Consider the function $ int_0^x [t]dt $ where $ x>0 $ and [t] denotes largest integer less than or equal...












1












$begingroup$


I have done the following:
$ int_0^x [t]dt $ can be divided as
$ int_0^1 [t]dt + int_1^2 [t]dt + int_2^3 [t]dt + ... + int_{x-1}^x [t]dt$

= $ int_0^1 0dt +int_1^2 1dt +....$

= $ 1 + 2 + 3 + ....{(x-1)} $

= $ frac{x(x-1)}{2} $

Therefore, $ f(x) = frac{x(x-1)}{2} $

If we use this expression to determine the continuity or differentiability of the f(x) then it is continuous and differentiable in [0,x]

However, if we use the original expression $ f(x)= int_0^x [t]dt $ to determine the nature of the function with the fundamental formulae,
then the results show that the function is continuous but not differentiable. Where is the fallacy?






integration functions continuity







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  • 3




    $begingroup$
    If $x$ is not an integer, the last integral should be $int_{lfloor xrfloor}^x [t]dt$, which is equal to $(x-lfloor xrfloor)lfloor xrfloor$.
    $endgroup$
    – CY Aries
    Jan 9 at 3:01


















1












$begingroup$


I have done the following:
$ int_0^x [t]dt $ can be divided as
$ int_0^1 [t]dt + int_1^2 [t]dt + int_2^3 [t]dt + ... + int_{x-1}^x [t]dt$

= $ int_0^1 0dt +int_1^2 1dt +....$

= $ 1 + 2 + 3 + ....{(x-1)} $

= $ frac{x(x-1)}{2} $

Therefore, $ f(x) = frac{x(x-1)}{2} $

If we use this expression to determine the continuity or differentiability of the f(x) then it is continuous and differentiable in [0,x]

However, if we use the original expression $ f(x)= int_0^x [t]dt $ to determine the nature of the function with the fundamental formulae,
then the results show that the function is continuous but not differentiable. Where is the fallacy?






integration functions continuity







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If $x$ is not an integer, the last integral should be $int_{lfloor xrfloor}^x [t]dt$, which is equal to $(x-lfloor xrfloor)lfloor xrfloor$.
    $endgroup$
    – CY Aries
    Jan 9 at 3:01
















1












1








1





$begingroup$


I have done the following:
$ int_0^x [t]dt $ can be divided as
$ int_0^1 [t]dt + int_1^2 [t]dt + int_2^3 [t]dt + ... + int_{x-1}^x [t]dt$

= $ int_0^1 0dt +int_1^2 1dt +....$

= $ 1 + 2 + 3 + ....{(x-1)} $

= $ frac{x(x-1)}{2} $

Therefore, $ f(x) = frac{x(x-1)}{2} $

If we use this expression to determine the continuity or differentiability of the f(x) then it is continuous and differentiable in [0,x]

However, if we use the original expression $ f(x)= int_0^x [t]dt $ to determine the nature of the function with the fundamental formulae,
then the results show that the function is continuous but not differentiable. Where is the fallacy?






integration functions continuity







share|cite|improve this question









$endgroup$




I have done the following:
$ int_0^x [t]dt $ can be divided as
$ int_0^1 [t]dt + int_1^2 [t]dt + int_2^3 [t]dt + ... + int_{x-1}^x [t]dt$

= $ int_0^1 0dt +int_1^2 1dt +....$

= $ 1 + 2 + 3 + ....{(x-1)} $

= $ frac{x(x-1)}{2} $

Therefore, $ f(x) = frac{x(x-1)}{2} $

If we use this expression to determine the continuity or differentiability of the f(x) then it is continuous and differentiable in [0,x]

However, if we use the original expression $ f(x)= int_0^x [t]dt $ to determine the nature of the function with the fundamental formulae,
then the results show that the function is continuous but not differentiable. Where is the fallacy?






integration functions continuity




integration functions continuity






share|cite|improve this question













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asked Jan 9 at 2:54









Anubhab GiriAnubhab Giri

345




345








  • 3




    $begingroup$
    If $x$ is not an integer, the last integral should be $int_{lfloor xrfloor}^x [t]dt$, which is equal to $(x-lfloor xrfloor)lfloor xrfloor$.
    $endgroup$
    – CY Aries
    Jan 9 at 3:01
















  • 3




    $begingroup$
    If $x$ is not an integer, the last integral should be $int_{lfloor xrfloor}^x [t]dt$, which is equal to $(x-lfloor xrfloor)lfloor xrfloor$.
    $endgroup$
    – CY Aries
    Jan 9 at 3:01










3




3




$begingroup$
If $x$ is not an integer, the last integral should be $int_{lfloor xrfloor}^x [t]dt$, which is equal to $(x-lfloor xrfloor)lfloor xrfloor$.
$endgroup$
– CY Aries
Jan 9 at 3:01






$begingroup$
If $x$ is not an integer, the last integral should be $int_{lfloor xrfloor}^x [t]dt$, which is equal to $(x-lfloor xrfloor)lfloor xrfloor$.
$endgroup$
– CY Aries
Jan 9 at 3:01












3 Answers
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Your function is shown in the graph below. Between two naturals $n$ and $n+1$ the graph has slope $n$, so it is differentiable. At the naturals it is not differentiable because the slope is different on the two sides. You have calculated the values at the naturals correctly, but it has values in between.



enter image description here






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    0












    $begingroup$

    Your division into several integrals appears to assume $x$ is an integer






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I believe an important issue is that your argument assumes that $x$ is integral, when that is not necessarily the case. This is why your $fleft(xright) = frac{xleft(x - 1right)}{2}$ is not true in all cases. In particular, the last term in the summation is provided by CY Aries in the comments to your question to properly be $left(x - lfloor x rfloorright)lfloor x rfloor$.






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        3 Answers
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        active

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        3 Answers
        3






        active

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        active

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        1












        $begingroup$

        Your function is shown in the graph below. Between two naturals $n$ and $n+1$ the graph has slope $n$, so it is differentiable. At the naturals it is not differentiable because the slope is different on the two sides. You have calculated the values at the naturals correctly, but it has values in between.



        enter image description here






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Your function is shown in the graph below. Between two naturals $n$ and $n+1$ the graph has slope $n$, so it is differentiable. At the naturals it is not differentiable because the slope is different on the two sides. You have calculated the values at the naturals correctly, but it has values in between.



          enter image description here






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Your function is shown in the graph below. Between two naturals $n$ and $n+1$ the graph has slope $n$, so it is differentiable. At the naturals it is not differentiable because the slope is different on the two sides. You have calculated the values at the naturals correctly, but it has values in between.



            enter image description here






            share|cite|improve this answer









            $endgroup$



            Your function is shown in the graph below. Between two naturals $n$ and $n+1$ the graph has slope $n$, so it is differentiable. At the naturals it is not differentiable because the slope is different on the two sides. You have calculated the values at the naturals correctly, but it has values in between.



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 4:06









            Ross MillikanRoss Millikan

            297k23198371




            297k23198371























                0












                $begingroup$

                Your division into several integrals appears to assume $x$ is an integer






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Your division into several integrals appears to assume $x$ is an integer






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Your division into several integrals appears to assume $x$ is an integer






                    share|cite|improve this answer









                    $endgroup$



                    Your division into several integrals appears to assume $x$ is an integer







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 9 at 3:00









                    MPWMPW

                    30.3k12157




                    30.3k12157























                        0












                        $begingroup$

                        I believe an important issue is that your argument assumes that $x$ is integral, when that is not necessarily the case. This is why your $fleft(xright) = frac{xleft(x - 1right)}{2}$ is not true in all cases. In particular, the last term in the summation is provided by CY Aries in the comments to your question to properly be $left(x - lfloor x rfloorright)lfloor x rfloor$.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          I believe an important issue is that your argument assumes that $x$ is integral, when that is not necessarily the case. This is why your $fleft(xright) = frac{xleft(x - 1right)}{2}$ is not true in all cases. In particular, the last term in the summation is provided by CY Aries in the comments to your question to properly be $left(x - lfloor x rfloorright)lfloor x rfloor$.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            I believe an important issue is that your argument assumes that $x$ is integral, when that is not necessarily the case. This is why your $fleft(xright) = frac{xleft(x - 1right)}{2}$ is not true in all cases. In particular, the last term in the summation is provided by CY Aries in the comments to your question to properly be $left(x - lfloor x rfloorright)lfloor x rfloor$.






                            share|cite|improve this answer











                            $endgroup$



                            I believe an important issue is that your argument assumes that $x$ is integral, when that is not necessarily the case. This is why your $fleft(xright) = frac{xleft(x - 1right)}{2}$ is not true in all cases. In particular, the last term in the summation is provided by CY Aries in the comments to your question to properly be $left(x - lfloor x rfloorright)lfloor x rfloor$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 9 at 6:16

























                            answered Jan 9 at 3:00









                            John OmielanJohn Omielan

                            3,1951214




                            3,1951214






























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