Motivation for the ring product rule $(a_1, a_2, a_3) \cdot (b_1, b_2, b_3) = (a_1 \cdot b_1, a_2 \cdot b_2, a_1 \cdot b_3 + a_3 \cdot b_2)$

In a lecture, our professor gave an example for a ring. He took it out of another source and mentioned that he does not know the motivation for the chosen operation.

Of course, it's likely that somebody just invented an arbitrary operation satisfying ring axioms. I'd still like to try my luck whether anyone here can decipher the operation and give any kind of motivation for that example.

On $\mathbb{R}^3$ define the operations $+$ and $\cdot$ by \begin{aligned} (a_1, a_2, a_3) + (b_1,b_2,b_3) &= (a_1+b_1,a_2+b_2,a_3+b_3) \\ (a_1, a_2, a_3) \cdot (b_1, b_2, b_3) &= (a_1 \cdot b_1, a_2 \cdot b_2, a_1 \cdot b_3 + a_3 \cdot b_2). \end{aligned} (The $+$ and $\cdot$ operations on the right side are the usual addition and multiplication from $\mathbb{R}$.) With those operations, one can confirm that $\left(\mathbb{R}^3, +, \cdot \right)$ is a ring.

• In case anyone is wondering, the multiplicative identity is $(1,1,0)$. – Alex Provost Dec 8 '17 at 19:00
• I think it's just an arbitrary thing. I can't see any motivating pattern here. If someone else does see something, that will be surprising and very interesting. Also, in case anyone is wondering, the multiplication is associative (I checked). – Zach Teitler Dec 8 '17 at 19:04
• I can't check at the moment, but is multiplication here commutative? Perhaps someone wanted to construct a noncommutative ring without referencing matricies? – Andrew Tawfeek Dec 8 '17 at 19:14
• The multiplication is not commutative, take $(1,0,0) \cdot (0,0,1) \neq (0,0,1) \cdot (1,0,0)$. – Qi Zhu Dec 8 '17 at 19:16
• mentioned that he does not know the motivation for the chosen operation Seriously?! If you show him the upper triangular matrix ring he will be rather abashed, then :) – rschwieb Dec 8 '17 at 21:36

This is just matrix multiplication in disguise. Specifically, if you identify $(a_1,a_2,a_3)$ with the matrix $\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}$, these operations are the usual matrix operations: $$\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}+\begin{pmatrix}b_1 & b_3 \\ 0 & b_2\end{pmatrix}=\begin{pmatrix}a_1+b_1 & a_3+b_3 \\ 0 & a_2+b_2\end{pmatrix}$$ $$\begin{pmatrix}a_1 & a_3 \\ 0 & a_2\end{pmatrix}\begin{pmatrix}b_1 & b_3 \\ 0 & b_2\end{pmatrix}=\begin{pmatrix}a_1b_1 & a_1b_3+a_3b_2 \\ 0 & a_2b_2\end{pmatrix}$$

• Oh of course! Nice and simple, thanks, the ring is much more motivated now! – Qi Zhu Dec 8 '17 at 21:32
• Unfortunately I realized that's exactly what I was describing a minute too late... – rschwieb Dec 8 '17 at 21:32

It is isomorphic to the ring of matrices

$$\left\{\begin{bmatrix}a_1&a_3\\0&a_2\end{bmatrix}\,\middle|\,a_1, a_2,a_3\in \mathbb R\right\}$$

It's a semiprimary ring whose Jacobson radical is the subset with $a_1=a_2=0$. The Jacobson radical is nilpotent, and $R/J(R)\cong\mathbb R\times\mathbb R$. Here is a list of more properties of such a ring.

This sort of ring is fairly famous, and has nice interpretations. One of them is that if you select a chain of subspaces $\{0\}<V<W<\mathbb R\times \mathbb R$ ($W$ of dimension $1$, $V$ of dimension $2$) then the linear transformations of $\mathbb R\times\mathbb R$ which stabilize this chain is isomorphic to this triangular matrix ring. That is, $\phi$ stabiliezes the chain if $\phi(V)\subseteq\phi(W)$.

Incidentally, you are always going to be able to extract some sort of matrix presentation for a multiplication like you are describing, because you can rely on it being a finite dimensional algebra. If it really is a valid ring multiplication, it's bilinear, and so you can work on figuring out what a logical 'basis' is and then deduce what it looks like with matrices.

• This should be the accepted and most upvoted answer. – Xam Dec 9 '17 at 5:54
• In your last paragraph, what exactly do you mean by "a multiplication like you are describing"? – Jack M Dec 9 '17 at 11:07
• @JackM Multiplication of $n$-tuples over a field, at least. Because that defines a representation with square matrices. – rschwieb Dec 9 '17 at 13:18
• @rschwieb Surely not any ring structure on $\mathbb F^n$ is isomorphic to a ring of $\mathbb F$-matrices. Maybe a ring structure in which addition is component wise, and multiplication is given by quadratic polynomials? – Jack M Dec 9 '17 at 13:29
• @JackM You're right, I should emphasize the pointwise addition. Remember I'm stipulating that the multiplication is actually a valid ring multiplication, not any rule with tuples. To rephrase, "if multiplication on elements of $F^n$ defines a valid ring structure, then the ring can be represented with $n\times n$ matrices." This is easy to see, because you just identify each tuple with the linear transformation it produces with the multiplication. – rschwieb Dec 9 '17 at 14:22


Orthogonally Diagonalize a matrix with nonreal eigenvectors

I am given the matrix A and asked to orthogonally diagonalize it.

$$A=\begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix}$$

While doing this I got $\lambda = 0,2.$ Then I found the eigenvectors corresponding to the eigenvalues to be $$\begin{pmatrix} i \\ 1 \\ \end{pmatrix}$$ and $$\begin{pmatrix} -i \\ 1 \\ \end{pmatrix},$$ respectively. While trying to divide each eigenvector by its norm I run into a problem, take $V_1$ for example $\|V_1\| = 0$ so I obviously can't divide by $V_1$ by its norm to make it orthogonal. My question is, is $A$ even orthogonally diagonizable at all? Or if the $\|V_1\| = 0$ does this mean that it is already orthogonal and I can just use my eigenvectors as is to generate the matrix $U$ such that $A=UDU^*?$

• $||V_1|| \neq 0$. Same goes for norm of $V_2$. You have to use absolute values around the squared entries. – layabout Jun 14 '18 at 18:47
• You mean to unitarily diagonalize it, no? – Mathematician 42 Jun 14 '18 at 18:48
• You are not working with a real inner product, but a Hermitian inner product, i.e. $\left\langle x,y \right\rangle=\sum_{i=1}^n x_i\overline{y_i}$. – Mathematician 42 Jun 14 '18 at 18:50
• The inner product is always positive definate. That is $\langle x,x\rangle \ge 0$ and only equals $0$ if $x = 0.$ (and the norm is $\sqrt {\langle x,x\rangle}$) To meet this criterion, we need a slightly different inner product definition when working with vectors over a complex field. $\langle x,y\rangle = x^\dagger y$ ($\dagger$ is the conjugate transpose) will do. So the norm $(i,1) = \sqrt {(-i\cdot i + 1\cdot 1)}$ – Doug M Jun 14 '18 at 18:52

The norm of a vector $v$ is

$$\|v\|^2 = (v^*)^Tv = \sum_k v_k^* v_k$$

where $^*$ denotes the complex conjugate. So in your case

$$\|v_1\|^2 = \pmatrix{i & 1}^* \pmatrix{i \\ 1} = \pmatrix{-i & 1} \pmatrix{i \\ 1} = 1 + 1 = 2$$

• Thank you very much. I had forgotten that there were special properties for cases like this. – Moseph Jun 14 '18 at 19:09
• @Moseph Happy to help – caverac Jun 14 '18 at 19:20

It happens that $\|(a,b)\|=\sqrt{|a|^2+|b|^2}$. Therefore, the norm of both vectors that you mentioned is $\sqrt2$.


# How to find the general solution of $\tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0$

Find the general solution of the equation.

\begin{eqnarray} \tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0\\ \end{eqnarray}

The answers in my textbook are $n\pi$ and $n\pi +\frac{\pi }{3}$.

Previously, I compute the similar questions by using the following operations :

\begin{eqnarray} \\\tan 3x&=&\cot 5x\\ \\\tan 3x&=&-\tan \left(\frac{\pi }{2}+5x\right)\\ \\\tan 3x&=&\tan \left(-\frac{\pi }{2}-5x\right)\\ \\3x&=&-\frac{\pi }{2}-5x\\ \\8x&=&n\pi -\frac{\pi }{2}\\ \\x&=&\frac{n\pi }{8}-\frac{\pi }{16}\\ \end{eqnarray}

But now there is a 3 in front of the second tan.

What should I do?

I do not know whether my method is correct. If you have any other methods, would you mind telling me the methods?

Update 1 : Found one of the answers, are my operations correct?

• Is your question correct? The question seems likely than $tan(x+\pi/3)=3tan(x-\pi/6)$ – MS.Kim Oct 6 '13 at 12:18
• Thx, the question now is correct. – Casper Oct 6 '13 at 12:24

If we assume that your question is to solve the equation $tan(x+\pi/3)+3tan(x-\pi/6)=0$. if we let $A=x+\pi/3, B=x-\pi/6$, then $A-B=\pi/2$, so we can transform the term $tanA$ to $tan(\pi/2+B)$.

and because $tan(\pi/2+B) = -cotB$, the equation will be $-cotB+3tanB=0$, $tan^2B=1/3$ so the general solution of $B=x-\pi/6=n\pi+\pi/6$ or $n\pi-\pi/6$

so $x=n\pi +\pi/3$ or $n\pi$

The solution is given as follows.

I don't know how to use spoiler for a block of code.

\documentclass[preview,border=12pt]{standalone}
\usepackage{amsmath}
\begin{document}
\abovedisplayskip=0pt\relax
\begin{gather*}
\tan (x+\frac{\pi}{3}) + 3\tan (x-\frac{\pi}{6})=0\\
\tan (x+\frac{\pi}{3}) = 3\tan (\frac{\pi}{6}-x)\\
\frac{\tan x + \tan \frac{\pi}{3}}{1-\tan x \tan \frac{\pi}{3}} = 3\frac{\tan \frac{\pi}{6} -\tan x}{1+\tan \frac{\pi}{6}\tan x}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = 3\frac{\frac{\sqrt 3}{3} -\tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\times\frac{3}{3}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ 3\sqrt 3 -9 \tan x}{3+ \sqrt 3\tan x}\\
(1- \sqrt 3\tan x) (3\sqrt 3 -9 \tan x) = (\tan x + \sqrt 3)(3+ \sqrt 3\tan x)\\
3\sqrt 3 -9\tan x -9\tan x +9\sqrt 3\tan^2x = 3\tan x +\sqrt 3\tan^2 x +3\sqrt 3 +3\tan x\\
8\sqrt 3\tan^2 x  -24 \tan x=0\\
\sqrt 3\tan^2 x -3\tan x=0\\
\tan x (\sqrt 3 \tan x -3)=0\\
\tan x =0 \text{ or } \sqrt 3 \tan x -3 =0\\
\tan x=0 \text{ or } \tan x =\sqrt 3\\
x=n\pi \text{ or } x=\frac{\pi}{3}+n\pi
\end{gather*}

\end{document}

• =口= This operation must take so much time when during exam...Is it the best way? – Casper Oct 6 '13 at 13:42
• @CasperLi: Not much even though you use LaTeX during the exam. :-) – kiss my armpit Oct 6 '13 at 13:44
• Are there any fast methods to input math/LaTeX? ~.~ – Casper Oct 6 '13 at 13:53
• @CasperLi: It depends on the text editor you are using. If there is a shortcut key for each macro then you can type much faster. But I think the important point is your typing speed. :-) – kiss my armpit Oct 6 '13 at 13:55


# Is "The empty set is a subset of any set" a convention?

Recently I learned that for any set A, we have $\varnothing\subset A$.

I found some explanation of why it holds.

$\varnothing\subset A$ means "for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set A". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set A). So $\varnothing\subset A$ holds.

What confused me was that, the following expression was also a vacuous truth.

For every object $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set A.

According to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set A) holds, so $\varnothing\not\subset A$ would be true, too.

Which one is correct? Or is it just a convention to let $\varnothing\subset A$?

• Maybe think of it like this: The empty set is like an empty bag. The set $A$ is like a bag with some stuff in it. It's possible to put your hand into either bag and take nothing out. Since everything you could take out of the empty bag $\varnothing$ was also in $A$, this means that $\varnothing \subset A$. This argument may need to be made rigorous, but I think the intuition is a good start. Oct 4, 2016 at 9:31
• It's a consequence of the following rule: any statement of the form 'if $x \in \emptyset$, then y' is true. That rule is itself a consequence of the agreement that "if P then Q" means "P is false or Q is true." On a philosophical level, I would distinguish this agreement from being a mere convention since it's an agreement about what reasoning itself means. But on a practical level, if you get annoyed thinking about these issues in any particular context, it's often better to shrug your shoulders, call it a convention, and move on! Oct 4, 2016 at 11:53
• It's the way math works... It is not a convention, but a theorem: we have an axiom asserting that there is a unique set $y$ such that : $\forall x \lnot (x \in y)$ and we call $y$ "the emptyset". Then we assume the def of set-inclusion : $A \subset B \leftrightarrow \forall x (x \in A \to x \in B)$ and we derive, by logical rules, that $\emptyset \subset A$, for any $A$. Oct 5, 2016 at 10:15
• It is true that for every $x$ and every set $A$, if $x$ belongs to the empty set, then $x$ both belongs to $A$ and $x$ does not belong to $A$ Oct 6, 2016 at 0:32
• “For every object $x$, if $x$ belongs to the empty set, then $x$ doesn’t belong to the set $A$.” This statement is vacuously true, indeed. However, it does not imply that $\varnothing\not\subset A$. For the latter to be true, there should actually exist some $x$ such that $x\in\varnothing$ but $x\notin A$. Oct 6, 2016 at 22:56

There’s no conflict: you’ve misinterpreted the second highlighted statement. What it actually says is that $\varnothing$ and $A$ have no element in common, i.e., that $\varnothing\cap A=\varnothing$. This is not the same as saying that $\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\varnothing\subseteq A$.

To expand on that a little, the statement $B\nsubseteq A$ does not say that if $x\in B$, then $x\notin A$; it says that there is at least one $x\in B$ that is not in $A$. This is certainly not true if $B=\varnothing$.

• @BlueRaja-DannyPflughoeft: Which is to say that he misinterpreted it: it does not mean what he understood it to mean. Oct 4, 2016 at 17:47
• @Searene You can derive anything from a contradiction. This is basic logic, and nothing wrong with it.
– orlp
Oct 5, 2016 at 7:17
• No you shouldn't say that 5=6 is true, but you can say that If 1=2, then 5=6 is true. The statement that x belongs to A is not necessarily true; what is true is precisely: for every object x, if x belongs to the empty set, then x also belongs to the set A, and that is the definition of the empty set being a subset of A. Oct 5, 2016 at 8:13
• @Searene You are making a fundamental error. You cannot derive that the conclusion is true!!! If you have a statement: $\forall x, x \in \emptyset \rightarrow \emptyset \not\subset A$ is true because $x \in \emptyset$ is false. This tells you nothing regarding $\emptyset \not\subset A$. In an implication $P \rightarrow Q$ if $P$ is false than the whole implication is true, but this does not say anything about the truth value of just $Q$. Oct 5, 2016 at 13:02
• ... vacuously true, but that doesn’t tell you anything about whether or not $\varnothing$ is a subset of $A$. The truth of $\forall x\,(x\in\varnothing\to x\in A)$, on the other hand, means by the definition of $\subseteq$ that $\varnothing\subseteq A$. Oct 5, 2016 at 17:25

From Halmos's Naive Set Theory:

A transcription:

The empty set is a subset of every set, or, in other words, $\emptyset \subset A$ for every $A$. To establish this, we might argue as follows. It is to be proved that every element in $\emptyset$ belongs to $A$; since there are no elements in $\emptyset$, the condition is automatically fulfilled. The reasoning is correct but perhaps unsatisfying. Since it is a typical example of a frequent phenomenon, a condition holding in the "vacuous" sense, a word of advice to the inexperienced reader might be in order. To prove that something is true about the empty set, prove that it cannot be false. How, for instance, could it be false that $\emptyset \subset A$? It could be false only if $\emptyset$ had an element that did not belong to $A$. Since $\emptyset$ has no elements at all, this is absurd. Conclusion: $\emptyset \subset A$ is not false, and therefore $\emptyset \subset A$ for every $A$.

• This paragraph shines with the rare virtue of sympathy with (and generosity of spirit toward) the uninitiated. +1. Oct 4, 2016 at 15:16
• I love this book. Oct 4, 2016 at 15:27
• With the text as opposed to with a picture of the text? Oct 4, 2016 at 19:41
• Okay, I don't see what search engines have to do with copyright violation issues, but in the meantime, I'd rather see the answer be accessibility compliant, and I guess it's your legal understanding that I'm responsible for any "issues" that might come up, so okay. Oct 4, 2016 at 19:48
• I like Halmos's second demonstration because it highlights a frequently used technique: For an implication to be false, there must be a witness to its falseness. That is, there is an element of some set that is a counterexample. Conversely, if it is impossible that there are counterexamples, then the implication is true. Oct 5, 2016 at 12:59

What confused me was that, the following expression was also a vacuous truth.

For every object of $$x$$, if $$x$$ belongs to the empty set, then $$x$$ doesn't belong to the set $$A$$.

As a complement (heh) to Brian Scott's (+1) answer, your argument shows that $$\varnothing \subset A^{c}$$, the complement of $$A$$. This statement is also (vacuously) true.

• So the empty set, being a subset of both the set $A$ and its complement, is a subset of the intersection of the set $A$ with its complement, which is the empty set. Oct 5, 2016 at 9:05
• I believe this is the key point. Oct 5, 2016 at 21:31
• @CarstenS: Be careful that in ZF set theory the complement of a set is never a set! Oct 7, 2016 at 14:17

Very subtle point:

"All x are not something" does not imply "Not all x are something".

The first may be vacuously true. The second one can not. If the x are vacuous then the second one has to be "vacuously false" as all x of nothing are any property so it is impossible for them not to be any property.

So "all elements of the empty set are not in $S$" does not imply "Not all elements of the empty set are in $S$" $\iff$ "It is not true that all elements of the empty set are in $S$" $\iff$ "There are some elements of the empty set that are not in $S$".

The first is vacuously true (and is equivalent to $\emptyset \subset S^c$ which is true) and the second set of equivalent statements are all equivalent to $\emptyset \not \subset S$ which is not true.

=====

The thing is what you say is absolutely correct for non empty sets.

More formally:

All elements $x$ in $S$ are not in $A$ $\implies$

$S \subset A^c$ $\implies$

$\color{red}{\text{There is an } x\in S \text{ where } x \not \in A}\implies$

It is not true that all $x \in S$ are also in $A$ $\implies$

$S \not \subset A$.

However the red line can only be concluded if $A$ is non-empty. If $A$ is empty the red line is simply false.

And without the red line there is simply no logic or means to jump from the line before to the line after:

All elements $x$ in $S$ are not in $A$ $\not\implies$

It is not true that all $x \in S$ are also in $A$.

That simply is not true for an empty $S$.

• You confused $A$ and $S$ at the end there. Oct 6, 2016 at 21:50

Every theory has axioms, which are some propositions held to be true without being proven from anything else, and are not provable from each other. Subsequent truths of the theory derived from the axioms are theorems.

The properly termed question is whether the empty set being a subset of every other set is axiom of set theory, or a theorem.

It depends on how "subset" is defined. If $A\subset B$ means that every element of $A$ is in $B$, it is not necessarily true that $\emptyset$ is a subset of anything, since it has no elements. In this case, $\emptyset \subset A$ can be added as an axiom. It doesn't conflict with anything, and simplifies all reasoning about subsets. Alternatively, if $A\subset B$ is defined as "$A$ has no elements that are not also in $B$", then we do not require the extra axiom for the $\emptyset$ case. If $A$ has no elements at all, it has no elements that are not in $B$.

Suppose that we use the first, positively termed definition of subset, and then adopt as an axiom not $\forall A:\emptyset \subset A$, but rather its negation: $\exists A:\emptyset \not\subset A$, or the outright proposition $\forall A:\emptyset \not\subset A$.

This is just going to cause problems. We can "do" set theory as before, but all the theorems will be uglified by having to avoid the special cases involving the empty set. In any derivation step in which we rely on a subset relation being true, or assert one, we will have to add the verbiage of an additional statement which asserts that the variable in question doesn't denote the empty set. This proposition then has to be carried in all the remaining derivations, unless something else makes it superfluous (some unrelated assurance from elsewhere that the set in question isn't empty).

Working with this clumsy subset definition that doesn't work with the empty set very well, someone is eventually going to have an epiphany and introduce a new subset-like relation which doesn't have these ugly problems: a new $A\ \mathbf{subset*}\ B$ binary relation which reduces exactly to $A\subset B$ when neither $A$ nor $B$ are $\emptyset$, and which, simply by definition, reduces to a truth whenever $A = \emptyset$, regardless of $B$. That person will then realize that all the existing work is simpler if this $\mathbf{subset*}$ operation is used in place of $\subset$.

At the end of the day it boils down to criteria like: is the system consistent (doesn't contradict itself), is it complete (does it capture the truths we want) and also is it convenient: are the rules configured so that we do not trip over unnecessary cases and superfluous logic.

This is a very mundane explanation (with finite sets). A subset is made of any combination of elements from the set. Suppose a set$S$ is made of three elements: $\{a,b,c\}$. Each element $a$, $b$ or $c$ can be, or not, in the combination. If all of them are not in the combination, they STILL form a combination of "absent elements", or $\emptyset$, a subset of $S$. They are the dual of the subset of "all elements", $\{a,b,c\}$.

In the same way that $\{a,b\}$ and $\{c\}$ are (complementary) subsets, $\emptyset$ and $\{a,b,c\}$ belong to the set of subsets.

What confused me was that, the following expression was also a vacuous truth.

For every object of $$x$$, if $$x$$ belongs to the empty set, then  $$x$$ doesn't belong to the set $$A$$.

The contrapositive of the above conditional is:

"For every $$x$$, if $$x$$ belongs to set $$A$$, then $$x$$ doesn't belong to the empty set" which is easy to understand as the empty set has no elements at all.

NOTE- $$p\rightarrow q$$ has same meaning as $$\lnot q\rightarrow \lnot p$$


# Help with separable equation

#### Kris

##### New member
( 4*x+1 )^2 dy/dx = 27*y^3

I'm trying to separate this into a separable equation. Does it matter which way I do it? I.e taking all xs to the left or all ys to the left or does it not matter as long as x and y are on different sides?

#### MarkFL

Staff member
As long as you have:

$$\displaystyle f(y)\,dy=g(x)\,dx$$ or $$\displaystyle f(x)\,dx=g(y)\,dy$$ it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.

#### Kris

##### New member
As long as you have:

$$\displaystyle f(y)\,dy=g(x)\,dx$$ or $$\displaystyle f(x)\,dx=g(y)\,dy$$ it does not matter. I tend to like this first form, with $y$ on the left and $x$ on the right, but that's just the way I was taught.
Hi Mark, I have tried separating the variables yet I can't figure out how to move the bracketed term across. Could you advise how to do this and show what the final solution would be?

Also how do I thank you as I can't find the button

#### MarkFL

Staff member
I will help you separate the variables, and then guide you to get the solution...you will get more from the problem that way.

We are given:

$$\displaystyle (4x+1)^2\frac{dy}{dx}=27y^3$$

See what you get when you divide through by $$\displaystyle (4x+1)^2y^3$$ (bearing in mind that in doing so we are eliminating the trivial solution $y\equiv0$).

edit: You should see a Thanks link at the lower right of each post, except your own.

#### Kris

##### New member
Thanks I can get the answer from here just wasn't sure if you could divide through by the whole bracket

#### Kris

##### New member
dy/(27*y^3)=dx/((4*x+1)^2)

This is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?

At this point it is okay to integrate right?

edit: Do you know what the answer will be for the integration because my work sheet doesnt have it and id like to have it before i finish the question

#### MarkFL

Staff member
dy/(27*y^3)=dx/((4*x+1)^2)

This is my final separation can you please tell me if this is the correct result. Can this be simplified down even further?

At this point it is okay to integrate right?

edit: Do you know what the answer will be for the integration because my work sheet doesn't have it and I'd like to have it before I finish the question.
Yes, that's fine, although I would probably choose to write:

$$\displaystyle y^{-3}\,dy=27(4x+1)^{-2}\,dx$$

I would use a $u$-substitution on the right side:

$$\displaystyle u=4x+1\,\therefore\,du=4\,dx$$ and we have (after multiplying through by 4):

$$\displaystyle 4\int y^{-3}\,dy=27\int u^{-2}\,du$$

Now, we are just a couple of steps from the solution (and don't forget to include the trivial solution we eliminated when separating variables when you state the final solution, if this trivial solution is not included in the general solution for a suitable choice of the constant of integration).

#### Kris

##### New member
Ive got a solution at y = -54 + c but i dont think this is right

I went with -1/2*y^2 * y = -1/u * u? as my integrations and then rearranged

Is this the correct answer or have I integrated something wrong which leads to my poor solution?

#### MarkFL

Staff member
Using the power rule for integration, and using the form in my post above, you should get:

$$\displaystyle 4\left(\frac{y^{-2}}{-2} \right)=27\left(\frac{u^{-1}}{-1} \right)+C$$

Multiply through by -1 and simplify a bit:

$$\displaystyle 2y^{-2}=27u^{-1}+C$$

Note: the sign of the parameter $C$ does not change as it can be any real number, negative or positive.

At this point, I would rewrite using positive exponents, and combine terms on the right:

$$\displaystyle \frac{2}{y^2}=\frac{Cu+27}{u}$$

Invert both sides:

$$\displaystyle \frac{y^2}{2}=\frac{u}{Cu+27}$$

$$\displaystyle y^2=\frac{2u}{Cu+27}$$

Back-substitute for $u$:

$$\displaystyle y^2=\frac{2(4x+1)}{C(4x+1)+27}$$

This is the general solution, and the only way we can get the trivial solution is for:

$$\displaystyle 4x+1=0$$

but we eliminated that possibility during the separation of variables as well.

#### Kris

##### New member
Thanks so much I see where i went wrong because i tried to integrate and invert before I multiplied out. It makes much more sense to multiply and flip then turn into a positive rather than trying to do it all at once


https://www.thoughtco.com/associative-and-commutative-properties-difference-3126316 (accessed July 22, 2022). By definition, commutative property is applied on 2 numbers, but the result remains the same for 3 numbers as well. For that reason, it is important to understand the difference between the two. That is. In Mathematics, a commutative property states that if the position of integers are moved around or interchanged while performing addition or multiplication operations, then the answer remains the same. Your Mobile number and Email id will not be published. By the commutative property of multiplication, 3 6 = 6 3. nhmDTH%PLI%@hQdK( (1) and (2), as per the commutative property of multiplication, we get; Find which of the following is the commutative property of addition and multiplication. For example: 1+2 = 2+1 and 2 x 3 = 3 x 2. Math Glossary: Mathematics Terms and Definitions, A Kindergarten Lesson Plan for Teaching Addition and Subtraction, IEP Math Goals for Operations in the Primary Grades, Definition and Usage of Union in Mathematics, Parentheses, Braces, and Brackets in Math. Essentially, the order does not matter when adding or multiplying. Yes. As per commutative property of addition, 827 + 389 = 389 + 827. Similarly, we can rearrange the addends and write: Example 4: Ben bought 3 packets of 6 pens each. As we already discussed in the introduction, as per the commutative property or commutative law, when two numbers are added or multiplied together, then a change in their positions does not change the result. Neither of these properties are applicable to division. Take, for example, the arithmetic problem (6 3) 2 = 3 2 = 1; if we change the grouping of the parentheses, we have 6 (3 2) = 6 1 = 5, which changes the final result of the equation. Whereas associative property holds regardless of grouping of numbers. Some operations are non-commutative. Lets see. For instance: This equation is an example of the commutative property of addition of real numbers. Choose the set of numbers to make the statement true. In other words, the placement of parentheses does not matter when it comes to adding or multiplying. No matter how the values are grouped, the result of the equation will be 10: As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Example 2: Use 14 15 = 210, to find 15 14. "The Associative and Commutative Properties." So, the total number of pens that Ben bought = 3 6, So, the total number of pens that Ben bought = 6 3. It states that if we swipe the positions of the integers, the result will remain the same. The correct answer is Both sides are equal to 33.. The commutative property concerns the order of certain mathematical operations. The left side equals 44 and the right side equals 33. By equation 1 and 2, as per commutative property of addition, we get; Q.3: Prove that A.B = B.A, if A = 4 and B = 3. What is the associative property of addition (or multiplication)? Checkout JEE MAINS 2022 Question Paper Analysis : explains that order of terms doesnt matter while performing, Commutative property is applicable only for addition and multiplication processes. The distributive property of Multiplication states that multiplying a sum by a number is the same as multiplying each addend by the value and adding the products then. Remember, with the commutative property, the order of the numbers does not matter when adding and multiplying. We can tell the difference between the associative and the commutative property by asking the question, Are we changing the order of the elements, or are we changing the grouping of the elements? If the elements are being reordered, then the commutative property applies.

(2020, October 29). Did they buy an equal number of pens or not? Mia bought 6 packets of 3 pens each. When two numbers are multiplied together and if we interchange their positions, then the product of the two remains the same. The Associative property holds true for addition and multiplication. KB(|Q-SFt4E! Even if both have different numbers of bun packs with each having a different number of buns in them, they both bought an equal number of buns, because 3 4 = 4 3. According to the associative law, regardless of how the numbers are grouped, you can add or multiply them together, the answer will be the same. The word, Commutative, originated from the French word commute or commuter means to switch or move around, combined with the suffix -ative means tend to. The property holds for Addition and Multiplication, but not for subtraction and division. By eq. Which of the following represents the commutative property of addition? 7)\). If $$x=2$$, $$y=5$$, and $$z=1$$, which of the following is true about this equation: $$2x+4y+9z=9z+4y+2x$$. The mathematical operations, subtraction and division are the two non-commutative operations. Example 4: Use the commutative property of addition to write the equation, 3 + 5 + 9 = 17, in a different sequence of the addends. Example 3: Use 827 + 389 = 1,216 to find 389 + 827. The other major properties of addition and multiplication are: Now, observe the other properties as well here: Associative Property of Addition and Multiplication. Since, 14 15 = 210, so, 15 14 also equals 210. Taylor, Courtney. by Mometrix Test Preparation | This Page Last Updated: June 2, 2022. The operation is commutative because the order of the elements does not affect the result of the operation. Which statement best illustrates the commutative property? Q.2: Prove that a+ b = b+a if a = 10 and b = 9. The associative property states that the grouping of factors in an operation can be changed without affecting the outcome of the equation. }8E| This can be shown by the equation (a + b) + c = a + (b + c). So, there can be two categories of operations that obeys commutative property: Although the official use of commutative property began at the end of the 18th century, it was known even in the ancient era. ThoughtCo. The commutative property states that values can be moved or swapped when adding or multiplying, and the outcome will not change. So, if we swap the position of numbers in subtraction or division statements, it changes the entire problem.

Which operations do not follow commutative property? Even if both have different numbers of apples and peaches, they have an equal number of fruits, because 2 + 6 = 6 + 2. Use the commutative property to find the missing value: $$45+44+43=43+44+$$_____. Which of the following represents the commutative property of multiplication? The associative property says you can regroup multiplied terms in any way. Which of the following expressions will follow the commutative property? "The Associative and Commutative Properties." B.A., Mathematics, Physics, and Chemistry, Anderson University. Commutative property is applicable only for addition and multiplication processes. Do they have an equal number of marbles? =*jb 5;dtOu2T*~GL:E7$_Bd% N Therefore, the literal meaning of the word is tending to switch or move around. 77; by commutative property of multiplication, 36; by commutative property of multiplication. Unlike addition, in subtraction switching of orders of terms results in different answers. Commutative property cannot be applied to subtraction and division. The Associative and Commutative Properties. Simply put, the commutative property states that the factors in an equation can be rearranged freely without affecting the outcome of the equation. Example: 4 3 = 1 but 3 4 = -1which are two different integers. Use the commutative property to find the missing values: $$4+6+$$ ____$$=6+$$____ $$+8$$. When the associative property is used, elements are merely regrouped. These propertiesthe commutative and the associativeare very similar and can be easily mixed up. Solution: Options 1, 2 and 5 follow the commutative law. Beth has 6 packets of 78 marbles each. eC:C%L"HX'JyS7yS| F: lj. The correct answer is $$6+5=5+6$$. The grouping of the elements, as indicated by the parentheses, does not affect the result of the equation. Required fields are marked *. Also, the division does not follow the commutative law. The correct answer is $$4(37)=(43)7$$. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. What is the distributive property of multiplication? Example 5: Lisa has 78 red and 6 blue marbles. Thus, it means we can change the position or swap the numbers when adding or multiplying any two numbers. Since Lisa has 78 red and 6 blue marbles. By the distributive property of multiplication over addition, we mean that multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. The associative property applies to multiplication but not division, so divided terms cannot be regrouped. We've updated our Privacy Policy, which will go in to effect on September 1, 2022. To solve more problems on properties of math, download BYJUS The Learning App from Google Play Store and watch interactive videos. (Except 2 + 2 and 2 2. (% This property states that when three or more numbers are added (or multiplied), the sum (or the product) is the same regardless of the grouping of the addends (or the multiplicands). Retrieved from https://www.thoughtco.com/associative-and-commutative-properties-difference-3126316. In mathematics, commutative property or commutative law explains that order of terms doesnt matter while performing arithmetic operations. For which all operations does the associative property hold true? This can be expressed through the equation a + (b + c) = (a + b) + c. No matter which pair of values in the equation is added first, the result will be the same. 3u(CXOD^$? We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best.

The commutative property, therefore, concerns itself with the ordering of operations, including the addition and multiplication of real numbers, integers, and rational numbers. The above examples clearly show that the commutative property holds true for addition and multiplication but not for subtraction and division. So, Lisa and Beth dont have an equal number of marbles. Example 1: Which of the following obeys commutative law? Commutative property holds regardless of order of numbers while addition or multiplication. If we pay careful attention to the equation, though, we see that only the order of the elements has been changed, not the grouping. As per commutative property of multiplication, 15 14 = 14 15. Rewrite the expression $$45+6+19$$ using the commutative property. The left side equals 33 and the right side equals 44. The expression $$45+6+19$$ is equivalent to $$6+45+19$$, because changing the order that we add does not affect the result. What Is the Difference of Two Sets in Set Theory? For the associative property to apply, we would have to rearrange the grouping of the elements as well: By clicking Accept All Cookies, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts.

When two numbers are added together, then if we swap the positions of numbers, the sum of the two remains the same. Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra.". However, note that the presence of parentheses alone does not necessarily mean that the associative property applies. The correct answer is $$4+6+\mathbf8=6+\mathbf4+8$$. a (b + c) = (a b) + (a c) where a, b, and c are whole numbers. Example 1: Fill in the missing numbers using the commutative property. Taylor, Courtney. However, unlike the commutative property, the associative property can also apply to matrix multiplication and function composition. ThoughtCo, Oct. 29, 2020, thoughtco.com/associative-and-commutative-properties-difference-3126316. Like commutative property equations, associative property equations cannot contain the subtraction of real numbers. Even though the terms are listed in a different order, the left and right side of the equation are both equal to 33. (a + b) + c = a + (b + c)(a b) c = a (b c) where a, b, and c are whole numbers. For example, the numbers 2, 3, and 5 can be added together in any order without affecting the final result: The numbers can likewise be multiplied in any order without affecting the final result: Subtraction and division, however, are not operations that can be commutative because the order of operations is important. Commutative Property Definition with Examples. Rearranging multiplied terms is an example of the commutative property. The three numbers above cannot, for example, be subtracted in any order without affecting the final value: As a result, the commutative property can be expressed through the equations a + b = b + a and a x b = b x a. Learn More All content on this website is Copyright 2022, $$3(1)^{2}+5(1)(2)+(3)=5(1)(2)+3+3(1)^{2}$$. The Rules of Using Positive and Negative Integers, Facts About the Element Ruthenium (or Ru), Lesson Plan: Adding and Multiplying Decimals, Use BEDMAS to Remember the Order of Operations. That is. The correct answer is $$6+45+19$$. By non-commutative, we mean the switching of the order will give different results. This is because we can apply this property on two numbers out of 3 in various combinations. The associative property, on the other hand, concerns the grouping of elements in an operation. So, the total number of marbles with Lisa = 78 + 6, So, the total number of marbles with Beth = 6 78. Your Mobile number and Email id will not be published. As per the commutative property of multiplication, when we multiply two integers, the answer we get after multiplication will remain the same, even if the position of the integers are interchanged. Taylor, Courtney. The commutative property allows the addition or multiplication of numbers in any order.

According to the Distributive Property, if a, b, c are real numbers, then: There are certain other properties such as Identity property, closure property which are introduced for integers. The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. According to the commutative property of addition, when we add two integers, the answer will remain unchanged even if the position of the numbers are changed. Clearly, adding and multiplying two numbers gives different results.

That is. For a binary operationone that involves only two elementsthis can be shown by the equation a + b = b + a. Can you apply the commutative property of addition/multiplication to 3 numbers? If the elements are only being regrouped, then the associative property applies. The correct answer is 45. Since, 827 + 389 = 1,216, so, 389 + 827 also equals 1,216. So, mathematically commutative property for addition and multiplication looks like this: a + b = b + a; where a and b are any 2 whole numbers, a b = b a; where a and b are any 2 nonzero whole numbers. This is one of the major properties of integers. So, both Ben and Mia bought an equal number of pens. There are several mathematical properties that are used in statistics and probability; two of these, the commutative and associative properties, are generally associated with the basic arithmetic of integers, rationals, and real numbers, though they also show up in more advanced mathematics.

[O J"FUQ&pwODUfYQ$z(['z@PY3"BjSN<8$N(whCZOAfDda( GhA!D}0.Fdap u JXEm]^m> wO]l"gTkJHDMOE\?II!>bH4VEJ%+OTEeu\!2% Note that when the commutative property is used, elements in an equation are rearranged. NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions Class 11 Business Studies, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 8 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions For Class 6 Social Science, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, Line Segment : Construct A Copy Of Line Segment, Important Questions Class 9 Maths Chapter 4 Linear Equations Two Variables, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths. For example, take the equation 2 + 3 + 5. =i*s{/_WT8yp4x1lDI

No matter the order of the values in these equations, the results will always be the same. apxZ=vE This is one of the major, Commutative property is only applicable for two arithmetic operations: Addition and Multiplication, Changing the order of operands, does not change the result, Commutative property of addition: A + B = B + A, Commutative property of multiplication: A.B = B.A. Q8 E T"4', BgFC01 DCKTEsIyaR`@!


I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square
503^2 = 253009
and
50*506=25300
then tacking the 3^2 on the end gets 253009.
Tricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.
Any help would be appreciated.

2. Originally Posted by Enkie
I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square
503^2 = 253009
and
50*506=25300
then tacking the 3^2 on the end gets 253009.
Tricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.
Any help would be appreciated.
Let $n\in\mathbb{N}$ then $n=a_0+10a_1+\cdots+10^n a_n$ where $a_k\in\{0,\cdots,9\}$. You are claiming that $n^2=\left(10 a_1+\cdots +10^n a_n\right)\cdot\left(2a_0+10 a_1+\cdots 10^n a_n\right)+a_0^2$, or equivalently

$\left(a_0+10 a_1+\cdots+10^n a_n\right)^2-\left(10a_1+\cdots+10^n a_n\right)\cdot\left(2a_0+10a_1+\cdots+10^na_n\rig ht)-a_0^2$ $=\left(\left(a_0+10a_1+\cdots 10^n a_n\right)^2-a_0^2\right)-\left(10 a_1+\cdots+10^na_n\right)\cdot\left(2a_0+10a_1+\cd ots+10^n a_n\right)$ $=0$.

But,

$\left(a_0+10a_1+\cdots+10^na_n\right)^2-a_0^2=\left(a_0+10a_1+\cdots+10^na_n-a_0\right)$ $\cdot\left(a_0+10a_1+\cdots 10^na_n+a_0\right)$ $=\left(10a_1+\cdots+10^na_n\right)\cdot\left(2a_0+ 10a_1+\cdots+10^n a_n\right)$.

From where the conclusion immediately follows.

3. Hello, Enkie!

Another approach . . .

I was noticing today that you can calculate the square of a number by:
(1) taking the number you want to square and removing the ones digit,
(2) and multiplying by the number you want to square plus the ones digit,
(3) then square the ones digit and add it on to get the square.

Any integer $N> 9$ is of the form: . $N \:=\:10T + U$
. . where $U$ is the units digit and $T$ is the "rest of the number."

We note that: . $N^2 \:=\:(10T + U)^2 \:=\:100T^2 + 20TU + U^2$

(1) Remove the units digit and append a zero.
. . . We have: . $10T$

(2) Multiply by $N + U\!:\;\;10T(N + U) \;=\;10T(10T + U + U) \;=\;100T^2 + 20TU$

(3) Add $U^2\!:\;\;100T^2 + 20TU + U^2\quad \hdots$ and the result is $N^2$

4. Let us try with $1485$. We get :

$148 \times (1485 + 5) = 220520$

Now add on $5^2 = 25 \Rightarrow 22052025$

But $1485^2 = 2205225$.

I understand what you mean but you need to mention that when adding the square of the last digit, you actually overlap the previous result when the last digit is greater than three.


# How to calculate the total probability inside a slice of a bivariate normal distribution in R?

I have a bivariate normal distribution composed of the univariate normal distributions $X_1$ and $X_2$ with $\rho \approx 0.3$.

$$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right)$$

Is there a simple way to calculate in R the cumulative probability of $X_1$ being less than a value $z$ given a particular slice of $X_2$ (between two values $a,b$) given we know all the parameters $\mu_1, \mu_2, \sigma_1, \sigma_2, \rho$?

$P(X_1 < z | a < X_2 < b)$

Can the distribution function I am looking for match (or be approximated by) the distribution function of a univariate normal distribution (to use qnorm/pnorm)? Ideally this would be the case so I can perform the calculation with less dependencies on libraries (e.g. on a MySQL server).

This is the bivariate distribution I am using:

means <- c(79.55920, 52.29355)
variances <- c(268.8986, 770.0212)
rho <- 0.2821711

covariancePartOfMatrix <- sqrt(variances[1]) * sqrt(variances[2]) * rho
sigmaMatrix <- matrix(c(variances[1],covariancePartOfMatrix,covariancePartOfMatrix,variances[2]), byrow=T, ncol=2)

n <- 10000
dat <- MASS::mvrnorm(n=n, mu=means, Sigma=sigmaMatrix)

plot(dat)

This is my numerical attempt to get the correct result. However it uses generated data from the bivariate distribution and I'm not convinced it will give the correct result.

a <- 79.5
b <- 80.5
z <- 50

sliceOfDat <- subset(data.frame(dat), X1 > a, X1 < b)
estimatedMean <- mean(sliceOfDat[,c(2)])
estimatedDev <- sd(sliceOfDat[,c(2)])

estimatedPercentile <- pnorm(z, estimatedMean, estimatedDev)

### Edit - R implementation of solution based on whuber's answer

Here is an implementation of the accepted solution using integrate, compared against my original idea based on sampling. The accepted solution provides the expected output 0.5, whereas my original idea deviated by a significant amount (0.41). Update - See wheber's edit for a better implementation.

# Bivariate distribution parameters
means <- c(79.55920, 52.29355)
variances <- c(268.8986, 770.0212)
rho <- 0.2821711

# Generate sample data for bivariate distribution
n <- 10000

covariancePartOfMatrix <- sqrt(variances[1]) * sqrt(variances[2]) * rho
sigmaMatrix <- matrix(c(variances[1],covariancePartOfMatrix,covariancePartOfMatrix,variances[2]), byrow=T, ncol=2)
dat <- MASS::mvrnorm(n=n, mu=means, Sigma=sigmaMatrix)

# Input parameters to test the estimation
w = 79.55920

a <- w - 0.5
b <- w + 0.5
z <- 52.29355

# Univariate approximation using randomness
sliceOfDat <- subset(data.frame(dat), X1 > a, X1 < b)
estimatedMean <- mean(sliceOfDat[,c(2)])
estimatedDev <- sd(sliceOfDat[,c(2)])

estimatedPercentile <- pnorm(z, estimatedMean, estimatedDev)
# OUTPUT: 0.411

# Numerical approximation from exact solution
adaptedZ <- (z - means[2]) / sqrt(variances[2])
adaptedB <- (b - means[1]) / sqrt(variances[1])
adaptedA <- (a - means[1]) / sqrt(variances[1])

integrand <- function(x) pnorm((adaptedZ - rho * x) / sqrt(1 - rho * rho)) * dnorm(x)
# 0.0121, abs.error 1.348036e-16, "OK"
exactPercentile = exactSolutionCoeff * exactSolutionInteg$value # OUTPUT: 0.500  ## 1 Answer Yes, a Normal approximation works--but not in all cases. We need to do some analysis to identify when the approximation is a good one. ### Exact solution Re-express$(X_1,X_2)$in standardized units so that they have zero means and unit variances. Letting$\Phibe the standard Normal distribution function (its CDF), it is well known from the theory of ordinary least squares regression that $$\Pr(X_1 \le z\,|\, X_2 = x) = \Phi\left(\frac{z - \rho x}{\sqrt{1-\rho^2}}\right).$$ The desired probability then can be obtained by integrating: \eqalign{\Pr(X_1 \le z\,|\, a \lt X_2 \le b) &= \frac{1}{\Phi(b)-\Phi(a)}\int_a^b \Pr(X_1\le z\,|\, X_2=x) \phi(x)\,dx \\&= \frac{1}{\Phi(b)-\Phi(a)}\int_a^b \Phi\left(\frac{z - \rho x}{\sqrt{1-\rho^2}}\right) \phi(x)\,dx.} This appears to require numerical integration (although the result for(a,b)=\mathbb{R}$is obtainable in closed form: see How can I calculate$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$). ### Approximating the distribution This expression can be differentiated under the integral sign (with respect to$z$) to obtain the PDF, $$f(z\,|\, a \lt X_2 \le b) = \phi(z)\ \frac{\Phi\left(\frac{b-\rho z}{\sqrt{1-\rho^2}}\right) - \Phi\left(\frac{a-\rho z}{\sqrt{1-\rho^2}}\right)}{\Phi(b) - \Phi(a)}.$$ This exhibits the PDF as product of the standard Normal PDF$\phi$and a "correction". When$b-a$is small compared to$\sqrt{1-\rho^2}$(specifically, when$(b-a)^2 \ll 1-\rho^2$), we might approximate the difference in the numerator with the first derivative: $$\Phi\left(\frac{b-\rho z}{\sqrt{1-\rho^2}}\right) - \Phi\left(\frac{a-\rho z}{\sqrt{1-\rho^2}}\right)\approx \phi\left(\frac{(a+b)/2-\rho z}{\sqrt{1-\rho^2}}\right)\frac{b-a}{\sqrt{1-\rho^2}}.$$ The error in this approximation is uniformly bounded (across all values of$z$) because the second derivative of$\Phi$is bounded. With this approximation, and completing the square, we obtain $$f(z\,|\, a \lt X_2 \le b) \approx \phi\left(z; \rho(a+b)/2, \sqrt{1-\rho^2}\right) \frac{(b-a)\exp\left(-(a+b)^2/8\right)}{(\Phi(b)-\Phi(a))\sqrt{2\pi}}.$$ ($\phi(*; \mu, \sigma)$denotes the PDF of a Normal distribution of mean$\mu$and standard deviation$\sigma$.) Moreover, the right hand factor (which does not depend on$z$) must be very close to$1$, because the$\phi$term integrates to unity, whence $$f(z\,|\, a \lt X_2 \le b) \approx \phi\left(z; \rho(a+b)/2, \sqrt{1-\rho^2}\right).$$ All this makes very good sense: the conditional distribution of$X_1$is approximated by its conditional distribution at the midpoint of the interval,$(a+b)/2$, where it has mean$\rho(a+b)/2$and standard deviation$\sqrt{1-\rho^2}$. The error is proportional to the width of the interval$b-a$and to an expression dominated by$\exp(-(a+b)^2/8)$, which becomes important only when both$a$and$b$are out in the same tail. Therefore, this Normal approximation works for narrow slices not too far into the tails of the bivariate distribution. Moreover, the difference between $$\frac{(b-a)\exp\left(-(a+b)^2/8\right)}{(\Phi(b)-\Phi(a))\sqrt{2\pi}}$$ and$1$serves as an excellent check of the quality of the approximation. ### Edit To check these conclusions, I simulated data in R for various values of$b$and$\rho$($a=-3$in all cases), drew their empirical density, and superimposed on that the theoretical (blue) and approximate (red) densities for comparison. (You cannot see the density plots because the theoretical plots fit over them almost perfectly.) As$|\rho|$gets close to$1$, the approximation grows poorer: this deserves further study. Clear the approximation is excellent for sufficiently small values of$b-a$. # # Numerical integration, to give a correct value. # f <- function(z, a, b, rho, value.only=FALSE, ...) { g <- function(x) pnorm((z - rho*x)/sqrt(1-rho^2)) * dnorm(x) / (pnorm(b) - pnorm(a)) u <- integrate(g, a, b, ...) if (value.only) return(u$value) else return(u)
}
#
# Set up the problem.
#
a <- -3  # Left endpoint
n <- 1e5 # Simulation size
par(mfrow=c(2,3))
for (rho in c(1/4, -2/3)) {
for (b in c(-2.5, -2, -1.5)) {
z <- seq((a-3)*rho, (b+3)*rho, length.out=101)
#
# Check the approximation (v needs to be small).
#
v <- (b-a) * exp(-(a+b)^2/8) / (pnorm(b) - pnorm(a)) / sqrt(2*pi) - 1
#
# Simulate some values of (x1, x2).
#
x.2 <- qnorm(runif(n, pnorm(a), pnorm(b)))  # Normal between a and b
x.1 <- rho*x.2 + rnorm(n, sd=sqrt(1-rho^2))
#
# Compare the simulated distribution to the theoretical and approximate
# densities.
#
x.hat <- density(x.1)
plot(x.hat, type="l", lwd=2,
main="Simulated, True, and Approximate",
sub=paste0("a=", round(a,2), ", b=", round(b, 2), ", rho=", round(rho, 2),
"; v=", round(v,3)),
xlab="X.1", ylab="Density")

# Theoretical
curve(dnorm(x) * (pnorm((b-rho*x)/sqrt(1-rho^2)) - pnorm((a-rho*x)/sqrt(1-rho^2))) /
(pnorm(b) - pnorm(a)), lwd=2, col="Blue", add=TRUE)

# Approximate
curve(dnorm(x, rho*(a+b)/2, sqrt(1-rho^2)), col="Red", lwd=2, add=TRUE)
}
}

• Thank you for the detailed answer. I had found that the normal approximation stopped producing the figures I expected when z was high or low. Your examination of the numerical analysis corresponds exactly to what I noticed varying a, b, and z, and makes me confident that I can get a better result by using the exact solution that you provided. – Michael Clark May 20 '16 at 11:54
• Would it be possible to add a little more detail on how to calculate the exact solution without numerical integration? An example line in R would be perfect. The part I am most confused about is how to convert the limits and sign of x in the integral ∫baΦ(z−ρx1−ρ2−−−−−√)ϕ(x)dx into the form ∫∞−∞Φ(w−ab)ϕ(w)dw∫−∞∞Φ(w−ab)ϕ(w)dw) so that I can use the result proved in the question that you linked. – Michael Clark May 20 '16 at 11:58
• Unfortunately there is no such conversion. Numerical integration really is needed. You could also pursue the development of the approximation further by using higher-order approximations to the $\Phi$ term in the integrand. Intuitively, the midpoint $(a+b)/2$ ought to be replaced by a point closer to the smaller of $a$ and $b$ (in size), because that's where most of the $X_2$ probability is. – whuber May 20 '16 at 12:26
• I've added a R example based on the exact solution -- using the mean & variance, the approximate solution was 0.41 and the exact 0.5. I am surprised how far the approximate solution is out in that case. Thanks again! – Michael Clark May 20 '16 at 20:17
• I saw that code, but it's not clear how it works or whether it's correct. At the very least you should compute the check value (at the end of my original post), look at $|b-a|/\sqrt{1-\rho^2}$ (in standardized units), and inspect the output of integrate to make sure it's not running into accuracy problems. – whuber May 20 '16 at 20:20


# How would I simplify this summation: $\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j$

$$\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j$$

I cant really get my head around on how to simplify these sigma notations, any help would be appreciated.

Thanks

• Just try it for modest $n$! Suppose $n=1$. Then try $n=2$. I expect a pattern will emerge.
– lulu
Oct 23 '16 at 14:05
• You could replace $j$ by $i$ and simplify. Oct 23 '16 at 14:08

You can “reorganize” the first summation: $$\sum_{i=1}^n(i+1)=\sum_{i=1}^n i+\sum_{i=1}^n1$$ Since $$\sum_{i=1}^n i = \sum_{j=1}^n j$$ you remain with $$\sum_{i=1}^n 1 = n$$

Another variation which might be helpful.

\begin{align*} \sum_{i=1}^n (i + 1) - \sum_{j=1}^n j&=\sum_{i=1}^n (i + 1) - \sum_{i=1}^n i\\ &=\sum_{i=1}^n (i + 1-i)\\ &=\sum_{i=1}^n 1\\ &=n \end{align*}

You have

$$\sum_{i=1}^n (i+1)=\sum_{i=2}^{n+1} i$$

so

$$\sum_{i=1}^n (i+1)-\sum_{j=1}^n j=\sum_{i=2}^{n+1} i-\sum_{j=1}^n j=n+1+\sum_{i=1}^n (i-i)+1=n+1-1=n.$$

(Too long for a comment)

As @lulu said you can investigate the behavior of the sum by hand but you can use a bit of algebra before to simplify something. By example observe that $$\sum_{k=1}^n (k+1)=\left(\sum_{k=1}^n k\right)+\left(\sum_{k=1}^n 1\right)=\left(\sum_{k=1}^n k\right) + n$$ and $$\sum_{k=1}^n k=\sum_{j=1}^n j=1+2+3+\cdots+n$$

Observe too that a summation can be written as

$$\sum_{k=1}^n k=\sum_{1\le k\le n}k$$

Then you can manipulate easily the inequality $1\le k\le n$ if you need to change the variable $k$ by, for example, $h=k+2$

$$1\le k\le n\iff 1+2\le k+2\le n+2\iff 3\le h\le n+2$$

Then you can rewrite

$$\sum_{k=1}^n k=\sum_{1\le k\le n}k=\sum_{3\le h\le n+2}(h-2)=\sum_{h=3}^{n+2}(h-2)$$


# Thread: Calculate the velocity of the brick using an assumption & draw the displacement-time graph

1. Calculate the velocity of the brick at the bottom of the inclined plane.

& Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)

Workings:

I am unable to think on the first two but I think the displacement and time graph should be something like ,

Many Thanks

2. To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.

So, if we equate the initial potential energy to the final kinetic energy, we have:

$\displaystyle mgh=\frac{1}{2}mv^2$

What do you get upon solving for $v$?

Originally Posted by MarkFL
To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.

So, if we equate the initial potential energy to the final kinetic energy, we have:

$\displaystyle mgh=\frac{1}{2}mv^2$

What do you get upon solving for $v$?
Thank you very much MarkFL

$\displaystyle mgh=\frac{1}{2}mv^2$

$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}mv^2$

$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$

$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$

$\displaystyle 100 J= v^2$

$\displaystyle 10 J= v$

Correct ?

4. I would solve the equation first, and then plug in the numbers:

$\displaystyle mgh=\frac{1}{2}mv^2$

$\displaystyle v=\sqrt{2gh}$

Okay, at this point we can plug in:

$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$

$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$

Originally Posted by MarkFL
I would solve the equation first, and then plug in the numbers:

$\displaystyle mgh=\frac{1}{2}mv^2$

$\displaystyle v=\sqrt{2gh}$

Okay, at this point we can plug in:

$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$

$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
Thank you very much

Originally Posted by mathlearn

Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)

Is the above drawn graph correct?

6. For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?

7. Originally Posted by mathlearn

Is the above drawn graph correct?
Hey mathlearn!

Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):

On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.

However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?

8. Originally Posted by mathlearn
$\displaystyle 100 J= v^2$

$\displaystyle 10 J= v$
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)

-Dan

Originally Posted by MarkFL
For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?

This kind of a curve with constant acceleration and with deceleration.

Originally Posted by I like Serena
Hey mathlearn!

Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):

On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.

However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?
The displacement time graph would look like the above graph

Originally Posted by topsquark
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)

-Dan
Thanks for the catch

10. Originally Posted by mathlearn
This kind of a curve with constant acceleration and with deceleration...
Why do you have the brick returning to zero displacement?


Home > Error Bound > Taylor Series Maximum Error

# Taylor Series Maximum Error

## Contents

Mathispower4u 48,779 views 9:00 Error or Remainder of a Taylor Polynomial Approximation - Duration: 11:27. Your cache administrator is webmaster. So for example, if someone were to ask you, or if you wanted to visualize. Alternating series error bound For a decreasing, alternating series, it is easy to get a bound on the error : In other words, the error is bounded by the next term http://accessdtv.com/error-bound/taylor-series-approximation-maximum-error.html

And this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be the same thing as f of a. Another use is for approximating values for definite integrals, especially when the exact antiderivative of the function cannot be found. But how many terms are enough? To see why the alternating bound holds, note that each successive term in the series overshoots the true value of the series. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation

## Taylor Polynomial Error Bound

So it's really just going to be, I'll do it in the same colors, it's going to be f of x minus P of x. It is going to be equal to zero. Suppose you needed to find .

Especially as we go further and further from where we are centered. >From where are approximation is centered. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an Nth degree polynomial centered at a. You can get a different bound with a different interval. What Is Error Bound I'll write two factorial.

Khan Academy 54,407 views 9:18 Loading more suggestions... Lagrange Error Bound Formula fall-2010-math-2300-005 lectures © 2011 Jason B. One way to get an approximation is to add up some number of terms and then stop.

If x is sufficiently small, this gives a decent error bound.

solution Practice B03 Solution video by PatrickJMT Close Practice B03 like? 6 Practice B04 Determine an upper bound on the error for a 4th degree Maclaurin polynomial of $$f(x)=\cos(x)$$ at $$\cos(0.1)$$. Taylor Polynomial Approximation Calculator What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b. Thus, we have What is the worst case scenario? If one adds up the first terms, then by the integral bound, the error satisfies Setting gives that , so .

## Lagrange Error Bound Formula

And we see that right over here. http://calculus.seas.upenn.edu/?n=Main.ApproximationAndError Therefore, one can think of the Taylor remainder theorem as a generalization of the Mean value theorem. Taylor Polynomial Error Bound Let's embark on a journey to find a bound for the error of a Taylor polynomial approximation. Lagrange Error Bound Calculator Similarly, you can find values of trigonometric functions.

with an error of at most .139*10^-8, or good to seven decimal places. this contact form And it's going to look like this. Close Yeah, keep it Undo Close This video is unavailable. Lagrange's formula for this remainder term is $$\displaystyle{ R_n(x) = \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!} }$$ This looks very similar to the equation for the Taylor series terms . . . Lagrange Error Bound Problems

That is the motivation for this module. However, only you can decide what will actually help you learn. Instead, use Taylor polynomials to find a numerical approximation. http://accessdtv.com/error-bound/taylor-maximum-error.html Autoplay When autoplay is enabled, a suggested video will automatically play next.

Please try the request again. Lagrange Error Bound Khan Academy If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. And what I wanna do is I wanna approximate f of x with a Taylor polynomial centered around x is equal to a.

## Return to the Power Series starting page Representing functions as power series A list of common Maclaurin series Taylor Series Copyright © 1996 Department of Mathematics, Oregon State University If you

Solving for gives for some if and if , which is precisely the statement of the Mean value theorem. solution Practice B02 Solution video by PatrickJMT Close Practice B02 like? 8 Practice B03 Use the 2nd order Maclaurin polynomial of $$e^x$$ to estimate $$e^{0.3}$$ and find an upper bound on The following theorem tells us how to bound this error. Alternating Series Error Bound patrickJMT 128,850 views 10:48 Calculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials - Duration: 1:34:10.

Links and banners on this page are affiliate links. You may want to simply skip to the examples. So it might look something like this. Check This Out A Taylor polynomial takes more into consideration.

Professor Leonard 99,296 views 3:01:45 Taylor Polynomials - Duration: 18:06. Krista King 59,295 views 8:23 Lec 38 | MIT 18.01 Single Variable Calculus, Fall 2007 - Duration: 47:31. Now, if we're looking for the worst possible value that this error can be on the given interval (this is usually what we're interested in finding) then we find the maximum So what I wanna do is define a remainder function.

The N plus oneth derivative of our error function or our remainder function, we could call it, is equal to the N plus oneth derivative of our function. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... This is going to be equal to zero. The square root of e sin(0.1) The integral, from 0 to 1/2, of exp(x^2) dx We cannot find the value of exp(x) directly, except for a very few values of x.

So, what is the value of $$z$$? $$z$$ takes on a value between $$a$$ and $$x$$, but, and here's the key, we don't know exactly what that value is. And that's the whole point of where I'm going with this video and probably the next video, is we're gonna try to bound it so we know how good of an So this is all review, I have this polynomial that's approximating this function. video by Dr Chris Tisdell Search 17Calculus Loading Practice Problems Instructions: For the questions related to finding an upper bound on the error, there are many (in fact, infinite) correct answers.

The following example should help to make this idea clear, using the sixth-degree Taylor polynomial for cos x: Suppose that you use this polynomial to approximate cos 1: How accurate is View Edit History Print Single Variable Multi Variable Main Approximation And Error < Taylor series redux | Home Page | Calculus > Given a series that is known to converge but Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do In general, if you take an N plus oneth derivative of an Nth degree polynomial, and you could prove it for yourself, you could even prove it generally but I think

Uploaded on Nov 11, 2011In this video we use Taylor's inequality to approximate the error in a 3rd degree taylor approximation. There is a slightly different form which gives a bound on the error: Taylor error bound where is the maximum value of over all between 0 and , inclusive. So, the first place where your original function and the Taylor polynomial differ is in the st derivative. But you'll see this often, this is E for error.