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Find Maximum of any discrete function (not necessarily a PDF)
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraWhy $2x$? Can't it be $x$?maximum area of a rectangle inscribed in a semi - circle with radius r.Finding the max/min turning points of $y=sin x$Can I calculate a fractional sum with functional equations and/or infinite series?Finding maximum of discrete functionDiscrete function: finding a maximumMaximum of a discrete function using derivativesWhat is the correct parametrisation of this curve, and why is it so?Finding the maximum value of summation of $sin$ series.Optimization for Exponential Bacteria Growth Rate
$begingroup$
How can we find the maximum of any discrete function, say
$$
f(n)=frac(n+1)^22^n,quad nin mathbbN
$$
that is not the PDF of any distribution? (This query is unrelated to statistics.)
By generating observations, we can obtain the maximum to be $9/4$ when $n=2$. However, other less rigourous methods using inequalities give poor estimates for this maximum point.
Wrong Approach 1 (Binomial Expansion):
$$2^n = (1+1)^nge beginpmatrixn\2endpmatriximplies frac(1+n)^2(1+1)^nle frac2(1+n)^2n(n-1)=frac2left(1+frac1n right)^21-frac1nle 2 quad(nto infty )$$
Wrong Approach 2 (Bernoulli Inequality):
$$
2^n=(1+1)^nge 1+n(1)implies frac(1+n)^2(1+1)^nlefrac(1+n)^21+n=1+nto infty quad (nto infty )
$$
How can we find the exact maximum point? If we try to differentiate with respect to $n$, we can get $napprox 1.88$, then we can try $n=1$ and $n=2$, but differentiation is just the wrong approach for a discrete function.
calculus sequences-and-series maxima-minima discrete-optimization
edited Apr 20 at 9:34
man on laptop
1
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
$endgroup$
add a comment |
$begingroup$
How can we find the maximum of any discrete function, say
$$
f(n)=frac(n+1)^22^n,quad nin mathbbN
$$
that is not the PDF of any distribution? (This query is unrelated to statistics.)
By generating observations, we can obtain the maximum to be $9/4$ when $n=2$. However, other less rigourous methods using inequalities give poor estimates for this maximum point.
Wrong Approach 1 (Binomial Expansion):
$$2^n = (1+1)^nge beginpmatrixn\2endpmatriximplies frac(1+n)^2(1+1)^nle frac2(1+n)^2n(n-1)=frac2left(1+frac1n right)^21-frac1nle 2 quad(nto infty )$$
Wrong Approach 2 (Bernoulli Inequality):
$$
2^n=(1+1)^nge 1+n(1)implies frac(1+n)^2(1+1)^nlefrac(1+n)^21+n=1+nto infty quad (nto infty )
$$
How can we find the exact maximum point? If we try to differentiate with respect to $n$, we can get $napprox 1.88$, then we can try $n=1$ and $n=2$, but differentiation is just the wrong approach for a discrete function.
calculus sequences-and-series maxima-minima discrete-optimization
edited Apr 20 at 9:34
man on laptop
1
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
$endgroup$
4
$begingroup$
Why do you say that differentiation is the wrong approach? That is the best approach for this question.
$endgroup$
– Kavi Rama Murthy
Apr 20 at 5:06
2
$begingroup$
Extend this function to positive real numbers, then use derivatives to show that it grows up to some point and then decreases (i.e. it is unimodal). The discrete maximum occurs at one of the immediate integer neighbors of the continuous maximum point, because it is less than those values on either side, by monotonicity. This works for any unimodal function.
$endgroup$
– Conifold
Apr 20 at 5:26
$begingroup$
@Conifold : I think it is generally not advisable to differentiate discrete distributions because many complications can occur - the least of which we need to check for differentiability. For instance, consider $f(n) = |n^2 - 2|$, where $n$ is any natural number. It has minimum value at $n=1$, but it is not differentiable at $x=sqrt2$. The function itself is not easy to differentiate.
$endgroup$
– NetUser5y62
Apr 20 at 5:53
4
$begingroup$
The question is whether the discrete function extends to a (piecewise) differentiable function. If so, the use of calculus is highly effective, and hence advisable. The points of non-differentiability should be treated as additional critical points and tested for potential maxima/minima along with zeros of the derivative on each interval between them. In your example you have three such intervals separated by $pmsqrt2$. Note that you will have to do the same when investigating maxima/minima of the continuous function $f(x) = |x^2 - 2|$, it is not the discreteness that makes the difference.
$endgroup$
– Conifold
Apr 20 at 6:26
add a comment |
$begingroup$
How can we find the maximum of any discrete function, say
$$
f(n)=frac(n+1)^22^n,quad nin mathbbN
$$
that is not the PDF of any distribution? (This query is unrelated to statistics.)
By generating observations, we can obtain the maximum to be $9/4$ when $n=2$. However, other less rigourous methods using inequalities give poor estimates for this maximum point.
Wrong Approach 1 (Binomial Expansion):
$$2^n = (1+1)^nge beginpmatrixn\2endpmatriximplies frac(1+n)^2(1+1)^nle frac2(1+n)^2n(n-1)=frac2left(1+frac1n right)^21-frac1nle 2 quad(nto infty )$$
Wrong Approach 2 (Bernoulli Inequality):
$$
2^n=(1+1)^nge 1+n(1)implies frac(1+n)^2(1+1)^nlefrac(1+n)^21+n=1+nto infty quad (nto infty )
$$
How can we find the exact maximum point? If we try to differentiate with respect to $n$, we can get $napprox 1.88$, then we can try $n=1$ and $n=2$, but differentiation is just the wrong approach for a discrete function.
calculus sequences-and-series maxima-minima discrete-optimization
edited Apr 20 at 9:34
man on laptop
1
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
$endgroup$
How can we find the maximum of any discrete function, say
$$
f(n)=frac(n+1)^22^n,quad nin mathbbN
$$
that is not the PDF of any distribution? (This query is unrelated to statistics.)
By generating observations, we can obtain the maximum to be $9/4$ when $n=2$. However, other less rigourous methods using inequalities give poor estimates for this maximum point.
Wrong Approach 1 (Binomial Expansion):
$$2^n = (1+1)^nge beginpmatrixn\2endpmatriximplies frac(1+n)^2(1+1)^nle frac2(1+n)^2n(n-1)=frac2left(1+frac1n right)^21-frac1nle 2 quad(nto infty )$$
Wrong Approach 2 (Bernoulli Inequality):
$$
2^n=(1+1)^nge 1+n(1)implies frac(1+n)^2(1+1)^nlefrac(1+n)^21+n=1+nto infty quad (nto infty )
$$
How can we find the exact maximum point? If we try to differentiate with respect to $n$, we can get $napprox 1.88$, then we can try $n=1$ and $n=2$, but differentiation is just the wrong approach for a discrete function.
calculus sequences-and-series maxima-minima discrete-optimization
calculus sequences-and-series maxima-minima discrete-optimization
edited Apr 20 at 9:34
man on laptop
1
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
edited Apr 20 at 9:34
man on laptop
1
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
edited Apr 20 at 9:34
man on laptop
1
edited Apr 20 at 9:34
man on laptop
1
edited Apr 20 at 9:34
man on laptop
1
1
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
asked Apr 20 at 4:42
NetUser5y62NetUser5y62
559215
559215
4
$begingroup$
Why do you say that differentiation is the wrong approach? That is the best approach for this question.
$endgroup$
– Kavi Rama Murthy
Apr 20 at 5:06
2
$begingroup$
Extend this function to positive real numbers, then use derivatives to show that it grows up to some point and then decreases (i.e. it is unimodal). The discrete maximum occurs at one of the immediate integer neighbors of the continuous maximum point, because it is less than those values on either side, by monotonicity. This works for any unimodal function.
$endgroup$
– Conifold
Apr 20 at 5:26
$begingroup$
@Conifold : I think it is generally not advisable to differentiate discrete distributions because many complications can occur - the least of which we need to check for differentiability. For instance, consider $f(n) = |n^2 - 2|$, where $n$ is any natural number. It has minimum value at $n=1$, but it is not differentiable at $x=sqrt2$. The function itself is not easy to differentiate.
$endgroup$
– NetUser5y62
Apr 20 at 5:53
4
$begingroup$
The question is whether the discrete function extends to a (piecewise) differentiable function. If so, the use of calculus is highly effective, and hence advisable. The points of non-differentiability should be treated as additional critical points and tested for potential maxima/minima along with zeros of the derivative on each interval between them. In your example you have three such intervals separated by $pmsqrt2$. Note that you will have to do the same when investigating maxima/minima of the continuous function $f(x) = |x^2 - 2|$, it is not the discreteness that makes the difference.
$endgroup$
– Conifold
Apr 20 at 6:26
add a comment |
4
$begingroup$
Why do you say that differentiation is the wrong approach? That is the best approach for this question.
$endgroup$
– Kavi Rama Murthy
Apr 20 at 5:06
2
$begingroup$
Extend this function to positive real numbers, then use derivatives to show that it grows up to some point and then decreases (i.e. it is unimodal). The discrete maximum occurs at one of the immediate integer neighbors of the continuous maximum point, because it is less than those values on either side, by monotonicity. This works for any unimodal function.
$endgroup$
– Conifold
Apr 20 at 5:26
$begingroup$
@Conifold : I think it is generally not advisable to differentiate discrete distributions because many complications can occur - the least of which we need to check for differentiability. For instance, consider $f(n) = |n^2 - 2|$, where $n$ is any natural number. It has minimum value at $n=1$, but it is not differentiable at $x=sqrt2$. The function itself is not easy to differentiate.
$endgroup$
– NetUser5y62
Apr 20 at 5:53
4
$begingroup$
The question is whether the discrete function extends to a (piecewise) differentiable function. If so, the use of calculus is highly effective, and hence advisable. The points of non-differentiability should be treated as additional critical points and tested for potential maxima/minima along with zeros of the derivative on each interval between them. In your example you have three such intervals separated by $pmsqrt2$. Note that you will have to do the same when investigating maxima/minima of the continuous function $f(x) = |x^2 - 2|$, it is not the discreteness that makes the difference.
$endgroup$
– Conifold
Apr 20 at 6:26
4
4
$begingroup$
Why do you say that differentiation is the wrong approach? That is the best approach for this question.
$endgroup$
– Kavi Rama Murthy
Apr 20 at 5:06
$begingroup$
Why do you say that differentiation is the wrong approach? That is the best approach for this question.
$endgroup$
– Kavi Rama Murthy
Apr 20 at 5:06
2
2
$begingroup$
Extend this function to positive real numbers, then use derivatives to show that it grows up to some point and then decreases (i.e. it is unimodal). The discrete maximum occurs at one of the immediate integer neighbors of the continuous maximum point, because it is less than those values on either side, by monotonicity. This works for any unimodal function.
$endgroup$
– Conifold
Apr 20 at 5:26
$begingroup$
Extend this function to positive real numbers, then use derivatives to show that it grows up to some point and then decreases (i.e. it is unimodal). The discrete maximum occurs at one of the immediate integer neighbors of the continuous maximum point, because it is less than those values on either side, by monotonicity. This works for any unimodal function.
$endgroup$
– Conifold
Apr 20 at 5:26
$begingroup$
@Conifold : I think it is generally not advisable to differentiate discrete distributions because many complications can occur - the least of which we need to check for differentiability. For instance, consider $f(n) = |n^2 - 2|$, where $n$ is any natural number. It has minimum value at $n=1$, but it is not differentiable at $x=sqrt2$. The function itself is not easy to differentiate.
$endgroup$
– NetUser5y62
Apr 20 at 5:53
$begingroup$
@Conifold : I think it is generally not advisable to differentiate discrete distributions because many complications can occur - the least of which we need to check for differentiability. For instance, consider $f(n) = |n^2 - 2|$, where $n$ is any natural number. It has minimum value at $n=1$, but it is not differentiable at $x=sqrt2$. The function itself is not easy to differentiate.
$endgroup$
– NetUser5y62
Apr 20 at 5:53
4
4
$begingroup$
The question is whether the discrete function extends to a (piecewise) differentiable function. If so, the use of calculus is highly effective, and hence advisable. The points of non-differentiability should be treated as additional critical points and tested for potential maxima/minima along with zeros of the derivative on each interval between them. In your example you have three such intervals separated by $pmsqrt2$. Note that you will have to do the same when investigating maxima/minima of the continuous function $f(x) = |x^2 - 2|$, it is not the discreteness that makes the difference.
$endgroup$
– Conifold
Apr 20 at 6:26
$begingroup$
The question is whether the discrete function extends to a (piecewise) differentiable function. If so, the use of calculus is highly effective, and hence advisable. The points of non-differentiability should be treated as additional critical points and tested for potential maxima/minima along with zeros of the derivative on each interval between them. In your example you have three such intervals separated by $pmsqrt2$. Note that you will have to do the same when investigating maxima/minima of the continuous function $f(x) = |x^2 - 2|$, it is not the discreteness that makes the difference.
$endgroup$
– Conifold
Apr 20 at 6:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Consider the ratio of successive terms $dfracf(n)f(n-1)$, for $nge 1$. (This ratio here equals $dfrac(n+1)^22n^2$.) Try and find for which values of $n$ we have $colorbluedfracf(n)f(n-1) ge 1$. Can you see how to use this information to find which $n$ maximises $f(n)$?
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
$endgroup$
add a comment |
$begingroup$
Hint: Prove that $$frac(n+1)^22^nle frac94$$ The equal sign holds if $$n=2$$ This is equivalent to $$(n+1)^2le 9cdot 2^n-2$$. You can prove this by induction.
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
1
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
votes
active
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votes
$begingroup$
Hint: Consider the ratio of successive terms $dfracf(n)f(n-1)$, for $nge 1$. (This ratio here equals $dfrac(n+1)^22n^2$.) Try and find for which values of $n$ we have $colorbluedfracf(n)f(n-1) ge 1$. Can you see how to use this information to find which $n$ maximises $f(n)$?
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
$endgroup$
add a comment |
$begingroup$
Hint: Consider the ratio of successive terms $dfracf(n)f(n-1)$, for $nge 1$. (This ratio here equals $dfrac(n+1)^22n^2$.) Try and find for which values of $n$ we have $colorbluedfracf(n)f(n-1) ge 1$. Can you see how to use this information to find which $n$ maximises $f(n)$?
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
$endgroup$
add a comment |
$begingroup$
Hint: Consider the ratio of successive terms $dfracf(n)f(n-1)$, for $nge 1$. (This ratio here equals $dfrac(n+1)^22n^2$.) Try and find for which values of $n$ we have $colorbluedfracf(n)f(n-1) ge 1$. Can you see how to use this information to find which $n$ maximises $f(n)$?
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
$endgroup$
Hint: Consider the ratio of successive terms $dfracf(n)f(n-1)$, for $nge 1$. (This ratio here equals $dfrac(n+1)^22n^2$.) Try and find for which values of $n$ we have $colorbluedfracf(n)f(n-1) ge 1$. Can you see how to use this information to find which $n$ maximises $f(n)$?
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
answered Apr 20 at 6:58
Minus One-TwelfthMinus One-Twelfth
3,653513
3,653513
add a comment |
add a comment |
$begingroup$
Hint: Prove that $$frac(n+1)^22^nle frac94$$ The equal sign holds if $$n=2$$ This is equivalent to $$(n+1)^2le 9cdot 2^n-2$$. You can prove this by induction.
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
1
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
add a comment |
$begingroup$
Hint: Prove that $$frac(n+1)^22^nle frac94$$ The equal sign holds if $$n=2$$ This is equivalent to $$(n+1)^2le 9cdot 2^n-2$$. You can prove this by induction.
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
1
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
add a comment |
$begingroup$
Hint: Prove that $$frac(n+1)^22^nle frac94$$ The equal sign holds if $$n=2$$ This is equivalent to $$(n+1)^2le 9cdot 2^n-2$$. You can prove this by induction.
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
Hint: Prove that $$frac(n+1)^22^nle frac94$$ The equal sign holds if $$n=2$$ This is equivalent to $$(n+1)^2le 9cdot 2^n-2$$. You can prove this by induction.
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
edited Apr 20 at 6:42
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Apr 20 at 5:36
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
79.6k42867
1
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
add a comment |
1
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
1
1
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
$begingroup$
I am afraid this is precisely the result I am unable to prove. Moreover, we are guessing the upper bound and then trying to prove it. We cannot guess the bound for more complicated expressions
$endgroup$
– NetUser5y62
Apr 20 at 6:12
add a comment |
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4
$begingroup$
Why do you say that differentiation is the wrong approach? That is the best approach for this question.
$endgroup$
– Kavi Rama Murthy
Apr 20 at 5:06
2
$begingroup$
Extend this function to positive real numbers, then use derivatives to show that it grows up to some point and then decreases (i.e. it is unimodal). The discrete maximum occurs at one of the immediate integer neighbors of the continuous maximum point, because it is less than those values on either side, by monotonicity. This works for any unimodal function.
$endgroup$
– Conifold
Apr 20 at 5:26
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@Conifold : I think it is generally not advisable to differentiate discrete distributions because many complications can occur - the least of which we need to check for differentiability. For instance, consider $f(n) = |n^2 - 2|$, where $n$ is any natural number. It has minimum value at $n=1$, but it is not differentiable at $x=sqrt2$. The function itself is not easy to differentiate.
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– NetUser5y62
Apr 20 at 5:53
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The question is whether the discrete function extends to a (piecewise) differentiable function. If so, the use of calculus is highly effective, and hence advisable. The points of non-differentiability should be treated as additional critical points and tested for potential maxima/minima along with zeros of the derivative on each interval between them. In your example you have three such intervals separated by $pmsqrt2$. Note that you will have to do the same when investigating maxima/minima of the continuous function $f(x) = |x^2 - 2|$, it is not the discreteness that makes the difference.
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– Conifold
Apr 20 at 6:26