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...and looks like somebody else did use Binet... :D – J. M. Nov 23 '10 at 7:20
@J.M. : I didn't see your comment when I wrote the answer :). – user17762 Nov 23 '10 at 7:23
am i interpreting this properly if i write: $\frac{\Phi^{n+m} - (1-\Phi)^{n+m}}{\sqrt{5}} = \frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}} * \frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}} + \frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}} * \frac{\Phi^{1+m} - (1-\Phi)^{1+m}}{\sqrt{5}}$ $\frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}}(\frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}} + \frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}} * F_1)$ $\frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}}(\frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}}+\frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}})$ ? what can i do at the end? – ghshtalt Nov 23 '10 at 7:41
@user3711: Use $\phi^2 = 1 + \phi$ to simplify. Or as the others have stated you can try induction. – user17762 Nov 23 '10 at 7:45
i'm sorry, but i'm having trouble figuring out how to use that to simplify.... – ghshtalt Nov 23 '10 at 8:08

These can almost always be solved by induction. (In general sequences that are defined by recursion go often well together with induction.)

For this formula we can do the following: Induct on $m$. If $m=1$, then $F_{n+m} = F_{n}F_{m-1} + F_{n-1}F_m = F_{n}F_{0} + F_{n-1}F_1 = F_n + F_{n-1} = F_{n+1}$.

Suppose then that the result holds for $m$ and smaller. Now $$F_{n+m+1} = F_{n+m} + F_{n+m-1} = F_{n}F_{m-1} + F_{n-1}F_m + F_{n}F_{m-2} + F_{n-1}F_{m-1}$$ $$= F_{n}(F_{m-1} + F_{m-2}) + F_{n-1}(F_m + F_{m-1}) = F_{n}F_{m} + F_{n-1}F_{m+1}.$$

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# Is the binomial theorem actually more efficient than just distributing

Is the binomial theorem actually more efficient than just distributing a given binomial. I believe it is more confusing for me to remember and work out using the binomial theorem as a guide, than to just distribute when given a problem like: $$(x-4)^6$$ I think distributing that would be equally as painful as using the binomial theorem. Am I alone or is there actually a reason to practice the painful procedures?

• Well if you distribute without adding equal terms, you will end up with $2^6$ terms, while the theorem gives you the right $6$ terms. – Phicar Jan 7 '17 at 6:26
• What about using Pascal's triangle to compute the binomial coefficients? – Fabio Somenzi Jan 7 '17 at 6:28
• What about $(x-1)^{10}$? Binomial theorem gives it by hand in under a minute. How long would you take to expand that by hand? The theorem is also much more conducive to algorithms, meaning it's strength in both computation and proof is far greater than direct distribution. – Nij Jan 7 '17 at 6:29
• @Phicar $7$ terms, not $6$. – Marc van Leeuwen Jan 7 '17 at 17:08
• @Nij Actually the OP's method of repeatedly multiplying can be more efficient than the binomial theorem for large polynomials of certain types. There is much symbolic computation literature on this topic, e.g.see R. J. Fateman, Polynomial Multiplication, Powers and Asymptotic Analysis: Some Comments, SIAM J. Comput., 3(3),(1974),196–213. – Bill Dubuque Jan 8 '17 at 5:46

The binomial theorem allows you to write out the expansion of your polynomial immediately.

It also allows you to answer such questions as "What is the coefficient of $x^{20}$ in $(1+x)^{100}$?"

Its generalisation to non-integer exponents allows you to get the expansion of $(1-x)^{-1/2}$.

It is a good thing.

• (+1) And needing "just one term" isn't an idle exercise, either: anyone having taken calculus should recognize the importance of: "What's the coefficient of $x^{n - 1}$ in $(x + h)^n - x^n$?" – pjs36 Jan 7 '17 at 6:49
• "It is a good thing" - Well said! :) (+1) – Hypergeometricx Jan 7 '17 at 15:50

There are all sorts of reasons. First of all, you'll learn when you've moved forward a bit that the binomial theorem can actually be applied in cases of non-integer exponents, by defining $\binom{n}{k} = \dfrac{n^{\underline{k}}}{k!}$, where the underline represents a falling power (i.e., $n(n-1)(n-2)\cdots$). Another that comes to mind immediately is the series expansion of $e$...

$$e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \lim_{n\to\infty}\left(1 + n\frac{x}{n} + \frac{n^\underline{2}}{2!}\frac{x^2}{n^2} + \frac{n^\underline{3}}{3!}\frac{x^3}{n^3} + \cdots + \frac{x^n}{n^n}\right)$$

As $n$ gets very large, for fixed $k$, $n^\underline{k}$ approaches $n^k$, and $\frac{x^n}{n^n}$ vanishes since $x$ is fixed, so you end up with the familiar series...

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$$

• Very nice link to expansion for $e^x$. (+1) – Hypergeometricx Jan 7 '17 at 15:51
• Weird. I don't see the underline in the definition of $\binom nk$. Am I the only one with this problem? I'm on Chromium 55.0.2883.87 (64-bit). – rubik Jan 7 '17 at 21:55
• @rubik I had the same problem -- edited the post to make it displaystyle so it would show up. I think we should bring up this on meta as a bug with mathjax, would you like to do it or should I? – 6005 Jan 7 '17 at 22:54
• The frac displays fine in my mobile Chrome (both with dfrac and frac). – Martin Argerami Jan 8 '17 at 6:35
• @6005 You're right, done. Here's the discussion. – rubik Jan 8 '17 at 9:05

If you're not seeing the utility of the binomial theorem, then I think you're missing an important observation:

The coefficients of every term can be calculated extremely efficiently (your example of $(x-4)^6$ is easy to expand by hand with the binomial theorem, without, it's doable but very tedious).

A useful shorthand for lower powers of the exponent is to remember the first few rows of Pascal's Triangle along with the rule for adding more if you need them. Example of first few rows:

$$1$$ $$1\qquad 1$$ $$1\qquad 2\qquad 1$$ $$1\qquad 3\qquad 3\qquad 1$$ $$1\qquad 4\qquad 6\qquad 4\qquad 1$$ $$1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1$$ $$1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1$$

The last row gives the coefficients of $x^k\cdot(-4)^{(6-k)}$ for $k$ running from $0$ to $6$. Computing $(-4)^{(6-k)}$ is easy enough by hand, and multiplying it by the relevant entry in the last row above is also fairly easy (give it a try).

One more thing: The binomial theorem admits an interesting generalization to powers of general polynomials of a given length sometimes called the multinomial theorem. (Challenge: try investigating the coefficients of the first few powers of $(x+y+z)^n$ and perhaps you can figure out the pattern for trinomials? How does it relate to the pattern for binomials? Can you generalize it to general polynomials?)

Marty Cohen and Phicar answered your question correctly. In mathematics, as we always use small numbers, It could be feasible to think this. Why learn the binomial theorem to compute $(a+b)^2$? But what can you say about $(3a+5b)^{987658}$? The binomial theorem is way more efficient in this case and the actual number of cases in which it wouldn't be efficient is very small.

The binomial theorem gives you a powerful tool: The computation for any exponent in - perhaps - the fastest way possible in which it is only not effective for a finite amount of exponents, but its the best possible for an endless amount of exponents.

This is one trap in mathematics: We implicitly assume that the mathematical tools are created to treat familiar cases, but they are created to treat all cases in the neatest way possible, they will be cumbersome for some simple cases (Example: Integrating $f(x)=mx$ in the interval $[0,a]$. This is just a triangle, there is no need for integral calculus here). I remember of me asking about why we need analysis if calculus seems to work well, but I was supposing (without realizing) that calculus is used only on familiar cases, but we want to be useful also for very imaginative cases!

• Just a small note to support the statement that it's as fast as it gets: using the recursion ${n\choose k+1} = {n\choose k}\frac{n-k}{k+1}$ with ${n\choose 0} = 1$ and the symmetry ${n\choose k} = {n\choose n-k}$ then we can compute all the $n$ binomial coefficients ${n\choose k}$ for $k=1,2,\ldots,n$ with only $n$ operations (multiplications and divisions). That is hard to beat! This can be compared with $\mathcal{O}(2^n)$ operations needed to distribute $(a+b)^n$. – Winther Jan 8 '17 at 3:28
• Excelent addendum, @Winther! – Billy Rubina Jan 8 '17 at 13:37

To add to other answers, in higher-level math it's important to be able to generalize the behavior of $(a + b)^n$ when $n$ is a variable. The work that you're doing now is mostly just training exercises for being familiar with, and having intuition for, the general binomial theorem.

For example, there's a really key moment early on in calculus when it's critical to understand that, for all $n$: $(x + h)^n = x^n + nx^{n-1}h....$ plus a bunch of other stuff with higher powers of $h$ in it, but whose exact numerical coefficients we can ignore.


# Evaluate $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$ [duplicate]

This question already has an answer here:

I need to evaluate, for a certain worded question:

If n is even $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n}$$ If n is odd $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n-1}$$

I havent been able to reduce this using the binomial expansion.

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## marked as duplicate by Grigory M, Sasha, Daniel Fischer♦, hardmath, DanJan 16 '14 at 22:19

unrelated: why is the first binomial coefficient in the title bigger than the others? –  kuch nahi Jun 2 '11 at 15:35
–  Grigory M Jun 2 '11 at 15:37
Just want to add that the original question was to find the number of binary sequences of length n that contain an even number of ones. Using a different argument the answer I got is $2^{n-1}$ –  kuch nahi Jun 2 '11 at 15:40
Hint: Do some computations for small values of $n$. You will see a pattern. When $n$ is odd, the pattern is most obviously proved by symmetry. When $n$ is even, look at Pascal's triangle. –  Aaron Jun 2 '11 at 15:41

Binomial expansion allows one to find both $\binom n0+\binom n1+\binom n2+\dots=(1+1)^n$ and $(\binom n0-\binom n1+\binom n2+\dots)=(1-1)^n$. Combining these two yields desired result: $\binom n0+\binom n2+\binom n4+\dots=\frac12\bigl((1+1)^n+(1-1)^n\bigr)=2^{n-1}$.

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I remember I've done this before. Thanks a ton! –  kuch nahi Jun 2 '11 at 15:45
The first version was helpful enough. Anyhow, thanks again. –  kuch nahi Jun 2 '11 at 15:54

The given sums count the number of even-cardinality subsets of an $n$-element set. I'll provide a simple argument that shows that the value of that sum is $2^{n-1}$.

Fix $x \in X$ (where $|X|=n$). The map $f \colon \mathcal{P}(X) \to \mathcal{P}(X)$ (where $\mathcal{P}(X)$ denotes the power set of $X$) defined by $$f(Y) = \begin{cases} Y \cup \{x\} &\text{if } x \notin Y \\ Y - \{x\} &\text{if } x \in Y \end{cases}$$

is a bijection of $\mathcal{P}(X)$ that takes even-cardinality subsets to odd-cardinality subsets and vice-versa. Thus there are the same number of even-cardinality subsets as odd-cardinality subsets. Since there are $2^{n}$ total subsets, there must be $2^{n}/2 = 2^{n-1}$ even-cardinality subsets.

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The sum of all binomial coefficients ${n \choose 0} + \cdots {n \choose n}$ is $2^n$. This is a well-known equation and the slick proof is to consider the binomial theorem for $(1+1)^n$.

Related is the binomial theorem applied to $(1-1)^n$ giving the sum

$${n \choose 0} - {n \choose 1} + {n \choose 2} - \cdots \pm {n \choose n} = 0.$$

The latter inequality is easy to see when you add up the $n$th row of Pascal's triangle with alternating signs and $n$ is odd. For example: $1-5+10-10+5-1 = 0$ is obvious from the symmetry of the entries.

To obtain the answer for every-other term in the first sum, you can add these two formulas together and the negative entries will cancel with the positive ones while the positive ones will double:

$$(1+1)^n + (1-1)^n = 2{n \choose 0} + 2{n \choose 2} + \cdots = 2^n + 0.$$

Now to finish, divide by two. To get the odd entries (the second part of your question), notice that $(1-1)^n = -(1-1)^n$ and use the technique above. Or you could just take the sum for all entries and subtract what we just obtained for the even entries.

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Let $X$ be a binomial$(n,p)$ random variable with $p=1/2$. If $n$ is even, then the probability that $X$ is even is equal to $$\sum\limits_{k = 0,2, \ldots ,n} {{n \choose k}\frac{1}{{2^n }}} .$$ On the other hand, this probability is obviously equal to $1/2$ (consider $X$ as the sum of $n$ independent random variables taking the values $0$ and $1$ with probability $1/2$ each). Hence, $$\sum\limits_{k = 0,2, \ldots ,n} {{n \choose k}} = 2^n \frac{1}{2} = 2^{n - 1} .$$

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+1: I'm always a fan of the probabilistic approach. :) –  Mike Spivey Jun 2 '11 at 16:37
@Mike - Thanks! –  Shai Covo Jun 2 '11 at 18:24

Here's a more general take. Sums of the form $\sum_k \binom{n}{2k} f(k)$ are sometimes called aerated binomial sums. My paper "Combinatorial Sums and Finite Differences" (Discrete Mathematics, 307 (24): 3130-3146, 2007) proves some general results about these aerated binomial sums. (See Section 4.) The case $f(k) = 1$ (as in the OP's question) falls out nicely.

Suppose you are interested, for some function $f(k)$, in the binomial sum

$$B(n) = \sum_{k=0}^n \binom{n}{2k} f(k).$$

Then, taking the finite difference $\Delta f(k) = f(k+1) - f(k)$, denote $A(n)$ by

$$A(n) = \sum_{k=0}^n \binom{n}{2k} \Delta f(k).$$

In the paper I prove that $B(n)$ and $A(n)$ are related via $$B(n) = 2^{n-1} f(0) + 2^n \sum_{k=2}^n \frac{A(k-2)}{2^k} + \frac{f(0)}{2}[n=0],$$ where $[n=0]$ evaluates to $1$ if $n=0$ and $0$ otherwise.

Since $f(k) = 1$ in the OP's question, $\Delta f(k) = 0$, and so $A(n) = 0$ for all $n$. Thus $$\sum_{k=0}^n \binom{n}{2k} = 2^{n-1} + \frac{1}{2}[n=0].$$

More generally, this approach can be used to prove that, for $m \geq 1$, $$\sum_k \binom{n}{2k} k^{\underline{m}} = n(n-m-1)^{\underline{m-1}} \, 2^{n-2m-1} [n \geq m+1],$$ and $$\sum_k \binom{n}{2k} k^m = n \sum_j \left\{ m \atop j\right\} \binom{n-j-1}{j-1} (j-1)! \, 2^{n-2j-1},$$ where $\left\{ m \atop j\right\}$ is a Stirling number of the second kind. (While the second equation swaps one sum for another, there are no more than $m$ terms in the right-hand side, and so it is useful when $m$ is small.)

The approach can also be extended to sums of the form $\sum_k \binom{n}{ak+b} f(k)$. (And the case $a = 2, b = 1$ is considered explicitly in the paper.) Using higher values of $a$ and $b$ would result in increasingly complicated expressions, though.

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I think it might help to look at this in terms of the rows on pascal's triangle. The sum of a given row is 2^n. All rows are symmetric about the middle of the row. As this sounds like it could be HW, i'll leave it to you to figure out how each of those relate to the sum of the row.

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I wish this were homework. As that would imply I am taking a course and have access to an instructor. –  kuch nahi Jun 2 '11 at 15:43
Ok, so they are both exactly half the of the row (they get every element once as opposed to twice). Since the sum of a row is 2^n, this gives 2^(n-1) as the answer. –  soandos Jun 2 '11 at 15:45
yup, got it. Thanks! +1 –  kuch nahi Jun 2 '11 at 15:54