Is f1+f2 unimodal if f1 and f2 is monotonic?
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I have recently encountered one programming problem which was reduced to find minimum value of function.Function f was sum of two functions f1 and f2 and f1 is strictly increasing and f2 is strictly decreasing. How can I prove that f1+f2 is unimodal from given information? I tried to think intuitively by considering slopes of f1 and f2 but I am not able to prove above fact. Can anyone give some approach or explanation? Thanks.
total-unimodularity
asked Jan 8 at 17:32
Parth PatelParth Patel
367
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add a comment |
$begingroup$
I have recently encountered one programming problem which was reduced to find minimum value of function.Function f was sum of two functions f1 and f2 and f1 is strictly increasing and f2 is strictly decreasing. How can I prove that f1+f2 is unimodal from given information? I tried to think intuitively by considering slopes of f1 and f2 but I am not able to prove above fact. Can anyone give some approach or explanation? Thanks.
total-unimodularity
asked Jan 8 at 17:32
Parth PatelParth Patel
367
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1
$begingroup$
Do you mean one was increasing and one was decreasing? If they are both increasing, so is the sum, so the maximum comes at the top end of the allowable interval.
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– Ross Millikan
Jan 8 at 17:44
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edited. Thanks for pointing out
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– Parth Patel
Jan 8 at 17:49
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I don't think this is true. It's easy to construct a counter-example, using piecewise linear functions.
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– verret
Jan 8 at 20:25
add a comment |
$begingroup$
I have recently encountered one programming problem which was reduced to find minimum value of function.Function f was sum of two functions f1 and f2 and f1 is strictly increasing and f2 is strictly decreasing. How can I prove that f1+f2 is unimodal from given information? I tried to think intuitively by considering slopes of f1 and f2 but I am not able to prove above fact. Can anyone give some approach or explanation? Thanks.
total-unimodularity
asked Jan 8 at 17:32
Parth PatelParth Patel
367
$endgroup$
I have recently encountered one programming problem which was reduced to find minimum value of function.Function f was sum of two functions f1 and f2 and f1 is strictly increasing and f2 is strictly decreasing. How can I prove that f1+f2 is unimodal from given information? I tried to think intuitively by considering slopes of f1 and f2 but I am not able to prove above fact. Can anyone give some approach or explanation? Thanks.
total-unimodularity
total-unimodularity
asked Jan 8 at 17:32
Parth PatelParth Patel
367
asked Jan 8 at 17:32
Parth PatelParth Patel
367
edited Jan 8 at 17:49
Parth Patel
asked Jan 8 at 17:32
Parth PatelParth Patel
367
asked Jan 8 at 17:32
Parth PatelParth Patel
367
asked Jan 8 at 17:32
Parth PatelParth Patel
367
367
1
$begingroup$
Do you mean one was increasing and one was decreasing? If they are both increasing, so is the sum, so the maximum comes at the top end of the allowable interval.
$endgroup$
– Ross Millikan
Jan 8 at 17:44
$begingroup$
edited. Thanks for pointing out
$endgroup$
– Parth Patel
Jan 8 at 17:49
$begingroup$
I don't think this is true. It's easy to construct a counter-example, using piecewise linear functions.
$endgroup$
– verret
Jan 8 at 20:25
add a comment |
1
$begingroup$
Do you mean one was increasing and one was decreasing? If they are both increasing, so is the sum, so the maximum comes at the top end of the allowable interval.
$endgroup$
– Ross Millikan
Jan 8 at 17:44
$begingroup$
edited. Thanks for pointing out
$endgroup$
– Parth Patel
Jan 8 at 17:49
$begingroup$
I don't think this is true. It's easy to construct a counter-example, using piecewise linear functions.
$endgroup$
– verret
Jan 8 at 20:25
1
1
$begingroup$
Do you mean one was increasing and one was decreasing? If they are both increasing, so is the sum, so the maximum comes at the top end of the allowable interval.
$endgroup$
– Ross Millikan
Jan 8 at 17:44
$begingroup$
Do you mean one was increasing and one was decreasing? If they are both increasing, so is the sum, so the maximum comes at the top end of the allowable interval.
$endgroup$
– Ross Millikan
Jan 8 at 17:44
$begingroup$
edited. Thanks for pointing out
$endgroup$
– Parth Patel
Jan 8 at 17:49
$begingroup$
edited. Thanks for pointing out
$endgroup$
– Parth Patel
Jan 8 at 17:49
$begingroup$
I don't think this is true. It's easy to construct a counter-example, using piecewise linear functions.
$endgroup$
– verret
Jan 8 at 20:25
$begingroup$
I don't think this is true. It's easy to construct a counter-example, using piecewise linear functions.
$endgroup$
– verret
Jan 8 at 20:25
add a comment |
2 Answers
2
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You can't prove it because it isn't true. $f1$ could increase rapidly for a while as $f2$ is slowly decreasing, leading the sum to increase, then $f2$ could decrease rapidly for a while while $f1$ is slowly increasing, making the sum decrease. Repeat this and you can have as many peaks in the sum as you want. An example is below. $f1$ is in blue, $f2$ is in red, $f1+f2$ is in yellow and is bimodal
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
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add a comment |
$begingroup$
If $f_1$ is an increasing and $f_2$ is decreasing it is not necessarily true that $g(x):= f_1(x) + f_2(x)$ is unimodal. Consider the following example: $f_1(x) = x$ when $x<1$, $f(x) = 4x$ when $xgeq 1$, and $f_2(x) = -x^2$. Then $g(x)=x-x^2$ when $x<1$ and $g(x) = 4x-x^2$ when $xgeq 1$. The function $g(x)$ has two local maxima at $(1/2,1/4)$ and $(2, 4)$.
answered Jan 8 at 18:00
irchansirchans
1,22949
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
You can't prove it because it isn't true. $f1$ could increase rapidly for a while as $f2$ is slowly decreasing, leading the sum to increase, then $f2$ could decrease rapidly for a while while $f1$ is slowly increasing, making the sum decrease. Repeat this and you can have as many peaks in the sum as you want. An example is below. $f1$ is in blue, $f2$ is in red, $f1+f2$ is in yellow and is bimodal
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
You can't prove it because it isn't true. $f1$ could increase rapidly for a while as $f2$ is slowly decreasing, leading the sum to increase, then $f2$ could decrease rapidly for a while while $f1$ is slowly increasing, making the sum decrease. Repeat this and you can have as many peaks in the sum as you want. An example is below. $f1$ is in blue, $f2$ is in red, $f1+f2$ is in yellow and is bimodal
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
You can't prove it because it isn't true. $f1$ could increase rapidly for a while as $f2$ is slowly decreasing, leading the sum to increase, then $f2$ could decrease rapidly for a while while $f1$ is slowly increasing, making the sum decrease. Repeat this and you can have as many peaks in the sum as you want. An example is below. $f1$ is in blue, $f2$ is in red, $f1+f2$ is in yellow and is bimodal
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
$endgroup$
You can't prove it because it isn't true. $f1$ could increase rapidly for a while as $f2$ is slowly decreasing, leading the sum to increase, then $f2$ could decrease rapidly for a while while $f1$ is slowly increasing, making the sum decrease. Repeat this and you can have as many peaks in the sum as you want. An example is below. $f1$ is in blue, $f2$ is in red, $f1+f2$ is in yellow and is bimodal
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
answered Jan 8 at 18:00
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
$begingroup$
If $f_1$ is an increasing and $f_2$ is decreasing it is not necessarily true that $g(x):= f_1(x) + f_2(x)$ is unimodal. Consider the following example: $f_1(x) = x$ when $x<1$, $f(x) = 4x$ when $xgeq 1$, and $f_2(x) = -x^2$. Then $g(x)=x-x^2$ when $x<1$ and $g(x) = 4x-x^2$ when $xgeq 1$. The function $g(x)$ has two local maxima at $(1/2,1/4)$ and $(2, 4)$.
answered Jan 8 at 18:00
irchansirchans
1,22949
$endgroup$
add a comment |
$begingroup$
If $f_1$ is an increasing and $f_2$ is decreasing it is not necessarily true that $g(x):= f_1(x) + f_2(x)$ is unimodal. Consider the following example: $f_1(x) = x$ when $x<1$, $f(x) = 4x$ when $xgeq 1$, and $f_2(x) = -x^2$. Then $g(x)=x-x^2$ when $x<1$ and $g(x) = 4x-x^2$ when $xgeq 1$. The function $g(x)$ has two local maxima at $(1/2,1/4)$ and $(2, 4)$.
answered Jan 8 at 18:00
irchansirchans
1,22949
$endgroup$
add a comment |
$begingroup$
If $f_1$ is an increasing and $f_2$ is decreasing it is not necessarily true that $g(x):= f_1(x) + f_2(x)$ is unimodal. Consider the following example: $f_1(x) = x$ when $x<1$, $f(x) = 4x$ when $xgeq 1$, and $f_2(x) = -x^2$. Then $g(x)=x-x^2$ when $x<1$ and $g(x) = 4x-x^2$ when $xgeq 1$. The function $g(x)$ has two local maxima at $(1/2,1/4)$ and $(2, 4)$.
answered Jan 8 at 18:00
irchansirchans
1,22949
$endgroup$
If $f_1$ is an increasing and $f_2$ is decreasing it is not necessarily true that $g(x):= f_1(x) + f_2(x)$ is unimodal. Consider the following example: $f_1(x) = x$ when $x<1$, $f(x) = 4x$ when $xgeq 1$, and $f_2(x) = -x^2$. Then $g(x)=x-x^2$ when $x<1$ and $g(x) = 4x-x^2$ when $xgeq 1$. The function $g(x)$ has two local maxima at $(1/2,1/4)$ and $(2, 4)$.
answered Jan 8 at 18:00
irchansirchans
1,22949
answered Jan 8 at 18:00
irchansirchans
1,22949
answered Jan 8 at 18:00
irchansirchans
1,22949
answered Jan 8 at 18:00
irchansirchans
1,22949
1,22949
add a comment |
add a comment |
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1
$begingroup$
Do you mean one was increasing and one was decreasing? If they are both increasing, so is the sum, so the maximum comes at the top end of the allowable interval.
$endgroup$
– Ross Millikan
Jan 8 at 17:44
$begingroup$
edited. Thanks for pointing out
$endgroup$
– Parth Patel
Jan 8 at 17:49
$begingroup$
I don't think this is true. It's easy to construct a counter-example, using piecewise linear functions.
$endgroup$
– verret
Jan 8 at 20:25