bernoulli's theorem experiment calculation

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Experiment #2: Bernoulli’s Theorem Demonstration

1. introduction.

Energy presents in the form of pressure, velocity, and elevation in fluids with no energy exchange due to viscous dissipation, heat transfer, or shaft work (pump or some other device). The relationship among these three forms of energy was first stated by Daniel Bernoulli (1700-1782), based upon the conservation of energy principle. Bernoulli’s theorem pertaining to a flow streamline is based on three assumptions: steady flow, incompressible fluid, and no losses from the fluid friction. The validity of Bernoulli’s equation will be examined in this experiment.

2. Practical Application

Bernoulli’s theorem provides a mathematical means to understanding the mechanics of fluids. It has many real-world applications, ranging from understanding the aerodynamics of an airplane; calculating wind load on buildings; designing water supply and sewer networks; measuring flow using devices such as weirs, Parshall flumes, and venturimeters; and estimating seepage through soil, etc. Although the expression for Bernoulli’s theorem is simple, the principle involved in the equation plays vital roles in the technological advancements designed to improve the quality of human life.

3. Objective

The objective of this experiment is to investigate the validity of the Bernoulli equation when it is applied to a steady flow of water through a tapered duct.

In this experiment, the validity of Bernoulli’s equation will be verified with the use of a tapered duct (venturi system) connected with manometers to measure the pressure head and total head at known points along the flow.

5. Equipment

The following equipment is required to complete the demonstration of the Bernoulli equation experiment:

• F1-10 hydraulics bench,
• F1-15 Bernoulli’s apparatus test equipment, and
• A stopwatch for timing the flow measurement.

6. Equipment Description

The Bernoulli test apparatus consists of a tapered duct (venturi), a series of manometers tapped into the venturi to measure the pressure head, and a hypodermic probe that can be traversed along the center of the test section to measure the total head. The test section is a circular duct of varying diameter with a 14° inclined angle on one side and a 21° inclined angle on other side. Series of side hole pressure tappings are provided to connect manometers to the test section (Figure 2.1).

Manometers allow the simultaneous measurement of the pressure heads at all of the six sections along the duct. The dimensions of the test section, the tapping positions, and the test section diameters are shown in Figure 2.2. The test section incorporates two unions, one at either end, to facilitate reversal for convergent or divergent testing. A probe is provided to measure the total pressure head along the test section by positioning it at any section of the duct. This probe may be moved after slackening the gland nut, which should be re-tightened by hand. To prevent damage, the probe should be fully inserted during transport/storage. The pressure tappings are connected to manometers that are mounted on a baseboard. The flow through the test section can be adjusted by the apparatus control valve or the bench control valve [2].

Bernoulli’s theorem assumes that the flow is frictionless, steady, and incompressible. These assumptions are also based on the laws of conservation of mass and energy.  Thus, the input mass and energy for a given control volume are equal to the output mass and energy:

These two laws and the definition of work and pressure are the basis for Bernoulli’s theorem and can be expressed as follows for any two points located on the same streamline in the flow:

P: pressure,

g: acceleration due to gravity,

v : fluid velocity, and

z: vertical elevation of the fluid.

In this experiment, since the duct is horizontal, the difference in height can be disregarded, i.e., z 1 =z 2

The hydrostatic pressure (P) along the flow is measured by manometers tapped into the duct. The pressure head (h), thus, is calculated as:

Therefore, Bernoulli’s equation for the test section can be written as:

The total head (h t ) may be measured by the traversing hypodermic probe. This probe is inserted into the duct with its end-hole facing the flow so that the flow becomes stagnant locally at this end; thus:

The conservation of energy or the Bernoulli’s equation can be expressed as:

The flow velocity is measured by collecting a volume of the fluid (V) over a time period (t). The flow rate is calculated as:

The velocity of flow at any section of the duct with a cross-sectional area of  is determined as:

For an incompressible fluid, conservation of mass through the test section should be also satisfied (Equation 1a), i.e.:

8. Experimental Procedure

• Place the apparatus on the hydraulics bench, and ensure that the outflow tube is positioned above the volumetric tank to facilitate timed volume collections.
• Level the apparatus base by adjusting its feet. (A sprit level is attached to the base for this purpose.) For accurate height measurement from the manometers, the apparatus must be horizontal.
• Install the test section with the 14° tapered section converging in the flow direction. If the test section needs to be reversed, the total head probe must be retracted before releasing the mounting couplings.
• Connect the apparatus inlet to the bench flow supply, close the bench valve and the apparatus flow control valve, and start the pump. Gradually open the bench valve to fill the test section with water.
• Close both the bench valve and the apparatus flow control valve.
• Remove the cap from the air valve, connect a small tube from the air valve to the volumetric tank, and open the air bleed screw.
• Open the bench valve and allow flow through the manometers to purge all air from them, then tighten the air bleed screw and partly open the bench valve and the apparatus flow control valve.
• Open the air bleed screw slightly to allow air to enter the top of the manometers (you may need to adjust both valves to achieve this), and re-tighten the screw when the manometer levels reach a convenient height. The maximum flow will be determined by having a maximum (h 1 ) and minimum (h 5 ) manometer readings on the baseboard.

If needed, the manometer levels can be adjusted by using an air pump to pressurize them. This can be accomplished by attaching the hand pump tube to the air bleed valve, opening the screw, and pumping air into the manometers.  Close the screw, after pumping, to retain the pressure in the system.

• Take readings of manometers h 1 to h 6 when the water level in the manometers is steady. The total pressure probe should be retracted from the test section during this reading.
• Measure the total head by traversing the total pressure probe along the test section from h 1 to h 6 .
• Measure the flow rate by a timed volume collection. To do that, close the ball valve and use a stopwatch to measure the time it takes to accumulate a known volume of fluid in the tank, which is read from the sight glass. You should collect fluid for at least one minute to minimize timing errors. You may repeat the flow measurement twice to check for repeatability. Be sure that the total pressure probe is retracted from the test section during this measurement.
• Reduce the flow rate to give the head difference of about 50 mm between manometers 1 and 5 (h 1 -h 5 ). This is the minimum flow experiment. Measure the pressure head, total head, and flow.
• Repeat the process for one more flow rate, with the (h 1 -h 5 ) difference approximately halfway between those obtained for the minimum and maximum flows. This is the average flow experiment.
• Reverse the test section (with the 21° tapered section converging in the flow direction) in order to observe the effects of a more rapidly converging section. Ensure that the total pressure probe is fully withdrawn from the test section, but not pulled out of its guide in the downstream coupling. Unscrew the two couplings, remove the test section and reverse it, then re-assemble it by tightening the couplings.
• Perform three sets of flow, and conduct pressure and flow measurements as above.

9. Results and Calculations

Please visit this link for accessing excel workbook for this experiment.

9.1. Results

Enter the test results into the Raw Data Tables.

Raw Data Table

  Raw Data Table

9.2 calculations.

For each set of measurements, calculate the flow rate; flow velocity, velocity head, and total head,  (pressure head+ velocity head).  Record your calculations in the Result Table.

Result Table

Use the template provided to prepare your lab report for this experiment. Your report should include the following:

• Table(s) of raw data
• Table(s) of results
• For each test, plot the total head (calculated and measured), pressure head, and velocity head (y-axis) vs. distance into duct (x-axis) from manometer 1 to 6, a total of six graphs. Connect the data points to observe the trend in each graph. Note that the flow direction in duct Position 1 is from manometer 1 to 6; in Position 2, it is from manometer 6 to 1.
• Comment on the validity of Bernoulli’s equation when the flow converges and diverges along the duct.
• Comment on the comparison of the calculated and measured total heads in this experiment.
• energy loss and how it is shown by the results of this experiment, and

Applied Fluid Mechanics Lab Manual Copyright © 2019 by Habib Ahmari and Shah Md Imran Kabir is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Experimental verification of bernoulli’s theorem.

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The procedure of Laboratory experiment to verify Bernoulli’s theorem, required apparatus and calculations to be done are explained in this article.

Bernoulli’s theorem states - For a continuous, steady and frictionless flow the total head (which is the sum of pressure head, velocity head and elevation head) at any section remains constant.

Total Head = pressure head + velocity head + elevation head

H = P/w + V 2 / 2g + Z

According to Bernoulli’ theorem, Total head at any two sections is expressed as :

Apparatus required

Apparatus required to conduct Bernoulli’s experiment are :

• Supply tank
• Tapered inclined pipe with piezometer tubes at different points
• Measuring tank

Test Procedure

Test procedure to verify Bernoulli’s experiment is as follows :

• Open the inlet valve and allow the water to flow from the supply tank to the receiving tank through a tapered inclined pipe.
• Adjust the flow using an outlet valve to make the head constant in the supply tank. At the constant head, head causing inflow and outflow are equal.
• After adjusting the flow, Note down the readings of the water level of each piezometer tube which are nothing but pressure heads at different points of tapered tube.

• Compute the area of cross-section of tapered pipe at points where piezometer tubes are located.
• Now, take the stopwatch and measure the height of water collected for a particular time interval.
• Also, note down the measuring tank dimensions.
• Repeat the same procedure for different discharges for at least two more times.

Observations

Pressure head or Piezometer readings, P/w =

Area of cross section of tapered pipe under piezometer tubes = A 1 , A 2 , A 3 …….

Area of Measuring tank, A =

Height of water collected for “t” interval of time =

Elevation head, Z =

Calculations

Volume of water Collected in measuring tank (V) = Area of measuring tank (A) X height of water collected for “t” interval of time (h)

Discharge Q = volume / time = V/t

Velocity of flow, v = Discharge / Area of cross section of tapered pipe

Velocities under each piezometer tube are v 1 = Q/A 1 ,

v 2 = Q/A 2 ,

v 3 = Q/A 3 ………

Velocity head = v 2 / 2g

Total Head = P/w + V 2 / 2g + Z

Hence, Bernoulli’s theorem is proved and It can be expressed as,

Sadanandam Anupoju

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14.6 Bernoulli’s Equation

Learning objectives.

By the end of this section, you will be able to:

• Explain the terms in Bernoulli’s equation
• Explain how Bernoulli’s equation is related to the conservation of energy
• Describe how to derive Bernoulli’s principle from Bernoulli’s equation
• Perform calculations using Bernoulli’s principle
• Describe some applications of Bernoulli’s principle

As we showed in Figure 14.27 , when a fluid flows into a narrower channel, its speed increases. That means its kinetic energy also increases. The increased kinetic energy comes from the net work done on the fluid to push it into the channel. Also, if the fluid changes vertical position, work is done on the fluid by the gravitational force.

A pressure difference occurs when the channel narrows. This pressure difference results in a net force on the fluid because the pressure times the area equals the force, and this net force does work. Recall the work-energy theorem,

The net work done increases the fluid’s kinetic energy. As a result, the pressure drops in a rapidly moving fluid whether or not the fluid is confined to a tube.

There are many common examples of pressure dropping in rapidly moving fluids. For instance, shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The reason is that the high-velocity stream of water and air creates a region of lower pressure inside the shower, whereas the pressure on the other side remains at the standard atmospheric pressure. This pressure difference results in a net force, pushing the curtain inward. Similarly, when a car passes a truck on the highway, the two vehicles seem to pull toward each other. The reason is the same: The high velocity of the air between the car and the truck creates a region of lower pressure between the vehicles, and they are pushed together by greater pressure on the outside ( Figure 14.29 ). This effect was observed as far back as the mid-1800s, when it was found that trains passing in opposite directions tipped precariously toward one another.

Energy Conservation and Bernoulli’s Equation

The application of the principle of conservation of energy to frictionless laminar flow leads to a very useful relation between pressure and flow speed in a fluid. This relation is called Bernoulli’s equation , named after Daniel Bernoulli (1700–1782), who published his studies on fluid motion in his book Hydrodynamica (1738).

Consider an incompressible fluid flowing through a pipe that has a varying diameter and height, as shown in Figure 14.30 . Subscripts 1 and 2 in the figure denote two locations along the pipe and illustrate the relationships between the areas of the cross sections A , the speed of flow v , the height from ground y , and the pressure p at each point. We assume here that the density at the two points is the same—therefore, density is denoted by ρ ρ without any subscripts—and since the fluid is incompressible, the shaded volumes must be equal.

We also assume that there are no viscous forces in the fluid, so the energy of any part of the fluid will be conserved. To derive Bernoulli’s equation, we first calculate the work that was done on the fluid:

The work done was due to the conservative force of gravity and the change in the kinetic energy of the fluid. The change in the kinetic energy of the fluid is equal to

The change in potential energy is

The energy equation then becomes

Rearranging the equation gives Bernoulli’s equation:

This relation states that the mechanical energy of any part of the fluid changes as a result of the work done by the fluid external to that part, due to varying pressure along the way. Since the two points were chosen arbitrarily, we can write Bernoulli’s equation more generally as a conservation principle along the flow.

Bernoulli’s Equation

For an incompressible, frictionless fluid, the combination of pressure and the sum of kinetic and potential energy densities is constant not only over time, but also along a streamline:

A special note must be made here of the fact that in a dynamic situation, the pressures at the same height in different parts of the fluid may be different if they have different speeds of flow.

Analyzing Bernoulli’s Equation

According to Bernoulli’s equation, if we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.

The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, let us consider some specific situations that simplify and illustrate its use and meaning.

Bernoulli’s equation for static fluids

First consider the very simple situation where the fluid is static—that is, v 1 = v 2 = 0 . v 1 = v 2 = 0 . Bernoulli’s equation in that case is

We can further simplify the equation by setting h 2 = 0 . h 2 = 0 . (Any height can be chosen for a reference height of zero, as is often done for other situations involving gravitational force, making all other heights relative.) In this case, we get

This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by h 1 h 1 , and consequently, p 2 p 2 is greater than p 1 p 1 by an amount ρ g h 1 ρ g h 1 . In the very simplest case, p 1 p 1 is zero at the top of the fluid, and we get the familiar relationship p = ρ g h p = ρ g h . ( Recall that p = ρ g h ( Recall that p = ρ g h and Δ U g = − m g h . ) Δ U g = − m g h . ) Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid is ρ g h ρ g h . Although we introduce Bernoulli’s equation for fluid motion, it includes much of what we studied for static fluids earlier.

Bernoulli’s principle

Suppose a fluid is moving but its depth is constant—that is, h 1 = h 2 h 1 = h 2 . Under this condition, Bernoulli’s equation becomes

Situations in which fluid flows at a constant depth are so common that this equation is often also called Bernoulli’s principle , which is simply Bernoulli’s equation for fluids at constant depth. (Note again that this applies to a small volume of fluid as we follow it along its path.) Bernoulli’s principle reinforces the fact that pressure drops as speed increases in a moving fluid: If v 2 v 2 is greater than v 1 v 1 in the equation, then p 2 p 2 must be less than p 1 p 1 for the equality to hold.

Example 14.6

Calculating pressure.

Substituting known values,

Significance

Applications of bernoulli’s principle.

Many devices and situations occur in which fluid flows at a constant height and thus can be analyzed with Bernoulli’s principle.

Entrainment

People have long put the Bernoulli principle to work by using reduced pressure in high-velocity fluids to move things about. With a higher pressure on the outside, the high-velocity fluid forces other fluids into the stream. This process is called entrainment . Entrainment devices have been in use since ancient times as pumps to raise water to small heights, as is necessary for draining swamps, fields, or other low-lying areas. Some other devices that use the concept of entrainment are shown in Figure 14.31 .

Velocity measurement

Figure 14.32 shows two devices that apply Bernoulli’s principle to measure fluid velocity. The manometer in part (a) is connected to two tubes that are small enough not to appreciably disturb the flow. The tube facing the oncoming fluid creates a dead spot having zero velocity ( v 1 = 0 v 1 = 0 ) in front of it, while fluid passing the other tube has velocity v 2 v 2 . This means that Bernoulli’s principle as stated in

Thus pressure p 2 p 2 over the second opening is reduced by 1 2 ρ v 2 2 1 2 ρ v 2 2 , so the fluid in the manometer rises by h on the side connected to the second opening, where

(Recall that the symbol ∝ ∝ means “proportional to.”) Solving for v 2 v 2 , we see that

Part (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air-speed indicators in aircraft.

A fire hose

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change.

Example 14.7

Calculating pressure: a fire hose nozzle.

where subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds v 1 v 1 and v 2 v 2 . Since Q = A 1 v 1 Q = A 1 v 1 , we get

Similarly, we find

This rather large speed is helpful in reaching the fire. Now, taking h 1 h 1 to be zero, we solve Bernoulli’s equation for p 2 p 2 :

Substituting known values yields

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Bernoulli’s theorem

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• OpenStax - College Physics 2e - Bernoulli’s Equation
• Khan Academy - What is Bernoulli 's equation?
• Physics LibreTexts - Bernoulli’s Equation

Bernoulli’s theorem , in fluid dynamics , relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar. First derived (1738) by the Swiss mathematician Daniel Bernoulli , the theorem states, in effect, that the total mechanical energy of the flowing fluid, comprising the energy associated with fluid pressure, the gravitational potential energy of elevation, and the kinetic energy of fluid motion, remains constant. Bernoulli’s theorem is the principle of energy conservation for ideal fluids in steady, or streamline , flow and is the basis for many engineering applications.

Bernoulli’s theorem implies, therefore, that if the fluid flows horizontally so that no change in gravitational potential energy occurs, then a decrease in fluid pressure is associated with an increase in fluid velocity. If the fluid is flowing through a horizontal pipe of varying cross-sectional area, for example, the fluid speeds up in constricted areas so that the pressure the fluid exerts is least where the cross section is smallest. This phenomenon is sometimes called the Venturi effect, after the Italian scientist G.B. Venturi (1746–1822), who first noted the effects of constricted channels on fluid flow.

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Bernoulli Experiment Lab Report

Aim of bernoulli experiment.

Aim of this experiment is to compare the behaviour of ideal fluid and real fluid using Bernoulli experiment.

1. Use the venture meter apparatus to perform Bernoulli experiment and the effect of area of the flow velocity and fluid pressure 2. Use the Bernoulli’s equation /  Bernoulli theorem  to compare the behaviour of ideal and real fluid

Introduction  Bernoulli Experiment

According to the Bernoulli’s experiment when area available for the fluid to flow decrease then flow of fluid increase and at the mean while time the fluid pressure or the fluid potential energy decreases.

Read about the Effect of Sluice Gate on the Flow of Fluid . 

This principle was name after the Daniel Bernoulli who first writes this principle in book named Hydrodynamic.  Following are some of the application of the Bernoulli experiment

Theory of Bernoulli's Theorem Experiment

Bernoulli experiment / principle state that the in a steady flowing fluid the sum of all the mechanical energies including kinetic energy, dynamic head, fluid pressure and potential energy should remain same at all the point of the flow. 

So if any type of energy increase like if kinetic energy increase then the other type of the energy like potential energy, pressure will decrease t o make the final sum same as before.  According to the B ernoulli's theorem experiment or can say  Bernoulli equation a flowing fluid have three things

• Pressure head
• Kinetic Energy
• Potential Energy

So we have P+  1/2×ρ×v^2+ ρgh=C P/ρg+  1/2×v^2/g+h=C According to the law of conservation of energy, energies at the input should be equal to the output so P_1/ρg+  (V_1^2)/2g+h=  P_n/ρg+  (V_n^2)/2g+h In the above equation  P = fluid pressure V = flow velocity Z = height ρ = density

From Bernoulli’s experiment / principle it can be stated that the density and pressure are inversely proportional to each other’s means high density fluid will apply more pressure while moving than the low density fluids. 

In the horizontal pipe where the inlet and outlet of the are at same height, the z quantity can be removed to give the above mention equation of Bernoulli’s principle a new look from where we can calculate the height at any point of the flow if we have the initial height of flow and velocity at respective positions. 

P_1/ρg+  (V_1^2)/2g=  P_n/ρg+  (V_n^2)/2g P_1/ρg=h1 and  P_n/ρg=hn  h_1+  (V_1^2)/2g= h_n+  (V_n^2)/2g h_n= h_1-[  (v_n^2)/2g-  (v_1^2)/2g]

Apparatus for Bernoulli's theorem experiment

• Venture meter
• Supply Hoses
• Measuring Tank

Procedure for Bernoulli experiment

Following is a complete procedure for performing the Bernoulli experiment

Graphs of  Bernoulli's theorem lab report

Following are the main Graphs of  Bernoulli's theorem lab report

Discussion of Bernoulli theorem lab report

Based on the calculation above and graph papers presented a detailed discussion of Bernoulli theorem lab report is as follow

• From the calculation it is very clear that with decrease in area of the flow velocity increase and pressure decrease • As shown in graph of all three test that the decrease in area of flow decrease the height of water in manometer column means they are directly proportional to each other • Difference in the theoretical and measured value it can be said that water is not an ideal fluid  • Height of water in the final column was not equal to the initial values which show that there are friction losses in water particle • This type of information is very use full in the case if nozzles, jets and diffusers

Conclusion of Bernoulli's theorem

This is because of the friction losses in the real fluid; ideal fluid does not have friction losses. From the experiment it can be conclude that with decrease in area of flow there is an increase in velocity and decrease in the flow pressure of the fluid.   

References for bernoulli equation lab report

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Experimentals: Examples of Bernoulli's theorem

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RUBEN: Good evening and welcome to This Is Your Theorem. Tonight we honour the great Swiss scientist Daniel Bernoulli who can't be with us tonight because he's dead. Bernoulli died in 1782, but he was some genius, as many of us know.

BERNIE: Bernoulli's theorem explains the strange behaviour we see when gases and liquids flow, some examples of which we'll see tonight. Wow.

RUBEN: So without any further do, Daniel Bernoulli, this is your theorem.

BERNIE: Bernoulli was educated in Switzerland, a tiny country where people wear clothes like this, eat cheese like this and where pocket knives have more accessories than you can poke a stick at. In 1725, Bernoulli moved to St Petersburg. And while his friends were back home yodelling, it was in St Petersburg that Bernoulli made his breakthrough.

RUBEN: No-one can be sure, but perhaps it was one of these that sparked Bernoulli's curiosity.

BERNIE: The water pouring down ought to keep an egg at the bottom of a glass, but instead it rises to the surface. Freaky, hey?

RUBEN: One can see why Bernoulli was intrigued, but it's not just his seminal work on eggs and water that made him famous. Bernoulli's work also had serious implications for balloons. Commonsense suggest that blowing between these balloons would cause them to move apart. But instead the opposite happens and they come together - bizarre.

BERNIE: Bernoulli discovered that the pressure of moving air is lower than the pressure of air that's still. The faster air moves, the lower its pressure. It might sound like regular physics mumbo jumbo, but Bernoulli's theorem is great because things always do the opposite of what you'd expect.

RUBEN: Like the liquid in this straw. When you blow across the top of a straw you'd expect the air to go down the straw and push the cordial down. But Bernoulli's theorem says that the air that's blown across the top is at a lower pressure. So it acts as though you were sucking on the straw. And look what happens - cordial is rising up the straw. That balloon - what a genius. But its brilliance didn't stop there.

BERNIE: If we put a balloon in the air stream of this hair dryer you might expect the balloon to go flying up and then fall back to the ground. But instead the balloon is trapped by the fast-moving air from the hair dryer, which, according to Bernoulli's theorem, is at a lower pressure than the still air around it. Once again, the balloon does the opposite of what you'd expect.

RUBEN: Here's a challenge Bernoulli would never have fallen for. Try blowing a ping-pong ball out of a funnel. You'll find it's impossible. Why? Because fast-moving air surrounds the ball and fast-moving air has lower pressure. The ball is trapped. And, look at this, the balls stays trapped even if you hold the funnel upside-down. Once again, the opposite of what you'd expect.

BERNIE: Yes, Bernoulli's work has changed lives the world over. Thanks to Bernoulli's theorem...

RUBEN: ..eggs can impress other breakfast foods by floating in running water...

BERNIE: ..hair dryers and balloons can live together in harmony...

RUBEN: ..and the humble ping-pong ball can resist the mightiest set of lungs.

BERNIE: (In falsetto) You can huff and puff, but you'll never blow me out.

RUBEN: So Daniel Bernoulli, this is your theorem.

SUBJECTS:   Science

YEARS:  7–8, 9–10

Have you ever wondered how a yacht sails into the wind?

Watch as the Experimentals team works through practical demonstrations of Bernoulli's theorem.

This science experiment will generate a few surprises as you learn how gases and liquids change their behaviour as they begin to flow.

Things to think about

• 1. Imagine that you are riding a bike. What will you feel if a truck passes close to you? Will you be pushed away or drawn towards the truck? Why do you think this happens?
• 2. The floating egg and the touching balloons are demonstrations of Bernoulli's theorem. What does this theorem say about pressure in moving gas or liquid?
• 3. Choose one of the examples used in the clip and make a diagram showing air movement and pressure zones to help explain what you saw. Using Bernoulli's theorem, think again about what happens when a truck passes close to you on your bike. Draw a diagram to show the pressure zones. Think of some practical ways that Bernoulli's theorem could be useful in our everyday lives.
• 4. Investigate how scientists and engineers have used Bernoulli's theorem as they developed new technologies. How might our lives be different without Bernoulli's theorem?

Date of broadcast: 10 Jun 2007

Metadata © Australian Broadcasting Corporation and Education Services Australia Ltd 2012 (except where otherwise indicated). Digital content © Australian Broadcasting Corporation (except where otherwise indicated). Video © Australian Broadcasting Corporation (except where otherwise indicated). All images copyright their respective owners. Text © Australian Broadcasting Corporation and Education Services Australia is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License (CC BY-SA 4.0).

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August 16, 2024

Bernoulli numbers and the harmonic oscillator, posted by john baez.

I keep wanting to understand Bernoulli numbers more deeply, and people keep telling me stuff that’s fancy when I want to understand things simply . But let me try again.

The Bernoulli numbers can be defined like this:

x e x − 1 = B 0 + B 1 x + B 2 x 2 2 ! + B 3 x 3 3 ! + ⋯ \frac{x}{e^x - 1} = B_0 + B_1 x + B_2 \frac{x^2}{2!} + B_3 \frac{x^3}{3!} + \cdots

and if you grind them out, you get

B 0 = 1 B 1 = − 1 2 B 2 = 1 6 B 3 = 0 B 4 = − 1 30 \begin{array}{lcr} B_0 &=& 1 \\ B_1 &=& -\frac{1}{2} \\ B_2 &=& \frac{1}{6} \\ B_3 &=& 0 \\ B_4 &=& -\frac{1}{30} \end{array}

and so on. The pattern is quite strange.

Bernoulli numbers are connected to hundreds of interesting things. For example if you want to figure out a sum like

1 10 + ⋯ + 1000 10 1^{10} + \cdots + 1000^{10}

you can use Bernoulli numbers — indeed Jakob Bernoulli boasted

It took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum 91,409,924,241,424,243,424,241,924,242,500.

For some more mysterious appearances of the Bernoulli numbers, see:

• John Baez, Bernoulli numbers and the J J -homomorphism .

But where the hell did this function x / ( e x − 1 ) x/(e^x - 1) come from?

If D D means derivative:

( D f ) ( x ) = f ′ ( x ) (D f)(x) = f'(x)

then e D − 1 e^D - 1 is a so-called ‘difference operator’:

( ( e D − 1 ) f ) ( x ) = f ( x + 1 ) − f ( x ) ((e^D - 1)f)(x) = f(x+1) - f(x)

which you can show using the Taylor series for f f if f f is an entire function. So D / ( e D − 1 ) D/(e^D - 1) is about derivatives versus differences, and its inverse is about integrals versus sums. This lets you reduce sums like the one above to integrals… if you know your Bernoulli numbers. For details try this:

• John Baez, Bernoulli numbers .

But x / ( e x − 1 ) x/(e^x - 1) also shows up when you compute the expected energy of a quantum harmonic oscillator in thermal equilibrium!

Let’s work in units where Planck’s constant and Boltzmann’s constant are 1 1 . Say we have a quantum harmonic oscillator whose allowed energies are 0 , 1 , 2 , 3 , … 0, 1, 2, 3, \dots etcetera. (Sometimes people add 1 2 \frac{1}{2} to each of these numbers, but let’s not.) If we compute this oscillator’s average or ‘expected’ energy at temperature T T , and divide it by T T , we get

x e x − 1 \frac{x}{e^x - 1}

x = 1 T x = \frac{1}{T}

So the quantum harmonic oscillator secretly knows about Bernoulli numbers.

What does this fact really mean??? I don’t know. I once read a book called Triangle of Thought about a conversation between Alain Connes and two other mathematicians, and he vaguely alluded to a fact of this sort, and said it was important.

I forget exactly what Connes said, so I imagine it’s something about how the Todd class in algebraic topology, which is usually defined using the function x / ( 1 − e − x ) x/(1 - e^{-x}) (whose power series also gives Bernoulli numbers), can be understood using the harmonic oscillator—perhaps because it appears in the Riemann–Roch theorem , which can probably be proved using ideas from quantum field theory (since it’s a special case of the Atiyah–Singer index theorem , which has a quantum proof ). But all this erudite stuff is probably a complicated spinoff of the basic ideas, and I’d like to understand the basic ideas first.

By the way, here you can see a calculation of the expected energy of a quantum harmonic oscillator:

• Harmonic oscillator statistics .

And here’s a little sanity check which is rather revealing. I said the expected energy divided by temperature is

where x = 1 / T x = 1/T . But in the limit T → + ∞ T \to +\infty the quantum harmonic oscillator should reduce to the classical harmonic oscillator. For that , the expected energy divided by temperature is just 1, since the oscillator has 2 degrees of freedom (position and momentum), and the equipartition theorem , which holds for classical systems with quadratic Hamiltonians, says we should get 1/2 times the number of degrees of freedom. And indeed, it works just as we’d expect:

lim x → 0 x e x − 1 = 1 \lim_{x \to 0} \frac{x}{e^x - 1} = 1

This limit is also, by definition, the zeroth Bernoulli number. So the zeroth Bernoulli number is telling us the energy over temperature of a quantum harmonic oscillator in the high-temperature limit. The rest of the Bernoulli numbers are telling us all the ‘low-temperature corrections’ to the oscillator’s energy over temperature:

where x = 1 / T x = 1/T .

I hope that if I think about it a bit harder, I’ll see that we’re using an analogy:

derivative operator D D : difference operator e D − 1 e^D - 1 :: classical harmonic oscillator : quantum harmonic oscillator

and that Bernoulli numbers arise from comparing the derivative and the difference operator — or comparing the classical harmonic oscillator with its continuous energy spectrum, to the quantum oscillator, with its discretely spaced energy spectrum — or more poetically, the continuous and the discrete!

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/3552

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Re: bernoulli numbers and the harmonic oscillator.

Misc. comments:

You might be interested in “Interactions between Lie theory and algebraic geometry” by Shilin Yu, tying together the GRR and the BCH formula.

The Todd op (mod signs) and its inverse are usually presented as differential ops, but, in a new MathOverflow question (not well received), I present the Todd op and its inverse also in terms purely of the finite difference op, which makes a direct connection to the Helmut Hasse formula for the Hurwitz zeta function and the Bernoulli function that Milnor presented in one of his papers. (The finite diff rep actually allows one of his divergent series derived from the diff op rep and related to the Harer-Zagier formula to be derived as a convergent series, but it is the asymptotic series that is desired.)

The Hurwitz ζ ( s , z ) \zeta(s,z) is related to the Bernoulli function via the derivative B s ( z ) = − ∂ z ζ ( − s , z ) = − s ζ ( − s + 1 , z ) B_s(z)=-\partial_z \zeta(-s,z) = -s\zeta(-s+1,z) , which evaluates as the Bernoulli polynomial B n ( x ) B_n(x) for a positive integer n = s n=s . Alternatively, a Mellin transform gives the same result. The Hurwitz zeta and the Barnes zeta function (which can be expanded in terms of the Hurwtiz zeta) are related to the Casimir effect, Bose-Eisntein condensation, and solutions for the Schrodinger equation of atoms in a harmonic oscillator potential in “Basic zeta functions and some applications in physics” by Klaus Kirsten.

Shilin Yu’s article Interactions between Lie theory and algebraic geometry is interesting, but it’s a personal research statement, so it doesn’t try to explain the connection Kontsevich discovered between the Duflo isomorphism and the Hochschild–Kostant–Rosenberg isomorphism, much less what I’m really interested in: why both these isomorphisms involve an expression that resembles the inverse of the Todd class, namely the ‘Duflo element’

J ( T ) = det ( 1 − e − T T ) J(T) = \det\left(\frac{1- e^{-T}}{T}\right)

of a linear transformation T T , or in fact actually the square root of this Duflo element. A much more expository account is here:

• Damien Calaque and Carlo A. Rossi, Lectures on Duflo isomorphisms in Lie algebras and complex geometry .

This is really nice. However, it still starts by introducing the formula for the Duflo element or its close relative

det ( sinh ( T / 2 ) T / 2 ) \det\left(\frac{\mathrm{sinh}(T/2)}{T/2}\right)

by fiat rather than by motivating it. It then explains a lot of other interesting math before explaining Kontsevich’s work.

Right now I’m trying to go down to the basic ideas, not up into the applications. So what I’d really enjoy is someone’s simple explanation of why you’d come up an expression like

det ( 1 − e − T T ) \det\left(\frac{1- e^{-T}}{T}\right)

x 1 − e − x \frac{x}{1 - e^{-x}}

if you didn’t already know about it. There are probably a number of good answers. I gave two in my post here.

I enjoyed this one:

• F. Hirzebruch, The signature theorem: reminiscences and recreation .

He says the only power series f ( x ) f(x) for which f ( 0 ) = 1 f(0) = 1 and the coefficient of x n x^n in f ( x ) n + 1 f(x)^{n+1} is always 1 1 is

f ( x ) = x 1 − e − x f(x) = \frac{x}{1 - e^{-x}}

And he explains why this funny condition on coefficients is necessary for the Todd class of ℂ P n \mathbb{C}\mathrm{P}^n to evaluate to 1 1 on the fundamental class of ℂ P n \mathbb{C}\mathrm{P}^n .

However, this still feels a bit ‘fancy’ to me. So I would like to connect this to the fact I presented here: at temperature T T , the ratio of the expected energy of the quantum harmonic oscillator with energy levels 0 , 1 , 2 , … 0,1,2,\dots to the expected energy of the classical harmonic oscillator with the same frequency is

I guess one concrete way forward is to think about the meaning of the powers of f ( x ) f(x) , and their coefficients, from a physics viewpoint.

In the book Concrete Mathematics they hand-wave the Euler summation formula based on the operator equation you mentioned elsewhere,

Δ = exp ( D ) − 1 \Delta = \exp(D) - 1

where D = d d x D = \frac{d}{d x} . Inverting, we get

∑ = 1 exp ( D ) − 1 = 1 D ( D exp ( D ) − 1 ) = 1 D + … \sum = \frac{1}{\exp(D)-1} = \frac{1}{D} (\frac{D}{\exp(D)-1}) = \frac{1}{D} + \ldots

the first term of which is ∫ \int . That is, a sum equals an integral plus Bernoulli correction terms. Anyway that’s how I motivate x / ( 1 − exp ( − x ) ) x/(1-\exp(-x)) to people.

I feel like there should be a generalization in which one compares two complex-oriented cohomology theories, involving a ratio of something involving their formal group laws, where the x / ( 1 − exp ( − x ) ) x/(1-\exp(-x)) case is about H H vs. K K . More specifically, let x x be a weight for a torus action on a line L L , and consider the class of 0 → ∈ L \vec 0 \in L as an element of equivariant cohomology/ K K -theory. I would write those as x x , 1 − exp ( − x ) 1-\exp(-x) respectively, though they live in different enough rings that it’s tricky to think about dividing one by the other.

Allen wrote:

I feel like there should be a generalization in which one compares two complex-oriented cohomology theories, involving a ratio of something involving their formal group laws…

That reminds me of this paper that Jack Morava pointed me to when I told him I was struggling to understand Bernoulli numbers:

• Haynes Miller, Universal Bernoulli numbers and the S 1 S^1 -transfer , Canadian Mathematical Society Conference Proceedings 2 (1982), 437–449.

He defines generalized Bernoulli numbers for any formal group F F over a commutative algebra A A over ℚ \mathbb{Q} . Apparently there’s a unique isomorphism of formal groups

log F : F → G a \log_F: F \to G_a

where G a G_a is the additive formal group of A A , and this has an inverse

exp F : G a → F \exp_F: G_a \to F

Then he defines ‘Bernoulli numbers’ B n ( F ) ∈ A B_n(F) \in A by

T exp F ( T ) = ∑ n = 0 ∞ B n ( F ) n ! T n \frac{T}{\exp_F(T)} = \sum_{n = 0}^\infty \frac{B_n(F)}{n!} T^n

When F = G a F = G_a these Bernoulli numbers are all zero except for B 0 ( F ) = 1 B_0(F) = 1 , since exp F \exp_F is the identity. When F = G m F = G_m is the multiplicative formal group of A A we get the usual Bernoulli numbers since then exp F ( T ) = 1 − e − T \exp_F(T) = 1 - e^{-T} .

So this sounds less general than what you’re dreaming of, but still hard for me to understand! For example, I don’t even see why exp G m ( T ) = 1 − e − T \exp_{G_m}(T) = 1 - e^{-T} . But it sounds like what I need is sitting in front of me. I think I need to just buckle down, learn some stuff and think about it.

A convention in the FG literature is to write the completion of a nice group law F F with identity element e e in terms of a coordinate near the identity, for example

( 0 − x ) + 𝔾 a ( 0 − y ) = 0 − ( x + y ) = 0 − x + 𝔾 ^ a y , (0-x)+_{\mathbb{G}_a}(0-y) = 0-(x+y) = 0-x+_{\hat{\mathbb{G}}_a}y,

( 1 − x ) + 𝔾 m ( 1 − y ) = 1 − ( x + y − x y ) = 1 − x + 𝔾 ^ m y , (1-x)+_{\mathbb{G}_m}(1-y) = 1-(x+y-x y) = 1-x+_{\hat{\mathbb{G}}_m}y ,

and similarly

− log 𝔾 ^ m ( x ) = log 𝔾 m ( 1 − x ) . - \log_{\hat{\mathbb{G}}_m}(x) = \log_{\mathbb {G}_m}(1-x) .

With this convention, Miller’s account of the generalized Hirzebruch genus may be exactly what Allen K is asking for. The issue is that the logarithm of a formal group law is a priori only defined over ℚ \mathbb{Q} , e.g.

log 𝔾 ^ m ( x ) = − ∑ n ≥ 1 x n n . \log_{\hat{\mathbb{G}}_m}(x) = -\sum_{n \geq 1} \frac{x^n}{n} .

Thanks to John for help posting the note above. In my haste I forgot to say that (good multiplicative) cohomology theories over ℚ \mathbb{Q} -algebras are, like cats in the dark, all isomorphic, so the ratio in the Miller’s formula makes sense, cf eg

https://ncatlab.org/nlab/show/Chern-Dold+character

There are other interesting number arrays and associated combinatorics related to the deformed Todd operator variant of 1 1 − te x \frac{1}{1-te^x} (see OEIS A131758 for a ref). A Taylor series expansion about x = 0 x=0 gives rational functions with the Eulerian polynomials (see OEIS A008292 & A173018) in the variable t t in the numerator, and, when these are expanded as a power series, the powers of the integers appear; e.g., for x 4 / 4 ! x^4/4! the coefficient is ( t 4 + 11 t 3 + 11 t 2 + t ) / ( 1 − t ) 5 = t + 2 4 t 2 + 3 4 t 3 + 5 4 t 4 + ⋯ (t^4+11t^3+11t^2+ t)/ (1-t)^5 = t + 2^4 t^2 + 3^4 t^3 + 5^4 t^4 + \cdots . The Eulerian polynomials are related to a slew of other number arrays and functions important in analysis and number theory as noted in the OEIS entries above, including, of course, the Bose-Einstein and Fermi-Dirac distributions.

The definition of the Bernoulli numbers that sticks in my head is the one in Pierre Cartier’s paper Mathemagics : they are defined by the equation

B n = ( B + 1 ) n B^n = (B + 1)^n

for n ≥ 2 n \geq 2 , and by B 0 = 1 B^0 = 1 . Then you change superscripts to subscripts.

This means the following. We start with B 0 = B 0 = 1 B_0 = B^0 = 1 . Then the equation above with n = 2 n = 2 says

B 2 = B 2 + 2 B 1 + 1 . B^2 = B^2 + 2B^1 + 1.

Cancelling gives

0 = 2 B 1 + 1 0 = 2B^1 + 1

and so B 1 = B 1 = − 1 / 2 B_1 = B^1 = -1/2 .

Next, take n = 3 n = 3 . The equation above says

B 3 = B 3 + 3 B 2 + 3 B 1 + 1 . B^3 = B^3 + 3B^2 + 3B^1 + 1.

Cancelling and changing superscripts to subscripts gives

0 = 3 B 2 + 3 B 1 + 1 . 0 = 3B_2 + 3B_1 + 1.

Then substituting in the value B 1 = − 1 / 2 B_1 = -1/2 that we just computed gives B 2 = 1 / 6 B_2 = 1/6 .

“And so on”. Clearly this is formal nonsense, but it’s formal nonsense that works.

This is very pretty, and also quite practical if you’re wanting to compute some Bernoulli numbers. It seems memorable, yet in fact I’d forgotten it— even though 21 years ago I gave a homework problem where I led students through how to derive it from

x e x − 1 = ∑ n = 0 ∞ B n x n n ! \frac{x}{e^x - 1} = \sum_{n = 0}^\infty B_n \frac{x^n}{n!}

(It’s problem 8 here .)

Is this stuff about acting like subscripts are powers the ‘umbral calculus’? I can never remember what the umbral calculus is, though I remember Rota trying to rescue it from the shadow of disrepute.

Alas, Pierre Cartier died on Saturday, August 17th, 2024.

Anyone who never met him should watch this interview of him, in French with English subtitles. It really gets across his personality. For some of his thoughts related to “mathemagics”—that is, math using mysterious formal manipulations—go to 48:32 . Great expression on his face when he describes how Bourbaki’s negative attitude to mathematical physics led them to neglect two of Weyl’s most important books. Then he turns to Boole’s algebraic approach to logic, then Dirac, then Heaviside….

Alas. He was a great man, with far more influence on the history of math than many another more well-known figure. His account of Grothendieck (as a person) is for example a non-mathematical example of the depth and lucidity of his thinking.

He had a sense of humor and a sense of irony. This really hurts.

Oh, that’s a great loss. I met him a few times at the IHES and conferences, and he seemed like one of those guys who would go on forever, in unbelievably robust health. His knowledge was amazing, and although I didn’t know him well, I had the strong impression that he loved to share it: he was always holding forth, in French or English, always with a little audience listening with rapt attention. The interview John links gives a flavour of that.

Let’s look at some really simple umbral Sheffer calculus (courtesy of Blissard–the gnarly stuff was used by Sylvester and Cayley in their explorations of invariant theory). Let the umbral substitution op and its action be given by a differential op as

( a . ) n = S x → a . x n = e a . ∂ x = 0 x n (a.)^n = S_{x \to a.} x^n = e^{a.\partial_{x=0}}x^n

= ∑ k ≥ 0 ( a . ) k ∂ x = 0 k k ! x n = ∑ k ≥ 0 a k ∂ x = 0 k k ! x n = a n . = \sum_{k \geq 0} (a.)^k \frac{\partial_{x=0}^k}{k!} x^n = \sum_{k \geq 0} a_k \frac{\partial_{x=0}^k}{k!} x^n = a_n.

Then we can easily rigorously derive a fundamental identity of the finite difference calculus

( a . ) n = a n = S x → a . x n = e a . ∂ x = 0 x n = e a . ∂ x = 0 ( 1 − ( 1 − x ) ) n (a.)^n = a_n = S_{x \to a.} x^n = e^{a.\partial_{x=0}}x^n = e^{a.\partial_{x=0}}(1-(1-x))^n

= e a . ∂ x = 0 ∑ k = 0 n ( − 1 ) k ( n k ) ∑ j = 0 k ( − 1 ) j ( k j ) x j = e^{a.\partial_{x=0}} \sum_{k=0}^n (-1)^k\binom{n}{k} \sum_{j=0}^k (-1)^j \binom{k}{j}x^j

= ∑ k = 0 n ( − 1 ) k ( n k ) ∑ j = 0 k ( − 1 ) j ( k j ) a j = ( 1 − ( 1 − a . ) ) n , =\sum_{k=0}^n (-1)^k\binom{n}{k} \sum_{j=0}^k (-1)^j \binom{k}{j} a_j= (1-(1-a.))^n,

and this suggests another useful rep for umbral substitution

a n = ( a . ) n = S x → a . x n = e − ( 1 − a . ) ∂ x = 1 x n = ( 1 − ( 1 − a . ) ) n a_n = (a.)^n = S_{x \to a.} x^n = e^{-(1-a.)\partial_{x=1}}x^n = (1-(1-a.))^n

that can lead to interpretation for complex s s of

( a . ) s = a s = ( 1 − ( 1 − a . ) ) s = e − ( 1 − a . ) ∂ x = 1 x s (a.)^s = a_s = (1-(1-a.))^s =e^{-(1-a.)\partial_{x=1}}x^s

= ∑ k ≥ 0 ( − 1 ) k ( s k ) ∑ j = 0 k ( − 1 ) j ( k j ) a j = \sum_{k \geq 0} (-1)^k\binom{s}{k} \sum_{j=0}^k (-1)^j \binom{k}{j} a_j

as a Newton series, which is intimately related to Mellin transform interpolation and its analytic continuation. (With a j = ( x + j ) α a_j = (x+j)^{\alpha} and s = b . s = b. , the Bernoulli number umbra, the Newton series gives a variant of the Helmut Hasse formula for the Hurwitz zeta function, the Bernoulli function B α ( x ) = − α ζ ( − α + 1 , x ) B_\alpha(x) = -\alpha \; \zeta(-\alpha+1,x) of Milnor.)

The sequence of Bernoulli polynomials is an iconic Appell-Sheffer polynomial sequence (ASPS). Given any sequence of real or complex numbers, say a n a_n , with a 0 = 1 a_0 =1 , an ASPS can be defined by umbral translation, or binomial convolution, via

A n ( x ) = T x → ( x + a . ) x n = ( a . + x ) n = e a . ∂ x x n A_n(x) = T_{x \to (x+a.)} x^n = (a.+x)^n = e^{a.\partial_x}x^n

= ∑ k = 0 n ( n k ) a k x n − k . =\sum_{k=0}^n \binom{n}{k} a_k x^{n-k} .

The Bernoulli polynomials and numbers B n ( x ) B_n(x) and b n b_n can be generated by

B n ( x ) = ( b . + x ) n = e b . ∂ x x n = ∂ x e ∂ x − 1 x n , B_n(x) = (b. +x)^n = e^{b.\partial_x} x^n = \frac{\partial_x}{e^{\partial_x}-1} x^n,

so the associated e.g.f.s are

e B . ( x ) t = e ( b . + x ) t = e b . ∂ x e xt = e b . t e xt = t e t − 1 e xt . e^{B.(x)t} = e^{(b.+x)t} = e^{b.\partial_x}e^{xt} =e^{b.t}e^{xt} = \frac{t}{e^t-1}e^{xt} .

e − B . ( 1 ) t = e − ( b . + 1 ) t = e − b . t e − t = − t e − t − 1 e − t = t e t − 1 = e b . t , e^{-B.(1)t} = e^{-(b.+1)t} = e^{-b.t}e^{-t} = \frac{-t}{e^{-t}-1}e^{-t} =\frac{t}{e^{t}-1} = e^{b.t},

( − B . ( 1 ) ) n = ( − 1 ) n ( b . + 1 ) n = ( b . ) n = b n , (-B.(1))^n = (-1)^n (b.+1)^n = (b.)^n = b_n,

the recursion relation noted by Cartier in Tom Leinster’s comments.

In these ways, the Bernoulli-Hirzebruch-Todd operator arises naturally in the umbral Sheffer operator calculus along with the associated inverse ops and the raising ops, which also involve the Bernoulli numbers, or more accurately the Riemann zeta numbers ζ ( − n ) \zeta(-n) .

Of course, there are direct formulas for the Bernoulli numbers involving the Stirling numbers of the second kind or the face counts of the permutahedra, which are easily derived from the umbral Sheffer calculus, and these you find in toric topology and soliton solutions of the KdV equation and associated quadratic Riccati equation. In fact, from my formulas in OEIS A008292 on the Eulerian polynomials, you can discern a relation of the Bernoulli numbers to a formal group law and generalized Hirzebruch-Todd classes as John discusses. Expand as a Taylor series the shifted reciprocal of the bivariate e.g.f. for the Eulerian polynomials A(x,a,b)= (e^(ax)-e^(bx))/(a e^(bx)-b e^(ax)), that is , x * (a e^(b x)-b e^(a x)) / (e^(a x)-e^(b x)), using Wolfram Alpha and you will find the Bernoulli numbers. See the links adjacent to the OEIS formulas for refs on the associated generalized cohomology theory and toric topology (related to the convex polytopes the associahedra, permutahedra, and stellahedra).

For a deeper investigation of the applications of umbral calculus to topology, see “Symbolic calculus: A 19th century approach to MU and BP” by Nigel Ray in Homotopy Theory: Proceedings of the Durham Symposium 1983, London Mmathematical Society Lecture Notes Series, 117.

See the two MSE questions “ What’s umbral calculus about? ” and “ Question about “baffling” umbral calculus result ” and the MO-Q “ How are Sheffer polynomials related to Lie theory? ” for more introductory info and links.

The relationships among the Newton series of the finite difference calculus; the compositional inverse pair h ( x ) = e x − 1 h(x) = e^x-1 and h − 1 ( x ) = ln ( 1 + x ) h^{-1}(x)=\ln(1+x) ; the umbral inverse pair of binomial Sheffer Stirling polynomials of the second and first kinds, a.k.a. the Bell/Touchard/exponential and the falling factorial polynomials–core sequences in combinatorics; and the Appell Sheffer Bernoulli polynomials and their umbral inverse polynomials are perhaps most easily and elegantly revealed with the umbral Sheffer calculus. I can post a new short note on these in my web blog if you are interested. They reflect the relations among the basic Lie group operations of translation and multiplicative and compositional inversions.

Euler-Maclaurin and Riemann-Roch

https://mathoverflow.net/questions/10667/

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