## Probability Puzzles (Part 3)

Here’s a puzzle based on something interesting that I learned from Greg Egan. I’ve dramatized it a bit.

Traditional Tom and Liberal Lisa are dating. They discuss their plans for having children:

Tom: I plan to keep having kids until I get two sons in a row.

Lisa: What?! That’s absurd. Why?

Tom: I want two to run my store when I get old.

Lisa: Even ignoring your insulting assumption that only boys can manage your shop, why in the world do you need two in a row?

Tom: From my own childhood, I’ve learned there’s a special bond between sons who are next to each other in age. They play together, they grow up together… they can run my shop together.

Lisa: Hmm. Well, then maybe I should have children until I have a girl followed directly by a boy!

Tom: What?!

Lisa: Well, I’ve observed that something special happens when a boy has an older sister, with no intervening siblings. They play together, they grow up together… and maybe he learns not to be such a sexist pig!

They decide they are incompatible, so they split up and each one separately tries to find a mate who will go along with their reproductive plan.

Now for some puzzles. In these puzzles, assume that each time someone has a child, they have a 50% chance of having either a daughter or a son. Also assume each event is independent: that is, the gender of any children so far has no effect on that of later ones. Also ignore twins and other tricky issues.

Puzzle 1. If Tom carries out his plan of having children until he has two consecutive sons, and then stops, what is the expected number of children he will have?

Puzzle 2. If Lisa carries out her plan of having children until she has a daughter followed directly by a son, and then stops, what is the expected number of children she will have?

Puzzle 3: Which is greater, Tom’s expected number of children or Lisa’s? Or are they equal?

For maximum benefit, try to answer Puzzle 3 before doing the calculations required to answer Puzzles 1 or 2.

### 64 Responses to Probability Puzzles (Part 3)

1. It’s late here right now so for now only preliminary quick guesses.
If you wanna have a more formal go, maybe don’t look at this until you want a hint or a distraction, depending on whether my guesses are right.

Puzzle 3 guess:
Since any particular two-birth-sequence is equally likely (1/4 each), I’d guess that both scenarios are equally likely.

Puzzle 1:
Possible sequences:
SS (1/4)
DSS (1/8)
DDSS (1/16)
SDSS (1/16)
DDDSS (1/32)
DSDSS (1/32)
SDDSS (1/32)
DDDDSS (1/64)
DDSDSS (1/64)
DSDDSS (1/64)
SDDDSS (1/64)
SDSDSS (1/64)
DDDDDSS (1/128)
DDDSDSS (1/128)
DDSDDSS (1/128)
DSDDDSS (1/128)
SDDDDSS (1/128)
DSDSDSS (1/128)
SDDSDSS (1/128)
SDSDDSS (1/128)
This looks suspiciously like $\sum_{n=1}^\infty \frac{\text{fib}\left(n\right)}{2^{n+1}}$
There is probably some nice combinatorial argument to make to prove this.
This sum – I didn’t prove but just quickly looked up – sums up to 1, so that’s good.
The expectation value version of that sum just adds an n:
$\sum_{n=1}^\infty n \frac{\text{fib}\left(n\right)}{2^{n+1}}$
And that appears to sum to 5.

Puzzle 2:
Possible sequences:
DS (1/4)
SDS (1/8)
DDS (1/8)
SSDS (1/16)
SDDS (1/16)
DDDS (1/16)
SSSDS (1/32)
SSDDS (1/32)
SDDDS (1/32)
DDDDS (1/32)
SSSSDS (1/64)
SSSDDS (1/64)
SSDDDS (1/64)
SDDDDS (1/64)
DDDDDS (1/64)

The series appears to be

$\displaystyle{ \sum_{n=1}^\infty \frac{n}{2^{n+1}} }$

which also sums to 1.

And the expectation value

$\displaystyle{ \sum_{n=1}^\infty \frac{n^2}{2^{n+1}} }$

goes to 3. So Lisa will actually most likely have fewer kids than Tom. Therefore my above prediction was wrong.

Furthermore, if I did this right, Lisa can expect her value to hover around 3 by 2, whereas for Tom that uncertainty sits around 4.69.
So in a fairly pessimistic (assuming you want to get there asap) outcome would be 5 trials for Lisa but 10 for Tom.

(I used that $\sigma^2 = \langle-^2\rangle$ but I’m tired and so possibly error-prone.)

• Michael Nelson says:

Here’s how fibonacci shows up:

You can count the number of ways of Tom winning via a string of n turns by breaking it up into two cases:

…….DSS
….DSDSS

The first case you counted already in the previous string of length n-1. And the second case you counted already from the string before that of length n-2.

• This is how I also ended up solving it, getting the Fibonnaci sequence immediately by using a construction approach to compute the number of binary strings (1 == boy, 0 == girl) that had the constraint of ending in ’11’ but with no other ’11’ in them. The construction as you show has the two ways to generate N+1: extend all the length N cases with a ‘0’ in the front and extend the N-1 cases adding ’10’ to the front. This means the number of ways to generate an appropriate string of length N+1 is Fib(N) since Count(N+1) = Count(N -1) + Count(N). And the total strings of length N+1 is $2^{N+1}$. Dividing gives you the probability of each string length (number of kids) and the expectation is as was shown above as well. And of course phi (the Golden Mean) shows up in the solution of the characteristic of the Fibonnaci recurrence so ultimately I ended up using it to solve this. Pretty cool!

2. It’s much quicker, Kram, to model each problem as a state machine.

Let $T_0$ and $T_1$ be Tom’s expectation of future children after zero and one son respectively. We can say immediately that $T_0 = \frac{1}{2}(s+T_1) + \frac{1}{2}(d+T_0)$ and $T_1 = \frac{1}{2}(s) + \frac{1}{2}(d+T_0)$. The solution, if I haven’t blundered, is $T_0 = 3d+3s$.

For Lisa: $L_0$ is her expectation before her first daughter; $L_1$ is her expectation after any daughter. $L_0 = \frac{1}{2}(s+L_0) + \frac{1}{2}(d+L_1)$ and $L_1 = \frac{1}{2}(d+L_1) + \frac{1}{2}(s)$. Solution: $L_0 = 2d+2s$.

Lisa can expect two sons and two daughters; Tom can expect three daughters and three sons.

The qualitative difference is that for Tom a daughter after a son puts him back a step, while for Lisa a child of the sex not hoped for leaves her in the same position.

• John Baez says:

This looks impressively efficient, Anton! But what does an equation like

$T_0 = \frac{1}{2}(s+T_1) + \frac{1}{2}(d+T_0)$

actually mean? I guess I’m lacking the necessary familiarity with state machines to know what type of entities these variables represent, and what addition means. I know what a state machine is; I’m just not familiar with this kind of equation.

I can guess that $T_n$ is allowed to be any expression of the form $a s + b d$ where $a,b \ge 0$, and

$T_n = a s + b d$

means “after having had $n$ sons, Tom expects in the future to have a total of $a$ sons and $b$ daughters”.

But then I don’t know why

$T_0 = \frac{1}{2}(s+T_1) + \frac{1}{2}(d+T_0)$

should be true. It seems to say “after having no sons, Tom expects in the future to have the same thing as if he had a 50% chance of having one more son than what he expected after having one son, and a 50% chance of having one more daughter than what he expected after having no sons”.

The horrible quoted phrase here certainly shows why a more concise notation is a good thing! It reminds me of what happens when you try to state the quadratic equation in ‘plain English’ without variables.

But the problem is, I don’t see why it should be true.

• My subscripts mean only that state 0 must occur before state 1 can occur. I could also say explicitly $T_2 = L_2 = 0$; does that make anything clearer?

My equations are the invariants of the tree of futures. The one that you query can be read as: “From state 0, the branches of Tom’s future are: with probability 1/2, he has a son and moves to state 1; and with probability 1/2, he has a daughter and remains in state 0.” As long as his sequence of daughters continues, his expectation of future children remains the same, so the definition of T0 is recursive. From state 1 (one son) Tom cannot remain in state 1; he must either terminate or go back to 0. After a daughter, previous sons are irrelevant to Tom’s future tree, which is why there are not more states in this model.

• Chris Upshaw says:

Shouldn’t that be s * T1 then, if its a parallel notation to sum of products datatype notation? Its certainly not the same operation as the middle ‘+’.

• That’s cos I don’t know nothin bout notating no datatypes.

• John Baez says:

I found Anton’s trick very hard to understand at first, mainly because the variables and operations in his equations had not been explained (and I didn’t already know this trick). Like Greg, I found XJY’s comment here helpful. Even then, I had to hit myself over the head with a rock a few times before I really understood what was going on.

I’ve spent a lot of time thinking about other aspects of Markov processes, but I never thought about this particular trick. It’s very nice, but I think I need to digest it and blend it with other things I know, like generating functions and—yes—recursively defined data types.

• Anton,

This is a really cool solution! Not only do you compute Lisa’s expected # of kids, you compute how many sons and daughters she can expect. I have never seen this concept of expressing the invariants of the tree of futures before. Do you have a good reference for learning more about this? Is it standard CS?

Thanks for posting this.

• Rob, sorry, I have no idea where I got the, er, idea.

“Invariant” may be a misnomer; thinking about it further, I reckon the analogy with the loop invariants of CS is not so strong.

I may have misunderstood the problem but it seems to me that as soon as Tom or Lisa reach their objective they STOP. Then Tom can have 2 sons in a row and it is over. On the other hand if they have a daughter Lisa can get a son and they stop, Tom can never get two sons.

• John Baez says:

Wow, you assumed that Tom and Lisa are married, and that their plans affect each other’s activities!

I never thought that: I was imagining two people at a bar, who obviously hate each other, each discussing their own separate plans. But now I see that there’s an ambiguity in the phrase “their plans for having children”. I’ll fix that!

• John Baez says:

Upon rereading what I wrote, I see there’s no real ambiguity in the puzzles as stated. One asks what will happen if Tom carries out his plan. The other asks what will happen if Lisa carries out her plan. There’s nothing about what will happen if both of them try to carry out their conflicting plans.

Nonetheless I added some stuff to make it crystal clear that Tom and Lisa are not going to be having children together.

4. jamievicary says:

I deliberately haven’t worked this out formally, because it’s more fun to see if intuition leads to a right answer. Write their progeny as a string in D and S. It’s ‘easy’ for a string of length n to lack a SS substring, as it could have no Ss, or just one S, or just two Ss spaced at least one apart, etc. But to lack a DS substring is relatively ‘hard’: you have to have all the sons before all the daughters. So, holding out for two sons in a row will take longer than waiting for a daughter followed by a son.

I guess the reason it’s a bit paradoxical is that for any two adjacent children, SS is just as likely as DS. I suppose the reason is that combinatorially, like-gendered children are indistinguishable. In reality this is not the case. They should just ask for “two children adjacent in age, both with the temperament to run a shop”!

(By the way, I found this post cognitively difficult to write, since on parenting forums DD and DS often meen ‘darling son’ and ‘darling daughter’.)

• John Baez says:

Jamie wrote:

I guess the reason it’s a bit paradoxical is that for any two adjacent children, SS is just as likely as DS.

Right, that was supposed to make the problem cognitively difficult.

(By the way, I found this post cognitively difficult to write, since on parenting forums DD and DS often meen ‘darling son’ and ‘darling daughter’.)

I never knew! I hadn’t expected that to cause cognitive difficulties. Do SD and SS mean ‘stupid daughter’ and ‘stupid son’?

(I would probably be banned from the forum for that joke.)

• Tamfang says:

or Silly or Sweet

• Charles Jackson says:

Well, you did make the problem cognitively difficult for me.

I did #3 first. I thought P(SS) = P(DS) => Expected births to first SS is same as for DS. WRONG—but

I did 2 relatively quickly with something like Anton’s approach above.

Then I looked at the answers in the comments. Oh, oh. I thought about it for awhile and I think I see an easy explanation for the answer.

Consider the outcomes with up to three children. Make a little truth table-like figure with all 8 possible sequences of combinations of boy and girl babies. See below. Notice that there are 4 occurrences of the sequence DS in the table—each of which is the first occurrence of DS in the sequence containing it. (capitalized in table below)

Birth Order
1 2 3
d d d
d D S
D S d
D S s
s d d
s D S
s s d
s s s

Now, search for SS, but only the 3 that are the first occurence in time.

d d d
d d s
d s d
d S S
s d d
s d s
S S d
S S s

The difference (only 3 SS versus 4 DS) arises from the second SS that is a subsequence of the SSS sequence
d d d
d d s
d s d
d s s
s d d
s d s
s s d
s S S

So, if the parties were restricted to 3 or fewer babies, Lisa would have a 50% chance of getting DS, but Tom would only have a 3/8 chance of getting SS.

Of course, if they both had three babies, Tom would have a 1/8 chance of getting SS twice, but Lisa would have zero chance of getting DS twice.

At the birth of the second baby they each have a 25% chance of getting their desired outcome. On the third baby, Lisa would also have a 25% chance but Tom would only have a 12.5% chance.

Lisa’s probability of terminating the process at 3, 4, or 5 is higher than the probability for Tom. This pushes up the expected number for him relative to her.

Chuck
PS Thanks for your blog. I only read some of the posts but always enjoy those I complete. No doubt if I had more time and patience I get even more enjoyment from it.

• John Baez says:

Nice. I’m glad you’re enjoying this blog!

• Jenny Meyer says:

The DD and DS tags are just a convention for anonymizing your family members. The D can be heavily ironic in context.

(The picture of the Duggars was enough to make this too cognitively difficult for me.)

5. Michael Nelson says:

I think I understand. If we think of this as a game, where Lisa needs to get the sequence DS and Tom needs to get the sequence SS, then Lisa will certainly have the advantage. As soon as a D shows up, Tom will lose. On the other hand, if a S shows up, Lisa can still win. So Lisa has a greater chance of winning this game.

• John Baez says:

I’m not assuming that Tom and Lisa are having children with each other: they are two separate people, each pursuing their own separate agenda with their own spouse, each assuming their spouse will knuckle under and go along with their insane plan.

Nonetheless you can still think of it as a game, where we imagine a random stream of babies being produced and both Tom and Lisa saying when they, personally, would prefer the stream to stop.

• Michael Nelson says:

Yes, this is what I meant. You phrased it better for me. I think I have a better way to think about this now. The number of ways that Lisa can win at the nth spot is just n. For example:

DDDDDS
SDDDDS
SSDDDS Lisa wins at 5th spot
SSSDDS
SSSSDS

DDDDDDS
SDDDDDS
SSDDDDS
SSSDDDS Lisa wins at 6th spot
SSSSDDS
SSSSSDS

The number of ways that Tom can win at the nth spot involves the fibonacci numbers though. For example:

DDDDSS
SDDDSS
DSDDSS Tom wins at 5th spot
DDSDSS
SDSDSS

DDDDDSS
SDDDDSS
DSDDDSS
DDSDDSS Tom wins at 6th spot
SDSDDSS
SDDSDSS
DSDSDSS

DDDDDDSS
SDDDDDSS
DSDDDDSS
DDSDDDSS Add D just before SS in 6th spot wins
SDSDDDSS
SDDSDDSS
DSDSDDSS

plus Tom wins at 7th spot

SDSDSDSS
DDSDSDSS
DSDDSDSS Replace the last S in 5th spot wins with DSS
SDDDSDSS
DDDDSDSS

That’s pretty cool that the fibonacci numbers are involved.

6. Dave Doty says:

I worked it out like this. I am nagged by the possibility that there is a more elegant simple way to do it.

Each is picking a string from the alphabet $\{ s,d \}$. Both have a list of “allowed” strings of length $n$ that indicate they must keep going.

Lisa omits from this list all strings with the substring $sd$, and Tom omits all strings with the substring $ss$. Already we can see that Lisa has fewer allowed strings than Tom, and will therefore wait a shorter time before stopping, because to remain in an allowed string, she can only transition once (from $s$ to $d$), whereas Tom can transition between $s$ and $d$ an arbitrary number of times.

Lisa’s allowed strings match the regular expression $s^*d^*$; e.g., for length 5, her 6 allowed strings are

$sssss, ssssd, sssdd, ssddd, sdddd, ddddd$

In general, Lisa has $n+1$ allowed strings of length $n$, so probability $\frac{n+1}{2^n}$ that a given length $n$ string will be allowed. If the string is $s^*$, then either of $s$ or $d$ is allowed next, whereas if it ends in a $d$, only a $d$ is allowed next. So the probability that Lisa ends on length $n$ is the probability that she does not end before (i.e., the length $n-1$ prefix is allowed) times the probability that the whole length $n$ string is disallowed, conditioned on length $n-1$ prefix being allowed. This is

$\displaystyle{ \frac{1}{2^{n-1}} \cdot 0}$

(the probability the string is $s^{n-1}$, times probability 0 since she will not stop after $s^{n-1}$), plus

$\displaystyle{ \frac{n}{2^{n-1}} \cdot \frac{1}{2} }$

(the probability the string is $s$‘s followed by at least one $d$, times the probability that the next symbol is $s$).

So Lisa’s probability of stopping after $n$ children is $\frac{n}{2^n}$.

Thus her expected stopping time is

$\displaystyle{ \sum_{n=2}^\infty n \cdot \frac{n}{2^n} = \sum_{n=2}^\infty \frac{n^2}{2^n} = \frac{11}{2} }$

Tom’s allowed strings omit the substring $ss$. We can count recursively how many strings of length $n$ satisfy this, denoted by $T_n$. If the first symbol is $d$, then all allowed strings of length $n-1$ could follow, of which there are $T_{n-1}$. If the first symbol is $s$, then the next symbol must be $d$, and any allowed string of length $n-2$ could follow, of which there are $T_{n-2}$. So $T_n = T_{n-1} + T_{n-2}$, the $n$‘th Fibonacci number, which is

$\displaystyle{ \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}} }$

So Tom’s probability of stopping after $n$ children is the probability that the last three symbols are $dss$ (if it were $sss$ he would have stopped at the second $s$) the prefix of length $n-3$ is allowed, which is

$\displaystyle{ \frac{T_n}{2^n} = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^{2n} \sqrt{5}} }$

Thus his expected stopping time is

$\displaystyle{ \sum_{n=2}^\infty n \cdot \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^{2n} \sqrt{5}} = \frac{19}{2} }$

• John Baez says:

Wow, what a tour de force! You are not getting the officially approved correct answers, but the officially approved correct answers do involve Fibonacci numbers, at least in the most direct approach.

7. Bruce Smith says:

I carefully solved each puzzle before reading beyond it (in case of spoilers), so puzzle 3 was not very interesting! (I had thought of it while starting to work on puzzle 2, but didn’t have the benefit of your stating it was worthwhile to work on it first, so I didn’t do so.)

Another mistake — I initially assumed Tom and Lisa were married and were discussing their joint plan for having kids! This leads to puzzle 4 — suppose they do marry (before having any kids) and agree to have kids until both their criteria are satisfied…

• John Baez says:

I think it’s too late for me to make Puzzle 3 into Puzzle 1 without causing massive confusion.

I have added verbiage to the puzzle to make it clear that Tom and Lisa are not married and never will be.

Nonetheless, Puzzle 4 is interesting, and will doubtless lead to an even larger expected family.

• For that case I get seven expected children, again equally divided.

Next question: For each scenario, what are the variances? Do the variance in daughters and the variance in sons add up to the variance in total children? I don’t know how to figure these.

• Unless I made an error, I already did the variances – or, well, the standard deviations which are just the square roots of the variances – in my reply (first one). However, all the other contributions here are much more fleshed out and nice. My approach was really just trying a couple and then guessing. It seems like the results were right, however it’s nice to read in the others’ replies why they are right.

To get the variance, if you have all the sub-probabilities, you just do:
Probability: $P=\sum_{n \in \mathbb{N}} p\left(n\right) = 1$

Expectation: $E = \sum_{n \in \mathbb{N}} n p\left(n\right)$

Variance: $\sigma^2 = \sum_{n \in \mathbb{N}} n^2 p\left(n\right) - E^2$

Standard Deviation: $\sigma = \sqrt{\sigma^2}$

For some reason, in my original reply, the $\LaTeX$ didn’t work right. Looks like the fixed version still isn’t right though. Hopefully the above works.

• John Baez says:

The LaTeX at the very end of this comment of yours didn’t work, and it was pretty cryptic, so I guessed you were trying to say this:

$\sigma^2 = \langle-^2\rangle$

This means ‘the square of the standard deviation is the expected value of the square of the quantity in question’. (The little dash is a standard symbol for ‘the quantity in question’: a blank to be filled in.) But this is only true for quantities whose mean is zero, which isn’t the case in this problem. So, I must have guessed your meaning incorrectly. I should have written

$\sigma^2 = \langle-^2\rangle - \langle - \rangle^2$

But probably you weren’t trying to use the ‘little dash’ notation, which gets a bit scary in formulas that also contain minus signs!

8. Ilya Surdin says:

Puzzle 1:
suppose d is the expectation value we’re looking for.
then d=0.5(d+1)+0.5(0.5*2+0.5(d+2)) [if a daughter is born first, we’re back to point 1 again, if a son is born first then if another son is born the value is 2, and if a daughter is born second we’re back to square 1].
so we get d = 0.75d +1.5 -> d=6

Puzzle 2:
suppose e is the expectation value we’re looking for.
e = 0.5(e+1) + 0.5f,

where summand 1 is if a son is born, and summand 2 is if a daughter is born, we only need to get a son (ds, dds, ddds, all options are good)

f = 2 [ f=0.51+0.5(f+1) ]

so e = 0.5e + 0.5 +1 -> e=3.

Puzzle 3:
At first sight, I guessed they will be equal.

9. Graham says:

Here’s a solution for Tom’s case using generating functions. Let $T$ be a random variable representing the number of children Tom has. Let $f(s) = \sum p_i s^i$ be the generating function for T, where $p_i = \Pr(T=i)$. The first few $p_i$ are

$p_0 = p_1 = 0, p_2=1/4, p_3=1/8.$

For $i \ge 4$, the sequence must end DSS, and there must be no SS occurring in the first $i-3$. So

$p_4 = (1/8)(1-p_1),$

$p_5 = (1/8)(1-p_1-p_2),$

and so on.

From this it follows that

$p_i = p_{i-1} - (1/8)p_{i-3}$

for $i \ge 4$.

Then by comparing $f(s), sf(s), s^3f(s)$, it is possible to calculate that

$\displaystyle{ f(s) = \frac{-s^2}{s^2+2s-4}. }$

Then

$E(X) = f'(1) = 6$

and

$E(X(X-1)) = f''(1) = 52$

from which $\mathrm{var}(X) = 52 + 6 -6^2 = 22$.

10. “In one word he told me secret of success in mathematics: Generalize!” What if the sex ratio at birth is not 1:1? I’ll get right on that — later.

• John Baez says:

Puzzle 5. For what sex ratio at birth would Tom and Lisa have an equal expected number of children?

(Obviously it needs to be more sons than daughters.)

• Greg Egan says:

John wrote:

Puzzle 5. For what sex ratio at birth would Tom and Lisa have an equal expected number of children?

$\displaystyle{ p_{son} = \frac{\sqrt{5}-1}{2} }$

or:

$\displaystyle{\frac{p_{son}}{p_{daughter}} = \frac{\sqrt{5}+1}{2} }$

To prove this (stealing from and generalising an answer by “XJY”, which is a slightly more coarse-grained version of Anton Sherwood’s approach), we can write the expectation values for the number of additional children that Tom and Lisa will each have after their latest child was a daughter or a son (who didn’t yet complete their target sequence) as:

$T_d = p_{son} (1+T_s) + (1-p_{son}) (1+T_d)$
$T_s = p_{son} +(1-p_{son}) (1+T_d)$
$L_d = p_{son} + (1-p_{son}) (1+L_d)$
$L_s = p_{son} (1+L_s) + (1-p_{son})(1+L_d)$

These are solved by:

$\displaystyle{T_d = \frac{1+p_{son}}{p_{son}^2} }$

$\displaystyle{T_s = \frac{1}{p_{son}^2} }$

$\displaystyle{L_d = \frac{1}{p_{son}} }$

$\displaystyle{L_s = \frac{1}{p_{son}(1-p_{son})} }$

But $T_d$ and $L_s$ are also the expectation values for the number of children each will have in total, because they have 0 out of 2 elements of their target sequence.

So if we solve:

$L_s = T_d$

the positive solution for $p_{son}$ is:

$\displaystyle{ p_{son} = \frac{\sqrt{5}-1}{2} }$

and

$L_s = T_d = 2+\sqrt{5}$

11. Bruce Smith says:

It turns out these puzzles show up (for numbers 1 and 3 sought by Alice and Bob) in this (otherwise mostly unrelated) blog post about primes:

https://rjlipton.wordpress.com/2016/03/26/bias-in-the-primes/

• John Baez says:

• Erica Klarreich, Mathematicians discover prime conspiracy, Quanta, 13 March 2016.

Soundararajan was drawn to study consecutive primes after hearing a lecture at Stanford by the mathematician Tadashi Tokieda, of the University of Cambridge, in which he mentioned a counterintuitive property of coin-tossing: If Alice tosses a coin until she sees a head followed by a tail, and Bob tosses a coin until he sees two heads in a row, then on average, Alice will require four tosses while Bob will require six tosses (try this at home!), even though head-tail and head-head have an equal chance of appearing after two coin tosses.

I decided to dramatize this using children instead of coin tosses.

By the way, I wrote about the ‘prime conspiracy’ here:

• John Baez, Weirdness in the primes, _The n-Category Café 15 March 2016.

There’s a bit of math here that’s not in the Quanta article.

12. Rob says:

Nice Puzzle.

from “Mathematical Plums” edited by Ross Honsberger, chap 5 some surprises in probability, “Suppose a fair coin is tossed repeatedly and an indefinitely long sequence of H’s and T’s is generated…. what a surprise to learn that it is twice as likely that the triple TTH will be encountered ahead of the triple THT…” I love surprises in probability, and I find that most things about probability surprise me! The chapter also talks about something else you may like if you haven’t seen it, “Efron’s Dice”.

By the way, have you read this paper, http://www.emis.ams.org/journals/TAC/volumes/26/4/26-04.pdf, Commutative Monads as a Theory of Distributions. I gather Prof. Kock is developing the theory to study probability theory. I can’t really understand it, but you may find it interesting.

• John Baez says:

Probability theory is often counterintuitive to humans, which must have something to do with why humans mess up so often.

I haven’t read that paper. I’m pretty good friends with Anders Koch’s son Joachim, who also does category theory related work. But this paper is by Anders. I find it pretty hard going, at first because I’ve never understood strengths as much as I’d like, even though I did help come up with the more conceptual definition of them given on the link. (The more common definition, which involves writing down some commutative diagrams, always bothered me.)

• Rob says:

Say the first child is a G. Then we are guarateed that GB will occur prior to to BB. (cause for BB to happen, we must first have had GB)

Now say the first child is a B. Then either the second child is a B, in which case BB happens first, or it is a G, in which case, again, GB will happen before BB.

So we see that GB is 3x more likely to occur in any given familiy than BB. Thus I would expect that the expected value of the second random varable is less than the expected value of the first. Hah! well probability is so amazing, I expect that I am wrong! Now I will read through the comments where I expect to find the right answer.

Also, if you don’t understand that paper, I never will! But I think it is really interesting so I will keep reading it. BTW I think your notes on Cat Theory are great. Thanks to all your students for putting that together.

13. arch1 says:

Puzzle 3: Represent all possible birth sequences as an infinite binary tree (at the risk of scaring you we’ll pretend that childbearing doesn’t stop when the goal is reached). Yes, at each level of the tree starting with the 2nd, ¼ of the arcs are DS arcs and ¼ are SS arcs. But the symmetry breaks when we look at the correlation between adjacent levels: While an SS arc can immediately follow an SS arc, a DS arc can’t immediately follow a DS arc (as that would require the shared middle node to be both an S and a D).

Upshot: Half of the SS arcs “hide behind” an immediate SS parent, but DS-DS repulsion prevents DS arcs from ever doing this. This makes a DS arc likelier than an SS arc to be the first arc of its kind on a path from the root (put differently, it causes DS arcs to more efficiently “cover” the tree as viewed from the root). So the expected number of levels before encountering a DS arc is less than for an SS arc.

Greg and John, I find these probability paradoxes very instructive and great for my humility. I hope you keep them coming.

• John Baez says:

Nice answer! Yes, probability puzzles are great for ones humility—maybe the right verb is “humiliating”?

There should be a book of probability puzzles, similar to Mark Levi’s excellent physics puzzle book Why Cats Land on Their Feet: And 76 Other Physical Paradoxes and Puzzles. But I don’t know it!

If anyone here hasn’t tried Probability Puzzles (Part 1) and Probability Puzzles (Part 2), now is the time! They’re all about boys and girls, oddly enough. The first one is particularly devastating, or infuriating.

• arch1 says:

(chuckle) I actually tend to find them more invigorating than humiliating, partly for noble reasons but mostly I think because I can’t wait to spring them on others.

Ironically the closest I’ve come to humiliation recently was last week when I, er, “judged” a high school science fair project which explained how cats can land on their feet using gauge theory.

• cclingen says:

High school?! Cats landing their feet?! Gauge theory?! Where can I learn more?

• John Baez says:

Hi, Charlie! It’s probably good to start here:

• Wikipedia, Falling cat problem.

but then go to this paper by a friend of mine who is an expert on classical mechanics at U. C. Santa Cruz:

• Richard Montgomery, Gauge theory of the falling cat, in M. J. Enos, Dynamics and Control of Mechanical Systems, American Mathematical Society, pp. 193–218.

• arch1 says:

Nowhere as far as I know (certainly not me as I gleaned only a scant understanding). Maybe a paper will come out later; meanwhile I’d prefer to let the student pursue things in his own way.

• Re Prob Puzzle books. I just bought “Fifty Challenging Problems in Probability”. It looks like it has some really cool stuff in it. And best of all it was cheap!

• Greg Egan says:

The counterintuitive aspects of real statistical phenomena might have put an innocent man in prison for 38 years:

An Unusual Pattern

Like the statisticians who discuss the case, I’m not claiming any certainty that Geen is innocent, but on the face of it the statistical basis of the prosecution’s argument appears alarmingly wrong-headed.

• That is creepy. As teachers we need to do a better job at explaining statistics. Examples like the above and the OP help a lot.

14. pauljurczak says:

I hate to be a party pooper, but using infinite series in the top answer is yet another reason why lies and statistics are often mentioned in the same sentence. Human female can only have a limited number of children, much smaller than 100, which is much smaller than infinity.

This is in general a problem of using mathematics to model physical “reality”, which always requires certain assumptions resulting in departure from “reality”. In this case, the assumption is immortality or more specifically, unlimited childbearing capability.

P.S. Don’t take this post too seriously ;-)

• John Baez says:

I won’t take it too seriously—just more seriously than you expected.

Puzzle 7: How do the answers for Tom’s and Lisa’s expected number of children change if we impose a ‘cutoff’, saying that a woman can have no more than 10 children?

• …Or if each pregnancy has probability q of leaving the mother unable to bear further children?

• Charles Jackson says:

Note that that there is about a 1% that Tom will have to have 24 or more children before the desired SS outcome occurs.

15. Edon says:

These processes can be seen as absorbing Markov chains. Let $T$ be the block in the transition matrix that holds the transitions from transient states to other transient states. It is a basic property that $(I - T)$ is invertible and that the $(i,j)$ entry of $M=(I - T)^{-1}$ gives the expected number of times the chain is in state $j$ before it is absorbed, given that we start from state $i$. Summing the entries in the appropriate row yields the answers to the puzzles.

16. For some reason there isn’t a reply button below that latest reply of you, John, but yeah, that was what I was going for, although indeed not with the ‘little dash’ notation, but that works too. With just one minus sign it’s not that bad. Thanks for the fix. There must have been more wrong with it than just $\langle \cdot \rangle$ if it was so cryptic. I’m not sure what I typed anymore but while typing it that was the only part I didn’t know whether it’d work. The rest I was pretty sure of. But oh well. I thought \left{<} would work for \langle but you guessed that part right.

• John Baez says:

Kram wrote:

For some reason there isn’t a reply button below that latest reply of you…

Unfortunately if you tell this blog that you don’t want comments to get indented more than 4 times (because they become absurdly skinny), it implements this by eliminating the ‘Reply’ button after a comment that’s been indented 4 times.

What you should do in this case is reply not to the comment but to the comment it’s commenting on! In fact it’s often wise to do this. This keeps the comments from getting too skinny. I can tell who is a really expert blog-commenter by whether they do this.

I thought \left{<} would work for \langle but you guessed that part right.

I’ve never seen \left{<} in LaTeX; I think LaTeX is pretty careful about the difference between the less than symbol $<$ and the left angle bracket $\langle$.

But there’s an extra problem with LaTeX in most blogs! In HTML, < is an escape symbol, so you to get a less than symbol in LaTeX you have to use the HTML symbol < rather than <. It’s very bizarre to be using HTML commands inside LaTeX equations, but that’s how it works! That’s true not only for this rather brain-dead WordPress blog but also some others I use.

17. Tom says:

Puzzle 6: Why did John use the name of his wife for the female character in this drama?

• John Baez says:

Answer: because it was the first woman’s name beginning with L that I thought of.

18. David Cinabro says:

This one got in my head, and I had to try to solve it.

First Puzzle 3. Two children SS, SD, DS, DD. Tom and Lisa both stop 1 out of 4. Three Children for Tom SDS, SDD, DSS, DSD, DDS, DDD. Tom stops 1 out of 6. Three Children for Lisa SSS, SSD, SDS, SDD, DDS, DDD. Lisa stops 2 out of 6. Clearly the expected number of children is different and Lisa will have a smaller number than Tom.

For Tom’s expected number of children, he has 1/4 chance stopping at 2, 1/6 stopping at 3, 2/10 stopping at 4, 3/16 stopping at 5, 5/26 stopping at 6, etc. The stopping chance numerators are Fibonacci seeded by 1 and 1, and the denominators are Fibonacci seeded by 4 and 6. The expected number of children is then 2(1/4) + 3(1-1/4)1/6 + 4(1-1/4)(1-1/6)3/16+…. ie he stops at 2 with a probability of 1/4, at 3 with a probability of (1-1/4)*1/6, etc. Perhaps there is a clever way to write that, but I wrote a little program to tell me that it goes to 6.

Lisa has 1/4 chance stopping at 2, 2/6 stopping at 3, 3/8 stopping at 4, etc. The stopping chance numerators are 1,2,3,etc. and the denominators at 4,6,8, etc. The same sort of calculation gives the expected number of children 21/4 + 3(1-1/4)2/6+4(1-1/4)(1-2/6)*3/8+… This one is certainly even simpler to work out in closed from , but again the program tells me that it sums to 4.

19. Michael Nelson says:

By the way, you can prove the classic result involving pascal’s triangle and the fibonacci numbers by partitioning Tom’s string of n children into these cases:

Tom has a total of 3 sons
Tom has a total of 4 sons

For those who aren’t familiar, the relation between pascal’s triangle and the fibonacci numbers involves summing over diagonals, like in this image:

In other words, the nth fibonacci number equals (n-1 choose 0) plus (n-2 choose 1) plus …

20. While on the subject of what is essentially run theory, mathematician Peter Donnelly has given a TED talk on the subject of fooling juries with statistics. His talk is based around the infamous English criminal case of Sally Clark. To illuminate the issues he poses this problem:
you toss an unbiased coin many times and count the number of tosses until you get the pattern ”HTH”. You then calculate the average. You repeat the experiment but this time you look for the pattern ”HTT”. The question is: ”Is the average number of trials required to get ”HTH” greater than, less than or equal to the average number of trials required to get ”HTT”? ” Donnelly has tried this out on a wide range of audiences including professional mathematicians and the audience not surprisingly says the number of trials were equal. The expected number of trials to get ”HTH” is 10 while the expected number of trials to get ”HTT” is 8.
My paper ( $\url{http://www.gotohaggstrom.com/Fooling%20juries%20with%20statistics.pdf}$ ) explains the generating function proof of Donnelly’s claim. This paper also sets out the background to the Sally Clark case. A 2013 English Court of Appeal case has had the effect of banning Bayesian reasoning in court cases. This shows that the “real” world of the justice system can intersect in bizarre ways with probability theory.

21. Puzzle 1.
We are looking at infinite sequence of coin flips (the coin has two sides: S and D) and waiting for first occurrence of SS.
There are essentially 3 different states:
s0: current sequence ends with D or is empty (initial state)
s1: current sequence ends with exactly one S
s2: current sequence ends with SS (end state)

Let ei = expected number of flips to get from si to s2.
Then we have the following system of linear equations:
e2 = 0
e1 = (1/2)(1+e2) + (1/2)(1+e0)
e0 = (1/2)(1+e0) + (1/2)(1+e1)

Solving it gives the answer to the puzzle: e0 = 6.

Puzzle 2.
We just need to wait for the first daughter, then wait for the first son. Since expected number of trials until first success equals 1/p, where p is the probability of success, the answer is obviously 2+2=4.

Puzzle 3.
The intuition is, that Tom has to make a “step back” sometimes but not Lisa, therefore she accomplishes her goal sooner.

22. Check out this interesting, related puzzle and discussion.

https://www.klittlepage.com/2013/11/20/dynasty-meets-the-central-limit-theorem/

I just came across it. It has some surprising conclusions.


# Find The Indicated Nth Roots Of A

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From this form a K-normal form C, can be deduced. 5) Complete the following table for!!!=7!. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. From this form a K-normal form C, can be deduced. In Exercises 48 and 49, find the modulus and the argument, and graph the complex number. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Finding real nth roots of a. Improve your math knowledge with free questions in "Nth roots" and thousands of other math skills. Numerical methods for ﬁnding the roots of a function The roots of a function f(x) are deﬁned as the values for which the value of the function becomes equal to zero. FINDING NTH ROOTS Find the indicated real nth root(s) of a. Given two numbers N and A, find N-th root of A. Round your answer to two decimal places when appropriate. 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When I try to find the roots of the same equation in Mathematica, I receive various errors. Geometric mean function for python. The solutions of the equation x^n=1, where n is a positive integer, are called the nth roots of 1 or "nth roots of unity. Homework Help With Finding Nth Roots And Rational Expressions. Perhaps, then, it is this calculator: Let's say you want to take the fifth root of 2. Asking for unreasonable roots, for example, show_nth_root < 1000000 >(2. This is a polynomial equation of degree n. (26 replies) In PythonWin I'm running a program to find the 13th root (say) of millions of hundred-digit numbers. Find nth roots of complex numbers Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Finding nth roots of Complex Numbers. However, there is still one basic procedure that is missing from the algebra of complex numbers. Both X and N must be real scalars or arrays of the same size. In mathematics, an nth root of a number x, where n is usually assumed to be a positive integer, is a number r which, when raised to the power n yields x: =, where n is the degree of the root. It uses Math. Homework Help On Finding Nth Roots And Rational Exponents. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. The math workaround described above works really well with pretty good accuracy. Square root of 64 is 8 because 8 times 8 is 64 Cube root of 27 is 3 because 3 times 3 times 3 = 27 fourth root of 16 is 2 because 2 times 2 times 2 times 2 = 16 Sometimes, you may get a real number when looking for the square root. The nth roots of a complex number For a positive integer n=1, 2, 3, … , a complex number w „ 0 has n different com-plex roots z. 1/2 Evaluate the expression. 24 The nth root of a number b is designated as nb,. EXAMPLE 1 Find nth roots Find the indicated real nth root (s) of a. is the radius to use. NaN = not a number To clear the entry boxes click "Reset". The most common root is the square root. It arises in a wide variety of practical applications in physics, chemistry, biosciences, engineering, etc. n = 2, a = 100 24. Find nth roots of complex numbers Contact If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Drag the slider to change the number of roots. The pictures. -polar graph of roots. If bn = a, then b is an nth root of a. An th root of unity in a finite field is an element satisfying , where is an integer. 0 root of 125. 5 or 10 1/2 and the cube root as 10 0. Find a number x = ξ such that f(ξ) = 0. Since CP is similar to A, a matrix Y E Mm K exists such that (Y-1CPY)= Y-iCPY = A. To calculate any root of a number use our Nth root calculator. How to find the nth root of a complex number. This online calculator for fifth roots is set up specifically to calculate 5th root. In effect we must find the roots of the equation z^3 = 1 We could write this z^3-1 = 0 (z-1)(z^2 + z + 1) = 0 and then z = 1 gives one root, and other two are solutions of the quadratic z^2+z+1 = 0 (these roots will be complex). ^2 96 sqr root 3 in. “nth Root of a”. An applet allowing you to investigate how the th roots of a complex number (in blue) appear on an Argand diagram. nth Root of a-to-the-nth-Power. 5, 13, 33, 89, I got this by adding the two sequences above. Evaluate Question: “What number, when raised to the 3rd power gives us 27?“ B. For instance, √(4) produces a principle root of 2. If all equations and starting values are real, then FindRoot will search only for real roots. I decided to take the input in the form of a Double. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. Relating to an unspecified ordinal number: ten to the nth power. Functional abstraction to find nth root of a number - Newton raphson. Created Date: 1/17/2019 10:36:18 AM. Finding nth roots of Complex Numbers. Find the indicated real nth root(s. (algorithm) Definition: This describes a "long hand" or manual method of calculating or extracting square roots. Rationalize one term denominators of rational expressions. The exponent of a number shows you how many times the number is to be used in a multiplication. 1 Evaluate nth Roots and Use Rational Exponents Goal p Evaluate nth roots and study rational exponents. Derivation from Newton's method. A root of degree 2 is called a square root, a root of degree 3 is called a cube root, a root of degree 4 is called a fourth root, and so forth. org - it looks like you don't have Java installed, please go to www. I know that if this were just normal numbers, I could find it u. In effect we must find the roots of the equation z^3 = 1 We could write this z^3-1 = 0 (z-1)(z^2 + z + 1) = 0 and then z = 1 gives one root, and other two are solutions of the quadratic z^2+z+1 = 0 (these roots will be complex). Using Bino's model of multiplication , you can compute the square root, cube root, or root of any positive number up to any decimal place for no-perfect cube number. Granted, then, you have a problem they currently have. What is the principal root of 121400? Rounded to two decimal places, the principal square root of 121400 is 348. Created Date: 1/14/2009 8:51:02 PM. The nth root calculator below will also provide a brute force rounded approximation of the principal nth root. She had invested$1500 more at 8%. Roots can be square roots, cube roots, fourth roots and so on. What is the fifty-first term?. Use nth roots in problem solving Animal Population The population P of a certain animal species after t months can be modeled by P = C(1. The for a different view). This online calculator is set up specifically to calculate 4th root. According to the Fundamental Theorem of Algebra, this equation has n solutions. Find indicated real nth roots of c. org - it looks like you don't have Java installed, please go to www. It uses Math. Homework Help On Finding Nth Roots And Rational Exponents. 5) Complete the following table for!!!=7!. First, we will define what square roots are and how you find the square root of a number. Looking for a primer on how to find the nth term of a geometric sequence? See how it's done with this free geometer's guide. The number that must be multiplied times itself n times to equal a given value. Discussion Suppose IDL is being used to find a real root of a number. Now I was thinking of adding the nth-Root of a Number. My current expression is -. 2 days ago · He said that 12 out of the 18 persons who had already indicated interest in the governorship election in the state in 2021 were encouragingly from Anambra South, adding that the expectation was that with time, aspirants from the other two senatorial zones would withdraw to join hands in finding an ideal person from the South to emerge as the. W HEN ONE THING DEPENDS on another, as for example the area of a circle depends on the radius -- in the sense that when the radius changes, the area will change -- then we say that the first is a "function" of the other. The number that must be multiplied times itself n times to equal a given value. 0 *Real nth Roots of a. Nth Root using the POWER Function. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Excel's powerful mathematical toolkit includes functions for square roots, cube roots, and even nth roots. This online calculator is set up specifically to calculate 4th root. 1 Evaluate nth Roots and Use Rational Exponents Goal p Evaluate nth roots and study rational exponents. Mathematical Connections Find the radius of the figure with the given volume. 32 = 32(cos0 + isin 0 ) in trig form. In the event you seek help on exam review or perhaps intermediate algebra syllabus, Polymathlove. Then, invoking the Intermediate Value Theorem, there is a root in the interval $[-2,-1]$. We find the the two roots by multiplying 2 by the square roots of unity. What is an “nth Root?” Extends the concept of square roots. It requires an initial guess, and then Newton-Raphson iterations are taken to improve that guess. To demonstrate that this simple method is fully applicable to roots of higher order, we shall attempt to find the 5th root of 2. What are the solution sets of the following 3 problems,the square root of x squared + 3x = x-3, the square root of 3x + 10 = x + 4 and finally the square root of 8x - 2 = the square root of 2x? Help much appriciated. Every real number has exactly one real. n5 3, a5 2 64 b. ^2 216 sqr root 3 in. In short, you can make use of the POWER function in Excel to find the nth root of any number. 1) the radicand has no perfect nth powers as factors 2) any denominator has been rationalized Example 4: Write the following Radicals in Simplest Form — Remember your perfect nth Roots u = 16. A formal mathematical definition might look something like: The nth roots of unity are the solutions to the equation x n = 1. Step 3: Write the answer using interval notation. ) ! n=5! a="32 b. Summing up: The following guidelines will be useful for finding the nth root of the perfect nth power of a two-digit or three-digit number. 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In fact, there are seven 7th roots of unity, and each gold disc in that picture is one of them. How do i find the indicated real nth root of:? Find the indicated real nth root(s) of a - Help? A) find indicated roots of the complex number b) express roots in standard form?. Given two numbers N and A, find N-th root of A. During my son’s math lesson today we got on the subject of , which I prefer to write as. The fundamental theorem of algebra can help you find imaginary roots. This function is pretty well behaved, so if you make a good guess about the solution you will get an answer, but if you make a bad guess, you may get the wrong root. Roots of unity have connections to many areas of mathematics, including the geometry of regular polygons, group theory, and number theory. This Algebra I: nth Root and Rational Exponents Worksheet is suitable for 8th - 11th Grade. Pow , mathematics. 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Improve your math knowledge with free questions in "Nth roots" and thousands of other math skills. Then we will apply similar ideas to define and evaluate nth roots. The resulting number from this expression means that if it’s raised to the nth power it equals the. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. One real root:. nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. The second root is usually called the "square root". Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. For example, to find the cube root of 8 in a cell, type the following:. Examples:. Use 27 cis (90 degress), which is the trigonometric from of 27i, to do the problem. June 19, 2019 July 22, 2019 Craig Barton. The finite field has prime order. I'm trying to find the n-th root of unity in a finite field that is given to me. 0 Votes 18 Views I am unable to prove that :. is the radius to use. Divide radicals that have the same index number. Instead you need to pass 1/10 as n√x = x(1/n). Unlike a square root function which is limited to nonnegative numbers, a cube root can use all real numbers because it is possible for three negatives to equal a negative. n th Root of an Integer Description Calculate the n th root of an integer. when x k +1 ≥ x k. In general, a root of degree n is called an nth root. Your program should ask the user for a number, and a root and print that root of the number. You can always tell FindRoot to search for complex roots by adding 0. If n = 3 , then it is called as Cube root. During my son’s math lesson today we got on the subject of , which I prefer to write as. An expression under a radical sign is called a radicand. Find the indicated real nth root(s) of a. Find the fourth roots of -16 and -16 is a complex number even though it is also real. n = 2, a = 100 24. The equation to calculating the nth term of the sequence is an+b; "a" is the fixed number that is being added to generate the series, and "b" gives the. n th Roots A square root of a number b , written b , is a solution of the equation x 2 = b. However, there is still one basic procedure that is missing from the algebra of complex numbers. 3 POLAR FORM AND DEMOIVRE’S THEOREM At this point you can add, subtract, multiply, and divide complex numbers. The Help Article also provides a convenient wrapper function. We get the nth root of 1 is the regular nth root of 1, which is just 1, times e to the i phi over n. De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem. Buy business plan online - best in texas, homework help on finding homework help us geography nth roots and rational exponents. , , , the six roots are () out of which , , and on the left s-plane are the roots of :. It need not be true that any of the fractions is actually a solution. Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. However, there is still one basic procedure that is missing from the algebra of complex numbers. (Buddy system! If there is no buddy, it must stay under the radical. Find the indicated real nth root(s) of a negative. De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem. n = 5, a = 0. Because n = 4 is even and a = 81 > 0, 81. 2 days ago · He said that 12 out of the 18 persons who had already indicated interest in the governorship election in the state in 2021 were encouragingly from Anambra South, adding that the expectation was that with time, aspirants from the other two senatorial zones would withdraw to join hands in finding an ideal person from the South to emerge as the. 3 POLAR FORM AND DEMOIVRE'S THEOREM 483 8. The 5th root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,204. Question 1: Simplify. When you’re giving a time-bound exam like SSC CGL, SSC CPO, Railways Group D, RPF & ALP this can drain you of your precious time. You can see there are many roots to this equation, and we want to be sure we get the n^ {th} root. Nth root calculator. V = 216 ft3 Find the real solution(s) of the equation. There might be some issues with this. IXL Learning Learning. Use nth roots in problem solving Animal Population The population P of a certain animal species after t months can be modeled by P = C(1. In mathematics, an nth root of a number x, is a number r which, when raised to the power n yields x. The number 0 (zero) has just one square root, 0 itself. The nth root calculator below will also provide a brute force rounded approximation of the principal nth root. Simplify the expression. There is no result accuracy argument of Nth_Root, because the iteration is supposed to be monotonically descending to the root when starts at A. When I try to find the roots of the same equation in Mathematica, I receive various errors. 1) the radicand has no perfect nth powers as factors 2) any denominator has been rationalized Example 4: Write the following Radicals in Simplest Form — Remember your perfect nth Roots u = 16. Example of How to Evaluate a “nth Root” Radical Expression. I assume you're saying that the calculator can do the four basic operations and square roots, but not arbitrary exponents. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. In particular, is called the primitive th root of unity. (I am aware that \sqrt as command for a general root is misleading) For the life of me I can't find the right syntax to get the input of n'th root of x. pl help Shambhu. Number of roots equal to n. find a formula for the nth partial sum of the series and use it to find the series sum if the series converges. n = 3, a = 216 b. Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. Suppose we wish to solve the equation z^3 = 2+2 i. Complex Numbers: nth Roots.


# Double Integral Set Up

The question was stated as follows,

Evaluate the following double integral;

$$\iint_R x^3y dA$$

where R is interior of triangle with vertices (0,0), (1,0), & (1,1) .

I thought for these types of double integrals I could have the limits in either of the two formats below;

$$\int_0^1\int_0^y x^3y dxdy$$ [1]

OR

$$\int_0^1\int_0^x x^3y dydx$$ [2]

However the answer sheet states that the answer for only the latter integral is correct.

How can I set up or visualize the problem in order to set it up in the correct way?

I did plot the points on a graph, and got my y=x "limit" from there, I just cannot understand why [2] is the correct integral, and how to go about making sure that in every double integral problem, I choose the correct integral set up.

All help is much appreciated :)

For the first one the bounds on x are wrong. The region is bounded by $y=x$ on the left and on the right by $x=1$. The bound on the left is a lower bound, so it should be

$$\int_0^1 \int_y^1 x^3ydxdy$$

• That makes sense! Thank you. – mnmakrets Jul 24 '15 at 14:10

You can draw the triangle.

Call the region inside of it $\Omega$.

Looking at lines of constant $y$, we'll have that $$\Omega = \{ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x \leq 1, y \leq x \leq 1 \}$$

Draw the triangle. Now think of it this way: if I fix $x$, what will be the extremes of integration for $y$? You'll get something that depends on $x$ and in particular $y$ goes from $0$ to $x$. Then integrate over the values taken on by $x$, and yo get the second form.

Alternatively, you can fix $y$ and notice that $x$ then varies from $y$ to $1$; then integrate over $y$ to get the correct form

$$\int_{0}^1 \int_y^1 x^3y dy dx$$

Usually you draw the area you want to integrate on, and draw the horizontal and vertical lines that corresponds to fixing $x$ or $y$. This helps visualize the problem and find the extremes of integration with ease

According fubini's theorem, you can either integrate parallel to the x-axis or the y-axis and hence there are two possibilities here:

$$\int_0^1 \int_0^x x^3y \ dy \ dx$$

or

$$\int_0^1 \int_y^1 x^3y \ dx \ dy$$

The best way to get to grips with these sort of questions is to practise more. For more, visit the following links:


# How do I obtain the enclosed area of this particular parametric plot?

I'm trying to find a way to obtain the enclosed area of this particular plot. Can someone show me how?

curveplot = ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]],
Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π},
PlotStyle -> {{Thickness[0.01], Darker[Purple]}},
AspectRatio -> Automatic, PlotRange -> All, AxesLabel -> {"x", "y"}]

## 6 Answers

You can get the curve in polynomial implicit form as below.

poly =
GroebnerBasis[{x^2 - ct, y^2 - st, ct^2 + st^2 - 1}, {x, y}, {ct,
st}][[1]]

(* Out[290]= -1 + x^4 + y^4 *)

To get the area, integrate the characteristic function for the interior of the region. That that's where the polynomial is nonpositive (just notice that it is negative at the origin, say).

area = Integrate[Boole[poly <= 0], {x, -2, 2}, {y, -2, 2}]

(* Out[292]= (2 Gamma[1/4] Gamma[5/4])/Sqrt[π] *)

N[area]

(* Out[293]= 3.7081493546 *)

There are other ways to do this if you cannot find an implicit form, but this seems most direct in this case.

--- edit ---

If you can just solve separately for x and y in terms of the parameter t then you can set up a region function. I do this below for the positive quadrant, and take advantage of symmetry to get the full area in approximate form.

reg = Function[{x, y},
If[And @@ {0 <= x <= 1, 0 <= y <= 1, y <= Sqrt[Sin[ArcCos[x^2]]]},
1, 0]];

approxarea = 4*NIntegrate[reg[x, y], {x, 0, 1}, {y, 0, 1}]

(* Out[321]= 3.70814937167 *)

One can actually recover the exact area from this by using Integrate instead of NIntegrate. But this seems like a viable approach in situations where the exact value might not be readily computed.

--- end edit ---

--- edit 2 ---

Here is a Monte Carlo method that does not rely on solving for anything. We extract the line segments, augment with a diagonal, and do some magic.

segs = Cases[curveplot, _Line, Infinity][[1, 1]];
segs = {Join[segs, N[Table[{j, 1 - j}, {j, 0, 1, 1/100}]]]};

I added an extra level of List due to requirements of some further code. First let's reform a line to take a look at this region.

Graphics[Apply[Line, segs]]

Now we create an in-out function, generate a bunch of random points in the unit square of the first quadrant, take a Monte-Carlo approximation of this area. Then multiply by 4 and add 2. Why? because that's what one always does-- it's like selecting "c" when we don't know the multiple choice answer. (Okay, we multiply by 4 to account for all quadrants, and add 2 because we have in effect excised a square of side length $\sqrt{2}$ from the full region.)

To create the in-out function I use code directly from here.

nbins = 100;
Timing[{{xmin, xmax}, {ymin, ymax}, segmentbins} =
polyToSegmentList[{segs[[1]]}, nbins];]

(* Out[414]= {0.040000, Null} *)

len = 100000;
pts = RandomReal[1, {len, 2}];
Timing[
inout = Map[pointInPolygon[#, segmentbins, xmin, xmax, ymin, ymax] &,
pts];]
approxarea = 4.*Length[Cases[inout, True]]/len + 2.

(* Out[419]= {2.750000, Null} *)

(* Out[420]= 3.7092 *)

I would imagine one could do a bit better by integrating just the unit square in the first quadrant with Method -> "QuasiMonteCarlo", sowing the points via the EvaluationMonitor option, and using those instead of the random set above. This will give a low-discrepancy sequence. Or generate such a set directly; bit offhand I don't know how to do that.

-- end edit 2 ---

• Thanks, I wanted to see if this checked out with a monte carlo estimate. – Black Milk Apr 2 '13 at 21:55

Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:

pts = First[Cases[
ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]},
{t, 0, 2 π}, Exclusions -> None, Method -> {MaxBend -> 1.},
PlotPoints -> 100] // Normal, Line[l_] :> l, ∞]];

PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2

PolygonSignedArea[pts]
3.7081447086368127

The value thus obtained is pretty close to the results in the other answers.

Surprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:

(* $VersionNumber < 10. *) GraphicsMeshMeshInit[]; PolygonArea[pts] 3.708144708636812 (*$VersionNumber >= 10. *)
GraphicsPolygonUtilsPolygonArea[pts]
3.708144708636812

But, what you really should know is that Lamé curves have been well studied, and there is in fact a closed form expression for the area of a Lamé curve. Given the Cartesian equation

$$\left|\frac{x}{a}\right|^r+\left|\frac{y}{b}\right|^r=1$$

the formula for the area of a Lamé curve (formula 5 here) is

$$A=\frac{4^{1-\tfrac1{r}}ab\sqrt\pi\;\Gamma\left(1+\tfrac1{r}\right)}{\Gamma\left(\tfrac1{r}+\tfrac12\right)}$$

In particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,

With[{a = 1, b = 1, r = 4},
N[(4^(1 - 1/r) a b Sqrt[π] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]]
3.7081493546027438369

This is the same as the answer Daniel obtained through more general methods.

• Never heard of that method before. Nice. (Yes, I upvoted..) – Daniel Lichtblau Apr 3 '13 at 15:42
• I've taught the "shoelace" method but never heard of the name. (+1) – Michael E2 Apr 3 '13 at 20:32
• @J.M. Thank you for the additional insight. – Black Milk Apr 3 '13 at 20:55
• @Michael, I must confess that I too have been using that method for a while now, but only learned it had a name on math.SE ... – J. M.'s technical difficulties Apr 4 '13 at 2:37
• You could just do PolygonArea@pts... no need to wrap it in Polygon – rm -rf Apr 25 '13 at 21:19

One can use one of the line integral forms of the area, derived from Green's Theorem: $$A = \frac12 \int_C x \; dy - y \; dx = \int_C x \; dy = - \int_C y \; dx$$ The first one is symmetric, which sometimes is an advantage.

c[t_] := {Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}

dA = 1/2 c'[t].Cross[c[t]]
(* complicated output *)

One problem with this parametrization are the derivatives of Abs and Sign. They are discontinuous at isolated points, and as far as the integral is concerned, it does not matter what value we assign them at the discontinuities. So we can simplify matters by substituting for them. We can also substitute 1 for Sign[x]^2, since x will be 0 only at few isolated points. Thus the differential is

dA = dA /. {Sign'[x_] :> 0, Abs'[x_] :> Sign[x]} /. {Sign[_]^2 :> 1} // Simplify
(* (Abs[Cos[t]] Cos[t] Sign[Cos[t]] + Abs[Sin[t]] Sign[Sin[t]] Sin[t]) /
(4 Sqrt[Abs[Cos[t]]] Sqrt[Abs[Sin[t]]]) *)

NIntegrate returns a small imaginary component

NIntegrate[dA, {t, 0, 2 π}]
(* 3.70815 - 1.97076*10^-10 I *)

Oddly, setting WorkingPrecision reduces the imaginary error, even if it is set to less than MachinePrecision (15.9546)

NIntegrate[dA, {t, 0, 2 π}, WorkingPrecision -> 10]
(* 3.708149355 + 0.*10^-21 I *)

Integrate returns an exact answer in this case:

Integrate[dA, {t, 0, 2 π}]
(* (3 Sqrt[2] π Gamma[5/4] + 4 Gamma[3/4] Gamma[5/4]^2) /
(2 Sqrt[π] Gamma[3/4]) *)

FullSimplify @ %
(* (Sqrt[π/2] Gamma[1/4])/Gamma[3/4] *)

N[%, 20]
(* 3.7081493546027438369 *)

In cases where it is not possible to get a closed form for the curve, you can use the image processing functions to get an approximation for the enclosed area. First, some small changes to the plot — get rid of the axes, labels and everything else that isn't needed, and set the aspect ratio to 1.

curveplot = ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]},
{t, 0, 2 π}, PlotStyle -> Thick, AspectRatio -> 1, Axes -> False, PlotRange -> {-1, 1}]

Next, close the holes in the curve using morphological operations:

im = Binarize@curveplot ~Opening~ 7 // DeleteSmallComponents

Depending on the curve, you'll have to tweak the parameters and be careful with DeleteSmallComponents, in case you have disconnected components in your plot (always do a visual check or ImageAdd[curveplot, im] to see if it's correct). Sometimes, this may even not be necessary if your curve is "nice".

Finally, use ComponentMeasurements to get the area of the enclosed space. The result is in sq. pixels, so I normalize by the total area (in sq. pixels) and multiply by the area of the plot range rectangle in the original plot (i.e., $(x_{max}-x_{min})(y_{max}-y_{min})$)

4 (1 /. ComponentMeasurements[im, "Area"]) / Times @@ ImageDimensions@im
(* 3.66738 *)

which looks about right, since your curve fits in a square of side 2 and is close to Daniel's answer of 3.708. You can get a closer approximation if you increase the ImageSize in the original plot. Using ImageSize -> 2000 gives me 3.70129 as the area (but note that the image processing steps will take longer to compute).

• Exclusions -> None in ParametricPlot[] should close up the artifact holes nicely. – J. M.'s technical difficulties Apr 3 '13 at 15:27

Integrating InterpolatingFunction.

p = Table[{
Sign[Cos@t] Sqrt[Abs@Cos@t],
Sign[Sin@t] Sqrt[Abs@Sin@t]}, {t, 0, Pi, Pi/100.}];
f = Interpolation[p];
2*NIntegrate[f@t, {t, -1, 1}]
3.70771

As a modified version of Michael E2's answer:

I tried to rewrite your original curve as below for $$t\in \left[0,\tfrac{\pi}{2}\right]$$, in order to make sure the derivatives of the parametric form can be obtained easily by avoiding Abs or Sign:

ncurve = {(Cos[t]^2 )^(1/4), (Sin[t]^2)^(1/4)}

Then the result of the closed area can be obtained by applying Green's theorem to the curve and by using the symmetry of the curve:

area = 4*Integrate[ncurve[[1]] D[ncurve[[2]], t], {t, 0, Pi/2}]

which gives:

$$\dfrac{8\; \Gamma^2 \left(\tfrac{5}{4}\right)}{\sqrt{\pi }}$$

and the numerical result is therefore:

area // N[#, 20] &
`

3.7081493546027438369


# What is the difference between ($\tan x \sec^2x$) and ($\sin x/\cos^3x$)? Why is the answer to the integration different?

$$\int \:\frac{\left(\sin x+\tan x\right)}{3\cos^2x}dx$$

I know I have to split the equation into

$$\frac{1}{3}\int \:\left(\:\frac{\sin x}{\cos x}\right)\left(\frac{1}{\cos x}\right)dx+\frac{1}{3}\int \:\left(\:\tan x\right)\left(\frac{1}{\cos^2x}\right)dx$$

I know that for the first part, it is $$\frac{1}{3}\int \tan x\sec xdx$$ which is $$\sec x$$.

However, for the second part, wouldn't it be $$\frac{1}{3}\int \tan x \sec^2xdx$$

If I used $$u=\tan x$$ then $$du=\sec^2xdx$$ so wouldn't the answer be $$\frac{1}{6}\tan^2x$$

However, the book is saying that the second part is supposed to be $$\frac{1}{6}\sec^2x$$ because I was supposed to convert the second part into $$\frac{1}{3}\int \frac{\sin x}{\cos^3x}dx$$ and let $$u=\cos x$$

What I am doing wrong? Why can't it be $$\tan \sec^2x$$ instead of $$\sin x/\cos^3x$$?

• Remember $\sec^2{\theta}=1+\tan^2{\theta}$ – user41736 Mar 4 '16 at 6:49

Recall that

$$1 + \tan^2 x = \sec^2 x$$

or, since I dislike the secant,

$$\frac{1}{\cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\cos^2 x} = 1+\tan^2 x.$$

Therefore, your answer and the book's answer only differ by a constant.

Both the answers differ by a constant(1/6). So both the answers and both the methods are correct. The constant of integration takes care of the constant(1/6).

Both are right as when you differentiate $1/6tan^2x$ with chain rule you get $\frac{1}{6}.2tanx.sec^2x=1/3.tanx.sec^2x$ also differentiating $1/6sec^2x$ you get the same answer . so both are right.

• Note the constants can be $c,c'$ as integration gives a family of curves and an exact constant gives a specific curve – Archis Welankar Mar 4 '16 at 6:48


## Section13.8Rational Exponents

The power to a power rule of exponents relates that $(x^m)^n=x^{mn}\text{.}$ This rule is fairly intuitive when both exponents are positive. For example, in the expression $(x^4)^3$ there are three factors of $x^4\text{,}$ each of which contains four factors of $x\text{,}$ so all together there are four factors of $x\text{,}$ three times, i.e. $3 \cdot 4$ factor of $x\text{.}$

While the power to a power rule is less intuitive once you move away from positive integer exponents, the rule remains the same regardless of the nature of the exponents. For example:

\begin{align*} (x^{1/3})^3\amp=x^{\frac{1}{3} \cdot 3}\\ \amp=x^{1}\\ \amp=x \end{align*}

But we already have a name for the expression that when cubed results in $x\text{,}$ and that name is $\sqrt[3]{x}$ (the cube root of $x$). So it must be the case that $x^{1/3}=\sqrt[3]{x}\text{.}$ In general, is $n$ is any positive integer, then:

\begin{equation*} x^{1/n}=\sqrt[n]{x} \end{equation*}

and more generally,

\begin{equation*} x^{m/n}=\sqrt[n]{x^m}\text{.} \end{equation*}

Several examples are shown below.

###### Example13.8.1.

Express $y^{7/5}$ as an equivalent radical expression

Solution
\begin{equation*} y^{7/5}=\sqrt[5]{y^7} \end{equation*}
###### Example13.8.2.

Express $\sqrt[3]{w^{12}}$ using an equivalent exponential expression

Solution
\begin{align*} \sqrt[3]{w^{12}}\amp=w^{12/3}\\ \amp=w^4 \end{align*}
###### Example13.8.3.

Express $\sqrt{x^9}$ using an equivalent exponential expression

Solution
\begin{equation*} \sqrt{x^9}=x^{9/2} \end{equation*}

You can use Figure 13.8.4 to explore this definition some more.

As long as both the numerator and denominator of a rational exponent are fairly small positive numbers, it is fairly easy to evaluate expressions that include rational exponents using the rule $x^{m/n}=\sqrt[n]{x^m}\text{.}$

###### Example13.8.5.

Evaluate $16^{1/2}\text{.}$

Solution
\begin{align*} 16^{1/2}\amp=\sqrt{16}\\ \amp=4 \end{align*}
###### Example13.8.6.

Evaluate $8^{2/3}\text{.}$

Solution
\begin{align*} 8^{2/3}\amp=\sqrt[3]{8^2}\\ \amp=\sqrt[3]{64}\\ \amp=4 \end{align*}
###### Example13.8.7.

Evaluate $100^{3/2}\text{.}$

Solution
\begin{align*} 100^{3/2}\amp=\sqrt{100^3}\\ \amp=\sqrt{1000000}\\ \amp=1000 \end{align*}

When the numerator of the rational exponent is large, the rule $x^{m/n}=\sqrt[n]{x^m}$ can become quite cumbersome. Consider, for example, evaluating $9^{5/2}\text{.}$ If we try to use the standard form we hit a brick wall. First, it's not trivial to calculate that $9^5=59,049$ (reality check ... I grabbed my calculator). Now that I have the value of 59,049, I have to determine its square root. Oh my!

Fortunately for us, the application of the exponent and the application of the radical can be done in either order. That is:

\begin{equation*} a^{m/n}=\sqrt[n]{x^m} \text{ and } a^{m/n}=(\sqrt[n]{x})^m \end{equation*}
###### Example13.8.8.

Using the second option, evaluate $9^{5/2}\text{.}$

Solution
\begin{align*} 9^{5/2}\amp=(\sqrt{9})^5\\ \amp=3^5\\ \amp=243 \end{align*}
###### Example13.8.9.

Using the second option, evaluate $16^{7/4}\text{.}$

Solution
\begin{align*} 16^{7/4}\amp=(\sqrt[4]{16})^7\\ \amp=2^7\\ \amp=128 \end{align*}

Rational exponents are allowed to be negative. If that's the case, you probably want to deal with the negative aspect of the exponent before taking on the fractional aspect.

###### Example13.8.10.

Evaluate $27^{-2/3}\text{.}$

Solution
\begin{align*} 27^{-2/3}\amp=\frac{1}{27^{2/3}}\\ \amp=\frac{1}{(\sqrt[3]{27})^2}\\ \amp=\frac{1}{3^2}\\ \amp=\frac{1}{9} \end{align*}

Sometimes radical expressions can be simplified after first rewriting the expressions using rational exponents and applying the appropriate rules of exponents. If the resultant expression still has a rational exponent, it is standard to convert back to radical notation. Several examples follow.

###### Example13.8.11.

Use rational exponents to simplify $\text{.}$ Where appropriate, your final result should be converted back to radical form.

Solution
\begin{align*} \sqrt[3]{y^2} \cdot \sqrt[6]{y}\amp=y^{2/3}y^{1/6}\\ \amp=y^{2/3+1/6}\\ \amp=y^{5/6}\\ \amp=\sqrt[6]{y^5} \end{align*}
###### Example13.8.12.

Use rational exponents to simplify $\sqrt[8]{t^4}\text{.}$ Where appropriate, your final result should be converted back to radical form.

Solution
\begin{align*} \sqrt[8]{t^4}\amp=t^{4/8}\\ \amp=t^{1/2}\\ \amp=\sqrt{t} \end{align*}
###### Example13.8.13.

Use rational exponents to simplify $\sqrt[10]{\sqrt{5^{40}}}\text{.}$ Where appropriate, your final result should be converted back to radical form.

Solution
\begin{align*} \sqrt[10]{\sqrt{5^{40}}}\amp=\sqrt[10]{5^{40/2}}\\ \amp=\sqrt[10]{5^{20}}\\ \amp=5^{20/10}\\ \amp=5^2\\ \amp=25 \end{align*}

### ExercisesExercises

Convert each exponential expression to a radical expression and each radical expression to an exponential expression. When converting to a rational exponent, reduce the exponent if possible. Assume that all variables represent positive values.

###### 1.

$x^{1/3}$

Solution

$x^{1/3}=\sqrt[3]{x}$

###### 2.

$y^{5/4}$

Solution

$y^{5/4}=\sqrt[4]{y^5}$

###### 3.

$z^{2/5}$

Solution

$z^{2/5}=\sqrt[5]{z^2}$

###### 4.

$\sqrt[11]{x^5}$

Solution

$\sqrt[11]{x^5}=x^{5/11}$

###### 5.

$\sqrt[4]{y^{20}}$

Solution

\begin{aligned}[t] \sqrt[4]{y^{20}}\amp=y^{20/4}\\ \amp=y^5 \end{aligned}

###### 6.

$\sqrt[15]{t^3}$

Solution

\begin{aligned}[t] \sqrt[15]{t^3}\amp=t^{3/15}\\ \amp=t^{1/5} \end{aligned}

Determine the value of each expression.

###### 7.

$4^{1/2}$

Solution

\begin{aligned}[t] 4^{1/2}\amp=\sqrt{4}\\ \amp=2 \end{aligned}

###### 8.

$27^{-1/3}$

Solution

\begin{aligned}[t] 27^{-1/3}\amp=\frac{1}{27^{1/3}}\\ \amp=\frac{1}{\sqrt[3]{27}}\\ \amp=\frac{1}{3} \end{aligned}

###### 9.

$\left(\frac{4}{9}\right)^{-1/2}$

Solution

\begin{aligned}[t] \left(\frac{4}{9}\right)^{-1/2}\amp=\left(\frac{9}{4}\right)^{1/2}\\ \amp=\sqrt{\frac{9}{4}}\\ \amp=\frac{3}{2} \end{aligned}

###### 10.

$8^{7/3}$

Solution

\begin{aligned}[t] 8^{7/3}\amp=(\sqrt[3]{8})^7\\ \amp=2^7\\ \amp=128 \end{aligned}

###### 11.

$100^{5/2}$

Solution

\begin{aligned}[t] 100^{5/2}\amp=(\sqrt{100})^5\\ \amp=10^5\\ \amp=100,000 \end{aligned}

###### 12.

$16^{-9/4}$

Solution

\begin{aligned}[t] 16^{-9/4}\amp=\frac{1}{16^{9/4}}\\ \amp=\frac{1}{(\sqrt[4]{16})^9}\\ \amp=\frac{1}{2^9}\\ \amp=\frac{1}{512} \end{aligned}

Simplify each radical expression after first rewriting the expression in exponential form. Assume that all variables represent positive values. Where appropriate, your final result should be converted back to radical form.

###### 13.

$\sqrt[5]{t^{20}}$

Solution

\begin{aligned}[t] \sqrt[5]{t^{20}}\amp=t^{20/5}\\ \amp=t^4 \end{aligned}

###### 14.

$6\sqrt[33]{x^{77}}$

Solution

\begin{aligned}[t] 6\sqrt[33]{x^{77}}\amp=6x^{77/33}\\ \amp=6x^{7/3}\\ \amp=6\sqrt[3]{x^7} \end{aligned}

###### 15.

$(\sqrt{3})^{10}$

Solution

\begin{aligned}[t] (\sqrt{3})^{10}\amp=3^{10/2}\\ \amp=3^5\\ \amp=243 \end{aligned}

###### 16.

$\sqrt[4]{9^2}$

Solution

\begin{aligned}[t] \sqrt[4]{9^2}\amp=9^{2/4}\\ \amp=9^{1/2}\\ \amp=\sqrt{9}\\ \amp=3 \end{aligned}

###### 17.

$\sqrt{w}\sqrt[4]{w}$

Solution

\begin{aligned}[t] \sqrt{w}\sqrt[4]{w}\amp=w^{1/2}w^{1/4}\\ \amp=w^{3/4}\\ \amp=\sqrt[4]{w^3} \end{aligned}

###### 18.

$\sqrt[7]{x^6}\sqrt[7]{x}$

Solution

\begin{aligned}[t] \sqrt[7]{x^6}\sqrt[7]{x}\amp=x^{6/7}x^{1/7}\\ \amp=x^1\\ \amp=x \end{aligned}

###### 19.

$(\sqrt[12]{x^7y^{16}})^{36}$

Solution

\begin{aligned}[t] (\sqrt[12]{x^7y^{15}})^{36}\amp=(x^7y^{16})^{36/12}\\ \amp=(x^7y^{16})^3\\ \amp=x^{21}y^{48} \end{aligned}

###### 20.

$\sqrt[5]{\sqrt[3]{x^{15}}}$

Solution

\begin{aligned}[t] \sqrt[15]{\sqrt[3]{x^{15}}}\amp=\sqrt[15]{x^{15/3}}\\ \amp=\sqrt[15]{x^5}\\ \amp=x^{5/15}\\ \amp=x^{1/3}\\ \amp=\sqrt[3]{x} \end{aligned}


IntMath Home » Forum home » Exponents and Radicals » exponential functions

# exponential functions [Solved!]

### My question

In expressions containing raised variables that are being divided, why do the variables not cancel each other out to a reciprocal of one? for example x^n/ x should be 1^n or is it still x^n?

### Relevant page

factoring polynomials with exponents - Google Search

### What I've done so far

(x^n/ x^n-2)^3 does this equal 1 or x^6?

X

In expressions containing raised variables that are being divided, why do the variables not cancel each other out to a reciprocal of one? for example x^n/ x should be 1^n or is it still x^n?
Relevant page

factoring polynomials with exponents - Google Search

What I've done so far

(x^n/ x^n-2)^3 does this equal 1 or x^6?

## Re: exponential functions

Hello Sam

It's often best to try some real numbers in algebraic expressions first, to see what is going on.

Let's try 10^4.

This means 10 xx 10 xx 10 xx 10

If we divide this by 10, we would have

(10 xx 10 xx 10 xx 10)/10

We cancel one of the 10s on top with the 10 on bottom, to give:

10 xx 10 xx 10

So what we've done is

(10^4)/10 = 10^3

If you do this for several other indices, you'll hopefully conclude that:

(10^n)/10 = 10^(n-1)

In general, for your first question, we'd have:

x^n/ x = x^(n-1)

For your other question, (x^n/ x^n-2)^3, I'm not sure if you mean (x^n/ x^n-2)^3 or (x^n/ (x^n-2))^3?

You are encouraged to use the math entry system which makes it easier to express your questions and follow your working.

X

Hello Sam

It's often best to try some real numbers in algebraic expressions first, to see what is going on.

Let's try 10^4.

This means 10 xx 10 xx 10 xx 10

If we divide this by 10, we would have

(10 xx 10 xx 10 xx 10)/10

We cancel one of the 10s on top with the 10 on bottom, to give:

10 xx 10 xx 10

So what we've done is

(10^4)/10 = 10^3

If you do this for several other indices, you'll hopefully conclude that:

(10^n)/10 = 10^(n-1)

In general, for your first question, we'd have:

x^n/ x = x^(n-1)

For your other question, (x^n/ x^n-2)^3, I'm not sure if you mean (x^n/ x^n-2)^3 or (x^n/ (x^n-2))^3?

You are encouraged to use the math entry system which makes it easier to express your questions and follow your working.

## Re: exponential functions

Sam didn't reply.

If his second question meant (x^n/ x^n-2)^3, the the result is (1-2)^3=(-1)^3=-1

If he meant (x^n/ (x^n-2))^3, then there is no "nice" simplification.

X

Sam didn't reply.

If his second question meant (x^n/ x^n-2)^3, the the result is (1-2)^3=(-1)^3=-1

If he meant (x^n/ (x^n-2))^3, then there is no "nice" simplification.

## Reply

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Height of a tetrahedron

How do I calculate: The height of a regular tetrahedron, side length 1.

Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.

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I'm curious. If you mind telling us, what was the purpose of the interview? What were the other mathematically relevant questions asked? –  user02138 Nov 20 '10 at 15:25
It's an old question for and Economics and Management interview at Oxford. –  Patrick Beardmore Nov 20 '10 at 16:43

The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.

Here's a view of the geometry:

and here's a view of the bottom face:

In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.

Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.

Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):

$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$

Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated here.

From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies

$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$

Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned here.

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It's not homework, it was an interview question I struggled with. I'd love to know the answer. –  Patrick Beardmore Nov 20 '10 at 14:57
Good to know. I'll edit with an explicit answer. –  Ｊ. Ｍ. Nov 20 '10 at 15:00
Thank-you very much. P.S: how did you make your lovely 3d shape diagrams? –  Patrick Beardmore Nov 20 '10 at 16:43
@Patrick: I used Mathematica 5.2 (yes, it's an old copy :) ) for these diagrams. –  Ｊ. Ｍ. Nov 20 '10 at 21:33

Consider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\frac{2}{\sqrt 3}$, and since the edges of this tetrahedron have length $\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \sqrt{\frac{2}{3}}$.

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Note that this way of doing this also gives that the distance from the centers of opposite sides of the tetrahedron is always ${1 \over \sqrt{2}}$ times the side length. –  Zarrax Nov 20 '10 at 15:26

You can also use trig based on the dihedral angle between two faces of the tetrahedron.

Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,

$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$

$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.

As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:

$$\begin{eqnarray} |OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\ s^2 &=& \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\ s^2 &=& \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\ 1 &=& \frac{3}{2} - \frac{3}{2} \cos{M} \\ \frac{-1}{2} &=& - \frac{3}{2} \cos{M} \\ \frac{1}{3} &=& \cos{M} \;\;\; (**)\\ \Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&=& \sin{M} \end{eqnarray}$$

Therefore,

$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$

(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):

$$\begin{eqnarray} \cos\left({\text{angle between two sides of a regular triangle}}\right) &=& \frac{1}{2}\\ \cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &=& \frac{1}{3}\\ \cos\left({\text{angle between two facets of a regular n-simplex}}\right) &=& \frac{1}{n} \end{eqnarray}$$

(Who would've suspected, upon first encountering it, that the "$2$" in "$\cos{60^{\circ}}=\frac{1}{2}$" was actually a reference to the dimension of the triangle?)

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Thank-you v. much! What's a simplex? –  Patrick Beardmore Nov 20 '10 at 17:14
A simplex is the analog of "triangle" in any dimension: the simplest possible shape. It's what you get when you join $(n+1)$ points in $n$-dimensional space. A triangle is a "$2$-simplex" ($3$ points in $2$ dimensions); a tetrahedron is a "$3$-simplex" ($4$ points in $3$ dimensions); going upward in dimensions, one generally just says "$4$-", "$5$-", ..., "$n$-simplex"; going downward, one can say the line segment is a "1-simplex" (determined by $2$ points) and the point is a "0-simplex" ($1$ point!). See en.wikipedia.org/wiki/Simplex and mathworld.wolfram.com/Simplex.html –  Blue Nov 20 '10 at 17:30
Very nice! I never knew about this pattern until now, thanks! :D –  Ｊ. Ｍ. Nov 20 '10 at 21:37