- #1
PFStudent
- 170
- 0
Hey,
1. Homework Statement .
I wrote my own proof for the below, I was wondering if you guys could take a look at it and give me some feedback please. Particularly, I would like to know if this proof is rigorous enough.
Lemma 2.4 - Suppose f has a second derivative everywhere, and that,
[tex]
{{f} + {f^{\prime\prime}}} = {0}
[/tex]
[tex]
{f(0)} = {0}
[/tex]
[tex]
{f^{\prime\prime}(0)}} = {0}
[/tex]
Then,
[tex]
{f(x)} = {0}
[/tex]
2. Homework Equations .
Knowledge of Calculus.
3. The Attempt at a Solution .
Let,
[tex]
{{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} = {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}
[/tex]
Prove that,
[tex]
{f} = {0}
[/tex]
Proof,
[tex]
{{f}+{f^{\prime\prime}}} = {0}
[/tex]
[tex]
{{\left({{f}^{\prime}}\right)}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {\left({0}\right)}{\left({{f}^{\prime}}\right)}
[/tex]
[tex]
{{{f}^{\prime}}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {0}
[/tex]
[tex]
{{\int_{}^{}}{\left[{{{f}^{\prime}}{\left({{f}{+{{f}^{\prime\prime}}}}\right)}}\right]}{dx}} = {{\int_{}^{}}{\left[{0}\right]}{dx}}
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,
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Let
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{0} = {a}
[/tex]
[tex]
{{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}
[/tex]
Let,
[tex]
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{u} = {f}
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{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
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{v} = {{f}^{\prime}}
[/tex]
[tex]
{du} = {{{f}^{\prime}}{dx}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{dv} = {{{f}^{\prime\prime}}{dx}}
[/tex]
[tex]
{{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}
[/tex]
[tex]
{{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{C}}\right)}
[/tex]
Let,
[tex]
{{{C}_{1}}+{{C}_{2}}} = {C}
[/tex]
[tex]
{{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{\left({C}\right)}} = {{{\left({0}\right)}{x}}+{C}}
[/tex]
[tex]
{{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}
[/tex]
[tex]
{{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}
[/tex]
[tex]
{{{\left({f(x)}\right)}^{2}}{+}{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {0}
[/tex]
, for all [itex]{x}[/itex].
This implies that [tex]{{f(x)} = {0}}[/tex] for all [itex]{x}[/itex].
[tex]
{\therefore}
[/tex]
[tex]
{f} = {0}
[/tex]
Thanks,
-PFStudent
EDIT: Thanks for the edit HallsofIvy.
1. Homework Statement .
I wrote my own proof for the below, I was wondering if you guys could take a look at it and give me some feedback please. Particularly, I would like to know if this proof is rigorous enough.
Lemma 2.4 - Suppose f has a second derivative everywhere, and that,
[tex]
{{f} + {f^{\prime\prime}}} = {0}
[/tex]
[tex]
{f(0)} = {0}
[/tex]
[tex]
{f^{\prime\prime}(0)}} = {0}
[/tex]
Then,
[tex]
{f(x)} = {0}
[/tex]
2. Homework Equations .
Knowledge of Calculus.
3. The Attempt at a Solution .
Let,
[tex]
{{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} = {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}
[/tex]
Prove that,
[tex]
{f} = {0}
[/tex]
Proof,
[tex]
{{f}+{f^{\prime\prime}}} = {0}
[/tex]
[tex]
{{\left({{f}^{\prime}}\right)}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {\left({0}\right)}{\left({{f}^{\prime}}\right)}
[/tex]
[tex]
{{{f}^{\prime}}{\left({{f}+{{f}^{\prime\prime}}}\right)}} = {0}
[/tex]
[tex]
{{\int_{}^{}}{\left[{{{f}^{\prime}}{\left({{f}{+{{f}^{\prime\prime}}}}\right)}}\right]}{dx}} = {{\int_{}^{}}{\left[{0}\right]}{dx}}
{\textcolor{white}{.}}
,
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
Let
{\textcolor{white}{.}}
{0} = {a}
[/tex]
[tex]
{{{\int_{}^{}}}{f}{{f}^{\prime}}{dx}+{{\int_{}^{}}{{f}^{\prime}}{{f}^{\prime\prime}}{dx}}} = {{{\int_{}^{}}}{\left({a}\right)}{dx}}
[/tex]
Let,
[tex]
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{u} = {f}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{v} = {{f}^{\prime}}
[/tex]
[tex]
{du} = {{{f}^{\prime}}{dx}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{\textcolor{white}{.}}
{dv} = {{{f}^{\prime\prime}}{dx}}
[/tex]
[tex]
{{{\int_{}^{}}{\left({u}\right)}{\left({du}\right)}}+{{\int_{}^{}}{\left({v}\right)}{\left({dv}\right)}}} = {{\int_{}^{}}{a}{dx}}
[/tex]
[tex]
{{\left({{\frac{{u}^{2}}{2}}+{{C}_{1}}}\right)}+{\left({{\frac{{v}^{2}}{2}}+{{C}_{2}}}\right)}} = {\left({{ax}+{C}}\right)}
[/tex]
Let,
[tex]
{{{C}_{1}}+{{C}_{2}}} = {C}
[/tex]
[tex]
{{{\frac{1}{2}}{\left({{{\left({f}\right)}^{2}}+{{\left({{f}^{\prime}}\right)}^{2}}}\right)}}+{\left({C}\right)}} = {{{\left({0}\right)}{x}}+{C}}
[/tex]
[tex]
{{\frac{1}{2}}{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}
[/tex]
[tex]
{{{\left({{{f}^{2}}+{{{f}^{\prime}}^{2}}}\right)}}} = {0}
[/tex]
[tex]
{{{\left({f(x)}\right)}^{2}}{+}{{\left({{{f}^{\prime}}{\left({x}\right)}}\right)}^{2}}} = {0}
[/tex]
, for all [itex]{x}[/itex].
This implies that [tex]{{f(x)} = {0}}[/tex] for all [itex]{x}[/itex].
[tex]
{\therefore}
[/tex]
[tex]
{f} = {0}
[/tex]
Thanks,
-PFStudent
EDIT: Thanks for the edit HallsofIvy.
Last edited: