The Math Thread

[Tannhauser]

...for everything math related: stuff you find interesting/fascinating, problems you have solved or are trying to solve, philosophical/theoretical/practical inquiries regarding math, etc.


Mathematics is a game played according to certain simple rules with meaningless marks on paper. --Hilbert

 

[Tannhauser]

I'll kick things off with this small brain teaser which I have asked quite a lot of people (mostly math students -- only one guy answered it correctly and confidently):

There is a town with two hospitals: one big and one small (just for intuition, let's say the big one contains 10 times as many patients as the small one). At a particular day, there were born 60% girls in one of the hospitals. Which one was it most likely?

 

[QuickTwist]

Smaller sample size. I'd say the smaller one.

 

[Feather]

The size of the hospital doesnt dictate boy or girl ratio as numbers get large. Except the small one would need at least x people to get to some margin close to 60 percent. So the real mathmatical question to solve is in x number of samples @ assumed 50/50 what is the odds of getting with in some margin 60 percent girls. Which once I get home get formula I think. Then if the bigger hospital is z times bigger you increase x and find the odds for the bigger hospital. Then you use the bayes theorem now that you have both odds to get final answer.

This reminds me of the problem in a prob and stat text book about 2 assembly lines each had some odds of defect what is the odds a defected product came from a certain assembly line. I think it's some extended version of bayes theorem I would have to check. But I know were to look.

 

[Tannhauser]

@Feather
I didn't quite follow why the size of the hospital did not matter. It is possible the question is slightly unprecise, but the point was, as QuickTwist wrote, that the smaller one will have a smaller sample. If we instead flipped coins for example, it is clearly more probable you will have 60% heads if you flip 10 times, compared to if you flip, say, 10000 times.

I think the picture in the wiki article on Law of Large numbers illustrates this well: https://en.wikipedia.org/wiki/Law_of_large_numbers

 

[Feather]

Ya I was being confusing. I was trying to think how to address the problem with math so I was stating assumptions as needed but I was on phone and it was shitty.

Assumptions:
1. Each hospital has the same odds of a boy or girl no matter its size or location or the type of patients that go there.
2. Those odds are 50/50

Sidetrack thought
{I noticed that say its hard to get 60% when only 5 people had kids that day. That lead my mind down a path of how to mathematically show that with number theory how many sets of trials could there even be that can get exactly 60%. Then my mind started to think how to define some range around a percentage and how many sets of trials is needed to get with in that range a certain percent of the time. [these thoughts are needed to solve the problem to the tooth but, overkill from the general idea of the problem]} leaving this thought by stating 60% or greater:

Now that we have 50/50 established by assumptions:

Hospital A = x babies
Hospital B = y babies

2^x= Number of combinations of babies Ex. 2^10=1024

P(>.6*x Girls) = 1 - [P(.6*x-1) + P((.6*x-1)-1) + Continue to P(0)]

{to find one of these terms a computer algorithm would have to figure out say for P(3) term how many sets have 3 girls. Kind of like for a coin flip figuring out 3 flips S{HHH,HHT,HTT,HTH,TTT,THH,TTH,THT} by a pattern. Then that number is divided by 2^x}

NOW: we would have the odds of getting 60% or greater girls from Hospital A
We could do the same thing for Hospital B.

THEN: once you have the odds for A & B you use the Bayes theorem to find out the exact odds say 60% or greater came from A rather than B.

 

[QuickTwist]

Does this mean I passed the logic test?

I'll take door number 2, its always door number 2.

 

[Feather]

Ya smaller sample would me more likely.

10 coin flips has 1024 outcomes each outcomes has the same odds .097 percent
20 coin flips has 1,048,576 outcomes each outcomes has same odds .000095 percent

As there is more outcomes there is a greater meat toward the center were you get fewer outcomes toward the edge say 8 heads compared to the total outcomes. So the smaller outcomes has less arrangements in the center closer to 5 heads. By arrangement I mean {HTHHTTHTHT} for ten flips.

 

[Tannhauser]

Another way of thinking about it is:

If there are born N babies and G is the no. of girls, the ratio we are looking at is G/N. The variable G is a sum of Bernoulli trials with p=0.5, so it has the Binomial distribution with mean N*0.5.

If we say G and N are the no. of girls and total no. of births in the small hospital respectively, while G' and N' are the corresponding variables for the big one (with N' = 10*N), we get that the variance of the ratio for the small one is

Var(G/N) = Var(G)/N^2 = N(1/4) / N^2 = 1/(4*N)

while for the big one gets Var(G'/N') = 1/(4N') = 1/(40*N)

hence if the small one has 1/10 the number of births of the big one, it will have 10 times as high variance of the ratio of girls.

 

[Coolydudey]

dude get real with the maths teasers:

The sume of some (not necessarily distinct) natural numbers (i.e. positive integers) is 100. How large can their product be?

 

[Tannhauser]

@Coolydudey
Good one. I'd start with guessing that it is 4^(25). It seems that 4 is the only integer that does not decompose into a sum of integers whose product exceeds itself. I.e. we have 2+2=4, 3+1=4, with 2*2=4 and 3*1 = 3 (if we tried for example 6 we get 6=3+3 but 3*3=9).

So when you have (4*...*4)*4 where there are 24 fours inside the parentheses, you cannot increase the product by decomposing the outside 4 into other factors.

 

[Tannhauser]

Here is, by the way, one I got at a recent job interview:

You have a group of people. 80% of them like dogs, 60% of them like cats. What can you say about how many like both dogs and cats?

 

[Inquisitor]

Here is, by the way, one I got at a recent job interview:

You have a group of people. 80% of them like dogs, 60% of them like cats. What can you say about how many like both dogs and cats?

40%?

 

[Cheeseumpuffs]

0 0 0 0 0 0 0 0 0 0 ------People
0 0 0 0 0 0 0 0 ---------Like Dogs
0 0 0 0 0 0 -------------Like Cats

0 0 0 0 0 0 0 0 0 0 ------ People
0 0 0 0 0 0 0 0 ---------- Like Dogs
-------0 0 0 0 0 0 ------ Like Cats
(Edit: argh, formatting is dumb)

At least 40% like both but no more than 60%
Although I'm not confident, as this seems like far too simple an answer. I feel like I'm skipping over something important.


Assuming the percentages are applied randomly you can take the 60% to apply to both dog-likers and non-dog-likers and multiply both groups by 60%
So you get:
Both-likers = (cat-likers)(dog-likers) = (.6)(.8) = .48 = 48%
Dog-only-likers = (dog-likers)(non-cat-likers) = (.8)(.4) = .32 = 32%
Cat-only-likers = (non-dog-likers)(cat-likers) = (.2)(.6) = .12 = 12%
Curmudgeons = (non-dog-likers)(non-cat-likers) = (.2)(.4) = .08 = 8%

But that's only if the percentages are applied evenly over the group. In reality I'd say that animal lovers are going to be animal lovers and you should expect the cat-likers group and the dog-likers group to overlap more. My personal estimate would be somewhere upwards of 50%.

 

[ummidk]

Atleast 40% like both? Anyway to prove more?

@Coolydudey
Good one. I'd start with guessing that it is 4^(25). It seems that 4 is the only integer that does not decompose into a sum of integers whose product exceeds itself. I.e. we have 2+2=4, 3+1=4, with 2*2=4 and 3*1 = 3 (if we tried for example 6 we get 6=3+3 but 3*3=9).

So when you have (4*...*4)*4 where there are 24 fours inside the parentheses, you cannot increase the product by decomposing the outside 4 into other factors.

I was gonna say 2^50(4^25), but you took it a step further, kudos

 

[Tannhauser]

@Inquisitor
No, sir. Also, a hint: one has to think slightly outside the box and look carefully at the question.

Edit: (after seeing Cheese's and ummidk's answers) yup, the answer is that it lies in the range [0.4, 0.6]

 

[Cheeseumpuffs]

look carefully at the question.

Ugh, this isn't some trick question bullshit is it?

You could say that at least 60% like dogs and cats, maybe 100%.

It depends how you apply the "and"

If you're looking for people who like (dogs *and* cats) then it's as I said in my previous post.

If you're looking for (people who like dogs) *and* (people who like cats) then it can be anywhere upwards of 60%

 

[Tannhauser]

@Cheese yeah, what I meant was that it is not asking for a number but some information about the number. My initial response at the interview was that you cannot tell what the number is They told me to read the question once more..

 

[ummidk]

Here is, by the way, one I got at a recent job interview:

You have a group of people. 80% of them like dogs, 60% of them like cats. What can you say about how many like both dogs and cats?

60% of them (dog lovers?) like cats? Therefore 48%?

 

[Tannhauser]

maybe if I would have written "60% of them dog lovers like cats"

 

[onesteptwostep]

40% to 60% like both cats and dogs.

What I did was add the number of people who did not like dogs (20%) with the people who did not like cats (40%) added them (20% + 40% = 60%) and then subtracted that from the whole (100% - 60% = 40%). 40% happens when there's the max amount of people who like only dogs or only like cats.

The 60% just comes from the possibility of the 60% cat lovers also liking dogs, which are 80%.

So 40% to 60%.

 

[Inquisitor]

Here is, by the way, one I got at a recent job interview:

You have a group of people. 80% of them like dogs, 60% of them like cats. What can you say about how many like both dogs and cats?

Oh haha, I see now.

Say there's 100 people. 80 will like dogs. 20 won't like dogs. 60% like cats, and 40% don't like cats.

60% of the 80 that like dogs also like cats. Answer is 48%.

It's impossible to say. 80% of the people like dogs, but 60% of the dogs like cats. There's no way to tell what percent of the people like cats, only that 20% don't like dogs. So...up to 80% of the people like both dogs and cats. Assuming dogs like themselves, 60% of dogs like cats.

 

[ummidk]

Many things could be said about the number, but it ranging from 40-60% seems to be the most meaningful, what am i missing?)

 

[Coolydudey]

How many subsets of {1,2,...,n} are there of even size?

 

[Cheeseumpuffs]

How many subsets of {1,2,...,n} are there of even size?

Taking n=1 we see that there are 0 subsets of even size.
n=2 yields 1
n=3 yields 3
n=4 yields 6
n=5 yields 10
n=6 yields 15
. . .

This gives us a sequence S in which the ith term (Si) is equal to the (i-1)th term (S(i-1)) plus (i-1)
or:
Sn = S(n-1)+(n-1)
(can't figure out subscript but imagine the stuff paired with the S's are subscript)

There's probably another step somewhere in there which would replace the S(n-1) so that the sequence isn't recursive, but I just got home from work and I'm tired and haven't had basically anything to eat today so I'm done here.

Oh wait, no I'm not I figured it out

S1 = 0 = 1(0) = 1(0/2)
S2 = 1 = 2(1/2) = 2(1/2)
S3 = 3 = 3(1) = 3(2/2)
S4 = 6 = 4(3/2) = 4(3/2)
S5 = 10 = 5(2) = 5(4/2)
S6 = 15 = 6(5/2) = 6(5/2)
. . .
Sn = n((n-1)/2)

Booyah motherfuckers.

 

[Coolydudey]

Taking n=1 we see that there are 0 subsets of even size.
n=2 yields 1
n=3 yields 3
n=4 yields 6
n=5 yields 10
n=6 yields 15
. . .

This gives us a sequence S in which the ith term (Si) is equal to the (i-1)th term (S(i-1)) plus (i-1)
or:
Sn = S(n-1)+(n-1)
(can't figure out subscript but imagine the stuff paired with the S's are subscript)

There's probably another step somewhere in there which would replace the S(n-1) so that the sequence isn't recursive, but I just got home from work and I'm tired and haven't had basically anything to eat today so I'm done here.

Oh wait, no I'm not I figured it out

S1 = 0 = 1(0) = 1(0/2)
S2 = 1 = 2(1/2) = 2(1/2)
S3 = 3 = 3(1) = 3(2/2)
S4 = 6 = 4(3/2) = 4(3/2)
S5 = 10 = 5(2) = 5(4/2)
S6 = 15 = 6(5/2) = 6(5/2)
. . .
Sn = n((n-1)/2)

Booyah motherfuckers.

Cheese I'm afraid you'll find that you got the number of subsets wrong for every n. First, the empty set is a subset (mathematical formality, but natural).

So:
n=1 : 1
n=2 : 2
n=3 : 4
n=4 : 8
etc.

The pattern is obvious, namely 2^(n-1)
It is worth knowing the following standard argument: the number of subsets of {1,2,...n} is 2^n. Indeed, all subsets can be constructed as follows:
is 1 in it? 2 choices
is 2 in it? 2 choices
etc.
In total 2^n choices of subsets.

You can modify this standard argument in a very pretty way when n is odd, i.e. how many even size subsets of {1,2,...n} are there when n is odd?

 

[Cheeseumpuffs]

Cheese I'm afraid you'll find that you got the number of subsets wrong for every n. First, the empty set is a subset (mathematical formality, but natural).

So:
n=1 : 1
n=2 : 2
n=3 : 4
n=4 : 8
etc.

The pattern is obvious, namely 2^(n-1)
It is worth knowing the following standard argument: the number of subsets of {1,2,...n} is 2^n. Indeed, all subsets can be constructed as follows:
is 1 in it? 2 choices
is 2 in it? 2 choices
etc.
In total 2^n choices of subsets.

You can modify this standard argument in a very pretty way when n is odd, i.e. how many even size subsets of {1,2,...n} are there when n is odd?

Ah fuck, I forgot the empty set. Fuck.





Fuck.

 

[Coolydudey]

Ah fuck, I forgot the empty set. Fuck.





Fuck.

Well I'll give you the argument:

if n is odd, every even sized subset of {1,2,...,n} leaves behind an odd-sized subset. So the number of even sized subsets=the number of odd sized subsets, and as the total number of subsets is 2^n, the number in question is 2^n-1.

 

[Rudolph Mondal]

dude get real with the maths teasers:

The sume of some (not necessarily distinct) natural numbers (i.e. positive integers) is 100. How large can their product be?

I realized no one has answered this so I will.

This should be right...but could be wrong.

I'd use lagrange multipliers first whereupon I'd realize that all the numbers adding up to 100 have to be the same to maximize the product.

Then I'd use unconstrained optimization by expressing x (the integer we're concerned with) in terms of n (the number of times it has to appear).

I think the answer would be 2/4.

EDIT: I think the distinct case would be a lot harder.

 

[Tannhauser]

I think the distinct case would be a lot harder.

I wrote a quick and dirty simulation based on random numbers to do that. The result:

max product = 21794572800
sequence = [2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

So it's basically all the smallest integers excluding 4. It is probably some amusing relation to the fact that 2+2=2*2, which I am too lazy to investigate right now.

 

[groovytaxi98]

I pretty much universally hate everything math, unless its basic level algebra or geometry. :P

But I do enjoy doing math tricks... for example, I love proving that one number is a completely different number... of course this seems cool until it is realized that it's all just converting bases. lol.

equations I use depend on how many variables are in the final converted number.
Two Digits: Y=mx+b
Three Digits: y= ax2 + bx + c
etc....

I could say that 43 = 234, by using...
(original number) = (1st digit of converted number)x^2 + (2nd digit of cnvtd nmbr)x + (3rd digit of cnvtd number)

43 = 2x^2 + 3x + 4
0 = 2x^2 + 3x - 39

then quadratic equationify the above:
1. (-3+/-(9-4(2)(-39)^1/2)/4
2. (-3+(17.664))/4
3. x= 3.66


Therefore, 43 equals 234. (*hiding the fact that this is only when 234 has a base of 3.66.)

 

[Irukanji]

There is a town with two hospitals: one big and one small (just for intuition, let's say the big one contains 10 times as many patients as the small one). At a particular day, there were born 60% girls in one of the hospitals. Which one was it most likely?

Well, knowing real life as I do, big hospitals are more likely to have maternity wards and small hospitals less likely. So I'd say the bigger one has a better chance of 60% girls.

However, you did say town, which to me conjures up an image of a small country town, 20 beds perhaps in the biggest hospital, and a small "doctors surgery" with only 2 beds(no overnight care, however). Thus, all births are performed at the big hospital.

But being the math thread and not the logic thread, and given a 1.05:1 ratio of males to females, the likelihood of 60% of the births being female isn't possible to calculate without the actual numbers.

But I read someones answer and the chance of the small hospital having 6 people give birth to females and 4 to males is better than 60 people giving birth to females and 40 to males...tfw logic goes the wrong direction. Assuming exactly 10:100 ratio, small:big.

Now you know why maths takes me forever

 

[QuickTwist]

dude get real with the maths teasers:

The sume of some (not necessarily distinct) natural numbers (i.e. positive integers) is 100. How large can their product be?

It would be 2^50 wouldn't it?

 

[Tannhauser]

Probability puzzle off the top of my head:

5 guys walk into a tavern and sit down at a round table with 5 seats. They are all of different age. What is the probability they will sit in order, according to age?

 

[Cheeseumpuffs]

Youngest guy sits down - 1
Next youngest sits next to him - 1/4
Next sits next to him - 1/3
Next - 1/2
Oldest takes the last chair - 1

1(1/4)(1/3)(1/2)(1/1) = 1/(4!) = 1/24

 

[Tannhauser]

Youngest guy sits down - 1
Next youngest sits next to him - 1/4
Next sits next to him - 1/3
Next - 1/2
Oldest takes the last chair - 1

1(1/4)(1/3)(1/2)(1/1) = 1/(4!) = 1/24

You're missing a piece of the puzzle

 

[Ex-User (9086)]

(5 * 4) / (5 * 5!) = 4/120 = 1/30

 

[Tannhauser]

Still ain't right I'm afraid.

 

[Ex-User (9086)]

Forgot about the redundancy of solutions in a round arrangement.

So this should be like this:

(2 * 5) / (5!) = 10/120 = 1/12

There are 2 solutions for each starting seat and there are 5! possible arrangements of elements in a 5 element set with 5 distinct pieces.

It's been a long time since I did combinatorics, I had to solve a few examples before I realised what was wrong with this one.

 

[Tannhauser]

Exactly.

 

[Tannhauser]

In the street where you live, there is a bar. There are 5 blocks between your house and the bar. At the moment you are standing between them, with 3 blocks to the bar and 2 blocks to home. Every time you walk 1 block, you flip a coin to decide the direction you go: if it is heads, you walk towards the bar, if it is tails you go towards home (and of course, you also flip the coin to decide your initial direction).

What is the probability you will end up in the bar?


Illustration: you are standing at point X:

B-----|------|------X------|------H

 

[Tannhauser]

No takers?

One can start with defining one's current position as 0, the bar as -3 and home as 2. Then define the probability as function of a variable position x like this: p(x) = P( B before H | position = x). So we are looking for p(0).

Since the probability is only dependent on the distance to the goal, it satisfies

p(x) = 0.5*p(x-1) + 0.5*p(x+1)

 

[Infinitatis]

No takers?

One can start with defining one's current position as 0, the bar as -3 and home as 2. Then define the probability as function of a variable position x like this: p(x) = P( B before H | position = x). So we are looking for p(0).

Since the probability is only dependent on the distance to the goal, it satisfies

p(x) = 0.5*p(x-1) + 0.5*p(x+1)

Perhaps I'm misinterpreting this (which isn't unlikely ), but wouldn't the probability of making it to the bar given a series of coin flips depend on how you get to a position, and not just the position itself? You could have an infinant number of flips before you reach either location. You stop flipping once you've reached home or the bar (i.e. once you've netted two tails or three heads).

Obviously the answer will be less than 1/2, since the flipper is already closer to home. My best estimate is that since you must net three heads, the probability is (1/2)³ = 1/8, but surely that's too simple. If that's not the case, the probability should be between 1/8 and 1/2, but I'm at a loss as to how one would go about calculating this.

 

[Infinitatis]

No takers?

One can start with defining one's current position as 0, the bar as -3 and home as 2. Then define the probability as function of a variable position x like this: p(x) = P( B before H | position = x). So we are looking for p(0).

Since the probability is only dependent on the distance to the goal, it satisfies

p(x) = 0.5*p(x-1) + 0.5*p(x+1)

I think that this problem requires the use of summation, or at least something derived from such, but I would not know where to start.

[H] — | — x — | — | —
0 1 2 3 4 5

Since the probability is solely dependent upon the starting position, I've been thinking about what is known, without mathematics:
– The probability that he makes it to the bar given that he starts at point 0 (home) is 0.
– The probability that he makes it to the bar given that he starts at point 5 (the bar) is 1.
– If he was halfway between both his home and the bar (say halfway between point 0 and 6: 3), the probability that he makes it to the bar is 1/2.
– The probability of reaching the bar before home decreases as x decreases and increases as x increases.
–It is possible to make it to the bar before home but not necessary (i.e. the probability is greater than zero but less than 1).
–The probability is less than one half because he is further from the bar than from home.
– Thus the probability of reaching the bar before home must be greater than 0 and less than 1/2.

Since the probability for a success (heads) or failure (tails) are the same, 1/2, it wouldn't surprise me if the probability is simply 2/5.

 

[Infinitatis]

Buffon's Needle Problem

You have a needle that is one inch long. On the floor, many parallel lines are drawn spaced one inch apart like so.

.....
—————————————————
—————————————————
—————————————————
—————————————————
—————————————————
—————————————————
.....

What is the probability that if you were to toss the needle onto the floor that the needle would cross one of the lines?

For those of you familiar with this famous problem, here's a question to take this a step further:

Given the same situation except with an additional set of parallel lines running perpendicular to the horizontal lines, making a grid, what is the probability that a needle crosses a line?

 

[Tannhauser]

I think that this problem requires the use of summation, or at least something derived from such, but I would not know where to start.

[H] — | — x — | — | —
0 1 2 3 4 5

Since the probability is solely dependent upon the starting position, I've been thinking about what is known, without mathematics:
– The probability that he makes it to the bar given that he starts at point 0 (home) is 0.
– The probability that he makes it to the bar given that he starts at point 5 (the bar) is 1.
– If he was halfway between both his home and the bar (say halfway between point 0 and 6: 3), the probability that he makes it to the bar is 1/2.
– The probability of reaching the bar before home decreases as x decreases and increases as x increases.
–It is possible to make it to the bar before home but not necessary (i.e. the probability is greater than zero but less than 1).
–The probability is less than one half because he is further from the bar than from home.
– Thus the probability of reaching the bar before home must be greater than 0 and less than 1/2.

Since the probability for a success (heads) or failure (tails) are the same, 1/2, it wouldn't surprise me if the probability is simply 2/5.

You are right about all of these deductions. Maybe there is an easier way of dealing with the problem, but continuing with the relationship

p(x) = 0.5*p(x-1) + 0.5*p(x+1)

one can see that this is a recurrence relation on the form

p_n = 2*p_(n-1) - p_(n-2)

From here one can use standard techniques for solving recurrence relations . In particular, the characteristic equation x^2 -2x +1 = 0 has a single root r =1 so the general solution of the recurrence is

p_n = C + D*n

and then using the conditions p(-3) = 1 and p(2) = 0 we get the particular solution

p_n = 2/5 - n/5

so, as you guessed, p_0 = 2/5

The handy thing about this type of solution is that one can easily substitute the fair coin with an unfair coin which has, say probability p of landing heads.

 

[Infinitatis]

You are right about all of these deductions. Maybe there is an easier way of dealing with the problem, but continuing with the relationship

p(x) = 0.5*p(x-1) + 0.5*p(x+1)

one can see that this is a recurrence relation on the form

p_n = 2*p_(n-1) - p_(n-2)

Ah, I see. I hadn't even realized that the formula you gave for p(x) is defined recursively. This makes a lot more sense now. Thanks for the explanation.

 

[Tannhauser]

Buffon's Needle Problem

You have a needle that is one inch long. On the floor, many parallel lines are drawn spaced one inch apart like so.

.....
—————————————————
—————————————————
—————————————————
—————————————————
—————————————————
—————————————————
.....

What is the probability that if you were to toss the needle onto the floor that the needle would cross one of the lines?

I've heard about this problem but never looked into it, so Im gonna try to freestyle it:

If we start with a discrete consideration of the problem to make things easier, one can say that

P(hit line) = \sum P(hit line | angle)*P(angle)

where "angle" is angle with which the needle falls, and \sum is summing over all possible angles.

Looking at P(hit line | angle): given an angle "a", this probability will be the proportion of distances to one of the lines such that the needle touches a line, to 1, the distance between the lines. Doing a quick sketch, I figured this is simply sin(a).

Then, since all angles are equally likely, P(angle) = 1/pi

So doing the continuous version of the problem, doing integration instead of summation, it's gonna be integrating sin(x)/pi from 0 to pi, which results in 2/pi

 

[Infinitatis]

So doing the continuous version of the problem, doing integration instead of summation, it's gonna be integrating sin(x)/pi from 0 to pi, which results in 2/pi

That's exactly correct. Nice job.

 

[Infinitatis]

This one's an interesting one. Find the indefinite integral.