Post Math Challenges/Questions

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>>8809565
Last time I posted this one, /sci/ spurged out.

>>8809565
sum all the prime numbers less than 2 million

what about solving some introductory physics

>>8809608
try looking in your 10th grade physics textbook.

>>8809565
Here's a quickie: derive the conservation of energy from first principles

Let [math]f: [0,\infty) \to \mathbb R[/math] a continuous surjection. Prove that f assumes each value infinitely many times.

>>8810926
Wow, that one is cool. Immediately tried to find a counterexample, because my intuition said it's wrong.

Here:

[math]
\begin{align*}
x^{x^{x^{.^{.^{.}}}}} &= 2 \\
x^2 &= 2 \\
x &= \sqrt{2}
\end{align*}
[/math]

and also

[math]
\begin{align*}
x^{x^{x^{.^{.^{.}}}}} &= 4 \\
x^4 &= 4 \\
x &= \sqrt[4]{4} \\
x &= \sqrt{2}
\end{align*}
[/math]

but then:

[math]
2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}} = 4
[/math]

Whats wrong here?

>>8810980
put that last one in a calculator

Given a positive integer n, let M(n) be the largest integer m such that:

mC(n-1) > (m-1)C(n)

Evaluate the limit as n approaches infinity of M(n)/n

>>8810980
>Whats wrong here?

That you never proved that any of those things make sense and you are just basically saying
"Algebra is magic".

Take a 3 digit number A, reverse its digits to get B.
If A > B then X = A - B
If A < B then X = B - A

Reverse the digits of X (If its a 2 digit number put a zero on the front i.e. 099) to get Y

Show that X + Y always equals the same number regardless of A

>>8810926
log(x) is a counterexample, right?

>>8810926
let there be a value [math]y \in \mathbb{R} [/math] which [math]f[/math] assumes only finitely many times, let [math]x_0[/math] be the largest point such that [math]f(x_0) = y [/math]. now [math]f( (x_0,\infty) )[/math] is a connected set lying in [math]\mathbb{R} \setminus \{ y \}[/math] and WLOG is therefore fully contained in [math] (y,\infty)[/math]. this implies that [math]f([0,x_0]) [/math], a compact set, contains all [math](-\infty,y)[/math]. contradiction.

amirite?

>>8811154
I also first thought of log(x), but it's domain is (0,infty) not [0,infty)

>>8811112
No, no, no, it was supposed to be a math challenge. One of them is correct, the challenge is to find out why the other one is not. (Yes you can use a calculator but where is the fun?)

>>8811165
Looks correct, well done

>>8811139
That one's pretty cool. Only times X + Y don't equal 1,089 are when the first and last of the 3 numbers are the same

>>8811171
Oh yeah, I forgot that detail in the question

It's a simple enough question to state that my normie Dad even enjoyed hearing the solution

>>8811139
A = a100 + b10 + c10
B = c100 + b10 + a
A > B iff a > c
A < B iff a < c
suppose a > c
X = A - B = (a-c)100 + (c-a)
but (c-a) is not a digit so we have
X = (a-c)100 - (a-c)
X = (a-c)99

Now there are 9 possibilities
X = 099,198,297,396,495,594,693,792,891
Y = 990,891,792,693,594,495,396,297,198

Then notice that:

X + Y = k*99 + (11-k)99 = 99k + 11*99 - 99k = 11*99

Nice.
I am taking an elementary course this semester. Did I do well daddy?

>>8811180
Exactly what I did to solve it

Find a 9 digit number such that the first two digits taken together are a multiple of 2, the first 3 digits are a multiple of 3 and so on, so the whole 9 digit number is a multiple of 9. There's only one solution

>>8810980
If [math] x^{x^{x^{.^{.^{.}}}}} = 2 [/math] then [math] x = \sqrt{2} [/math] is a correct statement.

If [math] x^{x^{x^{.^{.^{.}}}}} = 4 [/math] then [math] x = \sqrt{2} [/math] is also a correct statement.


If [math] x^{x^{x^{.^{.^{.}}}}} = 2 [/math] AND [math] x^{x^{x^{.^{.^{.}}}}} = 4 [/math] then [math]2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}} = 4 [/math] is also a correct statement.


The only "issue" is that [math]2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}} = 4 [/math] is wrong but none of the calculations show otherwise.

>>8811184
Let this number be
qwertyuio
(, means or)
If the first two digits are a multiple of 2 then
w = 0,2,4,6,8
if the first 3 digits are a multiple of 3 then
q + w + e = 3k
r10 + e = 4k
t=0,5
y=0,2,4,6,8
r + t + y = 3k
u = y = i = 0
q+w+e+r+t+y+u+i+o = 9k

r=1,e=6
w=0
q=6,3

THE ANSWER IS:
306450963

Not gonna lie there m8. I didn't even use the answers I originally wrote out. But after I had like 6 digits secured I started trial and erroring.
Great problem m8.

>>8811231
Ehh that defo works, but I forgot to mention that you have you use all the digits 1,2,3,4,5,6,7,8,9. That is, no 0 allowed. With 0 allowed there are many solutions but without it there is only one

I also remember another digits based problem but I've forgotten the details so I'll find that in a bit

>>8811247
Found it again

Find a 5 digit number A such that when you reverse its digits to get a new number B = 4 * A

>>8811261
Fuck should be a 4 digit number

>>8811247
Ah. I don't sure if that makes it easier or harder. With that then it is guaranteed that the 5th digit is 5, for example.
I am now busier so I'll see if I revisit this later.

Here is one my prof left for homework yeasterday:
Find digits x,y,z such that 5, 9 and 11 divide 2x1642y032z

>>8811265
28164280320

>>8811265
>>8811289
28164230325 works too.

The equations are:
z = 0 OR z = 5
x + y + z = 7 OR x + y + z = 16 OR x + y + z = 25
x = y + z OR x + 11 = y + z

Let [math]X, Y \in \mathcal M_{n-1,1}(\mathbb R)[/math] and [eqn]M = \begin{pmatrix} 0 & X^T \\ Y & 0_{n-1}\end{pmatrix}[/eqn]
Give a necessary and sufficient condition for M to be diagonalizable.

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You have four numbers: 2,5,8,9.
You have to use these numbers to get 52, using the four basic operations (+,-,*,/).
You can only use each number once and you can use as many brackets as you want.

>>8809568
Answer is 0.867.

>>8809565
My favourite one from a calculus class is:

Given the longitude and latitude of two cities and the radius of the earth, find the distance between the two cities.

Pretty simple but fun application!

>>8811955
>Give a necessary and sufficient condition for M to be diagonalizable.

A necessary and sufficient condition is that M has n linearly independent eigenvectors.

>>8811997
8*(9-5/2)

>>8809565
HAHAHAHAHAHAHAHAHA yeah we're turning into third world nations. Murica got its niggers and massive corruption /mind control/indoctrination and stupid fat people. And we've got muslims to fuck our shit up

>>8811036
(3+sqrt(5))/2

>>8812028
kek
why not just cut to the chase

>A necessary and sufficient condition for M to be diagonalizable is that M is diagonalizable

>>8812050
>>8812028
wew lads, a condition on X and Y (that can be checked in linear time if you're going to tell me that the condition was also a necessary and sufficient condition on X and Y)

Here's another one:
Let [math](a_n)[/math] and [math](b_n)[/math] be two real sequences such that [math](a_n + b_n) \rightarrow 0 [/math] and [math](e^{a_n} + e^{b_n}) \to 2[/math]. Prove that both sequences converge.

>>8812650
Suppose that [math](a_n)[/math] doesn't converge then one of two cases must happen:

Case 1: [math](a_n)[/math] has an unbounded subsequence [math](a_{n_k})[/math].
Without loss of generality suppose that [math] \lim_{k \to \infty} a_{n_k} = \infty [/math]. Then we must have [math] \lim_{k \to \infty} b_{n_k} = - \infty [/math] but then [math]\lim_{k \to \infty} (e^{a_{n_k}} + e^{b_{n_k}}) = \infty [/math] which is a contradiction.

Case 2: [math](a_n)[/math] has a subsequence [math](a_{n_k})[/math] that converges against a [math] c \neq 0 [/math].
Then [math] \lim_{k \to \infty} b_{n_k} = -c [/math] and [math]\lim_{k \to \infty} (e^{a_{n_k}} + e^{b_{n_k}}) = e^c + e^{-c} [/math] but [math] e^c + e^{-c} \neq 2 [/math] for [math]c \neq 0 [/math] so we have another contradiction.


Therefore out assumption must be wrong and [math](a_n)[/math] converges to [math]0 [/math]. Then [math](b_n)[/math] must naturally converge to [math]0 [/math] too.

Represent 0.dddddddd... As a fraction where d exists in {1,2,3,4,5,6,7,8,9,0}

>>8813078
d/9

>>8809565
x and y are reals.
1) Study f(x, y) = (x + y, xy).

2) We have the system (u = x + y, v = xy). Find x and as functions of u and v.

>>8813084
Prove it

>>8809565
This one is ez.

Let S be Shalaquandra's ass and B the blank space between her ass and the radius

r = r

>>8810980
You have a radius of convergence for the infinite power tower. It is clear that the square root of 2 is in the radius of convergence, but 4 is not.

>>8813141
[math]
0.ddd... = d\cdot \sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k = d\cdot \left(\frac{1}{1-\frac{1}{10}} -1 \right) = \frac{d}{9}
[/math]

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>>8809570
That thread was stupid

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>>8809570
>>8813225
Where exactly is the difficulty supposed to be?

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Requires calculus

>>8813259
[math] \int_0^{10} \left( \int_0^{10 cos(x/10)} y dy \right) dx = 25(5sin(2)+10)[/math]

>>8813287
I forgot to divide by the area of the quarter of circle

>>8812713
Great, that works.
Here's one in algebra.
Let M be a matrix with real coefficients. What is, in terms of M, the least number of coefficients you need to change to make it invertible ?

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what is the total area of the sum of all the grey circles?

>>8813297
HA what kind of nonsens is that! that has no aplication in the real world! dont waste your time!

>>8813297
Infinity.

>>8813291
wat

>>8813290
>>8813287
not quite

>>8813259
For people having a hard time
[math]
\bar{y} = (\frac{M_{xz}}{M}) = \frac{\int\int y dA}{\int\int 1 dA}= \frac{\int_{x=-10}^{x=10}\int_{y=0}^{y=\pm\sqrt{100-x^{2}}} ydydx}{\frac{\pi r^{2}}{2}}
[/math]

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To all the analysts:
How can I express delta as a function of epsilon without including c? I can't seem to figure it out. Any hints?

>>8813826
[eqn] \delta_\varepsilon = \sup_x |g'(x)| \varepsilon [/eqn]

You can do this for all [math]C^1[/math] functions. Just look at the fundamental theorem of Calculus.

>>8814040
are you saying all C^1 functions are uniformly continuous? because that's not true

>>8814046
The ones with [math] \sup_x |g'(x)| < \infty [/math] are.

>>8814058
right so nowhere near all C^1 functions... not even x^2

>>8814040
It's actually
[eqn] \sup_x |g'(x)| \delta_\varepsilon < \varepsilon [/eqn]

[eqn] \left| g(x) - g(c) \right| = \left| \int_c^x g'(t) dt \right| \leq \|g'\|_{\infty} |x - c| < \varepsilon [/eqn]

>>8813291
if M is an n*n matrix, then you need n changes. or more precisely, you need k changes where k is the algebraic multiplicity of 0 as the eigenvalue of M.

>>8814075
I would have said geometric multiplicity,(that is, n-rk(M)), maybe that's what you meant ? Take the matrix [math]M = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}[/math]. You need only change the bottom right coefficient to make it invertible but 0 has algebraic multiplicity 2.
In any case I expected a proof.

>>8813335
Given an n*n matrix, would you agree that you can make it invertible by changing all the coefficients ? (just replace each coefficient by the correspondent coefficient in the identity matrix). Now i'm asking the minimum number of changes you need to do this.

>>8814122
Can't you just take any value [math]a [/math] that is not in the spectrum and then subtract [math]a [/math] from the diagonal?

>>8814147
Yup, that means changing n coefficients (one for each on the diagonal), but you can do better than that

>>8814122
no, I meant algebraic, but it's wrong, should be geometric. I didn't realize that changing one entry could change more than one eigenvalues

>>8813297
[math].86997R^2[/math]

>>8809568
I get 1.758

>>8813297
I think a good first question might be "why is this construction even possible ?"

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>>8815404
This is what Wolfram tells me

>>8812046
Show the werk son.