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## Abstract

This study explored whether repeated infusions of intravenous anti-D could allow adults with recently diagnosed immune thrombocytopenic purpura (ITP) who had failed an initial steroid course to postpone and ultimately avoid splenectomy. Twenty-eight Rh+, nonsplenectomized adults with ITP diagnosed within 1 to 11 months and platelet counts 30 × 109/L (30 000/μL) or below were enrolled. Anti-D was infused whenever the platelet count decreased to 30 × 109/L (30 000/μL) or below. “Response” was defined as a platelet increase of more than 20 × 109/L (20 000/μL) to more than 30 × 109/L (30 000/μL) within 7 days of treatment. Patients were a median 3.5 months from ITP diagnosis at enrollment and had received a median of 2 previous therapies, including prednisone in 26 of 28 cases. They were followed for a median 26 months. A total of 93% responded to their initial infusion of anti-D, and 68% repeatedly responded with counts maintained above 30 × 109/L (30 000/μL) using anti-D alone. Currently, 12 (43%) of 28 patients have been off all treatment for more than 6 months without undergoing splenectomy, 6 maintaining counts above 100 × 109/L (100 000/μL). Seven continue on treatment, 8 underwent splenectomy, and 1 was lost to follow-up at 10 months. One patient discontinued anti-D because of toxicity. Patients with platelet counts at least 14 × 109/L (14 000/μL) at enrollment were more likely to discontinue treatment (P < .05). Anti-D was an effective maintenance treatment for two thirds of Rh+, nonsplenectomized adults with ITP who had failed an initial steroid course. Intermittent infusions of intravenous anti-D allowed more than 40% of these adults to avoid splenectomy and to achieve stable platelet counts off all therapy, even after many months of treatment. Platelet count at study entry was the primary predictor of outcome.

## Introduction

Immune thrombocytopenic purpura (ITP) is an autoimmune bleeding disorder in which antiplatelet antibodies cause accelerated platelet destruction by the mononuclear phagocytic system. Patients suffering from persistent, severe thrombocytopenia are at risk for hemorrhage and have an increased mortality rate.1,2Many patients, however, maintain a normal quality of life at below normal platelet counts with or without treatment.

Unlike children, less than one third of adults with this disease will go into sustained remission after an initial short course of steroids.2-8 The American Society of Hematology practice guidelines for ITP in 1996 show that there is considerable variation in the recommendations of expert hematologists on the treatment of those patients who do not go into remission.2 Generally, the standard of practice for patients who cannot maintain an adequate platelet count on tapering steroids is to undergo splenectomy. In 60% to 70% of patients this restores a normal platelet count for at least 5 to 10 years.9-11 It allows those patients who respond to this procedure to avoid both the risks of low platelet counts and the toxicity of daily steroids. However, although both the site of platelet destruction and the response to intravenous immunoglobulin (IVIG) have been suggested to predict response to splenectomy, there is no universally acknowledged way to predict which patients will respond.12-16 Patients are often reluctant to be exposed to the inherent risks of surgery and the small but significant risk of overwhelming sepsis after splenectomy without a guarantee of success.17,18

Because the standard of practice has been to perform early splenectomy in steroid nonresponders, little is known of the natural history of those patients who do not go into early remission. This pilot study was designed to assess how many patients would improve their platelet counts to a level sufficient to avoid splenectomy if they were supported with intermittent infusions of IV anti-D to maintain their platelet count above 30 × 109/L (30 000/μL). This platelet count was chosen as the threshold for treatment so that patients would be protected from the risks of serious hemorrhage while minimizing unnecessary treatment. IV anti-D was chosen in preference to IVIG because it is substantially less expensive, has a much shorter infusion time, and is made from a smaller donor pool. It was selected in preference to prednisone because it does not have the toxicities associated with the prolonged use of daily steroids.

A secondary aim of this study was to assess whether any clinical or biologic variables could predict which patients would respond to anti-D treatment and, in particular, which patients would improve their counts sufficiently to discontinue therapy and avoid splenectomy.

## Patients and methods

Twenty-eight patients, including 7 men and 21 women, were enrolled between May 1996 and March 1999 from the 31 eligible patients referred for this institutional review board–approved study at the Platelet Disorders Center at the New York Presbyterian Hospital; 3 patients declined to participate. Patients were eligible if they had been diagnosed within the last year (but not within the last month) with primary ITP; had no history or findings suggestive of human immunodeficiency virus; had platelet counts 30 × 109/L (30 000/μL) or below at study entry; were Rh+; and had not undergone splenectomy. All eligible patients were invited to participate, including patients with profound thrombocytopenia and those with active bleeding. Patients were not excluded on the basis of previous treatment, and if on prednisone at the time of study this was quickly tapered. Patients also had to be appropriate candidates for splenectomy. After their initial treatment and 1-week follow-up at the New York Presbyterian Hospital, most patients were managed by their referring hematologists.

Patients were randomized for their initial treatment to receive either 50 μg/kg or 75 μg/kg of anti-D (WinRho SDF; Cangene, Winnipeg, MB). The data describing the acute platelet increases and hemoglobin decreases of these 2 doses have been reported separately.19 There was no distinction in the dose effect on the outcome of this study, and it is not therefore used or analyzed here. Subsequent to the initial infusion, patients received 50 to 75 μg/kg for each treatment, rounded off to the nearest multiple of 300 μg.

Treatment was administered whenever the platelet count was 30 × 109/L (30 000/μL) or below. If patients maintained a stable platelet count above 30 × 109/L (30 000/μL) for more than 4 weeks and then had a single count between 20 × 109/L to 30 × 109/L (20 000-30 000/μL) in the absence of bleeding symptoms, treatment might not be administered at the discretion of the investigator provided that a subsequent count was 30 × 109/L (30 000/μL) or more.

Complete blood counts with platelet counts were obtained on days 0, 1, 7, 14, and 21 of therapy for the first infusion and weekly thereafter. The interval between these counts was extended to every 2 to 4 weeks in patients maintaining stable platelet counts. Plasma levels of glycocalicin20 and soluble CD16 (sCD16) levels21 were measured using an ELISA technique as previously described, prior to the first infusion in 17 and 22 patients, respectively. Twenty-five patients were assessed for their CDE rhesus phenotype to estimate the number of D sites on their red cells.22

Patients could receive IV anti-D for up to 1 year as part of the study. They were seen as clinically indicated and at least monthly for the first 12 months and at 3- to 6-month intervals until March 2001, until they underwent splenectomy, or until they were lost to follow-up.

Patients were assessed for signs of bleeding at their clinic visits and completed weekly diaries of their bleeding symptoms during the study period. In addition, they answered a series of specific questions at baseline, 6, 12, and 18 months aimed at determining their change in quality of life during the period of the study.

The questions analyzed were as follows: (1) I am as healthy as anybody I know. (2) Compared to 1 year ago, how would you rate your health in general now? (3) During the past 4 weeks, how much of the time did you have a lot of energy? (4) During the past 4 weeks, how much of the time did you feel tired? (5) During the past 4 weeks, to what extent has your physical health or emotional problems interfered with your normal social activities with family and friends? (6) During the past 4 weeks, how much of the time did you feel full of pep?

### Response criteria

Acute response to anti-D was defined as a platelet increment of 20 × 109/L (20 000/μL) or more to a count above 30 × 109/L (30 000/μL) within 7 days of treatment. Responders were continued on intermittent IV anti-D as needed but were given the option of alternative therapy including splenectomy throughout the study. Patients who did not respond were offered alternative therapy including splenectomy. Patients were considered “off treatment” if they maintained a platelet count above 30 × 109/L (30 000/μL) without treatment for at least 6 months.

Patients were classified depending on their treatment status as of their last clinic visit. Group A includes patients who were considered off treatment. Group B includes patients who received treatment within 6 months of their last visit. This group included patients who (1) continued to receive anti-D; (2) received medication for their ITP other than anti-D, such as IVIG, prednisone, or danazol; and (3) received medication for other medical conditions that could have affected their platelet count. Group C includes patients who underwent splenectomy.

### Cost analysis

Cost analysis for the period of this study was performed assuming the average cost of anti-D as $80 per 300-μg vial and the average cost of splenectomy as$16 000.23 The actual number of vials used per patient was used in the calculations.

### Statistical analysis

The mean, median, SD, and range were used to describe baseline variables, including age at study entry, months of ITP diagnosis, platelet counts before and after treatment, duration of response to treatment, posttreatment hemoglobin change, and initial mean corpuscular volume (MCV). The differences between the 3 outcome groups for these parameters were analyzed using the Kruskal-Wallis test and Mann-Whitney U test. The paired t test was used to assess the difference between the number of infusions of anti-D received before and after 12 months on study and to compare the health survey questions at 6, 12, and 18 months when compared with baseline. Relationships between the biologic variables were assessed using the Spearman rank correlation coefficient.

## Results

Patients were a median age of 28.5 years (range, 18-63 years) and a median of 3.5 months (range, 1-11 months) from their diagnosis of ITP at study initiation. As of March 2001, there was a median 26 months (range, 6-54 months) follow-up on study. Prior to study entry, 26 of the 28 patients had received a course of steroids; 15 had received IVIG; 4, danazol; 3, high-dose dexamethasone; and 6, IV anti-D. Fourteen patients had previously received both IVIG and steroids. Only 1 patient had received no previous ITP therapy.

### Twelve-month interim outcome

Patients were classified into the outcome groups described in “Patients and methods.” The percentage of patients in each outcome group for the first 12 months of the study is shown in Figure1. In summary, 26 (93%) of 28 patients responded to their initial infusion of IV anti-D (increase 20 × 109/L [20 000/μL] or more to a level above 30 × 109/L [30 000/μL]). Nineteen (68%) repeatedly responded with their platelet counts being maintained above 30 × 109/L (30 000/μL) on intermittent infusions of IV anti-D alone. Three of these 19 discontinued anti-D before the 12-month follow-up point for reasons unrelated to their platelet count. One developed breast cancer and received chemotherapy (patient no. 11, Figure 2), 1 elected to undergo splenectomy after 9 months despite continuing to respond to anti-D (no. 24), and 1 was lost to follow-up at 10 months (no. 16). Five of the 19 improved sufficiently to discontinue treatment, and 11 were still receiving intermittent doses of IV anti-D at 12 months. Of the 9 patients who could not be maintained on IV anti-D alone, 1 could not tolerate IV anti-D due to persistent anemia and repeated chill reactions despite premedication (no. 10), 2 did not respond to their first infusion, and 6 stopped responding within the first 4 months. Four of the 8 nonresponders underwent splenectomy, and 4 chose alternative therapy for their ITP (described below).

Fig. 1.

A bar chart representing the percentage of patients in each outcome group for each month up to and including month 12.

The x-axis is time in months. The y-axis is the percentage of patients in each outcome group. Outcome groups are identified as follows: Patients who underwent splenectomy are represented in yellow; patients who received other medical therapy such as chemotherapy for breast cancer are shown in dark blue; patients who received other ITP therapy such as IVIG, prednisone, or danazol are shown in intermediate blue; patients who were still on intermittent anti-D are shown in light blue; and patients who came off treatment are shown in pink.

Fig. 1.

A bar chart representing the percentage of patients in each outcome group for each month up to and including month 12.

The x-axis is time in months. The y-axis is the percentage of patients in each outcome group. Outcome groups are identified as follows: Patients who underwent splenectomy are represented in yellow; patients who received other medical therapy such as chemotherapy for breast cancer are shown in dark blue; patients who received other ITP therapy such as IVIG, prednisone, or danazol are shown in intermediate blue; patients who were still on intermittent anti-D are shown in light blue; and patients who came off treatment are shown in pink.

Fig. 2.

All 28 patients from diagnosis to last platelet count and their month-by-month treatment for their ITP from diagnosis to last known platelet count.

The x-axis represents time in months; the y-axis is composed of individual patients divided according to outcome group into which they fall: group A, “off treatment”; group B, “receiving treatment (anti-D or other)”; and group C, those undergoing “splenectomy.” The letters represent ITP treatments (explained in the key). The number at the end of each line is the last known platelet count for each patient. I indicates anti-D treatment; –, no treatment; S, splenectomy; D, danazol; G, IVIG; P, prednisone; M, IV methylprednisolone; A, azathioprine; R, rituxan; E, dexamethasone; 1, IVIG for surgery; 2, lost to follow-up; 3, prednisone for SLE (not low platelets); 4, IVIG for obstetric reasons; 5, chemotherapy for breast cancer; T, toxicity; C, chemotherapy; X, radiotherapy; and *, intermittent counts less than 30 000/uL, no treatment given.

Fig. 2.

All 28 patients from diagnosis to last platelet count and their month-by-month treatment for their ITP from diagnosis to last known platelet count.

The x-axis represents time in months; the y-axis is composed of individual patients divided according to outcome group into which they fall: group A, “off treatment”; group B, “receiving treatment (anti-D or other)”; and group C, those undergoing “splenectomy.” The letters represent ITP treatments (explained in the key). The number at the end of each line is the last known platelet count for each patient. I indicates anti-D treatment; –, no treatment; S, splenectomy; D, danazol; G, IVIG; P, prednisone; M, IV methylprednisolone; A, azathioprine; R, rituxan; E, dexamethasone; 1, IVIG for surgery; 2, lost to follow-up; 3, prednisone for SLE (not low platelets); 4, IVIG for obstetric reasons; 5, chemotherapy for breast cancer; T, toxicity; C, chemotherapy; X, radiotherapy; and *, intermittent counts less than 30 000/uL, no treatment given.

### Long-term outcome

The long-term outcome (to March 2001) of all patients, including their month-by-month treatment from diagnosis until their last month of follow-up, is illustrated in Figure 2 and discussed by groups below. As of their last follow-up, 12 (43%) of the initial 28 patients have been off all treatment for at least 6 months with a stable platelet count above 30 × 109/L (30 000/μL) (group A). Seven (25%) continue to receive treatment (group B): 3 with IV anti-D, 1 with IVIG and then rituximab, and 1 with low-dose prednisone. One patient received prednisone for lupus, 1 received 3 infusions of IVIG, and 1 IV anti-D during pregnancy. Eight (29%) underwent splenectomy (group C). The patient lost to follow-up at 10 months, who had not received anti-D for 5 months as of his last visit, is not included in the analyses of long-term outcome. The background demographics for each outcome group, including months from diagnosis to study entry, age, gender, and previous treatments are summarized in Table1.

Table 1.

Summary of patient characteristics at enrollment

Outcome group Median no. of months from ITP diagnosis to enrollment (range) Median age at enrollment, y (range) Gender (F, M) Steroids* IVIG* Danazol*
Off treatment (n = 12) 4  (1-11) 29.5  (21-63) 8, 4 10 2
On treatment (n = 7) 2  (1-11) 29  (18-56) 6, 1
Splenectomy (n = 8) 2  (2-10) 25  (20-61) 7, 1
Outcome group Median no. of months from ITP diagnosis to enrollment (range) Median age at enrollment, y (range) Gender (F, M) Steroids* IVIG* Danazol*
Off treatment (n = 12) 4  (1-11) 29.5  (21-63) 8, 4 10 2
On treatment (n = 7) 2  (1-11) 29  (18-56) 6, 1
Splenectomy (n = 8) 2  (2-10) 25  (20-61) 7, 1
*

Number of patients in each outcome group receiving these treatments prior to enrollment.

#### Group A.

The 12 patients who are now off all treatment were enrolled a median of 4 months (range, 1-11 months) from diagnosis. They received a median of 11.5 months (range, 3-41 months) of treatment on study. The median time from diagnosis to their last treatment was 16 months (range, 7-45 months). Eight of the 12 continuously responded to intermittent anti-D and received it alone for a median of 8.5 months (range, 3-41 months) before stopping treatment altogether. Two of the 12 stopped responding to anti-D within 5 months of study initiation and transiently used either prednisone or danazol (patient nos. 9 and 12). One received chemotherapy for breast cancer and then temporarily resumed IV anti-D (no. 11), and 1 discontinued anti-D because of toxicity (no. 10) and received 12 months of IVIG before coming off all treatment.

As of the last follow-up visit, 6 of the 12 have maintained platelet counts above 100 × 109/L (100 000/μL), with 3 of these maintaining normal counts (> 150 × 109/L [150 000/μL]) for more than 4 months. Strikingly, these 6 patients were on study for a median 15 months (range 8-35 months) and had been diagnosed with ITP for a median of 22.5 months (range 13-38 months) before achieving platelet counts above 100 × 109/L (100 000/μL). They were treated for a median of 16 months (range 7-20 months) from diagnosis to last treatment, including a median of 9 months (range 5-14 months) on study. The platelet counts of these 6 patients, beginning the month after they stopped treatment, are shown in Figure 3A. These patients continued to improve their counts even after stopping treatment. The platelet counts for those off treatment with counts below 100 × 109/L (100 000/μL) are shown in Figure 3B (also beginning the month after stopping treatment); their counts appear to be stable.

Fig. 3.

Platelet counts.

(A) The platelet counts for the 6 patients who have been off treatment for more than 6 months and have platelet counts at least 100 × 109/L (100 000/μL) at their last recorded count. The x-axis is time in months, and the y-axis is the platelet count. The counts are plotted beginning with the month after their last treatment. (B) The platelet count for the 6 patients who have been off treatment for more than 6 months with platelet counts below 100 × 109/L (100 000/μL) at their last follow-up count. The x-axis is time in months, and the y-axis is the platelet count. The counts are plotted beginning with the month after their last treatment.

Fig. 3.

Platelet counts.

(A) The platelet counts for the 6 patients who have been off treatment for more than 6 months and have platelet counts at least 100 × 109/L (100 000/μL) at their last recorded count. The x-axis is time in months, and the y-axis is the platelet count. The counts are plotted beginning with the month after their last treatment. (B) The platelet count for the 6 patients who have been off treatment for more than 6 months with platelet counts below 100 × 109/L (100 000/μL) at their last follow-up count. The x-axis is time in months, and the y-axis is the platelet count. The counts are plotted beginning with the month after their last treatment.

#### Group B.

The 7 patients still receiving treatment were enrolled a median of 2 months (range 1-11 months) from diagnosis and have been followed since study initiation for a median of 25 months (range 10-37 months). Three patients still receive IV anti-D. These 3 received a median of 0.42 infusions (range 0.42-0.58 infusions) per month during the first 12 months of the study and significantly less frequent (median 0.27 infusions [range 0.11-0.32]) infusions per month during the 9- to 22-month additional follow-up period (P = .037). Two patients stopped responding to IV anti-D within the first 4 months of the study. One received IVIG for 20 months and is now maintaining counts above 60 × 109/L (60 000/μL) following a course of rituximab (patient no. 18). The other has a count of 105 × 109/L (105 000/μL) on 7.5 mg prednisone a day (no. 20). Two patients received medication for other medical conditions. One developed systemic lupus erythematosus 20 months after starting the study and received prednisone for symptoms other than her platelet count (no. 17). The other was infused 3 times with 400 mg/kg IVIG while pregnant to prevent fetal loss (no. 19) and then received IV anti-D once immediately prior to delivery.

#### Group C.

The 8 patients who underwent splenectomy were enrolled a median of 4 months (range 2-10 months) from diagnosis and have been followed for a median of 27.5 months (range 6-39 months). Four were early nonresponders to anti-D and underwent splenectomy a median of 3.5 month (range 2-7 months) from starting the study. Four were continually responsive to anti-D but decided to undergo splenectomy a median of 27 months (range 9-36 months) from starting the study. Follow-up platelet counts are available from 2 to 10 months after splenectomy (median 3 months) in these patients. Six (75%) of the 8 had counts above 150 × 109/L (150 000/μL) after splenectomy without other therapy.

### Biologic variables and acute response to IV anti-D

There was no correlation between glycocalicin levels, sCD16 levels, initial MCV or initial hemoglobin level, and platelet response to treatment or hemoglobin change after treatment. There was also no relationship between the number of D sites on the red cell as estimated by CDE typing and the day 7 platelet response or hemoglobin decrease following anti-D.

### Comparisons between the outcome groups

Comparisons between the groups are shown in Table2. The 12 patients in the “off treatment” category had significantly higher platelet counts at study entry than the 8 patients who underwent splenectomy (P < .05). Eleven of the 12 “off treatment” patients had platelet counts at least 14 × 109/L (14 000/μL) at study entry, compared with only 2 of the 8 patients who underwent splenectomy. Furthermore, all of the 8 patients who were unresponsive to anti-D within the first few months had platelet counts below 14 × 109/L (14 000/μL) at study entry.

Table 2.

Comparison of pretreatment and peritreatment varables in each outcome group

Outcome group No. of patients Median initial platelet count (range) Median platelet increment after first infusion* (range) Median duration of first infusion (range)
Off treatment 12 19.5  (4-29) 106.5  (6-366) 28  (10-45.62)
On treatment 12  (5-26) 126  (8-440) 21  (11.15-56.75)
Splenectomy 9  (4-21) 47  (3-273) 19.5  (15.5-42.5)
Outcome group No. of patients Median initial platelet count (range) Median platelet increment after first infusion* (range) Median duration of first infusion (range)
Off treatment 12 19.5  (4-29) 106.5  (6-366) 28  (10-45.62)
On treatment 12  (5-26) 126  (8-440) 21  (11.15-56.75)
Splenectomy 9  (4-21) 47  (3-273) 19.5  (15.5-42.5)
*

Platelet count day 7 − day 0 × 109/L.

Days between infusions no. 1 and 2.

P < .05 MWU.

There was no significant difference in pretreatment glycocalicin or sCD16 levels between the long-term outcome groups, possibly because the numbers in each group were small. There was no significant predictive value of initial MCV, initial hemoglobin, or platelet increment or duration of response to the first infusion on the long-term outcome. There was no significant difference between the outcome groups for either rhesus phenotype (CDE) or ABO blood group.

### Bleeding diaries

No patient had a severe bleeding episode throughout the 64 years of patient follow-up following study enrollment. Twenty-four of the patients complained of nonsevere bleeding, including 7 nose bleeds, 7 mouth bleeds, 10 petechiae, and 11 increased bruising. Six of the 15 females described menorrhagia. Bleeding symptoms tended to occur at lower platelet counts; however, occasional patients described petechiae, bruising, and nose bleeds at normal platelet counts.

### Quality of life survey

There was no change from baseline to 6, 12, or 18 months in the scores of questions 3, 4, and 6. There was an increased score (greater well being) for question 5 at 12 months (P < .05) and at 18 months (P < .01) when compared with baseline. There was also an increased score at 6 (P < .05), 12 (P < .01), and 18 (P < .01) months in question 2 when compared with baseline. Question 1 showed a decreased score (decreased feeling of well being) at 6 (P < .05) and 18 (P < .05) months when compared with baseline.

### Cost analysis

The cost of treating patients with IV anti-D for each outcome group is estimated in Table 3 using the number of vials of anti-D actually used by each patient and the costs listed in “Patients and methods.” Using these estimates, the cost of the anti-D used in all 28 patients plus the 8 splenectomies would be $370 400. In comparison, the cost of performing splenectomy in all 28 patients would have been$448 000. The cost of treating the 12 patients who were able to discontinue treatment for the study period was $87 280, or$229 per patient per month. The cost of treating those patients still on treatment was $68 480, or$389 per patient per month. The cost of treating the patients who underwent splenectomy, including the costs of both anti-D and splenectomy but not treatment after splenectomy, was $214 640, or$1,850 per patient per month. These estimates are an oversimplification because they do not include many other costs, such as the cost of other therapies, blood tests, outpatient visits, nursing time, or the cost of time lost from work.

Table 3.

Cost comparisons of anti-D and splenectomy

Outcome group Number of anti-D infusions on study Other ITP treatment on study3-150 Months of patient follow-up Total cost of anti-D and splenectomy per group Cost per patient Cost per patient per month
Off treatment (n = 12) 79 16 G, 13 mo P, 28 mo D 381 $87 280$7 273 $229 On treatment (n = 7) 54 22 G, 31 mo P, 3 mo A 176$68 480 $9 978$389
Splenectomy (n = 8) 64 8 G, 1 V, 2 M, 18 mo D, 14 P 1163-151 $214 640 ($86 6403-152)  $26 830 ($10 8303-152$1 850 ($7473-152
Sum 197  673 $370 400 ($242 4003-152
Outcome group Number of anti-D infusions on study Other ITP treatment on study3-150 Months of patient follow-up Total cost of anti-D and splenectomy per group Cost per patient Cost per patient per month
Off treatment (n = 12) 79 16 G, 13 mo P, 28 mo D 381 $87 280$7 273 $229 On treatment (n = 7) 54 22 G, 31 mo P, 3 mo A 176$68 480 $9 978$389
Splenectomy (n = 8) 64 8 G, 1 V, 2 M, 18 mo D, 14 P 1163-151 $214 640 ($86 6403-152)  $26 830 ($10 8303-152$1 850 ($7473-152
Sum 197  673 $370 400 ($242 4003-152

G indicates IVIG; P, prednisone; D, danazol; V, vincristine; M, methylprednisolone; and A, azathioprine.

F3-150

The number of IVIG, vincristine, or methylprednisolone treatments and the number of months of prednisone, danazol, or azathioprine per group.

F3-151

Follow-up until month of splenectomy.

F3-152

Cost of anti-D alone.

## Discussion

Studies assessing the long-term outcome of adults with ITP (Table4) suggest that between 9% and 32% of newly identified adults will go into remission either spontaneously or following a short course of steroids.3-8 Those patients who do not go into early remission tend to undergo splenectomy.

Table 4.

Summary of studies assessing the long-term outcome of patients with ITP

Year Country No. of patients treated with steroids Initial CR, steroids alone (%) CR maintained after discontinuing treatment (%) Late CR Total CR (%)
Thompson et al3 1972 US 57 13  (23) 13  (23) 16  (28)
DiFino et al4 1980 US 59 25  (43) 13  (22) 16  (27)
Pizzuto et al5 1984 South America 818 386  (47) 262  (32) — 262  (32)
JiJi et al6 1984 US 91 22  (24) 21  (23) 26  (29)
Stasi et al7 1995 Italy 121 52  (39) 11  (9) — 11  (9)
Ikkala et al8 1978 Finland 40 16  (40) 10  (25) — 10  (25)
Year Country No. of patients treated with steroids Initial CR, steroids alone (%) CR maintained after discontinuing treatment (%) Late CR Total CR (%)
Thompson et al3 1972 US 57 13  (23) 13  (23) 16  (28)
DiFino et al4 1980 US 59 25  (43) 13  (22) 16  (27)
Pizzuto et al5 1984 South America 818 386  (47) 262  (32) — 262  (32)
JiJi et al6 1984 US 91 22  (24) 21  (23) 26  (29)
Stasi et al7 1995 Italy 121 52  (39) 11  (9) — 11  (9)
Ikkala et al8 1978 Finland 40 16  (40) 10  (25) — 10  (25)

CR indicates complete response.

Several studies have reported anecdotal patients who achieve late remission after multiple courses of steroids.3,4,6 In general, however, the natural history of those adults who do not improve after an initial course of steroids and do not undergo splenectomy is unknown. This pilot study was designed to assess the 1- to 5-year outcome of patients with recently diagnosed ITP who did not improve after their initial treatment. The hemorrhagic complications associated with persistently low platelet counts were avoided by using palliative treatment with repeated infusions of IV anti-D.

The results were encouraging. Twelve (43%) of the 28 adults who participated in this study have been off all treatment and have maintained platelet counts above 30 × 109/L (30 000/μL) for more than 6 months. Six (21%) of these continued to improve after discontinuing therapy and now have counts above 100 × 109/L (100 000/μL) with 3 (11%) in complete remission. Furthermore, 71% of the 28 patients have so far avoided splenectomy, and those still receiving IV anti-D are receiving it significantly less frequently. Of particular note, those patients with counts above 100 × 109/L (100 000/μL) (Figure 3A) achieved this a median of 22.5 months (range 13-38 months) from diagnosis and received a median of 16 months (range 7-20 months) of treatment. Given that the 28 patients enrolled in this study had completed a median of 2 courses of treatment and were a median 3.5 months from diagnosis at study entry, it was surprising that almost half the patients were able to discontinue treatment and that some patients were still improving and achieving near normal platelet counts up to 38 months after diagnosis.

It is possible that a referral bias resulted in a disproportionate number of patients with less severe ITP entering the trial. However, 28 of 31 patients eligible for the study were enrolled; 15 had platelet counts below 15 × 109/L (15 000/μL) at study entry, and 14 had previously received both steroids and IVIG.

One of the disadvantages of this approach is that patients were seen more frequently for outpatient visits and often had lower platelet counts than if they had undergone successful splenectomy. In addition, patients with marginal platelet counts in the range of 30 × 109/L to 50 × 109/L (30 000-50 000/μL), although off treatment, could potentially have periodic crises. These crises might require additional treatment, including hospitalization, all of which could have been relieved by early splenectomy, if it was successful. However, no ITP-related hospitalizations occurred during the course of the study, and there were no episodes of serious bleeding symptoms in the 64 patient years of follow-up in this study; nor were there significant adverse events or decrease in quality of life. Furthermore, many patients avoided both surgery and the toxicity of long-term steroids by following this protocol. The rudimentary cost analysis in Table 3 shows that if a substantial proportion of patients can eventually discontinue treatment without undergoing splenectomy, estimated previously to be 47%,23 the use of intermittent IV anti-D, even for many months, may be cost-effective in comparison to all patients undergoing splenectomy.

Predicting which patients will improve over time would further improve the cost-effectiveness and give patients a better idea of what they could expect if they do not undergo early splenectomy. Patients with higher platelet counts at study entry, in particular those with platelet counts above 14 × 109/L (14 000/μL), appear to be more likely to improve. No other variables—ie, glycocalicin levels, sCD16 levels, or blood type—were predictive, although the sample sizes were small.

Overall, this study proposes that splenectomy, which is only successful in 60% to 70% of patients,9-11 may not be immediately necessary in all adults with recently diagnosed ITP who do not go into remission with initial steroids. It suggests that most adults with ITP have a tendency for their platelet counts to improve over time. It also shows that IV anti-D is a safe and effective treatment in approximately two thirds of Rh+, nonsplenectomized adults with ITP and can be used as a steroid-sparing agent. There is no reason to believe that anti-D is curative or that it affected the natural history of the ITP in these adults. Rather, it appeared to give patients time to improve on their own.

ITP is clearly a heterogeneous disease. Better understanding of the pathology of this disease and identification of good predictors of response to splenectomy and other therapies would help to define treatment strategies that would be most appropriate for individual patients.

We thank Howard Fleit at the State University of New York at Stony Brook for performing the sCD16 assays, Ginette Lanoix for technical assistance, and all the physicians and their staff whose patients participated in this study: Drs Bernard Bernhardt, Edward Amorosi, Roy Berger, Ellin Berman, Laurence Bilsky, Terry Davies, Richard Furman, Gerard Hellman, Gary Horbar, Alexandra Ikeguchi, Robert Jacobson, Shirley Levine, Neal Lewin, Stuart Lewis, Eva Levitan, William Lipera, Taibor Moskowitz, Vijay Roy, Eduardo Saponara, William Solomon, Richard Taubman, Zella Zeigler, and Robert Zielinski.

Supported in part by a clinical research grant from Cangene, Winnipeg, Canada; the ITP Society of the Children's Blood Foundation, New York, NY; and the Swiss National Foundation of Science, Baden, Switzerland.

B.M.R.W. is an employee of Cangene Corporation.

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A dose of 75 μg/kg/d of IV anti-D increases the platelet count more rapidly and for a longer period of time than 50 μg/kg/d in adults with immune thrombocytopenic purpura.
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## Author notes

Nichola Cooper/James Bussel, Division of Hematology/Oncology, P-695, Dept of Pediatrics, New York Presbyterian Hospital—Weill Medical College of Cornell University, 525 East 68th St, New York, NY 10021; e-mail: nicholacooper@yahoo.com;jbussel@med.cornell.edu.


text:
# Twisted K-theory in motivic homotopy theory

### Maria Yakerson (ETH)

10-Sep-2021, 19:00-20:00 (4 months ago)

Abstract: In this talk, we will speak about algebraic K-theory of vector bundles twisted by a Brauer class, and its place in motivic homotopy theory. In particular, we will discuss a new approach to the motivic spectral sequence for twisted K-theory, constructed earlier by Bruno Kahn and Marc Levine. The talk is based on joint work in progress, with Elden Elmanto and Denis Nardin.

algebraic geometry

Audience: researchers in the topic

( slides )

format: plain


text:
# Relating Energy to Wavelength in curved space

Consider a curved space, e.g. Schwarzschild: \begin{align*} ds^2 = -\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2 \end{align*} Now, the energy of a photon is $E = \hbar \omega$, and $|\mathbf{k}|= \frac{2\pi}{\lambda}$, but am I correct in assuming that $\omega \neq |\mathbf{k}|$?

Because if $k^\mu = (\omega,\mathbf{k})$ then $k_\mu k^\mu = 0$ implies that: \begin{align*} g_{tt} \omega^2 + g_{rr}(k^1)^2+g_{\theta\theta}(k^2)^2+g_{\phi\phi}(k^3)^2 =0 \end{align*} So basically, is it correct that the relationship between $\omega$ and $|\mathbf{k}|$ will vary in curved space? (And so relationships like $E = \frac{h}{\lambda}$ no longer hold?)

-

The main thing to note is that the definition of the wavenumber you cite above is dependent on the underlying function satisfying the standard wave equation, because any function that satisfies

$$\eta^{ab}\partial_{a}\partial_{b}\phi(x) = 0$$

Will have its Fourier transform, $\Phi(k) = \int d^{4}x e^{ik^{a}x_{a}}\phi(x)$ satisfy

$$k^{a}k_{a}\Phi(k) = 0$$

But this is no longer true, because for the case of curved spacetime, the wave equation is

$$g^{ab}\partial_{a}\partial_{b}\phi(x) - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi(x) = 0$$

And this will require some modification to the Fourier transform to work. More physically, you have effects like gravitational lensing that cause light to interact with the gravitational field, so you don't get simple straight-line propogation of monochromatic modes.

Note, however, that it is always possible to locally transform to a coordinate system where $\Gamma_{ab}{}^{c} =0$ and $g_{ab} = \eta_{ab}$, and there, you will be able to have a well defined wavenumber and frequency which satisfies $\omega^{2} = k^{i}k_{i}$. This just won't work outside of your local neighborhood.

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$\Gamma_{ab}{}^c$ can in general only be made 0 at a point, not in a neighborhood. The latter is possible iff space is flat in that neighborhood. –  Robin Ekman Jun 8 at 16:20
@RobinEkman: you're right, of course. In my mind, I always imagine this as taking some tolerance for the size of the christoffel symbols, and adjusting the size of the neighborhood so that the christoffel symbols are below this tolerance, and then imagine spacetime as "approximately/essentially" flat in this region -- so that my picture of the equivalence principle is a little more physical than just talking about a single point, which is obviously not realizable physically. –  Jerry Schirmer Jun 8 at 16:31
but I shouldn't use my personal language like that when communicating with the public. –  Jerry Schirmer Jun 8 at 16:43

I would think what you are saying is correct, the dispersion relation for light in vacuum $\omega = kc$ is derived by considering plane wave solutions to Maxwell's equations $\partial_{\mu} F^{\mu \nu}=0$. For curved spacetime, you would have to replace the usual derivative by a covariant derivative.

-


text:
February 18, 2022

Global Real Economic Uncertainty and COVID-19

Ranie Lin, Juan M. Londono, and Sai Ma1

Introduction

The COVID-19 pandemic led to unprecedented disruptions in supply, demand, and productivity, which have had cataclysmic health, social, and economic implications across the globe. In this note, we explore the large increase in global real economic uncertainty observed during the pandemic as a channel that explains or magnifies the economic implications of COVID-19. We use the methodology in Jurado, Ludvigson, and Ng (2015) to calculate an index for real economic uncertainty (REU), and we center the attention on the dynamics and the economic effects of REU during the COVID-19 pandemic. To understand why the economic effects of REU might have been different during the pandemic compared to other crises, we exploit a cross section of 30 countries, which represent 65% of total global GDP, and we extend the time series of REUs in Londono, Ma, and Wilson (2021) to April 2021.

We find that, though the increase in REU during the COVID-19 pandemic was unprecedented and almost synchronized across all countries in our sample, the drops in industrial production (IP) that followed were similar in magnitude to those observed during the 2008 global financial crisis (GFC). However, unlike during the GFC, countries that suffered the largest increases in REU did not necessarily experience the largest declines in IP. To explore this apparent disconnect between, on the one hand, the unprecedented increase in REU and the relatively more modest decrease in IP during the COVID-19 pandemic, and, on the other hand, the heterogeneity in the economic effects of REU across countries, we explore two channels unique to the pandemic that drive the economic impact of REU over this period. Our results suggest that countries that experienced the more severe COVID-19 outbreaks suffered the largest economic effects; that is, relative to countries that suffered less severe outbreaks, their domestic IP dropped more following an increase in REU. However, fiscal stimulus packages moderated these economic effects. Thus, countries where governments responded more actively and injected more money into their economies during the pandemic suffered less declines in domestic IP in response to changes in REU relative to countries with more moderate fiscal stimulus packages over this period.

Real Economic Uncertainty during the COVID-19 Pandemic

We calculate the monthly REU index for the following 30 countries: Australia, Austria, Belgium, Brazil, Canada, Denmark, Finland, France, Germany, Greece, Iceland, Ireland, Israel, Italy, Japan, Korea, Luxembourg, Mexico, the Netherlands, New Zealand, Norway, Poland, Portugal, Slovenia, Spain, Sweden, Switzerland, Turkey, the United Kingdom, and the United States.

Following the methodology in Jurado, Ludvigson, and Ng (2015), we measure REU as an aggregate of the standard deviations of the unforecastable component of a set of economic indicators. Specifically, for each economic indicator (indexed by j) for a particular country at time $t$, $v_{it}$, the $h$-period ahead uncertainty is calculated as

${U}_{{jt}}\left({h}\right)=\sqrt{\mathbb{E}{\underbrace{\left({y}_{{jt}+{h}}-\mathbb{E}\left[{y}_{{jt}+{h}}\middle|{I}_{t}\right]\right)}_{Forecasting\ errors}}^\mathbf{2}|{I}_{t}},$

where the forecast is made based on a dynamic factor model using all information currently available ($I_t$). Intuitively, uncertainty is higher when the volatility of the forecast error is high; that is, when it is harder to predict this indicator. For each country, aggregate uncertainty is calculated as the average of all individual economic indicator uncertainties. We use data from the OECD Main Economic Indicators to calculate REU for all 30 countries in our sample. Using the same data source for all countries allows us to guarantee the comparability of economic indicators across countries.2 As pointed out by Ng (2021), variations of many underlying economic indicators during the pandemic could be interpreted as outliers, which poses a challenge for calculating REU in this period. To address this issue, we follow Ng (2021) and adjust REUs for each country using the number of COVID-19 related hospitalizations and positive cases after February of 2020.3

Figure 1 shows the time series of the equal-weighted average of REUs across all countries in our sample and the 12-month-rolloing window cross-country average of monthly IP growth.

Figure 1. Global REU and IP growth

The increase in REU during the pandemic was, indeed, unprecedented, reaching a maximum value of around seven standard deviations above its historical mean between March and May of 2020. For comparison, the previous maximum average REU realization, which could still be characterized as an extreme event, was around the collapse of Lehman Brothers in 2008 at 1.2 standard deviations above its full-sample historical mean.4 Overall, as documented by Londono et al. (2021), an increase in global REU is followed by a decrease in IP. Interestingly, however, the drop in IP during the pandemic is only slightly higher than that observed during the GFC. There are, of course, large fluctuations in IP during crises, including large rebounds in the second and third quarters of 2020 for many countries in our sample. Nevertheless, the 12-month-rolling-window average allows us to look at longer-term trends in IP.

Comparing the relation between REU and IP growth across countries can help us further understand this apparent disconnect between the unprecedented increase in REU during the COVID-19 pandemic—almost 5 times that during the GFC—and the somewhat similar decline in IP during this episode compared to the GFC. Figure 2 compares the maximum decline in IP during the GFC and the COVID-19 pandemic for countries grouped into low, medium, or high REU bins. The pattern for the GFC seems to follow better the intuition that a larger increase in REU is followed by a larger decline in IP. However, this pattern seems to break during the pandemic, not only because the magnitudes of the largest IP drops are smaller during this period but also because countries that experienced the largest REU did not necessarily experience the largest drop in IP.

Severity of COVID-19 Outbreaks and Fiscal Stimulus

Measures to prevent the spread of the virus and its effects were also unprecedented. For instance, in an effort to prevent the spread of the virus, governments implemented confinement measures, and these measures not only varied as the pandemic evolved but were also largely heterogeneous across countries, ranging from limited restrictions to total lockdowns that lasted for months in many countries. These measures partially explain why the severity of the outbreak varied substantially across countries and at different points in time during the pandemic. Governments around the world also implemented unprecedented packages to stimulate demand in their economies. For example, in the United States, a 2.2 trillion U.S. dollar stimulus package (around 10 percent of GDP), known as the U.S. CARES Act, was signed into law in March 2020; Finland's fiscal package, which supported unemployment and various sectors of the economy was valued at 10 percent of Finnish GDP; Japan's Emergency Economic Measures were valued at 1.1 trillion U.S. dollars (21 percent of GDP).5

To explore whether these two factors could explain the different economic impact of REU during the pandemic, we estimate the following regression:

$(1) \ \ \ \ \ \ \Delta{ip}_{i,t+h} = \alpha + (\beta_1 + \beta_2{Severity}_{i,t} + \beta_3{Stimulus}_{i,t}){REU}_{i,t} + \gamma{C} + \varepsilon_{i,t},$

where $\Delta{ip}_{i,t+h}$ is the annualized percentage growth in country $i$'s IP between months $t$ and $t+h$ and ${REU}_{i,t}$ is the country's domestic REU index. The economic effect of REU can be time varying and country specific as a function of both the severity of the outbreak and the magnitude of the fiscal stimulus package. $C$ includes the following control variables: the foreign REU, which is calculated as the equally-weighted average of all other countries' REUs, the level of the severity of the outbreak, and the level of the stimulus package. The last two control variables allow for a direct effect of the severity of the pandemic and the stimulus packages on IP, in addition to affecting the size of the response to REU.

Table 1 shows the estimates of the coefficients in equation (1). Our results confirm those in Londono et al. (2021) and Jurado et al. (2015) that REU has significant and economically meaningful effects. In the absence of COVID-19 cases and fiscal stimulus, a one standard deviation increase in domestic REU is followed by a maximum drop in IP of around one percent after 3 months.

Table 1: Economic effects of REU, COVID-19 severity, and fiscal stimulus
H 1 month 3 months 6 months
REU ($\beta_1$) 4.008 -1.186* -0.798***
(4.034) (0.619) (0.262)
REU*Severity ($\beta_2$) -0.958 -1.500** -0.570***
(5.02) (0.562) (0.196)
REU*Stimulus ($\beta_3$) 1.290* 0.457*** 0.176***
(0.746) (0.112) (0.0366)
R2 (%) 3.1 12 10.2
N 12,450 12,396 12,396

Note: this table shows panel regression results of IP growth on REU and a set of control variables across multiple time horizons. The severity of the Covid-19 outbreak is measured for each country as monthly new cases per billion residents, and the data are obtained from the Our World in Data (OWID) databank. Stimulus responses come from the Oxford COVID-19 Government Response Tracker, which assigns a monthly numerical index between 0 and 100 for each country capturing the strength of the overall government response to the pandemic. For each $h$-month time horizon, IP growth is computed as the annualized percentage growth in IP from time $t$ to time $t + h$. Country fixed effects are included to control for other time-invariant characteristics, and standard errors are clustered at the country level. The period spans January 1980 to April 2021. For brevity, we omit the coefficients for the control variables.

We also find evidence supporting our hypotheses that these the severity of COVID outbreaks and the magnitude of fiscal stimulus packages affect the size of the response to REU during the COVID-19 pandemic. In particular, the coefficient associated with the interaction between REU and the severity of the outbreak is negative and significant, which suggests that more severe outbreaks are followed by larger drops in IP when uncertainty increases. In contrast, the interaction between REU and fiscal stimulus is positive and significant, consistent with these stimulus packages being successful at preventing a further economic deterioration following the increase in REU.

To understand the economic magnitude of how the severity of outbreaks and stimulus packages drive the economic effects of REU, figure 3 plots these economic effects for different levels of each variable. The panel to the left shows different scenarios for the severity of outbreaks in the absence of fiscal stimulus. For countries with average COVID-19 cases, an increase in REU is followed by a two-month-ahead decrease in IP of about 8 percent, much higher than the one percent for countries with zero cases (minimum).

Figure 3. Effect of REU on IP growth for different levels of COVID-19 cases and stimulus responses

The panel to the right shows the economic effects of REU under different stimulus package scenarios, assuming as a reference the extreme case of the country with the maximum COVID-19 cases, Poland. In the absence of stimulus, a country that had as high a number of cases, an increase in REU is followed by a two-month-ahead drop in IP of more than 60 percent (annualized). Fiscal stimulus does counteract the magnifying effect of the severity of outbreaks. In particular, even if COVID cases were extreme, an increase in REU is followed by a drop in IP of around 7 percent when the stimulus package is the maximum in our sample, which corresponds to Greece in February of 2021.

In sum, in this note, we extend the time series of real economic uncertainties (REUs) for 30 countries in Londono, Ma, and Wilson (2021) to investigate the dynamics and the economic effects of REU during the COVID-19 pandemic. We find evidence that, while REUs are almost 5 times larger during the pandemic than during the 2007-09 GFC, the declines in global IPs are of a similar magnitude in both episodes. We show that two factors present during the COVID-19 pandemic, namely the severity of COVID outbreaks and the magnitude of fiscal stimulus packages, counteract each other and help explain the economic effects of REU on IP during the pandemic.

References

Jurado, K., Ludvigson, S. C., Ng, S., 2015. Measuring uncertainty. American Economic Review 105, 1177–1216.

Londono, J. M., Ma, S., Wilson, B. A., 2021. The global transmission of real economic uncertainty. IFDP working papers 1317.

Ng, S., 2021. Modeling macroeconomic variations after covid-19. Tech. Rep., National Bureau of Economic Research.

Ritchie, H., Mathieu, E., Rodés-Guirao, L., Appel, C., Giattino, C., Ortiz-Ospina, E., Hasell, J., Macdonald, B., Beltekian, D., Roser, M., 2020. Coronavirus Pandemic (COVID-19). Published online at OurWorldInData.org. Retrieved from: 'https://ourworldindata.org/coronavirus' [Online Resource]

1. Ranie Lin is at Rice University, Juan M. Londono is a Senior Economic Project Manager and Sai Ma is a Senior Economist at the Division of international Finance of the Federal Reserve Board. The analysis and conclusions set forth are those of the authors and do not indicate concurrence by other members of the research staff or the Board of Governors. Return to text

2. See Londono, Ma, and Wilson (2021) for more details. Return to text

3. Data on cases and hospitalizations are obtained from Our World in Data (https://ourworldindata.org/covid-hospitalizations). Return to text

4. The sample mean is significantly affected by the COVID-19 period. REU is around 4 standard deviations above its mean during the GFC when the COVID-19 period is excluded. Return to text

5. Source: the OECD coronavirus country tracker (https://www.oecd.org/coronavirus/country-policy-tracker/). It's worth noting that other types of policy support such as monetary policy and liquidity policies are not considered because of limited and ununified measures in the international sample as well the focus on the real activities instead of financial markets in the paper. Return to text


text:
# Difference between different radial and centripetal accelerations:(Doubt)

In cylindrical coordinates I had derived the acceleration as :

for the first component of acceleration, what is the difference between radial and centripetal ?

• A guess: centripetal is from a point, radial is perpendicular to a line. – garyp Jan 7 at 16:54

The radial acceleration, $$\ddot r$$, is responsible to change the rate of change of the radius vector, $$\dot r$$, which in turn is responsible to change the magnitude $$r$$ of the radius vector, its length.
The centripetal acceleration $$-r \dot \theta^{2}$$ is what makes the particle describe some curvilinear path. The "minus" signal in front of the centripetal acceleration indicates it's inward.
If we consider only the radial component in the case when $$\dot{\theta} = 0$$, then it becomes clear that $$\ddot{r}$$ is a radial acceleration, i.e. since the angular component of velocity is $$0$$, the particle is only moving along a straight line through the origin, and $$\ddot{r}$$ is the acceleration along this line. $$\\$$ Now if we consider when $$\ddot{r} = 0$$, with $$\dot{r} = 0$$ also (incidentally, this is just the condition for circular motion around the origin), then $$-r{\dot{\theta}}^2$$ must be the source of what is making us turn.


text:
# Cylindrical Coordinates

#### emt

Joined Oct 27, 2015
12
Cylindrical Coordinates

* why is there ρ in graph (from where came)?

*what is difference between "ρ0" and "ρ" ?

#### Papabravo

Joined Feb 24, 2006
12,770
ρ is the radial coordinate. ρ0 is a particular constant value of the radial coordinate at which a differential length or differential area is computed.

$$\rho=\sqrt{x^2 + y^2}$$

As shown in the diagram the cylindrical coordinates are ρ, φ, and z

emt


text:
mersenneforum.org New Fermat factors
 Register FAQ Search Today's Posts Mark Forums Read

2018-08-08, 08:10   #265
GP2

Sep 2003

2,579 Posts

Quote:
 Originally Posted by R. Gerbicz Just interestingly, if m=k*2^(n+2)+1 | F_n (where k can be even) and 0100.
Among the currently known factors, the largest n for which it the above condition 0<k<2^(n+2) fails (k can be even) is:

42 45 111318179143061

Last fiddled with by GP2 on 2018-08-08 at 08:37

2018-08-08, 08:19   #266
ATH
Einyen

Dec 2003
Denmark

3×5×193 Posts

Quote:
 Originally Posted by GP2 Maybe you could use pfgw to verify that the known Fermat factors really are prime factors and not composite factors? But maybe not for the goal I wanted. Or am I misunderstanding?
Looking in the pfgwdoc.txt for -gos2 it does test the Fermat factors without testing primality.

Quote:
 -g Generalized Fermat factor testing The complete syntax is: -g[x][o][s][q]{#,#}{#,#}[_dump_search_patterns] Also allowed is -g[o][q]# and -gx[o][q]#,# to test single GF (or xGF) number This will test the numbers to see if they divide any fermat number. All prp's or primes (N-1 only) which are in the correct form of being a Fermat factor will go through the test. o: The o is needed when you only want to test for Fermat factors, and NOT test the numbers for probable primality. ([O]nly perform GF divisibility). s: The s is an 'undocumented' feature, which saves the residue of each b^(2^(n-1)) into a file, so that a re-run of the number will be much quicker. (CAUTION here, this function does NOT delete the files, so things can add up quickly). The -g (or -go) may be followed by the GF base you wish to check for. if no base is entered, then the default will be -g[o]{2,5}{2,12} So -go2 will ONLY check (no prp) for Fermat divisors of base 2. Ranges of bases can also be entered. The format is:
Quote:
 Originally Posted by GP2 Among the currently known factors, the largest n for which it fails is: 42 45 111318179143061
It does not fail? 111318179143061*2^45+1 = 222636358286122*2^44+1 is prime.

I just tested pfgw on this factor, you just have to write the factor as 111318179143061*2^45+1 in a text file.
111318179143061*2^45+1 is a Factor of F42!!!! (0.000000 seconds)

Last fiddled with by ATH on 2018-08-08 at 08:25

2018-08-08, 08:35   #267
GP2

Sep 2003

2,579 Posts

Quote:
 Originally Posted by ATH It does not fail? 111318179143061*2^45+1 = 222636358286122*2^44+1 is prime.
No, I meant it fails the condition of:

0<K<2^(n+2)

where unlike the usual "k", this K can be even, i.e., K = 2^(m-(n+2)) * k

I edited my previous post to make this clear.

Last fiddled with by GP2 on 2018-08-08 at 08:37

2018-08-08, 13:37   #268
GP2

Sep 2003

257910 Posts

Quote:
 Originally Posted by Batalov Try on the same platform and compare times: 1. download pfgw 2. put all factors in a file 3. run pfgw -N -k -l -gos2 file
I downloaded pfgw from Sourceforge, with precompiled executable pfgw64, it's version 3.8.3_20170121

I ran it on a c5d.large instance on AWS (one core, Skylake, 3.0 GHz Xeon), same platform as for the GMP program.

Here are the times:

Code:
7905*2^352281+1 is a Factor of F352279!!!! (39.253506 seconds)
vs. my GMP program = 957 sec (24 times slower)

11*2^960901+1 is a Factor of F960897!!!! (219.145612 seconds)
vs. my GMP program = 7587 sec (35 times slower)
vs. Gary pmfs = 47 minutes = 2820 sec (13 times slower) running on unknown system

193*2^3329782+1 is a Factor of F3329780!!!! (3428.316006 seconds)
vs. my GMP program = 124205 sec (36 times slower)
Very impressive.

It's amusing that pfgw gives times to the millionth of a second...

I'm a bit alarmed for Gary though...

2018-08-08, 13:44   #269
ET_
Banned

"Luigi"
Aug 2002
Team Italia

112338 Posts

Quote:
 Originally Posted by GP2 I downloaded pfgw from Sourceforge, with precompiled executable pfgw64, it's version 3.8.3_20170121 I ran it on a c5d.large instance on AWS (one core, Skylake, 3.0 GHz Xeon), same platform as for the GMP program. Here are the times: Code: 7905*2^352281+1 is a Factor of F352279!!!! (39.253506 seconds) vs. my GMP program = 957 sec (24 times slower) 11*2^960901+1 is a Factor of F960897!!!! (219.145612 seconds) vs. my GMP program = 7587 sec (35 times slower) vs. Gary pmfs = 47 minutes = 2820 sec (13 times slower) running on unknown system 193*2^3329782+1 is a Factor of F3329780!!!! (3428.316006 seconds) vs. my GMP program = 124205 sec (36 times slower) Very impressive. It's amusing that pfgw gives times to the millionth of a second... Running the precompiled statically-linked version pfgw64s made no difference. I'm a bit alarmed for Gary though...
Don't

PMFS is quite impressive for N below 30,000.

 2018-08-08, 16:02 #270 rogue     "Mark" Apr 2003 Between here and the 2×29×101 Posts Actually according to your table, the cutover from using pfms to gfndsieve+pfgw is around n = 5000.
2018-08-08, 16:06   #271
ET_
Banned

"Luigi"
Aug 2002
Team Italia

11×433 Posts

Quote:
 Originally Posted by rogue Actually according to your table, the cutover from using pfms to gfndsieve+pfgw is around n = 5000.
Sure, but pmfs can make great use of threads ;-)

Last fiddled with by ET_ on 2018-08-08 at 16:06

 2018-11-02, 02:30 #272 Gary     "Gary" Aug 2015 Texas 608 Posts A new factor from FermatSearch I would like to report the following new Fermat factor: ***** 19,789,270,693 * 2^2148 + 1 divides F2144 This was discovered today running pmfs on my Superdome X system.
2018-11-02, 10:40   #273
ET_
Banned

"Luigi"
Aug 2002
Team Italia

11·433 Posts

Quote:
 Originally Posted by Gary I would like to report the following new Fermat factor: ***** 19,789,270,693 * 2^2148 + 1 divides F2144 This was discovered today running pmfs on my Superdome X system.
Congratulations Gary!

 2018-12-19, 04:01 #274 Gary     "Gary" Aug 2015 Texas 4810 Posts One more new factor for 2018 from FermatSearch I am happy to report the following new Fermat factor: ***** 6,604,326,057 * 2^3347 + 1 divides F3345 This was discovered on Dec 13 running pmfs on my HPE Superdome X system. Wishing everyone a joyous holiday season and a Happy New Year! Last fiddled with by Gary on 2018-12-19 at 04:33
 2018-12-20, 12:13 #275 ET_ Banned     "Luigi" Aug 2002 Team Italia 11·433 Posts Two in 6 days! December 19th, 2018 New Fermat factor from FermatSearch! 1075441212722595 . 2135+1 is a Factor of F132!!! Peter Strasser discovered the seventh Fermat factor of this year! He used George Woltman's mmff program running on his home computer. Congratulations to Peter from FermatSearch, for his second factor! Moderators please modify the title: Gostin factor was found on Dec. 13th. - Last fiddled with by ET_ on 2018-12-21 at 08:43

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# CS Home Area Quotas

UTK-CS home areas have quotas from today. I was highly over quota and and had to spend a lot of time backing up by data and deleting the unused files.
Also learnt that the UNIX command for deleting a non-empty folder is
UNIX> \rm -fr dir_name
-f option for force.
-r option for recursive.
Cooked Chole today after many many days.

### 5 thoughts on “CS Home Area Quotas”

1. can anybody give me an idea how can i send free sms to a bsnl mobile withou have gsm or gprs connection, please comment urgently a girl from vadodra


text:
# Decimal Representation of Irrational Numbers

Go back to  'Irrational Numbers'

## Introduction:

Let's recap the decimal representation of rational numbers. A rational number can have two types of decimal representations (expansions):

• Terminating: It means that the decimal representation or expansion terminates after a certain number of digits.
• Non-terminating but repeating: It means that although the decimal representation has an infinite number of digits, there is a repetitive pattern to it.

When irrational numbers are written in decimal form, we get non-terminating, non-repeating representations. Let's try to understand it better.

## Non-terminating, Non-repeating Decimal Expansion:

We have already seen that any terminating or non-terminating repeating representation corresponds to a rational number. It can be shown that every non-terminating, non-repeating representation is an irrational number (though the proof is advanced). Here are two irrational numbers whose decimal representations have been given up to 10 digits:

$\sqrt 2 = 1.4142135623 \ldots$

$\pi = 3.1415926535 \ldots$

Non-terminating, Non-repeating Decimal Expansion means that although the decimal representation has an infinite number of digits, there is no pattern to it.

The ellipsis (the three dots at the end of each representation) tells us that the sequence of digits never ends. You can keep evaluating these numbers to more and more decimal digits – there will be no end to it.

What does this really mean? Does this mean that we do not know the exact value of irrational numbers as we do for rational numbers? No, that’s not correct! When we talk about an irrational number, say, $$\sqrt 2$$ or $$\pi$$, we do know the exact number (or quantity) we are talking about. For example, we can exactly construct a length of $$\sqrt 2$$ units, given a length of 1 unit. We know that the ratio of the circumference to the diameter in any circle is exactly equal to $$\pi$$.

Think: If we know these numbers exactly, why do their decimal representations seem less than exact?

⚡Tip: To understand it better, go through the exactness of decimal representations.

Numbers and Number Systems
Numbers and Number Systems
grade 9 | Questions Set 2
Numbers and Number Systems


text:
# *Special Lecture* Exponential convergence for parabolic optimal transport with boundary

-
Jun Kitagawa, Michigan State University
Fine Hall 224

*Special Lecture*

Please note the time and location

A solution of the classical optimal transport (or Monge-Kantorovich problem) can be constructed from the solution of an elliptic Monge-Amp{\`e}re equation. A natural approach is then to consider a parabolic version of this equation and try to find stationary solution as time goes to infinity.

In this talk, I will discuss an exponential convergence result for such solutions on domains with nonempty boundary. This is related to a Li-Yau Harnack type inequality, but differs from the classical case since the boundary is nonempty, our proof utilizes the optimal transport structure, and the pseudo-Riemannian optimal transport metric defined by Kim and McCann.

This talk is based on joint work with Farhan Abedin (MSU).


text:
#### Volume 8, issue 4 (2008)

 Recent Issues
 The Journal About the Journal Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Subscriptions Author Index To Appear Contacts ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
Yang–Mills theory over surfaces and the Atiyah–Segal theorem

### Daniel A Ramras

Algebraic & Geometric Topology 8 (2008) 2209–2251
##### Abstract

In this paper we explain how Morse theory for the Yang–Mills functional can be used to prove an analogue for surface groups of the Atiyah–Segal theorem. Classically, the Atiyah–Segal theorem relates the representation ring $R\left(\Gamma \right)$ of a compact Lie group $\Gamma$ to the complex $K$–theory of the classifying space $B\Gamma$. For infinite discrete groups, it is necessary to take into account deformations of representations, and with this in mind we replace the representation ring by Carlsson’s deformation $K$–theory spectrum ${K}^{def}\left(\Gamma \right)$ (the homotopy-theoretical analogue of $R\left(\Gamma \right)$). Our main theorem provides an isomorphism in homotopy ${K}_{\ast }^{def}\left({\pi }_{1}\Sigma \right)\cong {K}^{-\ast }\left(\Sigma \right)$ for all compact, aspherical surfaces $\Sigma$ and all $\ast >0$. Combining this result with work of Tyler Lawson, we obtain homotopy theoretical information about the stable moduli space of flat unitary connections over surfaces.

##### Keywords
Atiyah–Segal theorem, deformation $K$–theory, flat connection, Yang–Mills theory
##### Mathematical Subject Classification 2000
Primary: 55N15, 58E15
Secondary: 58D27, 19L41


text:
# Fahrenheit to Celsius Converter

Created by Purnima Singh, PhD
Reviewed by Kenneth Alambra
Last updated: Sep 22, 2022

Omni's Fahrenheit to Celsius converter allows you to convert the temperature values from Fahrenheit to Celsius temperature scales.

Continue reading to learn more about the different temperature scales, i.e., Fahrenheit and Celsius, and the formula for conversion between the two. You will also see how to convert Fahrenheit to Celsius using our converter.

## Fahrenheit and Celsius temperature scales

Both degree Celsius (°C) and degree Fahrenheit (°F) are units of temperature.

On the Celsius temperature scale, the freezing point of water at 1 atm pressure is 0 °C, and the boiling point of water is 100 °C. The difference between these two points is divided into 100 equal parts, each part representing 1 °C.

On the Fahrenheit temperature scale the freezing point of water is 32 °F, and the boiling point is 212 °F. The difference between these two fixed points is further divided into 180 equal parts, each division representing 1 °F.

## Formula for conversion from Fahrenheit to Celsius

Using the information mentioned in the previous section, we will write the formula to convert between Fahrenheit and Celsius temperature scales as:

\small \begin{align*} \frac{T_c - 0}{100} &= \frac{T_F - 32}{180} \\ \\ \frac{T_c}{5} &= \frac{T_F - 32}{9} \\ \end{align*}

where:

• $T_c$ - Temperature on Celsius scale; and
• $T_F$ - Temperature on Fahrenheit scale.

For example, using the formula for conversion from Fahrenheit to Celsius, we can show that 212 °F is equivalent to 100 °C:

\small \begin{align*} T_c&= \frac{212 - 32}{9} \times 5 \\ & = \frac{180}{9} \times 5 \\ & = 100 \degree \text{C} \end{align*}

## How to convert Fahrenheit to Celsius

Now let us see how we can convert Fahrenheit to Celsius using our converter:

1. Enter the temperature value on the Fahrenheit scale.

2. The converter will display the temperature on the Celsius scale.

3. You can also use this tool to convert from Celsius to Fahrenheit.

## Other temperature scale converters

If you want to convert to other temperature scales, make sure to check out our other temperature converters:

## FAQ

### How do I convert Fahrenheit to Celsius?

To convert from Fahrenheit to Celsius temperature scale, follow the given instructions:

1. Subtract 32 from the temperature value on the Fahrenheit scale.

2. Divide the result by 9 and multiply by 5.

3. Congrats! You have converted from Fahrenheit to Celsius temperature scale.

### Is there a point where Celsius and Fahrenheit are equal?

Yes, at -40, both Celsius and Fahrenheit are equal, as shown below:

1. Use the formula for conversion from Celsius and Fahrenheit, i.e., T(°C) = (T(°F) - 32)/9 × 5.

2. Substitute x in the above equation for both temperature scales, x(°C) = (x (°F) - 32)/9 × 5.

3. Solve for x, and you will get x = -40.

4. Hence both Celsius and Fahrenheit are equal at -40 °C and -40 °F.

Purnima Singh, PhD
💡 You can also convert from Celsius to Fahrenheit!
°F
°C
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text:
# Thread: help with the domain of the function

1. ## help with the domain of the function

can someone please explain to find the domain of this function.

I keep getting:

x^2(-2x+21)(-67x+60)
x^2(21/2)(60/67)

which i type in my homework & is wrong.
does anyone know what to do?

2. Originally Posted by lsnyder
can someone please explain to find the domain of this function.

I keep getting:

x^2(-2x+21)(-67x+60)
x^2(21/2)(60/67)

which i type in my homework & is wrong.
does anyone know what to do?
The only limit imposed on this function is that the denominator CANNOT be zero.

So I recommend you factorise the denominator and find out which values of x lead the denominator to be 0. The domain of the function is all values of x, apart from the 3 values of x that lead the denominator to be 0.

3. Isnt the domain of a function where you find what numbers you can use to that don't make the problem undefined?

I think you're supposed to start out by factoring the top. You can take x out and you have
x(2x^2-1x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(x-3/2)(x-4)(x-5)

(-infinity,4) u (4,5) u (5,infinity)

Sorry, I dont know latex yet.

4. Originally Posted by ninobrn99
Isnt the domain of a function where you find what numbers you can use to that don't make the problem undefined?

I think you're supposed to start out by factoring the top. You can take x out and you have
x(2x^2-1x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(x-3/2)(x-4)(x-5)

(-infinity,4) u (4,5) u (5,infinity)

Sorry, I dont know latex yet.
You don't have to factor the top. The top is irrelevant to the problem.

You just have to factorise the bottom and that gives you your limits. As ytou rightly said.

$\text{denominator} = \bigg(x - \frac{3}{2}\bigg)\bigg(x -4\bigg)\bigg(x-5\bigg)$

Hence $x \in \bigg(- \infty, \frac{3}{2} \bigg) \cup \bigg( \frac{3}{2}, 4 \bigg) \cup \bigg(4, 5 \bigg) \cup \bigg(5, \infty \bigg)$

5. Originally Posted by Mush
You don't have to factor the top. The top is irrelevant to the problem.

You just have to factorise the bottom and that gives you your limits. As ytou rightly said.

$\text{denominator} = \bigg(x - \frac{3}{2}\bigg)\bigg(x -4\bigg)\bigg(x-5\bigg)$

Hence $x \in \bigg(- \infty, \frac{3}{2} \bigg) \cup \bigg( \frac{3}{2}, 4 \bigg) \cup \bigg(4, 5 \bigg) \cup \bigg(5, \infty \bigg)$
yeah, sorry. Im just so used to factoring everything. Just creating more work for myself, but yea it is only the bottom.

6. can some one explain how to find the hole and vertical acytope please.

The root(s) of , in increasing order, is/are: , , .

(last one does not exist, I know so far.)

has one hole at the point: ( , ).

7. Originally Posted by lsnyder
can some one explain how to find the hole and vertical acytope please.

The root(s) of , in increasing order, is/are: , , .

(last one does not exist, I know so far.)

has one hole at the point: ( , ).
What do you mean by hole?

A vertical asymptote is when the curve trails off to plus or minus infinity at a certain point, because the function is not defined at that point.

Your function is not defined for $x = \frac{3}{2}, x = 4, x = 5$. So there will by 3 vertical asymptotes. Whether the curve tends towards these positively or negatively depends on a table of signs.

I have no idea what a 'hole' is though.

8. Just discovered what a hole is lol.

A hole is when the zero of the numerator is equal to the zero on the denominator.

So for this one we DO need to factorise both top and bottom, and that gives us:

$\frac{x(x+1)(2x-3)}{\big(x-\frac{3}{2}\big)(x-4)(x-5)}$

Which can be written:

$\frac{x(x+1)(2x-3)}{(2x-3)(x-4)(x-5)}$

And you can see that the function has a zero on the top and bottom when $x = \frac{3}{2}$. So that is the x co-ordinate of your hole.

But to be honest, this is a trivial matter. We already knew there would be a 'gap' in the function that point due to the fact that the VA is there. I see no use in this 'hole' business.

9. okay this might of been answered, but how do i find the other hole?

because once again, i don't get it


text:
# A couple basic conceptual questions (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### haydn

Question 1:

I was told here that in Case A, the torque is equal to zero. I'm not really seeing that. I know torque is equal to rFsin$$\theta$$. If I draw a vector from the rotation point (the very center of the spool) to the point where the force is being applied, I get that the force and r are perpendicular, so sin$$\theta$$=1. Where am I going wrong? I was told that the torque in case B is rF, which I can clearly see by the same reasoning.

Question 2:

For both cases, I was told the net force and net torque on the system is 0. The system is considered to be both of the objects. If someone could explain this I'd appreciate it.

Thanks.

#### tiny-tim

Homework Helper
"net"

I was told here that in Case A, the torque is equal to zero. I'm not really seeing that.

For both cases, I was told the net force and net torque on the system is 0. The system is considered to be both of the objects. If someone could explain this I'd appreciate it.
Hi haydn!

(i'm assuming that the friction is preventing the spool from moving)

If the spool isn't moving, then yes the net torque is zero.

The net torque is the sum of the torque from the ribbon and the torque (or moment … same thing ) from the friction force (and, technically the torque of the weight also, though if you take torque about the centre, then that will be zero )

Question 2 is similar … you use the net force or torque, which means adding all the forces or torques.

#### haydn

Thanks for the reply. I'm not sure I'm understanding your explanation completely. Please tell me if I'm right or wrong here.

For my first question, in case A, the net torque is 0 because the friction force and the force applied via the ribbon cancel each other out. So incorporating the equation t=rFsin$$\theta$$, you would plug the net force, 0 into the equation?

Then for case B on the second question though, there are two force components, friction and the force applied via the ribbon... wouldn't this result in a net force at some angle other than 90 degrees? I was told thought that for case B, torque is just equal to rF.

Thanks for the help.

#### tiny-tim

Homework Helper
I'm really not following what you're doing.

The question quite clearly tells you to start by drawing "an extended free body diagram for each case".

Have you done this? What does it show? What equations can you get from it?

#### haydn

Yes I drew free body diagrams. For case A I have a force going to the right via the ribbon and a friction force going to the left. I believe these cancel out for a net force of 0.

For case B I have a force going straight up via the ribbon and a friction force perpendicular to that, going to the left along the table. I don't get how the net torque in this case is equal to rF.

#### tiny-tim

Homework Helper
Yes I drew free body diagrams.
No you didn't …

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text:
# Determining angle to cut two boards

I am building a small roof and I need to determine how to find the distance/angle in that I need to cut two 2x4's in order to make them fit together perfectly. I've drawn a crude image below to illustrate.

Given that my base is 6', and I'd like to have the center of the roof 2' high from the base, what formula can I use to determine how long a and b will be as well as the angle that boards a and b will come together?

-
Quite possibly the single most applied question I've ever seen on this site! –  Steven Stadnicki Mar 26 '12 at 16:31
mycarpentry.com/roof-framing.html –  user996522 Mar 26 '12 at 16:33
Very nice question, clearly put. –  André Nicolas Mar 26 '12 at 17:06

If the base is $2c$ and the height is $h$, then the roof framing cross sections are shown in blue below. Note that the triangles with sides $ach$ & $bch$ are congruent (mirror images, reflections about $h$), so really $a=b$ have the same length. (We only confuse things by using two variable names for them!) Furthermore, this length is the hypotenuse of a right triangle with horizontal and vertical legs $c$ & $h$, respectively, and we can use its trigonometry to find its complementary interior acute angles, which I will call $\alpha$ and $\beta$.
Now the angle $\beta$ opposite $h$, the roof's angle, has slope (or pitch) $\frac23$, from which we get $\beta$ using the arctangent (inverse tangent) function (with your calculator's angle mode set to degrees rather than radians): $$\tan\beta=\frac{h}{c}=\frac23\qquad\implies\qquad\beta =\tan^{-1}\frac23=0.588\text{ rad}=33.69^\circ$$ and the angle you want (opposite $c$ inside each congruent triangle at the top) is $$\tan\alpha=\frac{c}{h}=\frac32\qquad\implies\qquad\alpha =\tan^{-1}\frac32=0.9828\text{ rad}=56.31^\circ.$$ That is, you need to cut an acute angle $\alpha=90^\circ-\beta$ on each piece of wood, to get a total angle of $2\alpha$ at the top of the roof. As to the lengths, they are given by the famous Pythagorean formula: \eqalign{ a^2=b^2&=h^2+c^2\\&=2^2+3^2\\&=4+9\\&=13 \\\\ a=b &=\sqrt{13} \approx 3.60555\text{ ft} \\ &\approx 3\text{ ft }7.26661\text{ in} \\ &\approx 3\text{ ft }7\tfrac4{15}\text{ in} \\ &\approx 3\text{ ft }7\tfrac14\text{ in} \\ &\approx 109.9\text{ cm} } The alternate length unit and fraction are slightly more precise, but perhaps less convenient to work with.
Assuming you want $a=b$, you can draw a right-angled triangle sides $3$ and $2$ and hypotenuse $a$. So $a = \sqrt{(3^2 + 2^2)}$. Then the angle between $a$ and the horizontal is $\arctan(2/3)$, and the angle between a and b is $2 \arctan(3/2)$


text:
# What is the degree of a polynomial in more than one variable?

I'm studying basic algebra and I'm a bit confuse about degree of equations. I read quadratic equation are equations of the second degree, meaning it contains at least one term that is squared. My question is if that squared term also can be something like $$2xy$$ (because the degree of the term is also $$2$$ (the sum of exponents of each variable): example $$2xy + x = 12$$ or can only be single variables with exponent $$x^2, y^2, \dotsc$$.

• Yes, if you have a 2-variable expression, xy is a quadratic term. The degrees of x and y are added, together they give 2. – user376343 Jun 4 '20 at 19:51
• A quadratic equation typically considers only one variable. If you have more than one variable it would be a root of a multivariable quadratic function. – Peter Foreman Jun 4 '20 at 19:52
• Is "quadratic" the common name in the case $xy$ ? Isn't it just a term with degree $2$ ? – Peter Jun 4 '20 at 19:52
• It's not unheard of. For instance $xy$ and $x^2-y^2$ are the same function up to a change of basis, and calling $x^2-y^2$ quadratic would be more common. – runway44 Jun 4 '20 at 21:53

For example, the degree of $$xy$$ is $$1+1=2$$ (since $$x$$ and $$y$$ are raised to the $$1$$st power.) Similarly, the degree of $$x^2y$$ is $$2+1=3$$, and the degree of $$xyz$$ is $$1+1+1=3$$.