• Yesterday, 06:59
Hello and welcome to MHB! (Wave) We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions....
1 replies | 33 view(s)
• Yesterday, 02:52
Thanks GJA ... Appreciate your help ... Peter
3 replies | 45 view(s)
• Yesterday, 01:55
MarkFL replied to a thread Domain...2 in Pre-Calculus
No factor in the denominator can be zero, and nothing under a square root radical can be negative...so this leads to: t-1\ne0 0<t What do...
1 replies | 27 view(s)
• Yesterday, 01:47
I would write: y=\frac{2x}{x-1}=\frac{2x-2+2}{x-1}=\frac{2(x-1)+2}{x-1}=2+\frac{2}{x-1} We see this will have a horizontal asymptote at $y=2$,...
1 replies | 29 view(s)
• Yesterday, 01:31
Let's look at a definition: |u|=\begin{cases}u, & 0\le u \\ -u, & u<0 \\ \end{cases} Can you see that we must have: 0\le|u| ?
2 replies | 46 view(s)
• March 19th, 2018, 22:43
I am reading "The Basics of Abstract Algebra" by Paul E. Bland ... ... I am currently focused on Chapter 3: Sets with Two Binary Operations:...
3 replies | 45 view(s)
• March 19th, 2018, 22:30
Yes, and as Wilmer was pointing out, you would likely want to use the binomial theorem. If you raise both sides to the 5th power, you get: ...
10 replies | 106 view(s)
• March 19th, 2018, 21:56
You're being asked to find an area, and in essence, you're doing so by adding up a bunch of vertical lines, the length of which are determined by the...
5 replies | 54 view(s)
• March 19th, 2018, 21:48
To follow up, we get: \d{x}{t}=\frac{\d{y}{t}\left(x-12y^2\right)}{9x^2-y} Plugging in the given values, we find: ...
4 replies | 52 view(s)
• March 19th, 2018, 21:07
That's already included in the "top curve minus the bottom curve." :)
5 replies | 54 view(s)
• March 19th, 2018, 18:56
Yepper! Get well soon...
7 replies | 75 view(s)
• March 19th, 2018, 17:37
Well, your 375,462.29 result is way off: should be 176,900.08 Calculation (i = .0425/12): 704.26* / i = 176900.08 Then you need the future value...
7 replies | 75 view(s)
• March 19th, 2018, 15:10
Will do; pleasure is all mine. All yours Mark...
10 replies | 106 view(s)
• March 19th, 2018, 14:12
Didn't "introduce" anything... YOU asked about raising to 5th power... Gave you an example. HOKAY?!
10 replies | 106 view(s)
• March 19th, 2018, 13:15
SIMPLE: total payments - amount borrowed You knew that...right??
7 replies | 75 view(s)
• March 19th, 2018, 13:08
Well, you'd have quite a few terms to manipulate; as example: (a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5
10 replies | 106 view(s)
• March 19th, 2018, 12:54
So that results in M = 9.12 You can tell right away that's not correct. So why lose your time and ask? Regarding this sentence in problem:...
6 replies | 72 view(s)
• March 19th, 2018, 09:46
Let's first look at the bounded region: And so the area is: A=\int_0^3 (-x^2+6x)-(x^2-2x)\,dx=2\int_0^3 -x^2+4x\,dx=2\left_0^3=2(18-9)=18 ...
5 replies | 54 view(s)
• March 19th, 2018, 08:38
To follow up, the washer method gives us: V=\pi\int_{x_1}^{x^2} R^2-r^2\,dx The volume of an arbitrary washer is: ...
3 replies | 82 view(s)
• March 19th, 2018, 04:16
Thanks castor28 ... Hmm ... beginning to understand what you are saying ... Still concerned and a bit confused ... Surely R/I is a ring...
6 replies | 64 view(s)
• March 19th, 2018, 03:53
Thanks for the help, castor28 ... But just a point of clarification ... You write: " ... ... The identity of $R/I_1$ is $I_1$ ... ... " ...
6 replies | 64 view(s)
• March 19th, 2018, 02:24
I would first observe that $1+\sqrt{5}$ is a root of: x^2-2x-4=0 And so, the coefficients of the expansion: (1+\sqrt{5})^n Can be found...
10 replies | 106 view(s)
• March 19th, 2018, 02:16
I am reading "The Basics of Abstract Algebra" by Paul E. Bland ... ... I am currently focused on Chapter 3: Sets with Two Binary Operations:...
6 replies | 64 view(s)
• March 19th, 2018, 02:02
2 replies | 36 view(s)
• March 19th, 2018, 02:00
2 replies | 42 view(s)
• March 18th, 2018, 23:25
1 replies | 105 view(s)
• March 18th, 2018, 23:03
We consider that both $x$ and $y$ are functions of time $t$, and so beginning with: 3x^3+4y^3=xy We apply on the left the power and chain...
4 replies | 52 view(s)
More Activity

### 72 Visitor Messages

1. Haha, that would be Mark. I like the way sounds, don't you?
2. Hey Balarka,

As far as I'm concerned, you are welcome to post such information in any thread no matter how old it is.
3. Hey, way to go!
4. Hey Balarka, thanks, glad you are enjoying the challenge! As for the origin of this particular problem, I'm not sure, I found it in https://www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf (page 67, no solutions) and thought it was interesting. It only says "Japan 1990" so I'm guessing it's from a contest of some sort, yes (the second part of the challenge is a custom addition of mine).

Your proof appears to be correct although I will need some time to look at it in more detail. My solution approaches the problem in a similar way as yours, but proceeds somewhat differently, though I have not fully checked mine either (I will post it in a week as per the forum rules, as well as a reference solution if I can find one)
5. Done. Let me know if you need to edit more things in the future.
6. Hey Balarka,

topsquark was purposely given a wrong number and he wanted me to spot for the error, which it turns out, I was wrong. I guess I just am not good at math. Hehehe...
7. You should be able to edit it now. Let me know if there is a problem.
8. Hm, I'm interested in sending you a private message. Can you open up one space ?
9. I haven't had time to play with it, but it *is* interesting. The failure of right-exactness is what interests me: I suspect this is akin to why in groups left-split implies right-split, but not vice versa. So I would be tempted to investigate what a sufficient condition on $G$ would be so that the map: $Aut_N(G) \to Aut_N(H)$ is onto.

I feel the center of $G$ will be relevant, here, so I would look at the two extremes: $G$ abelian, and $G$ with trivial center. I think outer automorphisms will be troublesome. On the other hand if $H$ is an extension of $N$, and $G$ is an extension of $H$, this might be fruitful (again, mimicking the field construction).

For "concrete categories", for an object $A$ we can always form the group $Aut(A)$ of all isomorphisms $A \to A$. I can't say right off-hand if we can do so for ANY category, but this might be true (if our category is a poset, the automorphism groups are all trivial). So "some" generalization is possible, I'm not sure "how wide".
10. Yes, you are correct. The first post in a thread may be edited within 2 hours, and all other posts within 24 hours.

However, if you run over the 24 hour limit, I will lift the limit for you temporarily since you are adding such great content.
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#### Basic Information

Date of Birth
January 12, 2000 (18)
Biography:
I know a thing or two about topology. Find number theory interesting but don't know much about it.
Location:
West Bengal, India
Interests:
Number Theory
Country Flag:
India

#### Signature

Some of my notes on number theoretic topics are on a few primality tests and on Hardy & Littlewood's result (incomplete). The other articles are on quintics : about a brief description of Kiepert algorithm and a short introduction on quintic-solving algorithms, respectively. I have also written up an introductory thread on Riemann Hypothesis

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MarkFL
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