How to compare $2^pi$ and $pi^2$ using calculus

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How to compare $2^pi$ and $pi^2$ using calculus

I guess $$f(x)=fracln xx$$ wont help here since $2 lt e lt pi$

calculus

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edited Jul 21 at 16:41

user223391

asked Jul 21 at 16:37

Umesh shankar

2,27611018

add a comment | 

up vote
6
down vote

favorite

4

How to compare $2^pi$ and $pi^2$ using calculus

I guess $$f(x)=fracln xx$$ wont help here since $2 lt e lt pi$

calculus

share|cite|improve this question

edited Jul 21 at 16:41

user223391

asked Jul 21 at 16:37

Umesh shankar

2,27611018

add a comment | 

up vote
6
down vote

favorite

4

up vote
6
down vote

favorite

4

4

How to compare $2^pi$ and $pi^2$ using calculus

I guess $$f(x)=fracln xx$$ wont help here since $2 lt e lt pi$

calculus

share|cite|improve this question

edited Jul 21 at 16:41

user223391

asked Jul 21 at 16:37

Umesh shankar

2,27611018

How to compare $2^pi$ and $pi^2$ using calculus

I guess $$f(x)=fracln xx$$ wont help here since $2 lt e lt pi$

calculus

share|cite|improve this question

edited Jul 21 at 16:41

user223391

asked Jul 21 at 16:37

Umesh shankar

2,27611018

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 16:41

user223391

edited Jul 21 at 16:41

user223391

edited Jul 21 at 16:41

user223391

asked Jul 21 at 16:37

Umesh shankar

2,27611018

asked Jul 21 at 16:37

Umesh shankar

2,27611018

asked Jul 21 at 16:37

Umesh shankar

2,27611018

2,27611018

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
10
down vote



accepted

$2$ and $pi$ are on differing sides of the maximum of $fraclog(x)x$ at $x=e$, so it is hard to compare them. However, since $4=2^2$, we have $fraclog(2)2=fraclog(4)4$, and we can compare
$$
fraclog(pi)pigtfraclog(4)4=fraclog(2)2
$$
because $fraclog(x)x$ is monotonically decreasing for $xge e$. Therefore,
$$
pi^2gt2^pi
$$

---

Further Result

$2$ and $4$ is not the only pair of unequal numbers we can compute for which $fraclog(x)x=fraclog(y)y$. The same is true for
$$
x=left(1+frac1tright)^tquadtextandquad y=left(1+frac1tright)^t+1
$$
where $tgt0$. $t=1$ gives $2$ and $4$.

share|cite|improve this answer

edited Jul 21 at 17:45

answered Jul 21 at 17:09

robjohn♦

258k26297612

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
  – Umesh shankar
  Jul 21 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
  – robjohn♦
  Jul 21 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
  – Umesh shankar
  Jul 21 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
  – robjohn♦
  Jul 21 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes yes true about to post my edit, thanks
  – Umesh shankar
  Jul 21 at 17:22
  

  
  
  
  

 | 
show 1 more comment

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
10
down vote



accepted

$2$ and $pi$ are on differing sides of the maximum of $fraclog(x)x$ at $x=e$, so it is hard to compare them. However, since $4=2^2$, we have $fraclog(2)2=fraclog(4)4$, and we can compare
$$
fraclog(pi)pigtfraclog(4)4=fraclog(2)2
$$
because $fraclog(x)x$ is monotonically decreasing for $xge e$. Therefore,
$$
pi^2gt2^pi
$$

---

Further Result

$2$ and $4$ is not the only pair of unequal numbers we can compute for which $fraclog(x)x=fraclog(y)y$. The same is true for
$$
x=left(1+frac1tright)^tquadtextandquad y=left(1+frac1tright)^t+1
$$
where $tgt0$. $t=1$ gives $2$ and $4$.

share|cite|improve this answer

edited Jul 21 at 17:45

answered Jul 21 at 17:09

robjohn♦

258k26297612

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
  – Umesh shankar
  Jul 21 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
  – robjohn♦
  Jul 21 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
  – Umesh shankar
  Jul 21 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
  – robjohn♦
  Jul 21 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes yes true about to post my edit, thanks
  – Umesh shankar
  Jul 21 at 17:22
  

  
  
  
  

 | 
show 1 more comment

up vote
10
down vote



accepted

$2$ and $pi$ are on differing sides of the maximum of $fraclog(x)x$ at $x=e$, so it is hard to compare them. However, since $4=2^2$, we have $fraclog(2)2=fraclog(4)4$, and we can compare
$$
fraclog(pi)pigtfraclog(4)4=fraclog(2)2
$$
because $fraclog(x)x$ is monotonically decreasing for $xge e$. Therefore,
$$
pi^2gt2^pi
$$

---

Further Result

$2$ and $4$ is not the only pair of unequal numbers we can compute for which $fraclog(x)x=fraclog(y)y$. The same is true for
$$
x=left(1+frac1tright)^tquadtextandquad y=left(1+frac1tright)^t+1
$$
where $tgt0$. $t=1$ gives $2$ and $4$.

share|cite|improve this answer

edited Jul 21 at 17:45

answered Jul 21 at 17:09

robjohn♦

258k26297612

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
  – Umesh shankar
  Jul 21 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
  – robjohn♦
  Jul 21 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
  – Umesh shankar
  Jul 21 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
  – robjohn♦
  Jul 21 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes yes true about to post my edit, thanks
  – Umesh shankar
  Jul 21 at 17:22
  

  
  
  
  

 | 
show 1 more comment

up vote
10
down vote



accepted

up vote
10
down vote



accepted

$2$ and $pi$ are on differing sides of the maximum of $fraclog(x)x$ at $x=e$, so it is hard to compare them. However, since $4=2^2$, we have $fraclog(2)2=fraclog(4)4$, and we can compare
$$
fraclog(pi)pigtfraclog(4)4=fraclog(2)2
$$
because $fraclog(x)x$ is monotonically decreasing for $xge e$. Therefore,
$$
pi^2gt2^pi
$$

---

Further Result

$2$ and $4$ is not the only pair of unequal numbers we can compute for which $fraclog(x)x=fraclog(y)y$. The same is true for
$$
x=left(1+frac1tright)^tquadtextandquad y=left(1+frac1tright)^t+1
$$
where $tgt0$. $t=1$ gives $2$ and $4$.

share|cite|improve this answer

edited Jul 21 at 17:45

answered Jul 21 at 17:09

robjohn♦

258k26297612

$2$ and $pi$ are on differing sides of the maximum of $fraclog(x)x$ at $x=e$, so it is hard to compare them. However, since $4=2^2$, we have $fraclog(2)2=fraclog(4)4$, and we can compare
$$
fraclog(pi)pigtfraclog(4)4=fraclog(2)2
$$
because $fraclog(x)x$ is monotonically decreasing for $xge e$. Therefore,
$$
pi^2gt2^pi
$$

---

Further Result

$2$ and $4$ is not the only pair of unequal numbers we can compute for which $fraclog(x)x=fraclog(y)y$. The same is true for
$$
x=left(1+frac1tright)^tquadtextandquad y=left(1+frac1tright)^t+1
$$
where $tgt0$. $t=1$ gives $2$ and $4$.

share|cite|improve this answer

edited Jul 21 at 17:45

answered Jul 21 at 17:09

robjohn♦

258k26297612

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 17:45

edited Jul 21 at 17:45

edited Jul 21 at 17:45

answered Jul 21 at 17:09

robjohn♦

258k26297612

answered Jul 21 at 17:09

robjohn♦

258k26297612

answered Jul 21 at 17:09

robjohn♦

258k26297612

258k26297612

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
  – Umesh shankar
  Jul 21 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
  – robjohn♦
  Jul 21 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
  – Umesh shankar
  Jul 21 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
  – robjohn♦
  Jul 21 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes yes true about to post my edit, thanks
  – Umesh shankar
  Jul 21 at 17:22
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
  – Umesh shankar
  Jul 21 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
  – robjohn♦
  Jul 21 at 17:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
  – Umesh shankar
  Jul 21 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
  – robjohn♦
  Jul 21 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes yes true about to post my edit, thanks
  – Umesh shankar
  Jul 21 at 17:22
  

  
  
  
  

1

1

Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
– Umesh shankar
Jul 21 at 17:13

Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^pi$ and $pi^1.9$
– Umesh shankar
Jul 21 at 17:13

@Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
– robjohn♦
Jul 21 at 17:15

@Umeshshankar: actually, for $xle e$, $fraclog(x)x$ is monotonically increasing, so $fraclog(1.9)1.9ltfraclog(2)2ltfraclog(pi)pi$. So...
– robjohn♦
Jul 21 at 17:15

Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
– Umesh shankar
Jul 21 at 17:19

Yes perfect, i think now we can compare $a^pi$ and $pi^a$ for any $a$
– Umesh shankar
Jul 21 at 17:19

@Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
– robjohn♦
Jul 21 at 17:21

@Umeshshankar: Not so fast! it doesn't work for $2.1^pi$ and $pi^2.1$.
– robjohn♦
Jul 21 at 17:21

Yes yes true about to post my edit, thanks
– Umesh shankar
Jul 21 at 17:22

Yes yes true about to post my edit, thanks
– Umesh shankar
Jul 21 at 17:22

 | 
show 1 more comment

 

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