how can i prove $sum_{t=1}^T a_t.n_t > sum_{t=1}^T a_t.n'_t $?
$begingroup$
in partitioning of numbers (ways of writing a positive integer as a sum of positive integers) .
suppose that$ P $and $Q$ are two partitions of $N in mathbb N$.
$n_t$:=the number of times that a number is used in partitioning $P$ .
$n′_t$:=the number of times that a number is used in partitioning $q$ .
if $ T$ be a largest number in every partiotioning , and $forall t : 1leq tleq T$ , $ $ $a_t =frac{N^{T-t} (T-1)! }{(t-1)!} $ .is this true?
if $n_T>n'_T$ then $sum_{t=1}^T a_t.n_t > sum_{t=1}^T a_t.n'_t $
In all the examples I tested, the inequality was correct .but i don't know how can i prove it.
my try:
according to the defenition of $a_t$ we have $a_T=1$. then we must to prove the following:
$n_T+sum_{t=1}^{T -1}a_t.n_t > n'_T +sum_{t=1}^{T -1} a_t.n'_t $
This sum is composed of two parts. The first part is right according to the assumption. I think No matter how the value of second part is big, this value does not go to the value of the first part.
Is it Right?
How do I start proof?
calculus summation proof-writing
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
$endgroup$
add a comment |
$begingroup$
in partitioning of numbers (ways of writing a positive integer as a sum of positive integers) .
suppose that$ P $and $Q$ are two partitions of $N in mathbb N$.
$n_t$:=the number of times that a number is used in partitioning $P$ .
$n′_t$:=the number of times that a number is used in partitioning $q$ .
if $ T$ be a largest number in every partiotioning , and $forall t : 1leq tleq T$ , $ $ $a_t =frac{N^{T-t} (T-1)! }{(t-1)!} $ .is this true?
if $n_T>n'_T$ then $sum_{t=1}^T a_t.n_t > sum_{t=1}^T a_t.n'_t $
In all the examples I tested, the inequality was correct .but i don't know how can i prove it.
my try:
according to the defenition of $a_t$ we have $a_T=1$. then we must to prove the following:
$n_T+sum_{t=1}^{T -1}a_t.n_t > n'_T +sum_{t=1}^{T -1} a_t.n'_t $
This sum is composed of two parts. The first part is right according to the assumption. I think No matter how the value of second part is big, this value does not go to the value of the first part.
Is it Right?
How do I start proof?
calculus summation proof-writing
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
$endgroup$
$begingroup$
Is $a_t$ the same thing as $w_t?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:11
$begingroup$
@saulspatz , yes , I forgot to correct it .I edited it. Thanks
$endgroup$
– yaodao vang
Dec 7 '18 at 13:12
$begingroup$
Is this supposed to say that if $n_tge n'_t$ for $1le tle T$ then the conclusion is true? What you have written is clearly false.
$endgroup$
– saulspatz
Dec 7 '18 at 13:17
$begingroup$
What if $n_1=1, n'_1=10^{100}$ Why should the conclusion follow just because $n_T>n'_T$ for some $T?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:26
$begingroup$
I don't understand. This whole question seems very confused to me.
$endgroup$
– saulspatz
Dec 7 '18 at 14:48
add a comment |
$begingroup$
in partitioning of numbers (ways of writing a positive integer as a sum of positive integers) .
suppose that$ P $and $Q$ are two partitions of $N in mathbb N$.
$n_t$:=the number of times that a number is used in partitioning $P$ .
$n′_t$:=the number of times that a number is used in partitioning $q$ .
if $ T$ be a largest number in every partiotioning , and $forall t : 1leq tleq T$ , $ $ $a_t =frac{N^{T-t} (T-1)! }{(t-1)!} $ .is this true?
if $n_T>n'_T$ then $sum_{t=1}^T a_t.n_t > sum_{t=1}^T a_t.n'_t $
In all the examples I tested, the inequality was correct .but i don't know how can i prove it.
my try:
according to the defenition of $a_t$ we have $a_T=1$. then we must to prove the following:
$n_T+sum_{t=1}^{T -1}a_t.n_t > n'_T +sum_{t=1}^{T -1} a_t.n'_t $
This sum is composed of two parts. The first part is right according to the assumption. I think No matter how the value of second part is big, this value does not go to the value of the first part.
Is it Right?
How do I start proof?
calculus summation proof-writing
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
$endgroup$
in partitioning of numbers (ways of writing a positive integer as a sum of positive integers) .
suppose that$ P $and $Q$ are two partitions of $N in mathbb N$.
$n_t$:=the number of times that a number is used in partitioning $P$ .
$n′_t$:=the number of times that a number is used in partitioning $q$ .
if $ T$ be a largest number in every partiotioning , and $forall t : 1leq tleq T$ , $ $ $a_t =frac{N^{T-t} (T-1)! }{(t-1)!} $ .is this true?
if $n_T>n'_T$ then $sum_{t=1}^T a_t.n_t > sum_{t=1}^T a_t.n'_t $
In all the examples I tested, the inequality was correct .but i don't know how can i prove it.
my try:
according to the defenition of $a_t$ we have $a_T=1$. then we must to prove the following:
$n_T+sum_{t=1}^{T -1}a_t.n_t > n'_T +sum_{t=1}^{T -1} a_t.n'_t $
This sum is composed of two parts. The first part is right according to the assumption. I think No matter how the value of second part is big, this value does not go to the value of the first part.
Is it Right?
How do I start proof?
calculus summation proof-writing
calculus summation proof-writing
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
edited Dec 7 '18 at 15:53
yaodao vang
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
asked Dec 7 '18 at 13:00
yaodao vangyaodao vang
686
686
$begingroup$
Is $a_t$ the same thing as $w_t?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:11
$begingroup$
@saulspatz , yes , I forgot to correct it .I edited it. Thanks
$endgroup$
– yaodao vang
Dec 7 '18 at 13:12
$begingroup$
Is this supposed to say that if $n_tge n'_t$ for $1le tle T$ then the conclusion is true? What you have written is clearly false.
$endgroup$
– saulspatz
Dec 7 '18 at 13:17
$begingroup$
What if $n_1=1, n'_1=10^{100}$ Why should the conclusion follow just because $n_T>n'_T$ for some $T?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:26
$begingroup$
I don't understand. This whole question seems very confused to me.
$endgroup$
– saulspatz
Dec 7 '18 at 14:48
add a comment |
$begingroup$
Is $a_t$ the same thing as $w_t?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:11
$begingroup$
@saulspatz , yes , I forgot to correct it .I edited it. Thanks
$endgroup$
– yaodao vang
Dec 7 '18 at 13:12
$begingroup$
Is this supposed to say that if $n_tge n'_t$ for $1le tle T$ then the conclusion is true? What you have written is clearly false.
$endgroup$
– saulspatz
Dec 7 '18 at 13:17
$begingroup$
What if $n_1=1, n'_1=10^{100}$ Why should the conclusion follow just because $n_T>n'_T$ for some $T?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:26
$begingroup$
I don't understand. This whole question seems very confused to me.
$endgroup$
– saulspatz
Dec 7 '18 at 14:48
$begingroup$
Is $a_t$ the same thing as $w_t?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:11
$begingroup$
Is $a_t$ the same thing as $w_t?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:11
$begingroup$
@saulspatz , yes , I forgot to correct it .I edited it. Thanks
$endgroup$
– yaodao vang
Dec 7 '18 at 13:12
$begingroup$
@saulspatz , yes , I forgot to correct it .I edited it. Thanks
$endgroup$
– yaodao vang
Dec 7 '18 at 13:12
$begingroup$
Is this supposed to say that if $n_tge n'_t$ for $1le tle T$ then the conclusion is true? What you have written is clearly false.
$endgroup$
– saulspatz
Dec 7 '18 at 13:17
$begingroup$
Is this supposed to say that if $n_tge n'_t$ for $1le tle T$ then the conclusion is true? What you have written is clearly false.
$endgroup$
– saulspatz
Dec 7 '18 at 13:17
$begingroup$
What if $n_1=1, n'_1=10^{100}$ Why should the conclusion follow just because $n_T>n'_T$ for some $T?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:26
$begingroup$
What if $n_1=1, n'_1=10^{100}$ Why should the conclusion follow just because $n_T>n'_T$ for some $T?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:26
$begingroup$
I don't understand. This whole question seems very confused to me.
$endgroup$
– saulspatz
Dec 7 '18 at 14:48
$begingroup$
I don't understand. This whole question seems very confused to me.
$endgroup$
– saulspatz
Dec 7 '18 at 14:48
add a comment |
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$begingroup$
Is $a_t$ the same thing as $w_t?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:11
$begingroup$
@saulspatz , yes , I forgot to correct it .I edited it. Thanks
$endgroup$
– yaodao vang
Dec 7 '18 at 13:12
$begingroup$
Is this supposed to say that if $n_tge n'_t$ for $1le tle T$ then the conclusion is true? What you have written is clearly false.
$endgroup$
– saulspatz
Dec 7 '18 at 13:17
$begingroup$
What if $n_1=1, n'_1=10^{100}$ Why should the conclusion follow just because $n_T>n'_T$ for some $T?$
$endgroup$
– saulspatz
Dec 7 '18 at 13:26
$begingroup$
I don't understand. This whole question seems very confused to me.
$endgroup$
– saulspatz
Dec 7 '18 at 14:48