How to turn recursion sum formula into an integral?

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I have a formula that looks a lot like integral:
$$
m(t) = lim_Delta t to 0 sum_i in Delta t, 2Delta t, ... ,t ( m (i - Delta t) + v(i)) , Delta t
$$
where $v(t) = textconst$, $m(0)$ is given.
Yet I wonder - how to turn it into an integral and is it possible when to calculate $m(t)$ I have to obtain $mleft(t- Delta tright)$ ?







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  • Without any context we can't help you. What is "Pr(t)"? What is "n(t)"? What is "v(t)"? Where is "i" used in the sum?
    – Somos
    Jul 22 at 13:43










  • They are functions that depend only on time
    – DuckQueen
    Jul 22 at 22:01










  • simplified a bit
    – DuckQueen
    Jul 23 at 10:35














up vote
0
down vote

favorite












I have a formula that looks a lot like integral:
$$
m(t) = lim_Delta t to 0 sum_i in Delta t, 2Delta t, ... ,t ( m (i - Delta t) + v(i)) , Delta t
$$
where $v(t) = textconst$, $m(0)$ is given.
Yet I wonder - how to turn it into an integral and is it possible when to calculate $m(t)$ I have to obtain $mleft(t- Delta tright)$ ?







share|cite|improve this question





















  • Without any context we can't help you. What is "Pr(t)"? What is "n(t)"? What is "v(t)"? Where is "i" used in the sum?
    – Somos
    Jul 22 at 13:43










  • They are functions that depend only on time
    – DuckQueen
    Jul 22 at 22:01










  • simplified a bit
    – DuckQueen
    Jul 23 at 10:35












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a formula that looks a lot like integral:
$$
m(t) = lim_Delta t to 0 sum_i in Delta t, 2Delta t, ... ,t ( m (i - Delta t) + v(i)) , Delta t
$$
where $v(t) = textconst$, $m(0)$ is given.
Yet I wonder - how to turn it into an integral and is it possible when to calculate $m(t)$ I have to obtain $mleft(t- Delta tright)$ ?







share|cite|improve this question













I have a formula that looks a lot like integral:
$$
m(t) = lim_Delta t to 0 sum_i in Delta t, 2Delta t, ... ,t ( m (i - Delta t) + v(i)) , Delta t
$$
where $v(t) = textconst$, $m(0)$ is given.
Yet I wonder - how to turn it into an integral and is it possible when to calculate $m(t)$ I have to obtain $mleft(t- Delta tright)$ ?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 11:54









Sangchul Lee

85.5k12155253




85.5k12155253









asked Jul 22 at 11:58









DuckQueen

1377




1377











  • Without any context we can't help you. What is "Pr(t)"? What is "n(t)"? What is "v(t)"? Where is "i" used in the sum?
    – Somos
    Jul 22 at 13:43










  • They are functions that depend only on time
    – DuckQueen
    Jul 22 at 22:01










  • simplified a bit
    – DuckQueen
    Jul 23 at 10:35
















  • Without any context we can't help you. What is "Pr(t)"? What is "n(t)"? What is "v(t)"? Where is "i" used in the sum?
    – Somos
    Jul 22 at 13:43










  • They are functions that depend only on time
    – DuckQueen
    Jul 22 at 22:01










  • simplified a bit
    – DuckQueen
    Jul 23 at 10:35















Without any context we can't help you. What is "Pr(t)"? What is "n(t)"? What is "v(t)"? Where is "i" used in the sum?
– Somos
Jul 22 at 13:43




Without any context we can't help you. What is "Pr(t)"? What is "n(t)"? What is "v(t)"? Where is "i" used in the sum?
– Somos
Jul 22 at 13:43












They are functions that depend only on time
– DuckQueen
Jul 22 at 22:01




They are functions that depend only on time
– DuckQueen
Jul 22 at 22:01












simplified a bit
– DuckQueen
Jul 23 at 10:35




simplified a bit
– DuckQueen
Jul 23 at 10:35










1 Answer
1






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0
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The right side of your formula is a limit. If that limit exists, then it is an integral as you suggested. $, m(t) = int_0^t m(x) + v(x), dx. ,$ Your question about $, m(t-Delta t) ,$ turns out not to be a problem because in the limit that is the same as $, m(t). ,$ Taking the derivative of both sides of the equation, $, m'(t) = m(t) + v(t) ,$ which is a simple differential equation with initial value $, m(0)=0. ,$






share|cite|improve this answer























  • but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
    – DuckQueen
    Jul 23 at 11:04











  • @DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
    – Somos
    Jul 29 at 0:40










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













The right side of your formula is a limit. If that limit exists, then it is an integral as you suggested. $, m(t) = int_0^t m(x) + v(x), dx. ,$ Your question about $, m(t-Delta t) ,$ turns out not to be a problem because in the limit that is the same as $, m(t). ,$ Taking the derivative of both sides of the equation, $, m'(t) = m(t) + v(t) ,$ which is a simple differential equation with initial value $, m(0)=0. ,$






share|cite|improve this answer























  • but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
    – DuckQueen
    Jul 23 at 11:04











  • @DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
    – Somos
    Jul 29 at 0:40














up vote
0
down vote













The right side of your formula is a limit. If that limit exists, then it is an integral as you suggested. $, m(t) = int_0^t m(x) + v(x), dx. ,$ Your question about $, m(t-Delta t) ,$ turns out not to be a problem because in the limit that is the same as $, m(t). ,$ Taking the derivative of both sides of the equation, $, m'(t) = m(t) + v(t) ,$ which is a simple differential equation with initial value $, m(0)=0. ,$






share|cite|improve this answer























  • but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
    – DuckQueen
    Jul 23 at 11:04











  • @DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
    – Somos
    Jul 29 at 0:40












up vote
0
down vote










up vote
0
down vote









The right side of your formula is a limit. If that limit exists, then it is an integral as you suggested. $, m(t) = int_0^t m(x) + v(x), dx. ,$ Your question about $, m(t-Delta t) ,$ turns out not to be a problem because in the limit that is the same as $, m(t). ,$ Taking the derivative of both sides of the equation, $, m'(t) = m(t) + v(t) ,$ which is a simple differential equation with initial value $, m(0)=0. ,$






share|cite|improve this answer















The right side of your formula is a limit. If that limit exists, then it is an integral as you suggested. $, m(t) = int_0^t m(x) + v(x), dx. ,$ Your question about $, m(t-Delta t) ,$ turns out not to be a problem because in the limit that is the same as $, m(t). ,$ Taking the derivative of both sides of the equation, $, m'(t) = m(t) + v(t) ,$ which is a simple differential equation with initial value $, m(0)=0. ,$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 29 at 0:37


























answered Jul 23 at 10:45









Somos

11.5k1933




11.5k1933











  • but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
    – DuckQueen
    Jul 23 at 11:04











  • @DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
    – Somos
    Jul 29 at 0:40
















  • but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
    – DuckQueen
    Jul 23 at 11:04











  • @DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
    – Somos
    Jul 29 at 0:40















but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
– DuckQueen
Jul 23 at 11:04





but how to show that $m(x)$ has $x$ that is less than $v(x)$ and thus $m(x)$ has to be acquired on previous integration step?
– DuckQueen
Jul 23 at 11:04













@DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
– Somos
Jul 29 at 0:40




@DuckQueen What does "m(x) has x that is less than v(x)" mean? What previous integration step? Your comment does not make sense to me. Put your sensible comment in the question itself if it is important to you.
– Somos
Jul 29 at 0:40












 

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