Basic Mathematics - 2020-07-03

Knight

10:27

@user21820 Hello Hello Hello !

user21820

@Knight Hello!

Knight

I want to know how the average value of the functions depends on the function's nature, it is given that $f \lt 0$ for all values in its domain.

Then, the average value of $f$ in the interval $[0,1]$ will be $$\int_{0}^{1} f(t) dt$$ but can I get something more, just a little more information?

@user21820

user21820

@Knight Like what kind of information?

Knight

@user21820 Something about it's greatest lower bound

user21820

Then no.

Just because it's negative doesn't tell you that it can't go below something.

You can easily construct a function that has an arbitrarily low spike.

Knight

10:38

@user21820 Okay.

But let's say it's $f(0)=f(1)$ then?

We have $ \frac{f(0) + f(1)}{2}$ as an average? (am I completely wrong here?)

So we have average as $f(0)$

Which is negative

user21820

11:04

Firstly, you didn't state any condition on f, so it could very well have a point discontinuity that is arbitrary. The Riemann integral is unaffected by a single point value.

Secondly, I don't know why you would think (f(0)+f(1))/2 has anything to do with the integral, since f(0) and f(1) tells you nothing about what f is in the middle.

I think you should try coming up with various simple examples of negative functions and see what kind of extremes you can get. @Knight

Knight

@user21820 $f$ is continuous and everywhere $f \lt 0$. Then for the interval $[0,1]$ we have $f(0)=f(1)$.

user21820

Even if it is continuous, try to design a function satisfying your conditions with a spike.

Knight

Can we say that the average of $f$ in the interval $[0,1]$ is $f(0)$?

user21820

Why should we want to say such a thing?

Everything must have a reason.

I don't see any reason to call f(0) "the average of f".

Knight

Because we want average. And we have $$\frac{f(0)+f(1)}{2}= \frac{2 f(0)}{2}= f(0)$$

user21820

11:14

You can delete a message if it accidentally posts twice.

@Knight Anyway what you wrote doesn't make sense at all. Don't bother using LaTeX; that doesn't affect the mathematics.

You're effectively saying that (1+3)/2 = 1.

Knight

But it is 2

(1+3)/2= 2

user21820

That's why I said what you wrote doesn't make sense at all.

Just read it. You wrote "(f(0)+f(1))/2 = f(0)".

Knight

$f(0)=f(1)$ is given

user21820

Oh you mean based on that assumption.

Fine, but it still has nothing to do with the "average of f".

It's just the average of f(0) and f(1).

Not the average of f.

The average of f is usually defined as the integral you wrote earlier.

So why do you think f(0) and f(1) has anything to do with it?

Knight

Average of $f$ in the interval [0,1]

That integral was a finer approximation for the average, this one is error-ful (but still) average

user21820

11:21

Then I don't understand why you at first asked what you can say about the integral. The answer is you can't say anything except that it is negative.

As I said, construct your own negative continuous function f on [0,1] with f(0) = f(1) with a spike having width w and height h. And compute the integral.

And see how you can change the integral arbitrarily (as long as negative) by changing w,h and the other minor parts of f.

Knight

Yes I see

user21820

If you pick a triangular spike, it's easy to compute the integral over the spike to be w·h/2.

Knight

Yes

user21820

I suspect you have a real question but you're not really asking it. However, you can also try to convince yourself that you can make the integral as close to zero as you want no matter how negative f(0) and f(1) are.

Just make those downward spikes at 0 and 1 as thin as needed.

user69608

11:44

any probability expert ?

user21820

@user69608 I'm not an expert, but you can ask and see!

user69608

@user21820 I am probability very hard math.stackexchange.com/questions/3742411/… like in this question even after the solution it took me around an hour to understand . how do i improve my ability in assuming cases and proceed.

user21820

@user69608 Well, probabilistic processes are always best thought of in terms of states.

So one of the comments is really a good one:

I suggest going by states. There are really only four "active" states, labeled by the value of the last toss and the coin you are going to flip next. You can't return to the starting state since, after the first toss, the game can always end on the next one. — lulu 24 hours ago

Learn to identify the critical information needed to fully describe the current state of the process.

Then you can more easily analyze the process and the probability of various outcomes or paths.

user69608

@user21820 can you elaborate on this "best thought of in terms of states"

user21820

Ask yourself, what minimum information do I need to know at each point in the process to determine the possible subsequent behaviour of the process?

Clearly it cannot be "nothing", because then you can't tell after the next flip whether you've gotten a double head or double tail or not.

You also need to know which coin you're supposed to toss next.

That's what drives lulu's answer.

Based on that minimum information, set up the process as something that moves from one state to another according to the local step (in this case a toss of the appropriate coin).

Every probabilistic process can be thought of in this way.

user69608

11:56

i kept thinking like this HH...,THH..TT...,HTHH...TT but could not proceed.what should have i have thought after that i proceed towards solution?

user21820

@user69608 Well, do as I said instead.

You can't learn how to do it without trying to do it for every problem you come across.

Analyzing information flow is important in every area of mathematics, not just probability. So you really do need to learn to do it.

Viewing a process as a state-to-state operation is an important method of abstraction. Here is another example of that:

8

A: Understanding matrix multiplication

Although the duplicate question has an excellent top-voted answer, it may be a bit too dense for beginners. Here's a simple but instructive example. Fibonacci numbers! We all know them, don't we? =) $\def\nn{\mathbb{N}}$ Let $F_0 = 0$ and $F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ for any $n \in \n...

user69608

@user21820 ok yeah thanks . like in the question"In a regular decagon find the probability that the two diagonal chosen at random will intersect inside the polygon?" how do i start thinking

user21820

@user69608 You start by removing all the unnecessary information. Such as the geometry setting.

Since all that matters is the circular order of the vertices chosen, the geometry setting is just a distraction. You must learn to identify and remove such useless information.

So that you can rewrite the problem in its cleanest form.

user69608

how does that help?

user21820

It helps by forcing you to actually determine what the problem is about. What the internal structure of the problem is.

Same with the fibonacci example. You should read that first if you have not done so.

Because otherwise you won't really understand the power of such abstract factoring of structure.

user69608

12:09

ok i will read that:)

Knight

@user21820 What does the frequency of a periodic function tells us?

Does it mean number of complete oscillation in a unit interval like $[0,1]$?

user21820

More or less yes.

Knight

Okay... but I think there lies a big reason for a person like you to say “more or less”. May I learn something from that?

user21820

Heheh yes it's that you are probably talking about a function with a nice obvious period, like sin,cos... But consider a function f on the reals where f(x) = 1 if x is rational but 0 if x is irrational. Then f(x) = f(x+r) for any rational r.

And it doesn't have a single period, nor a smallest one.

Neither is its frequency well-defined.

Knight

Yes and it may not oscillate at all! It can even be a concave up or down

user21820

12:20

I'll be away for a bit. See you later!

Knight

See ya

Arjun

@Knight, hi

@user21820, can I ask a question?

user69608

12:57

how to do the indefinite integration of logx *gif(x)*sgn(x)d(x-gif(sinx))

user21820

13:07

@user69608 What is gif??

@Arjun Yes you can, but you must explain what you tried.

user69608

greatest integer function

sgn is signum function

user21820

Oh, that's usually called "floor".

user69608

oh we call gif

user21820

I don't know who is "we", but the standard notation is "floor", not "gif".

And it doesn't make sense to have "d(x-gif(sinx))".

Arjun

@user21820 , sorry for the inconveniences in reading that but in our country, this system is followed

user21820

13:11

@Arjun Uh?? If you don't wish to write in terms of things that the rest of the world understands, then please don't ask me.

Arjun

@user21820 , I didn't write it sir

@user69608, wrote it

user21820

It doesn't matter who wrote it. If it doesn't make sense, it doesn't make sense.

user69608

@user21820 ok it was given integral is xlog(f(X))-g(X) and asked to find f(X) and g(X)

user21820

@user69608 Impossible.

State the question exactly as you were given it.

user69608

user21820

13:19

Ok that is a really strange question. I'm going to say that the integral (as written in the question) does not make sense unless you can precisely define what it means.

If you cannot do that, then you have to ask whoever gave you that question to do that. A question that one does not even understand is meaningless.

user69608

ok leave that may be something wrong.

Arjun

@user69608 , I'll give some answer in a few hours

user69608

Arjun

13:35

@user69608, are u in 12 th?

user69608

@

@Arjun yes

Arjun

Delhi?

user69608

no

Arjun

Where?

user69608

gujarat

Arjun

13:37

Nice, any institute?

user69608

resonance

Arjun

Uttam, ati uttam:-)

user21820

@Arjun There is no answer to a meaningless question. If you wish to give any so-called answer, you would first have to define what the integral means. If not, please do not say there is an answer.

@user69608 What did you manage to get for this question?

Arjun

@user21820 ,I will answer in terms which most probably only @user69608 will understand and these type of meaningless questions are often made meaningful in my country once a solution is found

user21820

@Arjun No. Please do not insist on your way, otherwise you are welcome to leave this room.

I teach proper mathematics, not nonsense, in this room.

Arjun

13:44

@user21820 , as you wish

user69608

@user21820 i also did not understand that question please lets leave that

in above functional equation question it is easy from option but how to prove that?

user21820

@user69608 Of course. I don't mean to be dismissive, but let me just say that I am very familiar with the rigorous mathematical foundation for the Riemann integral and Lebesgue integral (from the mathematical field called measure theory), and that particular expression does not fall into "standard mathematics", so its meaning must be defined by the author or whatever before it can be answered. Ok, let's move on!

@user69608 Solving functional equations tend to revolve around finding key facts about it via substitution and algebraic manipulation, and then using various properties like continuity to squeeze solutions out of that.

For example, it's easy to find f(0).

user69608

how?

user21820

Wait I made a mistake lol. f(0) can't be determined algebraically.

user69608

yeah

user21820

13:54

Well here is my thought process.

This is a multiplicative form of an averaging identity.

The additive form is f(x−y) + f(x+y) = 2·f(x).

So the natural thing we should want to do is to collect and equate the ratios:

user69608

how did u do?

user21820

f(x+y) / f(x) = f(x) / f(x−y).

But before we can do that we must prove that f is nonzero everywhere.

Can you do that?

user69608

yeah it should be non zero otherwise 1/f(X) will not be defined?

user21820

Yes that is why we must prove it. Otherwise we cannot do the division.

I'm asking you to try to figure out the proof.

It's not hard.

Try to find a substitution that tells you that f is nonzero everywhere.

user69608

like some rational number substitution x/ytype?

user21820

13:59

It simpler than that.

Just state some results from substituting.

user69608

f(0)=f^2(X)/f(2x)

user21820

You should not do the division because you haven't proven that it is nonzero.

Instead, just say f(0)·f(2x) = f(x)^2.

Don't write "f^2(x)" because it means "f(f(x))" in standard notation.

"f(0)·f(2x) = f(x)^2" unfortunately does not seem to tell you that f(x) or f(2x) is nonzero, because they could be both zero...

Find another substitution.

If you try a few, you will surely find a suitable one.

(I know because I already found it haha..)

user69608

f(0)f(2x)=f(x)^2

user21820

(There are some problems where there is difficulty in just finding substitutions, but this is not one of those hard ones.)

user69608

sorry - in rhs

user21820

14:04

@user69608 Yes that's what I wrote above to fix your attempt, but it doesn't seem to get anywhere. (There is a way to get information from it, but it's not worth the effort.)

Just try more substitutions.

user69608

ok trying

can u give a hint?

user21820

Well, basically, you tried the substitution (x,y) = (t,t).

Typical simple substitutions include making one or more variables zero.

And making them equal, like you tried.

If those don't yield anything, we move on to find substitutions of special constants that simplify.

user69608

letting x=0 y=y?

user21820

Yes!

It gives f(y)·f(−y) = f(0)^2 ≠ 0.

Can you see why f(y) must be nonzero?

user69608

because f(0) is non zero?

user21820

14:13

Well.. I suspect you're not familiar with proofs. Let me explain a bit more then.

The given equation implies that f(y)·f(−y) = f(0)^2 for every y in the domain of f.

We also are given that f(0) ≠ 0.

user69608

yeah they dont make us do like this.sadly

user21820

By basic properties of real/complex numbers, f(0)^2 ≠ 0 as well.

So f(y)·f(−y) ≠ 0.

user69608

yeah

user21820

Now consider the two possibilities: (1) f(y) = 0. (2) f(y) ≠ 0.

(1) is impossible, because then f(y)·f(−y) = 0·f(−y) = 0 by the basic properties.

Therefore the only possibility is (2).

And we have shown this for every y in the domain of f.

user69608

yes

user21820

14:16

So great, you now know the rigorous reasoning for why f is nonzero everywhere.

That allows us to freely divide by f(...) whenever we like.

So let's go back to what we wanted to manipulate.

user69608

but sadly in exams we have only around 3 minutes for answering such question

user21820

We can come to that issue later. In case you're wondering, I am particular about rigorous reasoning because in my experience many students do not truly understand what they are doing, and even many high-school teachers cannot correctly solve questions that are designed to test their understanding, including this apparently simple mistake of dividing by zero.

But better to start learning now than later.

user69608

yes you are right

user21820

We were at f(x+y) / f(x) = f(x) / f(x−y).

Substituting a constant for y, we see that the ratio of increase over each interval is constant.

For instance, f(x+1) / f(x) = f(x) / f(x−1), for every x.

user69608

yup

user21820

14:21

So automatically I know that I can squeeze all the rational points out of this, by using induction.

The question as stated was imprecise and didn't specify the domain of f, so let us assume it is R.

user69608

squeezing rational points means?

user21820

First prove that f(k) = f(1)^k for every integer k, by induction in each direction.

Next prove that f(1/k)^k = f(1) for every positive integer k.

Oops I forgot about f(0).

Sorry.

Prove that f(k) = (f(1)/f(0))^k for every integer k.

And (f(1/k)/f(0))^k = f(1)/f(0) for every positive integer k.

Hmmph.. maybe I should have defined a new function to get rid of that f(0).

But anyway, the idea is that f on all rational points is automatically fixed by f(1)/f(0).

user69608

f(2)=f(1)^2/f(0)

user21820

Sorry.

f(k)/f(0) = (f(1)/f(0))^k

I keep forgetting the f(0).

I should do less in my head and more on paper.

But I'm a bit lazy today as it is friday.

=D

Do you get them now?

user69608

oops no

user21820

14:28

Well you did get f(2)/f(0) = (f(1)/f(0))^2.

Same for f(3)/f(0).

The general case is by induction.

That's the first step. For the second step you just divide the interval [0,1] into k equal parts and apply "f(x+y) / f(x) = f(x) / f(x−y)" to each part.

user69608

f(3)=f(1)^3/f(0)^2

user21820

That is, use f(x+1/k) / f(x) = f(x) / f(x−1/k).

@user69608 Yes that's right, and it indeed follows what I claimed:

> f(k)/f(0) = (f(1)/f(0))^k

(After fixing my careless mistakes.)

user69608

yeha

user21820

So do you get the second step?

user69608

@user21820 how this?

user21820

14:33

You need to see the manipulated equation as stating that the ratio of two points separated by the same amount is constant.

Choosing separation interval of 1/k gives that equation.

And that is what you need to get "(f(1/k)/f(0))^k = f(1)/f(0) for every positive integer k".

user69608

oh yeah

user21820

It's essentially the same reasoning as with the other case. But to see that you should do it relies on experience.

user69608

i have very little experience rn

user21820

It's okay. Do more mathematics and you will get more experience.

I know to do this because of a certain theorem:

> If two continuous functions on R agree on every rational point, then they are identical on every point.

The above two steps are just the natural stepping stones to proving that f(r)/f(0) = (f(1)/f(0))^r for every rational r.

user69608

but how we get to the function like something after integration that e may be coming?

so should we try differentiating the functions?

user21820

14:41

The equation "f(x+y) / f(x) = f(x) / f(x−y)" and the first two steps tell us that it agrees with an exponential curve on the rational points. To be complete we have to check that f(1)/f(0) > 0, which we can obtain from f(1)/f(0) = f(1/2)^2 > 0. That implies that (f(1)/f(0))^x is continuous in x. Then we can apply the above theorem to conclude that f(x)/f(0) = (f(1)/f(0))^x for every real x.

The conclusion should explain why the exponential function comes into the picture.

user69608

oh yeah nice

user21820

However, options (1) and (2) don't look correct (I didn't bother checking yet) because it doesn't match the final conclusion I get.

user69608

user21820

Option (4) is also clearly wrong because we are free to set f(1) and f(0) and it will satisfy "f(x+y) / f(x) = f(x) / f(x−y)".

user69608

yup

user21820

14:44

So if we want to get the answer quickly, we mentally see that "f(x+y) / f(x) = f(x) / f(x−y)" gives us all the rational points, and that it matches an exponential function with base f(1)/f(0), and so (4) is out and (1) and (2) don't match. It just takes a bit of time to check (3).

I don't like time-pressure exams, but I understand that many students have no choice but to do well at those.

I prefer exams that test conceptual understanding, and multiple-choice cannot do it.

user69608

true that

user21820

@user69608 This question is false, even without doing any calculation, because if (p,q,r) is a solution then so is (p·k,q·k^2,r·k) for any positive integer k.

But if you want 'simplified form', then I got to think...

user69608

@user21820 why this (pk,qk^2,rk)?

user21820

It preserves the value of (p+sqrt(q))/r.

user69608

@user21820 how that sorry i am not getting

user21820

14:50

(p+sqrt(q))/r = (p·k+sqrt(q·k^2))/(r·k).

user69608

oh hmm

user21820

So there is no unique (p,q,r) that satisfies cos α = (p+sqrt(q))/r, and none of the answers are preserved if we replace (p,q,r) by (p·k,q·k^2,r·k). That's why I said the question is false.

user69608

oh yes

can you now tell that decagon probability question?

user21820

Well, can you express the "intersects inside the polygon" condition in terms of the cyclic positions of the vertices?

user69608

@user21820 but if we neglect that is there a way out?

user21820

14:57

@user69608 I would need to think to know what is the minimal value of cos α. I am a bit rusty at such kinds of problems because most of them are just based on some tricks.

user69608

are you free from work any?

user21820

I'm free now, but was thinking about that problem. I know some high-power tools that would most probably work, but that's surely not the intended solution.

@user69608 I solved the question if we want p,q,r to be as small as possible. Let a,b,c be complex numbers on the unit circle. Then the question is equivalent to asking for the minimum possible Re(a) if Re(a) + Re(b) + Re(c) = 1 and Im(a) + Im(b) + Im(c) = 1.

Well, that means we have 3 unit-length segments from (0,0) to (1,1) in the complex plane, and we want to make the first segment go as low as possible.

user69608

15:12

yeah

user21820

In that situation, the next two segments will be taut.

So you have essentially just two segments, first one length 1 and second one length 2.

I'm too lazy to compute the answer, so I'll leave that to you. =P

user69608

@user21820 how to do after this and how length 2?

user21820

3 unit-length segments.

For the first to go as low as possible, the second and third must be taut (straight).

user69608

@user21820 how to put this imagination into mathematical equation

user21820

@user69608 Uh... I can prove it, but it's not a short proof. There may be an algebraic way, but I'm too lazy to find it.

Just too bad that I know too many high-power tools.

user69608

15:19

@user21820 (:

Knight

@user21820 I wish everyone should think like you Madame.

My academic life was destroyed only because of people who thinks and compete for questions like that

user21820

@user69608: In case you're curious, here is a sketch. Since the unit circle C is compact, C^3 is compact too. Consider the set S of all triples (a,b,c) in C^3 that satisfy Re(a+b+c) = 1 and Im(a+b+c) = 1. These equations have closed graphs, so S is also closed. Since S is a closed subset of a compact metric space, S is compact as well. Thus by the extreme value theorem, every continuous function on S attains its minimum value. In particular ( (a,b,c) ↦ Re(a) ) is continuous on S. [cont]

[cont] So it attains its minimum at some point (x,y,z). If y ≠ z, you can fudge them a little bit (I'm handwaving a bit here) to make x go lower. If you draw out the cases you will see that it is true. It's just troublesome to do it properly. Since (x,y,z) is already optimal (i.e. Re(x) ≤ Re(a) for any other (a,b,c) in S), such fudging is impossible, so we must have y = z.

The reason I think like this immediately is because many many optimization problems can be shown to attain their extreme values by this kind of argument, and that means we can directly focus on the optimal solutions to prove something about them.

user69608

@user21820 i get cos alpha =(-1+-sqrt(7))/4

user21820

@user69608 Oh okay. Is that the official answer?

user69608

yeah fro minimum we would take - sign right?

@user21820 most probably this is correct

user21820

15:32

@user69608 Yes.

@Knight Unfortunately the world is very non-optimal. We just have to make the best of what we can.

user69608

can u give intuitive explaination of y=z?

user21820

The intuitive explanation is already what I said: the second and third segments must be taut.

Maybe you don't know the meaning of "taut".

It means "pulled tight".

I also made a silly mistake, to minimize Re(a) you don't want a to go as low as possible, you want it to go as leftward as possible. But the idea is the same. Just pull it leftward as far as it would go.

You can easily imagine that the second and third segments will be pulled straight.

This is just physics intuition.

user69608

can we think of it like vector addition somehow?

user21820

I'm not sure what you mean. I already said that you have 3 unit-length segments corresponding to those 3 complex numbers a,b,c.

a+b+c is of course the sum of a,b,c.

So it is like segments joined end to end.

I don't want to call them vectors because that concept is not relevant here.

It's just addition of complex numbers.

user69608

@user21820 it looks correct but somehow i am not able to satisfy myself

user21820

15:41

@user69608 Get 3 straws of the same length c, put a string through them to make them joined end to end, attach one end to (0,0) on the ground and the other end of the 3 straws to (c,c) on the ground, and pull the free end of the first straw leftwards.

It will make the other two straws straighten out.

If you're talking about the proof sketch I gave above, it is deliberately a handwaving proof, and I do not claim it to be rigorous.

user69608

yeah

that implies what?

user21820

"handwaving" is a term used by mathematicians to mean that they just "wave their hands about and say it can be proven but don't actually prove it".

user69608

oh nice

dont they doubt sometime on their handwaving proof

user21820

Of course.

In professional mathematics, we first develop handwaving proofs, and then spend most of our time trying to make it rigorous.

=D

Sometimes, it fails, and we find that our handwaving proof goes poof!

user69608

great!

@user21820 what does straighten up implies?

user21820

15:50

@user69608 It means the segments are pointing in exactly the same direction.

So the corresponding complex numbers are equal.

user69608

@user21820 nicely explained!!!

finally i got it

user21820

That's great!

Arjun

@user21820, sorry for my behaviour the last time, i know you should be mad at me, i am not saying you just try to forget everything or forgive me. i just wanted to apologize for my rude behaviour.

peace out ;)

user21820

@Arjun That's okay. I wasn't mad at you, and I accept your apology! Thank you for that!

Arjun

@user21820 , i respect your kindness. good night

user69608

15:58

@user21820 i sometime hesitate to ask silly doubts. do u ever get frustated for someone asking stupid doubts?

user21820

@Arjun Good night!

@user69608 I don't get frustrated with anyone who wants to learn.

What I do get frustrated with are people who do not want to learn.

user69608

@user21820 :):

Knight

@user21820 I need help with Fourier series. "The condition that a function is always positive we have to have $a_0$ to be greater than the sum of the modulus of all other Fourier coefficients"

But I don't understand how. Here is the source for that statement

math.stackexchange.com/questions/2168897/answer/submit

user21820

@Knight I think you posted the wrong link.

Knight

math.stackexchange.com/questions/2168897/answer/submit

I don't know why it's not working, try this one

1

Q: Function always positive: Fourier Series

I have a function $v : \mathbb{R} \rightarrow \mathbb{R}$ which describes the norm of the velocity of an object moving under gravitational forces. As the object describes a closed period path, hence the function $v$ is periodic, of period 1 let's say, and always positive, as it is the norm of the...

user21820

16:08

@Knight The post is just saying that if a[0] is greater than the sum of the absolute values of all the other coefficients, then the whole expression is positive.

Note that a[0] is the coefficient of cos(0) = 1.

So it is just the constant term.

Nothing special.

Knight

yes

user21820

What don't you understand, then?

Knight

He is saying
$$
a_0 \gt |a_1| + |a_2| \cdots + |b_1|+|b_2| \cdots
$$

user21820

cos and sin are both bounded by 1, so all the other terms are bounded by the absolute value of the coefficients.

Right, if that holds then the whole sum is positive.

Knight

Didn't get that

user21820

16:10

|c·cos(...)| ≤ ?

Knight

c-1

user21820

Nope. What if c = 0?

Knight

is it $c+1$ ?

user21820

What if c = −1?

Knight

-2

user21820

16:13

Hmm, don't guess.

Figure it out based on the properties of absolute value.

What do you know about absolute value?

Knight

I really don't know the answer, ask me a simpler question

Absolute values means the thing inside it must come out the positive

user21820

Hmm do you know that abs(x·y) = abs(x)·abs(y) for any complex numbers x,y?

Same for reals, but you should know the general one.

Knight

Yeah I know that product rule. And I think I can prove even this one $|x+y| \leq |x| +|y|$

user21820

Yes so use the product rule here.

|c·cos(x)| = ... ≤ ?

Knight

$|c \cdot \cos x| = |c||\cos x| \leq |c|$

because the greatest value of $|\cos x|$ is 1.

user21820

16:23

Right.

Does that answer your question about the linked post now?

All the other terms are bounded by the coefficients' absolute values.

Knight

@user21820 I still don't get how does that condition imply that our function will always be positive. Can you explain a little more?

user21820

@Knight Look at a finite sum first. The claim for a finite sum is that if a > |b|+|c| then |a| > |b·cos(x)| + |c·sin(y)| for any reals x,y.

Do you understand that?

Knight

Yes, multiplying by cos (x) and sin(x) made our inequality even stronger

user21820

Well actually the claim was that for any reals a,b,c, if a > |b|+|c| then |a| > b·cos(x) + c·sin(y) for any reals x,y.

I accidentally added the absolute value signs too early.

Knight

Yes, by removing the absolute signs and multiplying with cos and sin made our inequality stronger

user21820

16:29

Ah I see. You didn't get the final step.

Knight

:(

user21820

We want a + b·cos(x) + c·sin(y) > 0. That's equivalent to a > − ( b·cos(x) + c·sin(y) ).

And that's true if abs( b·cos(x) + c·sin(y) ) < a.

Now use the properties of abs to break that left-hand expression apart

abs( ... ) ≤ abs( b·cos(x) ) + abs( c·sin(y) ) ≤ abs(b) + abs(c).

So if abs(b) + abs(c) < a, we get what we want.

Knight

What we did after getting
$$
a \gt - \left( b \cos x + c\sin y \right)
$$ ??

Of course we know that $a$ is positive.

user21820

Hmm. I don't really understand your question. If abs(b) + abs(c) < a, then a > − ( b·cos(x) + c·sin(y) ), because abs( − ( b·cos(x) + c·sin(y) ) ) = ... ≤ ...

Knight

I think I should see it after some time for getting it

Thanks for help

user21820

16:43

Ok. Remember that if |x| < y, then x < y.

That's why we want to bound the absolute value of "− ( b·cos(x) + c·sin(y) )".

user69608

17:28

please give a hint @user21820

user69608

18:28

nvm its done!

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♦ $Mathematics$ Basic Mathematics

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♦ $Mathematics$ Basic Mathematics