Exam in Japan

[Anonymous No.8397096]

This question is what a famous cram school in Japan made in order to prepare students for the entrace exam of Tokyo University.
Can anyone make it?

[Anonymous No.8397110]

I took calc in highschool. I'm familiar with everything in this problem but not enough to solve it. So infuriating.

[Anonymous No.8397114]

Negative infinity

[Anonymous No.8397120]

>>8397114
My explanation:
Since its concave up, the integral will be negative. The second half just turns into 0 because the denominator approaches infinity. So infinity times a negative number is negative infinity

[Anonymous No.8397126]

>>8397096
Pull left out of the limit and you get a definition for the integral from 0 to 1. The answer is 0.

[Anonymous No.8397129]

>>8397096
Isnt the sum a reimann sum approaching the integral? Thus the answer should be zero.

[Anonymous No.8397130]

>>8397096
Is this the only question on the exam?
Lmao, I get that they're only highschool students but still, any that have read ahead or have an interest in math and have gotten through the first parts of calc 2 should be able to do this in a ~15 minutes.

Also negative infinity ez.

[Anonymous No.8397141]

To tell you the truth, there was a hint to solve the problem.
This is it.

[Anonymous No.8397142]

>>8397129
I take this back, didnt take the N outside the () into account

[Anonymous No.8397154]

>>8397120
You're full of shit
suppose f"(x) = e^x -> f'(x) = e^x -> f(x) = e^x
e^x>0 for all x in R

[Anonymous No.8397189]

>>8397096
0

[Anonymous No.8397227]

>>8397120
Why would the integral be negative though?
assume f(x) = x^2, than the integral from 0 to 1 isn't negative (1/3)

[Anonymous No.8397242]

[eqn] \left( \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=0}^{n-1} f(\frac{k}{n}) \right) \leq \frac{1}{24 n^2} \max_{\xi \in [0,1]} |f''(\xi)| [/eqn]
So

[eqn] \lim_{n \to \infty} n \left( \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=0}^{n-1} f(\frac{k}{n}) \right) = 0 [/eqn]

[Anonymous No.8397255]

>>8397096
>>8397129
>>8397242
>>8397227

The Riemann sum approaches the same integral. However, there are errors in the approximation. The errors are approximately a triangle (this is where we use the second derivative is positive) of area [math]\frac{1}{2}\epsilon(f(x+\epsilon)-f(x)) [/math] where [math]\epsilon=1/n[/math]. We have to add up all these errors to get the final answer. Notice that this is the derivative and by the Fundamental theorem of calculus, its integral is the value at the boundaries.

In other words, the limit becomes

[eqn] n\sum_{k=0}^{n-1} \frac{1}{2} \frac{f(k/n+1/n)-f(k/n)}{n} \to \frac{1}{n}\sum_{k=0}^{n-1} \frac{1}{2} \frac{f(k/n+1/n)-f(k/n)}{1/n} \\ \to \frac{1}{2}\int_0^1 f'(x)d x = \boxed{ \frac{f(1)-f(0)}{2} }[/eqn]

[Anonymous No.8397260]

>>8397114
>>8397130
Wrong. Counterexample: f = 0

sum 0..n-1 (2*f(k/n)+f'(k/n)*1/n)/2n <= Integral <= sum 0..n-1 (2*f(k/n)+f'([k+1]/n)*1/n)/2n
sum 0..n-1 (f'(k/n)*1/n)/2n <= Integral-sum <= sum 0..n-1 (f'([k+1]/n)*1/n)/2n
sum 0..n-1 (f'(k/n)*1/n)/2 <= n*Integral-sum <= sum 0..n-1 (f'([k+1]/n)*1/n)/2
lim n*(Integral-sum) -> Integral 0..1 f'(x) dx/2

Bichez

[Anonymous No.8397262]

>>8397255
>>8397260
Which is the same since of course Integral 0..1 f'(x) dx / 2 = (f(1)-f(0))/2 :)

[Anonymous No.8397268]

>>8397096
Should be (f(1)-f(0))/2.
See how the areas in between look like little triangles with width (1/n) and total height (f(1) - f(0))? They have total area aproximately (f(1)-f(0))/(2n) so the limit is (f(1)-f(0))/2 ish.

>>8397255
Oh you beat me to it. Anyway this trick is used in the derivation of Stirling's approximation.

[Anonymous No.8397270]

>>8397255
The errors aren't fucking triangles unless f''(x) = 0 but this contradicts f''(x)>0.

[Anonymous No.8397274]

>>8397270
They aren't exactly triangles, but they are close enough to triangles for the purposes of the limit. Uh, to show that is true the only way I know is tedious geometry.

[Anonymous No.8397284]

>>8397270
>>8397274

>The errors are APPROXIMATELY a triangle

The non-triangular part of the errors give rise to higher powers of [math]\epsilon[/math]. If you had included these the final answer would be of the form [math]\frac{f(1)-f(0)}{2}+\frac{a}{n} + \frac{b}{n^2} +... [/math] where [math]a,b[/math] are independent of [math]n[/math]. When you take the limit these go to zero.

[Anonymous No.8397286]

>>8397260
but thats wrong

assume [math] f(x)=0[/math]
then [math]f''(x)=0[/math]
but [math]f''(x)>0[/math]
[math]\therefore f(x)\neq0[/math]
[math]\square [/math]

[Anonymous No.8397293]

>>8397286
Ok ok

[Anonymous No.8397296]

>>8397255
>The errors are approximately a triangle (this is where we use the second derivative is positive)
The areas also form approximately a triangle when the second derivative is negative like in that picture >>8397268. In fact every differentiable function has that property.

Meanwhile you can see by examples that OP's statements doesn't work for all differentiable functions so clearly your "proof" is lacking something.

[Anonymous No.8397300]

>>8397141
Pretty sure this says that
f(x) = f'(a) amirite?
since f'a = fb - fa/b-a

so if f(x) = f'a in the original eq. then...
1/n times the derivative of the sums of f(k/n) will be equal to 0 as n -> inf.

soooooooo the answer isssss:
1
S->f(x)dx
0
?

[Anonymous No.8397305]

[eqn] \int_0^1 f(x) dx = \sum_{k=0}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x) dx [/eqn]

By Taylor we have for k=0, 1, ..., n-1:
[eqn] \frac{f(\frac{k}{n})}{n} + \frac{f'(\frac{k}{n})}{2 n^2} + \frac{1}{6 n^2} \min_{\xi \in [\frac{k}{n}, \frac{k+1}{n}]} f''(\xi) \leq \int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x) dx \leq \frac{f(\frac{k}{n})}{n} + \frac{f'(\frac{k}{n})}{2 n^2} + \frac{1}{6 n^3} \max_{\xi \in [\frac{k}{n}, \frac{k+1}{n}]} f''(\xi) [/eqn]

Therefore

[eqn] \sum_{k=0}^{n-1} \frac{f'(\frac{k}{n})}{2n} + \frac{1}{6 n} \min_{\xi \in [0, 1]} f''(\xi) \leq n \left( \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=0}^{n-1} f(\frac{k}{n}) \right) \leq \sum_{k=0}^{n-1} \frac{f'(\frac{k}{n})}{2n} + \frac{1}{6 n} \max_{\xi \in [0, 1]} f''(\xi) [/eqn]

So by squeeze theorem

[eqn] \lim_{n \to \infty} n \left( \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=0}^{n-1} f(\frac{k}{n}) \right) = \frac{f(1) - f(0)}{2} [/eqn]


It still doesn't use [math] f''(x) > 0 [/math] strangly enough. Maybe this is just a ruse.

[Anonymous No.8397320]

Yes!You are right!
{f(1)-f(0)}/2 is the answer

[Anonymous No.8397321]

>>8397305
if f'' > 0 than zero then f1-f0/2 < 0, correct?

[Anonymous No.8397327]

>>8397305
>>8397320
well done man.
Looks like it probably took you longer to write that out than it did to solve.

[Anonymous No.8397332]

brainlet here, honestly how do i study all of this? HS math was easy af to me but proofs and theorems never clicked with me

[Anonymous No.8397341]

>>8397096
>Exam in Japan
>Question is in English

[Anonymous No.8397342]

>>8397332
this is analysis. so, in my school we start with real analysis then move on to topology.

[Anonymous No.8397407]

>>8397341
バカ外人

[Anonymous No.8397445]

>>8397305

What logic did you follow to solve the problem? What methods (if any) do you use to solve these kinds of questions in general?

I understand your proof but I don't think I would have come up with such an answer, not fast enough at least.

>inb4 get out of /sci/ you undergrad brainlet

[Anonymous No.8397491]

>>8397445
I'm not him but this >>8397260 anon.
The only thing I've used is obvious inequalities about approximations and the integral itself. These
>[math]\sum_0^{n-1} \frac{2 f(\frac k n)+f'(\frac k n) \frac 1 n}{2 n} \leq \int_0^1 f(x) dx \leq \sum_0^{n-1} \frac{2 f(\frac k n)+f'(\frac {k+1} {n}) \frac 1 n}{2 n}[/math]
Could be understood both through pictures like this >>8397268 or through strict inequalities (using the non-decreasing character of f'(x)).

In fact the left-hand side is what the integral would look like if the [math]f'(x)[/math] stayed the same [math]f'(\frac k n)[/math] through whole [math][\frac k n, \frac{k+1}{n}][/math] intervals (instead of increasing) and the right-hand side is conversely what the integral would look like if the [math]f'(x)[/math] was [math]f'(\frac{k+1} {n})[/math] all the way on [math][\frac k n, \frac{k+1}{n}][/math] instead of increasing all this time to that value.

[Anonymous No.8397736]

>>8397305
Damn. I just realized I'm not cut out for a math major.

[Anonymous No.8397748]

>>8397120
>Since its concave up, the integral will be negative
So the integral of x^2 is negative?

>The second half just turns into 0 because the denominator approaches infinity
Proofs? Smells like the divergent harmonic series

[Anonymous No.8397764]

Is it negative infinity?
If we say the part in the parentheses is g(x), there is no g(x) that would make the expression (at least using non-transcendental functions) approach a value other than +- infinity and 0, and since f''(x)>0, the original function must be non-zero concave up, which means the only possible value left is negative infinity.
Last year IB student checking in by the way, so I'm probably wrong

[Anonymous No.8397772]

>>8397748
I had a brainfart. i was picturing a specific function.

[Anonymous No.8397776]

>>8397736

if you go to a regular state school, you won't even do anything this difficult

at worst it'd be a challenge problem on the homework, but your professor would end up explaining because no other brainlets in your class were able so solve it

[Anonymous No.8397786]

>>8397776
What kinda math is this? I'm in my 3rd year

[Anonymous No.8397787]

>>8397786
This is analysis.

You haven't done analysis by your third year in a math major?

Oh boy...

I would tell you to kill yourself, but actually kill your fucking faculty for fucks sake. Actually do it man, don't pussy out on me. Go buy a gun like a "freedom loving american" ;^) WINK WINK hehe and then show your professors some... freedom.

[Anonymous No.8397799]

>>8397787
I've take differential, integral, and multivariable calculus, linear algebra, differential equations. Is analysis advanced calculus?

[Anonymous No.8397803]

>>8397799
>analysis advanced calculus?

Calculus is analysis man. The squeeze theorem was proved in Calc I for me.

Please, PLEASE do not tell me you do not know what the squeeze theorem is.

The rest there should usually be in an actual analysis class so take analysis.

[Anonymous No.8398191]

>>8397305
Now that i see the solution it's actually not that complicated
But no way i would have found this myself

[Anonymous No.8398249]

-1/12

[Anonymous No.8398270]

Brainlet here

No chance of myself solving that, as I could never understand the underlying concepts and I couldn't even memorize the steps of blindly doing it, if the task just varies a bit from the example that I should have memorized.

That's brutal

[Anonymous No.8398328]

>>8397786
>What kinda math is this? I'm in my 3rd year

1.Where do you come from, which college/university?

2.Your major?

We learn that stuff in the 1st/2nd term of university here at Germany, where standards are high if you are studying any decent major like math, physics, (electrical/mechanical)engineering, informatics.

[Anonymous No.8398329]

>tfw engineer with no rigorous understanding of mathematics
I've always felt that we should have more math in our curriculum.
At least I can do triple integrals...

[Anonymous No.8398334]

I swear I have seen this somewhere where you prove how fast the riemann sum converges to the integral.

[Anonymous No.8398398]

>>8397305
would you get zero if you used f( (k+1/2)/n) ) in the sum instead? seems you would have to.

[Anonymous No.8398438]

>>8398328
I understand everything except for the lowercase Xi and the max above it. They never taught us this notation

[Anonymous No.8398442]

This is a very non-trivial question. Assume that the left hand approximations of the integral approach the value of the integral faster than n^1, such that the limit provided is 0. Now, assume the same thing for the right hand approximations. If we take limit n-> infinity n*(LHA - INT(f(x))) = 0, and limit n-> infinity n*(INT(f(x)) - RHA) = 0. According to the rules of adding limits, we get limit n-> infinity n*(LHA - INT(f(x))) + n*(INT(f(x)) - RHA) = 0, which simplifies to limit n-> infinity n*(LHA - RHA) = 0. However, this is not the case, as LHA - RHA simplifies to f(0)/n - f(1)/n, so limit n-> infinity n*(LHA - RHA) = limit n-> infinity n*(f(0)/n - f(1)/n) = f(0) - f(1), which is not necessarily 0, so it is not true that both the LHA and the RHA approach the value of the integral faster than linearly. Because of this, we know that at least one of the integrals lim n->infinity n*(INT(f(x)) - LHA) or lim n-> infinity n*(INT(f(x)) - RHA) is different than 0, but why? and how do we calculate the limit?