/sci/ - Science & Math

>>10760791
Which step? Also, use /sqt/ next time

>>10760804
2^(n-1) +2^(n-1)

>>10760815
2^(n - 2) < 2^(n - 1)
Add 2^(n - 1) to both sides

>>10760815
come on man, you really have to be more specific.
do you understand what the inductive step is? we assume that the claim is true for n-1 and n-2, i.e. [math]F(n-2) < 2^(n-2)[/math] and [math]F(n-1) < 2^(n-1)[/math].
Adding these together, [math]F(n-2) + F(n-1) < 2^(n-2) + 2^(n-1)[/math].
But [math]F(n) = F(n-2) + F(n-1)[/math].
On the other hand, [math]2^(n-2) < 2^(n-1)[/math], so [math]F(n) < 2^(n-2) + 2^(n-1) < 2^(n-1) + 2^(n-1) = 2*2^(n-1) = 2^n[/math].
This completes the inductive step.

>>10760834
aw fuck, i forgot my curly braces. jesus christ.
whatever, try to figure out the exponents.

>proof by induction

Stopped reading there

jesus, how much of a brainlet do you have to be to not understand the assertion 2^(n-2) + 2^(n-1) < 2^(n-1) + 2^(n-1) = 2^n

>>10760834
you can preview your tex by click the button on the top left or right of reply interface

>>10760815
a+a=2a

>>10760791
it's really easy OP.

try substituting 3 for n:

[math]
F(3) = F(n-2) + F(n - 1)
F(3) = F(1) + F(2)
[\math]

now we already know that

[math]
F(1) < 2^1
F(2) < 2^2
[\math]

so

[math]
F(1) + F(2) < 2^1 + 2^2 < 2^3
[\math]

so therefore

[math]
F(3) < 2^3
[\math]

this can be extended for all n by repeating the process for n = 4,5,6,7 ... etc since we always know that the conditions for any n hold for n-2 and n-1

>>10761512
fuck I fucked up my formatting. I hope that makes sense anyways

>>10761034
I know I can preview my tex you fucking moron, I DIDNT FUCKING THINK I WOULD BE GETTING IT WRONG NOW DID I??

>>10761537
but you got it wrong anyways didnt you?
retard

Holy shit, is your brain replaced with used donkey shit toilet paper? You deserve no response, brainlet bag of your grandma's shit, faggot hairless paralyzed cave monkey. They say thinking is human's great power, but your neurological network resembles that of a sewage system. Your entire humanity up to this point can be reproduced as an image of a long shit leaving a cow's asshole midway. Never post here again bitch, head over to /pol where you're accepted.

>>10760815
it only works with 2
look at it as 2(2^(n-1)) instead of 2^(n-1)+2^(n-1)
since x(x^n)=x^(n+1) you can turn 2(2^(n-1)) into 2^n

>>10760791
CS major detected

>>10761673
this, and not ironically

>>10760971
What's wrong with induction, if it gives the correct result?

>>10761673
>>10761757
/pol/ is very high IQ though

>>10760971
Why don't you prove why induction doesn't work fag?

>>10760971
>if it gives the correct result
Lmao faggot. The reasoning must be correct, not just the answer.

>>10761537
Tbh I think it’s a valid tip. I didn’t know you could preview Tex before reading their post.

>>10761776
no, no it is not. it's one of the stupidest places i've ever seen, anywhere.

>>10761809
this is you

>>10761812
those people are probably more intelligent than /pol/ simply because they are more likely to admit they are wrong.
For a place that constantly argues the opposing side appeals to emotion to form their arguments, they sure do "feel" something is right with unscientific conviction.

>>10761832
All I'm seeing in this post is cope

>>10761842
im not even american. watching your country devolve into this shitshow requires no coping, trust me, it's hilarious

I remember doing this bullshit for CS
you know how often I used any of this shit while in a job: 0
That's right, zero. CS is bullshit. They just care if they can pay you less since you're wet behind the ears then throw you away once you're old if you don't move into managment
fuck programming and other "coders" everyone who works that job is a piece of fat shit

>>10762010
here's a perfect example of dealing with this horseshit:

I haven't touched python in 10 fuckin years but I need an older module/script/whatevergayshituwantocallit
I go to install the older script

>RuntimeError: Python version >= 3.5 required.

but I have python 3.....

>brew install python3
>Warning: python 3.7.3 is already installed and up-to-date
>To reinstall 3.7.3, run `brew reinstall python`


now I can waste my time looking around to find out why, without getting started on what I want to do.
you know why? because some faggot can't write FATLAB bs so I want to turn it into python so it's hopefully somewhat more readable instead of single character variable names everywhere

now someone will pipe up and laugh and say "OMG WHY DIDNT U JUST KNOW ABOUT BLERF DE BLARF FORT TYPE THAT IN GOD UR SO DUMB!!!"

that's your fat cowerker who hasn't taken a shower in half a week that you have to sit next to listening to him typewritter away on some manual keyboard becuase it's what all the cool kids are using

what a waste of 10+ fucking years of my life the money was not worht it
BLARRRRRRRRRRRRR

all the jews in sf can die in a fire

>>10761756
.....

>>10761695
That's not the issue I just dont understand how he got to 2^(n-1) + 2^(n-1) from 2^(n-1) +2^(n-2)

>>10762192
Say b < a. Then a + b < a + a. You're fucking stupid, and no matter how much you browse /pol/, that's not going to change.

>>10761812

>>10762192
plug in small n's. That should help you out.

>>10763466
Dilate. This whole site is /pol/ territory.

I still dont get it... can someone write out the proof if its soo fucking easy.

>>10762192
Because n-1 is larger than n-2.

>>10764476

>>10764476
here you go

>>10761812
high iq counter

>>10761812

>>10763525
they look so determined. they are probably not afraid to put countless hours into posting racial-iq and anti-climate-alarmist threads on /sci/.

>>10761812
This post just proved his point. Being an automatically defensive and anti-sjw when confronted is exactly what the average sjw does when presented with a new idea

>>10763525
The white master race kek

>>10765030
Mr High IQ over here

>>10762192
You shouldn't worry too much about how he "got" it.
You should worry about whether it's true or not.

It's true that:
[math] 2^{n-2} < 2^{n-1} [/math]

Therefore it's true that:
[math] 2^{n-2} + 2^{n-1} < 2^{n-1} + 2^{n-1} [/math]

And [math] 2^{n-1} + 2^{n-1} [/math] can be rewritten as [math] 2^n [/math], because:
[math] 2^{n-1} + 2^{n-1} = 2 \cdot 2^{n-1} = 2^n [/math]

Tell me which part you don't understand.

>>10760791

The idea behind this proof is that we use the fact that if we have a sum [math]A+B[/math], we can say that [math]A+B \leq 2M[/math] where [math]M=max(A,B)[/math].

So, we construct another recurrence that we know is larger than the Fibonacci numbers at each step by replacing each step with this inequality. It happens that this recurrence is [math]A(n) = 2^n[/math]

>>10765760
Stop trolling

>>10765575
I dont understand how the 2^(n-1) happen s

>>10767358
jesus fuck. Are you even making an effort?

>>10760791
haha the proposition number is 4.20 ha ha!

>>10769260
Grow up

>>10761512
Not OP but thanks, I'm an ESL so understanding math in English is kinda hard soemtimes

>>10770896
Holy shit that image is tiny

>>10760791
I have a quick and dirty "proof" that would make an engineer blush.
F(n) = F(n-1) + F(n-2)
Subtract F(n-2) and divide by 2.
(F(n) - F(n-2))/2= F(n-1)/2
Write the difference equation as a differential equation.
F'(x) = F(x)/2
F(x) = e^(x/2) + C
F(0) = 1 implies C=0
1/2 < ln(2) implies F(x) = e^(x/2) < 2^x.

>>10767358
[eqn]2^{n-1}+2^{n-1}=1\dot2^{n-1}+1\dot2^{n-1}[/eqn]

Factor out the ones Anon and see what happens :) I often forget this stuff too.

>>10771735
Sorry this, [eqn]1 \cdot 2^{n-1} + 1 \cdot 2^{n-1} [/eqn]

Now factor out the ones

>>10771745
Jesus sorry I just woke up lol factor out 2^(n-1) God damn it