How to evaluate $f(1)$?












0














$$f(x+1)-f(x) = 3$$
$$f(19) = 70$$



I'm trying to evaluate $f(1)$ from given data. However, what I noticed so far is that the function is linear.



My attempt:



For $x=19$



$$f(19) = f(20)-3$$



This yields



$$70 = f(20)-3 implies f(20) = 73$$



Since my goal is to evaluate $f(1)$, it makes no literal sense to proceed from there. Could you assist me?






functions







share|cite|improve this question






















  • Hint $ $ Put $,n=19,$ in $,f(n) = f(1) + 3(n-1) $
    – Bill Dubuque
    Nov 21 at 21:28
















0














$$f(x+1)-f(x) = 3$$
$$f(19) = 70$$



I'm trying to evaluate $f(1)$ from given data. However, what I noticed so far is that the function is linear.



My attempt:



For $x=19$



$$f(19) = f(20)-3$$



This yields



$$70 = f(20)-3 implies f(20) = 73$$



Since my goal is to evaluate $f(1)$, it makes no literal sense to proceed from there. Could you assist me?






functions







share|cite|improve this question






















  • Hint $ $ Put $,n=19,$ in $,f(n) = f(1) + 3(n-1) $
    – Bill Dubuque
    Nov 21 at 21:28














0












0








0







$$f(x+1)-f(x) = 3$$
$$f(19) = 70$$



I'm trying to evaluate $f(1)$ from given data. However, what I noticed so far is that the function is linear.



My attempt:



For $x=19$



$$f(19) = f(20)-3$$



This yields



$$70 = f(20)-3 implies f(20) = 73$$



Since my goal is to evaluate $f(1)$, it makes no literal sense to proceed from there. Could you assist me?






functions







share|cite|improve this question













$$f(x+1)-f(x) = 3$$
$$f(19) = 70$$



I'm trying to evaluate $f(1)$ from given data. However, what I noticed so far is that the function is linear.



My attempt:



For $x=19$



$$f(19) = f(20)-3$$



This yields



$$70 = f(20)-3 implies f(20) = 73$$



Since my goal is to evaluate $f(1)$, it makes no literal sense to proceed from there. Could you assist me?






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 21 at 19:53









Enzo

996




996












  • Hint $ $ Put $,n=19,$ in $,f(n) = f(1) + 3(n-1) $
    – Bill Dubuque
    Nov 21 at 21:28


















  • Hint $ $ Put $,n=19,$ in $,f(n) = f(1) + 3(n-1) $
    – Bill Dubuque
    Nov 21 at 21:28
















Hint $ $ Put $,n=19,$ in $,f(n) = f(1) + 3(n-1) $
– Bill Dubuque
Nov 21 at 21:28




Hint $ $ Put $,n=19,$ in $,f(n) = f(1) + 3(n-1) $
– Bill Dubuque
Nov 21 at 21:28










2 Answers
2






active

oldest

votes


















2














Observe that
begin{align}
f(1) &= f(1) - f(2)+f(2)-f(3)+ldots-f(19)+f(19)\
&= sum_{k=1}^{18} (f(k)-f(k+1)) + f(19)\
&= -sum_{k=1}^{18} (f(k+1)-f(k)) + 70\
&= -sum_{k=1}^{18} 3 + 70\
&= -3 cdot 18 + 70 = 16.
end{align}






share|cite|improve this answer























  • It's unclear after the summation.
    – Enzo
    Nov 21 at 20:01










  • @Enzo I have added two extra steps. Let me know if it is still unclear.
    – MisterRiemann
    Nov 21 at 20:03










  • Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
    – Enzo
    Nov 21 at 20:04










  • @Enzo Could you be more precise? Which row of the computation are you not familiar with?
    – MisterRiemann
    Nov 21 at 20:05










  • How did you also obtain $3$?
    – Enzo
    Nov 21 at 20:32



















1














We conclude that $$f(x)=3x+a$$for some real $a$ and for $xinBbb N$. Also $$f(19)=70=57+ato a=13$$and we can obtain$$f(1)=3+13=16$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Observe that
    begin{align}
    f(1) &= f(1) - f(2)+f(2)-f(3)+ldots-f(19)+f(19)\
    &= sum_{k=1}^{18} (f(k)-f(k+1)) + f(19)\
    &= -sum_{k=1}^{18} (f(k+1)-f(k)) + 70\
    &= -sum_{k=1}^{18} 3 + 70\
    &= -3 cdot 18 + 70 = 16.
    end{align}






    share|cite|improve this answer























    • It's unclear after the summation.
      – Enzo
      Nov 21 at 20:01










    • @Enzo I have added two extra steps. Let me know if it is still unclear.
      – MisterRiemann
      Nov 21 at 20:03










    • Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
      – Enzo
      Nov 21 at 20:04










    • @Enzo Could you be more precise? Which row of the computation are you not familiar with?
      – MisterRiemann
      Nov 21 at 20:05










    • How did you also obtain $3$?
      – Enzo
      Nov 21 at 20:32
















    2














    Observe that
    begin{align}
    f(1) &= f(1) - f(2)+f(2)-f(3)+ldots-f(19)+f(19)\
    &= sum_{k=1}^{18} (f(k)-f(k+1)) + f(19)\
    &= -sum_{k=1}^{18} (f(k+1)-f(k)) + 70\
    &= -sum_{k=1}^{18} 3 + 70\
    &= -3 cdot 18 + 70 = 16.
    end{align}






    share|cite|improve this answer























    • It's unclear after the summation.
      – Enzo
      Nov 21 at 20:01










    • @Enzo I have added two extra steps. Let me know if it is still unclear.
      – MisterRiemann
      Nov 21 at 20:03










    • Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
      – Enzo
      Nov 21 at 20:04










    • @Enzo Could you be more precise? Which row of the computation are you not familiar with?
      – MisterRiemann
      Nov 21 at 20:05










    • How did you also obtain $3$?
      – Enzo
      Nov 21 at 20:32














    2












    2








    2






    Observe that
    begin{align}
    f(1) &= f(1) - f(2)+f(2)-f(3)+ldots-f(19)+f(19)\
    &= sum_{k=1}^{18} (f(k)-f(k+1)) + f(19)\
    &= -sum_{k=1}^{18} (f(k+1)-f(k)) + 70\
    &= -sum_{k=1}^{18} 3 + 70\
    &= -3 cdot 18 + 70 = 16.
    end{align}






    share|cite|improve this answer














    Observe that
    begin{align}
    f(1) &= f(1) - f(2)+f(2)-f(3)+ldots-f(19)+f(19)\
    &= sum_{k=1}^{18} (f(k)-f(k+1)) + f(19)\
    &= -sum_{k=1}^{18} (f(k+1)-f(k)) + 70\
    &= -sum_{k=1}^{18} 3 + 70\
    &= -3 cdot 18 + 70 = 16.
    end{align}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 21 at 20:03

























    answered Nov 21 at 19:57









    MisterRiemann

    5,7291624




    5,7291624












    • It's unclear after the summation.
      – Enzo
      Nov 21 at 20:01










    • @Enzo I have added two extra steps. Let me know if it is still unclear.
      – MisterRiemann
      Nov 21 at 20:03










    • Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
      – Enzo
      Nov 21 at 20:04










    • @Enzo Could you be more precise? Which row of the computation are you not familiar with?
      – MisterRiemann
      Nov 21 at 20:05










    • How did you also obtain $3$?
      – Enzo
      Nov 21 at 20:32


















    • It's unclear after the summation.
      – Enzo
      Nov 21 at 20:01










    • @Enzo I have added two extra steps. Let me know if it is still unclear.
      – MisterRiemann
      Nov 21 at 20:03










    • Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
      – Enzo
      Nov 21 at 20:04










    • @Enzo Could you be more precise? Which row of the computation are you not familiar with?
      – MisterRiemann
      Nov 21 at 20:05










    • How did you also obtain $3$?
      – Enzo
      Nov 21 at 20:32
















    It's unclear after the summation.
    – Enzo
    Nov 21 at 20:01




    It's unclear after the summation.
    – Enzo
    Nov 21 at 20:01












    @Enzo I have added two extra steps. Let me know if it is still unclear.
    – MisterRiemann
    Nov 21 at 20:03




    @Enzo I have added two extra steps. Let me know if it is still unclear.
    – MisterRiemann
    Nov 21 at 20:03












    Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
    – Enzo
    Nov 21 at 20:04




    Yes, unfortunalety It is. So, I think you need to update something with summation since im not familiar with it.
    – Enzo
    Nov 21 at 20:04












    @Enzo Could you be more precise? Which row of the computation are you not familiar with?
    – MisterRiemann
    Nov 21 at 20:05




    @Enzo Could you be more precise? Which row of the computation are you not familiar with?
    – MisterRiemann
    Nov 21 at 20:05












    How did you also obtain $3$?
    – Enzo
    Nov 21 at 20:32




    How did you also obtain $3$?
    – Enzo
    Nov 21 at 20:32











    1














    We conclude that $$f(x)=3x+a$$for some real $a$ and for $xinBbb N$. Also $$f(19)=70=57+ato a=13$$and we can obtain$$f(1)=3+13=16$$






    share|cite|improve this answer


























      1














      We conclude that $$f(x)=3x+a$$for some real $a$ and for $xinBbb N$. Also $$f(19)=70=57+ato a=13$$and we can obtain$$f(1)=3+13=16$$






      share|cite|improve this answer
























        1












        1








        1






        We conclude that $$f(x)=3x+a$$for some real $a$ and for $xinBbb N$. Also $$f(19)=70=57+ato a=13$$and we can obtain$$f(1)=3+13=16$$






        share|cite|improve this answer












        We conclude that $$f(x)=3x+a$$for some real $a$ and for $xinBbb N$. Also $$f(19)=70=57+ato a=13$$and we can obtain$$f(1)=3+13=16$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 19:57









        Mostafa Ayaz

        13.6k3836




        13.6k3836






























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