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# The Progression of Perovskite Light Emitting Diodes (LEDs) in our Future

Authors

J. Guzman, R. Schweyk, and L. Traore

Abstract

Over 1.5 billion incandescent and III-V Light Emitting Diodes (LEDs) are currently in use in the United States today. Whether used in headlamps, flashlights, or lamp lights, these LEDs are consuming up 136 billion kilowatt-hours on a yearly basis in the United States. However, there is a possible solution concerning the new progress of perovskite LEDs. If we were to replace all the light sources in the United States that are dominated by III-V LEDs and incandescent light bulbs with the current progression of perovskites, would the enegry savings outweigh the financial cost, and what would this look like for the future of perovskite LEDs. Limited to their operational stability, perovskites as they stand are not a viable replacement. However, with the rapid evolution of perovskite LEDs as well as their superior optoelectric properties, it is probable that in the near future they will play a fundamental role in our daily lives.

In order to understand the cost effciency of perovskites, one must consider the economies of scale. When a company mass produces perovskites, instead of considering each and every material created in a single perovskite, one must take into account buying in bulk quantities. Not only this, but one must examine the advantages in energy consumption for perovskites in comparison to III-V and incandescent LEDs.

In this investigation, we will analyze the viability of replacing III-V LEDs and incandescent light bulbs with next-gen perovskite LEDs with an emphasis on the current operation stability of perovskite LEDs, how the stability will improve in the next 5-10 years, and when they reach an operational stability that rivals or surpasses that of III-V and incandescent technologies. Overall, this report will demarcate when it will make financial sense to switch to perovskite LEDs on a national scale.

# Background and Motivation

To understand what perovskite LEDs (PeLEDs) are, one must become familiar with what is available on the market. As of now the majority of the world utilizes Incandescent lights or LED bulbs as light sources. A light-emitting diode (LED) is a semiconductor light source that emits light when current flows through it. Before understanding the physics governing PeLEDs, we need to understand the ambipolar charge transport comprised of both holes and electrons that lead to light emission within PeLEDs.

Electrons can be described as subatomic particles with a net negative charge. Their most influential role is in the process of bonding individual atoms, known as atomic bonding. Electrons can be found in every element as they occupy the outer orbitals of an element. It is also important to understand electron flow, or in other words, current. This is essential when considering a semiconductor, as when a voltage is applied to a semiconductor, the protons (positively charged subatomic particles) are held stationary whereas the electrons flow with some drift velocity. An abundance of these charge carriers creates majority carriers, whereas a scarcity of these charge carriers creates minority carriers. This in turn induces a current through an electric field. This will be key in this investigation regarding some of the issues that currently face perovskite LEDs, including ion migration and Joule heating. Although similar to an electron in magnitude, holes have a net positive electronic charge associated with them. In essence, a hole represents the absence of an electron. Both however are charge carriers that are necessary for the current flow in ambipolar-based semiconductors.

Electrons in the semiconductor recombine with holes as electron-hole pairs, releasing energy in the form of photons thus leading to light emission. While perovskite light-emitting diodes depend upon the perovskite active layer which is in the form of ABX3, in lead halide perovskites, A represents a cation, B represents lead, and X represents a halogen. In the most common lead halide perovskites, halogens tend to most regularly be chlorine, fluorine, iodine, and bromine, the cations usually represent cesium or methylammonium, and as stated above, B represents lead. For example, CsPbBr3 is a typical lead halide perovskite.

Perovskites were first invented in the early 1950s only emitting red, but later green emission was possible. However, the color blue was and still is challenging to create, being 150 times less stable than green and red. Blue’s instability limits a wide color palette for perovskites to display, therefore impeding perovskite commercialization. One of the particular advantages of PeLEDs over current generation LEDs is that PeLEDs emit light with very narrow emissions resulting in ultra-high color purity. Thus, Lead Halide Perovskites are actively researched for their light emission capabilities and their promise for next-generation lighting and display technologies.

Considering all these factors required in the progression of perovskite LEDs, each and every component is essential in creating the most basic perovskite. In this report, we will further examine the cost analysis as well as the plausibility of our proposal. Given their current operational stability issues, we will predict how far into the future that these perovskite LEDs will be a suitable replacement for all the current incandescent and III-V LEDs in the United States.

# Methods and Results

To analyze when the US can transition to perovskite LEDs rather than incandescent and III-V LEDs, one must consider the output power of these PeLEDs. White light is the current most often utilized color of light, consisting of a superposition of three colors: red, blue, and green light. Once one determines the amount of power each color individually consumes, we can sum up the wattages resulting in a final approximation of the amount of power white light would need to be at its best efficiency.

## I. Power Analysis

When considering the power of each individual light source, the equation P=IV will be necessary to keep in mind. Boiled down to four specific steps, at this point one can thoroughly investigate the necessary amount of power needed to allow for a certain color of light to be emitted. First, one must gather the peak EQE (external quantum efficiency), the ratio between the amount of photons leaving the system to the amount of electrons entering, of the color[] Once this is gathered, use the EQE to characterize how much power the PeLED is consuming at that given operating point using J-V-L graphs. Similarly, one must do the same and compare this voltage number to a graph correlating J (current density) to voltage. This is one of the most important steps in the process as J is measured most commonly in mA/cm^2 which will be relevant in the step that follows. When understanding the equation P=IV, P represents power, I represents current, where V represents voltage. Then, one must input the data gathered into this equation. Although plugging in one’s value for V is simple, current is a bit more complicated. Since we are given units in mA/cm^2, one must multiply this value by the device area each perovskite substrate is created on, leaving with the overall current. Only then, will one be able to complete the calculation and solve for power.

After a thorough investigation of these criteria, we can begin the exact calculations. Take for instance red perovskite LEDs. Having a peak EQE currently at 17.8%, the corresponding voltage at the best efficiency occurs at 2.2 V. Similarly, the corresponding J value results in 8mA/cm^2, but must be multiplied by an area of 2.25 cm^2 as explained above. This has a result of approximately 41mW. Now, we can apply these same steps to green and blue perovskite LEDs. Green perovskite LEDs have their peak EQE at 21.63% being the highest of the three. With a voltage of 3.5 V and a current density value of 5mA/cm^2 both at the best efficiency of the perovskite, this results in a total of 63 mW. Although blue perovskite LEDs are still in development as are the others, blue perovskite LEDs have a long way to go in terms of stability and EQE as it is a very difficult color to produce. Containing a peak EQE of 10.11%, a voltage most efficient at 3.5 V, and a J value most efficient at 15 mA/cm^2, the power results in 118 mW .

The data above is reasonable as one must consider the amount of energy each color consumes as a total. Utilizing the equation $E= \frac{h c}{\lambda}$, the colors with the largest wavelengths will result in the least energy, and understanding the equation $P= \frac{E}{t}$, the greater the energy, the greater the power. When analyzing each color with this information, blue has the smallest wavelength and therefore consumes the largest amount of power. In contrast to blue, red has the longest wavelength so, in turn, has the least amount of power, whereas green falls in between the two.

Since all of these perovskites are currently not at a competitive EQE or stability to replace incandescent and III-V LEDs, it is important to understand how much they need to improve. State-of-the-art green LEDs reside at an approximate 38.4% EQE whereas red LEDs reside at 35.48%. The reason why these values are at such a high percentage is for the very reason of optical lenses. Optical lenses are utilized to ultimately direct light in a single forward direction. EQE is only measured by the forward light emission of the perovskite LEDs. Keeping this into consideration, when perovskite LEDs do indeed reach this level, it will be the logical solution as a replacement. However, this now begs the question of when we will be able to replace all current incandescent and III-V LEDs with perovskite LEDs in the future, and what does the future of blue perovskite LEDs look like.

## II. PeLED Operational Lifetime and EQE Estimates:

In order to predict commercialization of PeLEDs, one must consider the following factors: blue’s current and predicted EQE, Red, green, and blues’ projected operational lifetime in 10 years and their operational lifetime using ion migration suppression methods.

### Blue EQE Prediction:

Currently, blue perovskite LEDs are not at their maximum external quantum efficiency. In comparison to green and red which have come close to reaching their maximum EQE, blue is still behind. Knowing that green and red’s projected maximum EQE is around 25%, blue is expected to reach this maximum of 25% as well. In order to predict when blue perovskite LEDs would reach their maximum EQE, this investigation first gathered blue EQE from the years 2016-2021. With the information collected this research created a graph displaying the points to form a line of best fit. To create the line one had to find the average between the points. This average made the equation of Y= -825.58/x +50. Note that when using this equation one must use the last two digits of the year for “X”, for example in 2016, use 16 for x. This equation grows asymptotically because blue’s EQE would eventually plateau and approach a maximum of 25%. This means that in the year 2034 (12 years), blues predicted EQE will reach 25 percent (view equation (A) for solution and figure (6) for growth).

In addition, all red, green, and blue perovskite LED colors are significantly behind the operational lifetime of an LED thus withholding perovskite commercialization. This however is temporary as advances and predictions can be made to show when perovskite LEDs will surpass LEDs. First, the research started by predicting each color’s operational stability within the next ten years. This would help understand how rapidly perovskites are truly growing. In order to do so, research gathered papers that reported each color’s operational stability throughout the years.

### Predicted Blue PeLED Operational Lifetime in 10 years:

Starting with blue, in 2019, blue’s operational stability was 14.5 minutes, 51 minutes in 2020, and 81.3 minutes in 2021. With this data, one can create a line of best fit to average the points, create a growth rate, and predict future operational stability. Once done, research was able to average the data to formulate an equation of Y=33.4x-619.1, note that to again predict the years following, one must use the last two digits of the year. With this information, one replaced 31 to our x, representing 2031(10 years from now). This means that by the year 2031, blue would reach a predicted operational lifetime of 7 hours (view equation (B) for solution and figure 7 for growth).

It is important to notice that blue’s operational lifetime growth is linear. This is extremely valuable since blue is the most challenging color to create due to its large bandgap which will be further explained once compared to red and green’s operational lifetime growth.

### Predicted Red PeLED Operational Lifetime in 10 years:

Just like blue, the investigation took similar steps to predict red perovskites’ operational stability in 10 years. Again, one collected red’s previous operational stability throughout the years using research papers to predict a growth rate. In 2017, red operated for 16 hours; it increased to 30 in 2018 and currently in 2021, 317 hours. This data is a bit different from blue since it grows exponentially. Therefore instead of using a line of best fit that is linear, research made it exponential. This has to do with the fact that red is a much easier color to accomplish than blue allowing it to advance faster. With that said, the red exponential equation is $y=0.0000273735 \times 2.6169^{x}$ found by the average between points. With this equation, one imputed the year 31, representative of 2031 into the x of the equation. This shows that by the year 2031 red is predicted to operate at 726,735 hours, an extremely noticeable difference between our current operational stability( view equation (C) for solution and figure 8 for growth).

### Predicted Green PeLED Operational Lifetime in 10 years:

Again the research repeated the same process, this time gathering green’s perovskite operational stability to help figure out what the operational lifetime would look like in 10 years. The research found that in 2016 the green perovskite’s operational stability was ten hours, 2 years later in 2018 the LEDs operated for 46 hours, and finally, in 2021 the perovskites operated to their current maximum of 208 hours. This allowed the creation of a line of best fit that was exponential, being the average of the points $y=0.0033 \times 1.169^{x}$. Then use the equation to input the last two digits of the year into x as 31, representing 2031(10 years from 2021). Results in green’s operational lifetime to be 39,105 hours (view equation (D) for solution and figure 9 for growth). This operation stability for the year 2031 is a bit smaller than red’s considering that red is the most efficient.

Knowing that the blue’s bandgap is the widest, it directly correlates to the growth rate. The bandgap denotes the minimum energy required to excite an electron into a state that allows it to conduct current in the conduction band. The valence band is the lower energy level, and if there is a gap between this level and the higher energy conduction band, energy must be added to allow electrons to flow. As described, green and red have smaller band gaps which cause longer operational stability and exponential growth. Blue however has a wider bandgap which causes shorter operational stability and linear growth. Blue requires more energy to emit a photon because of the higher energy band gaps ultimately compromising the efficiency of perovskites.

## III. PeLED Commercialization Viability

### Red PeLED Commercialization:

Now that sufficient equations for each color have been gathered, predictions of commercialization can be considered. In order to find when each color can accomplish commercialization, one must acknowledge that an average LED can last 6 ×106h. Using this information, one can use the red perovskite equation to input 6 ×106h as our “ Y “ representing the operational stability to help find “X” our year. Doing so one can use $0.0000273735*2.16953^{x}=10^(6)*6h$, this answer would result in 33~34, representative of the year 2034. This means that by the year 2034 the red perovskite should be able to compete with LEDs. (view equation (E) for solution)

### Green PeLED Commercialization:

One can repeat the process for the green perovskite where we utilize our equation to input our desired operation lifetime. By replacing our “Y“ as 6 ×106h simplified to $10^{6}*6 =.00336293*1.69116^{x}$ is equivalent to 40 representing 2040 years of commercialization. By the year 2040 green perovskites are estimated to commercialize (view equation (F) for solution)

### Blue PeLED Commercialization:

Lastly, for the blue perovskite, one must solve for the estimated year which was $(33.4x)-619=10^{6}*6h$ equal to the year

Figure 8 Displays the equation $y=0.0000273735 x 2.6169^{x}$ to help predict the operational lifetime of the red perovskite in 10 years. Figure 9 displays the equation $y=.0033 x 1.69116^{x}$. To help predict operational lifetime of the green perovskite in 10 years.

179659, 179659 representative of the year 181,659.25. This year appears different in comparison to the other perovskite predicted commercialization year in which they were in the years 2000. To convert the number as one did for red and green one added 2000 which resulted in blues predicted commercialization year to be 181,659.25. ( view equation (G) for solution)

Blue is immensely behind, however, this is without taking into account advances that will be made in the future. For one, one must consider the fact that these numbers do not utilize ion migration joule heating suppression methods in addition to new research and studies that will specifically help advance the blue perovskite.

## IV. Pathways to enhance PeLED performance and stability

### Ion Migration Suppression Strategies:

As of now, there are many methods utilized to counteract migration and joule heating. One being B-site engineering. B site engineering is doping B-site cation ions with metal ions like Mn2+. Traps in the mid-gap can prevent electrons from recombining with holes, ultimately limiting the perovskite as light cannot emit. Reducing traps allows electrons to fall freely without encountering any obstructions. The addition of Mn2+ reduces the trap density, which results in less ion migration. Mn2+ can lower defect density (the number of defects ) in perovskites, reducing ion migration and resulting in greater operational stability. With that said, a study that utilized this method caused blue perovskites to operate 1,440 times longer than their original undoped perovskites. This study also applied B-site engineering to red perovskites allowing them to operate for 305 times longer than the undoped perovskites.

The second method was using Precursor Solution Composition Optimization. There are three crystallinity orders in a perovskite (substrate). There is amorphous, which means there is no discernible order. There is crystalline, where everything is in order and visually is similar to a checkerboard. Lastly, Polycrystalline indicates that halide ions can be arranged and slightly distributed in tiny groupings. There are pockets of small groups in polycrystalline thin films that will be in order. However, grain boundaries will exist inside the limits of those groupings.

In a polycrystalline material, a grain boundary is a point where two grains, or crystallites, meet. Grain boundaries are two-dimensional defects in the crystal structure that reduce the material’s electrical and thermal conductivity. This indicates that there are flaws in the device that cause it to degrade. Halide ions, such as bromide, chloride, or iodide, can easily cross grain boundaries since they have the lowest activation energy.

These halide ions are at grain boundary cusps, and because they are not strongly attached to the group, they can be whisked away by the electric field, resulting in ion migration. Halide ions are being swept across and concentrated at some point due to the electric field being applied to the layer, which is producing the most segregation. Instead of employing a cesium bromide, we can use a different method. CsTFA-derived films have a flatter energy landscape (a more homogenous energy level distribution for charges), a more stable crystal structure, superior optical characteristics, and reduced ion migration as compared to the CsBr method.

As a result, such grain boundaries get passivated, or the lead halide ions become more difficult to respect within the magnetic field. This causes tighter films where there are no bubbles, (effectively a sheet compared to cesium bromide with defects), less ion migration occurs. A research group utilized Precursor Solution Composition Optimization which allowed for green perovskites to operate 17 times more efficiently.

### Applying Precursor Solution Composition Optimization to Green PeLEDs:

Our research can then apply these numbers to the current undoped operational lifetime. Precursor Solution Composition Optimization allows green perovskites to function 17 times more efficiently. Therefore one first had to find a number multiplied to 17 to then calculate to 10^(6)*6 h resulting in 352,941h. One can then calculate the year our “X” in which green will be able to reach 352,941h. Solving for the year 35 representative of in the year 2035, green will be able to commercialize ( view equation (H) for solution).

### Applying B-site Engineering to Red PeLEDs:

Applying B site engineering red perovskites will be able to operate 305 times longer in comparison to our original predictions. One must find the minimum number that can be multiplied with 305 (305 representing how many times more the perovskite would last being doped) to be equivalent to 10^(6)*6h.

Set of equations A-J is processes taken in order to predict blue EQE (A), Blue perovskite estimated operational lifetime in ten years(B), Red perovskite estimated operational lifetime in ten years (C), Green perovskite estimated operational lifetime in ten years (D), Predicted Red perovskite commercialization year (E), Green perovskite commercialization year (F), Blue perovskite commercialization year (G), Green perovskite estimated commercialization year using Precursor Solution composition optimization(H), Red perovskite estimated commercialization year using B-site Engineering(I) and Blue perovskite estimated commercialization year using B-site Engineering(J).

This number is 19,672h, now replacing it for “Y” representing operational lifetime, into our red perovskite equation. Let us solve for “X” resulting in 26 representing the year 2026. This method allowed for the reduction of 7 years, in comparison to not using any ion migration suppression methods ( view equation (I) for solution).

### Applying B-site Engineering to Blue PeLEDs:

The same process can then be repeated for the blue perovskite. Instead, however, ion suppression made the operational lifetime 1440 times longer. The research found the minimum number that could be multiplied to 1440 (1440 representing how many times more the perovskite would operate for, being doped) to be equivalent to $10^{6}*6h$. Resulting in 4166h, which was used to represent Y operational lifetime into our blue perovskite equation. Then research proceeded to solve for X as our estimated year. Doing so, formulated the number 143 representing the year 2143 when the blue perovskite would be able to commercialize (view equation (J) for solution). Although this is significantly sooner than the undoped blue perovskite commercialization prediction there are still other factors one must consider. Since perovskites are rapidly evolving there are still numerous studies and ion suppression methods that can be applied. With this in mind, new research can be applied to all color’s estimated commercialization year. Regardless of red and green being able to commercialize sooner than blue, blue is still a color with great improvements expected over the next few years.

## V. Modeling the Economies of Scale towards Mass Production

To evaluate the total cost of mass-producing perovskite LED materials in the project of replacing all light sources in the US, we would need to apply economies of scale to our calculations. Economies of scale is the total average cost savings obtained by an enterprise for a greater quantity of production. As production increases, the total average cost decreases, resulting in a lesser total cost than the sum of the price per unit. In the generic model, the variables P1 and P2 represent the cost of production, and the variables Q1 and Q2 represent the quantity of production, and with an increase in the quantity of production (Q value), there is a gradual decline in the cost of production.

Our model evaluating costs, considering the impacts of economies of scale, was developed from a quotation by the Stanford Congreve Lab for TPBi, providing numbers for price at different numbers of grams. From these numbers, we were able to gather initial price numbers and establish a trend. The numbers initially provided were $566.00 for 2 grams and$1341.00 for 5 grams, thus $283.00/g and$268.20/g respectively, while the cost for 1 gram is between $600 and$700. The development of a model for economies of scale necessitated the development an asymptotic graph model based on these numbers, given costs for 2 grams and 5 grams respectively, to calculate the price per gram for 10,000 grams and price for bulk production of 1 billion at this rate.

For an increase in gram quantity by a factor of 2.5, there is a 5.23 % decrease in cost. This number was achieved by subtracting 268.20 from 283, the respective price per gram for 2 grams and 5 grams, and evaluating that percentage of the resulting 14.8. By this model, we can calculate a projected $169.69/g for 10,000 grams, and a projected$1.70/g for 1 billion. The asymptotic model for TPBi prices for increasing gram quantities follows this trend. Having calculated our model for the TPBi layer, we can then apply it similarly to each layer, which would be expected to follow a similar price trend.

To apply these numbers, we must consider the fact that for TPBi, 10,000 grams amounts to 1,000,000 substrates. For the ITO layer, $250.00 is the approximate price for 100 substrates, and by an increase in substrate by a factor of 250, there is a 0.05% decrease in cost per substrate. For 1,000,000 substrates, the cost can be estimated to be$237.48, a mere fraction of a cent per substrate at that quantity. Similarly, an approximate $125.00 corresponds to the quantity of 100 substrates for the PEDOT layer, and by the same process, the total cost for 1,000,000 substrates can be estimated to be$118.74. For the CsPbBr3 layer, the original costs were found to be for $38.00 CsBr and for$10.64 for PbBr2 for 1 substrate, and once more by the same process, the cost for 1,000,000 substrates can be estimated to be $34.29 for CsBr and$9.60 for PbBr2. For the LiF layer, $1 corresponds to the price of LiF for 1 substrate, and for 1,000,000 substrates, the cost can be calculated to an approximate$0.90. Finally, $0.27 corresponds to the price for 1 substrate in the Aluminum layer, and the cost for 1,000,000 substrates can be similarly estimated to$0.24. These costs were originally gathered from Sigma-Aldrich and Ossila.

Finally, from our calculated value of 3 to 4 cents for the cost of 1 substrate for a perovskite LED, we can apply our model to find this number at mass production, which would amount to an approximate fifth of a cent for a gram increase by a factor of 2.5. At production of 1 billion, the price would be about 11 million USD.

## VI. Financial, Performance, and Energy Analysis of Transitioning to PeLED-based Lighting

After taking a cumulative approach to this research of perovskite LEDs, the ultimate question of if the price of perovskite LEDs is worth the energy consumption can finally be answered. Considering the equation $P = \frac{E}{t}$ where P equates to power, E equates to energy, and t equates to time, it is possible to calculate the energy of each color perovskite LED given power and time. Finally, one must combine all these values to find the total energy the US would have to use to power lighting with perovskite LEDs. Considering these perovskites will be at the level where they can compete with current LEDs, one must assume each color perovskite is at their maximum EQE of 25%. The calculations proceed as follows: $((0.063*1.1)+(0.041*1.4)+(0.118*2.5))*(60*60*6*360*328,200,000)$. By multiplying each power value by specific numbers to approximate them to be around 25% EQE, this would give us a better and more accurate representation of the future. Multiplying this by 3,600 gives our value in hours rather than seconds, and final multiplying by the amount per year for every person in the US. This equates to approximately $1.5*10^{15}$ kWh. Compared to the 136 billion kWh the US uses in energy consumption a year to fuel light emission, this does not seem to be a logical fit currently, but with joule heating methods along with the rapid improvements of these perovskite LEDs, they are sure to reach the stability and energy of regular LEDs in the future.

## VII. Cumulative Research Prediction of White Peled

Taking into account power analysis, operational stability, and economies of scale, research is sufficient to help predict how much power a white PeLed would utilize, how long the peled would operate for, and how much the product would cost the U.S as a whole. As previously stated the blue Peled has an operational lifetime of 81 min noticeably behind the green and red Peled operational lifetimes, therefore if one were to utilize multiple blue peled in one white Peled it would help increment lifetime. For example, to create a white PeLED one red, green, and blue peled is needed. However one could utilize ten blue PeLEDs to increment the lifetime by 10 times as long as only one blue peled is used at a time. Nevertheless, there are limitations such as the cost of having multiple PeLEDs in one unit or utilizing an extreme amount of power to operate. Therefore as all colors currently stand research would only utilize 173 blue LEDs. Particularly, 173 blue perovskites in one white Peled because this would allow for the operation to increase to 207 hours (This lifetime being close to green Peleds current operational stability which is at 208 hours). Research multiplied the power needed to light up a blue Peled by 173. Finally adding the power needed to operate one green and one red Peled. The power for this Peled would equate to 20.518 kW. According to calculations 173 blue PeLEDs, one red and one green in one unit to supply the U.S population would cost 128 million. As they currently stand, they would not be the best alternative.

However, research can also calculate the lifetime of the white Peled in 16 years. It is important to consider how the white Peled would change in the future. Particularly in 16 years, since the green perovskite has reached operational stability of 1 million. At this time the blue perovskite would operate for a total of 10 hours, and could potentially operate for 1,000 hours using 100 blue Peled in one unit. This amount of blue Peleds would take less energy as time progressed, the power calculating to a total of 11.9 kW to operate one white LED in the year 2037. The cost of the white Peleds is 127 million dollars to fully transition. This number is lower in cost for a longer operational lifetime and less power needed compared to where they currently stand. It is also important to consider that in the future there could be new advances to combat ion migration which could also improve power, operational lifetime, and cost.

# Challenges Encountered

We are currently facing many problems when considering the stability of PeLEDs. At the moment being unstable, the goal for the future of perovskite LEDs is to eventually increase the stability of each perovskite, however, there are some issues that need to be solved that are preventing us such as ion migration, joule heating, etc.

Before understanding the problems with ion migration, we first must understand the meaning of an electric field. An electric field is a field that physically surrounds electrically charged particles which allows for the repulsion and attraction of other electrically charged particles. Ion migration occurs in the perovskite layer, where either cations or anions, typically halide ions (anions) in lead halide perovskites, as seen in figure 3, approach either the negative or positive side of the electric field, which creates unevenness throughout the perovskite. Once they congregate near the terminals, any sort of light emission will be uneven which in turn compromises the performance of the device. This is solely due to the electric field.

Joule heating occurs in the electron transport layer, the perovskite layer, and the hole transport layer. Joule heating allows for and creates heat, which is produced by current flowing in the material. The problem with this is that it can raise the temperature of the material by up to 40 degrees celsius which also, in turn, allows for degradation of the material and a decrease in instability. Any material is optimized at 200k, which progressively regresses as temp increases, which decreases current since electrons are more scattered. No flow amounts to no current, which results in no light emission. This is solely based on a thermal effect.

Another contributing factor to perovskite degradation is bandgap width, this can cause problems with operational lifetime, particularly with the color blue. The reason why red is the most efficient color is due to its small band gap width. Blue for example has the widest band gap width out of the three colors, this means that there is more energy that must move an electron. As for red, which has the smallest band gap width, the electron does not need as much energy to the dropdown. With smaller band gaps, we are able to have greater stability for red and green. For blue, since the color has a wider bandgap, occasionally the electron gets trapped in the middle of the bandgap; this area is called the midgap. Here the perovskite cannot emit light, as light emission does not occur in the mid-gap, only when the electron falls. This is one of the major problems faced as band gap width is not something that can be changed. Unlike green and red which have longer operational lifetimes, blue currently faces this problem and will continue to face it. In addition, the traps in the mid-gap also affect the operational lifetime to not only the blue perovskite. Therefore jeopardizing particularly the blue perovskite lifetime.

Another challenge encountered was the lack of perfect accuracy for our model of economies of scale. The economies of scale model utilized to calculate the cost values at various quantities is an imperfect measure of estimates. The model is based on very small quantities of TPBi, thus it could have been made more accurate if costs for larger quantities had also been provided for comparison. Additionally, all estimates are based on the model for TPBi, so they are expected to have slightly more inaccurate estimates for the other materials. However, although it does not specifically account for such inaccuracies, the model is itself an estimation and only approximates the trend in cost change at larger scales.

# Conclusion

Whether keeping into account each individual color of perovskite LEDs or the culmination of them all together, they most undoubtedly will play a fundamental role in our future. Each aspect explained upon in the preceding sections of this paper show countless evidence of the rapid evolution of these perovskites when considering either the power of each color, their operational stability, or the price of each perovskite.

The power analysis played an equally important role in determining the reliability of each perovskite. Considering the equation P=IV, the peak EQEs of each color, applying J-V graphs, and combining them collectively to get a final value for power, all portray the importance of the power analysis. As the rise of each color’s EQE continues, so will the efficiency of perovskites, which perhaps may be the most important factor in transitioning from III-V and incandescent LEDs to perovskites. With joule heating methods, perovskites are improving at exponential rates being very promising to their future.

By utilizing previous perovskite operational lifetimes there was sufficient information to predict if Peleds are worth the transition. Starting with the prediction of EQE for the blue perovskite, where research proved to show growth is rapid. Green and red perovskites too have shown extreme n operational lifetime improvements for the future. While the blue perovskites do seem to struggle, there are advances such as B-site engineering and precursor solution composition optimization to combat their short operational stability. Most importantly, the current white Peleds would not be a reasonable transition with the blue perovskite being the largest setback. However, with time we can be sure to see white Peleds thrive.

A significant consideration to evaluating the cost of transitioning to lighting by perovskite LEDs was economies of scale. At the high number of production necessary for this replacement, the total financial estimate could not be made accurately by multiplying the price per unit by the intended number of units, as such a calculation would overestimate the actual cost. Thus, the reduction in increase was calculated based on the original prices gathered from Sigma-Aldrich and Ossila. The economies of scale assumed for these calculations was a continued 5.23 % decrease in cost per gram for a quantity increase by a factor of 2.5, and by the model we used, this amounts to the LED cost of a fraction of a cent for a gram increase by that trend. Currently, to supply the U.S population with next-generation lighting, it would cost \$128 million.

# References

1. Aneer Lamichhane, Nuggehalli M. Ravindra, Energy Gap-Refractive Index Relations in Perovskites, Materials, 10.3390/ma13081917, 13, 8, (1917), (2020).
2. ResearchGate. Nov. 2014, http://www.researchgate.net/publication/ 280845761_Nobel_Prize_in_Physics_The_birth_of_the_blue_LED. Accessed 10 Aug. 2021.
3. Congreve Lab.” Congreve Lab, congrevelab.stanford.edu/.
4. “External Quantum Efficiency.” External Quantum Efficiency – an Overview | ScienceDirect Topics, www.sciencedirect.com/topics/engineering/external-quantum-efficiency.
5. Jiang, Yuanzhi. “Spectra stable blue perovskite light-emitting diodes.” Nature Communications, 23 Apr. 2019, http://www.nature.com/articles/s41467-019-09794-7. Accessed 27 July 2021.
6. Ma, Dongxin. “Chloride Insertion–Immobilization Enables Bright, Narrowband, and Stable Blue-Emitting Perovskite Diodes.” ACS Publications, 9 Mar. 2020, pubs.acs.org/doi/abs/10.1021/jacs.9b12323. Accessed 27 July 2021.
7. Ren, Zhenwei. “High-Performance Blue Perovskite Light-Emitting Diodes Enabled by Efficient Energy Transfer between Coupled Quasi-2D Perovskite Layers.” Onlinelibrary, 2021, onlinelibrary.wiley.com/doi/epdf/10.1002/ adma.202005570. Accessed 27 July 2021.
8. Iopscience.16 Oct. 2017, iopscience.iop.org/article/10.1088/1361-6528/aa8b8b/ meta. Accessed 27 July 2021.
9. Dong, Qi. “Operational stability of perovskite light-emitting diodes.”Iopscience, 2020,iopscience.iop.org/article/10.1088/2515-7639/ab60c4/pdf. Accessed 27 July 2021.
10. Li, Hanming. “Efficient and Stable Red Perovskite Light-Emitting Diodes with Operational Stability>300 h.” Wiley Online Library, 9 Mar. 2021, onlinelibrary.wiley.com/doi/abs/10.1002/adma.202008820. Accessed 27 July 2021.
11. Wang, Yanan. “Ultrastable, Highly Luminescent Organic–Inorganic Perovskite–Polymer Composite Films.” Advanced Materials, 2016, lcd.creol.ucf.edu/Publications/2016/Advanced_Materials.pdf. Accessed 27 July 2021.
12. Lin, Kebin. “Perovskite light-emitting diodes with external quantum efficiency exceeding 20 per cent.” Nature, 10 Oct. 2018, http://www.nature.com/articles/ s41586-018-0575-3?WT.feed_name=subjects_physical-sciences. Accessed 27 July 2021.
13. Han, Boning. “Green Perovskite Light-Emitting Diodes with 200 Hours Stability and 16% Efficiency: Cross-Linking Strategy and Mechanism.” Wiley Online Library, 17 Apr. 2021, onlinelibrary.wiley.com/doi/abs/10.1002/adfm.202011003. Accessed 27 July 2021.
14. Band Gap.” Energy Education, energyeducation.ca/encyclopedia/Band_gap#:~:text=A%20band%20gap%20is%20the,it%20can%20participate%20in%20conduction. Accessed 10 Aug. 2021.
15. Dong, Qi. “Operational stability of perovskite light-emittingpercent diodes.” Iopscience, 2020, iopscience.iop.org/article/10.1088/2515-7639/ ab60c4/pdf. Accessed 27 July 2021.
16. Wikipedia. en.wikipedia.org/wiki/Grain_boundary. Accessed 10 Aug. 2021.
17. Xiao, Peng. “Advances in Perovskite Light-Emitting Diodes Possessing Improved Lifetime.” Nanomaterials, 4 Jan. 2021, http://www.mdpi.com/2079-4991/11/1/103/htm. Accessed 10 Aug. 2021.
18. Kenton, Will. “Economies of Scale.” Investopedia, Investopedia, 10 Apr. 2021, http://www.investopedia.com/terms/e/economiesofscale.asp. Accessed 22 July 2021.
19. “Economies of Scale – Definition, Types, Effects of Economies of Scale.” Corporate Finance Institute, 31 Jan. 2021, corporatefinanceinstitute.com/resources/knowledge/economics/economies-of-scale/. Accessed 9 Aug. 2021.


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# Finding the probability density function for IID rv [duplicate]

The question is as follows:

Suppose that X1 and X2 are independent, identically distributed exponential random variables. Determine the PDF for for X1 - X2.

I understand that because X1 and X2 are IID, they have the same parameter and the same distribution. In this case, it is an exponential distribution. I am also under the assumption that I will need to use some variant of the PDF of the regular distribution:

Other than this, I am not certain how to solve this. Thank you.

• do you know about moment-generating functions? – symplectomorphic Apr 30 '14 at 0:42
• – symplectomorphic Apr 30 '14 at 0:43
• I am familiar with moment generating functions but would like to see some more examples. Let me check the link out that you provided. Ty. – DaGr8Gatzby Apr 30 '14 at 0:45

format: plain


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# Name two gases, other than CO_2, that are given out during burning of fossil fuels and contribute towards acid rain formation.

22 views
in Physics
closed
Name two gases, other than CO_2, that are given out during burning of fossil fuels and contribute towards acid rain formation.

by (71.7k points)
selected by

Sulphur dioxide and Nitric oxide.


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# Math Help - integrals

1. ## integrals

is this integrals correct?

1/y+t^2 (dt) = arctan t

what is the correct integral of 1 + y^2?

thanks!

2. Originally Posted by poundedintodust
is this integrals correct?

1/y+t^2 (dt) = arctan t

what is the correct integral of 1 + y^2?

thanks!
$\int \frac{1}{1 + t^2}~dt = atn(t)$

$\int \frac{1}{y + t^2}~dt = \frac{atn(t\sqrt{y})}{\sqrt{y}}$

-Dan

3. Originally Posted by poundedintodust
is this integrals correct?

1/y+t^2 (dt) = arctan t

what is the correct integral of 1 + y^2?

thanks!
$\int \frac{1}{y} + t^2 dt = \frac{t}{y} + \frac{t^3}{3} + C$

$\int 1 + y^2 dy = y + \frac{y^3}{3} + C$

I believe Dan answered what you wanted, but I listed these just in case.

Nevertheless, do not forget adding a constant in for indefinite integrals! $+ C$!


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Browse Questions

# Which is incorrect about an ecotone?

$\begin{array}{1 1}(a)\;\text{High habitat diversity} \\ (b)\;\text{Low species diversity} \\(c)\;\text{Presence of unique species} \\(d)\;\text{High micro climatic variability}\end{array}$

Can you answer this question?

Ecotone is the region where 2 distinct habitats meet.
Hence (b) is the correct answer.
answered Mar 24, 2014


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diagrams-lib-1.4.2: Embedded domain-specific language for declarative graphics

Description

Tools for adjusting the length of parametric objects such as segments and trails.

Synopsis

# Documentation

adjust :: (N t ~ n, Sectionable t, HasArcLength t, Fractional n) => t -> AdjustOpts n -> t Source #

Adjust the length of a parametric object such as a segment or trail. The second parameter is an option record which controls how the adjustment should be performed; see AdjustOpts.

data AdjustOpts n Source #

How should a segment, trail, or path be adjusted?

Instances

 Fractional n => Default (AdjustOpts n) Source # Methods

Which method should be used for adjusting?

Which end(s) of the object should be adjusted?

adjEps :: Lens' (AdjustOpts n) n Source #

Tolerance to use when doing adjustment.

data AdjustMethod n Source #

What method should be used for adjusting a segment, trail, or path?

Constructors

 ByParam n Extend by the given parameter value (use a negative parameter to shrink) ByAbsolute n Extend by the given arc length (use a negative length to shrink) ToAbsolute n Extend or shrink to the given arc length

Instances

 Source # Methods

Which side of a segment, trail, or path should be adjusted?

Constructors

 Start Adjust only the beginning End Adjust only the end Both Adjust both sides equally

Instances

 Source # Methods Source # Methods Source # Methods Source # Methods Source # Methods Source # MethodsshowList :: [AdjustSide] -> ShowS # Source # Methods


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### FitsTable

A representation of a FITS binary table

#### Description:

The FitsTable class is a representation of a FITS binary table. It inherits from the Table class. The parent Table is used to hold the binary data of the main table, and a FitsChan (encapsulated within the FitsTable) is used to hold the FITS header.

Note - it is not recommended to use the FitsTable class to store very large tables.

FitsTables are primarily geared towards the needs of the " -TAB" algorithm defined in FITS-WCS paper 2, and so do not support all features of FITS binary tables. In particularly, they do not provide any equivalent to the following features of FITS binary tables: " heap" data (i.e. binary data following the main table), columns holding complex values, columns holding variable length arrays, scaled columns, column formats, columns holding bit values, 8-byte integer values or logical values.

#### Inheritance

The FitsTable class inherits from the Table class.

#### Attributes

The FitsTable class does not define any new attributes beyond those which are applicable to all Tables.

#### Functions

In addition to those routines applicable to all Tables, the following routines may also be applied to all FitsTables:


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# Riesz representation theorem

There are several well-known theorems in functional analysis known as the Riesz representation theorem. They are named in honour of Frigyes Riesz.

This article will describe his theorem concerning the dual of a Hilbert space, which is sometimes called the Fréchet-Riesz theorem. For the theorems relating linear functionals to measures, see Riesz–Markov–Kakutani representation theorem.

## The Hilbert space representation theorem

This theorem establishes an important connection between a Hilbert space and its (continuous) dual space. If the underlying field is the real numbers, the two are isometrically isomorphic; if the underlying field is the complex numbers, the two are isometrically anti-isomorphic. The (anti-) isomorphism is a particular, natural one as will be described next.

Let H be a Hilbert space, and let H* denote its dual space, consisting of all continuous linear functionals from H into the field R or C. If x is an element of H, then the function φx, for all y in H defined by

$\varphi_x(y) = \left\langle y , x \right\rangle,$

where $\langle\cdot,\cdot\rangle$ denotes the inner product of the Hilbert space, is an element of H*. The Riesz representation theorem states that every element of H* can be written uniquely in this form.

Theorem. The mapping $\Phi$: HH* defined by $\Phi(x)$ = $\varphi_x$ is an isometric (anti-) isomorphism, meaning that:

• $\Phi$ is bijective.
• The norms of $x$ and $\varphi_x$ agree: $\Vert x \Vert = \Vert\Phi(x)\Vert$.
• $\Phi$ is additive: $\Phi( x_1 + x_2 ) = \Phi( x_1 ) + \Phi( x_2 )$.
• If the base field is R, then $\Phi(\lambda x) = \lambda \Phi(x)$ for all real numbers λ.
• If the base field is C, then $\Phi(\lambda x) = \bar{\lambda} \Phi(x)$ for all complex numbers λ, where $\bar{\lambda}$ denotes the complex conjugation of λ.

The inverse map of $\Phi$ can be described as follows. Given a non-zero element $\varphi$ of H*, the orthogonal complement of the kernel of $\varphi$ is a one-dimensional subspace of H. Take a non-zero element z in that subspace, and set $x = \overline{\varphi(z)} \cdot z /{\left\Vert z \right\Vert}^2$. Then $\Phi(x)$ = $\varphi$.

Historically, the theorem is often attributed simultaneously to Riesz and Fréchet in 1907 (see references).

In the mathematical treatment of quantum mechanics, the theorem can be seen as a justification for the popular bra–ket notation. When the theorem holds, every ket $|\psi\rangle$ has a corresponding bra $\langle\psi|$, and the correspondence is unambiguous. cf. also Rigged Hilbert space

## References

• M. Fréchet (1907). Sur les ensembles de fonctions et les opérations linéaires. C. R. Acad. Sci. Paris 144, 1414–1416.
• F. Riesz (1907). Sur une espèce de géométrie analytique des systèmes de fonctions sommables. C. R. Acad. Sci. Paris 144, 1409–1411.
• F. Riesz (1909). Sur les opérations fonctionnelles linéaires. C. R. Acad. Sci. Paris 149, 974–977.
• P. Halmos Measure Theory, D. van Nostrand and Co., 1950.
• P. Halmos, A Hilbert Space Problem Book, Springer, New York 1982 (problem 3 contains version for vector spaces with coordinate systems).


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# 统计学入门 Introduction to Statistics MATH10282

0

where $K$ is any one-dimensional kernel. Then there is a single bandwidth parameter $h$. At a target value $x=\left(x_{1}, \ldots, x_{d}\right)^{T}$, the local sum of squares is given by
$$\sum_{i=1}^{n} w_{i}(x)\left(Y_{i}-a_{0}-\sum_{j=1}^{d} a_{j}\left(x_{i j}-x_{j}\right)\right)^{2}$$
where
$$w_{i}(x)=K\left(\left|x_{i}-x\right| / h\right)$$

where $K$ is any one-dimensional kernel. Then there is a single bandwidth parameter $h$. At a target value $x=\left(x_{1}, \ldots, x_{d}\right)^{T}$, the local sum of squares is given by
$$\sum_{i=1}^{n} w_{i}(x)\left(Y_{i}-a_{0}-\sum_{j=1}^{d} a_{j}\left(x_{i j}-x_{j}\right)\right)^{2}$$
where
$$w_{i}(x)=K\left(\left|x_{i}-x\right| / h\right)$$
The estimator is
$$\widehat{r}{n}(x)=\widehat{a}{0}$$

## MATH10282 COURSE NOTES ：

Generally one grows a very large tree, then the tree is pruned to form a subtree by collapsing regions together. The size of the tree is a tuning parameter chosen as follows. Let $N_{m}$ denote the number of points in a rectangle $R_{m}$ of a subtree $T$ and define
$$c_{m}=\frac{1}{N_{m}} \sum_{x_{i} \in R_{m}} Y_{i}, \quad Q_{m}(T)=\frac{1}{N_{m}} \sum_{x_{i} \in R_{m}}\left(Y_{i}-c_{m}\right)^{2}$$
Define the complexity of $T$ by
$$C_{\alpha}(T)=\sum_{m=1}^{|T|} N_{m} Q_{m}(T)+\alpha|T|$$

# 统计学入门|Introduction to Statistics代写 STAT 515

0

To determine the expected value of a chi-squared random variable, note first that for a standard normal random variable $Z$,
\begin{aligned} 1 &=\operatorname{Var}(Z) \ &=E\left[Z^{2}\right]-(E[Z])^{2} \ &=E\left[Z^{2}\right] \quad \text { since } E[Z]=0 \end{aligned}
Hence, $E\left[Z^{2}\right]=1$ and so
$$E\left[\sum_{i=1}^{n} Z_{i}^{2}\right]=\sum_{i=1}^{n} E\left[Z_{i}^{2}\right]=n$$

The expected value of a chi-squared random variable is equal to its number of degrees of freedom.

Suppose now that we have a sample $X_{1}, \ldots, X_{n}$ from a normal population having mean $\mu$ and variance $\sigma^{2}$. Consider the sample variance $S^{2}$ defined by
$$S^{2}=\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{n-1}$$

## STAT515 COURSE NOTES ：

If the population mean $\mu$ is known, then the appropriate estimator of the population variance $\sigma^{2}$ is
$$\frac{\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}}{n}$$
If the population mean $\mu$ is unknown, then the appropriate estimator of the population variance $\sigma^{2}$ is
$$S^{2}=\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{n-1}$$
$S^{2}$ is an unbiased estimator of $\sigma^{2}$, that is,
$$E\left[S^{2}\right]=\sigma^{2}$$


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# Tag Info

1

I am not sure what exactly you are asking. If you ask why $\det \prod_{k=1}^N M_k=1$ when $\det M_1,\ldots,\det M_N=1$, this is because the determinant function is multiplicative (i.e. $\det AB=(\det A)(\det B)$). If you are asking why a sympletic matrix has determinant $1$, this has been explained in the related Wikipedia article. If you are asking why ...

1

Use $det(AB...Z)=det(A)det(B)...det(Z)$ and the fact that $det(M_k)=1$.

0

Hint: Use elementary row operations to bring $C$ to its reduced row echelon form $R$, then use elementary column operations to bring $R$ to its reduced column echelon form $C'$.

0

Which definition of $A>0$ do you use? Let $V$ be arbitrary inner product space. We say a linear map $A$ is pos.definite, if $A=A^*$ and for $x\in V$, $x\neq 0$ we have $\langle x,Ax\rangle>0$. Since $(AB)^*=B^*A^*$, $I=A^*(A^{-1})^*=A(A^{-1})^*$, so the unicity of the inverse yields $(A^{-1})^*=A^{-1}$. Furthermore, $$\langle x, A^{-1} ... 3 There is a theorem, that says that self-adjoint maps are allways diagonalizable, so there an eigenbasis and you are done. Let's sketch the proof: Let \lambda\in \mathbb C be an eigenvalue of T, E_\lambda := \ker(\lambda- T) the eigenspace. Then E_\lambda is T-invariant, and by self-adjointness, its orthogonal complement E_\lambda^\bot is also: ... 0 First, a notational issue. The vector you call |w\rangle "double u" is actually called "omega" i.e. |\omega\rangle (backslash-omega is the \rm\TeX command). The vector |s'\rangle is determined as the only (up to an arbitrary overall normalization including the phase) linear combination of |\omega\rangle and |s\rangle – note that you use the ... 0 2:True: FF^n=0\\\implies\text{Im}(F^n)\subset\text{Kernel}(F)\subset\text{Kernel}(F^n)\\\implies\text{Rank}(F^n)\le\text{Nullity}(F^n) 1,3,4: False: F:\mathbb R^3\to\mathbb R^3 given by F=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix} 0 I'd recommend you check out the Mathematics section here for free online video courses offered by some of the top institutions. 7 Let C = AB - BA and assume that (C - I)^n = 0. Then the minimal polynomial m(x) of C divides (x - 1)^{n}. Thus m(x) = (x-1)^{k} for some 1 \leq k \leq n. This shows that the only eigenvalue of C (considered in \Bbb{C}) is 1, hence the characteristic polynomial of C is (x-1)^{n}. In particular, \operatorname{tr}(C) = n > 0. This ... 2 I explain the relationship between Jordan and rational forms in these recently written notes. Here are the main points: 1) Jordan form is only available when a transformation is split, i.e., when its minimal polynomial (equivalently, its characteristic polynomial) is a product of linear factors over the ground field. 2) In this case, the rational ... 4 Let's go step by step. We want an equation of the following form: ... 3 As @Julien pointed out, every square matrix admits a PLU decomposition, where P is a permutation matrix. We have: A = P \cdot L \cdot U, such that: A=\begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 0 & 1 ... 1 No Conclusion Possible eigenvalues of a non-singular real skew-symmetric matrix are purely imaginary. Consequently such matrices have no eigenvalue in \mathbb R i.e. the number of eigenvalues is 0. The roots of the characteristic polynomial of a n\times n real symmetric matrices are all real whence the number of eigenvalues is n. You may form a ... 4 Since you observed that P is annihilated by X^n-1, it only remains to check that no nonzero polynomial of degree \leq n-1 annihilates P to conclude that the minimimal polynomial is X^n-1. Since the latter has degree n and divides the characteristic polynomial which is also monic of degree n, it will also follow that the characteristic ... 2 To solve$$ x'=Ax+b $$use$$ x=e^{At}x_0-A^{-1}b $$where e^{At}x_0 is the solution to x'=Ax. Often times it is useful to look at the 1-dimensional analog:$$ x'=ax+b $$where the solution is$$ x=x_0e^{at}-b/a $$I just put that solution into matrix form and checked it out to make sure it worked. Low tech math. 2 If there exists a vector v such that Av=b then consider the variable y=x-v. One finds y'=Ay. If there is no such v then A is not invertible. If A is still diagonalizable, however, then the tricky part of b is the bit lying in the kernel of the matrix. Thus write b=b_0+c where Ab_0=0 but there is some Av=c solution. Try computing ... 0 One common generalization to the case of square matrices is to define$$A \ge 0 :\Leftrightarrow A \; \textbf{is} \; \textbf{positive} \; \textbf{semidefinite},$$that is, v^T A v \ge 0 for every v \in \mathbb{R}^n. (assuming you're working with real matrices). Some common inequalities carry over immediately, appropriately modulated: for example,$$x^2 ...

1

when you consider $\{ (1,-1),(0,1)\}$ as base of domain's T and you catch $\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}$ then $(-1,-1)=1(1,-1)+0(1,0)$ so : $T(1,-1)=\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}=(1,0,-1)$

0

Look at the laplace expansion of the determinant of $A+I$ (http://en.wikipedia.org/wiki/Laplace_expansion) Your matrix has the form $$A = d\left( \begin{matrix} k_{1,1} &\ldots &k_{1,n} \\ \vdots &\ddots &\vdots \\ k_{n,1} &\ldots &k_{n,n} \end{matrix}\right)$$ and by laplace expanding the determinant using the ...

5

Hint: Can $-1$ be an eigenvalue for $A$? Alternatively: What is the determinant of $A+I$ modulo $d$?

1

Hints: First, \begin{align} (3v+w)\times(v-2w) &=-7v\times w \end{align} and $$\|-7v\times w\|=7\|v\|\|w\|\left|\sin\left(\frac{2\pi}{3}\right)\right|$$ For $\langle w\times 5v, v-3w\rangle$, note that $w\times v$ is perpendicular to both $v$ and $w$.

2

Hints: a real symmetric matrix is diagonalizable (via an orthogonal matrix, but you don't even need that here) in $M_n(\mathbb{R})$ and the condition $A^k=Id$ implies that the eigenvalues of $A$, which are all real, belong to $\{\pm 1\}$. Finally, note that the result is true.

1

If $A \succ 0$, $S$ is positive semidefinite iff the block matrix is positive semidefinite.

1

If $B$ commutes with $A$, it leaves the eigenspaces of $A$ invariant. In this case, these are the one-dimensional subspaces generated by the vectors of the canonical basis $\{e_i\;;\; i=1,2,3\}$. So $Be_i=\lambda_ie_i$ for $i=1,2,3$. That is $B$ is diagonal. The converse is clear. Conclusion: the commutant of $A$ in $M_3(\mathbb{R})$ is the subalgebra of ...

0

A matrix $B$ commutes with $A$ iff $AB=BA$ iff $B=ABA^{-1}$; fill the entries of $B$ with unknowns, write out this equation, and then find an equivalent system of relations for the unknowns: $$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & ... 1 Seems about right to me. The point is that a matrix is unitary if and only if its columns form an orthonormal basis with respect to the complex Euclidean inner product. In fact, if U = [v_1 | \cdots | v_n] then (UU^\dagger)_{ij} = \langle v_i, v_j \rangle (or U^\dagger U depending on your convention). Thus your question really boils down to: can any ... 3 if rank(A^*A)=rank(A) it means rank(A)=n-dim(null(A)) and rank(A^*A)=n-dim(null(A^*A)) so dim( null(A))=dim(null(A^*A)) if exist x such that Ax\ne 0 and A^*Ax=0 then x\in null(A^*A)but but x not belong to null(A) and since null(A)\subset rank(A^*A) so$$dim( null(A))\ne dim(null(A^*A))$$contradiction! on the other hand : we know ... 0 Perhaps this is slightly off topic but it's worth explaining slightly more intuitively why you can't divide by a vector. The key idea is to think about information (or more precisely numbers of constraints/equations). A dot product takes two vectors, both with n numbers in them, and combines them in a particular way (related to projecting then onto each ... 1 I'd take a cross product, say, (P-Q)\times(R-Q) to get a vector orthogonal to the plane, normalize that to get a normal vector, then form the scalar product of Z-Q with that normal vector and take the absolute value to get the distance of Z from the plane. 5 Note that$$\mathrm{Tr}\left[(A-A^T)^2\right]=-2\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2.$$Therefore the identity can be rewritten as$$2(1-\mathrm{Tr}\,A)^2-\mathrm{Tr}\left[(A-A^T)^2\right]=8.$$Since \mathrm{Tr}\,AA^T=\mathrm{Tr}\,1=3, this is equivalent to$$\left(\mathrm{Tr}\,A\right)^2-\mathrm{Tr}\,A^2-2\mathrm{Tr}\,A=0.$$The left side of the last ... 3 Hint: what would it mean for a vector to be orthogonal to itself? If the problem says "every other vector" as in the title (but not the quoted text) you could instead consider what it means for a given vector \vec{u} to be orthogonal to multiples of itself, e.g. to -\vec{u}. 1 We know that G=GL_n(\mathbb{R}) is generated by elementary matrices. Therefore A is in Z(G) if and only it commutes with all elementary matrices. By checking this condition, you will see that A has to be diagonal. 7 The dot product \cdot is an unusual type of multiplication. It takes two vectors in and produces a scalar out. Imagine a mommy elephant and a daddy elephant giving birth to a giraffe. There's no such thing as division in this context. You need to do the proof by looking at components of the vectors. u=(u_1,u_2,\ldots, u_n), v=(v_1, v_2,\ldots, v_n), ... 1 Let A \in Z(GL_n(\mathbb R)). Let B be defined as b_{ii}=i and b_{ij}=0 if i\neq j (or any other diagonal matrix with pairwise distinct entries). Then since AB=BA, for all i \neq j we have$$\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^n b_{ik}a_{kj} \Rightarrow a_{ij}b_{jj}=b_{ii}a_{ij}\,.$$Since i \neq j we have b_{ii} \neq b_{jj} ... 0 Well, I found the answer. It is easy to prove that x_v \le \lambda x_u simply by writing down Ax=\lambda x. Now, since A is irreducible, there must exist a power k \le m such that state u is connected to state v by a path of length k. A^k represents a graph with paths of length k between the states of the A-graph. This matrix has ... 0 V_\mathbb{R} = V \otimes_\mathbb{Q} \mathbb{R} is a \mathbb{R}-vector space in "the natural way". You do not need to talk about basis elements of neither V nor \mathbb{R} to define the structure. Any element of of V_\mathbb{R} looks like a sum of terms v \otimes r, where v \in V and r \in \mathbb{R}. To define a \mathbb{R}-vector space ... 1 Since that (1,-1) is the first vector of \beta it has coordinates (1,0). When you use matrices to apply the linear transformation you are using the coordinates of the vector on a fixed basis. So you should write$$\begin{bmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}.$$If \{v_1,\ldots,v_n\} is a basis ... 1 It suffices to prove that if a v = 0_E (I take there's a misprint in you formula, and I'm simplifying notation a bit), and a \ne 0_K, then v = 0_E. Now if a \ne 0_K, then a has an inverse a^{-1} in the field K. Thus$$ 0_E = a^{-1} \cdot 0_E = a^{-1} (a v) = (a^{-1} a) v = 1 v = v. $$1 Given 3 vertices, these vertices define 2 vectors; the unknown endpoint corresponds to the sum of these 2 vectors. Example: (1,1) and (1,3) form the vector (0,2). (1,1) and (4,2) form the vector (3,1). Sum these vectors to get (3,3). Then add the vertex (1,1) back to get (4,4) as the formerly unknown corner. 2 The three possibilities come from adding the coordinates of two points and subtracting the third. So (4,0)=(4,2)+(1,1)-(1,3). There are three choices of which point to subtract. You are translating the point you subtract to the origin, then adding the vectors to the other two points to find the opposite one, then translating back. So you could look at ... 0 Of course! \mathbb{Q} is a field, therefore if there is solution to linear system, the solutions will be rational. You can check it by writing the explicit formulas for the solution -- for example, the ones you provided as help or Cramer's rule -- and notice that the formulas involves only additions and multiplications of rationals. So if your system has ... 1 Hints (if I guessed correctly your notation and part of your problem):$$\begin{pmatrix}5&3&7&0\\ 2&\!\!-4&6&5\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\binom 50\iff\begin{cases}5x_1+3x_2+7x_3=5\\{}\\2x_1-4x_2+6x_3+5x_4=0\end{cases}$$Two particular solutions to this system are ... 4 Let \mathcal{A} = \cup_{i=1}^n A_i and \chi_i : \mathcal{A} \to \mathbb{N} be the indicator function for A_i, i = 1,\ldots n. For any n real numbers x_1, \ldots, x_n, not all zero. Consider the function f: \mathcal{A} \to \mathbb{R} defined by:$$f(a) = \sum_{i=1}^{n} x_i \chi_i(a)$$We have:$$ \sum_{i=1}^{n}\sum_{j=1}^{n} a_{ij} x_i x_j = ...

2

A fun problem. I think that the following approach is natural (at least it is the first that occurred to me, YMMV). Let us consider the union $$A=\bigcup_{i=1}^nA_i.$$ I will work in the space $F_A$ of real valued functions from $A$ to $\mathbb{R}$. If you list the elements of $A$ like $$A=\{a_1,a_2,\ldots,a_m\},$$ you can identify the space $F_A$ with ...

0

Hint: Claim: A simetric matrix $X=(x_{ij})_{VV}$ (here $V=\{1,2,\ldots, k, \ldots, n\}$ ) is positive semidefinite if and only if all of its principal minors $det(x_{ij})_{UU}$ (here $\emptyset \neq U\subset V$) are nonnegative. Claim: A simetric matrix $\begin{pmatrix}A& C^T\\ C& D\end{pmatrix}$ is positive semidefinite if and only if $D$ and ...

0

Recall that adding a multiple or subtracting a multiple of one row does not change the value of the determinant, see, for example ProofWiki. Using this fact and Laplace expansion you get |A_4|= \begin{vmatrix} 1& 1& 1& 1\\ 1& 2& 2& 2\\ 1& 2& 3& 3\\ 1& 2& 3& 4 \end{vmatrix}= ... 3 Let's prove that a linear function is continuous. Recalling the definition: A function f:\Bbb{R}^n\to\Bbb{R}^k is continuous iff for each x\in\Bbb{R}^n and for each \epsilon>0 there exists a \delta>0 such that for all y\in\Bbb{R}^n with ||y-x||<\delta we have ||f(y)-f(x)||<\epsilon. Now if f is linear, then it takes the form ... 0 \def\rk{\mathop{\mathrm{rank}}}We have, as \rk A = \rk A^t holds for every matrix, that \begin{align*} \dim U &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{pmatrix}\\ &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 ... 1 \def\Mat{\mathrm{Mat}}\def\R{\mathbb R}Let A \in \Mat_{n+1,m}(\R) given byA = \begin{pmatrix} x_1 & x_2 & \cdots & x_m\\ 1 & 1 & \cdots & 1 \end{pmatrix}  and $b \in \R^{n+1}$ by $b = \binom{0_n}1$. Then, by Farkas, either (a) for some $\lambda\in\R_{\ge 0}^m$, we have $A\lambda = b$, which is equivalent to $\sum_{i=1}^m ... 1 You should know that the dot product is distributive.$\left<\mathbf{a}, \mathbf{b} + \mathbf{c} \right> = \left<\mathbf{a}, \mathbf{b} \right> + \left<\mathbf{a}, \mathbf{c} \right>\$ So you should be able to obtain two dot products from this expression. Each of them would be what is called "scalar triple products." How is the result of ...

Top 50 recent answers are included


text:
# To calculate cross elasticity of demand, does percentage change in quantity demanded have to be market demand instead of individual demand? [closed]

If I try to calculate cross elasticity of demand (XED or CED) with individual demand quantity, it seems to break down since even small increase when the base number is small is a big increase in percentage. e.g. 1 to 2 is a 100% change.

If you calculate XED you have two relative terms.

$XED=\frac{\Delta q_i}{q_i}\cdot \frac{p_j}{\Delta p_j}$

It is right that in your case $\frac{\Delta q_i}{q_i}=\frac{2-1}{1}=1$

Now it depends on the current price of good j and the change of the price how large XED is. From the relative change of the demand you cannot conlcude to the level of XED.


text:
Outlook: Helios Technologies Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Wait until speculative trend diminishes
Time series to forecast n: 24 Jan 2023 for (n+8 weeks)
Methodology : Modular Neural Network (Market Volatility Analysis)

## Abstract

Helios Technologies Inc. Common Stock prediction model is evaluated with Modular Neural Network (Market Volatility Analysis) and Multiple Regression1,2,3,4 and it is concluded that the HLIO stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes

## Key Points

1. Fundemental Analysis with Algorithmic Trading
2. Is now good time to invest?
3. How do predictive algorithms actually work?

## HLIO Target Price Prediction Modeling Methodology

We consider Helios Technologies Inc. Common Stock Decision Process with Modular Neural Network (Market Volatility Analysis) where A is the set of discrete actions of HLIO stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Multiple Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Market Volatility Analysis)) X S(n):→ (n+8 weeks) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$

n:Time series to forecast

p:Price signals of HLIO stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## HLIO Stock Forecast (Buy or Sell) for (n+8 weeks)

Sample Set: Neural Network
Stock/Index: HLIO Helios Technologies Inc. Common Stock
Time series to forecast n: 24 Jan 2023 for (n+8 weeks)

According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for Helios Technologies Inc. Common Stock

1. The fair value of a financial instrument at initial recognition is normally the transaction price (ie the fair value of the consideration given or received, see also paragraph B5.1.2A and IFRS 13). However, if part of the consideration given or received is for something other than the financial instrument, an entity shall measure the fair value of the financial instrument. For example, the fair value of a long-term loan or receivable that carries no interest can be measured as the present value of all future cash receipts discounted using the prevailing market rate(s) of interest for a similar instrument (similar as to currency, term, type of interest rate and other factors) with a similar credit rating. Any additional amount lent is an expense or a reduction of income unless it qualifies for recognition as some other type of asset.
2. If an entity previously accounted at cost (in accordance with IAS 39), for an investment in an equity instrument that does not have a quoted price in an active market for an identical instrument (ie a Level 1 input) (or for a derivative asset that is linked to and must be settled by delivery of such an equity instrument) it shall measure that instrument at fair value at the date of initial application. Any difference between the previous carrying amount and the fair value shall be recognised in the opening retained earnings (or other component of equity, as appropriate) of the reporting period that includes the date of initial application.
3. If, in applying paragraph 7.2.44, an entity reinstates a discontinued hedging relationship, the entity shall read references in paragraphs 6.9.11 and 6.9.12 to the date the alternative benchmark rate is designated as a noncontractually specified risk component for the first time as referring to the date of initial application of these amendments (ie the 24-month period for that alternative benchmark rate designated as a non-contractually specified risk component begins from the date of initial application of these amendments).
4. The fact that a derivative is in or out of the money when it is designated as a hedging instrument does not in itself mean that a qualitative assessment is inappropriate. It depends on the circumstances whether hedge ineffectiveness arising from that fact could have a magnitude that a qualitative assessment would not adequately capture.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

Helios Technologies Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Helios Technologies Inc. Common Stock prediction model is evaluated with Modular Neural Network (Market Volatility Analysis) and Multiple Regression1,2,3,4 and it is concluded that the HLIO stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Wait until speculative trend diminishes

### HLIO Helios Technologies Inc. Common Stock Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB2Baa2
Balance SheetBaa2Caa2
Leverage RatiosBa3Ba2
Cash FlowBaa2Baa2
Rates of Return and ProfitabilityB2B2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 87 out of 100 with 643 signals.

## References

1. L. Panait and S. Luke. Cooperative multi-agent learning: The state of the art. Autonomous Agents and Multi-Agent Systems, 11(3):387–434, 2005.
2. Tibshirani R. 1996. Regression shrinkage and selection via the lasso. J. R. Stat. Soc. B 58:267–88
3. Abadie A, Cattaneo MD. 2018. Econometric methods for program evaluation. Annu. Rev. Econ. 10:465–503
4. Çetinkaya, A., Zhang, Y.Z., Hao, Y.M. and Ma, X.Y., Is FFBC Stock Buy or Sell?(Stock Forecast). AC Investment Research Journal, 101(3).
5. Holland PW. 1986. Statistics and causal inference. J. Am. Stat. Assoc. 81:945–60
6. V. Borkar. A sensitivity formula for the risk-sensitive cost and the actor-critic algorithm. Systems & Control Letters, 44:339–346, 2001
7. R. Sutton and A. Barto. Introduction to reinforcement learning. MIT Press, 1998
Frequently Asked QuestionsQ: What is the prediction methodology for HLIO stock?
A: HLIO stock prediction methodology: We evaluate the prediction models Modular Neural Network (Market Volatility Analysis) and Multiple Regression
Q: Is HLIO stock a buy or sell?
A: The dominant strategy among neural network is to Wait until speculative trend diminishes HLIO Stock.
Q: Is Helios Technologies Inc. Common Stock stock a good investment?
A: The consensus rating for Helios Technologies Inc. Common Stock is Wait until speculative trend diminishes and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of HLIO stock?
A: The consensus rating for HLIO is Wait until speculative trend diminishes.
Q: What is the prediction period for HLIO stock?
A: The prediction period for HLIO is (n+8 weeks)

## People also ask

What are the top stocks to invest in right now?


text:
# NAG CL Interfacex04ndc

## 1Purpose

x04ndc converts an integer error code to its corresponding string error name. The integer error code is typically returned as part of the NAG error structure by a prior call to a NAG C Library function.

## 2Specification

 #include
 const char * x04ndc (int code)
The function may be called by the names: x04ndc or nag_code_to_error_name.

## 3Description

x04ndc takes as input the value of the integer error code returned in the NAG error structure and returns the corresponding string error name. This can be useful when you want to print the returned error code as a name. For example the NE_BAD_PARAM code has the value $70$. A call to x04ndc$\left(70\right)$ returns the string ‘NE_BAD_PARAM’.
x04ndc returns $0$ (null pointer) if the input value is not recognized as a valid NAG error code.

None.

## 5Arguments

1: $\mathbf{code}$int Input
On entry: the value of a NAG error code.

None.

Not applicable.

## 8Parallelism and Performance

x04ndc is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.

None.

## 10Example

This example produces error name strings corresponding to a number of integer error codes.

### 10.1Program Text

Program Text (x04ndce.c)

None.

### 10.3Program Results

Program Results (x04ndce.r)


text:
UPDATE: For an application of adaptive sparse grid code to a macro model, see here.

A major issue with macroeconomic models, and Bellman iteration in general, is the curse of dimensionality. We generally do not have a parametric form for the value function and so we must approximate its true value on some set of grid points and then interpolate for values off the grid. The problem is that the number of grid points increase exponentially with the dimension of the problem, i.e. $N^d$. Sparse grids, to an extent, mitigate this issue. They rely on the observation that certain points are more important in the interpolation of a function. I follow the exposition given in Brumm and Scheidegger. My code for the multi-dimensional adaptive sparse grid is based off of the sparse grid interpolation algorithm Spinterp by Klimke and Wohlmuth. I have updated their algorithm to make the grid selection adaptive. This will be made clearer below.

using Plots

##### Collection of Basis Functions

All interpolation uses a collection of basis functions and weights to approximate a given function.

$f(x) \approx \sum_i w_i \phi_i(x)$

In the standard sparse grid literature, our basis functions are given by the hat function.

function φ(x)
-1.0 <= x <= 1.0 && return 1 - abs(x)
return 0.0
end

baseX = linspace(-1.0, 1.0, 500)
y = [φ(j) for j in baseX ]
plot(baseX , y, label = "Basis Function" )

The original interval is then broken into subintervals, half of the original width, and the basis function is scaled to this new subinterval. So for level $2$ the new intervals are $0$ to $\frac{1}{2}$ and $latex \frac{1}{2}$ to $1$. The basis functions look as follows

# Collection of Basis Functions
s = 2 # Current step level
h = 2.0^(-s) # Grid size
x = linspace(0.0, 1.0, 400)
y = [[φ( (j - i*h)/h ) for j in x] for i = 1:2:2^s-1] # Note that the i's are all odd
plot(x, y, label = ["Basis Function 2,1" "Basis Function 2,3"])

##### Sparse Grid Interpolation

In order to demonstrate the technique it helps to look at the sample function used in Brumm/Scheidegger. The two things that you should note are that the function is smooth and that it disappears on the boundaries of the interval. The smoothness implies that the sparse grid is in a certain sense optimal. Because it disappears on the boundaries we can use a very simple sparse grid scheme. This scheme will need to be updated to the “Clenshaw-Curtis” scheme when we have nonzero values on the boundaries.

f = x -> x^2*sin(π*x)
yFunc = [f(j) for j in x]
plot(x, yFunc, label = "f")

The first approximation is simply the hat function, and we weight it by functions value at $0.5$. This is not a very precise approximation as we can see below.

s = 1
h = 2.0^(-s)
x_i = [i*h for i = 1:2:2^s-1]
α_i = [f(x_ij) - (f(x_ij - h) + f(x_ij + h))/2 for x_ij in x_i]
y_1 = sum([[φ( (j - x_ij)/h )*α_ij for j in x] for (x_ij, α_ij) in zip(x_i, α_i)])
plot(x, [y_1 yFunc], label = ["Approximation" "Function"])

At the next level, we correct our approximation at two points, namely $0.25$ and $0.75$. We want the approximation to be exact at three points now. Note that both of our new basis functions are $0.0$ at $0.5$, so we do not need to change our original weight (this is one of the advantages of sparse grid interpolation, all previously calculated weights stay fixed as we move to more precise estimates). Our first approximation estimated the value at $0.25$ to be

$\frac{f(0.0) + f(0.5)}{2}$

Using this, we can see that a weight of $w_2 = f(0.25) - \frac{f(0.0) + f(0.5)}{2}$ will make our second degree approximation exact at $0.25$. We can do a similar exercise for the point $0.75$.

s = 2
h = 2.0^(-s)
x_i = [i*h for i = 1:2:2^s-1]

α_i = [f(x_ij) - (f(x_ij - h) + f(x_ij + h))/2 for x_ij in x_i]
y_2 = sum([[φ( (j - x_ij)/h )*α_ij for j in x] for (x_ij, α_ij) in zip(x_i, α_i)])
y_approx = y_1 .+ y_2

plot(x,[y_approx yFunc], label = ["Approximation" "Function"])

The previous step can be generalized. At each degree we can simple average the values of $f$ at the points $x_{ij} + h$ and $x_{ij} - h$ and subtract this value from $f(x_{ij})$ to get our new weights. Then, adding together the approximation from all previous degrees will give our new approximation.

s = 3
h = 2.0^(-s)
x_i = [i*h for i = 1:2:2^s-1]

α_i = [f(x_ij) - (f(x_ij - h) + f(x_ij + h))/2 for x_ij in x_i]
y_3 = sum([[φ( (j - x_ij)/h )*α_ij for j in x] for (x_ij, α_ij) in zip(x_i, α_i)])
y_approx = y_1 .+ y_2 .+ y_3

plot(x,[y_approx yFunc], label = ["Approximation" "Function"])

This is already a fairly close approximation to a nonlinear function. Note that I have written out each iteration rather than put it into a formula to help elucidate the code, but we need to make this more compact to be practical.

When it comes to interpolating a function, not all points are created equal. In the case of a linear function, only two points would be needed to “approximate” the function on an interval. So while we could sample 100 points and interpolate between them to “approximate” our linear function, 98 of those points would be a wasted calculation. Adaptive sparse grids apply this same logic to our approximating functions. Each points contribution to the approximation is proportional to its weight. So points with larger weights are more important.

function chld(value::Float64, level::Int )
h = 2.0^(-level-1)
return [value - h, value + h]
end

function AdptGrid1D( tol::Float64, f::Function )
lst = Tuple[(0.5, f(0.5), 1, chld(0.5, 1))]
ln0 = length(lst)
ln1 = ln0 + 1
i = 1
@time while ln1 > ln0
ln0 = ln1
l = lst[i]
for c in l[4]
h = 2.0^(-l[3]-1)
α_i = f(c) - (f(c - h) + f(c + h))/2
abs(α_i) <= tol && push!( lst, (c, α_i, l[3]+1, chld(c, l[3]+1)) )
end
ln1 = length(lst)
i += 1
end
push!(lst, (1.0, f(1.0), 1, []), (0.0, f(0.0), 1, []));
return lst
end

Part of the problems with the previous example is that the function was well approximated by an uniformly spaced grid. To demonstrate the computational costs that an adaptive grid can afford we will look at a two-dimensional function

$f(x,y) = \frac{1}{\vert 0.5 - x^4 - y^4\vert + 0.1}$

Note that this function is also highly nonlinear, not possessing a derivative along the curver $0.5 = x^4 + y^4$. Because the sparse grid approximation works best for smooth function so approximating this function posses a challenge.

##### Two-Dimensional Adaptive Sparse Grid Using Spinterp

Klimke and Wohlmuth provide an algorithm for sparse grid interpolation called spinterp. Their formulation does not choose grid points in an adaptive manner so I had to change the basic algorithm to implement that feature. Additionally, their algorithm is defined for the region $[0,1]^d$ which is restrictive for most macroeconomic models. In anticipation of taking this to the standard growth model I have adapted their algorithm to work in a generic rectangular region. The code is not for the faint of heart, so I have included at the bottom of the page after the results. While I do not give an in-depth discussion of the algorithm, anyone interested can read the paper by Klimke and Wohlmuth. I found that it explains this complex procedure very well.

The below figure shows the points of evaluation for our interpolation. Note that the points cluster along the ridge $\vert 0.5 - x^4 - y^4\vert$ where the derivative is not defined. In smooth areas, such as the lower left quadrant, only a few points are used to approximate the function.

box = [0.0 1.0; 0.0 1.0] # Define region of interpolation
b = box[:,2] # Get upper bounds
a = box[:,1] # Get lower bounds
nobs = 100 # Choose number of observations to interpolate, 100^2 = 10,000
x = linspace(a[1], b[1], nobs) # Create mesh for x axis
y = linspace(a[2], b[2], nobs) # Create mesh for y axis
f = (x,y) -> 1/(abs(0.5 - x^4 - y^4) + 0.1) # Define function to interpolate
d = 2 # Define dimension of the problem
n = 15 # Define the degree of the sparse grid algorithm

@time Znd, ΔHnd = spvals(f, d, n, b, a, .01); # Generate weights and nodes for ϵ >= 0.01

# Plot nodes
initPtsx = [x[1] for x in vcat(vcat(ΔHnd...)...)]
initPtsy = [x[2] for x in vcat(vcat(ΔHnd...)...)]
scatter(initPtsx, initPtsy, labels = "", marker = 1)

Below is a plot of the approximation using $\epsilon_{tol} = 0.01$ and $n = 15$. You can see that the ridge is roughly approximated but the rest of the function appears very smooth. This approximation was done with 2,339 points.

nd = 15-d # Maximum dimension of the interpolation
interVals = spinterp(d, nd, Znd, vcat(Base.product((x-a[1])/(b[1]-a[1]),(y-a[2])/(b[2]-a[2]))...), ΔHnd, b, a )
# Note that x and y are transformed to be in the unit interval before interpolating

@time z = reshape(interVals, nobs, nobs)'

#using Plots
Plots.surface(x,y,z, aspect_ratio=:equal)

The adaptive sparse grid approximation might seem poor, so it is worth comparing to other methods of interpolation. First, consider a regular sparse grid of degree $11$. This requires $15,361$ points to be evaluated and gives a considerably worse approximation.

Next consider bilinear interpolation. I used a grid size of $50 \times 50$ and of $100 \times 100$ to have a comparable calculation to the sparse grid approximations. The $50 \times 50$ grid performs much better than the regular sparse grid (which is unsurprising given that sparse grids are only optimal for smooth functions) but worse than the adaptive sparse grid. The $100 \times 100$ grid is comparable to the adaptive sparse grid but at the cost of $8,000$ more function evaluations. This highlights an important trade-off inherent in calculating sparse grids. We incur a higher computational cost in calculating the grid points and the interpolated values while having a lower computational cost in function evaluations. In this example, the trade-off is not in favor of the adaptive sparse grid. It took my computer $8$ seconds to calculate the adaptive grid and plot the interpolated values for this function, while it took my computer less than a fifth of a second to calculate a $300 \times 300$ bilinear interpolation of the same function (which is considerably more precise). However, in most macro models, it will be the function evaluations that are costly as you need to perform some form of optimization at every grid point. In these cases, adaptive sparse grids tend to outperform bilinear interpolation.

Finally, here is a Chebyshev Polynomial interpolation, using $25$ points in each dimension and fitting with a polynomial of degree $17$. As expected, the polynomial interpolation performs poorly as the function is not smooth.

##### Adaptive Sparse Grid Julia Code

The code that I used for the above figures can be found here.

UPDATE: I have written some more efficient adaptive sparse grid code which can be found here.

##### References
• Brumm, J., Scheidegger, S. (2017). “Using Adaptive Sparse Grids to Solve High-Dimensional Dynamic Models”. (Econometrica, Vol. 85, No. 5 (September, 2017), 1575–1612)
• Klimke, A., Wohlmuth, B. (2005). “Algorithm 847: Spinterp: piecewise multilinear hierarchical sparse grid interpolation in MATLAB”. (ACM Transactions on Mathematical Software (TOMS), Volume 31 Issue 4, (December, 2005), 561-579)

## One thought on “Adaptive Sparse Grids”

1. Quasi-reg monkey says:

Thanks for this great introduction. I have been searching for some examples of adaptive sparse grids for a long time.

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