Post History
#4: Post edited
by
Snoopy
·
2024-04-04T21:04:18Z (9 months ago)
- The first question can be rephrased in a clear way as follows:
Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.- If one rewrites:
- $$
- A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
- $$
- where the exponent is an [affine function](https://en.wikipedia.org/wiki/Affine_transformation) $h(t)= at+b$, and one can easily check that
- $$
- h(\lambda t_1+(1-\lambda)t_2) = \lambda h(t_1)+(1-\lambda)h(t_2)
- $$
- Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.
- To see (2) as a particular case from convexity of $F$, one only needs to write
- $$
- F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
- $$
- with
- $$
- \lambda = \frac1p, t_1 = 0, t_2=1\ .
- $$
- The first question can be rephrased in a clear way as follows:
- > Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
- If one rewrites:
- $$
- A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
- $$
- where the exponent is an [affine function](https://en.wikipedia.org/wiki/Affine_transformation) $h(t)= at+b$, and one can easily check that
- $$
- h(\lambda t_1+(1-\lambda)t_2) = \lambda h(t_1)+(1-\lambda)h(t_2)
- $$
- Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.
- To see (2) as a particular case from convexity of $F$, one only needs to write
- $$
- F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
- $$
- with
- $$
- \lambda = \frac1p, t_1 = 0, t_2=1\ .
- $$
#3: Post edited
by
Snoopy
·
2024-01-14T15:42:11Z (11 months ago)
- The first question can be rephrased in a clear way as follows:
- Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
- If one rewrites:
- $$
- A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
- $$
where the exponent is an [affine function](https://en.wikipedia.org/wiki/Affine_transformation) $h(t)= at+b$. Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.- To see (2) as a particular case from convexity of $F$, one only needs to write
- $$
- F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
- $$
- with
- $$
- \lambda = \frac1p, t_1 = 0, t_2=1\ .
- $$
- The first question can be rephrased in a clear way as follows:
- Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
- If one rewrites:
- $$
- A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
- $$
- where the exponent is an [affine function](https://en.wikipedia.org/wiki/Affine_transformation) $h(t)= at+b$, and one can easily check that
- $$
- h(\lambda t_1+(1-\lambda)t_2) = \lambda h(t_1)+(1-\lambda)h(t_2)
- $$
- Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.
- To see (2) as a particular case from convexity of $F$, one only needs to write
- $$
- F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
- $$
- with
- $$
- \lambda = \frac1p, t_1 = 0, t_2=1\ .
- $$
#2: Post edited
by
Snoopy
·
2024-01-14T15:40:15Z (11 months ago)
- The first question can be rephrased in a clear way as follows:
- Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
- If one rewrites:
- $$
- A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
- $$
where the *exponent* is an affine function $h(t)= at+b$. Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.- To see (2) as a particular case from convexity of $F$, one only needs to write
- $$
- F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
- $$
- with
- $$
- \lambda = \frac1p, t_1 = 0, t_2=1\ .
- $$
- The first question can be rephrased in a clear way as follows:
- Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
- If one rewrites:
- $$
- A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
- $$
- where the exponent is an [affine function](https://en.wikipedia.org/wiki/Affine_transformation) $h(t)= at+b$. Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.
- To see (2) as a particular case from convexity of $F$, one only needs to write
- $$
- F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
- $$
- with
- $$
- \lambda = \frac1p, t_1 = 0, t_2=1\ .
- $$
#1: Initial revision
by
Snoopy
·
2024-01-14T15:39:39Z (11 months ago)
The first question can be rephrased in a clear way as follows:
Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
If one rewrites:
$$
A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)}
$$
where the *exponent* is an affine function $h(t)= at+b$. Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.
To see (2) as a particular case from convexity of $F$, one only needs to write
$$
F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2)
$$
with
$$
\lambda = \frac1p, t_1 = 0, t_2=1\ .
$$