Tag Archives: factoring

Infinite Ways to an Infinite Geometric Sum

One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.

STANDARD INFINITE GEOMETRIC SUM DERIVATION

My standard explanation starts with a generic infinite geometric series.

$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$  (1)

We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r.

The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r

$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$

$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$  (2)

This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.

$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$

$\displaystyle S=\frac{g}{1-r}$

STUDENT PREFERENCES

I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.

In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play.

K’s VARIATION

K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by $r^2$ also determine the summation formula?

I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.

Multiplying (1) by $r^2$ gives

$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)

and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.

$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$

$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$

$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$

That was cool, but this success meant that there were surely many more options.

EXTENDING

Why stop at multiplying by r or $r^2$?  Why not multiply both sides of (1) by a generic $r^N$ for any natural number N?   That would give

$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)

where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives

$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$

$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$  (5)

There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.

Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that

$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$.  (6)

Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$

In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice.

FOOTNOTES

1) RESTRICTING r:  Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.

For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.

The last case, $r=-1$, is more subtle.  For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.

2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group.

So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches.

Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity.

Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.

Probability, Polynomials, and Sicherman Dice

Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems.  The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics.  I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH.

BINOMIALS:  FROM POLYNOMIALS TO SAMPLE SPACES

Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students.  The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting:

$\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$

But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability?  One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.

The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times.  Each term is an event.  Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs.  The overall exponent in the expand command is the number of trials.  For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.

The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on.

FROM POLYNOMIALS TO PROBABILITY

Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function.  For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by

The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die.  The probabilities of the other four events in the sample space are also shown.  Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.

Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them.

TAKING POLYNOMIALS FROM ONE DIE TO MANY

Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH.  Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice.  My immediate thought was to apply my earlier ideas.  As noted in my initial post, the expansion approach above is not limited to binomial situations.  My first reflexive CAS command in Steve’s session before he share anything was this.

By writing the outcomes in words, the CAS interprets them as variables.  I got the entire sample space, but didn’t learn gain anything beyond a long polynomial.  The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives.  The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four.  Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum.  I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values.  Time to listen to the speaker.

He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent.  That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die.  Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.

NOTE:  Exponents are handled in TWO different ways here.  1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event.  Coefficients have the same meaning as before.

Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen.  That means the sum of the faces when you roll two dice is determined by the following.

Notice that the output is a single polynomial.  Therefore, the exponents are the values of individual cases.  For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc.  The most commonly occurring outcome is the term with the largest coefficient.  For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$.  That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice.

While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past.  For example, what is the most common sum when rolling 5 dice and what is the probability of that sum?  On my CAS, I entered this.

In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice.  As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18.  This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.

With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common.  Technology confirms intuition.

ROLLING DIFFERENT DICE SIMULTANEOUSLY

What is the distribution of sums when rolling a 4-sided and a 6-sided die together?  No problem.  Just multiply two different polynomials, one representative of each die.

The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.

A BEAUTIFUL EXTENSION–SICHERMAN DICE

My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice.  The only restriction he gave was that all of the faces of the new dice had to have positive values.  This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get

$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.

Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product?  That is, does this polynomial factor into other polynomials that could multiply to the same product?  A CAS factor command gives

Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice?  As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?

Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2.  Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$).  This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution.  Cool, but not what I wanted.  Now what?

Factorization gave four distinct sub-polynomials, each with multitude 2.  One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die.  That means there are $3^4=81$ different possible dice combinations.  I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.

What follows is the result of thinking about the problem for a while.  Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them.  Problem solving is a messy–but very rewarding–business.

SOLUTION

Here are my insights over time:

1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces).  Because Steve asked for dice with all positive face values.  That meant each desired die had to have at least one x to prevent non-positive face values.

2) My first attempt didn’t create 6-sided dice.  The sums of the coefficients of the sub-polynomials determined the number of sides.  That sum could also be found by substituting $x=1$ into the sub-polynomial.  I want 6-sided dice, so the final coefficients must add to 6.  The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6.  The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1.  The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor.

3) That leaves the two $\left( x^2-x+1 \right)$ factors.  They could split between the two dice or both could be on one die, leaving none on the other.  We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors.  To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice:  Two normal 6-sided dice.  If I want different dice, I have to load both of these factors on one die.

That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.

One die would have single faces of 8, 6, 5, 4, 3, and 1.  The other die would have one 4, two 3s, two 2s, and one 1.  And this is exactly the result of the famous(?) Sicherman Dice.

If a 0 face value was allowed, shift one factor of x from one polynomial to the other.  This can be done two ways.

The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.

Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other.  While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces.  One of these is

corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.

CONCLUSION:

There are other very interesting properties of Sicherman Dice, but this is already a very long post.  In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond.  And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations.

Enjoy.

Powers of 2

Yesterday, James Tanton posted a fun little problem on Twitter:

So, 2 is one more than $1=2^0$, and 8 is one less than 9=2^3\$, and Dr. Tanton wants to know if there are any other powers of two that are within one unit of a perfect square.

While this problem may not have any “real-life relevance”, it demonstrates what I describe as the power and creativity of mathematics.  Among the infinite number of powers of two, how can someone know for certain if any others are or are not within one unit of a perfect square?  No one will ever be able to see every number in the list of powers of two, but variables and mathematics give you the tools to deal with all possibilities at once.

For this problem, let D and N be positive integers.  Translated into mathematical language, Dr. Tanton’s problem is equivalent to asking if there are values of D and N for which $2^D=N^2 \pm 1$.  With a single equation in two unknowns, this is where observation and creativity come into play.  I suspect there may be more than one way to approach this, but my solution follows.  Don’t read any further if you want to solve this for yourself.

Because D and N are positive integers, the left side of $2^D=N^2 \pm 1$,  is always even.   That means $N^2$, and therefore N must be odd.

Because N is odd, I know $N=2k+1$ for some whole number k.  Rewriting our equation gives $2^D=(2k+1)^2 \pm 1$, and the right side equals either $4k^2+4k$ or $4k^2+4k+2$.

Factoring the first expression gives $2^D=4k^2+4K=4k(k+1)$.   Notice that this is the product of two consecutive integers, k and $k+1$, and therefore one of these factors (even though I don’t know which one) must be an odd number.  The only odd number that is a factor of a power of two is 1, so either $k=1$ or $k+1=1 \rightarrow k=0$.  Now, $k=1 \longrightarrow N=3 \longrightarrow D=3$ and $k=0 \longrightarrow N=1 \longrightarrow D=0$, the two solutions Dr. Tanton gave.  No other possibilities are possible from this expression, no matter how far down the list of powers of two you want to go.

But what about the other expression?  Factoring again gives $2^D=4k^2+4k+2=2 \cdot \left( 2k^2+2k+1 \right)$.  The expression in parentheses must be odd because its first two terms are both multiplied by 2 (making them even) and then one is added (making the overall sum odd).  Again, 1 is the only odd factor of a power of two, and this happens in this case only when $k=0 \longrightarrow N=1 \longrightarrow D=0$, repeating a solution from above.

Because no other algebraic solutions are possible, the two solutions Dr. Tanton gave in the problem statement are the only two times in the entire universe of perfect squares and powers of two where elements of those two lists are within a single unit of each other.

Math is sweet.


# Distinguishable objects into distinguishable boxes

How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them respectively?

$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ \\ &1 &2 &3 &4 &5 \end{gather}$

The lines represent the $5$ distinguishable boxes and the numbers below represent how many distinguishable objects each box must hold. I'm thinking I have $C\left(15,1\right)$ options for the first box then $C\left(14,2\right)$ for the second box, all the way to $C\left(5,5\right)$ for the fifth box. I multiply all those combinations together because of the product rule and I have no idea if that's the right answer.

• Good clear correct analysis. I would say that for each option for the first box there are $\dots$. Surely it is not true that you have no idea whether this is the right answer! – André Nicolas Aug 16 '13 at 6:24
• I don't see where you're going with that ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have reached the correct answer. – Kasper-34 Aug 16 '13 at 6:30
• I just meant it should be made clearer why we multiply. Note that if it is not specified which boxes contain $1,2,\dots,5$ then we need to multiply your answer by $5!$. – André Nicolas Aug 16 '13 at 6:33
• Well I'm multiplying because of the product rule. I think? I would multiply by $5!$ if I wasn't restricted, because I could put them in any order such as $5,3,1,2,4$? – Kasper-34 Aug 16 '13 at 6:41
• Okay, in that case I agree with you. The use of the word "respectively" makes me think they must be in the order $1,2,3,4,5$ and only that order. – Kasper-34 Aug 16 '13 at 7:10

Ways to put the labels $\{1,2,3,4,5\}$ on the boxes according as how many objects they contain: 5!. Then, as you correctly presumed,

$\binom{15}{1}$ ways to select an object for the one-object box;

$\binom{14}{2}$ ways to select two objects for the two-object box;

$\binom{12}{3}$ ways to select three objects for the three-object box;

$\binom{9}{4}$ ways to select four objects for the four-object box;

$\binom{5}{5}=1$ way to put the remaining five objects into the five-object box.

I think the answer is $$5!\binom{15}{1}\binom{14}{2}\binom{12}{3}\binom{9}{4}.$$

If the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the term $5!$ should be suppressed.

• I would say in this particular case since it used the word "respectively" there is only one way to order the boxes, which means we can leave out the multiplication of $5!$. – Kasper-34 Aug 16 '13 at 7:17


Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics.

Evaluate $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$. This is from a combinatorics textbook so I'd like a combinatorial proof. I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle.

Similar question that appears just before this question in the text: Combinatorics problem involving series summation

• I'm not sure whether a combinatoric proof can sum those fractions, but induction shows the sum is $1-1/(n+1)!$.
– J.G.
Jul 7 '18 at 5:36
• @J.G. I linked a similar problem maybe that will help... (didn't help me) Jul 7 '18 at 5:42

Consider a uniformly at random selected permutation of $\{1,2,\dots,n,n+1\}$.

The probability that $2$ appears before $1$ is $\frac{1}{(1+1)!}$

Given that this does not occur, then $1$ and $2$ appear in the correct order. The probability then that $3$ appears before at least one of $2$ or $1$ as well as $1$ and $2$ appearing in the correct order is $\frac{2}{(2+1)!}$.

Given that this does not occur, then $1,2,3$ all appear in the correct order. The probability then that $4$ appears before at least on of $3,2,1$ and $1,2,3$ all appearing in the correct order is $\frac{3}{(3+1)!}$...

...

Given that this does not occur, then $1,2,3,\dots,n$ all appear in the correct order. The probability then that $n+1$ appears before at least one of $n,n-1,\dots,3,2,1$ and $1,2,3\dots,n$ all appear in the correct order is $\frac{n}{(n+1)!}$

Given that this does not occur, then $1,2,3,\dots,n,n+1$ all appear in the correct order. This occurs with probability $\frac{1}{(n+1)!}$

Note that these are all mutually exclusive and exhaustive events, so they add up to equal $1$. Note further that the sum you are interested in is the sum of all of the events except the last one. We have then

$$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$

Rephrased, by multiplying the expression by $(n+1)!$, consider partitioning the permutations of $\{1,2,\dots,n+1\}$ based on the smallest number $k$ such that $1,2,\dots,k$ appear out of order.

That is to say, if $k$ is the smallest number such that $1,2,\dots,k$ appear out of order then $1,2,\dots,k-1$ must appear in order while $k$ does not appear after $1,2,\dots,k-1$. To count how many permutations satisfy this condition first pick the spaces that $1,2,\dots,k-1,k$ occupy simultaneously and then pick which of those positions $k$ occupies noting that it cannot be the last. $1,2,\dots,k-1$ appear in their normal order in the remaining selected positions. Then all other elements are distributed among the other spaces. This occurs in

$$\binom{n+1}{k}(k-1)(n+1-k)!=\frac{(n+1)!}{k!(n+1-k)!}(k-1)(n+1-k)!=\frac{k-1}{k!}(n+1)!$$

which you should recognize as following the sequence $0,\frac{1}{2!},\frac{2}{3!},\frac{3}{4!},\frac{4}{5!},\dots$ with the additional factor of $(n+1)!$ which we introduced earlier, otherwise mimicking the desired sum.

By including also the additional case of the identity permutation, the above forms a partition of the permutations of $\{1,2,\dots,n+1\}$. It follows that their respective totals add up to $(n+1)!$.

By removing the identity permutation as well as dividing by the factor of $(n+1)!$ that we introduced, this yields the desired identity

$$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$

• Hello can you clarify what you mean by the last part: "based on the smallest number k such that 1,2,…,k appear out of order." Jul 7 '18 at 7:12
• @helios321 Added more details. If you are just having difficulty understanding the phrasing I used, perhaps an example will help. $\color{red}{1}5\color{blue}{4}\color{red}{2}6\color{red}{3}$ is an example of a permutation where $\color{blue}{4}$ is the smallest number which occurs out of order since $\color{red}{123}$ appear in the correct order. By "appearing in the correct order" that is not to say they are adjacent, but merely that $1$ appears before $2$, that $2$ appears before $3$, etc... Jul 7 '18 at 16:19
• Thanks I figured it out now. Interesting it seems the expression multiplied by $(n+1)!$ is the same as in this question math.stackexchange.com/questions/2334537/…, both equal $(n+1)!-1$, but the other is counted by fixing the position. Jul 8 '18 at 0:23

Solution

Notice that$$\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.$$

Hence, $$\sum_{k=1}^n \frac{k}{(k+1)!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)=1-\frac{1}{(n+1)!}.$$

Using generating functions, which are widely used in combinatorics: $$a_n=\sum\limits_{k=1}^{n}\frac{k}{(k+1)!}$$ which is the same as $$a_n=a_{n-1}+\frac{n}{(n+1)!}$$ with generating function $$f(x)=\sum\limits_{n}\color{red}{a_n}x^n =a_0+\sum\limits_{n=1}\left(a_{n-1}+\frac{n}{(n+1)!}\right)x^n=\\ a_0+x\sum\limits_{n=1}a_{n-1}x^{n-1}+\sum\limits_{n=1}\frac{n}{(n+1)!}x^n=\\ a_0+xf(x)+\sum\limits_{n=1}\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n+1)!}x^n=\\ a_0+xf(x)+\left(\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{x}\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}=\\ a_0+xf(x)+\left(e^x-1-x\right)'-\frac{1}{x}\left(e^x-1-x\right)=\\ a_0+xf(x)+e^x-\frac{1}{x}\left(e^x-1\right)$$ or $$f(x)=\frac{a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}$$ since $a_0=0$ $$f(x)=\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\ \frac{1}{x}\left(\sum\limits_{n=0}x^n - \sum\limits_{n=0}\frac{x^n}{n!}\right)=\sum\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1)!}\right)}x^{n}$$ as a result $$a_n=1-\frac{1}{(n+1)!}, n\geq1$$

• I don't know what the down-voter was up to, but by all means this is a combinatorial proof! Jul 7 '18 at 12:31
• I'm not the downvoter, but usually your kind of proof is considered to be an algebraic one, whereas a combinatoric proof takes a set with known cardinality and constructs a bijection with another set showing the wanted formula or binomial identity. Jul 8 '18 at 14:02
• @MarkusScheuer or you assume that for a set of size $n$ the number of favourable cases is $a_n$ and then try to derive the recurrence for $n+1$ from $n$. And after, solve the recurrence with generating functions. A technique very often used in enumerative combinatorics. Jul 8 '18 at 14:17
• No doubt, this technique is great and I also appreciate it and use it often. The point is, this type of proof is usually not denoted as combinatorial proof. We find for instance in R. P. Stanleys Enumerative Combinatorics in the introductory chapter How to Count: A proof that shows that a certain set $S$ has a certain number $m$ of elements by constructing an explicit bijection between $S$ and some other set that is known to have $m$ elements is called a combinatorial proof or bijective proof. - As we can see the key term denoting a combinatorial proof is bijection. Jul 8 '18 at 14:43


Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. Okay, so that sounds kind of complicated. The triangle ABD is isosceles. It looks like the kites you see flying up in the sky. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. So it doesn't always look like the kite you fly. A Square is a Kite? 3. • diagonals which alwaysmeet at right angles. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Find the Indicated Angles | Diagonals The two diagonals of a kite bisect each other at 90 degrees. Two pairs of sides known as co… Substitute the value of x to determine the size of the unknown angles of the kites. The vertex angles of a kite are the angles formed by two congruent sides.. By definition, a kite is a polygon with four total sides (quadrilateral). You can drag any of the red vertices to change the size or shape of the kite. The longer and shorter diagonals divide the kite into two congruent and two isosceles triangles respectively. Browse through some of these worksheets for free! Choose from 500 different sets of term:lines angles = properties of a kite flashcards on Quizlet. Using these facts about the diagonals of a kite (such as how the diagonal bisects the vertex angles) and various properties of triangles, such as the triangle angle sum theorem or Corresponding Parts of Congruent Triangles are Congruent (CPCTC), it is possible … In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. Two disjoint pairs of consecutive sides are congruent by definition. $\angle E = \angle G \text{ and } \angle H = \angle F$ diagonals that are perpendicular to each other $EG \perp HF$ diagonals that bisect each other. The legs of the triangles are 10 inches and 17 inches, respectively. Yes! Stay Home , Stay Safe and keep learning!!! 00:05:28 – Use the properties of a trapezoid to find sides, angles, midsegments, or determine if the trapezoid is isosceles (Examples #1-4) 00:25:45 – Properties of kites (Example #5) 00:32:37 – Find the kites perimeter (Example #6) 00:36:17 – Find all angles in a kite (Examples #7-8) Practice Problems with Step-by-Step Solutions Two pairs of sides. Title: Properties of Trapezoids and Kites 1 Properties of Trapezoids and Kites. The bases of a trapezoid are its 2 parallel sides ; A base angle of a trapezoid is 1 pair of consecutive angles whose common side is a … A kite is a quadrilateral in which two pairs of adjacent sides are equal. Charlene puts together two isosceles triangles so that they share a base, creating a kite. The sketch below shows how to construct a kite. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. It can be viewed as a pair of congruent triangles with a common base. In this section, we will discuss kite and its theorems. E-learning is the future today. Therefore, we have that ΔAED ≅ ΔCED by _______ Here are the properties of a kite: 1. Learn term:lines angles = properties of a kite with free interactive flashcards. In this section, we will discuss kite and its theorems. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles… Apply the properties of the kite to find the vertex and non-vertex angles. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . What are the Properties of a Kite? The main diagonal of a kite bisects the other diagonal. 3. In a kite, the measures of the angles are 3x °, 75°, 90°, and 120°.Find the value of x.What are the measures of the angles that are congruent? 3. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. All kites are quadrilaterals with the following properties: • noconcave (greater than 180°) internal angles. The angles The problem. Apply the properties of the kite to find the vertex and non-vertex angles. Convex: All its interior angles measure less than 180°. By definition, a kite is a polygon with four total sides (quadrilateral). Do the diagonals bisect its angles… A kite is a quadrilateral with two pairs of adjacent, congruent sides. Properties of Kites. A kite is a quadrilateral with two pairs of adjacent, congruent sides. The main diagonal of a kite bisects the other diagonal. Sometimes one of those diagonals could be outside the shape; then you have a dart. Use this interactive to investigate the properties of a kite. 1. A kite is the combination of two isosceles triangles. The smaller diagonal of a kite … Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. Find the Indicated Angles | Vertex and Non-Vertex Angles. 2. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. One diagonal is the perpendicular bisector of the other. Solve for x | Find the Indicated Angles in a Kite. When all the angles are also 90° the Kite will be a Square. The two non-vertex angles are always congruent. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. 2. It looks like the kites you see flying up in the sky. In every kite, the diagonals intersect at 90 °. These sides are called as distinct consecutive pairs of equal length. The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. Kite. A Kite is a flat shape with straight sides. Metric formulas. • noparallel sides. Parallel, Perpendicular and Intersecting Lines. Use the appropriate properties and solve for x. Kite properties. Explanation: . Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . 4. Types of Kite. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. ... Properties of triangle. 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Fibonacci equality, proving it someway

$F_{2n} = F_n(F_n+2F_{n-1})$

$F_n$ is a nth Fibonacci number. I tried by induction but i didn't get anywhere

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An answer just by induction on $n$ of the equality $F_{2n}=F_n(F_{n-1}+F_{n+1})$ is as follows:

For $n=2$ we have $3=F_4=(1+3)\cdot 1=(F_1+F_3)F_2$.

To go from $n$ to $n+1$:

\begin{align} F_{2n+2}&=3F_{2n}-F_{2n-2}\\ &=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\ &=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\ &=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\ &=F_{n-1}F_{n+1}+3F_{n+1}F_n\\ &=F_{n+1}(3F_n+F_{n-1})\\ &=F_{n+1}(2F_n+F_{n+1})\\ &=F_{n+1}(F_n+F_{n+2}) \end{align}

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A combinatorial interpretation:

$F_n$ is the number of ways to tile a row of $(n-1)$ squares with $1\times 1$ blocks and $1\times 2$ blocks.

The left hand side is the number of ways to tile a $1\times (2n-1)$ block with $1\times 1$ and $1\times 2$ blocks. Consider the middle square (the $(n-1)$th square.)

Case 1: It is used in a $1\times 1$ block. Then, there are $F_{n}$ ways to tile each of the $1 \times (n-1)$ blocks on each side of the middle, so $F_n^2$ total.

Case 2: It is used in a $1\times 2$ block. This block contains the $(n-1)$th square and either the $n$th or the $(n-2)$th square. In either case, there are $F_{n-1}$ ways to tile the shorter side and $F_n$ ways to tile the longer side.

We thus have $F_{2n} = F_n^2 + 2F_nF_{n-1},$ as desired.

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I didn't have mentioned about your way of doing it. I can't fully understand your solution. The man said it's solutionable by induction –  matiit Oct 14 '12 at 15:27
Not useful to the OP, perhaps, but still very nice. –  Brian M. Scott Oct 14 '12 at 15:46

This is very related to the answer at Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

That question has the identity:

$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$

which can be modeled to your identity by letting $n=m$

$F_{2n} = F_{n-1}F_n + F_n F_{n+1}$

$F_{2n} = F_n(F_{n-1} + F_{n+1})$

Setting the expression inside the parenthesis to:

$F_{n-1} + F_{n+1} = F_{n-1}+ F_{n-1} + F_{n} = F_n + 2F_{n-1}$

We get

$F_{2n} = F_n(F_n+2F_{n-1})$

Which is your identity. So work backwards to that identity and use the proof at the linked question to prove your relation.

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Here’s a purely computational proof. Let $$A=\pmatrix{F_2&F_1\\F_1&F_0}=\pmatrix{1&1\\1&0}\;;$$ a straightforward induction shows that $$A^n=\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}$$ for all $n\ge 1$. Then

\begin{align*} \pmatrix{F_{m+n+1}&F_{m+n}\\F_{m+n}&F_{m+n-1}}&=A^{m+n}\\ &=A^mA^n\\\\ &=\pmatrix{F_{m+1}&F_m\\F_m&F_{m-1}}\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}\\\\ &=\pmatrix{F_{m+1}F_{n+1}+F_mF_n&F_{m+1}F_n+F_mF_{n-1}\\F_mF_{n+1}+F_{m-1}F_n&F_mF_n+F_{m-1}F_{n-1}}\;, \end{align*}

so $F_{m+n}=F_{m+1}F_n+F_mF_{n-1}$. Take $m=n$, and this becomes

$$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\left(F_{n+1}+F_{n-1}\right)=F_n\left(F_n+2F_{n-1}\right)\;.$$

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Exact duplicate of Zchpyvr's 21-minute prior answer (you've simply inlined the lemma that he links to). –  Bill Dubuque Oct 14 '12 at 19:06
@Bill: I’m not terribly surprised. However, I didn’t see his answer until after I posted, and when I did see it, I was busy and didn’t feel like chasing down the link. And quite frankly, I don’t consider it an exact duplicate for that very reason. –  Brian M. Scott Oct 14 '12 at 21:52
It is most certainly an exact duplicate. As I said, you've simply inlined the link in the other answer. Lacking anything new, it should be deleted for the sake of the readers. –  Bill Dubuque Oct 14 '12 at 21:58
@Bill: You have a rather inexact definition of exact. The fact that the information is right on the page is a difference, in convenience if nothing else. And deleting it does not serve the readers; slightly the reverse, if anything, for that same reason. The only person whom it might possibly ill serve is Zchpyvr, and I’ve upvoted his answer. And that’s the end of it as far as I’m concerned. –  Brian M. Scott Oct 14 '12 at 22:04
Posting duplicate answers potentially wastes many reader's time, since they may read two or more answers when it would have sufficed to read one. Not to mention that the abstraction in Zchpyvr's answer gained by calling the lemma by name (vs. value) only serves to make the answer more comprehensible. By your argument, every textbook should inline the proof of all lemma's so that they are "right on the page". That is, of course, absurd. –  Bill Dubuque Oct 14 '12 at 22:14

Another direct proof, using the fact that

$$F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}\tag1$$

where

$$\phi=\frac{1+\sqrt5}{2},\qquad1-\phi=\frac{1-\sqrt5}{2}=-\frac{1}{\phi}.$$

From $(1)$ we have

$$s_n\equiv\frac{F_{2n}}{F_n}=\frac{\phi^{2n}-(1-\phi)^{2n}}{\phi^n-(1-\phi)^n}=\phi^n+(1-\phi)^n$$

then

\begin{align} s_n-F_n&=\phi^n+(1-\phi)^n-\frac{\phi^n-(1-\phi)^n}{\sqrt5}=\\ &=\frac{1}{\sqrt5}\left[\sqrt5\phi^n+\sqrt5(1-\phi)^n-\phi^n+(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-(1-\sqrt5)\phi^n+(1+\sqrt5)(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-2(1-\phi)\phi^n+2\phi(1-\phi)^n\right]=\\ &=-\frac{2\phi(1-\phi)}{\sqrt5}\left[\phi^{n-1}-(1-\phi)^{n-1}\right]=2F_{n-1} \end{align}

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This can be done also using the fact that

$$F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\bar{\sigma}^n\right),$$

from which it is easy to get

$$F_{2n} = F_n L_n,$$

(where $L_n$ is the $n$-th Lucas number) and we have only to prove

$$L_n= F_n+2F_{n-1},$$

that is true since $\{L_n\}_{n\in\mathbb{N}}$ and $\{F_n+2F_{n-1}\}_{n\in\mathbb{N}}$ are sequences with the same characteristic polynomial ($x^2-x-1$) and the same starting values $L_1=F_1+2F_0=1$, $L_2=F_2+2F_1=3$.

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# Solutions to Congruence Modulo 50

1. Jan 12, 2013

### knowLittle

1. The problem statement, all variables and given/known data
Find all solutions to the equation $35x\equiv 10mod50$

3. The attempt at a solution

gcd( 35,50)= 5
So, there is a solution to this, since 5| 10. Also, there is a theorem that guarantees the existence of exactly 5 solutions.

Now, dividing $35x\equiv 10mod50$ over 5 gives:
$7x\equiv 2mod10$
Now, what multiple of 7 gives us $\equiv 2mod10$
{ 2, 12, 22, 32, 42,...} Here, we found 42 that is a multiple of 7 and satisfies $\equiv 2mod10$
We can write $7x\equiv 42mod10$ . Now, I divided the expression by 7 and got $x\equiv 6mod10$

Now, there is another theorem that tells me this
6+(50/5)t, t=0, 1, ..., 4
I get: 6, 16, 26, 36, 46
So, the solutions are
$x\equiv 6mod50$ , $x\equiv 16 mod50$, $x\equiv 26 mod50$, $x\equiv 36mod 50$, $x\equiv 46mod 50$
These are the 5 solutions of $35x\equiv 10mod50$

I found other solutions online:
So x = 6, 16, 22, 28, 34, 40, 46 modulo 50 are the
solutions to the congruence 35x ≡ 10 mod 50.

Am I incorrect?

Thank you.

2. Jan 12, 2013

### kru_

Your method is correct. You can check your own solutions to verify that they are correct.

35*6 = 210 which is congruent to 10 mod 50.
35*16 = 560 which is also congruent to 10 mod 50.
Similarly for the other solutions that you found.

You can verify that 22, 28, and 40 are not solutions.
35*22 = 770 which is 20 mod 50.
35*28 = 980, is 30 mod 50.
etc.

3. Jan 12, 2013

### knowLittle

I have read somewhere that division is not defined in modular arithmetic. Can someone tell me how this affect my solution?

@kru: This is puzzling, since I found those other solutions at a .edu site.

4. Jan 12, 2013

### HallsofIvy

Division is not necessarily defined in modular arithmetic because there may be "0 divisors". For example, in modulo 6, 2(3)= 6= 0 (mod 6). If we had an equation of the form 3x= 1 (mod 6) we can immediately check that 3(1)= 3, 3(2)= 6, 3(3)= 9= 3, 3(4)= 12= 0, and 3(5)= 15= 3 mod 6. There is NO x such that 3x= 1 and so we could not, for example, "divide by 3" to get "1/3" as an answer. If we are working "modulo" a prime number, that doesn't happen and we can define "division".

The way I would do "35x= 10 mod 50" is this. This is the same as saying 35x= 50n+ 10 for some integer n- a linear "diophantine equation". The first thing we can do divide through by 5 to 7x= 10n+ 2 or 7x- 10n= 2. Now 7 divides into 10 once with remainder 3: 3= 10- 7. 3 divides into 7 twice with remainder 1: 1= 7- 2(3). We can replace that "3" with 10- 7 from the first equation: 1= 7- 2(10- 7)= 3(7)- 2(10)= 1 (The "Euclidean Divison Algorithm"). Multiply through by 2 to get 6(7)- 4(10)= 2.

So one solution to 7x- 10n= 2 is x= 6, n= 4. It is possible to write out the "general solution" but since 6 itself is between 0 and 10, x= 6 satisfies 7(6)= 2 (mod 10) and so 35(6)= 210= 10 (mod 50).

5. Jan 12, 2013

### knowLittle

According to Wikipedia, Diophantine equations are written as follows:
ax + by = c
The Diphantine equation that you are really writing is this
35x-50n=10?

I understand everything, until you change the equation 1=7 -2(10-7)= 3(7)-2(10)=1. I understand that 21-20=1, but why changing from 7- 2(10-7) to 3(7)-2(10)?
Also, I am acquainted with Euclid's GCD algorithm:
Euclid(a,b)
if b==0
return a
else return Euclid (b, a mod b)

Is there a way to use it without having to trace it?

Is this all solutions for 35x $\equiv$ 10 mod 50? Also, is it correct that there has to be exactly 5 solutions, since the gcd of 35, 50 is 5?
Is my solution correct?


# Math Help - Complex numbers-finding real number pairs

1. ## Complex numbers-finding real number pairs

Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit

2. Multiply both sides by 1+pi

Then you get...

3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p

=> 3 + 5i = (q - 4p) + 4i + pqi

=> 3 + i = (q - 4p) + pqi

So you want to solve...

q - 4p = 3

pq = 1

3. ## found 1 solution but not the other

I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?

4. This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.

5. Originally Posted by Neverquit
I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?
I think you are misinterpreting your own answer. When you solve the system of equations, you will find that

$p=1/4 ~~\mbox{ or }~~ p=-1$

These are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$. So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$. Since $pq=1$, it's a pretty straightforward calculation:

$p=1/4 \implies q=4$

$p=-1 \implies q=-1$

Therefore, the solutions are:

First solution: $p=1/4 \mbox{ and } q=4$

Second solution: $p=-1 \mbox{ and } q=-1$

You might also write this as:

$(p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)$

But you definitely would not say that the solutions are $(1/4,-1)$ and $(4,-1)$.

6. Originally Posted by Neverquit
Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit
$\frac{3+5i}{1+pi} =q+4i$

Multiply top and bottom of LHS by $1-pi$ to give:

$\frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i$

Equate real and imaginary parts to get:

$3+5p=(1+p^2)q$

and:

$5-3p=4(1+p^2)$

Now the problem is to find all solutions to this pair of equations.

CB

7. Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?

Solving the equations I arrived at...

q - 4p = 3

pq = 1

gives

q=1/p,

So subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1.

Hence q = 4 and -1.

Solutions are (1/4,4) and (-1, -1)

Are these the only solutions? Why must you multiply the top and bottom by the conjugate?

Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?
Your solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem.

9. ## Using the conjugate

Originally Posted by Plato
This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$.
I think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained.

Deastar, your solution still gives the same answer in the text book so it must be correct.

...........

10. ## generalisation

Originally Posted by Neverquit
Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit
the basic idea behind questions of your type is equaslising real and imaginary parts.


# What is the probability that Jan and Jon are chosen?

Jan , Jon and $10$ other children are in a classroom. The principal of the school walks in and choose $3$ children at random. What is the probability that Jan and Jon are chosen?

My approach:

Including Jon and Jan, number of ways of selection is $1\cdot1\cdot{10\choose 1} = 10$.
Total way of selection is ${12\choose 3} = 220$

So, probability is $\frac{10}{220}= \frac{1}{22}.$

But when I solving as follows:

\begin{align*} P(\text{selection of Jon and Jan}) &= 1- P(\text{not selection of Jon and Jan}) \\ &= 1- \frac{{10\choose 3}}{{12\choose 3}}\\ &= \frac{5}{11}. \end{align*}

Which approach is correct and why alternative one is wrong?

Note: My previous post had some mistakes, so I deleted that

I will call $A$, $B$ the events selecting Jon, Jan respectively. You did not take the complement correctly \begin{align*} P(A\cap B) &= 1-\color{red}{P(\bar A \cup\bar B)}\tag 1 \\ &= 1-[P(\bar A)+P(\bar B)-P(\bar A\bar B)]\tag 2 \\ &=1-\left[\frac{\binom{11}{3}}{\binom{12}{3}}+\frac{\binom{11}{3}}{\binom{12}{3}}-\frac{\binom{10}{3}}{\binom{12}{3}}\right]\\ &=\frac{1}{22} \end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true by inclusion-exclusion. As you can see the two methods give the same value.

Your first approach gives the probability that Jon AND Jan are selected, the second gives the probability that Jon OR Jan are selected.

When you computed the probability of not selecting Jon AND Jan, you didn't include the situations where Jon was selected but not Jan, and vice versa.

Lets say the even Jon is selected is A, Jan is selected B.

Then $P(A \wedge B)=1-P(Not (A \wedge B))= 1- P(Not A \vee Not B)$

$P(Not A \vee Not B)= P(Not A)+ P(Not B)- P(Not A \wedge Not B)$

Here $\wedge$ means and, and $\vee$ means or. Think of a Venn diagram, if we look at the union of two circles, the total area is equal to the sum of the circles minus the intersection, because we counted that part twice.

$P(Not A)=1-P(A)=1-(1*\binom{11}{2})/220)=3/4$

or

$P(Not A)=\binom{11}{3}/220=3/4$

$P(Not B)=3/4$

$P(Not A \wedge Not B)= \binom{10}{3}/220=6/11$

Now $3/4+3/4-6/11=21/22$ So $1-21/22=1/22$


# Number of ways to place $k$ non-attacking rooks on a $100\times 100$ chess board

I need to show that the number of ways to place $k$ non-attacking rooks (no two share the same column or row) on a $100\times 100$ chessboard is $k!{100 \choose k}^2$.

When I try to formulate this equation I end up getting ${100 \choose k}^2$ because you need to choose $k$ columns from $100$ columns and $k$ rows from $100$ rows. I know this isn't correct because if you have $k=100$, there is more than just $1$ solution. However I don't know how to come up with the $k!$ part of the equation.

• – Shaun May 16 '17 at 14:32

First of all, congrats on realizing that the answer you got can't possibly be correct. It's always a good idea to test formulas against special cases, to see if they stand up.

One way to arrive at the correct answer is to view the placement of the rooks in two steps: First choose the $k$ rows that the rooks will go in, and then, going row by row, decide which column to place that row's rook in. The first rook has $100$ columns to choose from, the second will have $99$, the third $98$, and so on. The total is thus

$${100\choose k}100\cdot99\cdot98\cdots(100-(k-1))={100\choose k}{100!\over(100-k)!}={100\choose k}{100!\over(100-k)!k!}k!={100\choose k}^2k!$$

First, choose your $k$ rows and columns, as you said. Start by considering the configuration in which the rooks are successively placed in the legal square furthest to the top and to the left (so that the rooks go "diagonally down to the right").

From there, it suffices to note that any rearrangement of the rook-columns results in a new and valid configuration. Since there are $k!$ such rearrangements, there are $k!$ configurations for any particular choice of $k$ rows and $k$ columns.

Your problem is that $\binom{100}{k}^2$ only gives you the ways to choose the columns and the rows separately, without specifiying which row goes with which column. This is why you need to add the factor $k!$, which corresponds to the number of bijections between your $k$ rows and your $k$ columns, i.e. the number of ways to associate them together.

An alternative way of obtaining this is to consider that you first choose $\binom{100}{k}$ column where you will place your rooks, then choose for each column the row where you place a rook; this second step amonts to choosing $k$ rows with order, which you can do in $\frac{100!}{(100-k)!}=\binom{100}{k}k!$ ways. Multiplying the two numbers together gives you the result.

Regards User. If i may contribute, here is my view :

A 100 $\times$ 100 chess board can be viewed as a matrix of size 100 $\times$ 100. For example : let $$(i, j), \:\: \text{ with } \:\:\: i,j=1,2,... ,100$$ denotes the $i$th row and $j$th column of the board.

To solve your problem, the key is : the $k$ non-attacking roots is as same as no two $(i_{1}, j_{1})$ and $(i_{2}, j_{2})$ with $i_{1}=i_{2}$ or $j_{1}=j_{2}$. Two rooks with position $(1, 100)$ and $(91, 100)$, for example, does not satisfy the non-attacking roots condition.

To illustrate how to solve this, first, you could start with $k=1$.

• For k=1. Let the position of this particular rook is $(x_{1},y_{1})$. Then there are 100 possibilities for $x_{1}$, and 100 possibilities for $x_{2}$. So the number of possibilities is (100)(100) = $1! \binom{100}{1}^{2}$

• For k=2. For each of the two rooks, their positions are denoted with $(x_{1},y_{1})$ and $(x_{2}, y_{2})$. For the first one, There are 100 possibilities for each $x_{1}$ and $y_{1}$. For the 2nd rook, $x_{2}$ and $y_{2}$ each has 99 possibilities (since they can't be equal to the 'coordinates' of the 1st rook). So the number of possibilities to put 2 non-attacking distinguished rooks is $$(100^{2})(99^{2})$$ For your problem, they are not distinguished, so we have to divide this by 2 (exactly $2!$), because we can choose either rook to be the 1st or the 2nd. So, output : $$\frac{(100^{2})(99^{2})}{2!} = \frac{(100 \cdot 99)(100 \cdot 99)}{2!} = \frac{(100 \cdot 99)(100 \cdot 99)}{2!} \left(\frac{98! \cdot 2!}{98! \cdot 2!} \right)^{2} = 2! \binom{100}{2}^{2}$$

• For k=3 and above, you would be confident to try and continue this method.

Hope this will helps. Regards, Arief.


# Permutation representation argument validity

#### kalish

##### Member
Hello,
I would like to check if the work I have done for this problem is valid and accurate. Any input would be appreciated. Thank you.

**Problem statement:** Let $G$ be a group of order 150. Let $H$ be a subgroup of $G$ of order 25. Consider the action of $G$ on $G/H$ by left multiplication: $g*aH=gaH.$ Use the permutation representation of the action to show that $G$ is not simple.

**My attempt:** Let $S_6$ be the group of permutations on $G/H$. Then, the action of $G$ on $G/H$ defines a homomorphism $f:G \rightarrow S_6$. We know $|S_6| = 720.$ Since $|G|=150$ does not divide 720, and $f(G)$ is a subgroup of $S_6$, $f$ cannot be one-to-one. Thus, $\exists$ $g_1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \implies f(g_1g_2^{-1})=e$. Thus, $\ker(f) = \{g:f(g)=e\}$. Since $\ker(f)$ is a normal subgroup of $G$, we have found a normal subgroup of $G$. Also, since $f$ is non-trivial, then $\ker(f)$ is a proper normal subgroup of $G.$ Hence $G$ is not simple.

Any suggestions or corrections?

#### Deveno

##### Well-known member
MHB Math Scholar
I think it is clear that since $|G|$ does not divide 720, $\text{ker}(f)$ is a non-trivial normal subgroup of $G$, so there is no need to talk about the existence of $g_1,g_2$ or restate the definition of $\text{ker}(f)$.

I *do* think you should say WHY $f$ is not the trivial homomorphism. It's pretty simple, though:

Since $|H| < |G|$, we can take any $g \in G - H$, which takes (under the action) the coset $H$ to $gH \neq H$, so $f(G)$ contains at least one non-identity element: namely, $f(g)$.

#### Deveno

##### Well-known member
MHB Math Scholar
This is actually a special case of a theorem proved in Herstein, which goes as follows:

If $G$ is a finite group with a subgroup $H$ such that $|G| \not\mid ([G:H])!$ then $G$ contains a non-trivial proper normal subgroup containing $H$.

One obvious corollary is then that such a group $G$ cannot be simple.

You would be far better off adapting your proof to this more general one, which can be re-used in many more situations.

#### kalish

##### Member
Hi,
Which Herstein book is this from? I would like to explore further.

Thanks.

#### Deveno

##### Well-known member
MHB Math Scholar
His classic Topics In Algebra.

#### kalish

##### Member
That's what I found from my search as well. Do you have a copy of the book or know where I can find one?

#### kalish

##### Member
That sounds like a fantastic result. I cannot find the book anywhere though. Could you please reproduce the proof for me here, so that I could use it to study? I would really appreciate it.

#### Deveno

##### Well-known member
MHB Math Scholar
Theorem 2.G (p. 62, chapter 2):

If $G$ is a group, $H$ a subgroup of $G$, and $S$ is the set of all right cosets of $H$ in $G$, then there is a homomorphism $\theta$ of $G$ into $A(S)$, and the kernel of $\theta$ is the largest normal subgroup of $G$ which is contained in $H$.

(a few words about notation: Herstein uses $A(S)$ to stand for the group of all bijections on $S$...if $|S| = n$, then $A(S)$ is isomorphic to $S_n$. Herstein also writes his mappings on the RIGHT, as in $(x)\sigma$ instead of $\sigma(x)$, so that composition and multiplication are "in the same order", instead of reversed. For this reason, he uses right cosets and right-multiplication instead of the left cosets (and left-multiplication) one often sees used in other texts. He also denotes the index of $H$ in $G$ by $i(H)$ , instead of $[G:H]$ and denotes $|G|$ by $o(G)$).

Proof: Let $G$ be a group, $H$ a subgroup of $G$. Let $S$ be the set whose elements are right cosets of $H$ in $G$. That is, $S = \{Hg: g \in G\}$. $S$ need not be a group itself, in fact, it would be a group only if $H$ were a normal subgroup of $G$. However, we can make our group $G$ act on $S$ in the following natural way: for $g \in G$ let $t_g:S \to S$ be defined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove:

(1) $t_g \in A(S)$ for every $g \in G$
(2) $t_{gh} = t_gt_h$.

Thus the mapping $\theta: G \to A(S)$ defined by $\theta(g) = t_g$ is a homomorphism of $G$ into $A(S)$. Can one always say that $\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\theta$. If $g_0 \in K$, then $\theta(g_0) = t_{g_0}$ is the identity map on $S$, so that for every $X \in S, Xt_{g_0} = X$. Since every element of $S$ is a right coset of $H$ in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \in G$, and using the definition of $t_{g_0}$, namely, $Hat_{g_0} = Hag_0$, we arrive at the identity $Hag_0 = Ha$ for every $a \in G$. On the other hand if $b \in G$ is such that $Hxb = Hx$ for every $x \in G$, retracing our argument we could show that $b \in K$. Thus $K = \{b \in G|Hxb = Hx$ all $x \in G\}$. We claim that from this characterization of $K,\ K$ must be the largest normal subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup of $G$ which is contained in $H$, then $N$ must be contained in $K$. We wish to show this is the case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \subset H$, for if $b \in K, Hab = Ha$ for every $a \in G$, so in particular, $Hb = Heb = He = H$, whence $b \in H$. Finally, if $N$ is a normal subgroup of $G$ which is contained in $H$, if $n \in N,\ a \in G$, then $ana^{-1} \in N \subset H$, so that $Hana^{-1} = H$; thus $Han = Ha$ for all $a \in G$. Therefore, $n \in K$ by our characterization of $K$.

**********

Remarks following the proof:

The case $H = (e)$ just yields Cayley's Theorem (Theorem 2.f). If $H$ should happen to have no normal subgroup of $G$, other than $(e)$ in it, then $\theta$ must be an isomorphism of $G$ into $A(S)$....(some text omitted)....

We examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i(H)$ (that is, the number of right cosets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ in $G$. The mapping, $\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\theta(G)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\theta$ must be larger than $(e)$; this kernel being the largest normal subgroup of $G$ which is contained in $H$, we can conclude that $H$ contains a nontrivial normal subgroup of $G$.

However, the above argument has implications even when $i(H)!$ is not less than $o(G)$. If $o(G)$ does not divide $i(H)!$ then by invoking Lagrange's theorem we know that $A(S)$ can have no subgroup of order $o(G)$, hence no subgroup isomorphic to $G$. However $A(S)$ does contain $\theta(G)$, whence $\theta(G)$ cannot be isomorphic to $G$, that is, $\theta$ cannot be an isomorphism. But then, as above, $H$ must contain a nontrivial normal subgroup of $G$. We summarize this as:

Lemma 2.21 If $G$ is a finite group, and $H \neq G$ is a subgroup of $G$ such that $o(G) \not\mid i(H)!$, then $H$ must contain a nontrivial normal subgroup of $G$. In particular, $G$ cannot be simple.

(Note to the moderating staff: although this is an excerpt from a copyrighted work, I believe this sample falls under the province of "Fair Use" for the purpose of "Scholarly research and exposition", and is not intended for commercial gain or to circumvent existing copyright laws).

Last edited:


# Spring Gun

1. Mar 7, 2014

### Steven60

I have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework.
Thanks!

2. Mar 7, 2014

Seems to be purely a math problem. Perhaps sketch the system and write the relevant equations needed to determine this?

3. Mar 7, 2014

### UltrafastPED

This is a nice physics exercise - there are several physical considerations, and then some simple math.

You have two forces acting on the marble ... the spring force, which launches the marble, and gravity.

Once the marble leaves the launch tube it will have a constant "horizontal" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V.

The vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T.

Then the distance from the table to the point of contact will be D = V x T.

The time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2.

Thus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a "good" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant.

If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2.

But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W.

Thus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X.

Thus the hypothesis is true!

Thanks to dauto for noticing the mistake at the end! :-)

Last edited: Mar 7, 2014
4. Mar 7, 2014

### DrewD

The equations involved will be $d=vt$ for constant $v$ (and assuming that the initial point when exiting the spring gun is defined as 0 distance), $U_{spring}=\frac12k\Delta x^2$ and $K=\frac12mv^2$. $\Delta x$ is the amount the spring is compressed, and $v$ is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring gun. Solve for $v$ to find the relationship between $\Delta x$ and $d$.

5. Mar 7, 2014

### dauto

You made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true.

6. Mar 9, 2014

### Steven60

Thanks for your replys. I actually worked this out myself and actually did the same exact steps as UltrafastPED.


# Is this induction procedure correct? ($2^n<n!$)

I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I've been able to solve some of the form $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$.

Now, I was presented this, for $n \ge 4$:

$$2^n<n!$$

I tried to do it with similar logic as the one suggested there. This is what I did:

Prove it for $n = 4$: $$2^4 = 16$$ $$4! = 1\cdot2\cdot3\cdot4 = 24$$ $$16 < 24$$ Assume the following: $$2^n<n!$$ We want to prove the following for $n+1$: $$2^{n+1}<(n+1)!$$ This is how I proved it:

• So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
• By our assumption, we know that $2^n\cdot2 < n!\cdot2$
• This is because I just multiplied by $2$ on both sides.
• Then we'll be finished if we can show that $n! \cdot 2 < (n+1)!$
• Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$
• Since both sides have $n!$, I can cancel them out
• Now I have $2<(n+1)!$
• This is clearly true, since $n \ge 4$

Even though the procedure seems to be right, I wonder:

• In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more "careful"?
• Is this whole procedure valid at all? I ask because, well, I don't really know if it would be accepted in a test.
• Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.
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I think your proof is fine, if a bit long-winded. But with experience, you'll learn what bits to shorten without losing rigour. –  Harald Hanche-Olsen Nov 27 '12 at 9:51
You should say 2^{n+1}\lt 2\cdot n!$. But because$2\lt $n+1$, it follows that $2\cdot n!\lt (n+1)!$. Unfortunately, you are still writing proofs "backwards" in a logically incorrect way. –  André Nicolas Nov 27 '12 at 9:59
@André: It seems a bit harsh to call this a "logically incorrect way". It's OK as long as "Since both sides have $n!$, I can cancel them out" is interpreted as "dividing through by $n!$ leads to an equivalent inequality". It's true that the reverse order would be clearer, and doing things in this order is incorrect if the implications used only go in one direction, but that's not the case here so the proof is still OK, if suboptimally structured. –  joriki Nov 27 '12 at 10:04
@Omega, your proof is correct..."almost", since as Andre apparently meant, there's some lack of logical rigour in your last lines. You must show that the implications there are double, i.e.: $$n!\cdot 2<(n+1)!\Longleftrightarrow 2<\frac{(n+1)!}{n!}=n+1$$ and then noting the last inequality is trivially true as we're working with $\,n\geq 4\,$... Also, don't right "equivalent" when it should be "equal", as in "$\,2^{n+1}\,$ is "equivalent"(should be "equal"!) to $\,2^n\cdot 2$ –  DonAntonio Nov 27 '12 at 10:47
@joriki: A bit harsh, perhaps, but other posts by the OP had some worse instances that were pointed out. Since the actual understanding of the problem is good, it is worthwhile to vaccinate the OP against $A\to B\to C\to 0=0$, therefore $A$. –  André Nicolas Nov 27 '12 at 16:47

Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.

For example, the base case could be re-written as follows:

When $n = 4$, we have $2^4 = 16 < 24 = 4!$

Next, the inductive hypothesis and the subsequent manipulations:

Suppose that for $n \geq 4$ we have $2^n < n!$

Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$.

Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D.

Note: I am not making a judgment about whether your write-up or the one I have included here is "better." I'm only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).

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