# Representation of a matrix transformation

An example asks me to define $T: M_{2\times2}(F) \to M_{2\times2}$ by $T(A) = A^t$. Compute $[T]_\alpha$

$\alpha$ is the standard ordered basis of $2\times 2$ matrices.

To find the transformation I performed the transformation on all four elements of $\alpha$, for example $T(\begin{pmatrix} 1&0\\0&0\\\end{pmatrix}) = (\begin{pmatrix} 1&0\\0&0\\\end{pmatrix})$ and so on. Basically the matrices are all the same before and after the transformation. Accordingly, the textbook's answer is that you arrange all the results $A^t$ into a $4\times4$ matrix:

$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$$

I don't understand this what means, and I've been trying to. How can this be the transformation $T$ on $\alpha$ if we're not even able to multiply the basis by it? What I mean is, we can't multiply this transformation matrix by anything in $\alpha$ since one is a $4\times4$ and one is a $2\times 2$.

Also, why doesn't this mean that any arbitrary $2\times 2$ matrix can be raised to the power of $t$ by simply multiplying by this matrix? (As per the previous paragraph, it can't, but it should).

• I changed $T: M_{2\times2}(F) -> M_{2\times2}$ to $T: M_{2\times2}(F) \to M_{2\times2}$. $\TeX$ is not so crude that you have to resort to things like that. – Michael Hardy Oct 11 '12 at 23:43
• Thanks - I couldn't find the command for it. I'll know for next time. – CodyBugstein Oct 11 '12 at 23:57

You need to distinguish here between the space $\mathbb F^2$ that the $2\times2$ matrices operate on and the space of the $2\times2$ matrices themselves. The former is two-dimensional, the latter is four-dimensional.

A matrix is a representation of a linear transformation with respect to bases of the domain and target of the transformation. If the linear transformation is an endomorphism, a linear map from a space to itself, the corresponding matrix is square, and one typically uses the same basis for the space in its two roles as domain and target.

The standard basis of the space of $2\times2$ matrices that you're using has four basis elements and spans the four-dimensional space of $2\times2$ matrices. A linear transformation has been defined on that space – the four-dimensional space of $2\times2$ matrices, not the two-dimensional space that they operate on.

Representing that linear transformation as a matrix with respect to the standard basis (both for the domain and the target) yields a $4\times4$ matrix. That's not a mismatch because the vectors and matrices that will be multiplied by this matrix aren't elements of $\mathbb F^2$; they're the vectors of coefficients in an expansion of the matrix they represent in the standard basis.

As an example, the matrix

$$\pmatrix{2&3\\4&5}$$

has coefficients $2$, $3$, $4$ and $5$, respectively, in the standard basis, so with respect to that basis it's represented by the column vector

$$\pmatrix{2\\3\\4\\5}\;.$$

Its transpose

$$\pmatrix{2&4\\3&5}$$

is represented by the column vector

$$\pmatrix{2\\4\\3\\5}\;.$$

And sure enough we have

$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\pmatrix{2\\3\\4\\5}=\pmatrix{2\\4\\3\\5}\;.$$

Thus this matrix really does represent the linear transformation $T$, transposition, in this basis.

• So matrices can unravel to form vectors? I've played around with some and it makes sense. Thanks for the amazing answer. I still have a question though; shouldn't the transforming vector actually put your sample vector to the power of something? Isn't that the whole idea? – CodyBugstein Oct 11 '12 at 23:32
• @Imray: I think you've misunderstood the notation. $A^t$ is merely a notational device to denote the transpose of $A$; nothing is being raised to a power. I prefer the slightly different notation $A^\top$ (A^\top) to avoid the impression that the $t$ is a variable used as an exponent. – joriki Oct 11 '12 at 23:34
• @Imray: As an exercise, I suggest to write down the matrix of $T$ with respect to the basis $$\left\{\pmatrix{1&0\\0&1},\pmatrix{1&0\\0&-1},\pmatrix{0&1\\1&0},\pmatrix{0&1\\-1&0}\right\}\;,$$ and then perhaps use it to apply $T$ to a couple of matrices and see whether the transpose comes out right. – joriki Oct 11 '12 at 23:38
• yes right! I didn't realize that! Ok I've got it now, thank you very much for explaining it to me. – CodyBugstein Oct 11 '12 at 23:38
• @Imray: You're welcome! – joriki Oct 11 '12 at 23:40

Just to make sure our definition of matrix representation is the same: Let $\alpha = (a_1, \ldots, a_k)$ be a basis for $V$, and let $T: V \rightarrow V$ be a linear map. Then matrix representation $[T]_{\alpha}$ is defined as follows: $$[T]_{\alpha} = \begin{bmatrix} [T(a_1)]_{\alpha} & \ldots & [T(a_k)]_{\alpha} \end{bmatrix}$$

The matrix representation obeys the following formula: for every $v \in V$, $[T(v)]_{\alpha} = [T]_{\alpha} [v]_{\alpha}$, where $[v]_{\alpha}$ is the coordinate vector of $v$ with respect to the basis $\alpha$.

So now, to answer your question, it's true that $[T]_{\alpha}$ is a $4 \times 4$ matrix in your example, but you don't multiply it onto the $2 \times 2$ matrices; you multiply it onto the $2 \times 2$ matrices after they have been converted into column vectors via the coordinate mapping. Since $2 \times 2$ matrices form a $4$-dimensional space, then this coordinate mapping turns a $2 \times 2$ matrix into an element of $\mathbb{R}^4$.

• How does coordinate mapping work? The rows all extend downward to become one long column? Can all matrices with $n$ elements be multiplied by any matrix with $n$ columns? – CodyBugstein Oct 11 '12 at 23:35
• What confused me is how $[T]_\alpha$ could be square. But one needs to remember that the columns are in vector notation too (not in matrix notation!), so they are actually not $k \times k$ but $k^2 \times 1$ and so in the end $[T]_\alpha$ is $k^2 \times k^2$. – philmcole Apr 10 '18 at 21:20


# Show that $a_n=\left(\sum_{k=1}^n \frac{1}{k}\right)-\log(n)$ is decreasing

How I decided to show it is by Induction, and I don't really like what I have, so could you guys please tell me if my proof valid. Thanks.

Aim of the proof: Show that $$a_n>a_{n+1}$$ for all $n\geq1$.

Base case: $$a_1-a_2=\log(2)-\frac{1}{2}>0.$$ Hence base case holds. Assume, for some $k>1$, that the result holds, that is:$$a_k>a_{k+1}$$ Hence show that $$a_{k+1}>a_{k+2}.$$ Now we can rewrite $a_{k+1}>a_{k+2}$ as, $$\left(\sum_{k=1}^{k+1}\frac{1}{k+1}\right)-\log(k+1)>\left(\sum_{k=1}^{k+1} \frac{1}{k+1}\right)+\frac{1}{k+2}-\log(k+2),$$ Which is equivealnt of saying: $$\log\left(\frac{k+2}{k+1}\right)>\frac{1}{k+2}$$

For our problem $k\geq1$ and $\log(\frac{3}{2})>\frac{1}{3}.$

My last step is saying that as $n$ tends infinity, the limit of LHS of our last inequality is infinite and RHS is 0. Hence the result holds.

What I don't like is the fact that I don't use the inductive hypothesis at all. (Sorry for selling and grammar errors, English is not my native language)

• This does not answer your question, but do you know that $\lim_n a_n =\gamma$, the Euler-Mascheroni constant? You might want to google it, you will for sure find something about the non decreasing behaviour of $a_n$. – Paolo Intuito Nov 23 '17 at 21:53
• A fundamental step you make in your proof is incorrect: $\lim_{k \rightarrow \infty} (\frac{k+2}{k+1}) = 0$, not infinity, as you claim. (In fact, if this step were correct, then this alone would be a proof - the whole induction set up would be unnecessary.) – John Don Nov 23 '17 at 21:59
• Oh crap, yeah, the fraction gets closer to 1 eqch time. Thank you for noticing this John. How would I go about proving it then?, and yes, G.S. I know about that fact but couldnt find anything on the question I am trying to solve – Scavenger23 Nov 23 '17 at 22:02

You wish to show the inequality,

$$\frac{1}{n+1}-\ln(n+1)<-\ln n$$ or equivalently, $$\frac{1}{n+1}<\ln (1+\frac{1}{n})$$

Now we have $$x-\frac{x^2}{2}<\ln(1+x)$$ so just check that $$\frac{1}{n+1}<\frac{1}{n}-\frac{1}{2n^2}$$

Which is indeed the case as it is not hard to show.

• How do you know that your second to last line is true? I feel like I am missing something basic. – Scavenger23 Nov 23 '17 at 22:26
• Either, you note that the next term in the power series is positive, or you could use the mean value theorem. – Rene Schipperus Nov 23 '17 at 22:27
• Or more straightforward, move all the terms to the right, and differentiate, and show the derivative is positive, and thus the function is increasing. – Rene Schipperus Nov 23 '17 at 22:33
• I see now, didnt realise that you tailor expanded the log. Thanks for the help. – Scavenger23 Nov 23 '17 at 22:35

I like Rene Schipperus' method a lot - it is very simple. Here is an commonly used alternative:

Note that, the function $\frac1x$ is strictly decreasing on $[1, \infty)$, so, for all $n \in \mathbb{N}^{\, \ge 1}$,

\begin{align} \ln (n+1) - \ln n &= \int_{n}^{n+1} \frac1x \, dx\\ &> \int_n^{n+1} \frac1{n+1} \, dx\\ &= \frac1{n+1} \end{align}

As you have already noted, the solution follows.

• very good work. So essentially the method I wanted to use was correct, ? The question in the book never told me how to prove it. Induction seemed the way forward for me. – Scavenger23 Nov 23 '17 at 22:43
• @KuderaSebastian That seems reasonable... I suppose that the thing to notice was that, at some point, (e.g. in your inductive step) you would have to relate $\log(n)$ to some function of integers somehow (as this is what you have on the LHS). One way is to use the power series expansion (as Rene has done), and the other is to notice that the derivative of $\log(x)$ is in fact $\frac1x$ (which both I and Peter have used in our answers). – John Don Nov 23 '17 at 22:53
• Jon.Nice answer, short and clear. – Peter Szilas Nov 24 '17 at 9:06

Show that $a_n -a_{n+1} >0.$

$\Delta_n: = a_n - a_{n+1} =$

$-\dfrac{1}{n+1} + \log(n+1) -\log(n).$

Mean Value Theorem:

$f(x) := \log(x)$ then:

$\dfrac{f(x+1)-f(x)}{1} = f'(t),$ $x \lt t \lt (1+x).$

$\dfrac{\log(n+1) -\log(n)}{1} = \dfrac{1}{t}$, $n\lt t\lt (n+1).$

$\Delta_n = -\dfrac{1}{n+1} + \dfrac{1}{t}$ , where $t \lt (n+1).$

Hence $\Delta_n >0.$

• NIce... my answer is essentially this, but integrated, in a sense. – John Don Nov 23 '17 at 22:49


divmod() in Python and its application

PythonServer Side ProgrammingProgramming

The divmod() is part of python’s standard library which takes two numbers as parameters and gives the quotient and remainder of their division as a tuple. It is useful in many mathematical applications like checking for divisibility of numbers and establishing if a number is prime or not.

Syntax

Syntax: divmod(a, b)
a and b : b divides a
a and b are integers or floats

Examples

In the below example see the cases of both integers and floats. On the application of divmod() they give us a resulting tuple which is can also contain integers and float values.

# with integers
print("5 and 2 give:",divmod(5,2))
print("25 and 5 give:",divmod(25,5))

# with Floats
print("5.6 and 2 give:",divmod(5.6,2))
print("11.3 and 9.2 give:",divmod(11.3,9.2))

Output

Running the above code gives us the following result −

5 and 2 give: (2, 1)
25 and 5 give: (5, 0)
5.6 and 2 give: (2.0, 1.5999999999999996)
11.3 and 9.2 give: (1.0, 2.1000000000000014)

Using Zero

If the first argument is zero then we get (0,0). And If the second argument is zero then we get Zerodivision error as expected.

Example

# With first argument as zero
print("0 and 8 give:",divmod(0,8))

# With second argument as zero
print("8 and 0 give:",divmod(8,0))

Output

Running the above code gives us the following result −

0 and 8 give: (0, 0)
Traceback (most recent call last):
File "xxx.py", line 6, in
print("8 and 0 give:",divmod(8,0))
ZeroDivisionError: integer division or modulo by zero

Checking divisibility

If the second value of the tuple after division is 0 then we say that the first number is divisible by second. Else it is not divisible. The below example illustrates this.

Example

m = 12
n = 4
quotient,remainder = divmod(m,n)
print(quotient)
print(remainder)
if (remainder==0):
print(m,' is divisible by ',n)
else:
print(m,' is not divisible by ',n)

Output

Running the above code gives us the following result −

3
0
12 is divisible by 4

Checking if Number is Prime

We can use divmod() to keep track of the reminders it generates when we start dividing a number by each number starting with itself till 1. For a prime number the count of zero remainder will be only one as no number other than itself will divide it perfectly. If the count of zero remainder is greater than 1 then the number is not prime,.

Example

num = 11
a = num
# counter the number of remainders with value zero
count = 0
while a != 0:
q, r = divmod(num, a)
a -= 1
if r == 0:
count += 1
if count > 2:
print(num, 'is not Prime')
else:
print(num, 'is Prime')

Output

Running the above code gives us the following result −

11 is Prime
Published on 07-Aug-2019 08:32:04


# What is the best way to intuitively explain the relationship between the derivative and the integral?

This is my first post so bear with me, but something I've been thinking about lately is: Why didn't I ever question the relationship between the derivative and the integral when I was taking calculus?

Let me explain what I mean: In most courses, the derivative is introduced as the slope of a curve at a point, or the "instantaneous rate of change". Then the integral is introduced as the area under a curve. Then students are told that these two things "undo" each other (the fundamental theorem of calculus). So now I'm wondering, why didn't I ever question why these "undo" each other? It's not intuitive at all. For example, addition and subtraction, multiplication and division, logarithms and exponentials; all of these things I can intuitively understand why they "undo" each other. But how does something that represents the area under a curve "undo" something that represents an instantaneous rate of change? What is the connection between those two concepts? Is there a better way to explain these concepts that makes the fundamental theorem of calculus more intuitive?

Looking forward to hearing some of y'alls responses. Thanks!

• $[\cdots]$ why didn't I ever question why these "undo" each other? --- Your calculus experience is probably in the minority, because it is fairly standard (at least in U.S. introductory calculus courses, both high school and college) to make a big deal about the connection, usually at the end of the first semester, this being especially the case since the rise of "reform calculus" from the late 1980s on. That said, see right brain explanation and left brain explanation. – Dave L Renfro Apr 2 '20 at 5:26
• Which resources have you perused before registering and asking this question? Voting to close. – Rusty Core Apr 2 '20 at 21:01
• In countries, where physics is a mandatory course in grade school, the traditional layman explanation is through kinematics. Average speed v, by definition is displacement s over time t. Start with the simplest case of uniform motion, draw a graph in (t, s), it would be a line climbing up. Tangent at any point is at the same angle (similarity of right triangles), this is the derivative by definition, which is speed. Draw speed in (t, v). Area below it is displacement. Continue with uniform acceleration and then with no-uniform motion, slicing up the area under speed into thin vertical bands. – Rusty Core Apr 2 '20 at 21:14
• @BenCrowell I think what makes this a math ed question is wanting better intuition to explain the relationship. Also, the fact that many of us didn't really get it until we were teaching speaks volumes to how much this is an education issue. – Sue VanHattum Apr 4 '20 at 20:52
• @SueVanHattum Exactly - that's why I posted it here. The funny thing is that when I learned how to prove the FTC in real analysis, I didn't find either of the proofs too difficult. But the intuition behind the overall concept still wasn't there. Then recently I realized that if a student came to me and asked "Why do these two concepts undo each other?", my best response would be to show them the proof. That's why I figured this was a math ed question, because I'm specifically asking what the best way is to explain this to a student and make it more intuitive. – Brain Gainz Apr 5 '20 at 18:43

You start by noticing that the Riemann sums (multiplication followed by addition) and the difference quotients (subtraction followed by division) undo each other. Their limits -- the integral and the derivative -- still undo each other.

Added: The first sentence is a part of what is called “Discrete Calculus” https://en.wikipedia.org/wiki/Discrete_calculus

• Wow. I'm unsure whether this intuition can be made to correspond to some rigorous proof, but it's just solid enough to make me feel it gave me some insight into the mechanism of the theorem's truth and not mere word-soup hand-waving. And I think the maximal spacing in a partition (that goes to zero for the Riemann integral) even corresponds to the step size in the difference quotient in a suitable sense. – Vandermonde Apr 3 '20 at 1:23
• @Vandermonde You are correct. By the way, this will be in the third volume of my book Calculus Illustrated: amazon.com/dp/B082WKCYHY – Peter Saveliev Apr 3 '20 at 1:44
• Surprising how entirely abstract and formal mathematical thinking can be intuitive :-). Perhaps not for everybody, but even this engineer sees symmetries. – Peter - Reinstate Monica Apr 3 '20 at 10:21
• 1. Can you please exhibit the formula that you're hinting to? 2. Can you please clarify 1. which "Riemann sums" you mean? 3. Which "difference quotients"? 4. Which "limits"? – NNOX Apps Apr 4 '20 at 5:27
• If one considers the differential $\text{d}$ and the integral $\int$ as the fundamental concepts of calculus (as opposed to $\frac{\text{d}(-)}{\text{d}x}$ and $\int(-)\text{d} x$), then the wording is even more simple: difference and sum undo each other. (Actually Leibniz called $\int$ the sum, before Bernoulli suggested integral) – Michael Bächtold Apr 6 '20 at 15:58

(this is from my calculus notes, see page 233 of: http://www.supermath.info/OldschoolCalculusII.pdf)

• So in words: the change in area is equal to the height of the function times the change in x? That actually makes sense. – Brain Gainz Apr 3 '20 at 0:31
• @BrainGainz thanks! Incidentally, I had exactly your feeling after taking years of calculus and an Advanced Calculus course where I suspect we proved the FTC. Your feeling is not unusual in my estimation. – James S. Cook Apr 3 '20 at 2:52
• I understand the pedagogical need, but I cringe when I read (or say myself when teaching calculus) "arbitrary function" and the picture is that of a differentiable function. – Martin Argerami Apr 3 '20 at 10:36
• @BrainGainz yep, I have a few Topology lectures posted. I am that James Cook. – James S. Cook Apr 4 '20 at 21:57
• @StevenGubkin cool, if I ever get to teach Calculus I again, I should use that. So many options to play with... – James S. Cook Apr 6 '20 at 3:49

Derivative of integral is the original function: Let $$F$$ be the integral function of $$f$$, so that $$F(x)$$ is the area under the graph from zero to $$x$$. For small $$h>0$$ the difference $$F(x+h)-F(x)$$ is the area of a narrow vertical strip. The width is $$h$$ and height approximately $$f(x)$$. As $$h\to0$$, this means $$F'(x)=f(x)$$.

If you like thinking in terms of infinitesimals, write $$dF=F(x+dx)-F(x)=f(x)dx$$ and divide by the differential. This can be a cleaner way to think as the limit process is left implicit in a way.

Integral of derivative is the original function: The integral is roughly $$\int_0^x f'(x)dx = \sum_{k=1}^n f'(x_k)(x_k-x_{k-1})$$ and the derivative is roughly $$f'(x_k) = \frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}.$$ Combine these and you get a telescoping sum and the desired result.

This is the way it makes intuitive and graphical sense to me. The link is not immediate; that's why it's a great theorem.

My answer also comes from physics.

Say p(t) is the position of an object in the time t. For concreteness, suppose you are traking a truck and the truck is going forward on a road from A to B. (i.e., all derivatives are positive)

It is very natural to graph p(t), and derive it, arriving at v(t), the speed of the truck for each time. After all, the derivative is just taking a small time increment $$dt$$, calculating the corresponding $$ds$$ (a space increment) and dividing

However, when I look at the graph of v(t), I can also obtain (almost everything) from p(t). Say I am at the position 100 km in the time t=0. After a $$dt$$, I shall be at $$p(t+dt) = 100+dt\cdot v(t)$$. And I can do it again and again $$p(t+dt+dt+dt) = 100+dt\cdot v(t)+dt\cdot v(t+dt)+dt\cdot v(t+2\cdot dt)+dt\cdot v(t+3\cdot dt)$$ (notice that after some $$dt$$s, I use the updated speed)

But that (taking a limit) is an integral!

The only weird thing is my assuption that I am at position 100km at time t=0. The graph of v(t) cannot tell me that. Intuitively, it does tell me how fast I am going, and therefore can tell me how much how much I walked, but it cannot tell me at which specific part of the road I started. That is that weird constant +c that appears on the indefinite integral.

But then again, we dont use the indefinite integral much, do we? We are more interested in summing $$dt$$*v(t) for ranges. Say, from t=3 to t=10. That gives me the amount I walked between those two times. So, mercifully, the +c disappears in most use cases: to know how much I walked, there is no need to know where I started.

• I love how velocity and position helps to cement the relationship. In one class, my students asked why area below the axis should count as negative. We agreed that negative velocity meant going backwards, and therefore decreased your distance from start. (Distance being measured by area under velocity curve, a negative velocity curve has to have "negative area" to decrease the total distance.) – Sue VanHattum Apr 4 '20 at 20:47

When I first began teaching Calculus, I realized that I really didn't understand the Fundamental Theorem. So I looked for something that would help me have a deep understanding, that would also help me help students to see it.

I found a lovely project, which I have modified over the years. Here are links to the pdf and to a .doc version (in which the formulas are messed up, because google docs couldn't read them). If you use this, you'll want to change the part near the end where I reference the textbook.

• Wow, this is really great. I'll definitely save this. Thanks! – Brain Gainz Apr 3 '20 at 0:29

I gave a similar post to this one touching on this on Math.StackExchange.

Basically, the way I would go about it is to say that there is a very easy way by which one can think of at least Riemann integration (the usual definition given in a "most courses" calculus course) as an inverse of differentiation by construction: that is, the relationship between the two is not an "accident", but design, so that given one, you could fairly easily be led to the other.

And to do that, I'd first suggest getting rid of the whole "slope" vs "area" business altogether - if anything, those are best given as theorems to be proved, after you have other, free-standing definitions of a tangent line and an area which, by the way, can be done, but just too-often aren't, in favor of various hand-waving arguments.

Instead, the relevant idea is change: the derivative of a function $$f$$, i.e. $$f'$$, stands for a kind of "sensitivity" measure. Suppose that $$f$$ were like a kind of meter or instrument, with a knob attached to it, and a readout of some kind. The knob attached is the function's input argument, typically denoted $$x$$. The indicator on the readout is the return value, $$f(x)$$. If $$x$$ is set at some value of interest $$x_0$$, then likewise the readout will be at $$f(x_0)$$. Now suppose you "wiggle" the knob $$x$$ back and forth a little bit, and you see how the needle $$f(x)$$ responds to that small impulse. For a continuous function, the size of the output's wiggle will be smaller in absolute terms the smaller you make the input, but the proportionate size, i.e. how much it wiggles relative to how much you wiggle the input value, may not be. When we say $$f$$ is differentiable, what that means is that there exists at each point $$x_0$$ a proportionality factor $$f'(x_0)$$ such that

$$f(x \pm \underbrace{dx}_\mbox{"wiggle" in x}) \approx f(x_0) \pm \underbrace{[f'(x_0)\ dx]}_\mbox{"wiggle" in f(x)}$$

so that

$$\mbox{proportional "wiggle"} = \frac{\mbox{"wiggle" in f(x)}}{\mbox{"wiggle" in x}}$$

so long as the change $$dx$$ is suitably small

Integration, then, goes the other way. Suppose that I am given now, not the function $$f$$, but only its derivative, $$f'$$, and want to find $$f$$. First off, one should observe that since $$f'$$ only deals with changes in the input, to start, we actually need one more piece of information, and that is some sort of initial value, i.e. $$f(0)$$. Suppose this to be given as well.

Starting at $$f(0)$$, suppose we apply a small, but nonzero, change $$\Delta x$$ to the input, i.e. we ask, "given $$f(0)$$, what is $$f(0 + \Delta x)$$, to the best we can do?" Well, since we know $$f(0)$$ and $$f'$$, then since $$\Delta x$$ is a small change or "wiggle", we can say that approximately,

$$f(0 + \Delta x) \approx f(0) + [f'(0)\ \Delta x]$$

which is just what you see above.

Now, suppose we take another step of $$\Delta x$$. We're now going from $$x = 0 + \Delta x$$ to $$x = (0 + \Delta x) + \Delta x$$ (or $$2\ \Delta x$$, but I find writing it this way makes it clearer what is going on - "simpler" isn't necessarily "better"). At this second step, likewise, treating $$0 + \Delta x$$ as a prior input in and of itself, we have

$$f([0 + \Delta x] + \Delta x) \approx f([0 + \Delta x]) + [f'([0 + \Delta x])\ \Delta x]$$

which, combining with the previous expression, becomes

$$f([0 + \Delta x] + \Delta x) \approx f(0) + [f'(0)\ \Delta x] + [f'([0 + \Delta x])\ \Delta x]$$

and it is not hard to then continue this process so that we see for $$N$$ steps,

$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(0 + i[\Delta x])\ \Delta x$$

or, if we set $$x_i := 0 + i[\Delta x]$$, we can say more neatly as

$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x$$

and to be more general, if we take $$N$$ steps of suitable, perhaps different, sizes $$\Delta x_i$$ to reach from $$0$$ some fixed point $$x_0$$,

$$f(x_0) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$

and then we consider what happens as the steps become arbitrarily fine, at which point we hope - and need to prove - that

$$f(x_0) = f(0) + \left[\lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i\right]$$

which leads us to define this new operation, given by the limit on the right...

$$\int_{0}^{x_0} f'(x)\ dx := \lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$

• This does a great job of explaining how to think of $\int_a^b f'(x)\mathrm dx = f(b) - f(a)$, but, in suggesting that we think of it as a definition, I'm left to wonder how to define $\int_a^b F(x)\mathrm dx$ without proving some separate theorems about existence of anti-derivatives. – LSpice Apr 6 '20 at 0:04

Intuitively, the fundamental theorem of calculus states that "the total change is the sum of all the little changes". $$f'(x) dx$$ is a tiny change in the value of $$f$$. We sum up all these little changes to get the total change $$f(b) - f(a)$$.

I elaborated on this explanation here: https://math.stackexchange.com/a/1537836/40119.

• Very nice! The idea that $\mathrm dx$ is not punctuation, but (conceptually) an actual infinitesimal measure of $x$-length, so that $f'(x)\mathrm dx$ has units of (slope)($x$-length) = $y$-length, I think is probably the most important pre-proof part of understanding the connection. That is, I think that most people think of integration as "something you do to $f'(x)$", not "something you do to $f'(x)\mathrm dx$"—hence later difficulties with differential forms, and earlier difficulties with changes of variable. – LSpice Apr 6 '20 at 0:06

Well, for me, the nicest intuition comes from physics. If $$F=F(x)$$ is some force applied to an object $$O$$ casuing $$O$$ to move, where $$x$$ is its displacement, when the work $$W(x)$$ produced by that force from $$x_0=0$$ to $$x$$ metres is given by:

$$W(x):=\int_0^xF(s)ds.$$

Now, what if we ask what is the rate of change of that work with respect to displacement? At some certain point $$x$$, the work provided to or substracted by that object $$O$$ is determined by the force $$F$$ applied on it. So, one can intuitively expect that:

$$W'(x)=\left(\int_0^xF(s)ds\right)'=F'(x).$$

That is, the larger the force, the faster the energy flows from the one applying the force towoards the object. The less the force, the slower that energy flows.

If you want more clearly "undoing" operations, define "stacking" as the following:

Take some $$\Delta x$$. Chop the curve into rectangles of width $$\Delta x$$ and height $$f(x)$$. (There's some leeway as to take $$f(x)$$ at the left side, right side, middle, minimum, maximum, etc. The rest of my description will be right side.) Now take each rectangle and put the bottom of each rectangle at the top of the previous rectangle. For instance, if you have $$f(x) = x^2$$ and $$\Delta x = 0.1$$, you'd have a rectangle with lower left corner at the origin, and top left at $$(0.1, 0.01)$$. Then the next rectangle would have lower left corner at $$(0.1, 0.01)$$, and upper right at $$(0.2, 0.05)$$, and so on.

Define "unstacking" as follows:

Take some $$\Delta x$$. Chop the curve into rectangles of width $$\Delta x$$ and height $$f(x)$$. Now, take each rectangle and shift it vertically down the height of the previous rectangle. For instance, if $$f(x) = x^2$$, the first rectangle will have height $$0.01$$. The second one will have height $$0.04$$, so for the unstacked version, we move it down $$0.01$$, leaving its new height as $$0.03$$. Another way of phrasing it is for each $$x$$, take the rectangle whose bottom left corner is at $$(x, f(x))$$ and top right corner is $$(x+\Delta x, f(x+\Delta x))$$. Then create a chart out of all of those rectangle, with all of them moved down so their bottom is on the x-axis.

For "well-behaved" functions (I believe that uniform continuity is a sufficient condition), as $$\Delta x$$ goes to zero, stacking becomes integration, and unstacking become differentiation (with the proper scaling).


# Elementary matrix operations

I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.

I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.

The first is $$A= \begin{pmatrix} 1 & 1 & 0 & 2\\ 2 & 1 & 1 &-2\\ 2 & 2 & 2 & 1 \end{pmatrix},\quad \tilde{b}= \begin{pmatrix} 12\\0\\14 \end{pmatrix}$$ Proceeding with an augmented matrix, I have \begin{align*} \left( \begin{array}{rrrr|c} 1 & 1 & 0 & 2 & 12\\ 2 & 1 & 1 &-2 & 0\\ 2 & 2 & 2 & 1 & 14 \end{array} \right)&\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 &-1 & 1 &-6 & -24\\ 0 & 0 & 2 & -3 &-10 \end{array} \right)\\ &\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 & 1 &-1 & 6 & 24\\ 0 & 0 & 1 & -3/2 &-5 \end{array} \right) \end{align*} I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.

Solving this now, I have \begin{align*} x_4&=\alpha\in\mathbb{R}\\ x_3&=-5+\frac{3}{2}\alpha\\ x_2&=19-\frac{9}{2}\alpha\\ x_1&=-7+\frac{5}{2}\alpha \end{align*}

So from this, my solution will be $$\tilde{x}= \begin{pmatrix} -7+\frac{5}{2}\alpha\\ 19-\frac{9}{2}\alpha\\ -5+\frac{3}{2}\alpha\\ \alpha \end{pmatrix} = \begin{pmatrix} -7\\19\\-5\\0 \end{pmatrix}+\beta \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$

This does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.

Edit: My professor's solution is $$\tilde{x}= \begin{pmatrix} 3\\1\\1\\4 \end{pmatrix}+\alpha \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$

Can somebody please let me know what I've done incorrectly?

Thank you.

• I tried using WolframAlpha, but it didn't tell me what I did wrong, and it spat out something that looked different to both my solution and my professor's (although it did provide the same nullspace!). – Babe in the Woods Aug 10 '16 at 23:22
• Oh, he went away.. – Babe in the Woods Aug 10 '16 at 23:22
• Can you provide your professor's solution? – Aweygan Aug 10 '16 at 23:25
• My apologies, my professor's solution has been added in the question. – Babe in the Woods Aug 10 '16 at 23:29
• Your professor simply chose to set $x_4=2\alpha$ instead of $\alpha$ to avoid denominators in the general solution. – Bernard Aug 10 '16 at 23:33

Your solution is correct. You can obtain the professor's solution from yours via the substitution $\beta=\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.
This is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found.


# Counting ways when first position is covered with a tie

Six horses are entered in a race. If two horses are tied for first place and there are no ties among the other four horses, in how many ways can the six horses cross the finish line?

My approach:
We choose $$2$$ out of $$6$$ for the first place and $$4$$ out of the remaining $$4$$:
$$_6P_2\cdot _4P_4 = \frac{6!}{(6 -2)!}\cdot\frac{4!}{(4 - 4)!}=720$$

But the solution states $$360$$.

What am I doing wrong here?

• Also, $\binom 44= \frac{4!}{0!4!}$. But you don't want $\binom 44$ because you care about the order of the remaining $4$ horses. You want the number of ways to permute those $4$ horses, which is simply $4!$. Feb 26 at 23:31
• @RobertShore: Yes the order matters that is why I used $\binom{n}{k}=\frac{n!}{(n-k)!}$ which is $n\cdot(n-1)\cdots\cdot(n-k+1)$. Also it is still $720$
– Jim
Feb 26 at 23:35
• Your values for the combinations are wrong. ${6 \choose 2}=\frac {6!}{4!2!}=15$. You don't want $4 \choose 4$ because the order of the last four horses matters. That gives $15 \cdot 4!=360$ Feb 26 at 23:40
• When you write $6 \choose 2$ you are saying that order does not matter. I don't know a simple one for when order does matter. Some use $_6P_2$ for that. Feb 27 at 0:09
• There are $\binom{6}{2}$ to select the subset of two horses which finish in a tie and $4!$ ways for the remaining four horses to finish in the remaining four positions. Notice that the order of selection for the two horses which finish in a tie does not matter, which is why your answer $P(6, 2)P(4, 4)$ is twice the correct answer. Feb 27 at 23:28

That leaves 4 choices for the horse to come in third, three choices for the horse to come in fourth, two choice for the horse to come in fifth, and one for the horse to come in sixth. The number of ways this can happen is $$15(4)(3)(2)(1)= \frac{6!}{2}= 720/2= 360$$,


# Find the eigenvalues and corresponding eigen vectors of the matrix

Find the eigenvalues and corresponding eigen vectors of the matrix $$\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$$

The eigenvalue $$\lambda_1 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.

The eigenvalue $$\lambda_2 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.

The eigenvalue $$\lambda_3 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.

I'm kind of stuck after a certain point. Here is what I have so far

I do know that $$(A - \lambda I)X = 0$$

so $$\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$$ $$\implies \lambda_1 = -3, \lambda_2 = -1, \lambda_3 = 2$$ so I have the eigenvalues but how can I find the corresponding eigenvectors?

• Do you know how to find eigenvalues? This is an upper triangular matrix. What about it's characteristic polynomial? – Kushal Bhuyan Jul 14 '16 at 1:40
• In any good textbook, class, video lectures or lecture notes on linear algebra, solving linear systems should be covered before finding eigenvalues. In particular, the Gaussian elimination method first puts a linear system into "upper triangular" form, and then finds the unknowns starting from the last ones. – arctic tern Jul 14 '16 at 3:27

You need to solve the equations $(A-\lambda I)v=0$ for $v$ for each of the three eigenvalues $\lambda$.

For instance when $\lambda=2$ we're solving

$$\begin{pmatrix} -5 & 6 & -43 \\ 0 & -3 & 9 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

The last equation is $0=0$ so it's superfluous. So we have two equations in three unknowns. If $z=0$ then the second equation implies $y=0$ and then the first equation implies $x=0$ which gives the zero vector, and that's not very interesting. Otherwise if $z\ne0$, we can scale the vector $(x,y,z)$ (scaling preserves the property of being a solution to the above system) in order to make $z=1$. In which case the second equation gives $y=3$ and then the first equation gives $x=-5$. So the eigenvector is $(-5,3,1)$ up to scaling.

Your turn. Do $\lambda=-3$ and $\lambda=-1$. (Sophia did $\lambda=-3$ for you, so now most of the work is done for you. If we do the problem for you, at least learn what it is we're doing!)

• For $\lambda_2 = -1$ I get $\begin{bmatrix}-2&6&-43\\0&0&9\\0&0&3\end{bmatrix}$ ~ $\begin{bmatrix}1&-3&\frac{43}{2}\\0&0&9\\0&0&0\end{bmatrix}$ so I thought it would be (-3,0,0) ? – Yusha Jul 14 '16 at 3:27
• When I multiply that matrix by $(-3,0,0)^T$ I don't get the zero matrix. In that situation, the last equation is redundant. What does the second equation tell you about $z$? What does the first equation then reduce to saying about $x$ and $y$? (BTW I wouldn't bother with the fractions.) – arctic tern Jul 14 '16 at 3:30
• Wow, I'm an idiot! I see now that $z = 0 \implies 6y = 2x \implies (3,1,0)$ – Yusha Jul 14 '16 at 3:31

You start with the understanding of this formula: $(A-\lambda I)\vec x=0$, which is equivalent to $\det(A-\lambda I)=0$ $$\begin{vmatrix}-3-\lambda&6&-43\\0&-1-\lambda&9\\0&0&2-\lambda\end{vmatrix}=(-3-\lambda)(-1-\lambda)(2-\lambda)=0$$ Therefore, $\lambda_1=-3, \ \lambda_2=-1, \ \lambda_3=2$.
Let's do one example for eigenvectors:
Plug in the value of $\lambda$ into the augmented form of the matrix:
With $\lambda_1=-3$, $$\left[\begin{array}{ccc|c}-3-(-3)&6&-43&0\\0&-1-(-3)&9&0\\0&0&2-(-3)&0\end{array}\right]=\left[\begin{array}{ccc|c}0&6&-43&0\\0&2&9&0\\0&0&5&0\end{array}\right]$$ Solve this matrix and get $v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$
Now you can use similar approach to find the eigenvectors of the next two eigenvalues.

• @Yusha, double check your calculations. At least for $\lambda=-1$, I did not get $(-3,0,0)$ – Ron Jul 14 '16 at 2:18
• I'm lost, how are you getting (1,0,0). You have in your work (0,0,0). – Yusha Jul 14 '16 at 2:43
• Both answers are correct, since one is a scalar multiple of the other. But it is customary to use numbers are small as possible, so $(1,0,0)$ would be prefered. – imranfat Jul 14 '16 at 2:47
• Does (1,0,0) @imranfat come from the 2nd column after its been put in RREF? – Yusha Jul 14 '16 at 2:50
• RREF has to give infinite solutions, I looked at Sophia's work, realizing that both vectors serve as the same eigenvector. Your RREF must have all zero's in bottom row – imranfat Jul 14 '16 at 2:55

Hint: In an upper triangular matrix the characteristic polynomial is $(x_1-\lambda_1)^{\alpha_1}(x_2-\lambda_2)^{\alpha_2}\ldots(x_n-\lambda_n)^{\alpha_n}$, where $x_i$ are diagonal entries with $\alpha$'s as their multiplicities.

In your case the characteristic polynomial is $(3+\lambda)(1+\lambda)(2-\lambda)$

• Here is one: sosmath.com/matrix/eigen2/eigen2.html – imranfat Jul 14 '16 at 2:27
• Khan academy usually rocks: khanacademy.org/math/linear-algebra/alternate-bases/… – imranfat Jul 14 '16 at 2:28
• OK, so I assume that you DO know how to find eigenvalues theoretically, but somehow your algebra is letting you down? It would be helpful to see your work in attempting finding the eigenvalue, because a lot of times it is a small algebraic mistake... – imranfat Jul 14 '16 at 2:33
• Here is a worked out example of a 2 by 2 matrix: calvin.edu/~scofield/courses/m256/materials/eigenstuff.pdf In my view you should first be fluent with 2by2's before going for 3by3's... – imranfat Jul 14 '16 at 2:34
• It sounds like your exchange converged. Anyway, this conversation has been moved to chat. If you need to revisit the full exchange of comments, go to that chatroom. I left the posts with links to resources also here. – Jyrki Lahtonen Jul 14 '16 at 6:26


# By using De Moivre's Theorem, show that $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$ [closed]

First step is

$$\cos5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5$$

thanks

• Because it's easier than using using other trigonometric formulae. Note that the real parts of the expressions on each side of the equation must be equal. – user164587 Sep 26 '14 at 14:39
• @user164587 well i meant how to get it? – problematic Sep 26 '14 at 14:41
• It is de Moivre – Seub Sep 26 '14 at 14:41
• @vera sorry bad english – problematic Sep 26 '14 at 14:41

Hint: Expand $(\cos\theta + i\sin\theta)^5$ using binomial theorem. i.e. $(a+b)^n = \sum\limits_{k = 0}^n {n \choose k}a^k b^{n-k}$
Here, $a = \cos\theta, b=i\sin\theta$.

\begin{align}(\cos\theta+i\sin\theta)^5 &= \sum\limits_{k = 0}^5 {5\choose k}(\cos\theta)^k (i\sin\theta)^{5-k}\\ &= {5\choose0}(\cos\theta)^0(i\sin\theta)^5 + {5\choose1}(\cos\theta)^1(i\sin\theta)^4 + {5\choose2}(\cos\theta)^2(i\sin\theta)^3\\&\quad+ {5\choose3}(\cos\theta)^3(i\sin\theta)^2+{5\choose4}(\cos\theta)^4(i\sin\theta)^1+{5\choose5}(\cos\theta)^5(i\sin\theta)^0 \\&= (i\sin\theta)^5 + 5(\cos\theta)(i^4\sin^4\theta) + 10\cos^2\theta(i^3\sin^3\theta)\\&\quad+10\cos^3\theta(i^2\sin^2\theta)+5\cos^4\theta(i\sin\theta)+\cos^5\theta \\&=\cdots \end{align}

Then, for even powers of $\sin \theta$, use $\sin^2 \theta = 1-\cos^2 \theta$.

Finally, equate the real parts of $\cos(5\theta)+i\sin(5\theta) = (\cos\theta+i\sin\theta)^5$.

• am i doing it right ? lol $(_0^5)$$(cos^0 \theta)$$(i sin^5 \theta)$ – problematic Sep 26 '14 at 15:03
• Yes, and there will be $5$ more terms. – taninamdar Sep 26 '14 at 15:05
• actually i don't know how binomial theoram works, can you give an example and show me what $(_0^5)$$(cos^0 \theta)$$(sin^5 \theta)$ goes to, i will do the rest my self – problematic Sep 26 '14 at 15:06
• See the edit, ask if you need more help. – taninamdar Sep 26 '14 at 15:13
• thank you, how to get the first number? i mean it looks like $(_1^5)$ : 5x1, $(_2^5)$ 5x2 , then $(_3^5)$ onward? – problematic Sep 26 '14 at 15:22

The point is that $$\cos 5 \theta$$ is the even part of $$\exp(5 i \theta) = [\exp(i \theta)]^5 = (\cos \theta + i \sin \theta)^5,$$ where the second equality follows from de Moivre's Theorem. Then, by expanding the right-hand side using the Binomial Theorem, we can immediately write $\cos 5 \theta$ as a sum of products of powers of $\sin \theta$ and $\cos \theta$. Then, we can eliminate even powers of $\sin \theta$ using the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1.$$

• You're welcome, I hope you found it helpful. – Travis Willse Sep 26 '14 at 15:37
• yes, you showed me exp(5iθ)=[exp(iθ)]5=(cosθ+isinθ)5, i needed this to know why we have the first step, i can't choose 2 accepted answers... – problematic Sep 26 '14 at 15:40
• No problem, that's just a part of the site mechanics, the important thing is that you understand the concepts you asked about better than you did before. – Travis Willse Sep 26 '14 at 15:51

De Moivre's Theorem states $$\cos n\theta + \mathbb i \sin n\theta = \left( \cos \theta + \mathbb i \sin \theta \right)^n$$


×

# Proof

Show that, for any positive integer $$n\geq 1$$ it is true that: $\displaystyle \sum_{k=1}^{n} \sqrt{k} \geq \sqrt{\displaystyle \sum_{k=1}^{n} k}.$

Note by Hjalmar Orellana Soto
10 months, 3 weeks ago

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for $$n\geq 2$$ it can be easily proven by applying triangle inequality recursively

assume a,b,c are three terms in the seuence

we have $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ as $$\sqrt{a},\sqrt{b},\sqrt{a+b}$$ form sides of a right triangle

now we have $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{a+b}+\sqrt{c}\geq\sqrt{a+b+c}$$ and so on....

sum of each of the outer sides of the spiral form the LHS of the equation,and RHS is given by the line to centre

also note that for n=1 the line to centre is same as side, the only point of equality

- 10 months, 1 week ago

very nice solution, Anirudh

- 10 months, 1 week ago

thank you :)

- 10 months, 1 week ago

$$LHS^2 = 1+2+\cdots+k + 2\sum_{1\le x<y\le n}\sqrt{xy}\ge 1+2+\cdots+n = RHS^2$$ and equallity occurs when $$n=1$$

- 10 months, 3 weeks ago

In addition to the problem.... Is this necessary information for saying that the limit below converges? $\lim_{x\to \infty} \dfrac{\displaystyle \sqrt{\sum_{n=1}^{x}n}}{\displaystyle \sum_{n=1}^{x}\sqrt{n}}$

- 10 months, 3 weeks ago

(Assuming I'm interpreting you correctly,) the essence of what you're asking here is:

If $$a_n, b_n$$ are sequences of (positive) reals, such that $$a_n \geq b_n$$, does $$\lim \frac{ b_n } { a_n }$$ exist?
If yes, why?
If no, what is a counter example? What other assumptions do we need to add to guarantee a true statement?

Staff - 10 months, 3 weeks ago

Exactly that is what I ask, I was thinking of asking for the value of $\displaystyle \sum_{n=1}^{\infty} (\frac{\sqrt{\sum_{k=1}^{n}n}}{\sum_{k=1}^{n}\sqrt{n}})$ but I really don't know how to approach it

- 10 months, 3 weeks ago

The answer to "does $$\lim \frac{ b_n } { a_n }$$ exist?" is no for general sequences. Do you see an obvious counter example?

For your recent comment, are you looking for the summation, or for the limit of the term (as in the prior comment)?

Staff - 10 months, 3 weeks ago

The answer to the limit question is yes for this case, using the same argument which is used for $$\lim_{x\to \infty} \frac{x}{e^x}$$.... and I'm looking for the summation

- 10 months, 3 weeks ago

See How to ask for help to understand how to provide proper context of what you want. E.g. You can see that people quickly established the inequality, but that doesn't help with your concern.

Staff - 10 months, 3 weeks ago

The summation is something else, something I supposed and now I'm trying to approach, and I think the inequality may help for knowing if the summation converges... and then I'll try to approach the value of the summation, I don't know if I'm explaining right...

- 10 months, 3 weeks ago

Asymptotically, $$\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}$$ can be approximated by the integral $\int_0^n \sqrt{x} \ dx = \frac{2}{3} n^{3/2}.$ And of course, $\sqrt{1 + 2 + \dots + n} = \sqrt{\frac{n(n + 1)}{2}}.$ That should help answer your question.

- 10 months, 3 weeks ago

We want to prove that $\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \ge \sqrt{1 + 2 + \dots + n}.$ Squaring the left-hand side, we get a term of $$\sqrt{k} \cdot \sqrt{k} = k$$ for each $$1 \le k \le n$$. The inequality follows.

- 10 months, 3 weeks ago

×


# On the number of roots of the polynomial $x^3+Ax^2+1=0$

I have the following cubic equation

$$x^{3}+Ax^{2}+1=0$$

where $$A$$ is an arbitrary (real) number.

I know that either:

• The 3 roots will be real.
• One root will be real and the other two will be complex conjugates of each other.

I would like to find out

• For what value/values of A the roots change from 3 real roots to one real and two complex roots.
• The signs of each of the real roots (both when they are all real and when there is only one real root)

Is there an analytical way of finding this as a function of $$A$$ or the only option is to solve the cubic numerically?

• Google Cardano's method. – hamam_Abdallah Jan 6 at 19:44

If we can show that the value of $$A$$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $$y$$-location and the other at a zero or negative $$y$$-location, then we know that this value of $$A$$ is the border for all real solutions versus 1 real and 2 complex solutions.

Denote $$f(x) = x^3 + Ax^2 + 1.$$

$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$

We see that $$f(0)$$ is always either a local maximum or minimum and $$f(0) > 0.$$ The other zero is given by $$x = -2A/3 = \xi.$$

We need $$f(\xi) \leq 0.$$

$$f(\xi) = \frac{4}{27}A^3 + 1 \leq 0$$

solves for the values of $$A$$ where 3 real solutions are guaranteed.

If there are 3 real roots, then since $$x = 0$$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $$A \leq -3/\sqrt[3]{4}$$ and the location of the moving extreme is $$x=-2A/3 > 0,$$ then the remaining real root has to be positive.

If there's only 1 real root, then $$A > -3/\sqrt[3]{4}$$ and the moving extreme is located at $$x < \sqrt[3]{2}.$$ Thus the only real root must be negative.

• How do we find the signs of the real roots? By actually finding them or is there a shortcut? – JennyToy Jan 6 at 20:21
• @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit? – AEngineer Jan 6 at 20:45

hint

Put $$A=-\frac 32B$$ and $$f(x)=x^3-\frac 32Bx^2+1.$$

$$f'(x)=3x(x-B).$$

$$f(0)=1$$ to have three real roots, we need

$$B>0 \text{ and } f(B)<0.$$

• To find when there are 3 real roots or one only use the sign of the discriminant $\Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots. – user2175783 Jan 6 at 20:13
• @user2175783 If $A=-\frac{3}{2^{2/3}}$, then there are a negative zero and a double zero on the positive axis. – Hanno Jan 12 at 22:17

The stationary points of $$f(x)$$ lie at $$x=0$$ and at $$x=-\frac{2A}{3}$$ and the sign of $$f(0)f\left(-\tfrac{2A}{3}\right) = 1+\tfrac{4}{27}A^3$$ decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $$f(0)f\left(-\tfrac{2A}{3}\right)$$ only depends on the stationary values having the same sign or not.

Too late at the party. But it looks good, so let's enter anyway.

TL;DR $$\quad f(x)=x^3+Ax^2+1\,$$ always has a simple zero on the negative axis. The other two zeros are

• real, positive and distinct, if $$\,A<-1.88988\:=\:-\dfrac{3}{\sqrt[3]{4}}\:=\:A_1$$
• merging into a double zero at $$x=\sqrt[3]{2}$$, if $$\,A=A_1$$
• non-real (thus complex-conjugate), if $$\,A>A_1$$

The party-giver also asked if there is an analytical way of finding this as a function of $$A$$?
Let's take a step in this direction, concentrating on the case of "one real zero", and express the negative zero as a function of $$A$$:
Let $$\,d=\big(\frac A3\big)^3 + \big(\frac 12\big)^2$$, and assume $$d>0$$ which corresponds to $$\,A>A_1$$. Define $$r \:=\:\left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\,,$$ $$r\,$$ is observed to be strictly negative. Then \begin{align} r^2\:= & \;\left(-\frac 12 -\sqrt d\right)^\frac 23 +\, \left(-\frac 12 +\sqrt d\right)^\frac 23 -\frac23 A \\[2ex] r^3\:= & \; -\frac23 Ar + \left(-\frac 12 -\sqrt d\right) +\, \left(-\frac 12 +\sqrt d\right) \\ & \quad -\frac13 A \left\{ \left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\right\} \\ = & -Ar - 1 \end{align} Whence $$1+Ar+r^3=0\quad\Longleftrightarrow\quad\frac1{r^3}+A\frac1{r^2}+1 =0 \\[2ex] \Longrightarrow\quad x \:=\: \frac1r \:=\: -r^2-A \:=\: -\frac13 A -\left(\sqrt d + \frac 12\right)^\frac 23 -\, \left(\sqrt d -\frac12\right)^\frac 23$$ satisfies $$x^3+Ax^2+1=0$$.

Added in Edit: Driven by user2175783's comment as of below, I found Lecture 4 in Mathematical Omnibus: 30 Lectures on Classic Maths by D. Fuchs + S. Tabachnikov: It gives an extensive and readable account of how to explicitly solve the cubic and quartic equations.

Visual TL;DR

• Wouldnt it be possible to generalize the algebra for the negative root for $A<A_{1}$? – user2175783 Jan 13 at 0:14
• @user2175783 I did not look closer at that case. I'd say there's no ambiguity for the sign of $\sqrt d$ because $r$ is a symmetrical expression. But afterwards complex cube roots have to be considered which might involve some choice. – Hanno Jan 13 at 9:31

The discriminant of a cubic polynomial with real coefficients is $$0$$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $$-4 A^3 - 27$$.


# Find all triplets (x,y,z)

#### anemone

##### MHB POTW Director
Staff member
Find all triplets of positive integers $(x, y, z)$ such that $$\displaystyle \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.

Last edited:

##### Well-known member
Re: Find all triplets of x, y and z

Find all triplets of positive integers $(x, y, z)$ such that $$\displaystyle \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.
(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them

solved as

Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2
(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

#### anemone

##### MHB POTW Director
Staff member
Re: Find all triplets of x, y and z

(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them

solved as

Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2
(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

Wow! I forgot to thank you explicitly for participating and also for your unique way of attacking the problem. What's worst is that I forgot completely how I approached it two months ago and I promise you to add my reply once I solved it because I could only tell at this point that I solved it differently than what you did. Sorry, kaliprasad!


# [SOLVED]2 coins are thrown 20 times

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail

If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?

Could you give me a hint for (b) ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!!

2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail

If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Hey mathmari !!

Yep.

Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
There seems to be a $5$ missing.
It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.

Could you give me a hint for (b) ?
Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.

#### mathmari

##### Well-known member
MHB Site Helper
There seems to be a $5$ missing.
It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.
Oh yes.. So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$

Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.
So, we have $$P(X\geq 2)=1-P(X<2)=1-P(X=1)-P(X=0)=1-\binom{20}{1}\cdot \left (\frac{1}{4}\right )^1\cdot \left (1-\frac{1}{4}\right )^{20-1}-\binom{20}{0}\cdot \left (\frac{1}{4}\right )^0\cdot \left (1-\frac{1}{4}\right )^{20-0}$$ or not?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yep. All correct.

#### Wilmer

##### In Memoriam
So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$
Looks good.
Ran a simulation: got 202276 out of 1 million.

MHB Site Helper
Thank you!!


# Thread: Problem #8 - Number Theory

1. ## Problem #8 - Number Theory

I got this one from our archives as well.

Show that $n^4 + 4^n$ is not prime for $n \geq 2$. (n is a positive integer!)

This is a more of a Number Theory problem than algebraic. As my proof relies more on a College Algebra level rather than Number Theory I'm curious to see what you might come up with.

-Dan

2. ## Re: Problem #8 - Number Theory

all n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values

3. ## Re: Problem #8 - Number Theory

Originally Posted by muddywaters
all n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
if n is an odd integer >= 3 and n is not a multiple of 5 then n^4 + 4^n is a multiple of 5 and therefore not a prime
Proof:
n is 1,2,3,or 4 (mod 5)
n^4 is 1 (mod 5)
4^n is -1 (mod 5)
So n^4 + 4^n is congruent to 0 (mod 5)

How do we handle the case n=5, 15, 25, 35, 45, ...... ?

4. ## Re: Problem #8 - Number Theory

If $n\equiv 0\pmod{2}$, then $n^4+4^n \equiv 0\pmod{2}$.

If $n\equiv 1\pmod{2}$, then $n+1\equiv 0\pmod{2}$, so $2^{n+1}$ is a perfect square, and $n^4+4^n$ factors as
$(n^2+n\sqrt{2^{n+1}}+2^n)(n^2-n\sqrt{2^{n+1}} + 2^n)$

So, all that is left is to show that for odd $n>2$, both terms are greater than 1. Since $n^2+2^n>1$, the first term is guaranteed to be greater than 1. So, we just need to show the second term is greater than one.

Since $(n^2-1)^2+2^{2n} \ge 2^{2n} > 2^{n+1}$ (for all $n>1$), we have

\begin{align*}n^4-2n^2+1+4^n-2^{n+1} > 0 & \Rightarrow n^4-2n^2+1+4^n-2^{n+1}+n^2 2^{n+1} > n^2 2^{n+1} \\ & \Rightarrow (n^2+2^n-1)^2 > n^2 2^{n+1} \\ & \Rightarrow n^2 + 2^n - 1 > n\sqrt{2^{n+1}} \\ & \Rightarrow n^2-n\sqrt{2^{n+1}}+2^n > 1\end{align*}

5. ## Re: Problem #8 - Number Theory

If n even, obvious.
If n odd, 2n+1 is a perfect square and the factorization are really neat observations. Thanks.

But what's the point of the congruences?

6. ## Re: Problem #8 - Number Theory

Originally Posted by Hartlw
But what's the point of the congruences?
$n\equiv 0\pmod{2}$ is the same as saying $n$ is even.
$n\equiv 1\pmod{2}$ is the same as saying $n$ is odd.

I have been working with modular arithmetic long enough that it feels more natural for me to use congruences rather than the words even and odd. The meaning is the same either way.

7. ## Re: Problem #8 - Number Theory

You have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.

8. ## Re: Problem #8 - Number Theory

Originally Posted by Hartlw
You have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.
If $n=1$, then the factorization I gave is $1^4+4^1 = (5)(1)$. However, 5 is a prime number. The problem asks to prove that for $n\ge 2$, $n^4+4^n$ is NOT prime. So, to do that, it is not enough to show the factors are positive. You must show they are both greater than 1.

9. ## Re: Problem #8 - Number Theory

Hey, lots of action on this one! Thanks to all.

SlipEternal gets some extra kudos for checking the "trivial" factorization problem. (That one of the factors might be 1.) I did eventually find the proof online (which was pretty much identical to Slip's and my own) but didn't address the trivial factorization problem.

-Dan


What is the probability of getting three pairs that makes up seven when six dice are rolled?

Stumbled upon this question in a game of dice where a point is dealt for each pair that makes seven. In the game there were a total of six dice that were to be thrown up to three times. Each time a pair of dice made seven 1+6, 2+5 or 4+3), the pair(s) was taken out from the rest, a point was dealt and you got to roll the remaining dice. You can not use dice from the excluded pairs to make up new ones. You win the game by getting the most points/pairs in 3 rounds.

Example:

round one [1] [1] [1] [3] [5] [6] - one pair make 7 (1+6)

round two [1] [3] [4] [2] - one pair make 7 (3+4)

round one [2] [6] - no pairs make 7

points awarded: 2

Out of curiosity I have been trying to calculate the probabilities of getting three pairs of sevens in the first round?

• We have six dice and we throw them all together in the first round. Let's say the faces are $3,4,5,2,6,6$. Because there are two pairs that make $7$, we remove the pairs and continue with two dice. Round 2 and 3 continues this way. What is the definition of winning? Is it the situation that we're left with no dice at the end? – gunes Nov 10 '19 at 18:18
• and how is it possible to roll three dice following those rules? – carlo Nov 10 '19 at 18:39
• Your question is unclear. Your title refers to three dice while your body text refers to six. You describe a game where the number of dice change, altering the probabilities. At which point in that game are you calculating the chance of getting two dice that add to 7? – Glen_b Nov 10 '19 at 21:02
• Sorry if the original edit was confusing, I have tried to clarify. The game description is just for context. The question is in the title and reformulated in the last paragraph. – joho Nov 13 '19 at 14:28

Consider a sequence of $$2k$$ dice each with the possible values $$1, 2, \ldots, 2n.$$ (The question concerns $$k=n=3.$$) The possible pairs are $$\{1,n\},$$ $$\{2,2n-1\},$$ and so on, through $$\{n,n+1\}.$$ Denote such a pair by its smallest value $$i$$ and for each $$i$$ let $$k_i\gt 0$$ be the number of such pairs that can be located in the sequence. We need to count the number of equally probable sequences for which $$k_1+k_2+\cdots + k_n = k.$$

Such a sequence is determined by (a) which pairs occur in it and (b) where in the sequence of $$2k$$ values each such pair occurs. The permutation group on $$2k$$ elements acts on the set of such sequences. For all $$i,$$ the stabilizer of such a sequence permutes the $$k_i$$ values of $$i$$ among themselves and the $$k_i$$ values of $$2n+1-i$$ among themselves. Thus, applying the method described at https://stats.stackexchange.com/a/415878/919, the number of ways of producing a sequence designated by $$\mathrm{k}=(k_1,k_2,\ldots,k_n)$$ is

$$p(\mathrm{k}) = \frac{(2k)!}{(k_1!)^2(k_2!)^2\cdots(k_n!)^2}.$$

Thus, the chance is obtained by summing $$p(\mathrm{k})$$ over all possible $$\mathrm{k}$$ whose components sum to $$k$$ and dividing by the total number of sequences, $$(2n)^{2k}.$$

These possibilities correspond to the weak compositions of $$k$$ into $$n$$ parts, which number $$\binom{k+n-1}{n-1}.$$ However, the amount of calculation is smaller than this, because $$p(\mathrm{k})$$ does not depend on the order of the $$k_i.$$ We may therefore do the calculation for all $$k_1\ge k_2\ge \cdots \ge k_n \ge 0$$ (giving a partition of $$k$$), multiplying each by the number of distinct re-orderings of $$\mathrm k.$$ Such sequences correspond to the Ferrers diagrams for $$n$$ having at most $$k$$ rows. They are relatively easy to enumerate.

With $$k=n=3,$$ we have $$\binom{3+3-1}{3-1}=10$$ possibilities for $$\mathrm k,$$ but they fall into just three groups corresponding to the partitions $$3 = 2+1 = 1+1+1:$$

$$p(3,0,0)=p(0,3,0)=p(0,0,3) = \frac{6!}{3!^2} = 20;$$ $$p(2,1,0)=p(2,0,1)=p(1,2,0)=p(1,0,2)=p(0,2,1)=p(0,1,2) = \frac{6!}{2!^21!^2}= 180;$$ $$p(1,1,1) = \frac{6!}{1!^21!^21!^2} = 720.$$

$$\Pr(\text{three pairs}) = \frac{3\times 20 + 6\times 180 + 1\times 720}{6^6} = \frac{(5)(31)}{(2^4)(3^5)} \approx 3.9866\%.$$
• @Ron I haven't thoroughly checked my general answer, but I did check the specific answer with a brute force enumeration of all possibilities. At least one mistake in your post is that many sequences without three pairs will sum to 21, such as 5,5,5,4,1,1, which has no pairs at all. Thus you substantially overcount. Here is my implementation: n <- 3; k <- 3; X <- expand.grid(rep(list(seq_len(2*n)), 2*k)); p <- function(x, d=n) { x <- sort(x); sum((x + rev(x))[seq_len(d)] == 2*d+1) }; table(apply(X, 1, p)) – whuber Nov 13 '19 at 19:04
• @Ron If you would like to explore the situation in more detail, this code (which follows the creation of data frame X in my previous comment) breaks the population down into all possible pair patterns: pattern <- function(x, d=n) { x <- sort(x); paste0(x[which((x + rev(x))[seq_len(d)] == 2*d + 1)], collapse="") }; table(apply(X, 1, pattern)) – whuber Nov 13 '19 at 19:19


# Choosing a linear map $(\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ which is nonzero on half of a sequence of vectors

Let $v_1,\ldots,v_m \in (\mathbb{Z}/2\mathbb{Z})^n$ be nonzero vectors. Is it always possible to choose a linear map $f : (\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ such that $f$ is nonzero on at least half of the $v_i$, i.e. such that

$$|\{\text{i | 1 \leq i \leq m and f(v_i) \neq 0}\}| \geq \frac{1}{2}m?$$

My guess is that the answer is yes; at the very least, it is true for $m=1$ and when the $v_i$ enumerate all of the nonzero vectors.

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Why don't you write in words what you want in the body of the question? Desciphering an equation when a perfectly good English sentence might be considerably easier to understand is silly! –  Mariano Suárez-Alvarez May 10 '13 at 3:45
@MarianoSuárez-Alvarez : Sure thing. But isn't that a little pedantic? The equation isn't all that complicated... –  Monica May 10 '13 at 3:47
I don't disagree with @Mariano's comment, but let me just add that this is a very nice question. Where did it come from? –  Pete L. Clark May 10 '13 at 13:33
And I am someone who generally finds contest type problems quite boring (which is not to say that I know how to solve them). Looking back at the comments, Monica got a less warm welcome than she should have: $\mathbb{Z}/2$ could only mean one thing (unlike $\mathbb{Z}_2$!) and upon reflection I do disagree with Mariano's comment. I'm glad it turned out well in the end. –  Pete L. Clark May 10 '13 at 17:10
@PeteL.Clark : Thanks! This experience definitely provides me another data point concerning using gendered names on the internet (I usually don't). It also appears that one can order the Babai book from U of C; see cs.uchicago.edu/research/publications/combinatorics –  Monica May 10 '13 at 17:25

The statement is true and let us prove it by induction on $n$.

Denote $\mathbb{Z}/2\mathbb{Z}:=\mathbb{Z}_2:=\{0,1\}$. When $n=1$, the statement is clearly true. Now assume that when $n\le n_0$, the statement is true for every $m\ge 1$, and let us show the statement is true for $n:=n_0+1$ and every $m\ge 1$.

Consider $(\mathbb{Z}_2)^n$ as $(\mathbb{Z}_2)^{n-1}\times\mathbb{Z}_2$ and denote by $P$ and $Q$ the projections of $(\mathbb{Z}_2)^n$ to its first $n-1$ coordinates and its last coordinate respectively, i.e. $$P: (\mathbb{Z}_2)^n\mapsto (\mathbb{Z}_2)^{n-1},\quad(a_1,\dots,a_{n-1},a_n)\mapsto (a_1,\dots,a_{n-1}),$$ and $$Q: (\mathbb{Z}_2)^n\mapsto \mathbb{Z}_2,\quad(a_1,\dots,a_n)\mapsto a_n.$$

Up to a rearrangement of $v_1,\dots,v_m$, we may assume that there exists $0\le m'\le m$, such that $Qv_i\ne 0$ if and only if $m'<i\le m$. If $m'=0$, simply choose $f=Q$ and we are done, so let us assume that $m'\ge 1$. By definition, $Pv_1,\dots,Pv_{m'}\ne 0$. By the induction assumption on $n$, there exists a linear function $g:(\mathbb{Z}_2)^{n-1}\to \mathbb{Z}_2$, such that $$|\{ 1\le i\le m' \mid g(P v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{1}$$ For $j=0,1$, define $$f_j:(\mathbb{Z}_2)^n\to \mathbb{Z}_2,\quad v\mapsto g(Pv)+j\cdot Qv.\tag{2}$$ By definition, for $j=0,1$, $f_j$ is linear and $f_j(v_i)=g(P v_i)$ when $1\le m\le m'$. Then by $(1)$, for both $j=0$ and $j=1$, $$|\{ 1\le i\le m' \mid f_j(v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{3}$$ If $m'=m$, then we are done. Otherwise, note that by $(2)$ and the choice of $m'$, for $j=0,1$, $$f_j(v_i)=0 \iff g(Pv_i)=1-j\iff f_{1-j}(v_i)=1,\quad \forall\, m'<i\le m.$$ It follows that $$|\{ m'< i\le m \mid f_0(v_i) \neq 0\}|+|\{ m'< i\le m \mid f_1(v_i) \neq 0\}|=m-m'.\tag{4}$$ Combining $(3)$ and $(4)$, we can conclude that the statement is true for either $f=f_0$ or $f=f_1$, which completes the proof.

Remark: Thanks to PeteL.Clark's nice question in the comment, I realized that the conclusion can be generalized to arbitrary finite field with the same argument as above.

Claim: Let $\mathbb{F}_q$ be the finite field of $q$ elements and let $v_1,\dots,v_m$ be nonzero vectors in $(\mathbb{F}_q)^n$. Then there exists a linear function $f:(\mathbb{F}_q)^n\to\mathbb{F}_q$, such that $$|\{1\le i\le m\mid f(v_i)\ne 0\}|\ge\frac{(q-1)m}{q}.\tag{5}$$

Sketch of Proof: Define projections $P:(\mathbb{F}_q)^n\to (\mathbb{F}_q)^{n-1}$, $Q:(\mathbb{F}_q)^n\to\mathbb{F}_q$ similarly. Define $m'$ and rearrange $v_1,\dots,v_m$ similarly. By induction, we may assume that there exists a linear function $g:(\mathbb{F}_q)^{n-1}\to\mathbb{F}_q$, such that for $f_j=g\circ P+j\cdot Q$, $0\le j<q$, $$|\{1\le i\le m'\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)m'}{q}.\tag{6}$$ By definition of $m'$ and $f_j$, for every $m'<i\le m$, $f_0(v_i),\dots, f_{q-1}(v_i)$ are pairwise different, so one and only one of them is $0$. It follows that for some $0\le j<q$, $$|\{m'< i\le m\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}.\tag{7}$$ $(5)$ follows from $(6)$ and $(7)$ for $f=f_j$.

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This is just lovely. –  Kevin Carlson May 10 '13 at 9:00
@KevinCarlson: Thank you! The question itself is lovely. –  23rd May 10 '13 at 9:08
How does this generalize to other finite fields, I wonder? –  Pete L. Clark May 10 '13 at 13:34
@PeteL.Clark: A simple observation based on this argument could show that for finite field $\mathbb{F}_q$, a lower bound is $\frac{(q-1)m}{q}$. Sketch of proof: we can define $f_j$, $0\le j\le q-1$ similarly as in $(2)$. By induction, we can assume that $(3)$ holds when $\frac{m'}{2}$ is replaced by $\frac{(q-1)m'}{q}$. When $i>m'$, $f_j(v_i)$ are pairwise different, so exactly only one of them is $0$. Then we can conclude that there exists $0\le j\le q-1$, such that $|\{m'<i\le m:f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}$, which completes the induction. –  23rd May 10 '13 at 14:01
@PeteL.Clark: Sorry, please ignore my first reply and see the current one. –  23rd May 10 '13 at 14:04

My coauthors and I needed this result recently (to prove something about pseudo-Anosov dilatations), and I was googling to see if it was known. I'm amazed that it was asked here so recently!

I'm answering now to record a short alternative proof that we found. I'll prove the same more general result that Landscape proved, namely:

Claim : Let $\vec{v}_1,\ldots,\vec{v}_m \in \mathbb{F}_q^n$ be nonzero vectors (not necessarily distinct). Then there exists a linear map $f : \mathbb{F}_q^n \rightarrow \mathbb{F}_q$ such that $f(\vec{v}_i) = 1$ for at least $\frac{q-1}{q}$ of the $\vec{v}_i$, i.e. such that $\{\text{$i|1 \leq i \leq m$,$f(\vec{v}_i) \neq 0$}\}$ has cardinality at least $\frac{q-1}{q}m$.

Proof : Let $\Omega$ be the probability space consisting of all linear maps from $\mathbb{F}_q^n \rightarrow \mathbb{F}_q$, each given equal probability. Let $\mathcal{X} : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to the cardinality of the set $\{\text{$i|1 \leq i \leq m$,$f(\vec{v}_i) \neq 0$}\}$. We will prove that the expected value $E(\mathcal{X})$ of $\mathcal{X}$ is $\frac{q-1}{q} m$, which clearly implies that there exists some element $f \in \Omega$ such that $\mathcal{X}(f) \geq \frac{q-1}{q} m$.

To prove the desired claim, for $1 \leq i \leq m$ let $\mathcal{X}_i : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to $1$ if $f(\vec{v}_i) \neq 0$ and to $0$ if $f(\vec{v}_i) = 0$. Viewing $\vec{v}_i$ as an element of the double dual $(\mathbb{F}_q^n)^{\ast \ast}$, the kernel of $\vec{v}_i$ consists of exactly $\frac{1}{q}$ of the elements of $(\mathbb{F}_q^n)^{\ast}$. This implies $E(\mathcal{X}_i) = \frac{q-1}{q}$. Using linearity of expectation (which recall does not require that the random variables be independent), we get that $$E(\mathcal{X}) = E(\sum_{i=1}^m \mathcal{X}_i) = \sum_{i=1}^m E(\mathcal{X}_i) = \sum_{i=1}^m \frac{q-1}{q} = \frac{q-1}{q} m,$$ as desired.

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# How many $1$s are in the first $1023$ binary numbers?

How many $1$s are in the first $1023$ binary numbers?

I'm not to sure how to approach this question. An idea, formula, or solution is appreciated!

• Have you noticed that $1023 = 2^{10}-1$? Feb 26 '17 at 0:20
• What is the "first" binary number? 0 or 1? Feb 26 '17 at 0:22
• Worth clarifying that we are talking about positive, unsigned integers.
– J...
Feb 26 '17 at 2:09
• @J... actually, can an unsigned integer be positive? Feb 26 '17 at 5:56
• @Joffan Cheeky bastard :P Yes, unsigned integers are positive because all of them (other than 0) are greater than 0. Feb 26 '17 at 9:35

Assuming the "first" binary number is $1$ note that the first $1023$ binary numbers, plus $0$, are all the binary numbers you can write with exactly $10$ binary digits or bits (prepending $0$s to "short" numbers, as in $0000101010_2$). Between all of them, you then have $1024 \cdot 10=10240$ bits, and for symmetry reasons exactly half of those, $5120$, are $1$s.

• why 1024*10 digits? Feb 26 '17 at 0:28
• Ohh, the amount of 0's don't matter! I see! Feb 26 '17 at 0:29
• It does matter, because if we start at 0, then 1111111111 should not be counted among the the first 1023 binary numbers. The question did not say the first 1023 positive binary numbers. Feb 26 '17 at 2:23
• In computer science, typically indexing start at 0. I was biased . I should have read the comments below the question... The issue was raised before. Feb 26 '17 at 2:52
• @MarDev We're looking at all of the integers we can represent with 10 bits - for every integer there is exactly one other which has all of its bits inverted (the result of an XOR with 1111111111) and therefore the opposite amount of zeros and ones. Furthermore, this inversion always lies in the opposite half of the possible numbers (the operation is essentially equivalent to 1111111111 - n) so each of the first 2^9 numbers (half of the possible 2^10) there exists exactly one number in the second half with the opposite number of zeros and ones. Feb 26 '17 at 8:46

Hint: For how many of those numbers will the one's bit be a $1$ (in other words: how many of those numbers are odd)? For how many of them will the two's bit be a $1$? For how many of them will the four's bit be a $1$? And so on. Also, it will probably be advantageous to include $0$ (and thus look at a collection of $1024$ binary numbers) to make the counting a bit easier. Or, if $0$ is already included, include $1023$ initially, then correct for it when you're done counting.

• I'm struggling with that idea right now. I'm not entirely sure how to solve that. Is it like half half? So if there are x digits in total, half of them are 1? Feb 26 '17 at 0:27
• On average, yes, half of the bits will be a $1$. Or something. Depends on what you mean by "of the bits". Feb 26 '17 at 0:31

Any such number can be represented by a string of 10 0s and 1s. The number of such strings with $n$ ones is $10$ choose $n$. Thus, the number of ones which appear is $$\sum_{n=0}^{10} n{10 \choose n}=5\cdot 2^{10}.$$

As an addendum to Arthur's answer - if you count to, say, $2^{4} - 1 = 15$, you can easily figure out a pattern in the columns:

0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

So if you count to $2^{n} - 1$, you have $2^{n} * n$ digits ("rows * columns"), of which 50% are 1's.

For your example: Counting to $1023 =2^{10} - 1$, gets you $2^{10} * 10 * \frac{1}{2} = 5120$ binary 1's in total.

It's easy to calculate. There is a pattern with any number used as pow of tow:

You have 2^x, where x> 0 and you will have a binary number as:

2^0 = 1 (Dec) = 1 (Binary)

2^1 = 2 (Dec) = 10 (Binary)

2^2 = 4 (Dec) = 100 (Binary)

...

2^10 = 1024 (Dec) = 100 0000 0000(Binary)

If you pay attention you will notice this:

2^10 - 1 = 1023 (Dec) = 011 1111 1111(Binary)

...

2^2 - 1 = 3 (Dec) = 011 (Binary)

2^1 = 1(Dec) = 01 (Binary)

2^0 - 1 = 0 (Dec) = 0 (Binary)

In this case, your question, the Number of ones matches with the number that is used for pow.


# invertible matrix theorem

Let two inverses of A be B and C Services; Math; Blog ; About; Math Help; Invertible Matrix and It’s Properties. This section consists of a single important theorem containing many equivalent conditions for a matrix to be invertible. Invertible Matrix Theorem. Section 3.5 Matrix Inverses ¶ permalink Objectives. * The determinant of $A$ is nonzero. * $A$ has only nonzero eigenvalues. : An matrix is invertible if and only if has only the solution . structure theorem for completely bounded module maps. Invertible matrix 2 The transpose AT is an invertible matrix (hence rows of A are linearly independent, span Kn, and form a basis of Kn). Showing any of the following about an $n \times n$ matrix $A$ will also show that $A$ is invertible. (If one statement holds, all do; if one statement is false, all are false.) Proof: Let there be a matrix A of order n×n which is invertible. Any nonzero square matrix A is similar to a matrix all diagonal elements of which are nonzero. According to WolframAlpha, the invertible matrix theorem gives a series of equivalent conditions for an n×n square matrix if and only if any and all of the conditions hold. e. The columns of A form a linearly independent set. The Invertible Matrix Theorem Let A be a square n×n matrix. Let A be a square n by n matrix over a field K (for example the field R of real numbers). 1. Understand what it means for a square matrix to be invertible. Then a natural question is when we can solve Ax = y for x 2 Rm; given y 2 Rn (1:1) If A is a square matrix (m = n) and A has an inverse, then (1.1) holds if and only if x = A¡1y. We define invertible matrix and explain many of its properties. 1 The Invertible Matrix Theorem Let A be a square matrix of size N × N. The following statement are equivalent: • A is an invertible matrix. (d)Show that if Q is invertible, then rank(AQ) = rank(A) by applying problem 4(c) to rank(AQ)T. (e)Suppose that B is n … Then the following statements are equivalent. Yes. We will append two more criteria in Section 6.1. An invertible matrix is sometimes referred to as nonsingular or non-degenerate, and are commonly defined using real or complex numbers. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … Theorem 1. 5.The columns of A are linearly independent. The Invertible Matrix Theorem|a small part For an n n matrix A, the following statements are equivalent. A is column-equivalent to the n-by-n identity matrix In. Let A be an n n matrix. A is an invertible matrix. Usually, when a set is written as the span of one vector, it’s one dimensional. Let A 2R n. Then the following statements are equivalent. • The columns of A form a linearly independent set. The matrix A can be expressed as a finite product of elementary matrices. Theorem (The QR Factorization) If A is an mxn matrix with linearly independent columns, then A can be factored as A=QR, where Q is an mxn matrix whose columns form an orthonormal basis for Col A and R is an nxn upper triangular invertible matrix with positive entries on the main diagonal. Theorem1: Unique inverse is possessed by every invertible matrix. The following statements are equivalent, i.e., for any given matrix they are either all true or all false: A is invertible, i.e. Some theorems, such as the Neumann Series representation, not only assure us that a certain matrix is invertible, but give formulas for computing the inverse. I. row reduce to! 4. The uniqueness of the polar decomposition of an invertible matrix. W. Sandburg [8] and Wu and Desoer [ … A has an inverse or is nonsingular. Dave4Math » Linear Algebra » Invertible Matrix and It’s Properties. Skip to content. 6.The linear transformation T deﬁned by T(x) = Ax is one-to-one. Note that finding this matrix B is equivalent to solving a system of equations. The following statements are equivalent: A is invertible, i.e. A has an inverse, is nonsingular, or is nondegenerate. tem with an invertible matrix of coefﬁcients is consistent with a unique solution.Now, we turn our attention to properties of the inverse, and the Fundamental Theorem of Invert- ible Matrices. lie in the commutants of d and 59’. The number 0 is not an eigenvalue of A. A matrix that has no inverse is singular. Here’s the first one. The Invertible Matrix Theorem (Section 2.3, Theorem 8) has many equivalent conditions for a matrix to be invertible. Introduction and Deﬂnition. Furthermore, the following properties hold for an invertible matrix A: • for nonzero scalar k • For any invertible n×n matrices A and B. Learn about invertible transformations, and understand the relationship between invertible matrices and invertible transformations. Menu. A is row-equivalent to the n-by-n identity matrix In. 2 det(A) is non-zero.See previous slide 3 At is invertible.on assignment 1 4 The reduced row echelon form of A is the identity matrix. When the determinant value of square matrix I exactly zero the matrix is singular. 5. (c)Showthatif P isaninvertiblem ×m matrix, thenrank(PA) = rank(A) byapplying problems4(a)and4(b)toeachofPA andP−1(PA). its nullity is zero. • A has N pivot positions. A is invertible.. A .. A is row equivalent to the n×n identity matrix. The Invertible Matrix Theorem has a lot of equivalent statements of it, but let’s just talk about two of them. Matrix I exactly zero the matrix must be a square n by n matrix over a field K for! A $3\times 3$ matrix is detailed along with characterizations the span of one vector, ’! 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