# Products of compact spaces with countable tightness

In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.

The following theorems are the main results in this post.

Theorem 1
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is compact, then $X \times Y$ is countably tight.

Theorem 2
The product of finitely many compact countably tight spaces is countably tight.

Theorem 3
Suppose that $X_1, X_2, X_3, \cdots$ are countably many compact spaces such that each $X_i$ has at least two points. If each $X_i$ is a countably tight space, then the product space $\prod_{i=1}^\infty X_i$ is countably tight.

____________________________________________________________________

Finite products

Before proving Theorem 1 and Theorem 2, we prove the following results.

Theorem 4
Let $f:Y_1 \rightarrow Y_2$ be a continuous and closed map from the space $Y_1$ onto the space $Y_2$. Suppose that the space $Y_2$ is countably tight and that each fiber of the map $f$ is countably tight. Then the space $Y_1$ is countably tight.

Proof of Theorem 4
Let $x \in Y_1$ and $x \in \overline{A}$ where $A \subset Y_1$. We proceed to find a countable $W \subset Y_1$ such that $x \in \overline{W}$. Choose $y \in Y_2$ such that $y=f(x)$.

Let $M$ be the fiber of the map $f$ at the point $y$, i.e. $M=f^{-1}(y)$. By assumption, $M$ is countably tight. Call a point $w \in M$ countably reached by $A$ if there is some countable $C \subset A$ such that $w \in \overline{C}$. Let $G$ be the set of all points in $M$ that are countably reached by $A$. We claim that $x \in \overline{G}$.

Let $U \subset Y_1$ be open such that $x \in U$. Because the space $Y_1$ is regular, choose open $V \subset U$ such that $x \in V$ and $\overline{V} \subset U$. Then $V \cap A \ne \varnothing$. Furthermore, $x \in \overline{V \cap A}$. Let $C=f(V \cap A)$. By the continuity of $f$, we have $y \in \overline{C}$. Since $Y_2$ is countably tight, there exists some countable $D \subset C$ such that $y \in \overline{D}$. Choose a countable $E \subset V \cap A$ such that $f(E)=D$. It follows that $y \in \overline{f(E)}$.

We show that that $\overline{E} \cap M \ne \varnothing$. Since $E \subset \overline{E}$, we have $f(E) \subset f(\overline{E})$. Note that $f(\overline{E})$ is a closed set since $f$ is a closed map. Thus $\overline{f(E)} \subset f(\overline{E})$. As a result, $y \in f(\overline{E})$. Then $y=f(t)$ for some $t \in \overline{E}$. We have $t \in \overline{E} \cap M$.

By the definition of the set $G$, we have $\overline{E} \cap M \subset G$. Furthermore, $\overline{E} \cap M \subset \overline{V} \subset U$. Note that the arbitrary open neighborhood $U$ of $x$ contains points of $G$. This establishes the claim that $x \in \overline{G}$.

Since $M$ is a fiber of $f$, $M$ is countably tight by assumption. Choose some countable $T \subset G$ such that $x \in \overline{T}$. For each $t \in T$, choose a countable $W_t \subset A$ with $t \in \overline{W_t}$. Let $W=\bigcup_{t \in T} W_t$. Note that $W \subset A$ and $W$ is countable with $x \in \overline{W}$. This establishes the space $Y_1$ is countably tight at $x \in Y_1$. $\blacksquare$

Lemma 5
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a compact space, then $f$ is a closed map.

Proof of Lemma 5
Let $A$ be a closed subset of $X \times Y$. Suppose that $f(A)$ is not closed. Let $y \in \overline{f(A)}-f(A)$. It follows that no point of $X \times \left\{y \right\}$ belongs to $A$. For each $x \in X$, choose open subset $O_x$ of $X \times Y$ such that $(x,y) \in O_x$ and $O_x \cap A=\varnothing$. The set of all $O_x$ is an open cover of the compact space $X \times \left\{y \right\}$. Then there exist finitely many $O_x$ that cover $X \times \left\{y \right\}$, say $O_{x_i}$ for $i=1,2,\cdots,n$.

Let $W=\bigcup_{i=1}^n O_{x_i}$. We have $X \times \left\{y \right\} \subset W$. Since $X$ is compact, we can then use the Tube Lemma which implies that there exists open $G \subset Y$ such that $X \times \left\{y \right\} \subset X \times G \subset W$. It follows that $G \cap f(A) \ne \varnothing$. Choose $t \in G \cap f(A)$. Then for some $x \in X$, $(x,t) \in A$. Since $t \in G$, $(x,t) \in W$, implying that $W \cap A \ne \varnothing$, a contradiction. Thus $f(A)$ must be a closed set in $Y$. This completes the proof of the lemma. $\blacksquare$

Proof of Theorem 1
Let $X$ be the factor that is compact. let $f: X \times Y \rightarrow Y$ be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space $Y$ is countably tight by assumption. Each fiber of the projection map $f$ is of the form $X \times \left\{y \right\}$ where $y \in Y$, which is countably tight. Then use Theorem 4 to establish that $X \times Y$ is countably tight. $\blacksquare$

Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. $\blacksquare$

____________________________________________________________________

Countable products

Our proof to establish that the product space $\prod_{i=1}^\infty X_i$ is countably tight is an indirect one and makes use of two non-trivial results. We first show that $\omega_1 \times \prod_{i=1}^\infty X_i$ is a closed subspace of a $\Sigma$-product that is normal. It follows from another result that the second factor $\prod_{i=1}^\infty X_i$ is countably tight. We now present all the necessary definitions and theorems.

Consider a product space $Y=\prod_{\alpha<\kappa} Y_\alpha$ where $\kappa$ is an infinite cardinal number. Fix a point $p \in Y$. The $\Sigma$-product of the spaces $Y_\alpha$ with $p$ as the base point is the following subspace of the product space $Y=\prod_{\alpha<\kappa} Y_\alpha$:

$\displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}$

The definition of the space $\Sigma_{\alpha<\kappa} Y_\alpha$ depends on the base point $p$. The discussion here is on properties of $\Sigma_{\alpha<\kappa} Y_\alpha$ that hold regardless of the choice of base point. If the factor spaces are indexed by a set $A$, the notation is $\Sigma_{\alpha \in A} Y_\alpha$.

If all factors $Y_\alpha$ are identical, say $Y_\alpha=Z$ for all $\alpha$, then we use the notation $\Sigma_{\alpha<\kappa} Z$ to denote the $\Sigma$-product. Once useful fact is that if there are $\omega_1$ many factors and each factor has at least 2 points, then the space $\omega_1$ can be embedded as a closed subspace of the $\Sigma$-product.

Theorem 6
For each $\alpha<\omega_1$, let $Y_\alpha$ be a space with at least two points. Then $\Sigma_{\alpha<\omega_1} Y_\alpha$ contains $\omega_1$ as a closed subspace. See Exercise 3 in this previous post.

Now we discuss normality of $\Sigma$-products. This previous post shows that if each factor is a separable metric space, then the $\Sigma$-product is normal. It is also well known that if each factor is a metric space, the $\Sigma$-product is normal. The following theorem handles the case where each factor is a compact space.

Theorem 7
For each $\alpha<\kappa$, let $Y_\alpha$ be a compact space. Then the $\Sigma$-product $\Sigma_{\alpha<\kappa} Y_\alpha$ is normal if and only if each factor $Y_\alpha$ is countably tight.

Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the $\Sigma$-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.

Theorem 8
Let $Y$ be a compact space. Then the product space $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight.

We now prove Theorem 3.

Proof of Theorem 3
Let $\omega_1=\cup \left\{A_n: n \in \omega \right\}$, where for each $n$, $\lvert A_n \lvert=\omega_1$ and that $A_n \cap A_m=\varnothing$ if $n \ne m$. For each $n=1,2,3,\cdots$, let $S_n=\Sigma_{\alpha \in A_n} X_n$. By Theorem 7, each $S_n$ is normal. Let $S_0=\Sigma_{\alpha \in A_0} X_1$, which is also normal. By Theorem 6, the space $\omega_1$ of countable ordinals is a closed subspace of $S_0$. Let $T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots$. We have the following derivation.

\displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned}

Recall that $\omega_1=\cup \left\{A_n: n \in \omega \right\}$. The space $W=\Sigma_{\alpha<\omega_1} W_\alpha$ is defined such that for each $n \ge 1$ and for each $\alpha \in A_n$, $W_\alpha=X_n$. Furthermore, for $n=0$, for each $\alpha \in A_0$, let $W_\alpha=X_1$. Thus $W$ is a $\Sigma$-product of compact countably tight spaces and is thus normal by Theorem 7. The space $T=\omega_1 \times \prod_{n=1}^\infty X_n$ is a closed subspace of the normal space $W$. By Theorem 8, the product space $\prod_{n=1}^\infty X_n$ must be countably tight. $\blacksquare$

____________________________________________________________________

Remarks

Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let $X=X_1 \times X_2 \times \cdots \times X_n$.

\displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset X_1 \times X_2 \times \cdots \times X_n \times X_{n+1} \times X_{n+2} \times \cdots \end{aligned}

In the above derivation, $t_m$ is a point of $X_m$ for all $m >n$. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.

____________________________________________________________________

Exercise

Exercise 1
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a countably compact space and $Y$ is a Frechet space, then $f$ is a closed map.

Exercise 2
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is a countably compact space and the other space is a Frechet space, then $X \times Y$ is countably tight.

Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.

____________________________________________________________________

Reference

1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

____________________________________________________________________
$\copyright \ 2015 \text{ by Dan Ma}$


# Rocket Ship Question

1. Nov 6, 2007

### Submission1

We are having a debate here at work and we need an answer from some smart folks (that's you).

If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.

Thanks,
Sub.

2. Nov 6, 2007

### mgb_phys

Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.

You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.

Last edited: Nov 6, 2007
3. Nov 6, 2007

### Submission1

Cool. Thanks.

4. Nov 6, 2007

### chronon

But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.

5. Nov 6, 2007

### mgb_phys

Does that apply if it's accelerating ?

6. Nov 6, 2007

### D H

Staff Emeritus
This topic ws recently discussed in this thread.

The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,

$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$

Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is

$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$

In other words, one must more than double the quantity of fuel to double a rocket's final velocity.

7. Nov 6, 2007

### mgb_phys

Thanks - I'm away for a week and look what I miss.

8. Nov 6, 2007

### chronon

Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.

9. Nov 6, 2007

### D H

Staff Emeritus
A little more detail regarding previous post:

The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.

Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.

However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.

Last edited: Nov 6, 2007
10. Nov 6, 2007

### D H

Staff Emeritus
Indeed. We just cross-posted.

11. Nov 6, 2007

### rcgldr

Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).

Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.

12. Nov 6, 2007

### Staff: Mentor

From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.

13. Nov 6, 2007

### TVP45

You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.

14. Nov 6, 2007

### D H

Staff Emeritus
That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.

Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.

The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.

15. Nov 8, 2007

### alvaros

So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.

( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )

16. Nov 8, 2007

### TVP45

A kind of new thought on this thread....
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.

Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.

Is there another way?

17. Nov 9, 2007

### alvaros

Im very sorry, but I dont understand your english.

18. Nov 9, 2007

### TVP45

Alvaros,

19. Nov 10, 2007

### alvaros

I think my example is clear enough:

It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )

I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).

20. Nov 10, 2007

### Staff: Mentor

Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.

I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:

Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.

Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.

"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.

Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).

Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.

Last edited: Nov 10, 2007


# Rocket Ship Question

1. Nov 6, 2007

### Submission1

We are having a debate here at work and we need an answer from some smart folks (that's you).

If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.

Thanks,
Sub.

2. Nov 6, 2007

### mgb_phys

Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.

You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.

Last edited: Nov 6, 2007
3. Nov 6, 2007

### Submission1

Cool. Thanks.

4. Nov 6, 2007

### chronon

But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.

5. Nov 6, 2007

### mgb_phys

Does that apply if it's accelerating ?

6. Nov 6, 2007

### D H

Staff Emeritus
This topic ws recently discussed in this thread.

The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,

$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$

Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is

$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$

In other words, one must more than double the quantity of fuel to double a rocket's final velocity.

7. Nov 6, 2007

### mgb_phys

Thanks - I'm away for a week and look what I miss.

8. Nov 6, 2007

### chronon

Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.

9. Nov 6, 2007

### D H

Staff Emeritus
A little more detail regarding previous post:

The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.

Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.

However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.

Last edited: Nov 6, 2007
10. Nov 6, 2007

### D H

Staff Emeritus
Indeed. We just cross-posted.

11. Nov 6, 2007

### rcgldr

Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).

Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.

12. Nov 6, 2007

### Staff: Mentor

From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.

13. Nov 6, 2007

### TVP45

You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.

14. Nov 6, 2007

### D H

Staff Emeritus
That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.

Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.

The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.

15. Nov 8, 2007

### alvaros

So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.

( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )

16. Nov 8, 2007

### TVP45

A kind of new thought on this thread....
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.

Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.

Is there another way?

17. Nov 9, 2007

### alvaros

Im very sorry, but I dont understand your english.

18. Nov 9, 2007

### TVP45

Alvaros,

19. Nov 10, 2007

### alvaros

I think my example is clear enough:

It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )

I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).

20. Nov 10, 2007

### Staff: Mentor

Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.

I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:

Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.

Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.

"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.

Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).

Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.

Last edited: Nov 10, 2007
21. Nov 10, 2007

### TVP45

DaleSpam,
This is the sort of example I was asking about. If I were a high school teacher (thank G-d I'm not; that's a hard job), how would I present this example so that students can grasp it? I'm not being in any way critical, any more than I was of Alvaros; I am simply curious about how this can be developed conceptually.

22. Nov 10, 2007

### Staff: Mentor

If I were a teacher in a high-school level physics course I would not even mention conservation of energy in this problem. I would stick entirely with conservation of momentum. You can solve for everything of interest that way, and the conservation of momentum principle does what a good conservation principle should: it simplifies things.

As a teacher I would not introduce conservation of energy into the problem since it does nothing to simplify the problem. If a student asked I would tell them that energy is conserved, but it doesn't make the problem any easier. I think students would understand that.

23. Nov 11, 2007

### TVP45

At first blush, I agree. Momentum is much easier to deal with than energy. Two questions occur to me:
(1) Can a student grasp the concept of momentum without being sidetracked by preconceptions about energy?
(2) Is the calculation of change of momentum rather than absolute momentum sensible?

24. Nov 11, 2007

### alvaros

DaleSpam:
Wrong, the change in kinetic energy on the fuel is 1000 times the change in kinetic energy on the rocket.

Momentum is an abstract concept, its much easier the 3rd Newtons law: the burning fuel pushes the rocket and the rocket pushes the fuel fuel <-> rocket
( you see the double arrow <-> , all forces ( ? ) are double arrow )
( It seems, as I read here, that the conservation of momentum is more "universal" than 3rd Newtons law, but, in this case, both laws are the same and true )

But, at the end, there isnt any clear answer to the problem.

25. Nov 11, 2007

### D H

Staff Emeritus
Invoking Newton's third law on this problem is in a sense more ad-hoc and more abstract than using conservation of momentum with regard to this problem. The concept of force (Newton's second law) is very abstract; anything change to an object's momentum involves some force. Applying conservation of momentum is no less abstract than applying Newton's third law and yields a deeper answer.

Put the spaceship in deep space, far from any massive object, so that there are no measurable external forces acting on the vehicle. Now, what exactly is the force that is making the spaceship accelerate? By assumption, there are no external forces. You can say posit some ad-hoc force $F$ that results from burning the fuel and get an answer via Newton's second law. Note well: Since the rocket's mass is changing, the simpler form $F=ma$ is not valid here. We have to use $F=dp/dt$ instead.

$$F = \frac{dp_r} {dt} = \frac{d}{dt} ( m_r \, v_r ) = \dot m_r v_r} + m_r \, a_r$$

Solving for the rocket's acceleration,

$$a_r = \frac {F} {m_r} - \frac {\dot m_r} {m_r} v_r}$$

This is not very satisying. What exactly is this force? It's purely ad-hoc for one thing. Moreover, it will turn out that the force is frame-dependent.

Here is how things turn out from conservation of momentum point of view. For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things $\Delta v$. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.

Nomenclature:
$$\vec v_r(t)$$ Rocket velocity at time $t$, inertial observer
$$\vec v_e(t)$$ Rocket exhaust velocity at time $t$, relative to vehicle velocity
$$m_r(t)$$ Rocket mass at time $t$
$$\dot m_f(t)$$ Rate at which rocket consumes fuel at$t$

Over a small time interval $\Delta t$, the rocket will eject a small mass of fuel $\dot m_f(t)\,\Delta t$. At the start of the interval, the rocket has mass $m_r(t)$, velocity $\vec v_r(t)$, and momentum $m_r(t) \, \vec v_r(t)$. At the end of the interval, the rocket has mass, velocity, and linear momentum

$$m_r(t+\Delta t) = m_r(t)-\dot m_f(t)\,\Delta t$$
$$\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)$$
$$\vec p_r(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))$$

The bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are

$$\Delta m_e(t) = \dot m_f(t)\,\Delta t$$
$$\vec v_{e_{\text{inertial}}}) = \vec v_r(t)+\vec v_e(t)$$
$\Delta \vec p_e(t+\Delta t) = \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))[/tex] The momentum of the rocket+exhaust at the end of the time interval is thus $$\vec p_{r+e}(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) + \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))$$ Dropping the second-order term [itex]\Delta t \Delta \vec v_r(t)$ and simplifying,

$$\vec p_{r+e}(t+\Delta t) = m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t)$$

Assuming no external forces act on the rocket and the ejected fuel during this time interval, the rocket and the ejected fuel form a closed system. Momentum is conserved in a closed system, so $\vec p_{r+e}(t+\Delta t)=\vec p_r(t)[/tex]: $$m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)$$ or $$m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = 0$$ Dividing by [itex]\Delta t$ and taking the limit $\Delta t \to 0$,

$$\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_f(t)} {m_r(t)} \, \vec v_e(t)$$

This is the equation for the acceleration of the rocket at time $t$.

By conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the fuel consumption rate: $\dot m_r(t) = -\dot m_f(t)$. If the relative exhaust velocity is a constant vector, both the left and right hand sides of the above acceleration equation are integrable. Integrating from some initial time $t_0$ to some final time $t_1$ yields

$$\vec v_r(t_1) - \vec v_r(t_0) = \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \vec v_e$$

This is the Tsiolokovsky rocket equation. Since the final mass is smaller than the initial mass, the logarithm will be negative. The change in velocity is directed against the exhaust direction.

Last edited: Nov 11, 2007


The difference between theoretical complexity and practical efficiency

If I have this pseudocode:

for i=0 to n/2 do
for j=0 to n/2 do
... do anything ....

The number of iterations is $n^2/4$.

What is the complexity of this program? Is it $O(n^2)$?

What is the intuition formal/informal for which is that complexity?

Next, if i have this other pseudocode :

for i=0 to n do
for j=0 to n do
... do anything ....

The complexity is again $O(n^2)$ -- is that correct?

Is there any difference between efficiency in practice and theoretical complexity in this case? Is one of these faster than the other?

• Rule 3 from Rob Pike, as cited in The Art of Unix Programming: "Rule 3. Fancy algorithms are slow when n is small, and n is usually small. Fancy algorithms have big constants. Until you know that n is frequently going to be big, don't get fancy." This is because of the difference between theoretical complexity and practical efficiency, which is ably defined in the answers below. – Wildcard Mar 21 '16 at 3:49
• @Wildcard That seems like a good rule of thumb. Benchmarking several contender algorithms would be even better. – G. Bach Mar 21 '16 at 12:47
• @Wildcard Very interesting quote. It is always interesting to think about the complexity when you're designing/implementing procedures but I usually don't break my head over them untill the procedure in question seems to be a bottleneck. – Auberon Mar 21 '16 at 16:18

Big O Formally

$O(f(n)) = \{g(n) | \exists c > 0, \exists n_0 > 0, \forall n > n_0 : 0 \leq g(n) \leq c*f(n)\}$

Big O Informally

$O(f(n))$ is the set of functions that grow slower (or equally fast) than $f(n)$. This means that whatever function you pick from $O(f(n))$, let's name her $g(n)$, and you pick a sufficiently large $n$, $f(n)$ will be larger than $g(n)$. Or, in even other words, $f(n)$ will eventually surpass any function in $O(f(n))$ when $n$ grows bigger. $n$ is the input size of your algorithm.

As an example. Let's pick $f(n) = n^2$.

We can say that $f(n) \in O(n^3)$. Note that, over time, we allowed notation abuse so almost everyone writes $f(n) = O(n^3)$ now.

Normally when we evaluate algorithms, we look at their order. This way, we can predict how the running time of the algorithm will increase when the input size ($n$) increases.

In your example, both algorithms are of order $O(n^2)$. So, their running time will increase the same way (quadratically) as $n$ increases. Your second algorithm will be four times slower than the first one, but that is something we're usually not interested in **theoretically*. Algorithms with the same order can have different factors ($c$ in the formal definition) or different lower order terms in the function that evaluates the number of steps. But usually that's because of implementation details and we're not interested in that.

If you have a algorithm that runs in $O(log(n))$ time we can safely say it will be faster than $O(n)$ [1] because $log(n) = O(n)$. No matter how bad the $O(log(n))$ algorithm is implemented, no matter how much overhead you put in the algorithm, it will always [1] be faster than the $O(n)$ algorithm. Even if the number of steps in the algorithms are $9999*log(n) = O(log(n))$ and $0.0001*n = O(n)$, the $O(log(n))$ algorithm will eventually be faster [1].

But, maybe, in your application, $n$ is always low and will never be sufficiently large enough so that the $O(log(n))$ algorithm will be faster in practice. So using the $O(n)$ algorithm will result in faster running times, despite $n = O(log(n))$

[1] if $n$ is sufficiently large.

• An $O(\log n)$ isn't always faster than an $O(n)$ one. It depends on the value of $n$ and on the hidden constants. A case in point is fast matrix multiplication, which isn't fast at all in practice. – Yuval Filmus Mar 21 '16 at 0:07
• That's why I added: If n is sufficiently large. In the case you mention, if the matrices grow sufficiently large, $O(log(n))$ WILL be faster, by definition – Auberon Mar 21 '16 at 0:09
• Isn't everything you're saying already in my answer? – Auberon Mar 21 '16 at 0:11
• The OP asks about the difference between theory and practice, and you're ignoring that. Asymptotic notation is usually, but not always, useful in practice. Improving the running time fourfold can make a huge difference in practice, but asymptotic notation is completely oblivious to it. – Yuval Filmus Mar 21 '16 at 0:14

Auberon provided a very good explanation of the Big O. I will try to explain what that means for you.

First of you are right. The first code example has $\frac{n^2}{4}$ iterations, but is still in the complexity class O(n^2).

Why?

The thing about complexity classes is, that we assume that doing something several times is not that bad for runtime.

Imagine a Algorithm with complexity O(2^n) running for n=3 and taking 1 second.

We could run this 10 times, and still expect an answer after about 10 seconds.

Now Imagine increasing n by 10. The Programm will take 2^10=1024 seconds.

So the computer scientists basically said: "Man, leading factors are annoying. Let's just assume n grows to infinity and compare functions there"

Which lead to the Big O.

In Practice

In Practice it is very well possible that a solution with much worse complexity runs much faster (for "small" inputs). It is easy to imagina a Programm that runs in O(n) but needs 10^10*n iterations.

It is O(n) but even an O(2^n) solution could be faster for small n.

In summary:

The Big O is a useful tool. But you still have to think about how fast your algorithm is in practice, since it only describes how much more time it will need if the input grows.

If I have this pseudocode:

for i=0 to n/2 do
for j=0 to n/2 do
... do anything ....

The number of iteration is $n^2/4$.

Actually, it's $(1+n/2)^2 = n^2/4 + n + 1$.

What is the complexity of this program? Is correct $O(n^2)$?

That depends entirely on what "do anything" is. For example, if "do anything" is replaced by the line

for k=0 to 2^n do {}

then the running time is $\Theta(n^22^n)$.


# Find local max, min, concavity, and inflection points

For the function $f(x)=\frac{x^2}{x^2+3}$ Find the intervals on which f(x) is increasing or decreasing.

Find the points of local maximum and minimum of f(x).

Find the intervals of concavity and the inflection points of f(x).

$f′(x)=\frac{6x}{(x^2+3)^2}$

$f''(x)=\frac{-18(x^2-1)}{(x^2+3)^3}$

• I know the critcal point is $x = 0$ if you plug that in to $f(x)$ and $f'(x)$ you get $0$ if you plug it into $f''(x)$ you get $f''(x) = \frac{2}{3}$ does this mean that the local min is $0$ at $x = 0$ ? – Csci319 Oct 31 '14 at 17:13
• Indeed it does. That method is called the "second derivative test" if you want to Google it. – GFauxPas Oct 31 '14 at 17:17
• but does that mean there is no max? how would I find the max? – Csci319 Oct 31 '14 at 17:47
• If the domain of your function is all real numbers, then the function has no maximum. – GFauxPas Oct 31 '14 at 18:05

## 1 Answer

Sign analysis of $f'$

$f'(x)=0$ when $x=0$ so partition the number line as $(-\infty,0)\cup(0,\infty)$.

$(-\infty,0)$: Pick a test point in this interval, say $x=-1$. Note $f'(-1)<0$ so $f'(x)<0$ on this interval. Hence $f$ is decreasing on this interval.

$(0,\infty)$: Pick a test point, say $x=1$. Then $f'(1)>0$ so $f'(x)>0$ on this interval. Hence $f$ is increasing on this interval.

Since $f$ went from decreasing to increasing at $x=0$, a local min occurs at $x=0$.

Sign analysis of $f''$

$f''(x)=0$ when $x=\pm 1$ so partition the number line as $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.

$(-\infty,-1)$: Pick a test point in this interval, say $x=-2$. Then $f''(-2)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval

$(-1,1)$: Pick a test point in this interval, say $x=0$. Then $f''(0)>0\implies f''(x)>0\implies f\text{ concave up}$ on this interval

$(1,\infty)$: Pick a test point in this interval, say $x=1$. Then $f''(1)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval

Since $f$ changed concavity at $x=\pm 1$, these are inflection points.

Indeed a plot of $y=f(x)$ bears out the information above:


# $\dim_K\operatorname{Im}(T^2) =\dim_K\operatorname{Im}(T) \implies \operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$

I need some help here...

Let $V$ be a vector space over a field $K$, and $T: V \to W$ a linear transformation. Prove that if $$dim_KV \lt \infty \ \ \text{and} \ \ \dim_K\operatorname{Im}(T^2) = \dim_K\operatorname{Im}(T)$$ then $$\operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$$

This is what I made so far.

Let $x \in \operatorname{Im}(T) \cap \operatorname{Ker}(T)$,where T is linear operator. Then $T(x)=0$ and $x=T(v)$ for some $v \in V$

$0 = T(x) = T(T(v)) = T^2(v) \implies v\in \operatorname{Ker}(T^2)$

My intention is to reach $x=0$, but I'm stuck at this point. I also deduced, via dimension theorem, that $\dim_K \operatorname{Ker}(T) = \dim_K \operatorname{Ker}(T^2)$ but I don't know where it can help. Every hint is appreciated.

• You have $\ker(T) \subset \ker(T^2)$. Now use the dimension. – Maik Pickl Apr 4 '16 at 19:49

The rank-nullity theorem says \begin{align} \dim V&=\dim\operatorname{Im}(T)+\dim\operatorname{Ker}(T) \\ \dim V&=\dim\operatorname{Im}(T^2)+\dim\operatorname{Ker}(T^2) \end{align} Next use the fact that $$\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$$ to conclude that $$\operatorname{Ker}(T)=\operatorname{Ker}(T^2)$$
• Oh, I did not know that $\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$. – Карпатський Apr 4 '16 at 19:59
• @Карпатський If $T(v)=0$, then $T^2(v)=0$. ;-) – egreg Apr 4 '16 at 20:00
• @egreg I haven't done linear algebra for a while, and I'm trying to refresh my memory. How exactly can we conclude that $\operatorname{Ker(T)}=\operatorname{Ker(T^2)}$? Does it follow from the fact that their dimensions are equal and the fact that the one is subset of the other implies that they live in the same space? – K.Power Apr 7 '16 at 23:19
• @K.Power Since $\ker(T)\subseteq\ker(T^2)$, when we show the former has the same dimension as the latter, we conclude they're equal (take a basis of $\ker(T)$: it has the number of elements needed to be a basis of $\ker(T^2)$). – egreg Apr 7 '16 at 23:44


# What is wrong with my solution? $\int \cos^2 x \tan^3x dx$

I am trying to do this problem completely on my own but I can not get a proper answer for some reason

\begin{align} \int \cos^2 x \tan^3x dx &=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \sin^3 x}{ \cos x}dx\\ &=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\ &=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\ &=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\ &=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\ &=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\ &=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\ &=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\ &=\ln|\sec x| + \frac{\cos 2x}{4} + C \end{align}

This is the wrong answer, I have went through and back and it all seems correct to me.

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Alright, now I would like to know why the first integral gives $\ln \sec x$ instead of $-\ln \cos x$. – Phira Jun 8 '12 at 23:07
I think that was just an error in typing out the question – user138246 Jun 8 '12 at 23:08
@Phira: Because you can bring the $-1$ up. – Eugene Jun 8 '12 at 23:16
Same function, different expression. Note that $$\ln|\sec x| = \ln\left|\frac{1}{\cos x}\right| = \ln\left(|\cos x|^{-1}\right) = -\ln|\cos x|.$$ And $$\cos 2x = \cos^2x - \sin^2x = \cos^2x-(1-\cos^2x) = 2\cos^2x - 1$$so $$\frac{\cos 2x}{4}+C = \frac{2\cos^2x}{4}-\frac{1}{4}+C = \frac{1}{2}\cos^2x + C'.$$ – Arturo Magidin Jun 8 '12 at 23:16
You're making good progress I see. – Eugene Jun 8 '12 at 23:27

A very simple way to check if the answer to an indefinite integral is correct is to differentiate the answer. If you get the original function, your answer is correct, and is equal, up to a constant, with any other solutions.

We have \begin{align*} \frac{d}{dx}\left(\ln|\sec x| + \frac{\cos 2x}{4} + C\right) &= \frac{1}{\sec x}(\sec x)' + \frac{1}{4}(-\sin(2x))(2x)' + 0\\ &= \frac{\sec x\tan x}{\sec x} - \frac{1}{2}\sin(2x)\\ &= \tan x - \frac{1}{2}\left(2\sin x\cos x\right)\\ &= \frac{\sin x}{\cos x} - \sin x\cos x\\ &= \frac{ \sin x - \sin x\cos^2 x}{\cos x}\\ &= \frac{\sin x(1 - \cos^2 x)}{\cos x}\\ &= \frac{\sin^3 x}{\cos x}\\ &= \frac{\sin ^3 x \cos^2 x}{\cos^3x}\\ &= \frac{\sin^3 x}{\cos^3 x}\cos^2 x\\ &= \tan^3 x \cos^2 x. \end{align*} So your answer is right.

This happens a lot with integrals of trigonometric identities, because there are a lot of very different-looking expressions that are equal "up to a constant". So the answer to $$\int\sin x\cos x\,dx$$ can be either $\sin^2 x + C$ or $-\cos^2x + C$; both correct, though they look different, because they differ by a constant: $-\cos^2x + C = 1-\cos^2x + (C-1) = \sin^2x + (C-1)$.

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@Always love your answers Prof. – Babak S. Jun 9 '12 at 2:35

$${\ln |\sec x| + \frac{{\cos 2x}}{4} + C}$$

$${\ln \left|\frac 1 {\cos x}\right| + \frac{{1+\cos 2x}}{4} + C}-\frac 1 4$$

$${-\ln \left| {\cos x}\right| + \frac 1 2\frac{{1+\cos 2x}}{2} + K}$$

$${-\ln \left| {\cos x}\right| + \frac 1 2 \cos ^2 x + K}$$

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this kind of thing makes me feel like Calculus II exams and assignments must be hell to mark. – crf Jun 8 '12 at 23:56
Everything is hell to mark. But one can often structure exam questions to minimize these issues. – André Nicolas Jun 9 '12 at 0:28

You have been simplifying things up until line 6 and then kind of turned back into complications. It would be natural to notice that $$\sin x dx = -d\left(\cos x\right)$$ Then the integral looks as follows: $$I=-\int\frac{1-\cos^2x}{\cos x}d\left(\cos x\right)$$ So it appears that $\cos x$ plays the role of a variable in its own right, so why not let for example $t=\cos x$. Now $$I=-\int\frac{1-t^2}{t}dt=-\int\frac{dt}{t}+\int tdt=-\ln t+\frac{t^2}{2}+C$$ Now plug $\cos x$ back into place. Recognising distinct "reusable" blocks within the expression is the most natural way to arrive at useful substitutions.

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The calculation is correct. There are many alternate forms of the integral, because of the endlessly many trigonometric identities.

If you differentiate the expression you got and simplify, you will see that you are right.

The answer you saw is likely also right. What was it?

Added: Since $\sec x=\frac{1}{\cos x}$, we have $\ln(|\sec x|)=-\ln(|\cos x|)$.

Also, since $\cos 2x=2\cos^2 x -1$, we have $\frac{\cos 2x}{4}=\frac{\cos^2 x}{2}-\frac{1}{4}$.

So your answer and the book answer differ by a constant. That's taken care of by the arbitrary constant of integration. As a simpler example, $\int 2x\, dx=x^2+C$ and $\int 2x\, dx=(x^2+17)+C$ are both right.

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$\frac{1}{2}cos^2 x - ln|cosx| + C$ – user138246 Jun 8 '12 at 23:18

Notice that:

$$\frac{1}{2} \cos^2 x = \frac{1}{2} \left(\frac{1}{2} + \frac{1}{2} \cos 2x\right) = \frac{1}{4} + \frac{1}{4} \cos 2x$$

And:

$$-\ln|\cos x| = \ln|(\cos x)^{-1}| = \ln|\sec x|$$

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∫tanxdx=ln|secx|=-ln|cosx|+ k since -ln |cos x|+k= ln|(cos x)^-1|+k = ln|sec x| +k

(cos2x)/4= 1/2cos^2(x) - 1/4

k-1/4 = new constant C and together you have the solution.

Your answer seems to be correct it is just manipulation or different approach to trig identities.

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A simple $u$-substitution will work even better, IMO. ${\cos^2\theta\tan^3\theta }$ yields ${\frac {\sin^3}{\cos\theta}}$ So starting from there $${\int\frac{\sin^2\theta}{\cos\theta}\sin\theta d\theta}$$ then $${\int\frac{1-\cos^2\theta}{\cos\theta}\sin\theta d\theta}$$ Let $u={\cos \theta}$ then ${du=-\sin\theta d\theta}$ $${-\int\frac {1-u^2}{u}du}$$ let ${v=1-u^2 }$ ${dv=-2u}$

let ${dw=1/u}$, ${w=\ln u}$

thus $${\ln u(1-u^2)+\int 2u\ln u}\,du$$ then

let ${v=\ln u}$, ${dv=u^{-1}}$

let ${dw=2u}$, ${w=u^2}$

then \begin{align*} {\ln u u^2-\int \frac{u^2}{u}\,du}\\{\ln u u^2-\int u\,du}\\ {\ln u(1-u^2)+\ln u u^2-1/2u^2} \end{align*} then \begin{align*}{\ln u(1-u^2+u^2)-1/2u^2}\\ {-(\ln u-1/2u^2)} \end{align*} substituting back yields $${-\ln\cos\theta+1/2\cos^2\theta+c}$$

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# A (small) theatre contains 10 seats. The seats are numbered from 1 to 10,

A (small) theatre contains 10 seats. The seats are numbered from 1 to 10, but not in order. Alice, Bob and Carol have tickets for seats 1, 2 and 3 respectively, and Damien, Eve, Francis, and Gertrude each have tickets without an assigned seat. Unfortunately they all arrive late and the lights are dim, so they can’t see what seat they are sitting in.

a) In how many ways can they be seated?

b) In how many ways can they be seated such that Alice is in her correct seat (eg, seat number 1)?

c) In how many ways can they be seated such that Alice, Bob and Carol are all in their correct seats?

d) In how many ways can they be seated such that none of Alice, Bob, Carol are in their correct seats?

What I did for part A is The first person has 10 seats to choose from, then the second person has 9 and so on. I got $10\times9\times8\times7\times6\times5\times4$ as the answer.

For part B I said alice has 1 seat to choose from and then did the rest like part A so I got $1\times9\times8\times7\times6\times5\times4$.

For C I did the same as B but with 3 people and got $1\times1\times1\times7\times6\times5\times4$ but this doesn't seem right to me.

A bit confused on how to do the rest.

"For c) I did the same as b) but with 3 people and got $7^{\underline{4}}=7\cdot6\cdot5\cdot4$ but this doesn't seem right to me"

Your first three parts are in fact all correct (though they suffer from a lack of convenient notation. If there were 50 people and 80 seats you don't want to waste time writing out all of the numbers in the product. Recommend using one of the following falling factorial notations $(7)_4, 7^{\underline{4}}, \frac{7!}{3!}, ~_7P_4, P(7,4),\dots$ but whichever you use make sure your audience understands the notation)

(Note on falling factorial notation: in all of the above, $(n)_r=n^{\underline{r}}=P(n,r)=\dots=\underbrace{n(n-1)(n-2)\cdots(n-r+2)(n-r+1)}_{r~\text{terms in product starting with}~n}$)

"A bit confused on how to do the rest"

The first three parts are most of the calculations needed to complete the fourth part. Recognize that by symmetry the answer to part b) is also the answer to the question of how many ways Bob is seated in his correct seat as well as the answer to how many ways Carol is in her correct seat.

Use inclusion-exclusion to figure out answer to "In how many ways can they be seated such that at least one of them is in their correct seat." Letting $A,B,C$ represent the events that Alice, Bob, and Carol made it to their correct seats respectively, and letting $\Omega$ represent the sample space i.e. the number of ways in which we don't care who did or didn't make it to their seat, we are tasked with finding the number of ways that none of them made it to their seat. I.e. $|A^c\cap B^c\cap C^c|$

Note that $|A^c\cap B^c\cap C^c| = |\Omega\setminus(A\cup B\cup C)|=|\Omega|-|A\cup B\cup C|$

$=|\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$

You've already found $|\Omega|,|A|,|A\cap B\cap C|$ and by symmetry found $|B|$ and $|C|$. The only information you are missing is $|A\cap B|,|A\cap C|,|B\cap C|$. A similar symmetry argument will reduce the effort it takes to find those.

• so for part d |A∩B|,|A∩C|,|B∩C|, would all be the same and would be calculated by adding P(9,4) to itself. Then I would put everything in the equation you provided? – KGT Nov 7 '16 at 17:52
• @KGT $P(n,r) = \underbrace{n(n-1)(n-2)(n-3)\cdots(n-r+3)(n-r+2)(n-r+1)}_{r~\text{total terms in product starting with}~n}$. I don't think you want to be using $P(9,4)$ – JMoravitz Nov 7 '16 at 17:55
• @KGT you did it correctly in the first three parts. Reworded using the $P$ notation, your answers were $P(10,7),P(9,6)$ and $P(7,4)$ respectively. If we had included the extra question in the middle of those of finding $|A\cap B|$, i.e. alice and bob get their correct seat and then we seat everyone else, how much would that be? – JMoravitz Nov 7 '16 at 18:00
• Oh ok. My mistake. I thought that meant P(9,4) meant 9x8x7x6x5x4. What I should use is P(9,6) correct? – KGT Nov 7 '16 at 18:01
• $P(9,6)$ was the answer to if we only cared about alice getting her correct seat but we didn't care about bob or carol getting their correct seats. If we cared about alice and bob, should that nine have been in the product? – JMoravitz Nov 7 '16 at 18:02


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# How to calculate cumulative cash flow from a Cash Flow table indexed by year

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I am a relative neophyte in using indexed parameters. I want to calculate a cumulative cash flow by year from a table of cash flows, in order to calculate when a project will break even. I am looking for the syntax of a call on each row of my cash flow list to add it to the previous row of my cumulative cash flow list. Once I have that, I also want to calculate a similar list of discounted cumulative cash flows to understand the effect of discounting the cash flow. Is it worthwhile to create a second index with cash flow, cumulative cash flow, and discounted cash flow? Or is there an even more clever way to do this using features I am unaware of?

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You are looking for the Cumulate function, which takes your cash flow by Year and returns the cumulated cash flow by Year.  The syntax is:

Cumulate( Cash_flow, Year )

To discount a cash flow, you just multiply Cash_flow by the discount factor for each year, like this:

Cash_flow * (1 - Discount_rate)^(@Year-1)

Here I've used @Year to number the years starting at 1. By using @Year-1, it means that the first year won't be discounted, whereas if you raise to the power of @Year you would discount the first year.  Think about which convention you want.  You can then pass this to the first parameter of Cumulate.

You want to compare the undiscounted case to the discounted case. A cool thing is that you can do this without duplicating all the logic for each case. The undiscounted case is simply the discounted case when the discount rate is 0. So, just make Discount_rate a Choice, i.e., between 0% and 7%, and then all your subsequent logic that uses discounted cash flow will work for either, or indeed for both at the same time (by selecting ALL for the choice), giving you the ability to compare any downstream variable without having to do anything additional.

In a short while, I will attach a model that demonstrates all of this and adds in the calculation for the breakeven year.

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This looks great! And the tip on 0% discount is excellent. I look forward to your example, and hope to get back to this work this week.

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This approach worked great.  Now I can get to my reason for doing this.  I would like to define a variable that finds the year when cumulative cash flow becomes positive, return the year, and add to that the value of cumulative cash flow for that year divided by the cash flow for that year, which gives me a number of years (with a fraction) to break even. Still looking to figure out how to do this.

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Sorry about the delay with posting the model that I had promised.  I hit a technical problem. Anyway, here it is:

Cash flow.ana

It includes a variable that finds the breakeven year and a fraction of the breakeven year. I'm not sure if my fraction of the breakeven year is exactly what you just described, but hopefully you can tweak it if not.

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I got taken away on other things (including a short vacation) and am just getting back to this reply.  I had in the meantime discovered the Solar Panel model, which gets the breakeven year by using the CondMin function, but you have come at it a separate way. Either way, I am learning useful new functions! I will look at it and see if it gets the same answer I was coming to.  Thanks!

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# how to know when a particular proof is appropriate for the given problem?

The main trouble I am currently having in math is knowing when the use cases are appropriate in a proof. I see many videos where they seem to choose a strategy like proof by contrapositive or proof by contradiction, but never quite understand how they came to the conclusion to use that proof strategy.

Here are some examples I have come across and my own solutions to them using the recommended proofs

the product of two odd integers is odd

I used contradiction to solve it

• suppose the product of two odd integers is even
• (2k+1)(2k+1) = 2k
• 2(2k^2 + 2k) + 1 = 2k
• given k is an integer, (2k^2 + 2k) is an integer. Therefore, an even number cannot be an odd number

if x^2 is even, x is even

Based on the wikipedia article on contrapositive

• if x is odd, then x^2 is odd
• (2k+1)^2 == 2(2k^2 +2k) + 1
• therefore, if x^2 is even, then x is even

However, what I am wondering is, is there a general principle as to when to use specific strategies for proofs? Eg, if you specifically know a theory is false, do you choose a strategy accordingly? What determines which strategy will be most effective, or is it arbitrary?

• – Shaun Oct 30 '14 at 12:38
• For "perfect" proofs see the book of Aigner and Ziegler, Proofs from THE BOOK – Dietrich Burde Oct 30 '14 at 12:55

This skill comes with lots of practice. Generally speaking, direct proofs are used when the result is "positive-sounding". In your example, the result is, "the product of two odd integers is odd". This result is positive-sounding (as opposed to a result like, "there is no smallest positive real number").

For the result, "if $x^2$ is even, then $x$ is even," you could do either a direct proof or a proof by contrapositive. A lot of times it does not matter and there is usually more than one way to prove something. The direct proof, in this case, is favored over a proof by contrapositive simply because it is more, well, direct. Here is how the result would be proven using the contrapositive:

Proof:

Suppose $x$ is odd. We show that $x^2$ is odd. So $x=2k+1$ for some integer $k$. Thus $x^2=(2k+1)^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1.$ Now, since $(2k^2+2k)$ is an integer, it follows that $x^2$ is odd.

For a result like "there is no smallest positive real number", this would be difficult to prove using a direct proof. The result is better suited for a proof by contradiction. It would start like this: "Suppose there exists a smallest positive real number…"

Some other examples of negative sounding results that would be more suited for proof by contradiction:

• If a is an even integer and b is an odd integer, then 4 does not divide $a^2+2b^2$
• The integer 100 cannot be written as the sum of three integers, and odd number of which are odd
• The sum of a rational number and an irrational number is irrational.

[Source: Chartrand, G., Polimeni, A.D., Zhang, P., 2013. Mathematical Proofs: A Transition to Advanced Mathematics, 3rd ed. Boston: Pearson.]

• ah, exactly what I was looking for, very clear and understandable, thank you! – corvid Oct 30 '14 at 13:05

There are many ways to write down a proof. The question is, apart from correctness, which one is more elegant, more concise, more constructive, and last not least easier und thus clearer. In your case, for me the easiast way is to do first an obvious computation, i.e. $$(2k+1)(2l+1)=4kl+2k+2l+1.$$ Your first step here is wrong, because you assume that both integers are equal, i.e., both given by $2k+1$. This would be a special case.
Then, in the second step, we consider everything modulo $2$. The integer on the RHS is odd, because it is of the form $2m+1$, with $m=2kl+k+l$. This finishes the proof.

• theoretically, anything you can prove by contradiction can also be proved by contrapositive, in that sense? – corvid Oct 30 '14 at 12:54