Moment of Inertia of Beam Sections

How to Calculate the Moment of Inertia of a Beam Section(Second Moment of Area)

Before we find the moment of inertia (or second moment of area) of a beam section, its centroid (or center of mass) must be known. For instance, if the moment of inertia of the section about its horizontal (XX) axis was required then the vertical (y) centroid would be needed first (Please view our Tutorial on how to calculate the Centroid of a Beam Section).

Before we start, if you were looking for our Free Moment of Inertia Calculator please click the link to learn more. Consider the I-beam section below, which was also featured in our Centroid Tutorial. As shown below the section was split into 3 segments:

The Neutral Axis (NA)

The Neutral Axis (NA) or the horizontal XX axis is located at the centroid or center of mass. In our Centroid Tutorial, the centroid of this section was previously found to be 216.29 mm from the bottom of the section. Now to calculate the total moment of ienrtia of the section we need to use the "Parallel Axis Theorem":

[math] {I}_{total} = \sum{(\bar{I}_{i} + {A}_{i} {{d}_{i}}^{2})} \text{ where:}\\ \begin{align} \bar{I}_{i} &= \text{The moment of inertia of the individual segment about its own centroid axis}\\ {A}_{i} &= \text{The area of the individual segment}\\ {d}_{i} &= \text{The vertical distance from the centroid of the segment to the Netrual Axis (NA)} \end{align} [math]

It is widely known that the moment of inertia equation of a rectangle about its centroid axis is simply:
[math] \bar{I}=\frac{1}{12}b{h}^{3} \text{ where:}\\\\ \begin{align} b &= \text{The base or width of the rectangle}\\ h &= \text{The height of the rectangle} \end{align} [math]

The moment of inertia of other shapes are often stated in the front/back of textbooks, however the rectangular shape is very common for beam sections.

Now we have all the information we need to use the "Parallel Axis Theorem" and find the total moment of inertia of the I-beam section:

[math] \text{Segment 1:}\\ \begin{align} \bar{I}_{1} &= \tfrac{1}{12}(250)(38)^{3} = 1,143,166.667 {\text{ mm}}^{4}\\ {A}_{1} &= 250\times38 = 9500 {\text{ mm}}^{2}\\ {d}_{1} &= \left|{y}_{1} - \bar{y} \right| = \left|(38 +300 +\tfrac{38}{2}) - 216.29\right| = 140.71 \text{ mm}\\\\ \end{align} [math] [math] \text{Segment 2:}\\ \begin{align} \bar{I}_{2} &= \tfrac{1}{12}(25)(300)^{3} = 56,250,000 {\text{ mm}}^{4}\\ {A}_{2} &= 300\times25 = 7500 {\text{ mm}}^{2}\\ {d}_{2} &= \left|{y}_{2} - \bar{y}\right| = \left|(38 +\tfrac{300}{2}) - 216.29\right| = 28.29 \text{ mm}\\\\ \end{align} [math] [math] \text{Segment 3:}\\ \begin{align} \bar{I}_{3} &= \tfrac{1}{12}(150)(38)^{3} = 685,900 {\text{ mm}}^{4}\\ {A}_{3} &= 150\times38 = 5700 {\text{ mm}}^{2}\\ {d}_{3} &= \left|{y}_{3} - \bar{y}\right| = \left|\tfrac{38}{2} - 216.29\right| = 197.29 \text{ mm}\\\\ \end{align} [math] [math] \begin{align} \therefore {I}_{total} &= \sum{(\bar{I}_{i} + {A}_{i} {{d}_{i}}^{2})} \\ &= (\bar{I}_{1} + {A}_{1}{{d}_{1}}^{2}) + (\bar{I}_{2} + {A}_{2}{{d}_{2}}^{2}) + (\bar{I}_{3} + {A}_{3}{{d}_{3}}^{2})\\ &= (1,143,166.667 + 9500\times140.71^{2}) + (56,250,000 + 7500\times28.29^{2}) + (685,900 + 5700\times197.29^{2})\\ &= 474,037,947.7 {\text{ mm}}^{4}\\ {I}_{total} &= 4.74 \times 10^{8} {\text{ mm}}^{4} \end{align} [math]

So in summary, you can see the results (calculated from our Free Moment of Inertia Calculator) along with some other section properties to check your work.

Of course you don't need to do all these calculations manually because you can use our fantastic Free Moment of Inertia Calculator to find the important properties and information about beam sections.

• Owain

hi, some guidance on how to calculate Ixx when you have rivets in the bottom flange would be great. Do you remove the X-sectional area of holes, determine NA and Ixx? or, do not remove holes and determine NA, then calculate Ixx with holes removed?

• Kabila

when is 1/12 factor used in calculating the moment of inertia? and why?

• side-fish

Hi Kabila, I hope I’m not too late. But the 1/12 coefficient from the moment of inertia of a rectangle is derived from calculus by integrating y^2dA. Your upper limit is the extreme fiber and the lower limit is the axis of rotation. In the case of a rectangle, you will integrate the top and the bottom of the rectangle. Note that dA=xdy. The new equation becomes I = ∫xy^2dy with the upper limit as h and the lower limit as h/2 if you want to integrate the top half (or h/2-0 if you want the bottom). Since the neutral axis of the rectangle is exactly in the middle, you simply integrate say… the top half and multiply twice. The coefficient that you should get is 1/12. If the axis of rotation is shifted to say the bottom fiber, integrating will yield a coefficient of 1/3, though you could use the transfer formula I=Io+Ad^2 and still get the same answer.

• Shaun Murrin

This is a tutorial question I’m struggling a bit with axis to base the i value on..
“Based on calculations it has been determined that a universal steel column must have a minimum value of I of 38748 cm4. Select a column suitable for this scenario showing clearly why you have selected that particular column. “

• Glenn

Used your calculator to find for simple ‘i’ beam. Whilst we both find identical Zy, my Zx is out compared to yours. I looked at your formula to find the centroid and it shows the addition of web thickness to the vertical centroid. Beam is: TFw & BFw = 100; TFt & BFt = 10; Wh = 80, Wt = 6. Following your tute on centroid for simple I beam, Cx would be 56, however your app shows (correctly in my mind) Cx and Cy both as 50. Yet my Ix = 4.566 x 106
mm4 ; calculator shows 4.323 x 106
mm4 (Iy is same as mine 1.668 x 106
mm4 ).
Been over calcs left and right and can’t find error. Unless Cx as 56 is culprit. Any ideas?

• Hi Glenn. As your section is symmetric about both horizontal and vertical and your section depth and width are both 100mm, then the centroid is definitely 50mm in both directions. Recalculate your Cx following the tutorial (see the image attached below).

• Cristian

There is an error in the last picture, it shows the subscripts of some answers for the “x” axis, while there is the axis “z”, maybe it will lead to some misunderstandings. Anyways nice tutorial.

• Thank you. This has been amended.


# Math Help - 2 Problems involving integrals

1. ## 2 Problems involving integrals

Suppose $f(0)=0$ and $\int_0^2{f'(2t)e^{f(2t)}}dt=5$. Find the value of f(4).

So I first recognized that $\frac{1}{2}e^{f(2t)}$ should be the antiderivative of the integral by FTC so that $\frac{1}{2}(e^{f(4)}-e^{f(0)})=5$.

Since f(0)=0, the equation becomes: $e^{f(4)}-1=10$

Isolating and then taking the natural log of both sides yields: f(4)=ln11
Is that right?

My other question is this: evaluate $\lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt$ It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?

The first thing I notice is that the variable in the integrand isn't with respect to x. If I rewrote it through FTC, I'd put $\lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}$? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?

2. Originally Posted by VectorRun
Suppose $f(0)=0$ and $\int_0^2{f'(2t)e^{f(2t)}}dt=5$. Find the value of f(4).

So I first recognized that $\frac{1}{2}e^{f(2t)}$ should be the antiderivative of the integral by FTC so that $\frac{1}{2}(e^{f(4)}-e^{f(0)})=5$.

Since f(0)=0, the equation becomes: $e^{f(4)}-1=10$

Isolating and then taking the natural log of both sides yields: f(4)=ln11
Is that right?
Yes, that is correct.

My other question is this: evaluate $\lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt$ It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?
Exactly right! That is $\lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$ where $f(x)= \int_a^x e^{arctan(t)}dt$. And, by the Fundamental Theorem of Calculus, $\frac{d}{dx}\int_a^x f(t)dt= f(x)$.

The first thing I notice is that the variable in the integrand isn't with respect to x.
The variable in the integrand is a "dummy variable"- it could be anything without changing the integral.

If I rewrote it through FTC, I'd put $\lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}$? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?
No, its not f that is being differentiated, it is the integral so the derivative of the integral is the integrand, f.

3. Hello, VectorRun!

Your first solution is correct . . . ecept for that $\frac{1}{2}$

$\displaystyle \text{Suppose }f(0)=0\:\text{ and }\:\int_0^2\!f'(2t)e^{f(2t)}}dt\:=\:5$

$\text{Find the value of }f(4).$

$\text{So I first recognized that }\frac{1}{2}e^{f(2t)}\text{ is the antiderivative of the integral.}$
Um ... not quite.

$\displaystyle \text{We have: }\:\int^2_0 u'\,e^u\,dt \;\;\text{ which equals: }\;e^u\,\bigg]^2_0$

$\text{Hence: }\;e^{f(2t)} \,\bigg]^2_0 \;=\;5 \quad \hdots \;\text{ etc.}
$


# How to calculate the nth term and the sum in this series if the common difference between them isn't explicit?

I have this series comprised of

$$1,2,5,10,17,26,...$$,

and so on. So far i have found that they add up in intervals being odd numbers. But I don't know how to find, let's say the 16th term and the sum up to that term. What should I do?

Edit:

Although it is somewhat possible to calculate the sum up to the 16th term in the above equation, what if the series is of higher order like this?.

$$6,7,14,27,46,...$$

Is there a shortcut to calculate the sum up to the 16th term? Can this be done by hand or would be necessary to use software like Maple?.

• So, we have $a_{n+1}-a_n=2n-1$ with $a_1=1$. Can you solve this recurrence relation? – Prasun Biswas Sep 30 '17 at 8:31

While Opt's answer is correct, it can take a good deal of practice to come up with such accurate observations on an unknown series. A very good and straightforward method in such a case is the method of differences

It formally states:

any term, of a given number series, can be expressed as a polynomial of degree $n$, if the $n$ differences of the series is constant.

Let me illustrate with your series.

$$1,2,5,10,17,26$$

The first differences are $$1(=2-1);3(=5-2); 5(=10-5);7(=17-10);9(=26-17)$$.

The second differences are $$2(=3-1); 2(=5-3); 2(=7-5);2(=9-7)$$ Note that all the second differences are constant! This implies, by the method of differences statement, that the $r$-th term in your series can be represented as a quadratic polynomial (degree 2) i.e. $T_r = ar^2+br+c$ for some constants $a,b,c$

Now, all you need to do is equate $T_1 =1; T_2=2; T_3=5$ and solve for $a,b,c$.

Added on request: To solve the above, you only need to do write them out after putting in $n=1,2,3$:

$$a+b+c=1...E1$$ $$4a+2b+c=2...E2$$ $$9a+3b+c=5...E3$$

You have three variables and three equations, thus they can be easily solved by subtraction and elimination. $E2-E1$ and $E3-E2$ respectively gives two new sets of equations:

$$3a+b=1...E4$$ $$5a+b=3...E5$$

and then $E5-E4$ straightaway gives $a=1$. Putting this in E5 gives $b=-2$. Putting them in $E1$ gives $c=2$.
Thus, our $T_r=r^2-2r+2=1+(r-1)^2$

For summation of terms from 1 to 16, you need to do:

$$\sum_{r=1}^{n}T_r=\sum_{r=1}^{n}(r^2-2r+2)$$

for which you may easily use the summation identities for $r^2$ and $r$.

Hope that helps!

• Sorry but my level of algebra isn't very proficient to solve the latter part of your explanation, mind if you extend a bit?. According to the method you proposed the polynomial is in terms of x but how does this translates as 'nth' term?. I do understand the nth is the degree but what's the x?. How do I solve for a, b and c, Can you show it to me please?. – Chris Steinbeck Bell Sep 30 '17 at 9:35
• For the general case (when the $n$-th difference is constant), it is simpler to use the binomial polynomials $\dbinom xi$ as a basis for the vector space of polynomials. – Bernard Sep 30 '17 at 9:37
• @Bernard Can you be a bit explicit on how do the binomial coefficient replaces constants in the above equation?. Am I understanding correctly?. Gaurang Tandon, There is also missing the last part of my question which involves the sum from 1 to the 16th term in the series, how can I do that?. – Chris Steinbeck Bell Sep 30 '17 at 9:46
• I'm not sure I was clear enough: I meant the canonical basis $1,x,x^2,\dots$ is replaced with the basis $\dbinom x0,\dbinom x1, \dbinom x2, \dots$. It's the natural basis in this situation because $\Delta\dbinom xk=\dbinom x{k-1}$. – Bernard Sep 30 '17 at 9:50
• @ChrisSteinbeckBell That confusion about $n$ and $x$ was my mistake. Sorry for that. Now, I've added extensive elaboration. Hope that helps! – Gaurang Tandon Sep 30 '17 at 9:56

As Guarang Tandon says. You can create a table of differences between consecutive terms and then repeat that process on the list you just created.

\begin{array}{|l|c|ccccccc|} \hline \text{index} & n & 0 & 1 & 2 & 3 & 4 & 5 & \dots\\ \hline \text{sequence} & f_n & 1 & 2 & 5 & 10 & 17 & 26 & \dots \\ \text{first differences} & \Delta f_n && 1 & 3 & 5 & 7 & 9 & \dots \\ \text{second differences} & \Delta^2 f_n &&& 2 & 2 & 2 & 2 & \dots \\ \hline \end{array}

If you are lucky, at some point, all of the differences will be constant. (Of course this is an unverifiable claim. All we know for sure is that the first four second differences are constant. We can only assume that this fortuitous pattern continues.)

It can be proved that, if $\Delta^d f_n$ is constant, then then $f_n$ is a $d$th degree polynomial in $n$.

If the second differences are constant, then the first differences must be an arithmetic sequence. In this case, we see that $\Delta f_n = 2n+1$. It follows that $$f_{n+1} = f_n + 2n+1 \tag 1$$.

If the first differences are an arithmetic sequence, then $f_n = an^2 + bn + c$ for some real numbers $a, b,$ and $c$.

We can now use modified form of mathematical induction to find the values of $a,b,$ and $c$.

Since $f_0=1$, then $1 = a\cdot0^2 + b\cdot 0 + c = c$. So $f_n=an^2+bn+1$.

First we note that $$f_{n+1} = a(n+1)^2 + b(n+1) + 1 = f_n + 2an + (a+b).$$ Comparing that to $f_{n+1}= f_n+ 2n+1$, we see that $2a=2$ and $a+b=1$. Hence $a=1$ and $b=0$.

We conclude that $f_n=n^2+1$.

We know that $\sum_{k=0}^n k^2=\frac 16n(n+1)(2n+1)$.

Then \begin{align} \sum_{k=0}^n f_k &= \sum_{k=0}^n (k^2+1) \\ &= \sum_{k=0}^n k^2 + \sum_{k=0}^n 1 \\ &= \frac 16n(n+1)(2n+1) + (n+1) \\ &= \frac 16(n+1)(2n^2+n) + \frac 16(n+1)6 \\ &= \frac 16(n+1)(2n^2+n+6) \\ \end{align}

Check:

$\qquad \sum_{k=0}^5 f_k = 1+2+5+10+17+26=61$

$\qquad \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=5}=\frac 16(6)(61)=61$

So, since we started counting at $n=0$, the $16$th term is $f_{15} = 15^2+1=226$

and the sum of the first $16$ terms is $\sum_{k=0}^{15} f_k = \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=15}=1256$.

• That's a nice table! :D – Gaurang Tandon Oct 1 '17 at 2:52

That's $$1 + 0 \\ 1 + 1 \\ 1 + 4 \\ 1 +9\\ 1 +16\\ 1 +25 \\ \vdots$$ Therefore, $$a_n = 1+(n-1)^2,$$

implying that $a_{16} = 1+(15)^2 = 226$.

As is said in other answers, the method of differences is certainly a way to go. However, in the simultaneous equations, you can obtain $a$ immediately:

If the $d^{th}$ row of the difference table is some constant $q$, the coefficient of the leading term of the polynomial is $\dfrac{q}{d!}$.

So, with the difference table as calculated in other answers, we have the second row is a constant 2. This gives us $a=\dfrac{2}{2!}=1$ without resorting to simultaneous equations.

Now, if you wanted to, you could use simultaneous equations to determine the other coefficients or you could subtract $n^2$ from each term:

$$1-0^2, 2-1^2, 5-2^2, 10-3^2, 17-4^2, 26-5^2,...\\ \text{which is }1,1,1,1,1,...$$

From which, the polynomial is $\fbox{n²+1}$


# Find the number of distinct real roots

#### anemone

##### MHB POTW Director
Staff member
Let $f(x)=x^3-3x+1$. Find the number of distinct real roots of the equation $f(f(x))=0$.

#### jacks

##### Well-known member

I am getting Total no. of real solution is $= 7$

Is is Right or not

Thanks

#### mathbalarka

##### Well-known member
MHB Math Helper
Quite a nice bit of a problem you've got there. Here's my solution --

Finding the real roots of $$\displaystyle f(f(x)) = 0$$ is equivalent to finding the ones of $$\displaystyle f(x) = x_i$$ where $$\displaystyle x_i$$ for i = 1, 2, 3 are the (real) roots of f(x).

We see that the roots of $$\displaystyle f(x)$$ are all real as the discriminant is positive. An inspection of $$\displaystyle f(x) = x^3 - 3x + 1$$ can be considered at the intervals $$\displaystyle [-2, -1] \cup [-1, 0] \cup [0, 1] \cup [1, 2]$$ to find out that three roots lie precisely in between the first, third and the fourth interval.

Note that for the equation $$\displaystyle f(x) = x_i$$ to have all real root, the discriminant must be positive, implying the $$\displaystyle -2 \leq 1 - x_i \leq 2$$. The first root lies between [-2, -1], so $$\displaystyle 1 - x_1$$ lies in between 2 and 3 and hence the discriminant comes down to a negative value, so one real root here exists and the other twos are complex conjugates of each other in $$\displaystyle \mathbb{C}$$.

For the second root, it lies between 0 and 1, so $$\displaystyle 1 - x_2$$ lies in between 2 and -2 obviously enough. The same applies for the third root. So we have 3 real roots each for these two.

So, the total number of real roots are 1 + 3 + 3 = 7.

Balarka
.

Last edited by a moderator:

#### anemone

##### MHB POTW Director
Staff member

I am getting Total no. of real solution is $= 7$

Is is Right or not

Thanks
Yes, 7 is the correct answer!

#### anemone

##### MHB POTW Director
Staff member
Finding the real roots of $$\displaystyle f(f(x)) = 0$$ is equivalent to finding the ones of $$\displaystyle f(x) = x_i$$ where $$\displaystyle x_i$$ for i = 1, 2, 3 are the (real) roots of f(x).

.

Hey Balarka,

I really like it you mentioned that to solve for $x$ in $$\displaystyle f(f(x)) = 0$$ is equivalent to solve for $$\displaystyle f(x) = x_i$$ where $$\displaystyle x_i$$ for i = 1, 2, 3 are the (real) roots of f(x).

And if I may use this idea to solve for the problem, I found out the total number of real roots for $$\displaystyle f(f(x)) = 0$$ are 7 as well, and this method is so much better than my first approach and thank you, Balarka for showing me something that I don't know!

 Solving for the number of real roots of $$\displaystyle f(x) = x_1=root_1$$ means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_1$, as shown by the green line and we see that in this case we have only one real root for $$\displaystyle f(x) = x_1=root_1$$. Solving for the number of real roots of $$\displaystyle f(x) = x_2=root_2$$ means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_2$, as shown by the purple line and we see that in this case we have only one real root for $$\displaystyle f(x) = x_2=root_2$$. Solving for the number of real roots of $$\displaystyle f(x) = x_3=root_3$$ means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_3$, as shown by the orange line and we see that in this case we have only one real root for $$\displaystyle f(x) = x_3=root_3$$.

So there are a total of 7 intersection points and hence there are 7 real roots for $$\displaystyle f(f(x)) = 0$$!

Thanks!

#### mathbalarka

##### Well-known member
MHB Math Helper
anemone said:
this method is so much better than my first approach
May I see your approach on this problem?

Balarka
.

#### anemone

##### MHB POTW Director
Staff member
May I see your approach on this problem? Balarka .
My first approach revolves around checking the nature of the curve of the function of $f(f(x))$ from its first and second derivative expression. If we let $f(x)=x^3-3x+1$ and $g(x)=f(f(x))=(f(x))^3-3(f(x))+1$, we find

 $f(x)=x^3-3x+1$ $g(x)=f(f(x))=(f(x))^3-3(f(x))+1$ $f'(x)=3x^2-3=3(x^2-1)=3(x+1)(x-1)$ $\rightarrow f'(x)=0$ iff $3(x+1)(x-1)=0$ or $x=\pm1$. $g'(x)=3f(x)^2f'(x)-3f'(x)=3f'(x)((f(x))^2-1)$ $\rightarrow g'(x)=0$ iff $x=\pm1$, $x=\pm \sqrt{3}$, or $x=-2$ or $x=0$. $f''(x)=6x$ $g''(x)=6f(x)(f'(x))^2+3f''(x)((f(x))^2-1)$

I then made another table to determine the nature of the critical points as follows:
 $x$ -2 $-\sqrt{3}$ -1 0 1 $\sqrt{3}$ $f(x)$ -1 1 3 1 -1 1 $f'(x)$ 9 6 0 -3 0 6 $f''(x)$ -12 $-6\sqrt{3}$ -6 0 6 $6\sqrt{3}$ $g(x)$ 3 -1 19 -1 3 -1 $g'(x)$ 0 0 0 0 0 0 $g''(x)$ -486 216 -144 54 0 216

And plot a rough sketch to determine the number of real roots of the function of $f(f(x))=0$:

From the graph we can tell there are 7 real roots of the function of $f(f(x))=0$.

#### mathbalarka

##### Well-known member
MHB Math Helper
Hmm... very nice. I like the way you analyzed the behavior of f(f(x)), very well done. Mine may have been a little time consuming but this one is more representative.

PS I particularly like this solution.

#### anemone

##### MHB POTW Director
Staff member
Hmm... very nice. I like the way you analyzed the behavior of f(f(x)), very well done. Mine may have been a little time consuming but this one is more representative.

PS I particularly like this solution.
Thank you for the compliment, Balarka!

And I must give full credit to Opalg, a very nice mentor here for giving me idea (from another question) to solve the problem!


# Integral of $\int_{-\infty}^{\infty} \left(\frac{1}{\alpha + ix} + \frac{1}{\alpha - ix}\right)^2 \, dx$

I'm having trouble integrating

$$\int_{-\infty}^{\infty} \left(\frac{1}{\alpha + ix} + \frac{1}{\alpha - ix}\right)^2 \, dx$$

where $\alpha$ is a real number and $i = \sqrt{-1}$. I'm guessing that I would have to integrate it using methods on complex analysis wherein I would use a semicircle contour enclosing a pole ($x = i\alpha$ for example) so that I could apply the residue theorem. But is there any other way to evaluate this integral (like an ingenious substitution or differentiating under the integral sign)?

• Is the integrand really typed correctly? Should this be, e.g., $\frac{1}{a - ix} + \frac{1}{a + ix}$? – Travis Willse Feb 5 '16 at 13:49
• @Travis Good call, thanks for pointing that out. – Aldon Feb 5 '16 at 13:53
• Have you tried combining the fractions? – user170231 Feb 5 '16 at 13:54
• @user170231 Yes, you'd get $\left(\frac{2 \alpha}{\alpha^2 + x^2}\right)^2$ which is another integral that's hard to integrate, which I'm thinking one that can be solved, again, by a contour integral. – Aldon Feb 5 '16 at 13:59

Observe that the second summand is conjugate of the first: $$\frac1{\alpha-ix}=\overline{\frac1{\alpha+ix}}$$ thus their sum will be real. In particular $$\frac1{\alpha+ix}+\frac1{\alpha-ix}=\frac{2\alpha}{\alpha^2+x^2}{}$$ thus the integrand will be $$\frac{4\alpha^2}{(\alpha^2+x^2)^2}=\frac{4}{\alpha^2}\frac{1}{(1+(x/\alpha)^2)^2}$$ The substitution $x=\alpha\tan y$ leads to (suppose wlog $\alpha>0$) \begin{align*} \int_{-\infty}^{+\infty}\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\,dx &=\frac{4}{\alpha^2}\int_{-\infty}^{+\infty}\frac{dx}{(1+(x/\alpha)^2)^2}\\ &=\frac{4}{\alpha^2}\int_{-\pi/2}^{+\pi/2}\frac{1}{(1+\tan^2y)^2}\frac{\alpha}{\cos^2y}\,dy\\ &=\frac{4}{\alpha}\int_{-\pi/2}^{+\pi/2}\frac{1}{\frac{1}{\cos^4y}}\frac{1}{\cos^2y}\,dy\\ &=\frac{4}{\alpha}\int_{-\pi/2}^{+\pi/2}{\cos^2y}\,dy\\ &=\frac{4}{\alpha}\frac12\left[\cos y\sin y +y\right]_{-\pi/2}^{+\pi/2}\\ &=\frac{2\pi}{\alpha} \end{align*}

• I have done the part where you'll combine the fractions to form $\frac{4 \alpha^2}{(\alpha^2 + x^2)^2}$ but I have not thought of factoring out $\alpha^2$ from the denominator. I will try that out thanks! – Aldon Feb 5 '16 at 14:02
• I'm beginning to see that this can be done using $x = \alpha \tan \theta$ but I keep on getting extremely weird answers. – Aldon Feb 5 '16 at 15:10
• Nothing weird Aldon! I completed the answer. I hope it could be clearer now! :-) – Joe Feb 6 '16 at 1:06
• I realized a few hours ago that I had wrong bounds for my integral when I converted it through $x = \alpha \tan \theta$. Thanks a lot though! – Aldon Feb 6 '16 at 1:14
• You're welcome! :-) – Joe Feb 6 '16 at 1:22

The integral is even in $\alpha$, so assume that $\alpha\gt0$.

Real Analysis Approach

Substitute $x\mapsto\alpha x$, \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=\int_{-\infty}^\infty\left(\frac{2\alpha}{\alpha^2+x^2}\right)^2\mathrm{d}x\\ &=\frac4\alpha\int_{-\infty}^\infty\frac1{(1+x^2)^2}\mathrm{d}x\tag{1} \end{align} Furthermore \begin{align} \pi &=\int_{-\infty}^\infty\frac1{1+x^2}\,\mathrm{d}x\tag{2}\\ &=\int_{-\infty}^\infty\frac{2x^2}{(1+x^2)^2}\,\mathrm{d}x\tag{3}\\ &=\int_{-\infty}^\infty\frac{1+x^2}{(1+x^2)^2}\,\mathrm{d}x\tag{4}\\ &=\int_{-\infty}^\infty\frac2{(1+x^2)^2}\,\mathrm{d}x\tag{5}\\ \end{align} Explanation:
$(2)$: arctan integral
$(3)$: integration by parts on $(2)$
$(4)$: multiply the integrand in $(2)$ by $\frac{1+x^2}{1+x^2}$
$(5)$: $2$ times $(4)$ minus $(3)$

Plug $(5)$ into $(1)$ to get \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=\frac4\alpha\frac\pi2\\[3pt] &=\frac{2\pi}{\alpha}\tag{6} \end{align}

Complex Analysis Approach

Let $z=ix$ and $\gamma=[-iR,iR]\cup iRe^{i[0,\pi]}$ as $R\to\infty$.

Then \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=-i\int_\gamma\left(\frac1{\alpha+z}+\frac1{\alpha-z}\right)^2\mathrm{d}z\tag{7}\\ &=-i\int_\gamma\left(\frac1{(\alpha+z)^2}+\frac1{(\alpha-z)^2}\right)\mathrm{d}z\\ &-\frac i\alpha\int_\gamma\left(\color{#C00000}{\frac1{\alpha+z}}+\frac1{\alpha-z}\right)\mathrm{d}z\tag{8}\\ &=\frac{2\pi}{\alpha}\tag{9} \end{align} Explanation:
$(7)$: the integral along the arc where $|z|=R$ is $\lesssim\frac{4\pi}R$, which vanishes as $R\to\infty$.
$(8)$: $\left(\frac1{\alpha+z}+\frac1{\alpha-z}\right)^2 =\frac1{(\alpha+z)^2}+\frac1{(\alpha-z)^2}+\frac2{(\alpha+z)(\alpha-z)} =\frac1{(\alpha+z)^2}+\frac1{(\alpha-z)^2}+\frac1\alpha\left(\frac1{\alpha+z}+\frac1{\alpha-z}\right)$
$(9)$: only the $\color{#C00000}{\frac1{\alpha+z}}$ term has a pole with non-zero residue inside $\gamma$

• Why is $\left(\frac{1}{\alpha +z} + \frac{1}{\alpha - z}\right)^2 = \frac{1}{(\alpha + z)^2} + \frac{1}{(\alpha - z)^2}$? – Aldon Feb 6 '16 at 1:07
• you could also use the Fourier transform (harmonic analysis) approach. if $\alpha < 0$ : $$\frac{1}{\alpha+ix} + \frac{1}{\alpha-ix} = 2 \pi \int_{-\infty}^\infty e^{-2 i \pi x t} e^{2 \pi \alpha |t|} dt$$ so by the Parseval identity : $$\int_{-\infty}^\infty \left| \frac{1}{\alpha+ix} + \frac{1}{\alpha-ix} \right |^2 dx = \int_{-\infty}^\infty |2 \pi e^{2 \pi \alpha |t|}|^2 dt = 4 \pi^2 \int_{-\infty}^\infty e^{4 \pi \alpha |t|} dt = 2\frac{4 \pi^2 }{4 \pi |\alpha|} = \frac{2 \pi }{|\alpha|}$$ – reuns Feb 6 '16 at 1:45
• and if $\alpha > 0$ : $\frac{1}{\alpha+ix} + \frac{1}{\alpha-ix} = - \left(\frac{1}{-\alpha-ix} + \frac{1}{-\alpha+ix} \right)$ so the result is still $\frac{2 \pi }{|\alpha|}$ – reuns Feb 6 '16 at 1:45
• @user1952009: indeed, there are many ways to compute the integral. – robjohn Feb 6 '16 at 7:03
• @Aldon: I have expanded the explanation of the Complex Analysis Approach. Note the explanation of $(8)$. – robjohn Feb 6 '16 at 7:39


These are interactive guides that focus on … [TAGALOG] Grade 9 Math Lesson: HOW TO SOLVE QUADRATIC INEQUALITY? Type 1: Algebraically solving x^2\lt a^2. Each STEM Case level has an associated Handbook. Example: Solve the inequality x^2 \lt 64 When solving quadratic inequalities it is important to remember there are two roots. 6. 9th - 11th grade. Edit. The range of values that satisfy the inequality is between -8 and 8.This can be expressed as, Definition Is an inequality that contains a polynomial of degree 2 and can be written in any of the following forms. Since the original inequality was less than or equal to, the boundary points are included. The above is an equation (=) but sometimes we need to solve inequalities like these: It then becomes a negative 9 plus 3, it becomes negative 3, and negative 3 will also become a negative 3. Visual intuition of what a quadratic inequality means. Solving quadratic equations by factoring. An inequality can therefore be solved graphically using a graph or algebraically using a table of signs to determine where the function is positive and negative. Quadratic Inequalities It is an inequality of the form of ax2 + bx + c (<, >, ≤, ≥) 0, where a, b and c are real numbers and a ≠ 0 3. Save. More Lessons for Grade 9 Math Math Worksheets Videos, worksheets, games and activities to help Algebra and Grade 9 students learn how to solve quadratic inequalities. Supplemental Math High School Grade 9 Q1.indd Created Date: So the width must be between 1 m and 7 m (inclusive) and the length is 8−width. Quadratic inequalities can be of the following forms: \begin{align*} ax^2 + bx + c > 0 \\ ax^2 + bx + c \ge 0 \\ ax^2 + bx + c < 0 \\ ax^2 + bx + c \le 0\end{align*} To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the $$x$$-axis. jamespearce. Thanks to all of you who support me on Patreon. But I know (and can verify from the above graph) that this quadratic only touches the axis from below; it is never fully above the axis. Multiple grade appropriate versions, or levels, exist for each STEM Case. 1. 0. Step by step guide to solve Solving Quadratic Inequalities . To solve a quadratic inequality, find the roots of its corresponding equality. $1 per month helps!! Graphing Quadratic inequalities – Example 1: Sketch the graph of $$y>3x^2$$. Some of the worksheets for this concept are Quadratic inequalities date period, Solving quadratic inequalities, Solving quadratic inequalities 1, Solving quadratic inequalities l3s1, Graphing quadratic, Graphing and solving quadratic inequalities, Solving quadratic inequalities algebraically work, Inequalitiesmep pupil text 16.$ < – 3 , – 2 > \cup < 1 , 2 ]\$ Quadratic inequalities worksheets Quadratic inequalities (4.4 MiB, 620 hits) 0 times. If the question was solve x^2=a^2 we would simple take the square root of both sides so that x=\pm a.. quadratic inequalities, conduct a spin-o˛ Think-Pair-Share activity (Lyman, 1981). • Diagrams are NOT accurately drawn, unless otherwise indicated. Choose a testing point and check the solution section. :) https://www.patreon.com/patrickjmt !! Nature of the roots of a quadratic equations. YouMore Kwenturuan tungkol sa kung paano gawin ang solving quadratic … This entry was posted in MATH and tagged math , math solver , mathway . OBJECTIVES : •Define the interval correctly •Express the Quadratic Inequalities •Solve the Quadratic Inequalities through graphs and interval notation 3. Mathematics 9 Lesson 2: Quadratic Inequalities 1. Edit. These are imaginary answers and cannot be graphed on a real number line. Quadratic Inequalities Example: x2 + x – 6 = … We need to find solutions. Yes we have two inequalities, because 3 2 = 9 AND (−3) 2 = 9. Before we get to quadratic inequalities, let's just start graphing some functions and interpret them and then we'll slowly move to the inequalities. Worked example 16: Solving quadratic inequalities • Answer the questions in the spaces provided – there may be more space than you need. Use the zero product property to find the solutions to the equation x2 - 9 = 16. x = -5 or x = 5 Adam is using the equation (x)(x + 2) = 255 to find two consecutive odd integers with a product of 255. In this case, we have drawn the graph of inequality using a pink color. Then fill in the region either inside or outside of it, depending on the inequality. Sum and product of the roots of a quadratic equations Algebraic identities. Therefore, the inequality x 2 + 2 x + 5 < 0 has no real solutions. An inequality can therefore be solved graphically using a graph or algebraically using a table of signs to determine where the function is positive and negative. Solving quadratic equations by quadratic formula. Solving quadratic inequalities, quadratic equations and quadratic simultaneous equations. Solving quadratic equations by completing square. By using this website, you agree to our Cookie Policy. The quadratic formula provides an easy and fast way to solve quadratic equations. Welcome to the presentation on quadratic inequalities. To graph a quadratic inequality, start by graphing the quadratic parabola. Solving absolute value equations Solving Absolute value inequalities However, this inequality is an "or equal to" inequality, so the "equal" part counts as part of the solution. GCSE (1 – 9) Quadratic Inequalities Name: _____ Instructions • Use black ink or ball-point pen. ... 2014/11/03 آ 3.1-Quadratic Functions & Inequalities Quadratic. MHR • Pre-Calculus 11 Solutions Chapter 9 Page 1 of 84 Chapter 9 Linear and Quadratic Inequalities Section 9.1 Linear Inequalities in Two Variables Section 9.1 Page 472 Question 1 a) y < x + 3 Try (7, 10). You da real mvps! To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the $$x$$-axis. 0% average accuracy. The method of completing the square provides a way to derive a formula that can be used to solve any quadratic equation. I need to find where y = –x 2 + 6x – 9 is above the axis. And that represents the graph of the inequality. an hour ago. ax2 + bx + c > 0 ax2 + bx + c ≥ 0 ax2 + bx + c < 0 ax2 + bx + c ≤ 0 where a, b, and c are real numbers and a ≠ 0. Let's say I had f of x is equal to x squared plus x minus 6. Transcript of Grade 9: Mathematics Unit 2 Quadratic Functions. This is a quadratic inequality. GCSE Grade 9 Quadratic Inequalities, Quadratic Equations and Quadratic Simultaneous Equations. A Quadratic Equation in Standard Form ( a , b , and c can have any value, except that a can't be 0.) Since this statement is false, the region between -9 and 1 is not correct. Siyavula's open Mathematics Grade 10 textbook, chapter 4 on Equations and inequalities covering Solving quadratic equations All answers are included. The easiest way to solve these inequalities is to draw the graphs and read the solutions of them. The same basic concepts apply to quadratic inequalities like $$y x^2 -1$$ from digram 8. Quadratic Inequalities (Visual Explanation) How to solve a quadratic inequality. Quadratic Inequalities 2. Mathematics. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. $$.. The set of solutions is the union of solutions of these two systems. This is the same quadratic equation, but the inequality has been changed to$$ \red . Solving Quadratic Inequalities DRAFT. Displaying top 8 worksheets found for - Quadratic Inequalities. Grade mathematics: Quadratic Inequalities 1. Try (–7, 10). MathematicsLearners Material9Module 2:Quadratic FunctionsThis instructional material was collaboratively developed and reviewed byeducators from public and private schools, colleges, and/or universities. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. A quadratic inequality is one that can be written in one of the following standard forms: $$ax^2+bx+c>0, ax^2+bx+c<0, ax^2+bx+c≥0, ax^2+bx+c≤0$$ Solving a quadratic inequality is like solving equations. To solve a quadratic inequality we must determine which part of the graph of a quadratic function lies above or below the $$x$$-axis. Left Side Right Side Left Side Right Side y x + 3 y x + 3 = 10 = 7 + 3 = 10 = –7 + 3 = 10 = –4 • You … Check: Consider the standard form of the quadratic equation \(ax^2 + bx + … It can be used in conjunction with Module 1 Lesson 7 of the DepEd Grade 9 Learning Modules. In this non-linear system, users are free to take whatever path through the material best serves their needs. Since this quadratic is not factorable using rational numbers, the quadratic formula will be used to solve it. Free quadratic inequality calculator - solve quadratic inequalities step-by-step This website uses cookies to ensure you get the best experience. Grade 9 - Mathematics Topic 1.8 : Solving Quadratic Inequalities 2. by jamespearce. Negative 3 squared is positive 9, you have a negative out front, it becomes negative 9 plus 6, which is negative 3. an hour ago. Grade 9 – Algebra Solving Quadratic Inequalities – This activity provides practice in using the graph of a quadratic function to solve quadratic inequalities. DLM 2 – Unit 2: Quadratic equations Summative Test for Quadratic Equations EASE Module 3 Quadratic Functions EASE Module 2 Quadratic Equations Deriving formula and solving for surface areas of solids GRADE 10 – RATIONAL ROOT THEOREM MATH 9 Q1M2.2 - Solving Quadratic Equations MATH 9 Q1M1.1 - Quadratic Equations DIG DOWN MY ROOTS 1 The software Geogebra is used as graphing tool. • Answer all questions. Worked example 16: Solving quadratic inequalities These unique features make Virtual Nerd a viable alternative to private tutoring. Add 4 to both sides of each inequality: 1 ≤ W ≤ 7. So it must be the region on either side of those points. If I were to graph it, Let's try it with 3, as well-- if we put a 3 there, 3 squared is 9. Solve the quadratic inequality: x2+11x+10≥0x^2+11x+10\ge0x2+11x+10≥0 Solving Quadratic Inequalities DRAFT.


GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Dec 2018, 15:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
• ### GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST

December 14, 2018

December 14, 2018

09:00 AM PST

10:00 AM PST

10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.

# Which of the following is/are terminating decimal(s)?

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 22 Jul 2014
Posts: 123
Concentration: General Management, Finance
GMAT 1: 670 Q48 V34
WE: Engineering (Energy and Utilities)
Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

06 Aug 2014, 09:08
1
3
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:10) correct 33% (01:26) wrong based on 193 sessions

### HideShow timer Statistics

Which of the following is/are terminating decimal(s)?

I 299/(32^123)
II 189/(49^99)
III 127/(25^37)

A) I only
B) I and III
C) II and III
D) II only
E) I, II, III
Tutor
Joined: 20 Apr 2012
Posts: 99
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

06 Aug 2014, 10:24
10
6
Terminating decimal is a such decimal that has only a finite number of non-zero digits. That's why if you will write such decimal as a fraction you will have as a denominator 10,100, 1000, etc. Since the fraction can be simplified, you can have in the denominator at the end some product of 2 or 5.

For example:
$$0.4=4/10=2/5$$, 5 in the denominator
$$0.25=25/100=1/4$$, 4=2*2 in the denominator.

But anyway, if you have terminating decimal you can't have anything other 2 or 5 in prime factorization of denominator. So we have a rule:

The fraction expressed as a decimal will be the terminating decimal if it can be presented as $$\frac{a}{{2^n\cdot 5^m}}$$ where $$a$$ is an integer, $$m=0,1,2,3..$$. , and $$n=0,1,2,3...$$ .

Or

if a fraction has in denominator any prime factor different from 2 and 5, such fraction will be infinite decimal.

I. Has only 2 as prime factor, since $$32=2^5$$. Terminating
II. Has 7 as prime factor. Infinite
III. Has only 5 as prime factor. Terminating

Hope this helps!:)
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

##### General Discussion
Manager
Joined: 18 Jul 2013
Posts: 73
Location: Italy
GMAT 1: 600 Q42 V31
GMAT 2: 700 Q48 V38
GPA: 3.75
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

06 Aug 2014, 13:57
smyarga wrote:
Terminating decimal is a such decimal that has only a finite number of non-zero digits. That's why if you will write such decimal as a fraction you will have as a denominator 10,100, 1000, etc. Since the fraction can be simplified, you can have in the denominator at the end some product of 2 or 5.

For example:
$$0.4=4/10=2/5$$, 5 in the denominator
$$0.25=25/100=1/4$$, 4=2*2 in the denominator.

But anyway, if you have terminating decimal you can't have anything other 2 or 5 in prime factorization of denominator. So we have a rule:

The fraction expressed as a decimal will be the terminating decimal if it can be presented as $$\frac{a}{{2^n\cdot 5^m}}$$ where $$a$$ is an integer, $$m=0,1,2,3..$$. , and $$n=0,1,2,3...$$ .

Or

if a fraction has in denominator any prime factor different from 2 and 5, such fraction will be infinite decimal.

I. Has only 2 as prime factor, since $$32=2^5$$. Terminating
II. Has 7 as prime factor. Infinite
III. Has only 5 as prime factor. Terminating

Hope this helps!:)

Great explanation, i forgot the concept of terminating decimals.
Tutor
Joined: 20 Apr 2012
Posts: 99
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

06 Aug 2014, 14:08
3
Current Student
Joined: 22 Jul 2014
Posts: 123
Concentration: General Management, Finance
GMAT 1: 670 Q48 V34
WE: Engineering (Energy and Utilities)
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

06 Aug 2014, 23:36
Tutor
Joined: 20 Apr 2012
Posts: 99
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

07 Aug 2014, 00:41
1
alphonsa wrote:

I like very much this source 700-800-level-quant-problem-collection-detailed-solutions-137388.html
You can find there problems divided by topics with detailed explanation.
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

07 Aug 2014, 18:46
1
alphonsa wrote:
Which of the following is/are terminating decimal(s)?

I 299/(32^123)
II 189/(49^99)
III 127/(25^37)

A) I only
B) I and III
C) II and III
D) II only
E) I, II, III

I

$$\frac{299}{32^{123}} = \frac{299}{(2^5)^{123}} = \frac{299}{2^{(5*123)}}$$

$$= \frac{299}{2^{(5*123)}} * \frac{5^{(5*123)}}{5^{(5*123)}}$$

$$= \frac{299 * 5^{(5*123)}}{10^{(5*123)}}$$ >> Power of 10 in denominator, this is a terminating decimal

II

$$\frac{189}{49^{99}} = \frac{7*27}{7^{198}} = \frac{27}{7^{197}}$$ >> This is not a terminating decimal

III

$$\frac{127}{25^{37}} = \frac{127}{5^{74}} = \frac{127}{5^{74}} * \frac{2^{74}}{2^{74}}$$

$$= \frac{127 * 2^{74}}{10^{74}}$$ >> Power of 10 in denominator, this is a terminating decimal

_________________

Kindly press "+1 Kudos" to appreciate

Manager
Joined: 22 Feb 2009
Posts: 171
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

07 Aug 2014, 21:16
PareshGmat wrote:
alphonsa wrote:
Which of the following is/are terminating decimal(s)?

I 299/(32^123)
II 189/(49^99)
III 127/(25^37)

A) I only
B) I and III
C) II and III
D) II only
E) I, II, III

I

$$\frac{299}{32^{123}} = \frac{299}{(2^5)^{123}} = \frac{299}{2^{(5*123)}}$$

$$= \frac{299}{2^{(5*123)}} * \frac{5^{(5*123)}}{5^{(5*123)}}$$

$$= \frac{299 * 5^{(5*123)}}{10^{(5*123)}}$$ >> Power of 10 in denominator, this is a terminating decimal

II

$$\frac{189}{49^{99}} = \frac{7*27}{7^{198}} = \frac{27}{7^{197}}$$ >> This is not a terminating decimal

III

$$\frac{127}{25^{37}} = \frac{127}{5^{74}} = \frac{127}{5^{74}} * \frac{2^{74}}{2^{74}}$$

$$= \frac{127 * 2^{74}}{10^{74}}$$ >> Power of 10 in denominator, this is a terminating decimal

I could solve the problem by using rules of terminating number, but your solution is so interesting. Thanks for sharing!!! + 1 kudos
_________________

.........................................................................
+1 Kudos please, if you like my post

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1825
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

07 Aug 2014, 21:44
PareshGmat wrote:
alphonsa wrote:
Which of the following is/are terminating decimal(s)?

I 299/(32^123)
II 189/(49^99)
III 127/(25^37)

A) I only
B) I and III
C) II and III
D) II only
E) I, II, III

I

$$\frac{299}{32^{123}} = \frac{299}{(2^5)^{123}} = \frac{299}{2^{(5*123)}}$$

$$= \frac{299}{2^{(5*123)}} * \frac{5^{(5*123)}}{5^{(5*123)}}$$

$$= \frac{299 * 5^{(5*123)}}{10^{(5*123)}}$$ >> Power of 10 in denominator, this is a terminating decimal

II

$$\frac{189}{49^{99}} = \frac{7*27}{7^{198}} = \frac{27}{7^{197}}$$ >> This is not a terminating decimal

III

$$\frac{127}{25^{37}} = \frac{127}{5^{74}} = \frac{127}{5^{74}} * \frac{2^{74}}{2^{74}}$$

$$= \frac{127 * 2^{74}}{10^{74}}$$ >> Power of 10 in denominator, this is a terminating decimal

I could solve the problem by using rules of terminating number, but your solution is so interesting. Thanks for sharing!!! + 1 kudos

Thank you so much for calling the solution interesting

One thing to add. In such type of problems, just look out for powers of 2 and/or powers of 5 (because they only compose 10) in denominator

For any other number, its not possible (Subject to complete simplification of the term)
_________________

Kindly press "+1 Kudos" to appreciate

Intern
Joined: 12 Feb 2013
Posts: 7
Location: India
Concentration: Finance, Marketing
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

07 Aug 2014, 23:11
The fraction will have terminating decimal if and only if the denominator of the fraction is of the form (2^n)(5^m).

If you look at the denominators:
1) (32^123) => (2^(5*123))(5^0) => Terminating Decimal
2) (49^99) => Can't be expressed as (2^n)(5^m) => Non terminating Decimal
3) (25^37) => (2^0)(5^(2*37)) => Terminating Decimal

Hence 1 and 3 are terminating decimals => Choice [B]
Non-Human User
Joined: 09 Sep 2013
Posts: 9138
Re: Which of the following is/are terminating decimal(s)?  [#permalink]

### Show Tags

11 Nov 2017, 11:17
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Which of the following is/are terminating decimal(s)? &nbs [#permalink] 11 Nov 2017, 11:17
Display posts from previous: Sort by


# Painted faces on a cube

Here's another challenge I used to give to my students:

Let's begin with a bunch of little white cubes assembled into a big white cube. All the little white cubes are equal.
Then I decide to paint some of the faces of the big cube blue. Afterwards I break the big cube apart into the smaller ones.
Only $$24$$ of the smaller cubes remain completely white.

How many cubes formed the big one? How many big faces did I paint?

Usually I use this problem to show how much we can deduce with so little information.

It hinges on the following insight:

Don't break the cube apart but keep the small cubes packed together. If you paint some faces of the cube, and remove the painted layers, you have a cuboid where each dimension is at zero, one or two units shorter than the original cube.

That means that the $$24$$ unpainted cubes

form a three-dimensional cuboid with dimensions that differ by at most two. The only way to factor $$24$$ like that is as $$24=2\times3\times4$$. Therefore you started with a $$4\times4\times4$$ cube and removed three layers, two of them opposite each other.

• That is a GREAT approach! If I could I give you $10$ upvotes! :)
– Pspl
Jul 10 '20 at 8:10

In general, you can paint 6 sides of an N×N×N cube in 10 ways:

* 0 sides
* 1 side
* 2 opposite sides
* 3 adjacent sides (a corner)
* 3 sides, two of them opposite (a band)
* 4 sides except 2 opposite
* 4 sides except 2 adjacent
* 5 sides
* 6 sides

which leaves you with these numbers of unpainted sub-cubes, respectively:

* N×N×N
* N×N×(N–1)
* N×(N–1)×(N–1)
* N×N×(N–2)
* (N–1)×(N–1)×(N–1)
* N×(N–1)×(N–2)
* N×(N–2)×(N–2)
* (N–1)×(N–1)×(N–2)
* (N–1)×(N–2)×(N–2)
* (N–2)×(N–2)×(N–2)

Given a number of unpainted cubes,

one needs to find a factorisation that fits one of those expressions.

As Jaap Scherphuis shows, the value of 24 fits just one, which yields only 1 answer.

• Note that all but three of those expressions are definitely ambiguous. Jul 10 '20 at 9:11
• @JaapScherphuis Yes. That means solution needn't be unique. If one takes a cube 6×6×6 and removes one layer from all faces, or one removes one layer from three adjacent faces of a 5×5×5 cube, or one leaves a 4×4×4 cube intact, the number of unit cubes left will be the same. Jul 10 '20 at 17:25

My approach was to...

start off by bounding the possible dimensions of the big cube, and then find a valid permutation of face-painting from that subset. But after bounding the size, finding a permutation became unnecessary.

To determine...

a maximum bound: We can be certain that the inner cube of dimensions $$(N-2)$$ will remain untouched no matter how the big cube is painted, so that must hold an equal or fewer number of small cubes than how many are left untouched. $$(N-2)^3 \leq 24$$. Since $$3^3$$ is already $$27$$,

$$N<5$$ is a maximum bound.

Then to determine...

a minimum bound: If at least one face of the big cube is painted, then the largest number of untouched unit cubes possible is the total number of cubes, minus one face of cubes. $$N^3-N^2 \geq 24$$. Since $$3^3-3^2$$ is only $$18$$,

$$N>3$$ is a minimum bound.

Therefore...

$$N$$ must be greater than $$3$$ and less than $$5$$, so $$4$$ is the only possible answer. It is not necessary to determine exactly what pattern of sides were painted to be certain of the answer.

With a 4x4x4 cube, there are 56 outer cubes; 40/16 paint/blank. The most cubes you can paint with a face is 16, so you need more than 2 faces. meanwhile 4 faces painted leaves only 8 or 10 cubes blank. So 3 faces must be painted. Finding the permutation is still unnecessary for the total answer. But to be specific, 16 for one face, +12 for each single-adjacent space. So 3 faces painted in a U-shape on a 4x4x4 cube is the full description of the cube.

• I too started with finding the bounds. +1. But the second half of the question is "How many big faces did I paint?" You do need the pattern (subject to rotation&reflection) to determine that. Jul 12 '20 at 19:50
• True - I could have elaborated on the final step, but everyone else had already given accurate answers - finding the number of painted sides is trivial once the size of the cube is known. I was just trying to emphasize that this piece of information could be gathered through simple, broad deduction without having to guess and check through permutations. Jul 16 '20 at 19:09

You can represent the unpainted volume like this:

$$(n-x_1-x_2)(n-y_1-y_2)(n-z_1-z_2) = 24$$
where $$x_1,x_2,y_1,...$$ are either $$0$$ or $$1$$, representing whether the face was painted.

This means all we need to do is

Factorise $$24$$ into 3 factors such that any two factors are at most $$2$$ apart

We know that

The prime factorisation of $$24$$ is $$2\cdot 2\cdot 2\cdot 3$$

Which means you have the following combinations:

• $$2\cdot 3\cdot 4$$
• $$2\cdot 2\cdot 6$$

Only one of which works, meaning that:

$$(n-x_1-x_2)(n-y_1-y_2)(n-z_1-z_2)=2\cdot 3\cdot 4\\\therefore(n-1-1)(n-1-0)(n-0-0)=2\cdot 3\cdot 4$$
Since $$x_1+x_2+y_1+y_2+z_1+z_2$$ sides are painted, we know that $$3$$ sides were painted.
Since $$(n-0-0)=4$$, we know that the original cube side length must have been $$4$$.

I haven't looked at the other answers so I hope I haven't done anything too similar to anyone else :)

• Having looked at the top answer, I realise this is essentially just a longer version of that, but I think it adds a little bit more explanation so I'll keep it up :) Jul 11 '20 at 16:26

Here is my way of looking at it:

I started off by representing situations with my favorite variable ‘x’ (I’m more of a math kind of person). For example, let x be the length of a side of the larger cube, so $$x^3-x^2=24$$ represents a situation where 1 face is painted blue, and those blue squares are removed ($$x^3$$ is the whole cube volume, so subtracting $$x^2$$ gives you the leftover volume after removing one face).

Then we move on to 2 blue faces...

$$x^3-2x^2=24$$ represents painting 2 blue faces that are opposite each other so the faces aren’t touching. And $$x^3-2x^2+x=24$$ represents 2 blue faces touching each other. The ‘plus x’ accounts for the fact that counting $$x^2$$ twice when the sides are touching means you are counting one edge two times; $$2x^2$$ tells you how many blue faces there are for the small cubes, so adding x will tell you how many cubes have any blue.

And so on... keeping in mind how many touching faces there are, we find that:

$$x^3-3x^2+2x=24$$ is the only equation out of all possible numbers of blue faces where a solution for x is a whole number ($$x=4$$). Remember that x represents the length of a side of the cube, so if the above equation is the only one with a whole number x, it’s the only one with a whole number side length and the only solution that works. The equation $$x^3-3x+2x=24$$ represents 3 blue sides, with 2 over laps (see coefficients). So since $$x=4$$, this means the whole cube has volume $$4^3$$ since x is AGAIN, the length of a side of the big cube.

Therefore...

You used 64 smaller cubes and painted 3 faces so that there are 2 edges where the faces touch (as in 2 of the faces are opposite each other).


# Totally ramified extension of $\mathbb{Q}_{p}$ which is not of a form $\mathbb{Q}_{p}(\sqrt[n]{pu})$

It is known that a finite extension $$K/\mathbb{Q}_{p}$$ is totally ramified if and only if $$K = \mathbb{Q}_{p}(\alpha)$$ where $$\alpha$$ is a root of Eisenstein polynomial. Is there any totally ramified extension that is not of the form $$\mathbb{Q}_{p}(\sqrt[n]{pu})$$ for some $$u\in \mathbb{Z}_{p}^{\times}$$? Every degree 2 totally ramified extensions has this form, but I don't know whether this is also true for degree 3 or higher. Thanks in advance.

• Lately I’ve been convincing myself of things that were not true, but I have convinced myself just now that $\Bbb Q_2(\sqrt3\,)$ is not of the form you claim for quadratic extensions. Could you check? Apr 21, 2019 at 2:42
• Yes, this is a phenomenon of wild ramification, when the residue characteristic divides the degree of the extension. For instance, I believe that $X^3+3X+3$ over $\Bbb Q_3$ gives another example. It’s a matter of the Herbrand transition function, or if you like, the location of the “breaks” in the ramification filtration. Apr 21, 2019 at 5:10
• @Lubin If you find the time to flesh that out in an answer, I'm sure I won't be the only user who will appreciate it :-) Apr 21, 2019 at 7:12
• I’ll see what I can do, @JyrkiLahtonen. Got lots on my plate the next 48 hours. Apr 21, 2019 at 18:10
• Thanks, @Lubin. We are not in a hurry here. Apr 21, 2019 at 18:11

There is a general theorem that every tamely totally ramified extension of $$\mathbf Q_p$$ with degree $$n$$ has the form $$\mathbf Q_p(\sqrt[n]{\pi})$$ for some prime $$\pi$$ in $$\mathbf Z_p$$, so $$\pi = pu$$ for a unit $$u$$ in $$\mathbf Z_p$$. (There is a similar theorem over other local fields.) So if you want a totally ramified extension not of that form you need $$n$$ to be divisible by $$p$$.

Let's try $$n=p$$. Something we can say about extensions $$\mathbf Q_p(\sqrt[p]{pu})$$ for $$p>2$$ is that they are not Galois over $$\mathbf Q_p$$: a field $$K$$ containing a full set of roots of $$x^p - pu$$ must contain the nontrivial $$p$$th roots of unity, and those have degree $$p-1$$ over $$\mathbf Q_p$$ so $$[K:\mathbf Q_p]$$ is divisible by $$p-1$$. Therefore $$[K:\mathbf Q_p] \not= p$$ when $$p>2$$. Thus a Galois totally ramified extension of $$\mathbf Q_p$$ having degree $$p$$ can't have the form $$\mathbf Q_p(\sqrt[p]{pu})$$.

Every totally ramified abelian Galois extension of $$\mathbf Q_p$$ with degree divisible by $$p$$ contains a subextension with degree $$p$$ since the Galois group has a subgroup of index $$p$$: in an abelian group of order $$n$$ there is a subgroup of each order dividing $$n$$ and thus also a subgroup of each index dividing $$n$$ by using a subgroup of order equal to the complementary factor in $$n$$ of the desired index. Subextensions of totally ramified extensions are totally ramified and subextensions of abelian Galois extensions are abelian Galois extensions. Thus all we need to do now is find a totally ramified abelian Galois extension of $$\mathbf Q_p$$ with degree divisible by $$p$$ and inside of it there are extensions of degree $$p$$, all of which are examples of the kind being sought (not having the form $$\mathbf Q_p(\sqrt[n]{pu})$$).

The easiest choice is a cyclotomic extension: $$\mathbf Q_p(\zeta_{p^2})$$ where $$\zeta_{p^2}$$ is a root of unity of order $$p^2$$. This field has degree $$p^2-p$$ over $$\mathbf Q_p$$, with cyclic Galois group $$(\mathbf Z/p^2\mathbf Z)^\times$$, so the field contains a unique subextension with degree $$p$$ over $$\mathbf Q_p$$, namely the field fixed by the unique subgroup of the Galois group with order $$(p^2-p)/p = p-1$$. That subgroup is the solutions to $$a^{p-1} \equiv 1 \bmod p^2$$, and a generator of this extension over $$\mathbf Q_p$$ is $$\sum_{a^{p-1} = 1} \zeta_{p^2}^a$$ where the sum runs over all solutions of $$a^{p-1} \equiv 1 \bmod p^2$$.

Example when $$p=3$$: $$a^2 \equiv 1 \bmod 9$$ has solutions $$\pm 1 \bmod 9$$ and $$\zeta_{9} + \zeta_9^{-1}$$ has minimal polynomial $$f(x) = x^3 - 3x + 1$$. Then $$f(x-1) = x^3 - 3x^2 + 3$$ is Eisenstein at $$3$$; the polynomial $$f(x+1)$$ is not. I did my calculation of the minimal polynomial in $$\mathbf C$$, which is okay since a primitive $$p$$th-power root of unity has the same degree over $$\mathbf Q_p$$ as it does over $$\mathbf Q$$, so the structure of the intermediate fields in a $$p$$th-power cyclotomic extension over $$\mathbf Q_p$$ and over $$\mathbf Q$$ are the same.

Example when $$p=5$$: solutions to $$a^4 \equiv 1 \bmod 25$$ are $$1, 7, 18$$, and $$24$$, and $$\zeta_{25} + \zeta_{25}^7 + \zeta_{25}^{18} + \zeta_{25}^{24}$$ has minimal polynomial over $$\mathbf Q_5$$ equal to $$g(x) = x^5 - 10x^3 + 5x^2 + 10x + 1$$. (Note $$g(x-1) = x^5 - 5x^4 + 25x^2 - 25x + 5$$ is Eisenstein at 5; the polynomial $$g(x+1)$$ is not Eisenstein at $$5$$.)

• Thanks! So if I understood correctly, the two examples are Galois extension of degree $p$ which can't be of the form, and they are Galois because they are subextensions of cyclotomic extension (which is abelian). Is this right? Apr 23, 2019 at 20:13
• That is correct.
– KCd
Apr 23, 2019 at 20:28

In response to Jyrki Lahtonen’s request, I’ll make an attempt to describe what’s going on here.

The Hasse-Herbrand transition function is a concave polygonal real-valued function on $$\Bbb R$$ that encapsulates much (but not all) of the information that comes out of the study of the higher ramification of a separable extension of local fields. You can read all about the subject in Chapter IV of Serre’s Corps Locaux (translated as Local Fields). What you see below will look nothing like Serre’s treatment, however. The least of the differences is that the traditional coordinatization of the plane, as in Serre, places the vertex describing the tame part of a totally ramified extension at the origin. My coordinatization puts this vertex at $$(1,1)$$.

Part I is to describe the Newton Copolygon. I won’t relate it to the more familiar Polygon, but you will see the connection. Let $$f(X)=\sum_na_nX^n\in\mathfrak o[X]$$, where for specificity’s sake I will suppose that $$\mathfrak o$$ is the ring of integers in a finite extension $$k$$ of $$\Bbb Q_p$$, and that we are using the (additive) valuation $$v$$ on $$k$$ normalized so that $$v(p)=1$$. For each nonzero monomial $$a_nX^n$$, draw the half-plane $$\Pi_n$$ described in $$\Bbb R^2$$ as all points $$(\xi,\eta)$$ satisfying $$\eta\le n\xi+v(a_n)$$. Then form the convex set $$\bigcap_n\Pi_n$$. This is the copolygon, but I hope I won’t confuse matters too badly by calling the “copolygon function” the function $$v_f$$ whose graph is the boundary of the convex set just described. You see, for instance, that if $$f(X)=pX+pX^2+X^3$$ the copolygon’s boundary has only one vertex, at $$(\frac12,\frac32)$$, with slope $$3$$ to the left and slope $$1$$ to the right. You see without difficulty that as long as $$g$$ has no constant term, $$v_{f\circ g}=v_f\circ v_g$$.

Part II. Without saying what the “lower breaks” and “upper breaks” of the ramification filtration of the Galois group of a Galois extension $$K\supset k\supset\Bbb Q_p$$ are, I simply proclaim that the Herbrand function is the polygonal real-valued function $$\psi^K_k$$ whose only vertices are at each break point $$(\ell_i,u_i)$$. The lovely fact about the transition functions is that if $$L\supset K\supset k$$, then $$\psi^L_k=\psi^K_k\circ\psi^L_K$$. The transition function $$\psi^K_k$$ is an invariant of the extension, not depending on any choices.

Part III is to relate these two polygonal functions, though this is not the place to explain why they are connected. Though the traditional description of the transition function, as in Serre, always starts from a Galois group, you will notice that there is no mention of groups below. For simplicity, I’ll describe only $$\psi^k_{\Bbb Q_p}$$ for $$k$$ totally ramified over $$\Bbb Q_p$$, since that’s enough to answer Saewoo Lee’s question.

Let $$\mathfrak o$$ be the ring of integers of $$k$$, and $$\pi$$ a prime element (generator of the maximal ideal), and let $$F(X)$$ be the minimal $$\Bbb Q_p$$-polynomial for $$\pi$$. Form the polynomial $$f(X)=F(X+\pi)$$, so that $$f$$ has no constant term. Now take the copolygon function $$v_f$$ of this $$f$$, and stretch it horizontally by a factor of $$e^k_{\Bbb Q_p}=[k:\Bbb Q_p]$$, to get $$\psi^k_{\Bbb Q_p}$$. That is, $$\psi^k_{\Bbb Q_p}(\xi)=v_f(\xi\,/\,e)$$.

Let’s work out three examples, namely $$\Bbb Q_2(\sqrt{2u}\,)$$, $$\Bbb Q_2(\sqrt3\,)$$, and $$\Bbb Q_3(\rho)$$ where the minimal polynomial for $$\rho$$ is $$X^3-3X-3$$.

First, over $$\Bbb Q_2$$, a prime is $$\pi=\sqrt{2u}$$, minimal polynomial $$F(X)=X^2-2u$$, giving $$f(X)=X^2+2\pi X$$. The copolygon has unique vertex at $$(\frac32,3)$$, and the transition function has unique vertex at $$(3,3)$$. (The initial segment of $$\psi^K_k$$ will always have slope $$1$$.)

Second, over $$\Bbb Q_2$$, a choice for a prime of $$\Bbb Z_2[\sqrt3\,]$$ is $$\sqrt3-1$$, with minimal polynomial $$F(X)=X^2+2X-2$$, so that $$f(X)=X^2+2\pi X+2X=X^2+2(1+\pi)X$$. The polygon has its one vertex at $$(1,2)$$, so that $$\psi$$ has its vertex at $$(2,2)$$, enough to show that $$\Bbb Q_2(\sqrt3\,)$$ is not of form $$\Bbb Q_2(\sqrt{2u}\,)$$.

Third, over $$\Bbb Q_3$$ with $$F(X)=X^3-3X-3$$, we get $$f(X)=X^3+3\rho X^2+3\rho^2X-3X$$, in which only the monomials $$X^3$$ and $$3(\rho-1)X$$ count, so that the copolygon has its vertex at $$(\frac12,\frac32)$$, and the transition function’s vertex is at $$(\frac32,\frac32)$$.

I’ll leave it to you to show that the vertex of the transition function for $$\Bbb Q_3(\sqrt[3]{3u}\,)$$ is at $$(\frac52,\frac52)$$. (Don’t be surprised that these vertices don’t have integral coordinates. That’s guaranteed only for normal, abelian extensions, by Hasse-Arf, and the cubic extensions here are neither.)


# Ways of showing $\sum_\limits{n=1}^{\infty}\ln(1+1/n)$ to be divergent

Show that the following sum is divergent $$\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)$$

I thought to do this using Taylor series using the fact that $$\ln\left(1+\frac1n\right)=\frac1n+O\left(\frac1{n^2}\right)$$ Which then makes it clear that $$\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)\sim \sum_{n=1}^{\infty}\frac1n\longrightarrow \infty$$ But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told?

• $\ln(1+1/n) \sim 1/n$ as $n \to +\infty$ means that for any $\epsilon > 0$ there is a $N$ such that for every $n > N$ : $\frac{1-\epsilon}{n} < \ln(1+1/n) < \frac{1+\epsilon}{n}$ – reuns Jul 30 '16 at 21:56
• and no you can't write $\sum_n \ln(1+1/n) \sim \sum_n 1/n$ (it means nothing) but you can show that $\sum_{n < N} \ln(1+1/n) \sim \sum_{n < N} 1/n$ as $N \to +\infty$ – reuns Jul 30 '16 at 21:59
• Why not? People seem to have understood – qbert Jul 30 '16 at 22:00
• because $f(n) \sim g(n)$ as $n\to +\infty$ means something precise, namely that $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 1$ – reuns Jul 30 '16 at 22:01
• I really, really like this problem from a pedagogical perspective: as we can see below, it can be solved using such a wide variety of techniques in the standard precalculus / calculus toolbox, but it can't be solved by directly applying any of the convergence tests in the standard list given to students. In this way, I feel the sequences & series section of calc 2 is too algorithmic, with relatively little creativity required from students. This is the perfect sort of problem for cutting through that. Thanks for sharing @qbert – Kaj Hansen May 27 '19 at 5:58

Notice the following: $$\log\left(1+\frac{1}{n}\right)=\log\left(\frac{n+1}{n}\right)=\log(n+1)-\log(n)$$ Hence $$\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)=\log(n+1) \to \infty$$

• Duh ok thank you. Any response to my second question? – qbert Jul 30 '16 at 21:54
• I think Taylor series would be the hint, for me at least. – preferred_anon Jul 30 '16 at 21:58
• This is an efficient and compelling approach. +1 – Mark Viola Jul 30 '16 at 22:28

This is a special case of: Suppose $f(1)= 0, f'(1) > 0.$ Then $\sum f(1+1/n) = \infty.$

Proof: From the definition of the derivative (no Taylor necessary), we have

$$\frac{f(1+h)-f(1)}{h}= \frac{f(1+h)}{h} > \frac{f'(1)}{2}$$

for small $h>0.$ Thus

$$f(1+1/n) > \frac{f'(1)}{2}\cdot\frac{1}{n}$$ for large $n.$ By the comparison test we're done.

• Nice generalization! – Kaj Hansen Jul 31 '16 at 1:06
• Very cool. I think this is my favorite – qbert Jul 31 '16 at 1:32

Note that we have

\begin{align} \log\left(1+\frac1n\right)&=\int_n^{n+1}\frac{1}{t}\,dt\\\\ &\ge\frac1{n+1} \end{align}

and the harmonic series diverges.

But, suppose one forgoes that comparison and instead writes

\begin{align} \sum_{n=1}^{2^N-1}\int_n^{n+1} \frac{1}{t}\,dt&=\int_1^{2^N}\frac{1}{t}\,dt\\\\ &=\int_1^2 \frac{1}{t}\,dt+\int_{2}^{4}\frac{1}{t}\,dt+\dots+\int_{2^{N-1}}^{2^N}\frac{1}{t}\,dt\\\\ &\ge \frac12 (2-1)+\frac14 (4-2)+\dots +\frac{1}{2^N}(2^N-2^{N-1})\\\\ &=\frac{N}{2} \end{align}

which goes to $\infty$ as $N\to \infty$. And we are done! .

• I really like using integrals to estimate series. Any books you'd recommend. I was only introduced to this method in complex variables – qbert Jul 30 '16 at 22:42
• Pleased to hear that this was useful. I don't have a specific book to recommend. – Mark Viola Jul 30 '16 at 22:45
• And I am trying to prove your first inequality. It seems clear that it is true for any positive increasing function in theintegrand, is that true? – qbert Jul 30 '16 at 22:46
• Actually, the integrand is decreasing. Since $t\le n+1$, then $\frac1t \ge \frac{1}{n+1}$. Therefore, $$\int_n^{n+1}\frac{1}{t}\,dt\ge \frac{1}{n+1}\left((n+1)-n\right)$$ – Mark Viola Jul 30 '16 at 22:48

Applying a property of logarithms gives the equality $$\displaystyle \sum_{n=1}^\infty \ln(1 + 1/n) = \ln \Bigg( \prod_{n=1}^\infty (1 + 1/n) \Bigg)$$. Therefore, if $$\displaystyle \sum_{n=1}^\infty \ln(1 + 1/n)$$ converges, say to $$c \in \mathbb{R}^+$$, then $$\displaystyle \prod_{n=1}^\infty (1 + 1/n)$$ should converge to $$e^c$$.

Therein lies a contradiction: expanding this product yields a clearly divergent sum: the expansion will include a positive copy of $$1/n$$ for all $$n \in \mathbb{N}$$.

• Super clever. Thank you! – qbert Jul 30 '16 at 21:59
• Glad I could help! – Kaj Hansen Jul 30 '16 at 21:59

$$\sum_{n=1}^{m}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{m}(\log(n+1)-\log n)=\log(m+1)$$ The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series$$\sum_{n=1}^{\infty}2^{n}\ln(1+1/2^{n})$$ converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.

• +1. It's the straightforward way to deal with the question. – Felix Marin Aug 1 '16 at 4:11

In THIS ANSWER, I showed using elementary analysis only that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le\log(x)\le x-1\tag1$$

Letting $$x=1+\frac1n$$ in $$(1)$$ reveals

$$\frac1{n+1}\le\log\left(1+\frac1n\right)\le \frac1n\tag2$$

And using the left-hand side of $$(2)$$ shows that

$$\sum_{n=1}^N \log\left(1+\frac1n\right)\ge \sum_{n=1}^N \frac1{n+1}$$

Inasmuch as the harmonic series diverges, we see that the series of interest diverges also. And we are done!

"Sophisticated" does not mean "complicated". In my opinion, despite using more sophisticated ideas (asymptotic analysis), your proof is simpler than all of the other current answers — even the one expressing it as a telescoping series.

Incidentally, you possibly made an oversight: to complete the proof,

$$\sum_{n=1}^{\infty} O\left(\frac{1}{n^2} \right) = O\left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right) = O(1)$$

(also, a remark: for this argument to be valid, it's important that the $O$ on the left is uniform; e.g. the same 'hidden constant' works for all $n$)

• Sorry I'm not sure I follow/may have used notation incorrectly. Are you saying that the final identity is ok since $(1/n^p)/(1/n^2)=c$ or that we even need to worry about how this c varies with p? – qbert Jul 31 '16 at 1:35
• @qbert: The final identity holds because the sum is convergent. It's a constant! – user14972 Jul 31 '16 at 1:42
• Ok, got it. Thanks! Also, what subjects use big o frequently? – qbert Jul 31 '16 at 1:43
• @qbert: For people who use the notation, it can come up in just about any field of mathematics. Often, there are many different ways to say the same thing (albeit with varying amounts of facility), and people will use the ones they're most comfortable with. However, off the top of my head, "analysis of algorithms" is the only subject I can think of where practically everybody uses it. – user14972 Jul 31 '16 at 1:47
• @qbert: For example, in $5$-adic analysis, $3$ and $53$ are 'nearby' integers. Some people like to emphasize the arithmetic statement: $3 \equiv 53 \pmod{5^2}$. Some people like to measure the distance: $d(3, 53) = 5^{-2}$. Others prefer to express the that they are the same up to some amount of precision: $3 = 53 + O(5^2)$. And each of these ways of thinking themselves have several ways they could be notated. – user14972 Jul 31 '16 at 1:53

You can also use the Abel's summation $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\sum_{n=1}^{N}1\cdot\log\left(1+\frac{1}{n}\right)=N\log\left(1+\frac{1}{N}\right)+\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t\left(t+1\right)}dt$$ where $\left\lfloor t\right\rfloor$ is the floor function. Since $\left\lfloor t\right\rfloor =t+O\left(1\right)$ we have $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=N\log\left(1+\frac{1}{N}\right)+\log\left(N+1\right)+O\left(1\right)$$ and taking $N\rightarrow\infty$ we can see that the series diverges.


# Convergence of $\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$

Does this integral converge to any particular value? $$\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$$ If the answer is yes, how should I calculate its value?
I tried to use convergence tests but I failed due to the complexity of the integral itself.

• Please look here. It's almost the same. – Galc127 Feb 20 '16 at 17:51
• In general, $$\int_0^\infty\frac{x^n}{e^x-1}~dx~=~n!~\zeta(n+1),$$ and $$\int_0^\infty\frac{x^n}{e^x+1}~dx~=~n!~\eta(n+1).$$ See the Riemann $\zeta$ and Dirichlet $\eta$ function for more information. – Lucian Feb 21 '16 at 1:02

Sure it does.

Collect $e^x$ in the denominator and you will get

$$\int_0^{+\infty}\frac{x}{e^{x}(1 + e^{-x})}\ \text{d}x$$

Since the integral range is from $0$ to infinity, you can see the fraction in this way:

$$\int_0^{+\infty}x e^{-x}\frac{1}{1 + e^{-x}}\ \text{d}x$$

and you can make use of the geometric series for that fraction:

$$\frac{1}{1 + e^{-x}} = \frac{1}{1 - (-e^{-x})} = \sum_{k = 0}^{+\infty} (-e^{-x})^k$$

thence thou have

$$\sum_{k = 0}^{+\infty}(-1)^k \int_0^{+\infty} x e^{-x} (e^{-kx})\ \text{d}x$$

Namely

$$\sum_{k = 0}^{+\infty}(-1)^k \int_0^{+\infty} x e^{-x(1+k)}\ \text{d}x$$

This is trivial, you can do it by parts getting

$$\int_0^{+\infty} x e^{-x(1+k)}\ \text{d} = \frac{1}{(1+k)^2}$$

Thence you have

$$\sum_{k = 0}^{+\infty}(-1)^k \frac{1}{(1+k)^2} = \frac{\pi^2}{12}$$

Which is the result of the integration

If you need more explanations about the sum, just tell me!

HOW TO CALCULATE THAT SERIES

There is a very interesting trick to calculate that series. First of all, let's write it with some terms, explicitly:

$$\sum_{k = 0}^{+\infty}\frac{(-1)^k}{(1+k)^2} = \sum_{k = 1}^{+\infty} \frac{(-1)^{k+1}}{k^2} = -\ \sum_{k = 1}^{+\infty}\frac{(-1)^{k}}{k^2}$$

The first terms of the series are:

$$-\left(-1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} - \frac{1}{25} + \frac{1}{36} - \frac{1}{64} + \frac{1}{128} - \cdots\right)$$

namely

$$\left(1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36} + \frac{1}{64} - \frac{1}{128} + \cdots\right)$$

Now let's call that series $S$, and let's split it into even and odd terms:

$$S = \left(1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \cdots\right) - \left(\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + \cdots\right) ~~~~~ \to ~~~~~ S = A - B$$

Where obviously $A$ and $B$ are respectively the odd and even part.

now the cute trick

take $B$, and factorize out $\frac{1}{4}$:

$$B = \frac{1}{4}\left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \cdots\right)$$

Now the series in the bracket is a well known series, namely the sum of reciprocal squares, which is a particular case of the Zeta Riemann:

$$\zeta(s) = \sum_{k = 1}^{+\infty} \frac{1}{k^s}$$

which is, for $s = 2$

$$\zeta(2) = \sum_{k = 1}^{+\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$$

Thence we have:

$$B = \frac{\zeta(2)}{4} = \frac{\pi^2}{24}$$

Are you seeing where we want to go? But this is not enough since we don't know what $A$ is. To do that, we can again split $B$ into even and odd terms! But doing so, we will find again the initial $A$ and $B$ series:

$$B = \frac{1}{4}\left(\left[1 + \frac{1}{9} + \frac{1}{25} + \cdots\right] + \left[\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots\right]\right) ~~~ \to ~~~ B = \frac{1}{4}\left(A + B\right)$$

This means:

$$4B - B = A ~~~~~ \to ~~~~~ A = 3B$$

So

$$A = 3\cdot \frac{\pi^2}{24} = \frac{\pi^2}{8}$$

Not let's get back to the initial series $S$ we wanted to compute, and substitution this we get:

$$S = A - B = \frac{\pi^2}{8} - \frac{\pi^2}{24} = \frac{\pi^2}{12}$$

• Could you please explain more about the sum and how you got the final answer? – FreeMind Feb 20 '16 at 18:01
• @FreeMind DONE! ^^ – Turing Feb 20 '16 at 18:42

Comparison test:

$$\frac x{1+e^x}\le\frac x{e^x}\;$$

and you can directly do integration by parts in the rightmost function to check it converges.

It is possible to generalize the result.

Claim: $$I=\int_{0}^{\infty}\frac{x^{a}}{e^{x-b}+1}dx=-\Gamma\left(a+1\right)\textrm{Li}_{a+1}\left(-e^{-b}\right)$$ where $\textrm{Li}_{n}\left(x\right)$ is the polylogarithm and $a>0$.

Consider $$\left(-1\right)^{n}\int_{0}^{\infty}x^{a}e^{-n\left(x-b\right)}dx=\frac{\left(-1\right)^{n}e^{-nb}}{n^{a+1}}\int_{0}^{\infty}y^{a}e^{-y}dy=\frac{\Gamma\left(a+1\right)\left(-1\right)^{n}e^{-bn}}{n^{a+1}}$$ and recalling that $$\textrm{Li}_{k}\left(x\right)=\sum_{n\geq1}\frac{x^{n}}{n^{k}}$$ we have $$\Gamma\left(a+1\right)\textrm{Li}_{a+1}\left(-e^{-b}\right)=\sum_{n\geq1}\left(-1\right)^{n}\int_{0}^{\infty}x^{a}e^{-n\left(x-b\right)}dx=\int_{0}^{\infty}x^{a}\sum_{n\geq1}\left(-1\right)^{n}e^{-n\left(x-b\right)}dx=$$ $$=-\int_{0}^{\infty}\frac{x^{a}}{e^{x-b}+1}dx.$$ Maybe it's interesting to note that the function ${x^{a}}/{e^{x-b}+1}$ is the Fermi-Dirac distribution.


Antiderivative of $e^x/(1+2e^x)$

I know the solution is $$\dfrac{\ln\left(2\mathrm{e}^x+1\right)}{2}+C$$ However, the result I got was $$\dfrac{\ln\left(\mathrm{e}^x+0.5\right)}{2}+C$$ What I did was:

\begin{align} \int \frac {e^x}{1+2e^x}dx &= \int \frac {e^x}{2*(0.5+e^x)}dx \\ &= 0.5 \cdot \int \frac{e^x}{0.5 + e^x} dx \\ &= 0.5 \cdot (\ln|0.5 + e^x| + C). \end{align}

I know there are antiderivative calculators online that show the correct method step by step, but I can't understand what I did wrong. What's the mistake?

• I think OP choice's of presentation on a line was better than the actual edit, symbols get all packet in a small area now. Why override OP's style when the original LaTeX is ok ? – zwim Dec 3 '17 at 0:34
• I split the lines because it accidentally carried over to a second line, which is distinctly unoptimal. The choice to make three lines instead of two was made on the fly. – davidlowryduda Dec 3 '17 at 0:37
• @zwim Also (I’m not taking sides) multiple lines are better for viewing on mobile. – Chase Ryan Taylor Dec 3 '17 at 1:51

These answers are the same. Namely,

$$\ln(2e^x + 1) = \ln(2(e^x + 0.5)) = \ln(2) + \ln(e^x + 0.5).$$

The added $\ln 2$ is absorbed by the $+ C$, and so we see that the two answers are the same up to an additive constant. There is no error.

You can tell your answer is right by differentiating it and seeing what you get back. Indeed, $$\frac{d}{dx}\left(\frac{\ln(e^x+0.5)}{2}\right) =\frac{1}{2}\frac{e^x}{e^x+0.5} =\frac{e^x}{2e^x+1}.$$ So you are correct. As the other answers point out, the two answers are the same up to a constant.

Let $C = \ln 2$, a constant. Then they're equivalent.

What you did is correct; you just have a different value of $C$. This is why the "$+C$" is so important!

The others have answered your question. Note that one runs into this often also when dealing with trig functions.For example $\int sinx cosx dx= sin^2(x)/2 + C$ if one chooses to make the substitution $u= sinx$, and $\int sin(x)cosx dx= -cos^2(x)/2 + C$, if one chooses $u=cosx$. Both answers are correct, as of course their difference is a constant, since $sin^2(x)+cos^2(x) = 1.$


# Is the set $\{2, 3, 4\}$ open in some metric spaces and not open in others?

I just want to check my understanding. This is from Baby Rudin:

2.18 Definition Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$.

$(a)$ A neighborhood of $p$ is a set $N_r(p)$ consisting of all $q$ such that $d(p, q)<r$ for some $r>0$. The number $r$ is called the radius of $N_r( p)$

$(e)$ A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N \subset E$

$(f)$ $E$ is open if every point of $E$ is an interior point of $E$.

Suppose we have the metric space with set $X=\{1, 2, 3, 4, 5\}$ and distance function $d(x, y)=|x-y|$. Now $2$ is an interior point of $\{2, 3, 4\}$ because $N_{0.5}(2)=\{2\} \subset \{2, 3, 4\}$ (and a similar argument can be made for $3$ and $4$ as well.

But if our metric space is $\mathbb{R}$ with the same distance function, then $\{2, 3, 4\}$ is not open because no neighborhood of $2$ is a subset of $\{2, 3, 4\}$, so $2$ is not an interior point of $\{2, 3, 4\}$, right?

• Right. That's completely correct. – fleablood Sep 4 '17 at 16:01
• Why "no neighborhood of $2$ is a subset of $\{2,3,4\}$"? You are right, but you need to add some explanation. – Krish Sep 4 '17 at 16:01
• @Krish When we are dealing with the set $\mathbb{R}$, every neighborhood will contain numbers which are not integers. – Ovi Sep 4 '17 at 16:03
• @Krish: Because any neighborhood of $2$ (by the definition given) has the form $(2-r,2+r)$ for some $r>0.$ This will readily contain some non-integer rational number. – Cameron Buie Sep 4 '17 at 16:03
• @CameronBuie sorry!!! But I was just checking whether OP understood the reason clearly or not. (+1) for the question. – Krish Sep 4 '17 at 16:07

Another thing to consider is that even when the underlying space is the same, using a different metric may yield different open sets. Letting our metric space be $\Bbb R,$ but with the distance function $$\delta(x,y):=\begin{cases}0 & x=y\\1 & x\neq y,\end{cases}$$ we can show that (for example) $\{2\}$ is a neighborhood of $2$ with radius $\frac12,$ and by similar reasoning conclude that $\{2,3,4\}$ is once again open.


# Binomial ID $\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$

Could I get some help with proving this identity?

$$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}.$$

It has been checked in Matlab for various small $n,m$ and $p$. I have a proof for $m=0$ that involves Pascal's rule to split it into two sums that mostly cancel, but this does not work for all $m$. I have looked at induction on various letters (and looked at simultaneous induction on multiple letters) without any success.

I would appreciate any help, algebraic or combinatoric.

• Where does this identity come from? Dec 4, 2016 at 23:58
• They are the coefficients of an expansion of a solution to Laplace's equation in terms of solid spherical harmonics. The $n$ and $m$ are the separation constants that occur in for example the spherical harmonics $P_n^m(\cos\theta)\exp(im\phi)$ Dec 5, 2016 at 0:23

Such identities are often reduced to the Chu--Vandermonde's identity $\sum_{i+j=\ell} \binom{x}i\binom{y}j=\binom{x+y}\ell$ by using reflection formulae $\binom{x}k=\binom{x}{x-k}$, $\binom{x}k=(-1)^k\binom{k-x-1}k$.

In your case you may write $$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}= \sum_{k=m}^p \binom{-m-1}{k-m}\binom{n+p+1}{p-k}= \binom{n+p-m}{p-m}$$ as you need, so it is Chu--Vandermonde for $x=-m-1$, $y=n+p+1$, $\ell=p-m$.

• Right, with the same generatingfunctionological explanation: $(1+t)^x (1+t)^y = (1+t)^{x+y}$. Dec 5, 2016 at 4:32
• @Noam but why formal series $(1+t)^x=\sum \binom{x}{n} t^n$ satisfy this equality? It may be justified by some abstract nonsense like 'this is known for reals or for positive integers, but the coefficients are polynomials in $x,y$, thus it holds formally too.' This argument may be reduced to bit less abstract: 'both parts of CV are polynomials in $x,y$ of degree at most $\ell$, thus it suffices to check that their values agree for integers $x,y\geqslant 0, x+y\leqslant \ell$ (this triangle is an interpolating set for polynomials of degree at most $\ell$) , where identity is just obvious.' Dec 5, 2016 at 5:54
• I suppose you can also use a (formal) differential equation to prove this. Dec 5, 2016 at 14:25

It's a generating-function exercise. We have $$\sum_{k=m}^\infty (-1)^{k+m} {k \choose m} x^{k-m} = (1+x)^{-(m+1)},$$ and (with $j=p-k$) $$\sum_{j=0}^\infty {n+p+1 \choose n+(p-j)+1} x^j = \sum_{j=0}^\infty {n+p+1 \choose j} x^j = (1+x)^{n+p+1}.$$ The desired sum is the $x^{p-m}$ coefficient of the product $(1+x)^{n+p-m}$ of these two power series, which is ${n+p-m \choose p-m} = {n+p-m \choose n}$, QED.

• Presumably "well-known", but easier to redo than to track down in the literature. Dec 5, 2016 at 3:07

I wish to explain a modern (high tech) method called the Wilf-Zeilberger (WZ) technique which might help you (and anyone interested) with the present question and many others you encounter in the future. This will save you time from hunting the literature and comparing notes with the milliard hypergeometric formulas.

Zeilberger developed an accompanying algorithm package which is by now part of the symbolic softwares, Maple and Mathematica. In case you have access to Maple, you may download and run a freely available online copy of it from here.

Start with the discrete function $$F(p,k)=(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}\binom{n+p-m}{n}^{-1}.$$ The above-mentioned algorithm furnished the companion function $$G(p,k)=-\frac{F(p,k)(k-m)(n+k+1)}{(p+1-k)(n+p+1-m)}.$$ Check (preferable using a symbolic software) that $$F(p+1,k)-F(p,k)=G(p,k+1)-G(p,k).\tag1$$ Convention: $\binom{a}b=0$ if $b>a$ or $b<0$. Now, sum both sides of (1) over all integers $k$, i.e. $\sum_{-\infty}^{\infty}$. Notice that the RHS of (1) vanishes because $\sum_{k\in\mathbb{Z}}G(p,k+1)=\sum_{k\in\mathbb{Z}}G(p,k)$. If we denote $f(p):=\sum_{k=m}^pF(p,k)=\sum_{\mathbb{Z}}F(p,k)$ then the LHS of (1) implies $$f(p+1)=f(p).$$ But, if $p=0$ then $f(0)=1$ and hence $f(p)=1$ identically. That means (after rewriting) $$\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$$ as desired.

Mathematica says: $$(-1)^{2 m} \binom{n+p+1}{m+n+1} \, _2F_1(m+1,m-p;m+n+2;1)$$


# Cardinality of the collection of all compact metric spaces

I was wondering today, that if $\mathscr{M}$ is the collection of all sets that admit a metric generating a compact topology, then

1. Is $\mathscr{M}$ a set in ZFC?
2. If it is, what is the cardinality of $\mathscr{M}$?

The reason why I found $1.$ interesting is because in general, the collection of all sets is not a set in ZFC, and each set admits a metric space structure via discrete metric and discrete topology. However, only those sets that are finite are compact with this metric. So could compactness assumption restrict this collection to become a set?

And about $2.$, I figured that atleast under the equivalence relation of homeomorphism the quotient space should be a set with cardinality less or equal than $2^{\mathfrak{c}}$, since every compact metric space admits an embedding to $[0,1]^{\mathbb{N}}$, which is of the size $\mathfrak{c}$. So for each equivalence class we could pick a representative from the power-set of $[0,1]^{\mathbb{N}}$. But how about if we consider the relation of being isometric instead of homeomorphic? And finally, if we drop the quotient spaces entirely?

My motivation for these questions arose when I was reading about the Gromov-Hausdorff distance for compact metric spaces. Maybe there is a trivial answer that $\mathscr{M}$ is not a set.

• The collection of all one-element sets is not a set. – Hurkyl Mar 11 '13 at 15:38
• @Hurkyl. How would you proceed to prove it? – T. Eskin Mar 11 '13 at 15:42
• The union of its elements is the collection of all sets. – Hurkyl Mar 11 '13 at 15:43
• A related concept: Urysohn's universal space is a separable metric space containing an isometric copy of every separable metric space. – Martin Mar 11 '13 at 16:13
• @ThomasE. Let $A$ be the collection of one-element sets. Assume $A$ is a set. Then $\{A\}$ is a set by Pairing Axiom and is one-element, hence $\{A\}\in A$. Again by Pairing, $B:=\{A,\{A\}\}$ is a set. By Axiom of Foundation, there exists $C\in B$ with $C\cap B=\emptyset$. Clearly, $C\ne A$ because then $\{A\}\in C\cap B$. Also, $C\ne\{A\}$ because then $A\in C\cap B$. Contradiction! Therefore $A$ is not a set. – Hagen von Eitzen Mar 11 '13 at 16:32

For any $x$, $\{x\}$ with the metric $d(x,x)=0$ is a compact metric space, so it’s clear that $\mathscr{M}$ cannot be a set. However, it is true that $|X|\le 2^\omega$ for every $X\in\mathscr{M}$, so there is a set $\mathscr{M}_0$ of compact metric spaces such that every compact metric space is homeomorphic (indeed isometric) to one in $\mathscr{M}_0$.

Added: For each cardinal $\kappa\le 2^\omega$ let $$\mathscr{M}_\kappa=\big\{\langle\kappa,d\rangle:d\text{ is a metric on }\kappa\big\}\subseteq{}^{\kappa\times\kappa}\Bbb R\;;$$ this is clearly a set, as is $\mathscr{M}=\bigcup\{\mathscr{M}_\kappa:\kappa\le 2^\omega\text{ is a cardinal}\}$, and every compact metric space is isometric to some space in $\mathscr{M}$. (For $\kappa>1$ $\mathscr{M}_\kappa$ contains $2^\kappa$ isometric copies of each space, corresponding to permutations of $\kappa$, so you might want to choose representatives of the isometry classes.)

To answer your last question in the comments, a compact metric space $X$ is separable, and every compatible metric is continuous on $X\times X$, so it has at most $2^\omega$ compatible metrics. On the other hand, it $d$ is a compatible metric, then so is $\alpha d$ for every $\alpha>0$, so $X$ has $2^\omega$ compatible metrics (unless $X$ is the one-point space!).

• Thanks Brian. Since you also addressed the isometric part, I have one more question. I figured that the quotient space under the relation of being isometric should have more equivalence classes than the homeomorphism relation, since not all homeomorphic spaces are isometric. What would be the cardinality of this quotient space? (notice the edit in this comment) – T. Eskin Mar 11 '13 at 15:57
• @Thomas: Let $X$ be a fixed set of cardinality $\kappa\le 2^\omega$. The set of metrics on $X$ is a subset of the set of functions from $X\times X$ to $\Bbb R$, so we can form the set $$\big\{\langle X,d\rangle:d\text{ is a metric on }X\big\}\;.$$ Do this with an $X$ for each $\kappa\le 2^\omega$, and you have your $\mathscr{M}_0$. – Brian M. Scott Mar 11 '13 at 16:02
• @Thomas: I’ve incorporated a few more specifics in my answer, including an answer to your question about the relationship between the homeomorphism and isometry classes. – Brian M. Scott Mar 11 '13 at 18:15
• Thank you Brian, this is an excellent answer. Thanks for your effort. A further question for your added part: you probably use the notation $\mathscr{M}$ in a different way as before, where you noted that $\mathscr{M}$ was not a set. Also, is it easy to prove that for a compact metric space $(X,d)$ of cardinality $\kappa$ there exists an isometry $\phi:\kappa\to X$ for a suitable metric on $\kappa$? Do we just take the bijection $\phi:\kappa\to X$ given by the cardinality of $X$, and pull back the metric $d$ to $\kappa$ by defining $e(\alpha,\beta)=d(\phi(\alpha),\phi(\beta))$ ? – T. Eskin Mar 11 '13 at 21:40
• @Thomas: You’re welcome. Yes, I used $\mathscr{M}$ in two different ways. And yes, just use the bijection to transfer the metric; it really is that simple. (Nice when that happens.) – Brian M. Scott Mar 11 '13 at 21:42

Generally collections of "all possible" are classes, simply because there is a proper class of sets of cardinality $1$, all of which are compact metric spaces.

But indeed every compact metric space can be embedded into $[0,1]^\Bbb N$, therefore we only need to find out how many closed sets this space has, and since it is a second countable metric space it has exactly $\frak c$ many closed subsets, so there are at most $\frak c$ compact metric spaces up to homeomorphism.

• Thanks Asaf for this clarification. – T. Eskin Mar 11 '13 at 15:53


# Math Help - Need help figuring out this linear equations table

1. ## Need help figuring out this linear equations table

X. F(x)
4. 13
2. 7
0. 1
-1. -2

using the table, what is the value of f(-1)?

Write a linear function equation for this table of values.

Using this equation or the table itself, find the value of f(1)

I don't understand this.

2. Originally Posted by brianfisher1208
X. F(x)
4. 13
2. 7
0. 1
-1. -2

using the table, what is the value of f(-1)?

Write a linear function equation for this table of values.

Using this equation or the table itself, find the value of f(1)

I don't understand this.
K so the question is saying that there is a function $y=f(x)$ such that plugging in any x into the function will give you a value for y=f(x) as listed in the table.

Using this, we can think of the table as telling us this:

$f(4)=13$
$f(2)=7$
$f(0) = 1$
$f(-1) = -2$

Now as for making a function. What can we decide from the table? It looks like if x =0, f(x)=f(0)=1. So it looks like we need to have a constant that isn't multiplied by any x.

Next, lets looks at the values, how can we turn 2 into 7 and 4 into 13?

Looks like $f(x) = 3x + 1$. Does this make sense how we came to this? It can be a bit of a trial and error process.

Let's test it.

$f(2) = 3(2) + 1 = 6 + 1 = 7 \ \ \ \text{Check!}$

$f(4) = 3(4) + 1 = 12 +1 = 13 \ \ \ \text {Check!}$

$f(0) = 3(0) + 1 = 0 +1 = 1 \ \ \ \text {Check!}$

$f(-1) = 3(-1) + 1 = -3 +1 = -2 \ \ \ \text {Check!}$

Can you find $f(1)$ now?

3. Nice avatar, Kasper !

Otherwise, you might want to consider a more straightforward algebraïc way.
Your problem strongly suggests that a linear relationship exists between $x$ and $f(x)$. This is equivalent to saying that $f(x) = ax + b$, for some $a$ and $b$, $a \neq 0$.

1. Find $b$

You have some values of the function (actually, only two are required). You can say that :

$13 = 4a + b$

$7 = 2a + b$

We can slightly reformulate this :

$13 - b = 4a$

$7 - b = 2a$

And now, we get smart ! We divide both equations together :

$\frac{13 - b}{7 - b} = \frac{4a}{2a}$

Which is equivalent to :

$\frac{13 - b}{7 - b} = 2$

Now we simplify the fraction :

$13 - b = 2(7 - b)$

$13 - b = 14 - 2b$

$- b + 2b = 14 - 13$

$\boxed{b = 1}$

2. Find $a$

Now that we know $b$, this is easy : we know that $b = 1$ : we can use a pair of values $\left ( x, f(x) \right )$ :

$13 = 4a + b$

Since $b = 1$, we get :

$13 = 4a + 1$

$12 = 4a$

$3 = a$

$\boxed{a = 3}$

3. Conclusion

You are done : you have found the linear relationship between $x$ and $f(x)$ :

$\boxed{f(x) = 3x + 1}$

Let's check those results :

$f(4) = 4 \times 3 + 1 = 12 + 1 = 13$ -->

$f(2) = 2 \times 3 + 1 = 6 + 1 = 7$ -->

$f(0) = 0 \times 3 + 1 = 0 + 1 = 1$ -->

$f(-1) = (-1) \times 3 + 1 = -3 + 1 = -2$ -->

Now that we can find $f(x)$ from $x$ with a simple linear formula, let's answer the last part of the question : substitute $x = 1$ to find the value of $f(1)$ :

$f(1) = 1 \times 3 + 1 = 3 + 1 = 4$

Done !
_________________

Does it make sense ?

4. Hello, brianfisher1208!

. . . $\begin{array}{c|c}x & f(x) \\ \hline\text{-}1 & \text{-}2 \\ 0 & 1 \\ 2 & 7 \\ 4 & 13 \end{array}$

Using the table, what is the value of $f(-1)$ ?

Um . . . -2 ?

Write a linear function equation for this table of values.
A linear function has the form: . $f(x) \;=\;ax+b$
And we must determine $a$ and $b.$

We can use any two values from our table.

For example:

. . $\begin{array}{ccccccccccccc}f(2) = 7: && a(2) + b &=& 7 && \Rightarrow && 2a + b &=& 7 & [1] \\
f(4) = 13: && a(4) + b &=& 13 && \Rightarrow && 4a + b &=& 13 & [2] \end{array}$

Subtract [2] - [1]: . $2a \:=\:6 \quad\Rightarrow\quad a \:=\:3$

Substitute into [1]: . $2(3) + b \:=\:7 \quad\Rightarrow\quad b \:=\:1$

Therefore: . $\boxed{f(x)\;=\;3x+1}$

Using this equation or the table itself, find the value of $f(1).$

$f(1) \;=\;3(1) + 1 \;=\;4$

5. Interesting! As you can see, Brian, Soroban chose to solve by substitution a system of linear equations in order to find $a$ and $b$, while I decided to arrange the equations to allow a division that will get rid of an unknown, making solving easy. Two equivalent solutions to one problem (although mine was a bit longer).

Thanks Soroban, I didn't know the command \boxed, I was sort of deceived my function didn't appear in math font