/sci/ - Science & Math » Thread #12331808

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Anonymous Wed 11 Nov 22:39:34 2020 No.12331808 View Reply Original Report

Quoted By: >>12331810 >>12331813 >>12331821 >>12331822 >>12331850 >>12331893 >>12331899 >>12331968 >>12331981 >>12331994 >>12332031 >>12332179 >>12332358 >>12332408 >>12332801

what is the mathematically correct way to say 1/n gets smaller as n increases?

Anonymous

Anonymous Wed 11 Nov 2020 22:40:10 No.12331810 Report

Quoted By:

>>12331808
1/n gets smaller as n increases

Anonymous

Anonymous Wed 11 Nov 2020 22:40:49 No.12331813 Report

Quoted By: >>12331834

>>12331808
1/n [arrow pointing to the right] -? as n [arrow pointing to the right] ?

Anonymous

Anonymous Wed 11 Nov 2020 22:42:02 No.12331821 Report

Quoted By:

>>12331808
The real function f with f(n) = 1/n is strictly monotonically decreasing for positive n.

Anonymous

Anonymous Wed 11 Nov 2020 22:42:21 No.12331822 Report

Quoted By: >>12331834

>>12331808
1 '/n
_ ->

N

Anonymous

Anonymous Wed 11 Nov 2020 22:42:56 No.12331823 Report

Quoted By: >>12331834 >>12331838

1/n is a series converging to zero

Anonymous

Anonymous Wed 11 Nov 2020 22:45:37 No.12331834 Report

Quoted By: >>12331869 >>12332105 >>12332436

>>12331813
>>12331822
>>12331823
The state of /sci/
For all n > m, 1/n < 1/m

Anonymous

Anonymous Wed 11 Nov 2020 22:46:57 No.12331838 Report

Quoted By:

>>12331823
it diverges

Anonymous

Anonymous Wed 11 Nov 2020 22:52:18 No.12331850 Report

Quoted By:

>>12331808
$\frac{1}{n}$ is just a figure eight w/ an intersection at $\pm\infty$

Anonymous

Anonymous Wed 11 Nov 2020 22:56:53 No.12331869 Report

Quoted By:

>>12331834
n = 1
m = -1

Anonymous

Anonymous Wed 11 Nov 2020 23:03:24 No.12331893 Report

Quoted By:

>>12331808
where n > 0:
n > n + 1
1/n < 1/(n+1)
...
a>0
n + a > 0
n < n + a
1/n > 1/(n + a)
This only looks at the positive case. I didn't feel like thinking beyond that because I am tired. someone else can finish it.

Anonymous

Anonymous Wed 11 Nov 2020 23:04:50 No.12331899 Report

Quoted By: >>12332127 >>12332700

>>12331808
lim (1/n) = 0
n->?

Anonymous

Anonymous Wed 11 Nov 2020 23:19:57 No.12331968 Report

Quoted By:

>>12331808
for all n e N
(1/n) > (1/(n+1)
induction maybe?

Anonymous

Anonymous Wed 11 Nov 2020 23:22:26 No.12331981 Report

Quoted By:

>>12331808
monotonic decreasing

Anonymous

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Anonymous Wed 11 Nov 2020 23:23:50 No.12331989 Report

Quoted By:

Anonymous

Anonymous Wed 11 Nov 2020 23:25:10 No.12331994 Report

Quoted By: >>12332077 >>12332384

>>12331808
is the joke that no one on this board can say 'inversely proportional'?

Anonymous

Anonymous Wed 11 Nov 2020 23:32:49 No.12332031 Report

Quoted By:

>>12331808
The question that ended /sci/

Anonymous

Anonymous Wed 11 Nov 2020 23:44:35 No.12332077 Report

Quoted By: >>12332202

>>12331994
inversely proportional to what

Anonymous

Anonymous Wed 11 Nov 2020 23:51:15 No.12332105 Report

Quoted By: >>12332627

>>12331834
Yeah it was embarrassing until you came along
Except a small correction: for all n, m ? ?...

Anonymous

Anonymous Wed 11 Nov 2020 23:54:02 No.12332110 Report

Quoted By:

$f\colon{\mathbb N}\to{\mathbb N}$
$f(n):=\dfrac{1}{n}$

Theorem 1:
$\forall n. (n\in{\mathbb N})\implies\Big(\forall m. (m=n+1)\implies\big(m>n\implies f(n)>f(m)\big)\Big)$

Theorem 2:
$\forall n. (n\in{\mathbb N})\implies\Big(\forall m. (m\in{\mathbb N})\implies\big(m>n\implies f(n)>f(m)\big)\Big)$

Anonymous

Anonymous Wed 11 Nov 2020 23:57:40 No.12332124 Report

Quoted By: >>12332133

$f\colon{\mathbb N}\to{\mathbb Q}$
$f(n):=\dfrac{1}{n}$
where we identify n in N with its caonical image in Q.
Add another clause in the expressions below if you're naturals have 0, so that you don't divide by 0.

Theorem 1:
$\forall n. (n\in{\mathbb N})\implies\big(\forall m. (m=n+1)\implies f(n)>f(m)\big)$

Where n+1 is also written Sn. You might also just write f(n+1) directly, but I want to mirror the one below more.

Theorem 2:
$\forall n. (n\in{\mathbb N})\implies\Big(\forall m. (m\in{\mathbb N})\implies\big(m>n\implies f(n)>f(m)\big)\Big)$

Anonymous

Anonymous Wed 11 Nov 2020 23:58:18 No.12332127 Report

Quoted By: >>12332134 >>12332136

>>12331899
this

Anonymous

Anonymous Wed 11 Nov 2020 23:59:14 No.12332133 Report

Quoted By:

>>12332124

Where in
$m>n\implies f(n)>f(m)$
I actually use two different orderings.

Every time I look at it I find some other type casuality.

Anonymous

Anonymous Wed 11 Nov 2020 23:59:22 No.12332134 Report

Quoted By:

>>12332127
Is this bait?

Anonymous

Anonymous Thu 12 Nov 2020 00:01:07 No.12332136 Report

Quoted By: >>12332152

>>12332127
cos(n)/n going to 0 doesn't mean that it's even a monotone function

El Arcón

El Arcón Thu 12 Nov 2020 00:03:39 No.12332150 Report

Quoted By:

"inversely proportional"

Anonymous

Anonymous Thu 12 Nov 2020 00:04:34 No.12332152 Report

Quoted By: >>12332166 >>12332173

>>12332136
good thing topic is 1/n.

Anonymous

Anonymous Thu 12 Nov 2020 00:08:34 No.12332166 Report

Quoted By: >>12332183 >>12332375

>>12332152
The topic is expressing that 1/n is falling.
That lim 1/n is 0 isn't doing that job.

Anonymous

Anonymous Thu 12 Nov 2020 00:09:51 No.12332173 Report

Quoted By: >>12332180 >>12332183

>>12332152
It doesn't matter
Something converging to zero has nothing to do with something decreasing in a monotonous

Anonymous

Anonymous Thu 12 Nov 2020 00:10:59 No.12332179 Report

Quoted By:

>>12331808
it depends on context.

Anonymous

Anonymous Thu 12 Nov 2020 00:11:05 No.12332180 Report

Quoted By:

>>12332173
..way
But I'm guessing this is bait there can't be people so mathematically illiterate over here

Anonymous

Anonymous Thu 12 Nov 2020 00:11:48 No.12332183 Report

Quoted By:

>>12332166
except for n = inf

>>12332173
gurr durr

Anonymous

Anonymous Thu 12 Nov 2020 00:18:14 No.12332202 Report

Quoted By:

>>12332077
Yo mom

Anonymous

Anonymous Thu 12 Nov 2020 01:12:40 No.12332358 Report

Quoted By:

>>12331808
d(1/n)/dn<0

Anonymous

Anonymous Thu 12 Nov 2020 01:22:09 No.12332375 Report

Quoted By:

>>12332166
>1/n is falling.
>That lim 1/n is 0 isn't doing that job.
except it does. 1/n is falling so hard, it falls right through -inf, crashes back to +inf and continues falling

Anonymous

Anonymous Thu 12 Nov 2020 01:26:03 No.12332384 Report

Quoted By:

>>12331994
Define inversely proportional in a non-circular manner.

Anonymous

Anonymous Thu 12 Nov 2020 01:38:13 No.12332408 Report

Quoted By:

>>12331808
It's a sequence of RATIONAL NUMBERS, you dumbass. You can compare two subsequent RATIONAL NUMBERS using the definition. I.e., 1/(n+1)<1/n, since n+1>n.

Anonymous

Anonymous Thu 12 Nov 2020 01:50:10 No.12332436 Report

Quoted By: >>12332622

>>12331834
retard detected
what if n or m is 0
1/0 cant be greater than or less than anything
RETARD
FAGGOT
FUCK
TRANNY
NIGGER

Anonymous

Anonymous Thu 12 Nov 2020 03:28:52 No.12332599 Report

Quoted By:

$\forall n \in \mathbb{R} — \{0\} -n^{-2} < 0$

Anonymous

Anonymous Thu 12 Nov 2020 03:37:58 No.12332622 Report

Quoted By:

>>12332436
>1/0 cant be greater than or less than anything
it is both greater and less than EVERYTHING retard
lrn2 $\pm\infty$

Anonymous

Anonymous Thu 12 Nov 2020 03:41:35 No.12332627 Report

Quoted By:

>>12332105
True, but n,m \in \R^+ would probably be a little better.

Anonymous

Anonymous Thu 12 Nov 2020 03:59:14 No.12332653 Report

Quoted By:

1 looks like a penis, so 1/n is the state of the penis depending on n number of nuts, hence it gets smaller

Anonymous

Anonymous Thu 12 Nov 2020 04:09:11 No.12332671 Report

Quoted By:

The derivative of 1/n is negative

Anonymous

Anonymous Thu 12 Nov 2020 04:29:55 No.12332700 Report

Quoted By:

>>12331899 winrar
The rest of you need to drink your milk and go to bed.
....and cut back on the caffeine.

Anonymous

Anonymous Thu 12 Nov 2020 05:29:32 No.12332801 Report

Quoted By:

>>12331808
$<br />\dfrac{1/n_{k+1}}{1/n_k}<1 <br />$

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