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# Order is Preserved on Positive Reals by Squaring
## Theorem
Let $x, y \in \R: x > 0, y >0$.
Then:
$x < y \iff x^2 < y^2$
## Proof 1
### Necessary Condition
Assume $x < y$.
Then:
$\ds x < y$ $\implies$ $\ds x \times x < x \times y$ Real Number Ordering is Compatible with Multiplication $\ds x < y$ $\implies$ $\ds x \times y < y \times y$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds x^2 < y^2$ Real Number Ordering is Transitive
So:
$x < y \implies x^2 < y^2$
$\Box$
### Sufficient Condition
Assume $x^2 < y^2$.
Then:
$\ds x^2$ $<$ $\ds y^2$ $\ds \leadsto \ \$ $\ds 0$ $<$ $\ds y^2 - x^2$ Real Number Ordering is Compatible with Addition $\ds \leadsto \ \$ $\ds \paren {y - x} \paren {y + x}$ $>$ $\ds 0$ Difference of Two Squares $\ds \leadsto \ \$ $\ds \paren {y - x} \paren {y + x} \paren {y + x}^{-1}$ $>$ $\ds 0 \times \paren {y + x}^{-1}$ as $x + y > 0$ $\ds \leadsto \ \$ $\ds y - x$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds x$ $<$ $\ds y$
So:
$x^2 < y^2 \implies x < y$
$\blacksquare$
## Proof 2
From Real Numbers form Ordered Field, the real numbers form an ordered field.
The result follows from Order of Squares in Ordered Field.
$\blacksquare$
## Proof 3
From Real Numbers form Ordered Field, the real numbers form an ordered field.
By definition, an ordered field is a totally ordered ring without proper zero divisors.
The result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors.
$\blacksquare$
## Proof 4
### Necessary Condition
Let $x < y$.
Then:
$\ds x < y$ $\implies$ $\ds x \times x < x \times y$ Real Number Ordering is Compatible with Multiplication $\ds x < y$ $\implies$ $\ds x \times y < y \times y$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds x^2 < y^2$ Real Number Ordering is Transitive
So:
$x < y \implies x^2 < y^2$
$\Box$
### Sufficient Condition
Let $x^2 < y^2$.
Aiming for a contradiction, suppose $x \ge y$.
Then: | {
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### Sufficient Condition
Let $x^2 < y^2$.
Aiming for a contradiction, suppose $x \ge y$.
Then:
$\ds x \ge y$ $\implies$ $\ds x \times x \ge x \times y$ Real Number Ordering is Compatible with Multiplication $\ds x \ge y$ $\implies$ $\ds x \times y \ge y \times y$ Real Number Ordering is Compatible with Multiplication $\ds$ $\leadsto$ $\ds x^2 \ge y^2$ Real Number Ordering is Transitive
But this contradicts our assertion that $x^2 < y^2$.
Hence by Proof by Contradiction it follows that:
$x < y$
$\blacksquare$ | {
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# prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer
Prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer. For clarity: the denominator is the only part being squared.
My thought process: The numerator is the product of the first n even numbers and the product of the first n odd numbers. That is, $(2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with n! resulting in the following quotient: $$(2^n)(2n-1)(2n-3)/(n!).$$ To me this looks even thanks to the powers of 2. But I am not convinced for some reason.
Sorry for the poor notation, I don't know any coding languages, my apologies.
-
Try counting the powers of 2 in the numerator and denominator. After a quick bit of experimentation, it looks like it's probably sufficient to consider $n=4m+k$ for $k=0,1,2,3$ (i.e. there are four cases) and you could demonstrate that there are always more powers of 2 in the numerator than in the denominator. – Chris Taylor Nov 3 '11 at 9:40
Could you give me more insight as to how you came up with the four cases for n? I don't see the derivation for n as you have shown. – hdtv1104 Nov 3 '11 at 9:50
Did you try using induction? It seems like that shouldn't be too difficult... – Braindead Nov 3 '11 at 13:59
We have $$\displaystyle \frac{(2n)!}{(n!)^2} = \binom{2n}{n}$$ and of course for every way to choose a combination of $n$ objects from a total of $2n$ objects, there exists a complementary choice (the left over $n$ objects). Since the ways to pick $n$ objects from $2n$ comes in pairs, it follows that the total number is even.
Another proof: $$\frac{(2n)!}{(n!)^2} = \frac{ 2n \cdot (2n-1)! }{(n!)^2 } = 2 \cdot \frac{ n \cdot (2n-1)! }{ n\cdot (n-1)! \cdot n!} = 2 \cdot \frac{(2n-1)!}{n! (n-1)!} = 2 \binom{2n-1}{n}$$ | {
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This one can be written in another flavor (using $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ and the symmetric property): $$\frac{(2n)!}{(n!)^2} = \binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n} = 2\binom{2n-1}{n}.$$ Yet another: The sum of the $2n$-th row of Pascal's triangle is $2^{2n}$ which is even, and the sum of all the entries excluding the central coefficient is also even, because of the symmetric property $\displaystyle \binom{n}{k} = \binom{n}{n-k}$. Thus, the central coefficient must also be even.
We can squeeze this problem a bit further and learn something interesting with this fourth proof. This one does not require any knowledge of binomial coefficients. To investigate the prime factors of factorials we use the following identity: $$l = \sum_{k=1}^{\infty} \bigg\lfloor \frac{n}{p^k} \bigg\rfloor$$
where $p$ is a prime number and $l$ is the unique natural number such that $p^l \mid n!$ but $p^{l+1} \nmid n!.$ Basically this counts the multiplicity of the prime factor $p$ in $n!$.
The idea of the proof is this: Write $n! = 1\cdot 2 \cdot 3 \cdots (n-1) \cdot n$, and ask yourself where the factors of $p$ come from. Clearly you get one factor of $p$ from $p, 2p, 3p, \cdots$ so there are $\lfloor n/p \rfloor$ factors of $p$ from these. But that's not the whole story - when we checked over the multiples $p,2p,3p\cdots$ we didn't count the second factor of $p$ every time there was $p^2, 2p^2, 3p^2 \cdots$ so there comes another $\lfloor n/p^2 \rfloor$ to add. Wait, we didn't count another factor every time we missed cubes in $p^3, 2p^3,\cdots$, so we add another $\lfloor n/p^3 \rfloor$ and so on. Also, note that this may look like an infinite series, but eventually $p^k > n$ so all the terms eventually become $0.$ | {
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Anyway, back to the main goal. Using our identity, we see that there are $$\sum_{k=1}^{\infty} \bigg \lfloor \frac{2n}{2^k} \bigg \rfloor - 2 \bigg \lfloor \frac{n}{2^k} \bigg \rfloor$$ factors of $2$ in $\displaystyle \frac{(2n)!}{ (n!)^2}$, and we wish to show that this number is at least $1.$
With some easy casework we can see that $$\lfloor 2x \rfloor - 2 \lfloor x \rfloor = \left\{ \begin{array}{lr} 1 & : \{ x\} \geq \frac{1}{2} \\ 0 & : \{ x\} < \frac{1}{2} \end{array} \right.$$ where $\{ x \}$ denotes the fractional part of $x.$ Thus our problem is reduced to showing that there exists some $k\in \mathbb{N}$ such that $\displaystyle \frac{n}{2^k}$ has fractional part greater than or equal to $1/2.$ This is of course true, since if $m$ is the largest positive integer such that $\displaystyle \frac{n}{2^m} \geq 1$ then $\displaystyle \frac{1}{2} \leq \frac{n}{2^{m+1} } < 1.$ Hence, $\displaystyle \frac{(2n)!}{(n!)^2}$ is even.
-
Thanks for the combinatorics response to this proof. I see that it is easier to invoke the combinatorial interpretation but I myself could not see this interpretation as readily. What inspired the use of binomial? – hdtv1104 Nov 3 '11 at 9:53
Just a bit of practice I guess. Every time I see a ratio of factorials, I check whether it fits the form $\frac{n!}{k! (n-k)!} = \binom{n}{k}.$ Also, it was particularly easy for me here because this particular coefficient, $\binom{2n}{n}$, is quite special and comes up relatively often. – Ragib Zaman Nov 3 '11 at 9:55
Thanks a lot Ragib. – hdtv1104 Nov 3 '11 at 9:58
You're welcome! – Ragib Zaman Nov 3 '11 at 10:00
In fact, the number of factors of $2$ in $\binom{2n}{n}$ is equal to the number of $1$ bits in the binary representation of $n$. – robjohn Nov 14 '11 at 21:24
It is often fun to try to solve such problems using Lagrange's Theorem from finite group theory. That is, to prove that $a$ divides $b$ you try to find a group of cardinality $b$ with a subgroup of cardinality $a$. | {
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In this case, the symmetric group on $2n$ letters, $S_{2n}$, is an obvious choice for a group of cardinality $(2n)!$. Let $G\subset S_{2n}$ be the set of all permutations which map the set $\{1,\ldots, n\}$ onto $\{1,\ldots, n\}$ or $\{n+1,\ldots, 2n\}$. You can check that $G$ is a subgroup of $S_{2n}$.
To compute the cardinality $\lvert G \rvert$, one option is to observe that $G\cong S_n\wr \mathbb{Z}_2$, so $\lvert G\rvert = \lvert S_n\rvert^{\lvert \mathbb{Z}_2\rvert} \lvert \mathbb{Z}_2\rvert = 2(n!)^2$. Alternatively, choosing an element of $G$ corresponds to choosing two permutations of a set of size $n$ and a choice of whether to swap $\{1,\ldots,n\}$ with $\{n+1,\ldots, 2n\}$ or not, so $\lvert G\rvert = 2(n!)^2$.
Thus by Lagrange's Theorem $2(n!)^2$ divides $(2n)!$, or in other words, $\frac{(2n)!}{(n!)^2}$ is even.
-
By the binomial theorem, $$4^n=(1+1)^{2n}=\sum\limits_{k=0}^{2n}{2n\choose k}={2n\choose n}+\sum\limits_{k=0}^{n-1}{2n\choose k}+\sum\limits_{k=n+1}^{2n}{2n\choose k}.$$ Since Pascal's triangle is symmetrical, each term in the sum from $k=0$ to $n-1$ coincides with one term in the sum from $k=n+1$ to $2n$, hence these two sums are equal and $$\frac{(2n)!}{(n!)^2}={2n\choose n}=4^n-2\cdot\sum\limits_{k=0}^{n-1}{2n\choose k}$$ is even for every $n\geqslant1$.
-
I had this one in my answer =) But not as fleshed out. – Ragib Zaman Nov 14 '11 at 21:02
Oh yes, I missed it but I see it now. Sorry about that. – Did Nov 14 '11 at 21:08
Let's observe an example:
$(2\cdot 5)!=10\cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!$ , we may conclude that:
$(2n)!=2n\cdot(2n-1)\cdot(2n-2)....(n+1)n!$ so we get following expression :
$$\frac{2n\cdot(2n-1)\cdot(2n-2)....(n+1)}{n!}=\frac{2n\cdot(2n-1)\cdot2\cdot(n-1)\cdot(2n-3)\cdot 2 \cdot (n-2)....(n+1)}{n!}$$
$$=\frac{2\cdot2\cdot2...\cdot2\cdot n\cdot(n-1)\cdot(n-2)\cdot......\cdot 1\cdot(2n-1)\cdot(2n-3)\cdot......(n+1)}{n!}$$ | {
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$$=\frac{2\cdot2\cdot2...\cdot2 \cdot n!\cdot(2n-1)\cdot(2n-3)\cdot......\cdot (n+1)}{n!}=$$
$$=2\cdot2\cdot2...\cdot2 \cdot(2n-1)\cdot(2n-3)\cdot......\cdot (n+1)$$ so number is even.
-
But I'm sorry, this proof is false (still got 11 upvotes). – user144248 May 15 '14 at 21:32
The error here is at the third equality: The terms that have been divided by $2$ don't go all the way down to $1$, since there are only (about) $n/2$ such terms. In fact, for $n$ even, the last term that is of this form is $n + 2$, so the correct expression would be $$= \frac{2 \cdots 2 \cdot n \cdot (n - 1) \cdot ... \cdot \frac{n + 2}{2} \cdot (2n - 1) \cdots (n + 1)}{n!}$$ – user61527 May 16 '14 at 4:49
This answer has been moved from this question, which was closed because it was judged to be a duplicate of this question.
The number of factors of a prime $p$ in $n!$ is $$\nu_p(n)=\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\dots\tag{1}$$ $\left\lfloor\frac{n}{p}\right\rfloor$ counts the number of positive multiples of $p$ no greater than $n$, but only once. $\left\lfloor\frac{n}{p^2}\right\rfloor$ counts the multiples of $p^2$ a second time, and $\left\lfloor\frac{n}{p^3}\right\rfloor$ the positive multiples of $p^3$ a third time, etc. Applying $(1)$ to the base-$p$ representation of $n$ $$n=\sum_{i=0}^kn_ip^i\tag{2}$$ where $0\le n_i<p$, yields \begin{align} \nu_p(n) &=\sum_{i=0}^kn_i\left(1+p+p^2+\dots+p^{i-1}\right)\\ &=\sum_{i=0}^kn_i\left(\frac{p^i-1}{p-1}\right)\\ &=\frac{1}{p-1}\left(n-\sum_{i=0}^kn_i\right)\\ &=\frac{n-\sigma_p(n)}{p-1}\tag{3} \end{align} where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$. | {
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Therefore, the number of factors $p$ in $\displaystyle\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is \begin{align} &\nu_p(n)-\nu_p(k)-\nu_p(n-k)\\ &=\frac{1}{p-1}\left(n-\sigma_p(n)-k+\sigma_p(k)-(n-k)+\sigma(n-k)\right)\\ &=\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{4} \end{align} Using $(4)$ with $p=2$ says that the number of factors of $2$ in $\displaystyle\binom{2n}{n}$ is $\sigma_2(n)+\sigma_2(n)-\sigma_2(2n)$ which is simply $\sigma_2(n)$ since $\sigma_2(n)=\sigma_2(2n)$. Therefore, the number of factors of $2$ in $\displaystyle\binom{2n}{n}$ is the number of $1$-bits in the binary representation of $n$.
Thus, if $n>0$, then there is at least one $1$-bit in $n$ in binary, and so $2$ divides $\displaystyle\binom{2n}{n}$.
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The exact power of $2$ dividing $\binom{2n}{n}$ is $2$^(sum of digits in base $2$ representation of $n$).
The exact power of $3$ dividing the trinomial coefficient $\binom{3n}{n,n,n} = (3n)!/(n!)^3$ is $3$^(sum of digits in base $3$ representation of $n$).
The exact power of $p$ dividing the $p$-nomial coefficient $\binom{pn}{n,n,\dots , n} = (pn)!/(n!)^p$ is $p$^(sum of digits in base $p$ representation of $n$) for any prime $p$.
This is from the formula for the power of $p$ dividing $n!$, which is ($n$ - sum of digits of $n$ when written in base $p$)/($p-1$). There also are combinatorial and generating function proofs but they are more complicated.
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Math Help - Finding a Tangent Line without Calculus
1. Finding a Tangent Line without Calculus
I hope this is in the right place, I'm not in a hurry, just curious.
How can I find an equation for a line tangent to a point on a parabola without using calculus?
I just started playing with this this morning
The equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
Using calculus I found the equation to be y = 4x -18
How can I do this without calculus?
2. Hello, Ranger SVO!
You're curious . . . good for you!
While playing with parabolas in college (centuries ago),
. . I "discovered" some facts that might help you.
The equation I'm using is $y \:= \:x^2 - 4x - 2$
and I'm looking for the equation of the tangent line at point $(4,\,-2)$
Using calculus I found the equation to be: $y \:= \:4x -18$
How can I do this without calculus?
Code:
* | */
| /
* | *P(x,y)
* | */
---------***---o----------
| /↑
| /(½x,0)
| /
|/
o ←(0,-y)
/|
Given the parabola: . $y \:=\:ax^2$, the tangent at point $P(x,y)$
. . has an x-intercept of $\left(\frac{1}{2}x,\,0\right)$ and a y-intercept of $(0,\,-y)$
The tangent at $P(x,y)$ always has a point directly below the vertex
. . with the same vertical displacement (only downward).
Your parabola $y \:=\:x^2-4x-2$ has its vertex at $(2,-6).$
Your point $P(4,-2)$ is 4 units above the level of the vertex.
Hence, it has a point 4 units directly below the vertex: $Q(2,-10).$
Now you can write the equation through $P$ and $Q.$
3. Originally Posted by Ranger SVO
I hope this is in the right place, I'm not in a hurry, just curious.
How can I find an equation for a line tangent to a point on a parabola without using calculus?
I just started playing with this this morning
The equation I'm using is y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
Using calculus I found the equation to be y = 4x -18 | {
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Using calculus I found the equation to be y = 4x -18
How can I do this without calculus?
I have another approach.
It does not always work.
But it does sometimes.
In your example, the tangent line is the line that only intersects the parabola at one point. Thus, you need to find an equation of a line that one intersects at one point, i.e. solution set is only one element. That means you need the discrimanant to be zero.
4. Soroban, I like your explination. But first, at my age curiousity is the only thing that keeps me from vegetating.
Here is what I've been working on. If I draw a rectangle using the vertex and the point where I want the tagent line, I can see the secant line bisecting the rectangle. I can easily notice that the slope of the tangent is twice that of the secant.
Is this always true? If so why?
Perfect Hacker, does that mean that I can use the quadratic equation to find the slope of a line tangent to some point.
5. Hello again, Ranger SVO!
Here is what I've been working on.
If I draw a rectangle using the vertex and the point where I want the tangent line,
I can see the secant line bisecting the rectangle.
I can easily notice that the slope of the tangent is twice that of the secant.
Is this always true? If so why?
Given the parabola: $y \:=\:ax^2$ and any point on it: $P(p,\,ap^2)$
The slope of the secant $OP$ is: . $m_s \:=\:\frac{ap^2 - 0}{p - 0} \:=\:ap$
The derivative is: . $y' \:=\:2ax$
. . Hence, the slope of the tangent at $P$ is: $m_t\:=\:2ap$
Therefore, the slope of the tangent is always twice the slope of the secant.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you're really bored, verify these claims.
1) On the parabola $y \:=\:ax^2$, pick any two points: . $P(p,\,ap^2)$ and $Q(q,\,aq^2)$
Draw the chord $PQ.$
Locate the tangent parallel to $PQ.$
It is found at: . $x \:=\:\frac{p+q}{2}$ . . . halfway between $p$ and $q.$ | {
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It is found at: . $x \:=\:\frac{p+q}{2}$ . . . halfway between $p$ and $q.$
2) On the parabola $y \:=\:ax^2$, pick any three points: . $P(p,\,ap^2),\;Q(q,\,aq^2),\;R(r,\,ar^2)$
Find the area of $\Delta PQR.$
Consider the tangents to the parabola at $P,\,Q,\,R.$
. . They will intersect at $A,\,B,\,C.$
Find the area of $\Delta ABC.$
We find that: . $(\text{area }\Delta ABC) \:=\:\frac{1}{2}(\text{area }\Delta PQR)$
6. y = x^2 - 4x - 2 and I'm looking for the equation of the tangent line at point ( 4, -2)
We want a tangent line at $(4,-2)$.
To do that I assume you are familar with the point-slope formula:
$y-y_0=m(x-x_0)$
In this case,
$(x_0,y_0)=(4,-2)$.
Thus, the equation of the line (tangent) at that point is determined by,
$y+3=m(x-4)$ that means ( $y=m(x-4)-3$)
Where,
$m$
Is the slope to be determined.
As I said we need to intersection to be a single point.
That is,
$\left\{ \begin{array}{c}y=x^2-4x-2\\y=m(x-4)-3 \end{array} \right\}$
We want this to has a single solution.
Equate,
$x^2-4x-2=m(x-4)-3$
$x^2-4x-2=mx-4m-3$
$x^2+x(-4-m)+(1+4m)=0$
To have a single real solution we need the discrimanant to be zero,
$(-4-m)^2-4(1)(1+4m)=0$
Solve to get the value of $m$. | {
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# Cambridge Vector Calculus Lecture Notes | {
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Publisher: Dalhousie University 2007 Number of pages: 106. 1 Vector calculus According to classical physics, The vector triple product,. Contents 1. These can be proven with a calculation, but what was more interesting is that the converse. They can be found here. Actually, this text also discusses integration and vector calculus (Chapter 10), but I personally found Rudin's treatment of such hard to follow when I was first learning the subject. Therefore the normal vector we look for is. For example we will see that the existence of partial derivatives does not guarantee that the function itself is continuous (as it is the case for a function of one variable). Unfortunatelythenumberof. Scalar and vector fields. Description. Math 53 - Multivariable Calculus -- [4 units] Course Format: Three hours of lecture and three hours of discussion per week. Lecture Notes: 178 kb: Module-5 Linear and Quadratic Approximations,Newton and Picard Methods: Lecture 15 : Newton\\\'s method: Lecture Notes: 184 kb: Module-6 Definition of Integral: Lecture 16 : Integral from upper and lower sums: Lecture Notes: 330 kb: Module-6 Definition of Integral: Lecture 17 : Fundamental theorem of calculus: Lecture. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. Vector Calculus; Line Integrals; Computational Vector Analysis Topics; Stoke’s and Green’s Theorems; The Joy of X" by Steven Strogatz; Laplace Transforms; Series: Power and Fourier; Partial Differential Equations; Numerical Methods; Topics in Engineering Mathematics; Computational Fluid Dynamics. pdf Lecture notes. From E→ to Φ. Index of Math Terms. Vector calculus identities explained. in Calculus. • how vector fields relate to certain physical phenomena, and how to carry out calculations using vector operations and vector identities, • how to evaluate line and surface integrals and to interpret these concepts into physical applications, • how to use the Divergence Theorem, Green, and Stokes to evaluate. | {
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into physical applications, • how to use the Divergence Theorem, Green, and Stokes to evaluate. The main focus of this module is on multivariable calculus in 2 and 3 dimensions, and vector calculus. Cambridge Notes Below are the notes I took during lectures in Cambridge, as well as the example sheets. , please let me know by e-mail. Higher derivatives and product rules 128 54. https://ocw. Vector geometry / Gilbert de B. Course Format: There are four 50-minute lectures each week. That weight function is commonly the arc length of the curve, or—if you’re integrating over a vector field—the scalar product with a vector differential in the curve. A fraction of the price of most calculus books. than 10 dimensions. The way we did in class was to observe that this rectangular region is vertical to the -plane. Notes are applicalicable for both 1st and 2nd sem students of CBCS scheme. Operator notation Gradient. Andrew Steane's Lecture Courses. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Scalar Product: The "dot" product of two vectors is a scalar. MATH 226 (Calculus III) Fall 2016 39571R (MWF at noon in SOS B44) Final exam is Wednesday, December 7, 2-4 pm. Applied Geometry for Computer Graphics and ms excel 2007 user guide pdf CAD D. These notes, in my view, can be used as a short reference for an introductory course on tensor algebra and calculus. May 17, 2020 - Lecture Notes - Review of Vector Calculus Notes | EduRev is made by best teachers of. They consist largely of the material presented during the lectures, though we have taken the liberty of eshing them out in some places and of being more cursory here than in the lectures in other places. Applied Advanced Calculus Lecture Notes by Jan Vrbik. The first lecture of Cosmology II is on Monday, Oct 28th, and the first exercise session is on Friday, Nov 8th. 2) F (x; y z) = P x y z I + Q x y z J R x y z K: For example, the vector field (18. | {
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Friday, Nov 8th. 2) F (x; y z) = P x y z I + Q x y z J R x y z K: For example, the vector field (18. Chapter 18 Vector Calculus 282 x 18. called the Euclidean metric. In this lesson, we will use the Calculus Shell Method to find the volume of a solid of revolution. The notes are adapted to the structure of the course, which stretches over 9 weeks. Advanced Integer Programming, Advanced Linear Programming, Advanced Nonlinear Programming, Calculus I, Vector Calculus, and Differential Equations, Calculus II. Vector Calculus: This lecture Contain problem of GREEN’S THEOREM PROBLEM-1 (pdf link notes is available in this description) Find Online Engineering Mathematics Online Solutions Here. MATH 1920 Lecture Notes - Lecture 31: Vector Calculus, Divergence Theorem, 2D Computer Graphics. The magnitude of c~usatis es jc~uj= jcjj~uj. This document is a sketch of what occurs in lecture. Please do not share or redistribute these notes without permission. and differential calc. Welcome! Please have a look at the syllabus for the class, which has all of the essential information for the class. In ordinary differential and integral calculus, you have already seen how derivatives and integrals interrelate. Its most recent edition, the fourth, was published in 2013 by Pearson and in 2017 by Cambridge University Press. Calculus and Linear Algebra II. Official Class Description from Campusnet. Notes: Some of these lectures reference the TI-89 graphing calculator. " " Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics. We will define line integrals, which can be used to find work done by a force field in moving an object along a curve. These course notes are intended for students of all TU/e departments that wish to learn the basics of tensor calculus and differential geometry. The concepts of vector algebra are covered in the first 100+ pages in great clarity. ) that make it easy to talk about volumes rather than just vectors. in lectures in the rst year, e. This | {
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make it easy to talk about volumes rather than just vectors. in lectures in the rst year, e. This course is about vector calculus, and covers material that all engineers should know. This page contains the animations I have created for this course. • how vector fields relate to certain physical phenomena, and how to carry out calculations using vector operations and vector identities, • how to evaluate line and surface integrals and to interpret these concepts into physical applications, • how to use the Divergence Theorem, Green, and Stokes to evaluate. While some of the pages are proofread pretty well over the years, others were written just the night before class. 15 pm Class location: ENG1 0286 Office hours: Tuesday/Thursday 4. Dörrzapf, who lectured the course in 2005 and 2006. KAZDAN Harvard University Lecture Notes. Homework problems are from Marsden and Tromba, "Vector Calculus, 6th Edition. 1 Definition of nDimensional. the matrix calculus is relatively simply while the matrix algebra and matrix arithmetic is messy and more involved. Inan Prentice Hall, 1999. Download free VTU Notes in pdf format. Course Description. 1 Vector Fields 15. leaves on a river). Lecture 3 (January 29): Gradient descent, stochastic gradient descent, and the perceptron learning algorithm. Derivatives (1) To work with derivatives you have to know what a limit is, but to motivate why we are going to study limits let’s rst look at the two classical problems that gave rise to the notion of a derivative: the tangent to a curve, and the instantaneous velocity of a moving object. NO Lecture due to Exam II; Class time will be converted to optional Office hours, to review home work solutions and discuss exam Part IV: Boundary value problems (5 Lectures) 1/16. Fluid Mechanics Richard Fitzpatrick Professor of Physics The University of Texas at Austin. Content Engineering Mathematics 2 ma8251 Unit 2 Vector Calculus. MP201 { Vector Calculus & Fourier Analysis Problem Set 6 Due by 5pm on Friday, | {
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2 Vector Calculus. MP201 { Vector Calculus & Fourier Analysis Problem Set 6 Due by 5pm on Friday, 10 November 2017 (Please write your name and tutorial day on the front of your assignment. The course is divided into four modules. Lecture Notes for EE261 – The Fourier Transform and its Applications. 1 Vector fields 16. This note explains following topics: Ordinary Differential Equations, First-Order Differential Equations, Second Order Differential Equations, Third and Higher-Order Linear ODEs, Sets of Linear, First-Order, Constant-Coefficient ODEs,Power-Series Solution, Vector Analysis, Complex Analysis, Complex Analysis, Complex Functions. In this book Saint-Venant, a convinced atomist, presented forces as divorced from the metaphysical concept of cause and from the physiological concept of muscular effort. THE THREE STOOGES: DIV, GRAD, AND CURL You might remember the following theorems from vector calc. 15 Vector Calculus 15. 3 (Applications of Second-Order Equations). Curves in R3. Moiola, University of Reading 2 Vector calculus lecture notes, 2016-17 1 Fields and vector differential operators For simplicity, in these notes we only consider the 3-dimensional Euclidean space R3, and, from time to time, the plane R2. These course notes are intended for students of all TU/e departments that wish to learn the basics of tensor calculus and differential geometry. It is useful to think. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples Details on obtaining and updating the source of DAMTP examples (this is aimed at DAMTP Unix account holders only), and the list of course codes and titles referred to in these pages. Neville Harnew: Lecture Material. Sc Mathematics Notes of Calculus with Analytic Geometry Notes of Calculus with Analytic Geometry. and differential calc. After each video lecture a) redo the examples done in lecture b) do the applicable homework assigbment. edu office: APM 5256, Office hours: MW: | {
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done in lecture b) do the applicable homework assigbment. edu office: APM 5256, Office hours: MW: 3:30-4:30 and by appointment (just talk to me after class, or email) Teaching assistants: Tianhao Wang, email: [email protected] Apply the mathematical skills required in problem-solving related to vectors, vector calculus, multiple integration, and partial derivatives. This course will be based on a series of lecture notes which will be posted regularly throughout the semester. Press, Cambridge (2007) Click for. Sloughter). Curves in R3. Vector Calculus - Winter 2019 Lectures: MWF 2-3 in Pepper Canyon Hall 109 Instructor:Hans Wenzl email: [email protected] In vector calculus, the derivative of a vector function y with respect to a vector x whose components represent a space is known as the pushforward (or differential), or the Jacobian matrix. This primitive concept, familiar from undergraduate physics and mathematics, applies equally in general relativity. Problems set for the 6th edition. Chapter 12: Vectors and Geometry of Space. SC 107: Calculus, Where are you ? Fall 2016 "If you give a child a fish you feed him for a day, if you teach him fishing you feed him for a lifetime. A vector can also be represented mathematically in the form of an equation: v = vxˆ. redistributed. Willard Gibbs (1839--1903) for his students at Yale University. Calculus mainly involves differentiation and integration Differential Calculus Integral Calculus 𝑑𝑑=𝑓𝑓 𝑑𝑑𝑓𝑓 𝑑𝑑𝑥𝑥. Lecture Notes for Section 14. These notes, in my view, can be used as a short reference for an introductory course on tensor algebra and calculus. Cambridge Notes Below are the notes I took during lectures in Cambridge, as well as the example sheets. Learning vector calculus techniques is one of the major missions to be accomplished by physics undergraduates. : (0711) 685-66346. Acceleration and Force 131 56. Corpus ID: 63957860. 2 Notes • Homework # 1 is uploaded on the course webpage Vector Calculus. 5-4 Lecture 5: | {
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63957860. 2 Notes • Homework # 1 is uploaded on the course webpage Vector Calculus. 5-4 Lecture 5: Trajectory Optimization 5. Willard Gibbs Josiah Willard Gibbs , Edwin Bidwell Wilson Yale University Press , 1901 - Vector analysis - 436 pages. Welcome To the Vector Calculus (Math 202) Home Page (These lecture notes are from Fall 2002) Click below for the course syllabus and lecture notes: Example Course Syllabus. Between points k 1 and kyou had the estimate of the arc length as p (x k)2 + (y k)2, but here you need the whole vector from ~r k 1 to ~r kin order to evaluate. Andre Lukas Oxford, 2013 3. VECTOR ALGEBRA r r 2 3 O r1. Multivariable Calculus, Fall 2018 (ASU MAT 267): Multivariable Calculus part 1, 18pp: Vectors and 3-Dimensional Geometry. It pro vides a way to describe physical quantities in three-dimensional space and the way in which these quantities vary. 1(a) The Vector Differential Operator. Classical Electromagnetism: An intermediate level course Richard Fitzpatrick Associate Professor of Physics The University of Texas at Austin. Brief notes for 268. MAE5201 - Solid Mechanics Course Notes About These notes are for the personal use of students who are enrolled in or have taken MAE5201 at the University of Colorado Colorado Springs in the Spring 2017 semester. There is no need for parametric equations. Given on September 4 th. Verify this result when S is the sphere jx j = R and A = ( z; 0;0) in Cartesian coordinates. 3 Independence of Path 15. [email protected] While some of the pages are proofread pretty well over the years, others were written just the night before class. The content may be incomplete. Course Notes and General Information Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. Description: Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak | {
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theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong normalization, type inference, denotational semantics, complete partial. Calculus III is the last course in the Calculus I, II, III sequence. PHY2060) and at least have co-registered in a vector. Andre Lukas Oxford, 2013 3. Notes in analysis on metric and Banach spaces with a twist of topology. : In V3, 3 non-coplanar vectors are linearly independent; i. Five lectures for undergraduates on general relativity, by Jorge. MACLACHLAN MURPHY, IAN S. When applied to a function defined on a one-dimensional domain, it denotes its standard derivative as defined in calculus. Multiple Integrals and Vector Calculus - Lecture Notes. Course Description. It is useful to think. pdf file) viewer can be obtained from Adobe Once the Acrobat plugin has been downloaded and installed, file can be view simply by clicking on the corresponding link. Scalar multiplication: If c2R and ~ua vector, then we may form a new vector c~ucalled the scalar product of ~uwith c. are expected to attend all of the lectures. Here is an unordered list of online mathematics books, textbooks, monographs, lecture notes, and other mathematics related documents freely available on the web. Precise Definition of Limit 6. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. This book uses SI units (the mks convention) exclusively. AUTUMN 2012 Lectures TR 1:00-2:00, F 2:00-3:00, 26-100 Instructor James McKernan, 2-274, phone 253-4391, [email protected] Don't show me this again. Vector Calculus - Free download as Powerpoint Presentation (. Textbook: Calculus. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. Lecture Notes Labs Assignments Download | {
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O N araigh Lecture notes in Vector Calculus, September 2017. Lecture Notes Labs Assignments Download Course Materials; Users may find additional or updated materials at Professor Carter's 3. Boqing Gong (Deep Neural Networks for Computer Vision Applications) Lecture 19: Deformable Models and Image Segmentation. ) Lecture notes by Giovanni Leoni. The main focus of this module is on multivariable calculus in 2 and 3 dimensions, and vector calculus. CMS College (Autonomous) The CMS College, Kottayam, founded by the Church Missionary Society of England, is one of the oldest institutions of. Vector Calculus Lecture Notes Adolfo J. 1: Vector function of one variable-- a vector, each component of which is a – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. 2 Spans, Lines and Planes The span of a single vector vin Rnis the set of all scalar multiples of v: spanfvg= ftvjt2Rg:. Lecture 4: Wrap up of vector calculus: Poisson & Laplace equations; curl 4. Whittaker, G. There are 9 chapters, each of a size that it should be possible to cover in one week. The unique vector of length zero is denoted ~0 and satis es ~0 +~v= ~v. IV Vector Calculus In Many Variables. This unit is offered in Semester 1. Study With Me - Probability, Vector Calculus, Analysis and more Last minute Vector Calculus and other flo's study diary 12,748 views. Probability About these notes. nagar - 2018 Material offline reading, offline notes, free download in App, Engineering Class handwritten notes, exam notes, previous year questions, PDF free download. Here are a set of practice problems for the Vectors chapter of the Calculus II notes. A lot of these gaps will be filled in in this course, and more material will be covered. • The inverse of a matrix is defined such that when it operates on the original matrix, the result is the unit matrix. Student Evaluations (Fall 2019 151) Student Evaluations (Fall 2019 152) Syllabus. Since the course is an experimental one and the | {
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151) Student Evaluations (Fall 2019 152) Syllabus. Since the course is an experimental one and the notes written before the lectures are delivered, there will inevitably be some sloppiness, disorganization, Our subject matter is intermediate calculus and linear. CM111A - Calculus I Compact Lecture Notes ACC Coolen Department of Mathematics, King's College London Version of Sept 2011. Multiple Integrals and Vector Calculus Prof. (14259 views) A Course of Modern Analysis by E. In this course, you'll learn how to quantify such change with calculus on vector fields. We will define line integrals, which can be used to find work done by a force field in moving an object along a curve. If v is a smooth vector field on R3, divcurlv = 0. 4 Mar Review for test March 23 Test 2 March 27 4. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. Tensors revision questions. Watch out for typos!. Vector Products, Triple Scalar Products. ) Lecture Notes on Multivariable Calculus by Barbara Niethammer and Andrew Dancer. Vector Calculus (Green's Theorem, Stokes' Theorem, Divergence Theorem) For differentiation, you can use Principles of Mathematical Analysis by Rudin (Chapter 9). ; Author: James Byrnie Shaw; Category: Calculus; Length: 325 Pages; Year: 1922. Students should read the textbook before each class and review lecture notes after each class. Description. Calculus Book with Video Lecture Preliminaries, Limits and Continuity, Differentiation, Applications of Derivatives, Integration, Applications of Definite Integrals, Transcendental Functions, Techniques of Integration, Further Applications of Integration, Conic Sections and Polar Coordinates, Infinite Sequences and Series, Vectors and the Geometry of Space, Vector-Valued Functions and Motion. 2) Divergent of a constant vector is always zero Ex: then. Section 1-8 : Tangent, Normal and Binormal Vectors. Vector Calculus deals with calculus in two and three dimensions, and develops the theory of curves, | {
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Vector Calculus deals with calculus in two and three dimensions, and develops the theory of curves, vector functions and partial derivatives, two and three dimensional 2 integration, line integrals and curl and divergence. Lecture notes; Slides of final lecture; Solar-System Dynamics. Chapter 5 : Vectors. PHY2061 Enriched Physics 2 Lecture Notes Refresher Math and Physics Refresher This course assumes that you have studied Newtonian mechanics in a previous calculus-based physics course (i. They consist largely of the material presented during the lectures, though we have taken the liberty of eshing them out in some places and of being more cursory here than in the lectures in other places. With the exception of economics, all these courses run in the Easter term. The final grade will consist of 25% homework, 45% midterms, and 30% final exam. 01 Partial differentiation, multiple integrals, and topics in differential and integral vector calculus, including Green's theorem, Stokes's theorem, and Gauss's theorem for students with a background in linear algebra. Lecture Notes on Classical Mechanics (A Work in Progress) Daniel Arovas Department of Physics University of California, San Diego May 8, 2013. Bear in mind that course syllabuses evolve over time, and different lecturers structure their courses differently and choose their own notation conventions. Multiple integrals. nb) and MathML files. In these lecture notes we shall represent vectors and vector elds using bold fonts, e. Thomas CalculusUse for Calculus I,Calculus II and MV CalculusAcknowledgements. SYLLABUS: Syllabus contains general information on topics, exams, grade, course structure and policies. Tensors revision questions. The direction is correct since the right hand side of the formula is a constant multiple of v so the projection vector is in the direction of v as required. Vector Calculus ADD. For instance R ˆZ (0. 1 Vector Fields 15. Introduction to the min-max theory for minimal surfaces: | {
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For instance R ˆZ (0. 1 Vector Fields 15. Introduction to the min-max theory for minimal surfaces: Hand-written lecture notes for a topic course on the min-max theory of minimal surfaces in 2013. Some gave vector fields; some gave scalar fields. As in those notes, the figures are made with Anders Thorup’s spline macros. Many topics in the physical sciences can be analysed mathematically using the techniques of vector. Sample Exam 1. Winter 2020 Final Exam (10 am lecture): Final Exam. Lecture notes. This section is largely based on my undergraduate lecture notes from a course given by Dr. Description: Topics covered in these notes include the untyped lambda calculus, the Church-Rosser theorem, combinatory algebras, the simply-typed lambda calculus, the Curry-Howard isomorphism, weak and strong normalization, type inference, denotational semantics, complete partial. Notes: Some of these lectures reference the TI-89 graphing calculator. Anastassiou, I. Also, I taught math 53 this past summer and the course page has some problems and solutions that might help studying. Catalog Description: Change of variable in multiple integrals, Jacobian, Line integrals, Green's theorem. edu office: Neill 315, 509-335-2134 office hours: MWF 10:15 – 11:45 am, and by appointment Lecture 2 is conducted by Mark Schumaker in CUE 203, MWF 2:10-3:00 pm email: [email protected] Matthews, P. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. 2 Notation and Nomenclature De nition 1 Let a ij2R, i= 1,2,,m, j= 1,2,,n. Interpretation of ~x′(t) as the velocity vector 129 55. The dates by some of the lectures are the date of the most recent revision. Lecture notes; Slides of final lecture; Solar-System Dynamics. To connect multivariate calculus to other fields both within and without mathematics. Related documents. 1 (Second-Order Equations with Constant Coefficients) Lecture Notes for Section 15. Introduction to Electrodynamics. VECTOR CALCULUS1. None of | {
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Lecture Notes for Section 15. Introduction to Electrodynamics. VECTOR CALCULUS1. None of this is official. The lecture will be recorded and can be found online at Canvas. Consider a trajectory x: [0, T] !C for a configuration space with d. Lecture Note Ser. Cambridge Course Notes. Written by Ross. 6 Divergence Theorem 15. best course for vector calculus. • Recognize that flow velocity is a vector field, which can be a function of space or a function of space and time. Lecture Notes for College Physics I Contents 1 Vector Algebra 1 2 Kinematics of Two-Dimensional Motion 2 3 Projectile Motion 5 4 Newton’s Laws of Motion 8 5 Force Problems 12 6 Forces due to Friction and Uniform Circular Motion 16 7 Newton’s Law of Universal Gravitation 20 8 Work-Energy Theorem I 22 9 Work-Energy Theorem II 24. In this course, you'll learn how to quantify such change with calculus on vector fields. ms access pdf reader 2 Geometry Of Space Curves. Vector Calculus - Winter 2019 Lectures: MWF 2-3 in Pepper Canyon Hall 109 Instructor:Hans Wenzl email: [email protected] Multivariable calculus. Many topics we will cover are generalizations of one variable calculus, including differentiation and integration, but there are completely new phenomena, and. Classical Electromagnetism: An intermediate level course Richard Fitzpatrick Associate Professor of Physics The University of Texas at Austin. Week 12: Integral definition of gradient, divergence and curl. 194 References The following references were consulted during the preparation of these lecture notes. GOAL: In Multivariable Calculus we complete the Calculus sequence. MAT1005: CALCULUS II Vector Calculus. The notes are designed to be used in conjunction with a set of online homework exercises which help the students read the lecture notes and learn basic linear algebra skills. 2 6 LECTURE 1. The second inner cover contains the basic equations of electrodynamics, the accepted values of some fundamental constants, and the transformation | {
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of electrodynamics, the accepted values of some fundamental constants, and the transformation equations for spherical and. 30-6 pm TA: palghamol. (b) If n is a unit vector, fis changing at the rate rf(0;ˇ=2) n = ˇ 2 ni in the direction n. Instead of Vector Calculus, some universities might call this course Multivariable Calculus or Calculus Three. In other coordinate systems, the unit vectors are not the same everywhere. Currently the book can be found online here, but the link may change as time progresses. Please check out Dr. Many topics in the physical sciences can be analysed mathematically using the techniques of vector. Plane polar co. These are the lecture notes for my online Coursera course,Vector Calculus for Engineers. A line can always be written as fA+ u: 2Rgfor a unit vector u2R. Functions of many variables, with a focus on surfaces in three dimensions, partial derivatives, gradients, and directional derivatives. The norm or length of a vector is jjxjj= (xx) 1=2 = X. The alternate version Stewart/Clegg/Watson Calculus, 9e, will publish later this spring. In these lecture notes we shall represent vectors and vector elds using bold fonts, e. A list of resources can be found below. In the unit we develop the theory of vector fields, flows and differential forms mainly for R n but with a view towards manifolds, in particular surfaces in R 3. Partial derivatives. 3 Vector Calculus In the last part of the course, we will study vector elds, which are functions that assign a vector to each point in its domain, like the vector-valued func-tion F described above. May 28, 2018 - Calculus 2 help for high school + college students (in-class or online). ; Author: James Byrnie Shaw; Category: Calculus; Length: 325 Pages; Year: 1922. Gauss’ Theorem (Divergence Theorem) Consider a surface S with volume V. A line can always be written as fA+ u: 2Rgfor a unit vector u2R. 3 Warnings and Disclaimers Before proceeding with this interactive manual we stress the following: † | {
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3 Warnings and Disclaimers Before proceeding with this interactive manual we stress the following: † These Web pages are designed in order to help students as a source. The magnitude of c~usatis es jc~uj= jcjj~uj. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Preliminaries (vectors, dot product, cross product, planes, lines) Functions of Several Variables (graphing, limits, calculus) Cylindrical & Spherical Coordinates; Partial Derivatives (limits, chain rule) Gradients and Directional Derivatives; Planes and Linear Approximation; Extreme-Values of Real. The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed. secret-bases. Notes for Calculus III (Multivariable Calculus) The notes below follow closely the textbook Introduction to Linear Algebra, Fourth Edition by Gilbert Strang. This would be pictured by drawing the vector (1, 2, 1) in the opposite direction. You may write on both sides of the page. Probability About these notes. GEOS 4430 Lecture Notes: Well Testing Dr. Lecture 1 Vectors A vector has direction and magnitude and is written in these notes in bold e. Continuity 7. Lecture Notes on Multivariable Calculus Notes written by Barbara Niethammer and Andrew Dancer This de nition is more suitable for the multivariable case, where his now a vector, so it in lectures in the rst year, e. " " Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics. Lecture Notes in Classical Mechanics (80751) Raz Kupferman Institute of Mathematics The Hebrew University July 14, 2008. Many topics we will cover are generalizations of one variable calculus, including differentiation and integration, but there are completely new phenomena, and. These notes are for helpful for undergraduate level (BSc or BS). The unique vector of length zero is denoted ~0 and satis es ~0 +~v= ~v. Digital PDF 9. Derivatives 8. Real Analysis. Relevant undergraduate courses are (for relevant | {
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~v. Digital PDF 9. Derivatives 8. Real Analysis. Relevant undergraduate courses are (for relevant schedules, example sheets and exam questions, refer to the General Resources):. For a vector function over a surface, the surface integral is given by Phi = int_SF·da (3) = int_S(F·n^^)da (4) = int_Sf_xdydz+f_ydzdx+f_zdxdy, (5) where a·b is a dot product and n^^ is a unit normal vector. The course will be conducted in Hindi and notes will be provided in English. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. 3 Vector Calculus In the last part of the course, we will study vector elds, which are functions that assign a vector to each point in its domain, like the vector-valued func-tion F described above. 2 Notes • Homework # 1 is uploaded on the course webpage Vector Calculus. edu/courses/mathematics/18-01-single-variable. 15 Vector Calculus 15. Vitalii Konarovskyi IPSP Winter 2019-2020 October 14 { February 4. Title: Vector Calculus. Elements of Vector Calculus: Lecture 1-Self Assesment Quiz: Self Assessment Quiz: 5 kb: Elements of Vector Calculus: Lecture 2-Self Assessment Quiz: Self Assessment Quiz: 7 kb: Elements of Vector Calculus: Lecture 3-Self Assessment Quiz: Self Assessment Quiz: 10 kb: Elements of Vector Calculus: Lecture 4-Self Assessment Quiz: Self Assessment. The key to understanding tensor calculus at a deep level begins with understanding linear and multilinear functions between vector spaces. These are personal notes written mostly long time ago (say 97) and need not be correct nor understandable. Examples of parametrized curves 125 52. In this section we are going to introduce the concepts of the curl and the divergence of a vector. Winter 2020 Final Exam (11 am lecture): Final Exam. Find materials for this course in the pages linked along the left. Dimock Dept. • The inverse of a matrix is defined such that when it operates on the original matrix, the result is the unit matrix. Let's start with the | {
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that when it operates on the original matrix, the result is the unit matrix. Let's start with the curl. Also, −1(1,2,1) = (−1,−2,−1). Prerequisites: MATH-102 or MATH-102H or MATH-102X Terms Offered: Summer, Fall, Winter, Spring A study of polar coordinates, parametric equations, and the calculus of functions of several variables with an introduction to vector calculus. Freeman (2003). Use Firefox to download the files if you have problems. , April 10: Solar activity: Chromospheric and coronal heating. In this course, Prof. Thus, I have chosen to use symbolic notation. Schey Div, grad, curl and all that: an informal text on vector calculus. A sound knowledge of these topics is a vital prerequisite for almost all the later courses in applied mathematics and theoretical physics. secret-bases. Calculus I or needing a refresher in some of the early topics in calculus. Abstract:These are lecture notes for the Cambridge mathematics tripos Part IA Vector Calculus course. Lecture Notes: Chapter 10: PARAMETRIC EQUATIONS AND POLAR COORDINATES. Continuity 7. Stanford Engineering Everywhere; 2007; Jeffrey A. The Mathematica® examples are provided in two formats: Mathematica® notebook files (. Multivariable Calculus is the second course in the series, consisting of 26 videos, 4 Study Guides, and a set of Supplementary Notes. Rogawski) Average values of modular L-series via the relative trace formula, Pure Appl. 5) Linear Operators (notes, lecture) [add day] 9/05 (1. Multiple Integrals and Vector Calculus Prof. In particular, the material is presented to (i) develop a physical understanding of the mathematical concepts associated with tensor calculus and (ii) develop the. Stephen Gull at the University of Cambridge. The lecture schedule described below is applicable to LectureSection 2. Freeman and Co. Actually, there are a couple of applications, but they all come back to needing the first one. Lecture notes. The following video provides an outline of all the topics you would | {
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the first one. Lecture notes. The following video provides an outline of all the topics you would expect to see in a typical Multivariable Calculus class (i. Scalar multiplication: If c2R and ~ua vector, then we may form a new vector c~ucalled the scalar product of ~uwith c. We use cookies to help give you the best experience on our website. These notes concentrate on the third part, which is covered in v. The following are important identities involving derivatives and integrals in vector calculus. 30-6 pm TA: palghamol. Lectures with an N after the lecture number have been rewritten to reference the TI-nspire graphing calculator. Tangents 3. Cambridge Course Notes. This unit is offered in Semester 1. There is no central location for these, so we have collated some resources below. We will reinforce this point of view throughout the course. Express A entirely in the spherical polar basis (i. The electromagnetism lecture notes is a book to provide an introduction to Electromagnetism for Electrical and Electronics Engineers. MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. LECTURE NOTES 15 The Divergence & Curl of B G Ampere’s Law As we have discussed in the previous P435 Lecture Notes, for the case of an infinitely long straight wire carrying a steady (constant) line current I =Izˆ, G the macroscopic magnetic field associated with this system is given by: () 0 ˆ 2 I Br r μ ϕ π ⎛⎞ =⎜⎟ ⎝⎠ GG for. F or underlined. Lecture #2 VECTOR (OR LINEAR) SPACES Handout#2 One sheet of paper. 9 : Lecture 12 : June 8 (Thu) Vector fields; Line integrals : 16. ISBN -13-805326-X. Introduction These are my notes for the course Math 53: Multivariable Calculus, at UC Berkeley, in the summer of 2011. Notes are applicalicable for both 1st and 2nd sem students of CBCS scheme. The pushforward along a vector function f with respect to vector v in R n is given by d f ( v ) = ∂ f ∂ v d v. This course will be based on a series of lecture notes which will be posted regularly throughout | {
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v. This course will be based on a series of lecture notes which will be posted regularly throughout the semester. You may find the following textbooks references useful: Cambridge University Press, 2007. Lectures Notes: Reading guides: Ch 14: Intro to Partial Derivatives: 14. Calculus Revisited: Multivariable Calculus (Res. Gradient Griffiths: Chapter 1 – skip section 1. Mathematical Tripos Part IA: Vector Calculus (1997-2000) My Vector Calculus notes from Lent 2000 are available in pdf and postscript form. This would require us to take the derivative of a vector. to be d(x;y) = jjx yjj: This gives a metric on E. Week 10: Application of vector calculus in mechanics, lines, surface and volume integrals. 01 graphing notebook. These are personal notes written mostly long time ago (say 97) and need not be correct nor understandable. Derivatives as. This course is about vector calculus, and covers material that all engineers should know. We may rewrite Equation (1. Although we developed many different formulas, everything in Chapter 2 could be summarized in one rule: the operators $\ddpl{}{x}$, $\ddpl{}{y}$, and $\ddpl{}{z}$ are the three components of a vector operator $\FLPnabla$. The derivative of a vector function 127 53. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. Vectors form a linear algebra (i. Welcome! This is one of over 2,200 courses on OCW. Homework should take at least 6 hours, although this can vary quite a lot depending on your background and goals. (Math 151 & 152) Calculus Workshop I & II. Vector Calculus (Green's Theorem, Stokes' Theorem, Divergence Theorem) For differentiation, you can use Principles of Mathematical Analysis by Rudin (Chapter 9). After each video lecture a) redo the examples done in lecture b) do the applicable homework assigbment. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. WEATHERBURN, C. Exams and grades. than | {
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(differentiation and integration) are a prerequisite. WEATHERBURN, C. Exams and grades. than 10 dimensions. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples. The proper way to understand this is that in both cases, the derivative is a linear transformation. Calculus and Linear Algebra II. University of Cambridge - Part IA Natural Sciences. The kind of things that give you insight into what the ideas mean or how they were developed. The lectures on cector calculus follow the book Calculus III by Marsden, Jerrold E. F or underlined. Calculus I and II). {\displaystyle d\,\mathbf {f} (\mathbf {v} )={\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}d\,\mathbf {v}. Understanding Basic Calculus S. Gives precise and intuitive topological pictures of antisymmetric tensors and their algebra and calculus in three dimensions. Anastassiou, I. Notes Outline: Section 10. ^ Kelly, P. Vitalii Konarovskyi IPSP Winter 2019-2020 October 14 { February 4. Gradient Griffiths: Chapter 1 – skip section 1. We discuss the basic ideas behind k-means clustering and study the classical algorithm. Description: These lecture notes provide a comprehensive introduction to Electromagnetism, aimed at undergraduates. 2 Spans, Lines and Planes The span of a single vector vin Rnis the set of all scalar multiples of v: spanfvg= ftvjt2Rg:. A vector can also be represented mathematically in the form of an equation: v = vxˆ. The historical motivation for homology theory came from vector calculus. Motion in a Noninertial Reference Frame, February 2009, 32pp. Applied Advanced Calculus Lecture Notes by Jan Vrbik. My lecture notes (PDF). Acceleration and Force 131 56. 9) due on July 22 (Fri): Solutions, Solutions to practice problems : Lecture 11 : July 20 (Wed) Change of variables in multiple integrals; Overview of vector calculus; Vector fields : 15. Boqing Gong): Support Vector Machines for Computer Vision Applications; Lecture 14: Guest | {
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: 15. Boqing Gong): Support Vector Machines for Computer Vision Applications; Lecture 14: Guest Lecture (Dr. 1 Gradient-Directional Derivative. PHY2060) and at least have co-registered in a vector calculus course (Calc 3). Perform operations on vectors and vector-valued. Module Overview. IA Vector Calculus (Cambridge), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degree, lecture notes, physical sciences. As the set fe^ igforms a basis for R3, the vector A may be written as a linear combination of the e^ i: A= A 1e^ 1 + A 2e^ 2 + A 3e^ 3: (1. Here are a set of practice problems for the Vectors chapter of the Calculus II notes. These notes are only meant to be a study aid and a supplement to your own notes. It is sta ed all six periods every class day. Lecture notes for Math 417-517 Multivariable Calculus J. It is equally valuable for students who are learning calculus for the first time. 02 instructor. Dividing by dt, we obtain dA dt = 1 2 fl fl fl flr £ dr dt fl fl fl fl = jcj 2 Therefore, the physical interpretation of Eq. Sloughter). 9 : Homework 5 (covering 15. Vector Calculus When working with functions one also has to study calculus in addition to studying algebra. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. 2 Vector Components and Dummy Indices Let Abe a vector in R3. Vector Calculus 16. 0130415316. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. Vector Calculus ADD. Vector calculus. Fundamental Theorem of Calculus for curve integrals: Path independence and potential: Simplices and boundary: Path, surface, and volume integrals, etc. In this appendix I use the following notation. Multiple integrals. Vector Calculus Tutorials: Vector Calculus | {
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I use the following notation. Multiple integrals. Vector Calculus Tutorials: Vector Calculus (wikipedia) Linear Algebra Tutorials: Professor G. 13 CURL OF A VECTOR1. The Organic Chemistry Tutor 1,896,958 views. Differential vector calculus. Notes: Some of these lectures reference the TI-89 graphing calculator. Tumblr is a place to express yourself, discover yourself, and bond over the stuff you love. Course Notes and General Information Vector calculus is the normal language used in applied mathematics for solving problems in two and three dimensions. Functions of many variables, with a focus on surfaces in three dimensions, partial derivatives, gradients, and directional derivatives. D, an award-winning teacher and. Its purpose is to prepare students for more advanced mathematics courses, particularly courses in mathematical programming (MAT 419), advanced engineering mathematics (MAT 430),. 2 n! 1=2: We de ne the distance between two points xand yin E. I teach maths to Engineering students and most of them take a course on vector calculus. The text for the course is Vector Calculus, fourth edition, by Susan J. Lecture Notes Section 1: (PostScript , PDF ) Revision of Vector Calculus. Vector calculus identities explained. Vector Calculus (aka Advanced Multivariable Calculus) Math222. See also: Vector algebra relations. He graduated in 1816 and spent the next 27 years as a civil engineer. Vector Calculus Collapse menu 1 Analytic Geometry. Lecture notes: Course Information; Section I: Electricity. •“Vector Analysis and Cartesian Tensors”, Bourne and Kendall 1999 by Nelson. 6 Divergence Theorem 15. 18-007, MIT OCW). Chapter 2. May 2016 "Constructing solutions to linear fractional-order PDEs", departmental seminar, Cambridge Analysts Knowledge Exchange, Faculty of Mathematics, University of Cambridge, UK. To see why this is true, consider the function given by r(t) = f(t)i + g(t)j. 00 RESULTS AND GRADES Answer to exam question 1; 2 can be found in lecture notes; 3, 4, 5 | {
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+ g(t)j. 00 RESULTS AND GRADES Answer to exam question 1; 2 can be found in lecture notes; 3, 4, 5 You can give feedback in WebOodi after the last lecture. Examples of this include sections on the statistical mechanical theory. 1aCOURSE CONTENT: MAC 2313 is the third semester in the calculus sequence and it gives a thorough introduction to multi-variable calculus. The dates by some of the lectures are the date of the most recent revision. Introduction These are my notes for the course Math 53: Multivariable Calculus, at UC Berkeley, in the summer of 2011. Read ISL, Section 9–9. Motion in a Noninertial Reference Frame, February 2009, 32pp. Math 4013: Vector Calculus, Summer 1998. Dan Sloughter Calculus of several variables; James Cook's lecture notes; Tom Apostol's calculus books; Richard Hammack Book of proof; Illustrations. This is the first in a series of lecture notes on k-means clustering, its variants, and applications. Find materials for this course in the pages linked along the left. Tangents and the unit tangent vector 133 57. Here are the pdf files for the calculus sequence note packets. pdf file of this paper. In this section we are going to introduce the concepts of the curl and the divergence of a vector. In this appendix I use the following notation. Polar Co-ordinate Systems Here dV indicates a volume element and dAan area element. MAT 203 Lecture Notes - Lecture 1: Multivariable Calculus, Vector Calculus, Graphing Calculator Exam. Calculus I and II). Prerequisites are linear algebra and vector calculus at an introductory level. The first lecture of Cosmology II is on Monday, Oct 28th, and the first exercise session is on Friday, Nov 8th. Homework should take at least 6 hours, although this can vary quite a lot depending on your background and goals. These notes are send by Umer Asghar, we are very thankful to him for providing these notes. Office Hours: Instructor: Hans Wenzl: Email: Hans Wenzl MW: 2:30-3:30 and by appointment (just talk to me after | {
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Instructor: Hans Wenzl: Email: Hans Wenzl MW: 2:30-3:30 and by appointment (just talk to me after class, or email) TA for sections A01, A02: James Hall in AP&M 5748 Email. 61 Differentiation of Processes Let E be a flat space with translation space V. Vector Analysis: A Text-book for the Use of Students of Mathematics and Physics, Founded Upon the Lectures of J. MULTIVARIABLE CALCULUS 28-MATH2063 SPRING 2020. Calculus mainly involves differentiation and integration Differential Calculus Integral Calculus 𝑑𝑑=𝑓𝑓 𝑑𝑑𝑓𝑓 𝑑𝑑𝑥𝑥. This is the one we will use. Publisher: Dalhousie University 2007 Number of pages: 106. Consider a trajectory x: [0, T] !C for a configuration space with d. A vector A, of components A1, A2 and A3 in the basis f^e 1;^e 2;^e 3g, will interchangeably be written as a column or row vector, A = 0 @ A1 A2 A2 1 A= (A1;A2;A3. Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Inan Prentice Hall, 1999. pdf Lecture notes. Continuity 7. edu, office hours: M10-11:50am, F: 4-6pm, office APM 2313. IA Vector Calculus Lecture notes 2000. It also contains a list of links to other web pages with information on vector calculus. Math 263 Calculus III Pierce College MAP 1 Lecture 16 Sections 18. You must prepare the notes. Views assigns the unit normal vector cosθ i+sinθj to each point on the cylinder S. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. Tromba Vector Calculus. Math 290-1: Linear Algebra & Multivariable Calculus Northwestern University, Lecture Notes Written by Santiago Ca˜nez These are notes which provide a basic summary of each lecture for Math 290-1, the first quarter. However, beginners report various difficulties dealing with the index notation due to. Lokesh, Acharya Instt of Tech, B'lore. The derivative of a vector function 127 53. Press, Cambridge (2007) Click for. When we move from derivatives of one function to derivatives of many | {
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Cambridge (2007) Click for. When we move from derivatives of one function to derivatives of many functions, we move from the. Motivation • In multivariable calculus, students become very adept at computing quantities involving vector fields However, there is difficulty in connecting the abstract concept of a vector field. Note that the third component of the curl is for fixed z just the two dimensional vector field F~ = hP,Qi is Q x. You can also fill in what we must leave out by reading the. Most of us last saw calculus in school, but derivatives are a critical part of machine learning, particularly deep neural networks, which are trained by optimizing a loss function. Chapter 5 : Vectors. There is no need for parametric equations. Not lecture notes, but here are some good lecture courses: Calculus of Several Real Variables Integral and Vector Calculus Transform Calculus and its applications in Differential Equations Multivariable Calculus Integral equations, calculus of variations and its applications Calculus of One Real Variable Differential Calculus in Several Variables. Derivatives as. (This covers the differential calculus portion of this class. Publisher: Dalhousie University 2007 Number of pages: 106. EXAM I: Friday January 24th, Covers Text Sections (2nd ed) 11. Vector Calculus Summary. Necessary concepts from linear algebra and other mathematical disciplines necessary to understand the text are also covered. A lot of these gaps will be filled in in this course, and more material will be covered. 4, Special Issue: In memory of Armand Borel. Lecture 2 Differentiable functions of many variables: pdf. Scalar Product: The "dot" product of two vectors is a scalar. Find materials for this course in the pages linked along the left. Quickly find Science & Math course-specific resources across a variety of academic disciplines such as digital and interactive textbooks, lecture notes, quiz packs, videos, presentations and more. You can use these notes and exercises on | {
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lecture notes, quiz packs, videos, presentations and more. You can use these notes and exercises on Scaling Analysis to check your current level and guide your study. Two semesters of single variable calculus is a typical prerequisite. That is the purpose of the first two sections of this chapter. Vector Analysis: A Text-book for the Use of Students of Mathematics and Physics, Founded Upon the Lectures of J. I would also like to thank Harold S. Mathematics Examples, Lecture Notes and Specimen Exam Questions and Natural Sciences Tripos Mathematics examples Details on obtaining and updating the source of DAMTP examples (this is aimed at DAMTP Unix account holders only), and the list of course codes and titles referred to in these pages. Relevant undergraduate courses are (for relevant schedules, example sheets and exam questions, refer to the General Resources):. MATH 226 (Calculus III) Fall 2016 39571R (MWF at noon in SOS B44) Final exam is Wednesday, December 7, 2-4 pm. LECTURE NOTES 15 The Divergence & Curl of B G Ampere’s Law As we have discussed in the previous P435 Lecture Notes, for the case of an infinitely long straight wire carrying a steady (constant) line current I =Izˆ, G the macroscopic magnetic field associated with this system is given by: () 0 ˆ 2 I Br r μ ϕ π ⎛⎞ =⎜⎟ ⎝⎠ GG for. 2) F (x; y z) = P x y z I + Q x y z J R x y z K: For example, the vector field (18. Vector product, Scalar triple product and their interpretation in terms of area and volume respectively. Based on lecture notes by James McKernan and on lectures by Pavel Etingof. Please do let me know if you. On one side is the definition of a vector space from the notes. 1 Step function Last year: inhomogeneous second order, constant coefficient, ODEs of the form f 00(t) af 0(t) bf (t) g(t) for a limited collection of g(t)s. Multivariable Calculus part 3, 23pp: Multiple Integration. Given on September 5 th. Norton & Company. Notes of lectures on Multivariable Calculus G. • Notice that one | {
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September 5 th. Norton & Company. Notes of lectures on Multivariable Calculus G. • Notice that one cannot add a column-vector and a row-vector!. Before doing so, however, it is useful to think about how calculus can be used in trajectory optimization. 2 Laplacian and second order operators 8. Revision of vector algebra, scalar product, vector product 2. To indicate that an item is in a set, we use the 2symbol. ISBN: 0130414085 (F05). Line, surface and volume integrals, curvilinear co-ordinates 5. This course is about vector calculus, and covers material that all engineers should know. Use Firefox to download the files if you have problems. Find calculus course notes, answered questions, and calculus tutors 24/7. The content may be incomplete. Published by Prentice Hall. Rogawski) Average values of modular L-series via the relative trace formula, Pure Appl. Read a portion of the text. to be d(x;y) = jjx yjj: This gives a metric on E. Written as a companion to multivariable calculus texts, this contains careful and intuitive explanations of several of the ideas covered in this course. Stochastic calculus has very important application in sciences (biology or physics) as well as mathematical nance. The dates by some of the lectures are the date of the most recent revision. ) Lecture Notes on Multivariable Calculus by Barbara Niethammer and Andrew Dancer. derivative as limit of a ratio, integral as limit of a sum. Math 320-3: Lecture Notes Northwestern University, Spring 2015 Written by Santiago Canez~ These are lecture notes for Math 320-3, the third quarter of \Real Analysis", taught at North-western University in the spring of 2015. Math 290-1: Linear Algebra & Multivariable Calculus Northwestern University, Lecture Notes Written by Santiago Ca˜nez These are notes which provide a basic summary of each lecture for Math 290-1, the first quarter. Please do let me know if you. Selected and mentored by James Stewart, Daniel Clegg and Saleem Watson continue Stewart's legacy | {
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Selected and mentored by James Stewart, Daniel Clegg and Saleem Watson continue Stewart's legacy of providing students with the strongest foundation for a STEM future. Not lecture notes, but here are some good lecture courses: Calculus of Several Real Variables Integral and Vector Calculus Transform Calculus and its applications in Differential Equations Multivariable Calculus Integral equations, calculus of variations and its applications Calculus of One Real Variable Differential Calculus in Several Variables. Real Analysis. Notes for the calculus courses. These notes will contain much the same material as the lecture, but will not be an exact copy. ENGINEERING MATHEMATICS 2 MA8251 Unit 2 VECTOR CALCULUS Notes Pdf Free download. MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. Analysis I (2003) Source of notes: Prof Körner's site; Vector Calculus (2000) Download file. On page 40 of the lecture notes on Chapter 19 Section 1: The Idea of Flux Integral, we saw the example which requires us to find the equation of the plane for the rectangular region with corners. Bowen Mechanical Engineering Lectures on Differential Geometry, Prentice-Hall, Englewood Cliffs, New Jersey, 1964. The proper way to understand this is that in both cases, the derivative is a linear transformation. This extends knowledge developed in A-levels and in the Year 1 module Calculus on the differentiation and integration of functions of a single variable, and provides the necessary ground work for Years 2 and 3 modules, such as Curves and Surfaces, Linear PDEs, Fluid Mechanics. Math 223 Vector Calculus Author: Arlo Caine Practice Exam 1 Name: Solutions Directions: Read all questions carefully. Gauss’ Theorem (Divergence Theorem) Consider a surface S with volume V. vector calculus: m427l linear algebra: m340l GEOMETRY for the high school classroom The following notes were developed in collaboration with Gary Hamrick, Diane Radin and a group of Austin-Area high school teachers. University | {
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with Gary Hamrick, Diane Radin and a group of Austin-Area high school teachers. University College Dublin An Col aiste Ollscoile, Baile Atha Cliath School of Mathematics and Statistics Scoil na Matamaitice agus na Staitistic Vector Integral and Di erential Calculus (ACM 20150) Dr Lennon O N araigh Lecture notes in Vector Calculus, September 2017. | {
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# Proof writing: $\varepsilon$-$\delta$ proof of the product rule for limits --- any tips for cleaning this up?
I've written the following proof for the product rule for limits. Here $$\lim_{x \to c} f(x) = L$$ and $$\lim_{x \to c} g(x) = M$$. While it works, I find it to be clunky and I can't shake the feeling that we somehow can reduce the general case to the much (in my opinion) cleaner proof of the special case where $$L = M = 0$$. Anyhow, here is the proof:
Suppose first that $$L = M = 0$$. Fix an arbitrary $$\varepsilon > 0$$ and let $$\delta_1$$ and $$\delta_2$$ be such that $$f(x)$$ and $$g(x)$$ are in a $$\sqrt{\varepsilon}$$ neighborhood of $$0$$, respectively. Choose $$\delta = \min(\delta_1, \delta_2)$$. Then when $$|x-c| < \delta$$ we have $$|f(x)g(x)| = |f(x)||g(x)| < \left(\sqrt{\varepsilon}\right)^2 = \varepsilon.$$
Suppose instead $$M \neq 0$$. Fix an arbitrary $$\varepsilon > 0$$ and let $$\delta_1$$ be such that it satisfies $$|f(x) - L| < \varepsilon_1 = \frac{\varepsilon}{3|M|}$$. Moreover, let $$\delta_2$$ be such that it satisfies $$|g(x) - M| < \varepsilon_2 = \min(\frac{\varepsilon}{3\varepsilon_1}, \frac{\varepsilon}{3|L|})$$ where $$\frac{\varepsilon}{3|L|} = \infty$$ if $$L = 0$$. Choose $$\delta = \min(\delta_1, \delta_2)$$. Then if $$|x - c| < \delta$$, we have \begin{align*} |f(x)g(x) - LM| &= |f(x)g(x) - Lg(x) + Lg(x) - LM| \\ &= |g(x)(f(x) - L) + L(g(x) - M)| \\ &\leq |g(x)||(f(x) - L)| + |L||(g(x) - M)| \\ &\leq |M \pm \varepsilon_2|\varepsilon_1 + |L|\varepsilon_2 \\ &\leq |M|\varepsilon_1 + \varepsilon_1\varepsilon_2 + |L|\varepsilon_2 \\ &\leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} < \varepsilon. \end{align*} | {
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Sure, the general case can be reduced to the case $$L = M = 0$$ by subtracting $$L$$ and $$M$$ from $$f$$ and $$g$$, respectively. Once we know that $$\lim_{x\to c}f(x)g(x) = 0$$ whenever $$f(x),g(x)\to 0$$ as $$x\to c$$, we deduce $$\lim_{x\to c}(f(x)-L)(g(x)-M) = 0.$$ Using algebraic limit laws, this simplifies to $$\lim_{x\to c}f(x)g(x) = LM.$$
Added: The algebraic limit laws I have in mind are (here all the limits are presumed to exist).
• $$\displaystyle\lim_{x\to c}[F(x) + G(x)] = \lim_{x\to c}F(x) + \lim_{x\to c}G(x)$$.
• For any real $$\alpha$$, $$\displaystyle\lim_{x\to c}\alpha F(x) = \alpha\lim_{x\to c}F(x)$$.
As for cleaning up the direct proof in the general case, I would proceed like this.
Proof. By the triangle inequality, $$|f(x)g(x)-LM| \le |f(x)-L||g(x)| + |L||g(x)-M|.$$ $$g$$ is bounded near $$c$$ because the limit $$M$$ exists. Sending $$x\to c$$, the right-hand side tends to $$0$$, as desired. $$\square$$
• Exactly how do you go about simplifying it using "algebraic limit laws"? Jun 13 '21 at 0:23
• @Peatherfed: I added the laws I was thinking about. Once you have these in mind, to finish the proof of the general case using the special case when the limits are zero, use the usual algebraic laws like distributivity, commutativity, and so on, and then apply the two bulleted laws when applicable to get the final result. Hope this helps. Jun 13 '21 at 1:20
• I see, I was hoping to actually get that second law as a consequence of the theorem. I might go about it by first showing it for constants, which will be almost immediate, and then proceeding through the special case into the general case using the linearity of limits. As concerns your suggestion for the direct proof, I like it, although I was aiming for the type of proof where the delta is explicitly provided. I essentially use that g is bounded when going from the first to the second inequality. Jun 13 '21 at 1:31 | {
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Note that you don't really need to split in two cases, you can do it all at once: It's easy to see that $$g$$ is bounded near $$x=c$$ (because the limit exists), say $$|g(x)|\leq K$$, $$K>0$$, for $$|x-c|<\delta_{3}$$. Given $$\varepsilon>0$$ there exists $$\delta_{1},\delta_{2}>0$$ such that $$|x-c|<\delta_{1}\Rightarrow |f(x)-L|<\frac{\varepsilon}{2K}\text{ and }|x-c|<\delta_{2}\Rightarrow |g(x)-M|<\frac{\varepsilon}{2(|L|+1)}.$$ Now take $$\delta=\min\{\delta_{1},\delta_{2},\delta_{3}\}$$. For $$|x-c|<\delta$$ we have that \begin{align*} |f(x)g(x)-LM|&\leq |f(x)g(x)-Lg(x)+Lg(x)-LM|\\ &\leq |g(x)|\cdot |f(x)-L|+|L|\cdot |g(x)-M|\\ &< K\cdot\frac{\varepsilon}{2K}+\frac{|L|\cdot\varepsilon}{2(|L|+1)}\\ &<\varepsilon \end{align*} and we are done.
Some time ago I found a very clean proof in Landau's Differential Calculus book which showed the product of continuous functions is continous.
The style was a bit outdated but I managed to update it and the idea works just fine with limits.
The advantage of the approach is that you don't have to show the function is bounded in a neighbourhood of the point. Furthermore you don't have to break into cases where $$L$$ and $$M$$ are zero or not.
First notice that
$$\begin{array}{c} |(fg)(x)-LM|&=|f(x)g(x)-f(x)M-Lg(x)+LM+Lg(x)-LM+Mf(x)-ML|\\ &=|(f(x)-L)(g(x)-M)+L(g(x)-M)+M(f(x)-L)|\\ &\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L| \end{array}$$ for every $$x$$. Now, given $$\varepsilon>0$$ there is $$\delta_1>0$$ such that
$$0<|x-a|<\delta_1\implies |f(x)-L|<\min\left\{1, \frac{\varepsilon}{3(1+|M|)}\right\}$$
and there is $$\delta_2>0$$ such that:
$$0<|x-a|<\delta_2\implies |g(x)-M|<\frac{\varepsilon}{3(1+|L|)}.$$ Now notice,
$$\begin{array}{c} 1<1+|L|&\implies \frac{1}{1+|L|}<1\\ |L|<1+|L|&\implies \frac{|L|}{1+|L|}<1\\ |M|<1+|M|&\implies \frac{|M|}{1+|M|}<1. \end{array}$$ Using this, it follows that | {
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$$0<|x-a|<\min\{\delta_1, \delta_2\}$$ implies \begin{align*} |(fg)(x)-LM|&\leq |f(x)-L||g(x)-M|+|L||g(x)-M|+|M||f(x)-L|\\ &<1\cdot \frac{\varepsilon}{3(1+|L|)}+|L|\frac{\varepsilon}{3(1+|L|)}+|M|\frac{\varepsilon}{3(1+|M|)}\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=\varepsilon. \end{align*}
The same idea also applies when you're working with sequences, for instance.
It may be of interest to you that (with some abstract tools from model theory) one can choose to use nonstandard analysis instead of $$\epsilon-\delta$$ and work with infinitesimal numbers. I could imagine that the book "Nonstandard Analysis" by Martin Väth contains basic calculus rules proven within the framework of nonstandard analysis. | {
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# Moderate Permutation-Combination Solved QuestionAptitude Discussion
Q. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
✖ A. 420 ✖ B. 360 ✖ C. 320 ✔ D. 210
Solution:
Option(D) is correct
Total number of arrangements,
$= \dfrac{8!}{4!×2!×2!} = 420$
Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements $= \textbf{210}$
## (2) Comment(s)
Numan
()
I think the answer is obtained by 7!/4!1!1! since you cant change the order of the first 1 and 2
Priyanka
()
why the 1st 1 before the first 2 is same as first 2 before the first 1 | {
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# Square Matrices
So I'm studying a few special families of square matrices, the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices and I just had a few questions.
I know...
a diagonal matrix is if every nondiagonal entry is zero, $a_{ij}$=0 whenever $i$ doesn't equal $j$.
an upper triangular matrix is if all entries below the diagonal are zero, $a_{ij}=0$ whenever $i >j$.
a lower triangular matrix is if all entries above the diagonal are zero, $a_{ij}=0$ whenever $i < j$.
symmetric if $a_{ij}=a_{ji}$ for all $i$ and $j$.
But I was just wondering, can the diagonal matrices, upper triangular matrices, lower triangular matrices and symmetric matrices have the $0_{n\times n}$? Also I know the $I_{n\times n}$ matrix is in the diagonal matrices, but can it be in the other three types?
• Both the zero matrix and the identity matrix are diagonal, upper/lower triangular, and symmetric. Check the definitions. – Ethan Alwaise Dec 9 '16 at 4:55
• Just check the hypothesis – Learnmore Dec 9 '16 at 4:56
A diagonal matrix doesn't care what is on the diagonal; it only cares that the off-diagonal entries are $0$. So the zero matrix $\mathbf 0_{n \times n}$ is indeed diagonal.
Similarly, an upper triangular matrix only cares that the elements below the diagonal are $0$, so the zero matrix is upper triangular. For the same reason, it is also lower triangular.
It is also symmetric, because $a_{ij} = a_{ji} = 0$ for all $i,j$.
Similar reasoning applies to the identity matrix.
Take for example the definition of upper triangular matrix. It's a matrix which has $a_{ij}=0$ whenever $i>j$, but this doesn't mean that $a_{ij}\neq 0$ for $i\le j$, it's possible to have $a_{ij}=0$ if $i\le j$. Then both $O_{n\times n}$ and $I_{n\times n}$ are upper triangular matrices.
The same idea applies to the diagonal, lower triangular and symmetric matrices. | {
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# How do you find the area of the shaded area of this circle?
I need to find the area of the shaded area. The triangle is equilateral. So far, I have found the area of the triangle to be $\sqrt 3$, but I cannot figure out how to find the radius of the circle in order to find the area of the circle. Any advice would be appreciated.
• Is the triangle equilateral? – Ahmed S. Attaalla Apr 7 '17 at 3:06
• Yes it is equilateral. – Justin Lam Apr 7 '17 at 3:10
• Do you know how to find the areas of circles and triangles? Oops, sorry, I see that you do. My bad. Try drawing in a radius and looking for relationships. Hint: Choose your radius wisely. – Arby Apr 7 '17 at 3:11
• Lookup circular segment, and think what all that comes down to for an equilateral triangle. – dxiv Apr 7 '17 at 3:13
We're looking for the radius right? So let's draw them..
Ok. Now we have an isosceles triangle $30-30-120$. If $r$ is the radius then law of sines tells us,
$$\frac{2}{\sin 120}=\frac{r}{\sin 30}$$
So $r=2 \frac{\sin 30}{\sin 120}=2\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$. I think you can take it from here.
• Thanks for the help! I got about 1.15 feet squared as the radius. Now I can do the rest. – Justin Lam Apr 7 '17 at 3:22
• My final answer is about 2.5 feet squared. Is this correct? – Justin Lam Apr 7 '17 at 3:29
• You're welcome. That's about right. The exact answer is $\frac{4}{3}\pi-\sqrt{3}$ feet squared. @JustinLam – Ahmed S. Attaalla Apr 7 '17 at 3:44
• Ahmed's drawing: cos(30°) = 1/r. Hence r = 1/cos(30°). Using cos(30°) = (1/2)×(√3) we get r. – Peter Szilas Apr 7 '17 at 6:04
• @Ahmed. OK , here we go. – Peter Szilas Apr 8 '17 at 2:33 | {
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Let's label Ahmed's drawing: Triangle $ABC$, lower left $A$, then counterclockwise $B$, and $C$ (top). Let the center of the circle be $M$. Extend $CM$ to intersect $AB$ in $D$. Note length $AD$ $=$ length $DB$ $=1$, $MD$ being the perpendicular bisector of $AB$. Triangle $ADM$ is a right angled triangle. Angle $MAD = 30°$.
$$\cos (30°) = \frac{1}{r}$$
$$r = \frac{1}{\cos (30°)}$$
Using $\cos (30°) = \frac{\sqrt{3}}{2}$ we get $r$. | {
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# Why do we teach even and odd functions?
I've been either a student or an instructor in Precalculus or Calculus 1 at about 6 institutions now, and teaching the definition of even functions (where $f(-x) = f(x)$) and odd functions (where $f(-x) = -f(x)$) has been universal.
But why? I don't see how these concepts are so useful that they need to be in the courses that are taught to everyone. I don't see how they lay the stage for understanding calculus.
I mean, seeing how it works graphically is nifty. But it seems like a disproportionate emphasis is placed on these classifications, in every curriculum I've seen.
• it seems like a disproportionate emphasis is placed on these classifications It's extremely useful in applications. What level of emphasis is it that you consider disproportionate? If it's covered in the textbook and mentioned once in a while in examples, then it probably doesn't even need to have 5 minutes dedicated to it in class.
– user507
Sep 11 '17 at 22:36
• I like to teach them because of graphing, integration, understanding the difference between proof by example (not proof) and proof of a general case. (Also: If everyone at your school studies calculus, then you teach at a very different institution than mine.) Sep 12 '17 at 0:24
• @BenCrowell It made up a noticeable part of my A-level, with no application at all at the time. Sep 13 '17 at 6:25
• It's an example of symmetry (and anti-symmetry) - which is arguably the most useful concept with a practical application in the whole of mathematics IMO. Sep 14 '17 at 1:22
• Many topics in mathematics are not useful. And no one has claimed them to be as well. Again useful means what, can they be of use in future etc are significant questions though. Sep 14 '17 at 6:51 | {
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One of the major themes of precalculus is what I call “connecting geometry to algebra”. Being able to translate between an algebraic statement like $f(x)= f(-x)$, and the geometric statement that the graph of $f$ is symmetric about the vertical axis is a great instance of this. This is just one more way to practice reinforcing function concepts, and the connection with graphs.
• Even a triple connection geometry - algebra - calculus!
– Basj
Sep 13 '17 at 20:09
Here you can see that knowing if the function is even or odd can help you when you are integrating over the interval $[-a, a]$.
You can reduce really-hard-to-look-at integrals to zero just by knowing this. As an example, to calculate $E(Z)$ where $Z \sim N(0, 1)$, the standard normal distribution, you have:
$\displaystyle E(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ze^{-z^2/2}dz$
Which is immediately reduced to zero as the inner function in the integral is odd, and you are integrating over $(-\infty, \infty)$.
This may not be too difficult, but knowing this property about odd functions lets you generalize this to all the odd moments of the standard normal: $E(Z), E(Z^3), E(Z^5), ...$ all of them equal zero.
• You could improve this answer by being more explicit here as to why this is useful. eg mention Fourier series Sep 12 '17 at 6:15
• +1, As an engineer and purely from a practical point this was my first thought. For instance, calculating Fourier series is made much easier knowing about major shortcuts that can be made from the knowledge of odd and even functions. Sep 12 '17 at 15:32
Besides applicability in topics like integration and Fourier analysis, it also connects algebra to calculus at least in the way that multiplication of even/odd functions behaves like addition even/odd numbers:
• Multiplying two even functions gives an even function.
• Multiplying two odd functions gives an even function, too.
• Multiplying an even and an odd function gives an odd function. | {
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Also, you can decompose every function as a sum of an even and an odd function as $$f(x) = \frac{f(x)+f(-x)}2 + \frac{f(x)-f(-x)}2$$ (which is a very useful concept an the same as writing a matrix as the sum of a symmetric and an antisymmetric one as $A = \tfrac12(A+A^T) + \tfrac12(A-A^T)$).
• This also gives one (among many) justification for using the hyperbolic trig functions $\sinh x$ and $\cosh x$ as the odd and even parts respectively of $e^x$ Sep 12 '17 at 14:08
• +1 Not forgetting that this approach generalizes to order three, four,... symmetries via discrete Fourier transforms and, ultimately, to representation theory of groups. Sep 30 '17 at 6:13
Learning to think about functions abstractly should be one goal in precalculus, and function symmetry helps. Also suppose we carefully protected a student from knowing anything about function symmetry. Upon learning about flux in vector calculus, would this student be able to quickly see that the flux of the vector field ${\bf F}(x,y,z)= y^2{\bf j}$ through the unit sphere is 0?
• Indeed, when I teach multivariable calculus, I show students lots of tricks to simplify integrals. Eliminating terms you know will evaluate to zero after integration, halving the domain so that you can "plug in" zeroes and simplify fewer terms, etc. It all comes down to symmetry. Sep 16 '17 at 13:09
Even and odd parity are probably the simplest examples of function symmetries.
In applied mathematics, the general observation of function symmetries allows to simplify calculations (as stated by others) and to produce more meaningful graphs. In physics, symmetrical parts of a function are sometimes associated to different physical phenomena.
Two examples: | {
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Two examples:
1. If you have a function which is invariant under inversion (a more complex symmetry), that is, $$f(1/x) = f(x),\ x>0,$$ it is better to plot the function by using a logarithmic $x$ axis because $f(a^x)$ is even.
2. The example given by Dirk on the decomposition of a matrix in the sum of a symmetric and antisymmetric part is useful in circuit theory: the symmetric part of the impedance and admittance matrices is associated to the average power dissipated by an electrical network when subjected to sinusoidal excitations.
• This is a better answer than mine. The psychology of voting on these sites is mysterious. Sep 13 '17 at 19:28
To express functions as sum of even and odd functions
$$f(x) = f_{even}(x) + f_{odd}(x)$$
And look at the properties of their graphs.
Teach Fourier transform and say it becomes easy to compute when $f(x)$ is an odd function. And the integral becomes zero.
• I think you've missed the point of the question. It isn't about what the syllabus requires, it's about the longer term benefits. Sep 13 '17 at 6:23
• now like applications of it ?@JessicaB Sep 13 '17 at 8:01
I agree that it is disproportionate. Ben Crowell (in comments) says that is helpful for applications. Well, I have a very strong background in natural sciences and engineering, and it is not a big building block. Very few derivations in physics rely on it for instance.
I know this is (-1) controversial, but I think the appeal comes from the ease of doing tricky questions based on it. Sort of ETS style questions. A similar thing would be the vertical angle theorem (or whatever it is called, where a line crosses a couple parallel lines) is big on the SAT.
(-2) I also think that many pure math people prefer things that are definitional and classificational. And that in preference to being able to do multistep mechanical problems like a maximization or power series expansion. | {
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• Really? The difference of Dirichlet versus Neumann boundary conditions (difference between the electromagnetic field behavior near conductors versus insulators) seems to be quite important. The method of images is how a the beginning of electrostatics is understood, and plays a role in the explanation of the Meissner effect in superconductors. The entirety of of the field of crystallography is based on understanding symmetries, and by extension things like understanding diffraction patterns. The difference between even and odd functions also explain the difference in timbre between brass and.. Sep 13 '17 at 20:53
• ... woodwinds in music, as well as between different types of percussive instruments. And I am merely a mathematician who doesn't do too much natural sciences. I am sure specialists can come up with many more examples. Sep 13 '17 at 20:56
Well symmetry helps investigate functions. It's the first things I need to know about the function I deal with. It helps as noted, evaluating integrals that are otherwise difficult to integrate. And it helps visualizing functions. I was a bit disapointed by the few examples given in class ass aplications of these concepts.
As with many items in math that have a name, if we didn't name it, it would still exist.
When manipulating and equation that contained Cos(-x) we are able to observe that this term is the same as Cos(x). On the other hand, when we see Sin(-x) and want to manipulate to form a positive argument, it's equal to -Sin(x).
I don't feel that we spend that much time on the concept. It's introduced in algebra, a parabola possibly being even, a third degree possibly odd, and then again in Trig with the examples I gave. Your experience may be different, more focus than I've observed. | {
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Learning mathematics isn't always about applying things to further mathematics - it can also be about actually applying that knowledge to other areas. So let me bring you a completely different answer to your question, coming from the background of someone that does programming and some web design.
You can easily apply even/odd functions to things like tables: one row has a dark background, the other has a lighter background. A simple enough requirement, that a good programmer might understand not only on the surface, but more deeply such as yourself.
I apologize if this answer is not relevant to the mathematics community directly, but I felt the need (seeing this question in a suggested list) to remind people that math isn't just about math ;) | {
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# What is the least integer of additive dimension 4?
Say that $m$ is the additive dimension of $n\in\Bbb N$, and write $m=\operatorname{ad}n$, if $m$ is the greatest integer for which there is an irredundant $m$-element set $M\subset\Bbb N$ that provides a partition of $n$ uniquely: namely $$n=k_1+\cdots+k_r\quad\text{with}\quad\{k_i:i=1,...,r\}=M,$$ for some $r\in\Bbb N$, where no other partition of $n$ involves only elements of $M$.
As an illustration, take $n=34$. We can choose $M=\{7,8,12\}$ (for example) in this case, with $34=7+7+8+12$, and so $\operatorname{ad}34=3$, since no other partition of $34$ can be formed from the elements of $M$, while any $4$-element set providing a partition of $34$ has a proper subset that does so too. (The last claim entails some checking.)
Note that $\operatorname{ad}0=0$; $\operatorname{ad}n=1$ for $n=1,2,3,4,6$; $\operatorname{ad}n=2$ for $n=5$ and $n=7,...,16$; and $\operatorname{ad}17=3$ (take $M=\{4,6,7\}$). Thus $17$ is the least integer of additive dimension $3$.
What is the least integer of additive dimension $4$? Further, what is the asymptotic behaviour of $\operatorname{ad}n$ as $n$ becomes large?
(This question was earlier posted on MathStackExchange but was not answered.)
-
Earlier post was math.stackexchange.com/questions/661256/… – Gerry Myerson Apr 8 '14 at 12:38
I wrote a quick and probably impressively sub-optimal computer program to investigate this. It takes two seconds to ascertain that the least integer of additive dimension at least 4 is 49. The relevant partition is $15+14+12+8$.
(I say "at least 4" because I haven't checked that it doesn't in fact have additive dimension 5.)
-
More a bet than a guess: the next one is $129=16+24+28+30+31$, and the numbers are $(n-1)2^n+1$ :) – მამუკა ჯიბლაძე Apr 8 '14 at 12:18
Let me prove that $A_n=(n-1)2^n+1$ has additive dimension $n$, with corresponding set $M=\{2^n-2^i\colon 0\leq i\leq n-1\}$ and representation $A_n=\sum M$. | {
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Assume that $A_n=k_1+\dots+k_r$ is some representation with $k_i\in M$. Since $k_i< 2^n$ we have $r\geq n$. Let $k_i=2^n-2^{d_i}$; then $\sum_i 2^{d_i}=(2^n-1)+(r-n)2^n$.
Lemma. If $\sum_i 2^{d_i}\equiv -1\pmod{2^k}$, then $|\{i\colon d_i<k\}|\geq k$.
Proof. If we have some equal $d_i$'s, replace them by one number $d_i+1$ with no effect to the mentioned sum. When this process stops, we have exactly one $d_i$ equal to each of $0,1,\dots,k-1$. It remains to notice that $|\{i\colon d_i<n\}|$ does not increase during the process. The lemma is proved.
Finally, the lemma inductively yields that the sum of $k\leq n$ least elements in the list $2^{d_1},\dots,2^{d_r}$ does not exceed $2^k-1$. Thus $\sum_i2^{d_i}\leq 2^n-1+(r-n)2^{n-1}$. Comparing this bound with the value found above we conclude that $n=r$, and the $d_i$ form a permutation of $\{0,1,\dots,n-1\}$, as required.
ADDENDUM. On the other hand, the least number in uniquely representing set $M$ of cardinality $n$ is at least $2^{n-1}$, so the least number of aditive dimension $n$ should be greater than $n2^{n-1}$.
To prove this, set $a=\min M$, $M'=M\setminus\{a\}$, and let $S$ be the set of subset sums of $M'$. Clearly, all such sums are distinct (otherwise we would obtain a different representation by replacing one subsum by another). Moreover, all the sets $S+ka$, $k=0,1,\dots$, are disjoint; otherwise we would have $s'=s+ka$ for some $s,s'\in S$, and we would be able to replace $s'$ by $s+ka$. Since $|S|=2^{n-1}$, the density argument shows that $a\geq 2^{n-1}$.
Perhaps, this argument can be extended to other elements of $M$?
-
Possible OEIS candidate, whose description sounds promising: A115981
-
Theorem: For n, $M_n:=(n-1)2^n+1$ is an upper bound.
As an example take $n=4$. now let \begin{aligned} k_1 & := 8 & = 1000 b \\ k_2 & := 12 & = 1100 b \\ k_3 & := 14 & = 1110 b \\ k_4 & := 15 & = 1111 b \\ \end{aligned}
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The numbers on the right side are the binary representations of $k_i$
${k_1,k_2,k_3,k_4}$ is a unique partition for $M_4 = k_1+k_2+k_3+k_4 = 49$.
The cases 0 to 3 are already settled in the question. Proof for $n>3$ by induction:
Suppose there would be two different partitions
$$(*) a_1k1+...+a_nk_n = b_1k1+...+b_nk_n = M_n$$
All factors $a_i$ and $b_i$ are lower than $k_1=2^{n-1}$, since otherwise $a_i*k_i\ge2^{2(n-1)}$, which exceeds $M_n=(n-1)2^n+1$ for all $n>3$. Equation $(*)$ also hold modulo $k_1=2^{n-1}$. This yields the same equation in the induction hypothesis (just drop the first binary digit!). As all coefficient are lower than the modulus, we have $a_i=b_i$ for $i>=2$. Then equation $(*)$ yields $a_1=b_1$ qed.
Conjecture: $M_n$ is already optimal for all $n$.
Edit
This proof is flawed! The induction hypothesis doesn't provide uniqueness mod $2^{n-1}$. Sorry!
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# Proof verification: powers of a group element of finite order are distinct
I'm finally attempting to conquer D&F (3rd ed) and I want to build good proof habits and fix mistakes early on. Here is the exercise (Ch. 1 ex. 32) and the following is my proof.
Prove that for $x \in G$, where $G$ is a group and $x$ has finite order $n$, all of $1, x, x^2, \ldots, x^{n-1}$ are distinct and deduce that $|x| \leq |G|$.
Proof. Let $a,b \in [0,n-1]\cap \mathbb{N}$ such that $x^a=x^b$. Then $1 = x^ax^{-a}=x^bx^{-a} = x^{b-a}$. Since $1 \leq a \lt n$ and $1 \leq b \lt n$, $b-a\lt n$. So $x^{b-a}=1 \implies b-a=0$ and $a=b$. Since $a$ and $b$ were arbitrarily chosen, $x^c$ is distinct for all $c \in [0,n-1] \cap \mathbb{N}$. We know $x^c\in G$ for all $c \in [0, n-1]\cap \mathbb{N}$ by definition of group closure, and $|[0,n-1]\cap \mathbb{N}|=n$. So $G$ contains at least $n$ elements, and $|x|\leq |G|$.
Is this a convincing proof? I feel like there might be some circular logic, and perhaps not enough detail. I was thinking of establishing that the elements form a cyclic group, and show this group is a subgroup of $G$. Any tips or comments would be greatly appreciated.
• It looks good to me. Aug 2 '16 at 16:16
Almost.
You assert $b-a < n$. I think you want to assume $b \ge a$ (without loss of generality) from the start, to make $b-a$ nonnegative as well. Otherwise you need to handle the case when it's not. | {
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• isn't $b-a \lt n$ even if $b-a$ is negative? I was thinking about clarifying that but I didn't think it would change anything Aug 2 '16 at 16:23
• @m1cky22: true, but you're appealing to the definition of order of an element, which is usually nonnegative. Aug 2 '16 at 16:25
• @m1cky22 Yes, it's less than $n$. But when it's less than $0$ you are dealing with a negative power in what follows. That needs work. Aug 2 '16 at 16:26
• I see. So it's more about keeping the consistency/logic of the definition of order rather than "the proof breaks if it's negative." Thanks, I think I can finally move on to the next chapter! Aug 2 '16 at 16:27
• @m1cky22 Proofs are all about keeping consistent to definitions. Aug 2 '16 at 16:28
Logically, everything looks good, except for @Ethan's point. If you are worried about circularity, make sure you understand why your statement “$x^{b-a} = 1 \implies b-a =0$” is true.
I think notations like
Let $a,b\in[0,n-1] \cap \mathbb{N}$ such that $x^a = x^b$.
are a bit cumbersome. Personally, I would rewrite the first sentence as
Let $a$ and $b$ be integers such that $0 \leq a \leq b \leq n-1$ and $x^a = a^b$.
Or perhaps even
Suppose $x^a = x^b$ for some integers $a$ and $b$ with $0 \leq a\leq b\leq n-1$.
In essence, you want to show $x^a = x^b \implies a=b$, so writing the first sentence that way aligns the paragraph with that logical statement.
Having a higher density of symbols does not make a more mathematically sophisticated argument.
Your proof is perfectly fine. Good Job!
One remark: One prof of me uses the following convention: $[0:n] := [0, n] \cap \mathbb N$.
This is an elegant way to write such "discrete intervals" and makes proofs more readable. Just an idea of me. Maybe you want to use it :)
• I was looking for better notation and didn't know what people use! I didn't want to write $1 \leq a \lt n$ so I chose my route, but yours is definitely better. Thanks! Aug 2 '16 at 16:24 | {
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I got confused when you brought $c$ into the mix.
Here is your proof with a lot of the argument removed. I don't think I have removed too much and it may be easier to follow.
Suppose the powers of $x$ are not all distinct.
Then there exist $a,b \in [0,n-1]\cap \mathbb{N}$ such that $b>a$ and $x^b=x^a$.
This implies that $0\lt b-a<n$, and $1=x^ax^{-a}=x^bx^{-a} = x^{b-a}$.
Since the order of $x$ is $n$, $x^{b-a}=1$ implies $n\, |\, b-a$.
This is a contradiction, since no number greater than $0$ and less than $n$ can be divisible by $n$.
So the original supposition, that the powers of $x$ are not all distinct, leads to a contradiction and must therefore be false. | {
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# Arbitrary precision spline interpolation
The current implementation of Interpolation does not allow arbitrary precision spline interpolation. Yu-Sung Chang says here that "it is not hard to implement it manually for arbitrary precision using BSplineBasis."
I am new to splines and not familiar with BSplineBasis. What is the best way to implement arbitrary precision spline interpolation in Mathematica?
I am interested primarily in an analog of the Method -> "Spline" of Interpolation which I investigated in this answer (BTW, what is the name of such a spline and of this kind of parametrization?). Also, J. M. gives in this answer an interesting implementation of a spline with centripetal parametrization, but it is not clear how to use it for interpolation.
P.S. I need an implementation where the degree of a spline is arbitrary. At the moment I am more interested in even-degree splines.
• Yu-Sung's pointer is wolfram.com/xid/0c0rpcn4bku6m-dj3q8d, where there are two examples of such a thing (I haven't tested them). What are those example lacking? – Dr. belisarius Jan 10 '14 at 12:45
• That is an example of cubic B-spline interpolation. I need an implementation where the degree of spline is arbitrary. – Alexey Popkov Jan 10 '14 at 12:51
• Please see my question about B-spline Basis Function – xyz May 28 '15 at 8:12
The OP linked to an answer of mine for interpolating over general point sets; for constructing a single interpolating function, a slight modification of my procedure is needed. (In particular, you don't need centripetal or chord-length parametrization in this case.)
data = {{0, 0}, {1/10, 3/10}, {1/2, 3/5}, {1, -1/5}, {2, 3}, {3, -6/5}};
To review the problems with using the built-in Interpolation[], let's try building an interpolating quartic spline from data:
m = 4; (* spline order *)
sp1 = Interpolation[data, InterpolationOrder -> m, Method -> "Spline"];
Now, evaluate:
sp1[3/2]
0.26753591659167364 | {
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Now, evaluate:
sp1[3/2]
0.26753591659167364
Herein lies the problem: even though the contents of data and the argument fed to sp1 are exact, the output is a machine precision number. Yeesh!
So, we go back to basics and build the interpolating spline ourselves from BSplineBasis[]. First, we generate the knots. The clamped condition is the most appropriate for interpolation, so we generate the appropriate knot sequence like so:
n = Length[data];
{xa, ya} = Transpose[data]
knots = Join[ConstantArray[xa[[1]], m + 1],
If[m + 2 <= n, MovingAverage[ArrayPad[xa, -1], m], {}],
ConstantArray[xa[[-1]], m + 1]];
To generate the control points for the B-spline, we construct the relevant linear system and solve it with LinearSolve[]:
cp = LinearSolve[Outer[BSplineBasis[{m, knots}, #2, #1] &,
xa, Range[0, Length[data] - 1]], ya];
Note that all the entries of cp are exact.
At this juncture, you might think that we can use BSplineFunction[] on the control points and the knots, but alas, the function also only evaluates at machine precision. We are left with no choice but to use BSplineBasis[] once more:
sp2[x_] = cp.Table[BSplineBasis[{m, knots}, j - 1, x], {j, n}];
Try it out:
sp2[3/2]
25251843/94386740
N[%]
0.2675359165916738
As a verification that we were able to reproduce the control points used internally by Interpolation[], let's do a comparison:
sp1bs = Cases[sp1, _BSplineFunction, ∞][[1]];
sp1bs["ControlPoints"]
{0., 0.777830630128766, 1.3470257035045392, -6.670028631670085, 13.100332135107115, -1.2}
N[cp]
{0., 0.777830630128766, 1.347025703504539, -6.6700286316700845, 13.100332135107113, -1.2}
Certainly, I can't end this post without at least one picture, so:
Plot[{sp1[x], sp2[x]}, {x, 0, 3}]
Plot[sp1[x] - sp2[x], {x, 0, 3}, PlotRange -> All]
I leave the task of bundling everything in a routine as an exercise. If you want to learn more, Piegl and Tiller's The NURBS Book is the canonical reference. | {
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• It is good to have you back. – Mr.Wizard May 24 '15 at 10:31
• I was able to borrow a friend's computer with a Mathematica installation, so I went for it and answered some questions. It was fun while it lasted… :) – J. M.'s technical difficulties May 24 '15 at 10:45
• Did you try contacting the Stack Exchange team about the matter? I understand if you do not wish to pursue it but I think they might help. – Mr.Wizard May 24 '15 at 10:48
• Maybe I'll ask them sometime… thanks for the concern. :) – J. M.'s technical difficulties May 24 '15 at 10:51
• @J.M. May I ask you also about a generalization of the spline interpolation (at least) for the 2D case? In the linked answer it is said that "Multi-dimension is simply tensor product version." I do not see this as a simple thing but it would be very helpful to have such a generalization. May be a separate thread would be the best place for this. – Alexey Popkov May 26 '15 at 14:01
Here is my implementation of quadric spline arbitrary precision interpolation which is defined exactly as in Interpolation with options Method->"Spline", InterpolationOrder->2.
## Theoretical background
Quadric spline interpolation for n datapoints is defined as a Piecewise function which consists of n-2 parabolas which are spliced together in the middles of successive datapoints (with exceptions for first and last intervals) by two conditions: successive parabolas at these middle points must be equal and their first derivatives must be equal too. So we have 2(n-3) such conditions.
Let us designate jth datapoint as {x[j],y[j]} and define jth parabola as
s[j_, xx_] = y[j + 1] + b[j] (xx - x[j + 1]) + c[j] (xx - x[j + 1])^2;
This definition automatically makes jth parabola equal to y[j + 1] at x[j + 1]. We have to add also 2 boundary conditions:
s[1, x[1]] == y[1]
s[n - 2, x[n]] == y[n]
And 2(n-3) splicing conditions: | {
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s[1, x[1]] == y[1]
s[n - 2, x[n]] == y[n]
And 2(n-3) splicing conditions:
Table[{
s[j, (x[j + 1] + x[j + 2])/2] == s[j + 1, (x[j + 1] + x[j + 2])/2],
ds[j, (x[j + 1] + x[j + 2])/2] == ds[j + 1, (x[j + 1] + x[j + 2])/2]
}, {j, 1, n - 3}]
where ds is first derivative:
ds[j_, xx_] = D[s[j, xx], xx];
Now we combine everything together and convert into matrix form:
eqs = Flatten[{
s[1, x[1]] == y[1],
Table[{
s[j, (x[j + 1] + x[j + 2])/2] == s[j + 1, (x[j + 1] + x[j + 2])/2],
ds[j, (x[j + 1] + x[j + 2])/2] == ds[j + 1, (x[j + 1] + x[j + 2])/2]
}, {j, 1, n - 3}],
s[n - 2, x[n]] == y[n]}];
arrs = Simplify@
Normal@CoefficientArrays[eqs, Flatten@Array[{b[#], c[#]} &, n - 2]];
MatrixForm /@ (4 arrs)
Looking at the matrices it is easy to see that they have periodic structure and contain only elements of the forms x[j+1]-x[j] and y[j+1]-y[j] with some numerical coefficients. This periodic structures can be expressed as SparceArrays.
## Implementation
Assuming that data contains an array of {x[j],y[j]}, we define
Δx[i_] := Subtract @@ data[[{i + 1, i}, 1]];
Δy[i_] := Subtract @@ data[[{i + 1, i}, 2]];
Two matrices in the arrs can be expressed as follows:
bB = SparseArray[{1 -> -4 Δy[1], -1 -> 4 Δy[n - 1], i_ /; OddQ[i] :> 0, i_ /; EvenQ[i] :> 4 Δy[i/2 + 1]}, 2 (n - 2)];
mM = SparseArray[
Join[{{1, 1} -> -4 Δx[1], {1, 2} -> 4 Δx[1]^2, {-1, -1} -> 4 Δx[n - 1]^2, {-1, -2} -> 4 Δx[n - 1]},
Table[Band[{2 i, 2 i - 1}] -> {{2 Δx[i + 1], Δx[i + 1]^2, 2 Δx[i + 1], -Δx[i + 1]^2}, {4, 4 Δx[i + 1], -4, 4 Δx[i + 1]}}, {i, 1, n - 3}]], {2 (n - 2), 2 (n - 2)}];
Now for obtaining coefficients b[i], c[i] we can use LinearSolve:
bcM = LinearSolve[mM, bB];
Grouping coefficients with identical indexes together:
bcM = Partition[bcM, 2];
The intervals at which the parabolas are defined:
intervList =
Partition[
Join[{data[[1, 1]]},
MovingAverage[data[[2 ;; -2, 1]], 2], {data[[-1, 1]]}], 2, 1];
Now we compile all the spline data in one object: | {
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Now we compile all the spline data in one object:
splineData = Table[{data[[i + 1]], bcM[[i]], intervList[[i]]}, {i, n - 2}];
splineData contains everything that is needed to construct the spline. It contains elements of the form:
{{x[i + 1], y[i + 1]}, {b[i], c[i]}, {xmin, xmax}}
It is very handy to use splineData for exact numerical integration (see here).
To produce the explicit piecewise function we define the constructor (HornerForm is already applied and gives 27% speedup):
makeSpline[splineData_List, x_Symbol] :=
Piecewise[
Append[Table[{d[[1, 2]] - d[[1, 1]] d[[2, 1]] +
d[[1, 1]]^2 d[[2, 2]] + x (d[[2, 1]] + x d[[2, 2]] -
2 d[[1, 1]] d[[2, 2]]), #1 <= x <= #2 & @@ d[[3]]}, {d, splineData}],
{Indeterminate, True}]]
It can be used as follows:
spline[\[FormalX]_] = makeSpline[splineData, \[FormalX]];
Compiling the code for creation of splineData into one function:
ClearAll[toSplineData, makeSpline];
Options[toSplineData] = {Method -> Automatic};
toSplineData[data_, OptionsPattern[]] /; MatrixQ[data, NumberQ] :=
Module[{Δx, Δy, bB, mM, bcM, intervList,
n = Length[data], dataS = Sort[data]},
Δx[i_] :=
Subtract @@ dataS[[{i + 1, i}, 1]]; Δy[i_] :=
Subtract @@ dataS[[{i + 1, i}, 2]];
bB = SparseArray[{1 -> -4 Δy[1], -1 ->
4 Δy[n - 1], i_ /; OddQ[i] :> 0,
i_ /; EvenQ[i] :> 4 Δy[i/2 + 1]}, 2 (n - 2)];
mM = SparseArray[
Join[{{1, 1} -> -4 Δx[1], {1, 2} ->
4 Δx[1]^2, {-1, -1} ->
4 Δx[n - 1]^2, {-1, -2} ->
4 Δx[n - 1]},
Table[Band[{2 i,
2 i - 1}] -> {{2 Δx[i + 1], Δx[
i + 1]^2,
2 Δx[i + 1], -Δx[i + 1]^2}, {4,
4 Δx[i + 1], -4,
4 Δx[i + 1]}}, {i, 1, n - 3}]], {2 (n - 2),
2 (n - 2)}];
bcM = LinearSolve[mM, bB, Method -> OptionValue[Method]];
bcM = Partition[bcM, 2];
intervList =
Partition[
Join[{dataS[[1, 1]]},
MovingAverage[dataS[[2 ;; -2, 1]], 2], {dataS[[-1, 1]]}], 2, 1];
Table[{dataS[[i + 1]], bcM[[i]], intervList[[i]]}, {i, n - 2}]];
Now to construct the spline from data we can evaluate | {
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Now to construct the spline from data we can evaluate
spline[\[FormalX]_] = makeSpline[toSplineData[data], \[FormalX]];
Any suggestions and improvements are welcome!
## What is so special about quadric spline interpolation?
Quadric spline interpolation with splicing points in the middle of successive data points is locally invariant and weakly depends on boundary condition. (But the same is not true for quadric spline interpolation with splicings in data points.) It gives much lesser artifacts than "usual" cubic spline interpolation. Moreover, it seems that splines of higher degree do not give any advantages over such quadric splines. As opposed to cubic spline, data points are not special points at which polynomials are spliced, and it makes much easier exact integration in these points. | {
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• Could you give a demo of your solution? THANKS:) – xyz Oct 10 '15 at 10:38
• @Shutao As you can see from the definition for the parabolas (the first formula in my answer) from which the spline is formed, my solution is non-parametric and so (as well as Interpolation) it is not intended for interpolating an arbitrary dataset (like yours). – Alexey Popkov Oct 10 '15 at 11:40
• @Shutao My solution does not use BSplineBasis and is not intended to be compatible with BSplineFunction and BSplineCurve. But it has an advantage of being very simple and efficient, mathematically clear and very handy when one needs to perform exact integration of interpolated data (without using NIntegrate). Actually I developed my solution exactly for the latter kind of applications. – Alexey Popkov Oct 10 '15 at 11:40
• @Shutao While I appreciate your solution very much I should note that your answer (as well as belisarius' answer) does not answer the original question because the question is explicitly about creation of arbitrary-precision version of Interpolate (preferably with richer set of spline interpolation methods), not about construction of BSplineCurve from a set of points. – Alexey Popkov Oct 10 '15 at 11:40
• @Shutao It would be great if you add a non-parametric version of your solution as another answer: something like BSplineInterpolate[pts][x]: BSplineInterpolate would accept a list of points as the first argument and return something line InterpolatingFunction (in my answer makeSpline returns a Piecewise function object), the latter can accept the abscissa x as its argument and return the interpolated value of the ordinate y. – Alexey Popkov Oct 10 '15 at 11:41
The following is a shameful plug of J. M.'s answer that you already linked to show how to get the parametrization by using BSplineBasis[] with arbitrary precision and packed into a function: | {
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splineInterp[data_, order_, prec_] :=
Module[{parametrizeCurve, tvals, bas, ctrlpts, knots},
parametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] :=
FoldList[
Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]];
tvals = parametrizeCurve[data];
(*knots for interpolating B-spline*)
knots =
Join[
ConstantArray[0, order + 1],
ConstantArray[1, order + 1]];
(*basis function matrix*)
bas =
Table[
BSplineBasis[{order, knots}, j - 1, N[tvals[[i]], prec]],
{i, Length[data]}, {j, Length[data]}];
ctrlpts = LinearSolve[bas, data];
Return[
Sum[
ctrlpts[[i + 1]] BSplineBasis[{order, knots}, i, #],
{i, 0, Length[testData] - 1}] &]
]
Usage
SeedRandom[42];
testData = RandomReal[{0, 1}, {10, 2}, WorkingPrecision -> prec];
f = splineInterp[testData, 5, 50];
f[1/10]
ParametricPlot[
f[t], {t, 0, 1},
Axes -> None,
Epilog ->
{Directive[Green, PointSize[Large]], Point[testData]}, Frame -> True
]
(*
{0.06025513038208479326777055464103084790562316, \
0.5927955887343273319037301352154452480202510}
*)
Please refer to J. M.'s answer for the details.
### Explanation
The centripetal distribution:
The function parametrizeCurve complete this algorithm
parametrizeCurve[pts_ /; MatrixQ[pts, NumericQ], a : (_?NumericQ) : 1/2] :=
FoldList[
Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] | {
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• I do not see in your answer how can I use this as interpolation in a regular way: interpolation[x] where x is not a parametrization variable t but the actual independent variable on which the dependent variable y depends by y[x]. – Alexey Popkov Jan 10 '14 at 16:28
• @AlexeyPopkov, Please see my explanation – xyz May 29 '15 at 13:04
• @ShutaoTang Thank you for the answer to my comment, now I see that the second argument of parametrizeCurve may be arbitrary but must be lesser or equal to $1$. BTW it would be better for others if you would post this explanation under your own answer. :^) – Alexey Popkov May 29 '15 at 13:22
• @AlexeyPopkov, My pleasue. I used the chord-length method to generate the parameters in my answer, so I think it is suitable to place this explanation in belisarius's answer:-) – xyz May 29 '15 at 13:34 | {
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# Factoring a polynomial
I am trying to factor the following polynomial: $$4x^3 - 8x^2 -x + 2$$
I am trying to do the following: $4x^2(x - 2)-x+2$ but I am stuck.
edit: correction.
-
It might be better to write it as $4x^2(x-2)- (x-2)$ to take advantage of what you have done so far. – Geoff Robinson Jan 24 '12 at 19:04
It may be of value to note that there is a general formula for finding roots of 3rd degree polynomials. The formula looks complex but may be of help if you can't guess a root. See this reference for more details: en.wikipedia.org/wiki/Cubic_function – Emmad Kareem Jan 25 '12 at 20:57
You can factor as
$$4x^2(x-2)-(x-2)=4x^2(x-2)-(1)(x-2)$$
Then factor out the $x-2$ to get
$$(x-2)(4x^2-1).$$
But, you may further factor $4x^2-1$ to get
$$(x-2)(2x-1)(2x+1).$$
-
+1:nice explanation... – pedja Jan 24 '12 at 19:07
It wasn't "factored wrong"; it's just that it wasn't taken far enough. – Michael Hardy Jan 24 '12 at 19:19
I agree with Michael. Could you perhaps edit that out? I think it's a little bit impolite. Incomplete and wrong are two very different things. :) – 000 Jan 25 '12 at 1:08
If you look at the revisions, the original posting contained an error where $(x-2)$ was incorrectly $(x-2x)$. I was not being impolite. But, I'll edit it out since things have changed. – Joe Johnson 126 Jan 25 '12 at 20:02
I'll use your polynomial to illustrate a more general procedure for factoring polynomials with integer coefficients (and assuming it has at least one rational root):
First, guess a root of $4x^3-8x^2-x+2$. The so called "rational roots test" will be helpful here.
Eventually, you'll discover that $x=2$ is a root of $4x^3-8x^2-x+2$. This will imply that your polynomial has the form $$\tag{1}(x-2)(ax^2+bx+c),$$ for some constants $a, b, c$.
To find those constants, you could do one of two things (and maybe more) | {
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To find those constants, you could do one of two things (and maybe more)
1. perform the division $4x^3-8x^2-x+2\over x-2$.
2. expand (1) and set it equal to the original polynomial. Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.
Once you've figured out what $a,b$, and $c$ are, factor the quadratic.
Using method (2), we have:
$$4x^3-8x^2-x+2 = ax^3+(b-2a)x^2+(c-2b)x-2c$$
A moment's reflection reveals that $c=-1$; whence $b=0$; whence $a=4$. Thus \eqalign{ 4x^3-8x^2-x+2 &= (x-2)(4x^2-1)\cr &= (x-2)(2x+1)(2x-1).\cr }
Of course, the other answers are more suitable to your problem; but in the event that your polynomial doesn't factor nicely (such as for $x^3+6x^2+11x+6$), you might try using this approach.
-
Franco, you said you are stuck. Did you have an approach?
In general, for a cubic equation to be factored means the person who is asking this question is indirectly giving you a hint that there is at least one real root.
Your approach should be as follows: Step 1: first to try to see if x=0, 1 -1, 2 or -2 is a root. Chances are in some cases that one of these is a root. In that case, let us say you figured that $x=2$ is a root of $4x^3 - 8x^2 -x + 2$, then if you know factoring polynomials a little bit then you could immediately figure $4x^3 - 8x^2 -x + 2 = (x-2)(4x^2-1)$ which means the other factors can be found by the fact that $(4x^2-1^2) = (2x-1)(2x+1)$. If in some cases you cannot find the obvious root as described in Step 1, then Step 2: Observe if there is a pattern like $4x^3-8x^2$ has two coefficients $4$ and $8$, which has the same pattern as $-x+2$, i.e. $4x^3-8x^2 = 4x^2(x-2)$ and $-x+2 = -(x-2)$, which means you can combine those as $$(x-2)4x^2-(x-2) = (x-2)(4x^2-1)$$
And in some cases as here you could also do $(4x^3-x) - (8x^2-2)$ which gives you
$$x(4x^2-1)-2(4x^2-1) = (x-2)(4x^2-1) = (x-2)(2x-1)(2x+1)$$ | {
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$$x(4x^2-1)-2(4x^2-1) = (x-2)(4x^2-1) = (x-2)(2x-1)(2x+1)$$
Luckily with this question, Step 1 and Step 2 worked. What if you have a cubic equation that does not have obvious ways to factor.
Step 3: For a general cubic equation $ax^3+bx^2+cx+d=0$, apply the substitution
$$x = y - \frac{b}{3a}$$
then we get
$$a\left(y-\frac{b}{3a} \right)^3 + b\left(y-\frac{b}{3a} \right)^2+c\left(y-\frac{b}{3a} \right)+d = 0$$
which simplifies to
$$ay^3 + \left( c-\frac{b^2}{3a}\right)y+ \left(d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0$$
This is called a depressed cubic equation, because the square term is eliminated. It is much easier to use this and then find the roots. (back substitute to get the roots in terms of $x$)
For example $$2x^3-18x^2+46x-30=0$$
Substitute $x=y+3$ and simplify this cubic equation to
$2y^3-8y=0 \Rightarrow y=0,2,-2$ which then gives the roots as $x=1,3,$ and $5$.
-
$(4x^3-x)-(8x^2-2)=x(4x^2-1)-2(4x^2-1)=(x-2)(4x^2-1)=$
$=(x-2)(2x-1)(2x+1)$
-
+1: Yours didn't appear until after I sent mine. – Joe Johnson 126 Jan 24 '12 at 19:04 | {
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# Confused about proof that $\log(n!) = \Theta(n \log n)$
So I was able to show that:
$\log(n!) = O(n\log n)$ without any problems.
My question is when trying to prove that $\log (n!) = \Omega(n\log n)$.
I was able to show that:
\begin{align*} \log n! &= \log(1 \cdot 2 \cdot 3 \cdots n )\\ &= \log 1 + \log2 + \log3 + \dots + \log n \\ &= \log 1 + \dots + \log\tfrac{n}{2} + \dots + \log n\\ &\geq \log\tfrac{n}{2} + \log\big(\tfrac{n}{2} + 1\big) + \dots + \log n &&\text{(i.e., the larger half of the sum)}\\ &\geq \log\big(\tfrac{n}{2}\big) + \log\big(\tfrac{n}{2}\big) + \dots + \log\big(\tfrac{n}{2}\big)&&\text{(adding \tfrac{n}2 times)} \\ &= \log\big(\tfrac{n}{2} \cdot \tfrac{n}{2} \cdots \tfrac{n}{2}) &&\text{(\tfrac{n}{2} times)} \\ &= \log\Big(\tfrac{n}{2}^{\tfrac{n}{2}}\Big)\\ &= \tfrac{n}{2} log\big(\tfrac{n}{2}\big) &&\text{(by log exponent rule)} \end{align*}
Thus, $\log(n!) \geq \tfrac{n}{2}\log\big(\tfrac{n}{2}\big)$, so we conclude that $\log(n!) = \Omega(n\log n)$.
I don't understand how finding the lower bound of $\log(n!)$ is found by getting the larger half of the sum. Why is that chosen to find $\Omega(n\log n)$? I feel like it's probably something obvious but it's the only thing keeping me from fully grasping the proof. If someone can enlighten me, I would appreciate it!
• Concisely: 1/2 is the simplest number in the interval (0,1).
– user12859
Sep 26, 2015 at 9:31
• @RickyDemer So what? Splitting the sum as $\log 1 + \dots + \log cn + \dots + \log n$ for any $0<c<1$ would work just as well. Sep 26, 2015 at 10:17
Consider the sum $S = \log(1) + \dots + \log(n)$. We're going to break it into two parts: $S=T+U$, where
\begin{align*} T &= \log(1) + \log(2) + \dots + \log(n/2)\\ U &= \log(n/2+1) + \dots + \log(n-1) + \log(n). \end{align*}
Basically, $T$ has the first $n/2$ terms of $S$, and $U$ has the remaining $n/2$ terms.
Now we'll lower-bound each of them. Start with $T$. Each term in $T$ is at least $\log(1)$, so we get | {
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$$T \ge \log(1) + \log(1) + \dots + \log(1) = 0 + 0 + \dots 0,$$
so $T \ge 0$. Next look at $U$. Each term in $U$ is at least $\log(n/2)$, so we get
$$U \ge \log(n/2) + \log(n/2) + \dots + \log(n/2) = (n/2) \times \log(n/2).$$
Now $S = T+U$, so plugging in the lower bounds obtained above, it follows that
$$S =T+U\ge 0 + (n/2) \times \log(n/2).$$
This is exactly the result you wanted to prove. Also, $(n/2) \times \log(n/2)$ is $\Omega(n \log n)$, so this proves that the sum $S$ is $\Omega(n \log n)$.
• Ok I think I'm starting to understand, but first, what is stopping me from not breaking up the sum into two parts and just lower bounding each term in the sum? Like this: log(1) + log(1) + ... + log(1) n times. Sep 26, 2015 at 5:21
• @Ghost_Stark, you can do that. That gives you the lower bound $S \ge 0$, which is a valid lower bound, but not very useful. (The lower bound $S \ge -42$ is also true, but even less useful.) It's just like if I tell you I'm thinking of a number between 1 and 100, and you ask for a hint, and I tell you: well, it's larger than $0$. You probably won't be amused at my useless hint, as you already knew that. It's the same for what you want to do. There are many lower bounds that one can prove -- the trick is to find one that is strong enough to prove what you want to prove.
– D.W.
Sep 26, 2015 at 5:23
• Ah I understand now! Thanks for the answer and the explanation! Very helpful. I guess my problem was not understanding that you have to have a reasonable lower bound to look for. Sep 26, 2015 at 5:31
$\dfrac n2$ is chosen so that you have sufficiently many factors ($O(n)$) that are sufficiently large ($O(n)$), so that the product remains $\sim n^n$ while being a lower bound to $n!$
The intuition is that $\log x$ is slowly-growing, and consequently "most" of the terms will be around $\log n$ in size. | {
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More precisely, if there are $\Theta(n)$ terms that are all $\Theta(\log n)$ in size, then their sum will indeed be $\Theta(n \log n)$ and we can conclude $\log n! \in \Omega(n \log n)$.
Taking half of the terms is merely the simplest idea to describe and calculate, and fortunately it satisfies the needed conditions. | {
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“时间-频率”译为未确定词的双语例句
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time-frequency
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The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy... The equipment is aimed at manipulating five sectionalizing posts and provided with both, telecontrol and telesignaling. Time-frequency system is employed for the channel. Telesignaling from the five sectionalizing posts works continually and automatically in cyclic order. Self-checking duel information codes are adopted for each operation, in which the direct code is being compared with its inverse counterpart. The logic design is suitable for use, the circuitry is simple, and the equipment is reliable and easy in maintenance. 本文介绍了一套用于电气化铁道分区亭的电子装置。它是1对5的分散集中式系统,遥控与遥信综合于同一系统中。通道采用时间-频率划分。各被控站遥信为经常自动循环传送方式,信息码采用二次传送,正反码校核。逻辑设计合理,电路简单,工作可靠,维修方便。 126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge... 126 lateral geniculate neurones were examined on unanaesthetized immobili-zed cats.The temporal frequency tuning curves of single neurones were measuredby stimulating the cat's eye with a sinusoidally modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its | {
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modulated light spot,generated ona CRT and presented to each neurone's receptive field center or its surround.Theaverage discharging rate was used as an index to judge the sensitivity to different mo-dulating frequencies.By comparing the mean impulse rates responding to modula-ted light stimulations(modulated discharge rate)with those responding to unmodulatedlight(unmodulated discharge rate),two opposite effects were observed.The majorityof the cells(93.7)% showed modulation-excitatory curves in which the modulateddischarge rates were higher than the unmodulated ones.A minority of the cells(6.3%)showed modulation-inhibitory curves in which the modulated dischargerates were lower than the unmodulated ones.The modulation-excitatory curvescould further be grouped into two subtypes,the“band-pass filters”and the“low-pass filters”.Similarly,the modulation-inhibitory curves could also be dividedinto two subtypes,the“band-rejection filters”and the“low-rejection filters”.Onboth sides of the temporal frequency tuning curve subsidiary sidebands oppositein direction to the main part of the curves could often be seen,i.e.inhibitorysidebands flanking the modunation-excitatory tuning curve and excitatorysidebands flanking the modulation-inhibitory tuning curve.Most of the tuningcurves have only one peak.of the 110 curves measured the peak loci show anormal distribution which centered at 7 Hz.The tuning curves due to stimulationof different locations of the periphery were almost the same.But the tuningproperties of the receptive field centers were different from those of the periphe-ries either in shape or in band-width. 用示波器产生亮度受正弦波调制的小光点刺激清醒猫外膝体神经元的感受野,以不同调制频率下神经元放电的平均频率为指标,分析了126个细胞感受野中心的时间频率调谐特性,主要结果如下。(1)大多数(93.7%)细胞呈调制-兴奋型反应,即刺激光的时间调制在一定频率范围内使放电频率增加;少数(6.3%)细胞呈调制-抑制型反应,即在一定频率范围内,时间调制使放电减少。(2)根据调谐曲线的形状和通带宽度,调制-兴奋型反应包括带通滤波器和低通滤波器两种类型,其110个调谐曲线的峰值分布接近正态曲线,多数细胞对7Hz 的调谐最敏感。调制一抑制型反应包括带除滤波器和低除滤波器两类。(3)调制-兴奋型曲线的通带旁边较當出现抑制性侧带,调制-抑制型曲线出现兴奋性侧带。(4)感受野中心区与外周区的时间频率调谐曲线的带宽和形状有所不同。 This | {
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This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic... This paper makes to model and predict the measured data in Variant Sample Time of Hydrogen, Cesium and Rubidium atomic clocks, mainly according to the principle of ARIMA models of time series analysis method with the Bias function theory of noise Variance of measuring statistics of atomic time/frequency and method of spline functions. Some preliminary useful predicting models and conclusions obtained in variant noise process have pretical sense in optimal predictions and monitoring of the behaviour of atomic time/frequency and in developing new optimal filting control method to improve the performances of atomic clocks. 本文主要根据时间序列分析方法的ARIMA(autoregressive integraled moving averge)模型原理[1-6],结合原子时间频率测量统计学的噪声方差偏函数理论[7-9],以及适当采用样条函数方法[10],对氢,铯、铷等原子钟不同取样时间的测量数据进行建模和预报。对于不同噪声过程,初步得到了一些有用的预报模型和结论,这对于原子时间频率性能的最佳预报和监测(如跳相和跳频)以及研究新的最佳滤波控制方法以提高原子钟的性能指标都有实际意义。 << 更多相关文摘 | {
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# How many 3 digit numbers that the sum of their digits equals 12?
How many positive 3-digit numbers exist such that the sum if their digits equals 12?
A) 54
B) 61
C) 64
D) 65
E) 66
I believe the answer is E.
Online problems state that is a stars and bars problem, however using the (n-1,k-1)/n-1C k-1 and (n+k-1,n)/n+k-1 C n formulas do not yield any of the answer choices.
Using (n-1,k-1) I get 55 and using (n+k-1,n) I get 91. Fifty five appears to be the closest answer.
I do not know how to do the inclusion method to remove results such as 066.
• Please show us your calculations so that we can see what you did and where you ran into difficulties. – N. F. Taussig Jul 23 '16 at 12:12
• Online problems state that is a stars and bars problem, however using the n-1 C k-1 and n+k-1 C n formulas do not yield 54. I get 190 from n+k-1 C n and 139 from n-1 C k-1 – G_Derek007 Jul 23 '16 at 12:21
• Please edit your question to show us exactly what you did. I suspect you have not taken into account the restrictions on the hundreds digit, tens digit, and units digit. – N. F. Taussig Jul 23 '16 at 12:23
• I have changed the question to 12 however it should still be the same concept – G_Derek007 Jul 23 '16 at 12:29
• If I read your work correctly, you used the formula for solutions in the positive integers to obtain $\binom{12 - 1}{3 - 1} = \binom{11}{2} = 55$ and for solutions in the non-negative integers to obtain $\binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} = 91$. Is that what you did? – N. F. Taussig Jul 23 '16 at 12:36
A three-digit positive integer has the form $100h + 10t + u$, where the hundreds digit $h$ satisfies the inequalities $1 \leq h \leq 9$, the tens digit $t$ satisfies the inequalities $0 \leq t \leq 9$, and the units digit $u$ satisfies the inequalities $0 \leq u \leq 9$. Therefore, we wish to determine the number of solutions of the equation $$h + t + u = 12 \tag{1}$$ subject to these restrictions. | {
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If we let $h' = h - 1$, then $0 \leq h' \leq 8$. Substituting $h' + 1$ for $h$ in equation 1 yields \begin{align*} h' + 1 + t + u & = 12\\ h' + t + u & = 11 \tag{2} \end{align*} Equation 2 is an equation in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of eleven ones. There are $$\binom{11 + 2}{2} = \binom{13}{2}$$ such solutions since we must choose which two of the thirteen symbols (eleven ones and two addition signs) will be addition signs.
However, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$. We must exclude these.
Suppose $h' > 8$. Then $h'$ is an integer satisfying $h' \geq 9$. Let $h'' = h' - 9$. Then $h'' \geq 0$. Substituting $h'' + 9$ for $h'$ in equation 2 yields \begin{align*} h'' + 9 + t + u & = 11\\ h'' + t + u & = 2 \tag{3} \end{align*} Equation 3 is an equation in the non-negative integers. Since a particular solution corresponds to the placement of two addition signs in a row of two ones, it has $$\binom{2 + 2}{2} = \binom{4}{2}$$ solutions.
Suppose $t > 9$. Then $t$ is an integer satisfying $t \geq 10$. Let $t' = t - 10$. Substituting $t' + 10$ for $t$ in equation 2 yields \begin{align*} h' + t' + 10 + u & = 11\\ h' + t' + u & = 1 \tag{4} \end{align*} Equation 4 is an equation in the non-negative integers with $\binom{3}{2} = 3$ solutions, depending on which variable is equal to $1$. By symmetry, there are also three solutions in which $u' > 9$. No two of these restrictions cannot be violated simultaneously since $9 + 10 = 19 > 12$. Thus, the number of three-digit positive integers with digit sum $12$ is $$\binom{13}{2} - \binom{4}{2} - 2\binom{3}{2} = 66$$ | {
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• It became a lot more clearer thanks! Is it too much to ask if its possible to derive a general solution? – G_Derek007 Jul 23 '16 at 13:20
• I believe I derived one already to be sure though if the sum of the digits were 14 it would be 70? The formula is S=sum of digits D=number of digits (S+D-1,D-1) - (S+D-1-9,D-1) - 2*(S+D-1-9-1,D-1)it should work for 3 digit numbers, which is what I need – G_Derek007 Jul 23 '16 at 13:35
• @user295641 Yes, there are 70 3 digit numbers whose digits sum to 14. – PM 2Ring Jul 23 '16 at 13:38
• That would be a rather long answer. For $h + t + u = k$, where $1 \leq k \leq 9$, you can stop after finding the solution to equation 2 with $k$ replacing $12$. For $h + t + u = 10$, you can stop after equation $3$ with $10$ replacing $12$. For $h + t + u = k$, where $11 \leq k \leq 18$, the same procedure will work if you replace $12$ by $k$. For $18 \leq k \leq 27$, we require the Inclusion-Exclusion Principle because two or more restrictions can be violated simultaneously in equation 2. – N. F. Taussig Jul 23 '16 at 13:39
• @user295641 You have an off-by-one error in that formula; I think you mean: $(S+1, 2) - (S-8, 2) - 2*(S-9, 2)$. However, that formula only works for $9 \le S \le 19$. Note that there is a symmetry. Let $f(S, D)$ be the count of $D$ digit numbers that sum to $S$. Then $f(9D+1-S, D) = f(S, D)$ – PM 2Ring Jul 23 '16 at 14:09
Here's a less algebraic way to look at it. .
Start by putting $1$ into the first cell, ${\boxed 1}\Large\boxed.\boxed.$, so you now only need a sum of $11$,
with the constraints that you can't put $\ge9$ in the first cell, and $\ge10$ in the other two.
Apply stars and bars, subtracting solutions that violate the constraints.
Since it is not possible to violate the constraint in more than one cell, $$ans = \binom{13}2 - \binom{13-9}2 - 2\binom{13-10}2 = 66$$ | {
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Interesting property related to the sums of the remainders of integers
Cross-posted on Math Overflow too
Let us define, $$r(b)=\sum_{k=1}^{\lfloor \frac{b-1}{2} \rfloor} (b \bmod{k})$$ After playing around with the $$r(b)$$ function for sometime I noticed that $$r(b)$$ appreared to be more even than odd. So to see the difference between the number of even and odd terms of $$r(b)$$, I defined a function, $$z(x)=\sum_{n=1}^x(-1)^{r(n)}$$ When user Peter ran a program for computing values of $$z(x)$$ in PARI, I observed that for $$x\le 10^{10}$$, $$z(x)\gt 0$$. This suggests that there are always more even terms of $$r(n)$$ than odd terms for any $$x$$.
This leads to my two questions:
1. Is $$z(x)$$ always positive? If so, then how do we prove this?
2. Is $$|z(x)|$$ bounded by some maximum value? If so, then what is this maximum value? Till now the maximum value of $$|z(x)|$$ found was $$49$$ for $$x = 5424027859$$. I find it odd that $$|z(x)|$$ goes to these large values and then returns back to small values as small as $$1$$.
Edit:
With the help of user Vepir I was able to plot $$z(x)$$ and noticed that the function has a sinusoidal-fractal-like appearance and it seems to grow without bounds, although very very slowly.
Is there any reason for this sinusoidal and fractal nature? | {
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Is there any reason for this sinusoidal and fractal nature?
• The efficient calculation is based on the formula for $d(n):=r(n)-r(n-1)$ , which is $$d(n)=2n-1-\sigma(n)$$ for even $n$ and $$d(n)=\frac{3n-1}{2}-\sigma(n)$$ for odd $n$ and the fact that $\sigma(n)$ is odd if and only if $n=k^2$ or $n=2k^2$ for some positive integer $k$. – Peter Mar 19 at 21:50
• $r(n)$ and $r(n-1)$ have the same parity if and only if $n$ is of the form $k^2,2k^2$ or $4k+3$ for some positive integer $k$ (in the case $4k+3$ , $k=0$ is also possible). – Peter Mar 19 at 21:54
• If my calculation is correct, we have $z(31988856^2)=80$ and the function does not really feel like it would be bounded from above. – Peter Košinár Mar 29 at 2:02
• @PeterKošinár thanks for the calculation. Was $z(n)$ ever negative in that range? – Mathphile Mar 29 at 4:48
• @Mathphile No, I have not found any case where $z(n)$ would be negative or zero. It has reached $1$ on multiple occasions, though. (again, assuming my computation was correct). – Peter Košinár Mar 29 at 16:43
I can't prove/disprove your conjectures, but I proved a claim which might be useful to prove/disprove your conjectures.
This answer proves the following claim :
Claim :
\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align} where $$c(n)=\#\{x:\text{ 1\le x\le n, and x is either a square or twice a square}\}$$ and $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$ | {
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According to oeis.org/A071860, one has $$c(n)=\lfloor\sqrt{n}\rfloor+\left\lfloor\sqrt{\frac n2}\right\rfloor$$ which might make the problem easier to deal with.
Also, it follows from the above claim that $$z(8m+4)=z(8m+6)$$ for every non-negative integer $$m$$.
The claim follows from the following lemmas :
Lemma 1 : $$r(n)=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n}\sigma(k)$$
Lemma 2 : $$r(n)\stackrel{\text{mod 2}}\equiv\begin{cases}\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 0,2,3,5\pmod 8}\\\\1+\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 1,4,6,7\pmod 8}\end{cases}$$
Lemma 3 : $$\text{\sigma(n) is odd \iff n is either a square or twice a square}$$
Lemma 4 : $$\sum_{k=1}^{n}\sigma(k)\equiv c(n)\pmod 2$$where $$c(n)=\#\{x:\text{ 1\le x\le n, and x is either a square or twice a square}\}$$
Lemma 5 : If $$n\equiv 3\pmod 4$$, then $$c(n)=c(n-1)$$.
Lemma 6 : $$z(x)=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Lemma 7 :
\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align} | {
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where $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
Lemma 1 : $$r(n)=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n}\sigma(k)$$
Proof : For $$n=1$$, it is true. In the following, let us use $$r(n+1)-r(n)=\begin{cases}2n+1-\sigma(n+1)&\text{if n is odd}\\\\\frac{3n+2}{2}-\sigma(n+1)&\text{if n is even}\end{cases}$$
Suppose that it is true for $$n=2m+1$$. Then, we get \begin{align}&r(n+1) \\\\&=r(n)+2n+1-\sigma(n+1) \\\\&=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg) -\sum_{k=1}^{n}\sigma(n)+2n+1-\sigma(n+1) \\\\&=(2m+1)^2-\frac{m(m+1)}{2}+2(2m+1)+1-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+2)^2-\frac{m(m+1)}{2}-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+2)^2-\frac{1}{2}\bigg(\left\lfloor\frac{2m-1}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{2m-1}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(n+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)\end{align}
Suppose that it is true for $$n=2m$$. Then, we get \begin{align}&r(n+1) \\\\&=r(n)+\frac{3n+2}{2}-\sigma(n+1) \\\\&=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg) -\sum_{k=1}^{n}\sigma(k)+\frac{3n+2}{2}-\sigma(n+1) \\\\&=(2m)^2-\frac{m(m-1)}{2}+3m+1-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+1)^2-\frac{(m+1)m}{2}-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(2m+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{2m+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{2m+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k) \\\\&=(n+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)\end{align}
So, it is also true for $$n+1$$.$$\quad\square$$ | {
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So, it is also true for $$n+1$$.$$\quad\square$$
Lemma 2 : $$r(n)\stackrel{\text{mod 2}}\equiv\begin{cases}\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 0,2,3,5\pmod 8}\\\\1+\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if n\equiv 1,4,6,7\pmod 8}\end{cases}$$
Proof : It follows from Lemma 1 that$$r(8m)=64m^2-2m(4m-1)-\sum_{k=1}^{8m}\sigma(k)\equiv \sum_{k=1}^{8m}\sigma(k)\pmod 2$$
$$r(8m+1)=(8m+1)^2-2m(4m+1)-\sum_{k=1}^{8m+1}\sigma(k)\equiv 1+\sum_{k=1}^{8m+1}\sigma(k)\pmod 2$$
$$r(8m+2)=(8m+2)^2-2m(4m+1)-\sum_{k=1}^{8m+2}\sigma(k)\equiv \sum_{k=1}^{8m+2}\sigma(k)\pmod 2$$
$$r(8m+3)=(8m+3)^2-(2m+1)(4m+1)-\sum_{k=1}^{8m+3}\sigma(k)\equiv \sum_{k=1}^{8m+3}\sigma(k)\pmod 2$$
$$r(8m+4)=(8m+4)^2-(2m+1)(4m+1)-\sum_{k=1}^{8m+4}\sigma(k)\equiv 1+\sum_{k=1}^{8m+4}\sigma(k)\pmod 2$$
$$r(8m+5)=(8m+5)^2-(2m+1)(4m+3)-\sum_{k=1}^{8m+5}\sigma(k)\equiv \sum_{k=1}^{8m+5}\sigma(k)\pmod 2$$
$$r(8m+6)=(8m+6)^2-(2m+1)(4m+3)-\sum_{k=1}^{8m+6}\sigma(k)\equiv 1+\sum_{k=1}^{8m+6}\sigma(k)\pmod 2$$
$$r(8m+7)=(8m+7)^2-(2m+2)(4m+3)-\sum_{k=1}^{8m+7}\sigma(k)\equiv 1+\sum_{k=1}^{8m+7}\sigma(k)\pmod2$$ So, the claim follows.$$\quad\square$$
Lemma 3 : $$\text{\sigma(n) is odd \iff n is either a square or twice a square}$$
Proof : See here or here.
Lemma 4 : $$\sum_{k=1}^{n}\sigma(k)\equiv c(n)\pmod 2$$where $$c(n)=\#\{x:\text{ 1\le x\le n, and x is either a square or twice a square}\}$$
Proof : It follows from Lemma 3 that $$\sum_{k=1}^{n}\sigma(k)=\underbrace{\sum_{k\in A}\sigma(k)}_{\text{sum of odd numbers}}+\underbrace{\sum_{k\not\in A}\sigma(k)}_{\text{sum of even numbers = even}}\equiv \sum_{k\in A}\sigma(k)=c(n)\pmod 2$$ where $$A=\{n\ :\ \text{n is either a square or twice a square}\}$$.
Lemma 5 : If $$n\equiv 3\pmod 4$$, then $$c(n)=c(n-1)$$.
Proof : Since we have $$\text{(a square)}\equiv 0,1\pmod 4\qquad\text{and}\qquad \text{(twice a square)}\equiv 0,2\pmod 4$$ we see that if $$n\equiv 3\pmod 4$$, then $$n$$ is neither a square nor twice a square. | {
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Lemma 6 : $$z(x)=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Proof : It follows from Lemma $$1,2,3,4,5$$ that $$z(x)=\sum_{k=1}^{x}(-1)^{r(k)}=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{r(8k)}+\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{r(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{r(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{r(8k+3)}+\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{r(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{r(8k+5)}+\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{r(8k+6)}+\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{r(8k+7)}$$
$$=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+3)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+6)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
$$=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Lemma 7 : | {
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Lemma 7 :
\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m) \\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m) \\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m) \\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m) \\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m) \\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m) \\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align}
where $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
Proof : This immediately follows from Lemma 6.
Claim: For every positive integer $$b$$ we have $$r(b)\equiv r(b-1)\pmod{2}$$ if and only if either
1. $$b\equiv3\pmod{4}$$, or
2. $$b=k^2$$ for some integer $$k$$, or
3. $$b=2k^2$$ for some integer $$k$$.
The main ingredient in proving this claim is the following lemma:
Lemma: For every positive integer $$b$$ we have $$r(b)-r(b-1)=c(b)-\sigma(b),$$ where $$\sigma(b)$$ denotes the sum of all positive divisors of $$b$$, and $$c(b):=\begin{cases} 2b-1&\text{ if }\ b\equiv0\pmod{2},\\ \tfrac{3b-1}{2}&\text{ if }\ b\equiv1\pmod{2}\ \text{ and }\ b\neq3,\\ 4&\text{ if }\ b=3. \end{cases}$$
This reduces the question to a question on the parity of $$\sigma(b)$$. | {
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Proof. For every pair of positive integers $$b$$ and $$k$$ there exist unique nonnegative integers $$q(b,k)$$ and $$r(b,k)$$ such that $$r(b,k) and $$b=q(b,k)\cdot k+r(b,k).$$ This is simply dividing $$b$$ by $$k$$ with remainder $$r(b,k)$$. With this, your function $$r$$ can be expressed as $$r(b)=\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}r(b,k).$$ To find a more manageable form for $$r(b)$$, note that $$q(b,k)=\lfloor\frac bk\rfloor$$, so that $$\begin{eqnarray*} r(b)&=&\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}r(b,k) =\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\Big(b-q(b,k)k\Big)=\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k. \end{eqnarray*}$$ Then the difference of two consecutive terms can be simplified. If $$b$$ is even: $$\begin{eqnarray*} r(b)-r(b-1)&=&\left(\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\big\lfloor\tfrac{b-2}{2}\big\rfloor(b-1) -\sum_{k=1}^{\lfloor\frac{b-2}{2}\rfloor}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\left(\frac{b-2}{2}b -\sum_{k=1}^{\frac{b-2}{2}}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\left(\frac{b-2}{2}\right)(b-1) -\sum_{k=1}^{\frac{b-2}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\frac{b-2}{2}-\sum_{k=1}^{\frac{b-2}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k,\\ \end{eqnarray*}$$ and similarly if $$b$$ is odd: $$\begin{eqnarray*} r(b)-r(b-1)&=&\left(\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\big\lfloor\tfrac{b-2}{2}\big\rfloor(b-1) -\sum_{k=1}^{\lfloor\frac{b-2}{2}\rfloor}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\left(\frac{b-1}{2}b -\sum_{k=1}^{\frac{b-1}{2}}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\frac{b-3}{2}(b-1) -\sum_{k=1}^{\frac{b-3}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ | {
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-\left(\frac{b-3}{2}(b-1) -\sum_{k=1}^{\frac{b-3}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\frac{3}{2}(b-1)-\big\lfloor\tfrac{2b}{b-1}\big\rfloor\frac{b-1}{2}-\sum_{k=1}^{\frac{b-3}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k, \end{eqnarray*}$$ where the extra term $$-\big\lfloor\tfrac{2b}{b-1}\big\rfloor\frac{b-1}{2}$$ appears because the summation in $$r(b)$$ has one more term than the summation in $$r(b-1)$$, with $$k=\tfrac{b-1}{2}$$. For odd $$b>3$$ this further simplifies to $$r(b)-r(b-1)=\frac{b-1}{2}-\sum_{k=1}^{\frac{b-3}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k.$$ Now these inner summations have more familiar closed forms; to see this, note that $$\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor=\begin{cases} 0&\text{ if }\ k\nmid b\\ 1&\text{ if }\ k\mid b\\ \end{cases}.$$ So effectively these summations sum precisely the divisors of $$b$$, up to $$\lfloor\tfrac{b-2}{2}\rfloor$$. A routine check shows that the only divisors not counted are $$b$$, and if $$b$$ is even also $$\tfrac{b}{2}$$, and if $$b=3$$ also $$1$$. Then we can further simplify the differences for even $$b$$ as $$\begin{eqnarray*} r(b)-r(b-1)&=&\frac{b}{2}-1-\Big(\sigma(b)-b-\tfrac{b}{2}\Big)\\ &=&2b-1-\sigma(b), \end{eqnarray*}$$ and for odd $$b>3$$ as $$\begin{eqnarray*} r(b)-r(b-1)&=&\frac{b-1}{2}-\Big(\sigma(b)-b\Big)\\ &=&\frac{3b-1}{2}-\sigma(b), \end{eqnarray*}$$ and the latter is easily verified to also hold for $$b=3$$.$$\quad\square$$ | {
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