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Do summations always increment by one? How do you incrment by a negative number? Or any real number?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The sum of an uncountable number of positive numbersA good book on programming for numerical optimizationMath Courses involving clever integration techniquesSolving equations that contain summationsSumming standard deviations? or summing variances?Evaluation of an Infinite Converging SumQuestion involving summations and the Θ-notation of running timesClosed-form Solution of Log SumChanging summation in a power seriesWhy I cannot simplify these two summations?Finding the formula for the summation $sum_i=1^n (2i-1) = 1+3+5+…+(2n-1)$.Jumping-index summations
$begingroup$
Do summations always increment by one?
Having a more of background in programming than math. I am just learning about summations and I look at them as loops that increment by one.
If my assumption correct what an equivalent method / form to increment by values not equal to one?
summation
$endgroup$
add a comment |
$begingroup$
Do summations always increment by one?
Having a more of background in programming than math. I am just learning about summations and I look at them as loops that increment by one.
If my assumption correct what an equivalent method / form to increment by values not equal to one?
summation
$endgroup$
1
$begingroup$
Summations do not always have to be incremented by $1$. Take $$sum_ktext eventext, 0 leq k leq 10k = 0 + 2 + cdots + 10 $$ for example. The possibilities with what you can write for summations are endless, but by convention, summations are incremented by $1$ if you have an upper index and lower index, unless of course indicated otherwise.
$endgroup$
– Clarinetist
Jan 5 '16 at 2:39
add a comment |
$begingroup$
Do summations always increment by one?
Having a more of background in programming than math. I am just learning about summations and I look at them as loops that increment by one.
If my assumption correct what an equivalent method / form to increment by values not equal to one?
summation
$endgroup$
Do summations always increment by one?
Having a more of background in programming than math. I am just learning about summations and I look at them as loops that increment by one.
If my assumption correct what an equivalent method / form to increment by values not equal to one?
summation
summation
asked Jan 5 '16 at 2:33
user136952user136952
13418
13418
1
$begingroup$
Summations do not always have to be incremented by $1$. Take $$sum_ktext eventext, 0 leq k leq 10k = 0 + 2 + cdots + 10 $$ for example. The possibilities with what you can write for summations are endless, but by convention, summations are incremented by $1$ if you have an upper index and lower index, unless of course indicated otherwise.
$endgroup$
– Clarinetist
Jan 5 '16 at 2:39
add a comment |
1
$begingroup$
Summations do not always have to be incremented by $1$. Take $$sum_ktext eventext, 0 leq k leq 10k = 0 + 2 + cdots + 10 $$ for example. The possibilities with what you can write for summations are endless, but by convention, summations are incremented by $1$ if you have an upper index and lower index, unless of course indicated otherwise.
$endgroup$
– Clarinetist
Jan 5 '16 at 2:39
1
1
$begingroup$
Summations do not always have to be incremented by $1$. Take $$sum_ktext eventext, 0 leq k leq 10k = 0 + 2 + cdots + 10 $$ for example. The possibilities with what you can write for summations are endless, but by convention, summations are incremented by $1$ if you have an upper index and lower index, unless of course indicated otherwise.
$endgroup$
– Clarinetist
Jan 5 '16 at 2:39
$begingroup$
Summations do not always have to be incremented by $1$. Take $$sum_ktext eventext, 0 leq k leq 10k = 0 + 2 + cdots + 10 $$ for example. The possibilities with what you can write for summations are endless, but by convention, summations are incremented by $1$ if you have an upper index and lower index, unless of course indicated otherwise.
$endgroup$
– Clarinetist
Jan 5 '16 at 2:39
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Do summations always increment by one?
That depends on what you mean by summation.
E.g
$$
S = sum_i = 1^n s_i = sum_i in I s_i
$$
means a summation, where the indices are picked from $I = 1,ldots, n$.
The index set $I$ could be traversed by starting at $1$ and incrementing to $n$ in steps of $+1$. But that is just one way of $n!$ possible ways to sum the $s_i$ up.
Having a more of background in programming than math. I am just
learning about summations and I look at them as loops that increment
by one.
If my assumption correct what an equivalent method / form to increment
by values not equal to one?
You could start with $n$ and finish with $1$. Or you first use the odd indices ascending and the the even ones descending.
$$
S = sum_overseti in I i bmod 2 = 1 s_i +
sum_overseti in I i bmod 2 = 0 s_i
$$
It should lead to the same sum, as order does not matter to the finite index set.
$endgroup$
add a comment |
$begingroup$
In programming, many languages provide for an optional step in the loop. It defaults to $1$, but you can set it to something else. In math, it is similar. The default is to increment by $1$. If you want to add up all the multiples of $3$ from $3$ to $30$, you could write $$sum_i=1^103i$$, where $i$ increments by $1$ and the factor $3$ gets you the increment you really want. In programming you can do the same thing. You can write for i=1 to 10;k=3*i and use k instead of i for your computations. You can also write for i=3 to 30 step 3. In math you can write $$sum_ substack i=3\ iequiv 0 pmod 3^30i$$ to get a step of 3. It seems fairly rare in math, more common in programming, in my experience
$endgroup$
add a comment |
$begingroup$
This may at first seem like a problem, but once you start thinking about how to solve it the solution becomes obvious and attests to the beauty of simplicity in math.
Let's solve the problem of being able to increment by any real number: simply multiply $i$ by $p$, the desired real number.
Using negative numbers? Increment by a negative number, etc.
$endgroup$
add a comment |
$begingroup$
No, summations do not always increment by one. In fact, summations can be over many kinds of sets, possibly not even restricted to integers or countable cardinality. Some examples from Wikipedia:
An example of a proof involving a summation over an uncountable set can be seen here on SE: The sum of an uncountable number of positive numbers
$endgroup$
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
1
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
add a comment |
$begingroup$
Edit, okay, okay strait to it then (if ya want more examples check the edits), so let's commence with the pealing with Python!
$$
h left( x right) = x over 2
$$
$$
sum_i=1^end=10 hleft( i right)
$$
end=10
really should just be10
to conform with the normal but it's labeled for referencing
$sum$ is very analogous to sum()
, not much really has to be done, but to ensure everyone's on the same page, both, under the hood kinda look like...
def sum_of_list(l):
last_value = 0
for value in l:
last_value += value
return value
... and $h left( x right)$ is a bit like...
def h(x):
# According to lore, multiplication is often faster than division.
return x * 0.5
... though not quite because that does not sufficiently express infinities, but let's set that aside for a moment.
$i = textstart$ is very much like an iterator, especially when paired with a limit such as $textend$, above an bellow $sum$ respectively. So let's consider it as such for now.
$h left( i right)$ when combined with $sum$, eg $sum_h left( i right)$ is almost a function
call, it's not quite equivalent but let's not quibble.
To approximate
these behaviors with some code, here's what I came up with...
#!/usr/bin/env python3
from collections import Iterator
class Rolling_Summation(Iterator):
"""
Rolling_Summation is an __approximation__ of Sigma Notation
On each iteration of `Rolling_Summation` the result of `f(i.next())` are `yeild`ed
The current sum maybe inspected at any point via the `.current_sum` property
"""
def __init__(self, i = range(1, 11), f = lambda x: float(x)):
super(Rolling_Summation, self).__init__()
self.i = i
self.f = f
self.__current_sum = 0
self.__finished = False
@property
def current_sum(self):
return self.__current_sum
@property
def is_finished(self):
if bool(self.__finished) is True:
return True
return False
def throw(self, type = None, traceback = None):
raise StopIteration
def next(self):
try:
if hasattr(i, 'next'):
# Python 2
i_value = self.i.next()
else:
# Python 3
i_value = self.i.__next__()
except (StopIteration, GeneratorExit):
self.__finished = True
else:
f_value = self.f(i_value)
self.__current_sum += f_value
return f_value
if self.is_finished:
self.throw(GeneratorExit)
# Python 3 compatibility
__next__ = next
... along with some examples.
$$
sum_i=1^10
$$
results = 0
sum_iter = Rolling_Summation(iter(range(1, 11)), lambda x: float(x))
for r in sum_iter:
print("r -> 0".format(r))
results += r
print("results -> 0".format(results))
print("sum_iter.current_sum -> 0".format(sum_iter.current_sum))
The above Python code could also be expressed as...
sigma = sum([x for x in Rolling_Summation(iter(range(1, 11)), lambda x: float(x))])
... if you're one for one-liners.
Both sum_iter.current_sum
and sigma
from above should result in the same value of 55.0
How do you increment by a negative number?
Stepping by -1
starting at -1
and going to -10
...
$$
sum_i = -1^-10
$$
sum([x for x in Rolling_Summation(iter(range(-1, -11, -1)), lambda x: float(x))])
# -> -55.0
While that may not be technically correct I doubt you'll confuse anyone as to the intent or the results expected. However to be explicit one could also translate it as...
$$
n left( x right) = -1n
$$
$$
sum_i = n left( 1 right)^-10
$$
... if you're one that must be reminded to decrement, and also a hint to yet another answer to...
Or any real number?
$$
p left( x right) = n + 2
$$
$$
sum_i = p left( 0 right)^ge10
$$
sum([x for x in Rolling_Summation(iter(range(2, 12, 2)), lambda x: float(x))])
Now even with this class
y method it's just an approximation
, a close one if ya dig into what the code is really doing to emulate much of what sigma notation can relay, but eventually you'll find something that'll compute as an oscillating value, or find floating point errors accumulating (just as two examples), and things'll get messy.
Bonus examples
It's a good idea to second-guess one's self when going from one to the other; off by one errors can be a pain, eg matrix
vs list
indexing...
$$
M = beginpmatrix0, 1, 2, dots 9 endpmatrix
$$
$$
sum_i = 1^10 M_i
$$
M = [x for x in range(0, 9)]
# -> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
sum([x for x in Rolling_Summation(iter(range(0, len(M) - 1)), M.index)])
... note where those 0
s and 1
s differ above to save yourself future headaches...
$$
f left( x right) = begincases
1, & textif 2 mid x \
0, & textotherwise
endcases
$$
$$
sum_i = M_1^10 fleft( i right)
$$
def f(x):
if x % 2 == 0:
return 1
return 0
sum([x for x in Rolling_Summation(iter(M[0:10]), f)])
But this has all just been one interpretation for how to translate as many of the features from one to the other, hopefully all before attention spans have run-out.
$endgroup$
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Do summations always increment by one?
That depends on what you mean by summation.
E.g
$$
S = sum_i = 1^n s_i = sum_i in I s_i
$$
means a summation, where the indices are picked from $I = 1,ldots, n$.
The index set $I$ could be traversed by starting at $1$ and incrementing to $n$ in steps of $+1$. But that is just one way of $n!$ possible ways to sum the $s_i$ up.
Having a more of background in programming than math. I am just
learning about summations and I look at them as loops that increment
by one.
If my assumption correct what an equivalent method / form to increment
by values not equal to one?
You could start with $n$ and finish with $1$. Or you first use the odd indices ascending and the the even ones descending.
$$
S = sum_overseti in I i bmod 2 = 1 s_i +
sum_overseti in I i bmod 2 = 0 s_i
$$
It should lead to the same sum, as order does not matter to the finite index set.
$endgroup$
add a comment |
$begingroup$
Do summations always increment by one?
That depends on what you mean by summation.
E.g
$$
S = sum_i = 1^n s_i = sum_i in I s_i
$$
means a summation, where the indices are picked from $I = 1,ldots, n$.
The index set $I$ could be traversed by starting at $1$ and incrementing to $n$ in steps of $+1$. But that is just one way of $n!$ possible ways to sum the $s_i$ up.
Having a more of background in programming than math. I am just
learning about summations and I look at them as loops that increment
by one.
If my assumption correct what an equivalent method / form to increment
by values not equal to one?
You could start with $n$ and finish with $1$. Or you first use the odd indices ascending and the the even ones descending.
$$
S = sum_overseti in I i bmod 2 = 1 s_i +
sum_overseti in I i bmod 2 = 0 s_i
$$
It should lead to the same sum, as order does not matter to the finite index set.
$endgroup$
add a comment |
$begingroup$
Do summations always increment by one?
That depends on what you mean by summation.
E.g
$$
S = sum_i = 1^n s_i = sum_i in I s_i
$$
means a summation, where the indices are picked from $I = 1,ldots, n$.
The index set $I$ could be traversed by starting at $1$ and incrementing to $n$ in steps of $+1$. But that is just one way of $n!$ possible ways to sum the $s_i$ up.
Having a more of background in programming than math. I am just
learning about summations and I look at them as loops that increment
by one.
If my assumption correct what an equivalent method / form to increment
by values not equal to one?
You could start with $n$ and finish with $1$. Or you first use the odd indices ascending and the the even ones descending.
$$
S = sum_overseti in I i bmod 2 = 1 s_i +
sum_overseti in I i bmod 2 = 0 s_i
$$
It should lead to the same sum, as order does not matter to the finite index set.
$endgroup$
Do summations always increment by one?
That depends on what you mean by summation.
E.g
$$
S = sum_i = 1^n s_i = sum_i in I s_i
$$
means a summation, where the indices are picked from $I = 1,ldots, n$.
The index set $I$ could be traversed by starting at $1$ and incrementing to $n$ in steps of $+1$. But that is just one way of $n!$ possible ways to sum the $s_i$ up.
Having a more of background in programming than math. I am just
learning about summations and I look at them as loops that increment
by one.
If my assumption correct what an equivalent method / form to increment
by values not equal to one?
You could start with $n$ and finish with $1$. Or you first use the odd indices ascending and the the even ones descending.
$$
S = sum_overseti in I i bmod 2 = 1 s_i +
sum_overseti in I i bmod 2 = 0 s_i
$$
It should lead to the same sum, as order does not matter to the finite index set.
edited Jan 5 '16 at 2:57
answered Jan 5 '16 at 2:42
mvwmvw
31.6k22252
31.6k22252
add a comment |
add a comment |
$begingroup$
In programming, many languages provide for an optional step in the loop. It defaults to $1$, but you can set it to something else. In math, it is similar. The default is to increment by $1$. If you want to add up all the multiples of $3$ from $3$ to $30$, you could write $$sum_i=1^103i$$, where $i$ increments by $1$ and the factor $3$ gets you the increment you really want. In programming you can do the same thing. You can write for i=1 to 10;k=3*i and use k instead of i for your computations. You can also write for i=3 to 30 step 3. In math you can write $$sum_ substack i=3\ iequiv 0 pmod 3^30i$$ to get a step of 3. It seems fairly rare in math, more common in programming, in my experience
$endgroup$
add a comment |
$begingroup$
In programming, many languages provide for an optional step in the loop. It defaults to $1$, but you can set it to something else. In math, it is similar. The default is to increment by $1$. If you want to add up all the multiples of $3$ from $3$ to $30$, you could write $$sum_i=1^103i$$, where $i$ increments by $1$ and the factor $3$ gets you the increment you really want. In programming you can do the same thing. You can write for i=1 to 10;k=3*i and use k instead of i for your computations. You can also write for i=3 to 30 step 3. In math you can write $$sum_ substack i=3\ iequiv 0 pmod 3^30i$$ to get a step of 3. It seems fairly rare in math, more common in programming, in my experience
$endgroup$
add a comment |
$begingroup$
In programming, many languages provide for an optional step in the loop. It defaults to $1$, but you can set it to something else. In math, it is similar. The default is to increment by $1$. If you want to add up all the multiples of $3$ from $3$ to $30$, you could write $$sum_i=1^103i$$, where $i$ increments by $1$ and the factor $3$ gets you the increment you really want. In programming you can do the same thing. You can write for i=1 to 10;k=3*i and use k instead of i for your computations. You can also write for i=3 to 30 step 3. In math you can write $$sum_ substack i=3\ iequiv 0 pmod 3^30i$$ to get a step of 3. It seems fairly rare in math, more common in programming, in my experience
$endgroup$
In programming, many languages provide for an optional step in the loop. It defaults to $1$, but you can set it to something else. In math, it is similar. The default is to increment by $1$. If you want to add up all the multiples of $3$ from $3$ to $30$, you could write $$sum_i=1^103i$$, where $i$ increments by $1$ and the factor $3$ gets you the increment you really want. In programming you can do the same thing. You can write for i=1 to 10;k=3*i and use k instead of i for your computations. You can also write for i=3 to 30 step 3. In math you can write $$sum_ substack i=3\ iequiv 0 pmod 3^30i$$ to get a step of 3. It seems fairly rare in math, more common in programming, in my experience
answered Jan 5 '16 at 2:53
Ross MillikanRoss Millikan
302k24200375
302k24200375
add a comment |
add a comment |
$begingroup$
This may at first seem like a problem, but once you start thinking about how to solve it the solution becomes obvious and attests to the beauty of simplicity in math.
Let's solve the problem of being able to increment by any real number: simply multiply $i$ by $p$, the desired real number.
Using negative numbers? Increment by a negative number, etc.
$endgroup$
add a comment |
$begingroup$
This may at first seem like a problem, but once you start thinking about how to solve it the solution becomes obvious and attests to the beauty of simplicity in math.
Let's solve the problem of being able to increment by any real number: simply multiply $i$ by $p$, the desired real number.
Using negative numbers? Increment by a negative number, etc.
$endgroup$
add a comment |
$begingroup$
This may at first seem like a problem, but once you start thinking about how to solve it the solution becomes obvious and attests to the beauty of simplicity in math.
Let's solve the problem of being able to increment by any real number: simply multiply $i$ by $p$, the desired real number.
Using negative numbers? Increment by a negative number, etc.
$endgroup$
This may at first seem like a problem, but once you start thinking about how to solve it the solution becomes obvious and attests to the beauty of simplicity in math.
Let's solve the problem of being able to increment by any real number: simply multiply $i$ by $p$, the desired real number.
Using negative numbers? Increment by a negative number, etc.
answered Jan 5 '16 at 2:54
theideasmiththeideasmith
603417
603417
add a comment |
add a comment |
$begingroup$
No, summations do not always increment by one. In fact, summations can be over many kinds of sets, possibly not even restricted to integers or countable cardinality. Some examples from Wikipedia:
An example of a proof involving a summation over an uncountable set can be seen here on SE: The sum of an uncountable number of positive numbers
$endgroup$
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
1
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
add a comment |
$begingroup$
No, summations do not always increment by one. In fact, summations can be over many kinds of sets, possibly not even restricted to integers or countable cardinality. Some examples from Wikipedia:
An example of a proof involving a summation over an uncountable set can be seen here on SE: The sum of an uncountable number of positive numbers
$endgroup$
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
1
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
add a comment |
$begingroup$
No, summations do not always increment by one. In fact, summations can be over many kinds of sets, possibly not even restricted to integers or countable cardinality. Some examples from Wikipedia:
An example of a proof involving a summation over an uncountable set can be seen here on SE: The sum of an uncountable number of positive numbers
$endgroup$
No, summations do not always increment by one. In fact, summations can be over many kinds of sets, possibly not even restricted to integers or countable cardinality. Some examples from Wikipedia:
An example of a proof involving a summation over an uncountable set can be seen here on SE: The sum of an uncountable number of positive numbers
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Jan 5 '16 at 3:02
Daniel R. CollinsDaniel R. Collins
5,9251534
5,9251534
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
1
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
add a comment |
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
1
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
$begingroup$
Your answer is misleading. As the second answer to the question you linked to states quite clearly, one is allowed to write a summation only when (one has shown that) the specified criterion holds. Otherwise it is not even meaningful, such as "$sum_x in [-1,1] x$", which cannot be assigned any meaningful interpretation.
$endgroup$
– user21820
Jan 30 '16 at 5:34
1
1
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
$begingroup$
@user21820: I changed "any arbitrary set" to "many kinds of sets"; thanks for the suggestion for improvement.
$endgroup$
– Daniel R. Collins
Jan 30 '16 at 6:05
add a comment |
$begingroup$
Edit, okay, okay strait to it then (if ya want more examples check the edits), so let's commence with the pealing with Python!
$$
h left( x right) = x over 2
$$
$$
sum_i=1^end=10 hleft( i right)
$$
end=10
really should just be10
to conform with the normal but it's labeled for referencing
$sum$ is very analogous to sum()
, not much really has to be done, but to ensure everyone's on the same page, both, under the hood kinda look like...
def sum_of_list(l):
last_value = 0
for value in l:
last_value += value
return value
... and $h left( x right)$ is a bit like...
def h(x):
# According to lore, multiplication is often faster than division.
return x * 0.5
... though not quite because that does not sufficiently express infinities, but let's set that aside for a moment.
$i = textstart$ is very much like an iterator, especially when paired with a limit such as $textend$, above an bellow $sum$ respectively. So let's consider it as such for now.
$h left( i right)$ when combined with $sum$, eg $sum_h left( i right)$ is almost a function
call, it's not quite equivalent but let's not quibble.
To approximate
these behaviors with some code, here's what I came up with...
#!/usr/bin/env python3
from collections import Iterator
class Rolling_Summation(Iterator):
"""
Rolling_Summation is an __approximation__ of Sigma Notation
On each iteration of `Rolling_Summation` the result of `f(i.next())` are `yeild`ed
The current sum maybe inspected at any point via the `.current_sum` property
"""
def __init__(self, i = range(1, 11), f = lambda x: float(x)):
super(Rolling_Summation, self).__init__()
self.i = i
self.f = f
self.__current_sum = 0
self.__finished = False
@property
def current_sum(self):
return self.__current_sum
@property
def is_finished(self):
if bool(self.__finished) is True:
return True
return False
def throw(self, type = None, traceback = None):
raise StopIteration
def next(self):
try:
if hasattr(i, 'next'):
# Python 2
i_value = self.i.next()
else:
# Python 3
i_value = self.i.__next__()
except (StopIteration, GeneratorExit):
self.__finished = True
else:
f_value = self.f(i_value)
self.__current_sum += f_value
return f_value
if self.is_finished:
self.throw(GeneratorExit)
# Python 3 compatibility
__next__ = next
... along with some examples.
$$
sum_i=1^10
$$
results = 0
sum_iter = Rolling_Summation(iter(range(1, 11)), lambda x: float(x))
for r in sum_iter:
print("r -> 0".format(r))
results += r
print("results -> 0".format(results))
print("sum_iter.current_sum -> 0".format(sum_iter.current_sum))
The above Python code could also be expressed as...
sigma = sum([x for x in Rolling_Summation(iter(range(1, 11)), lambda x: float(x))])
... if you're one for one-liners.
Both sum_iter.current_sum
and sigma
from above should result in the same value of 55.0
How do you increment by a negative number?
Stepping by -1
starting at -1
and going to -10
...
$$
sum_i = -1^-10
$$
sum([x for x in Rolling_Summation(iter(range(-1, -11, -1)), lambda x: float(x))])
# -> -55.0
While that may not be technically correct I doubt you'll confuse anyone as to the intent or the results expected. However to be explicit one could also translate it as...
$$
n left( x right) = -1n
$$
$$
sum_i = n left( 1 right)^-10
$$
... if you're one that must be reminded to decrement, and also a hint to yet another answer to...
Or any real number?
$$
p left( x right) = n + 2
$$
$$
sum_i = p left( 0 right)^ge10
$$
sum([x for x in Rolling_Summation(iter(range(2, 12, 2)), lambda x: float(x))])
Now even with this class
y method it's just an approximation
, a close one if ya dig into what the code is really doing to emulate much of what sigma notation can relay, but eventually you'll find something that'll compute as an oscillating value, or find floating point errors accumulating (just as two examples), and things'll get messy.
Bonus examples
It's a good idea to second-guess one's self when going from one to the other; off by one errors can be a pain, eg matrix
vs list
indexing...
$$
M = beginpmatrix0, 1, 2, dots 9 endpmatrix
$$
$$
sum_i = 1^10 M_i
$$
M = [x for x in range(0, 9)]
# -> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
sum([x for x in Rolling_Summation(iter(range(0, len(M) - 1)), M.index)])
... note where those 0
s and 1
s differ above to save yourself future headaches...
$$
f left( x right) = begincases
1, & textif 2 mid x \
0, & textotherwise
endcases
$$
$$
sum_i = M_1^10 fleft( i right)
$$
def f(x):
if x % 2 == 0:
return 1
return 0
sum([x for x in Rolling_Summation(iter(M[0:10]), f)])
But this has all just been one interpretation for how to translate as many of the features from one to the other, hopefully all before attention spans have run-out.
$endgroup$
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
add a comment |
$begingroup$
Edit, okay, okay strait to it then (if ya want more examples check the edits), so let's commence with the pealing with Python!
$$
h left( x right) = x over 2
$$
$$
sum_i=1^end=10 hleft( i right)
$$
end=10
really should just be10
to conform with the normal but it's labeled for referencing
$sum$ is very analogous to sum()
, not much really has to be done, but to ensure everyone's on the same page, both, under the hood kinda look like...
def sum_of_list(l):
last_value = 0
for value in l:
last_value += value
return value
... and $h left( x right)$ is a bit like...
def h(x):
# According to lore, multiplication is often faster than division.
return x * 0.5
... though not quite because that does not sufficiently express infinities, but let's set that aside for a moment.
$i = textstart$ is very much like an iterator, especially when paired with a limit such as $textend$, above an bellow $sum$ respectively. So let's consider it as such for now.
$h left( i right)$ when combined with $sum$, eg $sum_h left( i right)$ is almost a function
call, it's not quite equivalent but let's not quibble.
To approximate
these behaviors with some code, here's what I came up with...
#!/usr/bin/env python3
from collections import Iterator
class Rolling_Summation(Iterator):
"""
Rolling_Summation is an __approximation__ of Sigma Notation
On each iteration of `Rolling_Summation` the result of `f(i.next())` are `yeild`ed
The current sum maybe inspected at any point via the `.current_sum` property
"""
def __init__(self, i = range(1, 11), f = lambda x: float(x)):
super(Rolling_Summation, self).__init__()
self.i = i
self.f = f
self.__current_sum = 0
self.__finished = False
@property
def current_sum(self):
return self.__current_sum
@property
def is_finished(self):
if bool(self.__finished) is True:
return True
return False
def throw(self, type = None, traceback = None):
raise StopIteration
def next(self):
try:
if hasattr(i, 'next'):
# Python 2
i_value = self.i.next()
else:
# Python 3
i_value = self.i.__next__()
except (StopIteration, GeneratorExit):
self.__finished = True
else:
f_value = self.f(i_value)
self.__current_sum += f_value
return f_value
if self.is_finished:
self.throw(GeneratorExit)
# Python 3 compatibility
__next__ = next
... along with some examples.
$$
sum_i=1^10
$$
results = 0
sum_iter = Rolling_Summation(iter(range(1, 11)), lambda x: float(x))
for r in sum_iter:
print("r -> 0".format(r))
results += r
print("results -> 0".format(results))
print("sum_iter.current_sum -> 0".format(sum_iter.current_sum))
The above Python code could also be expressed as...
sigma = sum([x for x in Rolling_Summation(iter(range(1, 11)), lambda x: float(x))])
... if you're one for one-liners.
Both sum_iter.current_sum
and sigma
from above should result in the same value of 55.0
How do you increment by a negative number?
Stepping by -1
starting at -1
and going to -10
...
$$
sum_i = -1^-10
$$
sum([x for x in Rolling_Summation(iter(range(-1, -11, -1)), lambda x: float(x))])
# -> -55.0
While that may not be technically correct I doubt you'll confuse anyone as to the intent or the results expected. However to be explicit one could also translate it as...
$$
n left( x right) = -1n
$$
$$
sum_i = n left( 1 right)^-10
$$
... if you're one that must be reminded to decrement, and also a hint to yet another answer to...
Or any real number?
$$
p left( x right) = n + 2
$$
$$
sum_i = p left( 0 right)^ge10
$$
sum([x for x in Rolling_Summation(iter(range(2, 12, 2)), lambda x: float(x))])
Now even with this class
y method it's just an approximation
, a close one if ya dig into what the code is really doing to emulate much of what sigma notation can relay, but eventually you'll find something that'll compute as an oscillating value, or find floating point errors accumulating (just as two examples), and things'll get messy.
Bonus examples
It's a good idea to second-guess one's self when going from one to the other; off by one errors can be a pain, eg matrix
vs list
indexing...
$$
M = beginpmatrix0, 1, 2, dots 9 endpmatrix
$$
$$
sum_i = 1^10 M_i
$$
M = [x for x in range(0, 9)]
# -> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
sum([x for x in Rolling_Summation(iter(range(0, len(M) - 1)), M.index)])
... note where those 0
s and 1
s differ above to save yourself future headaches...
$$
f left( x right) = begincases
1, & textif 2 mid x \
0, & textotherwise
endcases
$$
$$
sum_i = M_1^10 fleft( i right)
$$
def f(x):
if x % 2 == 0:
return 1
return 0
sum([x for x in Rolling_Summation(iter(M[0:10]), f)])
But this has all just been one interpretation for how to translate as many of the features from one to the other, hopefully all before attention spans have run-out.
$endgroup$
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
add a comment |
$begingroup$
Edit, okay, okay strait to it then (if ya want more examples check the edits), so let's commence with the pealing with Python!
$$
h left( x right) = x over 2
$$
$$
sum_i=1^end=10 hleft( i right)
$$
end=10
really should just be10
to conform with the normal but it's labeled for referencing
$sum$ is very analogous to sum()
, not much really has to be done, but to ensure everyone's on the same page, both, under the hood kinda look like...
def sum_of_list(l):
last_value = 0
for value in l:
last_value += value
return value
... and $h left( x right)$ is a bit like...
def h(x):
# According to lore, multiplication is often faster than division.
return x * 0.5
... though not quite because that does not sufficiently express infinities, but let's set that aside for a moment.
$i = textstart$ is very much like an iterator, especially when paired with a limit such as $textend$, above an bellow $sum$ respectively. So let's consider it as such for now.
$h left( i right)$ when combined with $sum$, eg $sum_h left( i right)$ is almost a function
call, it's not quite equivalent but let's not quibble.
To approximate
these behaviors with some code, here's what I came up with...
#!/usr/bin/env python3
from collections import Iterator
class Rolling_Summation(Iterator):
"""
Rolling_Summation is an __approximation__ of Sigma Notation
On each iteration of `Rolling_Summation` the result of `f(i.next())` are `yeild`ed
The current sum maybe inspected at any point via the `.current_sum` property
"""
def __init__(self, i = range(1, 11), f = lambda x: float(x)):
super(Rolling_Summation, self).__init__()
self.i = i
self.f = f
self.__current_sum = 0
self.__finished = False
@property
def current_sum(self):
return self.__current_sum
@property
def is_finished(self):
if bool(self.__finished) is True:
return True
return False
def throw(self, type = None, traceback = None):
raise StopIteration
def next(self):
try:
if hasattr(i, 'next'):
# Python 2
i_value = self.i.next()
else:
# Python 3
i_value = self.i.__next__()
except (StopIteration, GeneratorExit):
self.__finished = True
else:
f_value = self.f(i_value)
self.__current_sum += f_value
return f_value
if self.is_finished:
self.throw(GeneratorExit)
# Python 3 compatibility
__next__ = next
... along with some examples.
$$
sum_i=1^10
$$
results = 0
sum_iter = Rolling_Summation(iter(range(1, 11)), lambda x: float(x))
for r in sum_iter:
print("r -> 0".format(r))
results += r
print("results -> 0".format(results))
print("sum_iter.current_sum -> 0".format(sum_iter.current_sum))
The above Python code could also be expressed as...
sigma = sum([x for x in Rolling_Summation(iter(range(1, 11)), lambda x: float(x))])
... if you're one for one-liners.
Both sum_iter.current_sum
and sigma
from above should result in the same value of 55.0
How do you increment by a negative number?
Stepping by -1
starting at -1
and going to -10
...
$$
sum_i = -1^-10
$$
sum([x for x in Rolling_Summation(iter(range(-1, -11, -1)), lambda x: float(x))])
# -> -55.0
While that may not be technically correct I doubt you'll confuse anyone as to the intent or the results expected. However to be explicit one could also translate it as...
$$
n left( x right) = -1n
$$
$$
sum_i = n left( 1 right)^-10
$$
... if you're one that must be reminded to decrement, and also a hint to yet another answer to...
Or any real number?
$$
p left( x right) = n + 2
$$
$$
sum_i = p left( 0 right)^ge10
$$
sum([x for x in Rolling_Summation(iter(range(2, 12, 2)), lambda x: float(x))])
Now even with this class
y method it's just an approximation
, a close one if ya dig into what the code is really doing to emulate much of what sigma notation can relay, but eventually you'll find something that'll compute as an oscillating value, or find floating point errors accumulating (just as two examples), and things'll get messy.
Bonus examples
It's a good idea to second-guess one's self when going from one to the other; off by one errors can be a pain, eg matrix
vs list
indexing...
$$
M = beginpmatrix0, 1, 2, dots 9 endpmatrix
$$
$$
sum_i = 1^10 M_i
$$
M = [x for x in range(0, 9)]
# -> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
sum([x for x in Rolling_Summation(iter(range(0, len(M) - 1)), M.index)])
... note where those 0
s and 1
s differ above to save yourself future headaches...
$$
f left( x right) = begincases
1, & textif 2 mid x \
0, & textotherwise
endcases
$$
$$
sum_i = M_1^10 fleft( i right)
$$
def f(x):
if x % 2 == 0:
return 1
return 0
sum([x for x in Rolling_Summation(iter(M[0:10]), f)])
But this has all just been one interpretation for how to translate as many of the features from one to the other, hopefully all before attention spans have run-out.
$endgroup$
Edit, okay, okay strait to it then (if ya want more examples check the edits), so let's commence with the pealing with Python!
$$
h left( x right) = x over 2
$$
$$
sum_i=1^end=10 hleft( i right)
$$
end=10
really should just be10
to conform with the normal but it's labeled for referencing
$sum$ is very analogous to sum()
, not much really has to be done, but to ensure everyone's on the same page, both, under the hood kinda look like...
def sum_of_list(l):
last_value = 0
for value in l:
last_value += value
return value
... and $h left( x right)$ is a bit like...
def h(x):
# According to lore, multiplication is often faster than division.
return x * 0.5
... though not quite because that does not sufficiently express infinities, but let's set that aside for a moment.
$i = textstart$ is very much like an iterator, especially when paired with a limit such as $textend$, above an bellow $sum$ respectively. So let's consider it as such for now.
$h left( i right)$ when combined with $sum$, eg $sum_h left( i right)$ is almost a function
call, it's not quite equivalent but let's not quibble.
To approximate
these behaviors with some code, here's what I came up with...
#!/usr/bin/env python3
from collections import Iterator
class Rolling_Summation(Iterator):
"""
Rolling_Summation is an __approximation__ of Sigma Notation
On each iteration of `Rolling_Summation` the result of `f(i.next())` are `yeild`ed
The current sum maybe inspected at any point via the `.current_sum` property
"""
def __init__(self, i = range(1, 11), f = lambda x: float(x)):
super(Rolling_Summation, self).__init__()
self.i = i
self.f = f
self.__current_sum = 0
self.__finished = False
@property
def current_sum(self):
return self.__current_sum
@property
def is_finished(self):
if bool(self.__finished) is True:
return True
return False
def throw(self, type = None, traceback = None):
raise StopIteration
def next(self):
try:
if hasattr(i, 'next'):
# Python 2
i_value = self.i.next()
else:
# Python 3
i_value = self.i.__next__()
except (StopIteration, GeneratorExit):
self.__finished = True
else:
f_value = self.f(i_value)
self.__current_sum += f_value
return f_value
if self.is_finished:
self.throw(GeneratorExit)
# Python 3 compatibility
__next__ = next
... along with some examples.
$$
sum_i=1^10
$$
results = 0
sum_iter = Rolling_Summation(iter(range(1, 11)), lambda x: float(x))
for r in sum_iter:
print("r -> 0".format(r))
results += r
print("results -> 0".format(results))
print("sum_iter.current_sum -> 0".format(sum_iter.current_sum))
The above Python code could also be expressed as...
sigma = sum([x for x in Rolling_Summation(iter(range(1, 11)), lambda x: float(x))])
... if you're one for one-liners.
Both sum_iter.current_sum
and sigma
from above should result in the same value of 55.0
How do you increment by a negative number?
Stepping by -1
starting at -1
and going to -10
...
$$
sum_i = -1^-10
$$
sum([x for x in Rolling_Summation(iter(range(-1, -11, -1)), lambda x: float(x))])
# -> -55.0
While that may not be technically correct I doubt you'll confuse anyone as to the intent or the results expected. However to be explicit one could also translate it as...
$$
n left( x right) = -1n
$$
$$
sum_i = n left( 1 right)^-10
$$
... if you're one that must be reminded to decrement, and also a hint to yet another answer to...
Or any real number?
$$
p left( x right) = n + 2
$$
$$
sum_i = p left( 0 right)^ge10
$$
sum([x for x in Rolling_Summation(iter(range(2, 12, 2)), lambda x: float(x))])
Now even with this class
y method it's just an approximation
, a close one if ya dig into what the code is really doing to emulate much of what sigma notation can relay, but eventually you'll find something that'll compute as an oscillating value, or find floating point errors accumulating (just as two examples), and things'll get messy.
Bonus examples
It's a good idea to second-guess one's self when going from one to the other; off by one errors can be a pain, eg matrix
vs list
indexing...
$$
M = beginpmatrix0, 1, 2, dots 9 endpmatrix
$$
$$
sum_i = 1^10 M_i
$$
M = [x for x in range(0, 9)]
# -> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
sum([x for x in Rolling_Summation(iter(range(0, len(M) - 1)), M.index)])
... note where those 0
s and 1
s differ above to save yourself future headaches...
$$
f left( x right) = begincases
1, & textif 2 mid x \
0, & textotherwise
endcases
$$
$$
sum_i = M_1^10 fleft( i right)
$$
def f(x):
if x % 2 == 0:
return 1
return 0
sum([x for x in Rolling_Summation(iter(M[0:10]), f)])
But this has all just been one interpretation for how to translate as many of the features from one to the other, hopefully all before attention spans have run-out.
edited Mar 28 at 7:56
answered Mar 27 at 6:58
S0AndS0S0AndS0
1917
1917
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
add a comment |
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
preposteriously too long and way to much overkill.
$endgroup$
– fleablood
Mar 28 at 0:59
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
$begingroup$
Heh, yeah it was getting ridiculous. Perhaps the edits have now honed it to a finer point.
$endgroup$
– S0AndS0
Mar 28 at 2:20
add a comment |
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1
$begingroup$
Summations do not always have to be incremented by $1$. Take $$sum_ktext eventext, 0 leq k leq 10k = 0 + 2 + cdots + 10 $$ for example. The possibilities with what you can write for summations are endless, but by convention, summations are incremented by $1$ if you have an upper index and lower index, unless of course indicated otherwise.
$endgroup$
– Clarinetist
Jan 5 '16 at 2:39