Mixing
Calculating a Summation Involving 2 Variables
$begingroup$
Having not taken a math course for multiple years, I appear to have forgotten some bare basics. Unfortunately, Google has not taken me to a solid answer after much searching.
How do you solve an summation dependent on two variables?
More specifically, I am looking for the more formal mathematical term for what I am trying to accomplish (so I can google more information about it) and how to input the function like what I am trying to solve into Wolfram Alpha.
For instance, I want a summation of all terms where $$x + y < 10$$ where $$x$$ can take any value from 0 to infinity.
$$y$$ can also take on any value from 0 to infinity as long as the constrain above is satisfied.
$$result = sum( (10)^x * (10)^y )$$
So written out valid entries in the summation would include (x=1, y=1), (x=1 y=2). and so forth as long the sum is less than 10.
And result would look like
$$((10)^1 * (10)^1 ) + ((10)^1 * (10)^2 ) + ...$$
Also, how would you write the summation with a constraint? Would it look something like:
$$sum_{ x + y <10}$$ ?
algebra-precalculus summation
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
$endgroup$
add a comment |
$begingroup$
Having not taken a math course for multiple years, I appear to have forgotten some bare basics. Unfortunately, Google has not taken me to a solid answer after much searching.
How do you solve an summation dependent on two variables?
More specifically, I am looking for the more formal mathematical term for what I am trying to accomplish (so I can google more information about it) and how to input the function like what I am trying to solve into Wolfram Alpha.
For instance, I want a summation of all terms where $$x + y < 10$$ where $$x$$ can take any value from 0 to infinity.
$$y$$ can also take on any value from 0 to infinity as long as the constrain above is satisfied.
$$result = sum( (10)^x * (10)^y )$$
So written out valid entries in the summation would include (x=1, y=1), (x=1 y=2). and so forth as long the sum is less than 10.
And result would look like
$$((10)^1 * (10)^1 ) + ((10)^1 * (10)^2 ) + ...$$
Also, how would you write the summation with a constraint? Would it look something like:
$$sum_{ x + y <10}$$ ?
algebra-precalculus summation
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
$endgroup$
add a comment |
$begingroup$
Having not taken a math course for multiple years, I appear to have forgotten some bare basics. Unfortunately, Google has not taken me to a solid answer after much searching.
How do you solve an summation dependent on two variables?
More specifically, I am looking for the more formal mathematical term for what I am trying to accomplish (so I can google more information about it) and how to input the function like what I am trying to solve into Wolfram Alpha.
For instance, I want a summation of all terms where $$x + y < 10$$ where $$x$$ can take any value from 0 to infinity.
$$y$$ can also take on any value from 0 to infinity as long as the constrain above is satisfied.
$$result = sum( (10)^x * (10)^y )$$
So written out valid entries in the summation would include (x=1, y=1), (x=1 y=2). and so forth as long the sum is less than 10.
And result would look like
$$((10)^1 * (10)^1 ) + ((10)^1 * (10)^2 ) + ...$$
Also, how would you write the summation with a constraint? Would it look something like:
$$sum_{ x + y <10}$$ ?
algebra-precalculus summation
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
$endgroup$
Having not taken a math course for multiple years, I appear to have forgotten some bare basics. Unfortunately, Google has not taken me to a solid answer after much searching.
How do you solve an summation dependent on two variables?
More specifically, I am looking for the more formal mathematical term for what I am trying to accomplish (so I can google more information about it) and how to input the function like what I am trying to solve into Wolfram Alpha.
For instance, I want a summation of all terms where $$x + y < 10$$ where $$x$$ can take any value from 0 to infinity.
$$y$$ can also take on any value from 0 to infinity as long as the constrain above is satisfied.
$$result = sum( (10)^x * (10)^y )$$
So written out valid entries in the summation would include (x=1, y=1), (x=1 y=2). and so forth as long the sum is less than 10.
And result would look like
$$((10)^1 * (10)^1 ) + ((10)^1 * (10)^2 ) + ...$$
Also, how would you write the summation with a constraint? Would it look something like:
$$sum_{ x + y <10}$$ ?
algebra-precalculus summation
algebra-precalculus summation
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
edited Feb 9 '14 at 18:01
user1431282
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
asked Feb 9 '14 at 7:54
user1431282user1431282
1011
1011
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I assume you mean that $x, y$ are integers. Then the sum ovar all pair of integers
$(x,y)$ such that both are nonnegative and both are less than or equal to $10$ can be written in many ways:
$$
sum_{x=0}^{10} sum_{y=0}^{10} 10^{x+y}
$$
Into Wolphram alpha you can just type: sum (x=0 to 10) sum(y=0 to 10) 10^(x+y)
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
$endgroup$
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
add a comment |
$begingroup$
I found this question still without an answer while looking for double variable sumatories and I thought I would give a solution despite it won't be useful for the OP 4 years before.
What you are looking for is a double sumatory where the second variable depends on the first. We need to find the lower and upper bounds for the variables. From your text we can guess that the lower bound is 1 for both variables (x >= 1 and y >= 1). We also have the inequality equation x + y < 10 that can be turned into x + y <= 9 as both bounds in a sumatory are included and they belong to the natural numbers. From this equation we can isolate y to obtain an upper bound for the dependant variable (y <= 9 - x). We are just missing x upper bound. Replacing the minimum value of y in x + y < 10 we get that x + 1 < 10 --> x < 9 --> x <= 8.
Summing up:
1 <= x <= 8
1 <= y <= 9 - x
That would be translated into:
$$
sum_{x=1}^{8} sum_{y=1}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=1 to 8) sum (y=1 to 9-x) 10^(x+y)
The result of this operation is 8 765 432 100.
If the lower bounds would be 0 instead of 1 the rest would have to be recalculated:
0 <= x <= 9
0 <= y <= 9 - x
That would be translated into:
$$
sum_{x=0}^{9} sum_{y=0}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=0 to 9) sum (y=0 to 9-x) 10^(x+y)
The result of this operation is 10 987 654 321.
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume you mean that $x, y$ are integers. Then the sum ovar all pair of integers
$(x,y)$ such that both are nonnegative and both are less than or equal to $10$ can be written in many ways:
$$
sum_{x=0}^{10} sum_{y=0}^{10} 10^{x+y}
$$
Into Wolphram alpha you can just type: sum (x=0 to 10) sum(y=0 to 10) 10^(x+y)
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
$endgroup$
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
add a comment |
$begingroup$
I assume you mean that $x, y$ are integers. Then the sum ovar all pair of integers
$(x,y)$ such that both are nonnegative and both are less than or equal to $10$ can be written in many ways:
$$
sum_{x=0}^{10} sum_{y=0}^{10} 10^{x+y}
$$
Into Wolphram alpha you can just type: sum (x=0 to 10) sum(y=0 to 10) 10^(x+y)
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
$endgroup$
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
add a comment |
$begingroup$
I assume you mean that $x, y$ are integers. Then the sum ovar all pair of integers
$(x,y)$ such that both are nonnegative and both are less than or equal to $10$ can be written in many ways:
$$
sum_{x=0}^{10} sum_{y=0}^{10} 10^{x+y}
$$
Into Wolphram alpha you can just type: sum (x=0 to 10) sum(y=0 to 10) 10^(x+y)
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
$endgroup$
I assume you mean that $x, y$ are integers. Then the sum ovar all pair of integers
$(x,y)$ such that both are nonnegative and both are less than or equal to $10$ can be written in many ways:
$$
sum_{x=0}^{10} sum_{y=0}^{10} 10^{x+y}
$$
Into Wolphram alpha you can just type: sum (x=0 to 10) sum(y=0 to 10) 10^(x+y)
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
answered Feb 9 '14 at 11:13
kjetil b halvorsenkjetil b halvorsen
4,85042638
4,85042638
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
add a comment |
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
$begingroup$
Thank you for the reply. I was looking for something similar to that except for the case where x+y is < 10 not the case where x is from 0 to 10 and y is from 0 to 10.
$endgroup$
– user1431282
Feb 9 '14 at 17:59
add a comment |
$begingroup$
I found this question still without an answer while looking for double variable sumatories and I thought I would give a solution despite it won't be useful for the OP 4 years before.
What you are looking for is a double sumatory where the second variable depends on the first. We need to find the lower and upper bounds for the variables. From your text we can guess that the lower bound is 1 for both variables (x >= 1 and y >= 1). We also have the inequality equation x + y < 10 that can be turned into x + y <= 9 as both bounds in a sumatory are included and they belong to the natural numbers. From this equation we can isolate y to obtain an upper bound for the dependant variable (y <= 9 - x). We are just missing x upper bound. Replacing the minimum value of y in x + y < 10 we get that x + 1 < 10 --> x < 9 --> x <= 8.
Summing up:
1 <= x <= 8
1 <= y <= 9 - x
That would be translated into:
$$
sum_{x=1}^{8} sum_{y=1}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=1 to 8) sum (y=1 to 9-x) 10^(x+y)
The result of this operation is 8 765 432 100.
If the lower bounds would be 0 instead of 1 the rest would have to be recalculated:
0 <= x <= 9
0 <= y <= 9 - x
That would be translated into:
$$
sum_{x=0}^{9} sum_{y=0}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=0 to 9) sum (y=0 to 9-x) 10^(x+y)
The result of this operation is 10 987 654 321.
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
$endgroup$
add a comment |
$begingroup$
I found this question still without an answer while looking for double variable sumatories and I thought I would give a solution despite it won't be useful for the OP 4 years before.
What you are looking for is a double sumatory where the second variable depends on the first. We need to find the lower and upper bounds for the variables. From your text we can guess that the lower bound is 1 for both variables (x >= 1 and y >= 1). We also have the inequality equation x + y < 10 that can be turned into x + y <= 9 as both bounds in a sumatory are included and they belong to the natural numbers. From this equation we can isolate y to obtain an upper bound for the dependant variable (y <= 9 - x). We are just missing x upper bound. Replacing the minimum value of y in x + y < 10 we get that x + 1 < 10 --> x < 9 --> x <= 8.
Summing up:
1 <= x <= 8
1 <= y <= 9 - x
That would be translated into:
$$
sum_{x=1}^{8} sum_{y=1}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=1 to 8) sum (y=1 to 9-x) 10^(x+y)
The result of this operation is 8 765 432 100.
If the lower bounds would be 0 instead of 1 the rest would have to be recalculated:
0 <= x <= 9
0 <= y <= 9 - x
That would be translated into:
$$
sum_{x=0}^{9} sum_{y=0}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=0 to 9) sum (y=0 to 9-x) 10^(x+y)
The result of this operation is 10 987 654 321.
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
$endgroup$
add a comment |
$begingroup$
I found this question still without an answer while looking for double variable sumatories and I thought I would give a solution despite it won't be useful for the OP 4 years before.
What you are looking for is a double sumatory where the second variable depends on the first. We need to find the lower and upper bounds for the variables. From your text we can guess that the lower bound is 1 for both variables (x >= 1 and y >= 1). We also have the inequality equation x + y < 10 that can be turned into x + y <= 9 as both bounds in a sumatory are included and they belong to the natural numbers. From this equation we can isolate y to obtain an upper bound for the dependant variable (y <= 9 - x). We are just missing x upper bound. Replacing the minimum value of y in x + y < 10 we get that x + 1 < 10 --> x < 9 --> x <= 8.
Summing up:
1 <= x <= 8
1 <= y <= 9 - x
That would be translated into:
$$
sum_{x=1}^{8} sum_{y=1}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=1 to 8) sum (y=1 to 9-x) 10^(x+y)
The result of this operation is 8 765 432 100.
If the lower bounds would be 0 instead of 1 the rest would have to be recalculated:
0 <= x <= 9
0 <= y <= 9 - x
That would be translated into:
$$
sum_{x=0}^{9} sum_{y=0}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=0 to 9) sum (y=0 to 9-x) 10^(x+y)
The result of this operation is 10 987 654 321.
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
$endgroup$
I found this question still without an answer while looking for double variable sumatories and I thought I would give a solution despite it won't be useful for the OP 4 years before.
What you are looking for is a double sumatory where the second variable depends on the first. We need to find the lower and upper bounds for the variables. From your text we can guess that the lower bound is 1 for both variables (x >= 1 and y >= 1). We also have the inequality equation x + y < 10 that can be turned into x + y <= 9 as both bounds in a sumatory are included and they belong to the natural numbers. From this equation we can isolate y to obtain an upper bound for the dependant variable (y <= 9 - x). We are just missing x upper bound. Replacing the minimum value of y in x + y < 10 we get that x + 1 < 10 --> x < 9 --> x <= 8.
Summing up:
1 <= x <= 8
1 <= y <= 9 - x
That would be translated into:
$$
sum_{x=1}^{8} sum_{y=1}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=1 to 8) sum (y=1 to 9-x) 10^(x+y)
The result of this operation is 8 765 432 100.
If the lower bounds would be 0 instead of 1 the rest would have to be recalculated:
0 <= x <= 9
0 <= y <= 9 - x
That would be translated into:
$$
sum_{x=0}^{9} sum_{y=0}^{9-x} 10^{x+y}
$$
You could use the following code in Wolfram Alpha:
sum(x=0 to 9) sum (y=0 to 9-x) 10^(x+y)
The result of this operation is 10 987 654 321.
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
answered Mar 1 '18 at 13:33
AdirioAdirio
1011
1011
add a comment |
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