Prove that a vector x from L2[a,b] is in the subspace generated by f1,f2,f3












-1














Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra differential-equations vector-spaces







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    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    – Nick Peterson
    Jan 4 at 17:03
















-1














Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra differential-equations vector-spaces







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Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 2




    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    – Nick Peterson
    Jan 4 at 17:03














-1












-1








-1


1





Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra differential-equations vector-spaces







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New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra differential-equations vector-spaces




linear-algebra differential-equations vector-spaces






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edited Jan 4 at 16:51









Davide Giraudo

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asked Jan 4 at 16:48









Raluca lokRaluca lok

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  • 2




    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    – Nick Peterson
    Jan 4 at 17:03














  • 2




    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    – Nick Peterson
    Jan 4 at 17:03








2




2




Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
– Nick Peterson
Jan 4 at 17:03




Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
– Nick Peterson
Jan 4 at 17:03










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To begin, we note that each of the functions



$f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



satisfy the differential equation



$x''' - x'' + x' - x = 0, tag 2$



as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



$f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



$f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



$V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



we see that every element of this subspace $V$ obeys (2).



To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



$x = e^{rt}, tag 5$



so that



$x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



we find that



$r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



we may divide out $e^{rt}$:



$r^3 - r^2 + r - 1 = 0; tag 8$



we may factor the cubic on the left:



$r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



thus,



$(r^2 + 1)(r - 1) = 0, tag{10}$



whence



$r = 1, ; r = pm i; tag{11}$



we thus find three functions



$f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






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    1 Answer
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    To begin, we note that each of the functions



    $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



    satisfy the differential equation



    $x''' - x'' + x' - x = 0, tag 2$



    as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



    $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



    $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



    $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



    we see that every element of this subspace $V$ obeys (2).



    To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



    $x = e^{rt}, tag 5$



    so that



    $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



    we find that



    $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



    we may divide out $e^{rt}$:



    $r^3 - r^2 + r - 1 = 0; tag 8$



    we may factor the cubic on the left:



    $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



    thus,



    $(r^2 + 1)(r - 1) = 0, tag{10}$



    whence



    $r = 1, ; r = pm i; tag{11}$



    we thus find three functions



    $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



    all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






    share|cite|improve this answer




























      1














      To begin, we note that each of the functions



      $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



      satisfy the differential equation



      $x''' - x'' + x' - x = 0, tag 2$



      as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



      $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



      $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



      $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



      we see that every element of this subspace $V$ obeys (2).



      To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



      $x = e^{rt}, tag 5$



      so that



      $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



      we find that



      $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



      we may divide out $e^{rt}$:



      $r^3 - r^2 + r - 1 = 0; tag 8$



      we may factor the cubic on the left:



      $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



      thus,



      $(r^2 + 1)(r - 1) = 0, tag{10}$



      whence



      $r = 1, ; r = pm i; tag{11}$



      we thus find three functions



      $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



      all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






      share|cite|improve this answer


























        1












        1








        1






        To begin, we note that each of the functions



        $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



        satisfy the differential equation



        $x''' - x'' + x' - x = 0, tag 2$



        as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



        $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



        $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



        $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



        we see that every element of this subspace $V$ obeys (2).



        To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



        $x = e^{rt}, tag 5$



        so that



        $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



        we find that



        $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



        we may divide out $e^{rt}$:



        $r^3 - r^2 + r - 1 = 0; tag 8$



        we may factor the cubic on the left:



        $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



        thus,



        $(r^2 + 1)(r - 1) = 0, tag{10}$



        whence



        $r = 1, ; r = pm i; tag{11}$



        we thus find three functions



        $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



        all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






        share|cite|improve this answer














        To begin, we note that each of the functions



        $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



        satisfy the differential equation



        $x''' - x'' + x' - x = 0, tag 2$



        as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



        $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



        $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



        $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



        we see that every element of this subspace $V$ obeys (2).



        To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



        $x = e^{rt}, tag 5$



        so that



        $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



        we find that



        $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



        we may divide out $e^{rt}$:



        $r^3 - r^2 + r - 1 = 0; tag 8$



        we may factor the cubic on the left:



        $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



        thus,



        $(r^2 + 1)(r - 1) = 0, tag{10}$



        whence



        $r = 1, ; r = pm i; tag{11}$



        we thus find three functions



        $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



        all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 0:27

























        answered Jan 4 at 17:56









        Robert LewisRobert Lewis

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