Skip to main content
deleted 137 characters in body
Source Link
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

Comment: It appears that $$\ln(2g(t))=\sum_{k=1}^\infty\frac{t^{2k}}{2n(2n-1)}.$$ If this is true, then the positive answer to

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

Comment: It appears that $$\ln(2g(t))=\sum_{k=1}^\infty\frac{t^{2k}}{2n(2n-1)}.$$ If this is true, then the positive answer to

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

Post Undeleted by Iosif Pinelis
Post Deleted by Iosif Pinelis
Post Undeleted by Iosif Pinelis
Post Deleted by Iosif Pinelis
added 137 characters in body
Source Link
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

Comment: It appears that $$\ln(2g(t))=\sum_{k=1}^\infty\frac{t^{2k}}{2n(2n-1)}.$$ If this is true, then the positive answer to

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

Comment: It appears that $$\ln(2g(t))=\sum_{k=1}^\infty\frac{t^{2k}}{2n(2n-1)}.$$ If this is true, then the positive answer to

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

Source Link
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

Coefficients of certain Taylor series

For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.