# Explicit expression for recursive sums

Let $$t_1,t_2,\dots,t_k$$ be non-negative integers. Can the following sum $$f_k(t_1,t_2,\dots,t_k):=\sum_{j_1=0}^{t_1} \sum_{j_2=0}^{t_2+j_1} \sum_{j_3=0}^{t_2+j_2} \dots \sum_{j_k=0}^{t_k+j_{k-1}} 1$$ be explicitly expressed as a polynomial in $$t_1,t_2,\dots,t_k$$ or via known combinatorial entities?

We surely have a recurrence formula: $$f_{k+1}(t_1,t_2,\dots,t_{k+1}) = \sum_{j=0}^{t_1} f_k(j+t_2,\dots,t_{k+1}),$$ which does not seem to easily unroll.

Just in case, first few terms are $$\begin{split} f_0 &= 1,\\ f_1(t_1) &= 1+t_1,\\ f_2(t_1,t_2) &= (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2,\\ f_3(t_1,t_2,t_3) &= \left[ (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2 \right](1+t_3) + \frac{t_2(1+t_2)}2 + \frac{3t_2^2 + 6t_2 + 2}6t_1 + \frac{1+t_2}2t_1^2 + \frac16t_1^3. \end{split}$$

UPDATED. Billy Joe found that $$f_k(n,d,d,\dots, d) = \frac{n+1}k \binom{n+k(d+1)}{k-1}.$$ In particular, at $$f_k(1,1,\dots,1)$$ gives $$(k+1)$$-st Catalan number.

• The equal arguments case looks like it should also give the Ehrhart polynomial of some lattice polytope. May 9 at 2:20
• FWIW it seems that $f_k(1,2,3,\ldots,k-2,k-2,k)=a(k+1)$, where $a(k)$ is A107877. May 10 at 17:32
• @BillyJoe: Nice catch! In fact, it is stated there as conjecture by Benedict W. J. Irwin. May 10 at 17:49
• Are you sure about A016121? Apparently $f_k(n,d,d,\ldots,d,d)=\frac{n+1}{k}\binom{n+k(d+1)}{k-1}$. May 10 at 21:08
• Irwin's conjecture on A107877 is equivalent to the earlier comment by Joerg Arndt, Apr 30 2011, which doesn't qualify it as a conjecture. May 11 at 7:19

Claim: The iterated sum $$f_k(t_1,\ldots,t_k)$$ counts the number of elements the interval $$[\emptyset,\lambda]$$ of Young's lattice, where $$\lambda = (\lambda_1,\lambda_2,\ldots,\lambda_k)$$ is the partition determined by $$\lambda_{k-i+1} = t_1 + \cdots + t_i$$. Equivalently, the function $$f_k$$ counts the number of subdiagrams of $$\lambda$$.
For an arbitrary partition $$\lambda$$, we have $$|[\emptyset,\lambda]| = \text{det} \left[\binom{\lambda_i + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$ which is a result due to P. A. MacMahon. The answer to Exercise 149 in Chapter 3 of Stanley's Enumerative Combinatorics, volume 1, 2nd edition provides a good reference of references for this result, with various extensions and specializations, including some of the results mentioned in the comments. For a short visual proof using Lindström-Gessel-Viennot, see Ciucu - A short conceptual proof of Narayana's path-counting formula.
If the claim is true, MacMahon's result implies $$\sum_{j_1=0}^{t_1}\sum_{j_2=0}^{t_2+j_1}\cdots\sum_{j_k=0}^{t_k+j_{k-1}} = \text{det} \left[\binom{t_1 + \cdots + t_{k - i + 1} + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$ which implies $$f_k(t_1,\ldots,t_k)$$ is a polynomial in $$t_1,\ldots,t_k$$.
Note that $$f_k(t_1,\ldots,t_k)$$ counts the number of $$(j_1,\ldots,j_k)$$ such that $$0 \leq j_1 \leq t_1$$ and $$0 \leq j_{i+1} \leq j_i + t_{i+1}$$ for $$i \geq 1$$. To establish the claim, it suffices to find a bijection between the set of $$\mu \subseteq \lambda$$ and the set of tuples satisfying the above constraints.
Sketch: Map $$\mu \subseteq \lambda$$ to $$(j_1,\ldots,j_k)$$, where $$j_i = \lambda_{k-i+1} - \mu_{k-i+1}$$. The visual interpretation is that each $$j_i$$ measures the distance between the walls of the $$i$$-th row from the bottom of the Young diagrams (English convention) for $$\mu$$ and $$\lambda$$. The $$t_i$$ specify how many boxes are added to the diagram for $$\lambda$$ in moving from the $$(i-1)$$-st row from the bottom to the $$i$$-th row. The constraints express the fact that in going from bottom to top in the diagram, the distance between walls increases by at most $$t_i$$. For a more direct definition chase, note that $$\lambda_{k-i} - \lambda_{k-i+1} = t_{i+1}$$. Since $$\mu$$ is a partition, we have $$\mu_{k-i+1} - \mu_{k-i} \leq 0$$. Combining the definitions and inequalities gives $$j_{i+1} \leq j_i + t_{i+1}$$.
$$f_k(t_1,\dots,t_k)$$ is counting the number of integer points in the "Pitman-Stanley polytope" $$\Pi_k(t_1,\dots,t_k)$$ defined here. The notation $$N(\Pi_k(\mathbf{t}))$$ is used in this paper, which has the determinantal formula given by Hugh Denoncourt, as well as a combinatorial formula. The combinatorial formula is equivalent to $$f_k(t_1,\dots,t_k)=\sum_{\mathbf{h}\in K_k} \binom{t_1+h_1}{h_1} \prod_{i=2}^k \binom{t_i+h_i-1}{h_i},$$ where $$K_k := \{\mathbf{h}\in\mathbb{N}^k\colon \sum_{i=1}^j h_i\geq j\ \mathrm{for\ all}\ 1\leq j\leq k-1\ \mathrm{and}\ \sum_{i=1}^k h_i=k \}.$$ The set $$K_k$$ has a Catalan number $$C_k$$ of elements.


# 4.4 The mean value theorem  (Page 2/7)

 Page 2 / 7

Verify that the function $f\left(x\right)=2{x}^{2}-8x+6$ defined over the interval $\left[1,3\right]$ satisfies the conditions of Rolle’s theorem. Find all points $c$ guaranteed by Rolle’s theorem.

$c=2$

## The mean value theorem and its meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$

## Mean value theorem

Let $f$ be continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right).$ Then, there exists at least one point $c\in \left(a,b\right)$ such that

$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

## Proof

The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is

$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

and the line passes through the point $\left(a,f\left(a\right)\right),$ the equation of that line can be written as

$y=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right).$

Let $g\left(x\right)$ denote the vertical difference between the point $\left(x,f\left(x\right)\right)$ and the point $\left(x,y\right)$ on that line. Therefore,

$g\left(x\right)=f\left(x\right)-\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\right]\text{.}$

Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $\left[a,b\right],$ $g$ is also continuous over $\left[a,b\right].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since

$g\prime \left(x\right)=f\prime \left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a},$

we see that

$g\prime \left(c\right)=f\prime \left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

Since $g\prime \left(c\right)=0,$ we conclude that

$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$

In the next example, we show how the Mean Value Theorem can be applied to the function $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right].$ The method is the same for other functions, although sometimes with more interesting consequences.

## Verifying that the mean value theorem applies

For $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.

We know that $f\left(x\right)=\sqrt{x}$ is continuous over $\left[0,9\right]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\prime }\left(x\right)=\frac{1}{\left(2\sqrt{x}\right)}.$ The slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ is given by

$\frac{f\left(9\right)-f\left(0\right)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$

We want to find $c$ such that ${f}^{\prime }\left(c\right)=\frac{1}{3}.$ That is, we want to find $c$ such that

$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$

Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The


# Rate of change of radius and volume

They put a gas bubble in someone's eye. The volume of a gas bubble changes from $0.4$ $cc$ to $1.6$ $cc$ in $74$ hours. Assuming that the rate of change of the radius is constant, find

• (a) The rate at which the radius changes;
• (b) The rate at which the volume of the bubble is increasing at any volume $V$;
• (c) The rate at which the volume is increasing when the volume

is $1$ $cc$. (Note: The volume of a ball of radius $r$ is $\frac{4}{3}\pi r^3$. Assume the bubble is spherical.)

Explanation would be appreciated.

I did differentiate the $\frac{4}{3}\pi r^3$ with respect to r that is $4\pi r^2$ and then made that equal to $1.66-06$ which is the rate change of $V$. But I Don't know if I am doing it right.

-
Remember that if the radius grows by a factor of x then the volume grows by X^3. So if the volume grew 4 times then the radius grew by... – DannyDan Nov 7 '13 at 18:10

$v=\frac{4\pi r^3}{3}$

or

$r=(3v/4\pi)^{1/3}$

$\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$ substitute for $r$ we get

and letting $dr/dt=K$ (constant) then

$v^{-2/3}dv=4\pi K(3/(4\pi))^{2/3} dt$

$3v^{1/3} = 4\pi K(3/(4\pi))^{2/3} dt$

Integrating between limits $0.4$ and $1.6$ for V and $t=0$ to $74$, we get $K= \frac{1.6^{1/3}-.4^{1/3}}{74 (4\pi/3)^{1/3}}$ which works out to $3.628E-3 cm/hr$ as the rate of change of radius.

b) $dv/dt=(36\pi)^{1/3} KV^{2/3}$

c) substitute $v=1$ in above we get $(36\pi)^{1/3} K$

-
It would be alot easier for part 1 to just: $$K = \frac{\sqrt[3]{1.6*3/4/\pi}-\sqrt[3]{0.4*3/4/\pi}}{74} cm/hr$$ but they both get the same answer – kaine Nov 7 '13 at 19:12

Hint: what are the starting and ending radii? Let $t$ be the number of hours from the start. Now write an equation $r=$ some function of $t$. Use that to write another equation $V=$ some function of $t$ Now use these functions to answer the questions.

-

If the radius increases at a constant rate, $r = \alpha t+r_o$ for some constant $\alpha$.

this simplifies the questions to:

a) What are the values of $\alpha$ and $r_o$? (should be solvable from the data)

b) Write $V(t)$, solve for $\frac{dV}{dt}$, and write that in terms of just V.

c) Specifically plug $V=1 cc$ into the answer for b.

Note that we are hesitating to give a direct answer as this appears to be homework but I will gladly confirm any answers you come up with.

-


Find the equation of the quadratic given it's real roots $2-\sqrt 3$ and $2 + \sqrt3$ which passes through the point (1,-2)

2. if $r_1$ and $r_2$ are two roots of a quadratic ...

$f(x) = k(x - r_1)(x - r_2)$

$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]$

now, substitute $r_1 = 2-\sqrt{3}$ and $r_2 = 2 + \sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $f(1) = -2$ to find $k$

3. Originally Posted by skeeter
if $r_1$ and $r_2$ are two roots of a quadratic ...

$f(x) = k(x - r_1)(x - r_2)$

$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]$

now, substitute $r_1 = 2-\sqrt{3}$ and $r_2 = 2 + \sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $f(1) = -2$ to find $k$
I'm not sure exactly what you mean, can you please work it down further? I have never learned this formula.

4. Hello, euclid2!

There are several approaches to this problem.
. . Here's one of them . . .

Find the equation of the quadratic given its real roots $2-\sqrt 3$ and $2 + \sqrt3$
which passes through the point (1,-2)

If the quadratic, $ax^2 + bx + c$, has roots $p\text{ and }q$, then: . $p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}$

So we have: . $\begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc }\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array}$ . $\Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}$

. . The function (so far) is: . $f(x) \:=\:ax^2 -4ax + a$

Since (1,-2) is on the graph: . $a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2$

. . Hence: . $a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1$

Therefore: . $f(x)\;=\;x^2 - 4x + 1$

5. Originally Posted by Soroban
Hello, euclid2!

There are several approaches to this problem.
. . Here's one of them . . .

If the quadratic, $ax^2 + bx + c$, has roots $p\text{ and }q$, then: . $p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}$

So we have: . $\begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc }\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array}$ . $\Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}$

. . The function (so far) is: . $f(x) \:=\:ax^2 -4ax + a$

Since (1,-2) is on the graph: . $a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2$

. . Hence: . $a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1$

Therefore: . $f(x)\;=\;x^2 - 4x + 1$

Thank you very much, although can you edit this to show me where you are getting those numbers from? Thanks, again.

6. $f(x) = k(x - r_1)(x - r_2)$

$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]$

$r_1 + r_2 = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$

$r_1 \cdot r_2 = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$

$f(x) = k(x^2 - 4x + 1)$

since $f(1) = -2$

$-2 = k[(1)^2 - 4(1) + 1]$

$-2 = k(-2)$

$k = 1$

so ...

$f(x) = x^2 - 4x + 1$


# Maclaurin Series Approximation

For instance, we know that sin0 = 0, but what is sin0. Maclaurin Series Small Angles Approximation Exam Questions with Full Solutions 5. Maclaurin Series(approximation) Thread starter naspek; Start date Dec 10, 2009; Tags use the corresponding Maclaurin polynomial of degree 5 to approximate. For example, the Taylor Series for ex is given by:. The Taylor series for at is (By convention,. Taylor Series in MATLAB First, let's review our two main statements on Taylor polynomials with remainder. 7, exercise 9. Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(π/3). The second order Taylor approximation provides a parabolic function approximation while the third order provides a cubic function approximation. The general form of a Taylor series is, assuming the function and all its derivatives exist and are continuous on an interval centered at and containing. The infinite series expansion for f (x) about x = 0 becomes:. As we have seen, a general power series can be centered at a point other than zero, and the method that produces the Maclaurin series can also produce such series. Thus, The Remainder Term is z is a number between x and 3. It is often useful to designate the inﬁnite possibilities by what is called the Taylor Series. In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x = a. Using Taylor series to evaluate limits. In the next video, I'll do this with some actual functions just so it makes a little bit more sense. POLYNOMIAL APPROXIMATION OF FUNCTIONS: Linear and Quadratic Approximation, Taylor and Maclaurin Polynomials, Approximation with Taylor Polynomials (1hour) POWER SERIES: Definition, Center and Radius, Interval of Convergence, Endpoint Convergence, Operations with Power Series, Differentiating and Integrating Power Series (1hour) Week 2 4 hours. Maclaurin Series of f(x) = about x = up to order = Calculate: Computing Get this widget. The n th partial sum of the Taylor series for a function $$f$$ at $$a$$ is known as the n th Taylor polynomial. It first prompts the user to enter the number of terms in the Taylor series and the value of x. If f has a power series representation (expansion) at a,. (b) The Maclaurin series for g evaluated at x = L is an alternating series whose terms decrease in absolute 17 value to 0. We can improve this approximation of f(x) in two ways: Take more terms, increasing N. ” This becomes clearer in the expanded …. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we'd like. The Maclaurin Series: Approximations to f Near x = 0 If we let a Taylor polynomial keep going forever instead of cutting it off at a particular degree, we get a Taylor series. Thanks, Prasad. An approximation for the exponential function can be found using what is called a Maclaurin series: e x ≈ 1 + x 1 1 ! + x 2 2 ! + x 3 3 ! + … We will write a program to investigate the value of e and the exponential function. 01SC Single Variable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw. This means that the power series converges fastest when x is closest to 0. Here’s the formula for …. We'll focus on the Maclaurin right now. Consider the function of the form. FP2: Taylor's Series What does it mean to perform a Taylor expansion on T and V? Why does trig not work when using the 90 degree angle, i. Limits and Continuity Definition of Limit of a Function Properties of Limits Trigonometric Limits The Number e Natural Logarithms Indeterminate Forms Use of Infinitesimals L’Hopital’s Rule Continuity of Functions Discontinuous Functions Differentiation of Functions Definition of the Derivative Basic Differentiation Rules Derivatives of Power Functions Product Rule Quotient Rule Chain Rule. When the Maclaurin series approximates a function, the series values and the function values are very close near x = 0. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. We now take a particular case of Taylor Series, in the region near x = 0. Taylor Series. Math 115 Exam #2 1. As you can imagine each order of derivative gets larger which is great fun to work out. Actually Maclaurin Series is just a special form of Taylor Series. Representation of Functions as Power Series 10 2. eq = 60 - 53 x. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we'd like. How accurate is the approximation?. It is the source of formulas for expressing both sin x and cos x as infinite series. Yesterday we learned: Definition of an nth-degree Taylor polynomial:. How Good is Your Approximation? Whenever you approximate something you should be concerned about how good your approximation is. The Maclaurin series of a function up to order may be found using Series[f, x, 0, n]. x 2 2 x 2 n 3 f 8 e 2 x 8 e 2 8 f 3 x 3 8 6 x 3 4 3 x 3 e 2 x 1 2 x 2 x 2 4 3 x from MATH 270 at DeVry University, Chicago. Free Online Library: Efficient and accurate approximation of infinite series summation using asymptotic approximation and fast convergent series. A Taylor polynomial approximates the value of a function, and in many cases, it’s helpful to measure the accuracy of an approximation. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Find the Maclaurin polynomial of degree n = 4 for. Concrete examples in the physical science division and various engineering ﬁelds are used to paint the applications. Taylor Series Expansions In the previous section, we learned that any power series represents a function and that it is very easy to di¤erentiate or integrate a power series. A calculator for finding the expansion and form of the Taylor Series of a given function. Theorem 40 (Taylor's Theorem). Maclaurin Series Michael Penna, Indiana University – Purdue University, Indianapolis Objective In this project we investigate the approximation of a function by its Maclaurin series. The binomial function Remark: If m is a positive integer, then the binomial function f m is a polynomial, therefore the Taylor series is the same polynomial, hence the Taylor series has only the ﬁrst m +1 terms non-zero. all derivatives exist. We could use either the We could use either the Taylor remainder term, with n = 2 since this is really the series through the 2 nd degree term, or we can. My mathematics python's programs is a set of Maclaurin's series to compute some of the most important functions in calculus. Theorem (Taylor Polynomial Approximation). XXIV – Taylor and Maclaurin Series 1. Partial sums of a Maclaurin series provide polynomial approximations for the function. The sum of partial series can be used as an approximation of the whole series. Colin Maclaurin published a special case of the Taylor series in the 1700s. The function ex or exp(x) can be approximated using the Maclaurin Series (a specific type of Taylor Series) as follows (do ? to find factorial): Write Matlab code using a while loop to calculate a vector, macexp, that will hold each successive approximation to exp(x). If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x)…. As a simple example, you can create the number 10 from smaller numbers: 1 + 2 + 3 + 4. Take a function, pick a specific point, and. Questa pagina è stata modificata per l'ultima volta il 23 giu 2018 alle 18:49. We now take a particular case of Taylor Series, in the region near x = 0. These Taylor (and Maclaurin) polynomials are used to numerically approximate functions. Lec 89 - Sine Taylor Series at 0 (Maclaurin) Lec 90 - Taylor Series at 0 (Maclaurin) for e to the x. To find the Maclaurin Series simply set your Point to zero (0). make us an approximation of the series to fourth term, and also test the effect of large amplitude angles on the period. please help me. That is, approximation? The second partial sum (sum of the rst three terms) is Z 2 0. (e)Explain why di culties occur using the series in part (b) to approxi-mate erf(1). Multivariate approximation methods and applications to geophysics and geodesy. If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. 1 Taylor series 2. As you can imagine each order of derivative gets larger which is great fun to work out. Actually Maclaurin Series is just a special form of Taylor Series. FP2: Taylor's Series What does it mean to perform a Taylor expansion on T and V? Why does trig not work when using the 90 degree angle, i. 1 Things to Remember • Given a function, be able to ﬁnd its Taylor or Maclaurin’s series. A Maclaurin series is a power series that allows one to calculate an approximation of a function f(x) for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. A MacLaurin Polynomial is a special case of a Taylor Polynomial. Taylor and Maclaurin Series Definitions In this section, we consider a way to represent all functions which are ”suﬃciently nice” around some point. One important application of power series is to approximate a function using partial sums of its Taylor series. Taylor and maclaurian series Derivation for Maclaurin Series for ex Derive the Maclaurin series x2 x3 ex x = + + + + 2! 3! So if we want to find out how many. T is a Maclaurin series, plot these functions together to see how well this Taylor. x is the ﬁrst (non-zero) term in the Maclaurin series for sin(x), 0+x+0x2+···. Given f(x), we want a power series expansion of this function with respect to a chosen point xo, as follows: (1) (Translation: find the values of a 0 , a 1 , a 2 , … of this infinite series so that the equation holds. Lec 92 - Visualizing Taylor Series Approximations. Use the Taylor series for the function defined as to estimate the value of. How do you find the Maclaurin series of #f(x)=sin(x)# ? How do you use a Maclaurin series to find the derivative of a function? See all questions in Constructing a Maclaurin Series. How do we ﬁnd a quadratic approximation to a function y = f(x) and how accurate is this approximation? The secret to solving these problems is to notice that the equation of the tangent line showed up in our integration by parts in (1. We attribute much of the founding theory to Brook Taylor (1685-1731), Colin Maclaurin (1698-1746) and Joseph-Louis Lagrange (1736-1813). For example, the Taylor Series for ex is given by:. Taylor and maclaurian series Derivation for Maclaurin Series for ex Derive the Maclaurin series x2 x3 ex x = + + + + 2! 3! So if we want to find out how many. On [Series:: esss] makes Series generate a message in this case. This procedure is also called the expansion of the function around (or about) zero. The main idea is this: You did linear approximations in first semester calculus. It is often the case that a convenient expansion point is x 0 = 0, and series about this special expansion point are also called Maclaurin series. A calculator for finding the expansion and form of the Taylor Series of a given function. Maclaurin series are fast approximations of functions, and they offer more accurate function approximations than just linear ones. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. In addition, when n is not an integer an extension to the Binomial Theorem can be used to give a power series representation of the term. We illustrate with some examples. Example: sine function. Lec 94 - Visualizing Taylor Series for e^x. Approximation of e^x using Maclaurin Series in Python. Example: Approximation for ln(1+x) Leaving Cert 2005 Q8 b (ii) Use those terms to find an approximation for ln. Using a Table of Basic Power Series to Determine More Power Series - Part 2 Determine the Maclaurin Series and Polynomial for Function in the Form ax^2*sin(bx. ) {}For the Euler-Maclaurin summation formula, see 65B15 for: 78: 41-XX Approximations and expansions {}For all approximation theory in the complex domain, see 30E05 and 30E10; for all trigonometric approximation and interpolation, see 42A10 and 42A15; for numerical. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter …. The program is really simple: The user inputs a parameter x (x being an angle in radians) and a float ε, which is the precision of the value of cos(x). The diagram shows the Maclaurin series approximation to degree n for the exponential function. We would like to find an easier-to-compute approximation to f(x), to see why f this is also called the MacLaurin polynomial, (called a power series). Without further ado, here it is: The notation f(n) means "the nth derivative of f. You can hide/reveal the graph of appropriate series and change the value of the pivot in the Taylor series. If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x)…. This chapter examines methods of deriving approximate solutions to problems or of approximating exact solutions, which allow us to develop concise and precise estimates of quantities of interest when analyzing algorithms. This website and its content is subject to our Terms and Conditions. Use your answer to find a Maclaurin series for f'. In an open interval around x= a, f(x) ≈ f(a)+f′(a)(x−a) linear approximation • Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. Let us revise how to construct a program for Taylor Series. Suppose a set of standardized test scores are normally distributed with mean and standard deviation Use and the first six terms in the Maclaurin series for to approximate the probability that a randomly selected test score is between and Use the alternating series test to determine how accurate your approximation is. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) - so it's a mathematical. These notes discuss three important applications of Taylor series: 1. Clicking the Draw button plots the polynomial of the selected degree; clicking the Next button increments the degree to the next odd integer and plots the polynomial of that degree. Suppose a set of standardized test scores are normally distributed with mean and standard deviation Use and the first six terms in the Maclaurin series for to approximate the probability that a randomly selected test score is between and Use the alternating series test to determine how accurate your approximation is. Lec 95 - Polynomial approximation of functions (part 1). Produces the result Note that function must be in the integrable functions space or L 1 on selected Interval as we shown at theory sections. You won’t use an infinite series to calculate the approximation. ) When , the series is called a Maclaurin series. So, I'm trying to create a program that calculates cos(x) by using a Taylor approximation. Using Taylor polynomials to approximate functions. Use the ﬁrst two non-zero terms of an appropriate series to give an approximation of Z 1 0 sin we can replace x with t3 to get the Maclaurin series for cost3: 1. Problems on Taylor's Theorem. Taylor Polynomials Preview. Just calculate the values of the red bits and plug them into the Maclaurin series to give you the series expansion formula. Maclaurin expansion of B * (s), which involves four moments of the service-time distribution, gives a better approximation than the one involving two moments. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter …. 2 correct to five decimal places. Example: sine function. Taylor and Maclaurin Series We have learned how to construct power series representations of certain functions by relating them to geometric series, either directly, or indirectly through di erentiation or integration. In fact, e−p8(1). This website and its content is subject to our Terms and Conditions. The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. A Taylor series is a representation of a function using an infinite sum. 1 (1974), 287--289, MathSciNet. Keywords: The Taylor series, the Maclaurin series, polynomial and nonpolynomial approx-imation. 6 Two examples 3 Indeterminate forms 3. Practice Taylor/Maclaurin, receive helpful hints, take a quiz, improve your math skills. Let n 1 be an integer, and let a 2 R be a point. First, we deﬂne the Bernoulli numbers B2n. We now take a particular case of Taylor Series, in the region near x = 0. Textbook solution for Calculus (MindTap Course List) 11th Edition Ron Larson Chapter 9. i don't even understand this topic. Leary Find other works by these authors. Compute the Remainder Term for. Category: Maclaurin's series Approximation in series expansion. ” This becomes clearer in the expanded …. Then to ﬁnd our approximation, we need to ﬁnd n such that (. Lecture 61: Power Series Representation Of Functions; Lecture 62: What Is The Taylor Series? Lecture 63: What Is The Maclaurin Series? Lecture 64: Application Of The Maclaurin Series; Lecture 65: Find The Maclaurin Series For Sinx; Lecture 66: Find The Maclaurin Series For Cosx; Lecture 67: Maclaurin Series For A Binominal Expansion: 1. Maclaurin Series of f(x) = about x = up to order = Calculate: Computing Get this widget. Though, the computation of an infinite sum which give the value of a function in terms of the derivatives evaluated at a special case where x0 = 0,in contrast with Taylor series. We now take a particular case of Taylor Series, in the region near x = 0. In other words, f0gives us a linear approximation of f(x) near c: for small values of "2R, we have f(c+ ") ˇf(c) + "f0(c) But if f(x) has higher order derivatives, why stop with a linear approximation? Taylor series take this idea of linear approximation and extends it to higher order derivatives, giving us a better approximation of f(x) near c. 12 (1975), no. Taylor's expansion, and the related Maclaurin expansion discussed below, are used in approximations. Power Series, Taylor and Maclaurin Polynomials and Series Power Series The Basics De nition 1 (Power Series). Notice that this Taylor Series for e x e^{x} e x is different from the Maclaurin Series for e x e^{x} e x. If you want to use a diﬀerent center, then just take the results from this documentandreplacex with(x a) everywhere. The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. Maclaurin Series Michael Penna, Indiana University – Purdue University, Indianapolis Objective In this project we investigate the approximation of a function by its Maclaurin series. Mathematica Code For the Euler-Maclaurin Formula20 k is a slowly divergent series, inclusive. Rather than referring to it as such, we use the following. The Maclaurin series expansion for cos x is. The approximation for g using the first two nonzero terms of this series is Show that 120 this approximation differs from g L by less than 200. ) Use a power series to approximate. Algorithm for Computing Taylor Series. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. The Maclaurin series is the Taylor series at the point 0. How to extract derivative values from Taylor series Since the Taylor series of f based at x = b is X∞ n=0 f(n)(b) n! (x−b)n, we may think of the Taylor series as an encoding of all of the derivatives of f at x = b: that information. For example, the Taylor Series for ex is given by:. We can improve this approximation of f(x) in two ways: Take more terms, increasing N. dx dy y − =− x Proving DE By further differentiation of this result, or otherwise, find the Maclaurin’s series for. power series. ( n=2 amd x>=0 and x<=0,5) Our instructor showed just a basic example about taylor/maclaurin approximations and it has nothing to do with this one. We set an initial value of 1 to the sum of the series and define the first term, t= 1. 5)2n+1 2n+1. It is more of an exercise in differentiating using the chain rule to find the derivatives. Practice Taylor/Maclaurin, receive helpful hints, take a quiz, improve your math skills. Therefore we need to calculate the first 11 terms of the MacLaurin series (remember that the first term is when n = 0). Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. i don't even learn this thing yet. all derivatives exist. Taylor Series 1. • Section 8. Series detects certain essential singularities. Choose a web site to get translated content where available and see local events and offers. 3 The binomial expansion 2. Here we show better and better approximations for cos(x). MIT OpenCourseWare http://ocw. Hopefully by the end of this, you'll be getting the hang of these things so we can start calculating stuff with them. How do we calculate the Maclaurin series?. Actually Maclaurin Series is just a special form of Taylor Series. One important application of power series is to approximate a function using partial sums of its Taylor series. Lec 91 - Euler's Formula and Euler's Identity. (See why we want to do this in the Introduction. 2c: By applying the ratio test, find the radius of convergence for this Maclaurin series. Taylor's expansion, and the related Maclaurin expansion discussed below, are used in approximations. Taylor Series Generalize Tangent Lines as Approximation. Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(π/3). It is the source of formulas for expressing both sin x and cos x as infinite series. Alternatively, it could also be used for approximation purposes in evaluating the value of certain numeric entities. Hello again everybody Tom from everystepcalculus. Concrete examples in the physical science division and various engineering ﬁelds are used to paint the applications. Maclaurin Series Small Angles Approximation Exam Questions with Full Solutions 5. Example: Second-order Taylor series approximation (in gray) of a function around origin. Using Taylor series to find the sum of a series. Determine a Maclaurin Series approximation for f(x)=sin?(6x) where n = 6. If you make the polynomial approach a degree of infinity, your approximation becomes infinitely close to the real function and is perfectly equal to the real function. Generally, Maclaurin series expressions are more compact and will give good approximations even for values far from the origin if enough terms are used. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. It can be used to approximate integrals by finite sums, or conversely to evaluate finite sums and infinite series using integrals and the machinery of calculus. This says that if 7 x 9, the approximation in part (a) is accurate to within 0. However,canafunctionf(x. The Maclaurin series of a function up to order may be found using Series[f, x, 0, n]. Multivariate approximation methods and applications to geophysics and geodesy. Using Taylor series to evaluate limits. As mentioned in §E. We attribute much of the founding theory to Brook Taylor (1685-1731), Colin Maclaurin (1698-1746) and Joseph-Louis Lagrange (1736-1813). The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. As you can imagine each order of derivative gets larger which is great fun to work out. Lec 93 - Generalized Taylor Series Approximation. The representation of Taylor series reduces many mathematical proofs. Summing 11 terms in the second series gives erf(1) ˇ0:842700790029219. We now take a particular case of Taylor Series, in the region near x = 0. Some series converge only at a, and others converge on an interval (a - r, a + r). It is the source of formulas for expressing both sin x and cos x as infinite series. The Maclaurin's series for ln(1+x) could be used to approximate the natural logarithm ln(x). eq = 60 - 53 x. The series carries on to inﬁnity, and has general term (x−a)n n! f(n)(a). Step-by-step method for computing a Taylor series, with example of finding the Taylor series expansion of f(x) = (1-x)-1 about x = 0. Solution of Example 3 (from pages 736. Linear Approximation of Functions Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. It is common practice to use a finite number of terms of the series to approximate a function. Series can expand about the point x = ∞. Since I want the Remainder Term, I need to find an expression for the derivative. But Taylor and Maclaurin polynomials can only approximate functions. In the next video, I'll do this with some actual functions just so it makes a little bit more sense. Multivariate Taylor series is used in many optimization techniques. The ﬁrst of these is to under-stand how concepts that were discussed for ﬁnite series and integrals can be meaningfully. Example 13. 0 Share this post. I have to solved this eq and draw the graph. If f has a power series representation (expansion) at a,. Now write the Maclaurin series for ln(x+. Use your pocket calculator or MATLAB to determine the true value. There are many applications for expansions of common functions around x=0. 4C  Composite Maclaurin Series 2016 EXPORT. There are many sensible notions of what 'good approximation' could mean. Leary Find other works by these authors. Complete Solution Before starting this problem, note that the Taylor series expansion of any function about the point c = 0 is the same as finding its Maclaurin series expansion. Hopefully by the end of this, you'll be getting the hang of these things so we can start calculating stuff with them. Linear Approximation of Functions Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. My mathematics python's programs is a set of Maclaurin's series to compute some of the most important functions in calculus. If you make the polynomial approach a degree of infinity, your approximation becomes infinitely close to the real function and is perfectly equal to the real function. Toggle Main Navigation. 10 Problem 47E. 10: Taylor and Maclaurin Series 1. putationally efﬁcient method of approximation. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. Taylor Series. Taylor series are extremely powerful tools for approximating functions that can be difficult to compute otherwise, as well as evaluating infinite sums and integrals by recognizing Taylor series. We can use power series to create a function that has the same value as another function, and we can then use a limited number of terms as a way to compute approximate values for the original function within the interval of convergence. In an open interval around x= a, f(x) ≈ f(a)+f′(a)(x−a) linear approximation • Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. On [Series:: esss] makes Series generate a message in this case. Maclaurin and Taylor Series ··· the curve representing ex is a better and better approximation. Let n 1 be an integer, and let a 2 R be a point. So, you can start by assuming the Taylor series definition:. So, I'm trying to create a program that calculates cos(x) by using a Taylor approximation. At the age of 19 he was elected a professor of mathematics at Marischal College, Aberdeen, and two. Maclaurin series expansion eulers formula for pi show 10 more Any revision resources for A-Level AQA Maths? FP2: Taylor's Series Maclaurin and Taylor Series! I need help :( What is the purpose of the power/maclaurin/taylor series Series expansions of odd functions. power series, such as the Taylor series of a basic function. The first derivative of tan x is very simple as you can see. Theorem 40 (Taylor's Theorem). Taylor series expansions of hyperbolic functions, i. org ) Created Date: 8/7/2013 5:18:45 PM. 3) Comment at the bottom of the page. Applications of Taylor Series Recall that we used the linear approximation of a function in Calculus 1 to estimate the values of the function near a point a (assuming f was di erentiable at a): f(x) ˇf(a) + f0(a)(x a) for x near a: Now suppose that f(x) has in nitely many derivatives at a and f(x) equals the. The sum of partial series can be used as an approximation of the whole series. The exponential function is shown in red and the Maclaurin series approximation function is shown in blue. The goal of a Taylor expansion is to approximate function values. [3] (iii) By substituting x = 3 8. Maclaurin and Taylor Series. The Maclaurin series above is more than an approximation of e x, it is equal to e x on the interval of convergence (- , ). 01 Single Variable Calculus, Fall 2005 Prof. com, a free online graphing calculator. Taylor's theorem (actually discovered first by Gregory) states that any function satisfying certain conditions can be expressed as a Taylor series. The th term of a Maclaurin series of a function can be computed in the Wolfram Language using SeriesCoefficient[f, x, 0, n] and is given by the inverse Z-transform. NASA Technical Reports Server (NTRS) Munteanu, M. To avoid this, we can rst nd the Maclaurin Series for g(x) = (1+x)2=3,. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) - so it's a mathematical. To find the Maclaurin. The main idea is this: You did linear approximations in first semester calculus. The Maclaurin Series: Approximations to f Near x = 0 If we let a Taylor polynomial keep going forever instead of cutting it off at a particular degree, we get a Taylor series. Use the Taylor series for the function defined as to estimate the value of. In the West, the subject was formulated by the Scottish mathematician James Gregory and formally introduced by the English mathematician Brook Taylor in 1715. In specific, the type of Taylor series used is technically a Maclaurin series, since the representation is centered at a=0. You can now regrow the entire creature from that tiny sample. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we’d like. Find the Taylor series for at xa. the Maclaurin series of a function is centred at 0, or talk of the series expansion around 0. In [5] the convergence radius for Liapunov series was found in case of homogeneous equilibrium figures (Maclaurin ellipsoids). TAYLOR AND MACLAURIN SERIES 3 Note that cos(x) is an even function in the sense that cos( x) = cos(x) and this is re ected in its power series expansion that involves only even powers of x. Course Material Related to This Topic: Read lecture notes, section 3, pages 4–5. The first one is easy because tan 0 = 0. A Taylor series can be used to describe any function ƒ(x) that is a smooth function (or, in mathematical terms. As you can imagine each order of derivative gets larger which is great fun to work out. We would like to find an easier-to-compute approximation to f(x), to see why f this is also called the MacLaurin polynomial, (called a power series). We now take a particular case of Taylor Series, in the region near x = 0. Maclaurin Series of Sqrt(1+x) In this tutorial we shall derive the series expansion of $$\sqrt {1 + x}$$ by using Maclaurin's series expansion function. 7 Taylor and Maclaurin Series The conclusion we can draw from (5) and Example 1 is that if ex has a power series expansion at 0, then ∞ xn x e = n! n=0 So how can we determine whether ex does have a power series representation?. Maclaurin series and the general Taylor series centered at x = a. If we're evaluating the series at a point within its interval of convergence. The Maclaurin series for this function is known as the binomial series. A Maclaurin series is the expansion of the Taylor series of a function about zero. What you did was you created a linear function (a line) approximating a function by taking two things into consideration: The value of the function at a point, and the value of the derivative at the same point. XXIV – Taylor and Maclaurin Series 1. You can hide/reveal the graph of appropriate series and change the value of the pivot in the Taylor series. , I might be ( 17;19)) and let x 0 be a point in I, i. The program approximates the function cos(x) using a Taylor series approximation. It is often useful to designate the inﬁnite possibilities by what is called the Taylor Series. Towards Maclaurin. Thus, The Remainder Term is z is a number between x and 3. In [5] the convergence radius for Liapunov series was found in case of homogeneous equilibrium figures (Maclaurin ellipsoids). If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. By using this website, you agree to our Cookie Policy. Thanks, Prasad. To nd Taylor series for a function f(x), we must de-. If not, for what values of x can I use the series to approximate f(x)? 5. ) When , the series is called a Maclaurin series.


# How can I define a very large matrix efficiently?

Here's a 7x7 matrix:

(matrix = {{1, 1, 1, 1, 1, 1, 1}, {1, -1, 0, 0, 0, 0, 0}, {1,
1, -2, 0, 0, 0, 0}, {1, 1, 1, -3, 0, 0, 0}, {1, 1, 1, 1, -4, 0,
0}, {1, 1, 1, 1, 1, -5, 0}, {1, 1, 1, 1, 1,
1, -6}}) // MatrixForm

The matrix looks like this:

How can I define the 100x100 equivalent of this matrix efficiently without having to manually type out every element? The documentation gives some ideas, but although there's certainly an order in this matrix, it's not something that translates easily (at all?) to the commands used in the documentation.

size = 7;
Table[Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0], {i, size}, {j, size}]

• Worked very well, now I'll have to understand your syntax =) – Allure Jul 2 '19 at 3:41
• You might be interested in a tweak to your method explained in my answer. – Michael E2 Jul 2 '19 at 17:35

The question refers to efficiency, which might mean with respect to programming, performance, or both. My guess is programming, but highlighting the performance strengths of Mathematica should promote learning the "Mathematica style" as it is sometimes referred to.

Timings on various approaches for size = 100 (timing code given at the end):

• 0.008800 @kglr
• 0.007000 @Rohit Namjoshi, which can be improved to 0.00026
• 0.000510 @MikeY, which can be improved to 0.000041 (best of all)
• 0.000047 @MichaelE2, (was never under 0.000042 but usually 0.000046 to 0.000048)

My best solution:

size = 5;
matrix =
With[{r    = Range@size,
zero = DeveloperToPackedArray@{0}},
UnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]] -
];
matrix // MatrixForm

### Discussion

Table[] vs. Array[]. The solution of @RohitNamjoshi can be improved by using Array[]. This involves turning the expression e into a function, which takes only wrapping the expression with Function[{i, j}, e] and removing i and j from the iterators:

Array[Function[{i, j},
Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0]
], {size, size}]; // RepeatedTiming
(*  {0.00039, Null}  *)

It can be sped up a little by compiling the function. Compiling with or without CompilationTarget -> "C" affects Table[] and Array[] differently. Compiling speeds up Table[] (0.0039 sec.), but whether to C or not makes no perceptible difference. Compiling to C speeds up Array[] slightly, but compiling to WVM the option slows down Array[] (0.00091 sec.); I don't know why.

cf = Compile[{{i, _Integer}, {j, _Integer}},
Which[
i == 1, 1,
i > j, 1,
i == j, -i + 1,
True, 0], CompilationTarget -> "C"];

Array[cf, {size, size}]; // RepeatedTiming
(*  {0.00026, Null}  *)

Vectorization vs. Compiling. Many basic functions in Mathematica are vectorized: They work efficiently on packed arrays. Elementary functions, many linear algebra functions, some list manipulation functions (esp. functions operating on rectangular arrays) are particularly efficient on packed arrays. Other functions might "unpack" the array, which involves copying the packed array into the unpacked form. This done internally but there are ways to tell (On["Packing"] and DeveloperPackedArrayQ for instance); mainly it is invisible to the user except for a performance hit. Usually, but not always, the vectorized functions are more efficient than compiled functions. The solution of @MikeY takes advantage of this up until ones and the use of ArrayFlatten[], which unpacks its arguments to assemble the final matrix. The reason ArrayFlatten[] unpacks is because one of the inputs, ones, is not packed. It turns out that user-entered lists are not packed. If we pack it, we get a great improvement in performance:

With[{n = size - 1},
Block[{m1, m2, ones},
m1 = LowerTriangularize[ConstantArray[1, {n, n}], -1];
m2 = m1 - DiagonalMatrix[Range[n]];
ones = DeveloperToPackedArray@{ConstantArray[1, n]};
ArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]
]]; // RepeatedTiming
(*  {0.000041, Null}  *)

Using ArrayPad[m2, {{1, 0}, {1, 0}}, 1], as suggested by @kglr, is about the same speed (the timings fluctuate around each other), even though it eliminates a line of code (for ones). But ones is just 1 x 100 and rather small compared to the matrices. At size = 1000, ArrayFlatten[] took 0.014-0.015 sec., ArrayPad[] took 0.015-0.16 sec., and my method took 0.012-0.014 sec.

My own answer avoids unpacking and uses vectorized functions. Outer[Plus, array1, array2] and Outer[List, array1, array2] are extremely efficient special cases of Outer[] on packed arrays.

### Appendix: Timing code

sa[size]; // RepeatedTiming                  (* @kglr *)
Table[Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0],
{i, size}, {j, size}]; // RepeatedTiming  (* @Rohit Namjoshi *)
With[{n = size - 1},
Block[{m1, m2, ones},
m1 = LowerTriangularize[ConstantArray[1, {n, n}], -1];
m2 = m1 - DiagonalMatrix[Range[n]];
ones = {ConstantArray[1, n]};
ArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]
]]; // RepeatedTiming                      (* @MikeY *)
With[{r    = Range@size,
zero = DeveloperToPackedArray@{0}},
UnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]] -
]; // RepeatedTiming                      (* @Michael E2 *)

• Nice comparison, thanks. I chose my approach more for ease of understanding, or maybe to align with the underlying logic of the matrix, rather than performance. Nice to see it did alright on the performance front too. I learned something! – MikeY Jul 2 '19 at 17:41
• Thanks Michael. I think you meant "can be improved by using Array". – Rohit Namjoshi Jul 2 '19 at 17:50
• @RohitNamjoshi Thanks for pointing out the error. – Michael E2 Jul 2 '19 at 18:06
• @MikeY You could also use ones = ConstantArray[1, {1, n}] instead of ToPackedArray. – Michael E2 Jul 2 '19 at 18:14
• @MichaelE2, I tweaked my answer, thanks for the tip. – MikeY Jul 2 '19 at 19:12

You can also use SparseArray as follows

sa[n_] := SparseArray[{{i_, i_} /; i > 1 -> 1 - i, {i_, j_} /; 1 < i < j ->  0}, {n, n}, 1]

matrix2 = sa[7]
matrix2 == matrix

True

One more, for fun and education. Treating it like a block matrix, do the lower right first. This makes a lower triangular matrix of 1s, with the -1 in the command telling it to make the diagonal = zero.

m1=LowerTriangularize[ConstantArray[1, {6, 6}], -1]

$$\left( \begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 \\ \end{array} \right)$$

m2 = m1 - DiagonalMatrix[Range[6]]

$$\left( \begin{array}{cccccc} -1 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 \\ 1 & 1 & -3 & 0 & 0 & 0 \\ 1 & 1 & 1 & -4 & 0 & 0 \\ 1 & 1 & 1 & 1 & -5 & 0 \\ 1 & 1 & 1 & 1 & 1 & -6 \\ \end{array} \right)$$

Now the upper right, a $$1 \times 6$$ matrix of ones

ones = ConstantArray[1, {1, n}]

$$\left( \begin{array}{cccccc} 1 & 1 & 1 & 1 & 1 & 1 \\ \end{array} \right)$$

Use ArrayFlatten to get the final answer.

ArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]

$$\left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & -2 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & -3 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & -4 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & -5 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & -6 \\ \end{array} \right)$$

• you can also use ArrayPad[#, {{1, 0}, {1, 0}}, 1] & on m2 for the last step (+1) – kglr Jul 2 '19 at 14:38

As already suggested in the other answers, vectorization is a key to high performance when working with large numeric arrays. To further increase calculation speed it is useful to remove unnecessary intermediate steps, because memory operations on large arrays can be quite time consuming. Here is a vectorized function with very few intermediate steps:

mFast[n_] := LowerTriangularize[DiagonalMatrix[-Range[n]] + 1] + UnitVector[n, 1];

Timing measurement:

RepeatedTiming[mFast[100];]

{0.000032, Null}

This can be compared to MichaelE2's function:

mMichaelE2[size_] := With[{r = Range@size, zero = DeveloperToPackedArray@{0}},
UnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]]
`


# If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$

I am trying to teach myself mathematics (I have no access to a teacher), but I am not getting very far. I am just working through the exercises at the end of the book's chapter, but unfortunately there are no solutions.

Anyway, I am trying to prove

If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$.

But I do not know how to approach it. I thought of starting with something like $n = 6k$ for the multiple and that $m^3$ is crucial, but I do not know how that would help or where to go next. Does anyone have any hints or suggestions? Please do not post the whole proof because I want to solve it myself, thank you.

-
You might also want to show that if $n = m^5-m$ for some integer $m$, then $n$ is a multiple of 30. (This is a bit trickier but basically the same idea.) – Michael Lugo Jul 17 '12 at 0:11
@Mic $\rm\ mod\ 5\!:\ 0^5\!\equiv 0,\ (\pm1)^5\!\equiv \pm1,\ (\pm2)^5\!\equiv \pm2.\ \ mod\ 6\!:\ m^5\equiv m^3 m^2 \equiv m\ m^2\equiv m\$ by OP. $\ \ \$ – Gone Jul 17 '12 at 2:09
Bill, that is more clever than the solution I had in mind - I hadn't thought to exploit what was done in the original question! I was thinking of factoring as $m(m-1)(m+1)(m^2+1)$ and then arguing that at least one factor is divisible by each of 2, 3, and 5. – Michael Lugo Jul 17 '12 at 18:53

Lets start by factoring $n:$$n = m^3-m = m(m^2-1) = (m-1)m(m+1)$$ Note$(m-1),m$and$(m+1)$are three consecutive integers so (at least) one of these must be a multiple of$2$and one of these must be a multiple of$3$. -  Thank you - that is helpful (I can probably do something now with factors) – George Jul 17 '12 at 0:16 Glad to help. Yes: $$n = (2p)\times (3r)\times t$$ where$p,r$and$t$are integers :) – Quixotic Jul 17 '12 at 0:17 so I did this:$\frac{n}{6} = ( (m - 1)(\frac{m}{2})(\frac{(m + 1)}{3}))$so$n = 6((m - 1)(\frac{m}{2})(\frac{(m + 1)}{3}))$is that cheating? Or am I going the right way? – George Jul 17 '12 at 0:31 I don't think this is the right way since if this is valid you can probably show divisibility by any number. – Quixotic Jul 17 '12 at 0:33 After what I said before,$n = (2p)\times (3r)\times t = 6 (prt)$and you are done. – Quixotic Jul 17 '12 at 0:42 show 1 more comment Hint$\rm\ mod\ 6\!:\ 0^3\!\equiv 0,\ (\pm1)^3\!\equiv \pm1,\ (\pm2)^3\!\equiv \pm2,\ 3^3\!\equiv 3\:\Rightarrow\:n^3\equiv n\ \ $QED Note$ $It's easier via balanced residues$\{0,\, \pm1,\, \pm2,\, 3\}$vs.$\,\{0,1,2,3,4,5\},\,$by$\rm\:4\equiv -2,\:\rm 5\equiv -1.\:$It is not difficult to prove a generalized Euler-Fermat theorem, namely Theorem$\ $For naturals$\rm\: e,m,n\: $with$\rm\: e,m>1 \rm\qquad\qquad\ m\ |\ n^e-n\ $for all$\rm\:n\ \iff\ m\:$is squarefree and prime$\rm\: p\:|\:m\: \Rightarrow\: p\!-\!1\ |\ e\!-\!1 $Yours is the special case$\rm\:e={\bf\color{blue}3},\ m = 6 = {\bf\color{#C00}2}\cdot{\bf\color{#0A0}3}\:$is squarefree, and$\rm\, {\bf\color{#C00}2}\!-\!1,{\bf\color{#0A0}3}\!-\!1\:|\:{\bf\color{blue}3}-1.$- $$n = m^3 - m$$ $$= m (m^2 - 1)$$ $$= m (m - 1) (m + 1)$$ Now, let$m$be odd. So,$m = 2k \pm 1$and$m \pm 1$will be even, and vice versa. So, atl east one of the numbers from$m$,$m \pm 1$has to be divisible by$2$. Again, since multiples of three are at a separation of$2$numbers, so, again we have one of$m$,$m \pm 1$being divisible by$6$. - You could also show this by induction, noting$0^3-0$is a multiple of$6$for the base case, and then showing that$(m+1)^2-(m+1) - (m^3-m)$is a multiple of$6$for all$m$for the inductive step. (It is enough to cover the nonnegative case, because$(-m)^3-(-m)=-(m^3-m)$.) You can show that the new expression, which simplifies to$3m(m+1)$, is a multiple of$6$either by noting that one of$m$or$m+1$is even as in Quixotic's answer, or again using induction. Note that$3\cdot 0(0+1)$is a multiple of$6$for the base case, and then show that$3(m+1)(m+2)-3m(m+1)=6(m+1)$is a multiple of$6$for the inductive step. Applying a similar method to Michael Lugo's problem in the comments shows after$3$steps of taking differences (and checking base cases) that$m^5-m$is divisible by$30$. If$f(m)=m^5-m$,$g(m)=f(m+1)-f(m)$,$h(m)=g(m+1)-g(m)$, and$k(m)=h(m+1)-h(m)$, then$f(0)=g(0)=0$,$h(0)=30$, and$k(m)=150 + 180 m + 60 m^2\$.


# Two questions about counting strings

I saw this two questions that I don't fully understand:

Let $E$ be the number of strings in $\{0,1\}^{17}$ with an even number of 1's, and Let $O$ be the number of strings with an odd number of 1's. What is $E-O$

The answer in the book is 0, but there's no explanation. Also

What is the number of strings in $\{0,1,2,3\}^{17}$ with an even sum of coordinates

The answer is $\frac{4^{17}}{2}$, but also no explanation. Thanks

-

If you write down a string with 17 digits, each of which is either $0$ or $1$, then you will either have an even number of $1$s and an odd number of zeros; or an odd number of $1$s and an even number of $0$s. So $E+O$ is the total number of strings.

However, if you take all strings, and in every string you exchange the roles of $0$ and $1$ (turn all $1$s to $0$s and all $0$s to $1$s) then you get back all strings again; but every string that was in $E$ will now be in $O$, and every string that was in $O$ will now be in $E$. That is, this operations swaps the contents of $E$ and $O$. So the number of elements in $E$ must be exactly the same as the number of elements in $O$. That is, $E=O$, so $E-O=0$.

Now consider the strings of length $17$ whose entries are $0$, $1$, $2$, or $3$. Exactly half of them have an even sum, and half have an odd sum. Why? Because of all strings of length $17$ with a given $16$ digit start, half will add up to an odd number and half will add up to an even number (those that end in $0$ and $2$ will have the same parity, those that end in $1$ or $3$ will have the same parity as each other, and opposite the parity of those that end in $0$ and $2$). Since there are $4^{17}$ strings total, half of them will add up to an even sum for a total of $\frac{4^{17}}{2}$.

-
Can I conclude from the first question (with a one-to-one relation) that every set $A=\{1,...,n\}$ has exactly $\frac{2^n}{2}=2^{n-1}$ sub-sets of even/odd size? –  yotamoo Mar 14 '12 at 17:52
@yotamoo: You can use complements to conclude that if $n$ is odd (since going from $B\subseteq A$ to $A-B$ will swap odd-sized subsets with eve-sized subsets). For $n$ even, though, that argument does not work; if you have some other way of establishing a correspondence between odd-sized subsets and even-sized subsets in that case, then yes. –  Arturo Magidin Mar 14 '12 at 18:05

For the first question, we can alternatively directly count both E and O like so:

Generally, the number of strings in $\{0,1\}^{17}$ with $k$ $1$'s is $\displaystyle\binom {17}k$, as we're ranging over all possible spots to put the $k$ ones in. So E equals the sum of this over all even numbers, and similarly O is the sum over all odd numbers. We can calculate

E - O $= \displaystyle \sum_{i=0}^{8} \displaystyle \binom {17}{2i} - \sum_{i=0}^{8} \displaystyle \binom {17}{2i + 1} = \sum_{i=0}^{17} \displaystyle (-1)^i \binom {17}{i} = \sum_{i=0}^{17}\displaystyle 1^{n-i}(-1)^i \binom {17}{i} = (1-1)^{17} = 0^{17} = 0.$

However, this approach doesn't seem to lend itself to the next question directly, unless you notice that a sum is even iff the the number of $1$'s plus the number of $3$'s is even. In this case, you can consider $1$ and $3$ as one element, and $0$ and $2$ as one element (sorting them by parity). We're essentially looking for E here and so we can apply our result that E = O from above. Combined with the knowledge that E $+$ O = our total number of strings, $4^{17}$, we see that E = $\dfrac{4^{17}}{2}$.

A computational approach is helpful and possible, but the often superior option is to come up with what is called a combinatorial proof based on direct reasoning, as in Arturo or lhf's answers. As Stanley writes in enumerative combinatics,

Not only is the above combinatorial proof much shorter than our previous proof, but also it makes the reason for the simple answer completely transparent. It is often the case, as occurred here, that the first proof to come to mind turns out to be laborious and inelegant, but that the final answer suggests a simple combinatorial proof. (page 12)

-


# Math Help - Trigonometry problem.

1. ## Trigonometry problem.

Find x, when:

$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$

I have come this far, but here it stops:

$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$

$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$

$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$

$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$

Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?

2. Originally Posted by MatteNoob
Find x, when:

$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$

I have come this far, but here it stops:

$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$

$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$

$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$

$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$

Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?

Looking good!

now, let $\theta = \arctan \bigg( \frac 13 \bigg)$ (so what you are looking for is $- \sin \theta$ and $\cos \theta$ -- since $\tan (-x) = - \tan x,~ \sin (-x) = - \sin x \mbox{, and } \cos (-x) = \cos x$)

$\Rightarrow \tan \theta = \frac 13$

thus we can draw a right-triangle, with an acute angle $\theta$, where the opposite side is of length 1 and the adjacent side is of length 3, since this is how we define the tangent ratio. By Pythagoras' theorem, the hypotenuse is $\sqrt{10}$

Now, recall that $\text{sine } = \frac {\text{opposite}}{\text{hypotenuse}}$ and $\text{cosine } = \frac {\text{adjacent}}{\text{hypotenuse}}$

i leave it to you to finish up

3. Originally Posted by MatteNoob
Find x, when:

$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$

I have come this far, but here it stops:

$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$

$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$

$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$

$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$

Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?

As you noted $x=\sin\left(\frac{\pi}{3}+\arctan\left(\frac{-1}{3}\right)\right)$

And also you noted that $\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\sin\left(B\right)\cos\left(A\right)$

So we have

$x=\sin\left(\frac{\pi}{3}\right)\cos\left(\arctan\ left(\frac{-1}{3}\right)\right)+\sin\left(\arctan\left(\frac{-1}{3}\right)\right)\cos\left(\frac{\pi}{3}\right)$

Simplifying we get

$x=\frac{\sqrt{3}}{2}\cos\left(\arctan\left(\frac{-1}{3}\right)\right)+\frac{1}{2}\sin\left(\arctan\l eft(\frac{-1}{3}\right)\right)$

Now it can be shown that $\cos\left(\arctan(x)\right)=\frac{1}{\sqrt{1+x^2}}$

and $\sin\left(\arctan(x)\right)=\frac{x}{\sqrt{x^2+1}}$

So we have that

$x=\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{1+\left(\f rac{-1}{3}\right)^2}}+\frac{1}{2}\frac{\left(\frac{-1}{3}\right)}{\sqrt{1+\left(\frac{-1}{3}\right)^2}}$

$~=\frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{10}}{10}+\fr ac{1}{2}\cdot\frac{-\sqrt{10}}{20}$

$=\frac{\left(3\sqrt{3}-1\right)\sqrt{10}}{20}\quad\blacksquare$

4. To Jhevon and Mathstud28:

Thank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again

5. Originally Posted by MatteNoob
To Jhevon and Mathstud28:

Thank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again


Categories

# binomial coefficient latex

The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. }}{{k!\left( {n - k} \right)!}} In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. The possibility to insert operators and functions as you know them from mathematics is not possible for all things. }}{{k!\left( {n - k} \right)!}}. It is especially useful for reasoning about recursive methods in programming. Here's an equation: math \frac {n!} Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. n! For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Binomial coefficient, returned as a nonnegative scalar value. All combinations of v, returned as a matrix of the same type as v. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? Latex binomial coefficient Definition. So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… where A is the permutation, $$A_n^k = \frac{n!}{(n-k)! This article explains how to typeset them in LaTeX. k-combinations of n-element set. }}{{k!\left( {n - k} \right)!}} The combination (n r) (n r) is called a binomial coefficient. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. binomial Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. It will give me the energy and motivation to continue this development. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! The binomial coefficient is defined by the next expression: \[ \binom {n}{k} = \frac {n ! (−)!. On the other side, \textstyle will change the style of the fraction as if it were part of the text. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. Since binomial coefficients are quite common, TeX has the \choose control word for them. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Binomial coefficient, returned as a nonnegative scalar value. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. C — All combinations of v matrix. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." In Counting Principles, we studied combinations.In the shortcut to finding ${\left(x+y\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. (n - k)!} In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. Specially useful for continued fractions. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. All combinations of v, returned as a matrix of the same type as v. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". Open an example in Overleaf ( n - k )! It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , coefficient Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. Usually, you find the special input possibilities on the reference page of the function in the Details section. (n-k)!} In Counting Principles, we studied combinations. Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. In this video, you will learn how to write binomial coefficients in a LaTeX document. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. Identifying Binomial Coefficients. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is Using fractions and binomial coefficients in an expression is straightforward. (n - k)!} b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. Latex Thank you ! = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$, \frac{n!}{k! \boxed, How to write table in Latex ? 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