Problem in calculation of daily expense
$begingroup$
I have a simple question.
Total amount = 50 USD
-----------------------------
- Expense | Remaining |
-----------------------------
- 20 | 30 |
-----------------------------
- 15 | 15 |
-----------------------------
- 9 | 6 |
-----------------------------
- 6 | 0 |
=============================
- 50 | 51 |
=============================
As you can see at the end value is not the same. If we calculate Expense and Remaining both value has changed. There is one more in Remaining balance.
Can you please let me know what is the problem, which I'm doing wrong.
algebra-precalculus arithmetic
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
$endgroup$
add a comment |
$begingroup$
I have a simple question.
Total amount = 50 USD
-----------------------------
- Expense | Remaining |
-----------------------------
- 20 | 30 |
-----------------------------
- 15 | 15 |
-----------------------------
- 9 | 6 |
-----------------------------
- 6 | 0 |
=============================
- 50 | 51 |
=============================
As you can see at the end value is not the same. If we calculate Expense and Remaining both value has changed. There is one more in Remaining balance.
Can you please let me know what is the problem, which I'm doing wrong.
algebra-precalculus arithmetic
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
$endgroup$
1
$begingroup$
Why are you adding up the Remaining column? If you spend $5$ lots of $10$ USD instead you'll have $40, 30, 20, 10 $ and $0$ in the Remaining column and adding them certainly won't give you $50$ (or even be only $1$ out)
$endgroup$
– postmortes
Dec 26 '18 at 16:36
$begingroup$
What makes you think they should be the same? The way to check this is that the beginning balance, minus the total expenditures, equals the last remaining balance. What would the sum of the remaining balances ever relate to in the real world?
$endgroup$
– saulspatz
Dec 26 '18 at 16:38
$begingroup$
Why do you think something's wrong? There's simply no reason that the sum of the "remaining" amounts should be the sum of the expenditures.
$endgroup$
– David C. Ullrich
Dec 26 '18 at 16:38
add a comment |
$begingroup$
I have a simple question.
Total amount = 50 USD
-----------------------------
- Expense | Remaining |
-----------------------------
- 20 | 30 |
-----------------------------
- 15 | 15 |
-----------------------------
- 9 | 6 |
-----------------------------
- 6 | 0 |
=============================
- 50 | 51 |
=============================
As you can see at the end value is not the same. If we calculate Expense and Remaining both value has changed. There is one more in Remaining balance.
Can you please let me know what is the problem, which I'm doing wrong.
algebra-precalculus arithmetic
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
$endgroup$
I have a simple question.
Total amount = 50 USD
-----------------------------
- Expense | Remaining |
-----------------------------
- 20 | 30 |
-----------------------------
- 15 | 15 |
-----------------------------
- 9 | 6 |
-----------------------------
- 6 | 0 |
=============================
- 50 | 51 |
=============================
As you can see at the end value is not the same. If we calculate Expense and Remaining both value has changed. There is one more in Remaining balance.
Can you please let me know what is the problem, which I'm doing wrong.
algebra-precalculus arithmetic
algebra-precalculus arithmetic
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
edited Dec 26 '18 at 16:37
David C. Ullrich
61.7k44095
61.7k44095
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
asked Dec 26 '18 at 16:29
Azeem HaiderAzeem Haider
1032
1032
1
$begingroup$
Why are you adding up the Remaining column? If you spend $5$ lots of $10$ USD instead you'll have $40, 30, 20, 10 $ and $0$ in the Remaining column and adding them certainly won't give you $50$ (or even be only $1$ out)
$endgroup$
– postmortes
Dec 26 '18 at 16:36
$begingroup$
What makes you think they should be the same? The way to check this is that the beginning balance, minus the total expenditures, equals the last remaining balance. What would the sum of the remaining balances ever relate to in the real world?
$endgroup$
– saulspatz
Dec 26 '18 at 16:38
$begingroup$
Why do you think something's wrong? There's simply no reason that the sum of the "remaining" amounts should be the sum of the expenditures.
$endgroup$
– David C. Ullrich
Dec 26 '18 at 16:38
add a comment |
1
$begingroup$
Why are you adding up the Remaining column? If you spend $5$ lots of $10$ USD instead you'll have $40, 30, 20, 10 $ and $0$ in the Remaining column and adding them certainly won't give you $50$ (or even be only $1$ out)
$endgroup$
– postmortes
Dec 26 '18 at 16:36
$begingroup$
What makes you think they should be the same? The way to check this is that the beginning balance, minus the total expenditures, equals the last remaining balance. What would the sum of the remaining balances ever relate to in the real world?
$endgroup$
– saulspatz
Dec 26 '18 at 16:38
$begingroup$
Why do you think something's wrong? There's simply no reason that the sum of the "remaining" amounts should be the sum of the expenditures.
$endgroup$
– David C. Ullrich
Dec 26 '18 at 16:38
1
1
$begingroup$
Why are you adding up the Remaining column? If you spend $5$ lots of $10$ USD instead you'll have $40, 30, 20, 10 $ and $0$ in the Remaining column and adding them certainly won't give you $50$ (or even be only $1$ out)
$endgroup$
– postmortes
Dec 26 '18 at 16:36
$begingroup$
Why are you adding up the Remaining column? If you spend $5$ lots of $10$ USD instead you'll have $40, 30, 20, 10 $ and $0$ in the Remaining column and adding them certainly won't give you $50$ (or even be only $1$ out)
$endgroup$
– postmortes
Dec 26 '18 at 16:36
$begingroup$
What makes you think they should be the same? The way to check this is that the beginning balance, minus the total expenditures, equals the last remaining balance. What would the sum of the remaining balances ever relate to in the real world?
$endgroup$
– saulspatz
Dec 26 '18 at 16:38
$begingroup$
What makes you think they should be the same? The way to check this is that the beginning balance, minus the total expenditures, equals the last remaining balance. What would the sum of the remaining balances ever relate to in the real world?
$endgroup$
– saulspatz
Dec 26 '18 at 16:38
$begingroup$
Why do you think something's wrong? There's simply no reason that the sum of the "remaining" amounts should be the sum of the expenditures.
$endgroup$
– David C. Ullrich
Dec 26 '18 at 16:38
$begingroup$
Why do you think something's wrong? There's simply no reason that the sum of the "remaining" amounts should be the sum of the expenditures.
$endgroup$
– David C. Ullrich
Dec 26 '18 at 16:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The short answer is that althought the difference of the 2 sums is only 1, it is because the sum in the second column does not represent the same fact as the sum in the first column. In fact, as explained below, the sum in the second column has no meaning relevant to the issue at hand.
To find the "Total Expenses", we sum each expense as you did in column 1:
$E=-e1-e2-e3-e4 = -50$
Now, for the column you named "Remaining" the sum has no mathematical meaning or significance. Why? Let me explain. Let's go back to what is calculated in the sum "Remaining" column, the sum can be written as follows:
$S=50-e1 + [50-e1]-e2 + [50-e1-e2] -e3 + [(50-e1)-e2-e3]-e4$
$S=4*50 - 4*e1 - 3*e2 - 2*e3 - 1*e4 = $
$4*(50)+4*(-20)+3*(-15)+2*(-9)+1*(-6)=51$
Notice that the above formula has no meaning or significance. It clearly does not represent the total expense, so it is no wonder that it differs from the correct total expense.
The fact that the absolute difference is only by value of $1$ from the sum of the left column is by chance! Had you started with an amount of 200 and applied the same calculations, the first column sum would still give $-50$ but the right (bogus) sum would give $651$. I hope this is now clear.
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
$endgroup$
add a comment |
$begingroup$
Generally speaking, your expenses are $a_1, a_2, ldots, a_n$ and must add up to $sum_{k=1}^n a_k = T,$ your total amount. If you look at the sum in the other column, it is actually summing the partial sums of the first one, so it is $sum_{k=1}^{n-1} s_k$ where $s_k = sum_{i=1}^k a_i$. In other words, you get
$$
sum_{k=1}^{n-1} sum_{i=1}^k a_i
= sum_{i=1}^{n-1} (n-i)a_i
= n sum_{i=1}^{n-1} a_i - sum_{i=1}^{n-1}ia_i
= nT - sum_{i=1}^{n-1}ia_i,
$$
and there is absolutely no reason why that expression should equal $T$...
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
$endgroup$
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
$begingroup$
The short answer is that althought the difference of the 2 sums is only 1, it is because the sum in the second column does not represent the same fact as the sum in the first column. In fact, as explained below, the sum in the second column has no meaning relevant to the issue at hand.
To find the "Total Expenses", we sum each expense as you did in column 1:
$E=-e1-e2-e3-e4 = -50$
Now, for the column you named "Remaining" the sum has no mathematical meaning or significance. Why? Let me explain. Let's go back to what is calculated in the sum "Remaining" column, the sum can be written as follows:
$S=50-e1 + [50-e1]-e2 + [50-e1-e2] -e3 + [(50-e1)-e2-e3]-e4$
$S=4*50 - 4*e1 - 3*e2 - 2*e3 - 1*e4 = $
$4*(50)+4*(-20)+3*(-15)+2*(-9)+1*(-6)=51$
Notice that the above formula has no meaning or significance. It clearly does not represent the total expense, so it is no wonder that it differs from the correct total expense.
The fact that the absolute difference is only by value of $1$ from the sum of the left column is by chance! Had you started with an amount of 200 and applied the same calculations, the first column sum would still give $-50$ but the right (bogus) sum would give $651$. I hope this is now clear.
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
$endgroup$
add a comment |
$begingroup$
The short answer is that althought the difference of the 2 sums is only 1, it is because the sum in the second column does not represent the same fact as the sum in the first column. In fact, as explained below, the sum in the second column has no meaning relevant to the issue at hand.
To find the "Total Expenses", we sum each expense as you did in column 1:
$E=-e1-e2-e3-e4 = -50$
Now, for the column you named "Remaining" the sum has no mathematical meaning or significance. Why? Let me explain. Let's go back to what is calculated in the sum "Remaining" column, the sum can be written as follows:
$S=50-e1 + [50-e1]-e2 + [50-e1-e2] -e3 + [(50-e1)-e2-e3]-e4$
$S=4*50 - 4*e1 - 3*e2 - 2*e3 - 1*e4 = $
$4*(50)+4*(-20)+3*(-15)+2*(-9)+1*(-6)=51$
Notice that the above formula has no meaning or significance. It clearly does not represent the total expense, so it is no wonder that it differs from the correct total expense.
The fact that the absolute difference is only by value of $1$ from the sum of the left column is by chance! Had you started with an amount of 200 and applied the same calculations, the first column sum would still give $-50$ but the right (bogus) sum would give $651$. I hope this is now clear.
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
$endgroup$
add a comment |
$begingroup$
The short answer is that althought the difference of the 2 sums is only 1, it is because the sum in the second column does not represent the same fact as the sum in the first column. In fact, as explained below, the sum in the second column has no meaning relevant to the issue at hand.
To find the "Total Expenses", we sum each expense as you did in column 1:
$E=-e1-e2-e3-e4 = -50$
Now, for the column you named "Remaining" the sum has no mathematical meaning or significance. Why? Let me explain. Let's go back to what is calculated in the sum "Remaining" column, the sum can be written as follows:
$S=50-e1 + [50-e1]-e2 + [50-e1-e2] -e3 + [(50-e1)-e2-e3]-e4$
$S=4*50 - 4*e1 - 3*e2 - 2*e3 - 1*e4 = $
$4*(50)+4*(-20)+3*(-15)+2*(-9)+1*(-6)=51$
Notice that the above formula has no meaning or significance. It clearly does not represent the total expense, so it is no wonder that it differs from the correct total expense.
The fact that the absolute difference is only by value of $1$ from the sum of the left column is by chance! Had you started with an amount of 200 and applied the same calculations, the first column sum would still give $-50$ but the right (bogus) sum would give $651$. I hope this is now clear.
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
$endgroup$
The short answer is that althought the difference of the 2 sums is only 1, it is because the sum in the second column does not represent the same fact as the sum in the first column. In fact, as explained below, the sum in the second column has no meaning relevant to the issue at hand.
To find the "Total Expenses", we sum each expense as you did in column 1:
$E=-e1-e2-e3-e4 = -50$
Now, for the column you named "Remaining" the sum has no mathematical meaning or significance. Why? Let me explain. Let's go back to what is calculated in the sum "Remaining" column, the sum can be written as follows:
$S=50-e1 + [50-e1]-e2 + [50-e1-e2] -e3 + [(50-e1)-e2-e3]-e4$
$S=4*50 - 4*e1 - 3*e2 - 2*e3 - 1*e4 = $
$4*(50)+4*(-20)+3*(-15)+2*(-9)+1*(-6)=51$
Notice that the above formula has no meaning or significance. It clearly does not represent the total expense, so it is no wonder that it differs from the correct total expense.
The fact that the absolute difference is only by value of $1$ from the sum of the left column is by chance! Had you started with an amount of 200 and applied the same calculations, the first column sum would still give $-50$ but the right (bogus) sum would give $651$. I hope this is now clear.
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
edited Dec 27 '18 at 9:54
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
answered Dec 26 '18 at 20:45
NoChanceNoChance
3,76621321
3,76621321
add a comment |
add a comment |
$begingroup$
Generally speaking, your expenses are $a_1, a_2, ldots, a_n$ and must add up to $sum_{k=1}^n a_k = T,$ your total amount. If you look at the sum in the other column, it is actually summing the partial sums of the first one, so it is $sum_{k=1}^{n-1} s_k$ where $s_k = sum_{i=1}^k a_i$. In other words, you get
$$
sum_{k=1}^{n-1} sum_{i=1}^k a_i
= sum_{i=1}^{n-1} (n-i)a_i
= n sum_{i=1}^{n-1} a_i - sum_{i=1}^{n-1}ia_i
= nT - sum_{i=1}^{n-1}ia_i,
$$
and there is absolutely no reason why that expression should equal $T$...
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
$endgroup$
add a comment |
$begingroup$
Generally speaking, your expenses are $a_1, a_2, ldots, a_n$ and must add up to $sum_{k=1}^n a_k = T,$ your total amount. If you look at the sum in the other column, it is actually summing the partial sums of the first one, so it is $sum_{k=1}^{n-1} s_k$ where $s_k = sum_{i=1}^k a_i$. In other words, you get
$$
sum_{k=1}^{n-1} sum_{i=1}^k a_i
= sum_{i=1}^{n-1} (n-i)a_i
= n sum_{i=1}^{n-1} a_i - sum_{i=1}^{n-1}ia_i
= nT - sum_{i=1}^{n-1}ia_i,
$$
and there is absolutely no reason why that expression should equal $T$...
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
$endgroup$
add a comment |
$begingroup$
Generally speaking, your expenses are $a_1, a_2, ldots, a_n$ and must add up to $sum_{k=1}^n a_k = T,$ your total amount. If you look at the sum in the other column, it is actually summing the partial sums of the first one, so it is $sum_{k=1}^{n-1} s_k$ where $s_k = sum_{i=1}^k a_i$. In other words, you get
$$
sum_{k=1}^{n-1} sum_{i=1}^k a_i
= sum_{i=1}^{n-1} (n-i)a_i
= n sum_{i=1}^{n-1} a_i - sum_{i=1}^{n-1}ia_i
= nT - sum_{i=1}^{n-1}ia_i,
$$
and there is absolutely no reason why that expression should equal $T$...
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
$endgroup$
Generally speaking, your expenses are $a_1, a_2, ldots, a_n$ and must add up to $sum_{k=1}^n a_k = T,$ your total amount. If you look at the sum in the other column, it is actually summing the partial sums of the first one, so it is $sum_{k=1}^{n-1} s_k$ where $s_k = sum_{i=1}^k a_i$. In other words, you get
$$
sum_{k=1}^{n-1} sum_{i=1}^k a_i
= sum_{i=1}^{n-1} (n-i)a_i
= n sum_{i=1}^{n-1} a_i - sum_{i=1}^{n-1}ia_i
= nT - sum_{i=1}^{n-1}ia_i,
$$
and there is absolutely no reason why that expression should equal $T$...
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
answered Dec 26 '18 at 16:42
gt6989bgt6989b
35.8k22557
35.8k22557
add a comment |
add a comment |
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1
$begingroup$
Why are you adding up the Remaining column? If you spend $5$ lots of $10$ USD instead you'll have $40, 30, 20, 10 $ and $0$ in the Remaining column and adding them certainly won't give you $50$ (or even be only $1$ out)
$endgroup$
– postmortes
Dec 26 '18 at 16:36
$begingroup$
What makes you think they should be the same? The way to check this is that the beginning balance, minus the total expenditures, equals the last remaining balance. What would the sum of the remaining balances ever relate to in the real world?
$endgroup$
– saulspatz
Dec 26 '18 at 16:38
$begingroup$
Why do you think something's wrong? There's simply no reason that the sum of the "remaining" amounts should be the sum of the expenditures.
$endgroup$
– David C. Ullrich
Dec 26 '18 at 16:38