Calvin, if $f$ and $g$ are positive functions then I’m unable to find any counter-example of $$avg \left(\frac{f}{g}\right) = \frac{avg(f)}{avg(g)}$$ Where avg means the average of the function.
Think about it like this Arjun. The limit as $x$ tends to infinity is equal to the sign of the coefficient of $x^n$, as the $x^n$ term dominates the other terms, and $x^n$ is positive
Then the limit as $x$ tends to negative infinity is equal to the negation of the sign of the coefficient of $x^n$, as $x^n$ is negative
So we have a function that ranges between negative infinity and infinity
And also it's a polynomial, so we know it's continuous
Then just appeal to the intermediate value theorem
@Knight I have a proof by a scaling argument that a counterexample exists but its easier just to look at this explicit counterexample, where I have done the conversion to step functions for you desmos.com/calculator/q5vb6vq8rj
a wobbly blue line at the bottom appears iff the parameters form a counterexample for $\int ab \le \int a \int b$
the proof is fiddly because of the setup of the problem (it would be nice if it was just $\mathbb R$ instead of $[0,1]$, and if we allowed the function to take the value 0) but these are things that can be worked around
find the number of 7 digits numbers the sum of whose digit is even. i want to do this multinomial theorem. by $(x+x^2....x^9)(1+x+x^2....x^9)^6$ how to do further??
@CalvinKhor This example and Holder etc is way too complicated. Just take $f = g$ to be characteristic functions for the interval $[0, 1/2]$. $fg$ is still the same function as $f$, whose integral is $1/2$. $1/2$ is not less than $1/4$ last I heard.
Knight wants to divide so modify this by making it $\varepsilon$ instead of $0$ outside the interval $[0, 1/2]$
In a Venn diagram, then I can draw a circle, label it as the real number set, and divide them into equal areas, one represents THE imaginary number set and the other one represents THE real number set.
under the assumption that $\int$ means $\int_0^x$ so you cannot assume (2) holds cuz there can be an arbitrary constant, and because you can always change what $g$ is, you cannot assume what $f$ is either
so either the question doesnt expect you to notice that you can treat $x \log f(x)$ as a remainder term, or its not $\int := \int_0^x$
if $\int = \int_0^x$ and you were to charitably force $f(x) = x$ to be true then for $1\gg x > 0$, [\sin x] =0 and. $[x]\sgn x = 1$ so the integral is just $x\log x$, and then (3) is also true
so thats what I'll go with, either none are true, or (1) and (3)
type it up, add your thoughts and post it as a question?
also I have no idea what it means to add two planes
and their definition of linearly dependent is not so correct
Here i'll do you a favour
Let $A$ is set of all possible planes passing through four vertices of given cube. Find number of ways of selecting four planes from set $A$, which are linearly dependent and one common point. (If planes $P_1 = 0$, $P_{2}=0, P_{3}=0$ and $P_{4}=0$ can be writen as $a P_{1}+b P_{2}+c P_{3}+d P_{4}=0$, where all $a, b, c, d$ are not equal to zero, then we say planes $P_{1}, P_{2}, P_{3}, P_{4}$ are linearly dependent planes). Let OABC is a regular tetrahedron and $P$ is any point in space. If edge length of tetrahedron is 1 unit. find the least value of $2\left(\mathrm{PA}^{2}+\mathrm{PB}^{2}…
I suggest you ignore everything in the brackets lol it just needs to be burned to the ground so that you can rebuild from scratch
well, i still have no clue what linear dependence of planes are, but if its anything like linear dependence of vectors, then $\{ v_1, v_2, v_2 \}$ can be a linearly dependent set even if the coefficients $a,b,c$ that witness this have $c=0$. What you need is that not all of them are zero (as opposed to all of them are not zero)
@Thorgott its probably just some exercise printed for preperation for some exam of user6numbers's apparently called JEE. Perhaps there's a .doc of this floating about or something
before I did UK-based IGCSEs my local exams were mostly like this. In fact even during my IGCSEs i think my teacher prepared some horrible .doc files though I didn't have the vocabulary to complain about typesetting
i think i can do the exercise but its a lot of work, i have some stuff i'm trying to do too, how about we meet halfway and you tell us what you've tried for your exercise. Do you know what this set $A$ of all planes is?
or ask it as a question so that you can find someone who wants the karma or has more free time
Expand all summands a la $(x-y)^2=x^2-2xy+y^2$ to obtain $$ \vec a^2+\vec b^2+\vec c^2-2(\vec a+\vec b+\vec c)\vec p+4\vec p^2,$$ which is $$ \left(2\vec p-\frac{\vec a+\vec b+\vec c}2\right)^2 + \text{something}$$
Sanity check: Consider $(\mathbb{Z}/2\mathbb{Z})^2$. As a finite ring, this is Noetherian. The zero is $(0,0)$. The only unit is $(1,1)$. The non-zero, non-unit elements $(1,0),(0,1)$ are both idempotent, hence reducible, so there are no irreducible elements in this and no non-zero, non-unit element factors into irreducibles. Does this make sense?
so now I can be smartass by pointing out this exercise asking me to prove every non-zero, non-unit element in a Noetherian ring factors into irreducibles is wrong and then be doubly smartass by pointing out it's true for Noetherian domains
I needed to think about Euclidean rings once to prove that for any PID $R$, $\text{SL}_n(R) \to \text{SL}_n(R/I)$ is surjective for any proper ideal $I$ in $R$
Cuz $R/I$ is then a Euclidean ring and Euclidean rings are elementarily generated so you pullback elementary matrices
for domains, every non-zero non-unit factors into irreducibles iff the ring satisfy the ACC on principal ideals and, in a domain where every non-zero non-unit factors into irreducibles, such a factorization is unique iff all irreducibles are prime
