Basic Mathematics - 2019-12-11 (page 1 of 2)

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5:57 AM

@user21820 How are scientific laws tested and verified in the past? Say the law is expressed as a mathematical equation, how is the equation verified in all different possible values of the variables (there are infinitely many cases)? Is this even possible?

I want to know how are scientific laws or scientific theories are tested and verified, and what is the criteria of test that they are accurate and us human beings can come into a consensus that they (best) describe/explain our world.

I've asked in the Physics SE chat but I did not get much help. Can you help me?

@yh05 I thought I already said "at that time, every scientist just used common sense".

Obviously you cannot completely test all of infinitely many cases.

That's why you have to understand kolmogorov complexity if you want to have any hope of a sensible justification of accepting certain explanations of empirical data over others.

And sadly most physicists today do not know kolmogorov complexity properly, so they still have only 'common sense'. Some of them don't even care about common sense, and make up explanations that have zero explanatory power.

It is also important to know that there are meta-assumptions involved, even in using kolmogorov complexity. For example, you have to assume that you have some amount of free will otherwise it could be that your experiments are all dictated by some entity that wants you to see only certain results.

And you can only empirically justify a statement that you can freely test any point within its domain. If the domain is uncountable (like a real interval), then you need to assume something like continuity almost everywhere, so that you can justify that your random test cases actually represent the domain.

6:24 AM

@user21820 can I ask a doubt related to binomial Theorem question here.?

@yh05: Do you understand kolmogorov complexity? Why given any computable function f, if you have enough input-output pairs of f then f has the minimum program length among all programs that match those pairs?

@yuvrajsingh Yes. But "doubt" ≠ "question".

Yes. So I was solving this question. Where I need to find sum of the coefficient of binomial series.

$$(1+x)^2010=c_0+c_1x+...........c_{2010}x^2010$$

So I put the value $x=1$.

That's correct. What's your question then?

And I got the sum that is equal to$2^2010$.

@yuvrajsingh That's wrong. I don't know how you got that. ... You just edited your message. It's correct now, but don't edit messages like that.

6:30 AM

Why?.

Is that look correct now?.

OK.

But in next part question says I need to find, sum of coefficient of

Of terms who have difference of 3,like.

$c_2+c_5+c_8.......c_2009$.

Now I do not know how do I find sum of these.

Do you know how to get the sum of odd terms?

Yes.

How?

Total sum is $n(n+1)/2$.

I subtract the even sum from them.

What?

6:36 AM

You ask for integer only.

Oh that's not what I meant.

I meant do you know how to find c[1] + c[3] + c[5] + ... + c[2009]?

Whether it is wrong or right, but in my mind I M, thinking the above expression as A. P SUM.

Well you're going to have to figure out this first, before you go on to more complicated things like c[2] + c[5] + c[8] + ...

I know how to take sum, but thing which confuse me is that they are not numbers, they are terms.

Not numbers, it sound like a stupid question but yes?.

Yes it's a stupid thing to say they are not numbers. As you defined it, c[k] is a number, for each k in [0..2010].

What you want to say is that you are summing terms of a sequence rather than the indices.

Either way, you have to figure it out. You already know c[0] + c[1] + c[2] + ...

Try to see if you can do something similar to figure out c[1] + c[3] + c[5] + ... + c[2009].

6:46 AM

Is that correct sum of of odd term is $1/2(2^2010)$.

And sum of three difference term ($c_2+c_5.....$)is

$1/3(2^2010-1)$.

You are going to have to prove what you claim, otherwise you must say it is just a guess.

Is it correct?. Then I can share my thoughts of writing this.

I said what I said.

OK, so for the first part I put $x=-1$ in the equation.

So the right hand term become zero.

So it will automatically get doubled. And I get the first result.

Right?.

@user21820.

Right idea, but I want to see the working.

7:00 AM

Mean should I post like putting x, then equal to zero.

Full working.

If I'm not convinced that you know precisely what is going on, we won't go on.

OK so I have the equation.

$$(1+x)^2010=c_0+c_1x+...........c_{2010}x^2010$$

Now putting $x=-1$ we get.

0=$c_0-c_1+c_2-c_3...........$.

And I get the odd terms on rhs.

So I get $c_1+c_3.........._2009=c_0.........c_2010$.

OK so far? @user21820

Yes.

Good.

Ah, I have one more question?. @user21820

Why do you think you're done?

7:09 AM

Why?.

You made a claim about c[2] + c[5] + c[8] + ... Do you have a proof or is that just a wild guess?

I have.

So from starting like that.?

Claiming to have a proof without having one would be dishonest, so show it.

Sorry. But I actually I guessed by above result. @user21820

Next time don't lie.

7:12 AM

OK I promise.

I thought about this, but I was not able to go through it.

The method you used worked for odd/even terms because you have a sequence that cycles between 1 and −1.

That allows you to cancel out some terms and add to others.

The way to get only indices at multiples of 3 is to find a sequence that cycles with period 3.

The sequence must be of the form 1,w,w^2,w^3,w^4,... What should w be?

Cube root of unity?

Which one?

Through $x^3=1$.

Which root?

7:17 AM

$x=1$,$x=-1+root 3i/2$,

And $-1-\root3i/2$.

@user21820

The other two.

Okay so just pick one. Specifically, all you need is that w^3 = 1 and w ≠ 1.

Now you have a period-3 sequence. You need another. Consider 1,w^2,w^4,w^6,...

It reduces to ...

For the earlier problem you used 1,1,1,... and 1,−1,1,−1,.... For this problem use 1,1,1,... and the two sequences I just gave you, and you can obtain what you want.

One question.

How do you get to know this equation would have complex root?

@user21820

I don't get your question. I showed you above why you want a period-3 sequence, and that's all there is to it.

It turns out that you need a complex root of unity, but I didn't have to guess it.

Got it.

@user21820, one last question, with my sincere efforts.

$(128)^(128)^(128)$.is divided by 3.

So I write the above part as $(64)^2(64)^2(64^2)

So on the next part I again write 64 as $8^2$.

And the I get $8^4.2.(8)^4.2(8)^4.2

And later I write 8=(9-1).

Is the way of approaching this question is right?

Ah, some math Jax error.

7:38 AM

I don't know what your a^b^c means. Be precise and specify the order by brackets.

i.stack.imgur.com/7D64f.jpg this what I convert @user21820.

And the original question was in this form.

$128^128^128$.

And later I write 64=$8^2$.

And then 8=9-1

Because three will always divide 9.

But I can say remainder as 1,but problem it has same number in powers.

@user21820

14 mins ago, by user21820

I don't know what your a^b^c means. Be precise and specify the order by brackets.

Mathematics is about precision. True precision requires symbolic precision. Diagrams can never be precise.

They are good for intuitive reasoning and helping you think, but you must learn to be symbolically precise.

7:59 AM

Ah, OK

And, for the fourth or fifth time, don't assume I'm a "sir".

Question is {128^128}^128

OK so far?

@user21820

That makes no sense, because it can be trivially simplified.

1 hour later…

9:06 AM

If you don't get it, you need to revise the definition of exponentiation and the basic properties of them.

9:36 AM

Sorry there was an server error.

So let me quote what I have tried.

128=64.2 @user21820

And later I write 64=8.8

I didn't ask you what you tried. I said your question itself makes no sense.

Define exponentiation for me.

Why?

Because you apparently don't know exponentiation at all.

For natural, integer, and rational powers it's exactly what you'd expect; for arbitrary real powers it is the unique continuous extension of the usual exponential to the real numbers

@user21820

Don't tell me it's what I expect. If you cannot rigorously define it, then you need to learn how to do it.

I already know how to do it. I don't think you do.

9:43 AM

Thus what I have learn?

You have to try, then I will know what you know and what you don't. Define natural number exponentiation.

(Don't even think about integer/rational/real exponentiation yet, until you have solid grasp of natural exponentiation.)

Exponentiation is related to growth that is always proportional to how large the quantity has grown so far. We can have different starts and rates but we can imagine the growth as happening gradually and continuously, and imagining how that would interpolate is how we extend to reals

@user21820

That is not a definition. That is a long rambling that has nothing to do with mathematical exponentiation.

If you don't know something, be honest and say so. That is one key criterion for me to teach you.

Sorry, for arguing, I just want that, how much I correctly know.

@user21820

Given natural numbers m,n, define m^n in terms of addition and multiplication.

By "define" it means that once you write it down, it is clear to everyone what m^n means, not just to people who already know what it means.

If you really don't know how to do that, I can give you a hint.

9:53 AM

Yes?

Hint: Start with the case where n = 0. Define m^n in that case first, and then define m^n in other cases.

m^n, where n=0 where m^0.

??

Sorry, I didn't, t get you, how you want me to define m^n

You have to tell me what it means as if I know nothing about it.

Or, you have to tell your grandparents about it and make them understand.

(Assuming they aren't mathematically trained.)

9:59 AM

OK.

Let say I have m packets of biscuits.

I suspect this is the first time you're told to define something, so try your best.

And I have to distribute this to n students.

So that each student get at least 1 biscuit.

At least?

Are the packets identical?

I don't think you got the right idea.

Can I have another chance?

Please last.

Sure, but maybe in the interest of time I will just give you the start.

It seems you're attempting to give an example related to m^n. Whether or not it is a correct example, it still does not count as a definition, because you don't justify that the example situation always results in the same 'answer'. Who knows, maybe with some children the number of ways is different.

A definition goes like this:

10:07 AM

OK sir.

Define m^n for naturals m,n by the following: If n = 0 then m^n = 1, otherwise m^n = m^(n−1)·m.

I write it this way to make it clear that the definition is well-founded, meaning that it is not circular. Indeed, you can for any m,n expand the definition of m^n all the way until it 'terminates' at the base case.

Try expanding the definition of 2^5.

By the way, we must also stipulate precedence rules. ^ has higher precedence than ·, so m^(n−1)·m = (m^(n−1))·m.

The precedence rule is not really part of the mathematical structure, but it is necessary for human communication.

I can, t define m^n=2^5, =m^(n-1).m

@user21820

I don't understand what you're doing at all. I already gave you the entire definition. I asked you to use it. Perhaps you don't understand basic English?

The definition is for arbitrary m,n, and this includes for m = 2 and n = 5. Just apply it.

OK.

Ah, OK 2^5=can be written as 2^(2-1)5

10:26 AM

That is completely wrong, and not at all what my definition expands to.

Try again.

Copy my definition and substitute m,n with the appropriate values.

Mu answer can sound stupid but (n-1)m,where m=2 n=5,can,t be 2^5.

6 mins ago, by user21820

Copy my definition and substitute m,n with the appropriate values.

Sorry, now I understand what you were conveying.

I still want to see it done.

I assume that it is not necessary m, n are integer.

So 2^5,=2^(7/2-1)2

10:37 AM

Why do you insist on not following my instructions?

I wrote:

30 mins ago, by user21820

Define m^n for naturals m,n by the following: If n = 0 then m^n = 1, otherwise m^n = m^(n−1)·m.

And then:

11 mins ago, by user21820

Copy my definition and substitute m,n with the appropriate values.

Literally, press Ctrl+C and Ctrl+V and manually substitute m,n for appropriate values so that you can apply the definition to 2^5.

Frankly, I don't know what kind of mathematics teachers you have had all these years.

Define m^n for naturals m,n by the following: If n = 0 then m^n = 1, otherwise m^n = m^(n−1)·m.

And

2^5=

For this.

It m^n will be equal to 1.

Stop.

It is clear that you don't understand "substitute", but you didn't even say so.

→ 7 messages moved to Trash

Substitute means you go through the entire definition I gave you and replace each occurrence of m with the same value (chosen by you), and likewise for n.

Don't write anything else besides the definition with the substitutions made.

11:19 AM

Are you still there, pal? @yuvrajsingh

11:51 AM

@skull patrol I am a girl, I am using my brother account, my name is Alesha.

12:01 PM

m^n=1 from m not equal to zero, and n equal to zero.

otherwise m^n = m^(n−1)·m.

OK so far? @user21820

Now substituting the Value.

2^0=1

"2^0 = 1" is correct, but unfortunately is not what I asked you to do.

"m^n=1 from m not equal to zero, and n equal to zero." is different from what I said, so it is wrong.

Look, just copy and paste what I wrote exactly without changing a single word or symbol, and then systematically replace each "m" with some thing (say "2"), and each "n" with something else.

And I advise you not to reveal your real name online as far as possible. For your own safety. (There are some really bad people, even on Math SE.)

12:19 PM

@user21820 I haven, t said anything wrong, and neither I am scared of anyone, it is just that I am a student, and at last I didn, t get what you want me to do, you said define a m^n, so I did the same.. (please do not worry regarding my safety, I answered wrong because I could not understand what you want to convey by this.)

Although you have written it several times..

For example... one possible substitution results in:

> If 3 = 0 then 4^3 = 1, otherwise 4^3 = 4^(3−1)·4.

As I said, copy exactly and replace only the m,n. I don't see why you cannot do that.

I have now given you an example. Do exactly the same, but choose a different substitution so that the result is relevant to 2^5.

2^5=2^(5-1)2, and same m^n=1 for n=0

Why did you delete so many words and add so many words of your own? You really need to learn to follow instructions.

OK so I am rewriting them. By the way my name is Alesha.

We define 2^5, such that if 5=0, then 2^5=1, and 2^5=2(5-1)2.

For 5 not equal to zero.

Is that OK?.

@user21820

@yuvrajsingh It's better, but still wrong because you changed "otherwise" to "and". Those two words don't even mean the same thing!

If you mean to attach "for 5 not equal to zero" at the back, it is better, but you should do it properly with the right words.

12:30 PM

Now what I get from this?

> If 5 = 0 then 2^5 = 1. If 5 ≠ 0 then 2^5 = 2^(5−1)·2.

You don't get anything because you didn't copy it right. Please learn to copy properly. You changed the words and missed a "^" and a "·" and it became wrong.

OK.

But still I am confuse with my binomial question?

Try again. I want you to be able to attain 100% precision. Logical reasoning is not about trying to say things in your own way.

What next?

You haven't done what I told you to do, so we're not going on. Don't change a single word or symbol except for m,n in the definition.

12:36 PM

If 5= 0 then 2^5= 1, otherwise 2^5 = 2^(5-1)·2

@user21820

Or again?

Now it's perfect.

OK now my question.

No. Wait.

Why?

You don't understand exponentiation, and yet you want to go off to another question? Be patient. Follow my instructions if you actually want to learn mathematics. If not, you're welcome to go somewhere else.

3 mins ago, by yuvraj singh

If 5= 0 then 2^5= 1, otherwise 2^5 = 2^(5-1)·2

You know that 5 ≠ 0. So what you just wrote tells you that 2^5 = 2^(5−1)·2 = 2^4·2.

12:39 PM

OK, I will follow what you will say.

Now apply the definition again to tell you what is 2^4.

Meaning, find another substitution to get you information about 2^4.

OK so by the same definition you given above?

Yes. The definition I gave you is supposed to be a complete definition of m^n for naturals m,n. So you only need to use it and nothing else.

Your first substitution gave you information about 2^5. Find another to tell you about 2^4.

OK if 4=0,then 2^4=1 otherwise 2^4=2^(4-1).2

Excellent.

From what you just wrote, and the fact that 4 ≠ 0, we know that 2^4 = 2^(4−1)·2 = 2^3·2.

Do you see where this is going?

We now have 2^5 = 2^4·2 = (2^3·2)·2.

12:44 PM

Yes.

Can you continue all the way? What is the final substitution we need?

For 2^n.

I mean, you have substituted (m,n) by (2,5) and (2,4) so far. You will need also (2,3) and ... What does the final one give you?

be wise now and generalize

(take your time :)

2^5=2^4.2=(2^3.2).2

12:48 PM

Right, and continue?

=((2^2).2).2

Missing something.

Bracket?

No, it's simply wrong.

Try to be more careful.

Ah,

(2^2.2).2)2

12:50 PM

Right.

So what is the final end?

You can drop the brackets as long as you remember that ^ is done first before ·.

(2.2).2).2).2

No no that's not exactly the end.

Use the definition. Don't just guess.

If you don't use the definition, you cannot truly understand how it works.

OK that is simple 2.2.2.2.2

No. I said use the definition. I don't want to see anything that didn't result from only the definition.

You're going to have to drop your habit of guessing, if you want to have a true grasp of mathematics. It is more beautiful and precise than you have been taught.

So if 5 is equal to zero, then 2^5=1,otherwise 2^5=((2.2).2. 2).2

12:58 PM

No you do not have that. Since you cannot see that you failed to use the definition, I want to see every single line of working.

22 mins ago, by yuvraj singh

If 5= 0 then 2^5= 1, otherwise 2^5 = 2^(5-1)·2

18 mins ago, by user21820

You know that 5 ≠ 0. So what you just wrote tells you that 2^5 = 2^(5−1)·2 = 2^4·2.

15 mins ago, by yuvraj singh

OK if 4=0,then 2^4=1 otherwise 2^4=2^(4-1).2

15 mins ago, by user21820

From what you just wrote, and the fact that 4 ≠ 0, we know that 2^4 = 2^(4−1)·2 = 2^3·2.

14 mins ago, by user21820

We now have 2^5 = 2^4·2 = (2^3·2)·2.

But that is meaning less., to write them again.

These are the first few lines, write the rest all out. It is not meaningless, because you haven't understood it. It is also easy to write it out because you have Ctrl+C.

Do it, then you will appreciate why I tell you to do it.

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