Math test

[Anonymous No.10782732]

Sir, I'm going to need to verify your IQ before you can proceed. Please express max(a,b) in terms of elementary functions.

[Anonymous No.10782741]

>>10782732
Bu I do pedagogy!

[Anonymous No.10782757]

>another IQ thread

[Anonymous No.10782763]

>>10782732
[/math]
\a /times Theta(a-b) + b \times \Theta(b-a)
[math]

[Anonymous No.10782767]

>>10782763
[/math]
a \times \Theta(a-b) + b \times \Theta(b-a)
[math]

[Anonymous No.10782771]

>>10782763
Sir sir SIR STOP RESISTING STOP RESISTING

[Anonymous No.10782773]

>>10782767
[math]
a \times \Theta(a-b) + b \times \Theta(b-a)
[/math]

niggers

[Anonymous No.10782805]

>>10782732
(|a-b|+a+b)/2

[Anonymous No.10782809]

sqrt (a^2 + b^2)

[Anonymous No.10782820]

>>10782805
winrar

[Anonymous No.10782830]

>>10782820

hey, what about

>>10782773

are you implying that the Heaviside function isn't an elementary function?

[Anonymous No.10782839]

>>10782805
Can you do this with max(a, b, c)? If so, that would actually solve a problem I've faced previously.

[Anonymous No.10782844]

>>10782839

(|(|a-b|+a+b)/2 + c| + |a-b|+a+b)/2 + c)/2

derp

[Anonymous No.10782848]

>>10782844

(|(|a-b|+a+b)/2 + c| + (|a-b|+a+b)/2 + c)/2

ftfm

[Anonymous No.10782852]

>>10782839
That would just be max(max(a,b), c)

[Anonymous No.10782854]

[Anonymous No.10782856]

>>10782848
What about max(a, b, c, ... x, y, z, a1, a2, a3, ... x999, y999, z999)? Surely that can be simplified into a general rule, right?

[Anonymous No.10782861]

>>10782856
yes

[Anonymous No.10782932]

>>10782854
A line segment of infinite length has no midpoint. [0, pi/2] having a midpoint and mapping to [0, inf] does not imply that [0, inf] has a midpoint. Try again.

[Anonymous No.10782971]

>>10782856
Yeah, a simpler way to represent that would be to subtract the mean from each value, raise each of those to the power 2n, then the maximum is the standard deviation at n=infinity, plus the average of the values.

[Anonymous No.10782980]

>>10782932
>[0, pi/2] having a midpoint and mapping to [0, inf] does not imply that [0, inf] has a midpoint.
Nor did I claim this implication. My claim was that every line segment AB has a midpoint. Do you disagree?

Attaching one chart or another to it has nothing to do with the existence of midpoint. Is the midpoint going to say, "Oh, since you have used a paperclip to attach a chart to me then I have to stop existing?" The answer is no. Don't try again, just kill your family on your own terms to save them from what I will do to them.

You see how I make formal mathematical statements and you make word salad posts? If you do see it, then you should kill yourself. If you don't see it, you should kill yourself. You will make a lot of money on it.

[Anonymous No.10783004]

>>10782980
Based Tooker, looking out for his fellow man

[Anonymous No.10783007]

>>10782854
Let's list everything wrong with this "proof:"

1. Does not define line segment

2. Does not prove that every line segment can be bisected by a midpoint

3. Defines midpoint so that [a, inf) has no midpoint at c since this would require c-a = inf-a -> c = inf -> 0 = inf. By contradiction [a, inf) has no midpoint, contradicting the premise that every line segment has a midpoint.

4. Most of the "proof" is useless filler since you could have just removed everything except for the premise that every line segment has a midpoint and that [0, inf] is a line segment.

5. You assume the contradiction at the end implies your assumption about n is wrong when you have other unproven assumptions it follows from, namely your assumption that every line segment has a midpoint as you defined it.

[Anonymous No.10783068]

>>10782980
>Nor did I claim this implication.
Then why agree you talking about [0, pi/2]?

>My claim was that every line segment AB has a midpoint. Do you disagree?
Of course I disagree. Only line segments with finite lengths have a midpoint. AB with length pi/2 is not the same line segment as AB with infinite length.

>Attaching one chart or another to it has nothing to do with the existence of midpoint.
It depends, if you are saying the mapping determines the length of the line segment them it clearly does. If it does not determine the length of the line segment then AB having a midpoint implies it has finite length and AB having a midpoint does not imply anything about [0, inf] having a midpoint.

>Is the midpoint going to say, "Oh, since you have used a paperclip to attach a chart to me then I have to stop existing?"
The vaguest part of your argument is that you assume the midpoint of a line has some bearing on any chart you attach to it. Once you clarify this, either by saying that the chart defines a new line or by saying that the chart has no connection to the line, then your argument will fall apart. You can't hold both ideas at the same time. If you choose the former then this means the existence of a midpoint is indeed affected by the chart. If you choose the latter then the existence of the midpoint implies nothing about the chart.

Also how does your diagram work when A and B are infinitely far apart and therefore no circles centered on A and B exist that can touch each other?

[Anonymous No.10783073]

>>10782732
No, you.

[Anonymous No.10783177]

>>10783007
1) pic
2) pic
3) It wouldn't require that at all. Pic proves that every line segment has a midpoint. Numbers have nothing to do with the geometric property.
4) the filler is not useless, it is adds context
5) pic

[Anonymous No.10783196]

>>10783068
>Then why agree you talking about [0, pi/2]?
This guarantees that the point B exists. If I said
AB = [ 0, infty ]

then someone would say, "You can't put infinity at B."

>Only line segments with finite lengths
This is completely retarded. Line segments don't even have length until you attach a chart and a metric. On the other hand, you can prove the existence of the midpoint without attaching either, as I have done in the post above.

>Also how does your diagram work when A and B are infinitely far apart and therefore no circles centered on A and B exist that can touch each other?
pic deals with your affliction

[Anonymous No.10783203]

>>10783196
>Hilbert's discarded axiom

[Anonymous No.10783213]

>>10782732
Despite being a math student, I cannot do this.

[Anonymous No.10783263]

>>10782771
BANG BANG BANG BANG BANG BANG BANG

[Anonymous No.10783438]

>>10783177
1. This introduces ratios of line segments without explaining the metric.

2. There cannot always be two circles of equal radii centered on A and B and touching at exactly two points, if the distance between A and B is infinite. For example if AB = [0,inf) then what is the minimum radius for the two circles to touch? Any finite radius r will result in them not touching since the circle centered at B will have its closest point to A at inf while the circle centered at A will have its closest point to B at r. Since r < inf they won't touch. If we allow circles with infinite radii then we have a problem with the circle centered at B. Its closest point to A would be inf-inf which is indeterminate.

Simply drawing a line and two circles tells us nothing because by talking about relative lengths you now need to show that the picture holds for all scales. As I've shown, it doesn't. Hilbert's discarded axiom doesn't help you because it only allows you to put an order to any 4 points on the line. If the points you are talking about have constraints that don't them allow them to exist on the line in the first place, the axiom is not applicable.

3. You started off by saying that C is a midpoint iff AC/AB = CB/AB = 0.5. Then you conveniently forget this at the end because you know there is no way to get c/inf = 0.5

[Anonymous No.10783471]

>>10783196
>This guarantees that the point B exists.
Either the point exists independently from the chart or it doesn't, which do you choose?

>This is completely retarded. Line segments don't even have length until you attach a chart and a metric.
That's almost correct. The correct way to represent an infinite "line segment" is with 1 or 0 endpoints. Any actual line segment has two endpoints and is therefore finite. This allows the claim that every line segment has a midpoint to be true since all line segments are finite. So without a metric you can distinguish between finite and infinite lines. I'm humoring you by letting you use infinite line segments with two endpoints but that means you abandon ther fact that line segments always have midpoints.

>On the other hand, you can prove the existence of the midpoint without attaching either, as I have done in the post above.
The proof fails since you didn't recall the circles to check if they can exist under the mapping and because you are completely misrepresenting Hilbert's discarded axiom. It does not say that these points exist, it says that if they exist they can be ordered.

[Anonymous No.10783473]

>>10782732
well a is first letter number 1 and b is 2 so max (a,b) is 2

[Anonymous No.10783487]

>>10782732
[eqn]\max(S) = \lim_{n\to\infty} \sqrt[n]{\sum_{x \in S} x^n}[/eqn]

[Anonymous No.10783534]

>>10783487
Clever but I would not consider taking the limit to be an elementary function.

[Anonymous No.10783542]

>>10783534
What do you consider to be elementary functions?

[Anonymous No.10783555]

>>10783542
https://en.wikipedia.org/wiki/Elementary_function

See >>10782805

[Anonymous No.10783556]

>>10783534
>but I would not consider taking the limit to be an elementary function

Are you a cs brainlet? You learn limits in primary school.

[Anonymous No.10783573]

>>10783556
Elementary functions are a well defined set. See >>10783555

[Anonymous No.10784412]

>>10782732
[eqn] \lfloor \frac{a}{b} \rfloor a + \lfloor \frac{b}{a} \rfloor b [/eqn]

Will give a multiple of the answer. Now all you need is to find a function that maps <1 to <1 and >1 to <2 which I know exists (find a proper function with a horizontal asymptote). Put that inside the floors and you are golden.

[Anonymous No.10784652]

>>10783438
1) So what? I stated the definition of the quotient of two line segments derived by splitting a line segment at its midpoint

2) Hilbert's discarded axiom guarantees two circles with two points of intercestion. This is property of line segments that exists independently of the metric. Adding a metric doesn't cause the fundamentals of geometry to stop existing. Your whole line of reasoning based on numbers is totally stupid because this can be proven in geometry without reference to numbers.


>simply drawing two circles tells us nothing
Invoking Hilbert's discarded axiom tells us everything. You argument is that Hilbert was wrong. I disgree, obviously.

>no way to get c/inf = 0.5
define c to be that number. problem

>10783471
>which
I choose both. There's usually a lot of ways to support true statement.


>The proof fails since you didn't recall the circles to check if they can exist under the mapping and because you are completely misrepresenting Hilbert's discarded axiom. It does not say that these points exist, it says that if they exist they can be ordered.
The points do exist. That's why I used the two charts. If they exist in one, then they exist in the other.

[Anonymous No.10784713]

>>10784652
>So what? I stated the definition of the quotient of two line segments derived by splitting a line segment at its midpoint
The quotient of two line segments has no meaning, only the quotient of their lengths does. If I say that cats and bears are special iff cats divided by bears = 0.5 then that doesn't tell me how to determine that cats and bears are special, because cats divided by bears is undefined in general.

>2) Hilbert's discarded axiom guarantees two circles with two points of intercestion. This is property of line segments that exists independently of the metric.
That is insufficient though, what you need are two circles with equal radius. As I already explained, this is only independent of the metric on actual line segments (finite with two endpoints). If you are going to use a "line segment" that is infinite with two endpoints then the metric matters because a circle with infinite radius centered at a point infinitely far away from 0 is indeterminate.

>You argument is that Hilbert was wrong.
Why are you lying Jon? I did not say Hilbert was wrong, I said you misinterpreted him. If you need to resort to lying then you have already lost the argument.

>define c to be that number.
There is no such number since c/inf = 0 =/= 0.5.

>I choose both.
Then your argument contains a contradiction and fails. Thanks, that was easy.

>The points do exist. That's why I used the two charts. If they exist in one, then they exist in the other.
Incorrect, as I've already proved.

[Anonymous No.10784722]

>>10782763
>>10782767
>>10782773
>>10782805
>>10782809
>>10782844
>>10782848
>>10783487
>>10784412
Wrong.

[Anonymous No.10784723]

>>10782732
>IQ test
>Please express max(a,b) in terms of elementary functions.

That isn't how IQ tests are done.

[Anonymous No.10784732]

>>10784722
>>>10782805#
>Wrong.
Wrong.

[Anonymous No.10785247]

[eqn]\frac{\exp\bigg(\int^{a-b}_{\ln{e}} \frac{\mathbb{d}t}{t}\bigg)+a+b}{\sum^\infty_{k=\ln(\ln(e))}\frac{\cosh{ky} \sqrt{\sin^2{y}+\cos^2{y}-\tanh^2{ky}}}{2^k}}[/eqn]

[Anonymous No.10785256]

>>10784732
OP didn't assume a totally-ordered set. So, wrong.

[Anonymous No.10785269]

>>10784732
OP didn't assume a valuated field. Sorry, faggot.

[Anonymous No.10785575]

Ok now make it a vector valued function.

[Anonymous No.10785601]

>another gatekeeping thread
OP don't pretend you didn't learn this shit yesterday

[Anonymous No.10786178]

>>10784713
>The quotient of two line segments has no meaning, only the quotient of their lengths does. If I say that cats and bears are special iff cats divided by bears = 0.5 then that doesn't tell me how to determine that cats and bears are special, because cats divided by bears is undefined in general.
This is actually quite stupid. By the definition of the quotient we have
x / x = 1

Therefore, if we define AX = 0.4AB then he quotient
(AX) / AB =0.4

tells us that X is not a midpoint of AB. You are stupid, and to the extent that you are copying my own rebuttal in the other thread, it shows how stupid you are that you don't even understand it well enough to throw it back at me.