# The total number of subarrays

I want to count the number of subarrays for a vector (not combinations of elements).
Ex.

A[1,2,3]

It has 6 subarrays :

{1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}

I think that for a vector of N elements the total number of subarrays is N*(N+1)/2.
I am not able to prove it, can someone do it?

• Welcoming the Math.SE! Here we have a culture of showing what you've tried, as it helps both the responders to give better answers and you to get a better understanding! :) – frogeyedpeas Mar 17 '15 at 22:08
• Looks you forgot to list $\{3, 1\}$ – AgentS Mar 17 '15 at 22:13
• No! I didn't forget {3,1}. A subarray has to have contiguous elements – user72708 Mar 18 '15 at 13:56

Suppose that your vector is $\langle a_1,a_2,\ldots,a_n\rangle$. Imagine a virtual element $a_{n+1}$ at the end; it doesn’t matter what its value is. A subarray is completely determined by the index of its first element and the index of the element that immediately follows its last element. For example, the subarray $\langle a_3,\ldots,a_{n-2}\rangle$ is determined by the indices $3$ and $n-1$, the subarray $\langle a_k\rangle$ is determined by the indices $k$ and $k+1$, and the subarray $\langle a_2,\ldots,a_n\rangle$ is determined by the indices $2$ and $n+1$. Moreover, each pair of distinct indices from the set $\{1,2,\ldots,n+1\}$ uniquely determines a subarray. Thus, the number of subarrays is the number of pairs of distinct indices from the set $\{1,2,\ldots,n+1\}$, which is

$$\binom{n+1}2=\frac{n(n+1)}2\;.$$

• I am confused why you use the index of the element that immediately follows its last element instead of just the last element. I was thinking the answer should be n C 2, instead of n+ 1 C 2 that you mentioned. Please correct me wherever I am wrong. – Abhishek Bhatia Jul 7 '18 at 19:33
• @Abhishek , You are right, It is n C 2 where we did not calculate the single character substrings. There can be n single character substring . So the result is (n C 2 + n C 1) = (n+1) C 2 . – shuva Oct 31 '18 at 20:15

Consider an arbitrary array of N DISTINCT ELEMENTS (if the elements are the same then I am afraid the formula you are seeking to prove no longer works!).

Naturally there exists 1 array consisting of all the elements (indexed from 0 to N-1)

There exist 2 arrays consisting of N-1 consecutive elements (indexed from 0 to N-2)

and in general there are k arrays consisting of N-k+1 consecutive elements (indexed from 0 to N-k-1)

Proof:

We can access elements 0 ... N-k-1 as the first array, then 1 ... N-k+2 is the second array, and this goes on for all N-k+r until N-k+r = N-1 (ie until we have hit the end). The r that does us is can be solved for :

$$N-k+r = N-1 \rightarrow r -k = -1 \rightarrow r = k-1$$

And the list $$0 ... k-1$$ contains k elements within it

Thus we note that the total count of subarrays is

1 for N elements

2 for N-1 elements

3 for N-2 elements

.

.

.

N for 1 element

And the total sum must be:

$$1 + 2 + 3 ... N$$

Let us see if your formula works

if:

$$1 + 2 +3 ... N = \frac{1}{2}N(N+1)$$

then

$$1 + 2 + 3 ... N+1 = \frac{1}{2}(N+1)(N+2)$$

We verify:

$$\frac{1}{2}N(N+1) + N+1 = (N+1)(\frac{1}{2}N + 1) = (N+1)\frac{N+2}{2}$$

So you're formula does indeed work! Now we verify that for N = 1

$$\frac{1*(1+1)}{2} = 1$$

And therefore we can use the above logic to show that for any and ALL whole numbers N the formula works!

This calculation can be seen as an arithmetic series (i.e. the sum of the terms of an arithmetic sequence).

Assuming the input sequence: $$(a_0, a_1, \ldots, a_n)$$, we can count all subarrays as follows:

\begin{align} \; 1 & \; \text{subarray from} \; a_0 \; \text{to} \; a_{n-1}\\ + \; 1 &\; \text{subarray from} \; a_1 \; \text{to} \; a_{n-1}\\ & \; \ldots \\ + \; 1 & \; \text{subarray from} \; a_{n-1}\; \text{to} \; a_{n-1}\\ = & \; n \end{align}

$$+$$

\begin{align} \; 1 & \; \text{subarray from} \; a_0 \; \text{to} \; a_{n-2}\\ + \; 1 &\; \text{subarray from} \; a_1 \; \text{to} \; a_{n-2}\\ & \; \ldots \\ + \; 1 & \; \text{subarray from} \; a_{n-2}\; \text{to} \; a_{n-2}\\ = & \; n-1\\ \end{align}

$$+ \; \ldots$$

\begin{align} \; \; \; 1 & \; \text{subarray only containing} \; a_0\\ = & \; 1\\ \end{align}

which results in the arithmetic series: $$n + n-1 + … + 1$$.

The above can also be represented as $$\sum_{i=1}^{n}i\;$$ and adds up to $$n (n+1)/2$$.

Elaborating on Brian's answer, lets assume we count all the single character sub-strings. There can be n such single character sub-string. By using the Binomial Coefficient, $$\binom{n}{r}$$ notation, we can say n = $$\binom{n}{1}$$ .

$$total_1 = n = \binom{n}{1}$$

Now let's assume we count all the sub-string that are not single character. Because we have to choose the beginning and end of the sub-string regardless of the order, the count should be $$\binom{n}{2}$$.

$$total_* = \binom{n}{2}$$

Now the total number of sub-string,

$$total = total_1 + total_* = \binom{n}{1} + \binom{n}{2}$$

Now recall Pascal triangle and recurrence relation $$\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}$$. So we can write,

$$total = total_1 + total_* = \binom{n}{1} + \binom{n}{2} = \binom{n+1}{2} = n(n+1)/2$$

There we have the mathematical deduction. (I kind of feel we do not need the fancy recurrence relation to get the final answer though).

• Use LaTeX please. – Michael Rozenberg Oct 31 '18 at 20:31

Trying to explain in layman terms.

Let's say f(0) = 0

f(1) = 1
f(2) = 3
f(3) = 6
f(4) = 10
f(5) = 15

By observation, you can see that each result is just an addition of previous result and current number.

f(n) = n + f(n-1)

So with this formula let's expand f(5).

f(5)
=> 5 + f(4)
=> 5 + 4 + f(3)
=> 5 + 4 + 3 + f(2)
=> 5 + 4 + 3 + 2 + f(1)
=> 5 + 4 + 3 + 2 + 1 + f(0)
=> 5 + 4 + 3 + 2 + 1 + 0
===> This is equal to sum of n numbers = n(n+1)/2


Given the planar curve, find the equation of the plane

Tags:
1. Oct 15, 2016

dlacombe13

1. The problem statement, all variables and given/known data
$r(t) = < 2e^t - 5 , e^t +3t^2 , 4t^2 +1>$
Is a curve that lies within a plane. Find the equation of this plane.

3. The attempt at a solution
I am not sure if my approach is correct. These are my results:
$x=2e^t - 5$
$y = e^t +3t^2$
$z = 4t^2 + 1$

$z = 4t^2 + 1 ~~\Rightarrow~~ t = \sqrt{\frac{z-1}{4}}$

$x = 2e^\sqrt{\frac{z-1}{4}} - 5 ~~\Rightarrow~~ \frac{x+5}{2} =e^\sqrt{\frac{z-1}{4}}$
$y = e^\sqrt{\frac{z-1}{4}} + \frac{3z-3}{4}$

$y = \frac{x+5}{2} + \frac{3z-3}{4}$
$= \frac{x}{2} + \frac{5}{2} + \frac{3z}{4} - \frac{3}{4}$

$= \frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$

2. Oct 15, 2016

pasmith

One way of specifying a plane is as $\{ \lambda \mathbf{a} + \mu\mathbf{b} + \mathbf{c} : (\lambda , \mu) \in \mathbb{R}^2\}$ for given vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ where $\mathbf{a}$ and $\mathbf{b}$ are linearly independent. Then for non-constant functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ we have that $$\mathbf{r}(t) = f(t)\mathbf{a} + g(t)\mathbf{b} + \mathbf{c}$$ is a curve which lies on this plane.

3. Oct 15, 2016

Ray Vickson

Easier: we have $2y- x = 6 t^2 + 5$, which has eliminated the $e^t$ terms. Now to eliminate the $t^2$ terms, just add or subtract a suitable multiple of $z$. That will leave you with a constant, having no $t$ in it anywhere.

4. Oct 15, 2016

Staff: Mentor

"I am not sure if my approach is correct."
You can (and really should) check your final equation. To do this choose three t values in your equation for r(t). That will give you three vectors, the endpoints of which lie in the plane. These three points should satisfy the plane equation you ended with. If all three points work, then you can be 99% sure that your work is correct. (Subtracting a tiny amount for arithmetic errors you might make). If the three points don't satisfy the plane equation, that's a sign that you have done something wrong.

5. Oct 15, 2016

LCKurtz

I didn't check all your work, but look at your last 3 lines. You start out with $y =$ and end up with $= \frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$.

Does $y=$ that or is your final equation supposed to be $\frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$? If that is your answer then notice that $\vec r(0) = \langle -3,1,1\rangle$, so that point must be on the plane. It doesn't seem to satisfy that last equation. It may be a simple arithmetic mistake.

But one reason I didn't check your work is that I don't think eliminating the parameter is a sensible way to do the problem. I would suggest a different approach. All you need for a plane is a point and a normal vector. It's easy to get a point on the plane. Then notice that given the curve is planar, that means the tangent vectors to the curve must be in the plane. So you could calculate $\vec r'(0)\times \vec r'(1)$ to get a normal.

6. Oct 15, 2016

dlacombe13

Thank you all for your replies. I tried plugging in t=0,1,2 into each parameter, and plugged it into the equation for the plane. For all three points, I get: -1.75. -15/8 = -1.875.
Thank you all for you're replies, I tried checking my values and got close to -15/8, but I don't like the error, since I think my arithmetic is right. Also, my original thinking was exactly as LCKurtz said. However I ran into issues, but I think that is because I did it wrong. I used:
$T = \frac{r'(t)}{|r'(t)|}$
T x r(t)
But it didn't seem to work. So what you're saying is I just need to take find two random tangent points along the curve, which will give me two random vectors on the plane it is within, and cross them which will yield an orthogonal vector (n), and then just form the equation from it?

7. Oct 15, 2016

dlacombe13

Thanks everyone, I did get the equation finally, using r'(0) x r'(1) = <8,-16,12> and got the equation:
8x - 16y +12z +28 = 0
I verified it by plugging in points, and it works out. One last question before I go...
Why didn't my attempt at using T, the unit tangent vector work? I mean it is still a vector that is in the direction of the curve, and thus on the plane, right?

8. Oct 15, 2016

LCKurtz

Yes, it should have worked. But dividing by the magnitude is unnecessary, adds square roots, and makes the calculations more error prone, which apparently got you.


# Convergence in distribution to a degenerate distribution

This question came up based on a disagreement I had with a TA. This was the specific example:

Let $$X_{1},…,X_{n}X_{1},...,X_{n}$$ be an iid random sample from a population with pdf $$f(x)=3(1-x)^2, 0. The $$nthnth$$ order statistic is represented as $$X_{(n)}X_{(n)}$$.

Question: Find a constant $$vv$$ such that $$n^v(1-X_{(n)})n^v(1-X_{(n)})$$ converges in distribution.

I believe that the intent of the question was to prompt the invocation of Slutsky's theorem (or maybe not); however, to which distribution the presented snippet was supposed to converge was not specified. So I presented a lazy alternative answer as follows:

As the sample size approaches infinity, $$X_{(n)}X_{(n)}$$ becomes arbitrarily close to 1. Therefore, we can simply set $$v=0v=0$$ and the statement will converge to a degenerate distribution. In this case, the order statistic converges to 1, so with $$v=0v=0$$, the statement converges to 0. I later realized that it actually converges to 1 after the transformation, which is reflected in the final degenerate distribution written toward the bottom of the question. I was surprised when the TA said that 0 is not a random variable or a distribution, and my answer made no sense. I second guessed myself and went further:

The cdf of the original distribution is $$F(x)=x^3-3x^2+3xF(x)=x^3-3x^2+3x$$ Using this, I derived the pdf of the order statistic,
$$f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1} f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1}$$
And the cdf of the order statistic, $$f_{X_{(n)}}(x)=(x^3-3x^2+3x)^nf_{X_{(n)}}(x)=(x^3-3x^2+3x)^n$$

I then used this to graph the cdf of the order statistic with $$n=1, n=100, n=1,000,000,000n=1, n=100, n=1,000,000,000$$. Predictably, the graph showed the cdf getting thinner and steeper, until with huge samples it visually looks like a vertical line at $$X_{(n)}=1X_{(n)}=1$$.

I reiterated my argument, and showed the visualization: it is converging to a degenerate distribution. Again, I was rebuffed. After pestering the TA and asking the instructor for the course, I still have no answer for why my answer was wrong, but I can't keep bothering them about it.

Can someone here tell me if my argument is solid, and if not, tell me specifically what I've done that is in error?

EDIT: Here is the final formalism that I gave to tie my argument together. Following Casella and Berger's definition,

A sequence of random variables, $$X_{1}, X_{2},...,X_{1}, X_{2},...,$$ converges in distribution to a random variable $$XX$$ if $$\lim_{n \to \infty} F_{X_n}(x) = F_X(x)\lim_{n \to \infty} F_{X_n}(x) = F_X(x)$$ at all points $$xx$$ where $$F_X(x)F_X(x)$$ is continuous

Since $$\lim_{n \to \infty} F_{X_(n)}(x) = 1\lim_{n \to \infty} F_{X_(n)}(x) = 1$$
And we can define a degenerate variable $$YY$$ with the cdf
$$F_Y(x)=\begin{cases}1, & x\ge1\\ 0, & else \\ \end{cases}F_Y(x)=\begin{cases}1, & x\ge1\\ 0, & else \\ \end{cases}$$
We can say
$$\lim_{n \to \infty} F_{X_(n)}(x) = F_Y(x) \lim_{n \to \infty} F_{X_(n)}(x) = F_Y(x)$$

So by definition, $$X_{(n)}X_{(n)}$$ converges in distribution to the degenerate distribution of $$YY$$.

EDIT: Adding a bounty to this, as it has become important. I did not formally perform the transformation $$Y=1-X_(n)Y=1-X_(n)$$ above, but doing so results in $$YY$$ having a distribution that still converges to the same degenerate distribution above. I was now told that my answer does not prove convergence in distribution at all. Question is answered if someone can definitively prove either 1) my answer is correct, and show exactly how you would have gone about proving it formally and rigorously, or 2) my answer is wrong, exactly why it is wrong, and prove any correct answer that uses degenerate distributions or degenerate variables if such an answer exists.

Your answer is correct (assuming that you have accurately transcribed the question). The proof:

Let $$F_n(c)F_n(c)$$ be the cdf of $$(1 - X_{(n)})(1 - X_{(n)})$$, where $$X_{(n)}X_{(n)}$$ is the greatest element in a sample of size $$nn$$. Let $$F(c)F(c)$$ be the cdf for the constant 0 distribution.

For $$c < 0c < 0$$, of course $$F_n(c) = 0 = F(c)F_n(c) = 0 = F(c)$$.

For $$c > 1c > 1$$, of course $$F_n(c) = 1 = F(c)F_n(c) = 1 = F(c)$$.

For $$0 < c \le 10 < c \le 1$$:

\begin{align} F_n(c) &= P(1 - X_{(n)} \le c) \\ &= P(X_{(n)} \ge 1-c) \\ &= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\ &= 1 - P(X_1 < 1-c)^n \to 1 = F(c) \end{align} \begin{align} F_n(c) &= P(1 - X_{(n)} \le c) \\ &= P(X_{(n)} \ge 1-c) \\ &= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\ &= 1 - P(X_1 < 1-c)^n \to 1 = F(c) \end{align}

And the case $$c = 0c = 0$$ doesn't matter, because $$FF$$ isn't continuous at $$00$$.

If the people you are arguing with don't realise that convergence in distribution to a constant is a thing, you could point them to e.g. Wikipedia's Proofs of convergence of random variables article.


# Did Feynman mentally compute $\sqrt[3]{1729.03}$ by linear approximation?

In the biopic "infinity" about Richard Feynman. (12:54) He computes $\sqrt[3]{1729.03}$ by mental calculation. I guess that he uses linear approximation. That is, he observe that $1728=12^3$. Let $f(x)=\sqrt[3]{x}$. Then $f'(x)=\frac{1}{3\sqrt[3]{x^2}}$ and $f'(1728)=\frac{1}{3\sqrt[3]{1728^2}}=\frac{1}{3\cdot 12^2}$. Therefore, $$\sqrt[3]{1729.03}=f(1729.03)\approx f(1728)+f'(1728)(1729.03-1728)=12+\frac{1.03}{3\cdot 12^2}=12.002384\overline{259}.$$

Question 1. If he used the linear approximation, how did he compute $\frac{1.03}{3\cdot 12^2}=0.002384\overline{259}$ by a mental calculation?

Question 2. If he didn't use the linear approximation, what is another method he might have used?

• You're asking about how a mental computation was accomplished in a scene from a fictional movie...? (Just clarifying; I'm not willing to click a youtube link to get the content of the question.) – Andrew D. Hwang May 1 '16 at 18:04
• @AndrewD.Hwang Yes. Although it is a fictinal movie, I think that maybe the scenarist had a method to approximate $\sqrt[3]{1729.03}$. And I hope this scene can motivate my student to learn linear approximation. – bfhaha May 1 '16 at 18:12
• $1.03/3\approx.34333\approx.34332$ $.34332/144=.05722/24=.02861/12\approx.002384$. – Mark S. May 1 '16 at 18:30
• You should mention what value was computed so that the question is self-contained. – Bill Dubuque May 1 '16 at 18:41
• @Andrew D. Hwang It is mentioned in his book "Surely You're joking, Mr. Feynman!". Give it a shot. – user537100 Jul 1 '18 at 13:09

Feynman tells the story in one of his books of anecdotes.

http://www.ee.ryerson.ca/~elf/abacus/feynman.html

$12$ is a very good first approximation and the linear term of the series expansion suffices to get high precision.

$$\sqrt[3]{1728 + d} = 12\sqrt[3]{1+x} = 12 + 4x + O(x^2)$$

where $d = 1.03$ and $x = \frac{d}{1728}$ is, in Feynman's words, about 1 part in 2000, so that the error term is of order $10^{-6}$.

Feynman says that he computed $12 + \frac{4d}{1728}$ as the approximate value.

The number was 1729.03. I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03 is only one part in nearly 2000, and I had learned in calculus that for small fractions, the cube root's excess is one-third of the number's excess. So all I had to do is find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.

He describes that as though $d=1$ for this part of the calculation, so maybe $12 + \frac{1}{432}$ was what he actually computed. By "adding two more digits" (to 12.002) he seems to mean working out the division in the fraction. It could also mean adding (0.03)/432 as two more decimal digits of accuracy to $(12 + 432^{-1})$, which requires only a multiplication by 3 of an already computed quantity 1/432.

Feynman's method is the one that would have been immediate for anyone familiar with the binomial series and with $12^3 = 1728$. He said that knew the latter as ft^3/in^3 and other people might know it from the Ramanujan 1729 story. The other ingredient, as Feynman says in the story, was being good at integer division.

I upvoted the other answer as it comes from the other book of anecdotes, but one way to calculate this without Feynman's experience might be the following.

You can start with the linear approximation, and then you have to calculate $(\dfrac{1.03}{3})/12^2$. $\dfrac{1.03}{3}\approx .34333$, but that doesn't lend itself to division by $12^2$, so you can change the last digit to get $\dfrac{1.03}{3}\approx .34332$

(You know this will be helpful by the divisibility test for $3$, and can note it's also divisible by $2$ and $4$ if you're thinking ahead.)

Then $\dfrac{.34332}{12^2}=\dfrac{.05722}{12*2}=\dfrac{.02861}{12}\approx.002384$.

• "linear approximation" here means taking the linear part of $(12 + x)^3$ when solving $(12 + x)^3 = 12^3 + 1.03$. Then $3x(12^2) = 1.03$ and the rest is as in the answer. Thinking of it as 1 or 1.03 divided by 432, how to divide by 432 with minimum calculation? The error in approximating that 432 by 500 is (1/432 - 1/500) = (500- 432)/500*432 or about 68/200000. This gives 0.002 + 0.00034 = 0.00234 using only division by 2, 4*5=20, and 100-32=68. – zyx May 2 '16 at 2:49
• Thanks MarkS. You and zyx gave the best answer for me. But it can only has one answer. So please forgive me to use upvote instead of mark it as an answer. – bfhaha May 2 '16 at 5:12


Ask your own question, for FREE!
Mathematics
OpenStudy (waheguru):

is there a special rick or is it just trial and error to solve this porblem

OpenStudy (waheguru):

OpenStudy (waheguru):

it would take a really long time to find he asnwer by trial and answer

jimthompson5910 (jim_thompson5910):

Best way to do this is to break things down. Start with 100. Clearly 1+0 = 1 which is not 0, so that doesn't work. But 101 does since 1+0 = 1 which is the last digit. Now ask yourself: are there any more numbers in the range of 100 to 109 that work? The answer is no since 1+0 = 1 is unique to this set of numbers (ie there are no more numbers in this range that have a 1 as the last digit) So again, in the list of numbers from 100 to 109, the only upright number is 101 Similarly, for numbers in the range 110 to 119, the only upright number is 112 Keep going to find the third upright number to be 123 Etc, etc When you come upon numbers that start with 19, you won't find any upright numbers since 1+9 = 10, which is not a single digit. So , in the range of 100 to 199, there are 8 numbers that are upright numbers So from 200 to 299, there are 7 numbers that work. Why 7 and not 8? notice that for numbers that start with 29 add to 2+9 = 11, but that's not a single digit. Likewise, numbers starting with 28 add to 2+8 = 10, again not a single digit. However, the rest work. Keep going and you'll find that only 6 numbers work from 300 to 399 etc etc After all that, you'll have the following number counts: 8, 7, 6, 5, 4, 3, 2, 1, 0 Add them all up to get 36 So the answer is choice C

OpenStudy (anonymous):

Interesting question. Here's how I tackled it:Look at the first few 3 digit numbers.100101102103104105106107108109They all have the same first two digits, but different last digits. You'll notice that exactly one of these numbers is upright. You'll find the same thing for 110 through 119. I can keep going unless the sum of the first two digits is greater than 9. For example, any number with 67 as the first two digits can never be upright because 6+7 is a two digits number.So to start counting upright numbers, add up all the possible 2 digit combinations you can get. you can use 10 through 99, so that's 90 possible two digit pairs. Now you need to subtract off the pairs that don't work, like 67. 10, 11, 12, 13, ... , 18 all work, but 19 doesn't. That's 9 upright numbers. 20, 21, ... , 27 all work, but 28 and 29 don't. That's 8 more upright numbers. 90 works, but 91, 92, ... , 99 don't. That's just one more upright number. See the pattern? So find 1+2+3+4+5+6+7+8, and you have your answer.

jimthompson5910 (jim_thompson5910):

you have the right idea pmilano, but you fell into the trap that there's a number in the range of 190 to 199 when there isn't one

OpenStudy (anonymous):

By the way, the fast way to add the numbers 1 through 8 (or, more generally, 1 through n) is by using the formula (n)(n+1)/2 So 8(8+1)/2 = 36

OpenStudy (anonymous):

Thanks jim.

OpenStudy (anonymous):

my answer is D)45. There is actually a pattern if you try to list down the first numbers. Starting at 1__, there are 9 possible answers. As you move up to the next hundreds digit, you decrease that number by one, because you can't have a sum of more than 9. 9+8+7+...+2+1 = 45

OpenStudy (anonymous):

I actually realized that before I posted but didn't proof read my post well enough (hence my correct answer and incorrect counting).

jimthompson5910 (jim_thompson5910):

My bad, I was ignoring zero Start with the range 100-199 and moving to the next 200-299, etc, the following upright numbers are found in the tens digits listed below 0, 1, 2, 3, 4, 5, 6, 7, 8 0, 1, 2, 3, 4, 5, 6, 7 0, 1, 2, 3, 4, 5, 6 0, 1, 2, 3, 4, 5 0, 1, 2, 3, 4 0, 1, 2, 3 0, 1, 2 0, 1 0 So there are 45 here

OpenStudy (waheguru):

i am getting 36?

OpenStudy (anonymous):

Here is another method lol (incase you didnt have enough) When the last digit is 'n', you want to ask yourself how many solutions are there to: x + y = n where x and y are non-negative integers? For example, if n was 3, then these are the possible solutions for x and y: x = 0, y = 3 x = 1, y = 2 x = 2, y = 1 x = 3, y = 0 These correspond to the three digits numbers: 033 123 213 303 Of course we would discard the '033' case. Anywhos, the equation that gives you the number of solutions to "x + y = n" is: $\left(\begin{matrix}n+2-1 \\ 2-1\end{matrix}\right) = \left(\begin{matrix}n+1 \\ 1\end{matrix}\right)$ We need to do this for n = 1, 2, 3,....,9, while remembering to not count the case of the first digit being 0. So we end up with: $\left(\begin{matrix}2 \\ 1\end{matrix}\right)+\left(\begin{matrix}3 \\ 1\end{matrix}\right)+\left(\begin{matrix}4 \\ 1\end{matrix}\right)+\ldots +\left(\begin{matrix}10 \\ 1\end{matrix}\right)-9$$= 2+3+4+\ldots+10-9 = 1+2+3+\ldots+9 = 45$

OpenStudy (anonymous):

First set of numbers: 101,112,123,134,145,156,167,178,189 ---> 9 numbers. For 2__, that is one less than the 1 hundreds because u can only use 0-7 for the 2nd digit. The last possible number is 909, the only number in the 9 hundreds.

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# Integrating $\int \frac{-\sin x}{1+\cos x}\, dx$, I get $\ln(1 + \cos x)$. WolframAlpha gives $2 \ln(\cos \frac x 2)$. Is WA wrong?

So, I'm watching a tutorial on differential equations, where I encountered this little trick:

$$\int \frac{y'}{y}\, dx = \ln(y)$$ It seems perfectly logical and easy to justify, but something fishy happens to this integral:

$$\int \frac{-\sin x}{1+\cos x}\, dx$$

The trick gives $$\int \frac{-\sin x}{1+\cos x}\, dx = \ln(1 + \cos x)$$

while WolframAlpha gives $$\int \frac{-\sin x}{1+\cos x}\, dx = 2 \ln(\cos \frac x 2)$$.

You guys who know this stuff - does WolframAlpha mess up here or is it something I've missed? Taking the derivative of $$2 \ln(\cos \frac x 2)$$ gives me $$-\tan \frac x 2$$, so I don't see how WA may be right.

• Remember that $\cos{2x}=2\cos^2x-1$ – Don Thousand Aug 12 at 11:57
• They are different in terms of their symbols, but they give you exactly the same values. What you have is a trigonometric identity. – Pixel Aug 12 at 12:01
• As others have mentioned, the key here is to use a trig identity to bridge the two solutions. ... Often, the trickiest part of Calculus is remembering your Pre-Calculus. :) – Blue Aug 12 at 12:03
• For future reference, you can always check your (or WA's) answer by differentiating it and checking it against the expression you're integrating (the integrand). If they match up then you know your answer is correct (up to a constant), even if you don't know any trig! – Ben Aug 12 at 12:05

## 5 Answers

$$1+\cos x = 2\cos^2\frac{x}{2}$$

$$\ln (1+\cos x )=\ln( 2\cos^2\frac{x}{2})=\ln2+2\ln\cos\frac{x}{2}$$

And this $$\ln2$$ adds together with the arbitrary constant $$c$$ in indefinite integral and gets cancelled in definite integral.

• Thank you so much! This makes perfect sense! – Seigemann Aug 12 at 11:58
• You're welcome! – Ak19 Aug 12 at 11:58

As Ben suggested in comments, it's always good to check an integral by taking the derivative:

$$\dfrac d {dx} \ln(1+\cos x)=\dfrac{-\sin x}{1+\cos x}=\dfrac{-2\sin\dfrac x2 \cos \dfrac x2}{2\cos^2\dfrac x2}=\dfrac{-\sin\dfrac x2}{\cos \dfrac x2}=\dfrac d {dx} 2 \ln \cos \dfrac x2$$

• N.B. The expressions are undefined when $x=\pi$ – J. W. Tanner Aug 12 at 13:11

$$1+\cos \, x=2\cos^{2}(\frac x 2)$$ so $$\ln (1+\cos \, x)=2 \ln (\cos (\frac x 2))+\ln 2$$ . Also, $$-\tan (\frac x 2)$$ is same as $$-\frac {\sin \, x} {1+\cos \, x}$$ because $$\sin \,x =2 \sin (\frac x 2)\cos (\frac x 2)$$ and $$1+\cos \, x=2\cos^{2}(\frac x 2)$$.

$$2\ln\cos\frac x2=\ln\cos^2\frac x2=\ln\frac{1+\cos x}2=\ln(1+\cos x)-\ln2$$ by the half-angle formula, so your answer and WA's are the same up to the integration constant.

When you do the one integral $$I(x)$$ by different methods you get different expressions $$I_1(x),I_2(x),I_3(x),.....$$, however the difference between any two of these is a constant independent of $$x$$. For instance $$I(x)=\int \sin \cos x dx =\frac{1}{2}\int \sin 2x~ dx =-\frac{1}{4} \cos 2x +C_1 =I_1(x).$$

Next if you do integration by parts you get $$I=\sin ^2x -\int \sin x \cos \Rightarrow I=\frac{1}{2} \sin^2 x +C_2=I_2(x).$$

Further I you use a substitution $$\cos x =-t$$, then $$I(x)=-\frac{1}{2}\cos^2 x +C_3=I_3(x).$$

Now check that the difference between any two of $$I_1,I_2,I_3$$ is just a constant.

As pointed in other solutions the thing stated above is happening in your case as well.


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Real numbers, rational numbers. Expressing complex numbers in form $a+bi$. Each complex number corresponds to a point (a, b) in the complex plane. The square of an imaginary number bi is −b2. In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line. can in general assume complex values with nonzero real parts, but in a particular case of interest, the real part is identically zero. Practice: Parts of complex numbers. In the last example (113) the imaginary part is zero and we actually have a real number. Rational numbers are numbers that can be expressed as a fraction or part of a whole number. It only takes a minute to sign up. There are three common forms of representing a complex number z: Cartesian: z = a + bi JR is the set of numbers that can be used to measure a distance, or the negative of a number used to measure a distance. 5.1.2 The Reals as a Subset of the Complex Numbers Since the complex numbers were seen as an extension of the set of real numbers, it is natural to believe that R is a subset of C. Of course, to prove this subset Similarly, it is asked, is every real number is a complex number? A complex number is a number that can be written in the form a + b i a + bi a+bi, where a and b are real numbers and i is the imaginary unit defined by i 2 = − 1 i^2 = -1 i2=−1. Classification of Real Numbers Examples. The real numbers have the following important subsets: rational numbers, irrational numbers, integers, whole numbers, and natural numbers. (0,1) = (-1,0), which is purely real and equals to -1. Example 2 : Tell whether the given statement is true or false. Thus we can consider the complex number system as having embedded within it, as a subset the real number … Furthermore, each real number is in the set of complex numbers,, so that the real numbers are a … Choices: A. integers, rational numbers, real numbers B. whole numbers, integers, rational numbers, real numbers C. natural numbers, whole numbers, integer numbers, rational numbers, real numbers D. irrational numbers, real numbers Correct Answer: A Complex numbers are often graphed on a plane. What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? Some examples of irrational numbers are $$\sqrt{2},\pi,\sqrt[3]{5},$$ and for example $$\pi=3,1415926535\ldots$$ comes from the relationship between the length of a circle and its diameter. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Oh I suppose Russel has a definition where the real number 3 is the set of all things there are 3 of. A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself). How are Quaternions derived from Complex numbers or Real numbers? B. Â¿CuÃ¡les son los 10 mandamientos de la Biblia Reina Valera 1960? The real numbers are a subset of the complex numbers. So, $$i \times i = -1$$ $$\Rightarrow i = \sqrt{-1}$$. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Complex numbers introduction. We will now introduce the set of complex numbers. The complex numbers are a plane with an additional real axis to calculate square roots (and other even roots) of negative numbers. The set of complex numbers C with addition and multiplication as defined above is a field with additive and multiplicative identities (0,0) and (1,0).It extends the real numbers R via the isomorphism (x,0) = x. Suppose that (1) 1 2S: (2) If a natural number kis in S;then the natural number k+ 1 also is in S: Then S= N:That is, every natural number nbelongs to S: REMARK. There are several types of subsets of real numbers—numbers that can be expressed as a decimal. Thus we can consider the complex number system as having embedded within it, as a subset the real number system. If you're seeing this message, it means we're having trouble loading external resources on our website. These numbers are called irrational numbers, and $\sqrt{2}$, $\sqrt{3}$, $\pi$... belong to this set. Is there even such a set? Proof that π is irrational. Email. But no real number, when squared, is ever equal to a negative number--hence, we call i an imaginary number. Better user experience while having a small amount of content to show. Explain your choice. Complex numbers in the form a + bi can be graphed on a complex coordinate plane. Complex Numbers. x is called the real part and y is called the imaginary part. (The counting numbers are 1,2,3,....) All of these types of numbers are real numbers. We will addres s complex (or imaginary) numbers in the Quadratic Functions chapter. That is the adjacent surface to our 3D! Learn vocabulary, terms, and more with flashcards, games, and other study tools. The real numbers are all the numbers on the number line, where you group rational numbers with a so called dedekind cut (you can form this cut so that it result is irrational). Popular Trending (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0 i, which is a complex representation.) Real numbers 21.5 pi. Some ﬁxed point O is chosen to represent the complex number … In the 1760s, Johann Heinrich Lambert proved that the number π (pi) is irrational: that is, it cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer. Therefore we have: z = Re(z) + iIm(z). Why or why not? In complex analysis, the complex numbers are customarily represented by the symbol z, which can be separated into its real (x) and imaginary (y) parts: = + for example: z = 4 + 5i, where x and y are real numbers, and i is the imaginary unit.In this customary notation the complex number z corresponds to the point (x, y) in the Cartesian plane. A set S of real numbers is called bounded from above if there exists some real number k (not necessarily in S) such that k ≥ s for all s in S.The number k is called an upper bound of S.The terms bounded from below and lower bound are similarly defined.. A set S is bounded if it has both upper and lower bounds. The complex numbers form a COMPLETE system of numbers of which the real numbers form a subset. The conjugate of a complex number z= a+ biis created by changing the sign on the imaginary part: z = a bi: Thus the conjugate of 2 + iis 2 + i= 2 i; the conjugate of p 3 ˇiis p 3 ˇi= p 3 + ˇi.

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# Is every odd order skew-symmetric matrix singular?

We call a square matrix $$A$$ a skew-symmetric matrix if $$A=-A^T$$. A matrix is said to be singular if its determinant is zero. Is every odd order skew-symmetric matrix with complex entries singular?

## 3 Answers

Yes, that holds, since: $$\det A=\det{(-A^T)}=(-1)^{odd}\det{A^T}=-\det A,$$ from where we get $\det{A}=0$.

This is actually the case :

Suppose, $A$ is an $n\times n$-matrix.

We have $$\det(A)=\det(-A^T)=(-1)^n\cdot \det(A^T)=(-1)^n\cdot \det(A)$$

Since $n$ is odd, we can conclude $\ \det(A)=-\det(A)\$ implying $\ \det(A)=0\$

• But the problem here is that the entries are complex. Is it true for complex entries as well? Nov 18 '16 at 12:05
• Yes, it is also true for complex entries. The proof does not assum real entries. Nov 18 '16 at 12:47

Consider the example $$\begin{bmatrix} 0 & i & -3\\ -i & 0 & 2i\\ 3 & -2i & 0\\ \end{bmatrix}$$

• I mean a 3 by 3 matrix with rows (0 i -3), ( -i 0 2i) and (3 -2i 0). Nov 18 '16 at 12:21
• If this was intended to be a counterexample, then it is not! The determinant of this matrix is zero. May 6 at 15:03