I'm bored as fuck, Veeky Forums. Give me some math problems

Lim of X as X goes to of (X(sin(erf(x)))

A is 70 B is 60

>Lim of X as X goes to of

lol epic fail!!! XDDDD

I know the answer to my question I just asked but how can it be true for one solution if it's a quadratic equation

(x-a)^2 has one real solution a=0

the vertex is at (a,0)

Using the nth term test, determine the convergence or divergence of the sequence. If the sequence converges, find its limit.
[math]a_n=cos2nπ[/math]

By definition, it diverges because it doesn't =0 right? The answer key says it converges. Can someone explain this to me? I know it =1, so that means it converges onto that point..?

yes, I am retarded.

in fact isn't it true even for two complex solutions (assuming real coefficients on the polynomial)?

it converges because its just a constant sequence of 1s, every cos(2pi*n) is equal to 1 for integer n

it's not because it doesnt equal 0

Go to Project Euler and do some programmin.

>do some programming

No.

Not OP, but how do I evaluate integral like this?

aw fuck, I realize what I misunderstood now after reading this and some other material.
thanks for answering, user!

no one knows, thats why the picture says its unsolved

the thing on the right is zeta(3) (en.wikipedia.org/wiki/Apéry's_constant), which is some irrational number but not even known whether to be transcendental or not

special values of zeta are generally only known for even values

Lemma:
The sum 1 + 2 + ... + n is equal to n(n+1)/2

Use this fact to come up with a formula for

1^2 + 2^2 + ... + n^2

isn't the nth term test for series not sequences?

I see someone already helped but a sequence of 1s definitely converges by the definition of convergence but a series from that sequence definitely does not converge.

if you mean triple integrals in general, just work on the leftmost integral first, keeping all other variables as coefficients if need be.

It's quadratic not polynominal. Ax^2 + bx +c only

quadratics are polynomials...

for example x^2+1=(x+i)(x-i)
the average of the two solutions is [i+(-i)]/2=0, which is the first coordinate of the vertex (0,1)

Easy.
(n(n+1)/2)^2

That's an interesting conclusion, but prove to me that it's true.

I should add now that I read "using the nth term test," not "given the nth term.." (which is what was on the test).
That's the main thing that had me confused. Ignore my stupidity.

No worries. Such is the road to learning mathematics and combining it with reading comprehension. We've all been there.

I mean how do I find the value of this [eqn]\int_0^1 \int_0^1 \int_0^1 \frac{\mathrm{d} x\, \mathrm{d} y\, \mathrm{d}z}{1-xyz}[/eqn] integral itself if I want to compute zeta this way?

did you read the text in the picture? it says its an unsolved problem

come up with a way and you'd likely get published in one of the top 3 number theory journals

(cont.)
people have been trying to compute zeta(3) for ~300 years btw, so good luck

I was so sure it's some undergrad integral that I've read it backwards

Easy, use the fact that [math]\frac{1}{1-xyz}=\sum_{n=0}^\infty(xyz)^n[/math]
Ignore the people saying it is impossible.

give this man a fields medal

You are too kind!

you have to use a trig sub

...

1+1=2, which is proved simply by reflexivity in MLTT.

Fun problem in one of my problem sets.
Consider a simple closed continuously differentiable curve [math]\Gamma[/math] enclosing a compact convex domain [math]D[/math]. Prove that any inscribed triangle [math]ABC[/math] (ie. whose points are on the boundary [math]\Gamma[/math]) that maximizes the perimeter among inscribed triangles is a periodic billiard trajectory (ie. it satisfies Snell's law of reflection).
Sounds complicated but it's basically multivariable calc

whose vertices*

>Cant into abstraction

Induction?

Have fun. I got x=2

Try it

Too trivial to prove but still a beautiful fact.

okay now that is extremely long and is probably unnecessarily long. I guess you can assume f(n) is true, then for f(n +1) use the derivative and show that it works for all x > 0 and x E R. Since N+ is a subset of R+, then it must also satisfy.

Looks like you can just put x=2 in the equation, then prove it's the only solution by analytical methods

Expressions cannot have solutions :P

I think you mean real root

Kek, good one, I'll use it if I ever have to scare undergrads off:
Let [math]x \in \mathbb R[/math]. We have [math]x^2 - 4x + 7 = (x-2)^2 + 3 \ge 3[/math], and therefore [math]0 \le \frac{\pi}{x^2 - 4x + 7} \le \frac{\pi}{3}[/math] and finally [math]0 \le \sin\left(\frac{\pi}{x^2-4x + 7} \right) \le \frac{\sqrt 3}{2}[/math].
Now, assume [math]4\sin\left(\frac{\pi}{x^2-4x + 7}\right) = \sqrt 3 (\log_2 |x| + \log_{|x|} 2)[/math]. Since the right hand side has the same sign as [math]\log x[/math], we must have [math] \log |x| > 0[/math].
We then have [math]\frac{\log |x|}{\log 2} + \frac{\log 2}{\log |x|} \ge 2[/math], and therefore [math]4\sin\left(\frac{\pi}{x^2-4x + 7}\right) \le 2\sqrt 3 \le \sqrt 3 (\log_2 |x| + \log_{|x|} 2)[/math], putting everything together
Because of the assumption, each inequality is an equality and we then have [math]\frac{\log |x|}{\log 2} + \frac{\log 2}{\log |x|} = 2[/math], which implies [math] |x| = 2[/math], as well as [math] \sin\left(\frac{\pi}{x^2-4x + 7} \right) = \frac{\sqrt 3}{2}[/math], which implies [math] x = 2[/math], using the fact that [math]\sin[/math] is one-to-one on [math]\left[0,\frac{\pi}{2}\right][/math].
Conversely, it is easy to check that 2 is a solution.

A closed form expression for [math]S_n^d[/math] where

[eqn] S_n^d = \sum_{k=0}^n\binom{n,k}k^d[/eqn]

Here's a problem I gave my Calc1 students.

Evaluate the following exactly: [math]\displaystyle \int_0^{\pi} \frac{x \sin(x)}{1+\cos^2(x)} dx [/math]

Protip: Even wolframalpha can't do this; ask it and it'll give you a decimal approximation. But you can find an exact answer using nothing more than Calc1.

Correction:
[eqn] S_n^d = \sum_{k=0}^n\binom{n}{k}k^d[/eqn]

that is pretty good i am thoroughly impressed but im a retarded undergrad so...

how did you come up with the general idea to solve this? did you just mess around and stumble upon the solution, or was it obvious to you when you saw the problem what techniques you should try, or both?

solve

Well at first, I thought it was just a randomly typed problem like you sometimes see here but then I noticed the [math]\log_2 |x| + \log_{|x|} 2[/math] which I knew was bounded from below by 2 so I knew it couldn't have been completely random. Then I looked at the polynomial and saw that it was bounded from below by 3; then I checked how it bounded the sine from above and; suprise, the supremum of the left hand side was exactly the infimum of the right-hand side and equality happened exactly when x=2.
The thing is, it looked so complicated that it was either random, or carefully constructed so that you would not have to compute anything to solve it.
But it was a fun one, wish I would have come up with it when I was TA-ing

youtube.com/watch?v=Y30VF3cSIYQ

Here. This one was extra credit for week 1 of my Calc 2 class. Had to sleep on it a few days.

Sorry, bad pic. The problem starts at the words "the figure" and ends after the diagram.

Assuming the radius is 1 (which doesn't change anything since everything is homogenous):
The area of the angular sector is [math]\frac{\theta}{2}[/math]
The area of the triangle OPR is [math]\frac{1}{2}||\overrightarrow{OP}\times \overrightarrow{OR}|| = \frac{\sin \theta}{2}[/math]
Hence [math]A(\theta) = \frac{\theta - \sin \theta}{2}[/math].
We also have [math]B(\theta) = \frac{\sin(\theta)(1-\cos \theta)}{2}[/math].
Finally; a Taylor expansion around 0 gives [math]\frac{A(\theta)}{B(\theta)} = \frac{\theta - \sin(\theta)}{\sin(\theta)(1- \cos \theta)} = \frac{\theta^3/6 + o(\theta^3)}{(\theta + o(\theta))(\theta^2/2 + o(\theta^2))} = \frac{\theta^3/6+o(\theta^3)}{\theta^3/2 + o(\theta^3)} = \frac{1}{3} + o(1)[/math].
Hence, [math]\frac{A(\theta)}{B(\theta)} \underset{\theta \to 0^+}{\rightarrow} \frac{1}{3}[/math]

you are throwing an n-dice

compute the expected length of a sequence with non-repeating numbers for any n

He's experienced in solving shitty homeworks from his study. Also, he thinks.

Let K be some finite extension of k(x) - you can assume quadratic at first. I'd like to find some way of characterizing elements of K who have the same norm over k(x).
In case, the norm of f is defined as the product of f by all its conjugates.

Find a formula for solutions of a quadratic equation in char 2 (this is a good one!)

remember the stacking boxes problem that isn't quite np hard because of dynamic programming?
If you decide that each box can only be used once, does it become np hard?

0,acos(0.5+sqrt(2)/2)

Not calculus but this baby tier problem stumped me.

A box with mass slides down a ramp on a 30 degree incline. What is the velocity and acceleration at the bottom of the ramp?

There is no coefficient of friction, the mass is not given and you do not assume the system is frictionless. Also you cannot look up any coefficients of friction.

A neutrino is born in the center of the sun. Calculate the probabilities that it will pass through your thumbnail, assuming you're standing on earth at the equator.

(Actual PhD program question)

Use google for data

I forgot to mention the ramp is 4 meters long.

Prove that any compostition of shift (on a vector lying on the plane), rotation and similarity transformation of a plane has a fixed point (f(x) = x).

Protip: you can't.
I can tho.

It's false for n=3, to start with.
((3*4)/2)^2 is not equal to 14, which is 1+4+9.

Now try doing it with only your Calc 1 toolbox. No vectors. No series. Just derivs and l'Hopital.

I got x=2 as well.
The quadratic is greater than or equal to 3 for all x. Thus the LHS of the equation is smaller than or equal to 2sqrt(3). The content of the parentheses on the RHS is greater than two for all x not equal to 2, so x=2 is the only solution.

Here's another from Stewart's that looks kind of brain teaser-y.

Divide a circle into equal area pieces which are not congruent, using only chords.

A shift by a non-zero vector doesn't have a fixed point.