convex function $f(x,y)=e^{alpha x^2+2y^2}-1$
I have to study if $f(x,y)=e^{alpha x^2+2y^2}-1$ is convex or not in its domain.If I calculate the Hessian matrix I have problem to define for which $alpha$ is convex (only for $alpha =0$ the Hessian is semi-positive definite).
But is $f(x,y)=e^{alpha x^2+2y^2}$ ever a convex function?
real-analysis
asked Nov 27 '18 at 16:49
Giulia B.
415211
add a comment |
I have to study if $f(x,y)=e^{alpha x^2+2y^2}-1$ is convex or not in its domain.If I calculate the Hessian matrix I have problem to define for which $alpha$ is convex (only for $alpha =0$ the Hessian is semi-positive definite).
But is $f(x,y)=e^{alpha x^2+2y^2}$ ever a convex function?
real-analysis
asked Nov 27 '18 at 16:49
Giulia B.
415211
add a comment |
I have to study if $f(x,y)=e^{alpha x^2+2y^2}-1$ is convex or not in its domain.If I calculate the Hessian matrix I have problem to define for which $alpha$ is convex (only for $alpha =0$ the Hessian is semi-positive definite).
But is $f(x,y)=e^{alpha x^2+2y^2}$ ever a convex function?
real-analysis
asked Nov 27 '18 at 16:49
Giulia B.
415211
I have to study if $f(x,y)=e^{alpha x^2+2y^2}-1$ is convex or not in its domain.If I calculate the Hessian matrix I have problem to define for which $alpha$ is convex (only for $alpha =0$ the Hessian is semi-positive definite).
But is $f(x,y)=e^{alpha x^2+2y^2}$ ever a convex function?
real-analysis
real-analysis
asked Nov 27 '18 at 16:49
Giulia B.
415211
asked Nov 27 '18 at 16:49
Giulia B.
415211
asked Nov 27 '18 at 16:49
Giulia B.
415211
asked Nov 27 '18 at 16:49
Giulia B.
415211
asked Nov 27 '18 at 16:49
Giulia B.
415211
415211
add a comment |
add a comment |
1 Answer
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Hint: If f and g are convex functions and g is non-decreasing , then $h(x)=g(f(x))$ is convex. Taking $g(x)=e^x$ which is non-decreasing and convex, you have that $h(x)=e^{f(x)}$ is convex if $f(x)$ is convex. In your case you have to find a value of $alpha$ such that $f(x,y)=alpha x^2+2y^2$ is convex.
Hint 2: Remember that the sum of two convex (concave) functions is convex (concave).
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
add a comment |
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1 Answer
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Hint: If f and g are convex functions and g is non-decreasing , then $h(x)=g(f(x))$ is convex. Taking $g(x)=e^x$ which is non-decreasing and convex, you have that $h(x)=e^{f(x)}$ is convex if $f(x)$ is convex. In your case you have to find a value of $alpha$ such that $f(x,y)=alpha x^2+2y^2$ is convex.
Hint 2: Remember that the sum of two convex (concave) functions is convex (concave).
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
add a comment |
Hint: If f and g are convex functions and g is non-decreasing , then $h(x)=g(f(x))$ is convex. Taking $g(x)=e^x$ which is non-decreasing and convex, you have that $h(x)=e^{f(x)}$ is convex if $f(x)$ is convex. In your case you have to find a value of $alpha$ such that $f(x,y)=alpha x^2+2y^2$ is convex.
Hint 2: Remember that the sum of two convex (concave) functions is convex (concave).
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
add a comment |
Hint: If f and g are convex functions and g is non-decreasing , then $h(x)=g(f(x))$ is convex. Taking $g(x)=e^x$ which is non-decreasing and convex, you have that $h(x)=e^{f(x)}$ is convex if $f(x)$ is convex. In your case you have to find a value of $alpha$ such that $f(x,y)=alpha x^2+2y^2$ is convex.
Hint 2: Remember that the sum of two convex (concave) functions is convex (concave).
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
Hint: If f and g are convex functions and g is non-decreasing , then $h(x)=g(f(x))$ is convex. Taking $g(x)=e^x$ which is non-decreasing and convex, you have that $h(x)=e^{f(x)}$ is convex if $f(x)$ is convex. In your case you have to find a value of $alpha$ such that $f(x,y)=alpha x^2+2y^2$ is convex.
Hint 2: Remember that the sum of two convex (concave) functions is convex (concave).
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
answered Nov 27 '18 at 16:57
Ramiro Scorolli
655113
655113
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
add a comment |
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
why $g(x)=e^x $ is non-decreasing?
– Giulia B.
Nov 27 '18 at 20:10
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
The exponential function is monotonic increasing. Take the derivative and note that it's always positive.
– Ramiro Scorolli
Nov 27 '18 at 20:16
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
Can I use the same principle to study convexity of f(x,y)=$int_{0}^{x^2+y^4} e^{t^2}dt$?
– Giulia B.
Nov 27 '18 at 21:03
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
give a look to the "Leibniz rule".en.wikipedia.org/wiki/Leibniz_integral_rule
– Ramiro Scorolli
Nov 27 '18 at 21:15
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
I have to calculate the partial derivatives for the Hessian matrix?
– Giulia B.
Nov 27 '18 at 21:29
add a comment |
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