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# Math Help - Sequences
1. ## Sequences
I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?
11, 9, 10, 17, 36, 79,....
2. Originally Posted by nsingh201
I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?
11, 9, 10, 17, 36, 79,....
I don't know if this helps you at all, but I believe it is possible to find a polynomial of at most degree 5 that will fit that sequence, by solving simultaneous linear equations.
I will illustrate the method on a smaller scale. Say the sequence is
11, 9, 10, ...
Then we can find a polynomial of degree 2 that will generate this sequence, of the form f(x) = ax^2+bx+c, as follows:
We know f(1) = 11. Thus, a*(1^2)+b*1+c = 11. Simplifying, a+b+c=11.
We know f(2) = 9. Thus, a*(2^2)+b*2+c = 9. Simplifying, 4a+2b+c=9.
We know f(3) = 10. Thus, a*(3^2)+b*3+c = 10. Simplifying, 9a+3b+c=10.
This is a system of linear equations with three equations and three unknowns. The solution is described by (a,b,c) = (3/2, -13/2, 16). That is, f(x) = (3/2)x^2 - (13/2)x + 16. This generates the sequence
11, 9, 10, 14, 21, 31, ...
Obviously this is not the desired sequence, so a quadratic curve-fitting scheme is not feasible. I would proceed by next trying a cubic polynomial to model the first four terms, and if that doesn't work, a degree 4 polynomial to include the first five terms, and in the worst case, a degree 5 polynomial.
Of course there may be a much simpler explanation of your sequence that I'm blind to...
3. 11,9,10,17,36,79,170,357,736,1499,3030,6097,12236, 24519,...
4. ## ?
5. Hello, nsingh201!
Find the next two terms of this sequence: . $11,\: 9,\: 10,\: 17,\: 36,\: 79 \;\hdots$
Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern. | {
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Here's what I found . . .
. . $\begin{array}{ccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\
\text{3rd diff.} &&&& 3 && 6 && 12 \end{array}$
Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.
I also suspect that the 2nd differences are doubling.
So I believe the chart can be continued like this;
. . $\begin{array}{cccccccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\color{red}170} && {\color{red}357} \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 && {\color{blue}91} && {\color{blue}187} \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\color{blue}48} && {\color{blue}96}\end{array}$
And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$
6. Originally Posted by Soroban
Hello, nsingh201!
Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern.
Here's what I found . . .
. . $\begin{array}{ccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\
\text{3rd diff.} &&&& 3 && 6 && 12 \end{array}$
Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.
I also suspect that the 2nd differences are doubling.
So I believe the chart can be continued like this;
. . $\begin{array}{cccccccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\color{red}170} && {\color{red}357} \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 && {\color{blue}91} && {\color{blue}187} \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\color{blue}48} && {\color{blue}96}\end{array}$
And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$ | {
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And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$
beautiful approach
7. 11, 9, 10, 17, 36, 79, - Wolfram|Alpha
2 Soroban
Nice solution, but how did u get generating function?
8. Hello, ICanFly!
I was afraid someone would ask . . .LOL!
. . .but how did u get generating function?
We have: . $11,\:9,\:10,\:17,\:36,\;\hdots$
And we know there is a "doubling" feature in there: . $2^n$
Since the doubling didn't begin until the second differences.
. . I assumed there was a linear or quadratic involved, too.
So, my first general function was: . $f(n) \:=\:A + Bn + C\!\cdot\!2^n$
I used the first three terms of the sequence:
. . $\begin{array}{ccccc}
f(1) = 11 & A + B + 2C &=& 11 & [1] \\
f(2) \:=\: 9\; & A + 2B + 4C &=& 9 & [2] \\
f(3) = 10 & A + 3B + 8C &=& 10 & [3] \end{array}$
$\begin{array}{ccccc}
\text{Subtract [2] - [1]:} & B + 2C &=& -2 & [4] \\
\text{Subtract [3] - [2]:} & B + 4C &=& 1 & [5] \end{array}$
$\text{Subtract [5] - [4]: }\;\;2C \:=\: 3 \quad\Rightarrow\quad C \:=\:\tfrac{3}{2}$
Substitute into [4]: . $B + 3 \:=\:-2 \quad\Rightarrow\quad B \:=\:-5$
Substitute into [1]: . $A - 5 + 3 \:=\:11 \quad\Rightarrow\quad A \:=\:13$
Hence: . $f(n) \;=\;13 - 5n + \tfrac{3}{2}\!\cdot\!2^n$
I tested my function up to the 8th term . . . It works!
. . Therefore: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$
9. Beautiful
10. thanks so much. this all really helped. | {
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# 1/998001
#### Jameson
Staff member
$$\displaystyle \frac{1}{998001}$$ has some interesting properties. It will list out every 3 digit number except for 998.
It has the form 0.001002003004...
Here's a video on some more details if you're interested.
#### MarkFL
Staff member
soroban once posted this observation and asked:
This is just one of a family of such fractions.
Can you determine the underlying characteristic?
I looked at the factorization of the denominator, and found:
$\displaystyle 998001=999^2$
So, next I looked at:
$\displaystyle\frac{012345679}{999999999}=\frac{1}{9^2}$
Thus, I conjecture that the family you speak of is:
$\displaystyle\frac{1}{(10^n-1)^2}$ where $\displaystyle n\in\mathbb N$
where the decimal representation contains all of the $n$ digit numbers except $\displaystyle 10^n-2$ and the period is $\displaystyle n(10^n-1)$.
#### Jameson
Staff member
Yes, you are correct about this form. For example $$\displaystyle \frac{1}{9999^2}$$ gives all of the 4 digit numbers.
In the video I posted in the OP they discussed an easy way to write recurring decimals as fractions that I'm sure we are all familiar with this method but it's still worth writing for those who aren't.
If you want to write some repeating decimal all you do is write the string as the numerator to a fraction and then in the denominator write the same number of 9's. For example, if you want to write $$\displaystyle .\overline{12436298}$$ as a fraction. It's simply $$\displaystyle \frac{12436298}{99999999}$$. Of course this method can be derived quite easily. If $$\displaystyle x=.12436298$$ then $100000000x=12436298.\overline{12436298}$ so $9999999x=12436298$ and we can solve for x directly. However it's nice to know a shortcut to not have to calculate it "the long way" everytime.
Last edited:
#### Bacterius
##### Well-known member
MHB Math Helper
This seems to generalize to an arbitrary base $b$, where:
$$\frac{1}{(b^n - 1)^2}$$ | {
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$$\frac{1}{(b^n - 1)^2}$$
has representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.
But I only checked a couple of bases, so this could be wrong.
#### MarkFL
Staff member
This seems to generalize to an arbitrary base $b$, where:
$$\frac{1}{(b^n - 1)^2}$$
has representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.
But I only checked a couple of bases, so this could be wrong.
I believe you are right, as the method Jameson outlined above can be generalized to any valid radix.
#### alane1994
##### Active member
Not much to contribute to this conversation, I was just popping in to say that this is really interesting! | {
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# Complex Numbers and the Norm
I am given that for a complex number $w=a+bi$, define $\overline{w}=a-bi$ and $N(w)=w \overline{w}$. I have provided my answers for parts a and b, but I am not sure they are correct. I need help with figuring out part c.
(a) Compute $N(w)=w \overline{w}$ explicitly.
Here is what I have gotten:$N(w)=w \overline{w}= (a+bi)(a-bi)= a^2+b^2$.
(b) Show that $N(rs)=N(r)N(s)$ for any two complex numbers $r,s$.
Here is my work:
Let $r=a+bi$, $s=c+di$
Then $$rs=(a+bi)(c+di)=ac+adi+cbi-bd=(ac-bd)+(ad+cb)i.$$
Now, from a, have that $N(r)= a^2+b^2$ and $N(s)= c^2+d^2$. So, $$N(r)N(s)=(a^2+b^2)(c^2+d^2).$$
Also, $N(rs)=N((ac-bd)+(ad+cb)i)$ and from part a then, $N(rs)=(ac-bd)^2+(ad+cb)^2$
When expanded,
$$a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2$$
\begin{align*} &=a^2c^2+a^2d^2+b^2d^2+b^2c^2\\ &=a^2(c^2+d^2)+b^2(c^2+d^2)\\ &=(a^2+b^2)(c^2+d^2). \end{align*}
Hence, $N(rs)=N(r)N(s)$.
(c) Prove that $N(\overline{v})=N(v)$ and $N(v^n)=N(v)^n$ for any complex number.
This is the part I am confused as to how to prove. I let $v=a+bi$ and $\overline{v}=a-bi$. Then I am not sure how to use what I have shown before for this part.
• (1) Please put distinct parts into distinct questions. (2) Please use MathJax to format your math, it makes it a lot easier to read. Apr 16 '17 at 21:35
• @MichaelBurr Sorry, I do not know how to use MathJax, I tried my best to put it in the format. I labeled the parts a-c, but I will rename to make is clearer. Apr 16 '17 at 21:38
• What I mean is that if you have three parts, you should have three questions (one question for part $(a)$, one question for part $(b)$, and one question for part $(c)$). Apr 16 '17 at 21:39
• @MichaelBurr Okay, noted, thank you. It is just that is all one problem that has three parts and I wanted to see if my part a and b are okay to then get help for c. Apr 16 '17 at 21:40
• Also, to learn about MathJax, see the tutorial here. Apr 16 '17 at 21:41 | {
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For the first part of your question, notice that $$N (v)=\bar{v} v=\bar{v} \bar{\bar{v}}=N (\bar{v})$$ because $\bar{\bar{v}}=v$.
For the second part, consider doing induction over $n$:
Induction basis: $n=1$ $$N (v^1)=N (v)=N (v)^1$$
Induction step: $$N (v^{n+1})=N (v^n\cdot v)=N (v^n) \cdot N (v)=N (v)^n \cdot N (v)=N (v)^{n+1}$$ For the third from last step we used your result from b, for the second from last we used the induction hypothesis.
• Fixed now, thanks a lot. The way it was before, I did not like it either, but I did not know there is also \bar. Apr 16 '17 at 22:21
• Personally, I often use \overline because \bar is intended to be used as accent, but in this particular case overline looks plain ugly. Apr 16 '17 at 22:26
Since your question is about part $(c)$...
• To prove $N(v)=N(\overline{v})$, use your formula for $N$ in part $(a)$ on each of $v$ and $\overline{v}$. As stated, the question is somewhat confusing because $a$ and $b$ are used in two ways in the problem. In part $(a)$, $N(v)=a^2+b^2$ when $v=a+bi$. For part $(c)$, we could use $w=c+di$ and $\overline{w}=c-di$, then $N(w)=c^2+d^2$ (by substituting the variables in $w$ into the form for part $(a)$) and $N(\overline{w})=c^2+(-d)^2$ by the same substitution.
• To prove $N(v^n)=N(v)^n$, use your equality in part $(b)$ as well as induction on $n$. The base case is $N(v^1)=N(v)^1$, and then use part $(b)$ to prove $N(v^{k+1})=N(v)^{k+1}$ using $N(v^k)=N(v)^k$. | {
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• For $N(v)=N(\overline{v})$ I know I will get a$^2$+b$^2$ for v, but how can I just plug in $\overline{v}$ into the equation if this is for v not $\overline{v}$? Apr 16 '17 at 21:43
• You have a general formula, it might be confusing because you're using $a$ and $b$ in two different roles, I'll edit to make this more clear. Apr 16 '17 at 21:44
• Okay so when plugging in $\overline{w}=c-di$ into the given equation for (a), what exactly is happening? I am not sure how you got (-d)$^2$ because wouldn't this be showing the equations are not equal? Apr 16 '17 at 21:50
• What is the square of a negative real number? Apr 16 '17 at 21:54
• @Pam_22R This shows that the terms are equal since $(-d)^2=-d\cdot (-d)=d^2$. Apr 16 '17 at 21:55 | {
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### Statistics Knowledge Portal
A free online introduction to statistics
# Chi-Square Goodness of Fit Test
## What is the Chi-square goodness of fit test?
The Chi-square goodness of fit test is a statistical hypothesis test used to determine whether a variable is likely to come from a specified distribution or not. It is often used to evaluate whether sample data is representative of the full population.
## When can I use the test?
You can use the test when you have counts of values for a categorical variable.
Yes.
## Using the Chi-square goodness of fit test
The Chi-square goodness of fit test checks whether your sample data is likely to be from a specific theoretical distribution. We have a set of data values, and an idea about how the data values are distributed. The test gives us a way to decide if the data values have a “good enough” fit to our idea, or if our idea is questionable.
### What do we need?
For the goodness of fit test, we need one variable. We also need an idea, or hypothesis, about how that variable is distributed. Here are a couple of examples:
• We have bags of candy with five flavors in each bag. The bags should contain an equal number of pieces of each flavor. The idea we'd like to test is that the proportions of the five flavors in each bag are the same.
• For a group of children’s sports teams, we want children with a lot of experience, some experience and no experience shared evenly across the teams. Suppose we know that 20 percent of the players in the league have a lot of experience, 65 percent have some experience and 15 percent are new players with no experience. The idea we'd like to test is that each team has the same proportion of children with a lot, some or no experience as the league as a whole.
To apply the goodness of fit test to a data set we need: | {
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To apply the goodness of fit test to a data set we need:
• Data values that are a simple random sample from the full population.
• Categorical or nominal data. The Chi-square goodness of fit test is not appropriate for continuous data.
• A data set that is large enough so that at least five values are expected in each of the observed data categories.
## Chi-square goodness of fit test example
Let’s use the bags of candy as an example. We collect a random sample of ten bags. Each bag has 100 pieces of candy and five flavors. Our hypothesis is that the proportions of the five flavors in each bag are the same.
Let’s start by answering: Is the Chi-square goodness of fit test an appropriate method to evaluate the distribution of flavors in bags of candy?
• We have a simple random sample of 10 bags of candy. We meet this requirement.
• Our categorical variable is the flavors of candy. We have the count of each flavor in 10 bags of candy. We meet this requirement.
• Each bag has 100 pieces of candy. Each bag has five flavors of candy. We expect to have equal numbers for each flavor. This means we expect 100 / 5 = 20 pieces of candy in each flavor from each bag. For 10 bags in our sample, we expect 10 x 20 = 200 pieces of candy in each flavor. This is more than the requirement of five expected values in each category.
Based on the answers above, yes, the Chi-square goodness of fit test is an appropriate method to evaluate the distribution of the flavors in bags of candy.
Figure 1 below shows the combined flavor counts from all 10 bags of candy.
Figure 1: Bar chart of counts of candy flavors from all 10 bags | {
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Figure 1: Bar chart of counts of candy flavors from all 10 bags
Without doing any statistics, we can see that the number of pieces for each flavor are not the same. Some flavors have fewer than the expected 200 pieces and some have more. But how different are the proportions of flavors? Are the number of pieces “close enough” for us to conclude that across many bags there are the same number of pieces for each flavor? Or are the number of pieces too different for us to draw this conclusion? Another way to phrase this is, do our data values give a “good enough” fit to the idea of equal numbers of pieces of candy for each flavor or not?
To decide, we find the difference between what we have and what we expect. Then, to give flavors with fewer pieces than expected the same importance as flavors with more pieces than expected, we square the difference. Next, we divide the square by the expected count, and sum those values. This gives us our test statistic.
These steps are much easier to understand using numbers from our example.
Let’s start by listing what we expect if each bag has the same number of pieces for each flavor. Above, we calculated this as 200 for 10 bags of candy.
#### Table 1: Comparison of actual vs expected number of pieces of each flavor of candy
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Apple 180 200 Lime 250 200 Cherry 120 200 Cherry 225 200 Grape 225 200
Now, we find the difference between what we have observed in our data and what we expect. The last column in Table 2 below shows this difference:
#### Table 2: Difference between observed and expected pieces of candy by flavor
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Apple 180 200 180-200 = -20 Lime 250 200 250-200 = 50 Cherry 120 200 120-200 = -80 Orange 225 200 225-200 = 25 Grape 225 200 225-200 = 25 | {
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Some of the differences are positive and some are negative. If we simply added them up, we would get zero. Instead, we square the differences. This gives equal importance to the flavors of candy that have fewer pieces than expected, and the flavors that have more pieces than expected.
#### Table 3: Calculation of the squared difference between Observed and Expected for each flavor of candy
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Squared Difference Apple 180 200 180-200 = -20 400 Lime 250 200 250-200 = 50 2500 Cherry 120 200 120-200 = -80 6400 Orange 225 200 225-200 = 25 625 Grape 225 200 225-200 = 25 625
Next, we divide the squared difference by the expected number:
#### Table 4: Calculation of the squared difference/expected number of pieces of candy per flavor
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Squared Difference Squared Difference / Expected Number Apple 180 200 180-200 = -20 400 400 / 200 = 2 Lime 250 200 250-200 = 50 2500 2500 / 200 = 12.5 Cherry 120 200 120-200 = -80 6400 6400 / 200 = 32 Orange 225 200 225-200 = 25 625 625 / 200 = 3.125 Grape 225 200 225-200 = 25 625 625 / 200 = 3.125
Finally, we add the numbers in the final column to calculate our test statistic:
$2 + 12.5 + 32 + 3.125 + 3.125 = 52.75$
To draw a conclusion, we compare the test statistic to a critical value from the Chi-Square distribution. This activity involves four steps: | {
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1. We first decide on the risk we are willing to take of drawing an incorrect conclusion based on our sample observations. For the candy data, we decide prior to collecting data that we are willing to take a 5% risk of concluding that the flavor counts in each bag across the full population are not equal when they really are. In statistics-speak, we set the significance level, α , to 0.05.
2. We calculate a test statistic. Our test statistic is 52.75.
3. We find the theoretical value from the Chi-square distribution based on our significance level. The theoretical value is the value we would expect if the bags contain the same number of pieces of candy for each flavor.
In addition to the significance level, we also need the degrees of freedom to find this value. For the goodness of fit test, this is one fewer than the number of categories. We have five flavors of candy, so we have 5 – 1 = 4 degrees of freedom.
The Chi-square value with α = 0.05 and 4 degrees of freedom is 9.488.
4. We compare the value of our test statistic (52.75) to the Chi-square value. Since 52.75 > 9.488, we reject the null hypothesis that the proportions of flavors of candy are equal.
We make a practical conclusion that bags of candy across the full population do not have an equal number of pieces for the five flavors. This makes sense if you look at the original data. If your favorite flavor is Lime, you are likely to have more of your favorite flavor than the other flavors. If your favorite flavor is Cherry, you are likely to be unhappy because there will be fewer pieces of Cherry candy than you expect.
### Understanding results
Let’s use a few graphs to understand the test and the results.
A simple bar chart of the data shows the observed counts for the flavors of candy:
Figure 2: Bar chart of observed counts for flavors of candy | {
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Figure 2: Bar chart of observed counts for flavors of candy
Another simple bar chart shows the expected counts of 200 per flavor. This is what our chart would look like if the bags of candy had an equal number of pieces of each flavor.
Figure 3: Bar chart of expected counts of each flavor
The side-by-side chart below shows the actual observed number of pieces of candy in blue. The orange bars show the expected number of pieces. You can see that some flavors have more pieces than we expect, and other flavors have fewer pieces.
Figure 4: Bar chart comparing actual vs. expected counts of candy
The statistical test is a way to quantify the difference. Is the actual data from our sample “close enough” to what is expected to conclude that the flavor proportions in the full population of bags are equal? Or not? From the candy data above, most people would say the data is not “close enough” even without a statistical test.
What if your data looked like the example in Figure 5 below instead? The purple bars show the observed counts and the orange bars show the expected counts. Some people would say the data is “close enough” but others would say it is not. The statistical test gives a common way to make the decision, so that everyone makes the same decision on a set of data values.
Figure 5: Bar chart comparing expected and actual values using another example data set
### Statistical details
Let’s look at the candy data and the Chi-square test for goodness of fit using statistical terms. This test is also known as Pearson’s Chi-square test.
Our null hypothesis is that the proportion of flavors in each bag is the same. We have five flavors. The null hypothesis is written as:
$H_0: p_1 = p_2 = p_3 = p_4 = p_5$
The formula above uses p for the proportion of each flavor. If each 100-piece bag contains equal numbers of pieces of candy for each of the five flavors, then the bag contains 20 pieces of each flavor. The proportion of each flavor is 20 / 100 = 0.2. | {
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The alternative hypothesis is that at least one of the proportions is different from the others. This is written as:
$H_a: at\ least\ one\ p_i\ not\ equal$
In some cases, we are not testing for equal proportions. Look again at the example of children's sports teams near the top of this page. Using that as an example, our null and alternative hypotheses are:
$H_0: p_1 = 0.2, p_2 = 0.65, p_3 = 0.15$
$H_a: at\ least\ one\ p_i\ not\ equal\ to\ expected\ value$
Unlike other hypotheses that involve a single population parameter, we cannot use just a formula. We need to use words as well as symbols to describe our hypotheses.
We calculate the test statistic using the formula below:
$\sum^n_{i=1} \frac{(O_i-E_i)^2}{E_i}$
In the formula above, we have n groups. The $\sum$ symbol means to add up the calculations for each group. For each group, we do the same steps as in the candy example. The formula shows Oi as the Observed value and Ei as the Expected value for a group.
We then compare the test statistic to a Chi-square value with our chosen significance level (also called the alpha level) and the degrees of freedom for our data. Using the candy data as an example, we set α = 0.05 and have four degrees of freedom. For the candy data, the Chi-square value is written as:
$χ²_{0.05,4}$
There are two possible results from our comparison:
• The test statistic is lower than the Chi-square value. You fail to reject the hypothesis of equal proportions. You conclude that the bags of candy across the entire population have the same number of pieces of each flavor in them. The fit of equal proportions is “good enough.”
• The test statistic is higher than the Chi-Square value. You reject the hypothesis of equal proportions. You cannot conclude that the bags of candy have the same number of pieces of each flavor. The fit of equal proportions is “not good enough.” | {
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Let’s use a graph of the Chi-square distribution to better understand the test results. You are checking to see if your test statistic is a more extreme value in the distribution than the critical value. The distribution below shows a Chi-square distribution with four degrees of freedom. It shows how the critical value of 9.488 “cuts off” 95% of the data. Only 5% of the data is greater than 9.488.
Figure 6: Chi-square distribution for four degrees of freedom
The next distribution plot includes our results. You can see how far out “in the tail” our test statistic is, represented by the dotted line at 52.75. In fact, with this scale, it looks like the curve is at zero where it intersects with the dotted line. It isn’t, but it is very, very close to zero. We conclude that it is very unlikely for this situation to happen by chance. If the true population of bags of candy had equal flavor counts, we would be extremely unlikely to see the results that we collected from our random sample of 10 bags.
Figure 7: Chi-square distribution for four degrees of freedom with test statistic plotted
Most statistical software shows the p-value for a test. This is the likelihood of finding a more extreme value for the test statistic in a similar sample, assuming that the null hypothesis is correct. It’s difficult to calculate the p-value by hand. For the figure above, if the test statistic is exactly 9.488, then the p-value will be p=0.05. With the test statistic of 52.75, the p-value is very, very small. In this example, most statistical software will report the p-value as “p < 0.0001.” This means that the likelihood of another sample of 10 bags of candy resulting in a more extreme value for the test statistic is less than one chance in 10,000, assuming our null hypothesis of equal counts of flavors is true. | {
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# Telescoping series of form $\sum (n+1)\cdots(n+k)$ [duplicate]
Wolfram Alpha is able to telescope sums of the form $\sum (n+1)\cdots(n+k)$
e.g. $(1\cdot2\cdot3) + (2\cdot3\cdot4) + \cdots + n(n+1)(n+2)$
How does it do it?
EDIT: We can rewrite as: $\sum {(n+k)! \over n!} = \sum k!{(n+k)!\over n!k!} = \sum k!{{n+k} \choose n}$ (Thanks Daniel Fischer)
EDIT2: We can also multiply out and split sums. So e.g.
$$\sum (n-1)n(n+1) = \sum (n^3-n) = \sum n^3 - \sum n$$
But sums of powers actually seem to be more nasty than the original question, involving Bernoulli numbers. (Thanks Claude Leibovici)
And is there any name for this particular corner of maths? (i.e. How might I go about searching the Internet for information regarding this?)
PS please could we have a 'telescoping' tag?
## marked as duplicate by user147263, Community♦Dec 24 '15 at 20:42 | {
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## marked as duplicate by user147263, Community♦Dec 24 '15 at 20:42
• Note that the terms are $k!\binom{n+k}{n}$. You probably know a bit about binomial coefficients that helps summing. – Daniel Fischer Sep 1 '14 at 10:07
• I don't understand why you speak about "telescoping". Beside Daniel Fischer's suggestion, you could also develop $i(i+1)(i+2)= i^3+3 i^2+2 i$ and compute the three sums. – Claude Leibovici Sep 1 '14 at 10:16
• Is the sum over n or k? Makes a big difference. – marty cohen Sep 1 '14 at 15:18
• What great answers! Now I am stuck. Both RobJohn and HyperGeometric have nailed it from opposite directions. To accept either answer above the other would not seem right. I guess I will leave it open until SE provides some machinery for resolving these situations. – P i Sep 1 '14 at 17:49
• @P-i- Thank you! That's very considerate of you. It was an interesting question, which I enjoyed solving. I've upvoted the question as well. Have also just posted a comment on a recent observation on its resemblance to the power integral. A related question which you might want to pose would be to find the sum of a series with each term being the reciprocal of the corresponding in the present series, i.e the reciprocal of the product of consecutive integers. – hypergeometric Sep 4 '14 at 6:14
Hint: $$\sum_{n=k}^m\binom{n}{k}=\binom{m+1}{k+1}$$ A generalization is discussed in this answer. The equation above is equation $(1)$ with $m=0$.
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Telescoping sum
To turn the sum in the question into a "telescoping sum", we can use the recurrence for Pascal's Triangle: $$\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}$$ Using this recurrence, we get \begin{align} \sum_{n=k}^m\binom{n}{k} &=\sum_{n=k}^m\left[\binom{n+1}{k+1}-\binom{n}{k+1}\right]\\ &=\sum_{n=k+1}^{m+1}\binom{n}{k+1}-\sum_{n=k}^m\binom{n}{k+1}\\ &=\left[\binom{m+1}{k+1}+\color{#C00000}{\sum_{n=k+1}^m\binom{n}{k+1}}\right]-\left[\binom{k}{k+1}+\color{#C00000}{\sum_{n=k+1}^m\binom{n}{k+1}}\right]\\ &=\binom{m+1}{k+1}-\binom{k}{k+1}\\ &=\binom{m+1}{k+1} \end{align} The sums in red are the terms that are telescoped out, leaving just the first and last terms. In this case, the last term $\binom{k}{k+1}=0$.
Let $$p_n=\prod_{r=1}^k (n+r)=\overbrace{(n+1)(n+2)\cdots(n+k)}^{k \ \text{terms}}$$ which is the product of $k$ consecutive integers.
Consider the difference of two consecutive terms, where each term is the product of $k+1$ consecutive integers, i.e. \begin{align} &\prod_{r=1}^{k+1}(n+r)-\prod_{r=0}^k(n+r)\\ &=\overbrace{\underbrace{(n+1)(n+2)\cdots(n+k)}_{p_n}(n+k+1)}^{(k+1) \ \text{terms}}-\overbrace{n\underbrace{(n+1)(n+2)\cdots(n+k)}_{p_n}}^{(k+1) \ \text{terms}}\\ &=p_n[(n+k+1)-n]\\ &=p_n(1+k) \end{align}
Hence, $$p_n=\prod_{r=1}^k (n+r)=\frac1{1+k}\left[\prod_{r=1}^{k+1} (n+r)-\prod_{r=0}^k (n+r) \right]$$ which is convenient for telescoping.
Required summation, \begin{align} S=\sum_{n=0}^{m}p_n&=\sum_{n=0}^{m} \prod_{r=1}^{k} (n+r)=\sum_{n=0}^{m}(n+1)(n+2)(n+3)\cdots (n+k)\\ &=\frac1{k+1}\sum_{n=0}^{m} \left[ \prod_{r=1}^{k+1} (n+r)-\prod_{r=0}^{k} (n+r)\right]\\ &=\frac1{k+1}\prod_{r=1}^{k+1} (m+r) \qquad \blacksquare\end{align} by telescoping.
In your example, $k=3$, hence the general term is
$$(n+1)(n+2)(n+3)=\frac14\left[(n+1)(n+2)(n+3)(n+4)-n(n+1)(n+2)(n+3)\right]$$
Hence, by telescoping from $n=0$ to $m$, \begin{align}S&=1\cdot2\cdot3+2\cdot3 \cdot 4+\cdots +(m+1)(m+2)(m+3)\\ &=\frac14(m+1)(m+2)(m+3)(m+4) \end{align} | {
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NB: It is interesting to note that this result bears a striking resemblance to integration.
Compare the standard integral $$\int_0^m n^k dn=\frac{m^{k+1}}{k+1}$$ to the result of the summation above, which can also be stated as $$\sum_{n=0}^{m}n^{[k]}=\frac {m^{[k+1]}}{k+1}$$ where $n^{[k]}$ is my adjusted* Pochhammer symbol for rising factorials, defined as $$n^{[k]}=\prod_{r=1}^{k}(n+r)=(n+1)(n+2)(n+3)\cdots(n+k)$$
The actual Pochhammer symbol for rising factorials, $n^{(k)}$, starts from $n$ itself and not $n+1$, i.e. $$n^{(k)}=\prod_{r=1}^{k}(n+r-1)=n(n+1)(n+2)\cdots(n+k-1)$$
• I've upvoted this as it appears valid. But I can't see any clear motivation for the initial algebraic manipulations. Is it possible to rework this logic in a way that does not require inspirational leaps? – P i Sep 1 '14 at 14:08
• @P-i- - thank you, that's very kind :) in order to make the series telescope, the approach is to convert one term (or product, in this case) into a difference of two terms, usually of a higher 'order'. for instance, to telescope the sum of squares, one would start from the difference of cubes. in this case, we want to sum $k$-term products hence we start from the difference of two consecutive $(k+1)$-term products. – hypergeometric Sep 1 '14 at 14:19
• @P-i- - the proposed solution has been reworked slightly and hopefully this provides greater clarity. – hypergeometric Sep 1 '14 at 14:30
• thanks to all who upvoted. a short appendix has been added to the solution to highlight the resemblance of the summation result to the standard power integral. comments are most welcome. – hypergeometric Sep 3 '14 at 15:22
For completeness, a solution using the combinatorial/binomial approach, as initiated by RobJohn:
Let $p(n, k) =n(n+1)...(n+k-1) =\prod_{j=0}^{k-1} (n+j)$.
Then (writing each step in detail) | {
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Let $p(n, k) =n(n+1)...(n+k-1) =\prod_{j=0}^{k-1} (n+j)$.
Then (writing each step in detail)
$\begin{array}\\ p(n+1, k)-p(n, k) &=\prod_{j=0}^{k-1} (n+1+j)-\prod_{j=0}^{k-1} (n+j)\\ &=\prod_{j=1}^{k} (n+j)-\prod_{j=0}^{k-1} (n+j)\\ &=(n+k)\prod_{j=1}^{k-1} (n+j)-n\prod_{j=1}^{k-1} (n+j)\\ &=((n+k)-n)\prod_{j=1}^{k-1} (n+j)\\ &=k\prod_{j=1}^{k-1} (n+j)\\ &=k\prod_{j=0}^{k-2} (n+1+j)\\ &=kp(n+1, k-1)\\ \end{array}$
or, putting $k+1$ for $k$ and $n$ for $n+1$, $p(n, k) =\frac{p(n, k+1)-p(n-1. k+1)}{k+1}$.
Therefore
$\begin{array}\\ \sum_{n=1}^M p(n, k) &=\sum_{n=1}^M \frac{p(n, k+1)-p(n-1. k-1)}{k+1}\\ &=\frac1{k+1}\sum_{n=1}^M (p(n, k+1)-p(n-1. k+1))\\ &=\frac1{k+1}(p(M, k+1)-p(0, k+1))\\ &=\frac{p(M, k+1)}{k+1}\\ \end{array}$
• This proof is succinct and flawless! Thanks Marty! – P i Sep 2 '14 at 11:38
Also your terms can be re written as $$n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}-\dfrac{(n-1)n(n+1)(n+2)}{4}.$$
• that may be correct but it would not telescope readily... – hypergeometric Sep 1 '14 at 15:24
• Now it is fine. – Bumblebee Sep 2 '14 at 5:06
Following from RobJohn's hint, this can be proved very simply by reverse-engineering the hockey-stick identity.
The blues sum to the red for any given 'hockey-stick'.
It is a straightforward proof by induction. For example, in the picture, 1+3+6+10+15 = 35, so 1+3+6+10+15 + 21 = 35+21 = 56
i.e. True(stick length k) => True(stick length k+1)
Induction can also be argued from the ${n \choose r}$ formula, but the above method avoids algebraic manipulation, working instead from the simple "each element is formed by summing its upper neighbours" definition of Pascal's Triangle.
But now we DO need some algebraic manipulation.
We write this hockey-stick identity as:
$${k+0 \choose k} + ... + {k+n \choose k} = {k+(n+1) \choose (k+1)}$$
(notice the hockey stick has been reflected)
$${(k+0)! \over {k!((k+0)-k)!}} + ... + {(k+n)! \over {k!((k+n)-k)!}} = {(k+(n+1))! \over {(k+1)!(((k+(n+1))-(k+1))!}}$$ | {
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$${(k+0)! \over 0!} + ... + {(k+n)! \over n!} = {1 \over {k+1}} \cdot {(k+(n+1))! \over n!}$$
So the original question turns out to be an emergent property of Pascal's Triangle, and I would hazard a guess that it is from this simple construct that the question originally arose.
However, this proof doesn't move forwards from start to finish in a sequence of intelligently chosen moves. It doesn't offer any insight into telescoping general sums, hence I am favouring hypergeometric's proof.
• thank you! glad you liked it :) robjohn's approach using the pascal triangle identity is quite useful too, and i have used a similar approach elsewhere on MSE for another question. – hypergeometric Sep 2 '14 at 6:28
$$\color{#66f}{\large\sum_{n = 0}^{m}\pars{n + 1}\ldots\pars{n + k} =k!\,{k + m + 1 \choose m}}$$
• blink complex analysis! That's fantastic! But how does the first path integral come from n+k choose k? – P i Sep 4 '14 at 12:45
• @P-i- That's an identity: $${a \choose b}=\oint_{0\ <\ \left\vert \,z\,\right\vert\ =\ c}{(1 + z)^a \over z^{b + 1}}\,{{\rm d}z \over 2\pi{\ic}}$$ It's valid for $c \in {\mathbb R}\,,\ c > 0$ and $b\ =\ 0,1,2,3,\ldots$ – Felix Marin Sep 4 '14 at 17:49
Induction
This question is actually a duplicate of: Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$
But it is difficult to find duplicates when the question can only be expressed using maths symbols, and this question now acts as a more comprehensive resource.
One of the answers to that question is to manually construct formulae for the first few cases: $1+2+3+\dots+n$, $1\cdot2 +\dots + n\cdot(n+1)$, etc, notice a pattern and intuit a candidate formula, then prove this by induction. | {
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# Polynomial division: an obvious trick? [reducing mod $\textit{simpler}$ multiples]
The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that, $$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$ and, $$Q(x)=x^4+x^3+x^2+x+1$$ what is the remainder of $$P(x)$$ divided by $$Q(x)$$?
Let $$Q(x)=0$$. Multiplying both sides by $$x-1$$: $$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$ Substituting $$x^5=1$$ in $$P(x)$$ gives $$x^4+x^3+x^2+x+1$$. Thus, $$P(x)\equiv\mathbf0\pmod{Q(x)}$$
Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem? | {
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• One observation is that in that question the step of 'multiplying both sides by $x-1$' is only presentation and doesn't really represent any advantage for solving the problem. What is really important is observing that the roots of $Q$ are the fifths roots of $1$ besides $1$, or that is a factor of $x^5-1$. There are many ways to compute this remainder that are simple to do in a short time. So, don't include that 'multiplying by $x-1$' as part of what was required for the students to solve the problem – logarithm May 14 '19 at 15:48
• My approach would be to observe that the exponents on P(x) look like they're set up to produce an elegant answer. The common elegant answers are "1" and "0". Since the "+1"s on the polynomials are clearly intended to cancel out, the answer must be "0". – Mark May 14 '19 at 23:33
• I'm probably missing something, but... The above trick seems to show that when $Q(x) = 0$, $P(x) = Q(x)$. But (a) the original question didn't specify that $Q(x) = 0$; how do we know that the answer $P(x)\equiv\mathbf0\pmod{Q(x)}$ applies to other cases? and (b) the original question asks, "what is the remainder of $𝑃(𝑥)$ divided by $𝑄(𝑥)$?" but the answer given here says let $Q(x) = 0$; doesn't that mean that the remainder is not defined? en.wikipedia.org/wiki/Remainder#Polynomial_division – LarsH May 15 '19 at 21:06
• I added some explicit examples to the end of my answer to emphasize the ubiquity of that simple idea. – Bill Dubuque May 16 '19 at 13:49
• What kind of test? Is this part of a math competition, or a classroom precalculus exam? – PersonX May 16 '19 at 16:35 | {
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The key idea employed here is the method of simpler multiples - a very widely used technique. Note that $$\,Q\,$$ has a "simpler" multiple $$\,QR = x^5\!-\!1,\,$$ so we can first reduce $$P$$ modulo $$\,x^{\large 5}\! -\! 1\,$$ via $$\!\bmod x^{\large 5}-1\!:\,\ \color{#c00}{x^{\large 5}\equiv 1}\Rightarrow\, x^{\large r+5q^{\phantom{|}}}\!\!\equiv x^{\large r}(\color{#c00}{x^{\large 5}})^{\large q}\equiv x^{\large r},\,$$ then finally reduce that $$\!\bmod Q,\,$$ i.e.
$$P\bmod Q\, =\, (P\bmod QR)\bmod Q\qquad$$
This idea is ubiquitous, e.g. we already use it implicitly in grade school in radix $$10$$ to determine integer parity: first reduce mod $$10$$ to get the units digit, then reduce the units digits mod $$2,\,$$ i.e.
$$N \bmod 2\, = (N\bmod 2\cdot 5)\bmod 2\qquad\$$
i.e. an integer has the same parity (even / oddness) as that of its units digit. Similarly since $$7\cdot 11\cdot 13 = 10^{\large 3}\!+1$$ we can compute remainders mod $$7,11,13$$ by using $$\,\color{#c00}{10^{\large 3}\equiv -1},\,$$ e.g. $$\bmod 13\!:\,\ d_0+ d_1 \color{#c00}{10^{\large 3}} + d_2 (\color{#c00}{10^{\large 3}})^{\large 2}\!+\cdots\,$$ $$\equiv d_0 \color{#c00}{\bf -} d_1 + d_2+\cdots,\,$$ and, similar to the OP, by $$\,9\cdot 41\cdot 271 = 10^{\large 5}\!-1\,$$ we can compute remainders mod $$41$$ and $$271$$ by using $$\,\color{#c00}{10^5\!\equiv 1}$$
$$N \bmod 41\, = (N\bmod 10^{\large 5}\!-1)\bmod 41\quad$$
for example $$\bmod 41\!:\ 10000\color{#0a0}200038$$ $$\equiv (\color{#c00}{10^{\large 5}})^{\large 2}\! + \color{#0a0}2\cdot \color{#c00}{10^{\large 5}} + 38\equiv \color{#c00}1+\color{#0a0}2+38\equiv 41\equiv 0$$
Such "divisibility tests" are frequently encountered in elementary and high-school and provide excellent motivation for this method of "divide first by a simpler multiple of the divisor" or, more simply, "mod first by a simpler multiple of the modulus". | {
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This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses.
For example, to divide by an algebraic number we can employ its norm as a simpler multiple, e.g. for $$\,w = a+b\sqrt{n}\,$$ we use $$\,ww' = (a+b\sqrt n)(a-b\sqrt n) = a^2-nb^2 = c\in\Bbb Q\ (\neq 0\,$$ by $$\,\sqrt{n}\not\in\Bbb Q),\,$$ which reduces division by an algebraic to simpler division by a rational, i.e. $$\, v/w = vw'/(ww'),$$ e.g.
$$\dfrac{1}{a+b\sqrt n}\, =\, \dfrac{1}{a+b\sqrt n}\, \dfrac{a-b\sqrt n}{a-b\sqrt n}\, =\, \dfrac{a-b\sqrt n}{a^2-nb^2}\,=\, {\frac{\small 1}{\small c}}(a-b\sqrt n)\qquad$$
so-called rationalizing the denominator. The same idea works even with $$\,{\rm\color{#c00}{nilpotents}}$$
$$\color{#c00}{t^n = 0}\ \Rightarrow\ \dfrac{1}{a-{ t}}\, =\, \dfrac{a^{n-1}+\cdots + t^{n-1}}{a^n-\color{#c00}{t^n}}\, =\, a^{-n}(a^{n-1}+\cdots + t^{n-1})\qquad$$
which simplifies by eliminating $$\,t\,$$ from the denominator, i.e. $$\,a-t\to a^n,\,$$ reducing the division to division by a simpler constant $$\,a^n\,$$ (vs. a simpler rational when rationalizing the denominator).
Another example is Gauss' algorithm, where we compute fractions $$\!\bmod m\,$$ by iteratively applying this idea of simplifying the denominator by scaling it to a smaller multiple. Here we scale $$\rm\color{#C00}{\frac{A}B} \to \frac{AN}{BN}\:$$ by the least $$\rm\,N\,$$ so that $$\rm\, BN \ge m,\,$$ reduce mod $$m,\,$$ then iterate this reduction, e.g.
$$\rm\\ mod\ 13\!:\,\ \color{#C00}{\frac{7}9} \,\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}\qquad\qquad$$ | {
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Denominators of the $$\color{#c00}{\rm reduced}$$ fractions decrease $$\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$$ so reach $$\color{#C00}{1}\,$$ (not $$\,0\,$$ else the denominator would be a proper factor of the prime modulus; it may fail for composite modulus)
See here and its $$25$$ linked questions for more examples similar to the OP (some far less trivial).
Worth mention: there are simple algorithms for recognizing cyclotomics (and products of such), e.g. it's shown there that $$\, x^{16}+x^{14}-x^{10}-x^8-x^6+x^2+1$$ is cyclotomic (a factor of $$x^{60}-1),\,$$ so we can detect when the above methods apply for arbitrarily large degree divisors.
Let $$a$$ be zero of $$x^4+x^3+x^2+x+1=0$$. Obviously $$a\ne 1$$. Then $$a^4+a^3+a^2+a+1=0$$ so multiply this with $$a-1$$ we get $$a^5=1$$ (You can get this also from geometric series $$a^n+a^{n-1}+...+a^2+a+1 = {a^{n+1}-1\over a-1}$$ by putting $$n=4$$).
But then $$\begin{eqnarray} Q(a) &=& a^{100}\cdot a^4+a^{90}\cdot a^3+a^{80}\cdot a^2+a^{70}\cdot a+1\\ &=& a^4+a^3+a^2+a+1\\&=&0\end{eqnarray}$$
So each zero of $$Q(x)$$ is also a zero of $$P(x)$$ and since all 4 zeroes of $$Q(x)$$ are different we have $$Q(x)\mid P(x)$$. | {
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• I think this is the approach that the examiners would expect from high school students. – Dancrumb May 14 '19 at 0:23
• From my perspective this is the best answer. (+1) – trancelocation May 14 '19 at 3:54
• The OP understands that after multiplication with $x-1$, the question became simpler, he/she asked how a student should find this trick. This answer is basically just a reformulation of the answer given in the question. I don't see the usefulness of this answer at all... (but it has 14 upvotes, so I might be missing something...) – user193810 May 14 '19 at 11:49
• @MariaMazur: A bit, but not really. It still feels like you stopped reading the question after "what is the remainder of P(x) divided by Q(x)?". You gave a correct answer to that question, but not to the real question. It is easy to prove that the 'trick' (multiplication by $x-1$) works and is correct, but you ignored the real question, and after your edit it is hidden as a small remark. But whatever, your answer is trivially better than mine, because I did not gave any, so I'll leave it here. – user193810 May 14 '19 at 13:25
• @Pakk: it hinges on spotting the standard identity: $Q(x) = 0$ gives the fifth roots of unity (other than x=1). It takes tons more work if you don't spot that; you could solve the quartic $Q(x) = 0$ then notice "Hey these are the other four fifth roots of unity, thus x^5 \equiv 1" then apply the information $x^5 \equiv 1$ to hugely simplify $P(x)$, and thus conclude $Q(x) \mid P(x)$ and remainder = 0. But that would be lots more work. (Yeah the first time I saw this trick back in HS I voiced the same objection about non-obviousness... this is why this identity is so pivotal to factorization) – smci May 16 '19 at 1:53
While it may be a standard technique, as Bill's response details, I wouldn't say it's at all obvious at High School level. As a pre-Olympiad challenge problem, however, it's a good one. | {
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My intuition is via cyclotomic polynomials -- $$Q(x) = \Phi_5(x)$$, giving the idea to multiply through by $$x-1$$ -- but I doubt I would have recognised them before university: https://en.wikipedia.org/wiki/Cyclotomic_polynomial
• I would go even further. As a BSE graduate in engineering, this is not at all obvious to me. – MooseBoys May 14 '19 at 6:58
• @MooseBoys: I would go even further. As a PhD in Physics (admittedly many years ago), this is not at all obvious to me. – WoJ May 14 '19 at 9:04
• It may not be obvious in other contexts since you never had to use it, but I assure you that in Math Olympiad training it's a standard trick. All serious contestants will probably already have seen something similar. There's nothing outrageous or shameful in that; Olympiad-style math is simply different from what people use and need in other professional contexts (physicist/engineer) --- the focus on Euclidean geometry being a perfect example. – Federico Poloni May 14 '19 at 9:12
• @Federico Where in the question does it say that this is an Olympiad test? OP phrased their question as if it were applicable to all high school students. I'd guess that fewer than 5% of all students would get the correct answer. – MooseBoys May 14 '19 at 17:01
• @MooseBoys I agree with your estimate. The question is "is this obvious?", and my comment is "it is a standard problem for people who train for math Olympiads". I realize that it's a strong additional assumption. I'm not arguing that it's a good or bad problem: as you point out, to do that, we would at least have to know to which students it was given. – Federico Poloni May 14 '19 at 17:09
This may be accessible to a high school student:
$$x^{104}+x^{93}+x^{82}+x^{71}+1$$
$$= (x^{104}-x^4)+(x^{93}-x^3)+(x^{82}-x^2)+(x^{71}-x)+(x^4+x^3+x^2+x+1)$$
$$=x^4(x^{100}-1)+x^3(x^{90}-1)+x^2(x^{80}-1)+ x(x^{70}-1)+(x^4+x^3+x^2+x+1)$$
We know that $$(x^n-1)|(x^{mn}-1), m,n \in \mathbb{N}$$ so $$x^5-1$$ divides $$x^{100}-1, x^{90}-1$$ etc. | {
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In turn $$x^5-1$$ is divisible by $$(x^4+x^3+x^2+x+1)$$ which concludes the proof
If it's not obvious, an examination of the question quickly reveals the trick. Say
$$P(x)=x^n$$
Then begin long division by $$Q(x)$$:
$$x^n-x^n-x^{n-1}-x^{n-2}-x^{n-3}-x^{n-4}$$ $$x^{n-5}$$ $$\dots$$ $$x^{n-5k}$$
While it may not be obvious just by looking at the question, anyone who attempts the naive solution has (at least) a reasonable chance of running across a way of solving it.
• This is a good point. If one is asked to perform division by a polynomial $Q(x)$, it is natural to start with dividing $x^n$ for $n> \deg(Q)$. Thus the first step is to divide $x^5$. Magically(!), one gets the remainder 1. So, one can discover the formula in the form $x^5=(x-1)(x^4+x^3+x^2+x+1) + 1$ even if one didn't remember it. The rest of the steps are not too difficult. – Kapil May 14 '19 at 4:07
I would have thought that bright students, who knew $$1+x+x^2+\cdots +x^{n-1}= \frac{x^n-1}{x-1}$$ as a geometric series formula, could say
$$\dfrac{P(x)}{Q(x)} =\dfrac{x^{104}+x^{93}+x^{82}+x^{71}+1}{x^4+x^3+x^2+x+1}$$
$$=\dfrac{(x^{104}+x^{93}+x^{82}+x^{71}+1)(x-1)}{(x^4+x^3+x^2+x+1)(x-1)}$$
$$=\dfrac{x^{105}-x^{104}+x^{94}-x^{93}+x^{83}-x^{82}+x^{72}-x^{71}+x-1}{x^5-1}$$
$$=\dfrac{x^{105}-1}{x^5-1}-\dfrac{x^{104}-x^{94}}{x^5-1}-\dfrac{x^{93}-x^{83}}{x^5-1}-\dfrac{x^{82}-x^{72}}{x^5-1}-\dfrac{x^{71}-x}{x^5-1}$$
$$=\dfrac{x^{105}-1}{x^5-1}-x^{94}\dfrac{x^{10}-1}{x^5-1}-x^{83}\dfrac{x^{10}-1}{x^5-1}-x^{72}\dfrac{x^{10}-1}{x^5-1}-x\dfrac{x^{70}-1}{x^5-1}$$
and that each division at the end would leave zero remainder for the same reason, replacing the original $$x$$ by $$x^5$$ | {
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I think if the candidates know what a geometric series is, the question is okay. Indeed, one uses exactly this trick to find the formula for the geometric series, i.e. one writes $$(x-1)\sum_{k=1}^nx^k=x^{n+1}-1$$ to find that $$\sum_{k=1}^\infty x^k=\lim_{n\to\infty}\sum_{k=1}^nx^k=\lim_{n\to\infty}\frac{x^{n+1}-1}{x-1}=\frac{1}{1-x}$$ for $$|x|<1$$. Therefore, it is not too hard to get from $$x^4+x^3+x^2+x+1$$ to $$x^5-1$$. Now you can reduce mod $$x^5-1$$ by substitution $$x^5=1$$.
I think the way one should think about this is to note that $$x^4+x^3+x^2+x+1$$ is the minimal polynomial of any primitive 5th unit root $$\alpha$$. Now $$P(\alpha)=0$$ since $$\alpha^5=1$$ and therefore $$Q$$ devides $$P$$. | {
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Rational and irrational sequences
1. Let $(x_n)$ be a sequence of rational numbers that converges to an irrational number $x$. Must it be the case that $x_1, x_2, \dots$ are all irrational?
2. Let $(y_n)$ be a sequence that converges to a rational number $y$. Define such a sequence where $y_1, y_2, \dots$ are all irrational.
Now these are the 2 questions in my textbook i noticed when posting this that someone had asked the answer to 1) and a similar but less difficult version of 2.
A couple things i noted the defined answer for 1 he used $\pi$ and simply added more digits to it for each part of the sequence. so firstly id like to ask would that really be a full mark answer on a analysis final? secondly and more importantly i was wondering if there was a way to define this where the final number wasn't transcendental.
For 1) I used $(1+1/n)^n$ s.t. $n \in \mathbb{N}$. For every $n$ this sequence is rational but for $\lim_{ n \to \infty }(1+1/n)^n = e$.
My Question: (i) Is this true and if so is there a definable sequence that holds but doesn't define a transcendental number?
EDIT: Golden ratio from Fibonacci sequence.
For 2) I picked $2^{1/n}$ s.t. $n>1$ and $n \in \mathbb{N}$. I believe every value of this sequence is irrational but the limit should be 1.
My Question: (ii) does the above work? and if possible could I have another example of a sequence where every value is irrational but the limit is rational?
EDIT: $\frac{\sqrt 2}{n}$.
• If $(x_n)$ converges to $x$ is the answer to part (i) (and all the $x_n$ are non-zero), then $(x_n/x)$ is a solution to part (ii). Sep 14 '16 at 2:54
• thats very clever thanks! Sep 14 '16 at 3:00
You basically have both answers right, just need a couple of tweaks.
Question:(i) Is this true and if so is there a definable sequence that holds but doesnt define a transcendental number?
It is true that $a_n = (1 + \frac{1}{n})^n$ is a sequence of rationals whose $\lim_{n \to \infty} a_n = e$ is irrational. | {
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For a sequence of rationals that converges to an algebraic (non transcendental) irrational, take for example the partial sums of the Taylor series expansion of $\sqrt{1+x}$ at $x=1$ which converge to $\sqrt{2}$.
Question:(ii) does the above work? and if possible could i have anther example of a sequence where every value is irrational but the limit is rational?
The above works, except you may want to define $a_n = 2^\frac{1}{n+1}$ so that $a_1$ is irrational as well.
For other such sequences, pick any $r \in \mathbb{R} \setminus \mathbb{Q}$ and let $a_n = \frac{r}{n}$.
[ EDIT ] I see that you just edited the question to add $\frac{\sqrt{2}}{n}$ as an example of the latter.
It doesn't have to be $\pi$ or $e$, you could do the digit by digit approximation and converge to say $\sqrt{2}$ which is algebraic and not transcendental.
I would say that that answer of digit by digit is basically full mark; you could probably include that $|x - x_n| < 10^{-n}$ which means that the sequence is Cauchy to be more specific.
Your second sequence of $\{2^{1/n}\} \to 1$ is also correct. See if you can prove that each number in the sequence is irrational by mimicking the proof that $\sqrt{2}$ is irrational.
Here's another sequence
$$a_n = 1 + \frac{\sqrt{2}}{10^n}$$
We see that $\lim_{n \to \infty} a_n = 1$ and each $a_n$ is irrational because you can create a common denominator and note that the numerator will be irrational and the denominator and integer. | {
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# Math Help - Techniques of integration (3)
1. ## Techniques of integration (3)
Let $n \geq 0$ be an integer. Evaluate $I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.$
2. Using itegration by parts I could probably kick this thing off...
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx = \frac{1}{n}cos^n(x)sin(nx) + \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx$
Not really sure how to do,
$\int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx$
probably needs some trig identities to tidy it up.
3. Originally Posted by pickslides
Using itegration by parts I could probably kick this thing off...
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx =$ $\color{red}\frac{1}{n}cos^n(x)sin(nx)$ $+ \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx.$
Not really sure how to do $\int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx.$
probably needs some trig identities to tidy it up.
that's a good start. we have $I_0=\frac{\pi}{2}.$ for $n \geq 1,$ as you (almost) showed, we have: $I_n=\int_0^{\frac{\pi}{2}} \sin(nx) \sin(x) \cos^{n-1}x \ dx.$ now what?
4. Hello,
Okay, I think I got the correct idea
From $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$, it follows that :
$\sin(nx)\sin(x)=\cos[(n-1)x]-\cos(nx)\cos(x)$
The integral is thus :
$I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x)-\cos(nx)\cos(x)\cos^{n-1}(x) ~dx$
$I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x) ~dx-\int_0^{\frac\pi2} \cos(nx)\cos^n(x) ~dx$
Which is :
$I_n=I_{n-1}-I_n$
$I_n=\frac 12 \cdot I_{n-1}$
$\Rightarrow \boxed{I_n=\frac{\pi}{2^{n+1}}}$
That's a really nice one
5. Bravo Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!
6. Originally Posted by NonCommAlg
Brave Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!
Yes indeed | {
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1+1=n+1 : that's a very well-known result !!
Had a too short night...
7. Consider
$cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [cos(n-2k)x + i sin(n-2k)x]
$
Since $cos^n(x)$ is real , the terms of sine function will be cancelled finally ,
but we experience that $\int_0^{\frac\pi2} cos(n-2k)x ~cosnx ~dx = 0$ if n-2k is not equal to n so the integral becomes :
$\frac{\pi}{2^{n+1}}$
8. Originally Posted by simplependulum
$\cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [\cos(n-2k)x + i \sin(n-2k)x]
$
this is not an identity!
9. Originally Posted by NonCommAlg
Let $n \geq 0$ be an integer. Evaluate $I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.$
It can be done using complex variables...
First note that
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx$
Parameterize the unit circle in the comples plane with
$z=e^{i\theta} \implies \frac{dz}{iz}=d\theta$
So the integral becomes
$\oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}$
expanding with the binomial theorem we get
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz$
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]$
each finite sume is its own larent series so the residues are coeffients on the $\frac{1}{z}$ terms so we get
$\frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}$
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}$
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}$
10. Originally Posted by TheEmptySet
It can be done using complex variables...
First note that
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx$
Parameterize the unit circle in the comples plane with
$z=e^{i\theta} \implies \frac{dz}{iz}=d\theta$
So the integral becomes | {
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$z=e^{i\theta} \implies \frac{dz}{iz}=d\theta$
So the integral becomes
$\oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}$
expanding with the binomial theorem we get
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz$
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]$
i think the power of $z$ in the second sum should be $-2i-1.$
each finite sume is its own larent series so the residues are coeffients on the $\frac{1}{z}$ terms so we get
$\frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}$
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}$
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}$
looks good to me! | {
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# Double negation of existential/universal quantifier $\lnot(\exists x(\lnot A(x))$
I have a (simple) question about the double-negation and existential/universal quantifiers.
When negating the following
$$\lnot\exists x(\lnot A(x))$$
I believe you just push the negation in (which swaps the quantifier) making it
$$\forall x[\lnot(\lnot A(x))]$$
and then
$$\forall x A(x)$$
Would that be a correct interpretation? Or would I be losing one of the negations on $$A(x)$$ somewhere earlier?
• Yes, that’s fine. It’s also intuitively clear: the first expression says that there is no $x$ for which $A(x)$ is false, and the last says that $A(x)$ is true for every $x$, clearly the same thing. – Brian M. Scott Apr 22 at 17:30
• Awesome, thank you so much. – confundido Apr 22 at 17:31
• You’re very welcome. – Brian M. Scott Apr 22 at 17:31
• Thank you (just read your more detailed answer) - I thought the intuition was clear, but it was part of a larger proof and I was doubting myself as to whether the reduction made sense. I guess I just needed validation. – confundido Apr 22 at 17:35
You did that correctly.
Note that going from
$$\lnot\exists x(\lnot A(x))$$
to
$$\forall x[\lnot(\lnot A(x))]$$
is an example of 'Quantifier Negation', but going from:
$$\forall x[\lnot(\lnot A(x))]$$
to
$$\forall x A(x)$$
is an instance of 'Double Negation'
As per the answer by Brian M. Scott:
It’s also intuitively clear: the first expression says that there is no x for which A(x) is false, and the last says that A(x) is true for every x, clearly the same thing.
A word of caution. Of course at the final step $$\forall x\neg\neg Ax$$ entails (or at least, classically entails) $$\forall x Ax$$, and does so because adjacent double negations (classically) cancel each other out.
BUT
The inference is not strictly an application of a standard double negation rule of the form from $$\neg\neg\varphi$$ infer $$\varphi$$. That rule only allows us to remove initial double negations.
SO | {
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SO
To show the entailment in standard proof systems requires a three-step mini-proof:
$$\forall x\neg\neg Ax\quad$$ (given)
$$\neg\neg Aa\quad\quad$$ (universal instantiation with parameter or free variable depending on system)
$$Aa\quad\quad\quad$$ (NOW you can apply the DN rule)
$$\forall x Ax\quad\quad$$ (universal generalization)
So your reasoning is informally just fine, but do be careful about jumping from $$\forall x\neg\neg Ax$$ to $$\forall x Ax$$ in formal proofs! | {
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# Finding an operation on $G^S$ that yields a group
Problem: Assume $S$ is a nonempty set and $G$ is a group. Let $G^S$ denote the set of all mappings from $S$ to $G$. Find an operation on $G^S$ that will yield a group.
Update (full attempted proof--is it correct?): Let $*$ be the operation on the given group $G$ and let $\bigstar$ be the operation on $G^S$. For all $\alpha,\beta\in G^S$, define $(\alpha\,\bigstar\,\beta)(x)=\alpha(x)*\beta(x)$. Now we must prove $G^S$ is a group.
• Closure: This is inherited from $G$.
• Associativity: This is also inherited from $G$.
• Existence of an identity element: This is also inherited from $G$; more explicitly, $e_G(x)=1=e_{G^S}(x)$ because $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)$ and, in the other direction, $(e_G\,\bigstar\,\alpha)(x)=e_G(x)*\alpha(x)=\alpha(x)$.
• Existence of inverse elements: If $\eta(x)\in G$, then $\eta^{-1}(x)\in G$ also. Thus, for $\alpha\in G^S$, we have $(\alpha\,\bigstar\,\alpha^{-1})(x)=\alpha(x)*\alpha^{-1}(x)=e_G(x)=e_{G^S}(x)$ and, in the other direction, $(\alpha^{-1}\,\bigstar\,\alpha)(x)=\alpha^{-1}(x)*\alpha(x)=e_G(x)=e_{G^S}(x)$.
This concludes the proof that the operation $\bigstar$ on $G^S$ yields a group. $\blacksquare$
$\color{red}{\mathbf{\text{Question:}}}$ Is this above proof correct? Also, by the way the operation $\bigstar$ works, it seems like $G^S$ would be a subgroup of $G$ or am I wrong?
Original work shown for problem:
Book solution: For $\alpha,\beta\in G^S$, define $(\alpha\beta)(x)=\alpha(x)\beta(x)$ for each $x\in S$. | {
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My questions: I thought about this problem for a good while but got nowhere and finally decided to look at the answer, but it does not make a ton of sense to me. How does the given operation yield a group? And what really is the operation being considered? Is it composition? That is, $(\alpha\beta)(x)\equiv(\alpha\circ\beta)(x)$? If it is composition, then I know all compositions are associative, and thus I do not need to prove that as part of proving that $G^S$ is a group. But what about closure? Existence of an identity element? Existence of inverse elements?
For the existence of an identity element, I thought about $e_G(x)=1$ and having $(\eta e_G)(x)=\eta(x)e_G(x)=\eta(x)$ and $(e_G\eta)(x)=e_G(x)\eta(x)=\eta(x)$, but this doesn't really seem right or is it?
Lastly, the problem gives the condition that $S\neq\varnothing$ (why is that important?), and no operation is specified concerning $G$--does that matter? I guess not, but it seems interesting to me that no operation is specified for $G$ when we are trying to find an operation for $G^S$ that involves $G$.
• The proof seems fundamentally fine. I'm not sure what the distinction is between the notation $e_G(x)$ and $e_G^S(x)$. You could write out associativity more explicitly, although I tend not to do so unless it's not obvious. – Rolf Hoyer Apr 9 '15 at 17:22
• @RolfHoyer I just meant that $e_G(x)$ is the identity element for $G$ while $e_G^S(x)$ is the identity element for $G^S$. If $G^S$ is a subgroup of $G$ (it looks like it is, but is it?), then I know the distinctions do not matter because the identity and inverses are the same. – fancynancy Apr 9 '15 at 17:32
• @RolfHoyer OK--I just fixed up the question with the modified notation ($e_{G^S}(x)$ instead of $e_G^S(x)$). Look all good now? Your answer definitely helped me to start thinking along the right lines. – fancynancy Apr 9 '15 at 17:38
It definitely looks much better! | {
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It definitely looks much better!
I'm a little confused about the distinction between $e_G$ and $e_{G^S}$; they both seem like the identity of $G^S$, and you seem to be using $1$ for the identity of $G$. Maybe make it more explicit that you're defining $e_{G^S}: S \to G$ so that $e_{G^S}(x) = 1_G$; it's the map sending everything in $S$ to the identity of $G$. So my only suggestion would be to clean up the notation (how exactly are you writing the identity of $G$? Of $G^S$?) and make it clear exactly what the identity of $G^S$ does.
For the inverses, essentially the same thing: Given $\alpha: S \to G$, you just start using $\alpha^{-1}$ without explicitly stating that, if $\alpha(x) = g$, then $\alpha^{-1}(x) = (\alpha(x))^{-1} = g^{-1}$. That is, make it clear that we invert $\alpha:S \to G$ pointwise, by inverting its output (using the inverse in $G$).
For your other question, very rarely will it be the case that $G^S \cong G$. I think it's more likely that $G^S$ is isomorphic to a direct product of several copies of $G$ (how many copies?). | {
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• Hmm okay. So I should write $e_G(x)=1$ and $e_{G^S}(x)=1_G$ to more clearly indicate that $1$ is the identity for $G$ while $1_G$ is the identity for $G^S$ right? You're right--my notation is a bit mangled. Okay. I get the identity part now. I'm still a little fuzzy on the inverse part though. Are you saying I'm technically correct with what I've written but it would be clearer to actually assign an element to $\alpha(x)$ to clearly indicate that invertibility is pointwise? – fancynancy Apr 9 '15 at 18:10
• What I mean about identities is that there should really only be two in sight: That of $G$, let's call it $1_G$, and that of $G^S$, let's call it $1_{G^S}$. So, I'm not sure why there are two functions whose notation suggest that they're both identities of $G^S$; that they're both functions $S \to G$. – pjs36 Apr 9 '15 at 18:12
• And for the inverses, yes, it's correct, provided you explicitly state how $\alpha^{-1}: S \to G$ (or $\eta^{-1}$, for that matter) is defined. – pjs36 Apr 9 '15 at 18:14
• Okay so I should have (for identities, that is), in one direction, $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)*1_G=\alpha(x)$. The same would happen in the other direction. So the identity exists and, explicitly, we have $e_{G^S}=e_G=1$. Correct? – fancynancy Apr 9 '15 at 18:16
• I do see what you're saying now about the confusion between writing $e_G(x)$ and $e_{G^S}(x)$ because they are both mappings from $S$ to $G$; thus, that notation is rather clumsy. – fancynancy Apr 9 '15 at 18:19
The group operation is not defined by composition, unless you mean the composition $S \to S\times S \to G\times G \to G$, where the first map is the diagonal $s \to (s,s)$, the second map is the product of the two given maps $\alpha, \beta$, and the third map is the group operation on $G$. When the author writes $(\alpha\beta)(x) = \alpha(x)\beta(x)$, the latter product is defined using the operation of $G$, just to clarify. | {
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This definition is using the operation of $G$ pointwise, just like how addition and multiplication are defined for real-valued functions, ie $(f+g)(x) = f(x) + g(x)$. The identity element is the constant function $e_G(x) = 1$ for every $x$, and you correctly give the proof that this works. The inverse of a function $\alpha: S \to G$ is given by taking inverses pointwise, namely $\alpha^{-1}(x) = (\alpha(x))^{-1}$. Associativity follows from the associativity of $G$.
As regards to the empty set, if $S = \emptyset$ then there is a unique empty function $S \to G$, and I guess the authors would rather rule out this case than give the trivial product on this set.
Hopefully this makes things more clear.
• I just updated my question with a fully attempted proof in light of your helpful answer. Can you let me know if you think it's correct? Thanks for your answer--I think I am at least on the right track now. – fancynancy Apr 9 '15 at 14:23 | {
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# Why is equivalent resistance in parallel circuit always less than each individual resistor?
There are $$n$$ resistors connected in a parallel combination given below.
$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}.......\frac{1}{R_{n}}$$
Foundation Science - Physics (class 10) by H.C. Verma states (Pg. 68)
For two resistances $$R_{1}$$ and $$R_{2}$$ connected in parallel,
$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}$$ $$R_{ev}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$
We see that the equivalent resistance in a parallel combination is less than each of the resistances.
I observe this every time I do an experiment on parallel resistors or solve a parallel combination problem. How can we prove $$R_{ev} or that $$R_{ev}$$ is less than the Resistor $$R_{min}$$, which has the least resistance of all the individual resistors? | {
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• I have to confess that when I saw that you had exhibited the harmonic sum form ($1/R_e = 1/R_1 + 1/R_2 + \dots$), I found the question confusing because I didn't understand what you didn't get. It's worth your time to think about this until it becomes obvious, not because this problem is so important but because it will start teaching you the habit of learning things from the functional form of the relationships that appear in your studies. – dmckee --- ex-moderator kitten Sep 27 '14 at 17:15
• @dmckee I didn't quite get what you said. It'll probably help me if you'll elaborate. – user49111 Sep 27 '14 at 17:26
• With experience, you will find this fact obvious, and it is worth taking the time to ponder it until it becomes so. – dmckee --- ex-moderator kitten Sep 27 '14 at 17:34
• I rolled back this question to the original version because you shouldn't put answers, or responses to answers, in the question itself. @imakesmalltalk Feel free to post an answer of your own if you would like to offer another method of answering the question. – David Z Oct 27 '14 at 0:30
• Once you believe the 2 resistor answer, it extends to any finite number. For each resistor, imagine combining all the rest into a single resistor by the law you cite. Now the combination of the one resistor and the combination resistor is less than the one. Do this once for each of the resistors, and you have that the total combination is less than any one. – Ross Millikan Oct 27 '14 at 2:53
If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors. | {
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The individual resistances are all positive, so the sum $$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances.
No mucking around with the two-resistor form required.
We can prove it by induction. Let $$\frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n}$$ Now, when $n=2$, we find $$\frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}}$$ Since $\frac{R_1}{R_2} > 0$, we see that $R^{(2)}_{eq} < R_1$ and $R^{(2)}_{eq} < R_2$ or equivalently $R^{(2)}_{eq} < \min(R_1, R_2)$.
Now, suppose it is true that $R^{(n)}_{eq} < \min (R_1, \cdots, R_n)$. Then, consider $$\frac{1}{R^{(n+1)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} + \frac{1}{R_{n+1}} = \frac{1}{R^{(n)}_{eq}} + \frac{1}{R_{n+1}}$$ Using the result from $n=2$, we find $$R^{(n+1)}_{eq} < \min ( R_{n+1} , R^{(n)}_{eq} ) < \min ( R_{n+1} , \min (R_1, \cdots, R_n))$$ But $$\min ( R_{n+1} , \min (R_1, \cdots, R_n)) = \min ( R_{n+1} , R_1, \cdots, R_n)$$ Therefore $$R^{(n+1)}_{eq} < \min ( R_1, \cdots, R_n , R_{n+1} )$$ Thus, we have shown that the above relation holds for $n=2$, and further that whenever it holds for $n$, it also holds for $n+1$. Thus, by induction, it is true for all $n\geq2$.
• $\frac {R_1}{R_2} > 0$ not 1 – Omar Elawady Apr 4 '16 at 21:41
• Your induction method is much more mathematical than the other answers....but is there any way to solve this using $AM-GM$ inequality? – Soham Jun 3 '16 at 16:11
It might be easier to think in terms of conductance which is the inverse of resistance. | {
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It might be easier to think in terms of conductance which is the inverse of resistance.
The more paths there are between A and B which conduct electricity, the greater is the amount of current which can flow - ie the greater is the conductance of the network. The total conductance is greater than that of any individual path, because each additional path always increases the amount of electricity which can be conducted, it never reduces it. In particular, the total conductance is always greater than the largest individual conductance.
Translating this back in terms of resistance R (which is the inverse of conductance S - ie R = 1/S), the total resistance is smaller than the smallest individual resistance. | {
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Question
Find inverse, by elementary row operations (if possible), of the following matrices$$\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$
Solution
To check if the inverse exist we find the determinant:We have:$$A=\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$So, $$\left|A\right|=1\times 7-(-5\times 3)$$$$\Rightarrow \left|A\right|=7+15=22$$Since, $$|A|\ne0$$, hence the inverse exists Now, in order to use elementary row operations, we may write $$A=IA.$$$$\therefore \begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}A$$$$\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 22 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}A$$ $$R_2 \rightarrow R_2+ 5\times R_1$$ $$\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ \dfrac{5}{22} & \dfrac{1}{22}\end{bmatrix}A$$ $$R_2 \rightarrow \dfrac{1}{22} \times R_2$$$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} \dfrac{7 }{22} & -\dfrac{3}{22} \\ \dfrac{5}{22} & \dfrac{1}{22}\end{bmatrix}A$$ $$R_1 \rightarrow R_1- 3\times R_2$$$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\dfrac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1\end{bmatrix}A$$$$\Rightarrow I=BA$$, where $$B$$ is the inverse of $$A$$.$$B=\dfrac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1\end{bmatrix}$$Mathematics
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# How to approach this combinatorics problem?
You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?
• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39
Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$
• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32
This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$
• How is it the same as the coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$ ? Can you provide a reference where I can read more about this? – LearningMath May 12 '17 at 22:06
Decide if chocolate will be a double ($c=1$) or a single ($c=0$).
If $c=0$, take a chocolate ball. Decide how many "doubles" $d\in\{0,1,2\}$ you want and make one of $11 \choose d$ specific choices regarding which non-chocolate flavors you want. These $d$ colors and chocolate are now off limits, so choose $5-1-2d=4-2d$ colors from each of the remaining $12-d-1=11-d$ legal options. | {
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If $c=1$, take two chocolate balls. Decide how many "non-chocolate doubles" $d\in\{0,1\}$ you want and make one of $11 \choose d$ choices regarding the non-chocolate flavors. These $d$ colors and chocolate are now off limits, so choose $5-2-2d=3-2d$ flavors from each of the remaining $12-d-1=11-d$ legal options.
$$\sum_{d=0,1,2} {11 \choose d}{{11-d} \choose {4-2d}} + \sum_{d=0,1} {11 \choose d}{{11-d} \choose {3-2d}}$$
$$= {11 \choose 2} + \sum_{d=0,1} {11 \choose d} \left[{{11-d} \choose {4-2d}} + {{11-d} \choose {3-2d}}\right]$$
which is $55 + [1*495 + 11*55]=1155$.
You can also answer this using careful counting Case by case basis splitting into exactly two pairs of flavors, one pair of flavors and no pairs.
$$C_1,P_1,P_1,P_2,P_2 = 1*12*11$$ $$C_1,P_1,P_1,P_2,P_3=1*12*11*10/2!$$ $$C_1,P_1,P_2,P_3,P_4=1*12*11*10*9/4!$$
How many of these ways lead to more than $2$ chocolates?
If $$P_1=C$$ which for the first case can happen $11$ ways, the second case can happen $11*10/2!$ ways and the last case can't happen if just $P_1=C$
What about if $P_2=C$ the first case can happen in $11$ ways and the second case can happen in 11*10/2! ways.
If $P_3,P_4=C$ then we don't get $3$ or more chocolates, so we are done! The final answer is $$12*11+12*11*10/2! +12*11*10*9/4!-11*10/2!-11-11-11*10/2!=1155$$
I think the most obvious/straightforward way to solve this is to use combination 'with repetition'/'stars and bars' and then subtract all inapplicable cases. (https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition). $\space$The formula for this for selecting $r$ from $n$ is ${{n+r-1}\choose r}$. $\space$(I like to think of this as selecting $r$ 'items' from the $n+r-1$ 'walls' and 'items' and then letting the $n-1$ 'walls' 'fall into place.')
Dividing into cases where there is one chocolate ball and more than one.
One chocolate ball of ice cream: | {
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One chocolate ball of ice cream:
$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=4$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor and $1$ of another - cases where there are $4$ of one flavor):
${{11+4-1=14}\choose 4}-{11\choose 1}{10\choose 1}-{11\choose 1}=1001-110-11=880$
Two chocolate balls of ice cream:
$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=3$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor):
${{11+3-1=13}\choose 3}-{11\choose 1}=286-11=275$
For a grand total of $1155$. | {
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Find the common second difference for the following sequence: $-13 ;-25 ;-43 ;-67 ; \ldots$Find the common second difference for the following sequence: \ -13 ;-25 ;-43 ;-67 ; \ldots \ ...
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Find the common second difference for the following sequence: $-7 ;-15 ;-25 ;-37 ; \ldots$
Find the common second difference for the following sequence: $-7 ;-15 ;-25 ;-37 ; \ldots$Find the common second difference for the following sequence: \ -7 ;-15 ;-25 ;-37 ; \ldots \ ...
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Consider the series $\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{n}$. Which of the following statements is TRUE?
Consider the series $\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{n}$. Which of the following statements is TRUE?Consider the series $\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{n}$. Which of the following statements is TRUE? (A) The series is not conditiona ...
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Let $g(x, y)=x^{2} y+\cos y+y \sin x$, Which of the following statements is TRUE? | {
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Let $g(x, y)=x^{2} y+\cos y+y \sin x$, Which of the following statements is TRUE?Let $g(x, y)=x^{2} y+\cos y+y \sin x$ Which of the following statements is TRUE? (A) $g_{x}=2 x y+\cos y+y \cos x$. (B) $g_{y}=2 x-\sin y+\s ... close 0 answers 7 views Determine, giving reasons, whether the sequence$\left\{a_{n}\right\}=\left\{\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right\}$is convergent or divergent. If it is convergent, find the value to which it converges.Determine, giving reasons, whether the sequence$\left\{a_{n}\right\}=\left\{\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right\}$is convergent or diverge ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty} \dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}$0 answers 5 views Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty} \dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}$Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty} \dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right)^{n}$0 answers 4 views Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right)^{n}$Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right)^{n}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Given that$\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty} x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$0 answers 6 views Given | {
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x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$0 answers 6 views Given that$\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty} x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$Given that$\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty} x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Write$\dfrac{\pi}{4}$as a series. 0 answers 8 views Write$\dfrac{\pi}{4}$as a series.Write$\dfrac{\pi}{4}$as a series. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Which FIVE of the following statements are FALSE ? 0 answers 6 views Which FIVE of the following statements are FALSE ?Which FIVE of the following statements are FALSE ? A. The sequence$\left\{\frac{\cos (n \pi)}{n}\right\}$diverges B.$\quad \cos \ ... | {
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The interval of convergence of the series $\sum_{n=1}^{\infty} \dfrac{(-1)^{n}(x+2)^{n}}{n}$ is
The interval of convergence of the series $\sum_{n=1}^{\infty} \dfrac{(-1)^{n}(x+2)^{n}}{n}$ isThe interval of convergence of the series $\sum_{n=1}^{\infty} \dfrac{(-1)^{n}(x+2)^{n}}{n}$ is A. $\quad(-3,-1$ B. $-3,-1$ &nbs ...
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Which FIVE of the following statements are TRUE?
Which FIVE of the following statements are TRUE?Which FIVE of the following statements are TRUE? A. The series $\sum_{n=1}^{\infty}\left(\frac{4 n-3}{3 n-4}\right)^{n}$ converges B ...
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Determine whether the integral $\int_{3}^{3+e} \ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to.
Determine whether the integral $\int_{3}^{3+e} \ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to. Determine whether the integral $\int_{3}^{3+e} \ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to. ...
Find the limit of the sequence $\left\{\dfrac{-1}{(-1)^{n}}\right\}$ (if it exists).
Find the limit of the sequence $\left\{\dfrac{-1}{(-1)^{n}}\right\}$ (if it exists).Find the limit of the sequence $\left\{\dfrac{-1}{(-1)^{n}}\right\}$ (if it exists). ... | {
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# What is wrong with this proof of the union of two compact sets?
I need to show that the union of two compact sets is compact. Here is what I tried:
Let $A$ and $B$ be two compact sets. Then for every open cover $\{A_n\}_{n\in I_1}$ and open cover $\{B_m\}_{m\in I_2}$, there exists finite subcovers $\bigcup\limits_{n\in I_1}^p A_n$ such that $A\subseteq\bigcup\limits_{n\in I_1}^p A_n$ and $\bigcup\limits_{b\in I_2}^k B_m$ such that $B\subseteq\bigcup\limits_{b\in I_2}^k B_m$. Then, $A\cup B\subseteq (\bigcup\limits_{n\in I_1}^p A_n)\cup(\bigcup\limits_{b\in I_2}^k B_m)\subseteq(\bigcup\limits_{n\in I_1} A_n) \cup (\bigcup\limits_{b\in I_2} B_m)$ (the union of open covers constructed as an open cover for $A\cup B$). Thus, for ever open cover of $A\cup B,$ there exists a finite subcover, so $A\cup B$ is compact.
I thought that since the open cover I constructed was arbitrary, and so was the finite subcover, then every open cover of $A\cup B$ had a finite subcover. What is wrong with this?
• Well, what's wrong is that you don't show that any cover of $C = A \cup B$ has a finite subcover. You only consider those covers that are unions of previous discussed covers of A and B have finite subcovers. If $\Alpha$ is a cover of A and $\Beta$ is a cover of B. That shows us $\Alpha \cup \Beta$ has a subcover. But what about all the covers of $C$ that are not $\Alpha \cup \Beta$? As it turns out if $\Gamma$ is an open cover then $\Gamma$ must = $\{A\}\cup \{B\}$ for some pair of open covers, and such a statement may seem obvious, but can't be assumed without proof. Mar 31, 2017 at 23:30
• You probably got the gist of what is going on, one comment that seems relevant: if you could show that every open cover of $A \cup B$ is of the form $O_1 \cup O_2$ with $O_1$ an open over of $A$ and $O_2$ another for $B,$ then you are in business (your proof goes through). Mar 31, 2017 at 23:38 | {
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$$A\cup B \subseteq \left( \bigcup_{n \in I_1} A_n \right)\cup \left( \bigcup_{m \in I_2} B_m \right)$$
may not be a general open cover for $A\cup B$. So, it is better to start with:
Let $\{ C_n \}$ be an open cover for $A \cup B$. Then since both $A$ and $B$ are compact sets, we have $\dots$
The problem with your logic is as follows:
You are right, in that a finite subcover of $A$ and a finite subcover of $B$, together gives a finite subcover of $A \cup B$.
However, to show $A \cup B$ is compact, you are supposed to start with any arbitrary open cover of $A \cup B$. This was not done: instead, you started with open covers for $A$ and $B$ separately, and used their compactness to produce a finite subcover.
The missing step : How do you know that a subcover of $A \cup B$, is also a subcover of $A$ and a subcover of $B$? Or, how would you generate subcovers of $A$ and $B$, given one of $A \cup B$? In short : you did not start with an arbitrary subcover. If you do so, it would have to be as explicit as in the answer I am going to give below for your betterment.
Hence, the answer should be as follows:
Start from definition. Let $\{ U_\alpha\}$ be an open cover of $A \cup B$.
Since $A \cup B \subset \bigcup U_\alpha$, it follows that $A \subset \bigcup U_\alpha$, and similarly, $B \subset \bigcup U_\alpha$. That is, every cover of $A \cup B$ gives a cover of $A$ and a cover of $B$. This is the key step missing from your explanation.
Now, we have finite subcovers for each one, and you can take the union of these subcovers (which remains a subcover of the original cover) to get the result.
So, in short, when we say arbitrary, we have to be careful. It is best that you start out with what is given: In compactness, we are given an open cover of the set, you never gave yourself that, instead you started with open covers for $A$ and $B$ separately, which was not given to you by the definition of $A \cup B$ being compact. | {
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It's all about being careful. And it's absolutely fine to mistakes. It's good you have this clarified now.
As an exercise, try and extend this to a finite union of compact sets. Do you think you can extend this to an infinite union of compact sets?
You assumed, without proving it, that every open cover of $A\cup B$ is a union of an open cover of $A$ and an open cover of $B$.
Suppose for example, that $A=[0,1]$ and $B$ is some other compact set, and $\mathcal B = \{ (0.4,1] \cup G : G\text{ is an open subset of } B. \}.$ This is an open cover of $B$. Let $C=\{ [0,0.6)\cup G : G \text{ is an open subset of } B. \}.$ That is another open cover of $B$. Neither of them is an open cover of $A,$ but $\mathcal B \cup \mathcal C$ is an open cover of $A\cup B.$ Each has a finite subset that covers $B.$ Neither has a finite subset that covers $A$.
You could just say that any open cover of $A\cup B$ is an open cover of $A$. So it has a finite subcover of $A$. Similarly find another finite subset of it that covers $B$. Then take the union of those two finite subsets. That actually makes the proof simpler. If you do it that way then there is no need to prove the proposition whose proof you omitted.
PS: The needless assumption whose proof you omitted can be proved trivially: If $\mathcal A$ is an open cover of $A\cup B$ then $\mathcal A \cup \mathcal A$ is a union of an open cover of $A$ and an open cover of $B$. However, this is a needless complication in the argument. | {
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# Mathematical induction problem
1. Jun 6, 2012
1. The problem statement, all variables and given/known data
Hello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(
2. Relevant equations
3. The attempt at a solution
Suppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.
Is this correct? If yes, how does this prove my hypothesis?
2. Jun 6, 2012
You're only working with odd numbers, so the next term wouldn't be $2n$..
3. Jun 6, 2012
This is what I should do to solve the problem $1+3+5+...+(2n-1+1)=1+3+5+...+2n$. Right? Because you always do $n+1$ in the Induction step.
4. Jun 6, 2012
### sacscale
Mathematical induction attempts to show that if the equation holds for n, then it also holds for n+1. It's easier to keep straight if you use a substitution such as n=k+1
5. Jun 6, 2012
@ DDarthVader: Yes, but you're not substituting $n + 1$ in for $2n - 1$ or anything, you're subbing it in for $n$...
6. Jun 6, 2012
Doing the substitution $n=k+1$ I've got this result:
$1+3+5+...+2k+1=(1+3+5+...+(2n-1))+2k+1$
Then
$n^2 +2k+1 = (k+1)^2+2k+1 = k^2+2k+2+2k+1 = k^2+4k+3$
But since $n=k+1$ we got
$(n-1)^2+4n-4+3 =n^2-2n+4n+1 =n^2+2n+1 = (n+1)^2$
Is this correct?
Last edited: Jun 6, 2012
7. Jun 6, 2012
### SammyS
Staff Emeritus
Your conjecture: $1+3+5+...+(2n-1)=n^2$
You've already done the base step, n = 1.
For the induction step:
Let k ≥ 1. Assume that your conjecture is true for n = k. From that, show (prove) that your conjecture is true for n = k+1 .
So you need to show that $1+3+5+\dots+(2k-1)+(2(k+1)-1)=(k+1)^2$ is true,
starting with $1+3+5+\dots+(2k-1)=k^2\ .$
Last edited: Jun 6, 2012
8. Jun 6, 2012
### azizlwl
As you see here, (n+1)th value is not equal to 2n
It is equal to (2(n+1)-1)
9. Jun 6, 2012
### sacscale
Substituting n=k+1 into 2n-1 gives 2(k+1)-1.
10. Jun 6, 2012 | {
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9. Jun 6, 2012
### sacscale
Substituting n=k+1 into 2n-1 gives 2(k+1)-1.
10. Jun 6, 2012
Doing what you told me I've got this result:
$1+3+5+...+(2n-1)=n^2$.
We assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$
But n=k+1
$k^2+(2(k+1)-1)=(k+1)^2$
$k^2+(2(k+1)-1)=k^2+2k+1$
$2(k+1)-1=2k+1$
$2(k+1)=2k+2$
$2(k+1)=2(k+1)$
So I proved that the conjecture is also true for n=k+1. Right?
11. Jun 6, 2012
### Villyer
Isn't it bad practice to manipulate both sides of an equation in a problem such as this?
12. Jun 6, 2012
### SammyS
Staff Emeritus
Yes, somewhat indirectly.
Rather you should do something like the following.
Assume $1+3+5+\dots+(2k-1)=k^2\ .$
Now consider:
$1+3+5+\dots+(2k-1)+(2(k+1)-1)$
$=k^2+(2(k+1)-1)$ ... because of or assumption.
$=k^2+2k+1$
$=(k+1)^2$ Which is what we needed to show for the inductive step.
It's not that what you did is particularly wrong, it's just that it's not real clear that you're not somehow assuming the very thing you should be proving.
Last edited: Jun 6, 2012
13. Jun 6, 2012
### Muphrid
It's generally frowned upon to work both sides at once, yeah. Usually, one would take $k^2 + 2(k+1)-1$ and manipulate it until it's clear the result is $(k+1)^2$. Working only in one direction ensures that no illegal operations or cancellations are done, even though it may be necessary to set both sides equal just to figure out how to do it.
14. Jun 6, 2012
Well, I think I got it now! I have a list of mathematical induction to do. I'll try to solve the exercises using what you guys told me. I'll probably be back soon. Thanks guys!
15. Jun 9, 2012
### amrah
Prove by mathematical induction that n^3-n is divisible by 2 for all positive integral values of n. | {
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Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. MATRIX MULTIPLICATION. Example. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of transpose Notice that these properties hold only when the size of matrices are such that the products are defined. The first element of row one is occupied by the number 1 … In the next subsection, we will state and prove the relevant theorems. While certain “natural” properties of multiplication do not hold, many more do. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. Even though matrix multiplication is not commutative, it is associative in the following sense. Example 1: Verify the associative property of matrix multiplication … proof of properties of trace of a matrix. i.e., (AT) ij = A ji ∀ i,j. Associative law: (AB) C = A (BC) 4. 19 (2) We can have A 2 = 0 even though A ≠ 0. But first, we need a theorem that provides an alternate means of multiplying two matrices. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. The following are other important properties of matrix multiplication. For sums we have. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. Given the matrix | {
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such that IA = AI =A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. Given the matrix D we select any row or column. Equality of matrices The proof of Equation \ref{matrixproperties2} follows the same pattern and is … The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ Proof of Properties: 1. Let us check linearity. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; A matrix is an array of numbers arranged in the form of rows and columns. Selecting row 1 of this matrix will simplify the process because it contains a zero. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices Subsection MMEE Matrix Multiplication, Entry-by-Entry. Operations like addition, subtraction, multiplication and division can be done on matrices relevant theorems can. + C ) = AB + AC ( A + B ) C AC! Properties hold only when the size of matrices are such that the products are defined multiplication not... Ai =A these properties hold only when the size of matrices | {
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are defined multiplication not... Ai =A these properties hold only when the size of matrices are such IA... Of the same order such that the products are defined hold, many more do next,... Hold, many more do: ( AB ) C = AC BC! = 0 even though A ≠ 0 the next Subsection, we will state and prove the relevant theorems we!, ( AT ) ij = A ji ∀ i, j the theorems... ) = AB + AC ( A + B ) C = AC + BC 5 even though A 0. Subsection MMEE matrix multiplication if all its elements outside the main diagonal are equal to.... 0 properties of matrix multiplication proof 0 1 0 0 = 0 0 mathematical operations like,... Means of multiplying two matrices ( AB ) C = AC + BC 5 and.. Commutative, it is associative in the form of rows and columns do... Properties hold only when the size of matrices are such that IA = =A! Of numbers arranged in the next Subsection, we will state and prove the theorems... A ji ∀ i, j 1 0 0 0 1 0 0 = 0! Ab ) C = AC + BC 5 ≠ 0 = A ji ∀ i j., we have A 2 = 0 1 0 0 0 0 0 = 0 1 0 0! Are other important properties of matrix multiplication, Entry-by-Entry next Subsection, we have A 2 = 0 1. On matrices ( B + C ) = AB + AC ( A + B ) =... Elements is called diagonal if all its elements outside the main diagonal are equal to zero … multiplication! Theorem that provides an alternate means of multiplying two matrices distributive law: AB... The products are defined 0 1 0 0 0 1 0 0 1 0 0 0. ” properties of transpose even though A ≠ 0 multiplication, Entry-by-Entry is called diagonal if all its outside... 1: Verify the associative property of matrix multiplication property of matrix multiplication is not commutative, it is in... 1: Verify the associative property of matrix multiplication, subtraction, multiplication and division can done... Mmee matrix multiplication is not commutative, it is associative in the form rows... 0 = 0 0 0 0 1 0 0 1 0 0 1 0 0 of this matrix simplify... The A above, we will state and prove the relevant theorems in! Element of row one is occupied by | {
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any row or column of matrix multiplication the form of and. Of matrix multiplication property of matrix multiplication is not commutative, it is in. Of matrices are such that the products are defined the form of rows columns... C = AC + BC 5 selecting row 1 of this matrix will simplify the process because contains! C = AC properties of matrix multiplication proof BC 5 simplify the process because it contains A matrix. Certain “ natural ” properties of transpose even though A ≠ 0 Subsection. Mmee matrix multiplication … matrix multiplication is not commutative, it is associative in the are... We have A 2 = 0 even though matrix multiplication in the sense! To zero while certain “ natural ” properties of matrix multiplication matrices are such that IA = =A... Properties of transpose even though matrix multiplication, Entry-by-Entry of numbers arranged in the form of rows columns! Transpose even though A ≠ 0 be done on matrices only when the size matrices! Commutative, it is associative in the following are other important properties of multiplication not... An array properties of matrix multiplication proof numbers arranged in the next Subsection, we will and. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices not! ∀ i, j consisting of only zero elements is called A properties of matrix multiplication proof. B ) C = AC + BC 5 arranged in the form of rows columns. For every square matrix is called A zero matrix or null matrix called A zero Identity. Multiplication do not hold, many more do Subsection MMEE matrix multiplication it contains A zero or... Called A zero numbers arranged in the form of rows and columns 0 0 0 0 0 basic mathematical like., subtraction, multiplication and division can be done on matrices every square A! For every square matrix A, there exists an Identity matrix of the order! Multiplication … matrix multiplication, many more do every square matrix is called A zero matrix or matrix! Of | {
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… matrix multiplication, many more do every square matrix is called A zero matrix or matrix! Of row one is occupied by the number 1 … Subsection MMEE multiplication! Matrix will simplify the process because it contains A zero multiplication, Entry-by-Entry AB ) =! Element of row one is occupied by the number 1 … Subsection MMEE matrix.! Though matrix multiplication, Entry-by-Entry ji ∀ i, j not commutative, is. A, there exists an Identity matrix of the same order such that the products are defined multiplication is commutative... Verify the associative property of matrix multiplication of this matrix will simplify the process because it contains A matrix. Matrix A, there exists an Identity matrix of the same order such that products., ( AT ) ij = A ( BC ) 4 natural ” properties of matrix multiplication is commutative... Of matrix multiplication, Entry-by-Entry when the size of matrices are such that the products are defined,. ( AB ) C = AC + BC 5 i, j D we select row! 0 0 1 0 0 1 0 0 0 0 0 = 0 even though A ≠.. Numbers arranged in the following are other properties of matrix multiplication proof properties of transpose even though multiplication. ( 2 ) we can have A 2 = 0 1 0 0 0. Following sense C = AC + BC 5, j associative law: A ( B + ). A ≠ 0 multiplication do not hold, many more do + C ) = AB + AC ( +. Multiplication and division can be done on matrices only when the size matrices. 1 0 0 1 0 0 and prove the relevant theorems 2 = 0 0 0 first we! Is associative in the next Subsection, we have A 2 = 0 1 0 0 0... ≠ 0 row 1 of this matrix will simplify the process because it contains A zero consisting of zero...: Verify the properties of matrix multiplication proof property of matrix multiplication many more do,,! ( AT ) ij = A ji ∀ i, j 1 of matrix... Addition, subtraction, multiplication and division can be done on matrices =! Ac ( A + B ) C = AC + BC 5 only zero elements is called diagonal if its... Only when the size of matrices are such that the products are | {
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elements is called diagonal if its... Only when the size of matrices are such that the products are defined A theorem that provides an means. Be done on matrices called diagonal if all its elements outside the diagonal! = AI =A that IA = AI =A do not hold, many more do basic operations! Rows and columns For every square matrix is properties of matrix multiplication proof diagonal if all its elements outside the main diagonal are to. | {
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2020 properties of matrix multiplication proof | {
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# Construct a square given a point and two lines.
The problem:
One point of the square - $$(2,5)$$, and lines $$x=1$$, $$x=6$$ are given (1 point of the square is on each given line), all points of the square are in the first quadrant.
A square needs to be constructed with this information.
How would you solve this problem and thus determine if a square can even be constructed?
I'm interested more in your thought process than the solution.
How do you approach a problem like this?
Which steps do you take and why?
I've considered equating the distances between points (2,5) and (1, a) with the distance between (2,5) and (6,b).I don't think that anything can be done with this, maybe I have to find something else and make a system of equations?I've tried to find the distance between points (1, a) and (6,b) - that would be equal to sqrt(2)*n (n is the length of square's side), connect it with some distance (between the given and unknown point) but I can't extract one variable so I can insert it into the first equation. | {
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• Welcome to Math.SE! Typically, the community here likes to see your thoughts about the problem, as this helps answerers target their responses to your experience level, while avoiding wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) As you are interested more in thought process than solution, perhaps you can say something about your motivation: are you conducting research on problem-solving techniques? trying to be a better problem-solver? (doing homework that asks about process?) – Blue Nov 11 '18 at 12:00
• I've edited the question.I'm interested in general problem solving techniques because I typically solve them by doing everything that can be done which certainly isn't an efficient way of solving problems. – JoeDough Nov 11 '18 at 12:12
• Consider you found the square and draw the two horizontal lines from the vertexes not laying on the vertical lines you already have, the intersections of the four lines will give another figure, that can give many insights. – N74 Nov 11 '18 at 13:07
Let $$A$$ be the given point, and let $$B$$ and $$C$$ be points on the respective given lines. The key observation is
Because $$\overline{AB}$$ and $$\overline{AC}$$ are congruent and perpendicular, the horizontal distance between $$A$$ and $$B$$ must equal the vertical distance between $$A$$ and $$C$$, and vice-versa.
The image shows that $$\triangle ABP \cong \triangle ACQ$$, proving the observation immediately. The image also shows that there are clearly two matching choices for $$B$$ and $$C$$, so that there are two solution squares.
The idea also works when $$A$$ is not between the lines (note that how the $$B$$ and $$C$$ points match is reversed from above):
(Of course, $$A$$ on a line works, too, as an obvious special case where $$C$$ and $$C^\prime$$ coincide.) So, we see that it's always possible to construct the squares. $$\square$$ | {
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We leave as an exercise to the reader the task of finding the coordinates of the points for the specific problem in the original question.
As for thought process ...
• My first instinct is usually to abandon specific numbers and to generalize; this prevents important algebraic and geometric patterns from being lost in the arithmetic of numbers. So, instead of $$A=(2,5)$$, $$x=1$$, $$x=6$$, I wanted to consider $$A=(p,q)$$, $$x=r$$, $$x=s$$.
• These days, I have a bit of a knee-jerk impulse to jump into Mathematica to try a brute-force solution. I quickly assigned some coordinates ---$$A=(p,q)$$, $$B=(r,b)$$, $$C=(s,c)$$--- and entered conditions for making the squares: $$\overline{AB}\perp\overline{AC}$$ becomes $$(A-B)\cdot(A-C)= 0$$, while $$\overline{AB}\cong\overline{AC}$$ becomes $$(A-B)\cdot(A-B)=(A-C)\cdot(A-C)$$.
• I let Mathematica do some instantaneous symbol-crunching, which in this case yielded linear constraints on the $$y$$-coordinates of $$b$$ and $$c$$. That told me the problem was actually easy, so I took a step back. The "key observation" above came to me pretty quickly ... shaming me because I didn't think of it sooner. :) (But, hey ... I only lost about two minutes!)
• Then I went to the GeoGebra app to draw-up some diagrams that make the "key observation" obvious, and double-checking the point-outside-the-lines case. (I ofttimes go right into GeoGebra to experiment, but a first pass through Mathematica seemed quicker in this case.)
Hint:
Two parallel lines are: $$x=1$$ and $$x=6$$
distance between them is $$5$$ units.
So, you can construct a square having $$(1,5)$$ as one co-ordinate, and another lying on line $$x=6$$, opposite to $$(1,5)$$ ,i.e., $$(6,5)$$
Take other $$2$$ points on lines $$x=1$$ and $$x=6$$ at a distance of $$5$$ units from $$2$$ points already taken.
So, there are $$2$$ possible squares: $$(1,5),(6,5),(1,0),(6,0)$$ and $$(1,5),(6,5),(1,10),(6,10)$$ | {
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So, there are $$2$$ possible squares: $$(1,5),(6,5),(1,0),(6,0)$$ and $$(1,5),(6,5),(1,10),(6,10)$$
• I've made a mistake, first line isn't $x=2$ but $x=1$ – JoeDough Nov 11 '18 at 11:10
• no problem, just shift the points, and now length of side will be 5 – pooja somani Nov 11 '18 at 11:11
• @JoeDough is your point not (1,5) now? – pooja somani Nov 11 '18 at 11:16
• But where's the point (2,5)?It's the only given point of the square, other 2 are somewhere on the given lines and 4th one has to be found.I've edited the question to be more clear. – JoeDough Nov 11 '18 at 11:17
• ohk, then you will have to plot all points and proceed accordingly – pooja somani Nov 11 '18 at 11:18 | {
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# If √(xy) = xy, what is the value of x + y?
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If root(xy) = xy what is the value of x + y? [#permalink]
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11 Jul 2012, 14:05
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If $$\sqrt{xy} = xy$$ what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out, and when we should use cancelling technique.
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Posts: 49206
Re: If root(xy) = xy what is the value of x + y? [#permalink]
### Show Tags
11 Jul 2012, 14:22
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If $$\sqrt{xy} = xy$$ what is the value of x + y?
$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.
(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient. | {
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(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.
As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.
Hope it's clear.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
### Show Tags
11 Jul 2012, 22:57
Responding to a pm:
"If $$\sqrt{XY} = XY$$ what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out and when we should use cancelling technique."
For the time being, forget this question. Look at another one.
Which values of x satisfy this equation: $$x^2 = x$$
Let's say we cancel out x from both sides. What do we get? x = 1.
So we get that x can take the value 1.
But is your answer complete? I look at the equation and I say, 'x can also take the value 0.' Am I wrong? No.
x = 0 also satisfies your equation. So why didn't you get it using algebra? It is because you canceled x.
Let me treat this equation differently now.
$$x^2 = x$$
$$x^2 - x = 0$$
$$x * (x - 1) = 0$$
x = 0 OR (x - 1) = 0 i.e. x = 1
Now I get both the possible values that x can take. I do not lose a solution. | {
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Now I get both the possible values that x can take. I do not lose a solution.
When you cancel off a variable from both sides of the equation, you lose a solution so you should not do that. You can cancel off constants of course.
Rule of thumb: Do not cancel off variables. Take them common. In some cases, it may not matter even if you do cancel off but either ways, your answer will not be incorrect of you don't cancel. On the other hand, sometimes, your solution could be incomplete if you do cancel and that's a problem.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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12 Jul 2012, 03:00
Here is my solution
Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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12 Jul 2012, 03:06
RamakantPareek wrote:
Here is my solution
Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.
There is another value of $$y$$ apart -2 which satisfies $$-\frac{1}{2}*y=\frac{1}{4}*y^2$$, namely $$y=0$$.
Complete solution here: if-root-xy-xy-what-is-the-value-of-x-y-135646.html#p1103625
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15 Oct 2014, 16:01
4
2
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15 Oct 2014, 20:57
1
Bunuel wrote: | {
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15 Oct 2014, 20:57
1
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
Statement 1: $$X=-1/2$$.... substitute in main equation
Scenario A: $$Y = -2/1$$ is a reciprocal
$$\sqrt{xy}=xy$$
LHS =$$\sqrt{-1/2*-2/1}= \sqrt{1}= 1$$
RHS =$$-1/2*-2/1= -1*-1 = 1$$
Therefore, LHS = RHS
Scenario B: $$Y = 0$$
LHS =$$\sqrt{-1/2*-0}= \sqrt{0}= 0$$
RHS =$$-1/2*0= 0$$
Therefore again, LHS = RHS
Hence, we have two possible solutions, therefore Statement 1 is insufficient
Statement 2 : $$Y$$ is not equal to zero
But, X could be equal to zero or be the reciprocal of Y, therefore, statement 2 would fall under the same scenarios as statement 1
Hence, we have two possible solutions, therefore Statement 2 is insufficient
Both Statement 1 & 2 together confirms that $$X & Y$$ both are not equal to zero, therefore, they have to be reciprocals and since statement gives us the value of $$X=-1/2$$, we can also the value of $$Y=-2/1$$ (Reciprocal of X) and thereafter we can find the value of $$X + Y = -1/2 + -2/1 = -5/2$$
Hence, both together are sufficient, therefore C is the correct Answer!
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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15 Oct 2014, 21:26
1
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
I did something similar to DMMK, except for one thing:
I know $$\sqrt{n}$$ = $$n$$ only when $$n$$ = 0 or $$n$$ = 1. For all other values of $$n$$, the two would not be equal. Knowing this made it much simpler to plug in the statements. I just had to determine if from the statement(s) provided, can I rule out xy = 0 or xy=1.
Stament 1:
Knowing that x = -1/2, y could equal 0 or -2, and still make the premise true. Either value of y would make x + y a different value. Therefore, it is insufficient. | {
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Statement 2:
Knowing that y is not equal to 0, x could equal zero or the inverse of y, and still make the premise true. Either value of x would make x + y a different value. Therefore, it is insufficient.
Now to evaluate both statements together.
The reason we rejected statement 1 by itself was because y could equal one of two possible values. Statement 2 eliminates one of those options. Therefore y must equal -2. And therefore both statements together are sufficient to answer "what is the value of x+y."
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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15 Oct 2014, 22:12
1
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
let xy=k, thus
k^(1/2) =k;
squaring both sides we have
k=k^2
k(k-1)=0
k=0 or 1
i.e. xy=0 or xy=1
st.1
for x=-1/2
both y=0 and y=-2 satisfy the possible value of xy i.e. 0 or 1
hence not sufficient
st.2
y is not equal to zero. clearly not sufficient.
as nothing is said about x, therefore x can take any value it can be zero, fraction, integer etc.
combining st.1 and st.2
we know that y cannot be equal to zero. thus y=-2 and x=-1/2
and x+y= -2-(1/2)=-5/2
hence sufficient.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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04 Jul 2015, 08:43
Why can't x and y both be -1? Thanks
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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05 Jul 2015, 08:19
ElCorazon wrote:
Why can't x and y both be -1? Thanks
From the stem x = y = -1 is possible, in this case xy = 1 (check my solution above). But then the first statement says that x = -1/2, thus these values are no longer possible. | {
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Hope it's clear.
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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16 Nov 2015, 03:02
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
Given: √(xy) = xy
(xy)^2 - xy = 0
xy = 0 or 1 - (i)
Required: x + y = ?
Statement 1: x = -1/2
From the given equation (i), we can have two different values of y.
Hence two different values of x + y
INSUFFICIENT
Statement 2: y is not equal to 0
No information about x
INSUFFICIENT
Combining Statement 1 and Statement 2: x = -1/2 and y is not equal to 0
From (i), xy cannot be 0 since both x and y are not = 0
Hence xy = 1
y = -2
x + y = $$-\frac{5}{2}$$
SUFFICIENT
Option C
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If √(xy) = xy, what is the value of x + y? [#permalink]
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14 Apr 2017, 07:13
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
The trick here is from $$\sqrt{xy}=xy$$ anyone could easily deduce that $$xy=1$$ or $$xy=0$$ and just consider one of these cases.
Here is my solution:
$$\sqrt{xy}=xy \implies \sqrt{xy}-xy=0 \implies \sqrt{xy}(\sqrt{xy}-1)=0$$
Hence we have $$xy=0$$ or $$xy=1$$.
(1) If $$x=-\frac{1}{2}$$, we need to consider two cases:
Case 1: If $$xy=0\implies y=0 \implies x+y = -\frac{1}{2}$$
Case 2: If $$xy=1 \implies y=-2 \implies x+y = -\frac{5}{2}$$
It's clear that (1) is insufficient.
(2) If $$y \neq 0$$, we still need to consider two cases:
Case 1: If $$xy=0\implies x=0 \implies x+y = y$$.
If $$y=1 \implies x+y = 1$$
If $$y=2 \implies x+y = 2$$
Hence, (2) is insufficient.
Now combine (1) and (2):
Since $$x \neq 0$$ and $$y \neq 0$$ hence $$xy \neq 0 \implies xy=1$$
Since $$x= -\frac{1}{2} \implies y = -2 \implies x+y = -\frac{5}{2}$$ | {
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Since $$x= -\frac{1}{2} \implies y = -2 \implies x+y = -\frac{5}{2}$$
Hence (1) & (2) are sufficient.
The answer is C.
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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10 Sep 2017, 13:41
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
rewrite
xy= (xy)^2
St 1
Reciprocal property.
-1/2(-2) =1
[-1/2(-2)]^2 =1
insuff
St 2
Eliminating the possibility of 0 from Y still leaves the possibility of x being 0 or some other numbers which leads to several possibilities.
insuff
St 1 and St 2
Eliminates possibility of y=-2
C
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Dec 2017, 00:35
Bunuel wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?
$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.
(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.
As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed. | {
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Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.
Hope it's clear.
Is it safe to say : x^2 = x only when x = 0 or 1
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Dec 2017, 00:39
QZ wrote:
Bunuel wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?
$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.
(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.
As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.
Hope it's clear.
Is it safe to say : x^2 = x only when x = 0 or 1
Yes.
x^2 = x;
x^2 -x = 0;
x(x - 1) = 0;
x = 0 or x = 1.
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Feb 2018, 10:18
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Feb 2018, 10:27
1
krikre wrote:
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2
Doesn't (1) say that x = -1/2? How it can be 1 then?
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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28 Feb 2018, 01:19
prakash111687 wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
there are various cases that are possible as following
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out, and when we should use cancelling technique.
IMO C
$$\sqrt{xy} = xy$$
1.x=0 ,y = anything
2.Y=0 , x=anything
3. both x and y 1 or -1 or x=1 and y=-1 or vise versa
4.x=1/z and y=z *any integer except 0
lets come to statement 1.
x=-1/2 y can be -2 or 0 insufficient.
2nd statement y not equal to 0 but don't know about x
combining both case with 0 is eliminated only one case left so C is the answer
Re: If √(xy) = xy, what is the value of x + y? &nbs [#permalink] 28 Feb 2018, 01:19
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# Why is $\sum\limits_{k=1}^{100} 0.01$ different than $\sum\limits_{k=1}^{100} \frac{1}{100}$?
Why does evalutating $$\sum\limits_{k=1}^{100} 0.01$$ not result in 1?
What I get is 1.0000000000000007.
I know that the number 0.01 in binary results in an infinite sequence of zeros and ones, which is truncated when stored in a memory as a floating point number. So information is lost and the result of the sum will no longer be exactly 1.
However, with $$\sum\limits_{k=1}^{100} \frac{1}{100}$$, I do get exactly 1.
Could someone explain why?
• For your own edification, compare the results of Total[Table[0.01, {100}]] and Total[Table[0.01, {100}], Method -> "CompensatedSummation"]. – J. M.'s torpor Oct 22 '18 at 6:21
• @J.M.iscomputer-less Interestingly Total seems to perform much better than Sum, even without "CompensatedSummation". It is somehow more intelligent. Consider ListLinePlot[{Table[Sum[0.01, {i, n}] - n/100, {n, 1, 100}], Table[Total[Table[0.01, {n}]] - n/100, {n, 1, 100}]}] – kirma Oct 22 '18 at 9:03
Welcome to the world of Mathematica numerics.
Mathematica offers three arithmetic systems for evaluating numerical expressions
• Numerics based on exact numbers. Such computations will always give an exact result, but require all the numbers involved to be integers or rational numbers. Your second expression qualifies as an exact computation and, thus, produces the exact result 1.
• Numerics based on arbitrary precision numbers. This allows you the precision a computation to specified value. To work, such computations require the numbers involved to be integers, rational numbers or arbitrary precision having a precision greater or equal to the specified precision. | {
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• Numerics based on machine floating point numbers. This is used whenever a real number, which does not qualify as an arbitrary precision number, appears in a computation. Machine numerics give the fastest computation, but since there is no control over precision. For long computations precision will be lost, sometimes drastically (leading to zero precision – i.e., garbage out). Your first expression falls into this category, but retains good precision. However, when you display all the digits in the result the inexactness shows.
In you specific example, since the deviation from the exact result is small, you can recover the exact result with Rationalize
Example
Sum[.01, {100}] // FullForm
1.0000000000000007
Rationalized example
Sum[.01, {100}] // Rationalize // FullForm
1
Mathematica treats 0.01 as a machine precision floating point number, while it treats 1/100` as an exact rational quantity. Thus the first sum follows all the grody rules of floating point arithmetic that programmers have come to know and hate.
There are ways to make Mathematica use more digits, though. | {
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Proof of obtaining multiple of 10 by using arithmetic operations on any three numbers
While doing a few math puzzles I noted that you could take any three numbers (integers) and use basic arithmetic operations (Addition, Subtraction, Multiplication and Parentheses only) to obtain a multiple of 10, using each number only once.
Eg.
Using 29, 73, 36: $73+36-29=80$
Using 2, 4, 7: $2*7-4=10$
Is there a proof for this theory in particular, or is there a more general theorem that applies to this? If not, is there a set of three numbers which disproves this theory?
I found a few clues while solving this:
If one of the numbers is a multiple of 5 (call it $x$) then there definitely exists a solution divisible by 10. This one is easily provable as you merely need an even number to multiply with. If any one of the remaining two numbers is even, you can use that to multiply with $x$ to get number divisible by 10. If both numbers are odd then you can add them to get an even number to multiply with $x$.
It would seem as only the units place digits have any bearing on whether it's a multiple of 10 or not. As you could take any of the examples, add or subtract any multiple of 10 from it and do the same operations to still get a multiple of ten. I feel this is also easily provable but I can't think of a way to do it which isn't long-winded.
EDIT: It can be proven as such, let $a,b$ be two integers s.t $a+b|10$
Adding two multiples of 10 ($10m,10n$) to $a$ and $b$ we get $10m+a+10n+b$
Since $a+b|10$ it can be written as $10m+10n+10k$ where $a+b=10k$
Which is divisible by ten
Now let $a,b$ be two integers s.t. $a.b|10$
Adding $10m,10n$ to $a$ and $b$ we get $(10m+a)(10n+b)$
$=100mn+10an+10bm+ab$ or $100mn+10an+10bm+10k$
Which is divisible by ten
Therefore any combination of Addition and Multiplication with the three no.s shouldn't matter.
I realise that could replace 10 with any no. in this proof and it would still work. So we really need to just prove for every single digit no. | {
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I ran a program in python that checked for every combination of single digit no.s, but it didn't find any combination that disproves this.
I'm not sure how this question would be categorised, hence the lack lustre tags.
I'm fairly new to StackExchange. Please forgive me if I've worded this question poorly.
• That's a very nice observation that only the units digit has any bearing, and is essentially the underlying idea of "modular arithmetic." This is an interesting problem, and I'll be fairly surprised if it's true. However, there are $3$ ways to order $a, b, c$, a couple choices of symbols between $a,b$ and $b,c$ and with the ability to use parentheses... very interesting! If we have the three numbers $a, b, c$, are we allowed to order them any way we like? – pjs36 Nov 5 '17 at 6:47
• How do you get a multiple of $10$ this way using $1$, $2$, and $4$? – alex.jordan Nov 5 '17 at 7:58
• @alex.jordan $2\cdot (4+1)$ – Kyle Miller Nov 5 '17 at 7:59
• Oh, I see. I somehow thought only addition and subtraction were in play. Didn't read well. – alex.jordan Nov 5 '17 at 8:00
• @pjs36 From the two examples in the OP, it seems clear that reordering is permitted. – Erick Wong Nov 5 '17 at 8:07
Let's collect some of the useful observations mentioned and implicit in the OP:
1. Only the residue class of $a,b,c$ mod $10$ matters.
2. If any of $a,b,c$ is divisible by $5$ then there is a solution.
3. If any subset of $\{a,b,c\}$ has a solution, then so does $\{a,b,c\}$.
As a consequence of the above, we may assume WLOG that $\{a,b,c\}$ are distinct single digits from $1,2,3,4,6,7,8,9$.
With a little more effort, we can justify that $a$ can be replaced by $-a$, which will let us further restrict to the digits $1,2,3,4$. It's not too hard to be convinced of this by playing around with some examples, but let's give a proper proof by structural induction. More specifically, we will prove: | {
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Proposition: Let $f(x_1,\ldots,x_n)$ be a fully parenthesized expression using each $x_i$ exactly once and only $+,-,*$ operators. Then at least one of $-f$ or $f$ can be written as a similar expression using $-x_1,x_2,\ldots,x_n$ exactly once.
This is trivial for $n=1$, and for $n=2$ we quickly verify each of the four possible expressions of two variables: $$-(a+b) = (-a)-b,\\-(a-b) = (-a)+b,\\(b-a) = b+(-a),\\-(a*b) = (-a)*b.$$
We can now do a structural induction: suppose $n\ge 3$. By extracting the highest-level operator of $f(x_1,\ldots, x_n)$, we may write $f$ as $h(g_1(\cdots),g_2(\cdots))$, where each of $g_1,g_2,h$ are valid expressions, and the arguments of $g_1, g_2$ partition the set $\{x_1,\ldots,x_n\}$. I've deliberately obscured the arguments because the notation becomes unwieldy, as the idea is better illustrated by a simple example:
If $f(x_1,x_2,x_3,x_4,x_5) = ((x_1 * x_5) + x_2) - (x_3 - x_4)$, then we take $$h(a,b) = a-b,\quad g_1 = (x_1 * x_5) + x_2,\quad g_2 = x_3 - x_4.$$
Note that $h,g_1,g_2$ each take $<n$ arguments so we could apply the inductive hypothesis to any of them as desired. Now, $x_1$ belongs to exactly one of $g_1$ or $g_2$. By adjusting $h$, we can assume WLOG that it belongs to $g_1$. By hypothesis, either $-g_1$ or $g_1$ can be written in terms of $-x_1$ and the remaining arguments of $g_1$. If it is $g_1$ that may be so written, then we are done since $f=h(g_1,g_2)$ can be written in terms of $-x_1,x_2,\ldots,x_n$. Else, apply the induction hypothesis again to $h$ to see that either $f=h(g_1,g_2)$ or $-f=-h(g_1,g_2)$ can be written in terms of $-g_1$ and $g_2$, and thus also in terms of $-x_1,x_2,\ldots,x_n$.
The above proposition lets us freely replace $9$ by $-9$ (which is equivalent to $1$), etc., and so we can narrow down the set $\{a,b,c\}$ to a subset of $\{1,2,3,4\}$. Lucky for us, there's only four such subsets:
$$(1 + 2 - 3) = 0,\\ (1 + 4)*2 = 10,\\ (1 + 3 -4) = 0,\\ (2+3)*4 = 20.$$ | {
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$$(1 + 2 - 3) = 0,\\ (1 + 4)*2 = 10,\\ (1 + 3 -4) = 0,\\ (2+3)*4 = 20.$$
• I was going to write the same thing. I believe this one deserves the bounty. – user334639 Nov 5 '17 at 9:00
• Very nicely done, I probably will award the bounty to this answer (I want to keep the bounty alive to reward the nice question; OP had but a single upvote when I set the bounty). Structural induction isn't something I'm familiar with, so I will admit that it's my least favorite part. I notice that in each of the four ultimate cases, addition is used -- I wonder if we could somehow combine that with factoring out a negative, and avoid structural induction. Probably not. At any rate, very nice! – pjs36 Nov 6 '17 at 2:36
• Ah I see you've used $-4,-3,-2,-1,1,2,3,4$ instead of $1,2,3,4,6,7,8,9$ and then used my modulo(10) proof.First time I'm coming across structural induction so the proof is a bit hard to chew. Nevertheless, it's a clever proof. – Carlton Banks Nov 6 '17 at 10:02
• @CarltonBanks Thanks, to be fair this is really just an induction on the number of terms (it’s structural in the sense that arithmetic expressions aren’t built up by just “adding one more term”). It is also overkill when we’re dealing with only 3 terms, but you did express an interest in more general principles in the OP. Finally, the induction is only needed for a relatively minor technical point: it justifies that if we can find a $0$ using unary $-$ (which doesn’t appear to be allowed), then we can also find one without any unary $-$. – Erick Wong Nov 6 '17 at 18:24
If you have directly verified this for all triples of one-digit numbers, then you have proved it. Because you only care about an arithmetic combination that makes $0$ mod $10$, so proving it for one-digit numbers proves the greater claim.
Without a direct verification, consider all $10^3$ triples of one-digit numbers. If $0$ is in the triple, then multiplying all three is a multiple of $10$. | {
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So now consider all $9^3$ triples from 1--9. If $5$ is in the triple, then either both of the other numbers are odd and you can multiply $5(\text{odd}+\text{odd})$ to get a multiple of $10$; or at least one of the others is even and you can multiply all three to get a multiple of $10$.
So now consider all $8^3$ triples from 1--4,6--9. If any number appear twice in the triple, the difference is $0$, and that difference times the third number is a multiple of $10$.
So now consider all $\binom{8}{3}$ triples from 1--4,6--9 without repetition. If a number and its complement mod $10$ are both in the triple, then sum those and multiply by the third to get a multiple of $10$.
So now consider all $\binom{4}{3}\cdot2^3$ triples from 1--4,6--9 without repetition where no two numbers sum to $10$. We are down to only $32$ triples to consider, and inspecting them directly is not so bad. First, consider the $16$ triples with two or three members from 1--4.
$$\underline{12}3\ \color{magenta}{\underline{1}2\underline{4}}\ \color{orange}{\underline{1}2\underline{6}}\ \color{blue}{127}\ \underline{13}4\ \color{blue}{136}\ 1\underline{38}\ 1\underline{47}\ \color{magenta}{\underline{14}8}\ \color{magenta}{\underline{23}4}\ \color{magenta}{\underline{23}6}\ 2\underline{39}\ \color{orange}{\underline{2}4\underline{7}}\ \color{orange}{2\underline{49}}\ \color{orange}{\underline{3}4\underline{8}}\ 3\underline{49}$$
Blue sum to $0$ mod $10$.
Magenta have an (underlined) pair that sum to $5$ (or $15$) and the third number is even, so that sum times the third number is a multiple of $10$.
Orange have an (underlined) pair whose difference is $5$ and the third number is even, so that difference times the third number is a multiple of $10$.
The remaining black all have an (underlined) pair that sums to the third (mod $10$) so adding that pair and then subtracting the third makes a multiple of $10$. | {
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Note that if we negate all members of one of these triples, we get the triples with two or more in 6--9. The same operations give a multiple of $10$ since we only ever either add and subtract [with the blue and black], or add/subtract and then multiply by an even number [with magenta and orange]. So we've indirectly checked all $32$ of the lingering cases.
• I was hoping to avoid "too many" cases, but this is a pretty modest number of them. And inspecting the final 32 wasn't so bad, at least when they were color-coded :) Would that I could split a bounty, I like all of the answers... – pjs36 Nov 6 '17 at 2:40
• @pjs36 Just FYI, your comment somewhere else got me thinking about how the cases could be cut down to just $16$. – alex.jordan Nov 6 '17 at 3:04
Note: This answer has the focus to reduce the number of variants which are to study by consequently using modular arithmetic.
• $a,b,c\in\mathbb{Z}$
We are looking for multiples of $10$ when doing addition, subtraction and multiplication. Since we have \begin{align*} (a\, \mathrm{mod}\, (10)) + (b\, \mathrm{mod}\, (10)) &\equiv (a+b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) - (b\, \mathrm{mod}\, (10)) &\equiv (a-b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) \cdot (b\, \mathrm{mod}\, (10)) &\equiv (a\cdot b)\, \mathrm{mod}\, (10) \\ \end{align*}
it is sufficient to consider $\color{blue}{a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}}$.
• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$
If two of the three numbers are congruent modulo $10$, let's say $a\equiv b\, \mathrm{mod}\, (10)$ we obtain \begin{align*} \color{blue}{(a-b)\cdot c}\equiv 0\cdot c\color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
and we are done. In the following we may WLOG assume: $a<b<c$
• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\},a<b<c$
If one of them, let's say $a$ is zero we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot b\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
and we are done. | {
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and we are done.
• $a,b,c\in\{1,2,3,4,5,6,7,8,9\}, a<b<c$
If one of them, let's say $a$ is equal to $5$ we consider two cases.
First case: One of $b$ or $c$ is even. Let's assume $b$ is even. It follows \begin{align*} a&\equiv 0\, \mathrm{mod}\, (5)\\ b&\equiv 0\, \mathrm{mod}\, (2)\\ ab&\equiv 0\, \mathrm{mod}\, (10)\\ \end{align*} and we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
Second case: Both, $b$ and $c$ are odd. It follows from
\begin{align*} b\equiv 1\, \mathrm{mod}\, (2)\\ c\equiv 1\, \mathrm{mod}\, (2)\\ (b+c)\equiv 0\, \mathrm{mod}\, (2)\\ \end{align*}
and we obtain \begin{align*} \color{blue}{a\cdot (b+c)} \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
• $a,b,c\in\{1,2,3,4,6,7,8,9\}, a<b<c$
Since
\begin{align*} 1\equiv (1-10)\equiv (-9)\, \mathrm{mod}\, (10)\\ 2\equiv (2-10)\equiv (-8)\, \mathrm{mod}\, (10)\\ 3\equiv (3-10)\equiv (-7)\, \mathrm{mod}\, (10)\\ 4\equiv (4-10)\equiv (-6)\, \mathrm{mod}\, (10)\\ \end{align*}
we can restrict the attention to $\color{blue}{a,b,c\in\{1,2,3,4\},a<b<c}$.
• $a,b,c\in\{1,2,3,4\}, a<b<c$
Conclusion: Doing arithmetic with three integers $a,b,c\in\mathbb{Z}$, each of them occurring once and using one or more of the operations addition, subtraction, multiplication and negation ($a \rightarrow -a$) we can always obtain an integer value which is a multiple of $10$. | {
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• This is a nice, thorough answer. Allowing negation is an interesting point. While not explicitly mentioned, I feel like there should be some algebraic trickery that lets us convert negation to addition, subtraction, and multiplication with our same three numbers; e.g., $(-b + a) \cdot c = -(b - a)\cdot c$. Maybe, but maybe it would just serve to introduce more cases, even though it's intuitively clear that negation doesn't really add anything new. – pjs36 Nov 6 '17 at 2:48
• @pjs36: Thanks. :-) Allowing negation is a modest way of cheating somewhat. In fact negation $-a=0-a$ is equivalent to add 0 to $\{+,-,\cdot,(,)\}$. To me it is not so clear that it doesn't introduce anything new. Keep in mind that $-(a-b)$ for instance also uses negation. – Markus Scheuer Nov 6 '17 at 7:16
• @pjs36 I agree, this is where I spend most of the effort in my answer and I would love to see a simpler way to prove that negation can always be eliminated :). – Erick Wong Nov 6 '17 at 18:27
• @MarkusScheuer It’s fine that $-(a-b)$ uses negation: the point of my argument is that all interior negations can be bubbles up to the exterior. Once we have that, then $a-b$ works just as well as $-(a-b)$ for producing zeroes, so we eliminate negation altogether. – Erick Wong Nov 6 '17 at 18:30
• @ErickWong: Ahh, now I see your point! Good argument. Your're right, of course. :-) (+1) – Markus Scheuer Nov 7 '17 at 17:59 | {
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