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How to simplify the following summation?
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I doing some derivation for my work. In some textbook, I got the following simplification, $x = sum_k=0^L-1frac12p+2(k+1)$ to $x=frac2^L-1sum_k=0^L-1fracL!k+1prod_k=0^L-12p+2(k+1)$, where $p$ is constant.
I could get the denominator, but I'm not able to get how the numerator got modified? Is this true simplification using the method of mathematical induction or if its true how it can done?
Please clarify my doubt? Thanks.
induction
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I doing some derivation for my work. In some textbook, I got the following simplification, $x = sum_k=0^L-1frac12p+2(k+1)$ to $x=frac2^L-1sum_k=0^L-1fracL!k+1prod_k=0^L-12p+2(k+1)$, where $p$ is constant.
I could get the denominator, but I'm not able to get how the numerator got modified? Is this true simplification using the method of mathematical induction or if its true how it can done?
Please clarify my doubt? Thanks.
induction
Is it possible that you forgot some $p$ factor at the numerator of the simplified version of $x$?
– Davide Morgante
Jul 21 at 19:25
No sir. But you suggest modifications accordingly. Thanks, @DavideMorgante
– user470730
Jul 22 at 8:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I doing some derivation for my work. In some textbook, I got the following simplification, $x = sum_k=0^L-1frac12p+2(k+1)$ to $x=frac2^L-1sum_k=0^L-1fracL!k+1prod_k=0^L-12p+2(k+1)$, where $p$ is constant.
I could get the denominator, but I'm not able to get how the numerator got modified? Is this true simplification using the method of mathematical induction or if its true how it can done?
Please clarify my doubt? Thanks.
induction
I doing some derivation for my work. In some textbook, I got the following simplification, $x = sum_k=0^L-1frac12p+2(k+1)$ to $x=frac2^L-1sum_k=0^L-1fracL!k+1prod_k=0^L-12p+2(k+1)$, where $p$ is constant.
I could get the denominator, but I'm not able to get how the numerator got modified? Is this true simplification using the method of mathematical induction or if its true how it can done?
Please clarify my doubt? Thanks.
induction
asked Jul 21 at 18:53
user470730
Is it possible that you forgot some $p$ factor at the numerator of the simplified version of $x$?
– Davide Morgante
Jul 21 at 19:25
No sir. But you suggest modifications accordingly. Thanks, @DavideMorgante
– user470730
Jul 22 at 8:40
add a comment |Â
Is it possible that you forgot some $p$ factor at the numerator of the simplified version of $x$?
– Davide Morgante
Jul 21 at 19:25
No sir. But you suggest modifications accordingly. Thanks, @DavideMorgante
– user470730
Jul 22 at 8:40
Is it possible that you forgot some $p$ factor at the numerator of the simplified version of $x$?
– Davide Morgante
Jul 21 at 19:25
Is it possible that you forgot some $p$ factor at the numerator of the simplified version of $x$?
– Davide Morgante
Jul 21 at 19:25
No sir. But you suggest modifications accordingly. Thanks, @DavideMorgante
– user470730
Jul 22 at 8:40
No sir. But you suggest modifications accordingly. Thanks, @DavideMorgante
– user470730
Jul 22 at 8:40
add a comment |Â
2 Answers
2
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oldest
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0
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is this:
$$x=frac sum_k=0^L-1left(prod_ine k^L-1 2p + 2(i+1) right) prod_j=0^L-1 2p + 2(j+1)$$
useful?
answered Jul 21 at 19:22
Mattia Ghio
189
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
add a comment |Â
up vote
0
down vote
For me seems very unlikely that the simplification holds true, there's no dependence on $p$ at the numerator where clearly should be being the denominator just the product of all the denominators without any simplification.
Let us take it easy and try to prove it with a small amount of factors, like $L=3$. In this case we get
$$x = sum_k=0^2 frac12p+2(k+1) = frac12p+2+frac12p+4+frac12p+6 \ = frac(2p+4)(2p+6)+(2p+2)(2p+6)+(2p+2)(2p+4)(2p+2)(2p+4)(2p+6) \= frac3(4p^2)+2(2pcdot6)+2(2pcdot4)+2(2pcdot2)+(4cdot6)+(2cdot6)+(2cdot4)prod_k=0^2 2p+2(k+1) \=frac3(4p^2)+p(8+16+24)+(8+12+24)prod_k=0^2 2p+2(k+1)$$
From your formula, for $L=3$ we have $$2^2sum_k=0^2frac3!k-1 = 4left(3!+frac3!2+frac3!3right)$$
which clearly is not the same thing we got before
answered Jul 22 at 9:00
Davide Morgante
1,812220
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
is this:
$$x=frac sum_k=0^L-1left(prod_ine k^L-1 2p + 2(i+1) right) prod_j=0^L-1 2p + 2(j+1)$$
useful?
answered Jul 21 at 19:22
Mattia Ghio
189
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
add a comment |Â
up vote
0
down vote
is this:
$$x=frac sum_k=0^L-1left(prod_ine k^L-1 2p + 2(i+1) right) prod_j=0^L-1 2p + 2(j+1)$$
useful?
answered Jul 21 at 19:22
Mattia Ghio
189
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
add a comment |Â
up vote
0
down vote
up vote
0
down vote
is this:
$$x=frac sum_k=0^L-1left(prod_ine k^L-1 2p + 2(i+1) right) prod_j=0^L-1 2p + 2(j+1)$$
useful?
answered Jul 21 at 19:22
Mattia Ghio
189
is this:
$$x=frac sum_k=0^L-1left(prod_ine k^L-1 2p + 2(i+1) right) prod_j=0^L-1 2p + 2(j+1)$$
useful?
answered Jul 21 at 19:22
Mattia Ghio
189
answered Jul 21 at 19:22
Mattia Ghio
189
answered Jul 21 at 19:22
Mattia Ghio
189
answered Jul 21 at 19:22
Mattia Ghio
189
189
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
add a comment |Â
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
I have also got the same expression. Thanks, @Mattia Ghio
– user470730
Jul 22 at 8:38
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
Try with the logarithm function, it trasform productories in summation
– Mattia Ghio
Jul 22 at 8:53
add a comment |Â
up vote
0
down vote
For me seems very unlikely that the simplification holds true, there's no dependence on $p$ at the numerator where clearly should be being the denominator just the product of all the denominators without any simplification.
Let us take it easy and try to prove it with a small amount of factors, like $L=3$. In this case we get
$$x = sum_k=0^2 frac12p+2(k+1) = frac12p+2+frac12p+4+frac12p+6 \ = frac(2p+4)(2p+6)+(2p+2)(2p+6)+(2p+2)(2p+4)(2p+2)(2p+4)(2p+6) \= frac3(4p^2)+2(2pcdot6)+2(2pcdot4)+2(2pcdot2)+(4cdot6)+(2cdot6)+(2cdot4)prod_k=0^2 2p+2(k+1) \=frac3(4p^2)+p(8+16+24)+(8+12+24)prod_k=0^2 2p+2(k+1)$$
From your formula, for $L=3$ we have $$2^2sum_k=0^2frac3!k-1 = 4left(3!+frac3!2+frac3!3right)$$
which clearly is not the same thing we got before
answered Jul 22 at 9:00
Davide Morgante
1,812220
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
add a comment |Â
up vote
0
down vote
For me seems very unlikely that the simplification holds true, there's no dependence on $p$ at the numerator where clearly should be being the denominator just the product of all the denominators without any simplification.
Let us take it easy and try to prove it with a small amount of factors, like $L=3$. In this case we get
$$x = sum_k=0^2 frac12p+2(k+1) = frac12p+2+frac12p+4+frac12p+6 \ = frac(2p+4)(2p+6)+(2p+2)(2p+6)+(2p+2)(2p+4)(2p+2)(2p+4)(2p+6) \= frac3(4p^2)+2(2pcdot6)+2(2pcdot4)+2(2pcdot2)+(4cdot6)+(2cdot6)+(2cdot4)prod_k=0^2 2p+2(k+1) \=frac3(4p^2)+p(8+16+24)+(8+12+24)prod_k=0^2 2p+2(k+1)$$
From your formula, for $L=3$ we have $$2^2sum_k=0^2frac3!k-1 = 4left(3!+frac3!2+frac3!3right)$$
which clearly is not the same thing we got before
answered Jul 22 at 9:00
Davide Morgante
1,812220
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For me seems very unlikely that the simplification holds true, there's no dependence on $p$ at the numerator where clearly should be being the denominator just the product of all the denominators without any simplification.
Let us take it easy and try to prove it with a small amount of factors, like $L=3$. In this case we get
$$x = sum_k=0^2 frac12p+2(k+1) = frac12p+2+frac12p+4+frac12p+6 \ = frac(2p+4)(2p+6)+(2p+2)(2p+6)+(2p+2)(2p+4)(2p+2)(2p+4)(2p+6) \= frac3(4p^2)+2(2pcdot6)+2(2pcdot4)+2(2pcdot2)+(4cdot6)+(2cdot6)+(2cdot4)prod_k=0^2 2p+2(k+1) \=frac3(4p^2)+p(8+16+24)+(8+12+24)prod_k=0^2 2p+2(k+1)$$
From your formula, for $L=3$ we have $$2^2sum_k=0^2frac3!k-1 = 4left(3!+frac3!2+frac3!3right)$$
which clearly is not the same thing we got before
answered Jul 22 at 9:00
Davide Morgante
1,812220
For me seems very unlikely that the simplification holds true, there's no dependence on $p$ at the numerator where clearly should be being the denominator just the product of all the denominators without any simplification.
Let us take it easy and try to prove it with a small amount of factors, like $L=3$. In this case we get
$$x = sum_k=0^2 frac12p+2(k+1) = frac12p+2+frac12p+4+frac12p+6 \ = frac(2p+4)(2p+6)+(2p+2)(2p+6)+(2p+2)(2p+4)(2p+2)(2p+4)(2p+6) \= frac3(4p^2)+2(2pcdot6)+2(2pcdot4)+2(2pcdot2)+(4cdot6)+(2cdot6)+(2cdot4)prod_k=0^2 2p+2(k+1) \=frac3(4p^2)+p(8+16+24)+(8+12+24)prod_k=0^2 2p+2(k+1)$$
From your formula, for $L=3$ we have $$2^2sum_k=0^2frac3!k-1 = 4left(3!+frac3!2+frac3!3right)$$
which clearly is not the same thing we got before
answered Jul 22 at 9:00
Davide Morgante
1,812220
edited Jul 22 at 9:12
answered Jul 22 at 9:00
Davide Morgante
1,812220
answered Jul 22 at 9:00
Davide Morgante
1,812220
answered Jul 22 at 9:00
Davide Morgante
1,812220
1,812220
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
add a comment |Â
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
Yes, I agree with your explanation.
– user470730
Jul 23 at 4:21
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
If this was what were you looking for, you should accept the answer. Here on MSE this gives points to the user who answered as well as letting all other users know that this forum is closed!
– Davide Morgante
Jul 23 at 7:55
add a comment |Â
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Is it possible that you forgot some $p$ factor at the numerator of the simplified version of $x$?
– Davide Morgante
Jul 21 at 19:25
No sir. But you suggest modifications accordingly. Thanks, @DavideMorgante
– user470730
Jul 22 at 8:40