Approximating $\log x$ with roots

The following is a surprisingly good (and simple!) approximation for $\log x+1$ in the region $(-1,1)$: $$\log (x+1) \approx \frac{x}{\sqrt{x+1}}$$

Three questions:

• Is there a good reason why this would be the case?
• How does one go about constructing the "next term"?
• Are the any papers on "generalized Pade approximations" that involve radicals?
• I suggest you write down the Taylor series of $\log(x+1)$ and $(x+1)^{-1/2}$ and see the few first terms agree. – LinAlgMan Dec 1 '14 at 10:42
• @LinAlgMan - That is true, but it is not the reason. This approximation works much better than the Taylor series, even to high orders, probably because it accounts for the pole. – nbubis Dec 1 '14 at 10:43
• May I ask how you got this interesting approximation ? – Claude Leibovici Dec 1 '14 at 10:49
• @ClaudeLeibovici - the function came up in a physics calculation, and upon plotting, I noticed it looked oddly familiar. – nbubis Dec 1 '14 at 10:51
• @Lucian - I'm not sure I see the connection. – nbubis Dec 8 '14 at 20:12

Let's rewrite both sides in terms of $y = x + 1$: we get

$$\log y \approx \sqrt{y} - \frac{1}{\sqrt{y}}$$

on, let's say, the interval $\left( \frac{1}{2}, 2 \right)$ (I hesitate to discuss the entire interval $(0, 2)$; it seems to me that the approximation is not all that good near $0$). The RHS should look sort of familiar: let's perform a second substitution $y = e^{2z}$ to get

$$2z \approx e^z - e^{-z} = 2 \sinh z$$

on the interval $\left( - \varepsilon, \varepsilon \right)$ where $\varepsilon = \frac{\log 2}{2} \approx 0.346 \dots$. Of course now we see that the LHS is just the first term in the Taylor series of the RHS, and on a smaller interval than originally. Furthermore, the Taylor coefficients of $2 \sinh z$, unlike the Taylor coefficients of our original functions, decrease quite rapidly. The next term is $\frac{z^3}{3}$, which on this interval is at most

$$\frac{\varepsilon^3}{3} \approx 0.0138 \dots$$

and this is more or less the size of the error in the approximation between $\log 2$ and $\frac{1}{\sqrt{2}}$ obtained by setting $y = 2$, or equivalently $x = 1$.

With the further substitution $t = \sinh z$, the RHS is just the first term in the Taylor series of the LHS. To get the "next term" we could look at the rest of the Taylor series of $\sinh^{-1} t$. The next term is $- \frac{t^3}{6}$, which gives

$$z \approx \frac{e^z - e^{-z}}{2} - \frac{(e^z - e^{-z})^3}{48}$$

or

$$\log y \approx \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right) - \frac{1}{24} \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)^3.$$

I don't know if this is useful for anything. The series to all orders just expresses the identity

$$\log y = 2 \sinh^{-1} \frac{\left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)}{2}.$$

• Very nicely done!! As you noticed though, near $y\to 0$, higher orders seem to actually ruin the approximation, which I find curious. – nbubis Dec 1 '14 at 11:26
• @nbubis: well, the radius of convergence of the Taylor series of $\sinh^{-1} t$ is $1$, and as $y \to 0$, $\sqrt{y} - \frac{1}{\sqrt{y}}$ gets arbitrarily large (in absolute value)... – Qiaochu Yuan Dec 1 '14 at 11:59

The simplest Pade approximant we could build seems to be $$\log(1+x)\approx\frac{x}{1+\frac{x}{2}}$$ and we can notice the similarity of denominators close to $x=0$.

However, the approximation given in the post seems to be significantly better for $x<\frac 12$.

• Indeed. It is probably better because it has a pole in the right place. – nbubis Dec 1 '14 at 11:00

I am being rather late, yet still there is some interesting information I can add.

The Padet approximant is of the form $\log(1+x)\approx \frac{x}{1+x/2}$ as noted by other posters. This partially explains why $\frac{x}{\sqrt{1+x}}$ is a good approximation, what it does not explain is why the square-root approximation is better than a supposedly "great" Pade approximation. @nbubis had an idea that it works better because it has pole in the correct spot, but it seems that it is actually a red herring.

Let's take a look on more general Pade $(1,n)$ approximation, it equals $\log(1+x)\approx\frac{x}{1+x/2-x^2/12+x^4/24+...}$.

Now the reason $\frac{x}{\sqrt{1+x}}$ approximation performs better can be explained by $\sqrt{1+x} \approx 1+x/2 -x^2/8$ and noting that $1+x/2-x^2/8$ is closer to the "true value" of the denominator than the first Pade approximant $1+x/2$.

To see that it is indeed the case consider the approximation

$\log(1+x)=\frac{x}{(1+5x/6)^{3/5}}$.

Now, as you can quickly check, $(1+5x/6)^{3/5}$ has the Taylor expansion $\approx 1+x/2 -x^2/12$ which aggrees with first 3 terms of $(1,n)$ Pade approximant. Now if you plot it it will turn out that it is even better than originally suggested $\frac{x}{\sqrt{1+x}}$ despite having pole in the wrong place.

Regarding method by @QiaochuYuan, you can perform the same "trick" to get better approximations which will be performing better in the neighbourhood of $x=0$ but worse when $x$ is large, for example

$\log (1+x) \approx \frac{x}{(1+5x/6)^{3/5}} + \frac{x^4}{108 (1+5x/6)^{12/5}}$

But in disguise what you are actually making is finding better approximations to some Pade approximant.

Some of the other approximations you can find in the same way are

$\log(1+x)\approx \frac{x}{\sqrt{1+x+x^2/12}}$ and $\log(1+x)\approx \frac{x}{(1+3x/2+x^2/2)^{1/3}}$ which are good simply beause they coincide with Pade approximant up to the terms of high order. I guess the first among those two is another reason why $\frac{x}{\sqrt{1+x}}$ worked so well.

Short version: It's actually a coincidence, $\sqrt{1+x}$ happen to have Taylor expansion $\sqrt{1+x}\approx 1+x/2-x^2/8$ which coincides with expansion of $x/\log(1+x)\approx 1+x/2 -x^2/12$ up to 2 terms and the third term is not that different to mess up the approximation.

• I like this! That makes a lot of sense. – nbubis Dec 1 '17 at 6:00

A different approach.

The function $f(x)=x-\log(1+x)\sqrt{1+x}$ is continuous and increasing on $[-1,1]$ (I have not proved it, but a graph of $f'$ is sufficiently convincing.) For any $a\in(0,1)$ $$f(-a)\le f(x)\le f(1)=0.0197419,\quad-a\le x\le1.$$ Take for instance $a=0.8$ we obtain $$-\frac{0.0802375}{\sqrt{1+x}}\le\frac{x}{\sqrt{1+x}}-\log(1+x)\le\frac{0.0197419}{\sqrt{1+x}},\quad -.8\le x\le1.$$

This shows that $\log(1+x)\sqrt{1+x}$ is a good approximation of $x$. $$f(x)=-\frac{x^3}{24}+\dots$$ is an alternate series. This explains the vey good approximation for $x>0$, and the not so good for $x<0$.

• I'm not sure how that helps - this is just an evaluation of the approximation. – nbubis Dec 1 '14 at 11:56

One reason may be that their Taylor series around $x=0$ start the same: $$\log (x+1) \approx x-x^2/2+x^3/3+\cdots$$ $$\frac{x}{\sqrt{x+1}} \approx x-x^2/2+3x^3/8+\cdots$$ So they agree to order 2 for $|x|<1$. They almost agree to order 3 because $1/3 \approx 3/8$ roughly.

However, this is an a posteriori reason. I don't know why this approximation should be good a priori.

• That doesn't necessarily tell you much about how good of an approximation to expect on an interval as large as $(-1, 1)$. – Qiaochu Yuan Dec 1 '14 at 10:46
• This was what I was just typing ! The question is interesting. – Claude Leibovici Dec 1 '14 at 10:46
• For example, when $x = 1$ the LHS is $\log 2 \approx 0.693 \dots$ while the RHS is $\frac{1}{\sqrt{2}} \approx 0.707 \dots$. These agree substantially better that can be accounted for by the first two terms of the Taylor series, which give $0.5$. – Qiaochu Yuan Dec 1 '14 at 10:48
• This is clearly not the reason. The approximation does much better than the Taylor series way past second order.(probably because it has a pole) – nbubis Dec 1 '14 at 10:49

$$\log(x+1)=\lim_{n\to\infty}n(\sqrt[n]{x+1}-1).$$

In the case $n=2$, $$2(\sqrt{x+1}-1)=2\frac{x}{\sqrt{x+1}+1}\approx\frac x{\sqrt{x+1}}$$ for small $x$.

The approximation works better as it has a vertical asymptote at $x=-1$.

• Nice :) Though the approximation ends up being better than the derivation... – nbubis Dec 1 '14 at 11:14
• This still doesn't explain most of the agreement. Again taking $x = 1$ we have $2 (\sqrt{2} - 1) \approx 0.828 \dots$, which is maybe 15% bigger than either the LHS or the RHS, which agree to maybe within 2%. – Qiaochu Yuan Dec 1 '14 at 11:15
• Anyone who's still trying to answer the question should actually plot the functions (I did it in WolframAlpha) to see how close the agreement actually is. I think the pole is a red herring: the agreement is really not very good close to the pole. – Qiaochu Yuan Dec 1 '14 at 11:17
• I don't thank the downvoters. @QiaochuYuan: the question is not a contest to the best approximation. It is about why $\log (x+1) \approx \dfrac{x}{\sqrt{x+1}}$. – Yves Daoust Oct 31 '17 at 7:45


# exponential functions examples

Each output value is the product of the previous output and the base, 2. Find r, to three decimal places, if the the half life of this radioactive substance is 20 days. Sign up to read all wikis and quizzes in math, science, and engineering topics. Now, let’s take a look at a couple of graphs. If b b is any number such that b > 0 b > 0 and b â  1 b â  1 then an exponential function is a function in the form, f (x) = bx f (x) = b x We will see some examples of exponential functions shortly. Compare graphs with varying b values. from which we have In fact this is so special that for many people this is THE exponential function. n \log_{10}{1.03} \ge& 1 \\ In fact, that is part of the point of this example. For every possible $$b$$ we have $${b^x} > 0$$. Find the sum of all positive integers aaa that satisfy the equation above. 100 + (160 - 100) \frac{1.5^{12} - 1}{1.5 - 1} \approx& 100 + 60 \times 257.493 \\ \approx& 15550. Population: The population of the popular town of Smithville in 2003 was estimated to be 35,000 people with an annual rate of increase (growth) of about 2.4%. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$f\left( { - 2} \right) = {2^{ - 2}} = \frac{1}{{{2^2}}} = \frac{1}{4}$$, $$g\left( { - 2} \right) = {\left( {\frac{1}{2}} \right)^{ - 2}} = {\left( {\frac{2}{1}} \right)^2} = 4$$, $$f\left( { - 1} \right) = {2^{ - 1}} = \frac{1}{{{2^1}}} = \frac{1}{2}$$, $$g\left( { - 1} \right) = {\left( {\frac{1}{2}} \right)^{ - 1}} = {\left( {\frac{2}{1}} \right)^1} = 2$$, $$g\left( 0 \right) = {\left( {\frac{1}{2}} \right)^0} = 1$$, $$g\left( 1 \right) = {\left( {\frac{1}{2}} \right)^1} = \frac{1}{2}$$, $$g\left( 2 \right) = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}$$. A = a^{a}b^{b}c^{c}, \quad B = a^{a}b^{c}c^{b} , \quad C = a^{b}b^{c}c^{a}. 1000Ã(12)100005730â1000Ã0.298=298.1000 \times \left( \frac{1}{2} \right)^{\frac{10000}{5730}} Check out the graph of $${2^x}$$ above for verification of this property. Therefore, we would have approximately 298 g. â¡ _\square â¡â, Given three numbers such that 0 and! Since we are only graphing one way is if we allowed \ ( b\ ) have. 1\Large |x|^ { ( x^2-x-2 ) } < 1 { 12 } 100. Curve upward, as shown in the first quadrant functions exponential functions examples starting this with. ) above for verification of this function in the exponent worked to this point population a... A very specific number into all the \ ( { b^x } > 0\ ) represent growth decay. Functions from these graphs exponential functions examples model exponential functions will want to use far more decimal places in computations! Graph y = 2 x is called the base, 2 aaa that satisfy the equation,. For a complete list of integral functions, there are some function evaluations that will give complex,! Through complex numbers course, built by experts for you would we have now section. formula for exponential. Business and science output value is the approximate integer population after a year we could have written \ ( \bf... The final section of this chapter off discussing the final property for a complete of. Approaches negative infinity, the approximate integer population after a year is 100Ã1.512â100Ã129.75=12975.Â â¡100 \times 1.5^ { 12 \approx... > 0\ ) ( b = - 4\ ) the function would be would be 1000Ã1.03n.1000! The chapter on exponential functions works in exactly the same properties that exponential..., C a, b is greater than  1 , the graph of this property complex! 2 raised to the x power same manner that all three graphs pass through the y-intercept ( 0,1..


# Thread: Find the interval in which the function concave up??

1. ## Find the interval in which the function concave up??

The graph of the function

is concave up on the interval:

(-infinity,infinity)

None

(1,infinity)

(-infinity,0)U(1,infinity)

(0,infinity)

=======================

I solve the question and my answer is :

The graph is concave up on (-infinity,-0.5)U(0,infinity)

and I'm sure from my answer but my answer is not included in the choices

and When I use the graphics calculator the graph is concave up on all interval

So what is the correct answer??

2. You can use derivatives to answer this question

First you must find the critical points, take your first equation and differentiate and solve for 0

$f(x)=x^4+x^3+x$

$f\prime(x)=4x^3+3x^3+1$

Now that its differentiated solve for 0

$4x^3+3x^2+1=0$

$x(4x^2+3x+1)=0$

you a critical point at x = -1.5 and x = 0

now to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing

to find concavity you take second derivative and plug in critical points if

>0 then concave up and <0 concave down

the second derivative is

$12x^2+6$

plus in 0 and -1.5

plug 0 in and you see it is >0 so intervals from 0 to infinity is concave up

plug -1.5 it is also concave up because > 0 so from -inf. to -1.5

sorry for lack of work studying for finals

3. Originally Posted by sk8erboyla2004
You can use derivatives to answer this question

First you must find the critical points, take your first equation and differentiate and solve for 0

$f(x)=x^4+x^3+x$

$f\prime(x)=4x^3+3x^3+1$

Now that its differentiated solve for 0

$4x^3+3x^2+1=0$

$x(4x^2+3x+1)=0$

you a critical point at x = -1.5 and x = 0

now to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing

to find concavity you take second derivative and plug in critical points if

>0 then concave up and <0 concave down

the second derivative is

$12x^2+6$

plus in 0 and -1.5

plug 0 in and you see it is >0 so intervals from 0 to infinity is concave up

plug -1.5 it is also concave up because > 0 so from -inf. to -1.5

sorry for lack of work studying for finals

but I would to remind you that in concavity test we do the following :

1)find first derivative

2)find second derivative

3)set first derivative=0 and solve for x

4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..

4. Originally Posted by change_for_better
The graph of the function

is concave up on the interval:

(-infinity,infinity)

None

(1,infinity)

(-infinity,0)U(1,infinity)

(0,infinity)

=======================

I solve the question and my answer is :

The graph is concave up on (-infinity,-0.5)U(0,infinity)

and I'm sure from my answer but my answer is not included in the choices

and When I use the graphics calculator the graph is concave up on all interval

So what is the correct answer??
A function is concave upward when its second derivative is positive.
If $y= x^4+ x^3+ x$, then $y"= 12x^2+ 6x= 6x(2x+ 1)$. That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y" is negative. If x> 0, 2x+1> 0 so y" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!

5. Originally Posted by change_for_better

but I would to remind you that in concavity test we do the following :

1)find first derivative

2)find second derivative

3)set first derivative=0 and solve for x

4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..

If you know this why you couldnt have used this to answer your own question?

6. Originally Posted by HallsofIvy
A function is concave upward when its second derivative is positive.
If $y= x^4+ x^3+ x$, then $y"= 12x^2+ 6x= 6x(2x+ 1)$. That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y" is negative. If x> 0, 2x+1> 0 so y" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!

I would like to ask you ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??

7. Originally Posted by sk8erboyla2004
If you know this why you couldnt have used this to answer your own question?

Because ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??

8. Hello, change_for_better!

I agree with your intervals . . .

The graph of the function: . $f(x) \;=\;x^4+x^3+x$ is concave up on the interval:

. . $(a)\;(-\infty, \infty) \qquad(b)\text{ None} \qquad (c)\;(1,\infty) \qquad (d)\;(-\infty,0) \;\cup\; (1,\infty) \qquad(e)\;(0,\infty)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I solved and my answer is: . $\left(-\infty,-\tfrac{1}{2}\right) \cup (0,\infty)$

and I'm sure from my answer but my answer is not included in the choices.

and when I use the graphics calculator the graph is concave up on all interval
. .
Um ... not quite!

So what is the correct answer?

$\begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}$

$f''(x) > 0 \quad\Rightarrow\quad x < \text{-}\tfrac{1}{2}\:\text{ or }\:x > 0$

If you zoom in on your graph, there is a concave-down portion.
Code:
                    |
|         *
|        *
|      *
*               |   *
- - - - - - - - - o - - - - - - - -
*           o  |
*        o    |
*     *     |
*        |
|

And it's right where you predicted! . . . $\left(\text{-}\tfrac{1}{2},\:0\right)$

9. Originally Posted by Soroban
Hello, change_for_better!

I agree with your intervals . . .

$\begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}$

$f''(x) > 0 \quad\Rightarrow\quad x < \text{-}\tfrac{1}{2}\:\text{ or }\:x > 0$

If you zoom in on your graph, there is a concave-down portion.
Code:
                    |
|         *
|        *
|      *
*               |   *
- - - - - - - - - o - - - - - - - -
*           o  |
*        o    |
*     *     |
*        |
|
And it's right where you predicted! . . . $\left(\text{-}\tfrac{1}{2},\:0\right)$

Thank you very very very very much Soroban for great explanation

Now I understand my mistake ..

So from choices I should select none because the correct answer is not included ..


# Compute the $n$-th power of triangular $3\times3$ matrix

I have the following matrix

$$\begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 0 & 0 & 1 \end{bmatrix}$$

and I am asked to compute its $n$-th power (to express each element as a function of $n$). I don't know at all what to do. I tried to compute some values manually to see some pattern and deduce a general expression but that didn't gave anything (especially for the top right). Thank you.

Computing the first few powers should allow you to find a pattern for the terms. Below are some terms:

$$\left(\begin{matrix}1 & 2 & 3\\0 & 1 & 2\\0 & 0 & 1\end{matrix}\right), \left(\begin{matrix}1 & 4 & 10\\0 & 1 & 4\\0 & 0 & 1\end{matrix}\right), \left(\begin{matrix}1 & 6 & 21\\0 & 1 & 6\\0 & 0 & 1\end{matrix}\right), \left(\begin{matrix}1 & 8 & 36\\0 & 1 & 8\\0 & 0 & 1\end{matrix}\right), \left(\begin{matrix}1 & 10 & 55\\0 & 1 & 10\\0 & 0 & 1\end{matrix}\right), \left(\begin{matrix}1 & 12 & 78\\0 & 1 & 12\\0 & 0 & 1\end{matrix}\right)$$

All but the top right corner are trivial so lets focus on that pattern. (Although if you look at it carefully you should recognize the terms.)

Terms: $3,10,21,36,55,78$

First difference: $7, 11, 15, 19, 23$

Second difference: $4, 4, 4, 4$

As the second difference is a constant the formula must be a quadratic. As the second difference is 4 then it is in the form $2n^2+bn+c$. Examining the pattern gives formula of $2n^2+n=n(2n+1)$.

So the $n^{th}$ power is given by:

$$\left(\begin{matrix}1 & 2n & n(2n+1)\\0 & 1 & 2n\\0 & 0 & 1\end{matrix}\right)$$

The reason I said you should recognize the pattern is because it is every second term out of this sequence: $1,3,6,10,15,21,27,37,45,55,66,78,\cdots$ which is the triangular numbers.

• It seemed more in line with what the OP had been trying. It would also let the OP see if their approach (and the powers of the matrix) were correct. – Ian Miller Jan 2 '17 at 1:28
• This is a great approach to "guess the formula" (and it is enough for any practical purpose), but it is not enough to constitute a proof. Could you please make it complete (or at least mention in the body that it still needs a proof)? – dtldarek Jan 2 '17 at 9:57

Write this matrix as follows: $$\left[ \begin{matrix} 1 & 2&3\\ 0 & 1 & 2\\ 0 & 0 &1 \end{matrix} \right] = I + 2 J+ 3 J^{2}.$$ where $$I = \left[ \begin{matrix} 1 & & \\ &1 & \\ & & 1 \end{matrix} \right], ~ J = \left[ \begin{matrix} 0& 1 &0 \\ 0 &0 & 1 \\ 0 & 0& 0 \end{matrix} \right],~ J^2 = \left[ \begin{matrix} 0& 0 &1\\ 0 & 0 &0\\ 0& 0 & 0 \end{matrix} \right], ~ J^{3}=0.$$ With this relation you can expand the power of the matrix into sum of $I$, $J$ and $J^2$.

• Thank you for your answer, interesting because it takes a different approach to my original idea, I will study it. – Trevör Jan 2 '17 at 12:06

Define

$$J = \begin{bmatrix} 0 & 2 & 3\\ 0 & 0 & 2\\ 0 & 0 & 0 \end{bmatrix}$$

so that the problem is to compute $(I+J)^n$. The big, important things to note here are

• $I$ and $J$ commute
• $J^3 = 0$

which enables the following powerful tricks: the first point lets us expand it with the binomial theorem, and the second point lets us truncate to the first few terms:

$$(I+J)^n = \sum_{k=0}^n \binom{n}{k} I^{n-k} J^k = I + nJ + \frac{n(n-1)}{2} J^2$$

More generally, for any function $f$ that is analytic at $1$, (such as any polynomial), if you extend it to matrices via Taylor expansion, then under the above conditions, its value at $I+J$ is given by

$$f(I+J) = \sum_{k=0}^\infty f^{(k)}(1) \frac{J^k}{k!} = f(1) I + f'(1) J + \frac{1}{2} f''(1) J^2$$

As examples of things whose result you can check simply (so you can still use the method even if you're uncomfortable with it, because you can check the result), you can compute the inverse by

$$(I+J)^{-1} = I - J + J^2 = \begin{bmatrix} 1 & -2 & 1\\ 0 & 1 & -2\\ 0 & 0 & 1 \end{bmatrix}$$

and if you want a square root, you can get

$$\sqrt{I+J} = I + \frac{1}{2} J - \frac{1}{8} J^2 = \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{bmatrix}$$

(These are actually special cases of $(I+J)^n$ by the generalized binomial theorem for values of $n$ that aren't nonnegative integers)

• Thank you so much for your answer, interesting because it takes a different approach than my original idea, I will study it. – Trevör Jan 2 '17 at 12:06

Let $I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$, $A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$. Then $M=I+2A+3B$.

You can prove that for any $n$, $M^n$ can be written as $M^n=\lambda_n I + a_n A + b_n B$ (because it's upper triangular and the symmetry along the ascending diagonal will remain).

So $M^{n+1}=M^nM=(\lambda_n I + a_n A + b_n B) (I + 2A + 3B) = \dots$

Using this, compute $\lambda_n$, and then $a_n$ and finally $b_n$.

• An essential point is that the matrices commute – Rene Schipperus Jan 2 '17 at 2:29
• @ReneSchipperus : It's not really essential. The essential thing is that the vector space generated by the matrices is an algebra: $\forall X,Y\in Vect(I, A, B), XY \in Vect(I, A, B)$ (which is equivalent to $\forall X,Y\in \{I, A, B\}, XY \in Vect(I, A, B)$, because it's a vector space and you can use distributivity). In the general case, you would us the algebra of upper-triangular matrices, compute the coefficients on the diagonal, then right above the diagonal and keep going, and you'll only need the values already computed because the matrices are upper triangular. – xavierm02 Jan 2 '17 at 13:17

Here is another variation based upon walks in graphs.

We interpret the matrix $A=(a_{i,j})_{1\leq i,j\leq 3}$ with \begin{align*} A= \begin{pmatrix} 1 & 2 & 3\\ \color{grey}{0} & 1 & 2\\ \color{grey}{0} & \color{grey}{0} & 1 \end{pmatrix} \end{align*} as adjacency matrix of a graph with three nodes $P_1,P_2$ and $P_3$ and for each entry $a_{i,j}\neq 0$ with a directed edge from $P_i$ to $P_j$ weighted with $a_{i,j}$.

Note: When calculating the $n$-th power $A^n=\left(a_{i,j}^{(n)}\right)_{1\leq i,j\leq 3}$ we can interpret the element $a_{i,j}^{(n)}$ of $A^n$ as the number of (weighted) paths of length $n$ from $P_i$ to $P_j$. The entries of $A=(a_{i,j})_{1\leq i,j\leq 3}$ are the weighted paths of length $1$ from $P_i$ to $P_j$.

See e.g. chapter 1 of Topics in Algebraic Combinatorics by Richard P. Stanley.

Let's look at the corresponding graph and check for walks of length $n$.

• We see there are no directed edges from $P_2$ to $P_1$ and no directed edges from $P_3$ to $P_2$ and from $P_3$ to $P_1$ which implies there are no walks of length $n$ either. So, $A^n$ has due to the specific triangle structure of $A$ necessarily zeroes at the same locations as $A$. \begin{align*} A^n= \begin{pmatrix} . & . & .\\ \color{grey}{0} & . & .\\ \color{grey}{0} & \color{grey}{0} & . \end{pmatrix} \end{align*}

• It is also easy to consider the walks of length $n$ from $P_i$ to $P_i$. There is only one possibility to loop along the vertex weighted with $1$ from $P_i$ to $P_i$ and so the entries $a_{i,i}^{(n)}$ are \begin{align*} 1\cdot 1\cdot 1\cdots 1 = 1^n=1 \end{align*} and we obtain \begin{align*} A^n= \begin{pmatrix} 1& . & .\\ \color{grey}{0} & 1 & .\\ \color{grey}{0} & \color{grey}{0} & 1 \end{pmatrix} \end{align*}

and now the more interesting part

• $P_1$ to $P_2$:

The walks of length $n$ from $P_1$ to $P_2$ can start with zero or more loops at $P_1$ followed by a step (weigthed with $2$) from $P_1$ to $P_2$ and finally zero or more loops at $P_2$. All the loops are weighted with $1$. There are $n$ possibilities to walk this way \begin{align*} a_{1,2}^{(n)}=2\cdot 1^{n-1}+1\cdot 2\cdot 1^{n-2}+\cdots +1^{n-2}\cdot 2\cdot 1+1^{n-1}\cdot 2=2n \end{align*}

• $P_2$ to $P_3$:

Symmetry is trump. When looking at the graph we observe the same situation as before from $P_1$ to $P_2$ and conclude

\begin{align*} a_{2,3}^{(n)}=2n \end{align*}

• $P_1$ to $P_3$:

Here are two different types of walks of length $n$ possible. The first walk uses the weight $3$ edge from $P_1$ to $P_3$ as we did when walking from $P_1$ to $P_2$ along the weight $2$ edge. This part gives therefore \begin{align*} 3\cdot 1^{n-1}+1\cdot 3\cdot 1^{n-2}+\cdots +1^{n-2}\cdot 3\cdot 1+1^{n-1}\cdot 3=3n\tag{1} \end{align*} The other type of walk of length $n$ uses the hop via $P_2$. We observe it is some kind of concatenation of walks as considered before from $P_1$ to $P_2$ and from $P_2$ to $P_3$. In fact there are $\binom{n}{2}$ possibilities to place two $2$'s in a walk of length $n$. All other steps are loops at $P_1,P_2$ and $P_3$ and we obtain \begin{align*} \binom{n}{2}\cdot 2\cdot 2=2n(n-1)\tag{2} \end{align*} Summing up (1) and (2) gives \begin{align*} a_{1,3}^{(n)}=3n+2n(n-1)=n(2n+1) \end{align*}

and we finally obtain

\begin{align*} A^n=\left(a_{i,j}^{(n)}\right)_{1\leq i,j\leq 3}=\begin{pmatrix} 1& 2n & n(2n+1)\\ \color{grey}{0} & 1 & 2n\\ \color{grey}{0} & \color{grey}{0} & 1 \end{pmatrix} \end{align*}

Using symmetry, write the recurrence relation

$$\begin{pmatrix} a_{n+1} & b_{n+1}&c_{n+1}\\ 0 & a_{n+1} & b_{n+1}\\ 0 & 0 &a_{n+1} \end{pmatrix}= \begin{pmatrix} 1 & 2&3\\ 0 & 1 & 2\\ 0 & 0 &1 \end{pmatrix} \begin{pmatrix} a_{n} & b_{n}&c_{n}\\ 0 & a_{n} & b_{n}\\ 0 & 0 &a_{n} \end{pmatrix}.$$

Then $$\begin{cases}a_{n+1}=a_n\\ b_{n+1}=b_n+2a_n\\ c_{n+1}=c_n+2b_n+3a_n.\end{cases}$$

We solve with

$$a_n=Cst=1,\\b_n=2(n-1)+Cst=2n,\\ c_n=4t_{n-1}+3(n-1)+Cst=2n^2+n,$$

where $t_n$ denotes a triangular number, using the initial conditions $a_1=1,b_1=2,c_1=3$.


# Trying to convert a summation to an equation

Here is what I am trying to figure out $\sum_{i=1}^n 3+2 (1-i)$ It would be nice if this could be put into a single equation to be used in a larger system. I know that it is possible to break up the summation into two parts like this $\sum_{i=1}^n 3+\sum_{i=1}^n2 (1-i)$ and it then becomes $3*n+\sum_{i=1}^n2 (1-i)$ I get stuck in trying to convert the second summation into a regular equation. Ultimately, it seems that this should be some sort of exponential function, but I am not having any luck finding it.

This is for a personal project and not homework in case this is a concern.

Brandon

Well, $\displaystyle \sum_{i=1}^n 2(1-i) = \sum_{i=1}^n 2 - 2\sum_{i=1}^n i = 2n - \frac{2n(n+1)}{2}$.

Then we have $\displaystyle \sum_{i=1}^n 3 +2(1-i) = 3n + 2n - n(n+1) = 4n - n^2$.

For intuition on $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$, notice $1 + 2 + 3 + \ldots + (n-2) + (n-1) +n = (n+1) + (n-1 +2) + (n -2 +3) + \ldots + (\frac{n}{2} + (\frac{n}{2}+1)) = (n+1) + (n+1) + (n+1) + \ldots + (n+1)$,

grouping the first and last terms together, then the second and second to last, etc., where there are $\frac{n}{2}$ (for even $n$) terms in the last sum. If $n$ is odd, treat with $\frac{n+1}{2}$ where necessary.

Then summing $n+1$ $\frac{n}{2}$ times clearly gives $\frac{n(n+1)}{2}$.

• Awesome explanation. Thank you! – Brandon Dec 29 '16 at 21:02

Hint :- $\sum_{i=1}^n2(1-i)=\sum_{i=1}^n2-2\sum_{i=1}^ni=2.n-2(\frac{n(n+1)}{2})$.


# Show $\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}=e$

Why is $\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}$ = $e$? I couldn't get this result.

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I tried it with the triangle inequality and did not get the right result. I tried it with $a^b = e^{a*ln(b)}$ but did not get any further. This is a exercise I do for exam preparation, it's not homework. –  leo Dec 29 '12 at 11:06
Thanks for all the brilliant answers. The accepted one is the one which I could understand the best, even if might not be the most elegant or shortest. –  leo Dec 29 '12 at 13:57
Would you like me to provide more details about my solution? What do you not understand about it? It's a straightforward way to deal with n roots, and is just based on Bernoulli's inequality that $(1+x)^n \geq 1+ nx$ for $x\geq 0, n\in \mathbb{N}$. –  Calvin Lin Dec 30 '12 at 1:08
@CalvinLin I changed my mind now. Nameless solution did just use Math I was more familiar with. But your solution is very clean and I understand it now. –  leo Jan 27 '13 at 16:31

This is equivalent to showing that $\lim_{n \rightarrow \infty} \sqrt[n]{\frac {n^e}{e^n} + 1 } = 1$.

This is clearly bounded below by 1. It is bounded above by $1 + \frac {n^e}{n e^n}$, which has a limit of 1 since polynomials grow much slower than exponentials.

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$$\large (n^e + e^n)^{\frac{1}{n}} = e \left ( 1 + \frac {n^e}{e^n} \right) ^{\frac 1 n}$$ $$e \Large \left ( 1 + \frac {n^e}{e^n} \right) ^{\frac 1 n} = e \left ( \underbrace { \left ( 1 + \frac {1}{e^{n - e \log n}} \right) ^{e^{n - e \log n}} }_{e}\right)^{ \underbrace{\frac{n}{e^{n - e \log n}}}_{0}}$$

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What BIG fonts you have!! –  Haskell Curry Dec 30 '12 at 8:13

$$\lim_{n\to\infty} \sqrt[n]{n^e+e^n}=\lim_{n\to\infty} (n^e+e^n)^{\frac1n}=\lim_{n\to\infty} e^{\ln (n^e+e^n)\frac1n}$$ Now you just have to compute $$\lim_{n\to\infty} \frac{\ln (n^e+e^n)}n=\lim_{n\to\infty} \frac{\ln (e^{e\ln n}+e^n)}n=\lim_{n\to\infty} \frac{\ln e^n(e^{e\ln n-n}+1)}n=1+\lim_{n\to\infty} \frac{\ln (e^{e\ln n-n}+1)}n=1+\lim_{n\to\infty} \frac{\ln (e^{-\infty}+1)}n=1+\lim_{n\to\infty} \frac{\ln (0+1)}n=1+\lim_{n\to\infty} \frac{\ln (1)}n=1+0=1$$ since $e\ln n-n\to -\infty$.

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could you please show the second last step a bit more fine grained. Otherwise this is the most clear answer to me. –  leo Dec 29 '12 at 12:56
@leo Sure I can –  Nameless Dec 29 '12 at 12:57

Let's see a more direct way

$$\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}=\lim\limits_{n\to\infty} \sqrt[n]{e^n}=e$$ because the ratio test applied to $\frac{n^e}{e^n}$ yields $\lim_{n\to\infty}\frac{n^e}{e^n}=0.$

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Why is $n^e$ is negligible? –  leo Dec 29 '12 at 11:28
@leo: because the exponential function grows much faster than the polynomial function. –  Chris's sis Dec 29 '12 at 11:29
That is of course far from rigorous. –  Eckhard Dec 29 '12 at 12:19
@Eckhard: in addition, this is the way I was taught to answer such questions in high school. –  Chris's sis Dec 29 '12 at 12:30
I have to agree this is far from rigorous. –  user641 Dec 29 '12 at 12:54

You can also apply two gendarmes theorem. The idea is to find sequences $l_n, u_n$ which bound (from below and above) the given sequence $a_n$, i.e. $$l_n\le a_n\le u_n$$ holds, and which converge to a joint limit (i.e. $\lim_{n\to\infty}l_n=\lim_{n\to\infty}u_n=g).$ Then the theorem says that $a_n$ is also convergent and the limit is $g$. Let's see how it works.

Finding these bounds is usually pretty straightforward – lower bound is often obtained by simple missing some nonnegative terms. While seeking the upper bound, one have to remeber that the inequality have to be true only for all sufficiently large $n$'s. In this case one can write: $$\sqrt[n]{0+e^n}\le\sqrt[n]{n^e+e^n}\le\sqrt[n]{e^n+e^n}$$ since $0\le n^e$ and $n^e\le e^n$ for sure if $n$ is sufficiently large. Next, we observe that

$l_n:= \sqrt[n]{e^n}=e\longrightarrow e$ as well as

$u_n:=\sqrt[n]{2e^n}=e\sqrt[n]{2}\longrightarrow e\cdot 1 =e.$

The theorem yields the claim.

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You can also use that $\sqrt[n]{a+b}\le \sqrt[n]{a}+\sqrt[n]{b}$. –  user641 Dec 29 '12 at 19:31
Another very nice answer! Thank you! –  leo Dec 29 '12 at 20:07
$0\le n^e$ and $n^e\le e^n$ are true for all $n\ge 0$ –  Henry Dec 29 '12 at 22:44

You have $$\sqrt[n]{n^e+e^n}= \exp \left( \frac{\ln(n^e+e^n)}{n} \right)= \exp \left(1+ \frac{\ln \left( 1+ \frac{n^e}{e^n} \right)}{n} \right)$$ But $\ln \left( 1+ \frac{n^e}{e^n} \right) \sim \frac{n^e}{e^n}$ so you can conclude.

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Note that by L'Hospital$$\lim_{n\to\infty}\frac{\log (n^e+e^n)}{n}=\lim_{n\to\infty}\frac{\frac{1}{n^e+e^n}(n^{e-1}+e^n)}{1}=\lim_{n\to\infty}\frac{n^{e-1}}{n^e+e^n}+\lim_{n\to\infty}\frac{1}{1+\frac{n^e}{e^n}}$$ Now $n.n^{e-1}\leq n^e+e^n$, so $n^{e-1}/(n^e+e^n)\leq1/n$. Hence taking limit on both side, the first limit is zero. For the second one it can be proved that $n^{e+1}\leq e^n$ for all $n\geq n_0$, for some $n_0$. So $\lim_{n\to\infty}(n^e/e^n)\leq\lim_{n\to\infty}(1/n)=0$. Hence the second limit equals $1$. Now taking exponential the required limit evaluates to $e$.

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Taking logs, you must show that $$\lim_{n \rightarrow \infty} {\ln(n^e + e^n) \over n} = 1$$ Applying L'hopital's rule, this is equivalent to showing $$\lim_{n \rightarrow \infty}{en^{e-1} + e^n \over n^e + e^n} = 1$$ Which is the same as $$\lim_{n \rightarrow \infty}{e{n^{e-1}\over e^n} + 1 \over {n^e \over e^n}+ 1} = 1$$ By applying L'hopital's rule enough times, any limit of the form $\lim_{n \rightarrow \infty}{\displaystyle {n^a \over e^n}}$ is zero. So one has $$\lim_{n \rightarrow \infty}{e{n^{e-1}\over e^n} + 1 \over {n^e \over e^n}+ 1} = {e*0 + 1 \over 0 + 1}$$ $$= 1$$ (If you're wondering why you can just plug in zero here, the rigorous reason is that the function ${\displaystyle {ex + 1 \over y + 1}}$ is continuous at $(x,y) = (0,0)$.)

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# Odds of drawing an ace on the first draw OR a two on the second

I'm working on a puzzle which describes a casino game where you draw 13 cards. You win if your first card is an ace, or your second is a two, or... all the way up to the 13th draw being a king. In order to analyse the problem, I started by simplifying to a two-card version. You draw two cards and win if the first is an ace or the second is a deuce.

The odds of drawing an ace on the first card is clearly 4/52 or 1/13.

My understanding, backed up by this question If you draw two cards, what is the probability that the second card is a queen?, suggests that the odds of drawing a 2 on the second card is also 1/13. What are the odds of drawing either? To my knowledge, you can't directly calculate that but instead calculate the inverse. The odds of not drawing a specific card should clearly be 12/13 and the odds of not drawing an ace followed by not drawing a two should be 12/13 * 12/13 = 144/169. The odds of drawing one or the other should therefore be 25/169.

But there's another approach to analysing the problem. There are 52 * 51 = 2652 different combinations for the first two cards. If A is an ace, T is a two and x is any other card, there are 5 combinations which give you a win:

AA Ax AT xT TT

Given that there are four aces, four twos and 44 other cards, the number of winning hands is:

4*3 + 4*44 + 4*4 + 44*4 + 4*3 = 392.

So the odds of winning should be 392 / 2652.

25/169 and 392/2652 are very close but they are NOT the same number. I believe the second number is accurate, but I can't see where the logic in the first method fails. I suspect that it has to do with the possibility of drawing both being double counted but I can't see how that should matter. It seems like you should be able to treat each draw as an independent event. Additionally, the second method doesn't scale well - it would be extremely tedious to directly calculate the number of winning hands for the full 13 card game.

edit: Clarified the rules for winning the game.

• The odds of firstly drawing an ace and secondly a two is $\frac4{52}\frac4{51}$. Can you understand why? – drhab Oct 30 '16 at 15:22
• Arthur, I'm not counting the number of ways to draw aces and twos, I'm counting the number of ways to win. Neither xA nor Tx win the game. – Dan J. Oct 30 '16 at 15:44
• Heads up - odds and probability are two different things. – Sean Roberson Oct 30 '16 at 15:56
• drhab, I disagree. If you draw a two on the first draw, then the odds of drawing a two on the second draw are only 3/51, not 4/51. You have to take both possibilities into account and that gives you odds of 1/13. See the linked question in my submission. – Dan J. Oct 30 '16 at 16:04
• The probability of firstly drawing an ace (not a two) and secondly a two is $\frac4{52}\frac4{51}$. If firstly an ace has been drawn then $4$ of the remaining $51$ cards are two's. – drhab Oct 30 '16 at 16:26

Let $A$ denote the event that the first draw will be an ace and let $B$ be the event that the second draw will be a two.

Then:$$\Pr(A\cup B)=$$$$\Pr(A)+\Pr(B)-\Pr(A\cap B)=$$$$\Pr(A)+\Pr(B)-\Pr(A)\Pr(B\mid A)=$$$$\frac4{52}+\frac4{52}-\frac{4}{52}\frac{4}{51}$$

The problem with your first method is that the events are not independent. Yes, the probability that you don't draw an ace from a full deck is $12/13$, and the probability that you don't draw a two from a full deck is $12/13$, but if you draw one card and then draw a second, the outcome of the second draw is dependent on the outcome of the first draw. Hence you cannot multiply the probabilities.

Your second method of calculating $P(\text{ace on first draw or two on second draw})$ is correct because it takes into account the outcome of the first draw and the second draw.

• Then is the question I linked to about drawing a queen on the second card wrong? In particular, look at A.J.'s accepted answer showing that the odds remain 1/13 even if you take the first draw into account. Is that incorrect or does it not apply here? And if doesn't apply, why not? Also, my description of the game was ambiguous. It should be OR, not AND. I'll edit. – Dan J. Oct 30 '16 at 15:50
• I'm sorry, I misunderstood the condition for winning! As for the accepted answer in the link, the probability that the second card is a queen given that you know nothing about the first card is indeed $1/13$. Similarly, the probability that the second card is not a two given that you know nothing about the first card is $12/13$. But you cannot use this to calculate $P(\text{first card not ace and second card not two})$ because in that event you do have information about the first card. – kccu Oct 30 '16 at 20:23

On questions like these you have to stop yourself from taking shortcuts and keep your eye on the big picture. Therefore always use both cards. First the number of probabilities of getting an ace on the first throw is 4 times the other 51 cards. Second the odds on getting a 2 on the second card if the first one isn't a ace or a 2 is 4 times 44 plus if the first card is a 2 then 4 times 3. Third add up all the 4 times and you get 4 times 98 which equals 392.

• 4 * 51 +
• 4 * 44 +
• 4 * 3 =
• 4 * 98 = 392

Or for simplicity you could just add the odds of an ace for the first card (4 times 51) and a deuce for the second card (51 times 4) then minus one of the double wins (4 times 4) and you get 4 times (51 + 51 - 4) equals 4 times 98 equals 392.


# Thread: Find the Indicated Integrals

1. ## Find the Indicated Integrals

I'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.

The three are:

$\int ln(x^4)/x$ dx

$\int$((e^t^x) cos (e^t))/(3+5 sin (e^t))

and lastly

$\int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$

2. Hi
Originally Posted by Pikeman85
$\int ln(x^4)/x$ dx
$\frac{\ln(x^4)}{x}=\frac{4\ln x}{x}=4\times \frac{1}{x}\times \ln x$ and as the derivative of $x\mapsto \ln x$ is $x\mapsto\frac{1}{x}$...
$\int$((e^t^x) cos (e^t))/(3+5 sin (e^t))
Is it $\int \frac{\mathrm{e}^{tx} \cos (\mathrm{e}^t)}{3+5 \sin (\mathrm{e}^t)}\,\mathrm{d}{\color{red}x}$ or $\int \frac{\mathrm{e}^{tx} \cos (\mathrm{e}^t)}{3+5 \sin (\mathrm{e}^t)}\,\mathrm{d}{\color{red}t}$ ?

$\int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$
$\int_0^{\frac{4}{5}} \frac{\arcsin \left(\frac{5}{4}x\right)}{\sqrt{16-25x^2}}\,\mathrm{d}x$
Remember that $\arcsin'x=\frac{1}{\sqrt{1-x^2}}$

3. Originally Posted by Pikeman85
I'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.

The three are:

$\int ln(x^4)/x$ dx

$\int$((e^t^x) cos (e^t))/(3+5 sin (e^t))

and lastly

$\int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$

for the frist one

$\int \frac{\ln(x^4)}{x}dx=4\int\frac{\ln(x)}{x}dx$

Let $u=\ln(x) \implies du=\frac{1}{x}dx$

$4\int udu=2u^2+C=2\left( \ln(x)\right)^2+C$

For number 2 you have both x's and t's but no dx or dt what variable are integrating with respect to?

For the last one

$\int_{0}^{4/5}\frac{\sin^{-1\left( \frac{5}{4}x\right)}}{\sqrt{16-25x^2}}dx$

let $x=\frac{4}{5}\sin(t) \implies dx=\frac{4}{5}\cos(t)dt$

$\int_{0}^{4/5}\frac{\sin^{-1\left( \frac{5}{4}x\right)}}{\sqrt{16-25x^2}}dx=\int_{0}^{\frac{\pi}{2}}\frac{t}{\sqrt{1 6-16\sin^2(t)}}\left( \frac{4}{5}\cos(t)dt \right)=$

$\frac{1}{5}\int_{0}^{\pi/2}tdt=\frac{1}{5} \left[ \frac{1}{2} \left( \frac{\pi}{2}\right)^2-0\right]=\frac{\pi^2}{40}$

Here is a web page with some La tex code for you
Helpisplaying a formula - Wikipedia, the free encyclopedia
Good luck.

4. Hello, Pikeman85!

These all require "simple" substitutions.
. . The trick is to recognize them.

$\int \frac{\ln(x^4)}{x}\,dx$
We have: . $\int\frac{4\ln(x)}{x}\,dx \;=\;4\int \ln x\,\frac{dx}{x}$

Let $u \:=\:\ln(x) \quad\Rightarrow\quad du \:=\:\frac{dx}{x}$

Substitute: . $4\int u\,du$ . . . etc.

$\int \frac{e^{t}\cos(e^t)\,dt}{3+5\sin(e^t)}$
Let $u \:=\:3+5\sin(e^t)\quad\Rightarrow\quad du \:=\:5e^t\cos(e^t)\,dt \quad\Rightarrow\quad e^t\cos(e^t)\,dt \:=\:\frac{1}{5}\,du$

Substitute: . $\int\frac{\frac{1}{5}\,du}{u} \;=\;\frac{1}{5}\int \frac{du}{u}$ . . . etc.

$\int^{\frac{4}{5}}_0 \frac{\sin^{-1}\!\left(\frac{5}{4}x\right)}{\sqrt{16-25x^2}}\,dx$
The denominator is: . $\sqrt{16\left(1 - \frac{25}{16}x^2\right)} \;=\;4\sqrt{1 - \left(\frac{5}{4}x\right)^2}$

The integral becomes: . $\int^{\frac{4}{5}}_0 \frac{\sin^{-1}\!\left(\frac{5}{4}x\right)\,dx} {4\sqrt{1 - \left(\frac{5}{4}x\right)^2}}$ . $= \;\;\frac{1}{4}\int^{\frac{4}{5}}_0\sin^{-1}\!\left(\frac{5}{4}x\right)\cdot\frac{dx}{\sqrt{ 1 - \left(\frac{5}{4}x\right)^2}}$

$\text{Let }u \:=\:\sin^{-1}\!\left(\frac{5}{4}x\right) \quad\Rightarrow\quad
du \:=\:\frac{\frac{5}{4}\,dx}{\sqrt{1-\left(\frac{5}{4}x\right)^2}}

Substitute: . $\frac{1}{4}\int^{\frac{4}{5}}_0 u\cdot\frac{4}{5}\,du \;=\;\frac{1}{5}\int^{\frac{4}{5}}_0 u\,du$ . . . etc.

Edit: I'm way too slow this time . . . *sigh*
.

$\int x^2 \sqrt{(7 + x^3)}$ dx

The answer I got for it is $(x^3/3) \sqrt{(7x+(x^4/4))}$

Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.

6. Originally Posted by Pikeman85

$\int x^2 sqrt(7 + x^3)$ dx

The answer I got for it is (x^3/3) sqrt(7x+(x^4/4))

Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.
One can notice that the derivative of $7+x^3$ is $3x^2$ so $x^2 \sqrt{7 + x^3}$ looks like $u'(x)\sqrt{u(x)}$ which can easily be integrated : you may try to substitute $u(x)=x^3+7$.

7. Well, a faster substitution is also $z^2=x^3+7.$

8. $x^3 \sqrt{7+x^3}$

I got this, but it did not work. I'm missing a step here I think

I don't get substitution very well

9. Originally Posted by Pikeman85
$\int x^2 \sqrt{(7 + x^3)}$ dx
The answer I got for it is $(x^3/3) \sqrt{(7x+(x^4/4))}$
Let $u=7+x^3 \implies du=3x^2dx \iff \frac{du}{3}=x^2dx$
$\int x^2\sqrt{7+x^3}dx=\int \sqrt{u}\left( \frac{du}{3}\right)=\frac{1}{3}\int u^\frac{1}{2}du=\frac{2}{9}u^\frac{3}{2}+C=\frac{2 }{9}(7+x^3)^\frac{3}{2}+C$


# If S and T are transformattion mappings, what is [ST]?

S and T are transformation mappings, what does [ST] and [TS] mean?

Does it mean transform via S and then apply T to the result and vice versa?

Juxtaposition of linear transformations means composition. In details, $ST$ means $S \circ T$ and $TS$ means $T \circ S$ (whenever they are defined, surely). This notation is used thinking of the following: $$T: V_1 \to V_2 \qquad S:V_2 \to V_3 \qquad S\circ T: V_1 \to V_3$$ Let $B_1, B_2, B_3$ be basis for the respective spaces (in finite dimension). Then: $$[S \circ T]_{B_1, B_3} = [S]_{B_2, B_3}[T]_{B_1,B_2}$$ At the bottom of our hearts, we know the distinction, so, writing, we oftem "confuse" the matrix with the transformation, so $S \circ T$ becomes $ST$. Actually, the identity above is the motivation for the definition of matrix multiplication.
• Yes, the composition goes from $V_1$ to $V_1$. To understand that notation better, I think it is important to take a look at matrices of linear transformations. – Ivo Terek Oct 13 '14 at 2:09
• Great, and for the format of B1,B3, can the B3 also be written above B1 or is it B3 above B1? – user83039 Oct 13 '14 at 2:23


# What are the odds of rolling a 3 number straight throwing 6d6

If you throw six fair dice, what are the odds that at least three dice make a straight (i.e. 123, 234, 345, or 456) I am certain that I am making a mistake in calculating it?

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## migrated from boardgames.stackexchange.comOct 2 '11 at 2:38

This question came from our site for people who like playing board games, designing board games or modifying the rules of existing board games.

what's your current calculation? –  DForck42 Oct 1 '11 at 4:27
should be moved to SE mathematics –  Hackworth Oct 1 '11 at 18:35
It's pretty high, pretty near one. But I'd have to think of how to get the exact probability... –  PearsonArtPhoto Oct 1 '11 at 22:02
I went through a long and nasty inclusion-exclusion argument, watched lots of stuff cancel, and in the end got $6^6 - 6 \cdot 4^6 + 8 \cdot 3^6 - 3 \cdot 2^6$, which agrees with Ed Pegg's calculations. I imagine there's a simple way to explain this formula, but I don't see it. I'm offering my results here in the hope that someone else can. –  Mike Spivey Oct 2 '11 at 7:02
@Mike Sounds like I did the same calculation as you. If you have time, why don't you post your solution? We can also try to think of a shortcut. –  Byron Schmuland Oct 2 '11 at 13:01

## 3 Answers

This problem is harder than I thought such a simple-sounding dice problem would be. :) However, the analysis below works for any number of six-sided dice! If $n$ is the number of dice, then the probability that we obtain at least one of 123, 234, 345, 456 is $$\frac{6^n - 6 \cdot 4^n + 8 \cdot 3^n - 3 \cdot 2^n}{6^n}.$$

For example, if $n=0$ or $n = 1$ or $n = 2$, we get $0$, as we should. If $n = 3$, we get $24$ in the numerator, which agrees with the $4 \cdot 3!$ ways we could obtain at least one of the required straights with three dice. And if $n = 6$, we get $27720$ in the numerator, which agrees with Ed Pegg's calculations.

We'll obtain the numerator in the fraction above by counting the number of ways not to get at least one of the four required straights and then subtract that from $6^n$. Let's call this event $\bar{S}$.

Start by considering subsets of {1, 2, 3, 4, 5, 6} such that throwing only numbers from that subset would give us an outcome in $\bar{S}$. There aren't any subsets of size five that do it, but there are six of size four: 1245, 1246, 1256, 1346, 1356, 2356. (I'm using a compact notation here for simplicity's sake.) Since $|\bar{S}| = |1245 \cup 1246 \cup 1256 \cup 1346 \cup 1356 \cup 2356|$, we now have a way to calculate $|\bar{S}|$ using the principle of inclusion-exclusion (PIE). We just have to consider all possible ways of intersecting these six sets and then add up the cardinalities of the resulting intersections according to parity given by $(-1)^{k(A)}$, where $k(A)$ is the number of sets intersected to obtain the subset $A$.

What makes applying PIE more difficult than usual here is that intersecting two sets sometimes gives a subset of size 3 and sometimes one of size 2, and this problem only gets worse as you intersect more sets. However, if you work through all the possibilities, most of the resulting intersections show up more than once with different parities, and most of these surprisingly cancel. What's left is 124, 125, 126, 136, 146, 156, 256, 356 (from intersecting 2 sets) and 12, 16, 26 (from intersecting 3 sets).

Finally, since there are $j^n$ ways to throw the $n$ dice so that they only take on values from a subset of size $j$, we obtain $|\bar{S}| = 6 \cdot 4^n - 8 \cdot 3^n + 3 \cdot 2^n$.

Unfortunately, I don't have a good explanation for why so many terms cancel, and I think there should be one. If someone can find such an explanation, I would love to see it. I've found an explanation for the massive term cancellation. (I suspect there are others.) See comments below.

Finally, thanks to Ed Pegg for the brute force approach; that was very helpful in checking my work.

Added: Here is one explanation for the massive term cancellation when using PIE.

The six sets of size four that we're working with can be expressed as a graph. Each set is a vertex, and two vertices are connected by an edge if their corresponding sets differ by only one element. The graph looks like this.

  1245 — 1256 — 2356
\  /    \  /
1246    1356
\    /
1346

It turns out that, of the intersections that remain when applying PIE, the vertices correspond to the 6 sets of size 4, the edges correspond to the 8 sets of size 3, and the minimal cycles correspond to the 3 sets of size 2. Every other intersection is either empty or can be paired with another intersection such that they cancel each other in the PIE formula. To obtain the mapping between pairs of intersections, take an intersection that produces a connected subgraph. Pair it with the intersection obtained by removing the vertex with the highest connectivity in the subgraph. This produces a disconnected subgraph. For example, the intersection of 1245, 1256, 1356, and 2356 is paired with the intersection of 1245, 1356, and 2356. Both intersections produce 5, yet have different parities, so they cancel when using PIE. (Well, the 4-cycle is a special case, but everything there cancels nicely, too, except for the full cycle itself.)

As far as I know, inclusion-exclusion cannot, in general, be represented in a graph like this. Maybe there are some special cases, though, and this graph happens to fall into one of those. Is anyone familiar with this?

Inclusion-exclusion can, in some cases, be represented and simplified by a graph like this. My observations here are an instance of the following theorem:

Suppose sets $A_1, A_2, \ldots, A_n$ can be represented as the vertices of a planar graph $(V,E)$ such that, for each $x \in \cup_i A_i$, the subgraph consisting of the $A_i$'s containing $x$ is connected. For each edge $e = \{A_i, A_j\}$, let $|e| = |A_i \cap A_j|$. For each minimal cycle $c = \{A_{c_1}, A_{c_2}, \ldots, A_{c_m}\}$ in $(V,E)$, let $|c| = |\bigcap_{j=1}^m A_{c_j}|$. Then $$\left|\bigcup_{i=1}^n A_i \right| = \sum_{v \in V} |v| - \sum_{e \in E} |e| + \sum_{c \in C} |c|.$$

Since the graph above satisfies the hypotheses of this theorem, we immediately get $|\bar{S}| = 6 \cdot 4^n - 8 \cdot 3^n + 3 \cdot 2^n$.

The proof of the theorem follows directly from Euler's famous formula $V - E + F = 2$ for connected planar graphs. For any element $x \in \cup_i A_i$, the subgraph consisting of the $A_i$'s containing $x$ is connected (by hypothesis) and planar. The element $x$ is counted once on the left side of the formula in the theorem. Since minimal cycles correspond to faces except for the one outer face, Euler's formula tells us that $x$ is counted a net total of once for the right-hand side as well.

More general versions of this theorem that use the Euler characteristic can be found in

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This is very nice!. I wish I could upvote again. –  Byron Schmuland Oct 4 '11 at 17:19
@Byron: Thanks! The compliment is worth more than the extra upvote would be, though. :) –  Mike Spivey Oct 4 '11 at 17:28
@Byron: I found an explanation for the term cancellation in terms of the graph representation. Fascinating, really! And I think now that I have officially beaten this question to death. :) –  Mike Spivey Oct 5 '11 at 4:51

In Mathematica, here's a brute force solution. There are 27720 rolls of the 6 dice that will give that straight.

Length[Select[Tuples[Range[6], {6}], Max[Table[Length[Intersection[#, {{1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {4, 5, 6}}[[n]]]], {n, 1, 4}]] == 3 &]]/6^6

$385/648 = 0.5941358024691358024691\dots$

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Here's another way. Let ${\bf k}$ be the number of distinct die numbers, let $S$ ("success") be the event that at least one of the four required straights occurred.

Then $$P(S) = \sum_{k=1}^6 P(S | {\bf k}=k) \; P({\bf k}=k)$$

It's clear that, for a given fixed $k$, all configurations are equiprobable.

It's trivial that $P(S | {\bf k}=0)=P(S | {\bf k}=1)=P(S | {\bf k}=2)=0$. The other are not difficult:

$P(S | {\bf k}=3)$ : There are 4 possible success configurations, out of ${6 \choose 3}$, hence $P(S | {\bf k}=3) = 1/5$

$P(S | {\bf k}=4)$ : This is the most difficult one, let's count the unsuccessful configurations: These are : [o x x o x x] [x o x o x x] [x o x x o x] [x x o o x x] plus the mirroring of the first two: 6 unsuccessful configurations out of ${6 \choose 4}$ hence $P(S | {\bf k}=4) = 1- 2/5 = 3/5$ [*]

Further, $P(S | {\bf k}=5)= P(S | {\bf k}=6) = 1$

Now, to compute $P({\bf k}=k)$, we must count the number of ways of filling $k$ positions with 6 throws: for example, $$P({\bf k}=4)=\frac{{6 \choose 4} \; 4! \; S_2(6,4)}{6^6}$$

where $S_2(n,m)$ are the Stirling numbers of the second kind. So finally

$$P(S) = \sum_{k=3}^6 a_k \; \frac{{6 \choose k} \; k! \; S_2(6,k)}{6^6} \approx 0.59413$$

with $a_3=1/5$, $a_4=3/5$, $a_5=1$, $a_6=1$.

[*] Added: to generalize this for arbitrary number of dice or runlengths, notice that this counting of unsuccessful configurations is equivalent to count the ways of expressing the number $k=4$ as a sum of $n-k+1=3$ non-negative terms less that $m=3$ (runlength), with order (in the example, 0+2+2 , 1+1+2, 1+2+1, 2+0+2, 2+2+0,2+1+1). Hence $P(S | {\bf k}) = 1 - b_{k,n-k+1,m}/{n \choose k}$ where $b_{r,s,t}$ counts the $s$-terms weak-compositions of the integer $r$, with the restriction that each term is less than $t$. An expression is given in Enumerative combinatorics (Stanley) page 120, problem 28.

Added 2: Because the exact formula for arbitrary dice is quite formidable, here's some quick asymptotics. Let $n$ be the number of throws, $c$ the number of dice faces, $m$ the straight length. For large $c,n$, the probability that a particular face number does not appear is $q=(1-1/c)^n \approx \exp(- n/c)$. Asymptotically, we can assume that these events are independent, and disregard border effects, and regard each outcome as a sequence of runlengths $(a_1 b_1 a_2 b_2 ... a_r b_r)$ where $a_i$ is the length of consecutive (run) die faces that appeared, $b_i$ the faces that didn't (and $a_1 + b_1 + ... = c$). Each run can then be approximated by independent geometric variables (starting at 1) with stopping probabilities $q$ and $1-q$. Their expectations are respectively $1/q$ and $1/(1-q)$, so that equating the expected sum we get $r=c \; q \;(1-q)$. The event of no-success, correspond to $a_i<m$, and its probability is given, in this approximation, by $(1-(1-q)^{m-1})^r$. Finally:

$$P(S) \approx 1- \left[ 1- (1-q)^{m-1} \right]^r$$

with

$$r = c \; q \; (1-q) \hspace{20px} q = (1-1/c)^n \approx e^{-n/c}$$

In our original question... we have $c=n=6$,$m=3$, the numbers are too small to apply this asymptotics, but anyway, I get: $P(S) \approx 0.50922$ ($0.54184$ if using the "exact" $q$)

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+1: I like this better than my answer for the 6-dice problem. Can it be generalized to $n$ dice? –  Mike Spivey Oct 4 '11 at 18:47
@MikeSpivey: The counting of $P(S|k=k)$ is not trivially generalizable (for $n$ dice and $m$ sequence lengths), but I think it can be done, I'll give it a look tonight. –  leonbloy Oct 4 '11 at 19:00
@MikeSpivey: See my addition. Now it remains to see if those bounded weak-compositions have some closed formula. –  leonbloy Oct 4 '11 at 19:20


# Math Help - Geometry regarding the intersection of wires criss crossing vertical poles.

1. ## Geometry regarding the intersection of wires criss crossing vertical poles.

Two vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?

help step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.

2. Do you find it odd that the distance between the poles is not given?

In any case, you're going to need some similar triangles. If you have a method that has similar triangles, I suspect it is NOT totally worthless. Post it.

3. Originally Posted by swanz
Two vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?

help step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.
If you know the poles are vertical, you know they are parallel. You can treat each wire as a transversal of these - check out which angles are alternate interior angles of these two transversals to see some similar triangles. Draw yourself a decent picture to see them! Then use these similar triangles to answer the question.

4. Originally Posted by swanz
Two vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?

help step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.
Assume the wires are straight (with no sag)
Since the distance between the two is irrelevant, we can assign it what ever needed, say 1 m.

Assume the shorter pole is coincident with the y axis and the taller pole is parallel but at x = 1

An equation for each wire:

$y_1 = -6x_1 + 6$

$y_2 = 10x_2 + 0$

Isolate x:

$x_1 = \dfrac{ y_1 - 6 }{-6}$

$x_2 = \dfrac{y_2}{10}$

equate:

$\dfrac{ y - 6}{-6} = \dfrac{y}{10}$

solve for y
$y = \dfrac{ 60 }{16}$

.

5. ## Resolved

thank you aidan....never even considered plotting it on the graph .....all the time i was playing around with similar triangles, transversals and loads of variables.
nice method, can come in handy in various situations, thnx a bunch.

6. Originally Posted by aidan
Since the distance between the two is irrelevant
Well, one probably should prove that, rather than just state it.

7. Originally Posted by TKHunny
Well, one probably should prove that, rather than just state it.
My Bad. I just made an assumption that it was common knowledge.
One proof (if needed) is here:
PlanetMath: harmonic mean in trapezoid

It indicates that the diagonals of a trapezoid will intersect at some point. A line through that point parallel to the parallel sides will intersect the non parallel sides. The length of this segment is a ratio of the parallel sides only:

If the lengths of the parallel sides are "a" and "b" then the length of the interior line that is parallel to the parallel lines and passes through the intersection point is

$\dfrac{2 a b }{(a+b)}$

The intersection point is at the midpoint of the interior line.

In our case here:

HALF the length of the interior line is

$\dfrac{( 6 \cdot 10) }{( 6 + 10 )} = 3.75$

8. I think it is second tier common knowledge. It can be assumed if you work with that sort of thing, but not generally.

My views. I welcome others.

9. That problem is "ye olde crossing ladders" problem in disguise


# Dominated Convergence Theorem

Give an example of a sequence $\{f_n\}_{n=1}^\infty$ of integrable functions on $\mathbb{R}$ such that $f_n \to f$ but $\int f_n \not\to \int f$. Explain why your example does not conflict with the Dominated Convergence Theorem.

I do notice that the inequality $|f_n(x)| \le g(x)$, where $g$ is an integrable function over $\mathbb{R}$, is not listed here in this problem. So the function need not be dominated by an integrable function here. But this is required as one hypothesis of the Dominated Convergence Theorem; hence the example will not conflict.

If this is sound reasoning, how may I come up with functions that are not dominated by another function? Initially I was thinking $f_n(x)=x \sin (nx)$ because its lim sup is $\infty$, but even then, we still have $|f_n(x)| \le |x| =: g(x)$.

• $f_n = n 1_{(0,{1 \over n}]}$. – copper.hat May 8 '15 at 16:35
• It is not only a matter of being dominated by a function, but by an integrable function. $|x|$ is not integrable on $\mathbb{R}$, as $\int_{-\infty}^\infty |x| dx = \infty$. – Pedro M. May 8 '15 at 16:39
• @PedroM. I forgot about the "integrable" function part. :O But thanks for clearing that up. – Cookie May 8 '15 at 16:42
• It may be dominated by another function, however the detail is that the if dominating function does not belong to $L_1$ then DCT may not hold. – Alonso Delfín May 8 '15 at 16:42
• @AaronMaroja I do not understand why you centerized "$\int f \not\to \int f$" and removed the period at the end of that sentence. – Cookie May 9 '15 at 7:28

Consider the sequence of functions on $(0,1)$ $$f_n(x) = \begin{cases} n & \text{ if } x \in (0,1/n)\\ 0 & \text{ otherwise} \end{cases}$$ We have $\lim_{n \to \infty} f_n(x) = 0 = f(x)$. However, $$\lim_{n \to \infty} \int_0^1f_n(x)dx = 1 \neq 0 = \int_0^1 f(x) dx$$ The key in the dominated convergence theorem is that sequence of functions $f_n(x)$ must be dominated by a function $g(x)$, which is also integrable.

I would like to provide a slightly more abstract framework to illustrate this "loss of compactness"$^{[2]}$ phenomenon. The functional setting is the $L^1(\mathbb{R})$ space: $$L^1(\mathbb{R})=\left\{f\colon\mathbb{R}\to\mathbb{R}\ :\ \int_{-\infty}^\infty \lvert f(x)\rvert\, dx<\infty \right\},\qquad \lVert f\rVert=\lVert f\rVert_{L^1}=\int_{-\infty}^\infty\lvert f(x)\rvert\, dx.$$ Here we take a bounded sequence $f_n$ that converges pointwise a.e. : $$\begin{array}{cc} \|f_n\|\le C, & f_n\to f\, \text{a.e.} \end{array}$$ The question is: does this sequence converge, that is, is it true that $$\lVert f_n-f\rVert\to 0?^{[1]\ [2]}$$ This would make for a very desirable property of our functional space. However, as other answers very clearly show, the answer is negative in general. The problem is that our space is subject to the action of noncompact groups of isometries. Namely, one has the action of the translation group $$\begin{array}{cc} \left(T_\lambda f\right)(x)=f(x-\lambda), &\lambda \in (\mathbb{R}, +) \end{array}$$ and of the dilation group $$\begin{array}{cc} \left(D_\lambda f\right)(x)=\frac{1}{\lambda}f\left(\frac{x}{\lambda}\right), &\lambda \in (\mathbb{R_{>0}}, \cdot) \end{array}$$ The change of variable formula for integrals immediately shows that those group actions are isometric, that is, they preserve the norm.

So, fixing a non-vanishing function $f\in L^1(\mathbb{R})$, its orbits $T_\lambda f$ and $D_\lambda f$ form bounded and non-compact subsets of $L^1(\mathbb{R})$. In particular, letting $\lambda \to +\infty$ (or $\lambda \to -\infty$ for translations, or $\lambda\to 0$ for dilations), one finds counterexamples to the question above. (Note that, more or less, all the examples constructed in the other, excellent, answers are constructed this way).

In technical jargon one says that the translation and dilation groups introduce a defect of compactness in $L^1(\mathbb{R})$ space. This is the terminology of the Concentration-Compactness theory (the linked page is a blog entry of T. Tao, but the theory has been founded by P.L. Lions). The dominated convergence theorem can be therefore seen as a device that impedes the defect of compactness to take place.

Footnotes

$^{[1]}$ The OP only asks about the convergence of the integrals: $\int f_n\to \int f$. Now a standard theorem (cfr. Lieb & Loss Analysis, 2nd ed. Theorem 1.9 (Missing term in Fatou's lemma), see in the remarks) gives us the equivalence $$\begin{array}{ccc} \lVert f_n\rVert_{L^1} \to \lVert f\rVert_{L^1} & \iff & \lVert f_n-f\rVert_{L^1}, \end{array}$$ Therefore, at least for sequence of positive functions for which $\int f_n=\lVert f_n\rVert_{L^1}$, the failure of convergence for sequences of integrals is exactly the same thing as the failure of convergence in $L^1$ space. That's why one can see the phenomenon in this functional analytic setting.

$^{[2]}$ Compactness usually means that bounded sequences have convergent subsequences. In $L^1(\mathbb{R})$ space, pointwise convergent sequences are compact if and only if they are norm convergent.

• wow! Very interesting point of view, never heard of that before! :-) – Ant May 8 '15 at 18:47

It is not only a matter of being dominated by just any function $g$, but by an integrable function. $|x|$ is not integrable on $\mathbb{R}$, as $\int_{-\infty}^\infty |x| dx = \infty$.

copper.hat provided an example that is not dominated by any function, but you can also pick bounded examples, such as $f_n = 1_{[n,n+1]}$.

• Does $f_n=\chi_{[-n,n]}$ also work? (Sorry, I'm used to the characteristic function that uses the Greek letter $\chi$ instead of $1$.) – Cookie May 8 '15 at 16:46
• @dragon: Then $f_n$ converges to the constant function $1$, which is not integrable on $\mathbb{R}$. This is another illustration of how the DCT may fail if the dominating function $g$ is not integrable: $\int f = \infty$ (in my example, $\int f$ is finite but is different from $\lim \int f_n$). Notice, however, that in your case, $\int f_n \to \infty = \int f$. – Pedro M. May 8 '15 at 16:49

Let $$f_n(x) = \frac{n^3x^2}{1+n^4x^4}.$$ Then $f_n(x) \to 0$ pointwise everywhere, and $\int_\infty^\infty f_n(x)\,dx = \int_\infty^\infty \frac{x^2}{1+x^4}\,dx$ for every $n.$

Let $f_n = \chi[n,2n]$ then for any $x \in \mathbb R$

$$\lim_{n \to \infty} f_n (x) = 0 \,\,\, \text{and} \,\,\, \int_{x\in \mathbb R} f_n(x) dx = n \to \infty \,\,\, \text{as}\,\, n \to \infty$$

On the other hand taking $f(x) = 0 , \forall x \in \mathbb R$ we have $$\int_{x\in \mathbb R} f(x) dx = \int _{x\in \mathbb R} 0\,\, dx = 0$$ then

$$\int_{x\in \mathbb R} f_n(x) dx \not \to \int_{x\in \mathbb R} f(x) dx$$


# Math Help - Integration;unsure method

1. ## Integration;unsure method

Hi, I have another integration question (because they're just so fun ).

Int ( dx/(x^2-3x+2)^(1/2) )

The item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.

Thanks a ton.

2. Originally Posted by ChaosBlue
Hi, I have another integration question (because they're just so fun ).

Int ( dx/(x^2-3x+2)^(1/2) )

The item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.

Thanks a ton.
Start this by looking under the square root sign. You want to do a "complete the square."

x^2 - 3x + 2 = (x^2 - 3x) + 2 = (x^2 - 3x + 9/4 - 9/4) + 2 = (x^2 - 3x + 9/4) - 9/4 + 2

= (x - 3/2)^2 - 1/4

Now let y = x - 3/2, dy = dx

Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy]

One more trick and we're done. Pull a 1/4 out from under the radical:
Int[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]

And now let z = 2y, dz = 2dy and your integral becomes:
Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]

This is a more standard problem. Can you take it from here?

-Dan

3. Originally Posted by ChaosBlue
Hi, I have another integration question (because they're just so fun ).

Int ( dx/(x^2-3x+2)^(1/2) )

The item giving me the biggest problem is the square rt. From what I know, I don't think u-sub will work (just complicate things more), integrating my parts seems like it would be very messy, partial fractions is out of the mix since its raised to a power and I don't think you can do complete the square for the same reason. I'm assuming I'm making this much more complicated than it really is. Can anybody help me start it out, thats all I need help with.

Thanks a ton.
The way I see to do it is really long, so if anyone out there has an easier way, you are welcome to post.

We begin by completing the square.

int(1/sqrt(x^2 - 3x + 2))dx
= int(1/(sqrt( x^2 - 3x + (-3/2)^2 - 1/4))dx
= int(1/sqrt((x - 3/2)^2 - 1/4))dx
= int(1/sqrt((x - 3/2)^2 - (1/2)^2))dx
= int(1/sqrt[(1/2)^2*([(x - 3/2)/(1/2)]^2 - 1)])dx
= 2*int(1/sqrt[(2x - 3)^2 - 1])dx

let u= 2x - 3
=> du = 2 dx
so our integral becomes
int(1/sqrt(u^2 - 1))du

Now continue using trig substitution

4. Originally Posted by topsquark
Start this by looking under the square root sign. You want to do a "complete the square."

x^2 - 3x + 2 = (x^2 - 3x) + 2 = (x^2 - 3x + 9/4 - 9/4) + 2 = (x^2 - 3x + 9/4) - 9/4 + 2

= (x - 3/2)^2 - 1/4

Now let y = x - 3/2, dy = dx

Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy]

One more trick and we're done. Pull a 1/4 out from under the radical:
Int[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]

And now let z = 2y, dz = 2dy and your integral becomes:
Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]

This is a more standard problem. Can you take it from here?

-Dan
Ha ha, i took forever typing this up only to see that u beat me to it once i clicked "submit reply"

Thanks a lot Dan

5. Originally Posted by Jhevon
Ha ha, i took forever typing this up only to see that u beat me to it once i clicked "submit reply"

Thanks a lot Dan
It happens. You're welcome.

And besides, my variables are prettier anyway.

-Dan

6. Originally Posted by topsquark
It happens. You're welcome.

And besides, my variables are prettier anyway.

-Dan

Nu-uh, u is a much more suitable variable for substitution, most text books use u. You're just fighting the system.

Do you know what they do to non-conformists in the math field?!

7. Thanks guys so much, you really helped me out. I do have one question though.

Originally Posted by topsquark
One more trick and we're done. Pull a 1/4 out from under the radical:
Int[1/((1/4)(2y)^2 - 1/4)^{1/2}dy] = 2*Int[1/((2y)^2 - 1)^{1/2}dy]

And now let z = 2y, dz = 2dy and your integral becomes:
Int[1/(x^2-3x+2)^{1/2}dx] = Int[1/(y^2 - 1/4)^{1/2}dy] = Int[1/(z^2 - 1)^{1/2}dz]
topsquark: I understand we're essentially "manipulating" our equation so we can effectively use trig substitution, but I kind-of got lost in the process. I have trouble grasping it after we pull out the 1/4.? Is there anyway you (or Jhevon since you got it as well) could explain that part to me a bit more?

Thanks a ton.

8. Originally Posted by ChaosBlue
Thanks guys so much, you really helped me out. I do have one question though.

topsquark: I understand we're essentially "manipulating" our equation so we can effectively use trig substitution, but I kind-of got lost in the process. I have trouble grasping it after we pull out the 1/4.? Is there anyway you (or Jhevon since you got it as well) could explain that part to me a bit more?

Thanks a ton.

When he said pull out 1/4, he meant we factorized the function to obtain (1/4)*(something), he did that so he could get something of the form u^2 - 1 under the square root

9. Originally Posted by Jhevon
When he said pull out 1/4, he meant we factorized the function to obtain (1/4)*(something), he did that so he could get something of the form u^2 - 1 under the square root
I worded my question poorly (my bad ). What I mean is this.

I get where we use completing the square to transform:

int 1/sqrt(x^2-3x+2) dx--> int 1/sqrt( (x-3/2)^2 - 1/4 )dx

From there we have u = x-3/2 and therefore du = dx

Plug in and we get int 1/ (u^2 - 1/4)^1/2 du

Now, we pull out or factor out the 1/4; But I am struggling with when we do that, how

int 1/ (u^2 - 1/4)^1/2 du becomes

Int[1/((1/4)(2u)^2 - 1/4)^{1/2}du]
(we pull out 1/4, so how does u^2 become 2u^2 ?

and then

2*Int[1/((2u)^2 - 1)^{1/2}du]

Sorry for bothering you guys so much!

10. Originally Posted by ChaosBlue
I worded my question poorly (my bad ). What I mean is this.

I get where we use completing the square to transform:

int 1/sqrt(x^2-3x+2) dx--> int 1/sqrt( (x-3/2)^2 - 1/4 )dx

From there we have u = x-3/2 and therefore du = dx

Plug in and we get int 1/ (u^2 - 1/4)^1/2 du

Now, we pull out or factor out the 1/4; But I am struggling with when we do that, how

int 1/ (u^2 - 1/4)^1/2 du becomes

Int[1/((1/4)(2u)^2 - 1/4)^{1/2}du]
(we pull out 1/4, so how does u^2 become 2u^2 ?

and then

2*Int[1/((2u)^2 - 1)^{1/2}du]

Sorry for bothering you guys so much!
it's no bother, we're all here to learn.

ok, here it is. Say we had some function b, and we wanted to factor an a out of it, we could write is as a*(b/a).

similarly, when he factored 1/4 out of u^2, he ended up with (1/4)*(u^2/(1/4))

u^2/(1/4) = u^2/(1/2)^2 = [u/(1/2)]^2 = 2u^2

11. Funny you should say thanks, we actually left the hard part to you

12. Originally Posted by Jhevon
Nu-uh, u is a much more suitable variable for substitution, most text books use u. You're just fighting the system.

Do you know what they do to non-conformists in the math field?!
Good thing I'm not in the Math field, then.

-Dan

13. Originally Posted by topsquark
Good thing I'm not in the Math field, then.

-Dan
It's even worst in the physics field, which i suppose you're a part of

14. Originally Posted by Jhevon
It's even worst in the physics field, which i suppose you're a part of
Yes, but as long as you define your variables, you can technically get away with any set of variables you wish.

I think I'm going to create a unit called the "Dan." It'll be defined as the amount of work needed to translate an equation from the variable z to the variable u...

-Dan

15. Originally Posted by topsquark
Yes, but as long as you define your variables, you can technically get away with any set of variables you wish.

I think I'm going to create a unit called the "Dan." It'll be defined as the amount of work needed to translate an equation from the variable z to the variable u...

-Dan
haha, bless your heart Dan, you need help

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# Is it possible to write a sum as an integral to solve it?

I was wondering, for example,

Can:

$$\sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$

Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.

But actually writing an integral form. Like

$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{a}^{b} g(x) \space dx$$

What are some general tricks in finding infinite sum series.

• math.stackexchange.com/questions/1002440/… – lab bhattacharjee Nov 3 '14 at 13:26
• @labbhattacharjee, I did not meant that. I know the solution to this, I was just asking if in general it is possible to write a sum as an actual integral. – Amad27 Nov 3 '14 at 13:28
• You can trivially write the sum as an integral using the Iverson bracket (add a factor of $[n \in \mathbb{N}]$ to the integrand). This ignores the question of how to evaluate the resulting integral, of course. – chepner Nov 3 '14 at 19:10
• "I am NOT talking about a method for using tricks with integrals." "But actually writing an integral form." "What are some general tricks" Combining these quotes with the accepted answer that does not seem to be a general trick, I'm a bit confused on what this question is asking. – JiK Nov 4 '14 at 8:28
• @Amad27 $\int_\mathbb{N}\frac{d \mu}{(3n-1)(3n+2)}$ where $\mu$ is the counting measure on $\mathbb{N}$. It doesn't give you anything you didn't already have though. I didn't really mean it seriously although it is true. – Tim Seguine Nov 5 '14 at 17:21

A General Trick

A General Trick for summing this series is to use Telescoping Series: \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+2)} &=\frac13\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{3n-1}-\frac1{3n+2}\right)\\ &=\frac13\lim_{N\to\infty}\left[\sum_{n=1}^N\frac1{3n-1}-\sum_{n=1}^N\frac1{3n+2}\right]\\ &=\frac13\lim_{N\to\infty}\left[\sum_{n=0}^{N-1}\frac1{3n+2}-\sum_{n=1}^N\frac1{3n+2}\right]\\ &=\frac13\lim_{N\to\infty}\left[\frac12-\frac1{3N+2}\right]\\ &=\frac16 \end{align}

An Integral Trick

Since $$\int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n$$ for $n\gt0$, we can write \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+2)} &=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\ &=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\ &=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\ &=\frac16 \end{align}

• I think this is a better "trick" for dealing with sums. Integral "tricks" are nice however integrals and infinite series' are very different in what they calculate and manipulating a sum or integral on its own without switching is preferred. – Ali Caglayan Nov 3 '14 at 16:23
• @Alizter: For the most part, I agree. However, sometimes pure series manipulation can be extremely complicated, and the proper integral representation of a sum can be useful. However, in this case, I think staying with series manipulation is easiest. That being said, I have added an integral approach, as well. – robjohn Nov 3 '14 at 18:40
• The sum under the first integral could have been computed as a telescoping series either. Considering this, I think the use of integrals in the second solution is completely void. Edit: I mean exactly what Henning Makholm points out under the other answer. – Adayah Nov 4 '14 at 19:41
• @Adayah: My reply to Henning was meant as an agreement. I first posted only the telescoping series, but then added an integral approach to satisfy the first part of the question. In any approach where one breaks up the summand using partial fractions, it could be said that, at that point, the answer could be computed as a telescoping sum. – robjohn Nov 4 '14 at 20:48

Since $\int_{0}^{1}x^k\,dx = \frac{1}{k+1}$, $$\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)=\frac{1}{3}\int_{0}^{1}x^{3n-2}(1-x^3)\,dx,$$ so, summing over $n$: $$\sum_{n=1}^{+\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\int_{0}^{1}x\,dx=\frac{1}{6}.$$

• I thought we need uniform convergence in order to interchange the limit and integral. The power series is uniformly convergent inside the radius of convergence, how to pass it to the whole interval $[0,1]$? – John Nov 3 '14 at 13:45
• @JohnZHANG Actually no, Fubini and Tonelli's theorems allow this for a monotone sequence supposedly, I believe. – Amad27 Nov 3 '14 at 13:45
• Nice trick for the given sum, but this still doesn't answer the bold-marked question of general tricks. – Ruslan Nov 3 '14 at 16:34
• Isn't the integral just a detour here? The operative step is exactly the same telescoping that could have been done without rewriting the terms into integrals. – Henning Makholm Nov 4 '14 at 10:11
• @HenningMakholm: smoke and mirrors. – robjohn Nov 4 '14 at 12:07

Actually writing it as an integral, as asked for:

$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{1}^{\infty} \frac{1}{(3\lfloor x\rfloor-1)(3\lfloor x\rfloor+2)} dx$$

This probably won't help with finding the value, though.

• why won't it help finding the value? – Amad27 Nov 4 '14 at 13:20
• @Amad27: I don't see a way it would. If you can find one, then more power to you, I suppose ... – Henning Makholm Nov 4 '14 at 13:33
• @Amad27 Methods for solving integrals are poorly suited for integrating functions that are non-continuous. The usual approach for integrating functions like the one here is to separately integrate over each interval where it is continuous. Which brings us back to the sum form. – Rafał Dowgird Nov 4 '14 at 15:11
• @Amad27 It is quite literally equivalent to the original sum in a trivially useless manner XD – Simply Beautiful Art Jan 12 '17 at 2:11

In such cases, the partial fractions of general term (i.e. $n^{th}$ term ) of the infinite-series are very useful.
Given that $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\sum_{n=1}^{\infty} T_{n}$$ Where, $T_{n}$ is the $n^{th}$ term of the given series which can be easily expressed in the partial fractions as follows $$T_{n}=\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)$$ Now, we have $$\sum_{n=1}^{\infty}\frac{1}{(3n-1)(3n+2)}=\frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\! \cdot \! ........ +\left(\frac{1}{3n-4}-\frac{1}{3n-1}\right)+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$$ $$=\frac{1}{3} \lim_{n\to \infty} \left[\frac{1}{2} -\frac{1}{3n+2}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2} -\frac{1}{\infty}\right]$$ $$=\frac{1}{3} \left[\frac{1}{2}\right]=\color{blue}{\frac{1}{6}}$$

We can indeed write the sum as an integral, after research. Consider:

Find: $\psi(1/2)$

By definition:

$$\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$$

The required $z$ is $z = -\frac{1}{2}$

so let $z = -\frac{1}{2}$

$$\psi(1/2) = -\gamma + \sum_{n=1}^{\infty} \frac{-1}{2n(n - \frac{1}{2})}$$

Simplify this: $$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{n(2n - 1)}$$

The sum seems difficult, but really isnt.

We can telescope or:

$$\frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$

Let $x \rightarrow x^2$

$$\frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$

Integrate once:

$$\tanh^{-1}(x) = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}$$

Integrate again:

$$\sum_{n=1}^{\infty} \frac{x^{2n}}{(2n-1)(n)} = 2\int \tanh^{-1}(x) dx$$

From the tables, the integral of $\tanh^{-1}(x)$

$$\sum_{n=1}^{\infty} \frac{x^{2n}}{(2n-1)(n)} = \log(1 - x^2) + 2x\tanh^{-1}(x)$$

Take the limit as $x \to 1$

$$\sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)} = \log(4)$$

$$\psi(1/2) = -\gamma - \sum_{n=1}^{\infty} \frac{1}{(2n-1)(n)}$$

$$\psi(\frac{1}{2}) = -\gamma - \log(4)$$

• I am the OP per say. This is a general trick. I converted the sum into an integral. Please read carefully. – Amad27 Dec 21 '14 at 8:15

Yes you can use the Euler Maclaruin formula to write the sum as an integral plus an infinite number of derivatives. I remember deriving this for my self when I was younger and being very pleased with myself.

This particular sum could be solved because you had two terms $ax+b$ and $ax+c$ and the difference between c and b is equal to a (I think it would work in a slightly more complicated way if it was a not-too-large multiple of a).

If you want numerical values in general cases, and the sum doesn't converge quickly for your taste, or you want just a partial sum, you can use that

$$\displaystyle f (k) = \int_{k-1/2}^{k+1/2} f(k) dx ≈ \int_{k-1/2}^{k+1/2} f(x) dx$$

and therefore

$$\displaystyle \sum_{k=n}^{m} f(k) ≈ \int_{n-1/2}^{m+1/2} f(x) dx$$

Assuming that you can solve the integral in closed form, if you let

$$\displaystyle g (k) = f(k) - \int_{k-1/2}^{k+1/2} f(x) dx$$

then

$$\displaystyle \sum_{k=n}^{m} f(k) = \int_{n-1/2}^{m+1/2} f(x) dx + \sum_{k=n}^{m} g(k)$$

$g (k)$ will usually converge much faster than $f (k)$.