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/sci/ Science & Math

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Anonymous 2013-02-20 08:35:00 No.5547373

[Missing image file: url.jpg]

What is bigger, e^pi or pi^e? Why


>>
Anonymous 2013-02-20 08:37:09 No.5547376
pi^e = 1

e^pi > 1

>>
Anonymous 2013-02-20 08:40:43 No.5547381
>>5547376
e^i*pi = -1 if that's what you're thinking of but pi^e definitely doesn't equal 1 considering they're both non-zero.

>>
Anonymous 2013-02-20 08:43:48 No.5547385
>>5547373
if x>y, then y^x>x^y. So answer your own question.

>>
Anonymous 2013-02-20 08:46:55 No.5547389
>>5547385
proof?

>>
Anonymous 2013-02-20 08:52:42 No.5547394
>>5547385

x = 1, y = 0

x > y

0^1 = 0
1^0 = 1

HMMMMMMMM...

>>
Anonymous 2013-02-20 09:00:20 No.5547397
>>5547394
Fine, For all x,y>0

>>
Anonymous 2013-02-20 09:01:03 No.5547398
You guys know e is about 2.7, right?

>>
Anonymous 2013-02-20 09:01:35 No.5547400
>>5547373
e= 2.71828182846
? = 3.14159265359

?^e= 22.4591577184
e^?=23.1406926328

e^? > ?^e

>>
Anonymous 2013-02-20 09:03:59 No.5547403
>>5547397

Nope, not exactly.

Take x = 5 and y = 1

x > y

1^5 = 1
5^1 = 5

HMMMMMMM....

>>
Anonymous 2013-02-20 10:03:43 No.5547484

[Missing image file: pic-head-swap6[1].jpg]
>>5547376
>pi^e = 1

>>
Anonymous 2013-02-20 10:24:59 No.5547506
>>5547373

using log(x+dx)<log(x)+1/x*dx, because log(x) is concave

log(e^pi)=pi * log(e) = pi
log(pi^e)=e* log(pi)<e*(1+(pi-e)/e)=
e+(pi-e)/e

the question is pi>e+(pi-e)/e
pie>e^2+pi-e
pi(e-1)>e^2-e
pi>e, which is true

log(pi^e)<e+(pi-e)/e<pi=log(e^pi)

so e^pi> pi>e

>>
Anonymous 2013-02-20 10:25:31 No.5547507
>>5547506
e^pi > pi^e

>>
Anonymous 2013-02-20 10:28:33 No.5547509
>>5547403
for x>y where y > 1, y^x > x^y
for x>y where y < 1, y^x > x^y

>>
Anonymous 2013-02-20 10:34:58 No.5547519
>>5547509
2>1.1 1.1^2 < 2^1.1
1>0.5 0.5^1 < 1^0.5

LEL

>>
Anonymous 2013-02-20 13:47:11 No.5547853
>>5547506

For x>y>1:

log_y(x)<1+(x-y)/( log(y) y) because of concaveness.

log_y(y^x)=x
log_y(x^y)=y*log_y(x)<y+(x-y)/log(y)

y+(x-y)/log(y) < x iff log(y)>=0, so if y>=e.

So the correct statement would, that if x>y>=e, then x^y > y^x.

>>
Anonymous 2013-02-20 13:50:07 No.5547863

[Missing image file: 93e.png]
>>5547385
>>5547397
>>5547509
This guy
Pretty sure it's just an ek in disquise

>>
Anonymous 2013-02-20 20:53:13 No.5548980

[Missing image file: epie.png]
Is this right?
Apologies for shitty handwriting.

>>
Anonymous 2013-02-20 22:29:33 No.5549309
e^pi < 3^3 = 9
pi^e >3^2.5 > 9

Therefor e^pi < pi^e







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