The Chain Rule is used when we want to diﬀerentiate a function that may be regarded as a composition of one or more simpler functions. J'ai constaté que la version homologue française « règle de dérivation en chaîne » ou « règle de la chaîne » est quasiment inconnue des étudiants. While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. In the next section, we use the Chain Rule to justify another differentiation technique. Differentiation – The Chain Rule Two key rules we initially developed for our “toolbox” of differentiation rules were the power rule and the constant multiple rule. Let’s start out with the implicit differentiation that we saw in a Calculus I course. 5:24. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. There are many curves that we can draw in the plane that fail the "vertical line test.'' Example of tangent plane for particular function. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 5:20. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). There is a chain rule for functional derivatives. Yes. That material is here. As u = 3x − 2, du/ dx = 3, so. Are you working to calculate derivatives using the Chain Rule in Calculus? If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f ′(x) = (g h) (x) = (g′ h)(x)h′(x). But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. The chain rule in calculus is one way to simplify differentiation. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 The Derivative tells us the slope of a function at any point.. Young's Theorem. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. This discussion will focus on the Chain Rule of Differentiation. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}. Chain Rule: Problems and Solutions. Then differentiate the function. The inner function is g = x + 3. There is also another notation which can be easier to work with when using the Chain Rule. So when using the chain rule: chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. The chain rule is not limited to two functions. Answer to 2: Differentiate y = sin 5x. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Categories. Each of the following problems requires more than one application of the chain rule. Let u = 5x (therefore, y = sin u) so using the chain rule. The chain rule says that. 2.10. 16 questions: Product Rule, Quotient Rule and Chain Rule. What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. Derivative Rules. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). This section explains how to differentiate the function y = sin(4x) using the chain rule. SOLUTION 12 : Differentiate . Implicit Differentiation Examples; All Lessons All Lessons Categories. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Thus, ( There are four layers in this problem. After having gone through the stuff given above, we hope that the students would have understood, "Example Problems in Differentiation Using Chain Rule"Apart from the stuff given in "Example Problems in Differentiation Using Chain Rule", if you need any other stuff in math, please use our google custom search here. Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. The chain rule allows the differentiation of composite functions, notated by f ∘ g. For example take the composite function (x + 3) 2. Hence, the constant 4 just tags along'' during the differentiation process. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. Together these rules allow us to differentiate functions of the form ( T)= . Try the Course for Free. 2.13. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. If cancelling were allowed ( which it’s not! ) Hessian matrix. So all we need to do is to multiply dy /du by du/ dx. I want to make some remark concerning notations. Taught By. We may still be interested in finding slopes of tangent lines to the circle at various points. Proof of the Chain Rule • Given two functions f and g where g is diﬀerentiable at the point x and f is diﬀerentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Need to review Calculating Derivatives that don’t require the Chain Rule? This calculator calculates the derivative of a function and then simplifies it. 2.12. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Transcript. Now we have a special case of the chain rule. Mes collègues locuteurs natifs m'ont recommandé de … In what follows though, we will attempt to take a look what both of those. Chain rule for differentiation. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The rule takes advantage of the "compositeness" of a function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. This rule … Linear approximation. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. It is NOT necessary to use the product rule. ) Kirill Bukin. For example, if a composite function f( x) is defined as Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. Here are useful rules to help you work out the derivatives of many functions (with examples below). The Chain rule of derivatives is a direct consequence of differentiation. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. All functions are functions of real numbers that return real values. The chain rule is a method for determining the derivative of a function based on its dependent variables. 2.11. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. This unit illustrates this rule. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. 10:07. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The chain rule tells us how to find the derivative of a composite function. In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. For those that want a thorough testing of their basic differentiation using the standard rules. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for diﬀerentiating a function of another function. 10:40. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 10:34. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. Next: Problem set: Quotient rule and chain rule; Similar pages. Second-order derivatives. 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. , dy dy dx du . Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. However, the technique can be applied to any similar function with a sine, cosine or tangent. If x + 3 = u then the outer function becomes f = u 2. Let’s do a harder example of the chain rule. Associate Professor, Candidate of sciences (phys.-math.) Examples of product, quotient, and chain rules ... = x^2 \cdot ln \ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Composite functions the next section, we will discuss the basic formulas of.... Where we use the chain rule in calculus thus, ( there are four layers in tutorial! To take a look what both of those, and learn how to differentiate functions of the following requires! 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# Kernel of successive powers of a matrix

For any $n \times n$ matrix $A$, is it true that $\ker(A^{n+1}) = \ker(A^{n+2}) = \ker(A^{n+3}) = \dots$ ? If yes, what is the proof and is there a name to this theorem? If not, for what matrices will it be true? How can I find a counterexample in the latter case?

I know that powers of nilpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps.

But is it necessary that for all singular matrices, all the rank reduction (if it occurs) must be in the initial exponents itself? In other words, is it possible for some matrices to have $\ker(A^{k}) = \ker(A^{k+1}) < \ker(A^{k+1+m})$ for some $m,k > 0$?

• The title isn't supposed to replace the first line of your question. As for the question, the answer depends on how you quantify over $n$. – Git Gud Sep 10 '14 at 18:59
• Added the first line. Could you please explain what you mean by 'quantify over n'? – allrtaken Sep 10 '14 at 19:07
• Let $P(n)$ be expression in the title before. If you mean $\exists n\in \mathbb NP(n)$, then the statement is true. If you mean $\forall n\in \mathbb NP(n)$, then the statement is false. – Git Gud Sep 10 '14 at 19:12
• I meant n to be the dimension of the matrix. – allrtaken Sep 10 '14 at 19:21

This is true. To my knowledge, there is no name for this theorem.

You can think of this as a consequence of Jordan canonical form. In particular, we can always write $$A = S[N \oplus P]S^{-1}$$ Where $N$ is nilpotent and $P$ has full rank. It suffices to show that $N$ has order of nilpotence at most equal to $n$, and that $P$ never reduces in rank.

• I want to check if I understood this correctly. Does the matrix N comprise of Jordan blocks with eigenvalue 0, and P is the matrix comprising of Jordan blocks corresponding to the other eigenvalues? – allrtaken Sep 10 '14 at 19:21
• @allrtaken that's exactly right. It is useful to note that a matrix is nilpotent if and only if all of its (complex) eigenvalues are equal to $0$. – Omnomnomnom Sep 10 '14 at 20:25
• I am not sure whether this is the right place for asking this question, but can you suggest a good textbook that covers the theory around this? – allrtaken Sep 11 '14 at 18:26
• Most linear algebra texts geared towards advanced undergraduates or graduates cover Jordan Canonical form at some point. I, in particular, used Horn and Johnson. Axler's "Linear Algebra Done Right" might be another good bet. If you want some more ideas, you could always post another question on this site. – Omnomnomnom Sep 11 '14 at 19:01

An important observation to be made here is that the if for some $k$, we have $\ker(A^k) = \ker(A^{k+1})$, then $\forall j\geq 0, \ker(A^{k+j}) = \ker(A^k)$. To show this, it would be sufficient to show that $\ker(A^{k+2}) = \ker(A^{k+1})$, and the rest would follow from a simple inductive argument.

Note that, we have $\ker(A^{k+1}) \subseteq \ker(A^{k+2})$, and thus it is enough to show that $\ker(A^{k+1}) \supseteq \ker(A^{k+2})$.

For this, consider a vector $v$ such that $v \in \ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \in \ker(A^{k+1})$ because $A^{k+1}(Av) = 0$. Since $\ker(A^{k+1}) = \ker(A^k)$, we have $Av \in \ker(A^k)$. Thus, $A^{k}(Av) = 0$, and hence $A^{k+1}v = 0$, which implies that $v \in \ker(A^{k+1})$.

Clearly, $\ker(A^{k+2}) \subseteq \ker(A^{k+1})$, and thus $\ker(A^{k+2}) = \ker(A^{k+1})$


# How does the middle term of a quadratic $ax^2 + bx + c$ influence the graph of $y = x^2$?

Every parabola represented by the equation $$y = ax^2 + bx + c$$ can be obtained by stretching and translating the graph of $$y = x^2$$.

Therefore:

The sign of the leading coefficient, $$-a$$ or $$a$$, determines if the parabola opens up or down i.e.

The leading coefficient, $$a$$, also determines the amount of vertical stretch or compression of $$y = x^2$$ i.e.

The constant term, $$c$$, determines the vertical translation of $$y = x^2$$ i.e.

Now for $$bx$$. Initially, I thought it would determine the amount of horizontal translation since the constant term, $$c$$, already accounted for the vertical translation, but when I plugged in some quadratics the graph of $$y = x^2$$ translated both horizontally and vertically. Here are the graphs:

Seeing as the middle term, $$bx$$, does more than just horizontally translate, how do you describe its effect on $$y=x^2$$? Would it be accurate to say that it both horizontally and vertically translates the graph of $$y = x^2$$?

• +1 for beautiful graphs and your efforts too!! – StammeringMathematician Sep 27 '18 at 4:18
• @StammeringMathematician Thank you! I used this to make the graphs: desmos.com/calculator – Slecker Sep 27 '18 at 4:22
• +1 from me as well. This attitude should be highly encouraged here on MSE. – Ahmad Bazzi Sep 27 '18 at 4:34
• Slecker. Beautiful +. – Peter Szilas Sep 27 '18 at 9:17

Yes, it will effect both a horizontal and vertical translation, and you can see how much by completing the square. For example, $$x^2+3x=\left(x+\frac32\right)^2-\frac94$$

Compare that to your graph of $$y=x^2+3x$$. Of course, if the coefficient of the quadratic term is not $$1$$ things get a little more complicated, but you can always see what the graph the graph will look like by completing the square.

• It took me a while to realize that you transformed it into vertex form. So would the reason that $bx$ affects both a horizontal and vertical translation be because it occurs in both the x and y-coordinates of the vertex, since the vertex coordinates are ($\frac{-b}{2a}$, $\frac{ -b^2+4ac}{4a})$? – Slecker Sep 27 '18 at 4:46
• @Slecker I'm not familiar with the term "vertex form," but I would say that you are correct. – saulspatz Sep 27 '18 at 5:04

Look at $$2$$ Cartesian coordinate systems $$X,Y$$ and $$X',Y'$$.

Origin of $$X',Y$$' is located at $$(x_0,y_0)$$, $$X'$$-axis parallel $$X$$-axis , $$Y'$$-axis parallel $$Y$$-axis(a translation),i.e.

$$x= x_0+x'$$; $$y= y_0+ y'$$.

Set up your normal parabola in the $$X',Y'$$ coordinate system.

$$y'=ax'^2$$, vertex at $$(0',0')$$.

Revert to original $$x,y$$ coordinates .

$$y-y_0= a(x-x_0)^2$$ ;

$$y=ax^2 -2(ax_0)x +ax_0^2$$.

Compare with $$y =ax^2+bc +c$$:

$$b=-2ax_0$$.

Can you interpret?

• I'm having a hard time understanding what you mean by "Revert to original $x$, $y$ coordinates" and where the subsequent equation, $y-y_0 = a(x-x_0)^2$, came from. I think once I understand that I can interpret the rest of your answer. – Slecker Sep 27 '18 at 15:22
• Slecker.Draw two coordinate systems, x,y and another one ,call it x',y'.Say, you put the origin of the x',y' system at x_0=3, y_0=4.x'y' system has its origin at (x_0,y_0)=(3,4), ok?. put a normal parabola y'=ax'^2 in the x',y' system.x'=1; y'=a; everything in x'y'.Take any x' coordinate, say x'=7, what is the x value in the original system: x= 7+ 3= x' +x_0 ok? Likewise y= y'+y_0. Solve for x' and y' and plug into y'=ax'2, get (y-y_0)=a(x-x_0)^2, now you are back in the original system.Your b =-2ax_0, where x_0 is the x-coordinate of the vertex.Let me know if ok. – Peter Szilas Sep 27 '18 at 17:56
• Ah ok thanks for the clarification! – Slecker Sep 27 '18 at 18:26
• Slecker. If anything else, just say so:) – Peter Szilas Sep 27 '18 at 18:35


# Cone shaped related rates of change question

A container is in the shape of a cone of semi-vertical angle $30^\circ$, with it's vertex downwards.

Liquid flows into the container at ${{\sqrt {3\pi } } \over 4}{\rm{ }}c{m^{^3}}/s$

At the instant when the radius of the circular surface of the liquid is 5 cm, find the rate of increase of:

(a) The radius of the circular surface of the liquid

(b) The area of the circular surface of the liquid

My attempt:

${{dV} \over {dt}} = {{\sqrt {3\pi } } \over 4}$

(A) I need to find the rate at which the radius increases as "h" increases, so I have to find ${{dr} \over {dt}}$.

The equation for the volume of a cone is:

$V = {1 \over 3}\pi {r^2}h$

I now must form a function in terms of r for h. As we are asked to a compute when the radius is 5 we can form an equation using similar triangles, so:

\eqalign{ & {h \over x} = {r \over 5} \cr & x: \cr & \tan 30^\circ = {5 \over x} \cr & x = {5 \over {\tan 30^\circ }} \cr & x = 5\sqrt 3 \cr & so: \cr & h = r\sqrt 3 \cr}

\eqalign{ & V = {1 \over 3}\pi {r^2}(r\sqrt 3 ) \cr & V = \pi {r^3}{{\sqrt 3 } \over 3} \cr & {{dV} \over {dr}} = \pi {r^2}\sqrt 3 \cr & {{dr} \over {dt}} = {{dV} \over {dt}} \times {{dr} \over {dV}} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over 4} \times {1 \over {\pi {r^2}\sqrt 3 }} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over {\pi {r^2}4\sqrt 3 }} = {{\sqrt \pi } \over {\pi {r^2}4}} \cr}

\eqalign{ & r = 5: \cr & {{\sqrt \pi } \over {4(25)\pi }} = {{\sqrt \pi } \over {100\pi }} = 0.005641... \cr}

For part (A) the answer is stated as 0.01 cm/s, nowhere in the question have I been asked to round my answer, have I obtained the correct answer? I just want to make sure..

Part (b)

Part (B) requires that I calculate the rate of increase of the circular area of the liquid, so essentially ${{dA} \over {dt}}$.

Area of a circle is: $A = \pi {r^2}$

\eqalign{ & A = \pi {r^2} \cr & {{dA} \over {dr}} = 2\pi r \cr & {{dA} \over {dt}} = {{dr} \over {dt}} \times {{dA} \over {dr}} \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {4\pi {r^2}}} \times 2\pi r \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {2r}} \cr & r = 5: \cr & {{dA} \over {dt}} = 0.1\sqrt \pi {\rm{ c}}{{\rm{m}}^2}/s \cr}

However the answer in the book for b is ${{dA} \over {dt}} = 0.1\pi {\rm{ c}}{{\rm{m}}^2}/s$ (pi is not square rooted), Where have I gone wrong?

Furthermore I'd love it if any answerers could suggest how I could improve on how I've done things, and any tricks/tips that would make things easier for myself in the future.

Thank you.

• If you are going to use $x$, you should define it. In fact, you don't need to find $h$, but if you want to, $\frac rh=\tan 30^\circ=\frac {\sqrt 3} 3$, so $h=5\sqrt 3$, which you got. – Ross Millikan May 2 '13 at 22:37
• @RossMillikan, Oh okay, I'll bear that in mind, thank you! – seeker May 2 '13 at 22:54

"Liquid flows into the container at: $\;\displaystyle \dfrac{\sqrt{3}\,\pi}{4} = {{(\sqrt {3})\,\pi} \over 4}{\rm{ }}\text{ cm$^3$per sec}$"
This would explain both discrepancies, (between both the solutions you obtained, and the solutions of the text), since your work seems to be fine, and was clearly done carefully. Simply changing your final evaluations using $\pi$ instead of $\sqrt{\pi}$ will yield the solutions you are given.


1. watchmath

The term of the absolute series is $$|\arctan(1/n)|/n^2<\frac{\pi/2}{n^2}$$. But the series $$\sum 1/n^2$$ is convergent (since it is a p-series with p=2), so the absolute series converges. Then the series is absolutely convergent.

2. anonymous

how did you come up with pi/2/x^2

3. watchmath

the range of the arctan function is between -pi/2 and pi/2

4. anonymous

and why did you put over x^2

5. watchmath

well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.

6. anonymous

so would you always compare it to pi/2 divided by denominator given in the problem

7. anonymous

for tan

8. watchmath

I wouldn't say always. I just see the opportunity that if we compare with pi/2 then I can have some conclusion. It is possible for a problem to have an arctan bu we don't compare with the pi/2.

9. anonymous

and if there isnt anything in the denominator of the given problem you would compare it to pi/2

10. watchmath

No, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series $$\sum pi/2$$ is divergent. So comparing with pi/2 doesn't give me any conclusion.

11. anonymous

okay what would you comoare arcsin (1/n) to and 1-cos1/n to

12. watchmath

the term of the series only arcsin(1/n) ?

13. anonymous

yah

14. watchmath

Just to make sure. So your series is $$\sum_{n=1}^\infty \arcsin(1/n)$$? I can't think how to do this right away...

15. anonymous

yah

16. watchmath

ok, arcsin is an increasing function and for positive x we have x > sin x Apply the arcsin arcsin x > x It follows that arcsin(1/n) > 1/n But the harmonic series $$\sum 1/n$$ is divergent Hence $$\sum \arcsin(1/n)$$ is divergent as well.

17. anonymous

what if it was just sin

18. watchmath

sin what?

19. anonymous

if instead of arc sin it was sin would you still make the same comparison

20. watchmath

you mean sin(1/n) ?

21. anonymous

yah

22. watchmath

well we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet.

23. anonymous

oh okay

24. anonymous

also if you were given sigma 1-cos(1/n) what would you compare it to

25. watchmath

That is actually a nice problem. I'll post it as a new question so everybody can give a response (BTW it is convergent)


# Integration of secant

\begin{align} \int \sec x \, dx &= \int \cos x \left( \frac{1}{\cos^2x} \right) \, dx \\ &= \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\ & = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx \end{align}

I am stuck in here. Any help to integrate secant?

## 5 Answers

\begin{align*}\int\sec x\,\mathrm dx&=\int\frac1{\cos x}\,\mathrm dx\\&=\int\frac{\cos x}{\cos^2x}\,\mathrm dx\\&=\int\frac{\cos x}{1-\sin^2x}\,\mathrm dx.\end{align*} Now, doing $$\sin x=t$$ and $$\cos x\,\mathrm dx=\mathrm dt$$, you get $$\displaystyle\int\frac{\mathrm dt}{1-t^2}$$. But\begin{align*}\int\frac{\mathrm dt}{1-t^2}&=\frac12\int\frac1{1-t}+\frac1{1+t}\,\mathrm dt\\&=\frac12\left(-\log|1-t|+\log|1+t|\right)\\&=\frac12\log\left|\frac{1+t}{1-t}\right|\\&=\frac12\log\left|\frac{(1+t)^2}{1-t^2}\right|\\&=\log\left|\frac{1+t}{\sqrt{1-t^2}}\right|\\&=\log\left|\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right|\\&=\log\left|\frac1{\cos x}+\frac{\sin x}{\cos x}\right|\\&=\log|\sec x+\tan x|.\end{align*}

• What a tricky..! Thx – Beverlie Jun 29 '17 at 15:25
• I've been reading about the early history of calculus. For a long period in the 17th century this was a significant unsolved problem. – DanielWainfleet Aug 3 '17 at 17:56
• @DanielWainfleet I think I read something about that in the historical notes of Spivak's Calculus. – José Carlos Santos Aug 3 '17 at 17:59

An alternative method: The trick here is to multiply $\sec{x}$ by $\dfrac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}$, then substitute $u=\tan{x}+\sec{x}$ and $du=(\sec^2{x}+\tan{x}\sec{x})~dx$:

$$\int \sec{x}~dx=\int \sec{x}\cdot \frac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}~dx=\int \frac{\sec{x}\tan{x}+\sec^2{x}}{\tan{x}+\sec{x}}~dx=\int \frac{1}{u}~du=\cdots$$

Not obvious, though it is efficient.

After $\int \cos x \left(\frac{1}{1-\sin^2x}\right)dx$ use the transformation $z = \sin x$ and $dz = \cos x \, dx$.

Edit:

$$\int\frac{1}{1-u^2}\,du = \frac{1}{2}\int\frac{(1+u)+(1-u)}{(1+u)(1-u)} = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u}\,du$$

And use, $\int \frac{1}{u}\,du = \ln|u|$

• I'd got $\int \frac{1}{1-u^2}du$ what would be the next step? – Beverlie Jun 29 '17 at 15:22
• Use partial fraction method as in my edit. – Dhruv Kohli - expiTTp1z0 Jun 29 '17 at 15:26

Although the integral can be evaluated in a straightforward way using real analysis, I thought it might be instructive to present an approach based on complex analysis. To that end, we now proceed.

We use Euler's Formula, $e^{ix}=\cos(x)+i\sin(x)$, to write $\displaystyle \sec(x)=\frac2{e^{ix}+e^{-ix}}=\frac{2e^{ix}}{1+e^{i2x}}$. Then, we have

\begin{align} \int \sec(x)\,dx&=\int \frac2{e^{ix}+e^{-ix}}\\\\ &=\int \frac{2e^{ix}}{1+e^{i2x}}\,dx \\\\ &=-i2 \int \frac{1}{1+(e^{ix})^2}\,d(e^{ix})\\\\ &=-i2 \arctan(e^{ix})+C\tag 1\\\\ &=\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)+C\tag2\\\\ &=\log\left(-i\left(\frac{1+\sin(x)}{i\cos(x)}\right)\right)+C\tag3\\\\ &=\log(\sec(x)+\tan(x))+C'\tag4 \end{align}

NOTES:

In going from $(1)$ to $(2)$, we used the identity $\arctan(z)=i2\log\left(\frac{1-iz}{1+iz}\right)$

In going from $(2)$ to $(3)$, we multiplied the numerator and denominator of the argument of the logarithm function by $1-ie^{ix}$. Then, we used

$$\frac{1-ie^{ix}}{1+ie^{ix}}=\frac{-i2\cos(x)}{2(1-\sin(x))}=-i\frac{1+\sin(x)}{\cos(x)}$$

Finally, in going from $(3)$ to $(4)$, we absorbed the term $\log(-i)$ into the integration constant $C$ and labeled the new integration constant $C'=C+\log(-i)$.

Just to spell out Lord Shark the Unknown's suggestion, $$t=\tan\frac{x}{2}\implies\sec x=\frac{1+t^2}{1-t^2},\,dx=\frac{2dt}{1+t^2}\implies\int\sec xdx=\int\frac{2 dt}{1-t^2}.$$From that point on, the same partial-fractions treatment as in multiple other answers can be used. Admittedly the expression thus obtained for the antiderivative is $$\ln\left|\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right|+C$$ instead of $$\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C$$ or $$\ln|\sec x+\tan x|+C$$, but of course they're all the same thanks to suitable trigonometric identities (again, obtainable by writing things as functions of $$\tan\frac{x}{2}$$).


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Fibonacci Sequence is a wonderful series of numbers that could start with 0 or 1. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Try it again. If we have an infinite series, $$S = 1 + ax + (ax)^2 + (ax)^3 + \cdots,$$, with $|ax| < 1$, then its sum is given by, This means, if the sum of an infinite geometric series is finite, we can always have the following equality -, $$\frac{1}{1 - ax} = 1 + ax + (ax)^2 + (ax)^3 + \cdots = \sum_{n \ge 0} a^n x^n$$, Using this idea, we can write the expression of $F(x)$ as, $$F(x) = \frac{1}{(\alpha - \beta)}\left(\frac{1}{1-x\alpha} - \frac{1}{1-x\beta} \right) = \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \ge 0 } x^n \beta^n \right)$$, Recalling the original definition of $F(x)$, we can finally write the following equality, $$F(x) = \sum_{n \ge 0}F_n x^n = \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \ge 0 } x^n \beta^n \right),$$, and comparing the $n-$th terms on both sides, we get a nice result, $$F_n = \frac{1}{\sqrt{5}} \left(\alpha^n - \beta^n \right),$$, (This number $\alpha$ is also a very interesting number in itself. 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Where, φ is the Golden Ratio, which is approximately equal to the value 1.618. n is the nth term of the Fibonacci sequence To calculate each successive Fibonacci number in the Fibonacci series, use the formula where is th Fibonacci number in the sequence, and the first … Next, we multiply the last equation by $x_n$ to get, $$x^n \cdot F_{n+1} = x^n \cdot F_n + x^n \cdot F_{n-1},$$, $$\sum_{n \ge 1}x^n \cdot F_{n+1} = \sum_{n \ge 1} x^n \cdot F_n + \sum_{n \ge 1} x^n \cdot F_{n-1}$$, Let us first consider the left hand side -, $$\sum_{n \ge 1} x^n \cdot F_{n+1} = x \cdot F_2 + x^2 \cdot F_3 + \cdots$$, Now, we try to represent this expansion in terms of $F(x)$, by doing the following simple manipulations -, $$\frac{1}{x} \left( x^2 \cdot F_2 + x^3 \cdot F_3 + \cdots \right)$$, $$\frac{1}{x} \left(- x \cdot F_1 + x \cdot F_1 + x^2 \cdot F_2 + x^3 \cdot F_3 + \cdots \right)$$, Using the definition of $F(x)$, this expression can now be written as, $$\frac{1}{x} \left(- x \cdot F_1 + F(x)\right)$$, Therefore, using the fact that $F_1=1$, we can write the entire left hand side as, $$\sum_{n \ge 1} x^n \cdot F_{n+1} = x \cdot F_2 + x^2 \cdot F_3 + \cdots = \frac{F(x) - x}{x}$$, $$\sum_{n \ge 1}x^n \cdot F_n + \sum_{n \ge 1} x^n \cdot F_{n-1}.$$, $$\left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right ) + \left( x^2 \cdot F_1 + x^3 \cdot F_2 + \cdots \right)$$. This equation calculates numbers in the Fibonacci sequence (Fn) by adding together the previous number in the series (Fn-1) with the number previous to that (Fn-2). x(n-2) is the term before the last one. Male or Female ? The powers of phi are the negative powers of Phi. Fibonacci Number Formula. . Alternatively, you can choose F₁ = 1 and F₂ = 1 as the sequence starters. Generalized Fibonacci sequence is defined by recurrence relation F pF qF k with k k k t 12 F a F b 01,2, Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student By the above formula, the Fibonacci number can be calculated in . Computing Fibonacci number by exponentiation. Table of Contents. If F(n) represents the nth Fibonacci number, then: F(n) = (a^n - b^n)/(a - b) where a and b are the two roots of the quadratic equation x^2-x-1 = 0. Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. We can also use the derived formula below. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. Francis Niño Moncada on October 01, 2020: Jomar Kristoffer Besayte on October 01, 2020: Mary Kris Banaynal on September 22, 2020: Ace Victor A. Acena on September 22, 2020: Andrea Nicole Villa on September 22, 2020: Claudette Marie Bonagua on September 22, 2020: Shaira A. Golondrina on September 22, 2020: Diana Rose A. Orillana on September 22, 2020: Luis Gabriel Alidogan on September 22, 2020: Grace Ann G. Mohametano on September 22, 2020. Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. A sequence derived from this equation is often called a Lucas sequence, named for French mathematician Edouard Lucas. The rule for calculating the next number in the sequence is: x(n) = x(n-1) + x(n-2) x(n) is the next number in the sequence. to get the rest. Remember, to find any given number in the Fibonacci sequence, you simply add the two previous numbers in the sequence. The first two terms of the Fibonacci sequence is 0 followed by 1. To calculate each successive Fibonacci number in the Fibonacci series, use the formula where is th Fibonacci number in the sequence, and the first … Fibonacci initially came up with the sequence in order to model the population of rabbits. In his memoir in the theory of conjugate axis and the moment of inertia of bodies, he enumerated the principle which is known now as Binet's Theorem. The Fibonacci sequence is a series where the next term is the sum of pervious two terms. A Fibonacci spiral having an initial radius of 1 has a polar equation similar to that of other logarithmic spirals . F n – 1 and F n – 2 are the (n-1) th and (n – 2) th terms respectively Also Check: Fibonacci Calculator. Solution for 88. $$0, 1, 1, 2, 3, 5, 8, 13 ,21, 34, 55, \cdots$$, Any number in this sequence is the sum of the previous two numbers, and this pattern is mathematically written as. Fibonacci Sequence. This pattern turned out to have an interest and … The Fibonacci Sequence is a series of numbers. The characteristic equation is, with roots. Although Fibonacci only gave the sequence, he obviously knew that the nth number of his sequence was the sum of the two previous numbers (Scotta and Marketos). # Program to display the Fibonacci sequence up to n-th term nterms = int(input("How many terms? ")) Abstract. A Fibonacci spiral having an initial radius of 1 has a polar equation similar to that of other logarithmic spirals . The Fibonacci series is a very famous series in mathematics. A Closed Form of the Fibonacci Sequence Fold Unfold. Example 2: Find the 25th term of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... Answer: Since you're looking for the 25th term, n = 25. Fibonacci number is defined by: Obviously, Fibonacci sequence is a difference equation (in above example) and it could be written in: Matrix Form. Computing Fibonacci number by exponentiation. In this tutorial I will show you how to generate the Fibonacci sequence in Python using a few methods. You can use the Binet's formula in in finding the nth term of a Fibonacci sequence without the other terms. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Now, this expression is fairly easy to understand and quite sufficient to produce any Fibonacci number by plugging the required value of $n$. By taking out a factor of $x$ from the second expansion, we get, $$\left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right ) + x \left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right).$$, Using the definition of $F(x)$, this can finally be written as. This pattern turned out to have an interest and … Our job is to find an explicit form of the function, $F(x)$, such that the coefficients, $F_n$ are the Fibonacci numbers. I know that the relationship is that the "sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term", but I don't think that is worded right? If you got 4 correct answers: You made it! ( Using power of the matrix {{1,1},{1,0}} ) This another O(n) which relies on the fact that if we n times … If you got between 0 and 1 correct answer: You can do it next time. It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre. Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. The equation is a variation on Pell's, in that x^2 - ny^2 = +/- 4 instead of 1. Fibonacci number is defined by: Obviously, Fibonacci sequence is a difference equation (in above example) and it could be written in: Matrix Form. The problem yields the ‘Fibonacci sequence’: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377 . Let us define a function $F(x)$, such that it can be expanded in a power series like this, $$F(x) = \sum_{n \ge 0}x^n F_n = x \cdot F_1 + x^2 \cdot F_2 + \cdots$$. Following the same pattern, 3 is found by adding 1 and 2, 5 is found by adding 2 and 3 and so on. So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first - quite a task, even with a calculator! F n = F n-1 + F n-2. The Fibonacci formula is used to generate Fibonacci in a recursive sequence. Fibonacci Sequence. Therefore, by equating the left and the right hand sides, the original formula can be re-written in terms of $F(x)$ as, $$\frac{F(x) - x}{x} = F(x) + xF(x) ~~ \Longrightarrow ~~ F(x) = \frac{x}{1-x-x^2}$$, Let us now simplify this expression a bit more. Mar 12, 2018 - Explore Kantilal Parshotam's board "Fibonacci formula" on Pinterest. The formula to calculate the Fibonacci numbers using the Golden Ratio is: X n = [φ n – (1-φ) n]/√5. Fibonacci Sequence is popularized in Europe by Leonardo of Pisa, famously known as "Leonardo Fibonacci".Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. If you got between 2 and 3 correct answers: Maybe you just need more practice. If we expand the by taking in above example, then. From this we find the formula, valid for all, and one desired continuous extension is clearly the real part Derivation of Fibonacci sequence . Fibonacci Sequence. THE FIBONACCI SEQUENCE, SPIRALS AND THE GOLDEN MEAN. It is not hard to imagine that if we need a number that is far ahead into the sequence, we will have to do a lot of "back" calculations, which might be tedious. However, if I wanted the 100th term of this sequence, it would take lots of intermediate calculations with the recursive formula to get a result. . Fibonacci Formula. Observe the following Fibonacci series: Derivation of Fibonacci sequence . The standard formula for the Fibonacci numbers is due to a French mathematician named Binet. Example 1: Find the 10th term of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... Answer: Since you're looking for the 10th term, n = 10. In order to make use of this function, first we have to rearrange the original formula. The sequence starts like this: 0, 1, 1, 2, 3, 4, 8, 13, 21, 34 (Issues regarding the convergence and uniqueness of the series are beyond the scope of the article). This sequence of Fibonacci numbers arises all over mathematics and also in nature. Number Theory > Special Numbers > Fibonacci Numbers > Binet's Fibonacci Number Formula Binet's formula is a special case of the Binet form with , corresponding to the th Fibonacci … . To improve this 'Fibonacci sequence Calculator', please fill in questionnaire. And even more surprising is that we can calculate any Fibonacci Number using the Golden Ratio: x n = φn − (1−φ)n √5. In this article, we are going to discuss another formula to obtain any Fibonacci number in the sequence, which might (arguably) be easier to work with. You can calculate the Fibonacci Sequence by starting with 0 and 1 and adding the previous two numbers, but Binet's Formula can be used to calculate directly any term of the sequence. Subscribe to the newsletter to receive more stories mailed directly to your inbox, The methods of finding roots of a quadratic equations are quite easy and are very well understood. Using The Golden Ratio to Calculate Fibonacci Numbers. In reality, rabbits do not breed this… The first two terms of the Fibonacci sequence is 0 followed by 1. The Fibonacci numbers are the sequence of numbers defined by the linear recurrence equation (1) Fibonacci Series Formula. Forty years ago I discovered that the Fibonacci Sequence (1, 1, 2, 3, 5, 8, etc) can be generated from the second degree Diophantine equation 5k^2 -/+ 4 = m^2 where the -,+ is taken alternately. The Fibonacci sequence was defined in Section 11.1 by the equations fi = 1, f2= 1, fn= fn=1 + fn-2 n> 3 %3D %3D Show that each of the following… Get the best viral stories straight into your inbox! He died in Paris, France in 1856. So, for n>1, we have: f₀ = 0, f₁ = 1, Let’s start by talking about the iterative approach to implementing the Fibonacci series. To create the sequence, you should think of 0 … Fibonacci number - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum of the two previous numbers. The authors would like to thank Prof. Ayman Badawi for his fruitful suggestions. The Fibonacci sequence is one of the most famous formulas in mathematics. 1 Binet's Formula for the nth Fibonacci number We have only defined the nth Fibonacci number in terms of the two before it: the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th. In this paper, we present properties of Generalized Fibonacci sequences. where: a = (F₁ - F₀ψ) / √5 b = (φF₀ - F₁) / √5 F₀ is the first term of the sequence, F₁ is the second term of the sequence. But what if you are asked to find the 100th term of a Fibonacci sequence, are you going to add the Fibonacci numbers consecutively until you get the 100th term? THE FIBONACCI SEQUENCE, SPIRALS AND THE GOLDEN MEAN. Fibonacci sequence equation. If at all, its only drawback is that, if we want to know a particular number, $F_n$ in the sequence, we need two numbers $F_{n-1}$ and $F_{n-2}$ that came before it; that's just how this formula works. F n = n th term of the series. Another way to write the equation is: Therefore, phi = 0.618 and 1/Phi. There are all kinds of approaches available, like, Ptolemy was an ancient astronomer, geographer, and mathematician who lived from (c. AD 100 – c. 170). In the case of the Fibonacci sequence, the recurrence is, with initial conditions. We can see from the following table, that by plugging the values of $n$, we can directly find all Fibonacci numbers! I have been assigned to decribe the relationship between the photo (attached below). 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# What is the dimension of two subspaces?

Let $V$ be the vector space with a basis $x_1, \ldots, x_9$ and $$V_1 = \{(a,a,a,b,b,b,c,c,c): a,b,c \in \mathbb{C}\}, \\ V_2=\{(a,b,c,a,b,c,a,b,c): a,b,c \in \mathbb{C}\}.$$ Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \cap V_2$?

I first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \}$$ and $V_2$ is
$$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \}.$$ By solving the collection of the equations for $V_1$ and $V_2$, I obtain the solutions $k_1=\cdots =k_9$. So the dimension of $V_1 \cap V_2$ is one dimensional. Is this correct?

Thank you very much.

It is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then
• Since $(a,b,c,d,e,f,g,h,i)\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$
• Since $(a,a,a,d,d,d,g,g,g)\in V_2$, $d=g=a$.
Therefore $V_1\cap V_2=\left\{(a,a,a,a,a,a,a,a,a)\in\mathbb{C}^9\,\middle|\,a\in\mathbb C\right\}$, which is clearly $1$-dimensional.
Yes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$: $$\left( \begin{array}{ccccccccc} 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ \hline 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ 0&0&1&0&0&1&0&0&1 \end{array}\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\left( \begin{array}{ccccccccc} \mathbf{1}&0&0&1&0&0&1&0&0\\ 0&\mathbf{1}&0&0&1&0&0&1&0\\ 0&0&\mathbf{1}&0&0&1&0&0&1\\ 0&0&0&\mathbf{1}&1&1&0&0&0\\ 0&0&0&0&0&0&\mathbf{1}&1&1\\ 0&0&0&0&0&0&0&0&0 \end{array}\right)$$ Hence $\dim(V_1+V_2)=5$ and $$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$


# Evaluating a double integral in polar coordinates

#### skatenerd

##### Active member
I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.

#### Prove It

##### Well-known member
MHB Math Helper
Re: evaluating a double integral in polar coordinates

Since you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually $$\displaystyle \displaystyle 2\pi$$, which means your $$\displaystyle \displaystyle \theta$$ bounds are actually 0 to $$\displaystyle \displaystyle 2\pi$$.

#### chisigma

##### Well-known member
Re: evaluating a double integral in polar coordinates

I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral
$$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$
into an integral in polar coordinates and then evaluate it.
Changing to polar coordinates I got the integral
$$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$
and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$.
I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.
The bounds of the inner integral are 0 and a, not -a and a so that is...

$\displaystyle S= \int_{0}^{2\ \pi} \int_{0}^{a} r\ d r\ d \theta = \pi\ a^{2}$ (1)

Kind regards

$\chi$ $\sigma$

#### skatenerd

##### Active member
Re: evaluating a double integral in polar coordinates

Thanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it!


# What kind of combinatorial problem is this?

Is there a theory from which the following problem comes? Does this type of problem have a name?

Find the largest possible number of $k$-element sets consisting of points from some finite set and have pairwise singleton or empty intersections.

I hope that was clear. If not, here's an example for $k=3$:

Let the set of points be $S=\{1,2,3,4,5,6\}$. The most 3-element sets (with pairwise singleton or empty intersections) that can be constructed from $S$ is 4, such as $\{456,236,124,135\}$.

I made a table for $|S|=3,4,5,6,7,8,9$ and got $1,1,2,4,7,8,12$, respectively, hoping I could dig up some information from OEIS.

I read a little on Steiner systems, and although it feels like I'm in the neighborhood, I'm not confident...

Edit1: typos.

Edit2: Johnson graphs and (for $k=3$) Steiner Triple Systems (STS) seem close to what I'm looking for. The condition of "pairwise singleton or empty intersections" is equivalent to "every 2-subset of S occurs in at most one $k$-element set". STS require that every 2-subset of S occurs in exactly one $3$-element set".

Edit3: Thank you to everyone who replied! All of your comments helped me push through a barrier I was facing for some time.

-
Looks a bit like a Johnson graph. Perhaps looking at some of the more popular objects in finite geometry will get you graphs matching your own – muzzlator Feb 19 '13 at 20:00
Thanks for the tip. – sasha Feb 19 '13 at 20:36
Design theory is also relevant – mrf Feb 19 '13 at 23:30
The $k=3$ sequence seems to be oeis.org/A001839 . – Kevin Costello Feb 19 '13 at 23:51
In general, you are looking for the maximal code of length $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,4)$, found at oeis.org/A004037. – mjqxxxx Feb 20 '13 at 0:57

You exactly want to determine the clique number of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which is the graph having all the $k$-element subsets as its $\binom{n}{k}$ vertices, where any two vertices are connected if and only if their cut contains at most $s$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that all their pairwise cuts are singleton or emtpy. The size of such a maximum clique is the clique number $\omega(KG_{n,k,s})$.

Googling around a bit, I could not find an exact expression therefore.

However, this states that $\omega(KG_{n,k,0}) = \lfloor \frac n k \rfloor$. Since for $s > 0$ edges are never removed, this also gives a lower bound on $\omega(KG_{n,k,s})$ for any $s \geq 0$, thus, $\omega(KG_{n,k,1}) \geq \lfloor \frac n k \rfloor$. But this bound seems rather weak, since it does not respect any singleton edges at all. For your example above we get 1,1,1,2,2,2,3 as lower bounds on 1,1,2,4,7,7,12.

Further, here is an expression for the chromatic number $\chi(KG_{n,k,1})$, which gives an upper bound on the clique number, since $\chi(G) \geq \omega(G)$ for any graph $G$, where for perfect graphs equality holds. For $n$ written as $n = (k-1) s + r$ for some $0 \leq r < k-1$ and large enough $n > n_0(k)$, the bound is given as $\chi(KG_{n,k,1}) = (k-1)\binom{s}{2} + rs \geq \omega(KG_{n,k,1})$. Ignoring any details on $n_0(k)$, since I have no access to the paper, this gives 1,2,4,6,9,12,16 as upper bounds on 1,1,2,4,7,7,12, which seems to approximate quite well.

Perhaps one of the proofs behind the above results can be adopted to the clique number of $KG_{n,k,1}$?

edit: I just realized that you even want to find a maximum clique - well, that is computationally very hard in general, but perhaps things get easier for $KG_{n,k,1}$?

-

A family of subsets of some finite set is a hypergraph; the subsets themselves are the edges (or hyperedges) of the hypergraph. If all the edges have size $k$, then the hypergraph is k-uniform. (For instance, a $2$-uniform hypergraph is just an ordinary undirected graph.) If no pair of edges has more than one point in common, the hypergraph is called linear. So your question can be reframed as:

What is the maximal number of edges in a $k$-uniform linear hypergraph on $n$ vertices?

-

If you're interested in the asymptotic situation instead of what happens for specific $|S|$, then this has been studied a fair bit under the name of packing problems. More generally, we can ask the question of the size of the largest collection of $k$-element subsets of $\{1, \dots, n\}$ such that each pair of subsets has intersection of size strictly less than $r$.

Since each $k$-element subset contains $\binom{k}{r}$ $r$-element subsets, and each $r$ element subset is in at most $1$ $k$-element subset, we can pick at most $$\frac{\binom{n}{r}}{\binom{k}{r}}$$ subsets in our collection.

Erdos and Hanani conjectured in 1963 that this was asymptotically optimal: For fixed $k$ and $r$, as $n$ tends to infinity, there is a collection of size $(1+o(1))$ times the bound above (for the specific case $r=2$ that you mentioned, this was conjectured earlier by Bose). The conjecture remained open for more than $20$ years until Rodl introduced his so-called "nibble method" to prove it ("On a Packing and Covering Problem", not available online as far as I can tell).

Another term you might want to search under is partial Steiner systems.

-


# Math Help - 1+2+3+...+n

1. ## 1+2+3+...+n

Hi, can anyone tell me why

1+2+3+...+n=n(n+1)/2

I can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself.

2. ## Re: 1+2+3+...+n

I think I can explain it like that:

1, 2, 3, 4, 5, 6, ... n

This is an arithmetic progression with first term 1, last term n with a common difference of 1.

The formula for the sum of the first n numbers is given by:

$S_n = \dfrac{n}{2}\left(a+l\right)$

a = 1, l = n so you simplify to get:

$S_n = \dfrac{n(n+1)}{2}$

How? Let's take an example:

1, 2, 3, 4, 5, 6, 7

If you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get:
4, 4, 4, 4, 4, 4, 4

The sum is then the 4n = 4(7) = 28

But what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7.

Does that make it any clearer?

3. ## Re: 1+2+3+...+n

Do you know 'induction'?

4. ## Re: 1+2+3+...+n

Yes, that was very clear. Thank you!

5. ## Re: 1+2+3+...+n

Originally Posted by Siron
Do you know 'induction'?
No, I don't. Does that relate to this problem?

6. ## Re: 1+2+3+...+n

Originally Posted by TwoPlusTwo
No, I don't. Does that relate to this problem?
Take a look here:
Mathematical induction - Wikipedia, the free encyclopedia
Your exercice is used as an example.

7. ## Re: 1+2+3+...+n

Hello, TwoPlusTwo!

$\text{Can anyone tell me why: }\:1+2+3+\hdots +n\:=\:\frac{n(n+1)}{2}$

Here is a geometric demonstration of the rule (not a proof, mind you).

Consider the case: . $n = 5$

We have this array of objects:

. . $\begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\:\circ \\ \circ\;\circ\;\circ\;\circ\;\circ \end{array}$

Left-justify the objects:

. . $\circ$
. . $\circ\;\:\circ$
. . $\circ\;\circ\;\circ$
. . $\circ\;\circ\;\circ\;\:\circ$
. . $\circ\;\circ\;\circ\;\circ\;\circ$

Append an inverted copy of the array:

. . $\begin{array}{c}\circ\;\bullet\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\; \bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet\;\:\bullet \\ \circ\;\circ\; \circ\;\circ\;\circ\;\:\bullet \end{array}$

We see that the rectangle has: . $5 \times 6\:=\:30$ objects.

Therefore, the triangle has: . $\frac{5 \times 6}{2} \:=\:15$ objects.

8. ## Re: 1+2+3+...+n

There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.

Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1

and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)

If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.

9. ## Re: 1+2+3+...+n

Originally Posted by HallsofIvy
There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.

Here is how he was supposed to have done it: write
1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it
100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1

and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)

If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.
As usual, Soroban got in just before me. I need to learn to type faster!


Question

# Assertion :If $$bc+qr=ca+rp=ab+pq=-1$$, then $$\begin{vmatrix} ap & a & p \\ bq & b & q \\ cr & c & r \end{vmatrix}=0\quad (abc,pqr\neq 0)$$ Reason: If system of equations $${ a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0,\quad { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0,{ \quad a }_{ 3 }x+{ b }_{ 3 }y+{ c }_{ 3 }=0$$ has non-trivial solutions, $$\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0$$

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct

Solution

## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$   ...(1)$$ca+rp+1=0$$   ...(2)$$ab+pq+1=0$$  ...(3)Multiplying (1),(2) and (3) by ap, bq,cr respectively, we get$$\left( abc \right) p+\left( pqr \right) a+ap=0\\ \left( abc \right) q+\left( pqr \right) b+bq=0\\ \left( abc \right) r+\left( pqr \right) c+cr=0$$These equation are consistent,Hence $$\begin{vmatrix} p\quad & a\quad & ap \\ q\quad & b\quad & bq \\ r\quad & c\quad & cr \end{vmatrix}=0\Rightarrow \begin{vmatrix} p & q & r \\ a & b & c \\ ap & bq & cr \end{vmatrix}=0$$ ( interchanging rows into columns)  $$\Rightarrow \left( -1 \right) \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0\quad \left( { R }_{ 1 }{ \leftrightarrow R }_{ 2 } \right) \\ \Rightarrow \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0$$Mathematics

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Question

# PQ and RS are two parallel chords of a circle whose centre is O and radius is $$10$$ cm. If PQ $$= 16$$ cm and RS $$= 12$$ cm. Then  the distance between PQ and RS, if they lie.(i) on the same side of the centre O and(ii) on the opposite of the centre O  are respectively

A
8 cm & 14 cm
B
4 cm & 14 cm
C
2 cm & 14 cm
D
2 cm & 28 cm

Solution

## The correct option is C $$2$$ cm & $$14$$ cmWe join $$OQ$$ &  $$OS$$, drop perpendicular from O to $$PQ$$ &  $$RS$$.The perpendiculars meet $$PQ$$ &  $$RS$$ at M &  N respectively.Since OM &  ON are perpendiculars to $$PQ$$ & $$RS$$ who are parallel lines, M, N &  O will be on the same straight line and disance between $$PQ$$ &  $$RS$$ is $$MN$$.........(i) and $$\angle ONQ={ 90 }^{ o }=\angle OMQ$$......(ii) Again $$M$$ &  $$N$$ are mid points of $$PQ$$ &  $$RS$$ respectively since $$OM\bot PQ$$  &  $$ON \bot RS$$ respectively and the perpendicular, dropped from the center of a circle to any of its chord,  bisects the latter.So $$QM=\dfrac { 1 }{ 2 }$$PQ$$=\frac { 1 }{ 2 } \times 16$$ cm $$=8$$ cm and $$SN=\dfrac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 12$$ cm$$=6$$ cm.$$\therefore \Delta$$ ONQ &  $$\Delta$$ OMQ are right triangles with $$OS$$ &  $$OQ$$  as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (i) $$PQ$$ &  $$RS$$ are to the opposite side of the centre O.(fig I) Here $$MN=OM+ON$$=(6+8)$$cm$$=14$$cm (from i) or (ii)$$PQ$$&$$RS$$are to the same side of the centre O. (fig II) Here$$MN=ON-OM=(8-6 )$$cm$$=2$$cm. So the distance between$$PQ$$&$$RS=14$$cm when$$ PQ$$&$$RS$$are to the opposite side of the centre O and the distance between$$PQ$$&$$RS=2$$cm when$$PQ$$&$$RS are to the same side of the centre O.Ans- Option C.Maths

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# If matrix product $AB$ is a square, then is $BA$ a square matrix?

## Problem 263

Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.
Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.

## Definition/Hint.

Let $A$ be an $m\times n$ matrix.
This means that the matrix $A$ has $m$ rows and $n$ columns.

Let $B$ be an $r \times s$ matrix.
Then the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$.

Definition. A matrix $C$ is called a square matrix if the size of $C$ is $n\times n$ for some positive integer $n$.
(The number of rows and the number of columns are the same.)

## Proof.

We prove that the matrix product $BA$ is defined and it is a square matrix.

Let $A$ be an $m\times n$ matrix and $B$ be an $r\times s$ matrix.

Since the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\times s$.
Since $AB$ is a square matrix, we have $m=s$.

Thus the size of the matrix $B$ is $n \times m$.
From this, we see that the product $BA$ is defined and its size is $n \times n$, hence it is a square matrix.

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# Math Help - somebody please teach me to complete the square

1. ## somebody please teach me to complete the square

hello

could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square

for example how would you solve this excercise

4x^2+3y^2+8x-6y=0

=========================
this is whow i would do it

2 (2x^2 +4+4) 3(y^2-2y+2) = 0

ok I give up, i don't know how to do it, please help me
thank you.

2. Originally Posted by jhonwashington

4x^2+3y^2+8x-6y=0
First you need to have the squared coefficient free, that is, equal to 1.
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the number squared and subtract,
$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$4(x+1)^2+3(y-3)^2=31$

3. Originally Posted by ThePerfectHacker
...
You do this by factoring,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
...
Hello TPH,

it looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$

EB

4. Hello, jhonwashington!

This problem has particularly ugly numbers . . . I'll modify it.

This is the approach I've taught in my classes.

$4x^2 + 3y^2 + 8x - 6y\:=$ 5

We have: . $4x^2+8x + 3y^2-6y\:=\:5$ .

Factor "in groups": . $4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$

This is the complete-the-square step:
. . Take one-half of the coefficient of the linear term and square it.
. . "Add to both sides."

The coefficient of $x$ is $2.$
. . $\frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$

So we "add to both sides" . . but be careful!

We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$
. . . . . . . . .
We wrote $+1$ on the left side
. . . . . . . .
but it is multiplied by the leading 4.
. . . . . . .
So we actually added 4 to the left side.

Complete the square for the $y$-terms.
. . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$

"Add to both sides": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1 $) \;=\;9 \,+$ 3

Factor: . $4(x+1)^2 + 3(y-1)^2\;=\;12$

Divide by $12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$

Then we have: . $\frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$

The ellipse is centered at $(-1,1)$
Its semiminor axis (x- direction) is: $\sqrt{3}$
Its semimajor axis (y-direction) is: $2$

5. Thank you so much for the help ThePerfectHacker ,earboth and soroban,

just one last question, how do you guys factorize? for example how did

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help

6. Originally Posted by jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,

just one last question, how do you guys factorize? for example how did

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help
Note:
$(x + a)^2 = x^2 + 2ax + a^2$

-Dan

7. Originally Posted by jhonwashington

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help
Whenever you add the the half term squared Then you can always factor.

For example,
$x^2+10x$
Add subtract half term squared,
$x^2+10x+25-25$
Thus,
$(x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".