Spivak calculus, Sum of a polynomial's all derivatives, chapter 11 question 26 [duplicate]
This question already has an answer here:
Sum of derivatives of a polynomial
2 answers
The question asked to prove if $f$ is a polynomial function of degree $n$, and $fgeq0$, then $f+f'+f''+...+f^{(n)}geq0$
The answer given is:
The minimum of $f+f'+f''+...+f^{(n)}$ occurs at point where $0=f'+f''+...+f^{(n)}$[the next term vanishing, since $f$ has degree $n$]. Since $fgeq0$,this means that the minimum occurs at a point where $f+f'+f''+...+f^{(n)}geq0$
The question is so short, so is the answer given. Why the minimum occurs at $0=f'+f''+...+f^{(n)}$? Does this point always exist?
calculus derivatives polynomials
asked Nov 27 '18 at 9:30
user401653user401653
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marked as duplicate by gimusi calculus
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Nov 27 '18 at 9:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Sum of derivatives of a polynomial
2 answers
The question asked to prove if $f$ is a polynomial function of degree $n$, and $fgeq0$, then $f+f'+f''+...+f^{(n)}geq0$
The answer given is:
The minimum of $f+f'+f''+...+f^{(n)}$ occurs at point where $0=f'+f''+...+f^{(n)}$[the next term vanishing, since $f$ has degree $n$]. Since $fgeq0$,this means that the minimum occurs at a point where $f+f'+f''+...+f^{(n)}geq0$
The question is so short, so is the answer given. Why the minimum occurs at $0=f'+f''+...+f^{(n)}$? Does this point always exist?
calculus derivatives polynomials
asked Nov 27 '18 at 9:30
user401653user401653
566
marked as duplicate by gimusi calculus
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Nov 27 '18 at 9:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Sum of derivatives of a polynomial
2 answers
The question asked to prove if $f$ is a polynomial function of degree $n$, and $fgeq0$, then $f+f'+f''+...+f^{(n)}geq0$
The answer given is:
The minimum of $f+f'+f''+...+f^{(n)}$ occurs at point where $0=f'+f''+...+f^{(n)}$[the next term vanishing, since $f$ has degree $n$]. Since $fgeq0$,this means that the minimum occurs at a point where $f+f'+f''+...+f^{(n)}geq0$
The question is so short, so is the answer given. Why the minimum occurs at $0=f'+f''+...+f^{(n)}$? Does this point always exist?
calculus derivatives polynomials
asked Nov 27 '18 at 9:30
user401653user401653
566
This question already has an answer here:
Sum of derivatives of a polynomial
2 answers
The question asked to prove if $f$ is a polynomial function of degree $n$, and $fgeq0$, then $f+f'+f''+...+f^{(n)}geq0$
The answer given is:
The minimum of $f+f'+f''+...+f^{(n)}$ occurs at point where $0=f'+f''+...+f^{(n)}$[the next term vanishing, since $f$ has degree $n$]. Since $fgeq0$,this means that the minimum occurs at a point where $f+f'+f''+...+f^{(n)}geq0$
The question is so short, so is the answer given. Why the minimum occurs at $0=f'+f''+...+f^{(n)}$? Does this point always exist?
This question already has an answer here:
Sum of derivatives of a polynomial
2 answers
calculus derivatives polynomials
calculus derivatives polynomials
asked Nov 27 '18 at 9:30
user401653user401653
566
asked Nov 27 '18 at 9:30
user401653user401653
566
asked Nov 27 '18 at 9:30
user401653user401653
566
asked Nov 27 '18 at 9:30
user401653user401653
566
asked Nov 27 '18 at 9:30
user401653user401653
566
566
marked as duplicate by gimusi calculus
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Nov 27 '18 at 9:37
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Nov 27 '18 at 9:37
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2 Answers
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Any non-constant polynomial goes to $ infty$ in absolute value as $x to pm infty$. Hence the minimum is attained at some finite point. If a differentiable function $g$ has a minimum at $x$ then $g'(x)=0$. Take $g=f+f'+...+f^{(n)}$.
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
add a comment |
Yes, it exists. Since $fgeqslant0$, the degree of $f$ must be even and therefore $f$ has a minimum in $mathbb R$.
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Any non-constant polynomial goes to $ infty$ in absolute value as $x to pm infty$. Hence the minimum is attained at some finite point. If a differentiable function $g$ has a minimum at $x$ then $g'(x)=0$. Take $g=f+f'+...+f^{(n)}$.
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
add a comment |
Any non-constant polynomial goes to $ infty$ in absolute value as $x to pm infty$. Hence the minimum is attained at some finite point. If a differentiable function $g$ has a minimum at $x$ then $g'(x)=0$. Take $g=f+f'+...+f^{(n)}$.
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
add a comment |
Any non-constant polynomial goes to $ infty$ in absolute value as $x to pm infty$. Hence the minimum is attained at some finite point. If a differentiable function $g$ has a minimum at $x$ then $g'(x)=0$. Take $g=f+f'+...+f^{(n)}$.
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
Any non-constant polynomial goes to $ infty$ in absolute value as $x to pm infty$. Hence the minimum is attained at some finite point. If a differentiable function $g$ has a minimum at $x$ then $g'(x)=0$. Take $g=f+f'+...+f^{(n)}$.
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
answered Nov 27 '18 at 9:34
Kavi Rama MurthyKavi Rama Murthy
51.5k31855
51.5k31855
add a comment |
add a comment |
Yes, it exists. Since $fgeqslant0$, the degree of $f$ must be even and therefore $f$ has a minimum in $mathbb R$.
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
Yes, it exists. Since $fgeqslant0$, the degree of $f$ must be even and therefore $f$ has a minimum in $mathbb R$.
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
Yes, it exists. Since $fgeqslant0$, the degree of $f$ must be even and therefore $f$ has a minimum in $mathbb R$.
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
Yes, it exists. Since $fgeqslant0$, the degree of $f$ must be even and therefore $f$ has a minimum in $mathbb R$.
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 27 '18 at 9:35
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
add a comment |
add a comment |
