Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?












1














Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?



And is it possible to do it without using calculus?




$$2^x=x+1.$$




Here is my attempts:



$2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.



Now, I need to look at these intervals.



$$xin (-1,0]; [0,1]; [1, infty)$$



Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.



Problematic point is , $xin [0,1]$ or $xin [-1,0]$.



For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.






inequality proof-writing exponential-function real-numbers







share|cite|improve this question





























    1














    Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?



    And is it possible to do it without using calculus?




    $$2^x=x+1.$$




    Here is my attempts:



    $2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.



    Now, I need to look at these intervals.



    $$xin (-1,0]; [0,1]; [1, infty)$$



    Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.



    Problematic point is , $xin [0,1]$ or $xin [-1,0]$.



    For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.






    inequality proof-writing exponential-function real-numbers







    share|cite|improve this question



























      1












      1








      1


      1





      Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?



      And is it possible to do it without using calculus?




      $$2^x=x+1.$$




      Here is my attempts:



      $2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.



      Now, I need to look at these intervals.



      $$xin (-1,0]; [0,1]; [1, infty)$$



      Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.



      Problematic point is , $xin [0,1]$ or $xin [-1,0]$.



      For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.






      inequality proof-writing exponential-function real-numbers







      share|cite|improve this question















      Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?



      And is it possible to do it without using calculus?




      $$2^x=x+1.$$




      Here is my attempts:



      $2^x>0 Rightarrow x+1>0 Rightarrow x>-1$.



      Now, I need to look at these intervals.



      $$xin (-1,0]; [0,1]; [1, infty)$$



      Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $xtoinfty$, we get $2^x>>x$.



      Problematic point is , $xin [0,1]$ or $xin [-1,0]$.



      For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.






      inequality proof-writing exponential-function real-numbers




      inequality proof-writing exponential-function real-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 7 at 22:53

























      asked Dec 7 at 22:07









      Beginner

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          4 Answers
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          To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
          $$begin{align}
          2^x
          &=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
          &=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
          &<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
          &=1+x\
          end{align} $$
          Which shows that $2^xlt 1+x$ for $xin (0,1)$.






          share|cite|improve this answer





























            2














            With calculus:



            Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.

            Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.



            Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.






            share|cite|improve this answer

















            • 2




              In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
              – jmerry
              Dec 7 at 22:35



















            2














            Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,



            Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution



            Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.



            Thus the only remaining points to check are $xin {0, 1}$.






            share|cite|improve this answer





























              1














              Function $xmapsto 2^x$ is strictly convex.



              A strictly convex intersects a line in 0, 1 or 2 points.



              So there exists at most two solutions of the equation $2^x = x +1$.






              share|cite|improve this answer























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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3














                To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
                $$begin{align}
                2^x
                &=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
                &=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
                &<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
                &=1+x\
                end{align} $$
                Which shows that $2^xlt 1+x$ for $xin (0,1)$.






                share|cite|improve this answer


























                  3














                  To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
                  $$begin{align}
                  2^x
                  &=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
                  &=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
                  &<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
                  &=1+x\
                  end{align} $$
                  Which shows that $2^xlt 1+x$ for $xin (0,1)$.






                  share|cite|improve this answer
























                    3












                    3








                    3






                    To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
                    $$begin{align}
                    2^x
                    &=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
                    &=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
                    &<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
                    &=1+x\
                    end{align} $$
                    Which shows that $2^xlt 1+x$ for $xin (0,1)$.






                    share|cite|improve this answer












                    To prove that no solution exists in $(0,1),$ you may use the series representation for the exponential function (which can be derived without using calculus). If $xin (0,1)$, we have
                    $$begin{align}
                    2^x
                    &=1+frac{ln(2)}{1!}x+frac{ln^2(2)}{2!}x^2+...\
                    &=1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}x+...bigg)\
                    &<1+xbigg(frac{ln(2)}{1!}+frac{ln^2(2)}{2!}+...bigg)\
                    &=1+x\
                    end{align} $$
                    Which shows that $2^xlt 1+x$ for $xin (0,1)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 7 at 22:42









                    Frpzzd

                    21.6k839107




                    21.6k839107























                        2














                        With calculus:



                        Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.

                        Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.



                        Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.






                        share|cite|improve this answer

















                        • 2




                          In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
                          – jmerry
                          Dec 7 at 22:35
















                        2














                        With calculus:



                        Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.

                        Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.



                        Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.






                        share|cite|improve this answer

















                        • 2




                          In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
                          – jmerry
                          Dec 7 at 22:35














                        2












                        2








                        2






                        With calculus:



                        Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.

                        Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.



                        Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.






                        share|cite|improve this answer












                        With calculus:



                        Lemma: If a function $f$ is $n$ times differentiable on an interval $I$, and $f$ has $n+1$ distinct zeros on $I$, then the $n$th derivative of $f$ has a zero somewhere in that interval.

                        Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $f'$ between each pair of zeros of $f$, for a total of at least $n$, then a zero of $f''$ between each zero of $f'$, and so on.



                        Now, for this problem? Consider $f(x)=2^x-x-1$. The second derivative is $f''(x)=ln^2(2)cdot 2^x$, which is positive on all of $mathbb{R}$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $f$ - on all of $mathbb{R}$. We already know of two ($0$ and $1$), so there are no others.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 7 at 22:30









                        jmerry

                        1,14616




                        1,14616








                        • 2




                          In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
                          – jmerry
                          Dec 7 at 22:35














                        • 2




                          In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
                          – jmerry
                          Dec 7 at 22:35








                        2




                        2




                        In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
                        – jmerry
                        Dec 7 at 22:35




                        In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them.
                        – jmerry
                        Dec 7 at 22:35











                        2














                        Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,



                        Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution



                        Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.



                        Thus the only remaining points to check are $xin {0, 1}$.






                        share|cite|improve this answer


























                          2














                          Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,



                          Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution



                          Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.



                          Thus the only remaining points to check are $xin {0, 1}$.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,



                            Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution



                            Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.



                            Thus the only remaining points to check are $xin {0, 1}$.






                            share|cite|improve this answer












                            Just another way $:quad$ You have already ruled out $x leqslant -1$. For others,



                            Case $xin (0, 1)$ using Bernoulli's inequality, $(1+1)^x < 1+x$, so there is no solution



                            Case $x in (-1, 0)cup (1, infty)$, again with Bernoulli's inequality we have $(1+1)^x > 1+x$.



                            Thus the only remaining points to check are $xin {0, 1}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 at 18:57









                            Macavity

                            35.1k52453




                            35.1k52453























                                1














                                Function $xmapsto 2^x$ is strictly convex.



                                A strictly convex intersects a line in 0, 1 or 2 points.



                                So there exists at most two solutions of the equation $2^x = x +1$.






                                share|cite|improve this answer




























                                  1














                                  Function $xmapsto 2^x$ is strictly convex.



                                  A strictly convex intersects a line in 0, 1 or 2 points.



                                  So there exists at most two solutions of the equation $2^x = x +1$.






                                  share|cite|improve this answer


























                                    1












                                    1








                                    1






                                    Function $xmapsto 2^x$ is strictly convex.



                                    A strictly convex intersects a line in 0, 1 or 2 points.



                                    So there exists at most two solutions of the equation $2^x = x +1$.






                                    share|cite|improve this answer














                                    Function $xmapsto 2^x$ is strictly convex.



                                    A strictly convex intersects a line in 0, 1 or 2 points.



                                    So there exists at most two solutions of the equation $2^x = x +1$.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 7 at 23:06

























                                    answered Dec 7 at 23:01









                                    dallonsi

                                    1187




                                    1187






























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