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https://www.physicsforums.com/threads/lorentz-force-simplification.645133/ | 1,508,407,828,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823260.52/warc/CC-MAIN-20171019084246-20171019104246-00539.warc.gz | 959,827,561 | 14,949 | # Lorentz Force Simplification
1. Oct 18, 2012
### RobertL
Hey everyone, I'm trying to make sure I did this question right. Any ideas would be greatly appreciated. Thanks!
1. The problem statement, all variables and given/known data
The general force that bends the electron in the e/m experiment is known as the Lorentz force, F = q(E + v x B), where force F, electric field E, velocity v, and magnetic field B are vectors, and x stands for the vector product. Given that there is no electric field, and the velocity vector is at an 45° angle with respect to the magnetic field, the formula for the Lorentz force experienced by a particle carrying electric charge q is which of the following:
qvB/2
qvB
qvB/sqrt(2)
qv/B
2. Relevant equations
F = q(E + v x B)
3. The attempt at a solution
I thought, as it was at a 45 degree angle, the qvB/sqrt(2) would be accurate as a right triangle with a 45 degree angle creates a hypotenuse of sqrt(2)
2. Oct 18, 2012
### tiny-tim
Hey RobertL!
Forget about right triangles.
The cross product means sin of the angle, and so it's sin 45° which is 1/√2.
3. Oct 18, 2012
### jambaugh
Robert,
Your reasoning is correct as far as selecting the correct answer from an offered set but to get the most understanding of the material from this problem as possible, you should work on finding the answer for arbitrary numbers. If it were 33° instead can you write down the answer? Hence tiny-tim's comments. | 378 | 1,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-43 | longest | en | 0.94598 |
http://isabelle.in.tum.de/repos/isabelle/file/bdc1e2f0a86a/src/HOL/Power.thy | 1,600,659,860,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00563.warc.gz | 64,385,776 | 18,796 | src/HOL/Power.thy
author wenzelm Tue Sep 01 22:32:58 2015 +0200 (2015-09-01) changeset 61076 bdc1e2f0a86a parent 60974 6a6f15d8fbc4 child 61378 3e04c9ca001a permissions -rw-r--r--
eliminated \<Colon>;
``` 1 (* Title: HOL/Power.thy
```
``` 2 Author: Lawrence C Paulson, Cambridge University Computer Laboratory
```
``` 3 Copyright 1997 University of Cambridge
```
``` 4 *)
```
``` 5
```
``` 6 section \<open>Exponentiation\<close>
```
``` 7
```
``` 8 theory Power
```
``` 9 imports Num Equiv_Relations
```
``` 10 begin
```
``` 11
```
``` 12 context linordered_ring (* TODO: move *)
```
``` 13 begin
```
``` 14
```
``` 15 lemma sum_squares_ge_zero:
```
``` 16 "0 \<le> x * x + y * y"
```
``` 17 by (intro add_nonneg_nonneg zero_le_square)
```
``` 18
```
``` 19 lemma not_sum_squares_lt_zero:
```
``` 20 "\<not> x * x + y * y < 0"
```
``` 21 by (simp add: not_less sum_squares_ge_zero)
```
``` 22
```
``` 23 end
```
``` 24
```
``` 25 subsection \<open>Powers for Arbitrary Monoids\<close>
```
``` 26
```
``` 27 class power = one + times
```
``` 28 begin
```
``` 29
```
``` 30 primrec power :: "'a \<Rightarrow> nat \<Rightarrow> 'a" (infixr "^" 80) where
```
``` 31 power_0: "a ^ 0 = 1"
```
``` 32 | power_Suc: "a ^ Suc n = a * a ^ n"
```
``` 33
```
``` 34 notation (latex output)
```
``` 35 power ("(_\<^bsup>_\<^esup>)" [1000] 1000)
```
``` 36
```
``` 37 notation (HTML output)
```
``` 38 power ("(_\<^bsup>_\<^esup>)" [1000] 1000)
```
``` 39
```
``` 40 text \<open>Special syntax for squares.\<close>
```
``` 41
```
``` 42 abbreviation (xsymbols)
```
``` 43 power2 :: "'a \<Rightarrow> 'a" ("(_\<^sup>2)" [1000] 999) where
```
``` 44 "x\<^sup>2 \<equiv> x ^ 2"
```
``` 45
```
``` 46 notation (latex output)
```
``` 47 power2 ("(_\<^sup>2)" [1000] 999)
```
``` 48
```
``` 49 notation (HTML output)
```
``` 50 power2 ("(_\<^sup>2)" [1000] 999)
```
``` 51
```
``` 52 end
```
``` 53
```
``` 54 context monoid_mult
```
``` 55 begin
```
``` 56
```
``` 57 subclass power .
```
``` 58
```
``` 59 lemma power_one [simp]:
```
``` 60 "1 ^ n = 1"
```
``` 61 by (induct n) simp_all
```
``` 62
```
``` 63 lemma power_one_right [simp]:
```
``` 64 "a ^ 1 = a"
```
``` 65 by simp
```
``` 66
```
``` 67 lemma power_Suc0_right [simp]:
```
``` 68 "a ^ Suc 0 = a"
```
``` 69 by simp
```
``` 70
```
``` 71 lemma power_commutes:
```
``` 72 "a ^ n * a = a * a ^ n"
```
``` 73 by (induct n) (simp_all add: mult.assoc)
```
``` 74
```
``` 75 lemma power_Suc2:
```
``` 76 "a ^ Suc n = a ^ n * a"
```
``` 77 by (simp add: power_commutes)
```
``` 78
```
``` 79 lemma power_add:
```
``` 80 "a ^ (m + n) = a ^ m * a ^ n"
```
``` 81 by (induct m) (simp_all add: algebra_simps)
```
``` 82
```
``` 83 lemma power_mult:
```
``` 84 "a ^ (m * n) = (a ^ m) ^ n"
```
``` 85 by (induct n) (simp_all add: power_add)
```
``` 86
```
``` 87 lemma power2_eq_square: "a\<^sup>2 = a * a"
```
``` 88 by (simp add: numeral_2_eq_2)
```
``` 89
```
``` 90 lemma power3_eq_cube: "a ^ 3 = a * a * a"
```
``` 91 by (simp add: numeral_3_eq_3 mult.assoc)
```
``` 92
```
``` 93 lemma power_even_eq:
```
``` 94 "a ^ (2 * n) = (a ^ n)\<^sup>2"
```
``` 95 by (subst mult.commute) (simp add: power_mult)
```
``` 96
```
``` 97 lemma power_odd_eq:
```
``` 98 "a ^ Suc (2*n) = a * (a ^ n)\<^sup>2"
```
``` 99 by (simp add: power_even_eq)
```
``` 100
```
``` 101 lemma power_numeral_even:
```
``` 102 "z ^ numeral (Num.Bit0 w) = (let w = z ^ (numeral w) in w * w)"
```
``` 103 unfolding numeral_Bit0 power_add Let_def ..
```
``` 104
```
``` 105 lemma power_numeral_odd:
```
``` 106 "z ^ numeral (Num.Bit1 w) = (let w = z ^ (numeral w) in z * w * w)"
```
``` 107 unfolding numeral_Bit1 One_nat_def add_Suc_right add_0_right
```
``` 108 unfolding power_Suc power_add Let_def mult.assoc ..
```
``` 109
```
``` 110 lemma funpow_times_power:
```
``` 111 "(times x ^^ f x) = times (x ^ f x)"
```
``` 112 proof (induct "f x" arbitrary: f)
```
``` 113 case 0 then show ?case by (simp add: fun_eq_iff)
```
``` 114 next
```
``` 115 case (Suc n)
```
``` 116 def g \<equiv> "\<lambda>x. f x - 1"
```
``` 117 with Suc have "n = g x" by simp
```
``` 118 with Suc have "times x ^^ g x = times (x ^ g x)" by simp
```
``` 119 moreover from Suc g_def have "f x = g x + 1" by simp
```
``` 120 ultimately show ?case by (simp add: power_add funpow_add fun_eq_iff mult.assoc)
```
``` 121 qed
```
``` 122
```
``` 123 lemma power_commuting_commutes:
```
``` 124 assumes "x * y = y * x"
```
``` 125 shows "x ^ n * y = y * x ^n"
```
``` 126 proof (induct n)
```
``` 127 case (Suc n)
```
``` 128 have "x ^ Suc n * y = x ^ n * y * x"
```
``` 129 by (subst power_Suc2) (simp add: assms ac_simps)
```
``` 130 also have "\<dots> = y * x ^ Suc n"
```
``` 131 unfolding Suc power_Suc2
```
``` 132 by (simp add: ac_simps)
```
``` 133 finally show ?case .
```
``` 134 qed simp
```
``` 135
```
``` 136 end
```
``` 137
```
``` 138 context comm_monoid_mult
```
``` 139 begin
```
``` 140
```
``` 141 lemma power_mult_distrib [field_simps]:
```
``` 142 "(a * b) ^ n = (a ^ n) * (b ^ n)"
```
``` 143 by (induct n) (simp_all add: ac_simps)
```
``` 144
```
``` 145 end
```
``` 146
```
``` 147 text\<open>Extract constant factors from powers\<close>
```
``` 148 declare power_mult_distrib [where a = "numeral w" for w, simp]
```
``` 149 declare power_mult_distrib [where b = "numeral w" for w, simp]
```
``` 150
```
``` 151 lemma power_add_numeral [simp]:
```
``` 152 fixes a :: "'a :: monoid_mult"
```
``` 153 shows "a^numeral m * a^numeral n = a^numeral (m + n)"
```
``` 154 by (simp add: power_add [symmetric])
```
``` 155
```
``` 156 lemma power_add_numeral2 [simp]:
```
``` 157 fixes a :: "'a :: monoid_mult"
```
``` 158 shows "a^numeral m * (a^numeral n * b) = a^numeral (m + n) * b"
```
``` 159 by (simp add: mult.assoc [symmetric])
```
``` 160
```
``` 161 lemma power_mult_numeral [simp]:
```
``` 162 fixes a :: "'a :: monoid_mult"
```
``` 163 shows"(a^numeral m)^numeral n = a^numeral (m * n)"
```
``` 164 by (simp only: numeral_mult power_mult)
```
``` 165
```
``` 166 context semiring_numeral
```
``` 167 begin
```
``` 168
```
``` 169 lemma numeral_sqr: "numeral (Num.sqr k) = numeral k * numeral k"
```
``` 170 by (simp only: sqr_conv_mult numeral_mult)
```
``` 171
```
``` 172 lemma numeral_pow: "numeral (Num.pow k l) = numeral k ^ numeral l"
```
``` 173 by (induct l, simp_all only: numeral_class.numeral.simps pow.simps
```
``` 174 numeral_sqr numeral_mult power_add power_one_right)
```
``` 175
```
``` 176 lemma power_numeral [simp]: "numeral k ^ numeral l = numeral (Num.pow k l)"
```
``` 177 by (rule numeral_pow [symmetric])
```
``` 178
```
``` 179 end
```
``` 180
```
``` 181 context semiring_1
```
``` 182 begin
```
``` 183
```
``` 184 lemma of_nat_power:
```
``` 185 "of_nat (m ^ n) = of_nat m ^ n"
```
``` 186 by (induct n) (simp_all add: of_nat_mult)
```
``` 187
```
``` 188 lemma zero_power:
```
``` 189 "0 < n \<Longrightarrow> 0 ^ n = 0"
```
``` 190 by (cases n) simp_all
```
``` 191
```
``` 192 lemma power_zero_numeral [simp]:
```
``` 193 "0 ^ numeral k = 0"
```
``` 194 by (simp add: numeral_eq_Suc)
```
``` 195
```
``` 196 lemma zero_power2: "0\<^sup>2 = 0" (* delete? *)
```
``` 197 by (rule power_zero_numeral)
```
``` 198
```
``` 199 lemma one_power2: "1\<^sup>2 = 1" (* delete? *)
```
``` 200 by (rule power_one)
```
``` 201
```
``` 202 lemma power_0_Suc [simp]:
```
``` 203 "0 ^ Suc n = 0"
```
``` 204 by simp
```
``` 205
```
``` 206 text\<open>It looks plausible as a simprule, but its effect can be strange.\<close>
```
``` 207 lemma power_0_left:
```
``` 208 "0 ^ n = (if n = 0 then 1 else 0)"
```
``` 209 by (cases n) simp_all
```
``` 210
```
``` 211 end
```
``` 212
```
``` 213 context comm_semiring_1
```
``` 214 begin
```
``` 215
```
``` 216 text \<open>The divides relation\<close>
```
``` 217
```
``` 218 lemma le_imp_power_dvd:
```
``` 219 assumes "m \<le> n" shows "a ^ m dvd a ^ n"
```
``` 220 proof
```
``` 221 have "a ^ n = a ^ (m + (n - m))"
```
``` 222 using \<open>m \<le> n\<close> by simp
```
``` 223 also have "\<dots> = a ^ m * a ^ (n - m)"
```
``` 224 by (rule power_add)
```
``` 225 finally show "a ^ n = a ^ m * a ^ (n - m)" .
```
``` 226 qed
```
``` 227
```
``` 228 lemma power_le_dvd:
```
``` 229 "a ^ n dvd b \<Longrightarrow> m \<le> n \<Longrightarrow> a ^ m dvd b"
```
``` 230 by (rule dvd_trans [OF le_imp_power_dvd])
```
``` 231
```
``` 232 lemma dvd_power_same:
```
``` 233 "x dvd y \<Longrightarrow> x ^ n dvd y ^ n"
```
``` 234 by (induct n) (auto simp add: mult_dvd_mono)
```
``` 235
```
``` 236 lemma dvd_power_le:
```
``` 237 "x dvd y \<Longrightarrow> m \<ge> n \<Longrightarrow> x ^ n dvd y ^ m"
```
``` 238 by (rule power_le_dvd [OF dvd_power_same])
```
``` 239
```
``` 240 lemma dvd_power [simp]:
```
``` 241 assumes "n > (0::nat) \<or> x = 1"
```
``` 242 shows "x dvd (x ^ n)"
```
``` 243 using assms proof
```
``` 244 assume "0 < n"
```
``` 245 then have "x ^ n = x ^ Suc (n - 1)" by simp
```
``` 246 then show "x dvd (x ^ n)" by simp
```
``` 247 next
```
``` 248 assume "x = 1"
```
``` 249 then show "x dvd (x ^ n)" by simp
```
``` 250 qed
```
``` 251
```
``` 252 end
```
``` 253
```
``` 254 class semiring_1_no_zero_divisors = semiring_1 + semiring_no_zero_divisors
```
``` 255 begin
```
``` 256
```
``` 257 subclass power .
```
``` 258
```
``` 259 lemma power_eq_0_iff [simp]:
```
``` 260 "a ^ n = 0 \<longleftrightarrow> a = 0 \<and> n > 0"
```
``` 261 by (induct n) auto
```
``` 262
```
``` 263 lemma power_not_zero:
```
``` 264 "a \<noteq> 0 \<Longrightarrow> a ^ n \<noteq> 0"
```
``` 265 by (induct n) auto
```
``` 266
```
``` 267 lemma zero_eq_power2 [simp]:
```
``` 268 "a\<^sup>2 = 0 \<longleftrightarrow> a = 0"
```
``` 269 unfolding power2_eq_square by simp
```
``` 270
```
``` 271 end
```
``` 272
```
``` 273 context semidom
```
``` 274 begin
```
``` 275
```
``` 276 subclass semiring_1_no_zero_divisors ..
```
``` 277
```
``` 278 end
```
``` 279
```
``` 280 context ring_1
```
``` 281 begin
```
``` 282
```
``` 283 lemma power_minus:
```
``` 284 "(- a) ^ n = (- 1) ^ n * a ^ n"
```
``` 285 proof (induct n)
```
``` 286 case 0 show ?case by simp
```
``` 287 next
```
``` 288 case (Suc n) then show ?case
```
``` 289 by (simp del: power_Suc add: power_Suc2 mult.assoc)
```
``` 290 qed
```
``` 291
```
``` 292 lemma power_minus_Bit0:
```
``` 293 "(- x) ^ numeral (Num.Bit0 k) = x ^ numeral (Num.Bit0 k)"
```
``` 294 by (induct k, simp_all only: numeral_class.numeral.simps power_add
```
``` 295 power_one_right mult_minus_left mult_minus_right minus_minus)
```
``` 296
```
``` 297 lemma power_minus_Bit1:
```
``` 298 "(- x) ^ numeral (Num.Bit1 k) = - (x ^ numeral (Num.Bit1 k))"
```
``` 299 by (simp only: eval_nat_numeral(3) power_Suc power_minus_Bit0 mult_minus_left)
```
``` 300
```
``` 301 lemma power2_minus [simp]:
```
``` 302 "(- a)\<^sup>2 = a\<^sup>2"
```
``` 303 by (fact power_minus_Bit0)
```
``` 304
```
``` 305 lemma power_minus1_even [simp]:
```
``` 306 "(- 1) ^ (2*n) = 1"
```
``` 307 proof (induct n)
```
``` 308 case 0 show ?case by simp
```
``` 309 next
```
``` 310 case (Suc n) then show ?case by (simp add: power_add power2_eq_square)
```
``` 311 qed
```
``` 312
```
``` 313 lemma power_minus1_odd:
```
``` 314 "(- 1) ^ Suc (2*n) = -1"
```
``` 315 by simp
```
``` 316
```
``` 317 lemma power_minus_even [simp]:
```
``` 318 "(-a) ^ (2*n) = a ^ (2*n)"
```
``` 319 by (simp add: power_minus [of a])
```
``` 320
```
``` 321 end
```
``` 322
```
``` 323 context ring_1_no_zero_divisors
```
``` 324 begin
```
``` 325
```
``` 326 subclass semiring_1_no_zero_divisors ..
```
``` 327
```
``` 328 lemma power2_eq_1_iff:
```
``` 329 "a\<^sup>2 = 1 \<longleftrightarrow> a = 1 \<or> a = - 1"
```
``` 330 using square_eq_1_iff [of a] by (simp add: power2_eq_square)
```
``` 331
```
``` 332 end
```
``` 333
```
``` 334 context idom
```
``` 335 begin
```
``` 336
```
``` 337 lemma power2_eq_iff: "x\<^sup>2 = y\<^sup>2 \<longleftrightarrow> x = y \<or> x = - y"
```
``` 338 unfolding power2_eq_square by (rule square_eq_iff)
```
``` 339
```
``` 340 end
```
``` 341
```
``` 342 context algebraic_semidom
```
``` 343 begin
```
``` 344
```
``` 345 lemma div_power:
```
``` 346 assumes "b dvd a"
```
``` 347 shows "(a div b) ^ n = a ^ n div b ^ n"
```
``` 348 using assms by (induct n) (simp_all add: div_mult_div_if_dvd dvd_power_same)
```
``` 349
```
``` 350 end
```
``` 351
```
``` 352 context normalization_semidom
```
``` 353 begin
```
``` 354
```
``` 355 lemma normalize_power:
```
``` 356 "normalize (a ^ n) = normalize a ^ n"
```
``` 357 by (induct n) (simp_all add: normalize_mult)
```
``` 358
```
``` 359 lemma unit_factor_power:
```
``` 360 "unit_factor (a ^ n) = unit_factor a ^ n"
```
``` 361 by (induct n) (simp_all add: unit_factor_mult)
```
``` 362
```
``` 363 end
```
``` 364
```
``` 365 context division_ring
```
``` 366 begin
```
``` 367
```
``` 368 text\<open>Perhaps these should be simprules.\<close>
```
``` 369 lemma power_inverse [field_simps, divide_simps]:
```
``` 370 "inverse a ^ n = inverse (a ^ n)"
```
``` 371 proof (cases "a = 0")
```
``` 372 case True then show ?thesis by (simp add: power_0_left)
```
``` 373 next
```
``` 374 case False then have "inverse (a ^ n) = inverse a ^ n"
```
``` 375 by (induct n) (simp_all add: nonzero_inverse_mult_distrib power_commutes)
```
``` 376 then show ?thesis by simp
```
``` 377 qed
```
``` 378
```
``` 379 lemma power_one_over [field_simps, divide_simps]:
```
``` 380 "(1 / a) ^ n = 1 / a ^ n"
```
``` 381 using power_inverse [of a] by (simp add: divide_inverse)
```
``` 382
```
``` 383 end
```
``` 384
```
``` 385 context field
```
``` 386 begin
```
``` 387
```
``` 388 lemma power_diff:
```
``` 389 assumes nz: "a \<noteq> 0"
```
``` 390 shows "n \<le> m \<Longrightarrow> a ^ (m - n) = a ^ m / a ^ n"
```
``` 391 by (induct m n rule: diff_induct) (simp_all add: nz power_not_zero)
```
``` 392
```
``` 393 lemma power_divide [field_simps, divide_simps]:
```
``` 394 "(a / b) ^ n = a ^ n / b ^ n"
```
``` 395 by (induct n) simp_all
```
``` 396
```
``` 397 declare power_divide [where b = "numeral w" for w, simp]
```
``` 398
```
``` 399 end
```
``` 400
```
``` 401
```
``` 402 subsection \<open>Exponentiation on ordered types\<close>
```
``` 403
```
``` 404 context linordered_semidom
```
``` 405 begin
```
``` 406
```
``` 407 lemma zero_less_power [simp]:
```
``` 408 "0 < a \<Longrightarrow> 0 < a ^ n"
```
``` 409 by (induct n) simp_all
```
``` 410
```
``` 411 lemma zero_le_power [simp]:
```
``` 412 "0 \<le> a \<Longrightarrow> 0 \<le> a ^ n"
```
``` 413 by (induct n) simp_all
```
``` 414
```
``` 415 lemma power_mono:
```
``` 416 "a \<le> b \<Longrightarrow> 0 \<le> a \<Longrightarrow> a ^ n \<le> b ^ n"
```
``` 417 by (induct n) (auto intro: mult_mono order_trans [of 0 a b])
```
``` 418
```
``` 419 lemma one_le_power [simp]: "1 \<le> a \<Longrightarrow> 1 \<le> a ^ n"
```
``` 420 using power_mono [of 1 a n] by simp
```
``` 421
```
``` 422 lemma power_le_one: "\<lbrakk>0 \<le> a; a \<le> 1\<rbrakk> \<Longrightarrow> a ^ n \<le> 1"
```
``` 423 using power_mono [of a 1 n] by simp
```
``` 424
```
``` 425 lemma power_gt1_lemma:
```
``` 426 assumes gt1: "1 < a"
```
``` 427 shows "1 < a * a ^ n"
```
``` 428 proof -
```
``` 429 from gt1 have "0 \<le> a"
```
``` 430 by (fact order_trans [OF zero_le_one less_imp_le])
```
``` 431 have "1 * 1 < a * 1" using gt1 by simp
```
``` 432 also have "\<dots> \<le> a * a ^ n" using gt1
```
``` 433 by (simp only: mult_mono \<open>0 \<le> a\<close> one_le_power order_less_imp_le
```
``` 434 zero_le_one order_refl)
```
``` 435 finally show ?thesis by simp
```
``` 436 qed
```
``` 437
```
``` 438 lemma power_gt1:
```
``` 439 "1 < a \<Longrightarrow> 1 < a ^ Suc n"
```
``` 440 by (simp add: power_gt1_lemma)
```
``` 441
```
``` 442 lemma one_less_power [simp]:
```
``` 443 "1 < a \<Longrightarrow> 0 < n \<Longrightarrow> 1 < a ^ n"
```
``` 444 by (cases n) (simp_all add: power_gt1_lemma)
```
``` 445
```
``` 446 lemma power_le_imp_le_exp:
```
``` 447 assumes gt1: "1 < a"
```
``` 448 shows "a ^ m \<le> a ^ n \<Longrightarrow> m \<le> n"
```
``` 449 proof (induct m arbitrary: n)
```
``` 450 case 0
```
``` 451 show ?case by simp
```
``` 452 next
```
``` 453 case (Suc m)
```
``` 454 show ?case
```
``` 455 proof (cases n)
```
``` 456 case 0
```
``` 457 with Suc.prems Suc.hyps have "a * a ^ m \<le> 1" by simp
```
``` 458 with gt1 show ?thesis
```
``` 459 by (force simp only: power_gt1_lemma
```
``` 460 not_less [symmetric])
```
``` 461 next
```
``` 462 case (Suc n)
```
``` 463 with Suc.prems Suc.hyps show ?thesis
```
``` 464 by (force dest: mult_left_le_imp_le
```
``` 465 simp add: less_trans [OF zero_less_one gt1])
```
``` 466 qed
```
``` 467 qed
```
``` 468
```
``` 469 text\<open>Surely we can strengthen this? It holds for @{text "0<a<1"} too.\<close>
```
``` 470 lemma power_inject_exp [simp]:
```
``` 471 "1 < a \<Longrightarrow> a ^ m = a ^ n \<longleftrightarrow> m = n"
```
``` 472 by (force simp add: order_antisym power_le_imp_le_exp)
```
``` 473
```
``` 474 text\<open>Can relax the first premise to @{term "0<a"} in the case of the
```
``` 475 natural numbers.\<close>
```
``` 476 lemma power_less_imp_less_exp:
```
``` 477 "1 < a \<Longrightarrow> a ^ m < a ^ n \<Longrightarrow> m < n"
```
``` 478 by (simp add: order_less_le [of m n] less_le [of "a^m" "a^n"]
```
``` 479 power_le_imp_le_exp)
```
``` 480
```
``` 481 lemma power_strict_mono [rule_format]:
```
``` 482 "a < b \<Longrightarrow> 0 \<le> a \<Longrightarrow> 0 < n \<longrightarrow> a ^ n < b ^ n"
```
``` 483 by (induct n)
```
``` 484 (auto simp add: mult_strict_mono le_less_trans [of 0 a b])
```
``` 485
```
``` 486 text\<open>Lemma for @{text power_strict_decreasing}\<close>
```
``` 487 lemma power_Suc_less:
```
``` 488 "0 < a \<Longrightarrow> a < 1 \<Longrightarrow> a * a ^ n < a ^ n"
```
``` 489 by (induct n)
```
``` 490 (auto simp add: mult_strict_left_mono)
```
``` 491
```
``` 492 lemma power_strict_decreasing [rule_format]:
```
``` 493 "n < N \<Longrightarrow> 0 < a \<Longrightarrow> a < 1 \<longrightarrow> a ^ N < a ^ n"
```
``` 494 proof (induct N)
```
``` 495 case 0 then show ?case by simp
```
``` 496 next
```
``` 497 case (Suc N) then show ?case
```
``` 498 apply (auto simp add: power_Suc_less less_Suc_eq)
```
``` 499 apply (subgoal_tac "a * a^N < 1 * a^n")
```
``` 500 apply simp
```
``` 501 apply (rule mult_strict_mono) apply auto
```
``` 502 done
```
``` 503 qed
```
``` 504
```
``` 505 text\<open>Proof resembles that of @{text power_strict_decreasing}\<close>
```
``` 506 lemma power_decreasing [rule_format]:
```
``` 507 "n \<le> N \<Longrightarrow> 0 \<le> a \<Longrightarrow> a \<le> 1 \<longrightarrow> a ^ N \<le> a ^ n"
```
``` 508 proof (induct N)
```
``` 509 case 0 then show ?case by simp
```
``` 510 next
```
``` 511 case (Suc N) then show ?case
```
``` 512 apply (auto simp add: le_Suc_eq)
```
``` 513 apply (subgoal_tac "a * a^N \<le> 1 * a^n", simp)
```
``` 514 apply (rule mult_mono) apply auto
```
``` 515 done
```
``` 516 qed
```
``` 517
```
``` 518 lemma power_Suc_less_one:
```
``` 519 "0 < a \<Longrightarrow> a < 1 \<Longrightarrow> a ^ Suc n < 1"
```
``` 520 using power_strict_decreasing [of 0 "Suc n" a] by simp
```
``` 521
```
``` 522 text\<open>Proof again resembles that of @{text power_strict_decreasing}\<close>
```
``` 523 lemma power_increasing [rule_format]:
```
``` 524 "n \<le> N \<Longrightarrow> 1 \<le> a \<Longrightarrow> a ^ n \<le> a ^ N"
```
``` 525 proof (induct N)
```
``` 526 case 0 then show ?case by simp
```
``` 527 next
```
``` 528 case (Suc N) then show ?case
```
``` 529 apply (auto simp add: le_Suc_eq)
```
``` 530 apply (subgoal_tac "1 * a^n \<le> a * a^N", simp)
```
``` 531 apply (rule mult_mono) apply (auto simp add: order_trans [OF zero_le_one])
```
``` 532 done
```
``` 533 qed
```
``` 534
```
``` 535 text\<open>Lemma for @{text power_strict_increasing}\<close>
```
``` 536 lemma power_less_power_Suc:
```
``` 537 "1 < a \<Longrightarrow> a ^ n < a * a ^ n"
```
``` 538 by (induct n) (auto simp add: mult_strict_left_mono less_trans [OF zero_less_one])
```
``` 539
```
``` 540 lemma power_strict_increasing [rule_format]:
```
``` 541 "n < N \<Longrightarrow> 1 < a \<longrightarrow> a ^ n < a ^ N"
```
``` 542 proof (induct N)
```
``` 543 case 0 then show ?case by simp
```
``` 544 next
```
``` 545 case (Suc N) then show ?case
```
``` 546 apply (auto simp add: power_less_power_Suc less_Suc_eq)
```
``` 547 apply (subgoal_tac "1 * a^n < a * a^N", simp)
```
``` 548 apply (rule mult_strict_mono) apply (auto simp add: less_trans [OF zero_less_one] less_imp_le)
```
``` 549 done
```
``` 550 qed
```
``` 551
```
``` 552 lemma power_increasing_iff [simp]:
```
``` 553 "1 < b \<Longrightarrow> b ^ x \<le> b ^ y \<longleftrightarrow> x \<le> y"
```
``` 554 by (blast intro: power_le_imp_le_exp power_increasing less_imp_le)
```
``` 555
```
``` 556 lemma power_strict_increasing_iff [simp]:
```
``` 557 "1 < b \<Longrightarrow> b ^ x < b ^ y \<longleftrightarrow> x < y"
```
``` 558 by (blast intro: power_less_imp_less_exp power_strict_increasing)
```
``` 559
```
``` 560 lemma power_le_imp_le_base:
```
``` 561 assumes le: "a ^ Suc n \<le> b ^ Suc n"
```
``` 562 and ynonneg: "0 \<le> b"
```
``` 563 shows "a \<le> b"
```
``` 564 proof (rule ccontr)
```
``` 565 assume "~ a \<le> b"
```
``` 566 then have "b < a" by (simp only: linorder_not_le)
```
``` 567 then have "b ^ Suc n < a ^ Suc n"
```
``` 568 by (simp only: assms power_strict_mono)
```
``` 569 from le and this show False
```
``` 570 by (simp add: linorder_not_less [symmetric])
```
``` 571 qed
```
``` 572
```
``` 573 lemma power_less_imp_less_base:
```
``` 574 assumes less: "a ^ n < b ^ n"
```
``` 575 assumes nonneg: "0 \<le> b"
```
``` 576 shows "a < b"
```
``` 577 proof (rule contrapos_pp [OF less])
```
``` 578 assume "~ a < b"
```
``` 579 hence "b \<le> a" by (simp only: linorder_not_less)
```
``` 580 hence "b ^ n \<le> a ^ n" using nonneg by (rule power_mono)
```
``` 581 thus "\<not> a ^ n < b ^ n" by (simp only: linorder_not_less)
```
``` 582 qed
```
``` 583
```
``` 584 lemma power_inject_base:
```
``` 585 "a ^ Suc n = b ^ Suc n \<Longrightarrow> 0 \<le> a \<Longrightarrow> 0 \<le> b \<Longrightarrow> a = b"
```
``` 586 by (blast intro: power_le_imp_le_base antisym eq_refl sym)
```
``` 587
```
``` 588 lemma power_eq_imp_eq_base:
```
``` 589 "a ^ n = b ^ n \<Longrightarrow> 0 \<le> a \<Longrightarrow> 0 \<le> b \<Longrightarrow> 0 < n \<Longrightarrow> a = b"
```
``` 590 by (cases n) (simp_all del: power_Suc, rule power_inject_base)
```
``` 591
```
``` 592 lemma power2_le_imp_le:
```
``` 593 "x\<^sup>2 \<le> y\<^sup>2 \<Longrightarrow> 0 \<le> y \<Longrightarrow> x \<le> y"
```
``` 594 unfolding numeral_2_eq_2 by (rule power_le_imp_le_base)
```
``` 595
```
``` 596 lemma power2_less_imp_less:
```
``` 597 "x\<^sup>2 < y\<^sup>2 \<Longrightarrow> 0 \<le> y \<Longrightarrow> x < y"
```
``` 598 by (rule power_less_imp_less_base)
```
``` 599
```
``` 600 lemma power2_eq_imp_eq:
```
``` 601 "x\<^sup>2 = y\<^sup>2 \<Longrightarrow> 0 \<le> x \<Longrightarrow> 0 \<le> y \<Longrightarrow> x = y"
```
``` 602 unfolding numeral_2_eq_2 by (erule (2) power_eq_imp_eq_base) simp
```
``` 603
```
``` 604 end
```
``` 605
```
``` 606 context linordered_ring_strict
```
``` 607 begin
```
``` 608
```
``` 609 lemma sum_squares_eq_zero_iff:
```
``` 610 "x * x + y * y = 0 \<longleftrightarrow> x = 0 \<and> y = 0"
```
``` 611 by (simp add: add_nonneg_eq_0_iff)
```
``` 612
```
``` 613 lemma sum_squares_le_zero_iff:
```
``` 614 "x * x + y * y \<le> 0 \<longleftrightarrow> x = 0 \<and> y = 0"
```
``` 615 by (simp add: le_less not_sum_squares_lt_zero sum_squares_eq_zero_iff)
```
``` 616
```
``` 617 lemma sum_squares_gt_zero_iff:
```
``` 618 "0 < x * x + y * y \<longleftrightarrow> x \<noteq> 0 \<or> y \<noteq> 0"
```
``` 619 by (simp add: not_le [symmetric] sum_squares_le_zero_iff)
```
``` 620
```
``` 621 end
```
``` 622
```
``` 623 context linordered_idom
```
``` 624 begin
```
``` 625
```
``` 626 lemma power_abs:
```
``` 627 "abs (a ^ n) = abs a ^ n"
```
``` 628 by (induct n) (auto simp add: abs_mult)
```
``` 629
```
``` 630 lemma abs_power_minus [simp]:
```
``` 631 "abs ((-a) ^ n) = abs (a ^ n)"
```
``` 632 by (simp add: power_abs)
```
``` 633
```
``` 634 lemma zero_less_power_abs_iff [simp]:
```
``` 635 "0 < abs a ^ n \<longleftrightarrow> a \<noteq> 0 \<or> n = 0"
```
``` 636 proof (induct n)
```
``` 637 case 0 show ?case by simp
```
``` 638 next
```
``` 639 case (Suc n) show ?case by (auto simp add: Suc zero_less_mult_iff)
```
``` 640 qed
```
``` 641
```
``` 642 lemma zero_le_power_abs [simp]:
```
``` 643 "0 \<le> abs a ^ n"
```
``` 644 by (rule zero_le_power [OF abs_ge_zero])
```
``` 645
```
``` 646 lemma zero_le_power2 [simp]:
```
``` 647 "0 \<le> a\<^sup>2"
```
``` 648 by (simp add: power2_eq_square)
```
``` 649
```
``` 650 lemma zero_less_power2 [simp]:
```
``` 651 "0 < a\<^sup>2 \<longleftrightarrow> a \<noteq> 0"
```
``` 652 by (force simp add: power2_eq_square zero_less_mult_iff linorder_neq_iff)
```
``` 653
```
``` 654 lemma power2_less_0 [simp]:
```
``` 655 "\<not> a\<^sup>2 < 0"
```
``` 656 by (force simp add: power2_eq_square mult_less_0_iff)
```
``` 657
```
``` 658 lemma power2_less_eq_zero_iff [simp]:
```
``` 659 "a\<^sup>2 \<le> 0 \<longleftrightarrow> a = 0"
```
``` 660 by (simp add: le_less)
```
``` 661
```
``` 662 lemma abs_power2 [simp]:
```
``` 663 "abs (a\<^sup>2) = a\<^sup>2"
```
``` 664 by (simp add: power2_eq_square abs_mult abs_mult_self)
```
``` 665
```
``` 666 lemma power2_abs [simp]:
```
``` 667 "(abs a)\<^sup>2 = a\<^sup>2"
```
``` 668 by (simp add: power2_eq_square abs_mult_self)
```
``` 669
```
``` 670 lemma odd_power_less_zero:
```
``` 671 "a < 0 \<Longrightarrow> a ^ Suc (2*n) < 0"
```
``` 672 proof (induct n)
```
``` 673 case 0
```
``` 674 then show ?case by simp
```
``` 675 next
```
``` 676 case (Suc n)
```
``` 677 have "a ^ Suc (2 * Suc n) = (a*a) * a ^ Suc(2*n)"
```
``` 678 by (simp add: ac_simps power_add power2_eq_square)
```
``` 679 thus ?case
```
``` 680 by (simp del: power_Suc add: Suc mult_less_0_iff mult_neg_neg)
```
``` 681 qed
```
``` 682
```
``` 683 lemma odd_0_le_power_imp_0_le:
```
``` 684 "0 \<le> a ^ Suc (2*n) \<Longrightarrow> 0 \<le> a"
```
``` 685 using odd_power_less_zero [of a n]
```
``` 686 by (force simp add: linorder_not_less [symmetric])
```
``` 687
```
``` 688 lemma zero_le_even_power'[simp]:
```
``` 689 "0 \<le> a ^ (2*n)"
```
``` 690 proof (induct n)
```
``` 691 case 0
```
``` 692 show ?case by simp
```
``` 693 next
```
``` 694 case (Suc n)
```
``` 695 have "a ^ (2 * Suc n) = (a*a) * a ^ (2*n)"
```
``` 696 by (simp add: ac_simps power_add power2_eq_square)
```
``` 697 thus ?case
```
``` 698 by (simp add: Suc zero_le_mult_iff)
```
``` 699 qed
```
``` 700
```
``` 701 lemma sum_power2_ge_zero:
```
``` 702 "0 \<le> x\<^sup>2 + y\<^sup>2"
```
``` 703 by (intro add_nonneg_nonneg zero_le_power2)
```
``` 704
```
``` 705 lemma not_sum_power2_lt_zero:
```
``` 706 "\<not> x\<^sup>2 + y\<^sup>2 < 0"
```
``` 707 unfolding not_less by (rule sum_power2_ge_zero)
```
``` 708
```
``` 709 lemma sum_power2_eq_zero_iff:
```
``` 710 "x\<^sup>2 + y\<^sup>2 = 0 \<longleftrightarrow> x = 0 \<and> y = 0"
```
``` 711 unfolding power2_eq_square by (simp add: add_nonneg_eq_0_iff)
```
``` 712
```
``` 713 lemma sum_power2_le_zero_iff:
```
``` 714 "x\<^sup>2 + y\<^sup>2 \<le> 0 \<longleftrightarrow> x = 0 \<and> y = 0"
```
``` 715 by (simp add: le_less sum_power2_eq_zero_iff not_sum_power2_lt_zero)
```
``` 716
```
``` 717 lemma sum_power2_gt_zero_iff:
```
``` 718 "0 < x\<^sup>2 + y\<^sup>2 \<longleftrightarrow> x \<noteq> 0 \<or> y \<noteq> 0"
```
``` 719 unfolding not_le [symmetric] by (simp add: sum_power2_le_zero_iff)
```
``` 720
```
``` 721 lemma abs_le_square_iff:
```
``` 722 "\<bar>x\<bar> \<le> \<bar>y\<bar> \<longleftrightarrow> x\<^sup>2 \<le> y\<^sup>2"
```
``` 723 proof
```
``` 724 assume "\<bar>x\<bar> \<le> \<bar>y\<bar>"
```
``` 725 then have "\<bar>x\<bar>\<^sup>2 \<le> \<bar>y\<bar>\<^sup>2" by (rule power_mono, simp)
```
``` 726 then show "x\<^sup>2 \<le> y\<^sup>2" by simp
```
``` 727 next
```
``` 728 assume "x\<^sup>2 \<le> y\<^sup>2"
```
``` 729 then show "\<bar>x\<bar> \<le> \<bar>y\<bar>"
```
``` 730 by (auto intro!: power2_le_imp_le [OF _ abs_ge_zero])
```
``` 731 qed
```
``` 732
```
``` 733 lemma abs_square_le_1:"x\<^sup>2 \<le> 1 \<longleftrightarrow> abs(x) \<le> 1"
```
``` 734 using abs_le_square_iff [of x 1]
```
``` 735 by simp
```
``` 736
```
``` 737 lemma abs_square_eq_1: "x\<^sup>2 = 1 \<longleftrightarrow> abs(x) = 1"
```
``` 738 by (auto simp add: abs_if power2_eq_1_iff)
```
``` 739
```
``` 740 lemma abs_square_less_1: "x\<^sup>2 < 1 \<longleftrightarrow> abs(x) < 1"
```
``` 741 using abs_square_eq_1 [of x] abs_square_le_1 [of x]
```
``` 742 by (auto simp add: le_less)
```
``` 743
```
``` 744 end
```
``` 745
```
``` 746
```
``` 747 subsection \<open>Miscellaneous rules\<close>
```
``` 748
```
``` 749 lemma (in linordered_semidom) self_le_power:
```
``` 750 "1 \<le> a \<Longrightarrow> 0 < n \<Longrightarrow> a \<le> a ^ n"
```
``` 751 using power_increasing [of 1 n a] power_one_right [of a] by auto
```
``` 752
```
``` 753 lemma (in power) power_eq_if:
```
``` 754 "p ^ m = (if m=0 then 1 else p * (p ^ (m - 1)))"
```
``` 755 unfolding One_nat_def by (cases m) simp_all
```
``` 756
```
``` 757 lemma (in comm_semiring_1) power2_sum:
```
``` 758 "(x + y)\<^sup>2 = x\<^sup>2 + y\<^sup>2 + 2 * x * y"
```
``` 759 by (simp add: algebra_simps power2_eq_square mult_2_right)
```
``` 760
```
``` 761 lemma (in comm_ring_1) power2_diff:
```
``` 762 "(x - y)\<^sup>2 = x\<^sup>2 + y\<^sup>2 - 2 * x * y"
```
``` 763 by (simp add: algebra_simps power2_eq_square mult_2_right)
```
``` 764
```
``` 765 lemma (in comm_ring_1) power2_commute:
```
``` 766 "(x - y)\<^sup>2 = (y - x)\<^sup>2"
```
``` 767 by (simp add: algebra_simps power2_eq_square)
```
``` 768
```
``` 769
```
``` 770 text \<open>Simprules for comparisons where common factors can be cancelled.\<close>
```
``` 771
```
``` 772 lemmas zero_compare_simps =
```
``` 773 add_strict_increasing add_strict_increasing2 add_increasing
```
``` 774 zero_le_mult_iff zero_le_divide_iff
```
``` 775 zero_less_mult_iff zero_less_divide_iff
```
``` 776 mult_le_0_iff divide_le_0_iff
```
``` 777 mult_less_0_iff divide_less_0_iff
```
``` 778 zero_le_power2 power2_less_0
```
``` 779
```
``` 780
```
``` 781 subsection \<open>Exponentiation for the Natural Numbers\<close>
```
``` 782
```
``` 783 lemma nat_one_le_power [simp]:
```
``` 784 "Suc 0 \<le> i \<Longrightarrow> Suc 0 \<le> i ^ n"
```
``` 785 by (rule one_le_power [of i n, unfolded One_nat_def])
```
``` 786
```
``` 787 lemma nat_zero_less_power_iff [simp]:
```
``` 788 "x ^ n > 0 \<longleftrightarrow> x > (0::nat) \<or> n = 0"
```
``` 789 by (induct n) auto
```
``` 790
```
``` 791 lemma nat_power_eq_Suc_0_iff [simp]:
```
``` 792 "x ^ m = Suc 0 \<longleftrightarrow> m = 0 \<or> x = Suc 0"
```
``` 793 by (induct m) auto
```
``` 794
```
``` 795 lemma power_Suc_0 [simp]:
```
``` 796 "Suc 0 ^ n = Suc 0"
```
``` 797 by simp
```
``` 798
```
``` 799 text\<open>Valid for the naturals, but what if @{text"0<i<1"}?
```
``` 800 Premises cannot be weakened: consider the case where @{term "i=0"},
```
``` 801 @{term "m=1"} and @{term "n=0"}.\<close>
```
``` 802 lemma nat_power_less_imp_less:
```
``` 803 assumes nonneg: "0 < (i::nat)"
```
``` 804 assumes less: "i ^ m < i ^ n"
```
``` 805 shows "m < n"
```
``` 806 proof (cases "i = 1")
```
``` 807 case True with less power_one [where 'a = nat] show ?thesis by simp
```
``` 808 next
```
``` 809 case False with nonneg have "1 < i" by auto
```
``` 810 from power_strict_increasing_iff [OF this] less show ?thesis ..
```
``` 811 qed
```
``` 812
```
``` 813 lemma power_dvd_imp_le:
```
``` 814 "i ^ m dvd i ^ n \<Longrightarrow> (1::nat) < i \<Longrightarrow> m \<le> n"
```
``` 815 apply (rule power_le_imp_le_exp, assumption)
```
``` 816 apply (erule dvd_imp_le, simp)
```
``` 817 done
```
``` 818
```
``` 819 lemma power2_nat_le_eq_le:
```
``` 820 fixes m n :: nat
```
``` 821 shows "m\<^sup>2 \<le> n\<^sup>2 \<longleftrightarrow> m \<le> n"
```
``` 822 by (auto intro: power2_le_imp_le power_mono)
```
``` 823
```
``` 824 lemma power2_nat_le_imp_le:
```
``` 825 fixes m n :: nat
```
``` 826 assumes "m\<^sup>2 \<le> n"
```
``` 827 shows "m \<le> n"
```
``` 828 proof (cases m)
```
``` 829 case 0 then show ?thesis by simp
```
``` 830 next
```
``` 831 case (Suc k)
```
``` 832 show ?thesis
```
``` 833 proof (rule ccontr)
```
``` 834 assume "\<not> m \<le> n"
```
``` 835 then have "n < m" by simp
```
``` 836 with assms Suc show False
```
``` 837 by (simp add: power2_eq_square)
```
``` 838 qed
```
``` 839 qed
```
``` 840
```
``` 841 subsubsection \<open>Cardinality of the Powerset\<close>
```
``` 842
```
``` 843 lemma card_UNIV_bool [simp]: "card (UNIV :: bool set) = 2"
```
``` 844 unfolding UNIV_bool by simp
```
``` 845
```
``` 846 lemma card_Pow: "finite A \<Longrightarrow> card (Pow A) = 2 ^ card A"
```
``` 847 proof (induct rule: finite_induct)
```
``` 848 case empty
```
``` 849 show ?case by auto
```
``` 850 next
```
``` 851 case (insert x A)
```
``` 852 then have "inj_on (insert x) (Pow A)"
```
``` 853 unfolding inj_on_def by (blast elim!: equalityE)
```
``` 854 then have "card (Pow A) + card (insert x ` Pow A) = 2 * 2 ^ card A"
```
``` 855 by (simp add: mult_2 card_image Pow_insert insert.hyps)
```
``` 856 then show ?case using insert
```
``` 857 apply (simp add: Pow_insert)
```
``` 858 apply (subst card_Un_disjoint, auto)
```
``` 859 done
```
``` 860 qed
```
``` 861
```
``` 862
```
``` 863 subsubsection \<open>Generalized sum over a set\<close>
```
``` 864
```
``` 865 lemma setsum_zero_power [simp]:
```
``` 866 fixes c :: "nat \<Rightarrow> 'a::division_ring"
```
``` 867 shows "(\<Sum>i\<in>A. c i * 0^i) = (if finite A \<and> 0 \<in> A then c 0 else 0)"
```
``` 868 apply (cases "finite A")
```
``` 869 by (induction A rule: finite_induct) auto
```
``` 870
```
``` 871 lemma setsum_zero_power' [simp]:
```
``` 872 fixes c :: "nat \<Rightarrow> 'a::field"
```
``` 873 shows "(\<Sum>i\<in>A. c i * 0^i / d i) = (if finite A \<and> 0 \<in> A then c 0 / d 0 else 0)"
```
``` 874 using setsum_zero_power [of "\<lambda>i. c i / d i" A]
```
``` 875 by auto
```
``` 876
```
``` 877
```
``` 878 subsubsection \<open>Generalized product over a set\<close>
```
``` 879
```
``` 880 lemma setprod_constant: "finite A ==> (\<Prod>x\<in> A. (y::'a::{comm_monoid_mult})) = y^(card A)"
```
``` 881 apply (erule finite_induct)
```
``` 882 apply auto
```
``` 883 done
```
``` 884
```
``` 885 lemma setprod_power_distrib:
```
``` 886 fixes f :: "'a \<Rightarrow> 'b::comm_semiring_1"
```
``` 887 shows "setprod f A ^ n = setprod (\<lambda>x. (f x) ^ n) A"
```
``` 888 proof (cases "finite A")
```
``` 889 case True then show ?thesis
```
``` 890 by (induct A rule: finite_induct) (auto simp add: power_mult_distrib)
```
``` 891 next
```
``` 892 case False then show ?thesis
```
``` 893 by simp
```
``` 894 qed
```
``` 895
```
``` 896 lemma power_setsum:
```
``` 897 "c ^ (\<Sum>a\<in>A. f a) = (\<Prod>a\<in>A. c ^ f a)"
```
``` 898 by (induct A rule: infinite_finite_induct) (simp_all add: power_add)
```
``` 899
```
``` 900 lemma setprod_gen_delta:
```
``` 901 assumes fS: "finite S"
```
``` 902 shows "setprod (\<lambda>k. if k=a then b k else c) S = (if a \<in> S then (b a ::'a::comm_monoid_mult) * c^ (card S - 1) else c^ card S)"
```
``` 903 proof-
```
``` 904 let ?f = "(\<lambda>k. if k=a then b k else c)"
```
``` 905 {assume a: "a \<notin> S"
```
``` 906 hence "\<forall> k\<in> S. ?f k = c" by simp
```
``` 907 hence ?thesis using a setprod_constant[OF fS, of c] by simp }
```
``` 908 moreover
```
``` 909 {assume a: "a \<in> S"
```
``` 910 let ?A = "S - {a}"
```
``` 911 let ?B = "{a}"
```
``` 912 have eq: "S = ?A \<union> ?B" using a by blast
```
``` 913 have dj: "?A \<inter> ?B = {}" by simp
```
``` 914 from fS have fAB: "finite ?A" "finite ?B" by auto
```
``` 915 have fA0:"setprod ?f ?A = setprod (\<lambda>i. c) ?A"
```
``` 916 apply (rule setprod.cong) by auto
```
``` 917 have cA: "card ?A = card S - 1" using fS a by auto
```
``` 918 have fA1: "setprod ?f ?A = c ^ card ?A" unfolding fA0 apply (rule setprod_constant) using fS by auto
```
``` 919 have "setprod ?f ?A * setprod ?f ?B = setprod ?f S"
```
``` 920 using setprod.union_disjoint[OF fAB dj, of ?f, unfolded eq[symmetric]]
```
``` 921 by simp
```
``` 922 then have ?thesis using a cA
```
``` 923 by (simp add: fA1 field_simps cong add: setprod.cong cong del: if_weak_cong)}
```
``` 924 ultimately show ?thesis by blast
```
``` 925 qed
```
``` 926
```
``` 927 subsection \<open>Code generator tweak\<close>
```
``` 928
```
``` 929 code_identifier
```
``` 930 code_module Power \<rightharpoonup> (SML) Arith and (OCaml) Arith and (Haskell) Arith
```
``` 931
```
``` 932 end
``` | 14,840 | 39,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.535854 |
https://au.hisvoicetoday.org/3289-can-honeybees-learn-math-research-suggests-they-can.html | 1,624,504,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00399.warc.gz | 111,260,945 | 11,911 | # Can Honeybees Learn Math? Research Suggests They Can
In a new report, researchers from RMIT University say they have taught honeybees how to perform the arithmetic operations addition and subtraction, a remarkable feat for creatures whose brains have fewer than 1 million neurons.
## How Do You Teach a Honeybee?
The way we learn to perform arithmetic operations as children requires us to learn what the symbolic operators (+) and (-) represent. A (+) means adds two together to form a new number, while (-) means subtract one number from the other to get a new number.
Researchers used colored shaped to represent these same concepts in a way that a honeybee could see them and set up a Y-shaped box for the bees to navigate for sugar water. At the opening, there was a symbol indicating the operation to perform, either to add 1 or to subtract 1 from the number presented by the shape.
The bees would then choose either the left or right branch where two answers were represented. The correct answer contained high-sugar content liquid for the bees to gorge on and bring back to their hive. The incorrect answers contained a bitter liquid instead.
Over time, bees began to navigate more toward the correct answers, which researchers would switch between the branches to ensure the bees couldn’t learn where the sugar water was located. They also switched up the number to be added and subtracted, so that the bees couldn’t be using some other pattern to find the sugar water.
The only way they could do that is if they’d begun to understand what the shapes meant and correctly deduced the operation needed to access the sugar. Moreover, they began to actually perform the arithmetic.
According to the researchers, “During testing with a novel number, bees were correct in addition and subtraction of one element 64-72% of the time. The bee’s performance on tests was significantly different than what we would expect if bees were choosing randomly, called chance level performance (50% correct/incorrect).”
“Thus, our ‘bee school’ within the Y-maze allowed the bees to learn how to use arithmetic operators to add or subtract.”
## Two Levels of Processing Needed for Math
Why this is important is that arithmetic—even simple addition and subtraction—requires two levels of processing in the brain, one requiring the bees to understand the numerical values while the second requires the bees to work with the numbers mentally in their working memory to find the correct answer.
Moreover, the bees had to work with the numerical value to be added or subtracted when it was not visually present, so they had to recall the number they had seen previously. This required the bees to “abstract” the value they needed to add or subtract, a key signifier of symbolic thinking that is the bedrock of higher level intelligence.
“[O]ur findings show that the understanding of maths symbols as a language with operators is something that many brains can probably achieve, and helps explain how many human cultures independently developed numeracy skills,” they concluded. | 607 | 3,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-25 | latest | en | 0.962919 |
https://metanumbers.com/17231 | 1,632,296,597,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00371.warc.gz | 438,789,498 | 7,279 | # 17231 (number)
17,231 (seventeen thousand two hundred thirty-one) is an odd five-digits prime number following 17230 and preceding 17232. In scientific notation, it is written as 1.7231 × 104. The sum of its digits is 14. It has a total of 1 prime factor and 2 positive divisors. There are 17,230 positive integers (up to 17231) that are relatively prime to 17231.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 14
• Digital Root 5
## Name
Short name 17 thousand 231 seventeen thousand two hundred thirty-one
## Notation
Scientific notation 1.7231 × 104 17.231 × 103
## Prime Factorization of 17231
Prime Factorization 17231
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 17231 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 9.75447 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 17,231 is 17231. Since it has a total of 1 prime factor, 17,231 is a prime number.
## Divisors of 17231
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 17232 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 8616 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 131.267 Returns the nth root of the product of n divisors H(n) 1.99988 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 17,231 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 17,231) is 17,232, the average is 8,616.
## Other Arithmetic Functions (n = 17231)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 17230 Total number of positive integers not greater than n that are coprime to n λ(n) 17230 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1986 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 17,230 positive integers (less than 17,231) that are coprime with 17,231. And there are approximately 1,986 prime numbers less than or equal to 17,231.
## Divisibility of 17231
m n mod m 2 3 4 5 6 7 8 9 1 2 3 1 5 4 7 5
17,231 is not divisible by any number less than or equal to 9.
## Classification of 17231
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (17231)
Base System Value
2 Binary 100001101001111
3 Ternary 212122012
4 Quaternary 10031033
5 Quinary 1022411
6 Senary 211435
8 Octal 41517
10 Decimal 17231
12 Duodecimal 9b7b
20 Vigesimal 231b
36 Base36 dan
## Basic calculations (n = 17231)
### Multiplication
n×y
n×2 34462 51693 68924 86155
### Division
n÷y
n÷2 8615.5 5743.67 4307.75 3446.2
### Exponentiation
ny
n2 296907361 5116010737391 88153981015984321 1518981246886425835151
### Nth Root
y√n
2√n 131.267 25.8288 11.4572 7.03497
## 17231 as geometric shapes
### Circle
Diameter 34462 108266 9.32762e+08
### Sphere
Volume 2.14299e+13 3.73105e+09 108266
### Square
Length = n
Perimeter 68924 2.96907e+08 24368.3
### Cube
Length = n
Surface area 1.78144e+09 5.11601e+12 29845
### Equilateral Triangle
Length = n
Perimeter 51693 1.28565e+08 14922.5
### Triangular Pyramid
Length = n
Surface area 5.14259e+08 6.02928e+11 14069.1
## Cryptographic Hash Functions
md5 a9b67cca413479cd258a8b3687aa693e 04bd392f228f637be35570351338d6a1837d2aeb 794929e42589f3125db6f231232d471d89bc7631f3e760d9ef7ce3827053afa9 0ac48f5a841635977b2fca33cb9b84f4e2394fd25fcfa979060702f2ced0fb563acc32327a3a3492602d2035593b7a97165457530395a2000b220404742819d0 025761ca92d3a3b2065a018be3157ae661742489 | 1,427 | 4,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-39 | latest | en | 0.821585 |
https://mathpro.weebly.com/sample-quiz1.html | 1,524,433,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00021.warc.gz | 648,400,701 | 4,764 | ## Sample Quiz
1.(3)2 + b2 = (6)2 2. (5)2 + (7)2 = c2 3. a2 + (1)2 = (10)2 | 50 | 81 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-17 | longest | en | 0.169579 |
https://www.rcspoint.com/7pw6dv/333d3f-double-line-graphs-worksheets | 1,627,724,261,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00069.warc.gz | 1,015,133,507 | 8,193 | Try some of these worksheets for free! Bar Graphs Maths Class 5 Cbse Icse Youtube . Please update your bookmarks accordingly. Exercises to draw line graphs and double line graphs with a suitable scale; labeling the axes; giving a title for the graph and more are included in these printable worksheets for students of grade 2 through grade 6. Navigate through our free printable line graph worksheets to read, interpret, and draw a line graph from the data provided. Their schools of grammar and rhetoric attracted students from Rome itself. 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Double Line Graph Lesson Plans Worksheets Reviewed By Teachers. Feb 13, 2020 - Doubles Rap Printable - 25 Doubles Rap Printable , Blank Bar Graph Printable â sociallawbook More information Doubles Rap Printable Printable Line Graph Worksheets â Primalvape From Rome itself was pleased with Jamil 's story and rewarded him a! Students solve 5 short answer and graphing problems the double line graphs Sequencing worksheets graphing Comprehension!, think something about his worldly prospects and graphs Sequencing worksheets graphing Activities Comprehension worksheets school worksheets Grade... And rhetoric attracted students from Rome itself the classroom only Graph and analyze how the data from the and! Off the coverings and the blue broken line represents school 1, and the broken... How the data from the graphs to download/print, click on pop-out icon or print icon to Worksheet to or... Represents school 1, and the Princess sprang to her feet Comprehension Graph the given as... Mean, median, mode, and the Princess sprang to her feet off the coverings and Princess. Add selected questions to a set of data presented in a double line learning! True, think something about his worldly prospects Y^ Month Lesson Worksheet q1: the following line graphs the!, interpret and compare the data from the graphs and label the axes as well to print or.. Drawing line graphs analyze how the data changes over a period of time with help... Answering 4 questions about a given data set members then divided between the! & download or print using the browser document reader options the Princess sprang to her feet Month Lesson:. Including but NOT limited to Teacher web pages of two related sets of data presented in double... Worksheets filing cabinet to save their favorite worksheets this data activity, 6th graders read definitions! Kids name the graphs types of charts and graphs compare the average number of first graders in two different for... Set of 2 double line Graph worksheets are copyright material and are for... ; Scales count By 10sand 2s ; Grades 2-3 collection of line Graph - top!, click on pop-out icon or print icon to Worksheet to print or download given as... Is their story, O Commander of the Faithful 5th Grade Math for! Students solve 5 short answer and graphing problems understanding the equivalence of two related sets of data in! Represents school 1, and the blue broken line represents school 2 worksheets... School worksheets 5th Grade worksheets Sequencing worksheets graphing Activities Comprehension worksheets school worksheets 5th Grade worksheets Sequencing worksheets double line graphs worksheets Comprehension! Ample practice skills to analyze, interpret and compare the average number first... For 4th Grade was merely glad to be rid of him learning exercise, solve! Button before moving to another page school worksheets 5th Grade Math worksheets double line Graph worksheets have ample skills... The Faithful perfect and present Perfct Continuous, Long Division no Remainder 2 Digit Dividend draw a line.. Students from Rome itself to save their favorite worksheets 4th Grade was merely glad to be of! 1: Student will create double line graphs Worksheet is suitable for 6th Grade as... Age 8 ; Scales count By 10sand 2s ; Grades 2-3 Graph Plans... Read how to work with different types of charts and graphs definitions of mean, median,,. | 3,076 | 14,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.90862 |
https://www.physicsforums.com/threads/does-a-refl-anti-symm-relation-on-a-set-a-have-this-property.785355/ | 1,529,902,284,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00225.warc.gz | 886,551,155 | 15,007 | # Homework Help: Does a refl/anti-symm relation on a set A have this property?
1. Dec 2, 2014
### pandaBee
1. The problem statement, all variables and given/known data
Let $R$ be an ordered relation on a set $A$ that is reflexive and anti-symmetric.
If there is a chain of elements in $R$ that begins and ends with the same element, say the element $x \in A$ is it true that all the elements of $R$ sandwiched in between the ones beginning and ending in $x$ are equal to each other?
2. Relevant equations
For clarification Let the 'chain' of elements in $R$ be the following:
$(x,b_0), (b_0,b_1), ..., (b_n, x)$
let $Z = \{b_0, b_1, ..., b_n\}$
If $R$ is a reflexive and anti-symmetric ordered relation are all elements of $Z$ equivalent?
3. The attempt at a solution
I came up with a few antisymmetric, reflexive ordered relations and tested the validity of this argument and it passed in each one so I am guessing that the theorem above may be true since I have not found a counter-example yet. I just have no idea how to express it in a proof. The examples I came up with were:
The set $G=\{(x,y) \in P\text{ x }P~ |~ x \ \text{genes come from}\ y\}$ where $P$ is the set of all people,
The set $D=\{(x,y) \in ℤ\text{ x }ℤ ~|~x \ \text{divides}\ y\}$
And also the standard less than or equal to ordered relation.
In every case it was implied that if a element of the set under which the relation is ordered under begins and ends a chain of ordered pairs in the ordered relation, then all of the elements are equivalent to one another, as a consequence of anti-symmetry and reflexivity.
I'm unsure of how to start this proof. I've tried defining a set containing all the intermediate elements (in this case being the $b_j$'s in the set $Z$ and assuming that there is at least two elements that are not equal to one another in the set and trying to find a contradiction, but I ran into a dead end there as well. Any suggestions? Or perhaps someone can think of a counter example ruling out the theorem completely?
Last edited: Dec 2, 2014
2. Dec 3, 2014
### RUber
I think in general, you only need to show that it is true for any element y such that x R y R x implies that x=y.
I do not have a clear mental image of a rigorous proof, but I would start with the basics:
reflexive means that x R x for all x.
anti-symmetric means that x R y implies y ~R x.
So, x ~R x. If y R x and y ~R x, does this mean y = x?
Also, I think ordering is useful since there should not be anything between x and x in an ordered relation other than more x. | 687 | 2,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-26 | latest | en | 0.939765 |
https://mathspace.co/textbooks/syllabuses/Syllabus-408/topics/Topic-7232/subtopics/Subtopic-96642/?activeTab=interactive | 1,719,034,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00412.warc.gz | 336,173,904 | 52,624 | # Mixed operations with algebraic terms (incl negatives)
## Interactive practice questions
Simplify the expression:
$5m+9m+10m$5m+9m+10m
Easy
< 1min
Simplify the expression:
$4x-10x$4x10x
Easy
< 1min
Simplify the expression $q\times9\times p$q×9×p.
Easy
< 1min
Simplify the expression $\frac{20n}{5}$20n5.
Easy
< 1min | 119 | 328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.745988 |
http://numerical--analysis.blogspot.com/2010/02/faddeev-leverrier-method.html | 1,502,983,563,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103579.21/warc/CC-MAIN-20170817151157-20170817171157-00319.warc.gz | 310,372,174 | 13,449 | ## Monday, February 1, 2010
Let be an n × n matrix. The determination of eigenvalues and eigenvectors requires the solution of
(1)
where is the eigenvalue corresponding to the eigenvector . The values must satisfy the equation
(2) .
Hence is a root of an nth degree polynomial , which we write in the form
(3) .
The Faddeev-Leverrier algorithm is an efficient method for finding the coefficients of the polynomial . As an additional benefit, the inverse matrix is obtained at no extra computational expense.
Recall that the trace of the matrix , written , is
(4) .
The algorithm generates a sequence of matrices and uses their traces to compute the coefficients of ,
(5)
Then the characteristic polynomial is given by
(6) .
In addition, the inverse matrix is given by
(7) . | 182 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-34 | longest | en | 0.901001 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=29&t=31830 | 1,566,715,339,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323221.23/warc/CC-MAIN-20190825062944-20190825084944-00255.warc.gz | 544,442,441 | 12,712 | ## Formal charge?
Alondra Juarez section 1E
Posts: 47
Joined: Fri Sep 29, 2017 7:03 am
### Formal charge?
What is formal charge and how do we calculate it?
Kuldeep Gill 1H
Posts: 44
Joined: Fri Apr 06, 2018 11:02 am
Been upvoted: 1 time
### Re: Formal charge?
Hi, you take the number of valence electrons minus half the number of shared electrons minus the number of unshared electrons. This will give you the formal charge of that atom!
sylvie_1D
Posts: 31
Joined: Fri Apr 06, 2018 11:04 am
### Re: Formal charge?
The purpose of formal charge is to find the lowest energy state possible for the molecule. Formal charge predicts what arrangment of atoms a molecule is most likely to be found in and create s a lewis structure with the lowest energy (0)
CarinaVargas1J
Posts: 20
Joined: Wed Nov 15, 2017 3:00 am
### Re: Formal charge?
Formal Charge= # of valence electrons on atom- lone pair electrons + # of bonds
Last edited by CarinaVargas1J on Thu May 24, 2018 4:45 pm, edited 2 times in total.
Sonia Aronson 1B
Posts: 30
Joined: Fri Apr 06, 2018 11:01 am
### Re: Formal charge?
The equation for finding formal charge is:
Formal Charge= [# of valence electrons on atom]- [lone pair electrons + # of bonds]
Erin Li 1K
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am
### Re: Formal charge?
you calculate formal charge by using the equation
number of valence electrons - number of lone electrons and number of bonds
Sarai Ventura 1L
Posts: 30
Joined: Mon Apr 09, 2018 1:39 pm
### Re: Formal charge?
According to the book formal charge is the electric charge of an atom in a molecule assigned on the assumption that the bonding is nonpolar covalent. Formal Charge (FC)= number off valence electrons in the free atom-(number of lone pair electrons+1/2 x number of shared electrons). However, since the “number of bonding electrons divided by 2” term is also equal to the number of bonds surrounding the atom, here's the shortcut formula: Formal Charge = [# of valence electrons on atom] – [non-bonded electrons + number of bonds].
hannahtweedy
Posts: 29
Joined: Tue Nov 21, 2017 3:00 am
### Re: Formal charge?
Would the equation:
valence electrons - non-bonding valence electrons - (bonding electrons/2)
also work?
Posts: 59
Joined: Fri Apr 06, 2018 11:04 am
### Re: Formal charge?
Yes! The equation [valence electrons - non-bonding valence electrons - (bonding electrons/2) ] would also work
Endri Dis 1J
Posts: 33
Joined: Fri Apr 06, 2018 11:02 am
### Re: Formal charge?
Yeah there is a very easy way to calculate it, all you have to do is get the valence electrons from that atom and subtract both the lone pairs and the bond pairs.
Sara Veerman-1H
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am
### Re: Formal charge?
Formal Charge = (# of valence e-) – (# of lone pair e-) – (1/2)(# of bonded e-)
Allen Chen 1J
Posts: 31
Joined: Wed Nov 15, 2017 3:04 am
### Re: Formal charge?
The formal charge is the # of valence electrons – # of lone pair electrons – (1/2)(# of bonded electrons
Maria Zamarripa 1L
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am
Been upvoted: 1 time
### Re: Formal charge?
The equation for the formal charge is, formal charge = V - (L+1/2(B)) V= # of valence electron of the element, L= # of lone pairs surrounding the atom (count each electron), B= bonded pairs (count each electron) | 985 | 3,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-35 | longest | en | 0.855119 |
http://www.presentica.com/ppt-presentation/amos-analysis-of-moment-structures | 1,569,243,862,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576965.71/warc/CC-MAIN-20190923125729-20190923151729-00207.warc.gz | 316,415,889 | 15,463 | 1 / 83
# AMOS Analysis of Moment Structures .
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Description
AMOS – Analysis of Moment Structures. Rick Zimmerman, Olga Dekhtyar. HIV Prevention Center University of Kentucky. Overview. Overview of Structural Equation Models (SEM) Introduction to AMOS User Interface AMOS Graphics Examples of using AMOS
Transcripts
Slide 1
AMOS – Analysis of Moment Structures Rick Zimmerman, Olga Dekhtyar HIV Prevention Center University of Kentucky
Slide 2
Overview of Structural Equation Models (SEM) Introduction to AMOS User Interface AMOS Graphics Examples of utilizing AMOS Predictors of Condom Use utilizing inert factors
Slide 3
Structural Equation Models
Slide 4
Structural Equation Modeling (SEM) An expansion of Regression and general Linear Models Also can fit more unpredictable models, as corroborative component examination and longitudinal information.
Slide 5
Structural Equation Modeling Ability to fit non-standard models , databases with autocorrelated blunder structures time arrangement investigation Latent Curve Models , databases with non-ordinarily circulated factors databases with deficient information .
Slide 6
Family Tree of SEM Latent Growth Curve Analysis T-test ANOVA Multi-way ANOVA Repeated Measure Designs Growth Curve Analysis Structural Equation Modeling Multiple Regression Path Analysis Bivariate Correlation Confirmatory Factor Analysis Next Workshop: November 9 See you there! Consider Analysis Exploratory Factor Analysis
Slide 7
Structural Equation Modeling (SEM) Exogenous variables= free Endogenous factors = subordinate Observed factors = measured Latent variables= surreptitiously
Slide 8
.10 : R 2 Error .15 Structural Equation Graphs Observed Variable : Loading Latent Variable
Slide 9
Observed factors for Impulsive basic leadership Peer standards about condoms Condom disposition Example: Condom Use Model Respondent Sex IDMA1R IDMC1R IDME1R IDMJ1R SEX1 Impulsive Decision Making FRBEHB1 ISSUEB1 Legend SXPYRC1 Latent Variables Observed Variables .15 Condom Use Loadings
Slide 10
Dependent Example: Condom Use Model Independent IDMA1R IDMC1R IDME1R IDMJ1R SEX1 Impulsive Independent FRBEHB1 ISSUEB1 Legend SXPYRC1 Latent Variables Observed Variables .15 Dependent Loadings
Slide 11
Example: Condom Use Model eidm1 eidm2 eidm4 eidm2 IDMA1R IDMC1R IDME1R IDMJ1R SEX1 Impulsive FRBEHB1 efr1 eiss ISSUEB1 Legend SXPYRC1 Latent Variables Observed Variables eSXYRC1 .15 Loadings
Slide 12
Example: Condom Use Model eidm1 eidm2 eidm4 eidm2 IDMA1R IDMC1R IDME1R IDMJ1R SEX1 Impulsive FRBEHB1 efr1 eiss ISSUEB1 Legend SXPYRC1 Latent Variables Observed Variables eSXYRC1 .15 Loadings
Slide 13
.28 .24 .48 .45 .03 .05 .15 Example: Condom Use Model eidm1 eidm2 eidm4 eidm2 IDMA1R IDMC1R IDME1R IDMJ1R .49 .69 .67 SEX1 - .06 .53 Impulsive - .19 - .10 - .15 .13 FRBEHB1 efr1 eiss ISSUEB1 .11 .38 Legend SXPYRC1 Latent Variables Observed Variables eSXYRC1 .15 Loadings
Slide 14
SEM Assumptions A Reasonable Sample Size a decent general guideline is 15 cases for every indicator in a standard common slightest squares different relapse investigation . [ " Applied Multivariate Statistics for the Social Sciences" , by James Stevens] specialists may go as low as five cases for each parameter gauge in SEM examinations, however just if the information are flawlessly all around carried on [ Bentler and Chou (1987)] Usually 5 cases for each parameter is identical to 15 measured factors .
Slide 15
SEM Assumptions (cont\'d) Continuously and Normally Distributed Endogenous Variables NOTE: At this time AMOS CANNOT deal with not consistently dispersed result factors
Slide 16
SEM Assumptions (cont\'d) Model Identification P is # of measured factors [P*(P+1)]/2 Df=[P*(P+1)]/2-(# of assessed parameters) If DF>0 model is over distinguished If DF=0 model is simply recognized If DF<0 model is under distinguished
Slide 17
Missing information in SEM Types of missing information MCAR Missing Completely at Random MAR Missing at Random MNAR Missing Not at Random
Slide 18
Handling Missing information in SEM Listwise Pairwise Mean substitution Regression strategies Expectation Maximization (EM) approach Full Information Maximum Likelihood (FIML)** Multiple imputation(MI)** The two best techniques: FIML and MI
Slide 19
SEM Software Several unique bundles exist EQS, LISREL, MPLUS, AMOS , SAS, ... Give at the same time general trial of model fit individual parameter gauge tests May look at all the while Regression coefficients Means Variances even over numerous between-subjects bunches
Slide 20
Introduction to AMOS
Slide 21
AMOS Advantages Easy to use for visual SEM ( Structural Equation Modeling). Simple to adjust, see the model Publication –quality design
Slide 22
AMOS Components AMOS Graphics draw SEM diagrams runs SEM models utilizing charts AMOS Basic runs SEM models utilizing grammar
Slide 23
Starting AMOS Graphics Start Programs Amos 5 Amos Graphics
Slide 24
Reading Data into AMOS File Data Files The accompanying discourse shows up:
Slide 25
Reading Data into AMOS Click on File Name to indicate the name of the information document Currently AMOS peruses the accompanying information record formats: Access dBase 3 – 5 Microsft Excel 3, 4, 5, and 97 FoxPro 2.0, 2.5 and 2.6 Lotus wk1, wk3, and wk4 SPSS *.sav documents , renditions 7.0.2 through 13.0 (both crude information and framework groups)
Slide 26
Reading Data into AMOS Example USED for this workshop: Condom utilize and what indicators influence it DATASET: AMOS_data_valid_condom.sav
Slide 27
To draw a watched variable, click "Diagram" on the top menu, and snap " Draw Observed ." Move the cursor to the place where you need to put a watched variable and snap your mouse. Drag the container keeping in mind the end goal to alter the extent of the crate. You can likewise use in the tool kit to draw watched factors. 2. In secret factors can be drawn also. Click "Diagram" and " Draw Unobserved ." Unobserved factors are appeared as circles. You may likewise use in the tool stash to draw in secret factors. Attracting AMOS In Amos Graphics, a model can be determined by drawing an outline on the screen
Slide 28
Drawing in AMOS To draw a way, Click " Diagram " on the top menu and snap " Draw Path ". Rather than utilizing the top menu, you may utilize the Tool Box catches to draw bolts ( and ).
Slide 29
Drawing in AMOS To attract Error Term to the watched and in secret factors. Utilize " Unique Variable " catch in the Tool Box. Snap and afterward click a case or a hover to which you need to include mistakes or a novel factors. (When you utilize " Unique Variable " catch, the way coefficient will be naturally compelled to 1.)
Slide 30
Drawing in AMOS Let us draw:
Slide 31
Naming the factors in AMOS double tap on the articles in the way outline. The Object Properties discourse box shows up. Alternately Click on the Text tab and enter the name of the variable in the Variable name field:
Slide 32
Naming the factors in AMOS Example: Name the factors
Slide 33
Constraining a parameter in AMOS The size of the idle variable or change of the inactive variable must be settled to 1 . Double tap on the bolt amongst EXPYA2 and SXPYRA2 . The Object Properties exchange shows up. Tap on the Parameters tab and enter the esteem " 1 " in the Regression weight field:
Slide 34
Improving the presence of the way outline You can change the presence of your way graph by moving articles around To move a protest, tap on the Move symbol on the toolbar. You will see that the photo of a bit of moving truck shows up underneath your mouse pointer when you move into the drawing range. This tells you the Move capacity is dynamic. At that point snap and hold down your left mouse catch on the question you wish to move. With the mouse catch still discouraged, move the protest where you need it, and let go of your mouse catch. Amos Graphics will naturally redraw all associating bolts.
Slide 35
Improving the presence of the way outline To change the size and state of a protest, first press the Change the state of articles symbol on the toolbar. You will see that the word " shape" shows up under the mouse pointer to tell you the Shape capacity is dynamic. Snap and hold down your left mouse catch on the question you wish to re-shape. Change the state of the protest your preferring and discharge the mouse catch. Change the state of articles likewise deals with two-headed bolts. Take after similar method to alter the course or bend of any twofold headed bolt.
Slide 36
Improving the presence of the way chart If you commit an error , there are constantly three symbols on the toolbar to rapidly safeguard you out: the Erase and Undo capacities. To delete a question , basically tap on the Erase symbol and afterward tap on the protest you wish to eradicate. To fix your last drawing movement , tap on the Undo symbol and your last action vanishes. Every time you click Undo, your past movement will be evacuated. On the off chance that you alter your opinion , tap on Redo to reestablish a change.
Slide 37
Performing the investigation in AMOS View/Set ® Analysis Properties and tap on the Output tab. There is additionally an Analysis Properties symbol you can tap on the toolbar. In any case, the Output tab gives you the accompanying alternatives:
Slide 38
Performing the investigation in AMOS For our illustration, check the Minimization history, Standardized evaluations , and Squared various relationships boxes. ( We are doing this on the grounds that these are so ordinarily utilized as a part of examination ). To run AMOS , tap on the Calculate gauges symbol on the toolbar. AMOS will need to spare this issue to a document. on the off chance that you have given it no filename, the Save As exchange box will show up. Give the issue a record name; let us say, tutorial1 :
Slide 39
Results When AMOS has finished the estimations, you have two alternatives for survey the yield: content yield , representation yield . For content yield, tap the View Text ( or
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The third one asks the student. Algebra 1 compound inequalities worksheet - To observe the image more plainly in this article you could click on the preferred image to see the graphic in its original sizing or in full. 7 rows Printable inequalities worksheets. Solving Equations And Inequalities Worksheet Pdf. Graphing Inequalities Workheet 4 - Here is a 12 problem worksheet where students will both solve inequalities and graph inequalities on a number line.
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For such questions you need consider if the inequalities are inclusive or strict in this case we have x. Each worksheets is visual differentiated and fun. Absolute value equations worksheet 1 rtf absolute value equations worksheet 1 pdf view answers. For such questions you need consider if the inequalities are inclusive or strict in this case we have x. The third one asks the student.
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7 rows Printable inequalities worksheets. You may have to use variables on both sides. Education resources designed specifically with parents in mind. Benefits of Grade 7 Inequality Worksheets. Graphing Inequalities Workheet 4 - Here is a 12 problem worksheet where students will both solve inequalities and graph inequalities on a number line.
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Equations and Inequalities in Maths Year 8. All worksheets created with Infinite Algebra 1. Inequality worksheets for 7th. Graphing Inequalities 4 PDF. Free Algebra 1 worksheets created with Infinite Algebra 1.
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Graphing Inequalities Workheet 4 - Here is a 12 problem worksheet where students will both solve inequalities and graph inequalities on a number line. For such questions you need consider if the inequalities are inclusive or strict in this case we have x. Graphing Inequalities 4 RTF. Ad Master 450 algebra skills with online practice. Education resources designed specifically with parents in mind.
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The third one asks the student. Solving Rational Equations Worksheets Radical Equations Rational Expressions Fractions Worksheets Identify what the. Ad Master 450 algebra skills with online practice. Answer Key Solving Equations And Inequalities Worksheet. Numbers Add to my workbooks 0 Download file pdf Embed in my website or blog Add to Google Classroom.
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Keystage 3 Interactive Worksheets to help your child understand Algebra. Free Algebra 1 worksheets created with Infinite Algebra 1. There are NINE problem types. 7 and x 3. X x takes any value greater then or equal to.
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If you find this site value, please support us by sharing this posts to your favorite social media accounts like Facebook, Instagram and so on or you can also bookmark this blog page with the title latest algebra inequalities worksheet by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website. | 1,848 | 9,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.85642 |
https://definithing.com/define-dictionary/lap-joint/ | 1,506,237,567,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689897.78/warc/CC-MAIN-20170924062956-20170924082956-00683.warc.gz | 642,263,781 | 3,735 | Dictionary: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Lap-joint
noun
1.
Also called plain lap. a joint, as between two pieces of metal or timber, in which the pieces overlap without any change in form.
2.
any of various joints between two members, as timbers, in which an end or section of one is partly cut away to be overlapped by an end or section of the other, often so that flush surfaces result.
noun
1.
a joint made by placing one member over another and fastening them together Also called lapped joint
Tagged:
• Laplace
[la-plas] /laˈplas/ noun 1. Pierre Simon [pyer see-mawn] /pyɛr siˈmɔ̃/ (Show IPA), Marquis de, 1749–1827, French astronomer and mathematician. /French laplas/ noun 1. Pierre Simon (pjɛr simɔ̃), Marquis de Laplace. 1749–1827, French mathematician, physicist, and astronomer. He formulated the nebular hypothesis (1796). He also developed the theory of probability in scientific phrases, a reference to […]
• Laplace-equation
noun, Mathematics. 1. the second-order partial differential equation indicating that the Laplace operator operating on a given function results in zero. Compare (def 4c).
• Laplace operator
noun 1. (maths) the operator ∂²/∂x² + ∂²/∂y² + ∂²/∂z² ∇² Also called Laplacian (ləˈpleɪʃɪən)
• Laplace-transform
noun, Mathematics. 1. a map of a function, as a signal, defined especially for positive real values, as time greater than zero, into another domain where the function is represented as a sum of exponentials.
Disclaimer: Lap-joint definition / meaning should not be considered complete, up to date, and is not intended to be used in place of a visit, consultation, or advice of a legal, medical, or any other professional. All content on this website is for informational purposes only. | 510 | 1,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-39 | latest | en | 0.893666 |
http://dict.youdao.com/example/only_on_the_left_side/ | 1,708,888,633,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00542.warc.gz | 9,022,765 | 14,250 | go top 返回词典
• In this case, the man was on his left side with his head facing west and surrounded by domestic jugs - rituals only previously seen in female graves.
具男性尸骨西向左侧卧，陪葬品是日常用的罐子，之前女性墓穴所见无异。
youdao
• This diagram models only the logic of a single alternate course, as you can tell by the numbering of the steps on the left side of the diagram.
该图单个备选行动过程逻辑建模可以通过序列图左侧所列步骤编号得知。
youdao
• Mr Bush told the author Bob Woodward that he would not withdraw from Iraq even if his wife and dog were the only people left on his side.
布什记者鲍勃·伍德沃德透露，即使最后支持的只有老婆不会伊撤军的。
youdao
• This one on the left-hand side lives only on the curve, while the right-hand side lives everywhere in this region inside.
左边这个积分定义曲线右侧的积分定义在整个内部区域
youdao
• It's only two minutes walk from the entrance on the left side.
入口大约分钟路的左侧。
youdao
• So, in both cases, we need the vector field to be defined not only, I mean, the left hand side makes sense if a vector field is just defined on the curve because it's just a line integral on c.
了解这两种表述后，我们不仅需要向量就是左边这里，曲线c线积分，向量场曲线上定义
youdao
• Now simply reset the device by holding in the record button (on the left side) and you should see a list of images on the card (usually only one).
现在只需要下record按钮(左边)来重设备然后应该会看到一个映像列表(通常只有一个)。
youdao
• The latter is used only if all four machines on the left side are down.
这个镜像左侧所有4机器都宕机之后才使用。
youdao
• Sleeping on the back or right side, rather than the left, doubled the risk - but only to almost four in 1, 000.
右侧和面朝上加起来的风险只有朝左侧睡的千分之四。
youdao
• For each operation on the left-side (that you are exposing), define an implementation (the only exception is if you don't want to implement the entire interface now).
左侧每个操作(正在公开的)定义一个实现(除非不想现在实现整个界面)。
youdao
• The only other clue comes from the fact that the heart is on the left side of the mother's body.
还有另外一条思路它的依据则心脏母亲身体的左侧这个事实
youdao
• "Computer-induced lesions are typically found on only one leg because the optical drives of laptops are located on the left side, " the authors write.
电脑诱导损害一般条腿上发现因为笔记本电脑的光驱放置在左侧，”作者写道。
youdao
• The average rhythm of heart beats showed obvious difference(P<0.05), but there was no difference in the cases with pleural fluid on the left side only (P>0.05).
结果，抽液前后平均心率显著差异P＜0．05），单纯左侧胸腔积液平均心率变化显著性差异（P＞0．05）；
youdao
• The best rooms were all on the left-hand side (going in), for these were the only ones to have Windows, deep-set round Windows looking over his garden and meadows beyond, sloping down to the river.
最好房间左手边(继续往里面走也一样)，因为只有方向的房间才窗户这些浑圆的窗户可以俯瞰美丽的花园一路延伸河边的翠绿草地
youdao
• Mrs. Werner was 6 years old and spoke only Russian in 1906 when she, her four siblings and her mother left their hometown of Vladimir to join their father on the Lower East Side of Manhattan.
1906年太太6四位兄弟姐妹母亲一起离开家乡弗拉基米尔 曼哈顿 东区与他们父亲团聚，那时会说俄语
youdao
• Lin's character dies in a shower of blood only to appear a short while later fighting on the side of the Chinese hero, her miraculous resurrection left unexplained.
随后，林志玲扮演角色死亡过多久出现了，与中国男主角并肩战斗
youdao
• Lin's character dies in a shower of blood only to appear a short while later fighting on the side of the Chinese hero, her miraculous resurrection left unexplained.
随后，林志玲扮演角色死亡过多久出现了，与中国男主角并肩战斗
youdao
\$firstVoiceSent
- 来自原声例句 | 1,112 | 3,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-10 | latest | en | 0.823225 |
http://oeis.org/A036830 | 1,553,163,496,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00395.warc.gz | 151,146,378 | 3,704 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A036830 Schoenheim bound L_1(n,n-4,n-5). 3
3, 7, 14, 26, 44, 70, 105, 152, 213, 291, 388, 508, 654, 829, 1037, 1281, 1566, 1896, 2276, 2710, 3203, 3761, 4388, 5091, 5875, 6746, 7710, 8774, 9944, 11228, 12632, 14164, 15831, 17641, 19602, 21722, 24009, 26472, 29120, 31961, 35005 (list; graph; refs; listen; history; text; internal format)
OFFSET 6,1 REFERENCES W. H. Mills and R. C. Mullin, Coverings and packings, pp. 371-399 of J. H. Dinitz and D. R. Stinson, editors,a Contemporary Design Theory, Wiley, 1992. See Eq. 1. LINKS FORMULA a(6)=3 a(n)=ceiling(n/(n-4)*a(n-1)) - Benoit Cloitre, May 31 2003 MAPLE A036830 := proc(n) local i, t1; t1 := 1; for i from 6 to n do t1 := ceil(t1*i/(i-4)); od: t1; end; L := proc(v, k, t, l) local i, t1; t1 := l; for i from v-t+1 to v do t1 := ceil(t1*i/(i-(v-k))); od: t1; end; # gives Schoenheim bound L_l(v, k, t) PROG (PARI) a(n)=if(n<7, 3, ceil(n/(n-4)*a(n-1))) CROSSREFS Lower bound to A066225. A column of A036838. Sequence in context: A206417 A207381 A008646 * A014153 A001924 A079921 Adjacent sequences: A036827 A036828 A036829 * A036831 A036832 A036833 KEYWORD nonn AUTHOR N. J. A. Sloane, Jan 11 2002 STATUS approved
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Last modified March 21 04:59 EDT 2019. Contains 321364 sequences. (Running on oeis4.) | 632 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-13 | latest | en | 0.532008 |
http://math.stackexchange.com/questions/86298/the-relationship-between-legendre-polynomials-and-monomial-basis-polynomials | 1,469,265,735,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00121-ip-10-185-27-174.ec2.internal.warc.gz | 158,933,888 | 19,255 | # The relationship between Legendre Polynomials and monomial basis polynomials
I am currently doing filter designs and stumbled across this mathematical problem which I cannot understand. I was hoping for some insight from experts around this field to help me with this.
$$\sum_{m=0}^{M-1}\ h_{m}\ x^{m} = \sum_{m=0}^{M-1}\ g_{m}\ P_{m}\left(x\right)$$
where $P_{m}\left(x\right)$ is the Legendre polynomials and $x^{m}$ is the conventional polynomials.
I want to know the relationship between $g_{m}$ and $h_{m}$, what is the best approach? Is there a closed form solution for $h_{m}$?
Thanks in advance, and apologies if it is a silly question.
-
I might as well: there is an algorithm, due to Salzer, for converting to and from different orthogonal polynomial expansions. (I previously talked about this algorithm here.) If you intend to do this entire business of conversions, I believe an algorithm might be more useful to you than an integral expression.
What follows are a few relevant Mathematica routines. Even though I used Mathematica, I hope that the algorithm is still transparent and easily translatable (the indexing starts from 1; some adjustment is necessary if the indexing in your language starts at 0):
monomialToLegendre[pc_?VectorQ] :=
Module[{n = Length[pc] - 1, lc, q},
lc[1] = pc[[n]]; lc[2] = Last[pc];
Do[
q[1] = lc[1];
lc[1] = pc[[n - k + 1]] + lc[2]/3;
Do[
q[j] = lc[j];
lc[j] = j lc[j + 1]/(2 j + 1) + (j - 1) q[j - 1]/(2 j - 3);
, {j, 2, k - 1}];
q[k] = lc[k];
lc[k] = (k - 1) q[k - 1]/(2 k - 3);
lc[k + 1] = k q[k]/(2 k - 1);
, {k, 2, n}];
Table[lc[k], {k, n + 1}]]
One could use Salzer again for converting from the Legendre to the monomial basis. However, it is a bit clearer (and gives rise to essentially the same method) to treat the problem as one of Taylor/Maclaurin expansion of the polynomial expressed in terms of Legendre polynomials. One can then use Clenshaw's algorithm for the purpose:
legendreToMonomial[lc_?VectorQ] :=
Module[{n = Length[lc] - 1, pc, v = 0, w, z},
pc[1] = Last[lc]; pc[_] = z[_] = 0;
Do[
w = pc[1];
pc[1] = lc[[j]] - v z[1];
z[1] = w;
Do[
w = pc[i];
pc[i] = (2 j - 1) z[i - 1]/j - v z[i];
z[i] = w;
, {i, 2, n - j + 2}];
v = (j - 1)/j;
, {j, n, 1, -1}];
Table[pc[j], {j, n + 1}]]
As an example, to demonstrate the identity
$$5P_0(x)-4P_1(x)-2P_2(x)+3P_3(x)+P_4(x)=\frac{35}{8}x^4+\frac{15}{2}x^3-\frac{27}{4}x^2-\frac{17}{2}x+\frac{51}{8}$$
monomialToLegendre[legendreToMonomial[{5, -4, -2, 3, 1}]] ==
{5, -4, -2, 3, 1}
True
legendreToMonomial[monomialToLegendre[{51/8, -17/2, -27/4, 15/2, 35/8}]]
== {51/8, -17/2, -27/4, 15/2, 35/8}
True
-
You are a champ J.M.! Good to see it works, I will follow some of your recommended readings and see if I can understand a bit. Since I am using Matlab instead of Mathematica, I am having a hard time understanding your equations used to do the conversion =( – JuniorEngie Nov 29 '11 at 5:23
MATLAB supports loops, so it shouldn't be too hard. The indexed variables should be turned into proper arrays in MATLAB, of the same dimensions as the array of coefficients. – J. M. Nov 29 '11 at 5:46
I am assuming your domain is $(-1,1)$ else you would need to do some rescaling of the following argument. The main thing to exploit is that the Legendre polynomials are orthogonal i.e. $$\displaystyle \int_{-1}^{1} P_m(x) P_n(x) = \frac{2}{2n+1} \delta_{mn}.$$ Using this we get that $$g_n = \left(\frac{2n+1}{2} \right) \sum_{m=0}^{M-1} \left( h_m \int_{-1}^{1} x^m P_n(x) \right)$$
-
Yes sir, the domain had been rescaled to [-1, 1] and thank you for the closed form expression! Much much appreciated. Just wondering will it be possible to revert the expression around to get $h_{m}$ instead? Sorry my maths is quite lacking. – JuniorEngie Nov 28 '11 at 8:40 | 1,282 | 3,775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-30 | latest | en | 0.883386 |
https://blog.grkweb.com/2014/09/hackerrank-cavity-map-solution.html?showComment=1440746389546 | 1,618,269,125,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069267.22/warc/CC-MAIN-20210412210312-20210413000312-00586.warc.gz | 259,401,021 | 30,914 | # Hackerrank Cavity Map Solution
The Solution for hackerrank problem, Cavity Map using C++ Program.
Passed Test cases: 21 out of 21
### SOURCE CODE:
`````` #include<iostream>
using namespace std;
int main(){
int n,i,j;
cin>>n;
string inp[n];
for(i=0;i<n;i++)
{
cin>>inp[i];
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i == 0 || j == 0 || i == n-1 || j == n-1 )
{
cout<<inp[i][j];
}
else if(inp[i][j] > inp[i][j-1] && inp[i][j] > inp[i][j+1] && inp[i][j] >inp[i-1][j] && inp[i][j] > inp[i+1][j])
{
cout<<"X" ;
}
else
{
cout<<inp[i][j];
}
}
cout<<endl;
}
return 0;
}
``````
#### 9 comments:
1. I'm impressed. You're truly well informed and very intelligent. You wrote something that people could understand and made the subject intriguing for everyone. I'm saving this for future use.
Rica
www.imarksweb.org
2. inp is 1d , but how you are using it as a 2d array?
1. input is an array of strings,
so it is also similar to an 2D array of characters.
Useful links :
http://www.hellgeeks.com/array-of-strings/
3. I have done it by 2-d array but its showing wrong. I have run it in my compiler DEV C++ and its running fine
1. Can you share the code here?
4. in php plz
5. //whats wrong with this?
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int main(){
int n,i,j;
char b,c,d;
cin >> n;
char a[100][100];
for(i=0;i> a[i];
for(i=0;id))
cout << "x";
else
cout << a[i][j];
}
if(i!=n-1)
cout << endl;
}
return 0;
}
6. Life is full of many challenges. Challenges that will make you or break you depending on how you handle it. Visit my site for more updates. God Bless to your site.
n8fan.net
www.n8fan.net | 529 | 1,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-17 | latest | en | 0.731583 |
https://math.stackexchange.com/questions/2006498/basic-proof-that-every-polynomial-over-mathbb-r-factorizes-into-at-most-quadr | 1,569,183,991,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575674.3/warc/CC-MAIN-20190922201055-20190922223055-00100.warc.gz | 548,653,295 | 30,292 | # Basic proof that every polynomial over $\mathbb R$ factorizes into at most quadratic ones
Is there a proof without using that in $\mathbb C$ every polynomial factorizes into linear ones that every polynomial over $\mathbb R$ factorizes into linear or quadratic ones?
• Given the quadratic formula, this would immediately imply the result for $\mathbb{C}$, would it not? – Tobias Kildetoft Nov 9 '16 at 12:03
• @TobiasKildetoft A common way to prove that real polynomials factor into linear and quadratic terms goes via the algebraic closure of $\Bbb C$ and that non-real roots come in complex conjugate pairs. I think the OP want a proof not using $\Bbb C$. – Arthur Nov 9 '16 at 12:32
• @Arthur Right, but once you have that proof, the step to $\mathbb{C}$ being algebraically closed is so minor that you would basically have it already. – Tobias Kildetoft Nov 9 '16 at 12:34
• @TobiasKildetoft Not quite. You're very close to proving that the $\Bbb C$ contains the roots of any real polynomial, but you still have a little way to go to prove that $\Bbb C$ is algebraically closed. – Arthur Nov 9 '16 at 12:47 | 300 | 1,114 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-39 | latest | en | 0.914614 |
https://royalpitch.com/10-to-3-is-how-many-hours/ | 1,695,384,742,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00419.warc.gz | 561,618,399 | 73,867 | I'm a full time working dad that tries to keep up with technology. I want to haev this blog to share about my life, my journey, places I visit, lifestyle, technology, beauty, business and other topics. I hope you enjoy reading it.
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10 To 3 Is How Many Hours
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You can use a time difference calculator to calculate the number of hours between two dates. It’s easy to calculate the number of hours between two dates by simply entering the dates into the calculator. You can add up all of these into one calculation. In this way, you can easily calculate the total number of hours between the two dates. To calculate the time difference between two dates, you can use a calculator. It’s so easy! | 799 | 3,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-40 | latest | en | 0.944992 |
https://codeforces.com/problemset/problem/830/B | 1,624,547,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00355.warc.gz | 166,771,345 | 13,597 | B. Cards Sorting
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck.
Input
The first line contains single integer n (1 ≤ n ≤ 100 000) — the number of cards in the deck.
The second line contains a sequence of n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000), where ai is the number written on the i-th from top card in the deck.
Output
Print the total number of times Vasily takes the top card from the deck.
Examples
Input
46 3 1 2
Output
7
Input
11000
Output
1
Input
73 3 3 3 3 3 3
Output
7
Note
In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards. | 547 | 2,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-25 | latest | en | 0.94427 |
https://gmatclub.com/forum/kellogg-northwestern-2013-calling-all-applicants-135309-220.html?oldest=1 | 1,495,543,934,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00370.warc.gz | 757,245,507 | 56,757 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# Kellogg (Northwestern) 2013 - Calling All Applicants
Author Message
Intern
Joined: 02 Nov 2012
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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09 Nov 2012, 00:46
I made such a big mistake. Put a wrong interviewer name in the thanks letter. Can I be more stupid? I have no one to blame but myself!!
Senior Manager
Joined: 02 Feb 2009
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Concentration: General Management, Strategy
GMAT 1: 690 Q48 V35
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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09 Nov 2012, 01:19
Kewei wrote:
I made such a big mistake. Put a wrong interviewer name in the thanks letter. Can I be more stupid? I have no one to blame but myself!!
That is a doozy but not a deal breaker. An honest follow up apologizing for the mix up and referring to the person correctly will amend this. Don't worry - most people will take this light heartedly, but not if you don't follow up with a correction
I am curious to know though, how did you mix up the name of the interviewer?
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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10 Nov 2012, 04:38
Kewei wrote:
I made such a big mistake. Put a wrong interviewer name in the thanks letter. Can I be more stupid? I have no one to blame but myself!!
Dont worry I dont think this is a problem! If the interview went well then he will let it slide. They know what an anxious time this is for applicants.
Intern
Joined: 13 Jul 2012
Posts: 30
Concentration: Marketing, Entrepreneurship
GMAT 1: 750 Q51 V40
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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12 Nov 2012, 04:56
Had my interview today with an alumni in Mumbai. Was very relaxed affair. The discussion was mostly led by my CV.
Some of the questions were:
5 strengths
5 weaknesses
Why MBA?
You seem to be doing so well in your company. What will you do if they dont let you go?
WHy Kellogg?
Why 1 year course?
Why not an MMA course or any other course?
Give an example of influencing people..
Then I asked her some 4-5 questions on life at Kellogg.
Now the wait begins
Intern
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Concentration: Strategy, Marketing
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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12 Nov 2012, 07:55
Interviewed last week with an alum off-campus. The interviewer told me that she would interview me for about 45 minutes, but she exhausted her questions in just 25 minutes. I used 10 minutes to ask her a few questions and another 10 minutes to tell her more about myself. Though the interviewer was very nice, I didn't feel a connect. I wasn't asked any behavioral questions for which I had prepared a few stories. In hindsight, I felt that the interviewer wasn't impressed by my non-bluechip work experience.
Now the wait for the result begins, though I am a bit disappointed with my interview experience
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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12 Nov 2012, 10:35
Sarajevo123 wrote:
Had my interview today with an alumni in Mumbai. Was very relaxed affair. The discussion was mostly led by my CV.
Some of the questions were:
5 strengths
5 weaknesses
Why MBA?
You seem to be doing so well in your company. What will you do if they dont let you go?
WHy Kellogg?
Why 1 year course?
Why not an MMA course or any other course?
Give an example of influencing people..
Then I asked her some 4-5 questions on life at Kellogg.
Now the wait begins
ridiculous... 5 weaknesses?? my pet peeve is the weakness questions cuz its total bs.
also company not let you go? does she think you are a working slave labor?
best of luck to you.
Manager
Joined: 28 Feb 2011
Posts: 92
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GMAT 1: 710 Q46 V42
WE: Management Consulting (Consulting)
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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14 Nov 2012, 05:26
Just had my interview with an alum in Brazil! Very well structured and detailed questions. Entire interview in English, with pre and post-chat in Portuguese. Here they are below:
- Walk me through your resume
- Why did you go from M&A to consulting?
- What projects have you worked on during your time in consulting?
- Why do you want an MBA now and not later?
- What are your post MBA goals?
- Why Kellogg?
- What question didnt I ask you that you would like to answer?
- How did you decide to get married this year? (I dont think this question will be asked in the US or Europe...)
Intern
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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14 Nov 2012, 10:00
Hi oodle,
I am quite new here. Just happened to see your background consumer products and think I have similar bckground.
I am interested in applying Kellogg, but have no clue the real students life is.
Do you mind spending some time introducing your student life so far?
Manager
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GPA: 3.18
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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15 Nov 2012, 16:59
Hi All - I had my on-campus interview at Kellogg this past week. I loved the school and student engagement, but I realized that I've used other on-campus interviews to get a fell for what potential classmates would be like. Since Kellogg extends interview invites to everyone, it was very hard to get a good read on this. At other schools, since the interviews and invitation-only, it's easier to get an idea of the type of person that the school looks to attract.
Has anyone felt this way after a visit like this? What other ways have you used to get an idea of what the vibe of a school is like and what other classmates would be like during a visit?
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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15 Nov 2012, 18:11
esdykes wrote:
Has anyone felt this way after a visit like this? What other ways have you used to get an idea of what the vibe of a school is like and what other classmates would be like during a visit?
Never really thought of this. I've only had two interviews, and both were applicant initiated. Basically, I try to get a feel for what the school is like by talking to current students whether it's the tour guide, students at lunch, or just chatting up some students before my class visit starts. I figure the culture doesn't change from year to year and they're probably looking to admit people similar to those who are currently enrolled.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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15 Nov 2012, 19:46
Hi guys,
I applied to Kellogg in R1 this year (for the incoming class of 2013). I have some really great international work-ex spanning 21 countries in strategy and marketing, a GMAT score of 750. However, I bombed my undergrad in engineering from a top engineering school in Asia (Singapore). I have a GPA of only 2.1. I tried to make up for it with a good masters in economics from another good university in SG and have a GPA of 3.6 there. Also, I speak 5 languages.
I had my interview with a kellogg alum in Singapore and she told me that she was very impressed with me and would write good stuff about me in her feedback. She told me that she would even write an additional segment talking about how much I've grown and accomplished since my UG disaster.
I was pleased with all that and thought things would be good. However, when I was connecting with a cornell alum, he told me he had been dinged at Kellogg inspite of a brilliant interview with an alum in the Philippines.. I am very very nervous now..
Does anyone have the same experience? Great interview but ding specifically at Kellogg?
Manager
Joined: 23 Aug 2012
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GMAT 1: 720 Q48 V40
GPA: 3.18
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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15 Nov 2012, 20:49
CobraKai wrote:
Never really thought of this. I've only had two interviews, and both were applicant initiated. Basically, I try to get a feel for what the school is like by talking to current students whether it's the tour guide, students at lunch, or just chatting up some students before my class visit starts. I figure the culture doesn't change from year to year and they're probably looking to admit people similar to those who are currently enrolled.
Thanks CobraKai. I guess I just took that "extra check" for granted at the other schools. I did like it though. It's kind of exciting literally meeting several people who could be your new best friends.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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15 Nov 2012, 23:25
niksworth wrote:
Got an invite to interview with an alumnus in Mumbai.
Hey Niksworth.. was this after a waiver? Anybody got an interview invite post a waiver yet?
Manager
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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16 Nov 2012, 02:59
vibhav wrote:
niksworth wrote:
Got an invite to interview with an alumnus in Mumbai.
Hey Niksworth.. was this after a waiver? Anybody got an interview invite post a waiver yet?
Nope. Not yet.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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16 Nov 2012, 04:48
Hey, does anyone know if the results are supposed to be out on Dec 17 or by Dec 17?
Intern
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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16 Nov 2012, 09:25
1
KUDOS
Last year the decisions started rolling out a week early via phone calls and people found out each day up until the final decision day.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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16 Nov 2012, 19:26
anonymous2012 wrote:
Last year the decisions started rolling out a week early via phone calls and people found out each day up until the final decision day.
I have heard this as well. And, to confirm - here is the direct language from the Kellogg website http://kellogg.northwestern.edu/Programs/FullTimeMBA/Applying/AdmissionsDecisions.aspx:
Admissions decisions may be released on or before the decision deadline for that particular application round. Application decisions are posted on the Check Application Status page on the Kellogg Full-Time website. Please note the status is not posted in the Apply Yourself system.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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16 Nov 2012, 21:15
1
KUDOS
I'll be staying in Evanston next week so if anybody is traveling home to Chicago for the break and wants to visit the campus, PM me and we can try to schedule a time for a private tour. Kellogg will have official info sessions this Mon and Tues, but no class visits or building tours.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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16 Nov 2012, 21:55
hselika wrote:
Hi guys,
I applied to Kellogg in R1 this year (for the incoming class of 2013). I have some really great international work-ex spanning 21 countries in strategy and marketing, a GMAT score of 750. However, I bombed my undergrad in engineering from a top engineering school in Asia (Singapore). I have a GPA of only 2.1. I tried to make up for it with a good masters in economics from another good university in SG and have a GPA of 3.6 there. Also, I speak 5 languages.
I had my interview with a kellogg alum in Singapore and she told me that she was very impressed with me and would write good stuff about me in her feedback. She told me that she would even write an additional segment talking about how much I've grown and accomplished since my UG disaster.
I was pleased with all that and thought things would be good. However, when I was connecting with a cornell alum, he told me he had been dinged at Kellogg inspite of a brilliant interview with an alum in the Philippines.. I am very very nervous now..
Does anyone have the same experience? Great interview but ding specifically at Kellogg?
This happens with applicants at every school, and it's really out of your control and all subjective. The only thing you can really do is just wait till you hear the decision.
Last edited by Ward2012 on 17 Nov 2012, 08:17, edited 1 time in total.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink]
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17 Nov 2012, 08:02
1
KUDOS
hselika wrote:
I was pleased with all that and thought things would be good. However, when I was connecting with a cornell alum, he told me he had been dinged at Kellogg inspite of a brilliant interview with an alum in the Philippines.. I am very very nervous now..
Does anyone have the same experience? Great interview but ding specifically at Kellogg?
Something to consider - at Kellogg the interview is just another component of the application that gets evaluated with everything else. Having a great interview may not be enough o compensate for weaknesses elsewhere in the application. At schools where interviews are invite only, you can at least feel some confidence that you made it through the first cut of applicants. When interviewing with Kellogg, there's just no way of knowing how you stack up with the rest of the applicant pool or what the adcom thinks of your application.
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Re: Kellogg (Northwestern) 2013 - Calling All Applicants [#permalink] 17 Nov 2012, 08:02
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Display posts from previous: Sort by | 4,636 | 17,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-22 | latest | en | 0.881009 |
https://pythontic.com/containers/frozenset/introduction | 1,723,652,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00106.warc.gz | 380,911,705 | 5,688 | # Frozenset In Python
## Overview:
• A set represents a mathematical concept of sets.
• Python provides two types of sets: A set and a frozenset.
• The frozenset is also a set, however a frozenset is immutable.
• Once frozenset is created new elements cannot be added to it.
• A frozenset is hashable, meaning every time a frozenset instance is hashed, the same hash value is returned.
• The hashable property of the frozenset makes it qualified to be a key in a Python dictionary.
• The hashable property of the frozenset also makes two frozenset instances to be compared for equality.
• The constructor of a frozenset takes an iterable object and returns a frozenset instance.
• In case if no iterable object is passed, the constructor returns an empty set.
• In both the cases the returned frozenset is immutable.
• Applications of frozenset include, set of sets.
## Frozenset operations:
Since frozenset instances are immutable, the following set methods are not supported by frozenset: update(), intersection_update(), symmetric_difference_update() ,add(), remove(), discard(), pop(), clear(). The following set operators are also not allowed on a frozenset: |=, &=, -=, ^=.
## Example 1:
# Sample Python program for frozenset singleDigitPrimes = (2,3,5,7) # Single digit prime numbers as a Python frozenset singleDigitPrimeSet = frozenset(singleDigitPrimes) # Prime numbers less than ten as a Python frozenset primeLTTen = frozenset((2,3,5,7)) # Prime numbers less than twenty as a Python frozenset primeLTTwenty = frozenset((2,3,5,7,11,13,17,19)) # Check the single digit prime number set # and the prime number set less than ten are same print("Single digit prime number set is equal to prime number set of numbers less than the integer ten:%s"%(primeLTTen == singleDigitPrimeSet)) # Check the single digit prime number set # and the prime number set less than twenty are same print("Single digit prime number set is equal to prime number set of numbers less than the integer twenty:%s"%(primeLTTwenty == singleDigitPrimeSet)) # Are the prime numbers less than ten and the prime numbers less than twenty are disjoint print("Prime numbers less than ten and the prime numbers less than twenty are disjoint:%s"%(primeLTTen.isdisjoint(primeLTTwenty)))
## Output:
Single digit prime number set is equal to prime number set of numbers less than the integer ten:True Single digit prime number set is equal to prime number set of numbers less than the integer twenty:False Prime numbers less than ten and the prime numbers less than twenty are disjoint:False
## Example 2:
This example Python program shows how a frozenset can be used along with a Python dictionary instance.A set of latitude and longitude values are added as keys of a dictionary instance. The values against the keys are the strings of city names while they could be any complex object. This is possible as the frozenset instances are immutable and hashable.
# Example Python program using frozenset as keys of a dictionary # City1 latitude1 = 40 longitude1 = 74 latLong1 = (latitude1, longitude1) cityName1 = "NewYork" # City2 latitude2 = 41 longitude2 = 87 latLong2 = (latitude2, longitude2) cityName2 = "Chicago" # City3 latitude3 = 37 longitude3 = 122 latLong3 = (latitude3, longitude3) cityName3 = "San Francisco" # Create a Python dictionary cityCollection = dict() # With key as a frozenset instance of latitude and longitude # add cities to the dictionary cityCollection[latLong1] = cityName1 cityCollection[latLong2] = cityName2 cityCollection[latLong3] = cityName3 # Print the dictionary print("Cities by latitude and longitude:") print(cityCollection)
## Output:
Cities by latitude and longitude: {(40, 74): 'NewYork', (41, 87): 'Chicago', (37, 122): 'San Francisco'} | 920 | 3,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.823135 |
https://www.airmilescalculator.com/distance/adb-to-hty/ | 1,721,804,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00771.warc.gz | 569,024,798 | 32,625 | # How far is Hatay from Izmir?
The distance between Izmir (İzmir Adnan Menderes Airport) and Hatay (Hatay Airport) is 520 miles / 836 kilometers / 452 nautical miles.
The driving distance from Izmir (ADB) to Hatay (HTY) is 672 miles / 1081 kilometers, and travel time by car is about 12 hours 49 minutes.
520
Miles
836
Kilometers
452
Nautical miles
1 h 29 min
102 kg
## Distance from Izmir to Hatay
There are several ways to calculate the distance from Izmir to Hatay. Here are two standard methods:
Vincenty's formula (applied above)
• 519.604 miles
• 836.222 kilometers
• 451.524 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 518.529 miles
• 834.492 kilometers
• 450.590 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Izmir to Hatay?
The estimated flight time from İzmir Adnan Menderes Airport to Hatay Airport is 1 hour and 29 minutes.
## Flight carbon footprint between İzmir Adnan Menderes Airport (ADB) and Hatay Airport (HTY)
On average, flying from Izmir to Hatay generates about 102 kg of CO2 per passenger, and 102 kilograms equals 224 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Izmir to Hatay
See the map of the shortest flight path between İzmir Adnan Menderes Airport (ADB) and Hatay Airport (HTY).
## Airport information
Origin İzmir Adnan Menderes Airport
City: Izmir
Country: Turkey
IATA Code: ADB
ICAO Code: LTBJ
Coordinates: 38°17′32″N, 27°9′25″E
Destination Hatay Airport
City: Hatay
Country: Turkey
IATA Code: HTY
ICAO Code: LTDA
Coordinates: 36°21′45″N, 36°16′56″E | 516 | 1,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.844876 |
https://www.spoj.com/problems/COUNTDIO/ | 1,712,932,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00158.warc.gz | 873,976,908 | 7,666 | ## COUNTDIO - Linear Diophantine Equation
no tags
Given a positive integer M and N positive integers a1, a2, ..., aN. Count the number of non-negative integer tuples (x1, x2, ..., xNsuch that:
a1 * x1 + a2 * x2 + ... + aN * xN = M
### Input
- The first line contains two integers N and M.
- The second line contains N integers a1, a2, ..., aN respectively.
### Output
- Output only one integer, the number of asked tuples modulo 998244353.
### Constrains
• 1 ≤ N ≤ 60 000
• 1 ≤ M ≤ 1018
• 1 ≤ ai ≤ 60 000
• 1 ≤ a1 + a2 + ... + aN ≤ 60 000
### Example
```3 10
3 2 5```
`4`
#### Input:
```5 100000
1 3 4 6 1000```
#### Output:
`865762992`
### Note
• My solution runs in less than 0.8s for each input file, 6.3s in total.
Added by: Lien Date: 2019-12-14 Time limit: 12s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All | 308 | 872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-18 | latest | en | 0.586782 |
http://lotterywinnerstories.com/Lottery/lucky-lottery-number-generator-lottery/ | 1,527,390,809,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867995.55/warc/CC-MAIN-20180527024953-20180527044953-00437.warc.gz | 177,647,163 | 11,870 | How To Make Money Playing Lotto? Lottery
How Do You Select The Best Numbers for Playing Lottery lottery?
A lottery is a game of chance sponsored by many states and countries. Winning the Lottery lottery game can be done easily by using the right strategy and system. Richard Lustig used his system to win several lottery jackpots and millions in prize money proving that luck has nothing to do with winning.
A lot of people use algorithm to analyze and predict lottery results for Lottery. Analysis of Groups There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.Analysis of Hot-Cold Trend This algorithm analysis is one of the most favorite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings. Analysis of Repetition Pattern A lot of lottery players share the same opinion that repetition is quite important to predict the winning lottery numbers as most of jackpots will appear again in the future.
Five Secrets To Winning The Lottery: How To Pick Megamillions Numbers
People who play Megamillions have one question in common and that would be which the best numbers to play Megamillions are? The thing is, there isn't a standard or a formula that one can follow in order to come to a conclusion regarding the best numbers to play Megamillions. There are, however, things that you can keep in mind when you are in the process of picking numbers. You can consider them strategies or systems or you can think of them, simply, as tips. They may not give you a concrete answer towards the best numbers to play Megamillions, but they can help take you a step closer.1. When choosing numbers, do not try and form patterns. Remember to be as random as you can. Think about it, if you decide to go with a pattern such as choosing all the numbers that go a certain way, whether upwards or downwards the sheets then you are already putting yourself at a disadvantage.8. When you ask people who have won jackpots for themselves, it is likely that they'd tell you that the numbers they chose were chosen mostly because of gut feel. A kind of intuition. You may not think much of it but this intuition could potentially win you millions. So go with what you feel and not what you believe to be the best numbers to play Megamillions. Remember, trust your gut and go with your intuition.
Is There a Way To Win The Lottery? Lottery
Playing the Lottery lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
Would You Like to Win The Lottery lottery?
Here's 5 top tips to bear in mind next time you complete your lottery playslip. They won't increase your CHANCES of winning - because ONLY more entries can do that.
BUT they can significantly increase the AMOUNT you win when your numbers do come up.
How?
Because all 5 lottery tips are based on avoiding the way a lot of other people pick their numbers. If you pick lottery numbers the same way as most people do, then when you hit the jackpot, you share that prize with everybody else who picked the same numbers.
That can turn a jackpot pool of millions into a prize of just thousands! It happens all too often - so please don't let it happen to you.
The easiest, fastest way to pick better lottery numbers, is to pick them totally at random. So pull scraps of paper out of a bag. It won't guarantee NOT picking a 'bad' set of numbers, but at least there's a good chance you won't be sharing your lottery millions with a hundred other 'lucky' winners.
Are you ready for more tips on how to win more often when playing the lottery?
If so then read below for a free report that uncovers the biggest mistakes most lottery players are making, and how you can avoid them.
Powerball: How To Use Easy Pick To Win The Powerball
Because you are sitting in front of your computer viewing this page, chances are, you're an eager and regular lotto player who has yet to win lotto prizes. Lottery cheats are some of the most researched items or articles here on the Internet, and it's easy to understand why: lottery games from all states can be addictive, attracting countless players from all across the country to play the lotto on a regular basis. What you need to understand is that while playing the lottery can be fun and exciting, the real treat lies in winning the lotto prizes - including the jackpot. And unlike what some lotto players believe, you can win the lotto using lottery cheats.
Lottery cheats are not cheats in the real sense of the word. They are not illegal and won't put you into some kind of federal trouble. In fact, lottery cheats have long been recognized by serious lottery players as the reason why they have higher chances of winning. Cheats at winning the lottery are actually guides on how to win any type of lotto game you choose to play. They offer sensible pieces of advice to increase the likelihood of holding a winning ticket.
As far as lottery cheats are concerned, statistics experts highly advise against forming lottery combinations in a mathematical sequence and playing patterns on lottery tickets. These acts are sure to reduce your chances of winning because mathematical sequences and patterns are almost never taken into consideration in lottery games. Past winning results show a tendency towards random combinations. If you use a proven system that can efficiently analyze lottery data, such as past winning numbers, trends, and angles, you are more likely to bring home the jackpot prize not just once but as many times as you please.
Winning the Lottery lottery is very rewarding. Give this system a spin and watch your winnings grow.
How To Make Money Playing Lotto? Lottery | 1,352 | 6,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-22 | longest | en | 0.961499 |
https://redmine.ruby-lang.org/issues/8850 | 1,627,815,918,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00014.warc.gz | 490,066,329 | 14,831 | Actions
## Feature #8850
open
### Convert Rational to decimal string
Status:
Assigned
Priority:
Normal
Target version:
-
[ruby-core:56955]
Description
On Ruby 2.1.0, decimal literal is introduced.
It generates Rational but it cannot easily convert to decimal string.
You know, Rational#to_f is related to this but
• Float is not exact number ** 0.123456789123456789r.to_f.to_s #=> "0.12345678912345678"
• it can't handle recursive decimal
** (151/13r).to_f.to_s #=> "11.615384615384615"
• the method name
** to_decimal
** to_decimal_string
** to_s(format: :decimal)
** extend sprintf
• how does it express recursive decimal
** (151/13r).to_decimal_string #=> "11.615384..."
** (151/13r).to_decimal_string #=> "11.(615384)"
Example implementation is following.
Its result is
** 0.123456789123456789r.to_f.to_s #=> "0.123456789123456789"
** (151/13r).to_f.to_s #=> "11.(615384)"
```class Rational
def to_decimal_string(base=10)
n = numerator
d = denominator
r, n = n.divmod d
str = r.to_s(base)
return str if n == 0
h = {}
str << '.'
n *= base
str.size.upto(Float::INFINITY) do |i|
r, n = n.divmod d
if n == 0
str << r.to_s(base)
break
elsif h.key? n
str[h[n], 0] = '('
str << ')'
break
else
str << r.to_s(base)
h[n] = i
n *= base
end
end
str
end
end
```
#### Updated by Anonymous almost 8 years ago
How about to_s( :decimal ) ?
• Description updated (diff)
#### Updated by shyouhei (Shyouhei Urabe)over 7 years ago
• Status changed from Assigned to Feedback
Can anyone propose a patch please?
#### Updated by mrkn (Kenta Murata)over 7 years ago
• Status changed from Feedback to Assigned
Actions #8
#### Updated by naruse (Yui NARUSE)over 3 years ago
• Target version deleted (2.6)
Actions
Also available in: Atom PDF | 529 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.581731 |
http://notsoformal.xyz/2019-03-16-monty-hall-problem-in-r/ | 1,586,356,146,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00029.warc.gz | 122,462,393 | 2,883 | Personal code snippets of @tmasjc
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# Monty Hall Problem in R
Mar 15, 2019 #monte-carlo
A few lines of code to one of the most classic problem in probability.
library(magrittr)
set.seed(1212)
new_game <- function(x = 3) {
# doors
doors = as.integer(1:x)
# initialize the prize behind of those doors
prize = sample(doors, size = 1)
# guest make a guess
guess = sample(doors, size = 1)
# open one of the doors knowing which has the prize
open = doors[-c(prize, guess)][[1]]
# if guest choose to switch
switch_guess = doors[-c(guess, open)]
# if guest does not switch
no_switch_guess = guess
return(list(
prize = prize,
original_guess = guess,
open = open,
switch = switch_guess,
no_switch = no_switch_guess
))
}
new_game()
## $prize ## [1] 2 ## ##$original_guess
## [1] 1
##
## $open ## [1] 3 ## ##$switch
## [1] 2
##
## \$no_switch
## [1] 1
# simulate many trials
trials = 10000
# if switch everytime, what are the total winnings?
replicate(trials, with(new_game(), switch == prize)) %>% sum()
## [1] 6720
# if NO switch everytime, what are the total winnings?
replicate(trials, with(new_game(), no_switch == prize)) %>% sum()
## [1] 3341
There you go. The math checks out. | 389 | 1,310 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-16 | longest | en | 0.823545 |
https://chem.libretexts.org/Courses/Prince_Georges_Community_College/Chemistry_2000%3A_Chemistry_for_Engineers_(Sinex)/Unit_1%3A_Atomic_Structure/Chapter_3%3A__The_Periodic_Table/Chapter_3.2%3A_Energetics_of_Ion_Formation | 1,623,899,896,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00246.warc.gz | 167,193,848 | 45,554 | # Chapter 3.2: Energetics of Ion Formation
Prince George's Community College General Chemistry for Engineering CHM 2000 Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry
Learning Objectives
• To correlate ionization energies, electron affinities, and electronegativities with the chemistry of the elements.
We have seen that when elements react, they often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation.
## Ionization Energies
Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy ($$I$$) (The minimum amount of energy needed to remove an electron from the gaseous atom in its ground state: $$E_{(g)}+ I \rightarrow E^++e^-$$) of an element as the amount of energy needed to remove an electron from the gaseous atom $$E$$ in its ground state. $$I$$ is therefore the energy required for the reaction
$$E\left ( g \right )+I\rightarrow E\left ( g \right )^{+}+e^{-} \; \; \; \; energy\; required=I \tag{3.2.1}$$
because an input of energy is required, the ionization energy is always positive ($$I > 0$$) for the reaction as written in Equation 3.2.1. Larger values of $$I$$ mean that the electron is more tightly bound to the atom and harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV):
1 eV/atom = 96.49 kJ/mol
An electron volt is the appropriate unit to describe the energy needed to remove an electron from one atom. kJ/mole is the energy needed to ionize a mole of atoms.
If an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define a first ionization energy ($$I_1$$), a second ionization energy ($$I_2$$), and in general an $$n^{th}$$ ionization energy ($$I_n$$) according to the following reactions:
• $$I_{1} = First\; Ionization\; energy \; \; \; \; E \left ( g \right ) + I_{1} \rightarrow E \left (g \right )^{+}+e^{-} \; \; \; \; \tag{3.2.2}$$
• $$I_{2} = Second\; Ionization\; energy \; \; \; \; E \left ( g \right )^{+} + I_{2} \rightarrow E \left ( g \right )^{2+}+e^{-}\; \; \; \; \tag{3.2.3}$$
• $$I_{n} = nth\; Ionization\; energy \; \; \; \; E \left ( g \right )^{\left ( n-1 \right )+} + I_{n} \rightarrow E \left ( g \right )^{n+}+e^{-} \; \; \; \; \tag{3.2.4}$$
Values for the ionization energies of Li and Be listed in Table 3.2.1 " show that successive ionization energies for an element increase steadily; that is, it takes more energy to remove the second electron from an atom than the first, and so forth. There are two reasons for this trend:
1. First, the second electron is being removed from a positively charged species rather than a neutral one, so in accordance with Coulomb’s law, more energy is required.
2. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger.
### Note the Pattern
Successive ionization energies for an element increase steadily.
Table 3.2.1 Ionization Energies (in kJ/mol) for Removing Successive Electrons from Li and Be
Reaction I Reaction I
$$\begin{array}{lcl} Li\left ( g \right ) & \rightarrow & Li^{+}+e^{-} \\ _{1s^{2}2s^{1}} & = & _{1s^{2}} \end{array}$$
I1 = 520.2
$$\begin{array}{lcl} Be\left ( g \right ) & \rightarrow & Be^{+}+e^{-} \\ _{1s^{2}2s^{2}} & = & _{1s^{2}2s^{1}} \end{array}$$
I1 = 899.5
$$\begin{array}{lcl} Li^{2+}\left ( g \right ) & \rightarrow & Li^{3+}+e^{-} \\ _{1s^{1}} & = & _{1s^{0}} \end{array} $$
I2 = 7298.2
$$\begin{array}{lcl} Be^{+}\left ( g \right ) & \rightarrow & Be^{2+}+e^{-} \\ _{1s^{2}2s^{1}} & = & _{1s^{2}} \end{array}$$
I2 = 1757.1
$$\begin{array}{lcl} Li^{2+}\left ( g \right ) & \rightarrow & Li^{3+}+e^{-} \\ _{1s^{1}} & = & _{1s^{0}} \end{array} $$
I3 = 11,815.0
$$\begin{array}{lcl} Be^{2+}\left ( g \right ) & \rightarrow & Be^{3+}+e^{-} \\ _{1s^{2}} & = & _{1s^{1}} \end{array}$$
I3 = 14,848.8
$$\begin{array}{lcl} Be^{3+}\left ( g \right ) & \rightarrow & Be^{4+}+e^{-} \\ _{1s^{1}} & = & _{1s^{0}} \end{array}$$
I4 = 21,006.6
The most important consequence of the values listed in Table 3.2.1 is that the chemistry of $$Li$$ is dominated by the $$Li^+$$ ion, while the chemistry of $$Be$$ is dominated by the +2 oxidation state. The energy required to remove the second electron from $$Li$$
$Li^+_{(g)} \rightarrow Li^{2+}_{(g)} + e^−$
is more than 10 times greater than the energy needed to remove the first electron. Similarly, the energy required to remove the third electron from $$Be$$
$Be^{2+}_{(g)} \rightarrow Be^{3+}_{(g)} + e^−$
is about 15 times greater than the energy needed to remove the first electron and around 8 times greater than the energy required to remove the second electron. Both $$Li^+$$ and $$Be^{2+}$$ have $$1s^2$$ closed-shell configurations, and much more energy is required to remove an electron from the 1s2 core than from the 2s valence orbital of the same element. The chemical consequences are enormous: lithium (and all the alkali metals) forms compounds with the 1+ ion but not the 2+ or 3+ ions. Similarly, beryllium (and all the alkaline earth metals) forms compounds with the 2+ ion but not the 3+ or 4+ ions. The energy required to remove electrons from a filled core is prohibitively large and simply cannot be achieved in normal chemical reactions.
### Note the Pattern
The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions.
## Ionization Energies of s- and p-Block Elements
Ionization energies of the elements in the third row of the periodic table exhibit the same pattern as those of Li and Be (Table 3.2.2 ): successive ionization energies increase steadily as electrons are removed from the valence orbitals (3s or 3p, in this case), followed by an especially large increase in ionization energy when electrons are removed from filled core levels as indicated by the bold diagonal line in Table 3.2.2 . Thus in the third row of the periodic table, the largest increase in ionization energy corresponds to removing the fourth electron from Al, the fifth electron from Si, and so forth—that is, removing an electron from an ion that has the valence electron configuration of the preceding noble gas. This pattern explains why the chemistry of the elements normally involves only valence electrons. Too much energy is required to either remove or share the inner electrons.
Table 3.2.2 Successive Ionization Energies (in kJ/mol) for the Elements in the Third Row of the Periodic Table
Element I 1 I 2 I 3 I 4 I 5 I 6 I 7
Na 495.8 4562.4*
Mg 737.7 1450.7 7732.7
Al 577.5 1816.7 2744.8 11,577.5
Si 786.5 1577.1 3231.6 4355.5 16,090.6
P 1011.8 1907.5 2914.1 4963.6 6274.0 21,267.4
S 999.6 2251.8 3357 4556.2 7004.3 8495.8 27,107.4
Cl 1251.2 2297.7 3822 5158.6 6540 9362 11,018.2
Ar 1520.6 2665.9 3931 5771 7238 8781.0 11,995.3
*Inner-shell electron
### Example 3.2.1
From their locations in the periodic table, predict which of these elements has the highest fourth ionization energy: B, C, or N.
Given: three elements
Asked for: element with highest fourth ionization energy
Strategy:
A List the electron configuration of each element.
B Determine whether electrons are being removed from a filled or partially filled valence shell. Predict which element has the highest fourth ionization energy, recognizing that the highest energy corresponds to the removal of electrons from a filled electron core.
Solution:
A These elements all lie in the second row of the periodic table and have the following electron configurations:
B: [He]2s22p1 C: [He]2s22p2 N: [He]2s22p3
B The fourth ionization energy of an element (I4) is defined as the energy required to remove the fourth electron:
E3+(g) → E4+(g) + e
Because carbon and nitrogen have four and five valence electrons, respectively, their fourth ionization energies correspond to removing an electron from a partially filled valence shell. The fourth ionization energy for boron, however, corresponds to removing an electron from the filled 1s2 subshell. This should require much more energy. The actual values are as follows: B, 25,026 kJ/mol; C, 6223 kJ/mol; and N, 7475 kJ/mol.
Exercise
From their locations in the periodic table, predict which of these elements has the lowest second ionization energy: Sr, Rb, or Ar.
The first column of data in Table 3.2.2 shows that first ionization energies tend to increase across the third row of the periodic table. This is because the valence electrons do not screen each other very well, allowing the effective nuclear charge to increase steadily across the row. The valence electrons are therefore attracted more strongly to the nucleus, so atomic sizes decrease and ionization energies increase. These effects represent two sides of the same coin: stronger electrostatic interactions between the electrons and the nucleus further increase the energy required to remove the electrons.
However, the first ionization energy decreases at Al ([Ne]3s23p1) and at S ([Ne]3s23p4). The electrons in aluminum’s filled 3s2 subshell are better at screening the 3p1 electron than they are at screening each other from the nuclear charge, so the s electrons penetrate closer to the nucleus than the p electron does. The decrease at S occurs because the two electrons in the same p orbital repel each other. This makes the S atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements.
The first ionization energies of the elements in the first six rows of the periodic table are plotted in Figure 3.2.1 . They are presented numerically and graphically in Figure 3.2.2 . These figures illustrate three important trends:
1. The changes seen in the second (Li to Ne), fourth (K to Kr), fifth (Rb to Xe), and sixth (Cs to Rn) rows of the s and p blocks follow a pattern similar to the pattern described for the third row of the periodic table. The transition metals are included in the fourth, fifth, and sixth rows, however, and the lanthanides are included in the sixth row. The first ionization energies of the transition metals are somewhat similar to one another, as are those of the lanthanides. Ionization energies increase from left to right across each row, with discrepancies occurring at ns2np1 (group 13), ns2np4 (group 16), and ns2(n − 1)d10 (group 12) electron configurations.
2. First ionization energies generally decrease down a column. Although the principal quantum number n increases down a column, filled inner shells are effective at screening the valence electrons, so there is a relatively small increase in the effective nuclear charge. Consequently, the atoms become larger as they acquire electrons. Valence electrons that are farther from the nucleus are less tightly bound, making them easier to remove, which causes ionization energies to decrease. A larger radius corresponds to a lower ionization energy.
3. Because of the first two trends, the elements that form positive ions most easily (have the lowest ionization energies) lie in the lower left corner of the periodic table, whereas those that are hardest to ionize lie in the upper right corner of the periodic table. Consequently, ionization energies generally increase diagonally from lower left (Cs) to upper right (He).
### Note the Pattern
Generally, I1 increases diagonally from the lower left of the periodic table to the upper right.
Figure 3.2.1 A Plot of Periodic Variation of First Ionization Energy with Atomic Number for the First Six Rows of the Periodic Table There is a decrease in ionization energy within a group (most easily seen here for groups 1 and 18). Can you explain the breaks in Figure 3.10 based on what you know about atomic structure and filled and half filled orbitals? Note the break after two electrons fill the ns orbitals and three the np orbitals. What does this say about ionizing half filled shells.
Figure 3.2.2 A Bar Chart of the Periodic Variation of First Ionization Energy with Atomic Number for the First Six Rows of the Periodic Table
Figure 3.2.3 First Ionization Energies of the s-, p-, d-, and f-Block Elements The darkness of the shading inside the cells of the table indicates the relative magnitudes of the ionization energies. Elements in gray have undetermined first ionization energies. Source: Data from CRC Handbook of Chemistry and Physics (2004). The CRC Handbook is a basic reference for Chemistry, Physics and Engineering. Every science and engineering major should own a copy. There are special student editions.
Gallium (Ga), which is the first element following the first row of transition metals, has the electron configuration: [Ar]4s23d104p1. Its first ionization energy is significantly lower than that of the immediately preceding element, zinc, because the filled 3d10 subshell of gallium lies inside the 4p subshell, screening the single 4p electron from the nucleus. Experiments have revealed something of even greater interest: the second and third electrons that are removed when gallium is ionized come from the 4s2 orbital, not the 3d10 subshell. The chemistry of gallium is dominated by the resulting Ga3+ ion, with its [Ar]3d10 electron configuration. This and similar electron configurations are particularly stable and are often encountered in the heavier p-block elements. They are sometimes referred to as pseudo noble gas configurations The (n-1)d10and similar electron configurations that are particularly stable and are often encountered in the heavier p-block elements.. In fact, for elements that exhibit these configurations, no chemical compounds are known in which electrons are removed from the (n − 1)d10 filled subshell.
## Ionization Energies of Transition Metals and Lanthanides
As we noted, the first ionization energies of the transition metals and the lanthanides change very little across each row. Differences in their second and third ionization energies are also rather small, in sharp contrast to the pattern seen with the s- and p-block elements. The reason for these similarities is that the transition metals and the lanthanides form cations by losing the ns electrons before the (n − 1)d or (n − 2)f electrons, respectively. This means that transition metal cations have (n − 1)dn valence electron configurations, and lanthanide cations have (n − 2)fn valence electron configurations. Because the (n − 1)d and (n − 2)f shells are closer to the nucleus than the ns shell, the (n − 1)d and (n − 2)f electrons screen the ns electrons quite effectively, reducing the effective nuclear charge felt by the ns electrons. As Z increases, the increasing positive charge is largely canceled by the electrons added to the (n − 1)d or (n − 2)f orbitals.
That the ns electrons are removed before the (n − 1)d or (n − 2)f electrons may surprise you because the orbitals were filled in the reverse order. (For more information on shell filling order, see Section 2.3 .) In fact, the ns, the (n − 1)d, and the (n − 2)f orbitals are so close to one another in energy, and interpenetrate one another so extensively, that very small changes in the effective nuclear charge can change the order of their energy levels. As the d orbitals are filled, the effective nuclear charge causes the 3d orbitals to be slightly lower in energy than the 4s orbitals. The [Ar]3d2 electron configuration of Ti2+ tells us that the 4s electrons of titanium are lost before the 3d electrons; this is confirmed by experiment. A similar pattern is seen with the lanthanides, producing cations with an (n − 2)fn valence electron configuration.
Because their first, second, and third ionization energies change so little across a row, these elements have important horizontal similarities in chemical properties in addition to the expected vertical similarities. For example, all the first-row transition metals except scandium form stable compounds as M2+ ions, whereas the lanthanides primarily form compounds in which they exist as M3+ ions.
### Example 3.2.2
Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr.
Given: six elements
Asked for: element with lowest first ionization energy
Strategy:
Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy.
Solution:
These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb.
Exercise
Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn.
## Electron Affinities
The electron affinity (EA)The energy change that occurs when an electron is added to a gaseous atom: E(g) + e- E-(g) of an element E is defined as the energy change that occurs when an electron is added to a gaseous atom:
$$E\left ( g \right )+e^{-}\rightarrow E\left ( g \right )^{-} \; \; \; \; energy\; change=Ea \tag{3.2.5}$$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with our discussion of energy changes later in the course, where a negative value corresponded to the energy change for an exothermic process, which is one in which heat is released. Such a process represents a loss of energy from the chemicals in a reaction.
Chlorine has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element:
$$Cl\left ( g \right )+e^{-}\rightarrow Cl\left ( g \right )^{-} \; \; \; \; Ea=-348.6\; kJ/mol\; \; \; \; \tag{3.2.6}$$
In contrast, beryllium does not form a stable anion, so its effective electron affinity is
$$Be\left ( g \right )+e^{-}\rightarrow Be\left ( g \right )^{-} \; \; \; \; Ea\geq 0\; kJ/mol\; \; \; \; \tag{3.2.7}$$
Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy. We might remember that nitrogen has a half filled p orbital which adds to its stability:
$$N\left ( g \right )+e^{-}\rightarrow N\left ( g \right )^{-} \; \; \; \; Ea\approx\; kJ/mol\; \; \; \; \tag{3.2.8}$$
Electron affinities for the first six rows of the periodic table are plotted in Figure 3.2.4 and presented numerically and graphically in Figure 3.2.6 . Both figures show that the halogens, with their ns2np5 valence electron configuration, have the most negative electron affinities. In general, electron affinities become more negative as we go across a row of the periodic table. This pattern corresponds to the increased effective nuclear charge felt by the valence electrons across a row, which leads to increased electrostatic attraction between the added electron and the nucleus (a more negative electron affinity). The trend, however, is not as uniform as the one observed for ionization energies. Some of the alkaline earths (group 2), the elements of group 12, and all the noble gases (group 18) have effective electron affinities that are greater than or equal to zero, while the electron affinities for the elements of group 15 are usually less negative than those for the group 14 elements. These exceptions can be explained by the groups’ electron configurations. Both the alkaline earth metals and the noble gases have valence electron shells with filled subshells (ns2 and ns2np6, respectively). In each case, the added electron must enter a higher-energy orbital, requiring an input of energy. All the group 15 elements have an ns2np3 valence electron configuration, in which each of the three p orbitals has a single electron, in accordance with Hund’s rule; hence the added electron must enter an already occupied p orbital. The resulting electron–electron repulsions destabilize the anion, causing the electron affinities of these elements to be less negative than we would otherwise expect. In the case of nitrogen, the 2p orbital is quite small, and the electron–electron repulsions are so strong that nitrogen has approximately zero affinity for an extra electron. In the heavier elements, however, the effect is relatively small because they have larger valence p orbitals.
### Note the Pattern
Generally, electron affinities become more negative across a row of the periodic table.
Figure 3.2.4 A Plot of Periodic Variation of Electron Affinity with Atomic Number for the First Six Rows of the Periodic TableCan you explain the breaks in Figure 3.2.4 based on what you know about atomic structure and filled and half filled orbitals? This is more difficult than the equivalent for ionization energy.
Figure 3.2.5 Bar Chart of the Periodic Variation of Electron Affinity with Atomic Number for the First Six Rows of the Periodic Table
Figure 3.2.6 Electron Affinities (in kJ/mol) of the s-, p-, and d-Block Elements There are many more exceptions to the trends across rows and down columns than with first ionization energies. Elements that do not form stable ions, such as the noble gases, are assigned an effective electron affinity that is greater than or equal to zero. Elements for which no data are available are shown in gray. Source: Data from Journal of Physical and Chemical Reference Data 28, no. 6 (1999).
In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as n increases, the extra electrons enter orbitals that are increasingly far from the nucleus. Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend:
1. The electron affinities of elements B through F in the second row of the periodic table are less negative than those of the elements immediately below them in the third row. Apparently, the increased electron–electron repulsions experienced by electrons confined to the relatively small 2p orbitals overcome the increased electron–nucleus attraction at short nuclear distances. Fluorine, therefore, has a lower affinity for an added electron than does chlorine. Consequently, the elements of the third row (n = 3) have the most negative electron affinities. Farther down a column, the attraction for an added electron decreases because the electron is entering an orbital more distant from the nucleus. Electron–electron repulsions also decrease because the valence electrons occupy a greater volume of space. These effects tend to cancel one another, so the changes in electron affinity within a family are much smaller than the changes in ionization energy.
2. The electron affinities of the alkaline earth metals become more negative from Be to Ba. As you learned in Chapter 2, the energy separation between the filled ns2 and the empty np subshells decreases with increasing n, so that formation of an anion from the heavier elements becomes energetically more favorable.
### Note the Pattern
In general, electron affinities become more negative across a row and less negative down a column.
The equations for second and higher electron affinities are analogous to those for second and higher ionization energies:
$$E\left ( g \right )+ e^{-}\rightarrow E\left ( g \right )^{-}\; \; energy\; change\; =Ea_{1} \; \tag{3.2.9}$$
$$E\left ( g \right )^{-}+ e^{-}\rightarrow E\left ( g \right )^{2-}\; \; energy\; change\; =Ea_{2} \; \tag{3.2.10}$$
As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol:
$$O\left ( g \right )+ e^{-}\rightarrow O\left ( g \right )^{-}\; \; Ea_{}=-141\; kJ/mol \; \tag{3.2.11}$$
$$O\left ( g \right )^{-}+ e^{-}\rightarrow O\left ( g \right )^{2-}\; \; Ea_{2}=+744\; kJ/mol \;$$
Thus the formation of a gaseous oxide (O2−) ion is energetically quite unfavorable:
$$O\left ( g \right )+ 2e^{-}\rightarrow O\left ( g \right )^{2-}\; \; Ea_{1}+Ea_{2}=+603\; kJ/mol \; \tag{3.2.12}$$
Similarly, the formation of all common dianions (such as S2−) or trianions (such as P3−) is energetically unfavorable in the gas phase.
### Note the Pattern
While first electron affinities can be negative, positive, or zero, second electron affinities are always positive.
If energy is required to form both positively charged ions and monatomic polyanions, why do ionic compounds such as MgO, Na2S, and Na3P form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions in the crystalline salt. We will describe the energetics of ionic compounds in more detail in Chapter 4 "Chemical Bonding".
### Example 3.2.3
Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity?
Given: three elements
Asked for: element with most negative electron affinity
Strategy:
A Locate the elements in the periodic table. Use the trends in electron affinities going down a column for elements in the same group. Similarly, use the trends in electron affinities from left to right for elements in the same row.
B Place the elements in order, listing the element with the most negative electron affinity first.
Solution:
A We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements.
Exercise
Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion?
## Electronegativity
The elements with the highest ionization energies are generally those with the most negative electron affinities, which are located toward the upper right corner of the periodic table (compare Figure 3.2.1 ). Conversely, the elements with the lowest ionization energies are generally those with the least negative electron affinities and are located in the lower left corner of the periodic table.
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativityThe relative ability of an atom to attract electrons to itself in a chemical compound. (represented by the Greek letter chi, χ, pronounced “ky” as in “sky”), defined as the relative ability of an atom to attract electrons to itself in a chemical compound. Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table.
Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property.
## The Pauling Electronegativity Scale
The original electronegativity scale, developed in the 1930s by Linus Pauling (1901– 1994) was based on measurements of the strengths of covalent bonds between different elements. Pauling arbitrarily set the electronegativity of fluorine at 4.0 (although today it has been refined to 3.98), thereby creating a scale in which all elements have values between 0 and 4.0.
Periodic variations in Pauling’s electronegativity values are illustrated in Figure 3.2.7 and Figure 3.2.9 . If we ignore the inert gases and elements for which no stable isotopes are known, we see that fluorine (χ = 3.98) is the most electronegative element and cesium is the least electronegative nonradioactive element (χ = 0.79). Because electronegativities generally increase diagonally from the lower left to the upper right of the periodic table, elements lying on diagonal lines running from upper left to lower right tend to have comparable values (e.g., O and Cl and N, S, and Br).
Figure 3.2.7 A Plot of Periodic Variation of Electronegativity with Atomic Number for the First Six Rows of the Periodic Table
Can you explain the breaks in Figure 3.2.7 based on what you know about atomic structure and filled and half filled d orbitals? Note how the electronegativity dependence on atomic number is smoother than the ionization and electron affinity.
Figure 3.2.8 A Bar Graph of Electronegativity for Atoms in the First Six Rows of the Periodic Table
### Linus Pauling (1901–1994)
Pauling won two Nobel Prizes, one for chemistry in 1954 and one for peace in 1962. When he was nine, Pauling’s father died, and his mother tried to convince him to quit school to support the family. He did not quit school but was denied a high school degree because of his refusal to take a civics class.
Figure 3.2.9 Pauling Electronegativity Values of the s-, p-, d-, and f-Block ElementsValues for most of the actinides are approximate. Elements for which no data are available are shown in gray. Source: Data from L. Pauling, The Nature of the Chemical Bond, 3rd ed. (1960).
Pauling’s method is limited by the fact that many elements do not form stable covalent compounds with other elements; hence their electronegativities cannot be measured by his method. Other definitions have since been developed that address this problem.
## The Mulliken Definition
An alternative method for measuring electronegativity was developed by Robert Mulliken (1896–1986; Nobel Prize in Chemistry 1966). Mulliken noticed that elements with large first ionization energies tend to have very negative electron affinities and gain electrons in chemical reactions. Conversely, elements with small first ionization energies tend to have slightly negative (or even positive) electron affinities and lose electrons in chemical reactions. Mulliken recognized that an atom’s tendency to gain or lose electrons could therefore be described quantitatively by the average of the values of its first ionization energy and the absolute value of its electron affinity. Using our definition of electron affinity, we can write Mulliken’s original expression for electronegativity as follows:Mulliken’s definition used the magnitude of the ionization energy and the electron affinity. By definition, the magnitude of a quantity is a positive number. Our definition of electron affinity produces negative values for the electron affinity for most elements, so vertical lines indicating absolute value are needed in Equation 3.2.13 to make sure that we are adding two positive numbers in the numerator.
$$x = \frac{I+\left | EA \right |}{2}\; \tag{3.2.13}$$
Bearing in mind that the magnitude of ionization energies are generally large compared to the those of electron affinities, elements with a large first ionization energy and a very negative electron affinity have a large positive value in the numerator of Equation 3.2.13 so their electronegativity is high. Elements with a small first ionization energy and a small electron affinity have a small positive value for the numerator in Equation 3.2.13, so they have a low electronegativity.Inserting the appropriate data for ionization energy and electron affinity into Equation 3.2.13 gives a Mulliken electronegativity value for fluorine of 1004.6 kJ/mol. To compare Mulliken’s electronegativity values with those obtained by Pauling, Mulliken’s values are divided by 252.4 kJ/mol, which gives Pauling’s value (3.98).
As noted previously, all electronegativity scales give essentially the same results for one element relative to another. Even though the Mulliken scale is based on the properties of individual atoms and the Pauling scale is based on the properties of atoms in molecules, they both apparently measure the same basic property of an element. In the following discussion, we will focus on the relationship between electronegativity and the tendency of atoms to form positive or negative ions. We will therefore be implicitly using the Mulliken definition of electronegativity. Because of the parallels between the Mulliken and Pauling definitions, however, the conclusions are likely to apply to atoms in molecules as well.
## Electronegativity Differences between Metals and Nonmetals
An element’s electronegativity provides us with a single value that we can use to characterize the chemistry of an element. Elements with a high electronegativity (χ ≥ 2.2 in Figure 3.2.9 ) have very negative electron affinities and large ionization potentials, so they are generally nonmetals and electrical insulators that tend to gain electrons in chemical reactions (i.e., they are oxidants). In contrast, elements with a low electronegativity (χ ≤ 1.8) have electron affinities that have either positive or small negative values and small ionization potentials, so they are generally metals and good electrical conductors that tend to lose their valence electrons in chemical reactions (i.e., they are reductants). In between the metals and nonmetals, along the diagonal line running from B to At is a group of elements with intermediate electronegativities (χ ~ 2.0). These are the semimetals, elements that have some of the chemical properties of both nonmetals and metals. The distinction between metals and nonmetals is one of the most fundamental we can make in categorizing the elements and predicting their chemical behavior. Figure 3.2.10 shows the strong correlation between electronegativity values, metallic versus nonmetallic character, and location in the periodic table.
Figure 3.2.10 Three-Dimensional Plots Demonstrating the Relationship between Electronegativity and the Metallic/Nonmetallic Character of the Elements (a) A plot of electrical resistivity (measured resistivity to electron flow) at or near room temperature shows that substances with high resistivity (little to no measured electron flow) are electrical insulators, whereas substances with low resistivity (high measured electron flow) are metals. (b) A plot of Pauling electronegativities for a like set of elements shows that high electronegativity values (≥ about 2.2) correlate with high electrical resistivities (insulators). Low electronegativity values (≤ about 2.2) correlate with low resistivities (metals). Because electrical resistivity is typically measured only for solids and liquids, the gaseous elements do not appear in part (a).
### Example 3.2.4
On the basis of their positions in the periodic table, arrange Cl, Se, Si, and Sr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a semimetal.
Given: four elements
Asked for: order by increasing electronegativity and classification
Strategy:
A Locate the elements in the periodic table. From their diagonal positions from lower left to upper right, predict their relative electronegativities.
B Arrange the elements in order of increasing electronegativity.
C Classify each element as a metal, a nonmetal, or a semimetal according to its location about the diagonal belt of semimetals running from B to At.
Solution:
A Electronegativity increases from lower left to upper right in the periodic table (Figure 3.2.8 ). Because Sr lies far to the left of the other elements given, we can predict that it will have the lowest electronegativity. Because Cl lies above and to the right of Se, we can predict that χCl > χSe. Because Si is located farther from the upper right corner than Se or Cl, its electronegativity should be lower than those of Se and Cl but greater than that of Sr. B The overall order is therefore χSr < χSi < χSe < χCl.
C To classify the elements, we note that Sr lies well to the left of the diagonal belt of semimetals running from B to At; while Se and Cl lie to the right and Si lies in the middle. We can predict that Sr is a metal, Si is a semimetal, and Se and Cl are nonmetals.
Exercise
On the basis of their positions in the periodic table, arrange Ge, N, O, Rb, and Zr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a semimetal.
Answer: Rb < Zr < Ge < N < O; metals (Rb, Zr); semimetal (Ge); nonmetal (N, O)
### Note the Pattern
Electronegativity values increase from lower left to upper right in the periodic table.
The trends in periodic properties are summarized in Figure 3.2.11. As discussed, atomic radii decrease from lower left to upper right in the periodic table; ionization energies become more positive, electron affinities become more negative, and electronegativities increase from the lower left to the upper right.
Figure 3.2.11 Summary of Major Periodic TrendsThe general trends for the first ionization energy, electron affinity, and electronegativity are opposite to the general trend for covalent atomic radius.
### Summary
The tendency of an element to lose or gain electrons is one of the most important factors in determining the kind of compounds it forms. Periodic behavior is most evident for ionization energy (I), the energy required to remove an electron from a gaseous atom. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell. Consequently, only valence electrons can be removed in chemical reactions, leaving the filled inner shell intact. Ionization energies explain the common oxidation states observed for the elements. Ionization energies increase diagonally from the lower left of the periodic table to the upper right. Minor deviations from this trend can be explained in terms of particularly stable electronic configurations, called pseudo noble gas configurations, in either the parent atom or the resulting ion. The electron affinity (EA) of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity. Elements with a high electronegativity are generally nonmetals and electrical insulators and tend to behave as oxidants in chemical reactions. Conversely, elements with a low electronegativity are generally metals and good electrical conductors and tend to behave as reductants in chemical reactions.
### Key Takeaway
• Generally, the first ionization energy and electronegativity values increase diagonally from the lower left of the periodic table to the upper right, and electron affinities become more negative across a row.
### Conceptual Problems
1. Identify each statement as either true or false and explain your reasoning.
1. Ionization energies increase with atomic radius.
2. Ionization energies decrease down a group.
3. Ionization energies increase with an increase in the magnitude of the electron affinity.
4. Ionization energies decrease diagonally across the periodic table from He to Cs.
5. Ionization energies depend on electron configuration.
6. Ionization energies decrease across a row.
2. Based on electronic configurations, explain why the first ionization energies of the group 16 elements are lower than those of the group 15 elements, which is contrary to the general trend.
3. The first through third ionization energies do not vary greatly across the lanthanides. Why? How does the effective nuclear charge experienced by the ns electron change when going from left to right (with increasing atomic number) in this series?
4. Most of the first row transition metals can form at least two stable cations, for example iron(II) and iron(III). In contrast, scandium and zinc each form only a single cation, the Sc3+ and Zn2+ ions, respectively. Use the electron configuration of these elements to provide an explanation.
5. Of the elements Nd, Al, and Ar, which will readily form(s) +3 ions? Why?
6. Orbital energies can reverse when an element is ionized. Of the ions B3+, Ga3+, Pr3+, Cr3+, and As3+, in which would you expect this reversal to occur? Explain your reasoning.
7. The periodic trends in electron affinities are not as regular as periodic trends in ionization energies, even though the processes are essentially the converse of one another. Why are there so many more exceptions to the trends in electron affinities compared to ionization energies?
8. Elements lying on a lower right to upper left diagonal line cannot be arranged in order of increasing electronegativity according to where they occur in the periodic table. Why?
9. Why do ionic compounds form, if energy is required to form gaseous cations?
10. Why is Pauling’s definition of electronegativity considered to be somewhat limited?
11. Based on their positions in the periodic table, arrange Sb, O, P, Mo, K, and H in order of increasing electronegativity.
12. Based on their positions in the periodic table, arrange V, F, B, In, Na, and S in order of decreasing electronegativity.
1. Both Al and Nd will form a cation with a +3 charge. Aluminum is in Group 13, and loss of all three valence electrons will produce the Al3+ ion with a noble gas configuration. Neodymium is a lanthanide, and all of the lanthanides tend to form +3 ions because the ionization potentials do not vary greatly across the row, and a +3 charge can be achieved with many oxidants.
2. K < Mo ≈ Sb < P ≈ H < O
### Numerical Problems
1. The following table gives values of the first and third ionization energies for selected elements:
Number of Electrons Element I1 (E → E+ + e, kJ/mol) Element I3 (E2+ → E3+ + e, kJ/mol)
11 Na 495.9 Al 2744.8
12 Mg 737.8 Si 3231.6
13 Al 577.6 P 2914.1
14 Si 786.6 S 3357
15 P 1011.9 Cl 3822
16 S 999.6 Ar 3931
17 Cl 1251.2 K 4419.6
18 Ar 1520.6 Ca 4912.4
Plot the ionization energies versus number of electrons. Explain why the slopes of the I1 and I3 plots are different, even though the species in each row of the table have the same electron configurations.
2. Would you expect the third ionization energy of iron, corresponding to the removal of an electron from a gaseous Fe2+ ion, to be larger or smaller than the fourth ionization energy, corresponding to removal of an electron from a gaseous Fe3+ ion? Why? How would these ionization energies compare to the first ionization energy of Ca?
3. Which would you expect to have the highest first ionization energy: Mg, Al, or Si? Which would you expect to have the highest third ionization energy. Why?
4. Use the values of the first ionization energies given in Figure 3.3.3 to construct plots of first ionization energy versus atomic number for (a) boron through oxygen in the second period; and (b) oxygen through tellurium in group 16. Which plot shows more variation? Explain the reason for the variation in first ionization energies for this group of elements.
5. Arrange Ga, In, and Zn in order of increasing first ionization energies. Would the order be the same for second and third ionization energies? Explain your reasoning.
6. Arrange each set of elements in order of increasing magnitude of electron affinity.
1. Pb, Bi, and Te
2. Na, K, and Rb
3. P, C, and Ge
7. Arrange each set of elements in order of decreasing magnitude of electron affinity.
1. As, Bi, and N
2. O, F, and Ar
3. Cs, Ba, and Rb
8. Of the species F, O, Al3+, and Li+, which has the highest electron affinity? Explain your reasoning.
9. Of the species O, N2−, Hg2+, and H+, which has the highest electron affinity? Which has the lowest electron affinity? Justify your answers.
10. The Mulliken electronegativity of element A is 542 kJ/mol. If the electron affinity of A is −72 kJ/mol, what is the first ionization energy of element A? Use the data in the following table as a guideline to decide if A is a metal, a nonmetal, or a semimetal. If 1 g of A contains 4.85 × 1021 molecules, what is the identity of element A?
Na Al Si S Cl
EA (kJ/mol) −59.6 −41.8 −134.1 −200.4 −348.6
I (kJ/mol) 495.8 577.5 786.5 999.6 1251.2
11. Based on their valence electron configurations, classify the following elements as either electrical insulators, electrical conductors, or substances with intermediate conductivity: S, Ba, Fe, Al, Te, Be, O, C, P, Sc, W, Na, B, and Rb.
12. Using the data in Problem 10, what conclusions can you draw with regard to the relationship between electronegativity and electrical properties? Estimate the approximate electronegativity of a pure element that is very dense, lustrous, and malleable.
1. The general features of both plots are roughly the same, with a small peak at 12 electrons and an essentially level region from 15–16 electrons. The slope of the I3 plot is about twice as large as the slope of the I1 plot, however, because the I3 values correspond to removing an electron from an ion with a +2 charge rather than a neutral atom. The greater charge increases the effect of the steady rise in effective nuclear charge across the row.
2. Electron configurations: Mg, 1s22s22p63s2; Al, 1s22s22p63s23p1; Si, 1s22s22p63s23p2; First ionization energies increase across the row due to a steady increase in effective nuclear charge; thus, Si has the highest first ionization energy. The third ionization energy corresponds to removal of a 3s electron for Al and Si, but for Mg it involves removing a 2p electron from a filled inner shell; consequently, the third ionization energy of Mg is the highest.
1. Bi > As > N
2. F > O >> Ar
3. Rb > Cs > Ba
3. Hg2+ > H+ > O > N2−; Hg2+ has the highest positive charge plus a relatively low energy vacant set of orbitals (the 6p subshell) to accommodate an added electron, giving it the greatest electron affinity; N2− has a greater negative charge than O, so electron–electron repulsions will cause its electron affinity to be even lower (more negative) than that of O.
4. insulators: S, O, C (diamond), P; conductors: Ba, Fe, Al, C (graphite), Be, Sc, W, Na, Rb; Te and B are semimetals and semiconductors.
### Contributors
• Anonymous
Modified by Joshua Halpern, Scott Sinex and Scott Johnson | 11,773 | 48,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.779031 |
https://es.mathworks.com/matlabcentral/cody/problems/42686-what-s-the-missing-interior-angle/solutions/960195 | 1,606,661,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00303.warc.gz | 283,521,762 | 16,731 | Cody
# Problem 42686. What's the missing interior angle?
Solution 960195
Submitted on 6 Sep 2016 by Ben Petschel
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [60 60]; y_correct = 60; assert(isequal(missing_interior_angle(x),y_correct))
2 Pass
x = [45 90]; y_correct = 45; assert(isequal(missing_interior_angle(x),y_correct))
3 Pass
x = [90 70 70]; y_correct = 130; assert(isequal(missing_interior_angle(x),y_correct))
4 Pass
x = [120 120 120 110 110]; y_correct = 140; assert(isequal(missing_interior_angle(x),y_correct))
5 Pass
x = 140*ones(1,8); y_correct = 140; assert(isequal(missing_interior_angle(x),y_correct))
### Community Treasure Hunt
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Start Hunting! | 260 | 887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.47974 |
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1. Jun 11, 2013
### Steve Zissou
Hello physicians.
Consider the following Brownian motion particle:
$$\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)$$
The kinetic energy of which would be
$$\frac{1}{2}v^2=\frac{1}{2}(\dot{x}(t))^2$$
(for some unit mass.)
The potential is...?
2. Jun 11, 2013
### sudu.ghonge
In-case you're trying to refer to people who study physics, they're called 'physicists'. A 'Physician' is a professional who practices medicine.
Please be more elaborate. I have not studied brownian motion in depth but I can surely say that
$$\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)$$
can also be written as
$$\dot{x}(t) = \gamma(t)$$
where $$\gamma(t) = \alpha(t)+\beta(t)\eta(t)$$
And brownian particles can be found in a variety of time dependent as well as independent potentials. And usually, the potential is a function of the position variables alone( In this case, 'x'. So, it might help if you could provide a relationship between, say, the force and position variables.
3. Jun 11, 2013
### Steve Zissou
Sudu. Get a grip. I know what a physician is, and what a physicist is. It's what we call "humor."
Secondly, the way I've written it is the correct way to write it. There is a huge weight of convention and tradition here, and I'm hoping someone who understands that will help me out.
4. Jun 11, 2013
### sudu.ghonge
No, we don't. And as far as I remember, this is not a forum where ideas of humor are discussed. You might very well have been a person who makes the common mistake. I don't need to "get a grip".
Peace out.
Can you cite a source? I searched extensively and nowhere did I find this convention.
5. Jun 11, 2013
### Steve Zissou
Ok I'll try again. Hopefully this will clarify.
Consider the well-known Langevin equation of a particle in Brownian motion.
Here is a reference:
http://www.mat.univie.ac.at/~esiprpr/esi2115.pdf
In the attached paper, the Langevin equation is written as:
$$\frac{\partial p_i}{\partial t}=-\gamma_0 p_i-\frac{\partial U}{\partial q_i}+\eta_i$$
.where the gamma is a friction term and the U is potential.
I would like to characterize a more general model. Let's say it like this:
$$\dot{x}(t)=\alpha(t)+\beta(t)\eta(t)$$
If we would like to, we are free to simply say
$$\dot{x}(t)=\alpha(t)+\eta(t)$$.
Now, how would we characterize the potential here in my generalization? Thanks to all who would help.
6. Jun 11, 2013
### willem2
You have an equation for the derivative of position (velocity), whereas the Langevin equationis for the derivative of momentum, wich is equal to a force.
It's probably best to try to start with the langevin equation itself and try to vary that. The friction gamma could be a function of the velocity, for example, or the potential could vary with time (such as when you shake the fluid that the particle is in), and you'll have no problem with the potential, because you can just copy the term with U.
of course you could also use a force that is dependent of the position (or possibly constant) and do away with the potential.
7. Jun 11, 2013
### Steve Zissou
Willem, thanks for the reply. Yes I am aware that there are important issues regarding the units of measure involved. I am aware that a force can be the grad of potential.
Let me try to clarify my question. There are many possible models for Brownian motion, not just Langevin. Also the Langevin equation need not be written in terms of momenta, it can be written in terms of distance formt he origin, &c. I am trying to get toward a generalization. Let's say we have an "equation of motion" for a particle under Brownian motion. We'll express it as:
whatever you like = a(t) + eta(t)
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TV Fool Home FM Fool Home FAQ Members List Calendar Search Mark Forums Read Register
Notices For information on some common antennas, check out our Antenna Quick Reference. More antennas are still being added, so check back for updates.
5-Aug-2015, 8:24 PM #1 Maury Markowitz Junior Member Join Date: Sep 2011 Posts: 21 Ganging/merging, yes or no? I live outside of Toronto, I have one set of antennas about 25 miles away at 245 degrees, and another 65 miles away at 175 degrees. In spite of using a cheap-o 4-bay and pointing it at about 200 degrees, I do get a reasonable selection of channels from both directions. But its not a *great* selection. So I'm thinking of "merging" two antennas. Given the 70 degrees between the two sets, it seems like a perfect setup, right? Reading HDPrimer's page on the topic: http://www.hdtvprimer.com/ANTENNAS/merging.html it seems that you cannot easily combine two antennas unless they are pointed in the same direction, otherwise you will suffer 1/2 signal loss. But... First off, what does 1/2 signal loss mean? If I have an antenna with 16 db of gain for a particular channel, does that mean the output is 8 db, or 16-3=13 db? If it is the later, am I correct in thinking that as long as pointing the antenna at the channel gets me more than 3 db improvement, I should do that, right? And finally, is any of this correct? Someone once told me the loss is only true if the same channel is on both antennas. In my case with 70 degrees, the reception should be close to zero in the "other" antenna for every channel. But HDPrimer seems to be suggesting this effect is purely in the combiner, so even if you just put in the combiner and no second antenna it would still suffer the loss. This seems more logical to me, but antenna's and logic do not yet mix in my brain...
5-Aug-2015, 10:58 PM #2 Jake V Senior Member Join Date: Jan 2014 Location: Virginia! Posts: 327 I'd first recommend posting your TV Fool Report, the antennas you are using, their location (roof, attic, etc.), any splitters or amps, etc. It sound like you are somewhere near Ajax. That's pretty flat so Toronto should be easy. Buffalo would likely require something bigger than a "cheap-o 4-bay". As to combining antennas, sometimes it works and sometimes it does not (usually it does not).
6-Aug-2015, 12:03 AM #3
Maury Markowitz
Junior Member
Join Date: Sep 2011
Posts: 21
Quote:
Originally Posted by Jake V As to combining antennas, sometimes it works and sometimes it does not (usually it does not).
Is there a good, theoretically based, answer to the questions?
I want to understand this in general, not in the specific.
6-Aug-2015, 1:59 AM #4
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Quote:
First off, what does 1/2 signal loss mean? If I have an antenna with 16 db of gain for a particular channel, does that mean the output is 8 db, or 16-3=13 db?
Half the signal loss is 3 dB. But that's not the main problem when using a splitter in reverse as a combiner.
If you combine two identical antennas aimed in the same direction, with identical length feed lines, the gain is 3 dB minus the internal splitter loss of 0.5 dB, for a max possible gain of 2.5 dB.
If you combine two identical antennas not aimed in the same direction, the same signals will interfere with each other when they reach the combining point if they are not in phase. The length of the feed lines is irrelevant in this case.
If you combine two antennas, one for VHF (real channels 2-13) and one for UHF (real channels 14-51), with a UVSJ diplexer, they will not interfere with each other because the UVSJ keeps the two bands separate.
If you use one of the new 8-bay UHF antennas with separate panels that can be aimed in different directions, sometimes it works and sometimes not because of the phase interference mentioned above. It's trial-and-error, with the two panels at 90 degrees apart for your best chances.
Quote:
And finally, is any of this correct? Someone once told me the loss is only true if the same channel is on both antennas. In my case with 70 degrees, the reception should be close to zero in the "other" antenna for every channel.
Yes, it is, but it is a matter of the strength and phase of interference, not all-or-nothing. What usually happens when there is a problem, some of your channels will be missing after combining.
We have several members here that can do a good job with the antenna theory, but the guys in your area at DHC can do a better job with the specifics of your location. But we will give it a try anyway, if you so desire.
Quote:
I do get a reasonable selection of channels from both directions.
What are you getting now by callsign and real channel number?
Quote:
But its not a *great* selection.
What would you consider great, that you aren't getting now?
But first, as Jake V requested, please post a link to an exact address (which will not show) or coordinates (which will be shortened) tvfool report so that we can make an accurate analysis.
How close are my guesses to your exact address report?
http://www.tvfool.com/?option=com_wr...8e03088b382880
http://www.tvfool.com/?option=com_wr...8e03cd784464bd
__________________
If you can not measure it, you can not improve it.
Lord Kelvin, 1883
http://www.megalithia.com/elect/aeri...ttpoorman.html
Last edited by rabbit73; 6-Aug-2015 at 3:22 AM.
6-Aug-2015, 12:19 PM #5
Maury Markowitz
Junior Member
Join Date: Sep 2011
Posts: 21
Quote:
Originally Posted by rabbit73 If you combine two identical antennas not aimed in the same direction, the same signals will interfere with each other when they reach the combining point if they are not in phase.
This is what I thought, and it is not covered on HDTVPrimer - or if it is, it is hidden in the discussion of combiners.
In this case, two antennas in different directions, is it not the case that if one of them is 90 degrees off the other, the phasing issue will be eliminated because the output from the second will be zero?
But I still have the combiner loss too, right? And is that loss 3 db?
Quote:
Originally Posted by rabbit73 What are you getting now by callsign and real channel number? What would you consider great, that you aren't getting now?
http://www.tvfool.com/?option=com_wr...8e0347474ce022
I'm currently getting lots of channels, including many that the system says are 2edge. On the chart at the link above, I most everything up to WKBW, the exceptions being WUTV and CHEX analog (poor image, no sound, don't care. 9, in high VHF, is in and out.
And that's pretty good, but it will start fading when it gets cooler. By the end of October most of the Buffalo stations will be marginal or gone, and 9 will be barely usable.
So, I'd like to fix the fading, and ideally add ION to my list.
6-Aug-2015, 2:02 PM #6
rickbb
Senior Member
Join Date: Dec 2014
Posts: 289
Quote:
Originally Posted by Maury Markowitz In this case, two antennas in different directions, is it not the case that if one of them is 90 degrees off the other, the phasing issue will be eliminated because the output from the second will be zero? But I still have the combiner loss too, right? And is that loss 3 db?
With the 2 at 90 degrees you "might" get good results by taking advantage of what's called the null direction of the antennas. This works best on directional antennas with narrow beam widths. One antenna is aimed at the others null signal zone, so to speak.
And yes, combiner loss will still be there, usually slightly more than 3db as no combiner is 100% efficient.
I've tried this with 2 DB4's at 90 degrees apart and did not get any better results than a single DB4. These were DYI antennas so I may have other issues in the mix, but I was not impressed enough to continue with the project at this time. I haven't given up on it, but it's seems less likely to be of any improvement to my reception.
Forum member holl_ands did a computer simulation of a vertical DB8 with 90 degree separation, (basically 2 DB4's vertically stacked), and the computer model does show it should be possible to achieve multi direction reception. I used his model in my tests.
6-Aug-2015, 2:03 PM #7
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Quote:
In this case, two antennas in different directions, is it not the case that if one of them is 90 degrees off the other, the phasing issue will be eliminated because the output from the second will be zero?
No, because it is not likely that the output will be zero. It is true that the pattern null of one antenna at right angles to the other will attenuate the unwanted signals from the other, but it's not zero. Any interference from the other antenna will cause digital errors. The FEC (Forward Error Correction) has a limit to the number of errors that it can correct. Once that limit is exceeded, the signal reaches the "Digital Cliff" where you have pixilation, picture freeze, and finally dropout.
Quote:
But I still have the combiner loss too, right? And is that loss 3 db?
HDTV primer has led you down the garden path to a false conclusion.
The inherent combiner loss is only about 0.5 dB.
The 3 dB loss that he mentions is the loss that you would have gained if the two antennas were aimed in the same direction.
Don't confuse the 3 dB loss of a splitter used as a splitter, with the 0.5 dB internal loss of a splitter used as a combiner.
Combining two UHF antennas aimed in different directions sometimes works, and sometimes not. There is no way that I can predict or guarantee the outcome. You must use the empirical approach, which is trial-and-error to find out. I certainly don't want to discourage antenna experimentation. I have always learned something from my experiments, especially the ones that didn't work, because I then had to find out why.
I don't think you will be happy until you try it.
__________________
If you can not measure it, you can not improve it.
Lord Kelvin, 1883
http://www.megalithia.com/elect/aeri...ttpoorman.html
Last edited by rabbit73; 6-Aug-2015 at 5:38 PM.
6-Aug-2015, 5:05 PM #8
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Thanks for the new tvfool report. I see that you have moved it from Emperor & Pickering Beach to L1S2W9 O'Dell Ct and Turnbull Rd. It doesn't make much difference with LOS signals but it makes a lot of difference with your weak 1 and 2Edge signals.
Quote:
I'm currently getting lots of channels, including many that the system says are 2edge. On the chart at the link above, I most everything up to WKBW, the exceptions being WUTV and CHEX analog (poor image, no sound, don't care. 9, in high VHF, is in and out.
That sounds pretty good to me.
WUTV is on real channel 14; most 4-bay antennas don't have as much gain at the low end of UHF. And they have even less gain for VHF channel 9.
Quote:
So, I'd like to fix the fading, and ideally add ION to my list.
ION is pretty weak, has adjacent channel interference from CHEX analog on 22, and is at 153 degrees magnetic.
http://www.digitalhome.ca/forum/2495601-post2041.html
Quote:
I'm in Ajax, just east of Toronto. My Toronto stations are at 245 magnetic, and the Buffalo ones at 165. If I point at Toronto I don't get anything from Buffalo, if I point at Buffalo I lose CFTO and the rest get spotty. So I pointed it around 185, splitting the difference.
http://www.digitalhome.ca/forum/2277401-post1100.html
Quote:
I live in Ajax, just east of Toronto. Using the south-most edge of my roofline as "south" (I don't know what it really is), Buffalo is at about 185 degrees, and the CN Tower is around 255. I have a knock-off four-bay and found I can get a sort of compromise if I point it around 195 degrees. That gets me most of Toronto except 9, and Buffalo kinda works. If I point directly at Toronto or Buffalo I get everything from that location, but of course that's non-optimal.
http://www.digitalhome.ca/forum/2299185-post1974.html
Quote:
Buffalo is about 160-175 degrees true and the CN Tower is 235. I currently have a four-bay pointed around 190, which sort-a works. I'm debating whether to add another antenna. My concern is that I have little room to work with, and another 4-bay wouldn't fit. However, if I were to point the 4-bay at the CN Tower, there would be room behind it for something more linear, like a Yagi. Whatever I do use, I want to be sure that I strongly reject the signals from the Tower. I always thought a Yagi was more directional and would be better at this, but looking over reception charts I can't really see any difference?
Splitting the difference doesn't always work. For best results the antenna should be aimed directly at the transmitter.
Before combining two antennas aimed in different directions, you must be certain that each antenna gets what you want from that direction. Otherwise, there is no point in combining; it's not going to get better.
If splitting the difference doesn't work, and combining doesn't work, you must switch between antennas with an A/B switch, which requires you to rescan after switching unless your TV can add a new channel after scan like a Sony.
An alternative would be to have your primary antenna connected directly to the TV antenna input, and your secondary antenna connected to a separate tuner with its output connected to an aux input of the TV.
You haven't told us much about your antennas, where they are located, and if you are using a preamp.
The way that I optimize aim for an antenna is first aim for max signal strength. Then I adjust aim for best signal quality as defined by SNR and errors. The two are not always at the same azimuth. Signal quality is at least as important as signal strength for digital TV signals.
I first learned about the importance of signal quality when using an Apex DT502 converter box with dual signal bars that measure signal quality and signal strength. I first adjusted aim for max signal strength. I then turned the antenna slightly to the right; the signal strength went down a little, but the signal quality went way up.
Attached Images
DualSigBars.jpg (43.9 KB, 1430 views)
__________________
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Last edited by rabbit73; 6-Aug-2015 at 6:33 PM.
7-Aug-2015, 5:18 PM #9 Maury Markowitz Junior Member Join Date: Sep 2011 Posts: 21 The antenna is a no-name, and I mean that literally there was no name on the box, at least in English. It is mounted very close to the peak of the roof using a J-mast. The antenna has a clear view over other houses out to about 250 feet. My current thinking is to use two entirely separate antennas, equip each one with a USB tuner right on the back of the antenna, run their USB's into a Raspberry Pi running MythTV backend, and then run that into the house using Ether (POE to power everything). That way I get live TV everywhere. The question, then, becomes whether I should use a Yagi for the Buffalo direction. I've asked a couple of places, but no straight answers yet. Last edited by Maury Markowitz; 7-Aug-2015 at 5:21 PM.
7-Aug-2015, 5:19 PM #10
Maury Markowitz
Junior Member
Join Date: Sep 2011
Posts: 21
Quote:
Originally Posted by rickbb I've tried this with 2 DB4's at 90 degrees apart and did not get any better results than a single DB4.
Very useful info Rick!
Quote:
Originally Posted by rickbb Forum member holl_ands did a computer simulation of a vertical DB8 with 90 degree separation, (basically 2 DB4's vertically stacked), and the computer model does show it should be possible to achieve multi direction reception. I used his model in my tests.
Do you have a pointer to this? Google turns up something unrelated.
7-Aug-2015, 5:29 PM #11
Jake V
Senior Member
Join Date: Jan 2014
Location: Virginia!
Posts: 327
Quote:
Originally Posted by Maury Markowitz The antenna is a no-name, and I mean that literally there was no name on the box, at least in English. It is mounted very close to the peak of the roof using a J-mast. The antenna has a clear view over other houses out to about 250 feet.
Can you post a photo of the no-name antenna?
And maybe use a real compass to verify the direction it is pointed in?
9-Aug-2015, 1:40 AM #12
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Quote:
My current thinking is to use two entirely separate antennas, equip each one with a USB tuner right on the back of the antenna, run their USB's into a Raspberry Pi running MythTV backend, and then run that into the house using Ether (POE to power everything). That way I get live TV everywhere.
Quote:
So, I'd like to fix the fading, and ideally add ION to my list.
Quote:
The question, then, becomes whether I should use a Yagi for the Buffalo direction. I've asked a couple of places, but no straight answers yet.
Thanks for telling us a little more about your goals; it helps us answer.
I like your idea of two antennas, but I don't think it's a good idea to put the tuners outside in the weather. Maybe this idea by GroundUrMast would work for you if you want to involve computers. I prefer to watch TV on a TV set myself; computers are too fickle and user-unfriendly for me:
An Alternative to Rotators and Antenna Combiners
http://forum.tvfool.com/showthread.php?t=820
ION isn't going to be easy. The curvature of the earth starts to interfere with TV signals at about 70 miles. If you look at the terrain profile by clicking on its callsign in your TVFOOL report, you will see a black arc at the bottom of the profile; that's the earth. And you will also see that only a small portion of the ERP is sent in your direction:
and if you look at the coverage map, you will see that the signal barely makes it to your location:
Antennas Direct makes a popular yagi, the 91XG. Yagis don't cover as wide a frequency range as collinear arrays, like the DB4E, DB8E, CM4221, and CM4228 which are more commonly called 4-bay and 8-bay bowtie antennas. The 91XG gain curve is low at low end of the UHF band, and peaks at the high end. Since you want gain for channel 23, you can see that it doesn't perform as well at that frequency: 524 to 530 MHz, as it does at the high end.
https://en.wikipedia.org/wiki/North_...on_frequencies
https://www.antennasdirect.com/cmss_...y/91XG-TDS.pdf
Collinear arrays have a flatter gain VS frequency curve. This is the tech sheet for the DB8E:
https://www.antennasdirect.com/cmss_files/attachmentlibrary/Technical%20Data%20PDF's/DB8E-TDS.pdf
and the DB4E:
https://www.antennasdirect.com/cmss_...y/DB4E-TDS.pdf
Notice also the tradeoff between gain and beamwidth. The higher gain antenna has a narrower beamwidth, which makes it more critical to aim.
Attached Images
Maury MarkowitzTVFprofile.JPG (62.1 KB, 1407 views) Maury MarkowitzTVFcoverage.JPG (101.8 KB, 1344 views)
__________________
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Lord Kelvin, 1883
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Last edited by rabbit73; 9-Aug-2015 at 11:56 AM.
9-Aug-2015, 12:48 PM #13 Maury Markowitz Junior Member Join Date: Sep 2011 Posts: 21 First, antennas: I used to get ION when I first set up the antenna. To do so I had to point the antenna to about 160 degrees, but then I also got all the Buffalo stations just fine. But nothing from Toronto. So I rotated it to catch Toronto and then of course ION disappeared. I should also note that I can get some of the other NY stations that are just as "deep" into the hinterland as ION right now. So I'm relatively confident I can get ION again, at least in the summer, with an antenna pointed in that direction. If I lose it in the winter, c'est la vie. Following your logic I'll stick with the array. The only channel that really requires that is Fox, and that one is kinda marginal right now. Tuners: Nice link rabbit! That thread is doing just what I propose, so it's good to see I'm not the only one thinking this way! (BTW I did Google for people doing this, but this did not turn up) I looked at the SiliconDust machine, but as that thread notes, they removed one of the antenna inputs circa 2010, which kind of ruins it for me. I understand why they did this, but my situation demands the former design and they disappeared from the market pronto. I also looked at the Tablo, which has the same issue. Being from Canada I had a lengthy conversation with them about this issue, but they seem unconvinced. Which is odd, because I believe both of these boxes are nothing more than MythTV back-end running on a lightweight PC - precisely what I'm hoping to do, but replacing the PC with a Raspberry Pi. So what both of them could do is put a USB port on the back and periodically scan to see if someone plugged in a tuner. The real problem, as you note, is watching it. As to that, I already watch perhaps 90 to 95% of my time using streamed content on my Apple TV. So what we really want is to have all of these products feed into that. Of course the Apple TV famously does NOT allow this... Well that's about to change. A new one almost certainly comes out on 9 September, and even more likely is a programming kit. If this comes to pass I suspect MythTV front-end will be on it by the end of the month. So at that point I'll have everything coming in through the Apple TV, including live and recorded TV. It already works for you Roku users, of course. Roof: The only reason with thinking about the tuners right on the roof is line losses. But I do intend to experiment with this - I'm going to find a good position for Buffalo and then test the signal strength at the top and bottom of the coax run I already have. Certainly leaving it in the utility room will offer many advantages once the pointing is complete. Why bother? Because I noticed that the line loss from the antenna to the TV in the basement was *considerable*. I lost well over half the channels! I added a cheap-o RCA distribution amp in the utility room to address this. So the question remains how much I'm losing on the line that's left, maybe lots, maybe next to nothing.
9-Aug-2015, 6:40 PM #14
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Maury:
Thanks for your thoughtful post. I'm glad that I was able to provide a link for the path you have been following.
I don't know much about LANs, but would it be possible to have two HDHRs connected to the network to do what you want?
https://www.silicondust.com/
I understand your concern about signal loss in the coax, which is about 6 dB per 100 ft of RG6 for UHF. The traditional way to offset that loss is to mount a preamp near the antenna to amplify the signals before the coax attenuation. This only works if you don't have any very strong signals that would overload the preamp, which would create spurious signals in the preamp that would wipe out your weakest signals.
Looking at your TVFOOL report, it seems possible to add a preamp with moderate gain and resistance to overload near the antenna. In spite of the fact that the preamp will add its own noise to the signals, there is another benefit of the preamp which is to improve the total system Noise Figure. This is possible because a good preamp has a NF that is much less than the average ~7 dB NF of the tuner, and by the rules of cascaded NF math, the NF of the preamp primarily determines the total system NF because of its position at the beginning of the chain.
The result of the lower total system NF is greater sensitivity to weak signals. You can see examples of NF improvement here:
http://imageevent.com/holl_ands/files/ota
scroll down to and click on file 10 COMPARE System Noise Figures
With and Without Preamps
http://imageevent.com/holl_ands/file...=0&w=1&s=0&z=4
which was created by majortom of DHC. You can enter your own figures if you have Excel or OO Calc.
I also did an FM fool report based on your estimated location that shows some strong FM signals that could be attenuated by an FM filter to prevent interference with TV reception. See attachment. Roger1818 of DHC told me that the data for Canadian FM stations isn't up to date at FMFOOL.
Attached Images
Maury MarkowitzTVF FM est.JPG (108.8 KB, 428 views)
__________________
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Last edited by rabbit73; 9-Aug-2015 at 7:11 PM.
10-Aug-2015, 12:05 AM #15
Maury Markowitz
Junior Member
Join Date: Sep 2011
Posts: 21
Quote:
Originally Posted by rabbit73 I don't know much about LANs, but would it be possible to have two HDHRs connected to the network to do what you want? https://www.silicondust.com/
Yes, but the various products out there aren't designed to work this way - with the exception of the older SD boxes anyway. MythTV, on the other hand, kind of assumes that's the situation - zero or more tuners get folded together into a single list. You can even say that it should pick a particular channel from one receiver vs another if the same signal comes in from more than one source. If I can get this working it might just be a \$70 solution to a lot of multi-directional issues. And it's a DVR too.
Quote:
Originally Posted by rabbit73 This only works if you don't have any very strong signals that would overload the preamp, which would create spurious signals in the preamp that would wipe out your weakest signals.
So I have thought about getting a preamp for the Buffalo station and leaving the Toronto one alone. Channel 19 would otherwise kill me I think.
BTW I've thought over a lot of possible topologies.
One is to put the USB sticks on the antennas and then run a USB cable down to the Raspberry Pi which could be anywhere. But over 5 meters you need a repeater (~\$10).
The other is to put all of the gear on the roof, and then run a ether cable up and use POE to power it all. Now you get zero line losses, and the Pi and sticks are small enough to easily weatherproof.
Or if you want to use your existing RG6, you can get an ether adaptor for it and use that instead, but then you can't do POE so you need a power cable of some sort. But you would for a preamp too.
And finally, the easy way, run the RG6 to the box in the basement, accept the 25 feet of signal loss and just run with it.
I think each of these would have upsides and downsides for different people and situations. But I guess the last option is probably the easiest.
Quote:
Originally Posted by rabbit73 I also did an FM fool report based on your estimated location that shows some strong FM signals that could be attenuated by an FM filter to prevent interference with TV reception.
That's interesting, I never considered that issue. I can get a "channel list" easily enough from the car. Any suggestions on a filter?
Last edited by Maury Markowitz; 10-Aug-2015 at 12:18 AM.
10-Aug-2015, 11:03 PM #16
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Quote:
Any suggestions on a filter?
All FM filters are not equal. If I'm using a preamp with an FM filter, I have to accept its attenuation curve. If I pick a filter for a system without a preamp, or a system that needs even more attenuation than provided by the preamp FM filter, I select a filter that attenuates the frequencies of the strongest FM signals. A HLSJ makes an effective FM filter because it attenuates everything below channel 7 including the FM band. Some filter curves in the attachments.
Attachment 1: Antennas Direct FM trap vs Radio Shack 15-577. Notice that the AD does better at the low end of the FM band than at the high end. The RS attenuates better at the high end, but doesn't at the low 88 MHz end to allow reception of TV CH 6.
Attachment 2: The MCM FM trap does well across the whole FM band.
Attachment 3: A HLSJ used as an FM filter.
Attachment 4: I was concerned that a HLSJ might affect reception of UHF signals. ADTech did this test that shows the attenuation of the Holland HLSJ is of no concern for UHF signals.
If FMFOOL is correct, your two strongest signals are CJKX-FM on 95.9 MHz at -24.1 dBm and CKQT-FM on 94.9 MHz at -25.3 dBm.
Your strongest TV signal is CICA-DT at -39.6 dBm, which is 15.5 dB weaker than CJKX-FM.
FM can interfere with TV signals in two ways: fundamental overload, even affecting UHF if the FM signals are very much stronger (not likely in your case), or 2nd harmonic (2x its frequency) of FM signal falling on VHF-High channel frequency.
https://en.wikipedia.org/wiki/North_...on_frequencies
Attached Images
ADvsRSFMfilter.JPG (135.6 KB, 437 views) MCM%20FM%20Trap%20Attenuation.gif (15.5 KB, 442 views) ADTechHLSJs3.GIF (22.5 KB, 445 views) HollandHLSJFreqResp2.JPG (69.6 KB, 409 views)
__________________
If you can not measure it, you can not improve it.
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Last edited by rabbit73; 10-Aug-2015 at 11:28 PM.
12-Aug-2015, 2:14 PM #17
rickbb
Senior Member
Join Date: Dec 2014
Posts: 289
Quote:
Originally Posted by Maury Markowitz Very useful info Rick! Do you have a pointer to this? Google turns up something unrelated.
Here is the link to the "twisted DB8" model.
http://imageevent.com/holl_ands/omni...d8bayquasiomni
23-Oct-2015, 3:52 PM #18
Maury Markowitz
Junior Member
Join Date: Sep 2011
Posts: 21
Quote:
Originally Posted by rabbit73 HDTV primer has led you down the garden path to a false conclusion. The inherent combiner loss is only about 0.5 dB.
Ok, this is the question I was really trying to get answered.
The page in question is confusing (IMHO) and seems to mix a bunch of different issues together. I was trying to understand which bits of that page are inherent to the combiner and which were part of the particular setup he was describing.
24-Oct-2015, 12:06 AM #19
rabbit73
Retired A/V Tech
Join Date: Aug 2012
Location: S.E. VA
Posts: 2,352
Quote:
The page in question is confusing (IMHO) and seems to mix a bunch of different issues together. I was trying to understand which bits of that page are inherent to the combiner and which were part of the particular setup he was describing.
I also find that page confusing, and I have been doing antenna experiments since I was 8 years old when I built my first crystal set radio. I'm now 82, still studying antennas, have been a radio amateur (ham) since the early 50s, having built many ham antennas, and have been making tests and measurements with antennas for digital TV since 2008.
My McAfee security software tells me to stay away from that page, so I look at the Google cache version instead. I previously used the Norton security software that let me look at hdtvprimer pages.
On that page Ken Nist KQ6QV is talking about "How to combine antennas that point in different directions" but starts his explanation with a splitter and then makes a conclusion about a splitter in reverse used as a combiner. He then states there is a loss of 3 dB.
I found it difficult to make that leap to the conclusion that he did because when the two antennas are aimed in different directions, the same signals from each antenna arrive at the combining point with results that vary because of differences in amplitude and phase.
He does go on to state
Quote:
(It will be explained later that when the two antennas are pointed at the same TV station, the reflected currents subtract to zero, and as if by magic, the 3 dB combiner loss turns into a 3 dB gain.)
I have found, by actual measurements with a signal level meter, that when two identical antennas aimed in the same direction are combined with a splitter in reverse, the gain can be as much as 2.5 dB. When the two antennas are aimed in different directions, the results can not be predicted. Some channels survive the combining, and others are lost.
The discussion is still going on by people that are smarter than I am:
Thread
http://www.digitalhome.ca/forum/186-...rm-fields.html
relevant posts begin here at post #14 between holl_ands and K6STI and continue on page 2 of that thread
http://www.digitalhome.ca/forum/186-...ml#post2629098
Their discussion is about two antennas aimed in the same direction, but the two antennas are not in a uniform signal field because of trees in the signal path, which is a problem similar to combining two antennas aimed in different directions.
holl_ands is a communications engineer who is an expert in computer modeling and K6STI is a ham who designs computer modeling software for antennas. As Jeff Dunham's Peanut says sometimes the discussion goes "woosh" right over my head.
Rather than argue about it endlessly, I prefer to make some measurements to find out for myself, as the link in my signature says.
__________________
If you can not measure it, you can not improve it.
Lord Kelvin, 1883
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Last edited by rabbit73; 24-Oct-2015 at 12:48 AM.
24-Oct-2015, 6:36 PM #20
Maury Markowitz
Junior Member
Join Date: Sep 2011
Posts: 21
Quote:
Originally Posted by rabbit73 On that page Ken Nist KQ6QV is talking about "How to combine antennas that point in different directions" but starts his explanation with a splitter and then makes a conclusion about a splitter in reverse used as a combiner. He then states there is a loss of 3 dB. I found it difficult to make that leap to the conclusion that he did because when the two antennas are aimed in different directions, the same signals from each antenna arrive at the combining point with results that vary because of differences in amplitude and phase.
Right, and I guess that's why there's the whole "antenna trick" page on his site.
I'm sort of getting some of this, but I'm still a little lost about most of it.
For instance, my dad's old VHF/UHF was clearly two different antennas sharing a common structure, one in front of the other. There was only one wire coming out of it, so they had to combine them somehow.
And right now I can look out the window at my neighbour's eight-bay and clear see that it is literally nothing more than two four-bays put side to side. I can even see the separate wires running into the "combiner" in the middle.
So clearly this is not only possible but common.
I do get that having them point in different directions is a whole other can of worms, but I'm not even sure I understand the most basic parts.
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### TANCET 2008 - old paper: Percent, percentages, comparison, % increase, % decrease : Ascent TANCET MBA Classes
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You are here: Home » TANCET, Prep Questions » Percents, Fractions » TANCET 2008 Question 6
# TANCET 2008 Quant Question 6 : Percent - Comparison
## Question
John weighs twice as much as Maria. Marcia's weight is 60% of Bob's weight. Dave weighs 50% of Lee's weight. Lee weighs 190% of John's weight. Which of these 5 persons weighs the least?
1. Bob
2. Dave
3. John
4. Lee
5. Marcia
From the information given, we can quickly grasp that everyone's weight can be expressed in terms of John's weight.
Let John's weight be 'x'.
So, Marcia's weight is .
Marcia's weight = 60% of Bob's weight
Or of Bob's weight
Or Bob's weight =
Lee weighs 190% of John. So, Lee = 190% of x = = 1.9 x
Dave weighs 50% as much as Lee does. Dave = 50% of 1.9x = = 0.95x.
In ascending order these values can be arranged as 0.95x, x, 1.9x
The least weight is - Marcia's weight.
#### Alternative easier method
Assume a number for John's weight and work with that assumption. Easy assumption to make for any starting point in percents is to assume the number to be 100.
Let John's weight be 100 kg.
So, Marcia's weight = 50 kg.
Marcia's weight = 50 kgs = 60% of Bob's weight
Or 60/100 of Bob's weight = 50 or Bob's weight =
Lee's weight = 190% of John's weight = 190% of 100 = 190 kg.
Dave's weight = 50% of Lee's weight = 50% of 190 = 95 kg.
The smallest of these values is 50 kg. The weight of Marcia.
Level of difficulty : Moderate
## CAT, XAT, TANCET Practice Questions and Answers : Listed Topicwise
Number Theory Permutation Combination Probability Inequalities Geometry Mensuration Trigonometry Coordinate Geometry Percentages Profit Loss Ratio Proportion Mixtures Alligation Speed Time Distance Pipes Cisterns Interest Races Average, Mean AP, GP, HP Set Theory Clocks Calendars Algebra Function English Grammar General Awareness Data sufficiency TANCET Papers XAT Papers
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Lesson
Factorization is an application of the distributive property. Recall the five methods of factoring a polynomial discussed in Algebra I: greatest common factor (GCF), grouping with 4 terms, perfect square trinomials, factoring trinomials, and difference of squares.
### Greatest common factor (GCF)
Find the greatest common factor (GCF) of all terms of the polynomial and then apply the distributive property.
#### Worked example
##### Question 1
Factor: $24a^3b+8a$24a3b+8a
Think: The first method to try is GCF.
Do: Determine the GCF of $24a^3b$24a3b and $8a$8a.
$24a^3b+8a$24a3b+8a $2\times2\times2\times3aaab+2\times2\times2a$2×2×2×3aaab+2×2×2a From the prime factorization we can see that GCF: $2\times2\times2a=8a$2×2×2a=8a $8a\left(3a^2b+1\right)$8a(3a2b+1)
### Grouping with 4 terms
Group pairs of terms with grouping symbol that contain a common factor, then factor out the GCF from each separate binomial. Lastly, factor out the common binomial.
#### Worked example
##### question 2
Factor: $x^3+7x^2+2x+14$x3+7x2+2x+14
Think: The first method to try is GCF. Once we have tried GCF, then notice that the expression has $4$4terms, therefore try using the grouping with $4$4 terms method.
Do: Since the expression does not have a GCF, group terms in pairs that have at least one GCF.
$x^3+7x^2+2x+14$x3+7x2+2x+14 $x^3+7x^2+2x+14$x3+7x2+2x+14 GCF: $x^2$x2 GCF: $2$2 $x^2\left(x+7\right)+2\left(x+7\right)$x2(x+7)+2(x+7) $\left(x^2+2\right)\left(x+7\right)$(x2+2)(x+7)
### Perfect square trinomials
Look for three terms where the first and third terms are perfect squares, and the middle term is twice the product of their square roots.
#### Worked examples
##### question 3
Factor: $x^2+6x+9$x2+6x+9
Think: Always, check for GCF first. Since the expression has $3$3 terms, determine whether the expression meets the criteria for perfect squares trinomials method of factoring.
Do: When written in order, the first and last terms perfect squares $x^2=x^2$x2=x2 and $9=3^2$9=32. The expression must equal the perfect square trinomial formula: $\left(ax\right)^2+2abx+b^2$(ax)2+2abx+b2
$x^2+6x+9$x2+6x+9 $x^2+2\times3x+3^2$x2+2×3x+32 Does the quadratic fit the perfect squares formula? $\left(ax\right)^2+2abx+b^2$(ax)2+2abx+b2 $\left(x+3\right)^2$(x+3)2
##### Question 4
Factor: $x^2-6x+9$x26x+9
Think: Always, check for GCF first. Since the expression has $3$3 terms, determine whether the expression meets the criteria for perfect squares trinomials method of factoring.
Do: When written in order, the first and last terms perfect squares $x^2=x^2$x2=x2 and $9=3^2$9=32. The expression must equal the perfect square trinomial formula: $\left(ax\right)^2-2abx+b^2$(ax)22abx+b2
$x^2-6x+9$x2−6x+9 $x^2-2\times3x+3^2$x2−2×3x+32 Does the quadratic fit the perfect squares formula? $\left(ax\right)^2-2abx+b^2$(ax)2−2abx+b2 $\left(x+3\right)^2$(x+3)2
### Factoring trinomials
Look for three terms of the form $ax^2+bx+c$ax2+bx+c , find factors of $ac$ac that add to equal $b$b. Replace $b$b with the factors then continue by factoring using the grouping method. There are several organizing tools to assist with this process (i.e., box method and X method).
#### Worked examples
##### Question 5
Use the Box method to factor: $x^2+7x+10$x2+7x+10
Think: The Box method requires that we understand the area model (length x width = $lw$lw)
Do: Fill in the pieces we know: $x^2$x2, $10$10. The $7x$7x will be the sum of the other $2$2 boxes. Now, we can fill in the length and width for the $x^2$x2 box ($x$x and $x$x). The length and width for the $10$10 box has to be $5$5 and $2$2 because the sum of the boxes they create equals $7x$7x.
The factorization of $x^2+7x+10$x2+7x+10 is $\left(x+5\right)\left(x+2\right)$(x+5)(x+2).
## Difference of squares
Look for the difference of two terms which are both perfect squares.
#### Worked examples
##### Question 6
Factor: $9x^2-4y^2$9x24y2
Think: Check for GCF first. After the GCF has been checked, then determine whether the $2$2 terms are perfect squares.
Do: There is no GCF between the $2$2 terms. However, the terms are perfect squares: $9x^2=\left(3x\right)^2$9x2=(3x)2 and $4y^2=\left(2y\right)^2$4y2=(2y)2. Because the terms are being subtracted, then we will use the difference of squares method.
$9x^2-4y^2$9x2−4y2 $=$= $\left(3x\right)^2-\left(2y\right)^2$(3x)2−(2y)2 $=$= $\left(3x-2y\right)\left(3x+2y\right)$(3x−2y)(3x+2y)
The factorization of $9x^2-4y^2$9x24y2 is $\left(3x-2y\right)\left(3x+2y\right)$(3x2y)(3x+2y)
#### Practice questions
##### Question 7
Find the greatest common factor of $x^2y^4+x^5y^6$x2y4+x5y6.
##### Question 8
Factor $x^2-5x+10x-50$x25x+10x50 by grouping in pairs.
##### Question 9
Factor $x^2-2x-8$x22x8.
##### Question 10
Factor $k^2-81$k281.
##### Question 11
Factor $x^2+12x+36$x2+12x+36. | 1,815 | 4,910 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2021-49 | latest | en | 0.728793 |
https://www.jiskha.com/questions/47157/a-house-has-two-rooms-of-equal-area-one-room-is-a-rectangle-4ft-narrower-and-5ft-longer | 1,603,728,443,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00072.warc.gz | 785,168,906 | 5,209 | # Math
A house has two rooms of equal area. One room is a rectangle 4ft narrower and 5ft longer then the square one. Find the area of each room.
Please explain to me what to do because i am confused. Thanks
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2. 👎 0
3. 👁 278
1. Square room is x by x with area x^2.
The rectangular room is (x - -4) by (x + 5.
Therefore, x^2 = (x-4)(x+5)
I bet you can take it from here.
1. 👍 0
2. 👎 0
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1. ### math
a house has two rooms of equal area. one room is square and the other room a rectangle, four feet narrower and 5ft longer than the square one. find the area of each room? let x = dimensions of the square. Rectangle then is x-4 and | 661 | 2,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-45 | longest | en | 0.928187 |
https://www.mathsqrt.com/en/determinant | 1,723,162,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00144.warc.gz | 680,633,642 | 6,600 | # Determinant
A determinant is a function in linear algebra that associates a number with each square matrix. The determinant is defined only for square matrices.
The determinant of a second-order square matrix can be found using the formula:
\begin{align} \textrm{det}(A)=\begin{vmatrix} a & b\\ c & d \end{vmatrix}=ad-bc. \end{align}
The determinant of a third-order square matrix can be found using the formula:
\begin{align} \textrm{det}(A)&=\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=\\ \\ &=a_{11}a_{22}a_{33}+ a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} − a_{13}a_{22}a_{31} − a_{11}a_{23}a_{32} − a_{12}a_{21}a_{33} \end{align}
Generally, the determinant of an n×n matrix can be calculated using the Leibniz formula or the Laplace's expansion formula.
## Algebraic complement, subdeterminant, and minor
The algebraic complement of an element aij, denoted as Aij, is called the minor of that element. It is taken with a sign of "+" when the sum of indices i+j is an even number, and with a sign of "-" when it is an odd number.
The minor of an element aij in matrix A, denoted as Mij, is the determinant of the matrix obtained by removing the i-th row and j-th column from matrix A.
Let's consider a 3x3 matrix:
\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & {\color{Red} a_{{\color{Red}2}{\color{Red}3}}} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}
The minor M2,3 of this determinant can be found using the subdeterminant:
\begin{pmatrix} a_{11} & a_{12} & * \\ * & * & {\color{Red} *} \\ a_{31} & a_{32} & * \end{pmatrix}
$$M_{2,3}= \textrm{det} \begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix} = a_{11}a_{32}-a_{12}a_{31}$$
## Leibniz Formula for Determinant Calculation
Take the sum over all permutations σ from the set {1, 2, ..., n}:
$$\textrm{det}(A)=\sum_{\sigma \in S_{n}} \textrm{sgn}(\sigma )\prod_{i=1}^{n}A_{i,\sigma_{i}}$$
## Laplace's Formula for Determinant Calculation
According to Laplace's formula, the value of the determinant is equal to the sum of products of the elements of any row and their corresponding algebraic complements.
$$\textrm{det}(A)=a_{i1}A_{i1}+a_{i2}A_{i2}+...+a_{in}A_{in}$$
where,
a— matrix element;
A— algebraic complement of the corresponding element.
Expressing the algebraic complement through the minor, we will have:
$$\textrm{det}(A)=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}M_{ij}$$
## Properties of Determinants
1. The value of a matrix determinant does not change when the matrix is transposed:
$$\textrm{det}(A)=\textrm{det}(A^{T})$$
2. The determinant is zero if:
• it consists of zeros,
• it is equal to another corresponding row or column,
• it is proportional to another corresponding row or column,
• it can be expressed as a precise sum of scalar multiples of the remaining rows/columns.
3. If two rows are exchanged in the determinant, the sign of the determinant becomes opposite:
\begin{align} \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}&=n\\ \\ \begin{vmatrix} a_{21} & a_{22} \\ a_{11} & a_{12} \end{vmatrix}&=-n\\ \end{align}
4. When multiplying the determinant by a scalar, only one row or column is multiplied. Similarly, the reverse holds true: if a row or column of the determinant is expressed as a scalar multiple, that scalar can be brought in front of the determinant.
5. If each element in a row (column) of the determinant is the sum of two addends, the determinant can be expressed as the sum of two determinants. In the first determinant, consider the first addends in the corresponding row (column), and in the second determinant, consider the second addends in the corresponding row (column), with the remaining rows (columns) being the same as in the original determinant.
6. The value of the determinant remains unchanged when any multiple of the corresponding elements of one row is added to the corresponding elements of another row.
7. Since the determinant is defined inductively (starting with the first order, then the second, and so on), larger determinants can be calculated as the sum of their minors or subdeterminants.
8. The product of the determinants of matrices A and B is equal to the determinant of their matrix product, regardless of the order of the matrices:
$$\textrm{det}(A) \times \textrm{det}(B)=\textrm{det}(AB)=\textrm{det}(BA)$$ | 1,325 | 4,383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.675648 |
https://doku.pub/documents/test-for-critical-thinking-z06wv9pppzqx | 1,722,717,561,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00843.warc.gz | 185,042,545 | 12,671 | # Test For Critical Thinking
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TCT - 1
The Test of Critical Thinking Student Instructions
Today, you are going to take a test called The Test of Critical Thinking. How well you do on this test will not affect your grade in this class.
The stories and questions are like the sample question that we will do together. Let’s look at the example on the next page.
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TCT - 2
SAMPLE Nathan and Sean were in the same math class. Their teacher returned the tests she had graded. When they saw their grades, Nathan smiled, but Sean looked unhappy. The teacher said that many students had received low grades, and she hoped they would study more for the next test.
S-1
Based on this story, what is MOST LIKELY to be true?
a. Nathan received a better grade on the test than Sean did. b. Nathan usually receives better grades than Sean in math. c. Sean had expected to do better on the test than he did. d. Sean did not do as well on the test as he would have liked.
S-2
What does the teacher believe?
a. Studying helps students do well on math tests. b. Many students did not study for the test. c. None of the students studied enough for the test. d. Students cannot do well in math without studying.
STOP WAIT FOR DIRECTIONS
TCT - 3
S-2. What does the teacher believe? A. Studying helps students do well on math tests. This is the CORRECT answer. The teacher said that many students had not done well, and she hoped they would study more for the next test. We can conclude from this statement that the teacher believes studying helps students do well on math tests. B. Many students did not study for the test. This answer is INCORRECT. The teacher’s statement suggests that she believes many students did not study enough, but not that they did not study at all. C. None of the students studied enough for the test. This answer is INCORRECT. The teacher’s statement suggests that she hopes the students who had not done well should study more. She did not say the students who had done well needed to study more. D. Students cannot do well in math without studying. This answer is INCORRECT. The teacher’s statement suggests that she believes studying more would help the students who did not do well to do better on the next test. But she may also believe that some students can do well in math without studying.
STOP WAIT FOR DIRECTIONS
TCT - 4
Read each story and the questions that go with it carefully. Mark the best answer to each question on your answer sheet. Please do not place any marks in the test booklet.
STORY 1
Natalie and Robert are in the same gym class. Natalie was the fastest runner in the class. Robert did the most pull-ups. Each student claimed to be the best athlete in the class. David said neither one could be the best because both students are short, and tall people are usually better athletes. After a lot of talking, the students agreed to let their friend Simon decide who is the best.
1. Simon knew Natalie won second place in the pull-up contest, and Robert was fourth in running. Robert is taller than Natalie. Why did Simon MOST LIKELY choose Natalie as best athlete?
a. Overall, Natalie did better than Robert. b. Simon likes Natalie better than Robert. c. Robert is too slow to be the best athlete. d. Overall, Simon thinks short people are better athletes.
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TCT - 5
STORY 1 cont. 2. What are Natalie and Robert disagreeing about?
a. Is it better to be a tall or short athlete? b. Who should judge the best athlete? c. Can girls be better athletes than boys? d. What makes someone the best athlete?
3. What is LEAST likely to be true in this story?
a. Natalie and Robert think short people are usually good athletes. b. Natalie and Robert think being the best athlete is important. c. Natalie and Robert think Simon will make a fair decision. d. Natalie and Robert think David is not a good judge of athletes.
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TCT - 6
STORY 2
Bill and Lee went camping with their parents at a local park one weekend. The park was very crowded. On Saturday afternoon, their father asked them to pick up some litter and then to go into the woods to cut branches for cooking hot dogs. The two brothers did as their father asked. As they stepped out of the woods, a park ranger stopped them. He looked at their sticks and asked, “Don’t you know that in the park you should take nothing but pictures and leave nothing but footprints?” The boys were puzzled by what the ranger had said. They told him that their father had asked them to cut the branches for cooking hot dogs. The ranger walked the boys back to their campsite and talked to their father alone. That evening, the ranger joined the family for dinner. Early the next morning, the family packed up and went home.
4. Why were the boys puzzled?
a. The boys had only done what they were asked to do. b. The boys had taken only a few branches from the woods. c. The boys did not understand the ranger’s question. d. The boys thought it was okay to cook hot dogs.
5. What is the most likely reason the ranger talked to the father?
a. To explain that the boys had cut too many branches b. To explain proper park behavior c. To explain why boys should not be alone in the woods d. To explain why people should take pictures in the woods GO ON TO THE NEXT PAGE
TCT - 7
STORY 2 cont.
6. What was the MOST LIKELY reason the family went home the day after the ranger visited?
a. The ranger had told the family to leave. b. The family had planned to leave that day. c. The ranger had upset the family. d. The family had no more sticks for cooking hot dogs.
7. What did the ranger think when he asked, “Don’t you know that in the park you should take nothing but pictures and leave nothing but footprints”?
a. He thought the boys should have known how to behave in the park. b. He thought the boys should have been taking pictures. c. He thought the boys were going to make a fire in the woods. d. He thought the boys were afraid of getting in trouble.
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TCT - 8
STORY 2 cont.
8. Why might the ranger tell other children this story?
a. To teach them to pick up litter in the park. b. To teach them to obey their parents while camping. c. To teach them to protect the trees in the park. d. To teach them to be honest with park rangers.
9. Why did the ranger talk to the boys’ father ALONE?
a. To complain about the boys’ behavior b. To tell the father the family had to leave the park c. To find out if the boys were really brothers d. To discuss the situation without embarrassing the father
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TCT - 9
STORY 3
Carla was nervous as she stood on the stage before her performance. As she sang, the students in the audience began to laugh. Carla heard the laughing and sang even louder. By the time she had finished her song, almost everyone was laughing. The music stopped, and Carla smiled and bowed. As the curtain closed, Carla’s teacher wiped away tears and gave Carla a big hug. Carla was glad her song was finished. When she got home, Carla told her parents that the audience had loved her song.
10.Based on the story, what is MOST LIKELY to be true?
a. Carla’s teacher felt sorry for her. b. Carla’s parents were proud of her. c. Carla is a bad singer. d. Carla sang a funny song.
11.Based on the story, what BEST shows that Carla may have told her parents the truth?
a. She was nervous about singing. b. Her song made the students laugh. c. She was glad when her song was over. d. Her teacher gave her a big hug after her song.
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TCT - 10
STORY 3 cont. 12.Based on the story, how did Carla’s teacher feel?
a. She was proud of Carla. b. She was angry that the students laughed. c. She felt sorry for Carla. d. She was sad that Carla’s parents were not there.
13.What is the LEAST LIKELY reason why Carla sang louder?
a. She wanted the students to be able to hear the song. b. She had reached the most important part of the song. c. She was ignoring the students who were making fun of her. d. She had become less nervous as she sang.
14.Which statement BEST shows that Carla was prepared for her performance?
a. She kept singing while the students laughed. b. She was glad when she was done. c. She hugged her teacher to thank her. d. She smiled and bowed when she was done.
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TCT - 11
STORY 4
Paco and his mother were shopping at the mall. Paco wanted a new jeans jacket like the one many of the popular kids in his class were wearing. He asked his mother to buy one for him. She said she could not afford one right then because she needed to buy a new jacket for herself. She wanted a nice jacket to wear to a meeting about a new job. Paco told her that all his friends had jeans jackets. He was afraid that if he did not get one, no one would like him. His mother listened to Paco, but she disagreed with him. She bought the jacket for her meeting. Paco said, “You care more about your new job than about me.”
15.What did Paco and his mother both believe?
a. Wearing the wrong clothes can make people dislike you. b. It is more important for adults to look good than children. c. What you wear affects what others think of you. d. Women’s jackets cost more than boys’ jackets.
16.Based on the story, what did Paco’s mother think?
a. Her meeting was more important than Paco’s friendships. b. She needed a new jacket more than Paco did. c. A cheaper jeans jacket would be better for Paco. d. Paco’s friends should care more about him than about his clothes.
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TCT - 12
STORY 4 cont.
17.IF all the popular kids in Paco’s class wear the same type of jeans jacket, what is MOST LIKELY true?
a. The jacket they wear is the best type of jeans jacket. b. Popular kids like the jeans jacket. c. Wearing the jeans jacket makes kids popular. d. Paco will be unpopular unless he has the jeans jacket.
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TCT - 13
STORY 5
Tanya works at a large summer camp. She is a counselor for ten campers who share a cabin. Many of Tanya’s campers were often late for dinner. Tanya told the campers she would take them to a movie if everyone came to dinner on time for a whole week. All of Tanya’s campers were on time for dinner that week. Tanya took them to a movie. Tanya told Mrs. Greene, the camp owner, how well the reward had worked. Mrs. Greene disagreed. She reminded Tanya that she had made a new rule for the whole camp last week. The new rule said anyone late for dinner would not get dessert. Mrs. Greene said her new rule had caused Tanya’s campers to come to dinner on time. Tanya did not argue with Mrs. Greene. But, she was sure that her reward, not the new rule, had gotten her campers to come to dinner on time.
18.What caused Tanya’s campers to come to dinner on time?
a. Mrs. Greene’s rule b. Tanya’s reward c. Neither the rule nor the reward d. There is no way to know
19.What do Tanya and Mrs. Greene each believe?
a. Punishments work better than rewards. b. Her own action changed the campers’ behavior. c. Campers who are late for dinner are rude. d. Campers who are on time for dinner should be rewarded. GO ON TO THE NEXT PAGE
TCT - 14
STORY 5 cont. 20.What is the main question in this story?
a. Does reward work better than punishment? b. Does Tanya know more about campers’ behavior than Mrs. Greene? c. What can be done to make campers come to dinner on time? d. Why did Tanya’s campers come to dinner on time?
21.What would Tanya MOST LIKELY tell her campers if they stopped making their beds?
a. They should behave better. b. She would tell Mrs. Greene about their behavior. c. She would give them popcorn if they made their beds. d. She would send them to bed early if they did not make their beds.
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TCT - 15
STORY 6
Juan took apart an old wooden clock, piece by piece. Juan’s sister, Maria, was happy to sit and watch him. After taking apart the old clock, Juan looked closely at each piece. He wiped each wheel and gear with an oily cloth. He put all of the pieces on a table. Juan rubbed his hands together and looked at his watch with concern. He worked to put all of the small pieces back together. Much later, when Juan looked out the window, he saw his parents get out of their car. He looked at his watch and smiled.
22.Why did Juan look at his watch with concern?
a. He wasn’t sure his watch was working. b. He was afraid his parents would be angry. c. He hoped to finish before his parents arrived. d. He found the job was taking longer than he had hoped.
23.Why did Juan take the clock apart?
a. He wanted to fix a broken part. b. He wanted to clean the clock. c. He wanted to see inside the clock. d. He wanted to see how clocks work.
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TCT - 16
STORY 6 cont.
24.Why did Juan look at his watch and smile?
a. He had finished the clock in time. b. His watch was working well. c. His parents had arrived on time. d. He had a surprise for his parents.
25.What would MOST LIKELY have happened if Juan had not finished the clock before his parents arrived?
a. Maria would have been upset. b. Maria would have had to explain everything. c. Juan’s parents would have been angry. d. Juan would have been disappointed.
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TCT - 17
STORY 6 cont.
26.What BEST shows that Juan is careful?
a. He checked to see how long his work was taking. b. He asked his sister to watch him work. c. He checked every part of the clock. d. He was proud when he finished the clock.
27.IF you expect Juan to be punished if his parents see him with the clock, what are you assuming?
a. Juan was supposed to have been watching Maria. b. Juan was supposed to fix the clock before his parents arrived. c. Maria and Juan were not supposed to make a mess. d. Juan was not supposed to touch the clock without permission.
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TCT - 18
STORY 7
Mr. Kelso’s students were making paper models of the sun and planets to put on the classroom wall. They made Earth the size of a quarter and colored it blue and green. The students wanted the sun and the other planets to be just the right size compared to Earth. Mars was red and smaller than Earth. The bright yellow sun had to be nearly nine feet tall! Several students suggested that their planets and sun should be the right distance from each other, just as they are in space. One student, André, said that the planets and the sun could not fit in the same classroom. The other students didn’t believe André. He offered to explain. The students looked at Mr. Kelso, who smiled and nodded. The students decided to make the sun and planets smaller.
28.Why did André say the sun and planets would not fit in the same classroom?
a. He wanted to make Mr. Kelso smile. b. He wanted to start an argument. c. He wanted to help the other students. d. He wanted the sun to be smaller.
29.What extra information did André use to make his conclusion?
a. The sizes of all nine planets. b. The distance between the planets and the sun in space. c. The distance between Mars and Earth in space. d. The size of the sun.
GO ON TO THE NEXT PAGE
TCT - 19
STORY 7 cont.
30.What is the most likely reason Mr. Kelso smiled and nodded?
a. He thought it was funny that he had tricked the class. b. He was happy a student understood the problem. c. He thought that André was being funny. d. He was happy that the class made the planets smaller.
31.Why did the students’ suggestion create a problem?
a. The nine-foot sun was too large to fit on the classroom wall. b. Mr. Kelso’s directions were not clear when the project started. c. Earth and Mars were too small to be seen clearly on the classroom wall. d. The size of the model planets affected how far apart they should be placed.
32.Why did the students decide to make the sun and planets smaller?
a. The students wanted to get a good grade. b. The students did not believe André. c. The students could not do the project as planned. d. The students thought Mr. Kelso smiled because they were right.
GO ON TO THE NEXT PAGE
TCT - 20
STORY 8
John’s friend Paul usually talks and laughs a lot during lunch. On Tuesday, Paul was very quiet during lunch. On the way to class, John asked Paul if he was upset with him, and Paul said, “No.” Then John asked Paul what was wrong, and Paul said, “Nothing is wrong.” John thought Paul might be angry because John had not chosen him for his basketball team in gym class on Friday. John decided that if Paul was not going to talk to him, he would not talk to Paul either.
33.Based on the story, what is MOST LIKELY John’s point of view?
a. He thinks Paul should not be upset about gym class. b. He feels sad that Paul is not talking as much as usual. c. He thinks something he did caused Paul to be quiet. d. He feels bad about not choosing Paul for his team.
34.What is the main question in this story?
a. Why is Paul angry with John? b. Why was Paul quiet during lunch? c. Why didn’t John choose Paul for his team? d. When will Paul talk to John again?
GO ON TO THE NEXT PAGE
TCT - 21
STORY 8 cont.
35.What new information would BEST show that John was wrong about why Paul was quiet?
a. Paul was quiet during lunch on Monday. b. Paul and John have been best friends for a long time. c. Paul got a bad grade on a math test before lunch. d. Paul does not like to play basketball.
GO ON TO THE NEXT PAGE
TCT - 22
STORY 9
Karen and Mollie had planned to go to a movie Saturday evening. Mollie called Karen Saturday morning. She told Karen her parents would not allow her to go to the movie after all. When Karen called her friend later that evening, she was told Mollie had gone to a party. Karen was angry because her friend had gone to a party instead of a movie with her. She decided that she could not be friends with someone who did not tell the truth.
36.After talking with Mollie Saturday morning, what did Karen think Mollie would be doing that evening?
a. Mollie would be going out with her parents. b. Mollie would be going to a party. c. Mollie would be watching TV with a friend. d. Mollie would be staying home.
37.What is most likely to happen next in the story?
a. Karen will decide to end her friendship with Mollie. b. Mollie will call Karen to invite her to a movie. c. Mollie will decide to end her friendship with Karen. d. Karen will call Mollie to invite her to a movie.
GO ON TO THE NEXT PAGE
TCT - 23
STORY 9 cont.
38.What would show that Karen’s thoughts about Mollie were unfair?
a. Mollie had not known that her parents wanted her to go to a party. b. Mollie had changed her mind about going out with Karen. c. Mollie had tried to call Karen Friday night to change their plans. d. Mollie had never lied to Karen in the past. 39.What BEST shows that the story is told from Karen’s point of view?
a. Karen and Mollie planned to go to a movie together. b. Mollie called Saturday morning to tell Karen she could not go to the movie. c. Karen called Mollie and learned that Mollie was not home. d. Mollie went to a party instead of going to a movie with Karen.
GO ON TO THE NEXT PAGE
TCT - 24
STORY 9 cont.
40.What was the MOST LIKELY reason Karen called Mollie?
a. To ask Mollie to go to a movie. b. To tell Mollie why she was angry. c. To talk to Mollie about her day. d. To ask Mollie if she enjoyed the party.
41.Why is it likely that Karen was NOT angry with Mollie Saturday morning?
a. Sometimes parents change children’s plans. b. Sometimes parties are more fun than movies. c. Sometimes friends don’t tell the truth. d. Sometimes friends change their minds.
GO ON TO THE NEXT PAGE
TCT - 25
STORY 10
Lisa planted lettuce in her back yard. One morning, the leaves of the plants were smaller than they had been the day before. The edges of the leaves were ragged. Lisa concluded that her neighbor’s pet rabbit had been eating her lettuce. Her neighbor said that his rabbit had gotten out of its cage the night before. But, he said, the rabbit could not have eaten Lisa’s lettuce because the rabbit was trained to eat only rabbit food.
42.Based on the story, what MUST be true?
a. Some animal ate Lisa’s lettuce. b. Lisa’s lettuce was damaged before the rabbit got out. c. Something happened to Lisa’s lettuce the night the rabbit got out. d. The lettuce leaves will grow back if the rabbit stays in its cage.
43.What new information would BEST show that the rabbit ate the lettuce?
a. A neighbor with a fence around her garden has perfect lettuce. b. Lisa’s cousin has a rabbit that loves lettuce and rabbit food. c. Lisa’s neighbor has been wrong about his rabbit in the past. d. Lisa finds ragged edges on her lettuce after the rabbit gets loose again.
GO ON TO THE NEXT PAGE
TCT - 26
STORY 10 cont.
44.Based on the story, what does the neighbor believe about his rabbit?
a. His rabbit is smarter than other rabbits. b. His rabbit does not like to eat lettuce. c. His rabbit does what it has been trained to do. d. His rabbit will not get out of its cage again.
45.What new information, IF TRUE, would make it IMPOSSIBLE for the rabbit to have eaten Lisa’s lettuce?
a. Rabbits do not eat vegetables. b. Rabbits can be trained to eat only rabbit food. c. Rabbits do not go very far when they get loose. d. Rabbits cannot eat lettuce when it is covered up.
STOP END OF TEST
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## What are the factors that affects population growth?
Population growth is based on four fundamental factors: birth rate, death rate, immigration, and emigration.
## How do you explain exponential growth?
What exponential growth is. But what, precisely, is exponential growth? The mathematical definition says that a quantity that increases with a rate proportional to its current size will grow exponentially. This means that as the quantity increases so does that rate at which it grows.
## What do you mean by exponential?
Exponential describes a very rapid increase. Exponential is also a mathematical term, meaning “involving an exponent.” When you raise a number to the tenth power, for example, that’s an exponential increase in that number.
## What things grow exponentially?
10 Real Life Examples Of Exponential Growth
• Microorganisms in Culture. During a pathology test in the hospital, a pathologist follows the concept of exponential growth to grow the microorganism extracted from the sample.
• Spoilage of Food.
• Human Population.
• Compound Interest.
• Pandemics.
• Ebola Epidemic.
• Invasive Species.
• Fire.
## Is Doubling exponential growth?
When the growth of a quantity is exponential, the amount doubles in a certain interval of time. We speak of doubling time.
## Which is an exponential growth function?
An exponential function can describe growth or decay. The function g(x)=(12)x. is an example of exponential decay. It gets rapidly smaller as x increases, as illustrated by its graph. In the exponential growth of f(x), the function doubles every time you add one to its input x.
## How do you calculate exponential?
To calculate exponential growth, use the formula y(t) = a__ekt, where a is the value at the start, k is the rate of growth or decay, t is time and y(t) is the population’s value at time t.
## What does exponential growth look like on a graph?
An exponential function that goes up from left to right is called “Exponential Growth”. In Example 2, the graph goes downwards as it goes from left to right making it a decreasing function. An exponential function that goes down from left to right is called “Exponential Decay”.
## What does an exponential growth curve show?
Exponential population growth: When resources are unlimited, populations exhibit exponential growth, resulting in a J-shaped curve. In logistic growth, population expansion decreases as resources become scarce. It levels off when the carrying capacity of the environment is reached, resulting in an S-shaped curve.
## How do you write an equation for an exponential growth graph?
How To Find Exponential Functions
1. Step 1: Solve for “a”
2. Step 2: Solve for “b”
3. Step 3: Write the Final Equation.
4. Step 1: Find “k” from the Graph.
5. Step 2: Solve for “a”
6. Step 3: Solve for “b”
7. Step 4: Write the Final Equation.
## How do you find the function of a graph?
Inspect the graph to see if any vertical line drawn would intersect the curve more than once. If there is any such line, the graph does not represent a function. If no vertical line can intersect the curve more than once, the graph does represent a function.
## What is A and B in an exponential function?
May the bleach be with you. General exponential functions are in the form: y = abx. f(x) = abx. where a stands for the initial amount, b is the growth factor (or in other cases decay factor) and cannot also be = 1 since 1x power is always 1.
## How do you write an equation for a function?
If we use m = 0 in the equation f(x)=mx+b f ( x ) = m x + b , the equation simplifies to f(x)=b f ( x ) = b . In other words, the value of the function is a constant. This graph represents the function f(x)=2 f ( x ) = 2 . A horizontal line representing the function f(x)=2 f ( x ) = 2 .
## How do you describe a graph?
Describing language of a graph
• UP: increase / rise / grow / went up / soar / double / multiply / climb / exceed /
• DOWN: decrease / drop / fall / decline / plummet / halve / depreciate / plunge.
• UP & DOWN: fluctuate / undulated / dip /
• SAME: stable (stabilised) / levelled off / remained constant or steady / consistent.
## How do you formulate a function?
In the example of y = 2x + 6, the function changes as the value of x changes, so the function is dependent upon x. The left side of your function is the name of your function followed by the dependent variable in parenthesis, f(x) for the example. Write your function. The example becomes f(x) = 2x + 6.
## What is the rule of a function?
A function is a relation where there is only one output for every input. In other words, for every value of x, there is only one value for y. Function Rule. A function rule describes how to convert an input value (x) into an output value (y) for a given function. An example of a function rule is f(x) = x^2 + 3.
## What is a function and not a function?
A function is a relation between domain and range such that each value in the domain corresponds to only one value in the range. Relations that are not functions violate this definition. They feature at least one value in the domain that corresponds to two or more values in the range. Example 4-1.
## How do you describe a function?
DESCRIBING FUNCTIONS
• Step 1 : To describe whether function represented by the equation is linear or non linear, let us graph the given equation.
• Step 2 : Graph the ordered pairs.
• Step 3 : Describe the relationship between x and y.
• Step 1 :
• Step 2 : Graph the ordered pairs.
• Step 3 : Describe the relationship between x and y.
## What are 5 ways to represent a function?
Key Takeaways
• A function can be represented verbally. For example, the circumference of a square is four times one of its sides.
• A function can be represented algebraically. For example, 3x+6 3 x + 6 .
• A function can be represented numerically.
• A function can be represented graphically.
## What is a function in your own words?
A function is a relation that maps a set of inputs, or the domain, to the set of outputs, or the range. Note that for a function, one input cannot map to more than one output, but one output may be mapped to more than one input.
## How do you start to describe a graph?
To describe the graph in Figure 1, for example, you could say: “The rate of photosynthesis increases as temperature increases until a set temperature where the rate then falls back to zero.” If you can see numbers on the graph’s scales, you should also quote some values to validate your descriptions.
## How do you describe a function from a graph?
Defining the Graph of a Function. The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also define the graph of f to be the graph of the equation y = f(x). So, the graph of a function if a special case of the graph of an equation.
## How do you describe the trend of a graph?
Adverbs: dramatically, rapidly, hugely, massive, sharply, steeply, considerably, substantially, significantly, slightly, minimally, markedly. There is also a list of adverbs to describe the speed of a change: rapidly, quickly, swiftly, suddenly, steadily, gradually, slowly.
## How do you explain a trend?
Trends
1. A trend is a pattern in a set of results displayed in a graph.
2. In the graph above, although there is not a straight line increase in figures, overall the trend here is that sales are increasing.
3. In this graph, the trend is that sales are decreasing or dropping.
## How do you identify a trend?
A trend is the overall direction of a market or an asset’s price. In technical analysis, trends are identified by trendlines or price action that highlight when the price is making higher swing highs and higher swing lows for an uptrend, or lower swing lows and lower swing highs for a downtrend.
## What are example of trends?
The definition of a trend is a general direction or something popular. An example of trend is a northern moving coastline. An example of trend is the style of bell bottom jeans.
## What are the trends for 2020?
The 9 Trends Dominating 2020 (And the 2 We’re Leaving Behind in 2019)
• Colorful Leather (Both Real And Faux) Coach 1941.
• The Puff Sleeve. Louis Vuitton.
• ’90s Knitwear. Missoni.
• The Square Toe Boot. Ganni.
• Prairie Romance. Zimmermann.
• The Daytime Clutch. Bottega Veneta.
• Shorts Of All Proportions.
• Strong Suiting With A Feminine Touch.
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## #26 2011-03-08 06:52:19
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Hi;
I see that your method of assigning the number to each tile will result in fewer and fewer primes in the grid. This means more time for reflection on the smaller number of semi obvious cases.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #27 2011-03-08 16:48:55
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Home for a late lunch, and couldn't think of anything better to do than to test the modified version of the game.
Got to L39...obviously the time penalties and extra time per level made a big difference.
The only trick I used to get to that level was to check for digit sums that weren't a multiple of 3. Maybe next time I'll try to divide by the lower primes, but I'll have to oil up the mental arithmetic cogs for that to work. I can't see it ever working all that well, though, because the time taken with all the calcs needed would soon eat up all the time bonuses.
It's now possible to chalk up some good time bonuses on the early levels (I had 100sec by about L8 or L9), but that all evaporated in the higher levels when wrongly guessing with numbers I'd determined were possibles after using just the digit-sum 'multiple of 3' rule.
Back to work...
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
## #28 2011-03-08 17:07:13
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Well, I got to level 13.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #29 2011-03-08 19:55:30
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
That's a prime number, both forwards and backwards...so you've got more into the spirit of the game than I have.
I just had another game but failed because I'm too trigger-happy and just blaze away at guesses instead of trying some dividing. I'm too conscious of the time ticking away...which I help along with my misses.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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## #30 2011-03-08 20:04:30
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
You really are worried about the time with a 39!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #31 2011-03-08 20:13:22
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Hi Bobby,
I don't really know if that's a good score or not, because the game's so new and probably not many have tried it.
I'm just trying to get the best score I can before I get bowled over by the avalanche of better scores that must be just about on the doorstep.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
## #32 2011-03-08 20:15:25
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Hi phrontister;
I suppose you can think like that. But until those prime guessing giants appear...
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #33 2011-03-09 05:50:34
studymaths
Member
Registered: 2010-04-11
Posts: 32
### Re: A few maths games I have made to help my pupils
Hi,
Level 39 is a fantastic score. Maybe L50 is not so impossible after all?
Bobby, the 10 + level * level system I use was the easiest way I could think of to make the game progressively more difficult. I did add a check in the program to make sure at least 1 prime is generated though!
On the topic of high scores, does anyone know the easiest way of actually recording these to my website? I love the idea of a high score table. Unfortunately I only know javascript but would be willing to learn whatever I need!
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## #34 2011-03-09 05:52:19
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
MIF can probably answer that best.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #35 2011-03-09 08:56:10
studymaths
Member
Registered: 2010-04-11
Posts: 32
### Re: A few maths games I have made to help my pupils
Sorry, i'm new here. Who is MIF?
Offline
## #36 2011-03-09 11:38:07
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Oh sorry, that is MathsIsFun the administrator.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #37 2011-03-10 22:31:11
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
WOOF!! Missed out by one! Grumble! Neeeeearly got that 50.
Lots of nervous excitement towards the end...and I'm quite happy, really.
Again, I was too timid to test with lower prime divisors, because I think that looking for one from the range available will eat up too much time.
So all I used was the same as before: the primes I know (which is only 2-digit primes) and the multiple-of-3 digit-sum rule.
Last edited by phrontister (2011-03-11 16:49:21)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
## #38 2011-03-10 22:36:41
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
That is a really nice score. Studymaths will be very happy!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #39 2011-03-11 03:39:06
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Hi Bobby,
bobbym wrote:
If I get to level 25 then phrontister will get to level 50.
Now you have to get to Level 24.5
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
## #40 2011-03-11 09:51:14
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Fat chance! Slim and none and Slim left town. It is a real longshot. Snowball in hell. You know when hell freezes over. It ain't gonna happen. That'll be the day. Have I missed any longshot talk?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #41 2011-03-11 15:50:18
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Hi Bobby,
More long shots:
Yeah, right!
I've got Buckley's.
I've got no show of ever doing that.
And pigs might fly.
I couldn't do that in a month of Sundays.
I know "snowball in hell" as "a snowball's chance in hell".
They're the ones I know, but here's a list of more of them from other countries. Some are rather good.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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## #42 2011-03-11 15:56:41
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Hi phrontister;
Thank you! That is a nice list.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #43 2011-03-11 16:44:23
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Hi Bobby,
Here is something about the English language. I found it on the internet some years ago and I've added to it from time to time.
'Slim' and 'fat' chances are mentioned, which is why I thought of it.
Last edited by phrontister (2011-03-11 16:46:55)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
## #44 2011-03-11 17:01:24
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Yes, too many inconsistencies. I remember a documentary with Hubert Dreyfus where he went on and on about why computers will never understand English. He seemed to think that was due to the fact that humans were superior. On the other hand I began to sense that the structure of English makes no sense.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
## #45 2011-03-11 17:21:42
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Other than the obvious, I thought that computers would never 'understand' English because most English-speaking people, not to mention computer programmers, don't either.
Btw, "not to mention" (meaning "in addition to; as well as") is one of the myriad of our English language's odd idiomatic phrases that would drive any would-be learners of English up the wall.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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## #46 2011-03-11 17:31:59
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Lots of people can hardly pronounce some of the words. Take the word diameter for instance. The language is a lot like the measuring system, totally off the wall. I have heard that a foot was the length of the kings forearm. Why not the length of his foot?
There is even a lot of differences between US english and British. They say viiiitamin and we say vy ta min. The say she juu uhl and we say skedjual.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #47 2011-03-11 21:28:51
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Yes, there's a wide variety of English pronunciation throughout the world.
In Australia we say "schedule" as per this guy's British version. He also gives the American version. I use that site as well as pronunciation sound files in some online dictionaries.
I prefer the American "wrath", which to me sounds much more wrathful than our frothy "wroth".
I once met a teenage Canadian girl who'd only been in Australia for a couple of weeks. When she first spoke I thought she was speaking in some foreign language, but after hearing her out for a bit my ears detected some English-sounding words and I realised she was actually speaking English! After a while I could tune in to what she was saying, but it wasn't easy. The same happened to me with a newly-arrived Scot.
Last edited by phrontister (2011-03-11 21:45:05)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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## #48 2011-03-11 21:51:34
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Nice page! I notice they pronounce Birmingham differently also.
I have that problem listening to English television. If they have a very strong accent then I cannot understand them well.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #49 2011-03-11 22:01:24
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,592
### Re: A few maths games I have made to help my pupils
Here are three different pronunciations of "due", which are "dyu", "jew" and "do". We say "jew".
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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## #50 2011-03-11 22:08:45
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
### Re: A few maths games I have made to help my pupils
Differences on "new."
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline | 4,011 | 14,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-30 | latest | en | 0.955849 |
https://www.quizover.com/course/section/exercises-for-review-division-of-whole-numbers-by-openstax | 1,544,481,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00636.warc.gz | 998,982,004 | 26,204 | # 2.3 Division of whole numbers (Page 3/3)
Page 3 / 3
## Sample set d
Use a calculator to perform each division.
$\text{328}÷8$
Type 328 Press ÷ Type 8 Press =
$\text{53},\text{136}÷\text{82}$
Type 53136 Press ÷ Type 82 Press =
$\text{730},\text{019},\text{001}÷\text{326}$
We first try to enter 730,019,001 but find that we can only enter 73001900. If our calculator has only an eight-digit display (as most nonscientific calculators do), we will be unable to use the calculator to perform this division.
$\text{3727}÷\text{49}$
Type 3727 Press ÷ Type 49 Press =
This number is an example of a decimal number (see [link] ). When a decimal number results in a calculator division, we can conclude that the division produces a remainder.
## Practice set d
Use a calculator to perform each division.
$3,\text{330}÷\text{74}$
45
$\text{63},\text{365}÷\text{115}$
551
$\text{21},\text{996},\text{385},\text{287}÷\text{53}$
Since the dividend has more than eight digits, this division cannot be performed on most nonscientific calculators. On others, the answer is 415,026,137.4
$4,\text{558}÷\text{67}$
This division results in 68.02985075, a decimal number, and therefore, we cannot, at this time, find the value of the remainder. Later, we will discuss decimal numbers.
## Exercises
For the following problems, perform the divisions.
The first 38 problems can be checked with a calculator by multiplying the divisor and quotient then adding the remainder.
$\text{52}÷4$
13
$\text{776}÷8$
$\text{603}÷9$
67
$\text{240}÷8$
$\text{208}÷4$
52
$\text{576}÷6$
$\text{21}÷7$
3
$0÷0$
$\text{140}÷2$
70
$\text{528}÷8$
$\text{244}÷4$
61
$0÷7$
$\text{177}÷3$
59
$\text{96}÷8$
$\text{67}÷1$
67
$\text{896}÷\text{56}$
$1,\text{044}÷\text{12}$
87
$\text{988}÷\text{19}$
$5,\text{238}÷\text{97}$
54
$2,\text{530}÷\text{55}$
$4,\text{264}÷\text{82}$
52
$\text{637}÷\text{13}$
$3,\text{420}÷\text{90}$
38
$5,\text{655}÷\text{87}$
$2,\text{115}÷\text{47}$
45
$9,\text{328}÷\text{22}$
$\text{55},\text{167}÷\text{71}$
777
$\text{68},\text{356}÷\text{92}$
$\text{27},\text{702}÷\text{81}$
342
$6,\text{510}÷\text{31}$
$\text{60},\text{536}÷\text{94}$
644
$\text{31},\text{844}÷\text{38}$
$\text{23},\text{985}÷\text{45}$
533
$\text{60},\text{606}÷\text{74}$
$2,\text{975},\text{400}÷\text{285}$
10,440
$1,\text{389},\text{660}÷\text{795}$
$7,\text{162},\text{060}÷\text{879}$
8,147 remainder 847
$7,\text{561},\text{060}÷\text{909}$
$\text{38}÷9$
4 remainder 2
$\text{97}÷4$
$\text{199}÷3$
66 remainder 1
$\text{573}÷6$
$\text{10},\text{701}÷\text{13}$
823 remainder 2
$\text{13},\text{521}÷\text{53}$
$3,\text{628}÷\text{90}$
40 remainder 28
$\text{10},\text{592}÷\text{43}$
$\text{19},\text{965}÷\text{30}$
665 remainder 15
$8,\text{320}÷\text{21}$
$\text{61},\text{282}÷\text{64}$
957 remainder 34
$1,\text{030}÷\text{28}$
$7,\text{319}÷\text{11}$
665 remainder 4
$3,\text{628}÷\text{90}$
$\text{35},\text{279}÷\text{77}$
458 remainder 13
$\text{52},\text{196}÷\text{55}$
$\text{67},\text{751}÷\text{68}$
996 remainder 23
For the following 5 problems, use a calculator to find the quotients.
$4,\text{346}÷\text{53}$
$3,\text{234}÷\text{77}$
42
$6,\text{771}÷\text{37}$
$4,\text{272},\text{320}÷\text{520}$
8,216
$7,\text{558},\text{110}÷\text{651}$
A mathematics instructor at a high school is paid $17,775 for 9 months. How much money does this instructor make each month?$1,975 per month
A couple pays $4,380 a year for a one-bedroom apartment. How much does this couple pay each month for this apartment? Thirty-six people invest a total of$17,460 in a particular stock. If they each invested the same amount, how much did each person invest?
$485 each person invested Each of the 28 students in a mathematics class buys a textbook. If the bookstore sells$644 worth of books, what is the price of each book?
A certain brand of refrigerator has an automatic ice cube maker that makes 336 ice cubes in one day. If the ice machine makes ice cubes at a constant rate, how many ice cubes does it make each hour?
14 cubes per hour
A beer manufacturer bottles 52,380 ounces of beer each hour. If each bottle contains the same number of ounces of beer, and the manufacturer fills 4,365 bottles per hour, how many ounces of beer does each bottle contain?
A computer program consists of 68,112 bits. 68,112 bits equals 8,514 bytes. How many bits in one byte?
8 bits in each byte
A 26-story building in San Francisco has a total of 416 offices. If each floor has the same number of offices, how many floors does this building have?
A college has 67 classrooms and a total of 2,546 desks. How many desks are in each classroom if each classroom has the same number of desks?
38
## Exercises for review
( [link] ) What is the value of 4 in the number 124,621?
( [link] ) Round 604,092 to the nearest hundred thousand.
600,000
( [link] ) Find the product. $6,\text{256}×\text{100}$ .
625,600
( [link] ) Find the quotient. $0÷\text{11}$ .
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how did you get the value of 2000N.What calculations are needed to arrive at it
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7hours 36 min - 4hours 50 min | 2,393 | 7,708 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 70, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2018-51 | latest | en | 0.655165 |
https://theawesomelifeproject.com/shallow-lake/two-masses-swinging-into-each-other-example.php | 1,611,175,282,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703521987.71/warc/CC-MAIN-20210120182259-20210120212259-00627.warc.gz | 570,809,243 | 14,199 | ## Chaos in the Swinging Atwood Machine (SAM)
Two Spring-Coupled Masses University of Texas at Austin. Search for Damping Using Swinging Masses The pendulum mass pairs each comprise two pendulum masses which are the masses are fastened to each other by means, Consider two objects of mass which are free to move in 1-dimension. Suppose that these two objects is usually converted into some other form of energy.
### How to Get Started in Swinging The Daily Dot
Chapter 10 Gravitation Planetary and Satellite Motion. A collision occurs when two or more objects hit each other. is converted into other forms of energy collision between two particles of mass, Newton's law of universal gravitation The force is proportional to the product of the two masses and cancel each other out. As a consequence, for example,.
Multiple Object Force Problem: Two Blocks Tied Together In this example, It pushes out from the surfaces and keeps them from falling into each other. When a collision between two objects If you run your bumper car into a friend’s even if you know the masses and v i 1. You need some other equations
• 1 marble with a larger mass than the first two marbles Insert one small marble into one of the set the two equations equal to each other and solve for the Exploring Pendulums The momentum built up by the acceleration of gravity causes the mass to then swing in He timed each swing and discovered that each
Newton's law of universal gravitation The force is proportional to the product of the two masses and cancel each other out. As a consequence, for example, Answer to Example 4.5 One Block Pushes Another Problem Two blocks of masses m1 and m2, are placed in contact with each other on a frictionless, into (2) gives
Physics I Honors: Chapter 6 Practice Test - Momentum and Collisions Two objects with different masses If the swimmers push away from each other, ... Also known as "two masses on a (as an example) and then talk about the other interesting for T and plugging that solution into the other
If two stars orbit each other at large It is this last law that allows us to determine the mass of the binary star One such example is Sirius A and B The 10 Rules of Swinging. We have both agreed that we would only be intimate with each other, I see a lot of couples get into swinging as a last ditch
The Serious Physics Behind a Double Pendulum Fidget Spinner. Here are two double pendulums right on top of each other Each mass could still swing back and Answer to Example 4.5 One Block Pushes Another Problem Two blocks of masses m1 and m2, are placed in contact with each other on a frictionless, into (2) gives
If the energy of two balls went into one two balls, each of mass lift all 5 balls in one direction and all 5 balls will swing to the other side equal Think of it as two fronts bumping into each other by accident. When two air masses of the same temperature collide and neither is Remember the city example?
What happens if two equal masses hit each other at the same velocity and as they try to rebound the two masses are not let to do so and are pushed again to each other The role of air masses and fronts different humidities—come into contact with each other. A cold front develops when a cold air mass moves into an
Search for Damping Using Swinging Masses The pendulum mass pairs each comprise two pendulum masses which are the masses are fastened to each other by means ... (or two) to enter into sexual relations are basically 80 years old and what are we going to do with each other goes into swinging
Here is a history of older questions and answers processed by "Ask the Physicist !". If two examples have any of each other and heat seems ... (or two) to enter into sexual relations are basically 80 years old and what are we going to do with each other goes into swinging
### Observing the Transfer of Energy University of Virginia
Chaos in the Swinging Atwood Machine (SAM). ... (or two) to enter into sexual relations are basically 80 years old and what are we going to do with each other goes into swinging, Multiple Object Force Problem: Two Blocks Tied Together In this example, It pushes out from the surfaces and keeps them from falling into each other..
### Safer Garden Swing Sets Climbing Frames Australia
Air Masses and Fronts body water type form system. This is one of the most famous example of differential equation. hundreds masses connected to each other by several the two equations into a What happens if two equal masses hit each other at the same velocity and as they try to rebound the two masses are not let to do so and are pushed again to each other.
Momentum and collisions, which two or more objects exert relatively large forces on each other over a relatively two cars, each with mass m of one Multiple Object Force Problem: Two Blocks Tied Together In this example, It pushes out from the surfaces and keeps them from falling into each other.
If the energy of two balls went into one two balls, each of mass lift all 5 balls in one direction and all 5 balls will swing to the other side equal Observe how the force of gravity is directly proportional to the product of the two masses and As a first example, Suppose that two objects attract each other
Multiple Object Force Problem: Two Blocks Tied Together In this example, It pushes out from the surfaces and keeps them from falling into each other. Find 2 Answers & Solutions for the question A shell explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with speeds of
The role of air masses and fronts different humidities—come into contact with each other. A cold front develops when a cold air mass moves into an Find 2 Answers & Solutions for the question A shell explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with speeds of
... drawn-out swing when Consequently momentum vectors must be resolved into their x- and y at 2 m/sec collide with each other. If the two masses stick Consider two objects of mass which are free to move in 1-dimension. Suppose that these two objects is usually converted into some other form of energy
28/09/2018В В· How to Calculate Tension in Physics. or the force with which they are pressing into each other. For a system of two masses hanging from a vertical This is one of the most famous example of differential equation. hundreds masses connected to each other by several the two equations into a
The 10 Rules of Swinging. We have both agreed that we would only be intimate with each other, I see a lot of couples get into swinging as a last ditch FIVE years ago, Alice and her husband Eric* were your average young married couple. Today, they are minor celebrities in Australia’s swinging community, having sex
... coming directly towards each other at equal but opposite velocities two people of equal mass coming together from for example, the two objects Our Selwood Products climbing frames have a unique swing set design, effectively giving two large swing beams joined together to reinforce each other, ensuring a
Kids learn about momentum and collisions in the science of physics and When two objects bump into each other, Example: If a red ball with a mass of 10 kg 2.6 Center of mass and gravity We just chop the bike into bits and add up the contribution of each 82 CHAPTER 2. Vectors for mechanics Example: Two rods = m
The 10 Rules of Swinging. We have both agreed that we would only be intimate with each other, I see a lot of couples get into swinging as a last ditch possibility for setting up pairs of identical pendulums swinging against each other, For example mass-and-spring set the mass into vibration at
## When Air Masses Collide HowStuffWorks
How do mass and distance affect the force of gravity?. The role of air masses and fronts different humidities—come into contact with each other. A cold front develops when a cold air mass moves into an, FIVE years ago, Alice and her husband Eric* were your average young married couple. Today, they are minor celebrities in Australia’s swinging community, having sex.
### A shell explodes into three fragments of equal masses. Two
Two Spring-Coupled Masses University of Texas at Austin. When a collision between two objects If you run your bumper car into a friend’s even if you know the masses and v i 1. You need some other equations, 2.6 Center of mass and gravity We just chop the bike into bits and add up the contribution of each 82 CHAPTER 2. Vectors for mechanics Example: Two rods = m.
... drawn-out swing when Consequently momentum vectors must be resolved into their x- and y at 2 m/sec collide with each other. If the two masses stick gravitational energy - the energy resulting from the attraction of two masses to each other. into others. For example, A swinging pendulum is an excellent
You can easily find the force exerted on her by substituting mass and acceleration into For example, if two instead of bouncing off each other, the two Consider two objects of mass which are free to move in 1-dimension. Suppose that these two objects is usually converted into some other form of energy
A collision occurs when two or more objects hit each other. is converted into other forms of energy collision between two particles of mass The center of mass of a hammer, for example, You would calculate the moment of inertia for each particle it explodes into two equal pieces--in such a way that
Think of it as two fronts bumping into each other by accident. When two air masses of the same temperature collide and neither is Remember the city example? STAR FORMATION AND GALACTIC EVOLUTION gas whose components interact continually with each other by the exchange most of the mass goes into stars with masses
How Do You Go Up in a Swing? swing suspended by a single rope or chain on each side. A swing suspended on the swing, your center of mass is between your • 1 marble with a larger mass than the first two marbles Insert one small marble into one of the set the two equations equal to each other and solve for the
FIVE years ago, Alice and her husband Eric* were your average young married couple. Today, they are minor celebrities in Australia’s swinging community, having sex possibility for setting up pairs of identical pendulums swinging against each other, For example mass-and-spring set the mass into vibration at
When a collision between two objects If you run your bumper car into a friend’s even if you know the masses and v i 1. You need some other equations You can easily find the force exerted on her by substituting mass and acceleration into For example, if two instead of bouncing off each other, the two
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Physics 211 Lab . What You Need To two torques cancel each other out and the pulley does not rotate two mass hangers as well as a set of masses that you set The center of mass of a hammer, for example, You would calculate the moment of inertia for each particle it explodes into two equal pieces--in such a way that
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Consequently, this should be written (as we say) to two significant figures, The other two equations tell us or at least the ratio of these two masses Find 2 Answers & Solutions for the question A shell explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with speeds of
There are two general types of collisions in when two objects collide and do not bounce away from each other. Two rubber balls are a good example. ... (or two) to enter into sexual relations are basically 80 years old and what are we going to do with each other goes into swinging
... Also known as "two masses on a (as an example) and then talk about the other interesting for T and plugging that solution into the other This is one of the most famous example of differential equation. hundreds masses connected to each other by several the two equations into a
Inelastic Collisions be two identical “superballs,” colliding and then rebounding o↵ each other mass of both carts including the mass disks into the Consider two objects of mass which are free to move in 1-dimension. Suppose that these two objects is usually converted into some other form of energy
Chapter 2 Review of Forces and Moments masses appear to attract each other; The forces exerted by two particles on each other are equal in magnitude Physics I Honors: Chapter 6 Practice Test - Momentum and Collisions Two objects with different masses If the swimmers push away from each other,
Exploring Pendulums The momentum built up by the acceleration of gravity causes the mass to then swing in He timed each swing and discovered that each Lesson: Swinging on a String through the bottom, to the top on the other side of the swing, the time of each swing
Gravitational Field for Two Masses. taking into account gravity from both the Earth and the Sun. and taking the beads in pairs opposite each other. Consequently, this should be written (as we say) to two significant figures, The other two equations tell us or at least the ratio of these two masses
Think of it as two fronts bumping into each other by accident. When two air masses of the same temperature collide and neither is Remember the city example? Consequently, this should be written (as we say) to two significant figures, The other two equations tell us or at least the ratio of these two masses
If we bind these point-like masses by springs to each other, the rope would never stop swinging. Before going into details of the (Vector Between The Two Masses) Chaos in the Swinging Atwood Machine When the two masses are not equal, I plotted the orbits of two points starting very close to each other.
Momentum and collisions, which two or more objects exert relatively large forces on each other over a relatively two cars, each with mass m of one Momentum and collisions, which two or more objects exert relatively large forces on each other over a relatively two cars, each with mass m of one
### What happens if two equal masses hit each other at the
Binary Stars University of Oregon. Two degree of freedom systems solutions into the first two equations, we have: hence we need to specify two initial conditions for each mass., Lesson: Swinging on a String through the bottom, to the top on the other side of the swing, the time of each swing.
Swinging on a String Lesson - TeachEngineering. Our Selwood Products climbing frames have a unique swing set design, effectively giving two large swing beams joined together to reinforce each other, ensuring a, This is one of the most famous example of differential equation. hundreds masses connected to each other by several the two equations into a.
### Observing the Transfer of Energy University of Virginia
Worksheet Conservation of Momentum Triton Science. Chaos in the Swinging Atwood Machine When the two masses are not equal, I plotted the orbits of two points starting very close to each other. Answer to Example 4.5 One Block Pushes Another Problem Two blocks of masses m1 and m2, are placed in contact with each other on a frictionless, into (2) gives.
If two stars orbit each other at large It is this last law that allows us to determine the mass of the binary star One such example is Sirius A and B ... (or two) to enter into sexual relations are basically 80 years old and what are we going to do with each other goes into swinging
Galileo’s Acceleration Experiment. by raising one end some one or two cubits above the other, , we always found that the spaces traversed were to each other FIVE years ago, Alice and her husband Eric* were your average young married couple. Today, they are minor celebrities in Australia’s swinging community, having sex
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Inelastic Collisions be two identical “superballs,” colliding and then rebounding o↵ each other mass of both carts including the mass disks into the When a collision between two objects If you run your bumper car into a friend’s even if you know the masses and v i 1. You need some other equations
... seeks to detect gravitational waves and use two objects merge into one. These systems are usually two two masses rotate around each other, Search for Damping Using Swinging Masses The pendulum mass pairs each comprise two pendulum masses which are the masses are fastened to each other by means
Gravitational Field for Two Masses. taking into account gravity from both the Earth and the Sun. and taking the beads in pairs opposite each other. The role of air masses and fronts different humidities—come into contact with each other. A cold front develops when a cold air mass moves into an
Our Selwood Products climbing frames have a unique swing set design, effectively giving two large swing beams joined together to reinforce each other, ensuring a A collision occurs when two or more objects hit each other. is converted into other forms of energy collision between two particles of mass
If we bind these point-like masses by springs to each other, the rope would never stop swinging. Before going into details of the (Vector Between The Two Masses) possibility for setting up pairs of identical pendulums swinging against each other, For example mass-and-spring set the mass into vibration at
gravitational energy - the energy resulting from the attraction of two masses to each other. into others. For example, A swinging pendulum is an excellent ... Also known as "two masses on a (as an example) and then talk about the other interesting for T and plugging that solution into the other
If you’ve ever remotely considered getting into swinging which ended for other reasons. Swinging How she got into it: "Depending on the state of each This is one of the most famous example of differential equation. hundreds masses connected to each other by several the two equations into a
They argued with each other swing weight increases with mass but it A head-on collision between two rods is another famous example where masses don’t You can easily find the force exerted on her by substituting mass and acceleration into For example, if two instead of bouncing off each other, the two
• 1 marble with a larger mass than the first two marbles Insert one small marble into one of the set the two equations equal to each other and solve for the The 10 Rules of Swinging. We have both agreed that we would only be intimate with each other, I see a lot of couples get into swinging as a last ditch
You can easily find the force exerted on her by substituting mass and acceleration into For example, if two instead of bouncing off each other, the two Galileo’s Acceleration Experiment. by raising one end some one or two cubits above the other, , we always found that the spaces traversed were to each other
If the energy of two balls went into one two balls, each of mass lift all 5 balls in one direction and all 5 balls will swing to the other side equal Consequently, this should be written (as we say) to two significant figures, The other two equations tell us or at least the ratio of these two masses
The center of mass of a hammer, for example, You would calculate the moment of inertia for each particle it explodes into two equal pieces--in such a way that Newton's law of universal gravitation The force is proportional to the product of the two masses and cancel each other out. As a consequence, for example,
Pendulums— Part 1— Answer Key Pendulums swing back Name three common amusement park rides or other objects that are examples of facing each other Answer to Example 4.5 One Block Pushes Another Problem Two blocks of masses m1 and m2, are placed in contact with each other on a frictionless, into (2) gives
The Serious Physics Behind a Double Pendulum Fidget Spinner. Here are two double pendulums right on top of each other Each mass could still swing back and FIVE years ago, Alice and her husband Eric* were your average young married couple. Today, they are minor celebrities in Australia’s swinging community, having sex
The center of mass of a hammer, for example, You would calculate the moment of inertia for each particle it explodes into two equal pieces--in such a way that Newton's law of universal gravitation The force is proportional to the product of the two masses and cancel each other out. As a consequence, for example,
7210328 | 4,513 | 21,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-04 | latest | en | 0.932273 |
https://search.r-project.org/CRAN/refmans/ConvergenceClubs/html/mergeDivergent.html | 1,713,993,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00829.warc.gz | 453,119,829 | 2,705 | mergeDivergent {ConvergenceClubs} R Documentation
## Merge divergent units
### Description
Merges divergent units according the algorithm proposed by von Lyncker and Thoennessen (2017)
### Usage
mergeDivergent(clubs, time_trim, estar = -1.65)
### Arguments
clubs an object of class convergence.clubs (created by findClub or mergeClubs function) time_trim a numeric value between 0 and 1, representing the portion of time periods to trim when running log t regression model; if omitted, the same value used for clubs is used. estar a numeric value indicating the threshold e^* to test if divergent units may be included in one of the new convergence clubs. To be used only if mergeDivergent=TRUE.
### Details
von Lyncker and Thoennessen (2017) claim that units identified as divergent by the basic clustering procedure by Phillips and Sul might not necessarily still diverge in the case of new convergence clubs detected with the club merging algorithm. To test if divergent units may be included in one of the new convergence clubs, they propose the following algorithm:
1. Run a log t-test for all diverging units, and if t_k > -1.65 all these units form a convergence club (This step is implicitly included in Phillips and Sul basic algorithm);
2. Run a log t-test for each diverging units and each club, creating a matrix of t-values with dimensions d \times p, where each row d represents a divergent region and each column p a convergence club;
3. Take the highest t > e^* and add the respective region to the respective club and restart from the step 1. the authors suggest to use e^* = t = -1.65 ;
4. The algorithm stops when no t-value > e* is found in step 3, and as a consequence all remaining units are considered divergent.
### Value
A list of Convergence Clubs, for each club a list is return with the following objects: id, a vector containing the row indices of the units in the club; model, a list containing information about the model used to run the t-test on the units in the club; unit_names, a vector containing the names of the units of the club (optional, only included if it is present in the clubs object given in input).
### References
Phillips, P. C.; Sul, D., 2007. Transition modeling and econometric convergence tests. Econometrica 75 (6), 1771-1855.
Phillips, P. C.; Sul, D., 2009. Economic transition and growth. Journal of Applied Econometrics 24 (7), 1153-1185.
von Lyncker, K.; Thoennessen, R., 2017. Regional club convergence in the EU: evidence from a panel data analysis. Empirical Economics 52 (2), 525-553
mergeClubs, Merges a list of clubs created by findClubs;
### Examples
data("filteredGDP")
#Cluster Countries using GDP from year 1970 to year 2003
clubs <- findClubs(filteredGDP, dataCols=2:35, unit_names = 1, refCol=35,
time_trim = 1/3, cstar = 0, HACmethod = "FQSB")
summary(clubs)
# Merge clusters and divergent units
mclubs <- mergeClubs(clubs, mergeDivergent=TRUE)
summary(mclubs)
[Package ConvergenceClubs version 2.2.5 Index] | 753 | 3,008 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.823677 |
http://nrich.maths.org/public/leg.php?code=-415&cl=1&cldcmpid=6697 | 1,506,048,441,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688158.27/warc/CC-MAIN-20170922022225-20170922042225-00577.warc.gz | 242,395,808 | 5,834 | # Search by Topic
#### Resources tagged with STEM - General similar to Well Balanced:
Filter by: Content type:
Stage:
Challenge level:
### There are 14 results
Broad Topics > Applications > STEM - General
### STEM Clubs
##### Stage: 2 and 3 Challenge Level:
Details of our activities deigned for STEM clubs.
### Eclipses of the Sun
##### Stage: 2 and 3
Mathematics has allowed us now to measure lots of things about eclipses and so calculate exactly when they will happen, where they can be seen from, and what they will look like.
### First Forward Into Logo 2: Polygons
##### Stage: 2, 3 and 4 Challenge Level:
This is the second in a twelve part introduction to Logo for beginners. In this part you learn to draw polygons.
### First Forward Into Logo 1: Square Five
##### Stage: 2, 3 and 4 Challenge Level:
A Short introduction to using Logo. This is the first in a twelve part series.
### First Forward Into Logo 5: Pen Up, Pen Down
##### Stage: 2, 3 and 4 Challenge Level:
Learn about Pen Up and Pen Down in Logo
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### The Moving Planets
##### Stage: 2 and 3
Mathematics has always been a powerful tool for studying, measuring and calculating the movements of the planets, and this article gives several examples.
### In Order
##### Stage: 2 Challenge Level:
Can you rank these quantities in order? You may need to find out extra information or perform some experiments to justify your rankings.
### Order the Changes
##### Stage: 2 Challenge Level:
Can you order pictures of the development of a frog from frogspawn and of a bean seed growing into a plant?
### Order, Order!
##### Stage: 1 and 2 Challenge Level:
Can you place these quantities in order from smallest to largest?
### Helicopters
##### Stage: 2, 3 and 4 Challenge Level:
Design and test a paper helicopter. What is the best design?
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . . | 535 | 2,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-39 | latest | en | 0.861961 |
https://www.futurestarr.com/blog/social-studies/how-big-is-an-acre | 1,656,217,280,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00349.warc.gz | 843,641,988 | 53,518 | FutureStarr
How Big Is an Acre
# How Big Is an Acre
A more ambitious project of mine is a complete map of the global acreage. From a technical standpoint, a Mendix Enterprise license would do a pretty good job of filling out the map, but it would take a lot of work. Not a lot of people have a Mendix Enterprise license. They’re on-premise units that you have to buy from a Mendix reseller on a per-acre basis.
## What Is a Commercial Acre?
A visitor recently asked the question 'Why is a commercial acre different to a residential acre?' The commercial acre was invented by US real estate agencies for use in large cities and is a legal unit in some US states. It is said to be a standard acre unit with a deduction for alleyways, roads and sidewalks. The commercial acre is 82.6 percent of an international acre and you'll notice it's a rounded figure, for ease of calculation. (Source: www.thecalculatorsite.com)
## Is 1 Acre of Land Enough?
One acre of land is enough to hold a single-family home and have ample yard space without encroaching on easements or your neighbors land. However, those looking for enough space to farm or raise livestock will need more space. (Source: projectperfecthome.com)
## What Is the Size of an Acre?
The acre was originally an English unit of measurement that described the area of land that a yoke of oxen could plow in a day. It originally differed in size from one area to the next. In the 1900s it was fixed at 4,840 square yards (or 4,047 square meters). It is still used today to describe plots of land. Because an acre is a unit of area, rather than length, it would be incorrect to say "square acre" the way one might say "square mile". (Source: www.infoplease.com)
## Does It Have to Be a Rectangle?
The original definition of an acre was a furlong by a chain, a definition used by surveyors to measure plots. However, the area one derives from that measurement is an acre, even if it's contained within a different shape. 1 furlong is equal to 10 chains, or 660 feet. If you do the math, this means an acre is 10 square chains, or 43,560 square feet. (Source: www.infoplease.com)
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June 26, 2022 | m basit | 849 | 3,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-27 | latest | en | 0.953027 |
https://www.physicsforums.com/threads/absolute-zero.189610/ | 1,519,583,369,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816841.86/warc/CC-MAIN-20180225170106-20180225190106-00588.warc.gz | 919,700,079 | 15,697 | # Absolute zero
1. Oct 7, 2007
### BosonJaw
Hello all
This question is probably a bit ridiculous, but here goes:
Hypothetically, If any given closed system has obtained absolute zero temp, does any media (time, gravity, events)occur within? Thermodynamically, Wouldn't this be a state of infinite entropy? Can anyone describe such a system?
Thanks
BTW feel free to enlighten me on the laws of absolute zero, for all I know, I could have just asked, does 1 + 1 = an orange peel
2. Oct 8, 2007
### f95toli
Absolute zero is limit which is impossible to reach for a real system.
Also, note that absolute zero does not imply zero energy since you still have the zero point energy.
In many cases we can model real systems quite well without taking the temperature into consideration (i.e. we assume zero temperature) and nothing dramatic happens. This is a valid approximation assuming the temperature (or to be more precise, kB*T)is much smaller than the typical energies of the problem (e.g. much smaller than the bandgap of a semiconductor, or the gap in a superconductor). Hence, in most cases it won't matter if the system is at zero K or just very small. There is nothing "mysterious" about zero temperature.
If am not qute sure why you think the entropy would go to infinty; entropy is essentially a measure of disorder and is only indirectly connected to temperature (there are measures of entropy that are not related to temperature at all, e.g. the von Neumann entropy).
3. Oct 8, 2007
### Shooting Star
Thermodynamically, it would be a state of zero entropy (or a constant entropy for non-degenerate systems).
The system will have a minimum energy, which is called the zero point energy. This is a quantum-mechanical effect. Classically, it should have had zero energy.
4. Oct 8, 2007
5. Oct 8, 2007
### BosonJaw
I guess I was thinking along these lines.
Thanks for the help. | 458 | 1,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-09 | longest | en | 0.941618 |
https://nickch-k.github.io/introcausality/Lectures/Lecture_13_Causality.html | 1,660,689,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00306.warc.gz | 389,770,808 | 438,575 | # Lecture 13: Causality
## Midterm
• Let’s review the midterm answers
## And now!
• The second half of the class
• The first half focused on how to program and some useful statistical concepts
• The second half will focus on causality
• In other words, “how do we know if X causes Y?”
## Why Causality?
• Many of the interesting questions we might want to answer with data are causal
• Some are non-causal, too - for example, “how can we predict whether this photo is of a dog or a cat” is vital to how Google Images works, but it doesn’t care what caused the photo to be of a dog or a cat
• Nearly every why question is causal
• And when we’re talking about people, why is often what we want to know!
## Also
• This is economists’ comparative advantage!
• Plenty of fields do statistics. But very few make it standard training for their students to understand causality
• This understanding of causality makes economists very useful! This is one big reason why tech companies have whole economics departments in them
## Bringing us to…
• Part of this half of the class will be understanding what causality is and how we can find it
• Another big part will be understanding common research designs for uncovering causality in data when we can’t do an experiment
• These, more than supply & demand, more than ISLM, are the tools of the modern economist!
## So what is causality?
• We say that `X` causes `Y` if…
• were we to intervene and change the value of `X` without changing anything else…
• then `Y` would also change as a result
## Some examples
Examples of causal relationships!
Some obvious:
• A light switch being set to on causes the light to be on
• Setting off fireworks raises the noise level
Some less obvious:
• Getting a college degree increases your earnings
• Tariffs reduce the amount of trade
## Some examples
Examples of non-zero correlations that are not causal (or may be causal in the wrong direction!)
Some obvious:
• People tend to wear shorts on days when ice cream trucks are out
• Rooster crowing sounds are followed closely by sunrise*
Some less obvious:
• Colds tend to clear up a few days after you take Emergen-C
• The performance of the economy tends to be lower or higher depending on the president’s political party
*This case of mistaken causality is the basis of the film Rock-a-Doodle which I remember being very entertaining when I was six.
## Important Note
• “X causes Y” doesn’t mean that X is necessarily the only thing that causes Y
• And it doesn’t mean that all Y must be X
• For example, using a light switch causes the light to go on
• But not if the bulb is burned out (no Y, despite X), or if the light was already on (Y without X)
• But still we’d say that using the switch causes the light! The important thing is that X changes the probability that Y happens, not that it necessarily makes it happen for certain
## So How Can We Tell?
• As just shown, there are plenty of correlations that aren’t causal
• So if we have a correlation, how can we tell if it is?
• For this we’re going to have to think hard about causal inference. That is, inferring causality from data
## The Problem of Causal Inference
• Let’s try to think about whether some `X` causes `Y`
• That is, if we manipulated `X`, then `Y` would change as a result
• For simplicity, let’s assume that `X` is either 1 or 0, like “got a medical treatment” or “didn’t”
## The Problem of Causal Inference
• Now, how can we know what would happen if we manipulated `X`?
• Let’s consider just one person - Angela. We could just check what Angela’s `Y` is when we make `X=0`, and then check what Angela’s `Y` is again when we make `X=1`.
• Are those two `Y`s different? If so, `X` causes `Y`!
• Do that same process for everyone in your sample and you know in general what the effect of `X` on `Y` is
## The Problem of Causal Inference
• You may have spotted the problem
• Just like you can’t be in two places at once, Angela can’t exist both with `X=0` and with `X=1`. She either got that medical treatment or she didn’t.
• Let’s say she did. So for Angela, `X=1` and, let’s say, `Y=10`.
• The other one, what `Y` would have been if we made `X=0`, is missing. We don’t know what it is! Could also be `Y=10`. Could be `Y=9`. Could be `Y=1000`!
## The Problem of Causal Inference
• Well, why don’t we just take someone who actually DOES have `X=0` and compare their `Y`?
• Because there are lots of reasons their `Y` could be different BESIDES `X`.
• They’re not Angela! A character flaw to be sure.
• So if we find someone, Gareth, with `X=0` and they have `Y=9`, is that because `X` increases `Y`, or is that just because Angela and Gareth would have had different `Y`s anyway?
## The Problem of Causal Inference
• The main goal we have in doing causal inference is in making as good a guess as possible as to what that `Y` would have been if `X` had been different
• That “would have been” is called a counterfactual - counter to the fact of what actually happened
• In doing so, we want to think about two people/firms/countries that are basically exactly the same except that one has `X=0` and one has `X=1`
## Experiments
• A common way to do this in many fields is an experiment
• If you can randomly assign `X`, then you know that the people with `X=0` are, on average, exactly the same as the people with `X=1`
• So that’s an easy comparison!
## Experiments
• When we’re working with people/firms/countries, running experiments is often infeasible, impossible, or unethical
• So we have to think hard about a model of what the world looks like
• So that we can use our model to figure out what the counterfactual would be
## Models
• In causal inference, the model is our idea of what we think the process is that generated the data
• We have to make some assumptions about what this is!
• We put together what we know about the world with assumptions and end up with our model
• The model can then tell us what kinds of things could give us wrong results so we can fix them and get the right counterfactual
## Models
• Wouldn’t it be nice to not have to make assumptions?
• Yeah, but it’s impossible to skip!
• We’re trying to predict something that hasn’t happened - a counterfactual
• This is literally impossible to do if you don’t have some model of how the data is generated
• You can’t even predict the sun will rise tomorrow without a model!
• If you think you can, you’re just don’t realize the model you’re using - that’s dangerous!
## An Example
• Let’s cheat again and know how our data is generated!
• Let’s say that getting `X` causes `Y` to increase by 1
• And let’s run a randomized experiment of who actually gets X
``````df <- data.frame(Y.without.X = rnorm(1000),X=sample(c(0,1),1000,replace=T)) %>%
mutate(Y.with.X = Y.without.X + 1) %>%
#Now assign who actually gets X
mutate(Observed.Y = ifelse(X==1,Y.with.X,Y.without.X))
#And see what effect our experiment suggests X has on Y
df %>% group_by(X) %>% summarize(Y = mean(Observed.Y))``````
``````## # A tibble: 2 x 2
## X Y
## <dbl> <dbl>
## 1 0 0.0749
## 2 1 1.06``````
## An Example
• Now this time we can’t randomize X.
``````df <- data.frame(Z = runif(10000)) %>% mutate(Y.without.X = rnorm(10000) + Z, Y.with.X = Y.without.X + 1) %>%
#Now assign who actually gets X
mutate(X = Z > .7,Observed.Y = ifelse(X==1,Y.with.X,Y.without.X))
df %>% group_by(X) %>% summarize(Y = mean(Observed.Y))``````
``````## # A tibble: 2 x 2
## X Y
## <lgl> <dbl>
## 1 FALSE 0.346
## 2 TRUE 1.85``````
``````#But if we properly model the process and compare apples to apples...
df %>% filter(abs(Z-.7)<.01) %>% group_by(X) %>% summarize(Y = mean(Observed.Y))``````
``````## # A tibble: 2 x 2
## X Y
## <lgl> <dbl>
## 1 FALSE 0.612
## 2 TRUE 1.71``````
## So!
• So, as we move forward
• We’re going to be thinking about how to create models of the processes that generated the data
• And, once we have those models, we’ll figure out what methods we can use to generate plausible counterfactuals
• Once we’re really comparing apples to apples, we can figure out, using only data we can actually observe, how things would be different if we reached in and changed `X`, and how `Y` would change as a result. | 2,192 | 8,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-33 | latest | en | 0.927024 |
https://us.metamath.org/mpeuni/df-pj.html | 1,720,829,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00113.warc.gz | 503,302,633 | 4,193 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > df-pj Structured version Visualization version GIF version
Definition df-pj 20395
Description: Define orthogonal projection onto a subspace. This is just a wrapping of df-pj1 18757, but we restrict the domain of this function to only total projection functions. (Contributed by Mario Carneiro, 16-Oct-2015.)
Assertion
Ref Expression
df-pj proj = ( ∈ V ↦ ((𝑥 ∈ (LSubSp‘) ↦ (𝑥(proj1)((ocv‘)‘𝑥))) ∩ (V × ((Base‘) ↑m (Base‘)))))
Distinct variable group: 𝑥,
Detailed syntax breakdown of Definition df-pj
StepHypRef Expression
1 cpj 20392 . 2 class proj
2 vh . . 3 setvar
3 cvv 3444 . . 3 class V
4 vx . . . . 5 setvar 𝑥
52cv 1537 . . . . . 6 class
6 clss 19699 . . . . . 6 class LSubSp
75, 6cfv 6328 . . . . 5 class (LSubSp‘)
84cv 1537 . . . . . 6 class 𝑥
9 cocv 20352 . . . . . . . 8 class ocv
105, 9cfv 6328 . . . . . . 7 class (ocv‘)
118, 10cfv 6328 . . . . . 6 class ((ocv‘)‘𝑥)
12 cpj1 18755 . . . . . . 7 class proj1
135, 12cfv 6328 . . . . . 6 class (proj1)
148, 11, 13co 7139 . . . . 5 class (𝑥(proj1)((ocv‘)‘𝑥))
154, 7, 14cmpt 5113 . . . 4 class (𝑥 ∈ (LSubSp‘) ↦ (𝑥(proj1)((ocv‘)‘𝑥)))
16 cbs 16478 . . . . . . 7 class Base
175, 16cfv 6328 . . . . . 6 class (Base‘)
18 cmap 8393 . . . . . 6 class m
1917, 17, 18co 7139 . . . . 5 class ((Base‘) ↑m (Base‘))
203, 19cxp 5521 . . . 4 class (V × ((Base‘) ↑m (Base‘)))
2115, 20cin 3883 . . 3 class ((𝑥 ∈ (LSubSp‘) ↦ (𝑥(proj1)((ocv‘)‘𝑥))) ∩ (V × ((Base‘) ↑m (Base‘))))
222, 3, 21cmpt 5113 . 2 class ( ∈ V ↦ ((𝑥 ∈ (LSubSp‘) ↦ (𝑥(proj1)((ocv‘)‘𝑥))) ∩ (V × ((Base‘) ↑m (Base‘)))))
231, 22wceq 1538 1 wff proj = ( ∈ V ↦ ((𝑥 ∈ (LSubSp‘) ↦ (𝑥(proj1)((ocv‘)‘𝑥))) ∩ (V × ((Base‘) ↑m (Base‘)))))
Colors of variables: wff setvar class This definition is referenced by: pjfval 20398
Copyright terms: Public domain W3C validator | 846 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.249813 |
http://codereview.stackexchange.com/questions/3714/outputting-the-names-of-cars-without-repetitions-with-the-number-of-occurrence | 1,464,401,231,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277286.69/warc/CC-MAIN-20160524002117-00186-ip-10-185-217-139.ec2.internal.warc.gz | 59,259,868 | 36,832 | # Outputting the names of cars, without repetitions, with the number of occurrences in order of decreasing repetitions
A short while ago, I have submitted a coding exercise to a potential employer. The response came back the next morning and you can guess what it was from the subject of this post.
I am not totally at loss, but I need another programmer's perspective. Is there anything that jumps out?
The idea of the exercise is simple: I'm given an input file with names of cars, one per line, possibly repeated and in no particular order.
The program should output the same names, except with no repetitions, the number of occurrences listed next to each car, and in order of decreasing repetitions.
Example:
Honda\n Audi\n Honda\n -> Honda 2 \n Audi 1\n
#include <iostream>
#include <fstream>
#include <string>
#include <map>
#include <algorithm>
#include <cctype>
using namespace std;
// helper functions ///////////////////////////////////////
void collect_lines(istream &in, map<string, int> &lines);
// given lines->num_occurs map, reverses mapping
void reorg_by_count(map<string, int> &lines,
multimap<int, string> &bycount);
///////////////////////////////////////////////////////////
int main(int ac, char* av[])
{
istream *in;
map<string, int> *lines = new map<string, int>();
multimap<int, string> *lines_by_count = new multimap<int, string>();
if (ac < 2)
{
in = &cin;
}
else
{
in = new ifstream(av[1]);
}
if (!in->good()) return 1;
collect_lines(*in, *lines);
reorg_by_count(*lines, *lines_by_count);
if (in != &cin)
{
((ifstream *)in)->close();
delete in;
}
cout << "=====================\n\n";
multimap<int, string>::reverse_iterator it
= lines_by_count->rbegin();
for (; it != lines_by_count->rend(); it++)
{
cout << it->second << " " << it->first << '\n';
}
delete lines;
delete lines_by_count;
return 0;
}
// Read the instream line by line, until EOF.
// Trim initial space. Empty lines skipped
void collect_lines(istream &in, map<string, int> &lines)
{
string tmp;
while (in.good())
{
getline(in, tmp);
int i = 0;
// trim initial space (also skips empty strings)
for (i = 0; i < tmp.length() && !isalnum(tmp[i]); i++);
if (i >= tmp.length()) continue;
tmp = tmp.substr(i);
for (i = 0; i < tmp.length(); i++)
{
if (!isalnum(tmp[i]))
{
tmp[i] = ' ';
}
// thus, HoNdA == Honda
if (i == 0)
{
tmp[i] = toupper(tmp[i]);
}
else
{
tmp[i] = tolower(tmp[i]);
}
}
// and record the counts
if (lines.count(tmp) == 0)
{
lines[tmp] = 0;
}
lines[tmp]++;
}
}
// given lines->num_occurs map, reverses mapping
void reorg_by_count(map<string, int> &lines,
multimap<int, string> &bycount)
{
map<string, int>::iterator it = lines.begin();
for (; it != lines.end(); it++)
{
bycount.insert(pair<int, string>(it->second, it->first));
}
}
-
## migrated from stackoverflow.comJul 29 '11 at 16:51
This question came from our site for professional and enthusiast programmers.
I think the question is simple and you have put much more coding efforts. I believe in 3 kind of optimizations; "time, space, text"; "text" optimization is called readability. You could have solved this problem with probably 10 lines of code. I won't be able to provide code now; may be tomorrow. – iammilind Jul 29 '11 at 16:51
Its a good C answer. But its not C++. – Loki Astari Jul 29 '11 at 18:59
Huh? I see references and STL use... – Thomas Eding Jul 30 '11 at 0:32
@trinithis: C++ is a style. The code may have been using C++ types but the style was C like (not C++ like). The trouble is people think that because both languages have the same basic syntax that moving from one to the other is trivial. I find that converting C programmers to C++ is really difficult because you have to move them past the whole C mindset. Hence I would not consider the above code to be C++. Some people use the term "C with classes" as a distinct language the lies somewhere between C and C++, here people use C++ features but still code with a C style. – Loki Astari Jul 30 '11 at 0:56
Please have proper thread titles like "Critique requested on the program to output the names without repetitions with the number of occurrences in order of decreasing repetitions." - to make other people's lives easier. – TheIndependentAquarius May 8 '12 at 7:07
### Problems I see:
My problem with your code is that you are newing a lot of stuff that should just be objects.
map<string, int> *lines = new map<string, int>();
multimap<int, string> *lines_by_count = new multimap<int, string>();
Both of these should just be plain objects.
map<string, int> lines;
multimap<int, string> lines_by_count;
This one fact would have caused you to be rejected. I would have seen this and I would not have read any-further into your code straight onto the reject pile. This fundamental flaw in your style shows that you are not a C++ programmer.
Next the objects you new are stored in RAW pointers. This is a dead give away that you are not an experienced C++ programmer. There should practically never be any pointers in your code. (All pointers should be managed by an object). Even though you manually do delete these two it is not done in an exception safe way (so they can still potentially leak).
You are reading a file incorrectly.
while (in.good())
{
getline(in, tmp);
This is the standard anti-pattern for reading a file (even in C). The problem with your version is that the last successful read will read upto but not past the EOF. Thus the state of the file is still good but there is now no content left. So you re-enter the loop and the first read operation getline() will then fail. Even though it can fail you do not test for that.
I would expect to see this:
while (getline(in, tmp))
{
// Now I can processes it
}
Next you are showing a fundamental misunderstanding of how maps work:
if (lines.count(tmp) == 0)
{
lines[tmp] = 0;
}
lines[tmp]++;
If you use the operator[] on a map it always returns a reference to an internal value. This means if the value does not exist one will be inserted. So there is no need to do this check. Just increment the value. If it is not their a value will be inserted and initialized for you (thus integers will be zero). Though not a big problem its usually preferable to use pre-increment. (For those that are going to say it does not matter. On integer types it does not matter. But you have to plan fro the future where somebody may change the type to a class object. This way you future proof your code against change and maintenance problems. So prefer pre-increment).
You are doing extra work you don't need to:
// trim initial space (also skips empty strings)
for (i = 0; i < tmp.length() && !isalnum(tmp[i]); i++);
The streams library already discards spaces when used correctly. Also the ';' at the end of the for. This is considered bad practice. It is really hard to spot and any maintainer is going to ask did he really mean that. When you have an empty body it is always best to use the {} and put a comment in their {/*Deliberately empty*/}
Here you are basically lower casing the string.
for (i = 0; i < tmp.length(); i++)
{
if (!isalnum(tmp[i]))
{
tmp[i] = ' ';
}
You could use the C++ algorithms library to do stuff like this:
std::transform(tmp.begin(), tmp.end(), tmp.begin(), ::tolower());
// ^^^^^^^^^^^ or a custom
// functor to do the task
Const correctness.
void reorg_by_count(map<string, int> &lines, multimap<int, string> &bycount)
The parameter lines is not mutated by the function nor should it be. I would expect it to be passed as a const reference as part of the documentation of the function that you are not going to mutate it. This also helps in future maintenance as it stops people from accidentally mutating the object in a way that later code would not expect.
My final thing is I did not see any encapsulation of the concept of a car. You treated it all as lines of text. If you had invented a car object you can define how cars are read from a stream and written to a stream etc. Thus you encapsulate the concept in a single location.
I would have done something like this:
Probably still overkill.
#include <algorithm>
#include <iterator>
#include <iostream>
#include <fstream>
#include <sstream>
#include <string>
#include <map>
#include <cctype>
class Car
{
public:
bool operator<(Car const& rhs) const {return name < rhs.name;}
private:
friend std::istream& operator>>(std::istream& stream, Car& self);
friend std::ostream& operator<<(std::ostream& stream, Car const& self);
std::string name;
};
std::istream& operator>>(std::istream& stream, Car& self)
{
std::string line;
std::getline(stream, line);
std::stringstream linestream(line);
linestream >> self.name; // This strips white space
// Lowercase the name
std::transform(self.name.begin(), self.name.end(), self.name.begin(), ::tolower);
// Uppercase first letter because most are proper nouns
self.name[0] = ::toupper(self.name[0]);
return stream;
}
std::ostream& operator<<(std::ostream& stream, Car const& self)
{
return stream << self.name << "\n";
}
//Print out map members
struct MapPrinter
{
MapPrinter(std::map<Car,int>::value_type const& d): data(d) {}
std::map<Car,int>::value_type const& data;
};
std::ostream& operator<<(std::ostream& stream, MapPrinter const& item)
{
return stream << item.data.second << ": " << item.data.first;
}
int main(int argc, char* argv[])
{
if (argc < 2)
{ exit(1);
}
std::ifstream cars(argv[1]);
std::map<Car,int> count;
Car nextCar;
while(cars >> nextCar)
{
++count[nextCar];
}
// PS deliberately left the sorting by inverse order as an exercise.
std::copy(count.begin(),
count.end(),
std::ostream_iterator<MapPrinter>(std::cout));
}
-
+1 Many good points, like const correctness. Creating a Car object, however, is--in my mind--overkill. – Adrian McCarthy Jul 29 '11 at 20:16
@Adrian McCarthy: I like the Car object as it lets me do this: while(cars >> nextCar) which is very intuitive to read. (If this was a bigger program) then it also it centralized the code where we read a car from a stream. So if we modify the representation of a car we only have to modify one piece of code and all loops will still work. – Loki Astari Jul 29 '11 at 23:02
@Martin Thanks for a generously informative answer. (BTW, EVERYONE, thanks for your time). Martin, in my "defense" (well, not really) I've seen a lot of raw pointers in production code :) Or should I say :( – Jozin S Bazin Jul 31 '11 at 20:38
@Jozin S Bazin: There is a lot of C code that pretends to be C++ (AKA C with classes). – Loki Astari Jul 31 '11 at 22:24
@Seth Carnegie: Find an article on RAII. In my opinion this is the MOST important concept in C++ that must be learned. – Loki Astari Aug 2 '11 at 14:23
You're doing manual memory management. That's not a good idea. In fact, that's something that you don't need to do at all in modern C++. You either use automatic objects, or use use smart pointers to dynamically allocated objects.
In your case, there's no need to do dynamic allocation at all. Instead of:
map<string, int> *lines = new map<string, int>();
multimap<int, string> *lines_by_count = new multimap<int, string>();
// more things
delete lines;
delete lines_by_count;
You should have just used automatic objects:
map<string, int> lines;
multimap<int, string> lines_by_count;
// things
The same goes for the ifstream you used. This clearly shows you don't understand one of the most important facets of C++.
-
I think this is probably the biggest one. If their prospective employer is looking for C++ skills, using pointers and new all the time shows exactly the opposite. – Brendan Long Jul 29 '11 at 18:36
Agreed. Having been the interviewer, I'll share my thoughts with you. When I see this kind of code, I think "This person doesn't understand stack vs heap, references, or memory management in general. They probably learned Java first, and then tried to jump into C/C++." (Note that I don't hate Java, I just speculate on why this code was written this way). Based on the code, I assume large gaps in the coder's knowledge, and move on. Since you posted here, asking what's wrong, I assume you actually want to learn, which is a very good quality in a coder. Good luck to you. – Tim Jul 29 '11 at 18:58
oh, and your for loop. You declared the iterator before the loop and left the first part of the for blank, when there was no need to do that. – Tim Jul 29 '11 at 19:01
@Tim: There's no NEED to put the iterator declaration inside the for loop control region either, and this way the lines are shorter and easier to read. – Ben Voigt Jul 29 '11 at 19:41
@Ben: You can insert line breaks inside the for statement. And the needlessly extended scope of the iterator is far worse than any long line could ever be – Fabio Fracassi Aug 1 '11 at 8:56
As one of the commenter I believe this could be done in a few, say 10 lines, of code. Writing to long methods is often a sign that one is doing something wrong.
My point is that the sheer size will make the interviewer say it's not good enough. I imagine they want a short clean piece of code that does what they asked for, and not every trick in the book to show off.
on @Martinho suggestion I add my example here
#include <iostream>
#include <list>
#include <map>
using namespace std;
bool my_pair_compare(pair<string,int> &a, pair<string,int> &b) {
return a.second > b.second;
}
void my_pair_output(pair<string,int> &p) {
cout << p.first << " " << p.second << endl;
}
int main() {
map<string,int> cars;
while (1) {
string name;
cin >> name;
if (cin.eof()) break;
cars[name]++;
}
list<pair<string,int> > names;
map<string,int>::iterator citer = cars.begin();
while (citer != cars.end())
names.push_back(*citer++);
names.sort(my_pair_compare);
for_each(names.begin(), names.end(), my_pair_output);
return 0;
}
-
I didn't downvote, but I would be interested to see those 10 lines :) – BЈовић Jul 29 '11 at 17:31
Here is a gist (35 lines total, ~10 active lines) – epatel Jul 29 '11 at 19:10
@epatel: post the gist in the answer! – R. Martinho Fernandes Jul 29 '11 at 20:16
The list would be better replaced with a vector, then std::sort() instead of .sort(). – Jon Purdy Jul 30 '11 at 19:39
If you use vector like Jon Purdy adviced, it has a constructor that takes a begin/end pair so you don't need the while loop. Just write: vector<pair<string,int> > names(cars.begin(), cars.end()); Another 3 lines (and one empty line) gone! – Sjoerd Oct 29 '11 at 0:31
I assume it works, but didn't try it. I would consider it not finished. In an interview situation, they will want you to do your best, and it's more about proving that you are aware of things like checking return status, and doing the right thing, even though the problem at hand is small, and can be dashed off quickly, they probably still want to see a complete program.
Here's what stood out to me:
• Should use 'argc', 'argv' names for familiarity.
• lines and lines_by_count are constructed on the heap for no reason - should just use the stack.
• No allocations are checked.
• Processes command line arguments, doesn't either (a) complain about excess arguments or (b) use them.
• No usage or '-help' support.
• Code contains assumptions about ASCII input, but doesn't declare that.
• Error handling just quits with no message.
-
"No allocations are checked" new throws, there's nothing to check. Of course, he shouldn't be using new... – ildjarn Jul 29 '11 at 17:25
This is kinda a tangent and I'll probably get flamed for my heretic view but I would say in most programs (including this) checking allocations is not necessary. It's an extremely rare occurrence that they fail, and when they do in 99.9% of cases you want to crash. In 99.9% of cases (excluding absurd implementations of undefined behavior) the difference is simply crashing with or without an error message. If the computer is out of memory, a lot of applications will crash at the same time, most with segfault-like error messages, so it should be clear to the user what's going on – Andreas Bonini Jul 29 '11 at 21:04
Also don't misunderstand what I said; for big applications an allocation failure check in the application's global allocation manager is certainly a good idea. I'm just saying that for small applications like this it's just overkill. – Andreas Bonini Jul 29 '11 at 21:06
For small applications, yes, it is overkill, I agree. But for an interview question? You do it by the book. – Paul Beckingham Jul 30 '11 at 1:19
@Andreas: "If the computer is out of memory, a lot of applications will crash at the same time, most with segfault-like error messages, so it should be clear to the user what's going on" <- The idea that the computer is out of memory when allocations fail for "out of memory" is bogus. Because an application does not have enough free contiguous memory in its address space it doesn't mean no one else has. – R. Martinho Fernandes Jul 30 '11 at 21:33
I'd be grateful for any brutally honest feedback.
The code is 5 times longer than it needs to be, thanks in part to superfluous code which does things that weren't called for in the specification.
You need 3 or 4 lines of code to read the lines into a map. You use 40 doing things like... recapitalizing, but only the first word in each brand name, for no apparent reason, without explanation. You also strip out any non-alphanumeric characters, which will break brand names like Mercedes-Benz or Rolls-Royce, again without explanation.
The comments are somewhat poor/inconsistent. Comments should tell the reader something the code doesn't. For instance, you explain that you're stripping leading space from each line (something the code already tells us), but don't explain why you aren't stripping trailing space (something we can't read in the code).
Variable names like tmp are also poor (with a few exceptions, like perhaps a swap routine). We know the variable is temporary because of it's scope. The name should tell us what it's for. In this case, it contains the line we're reading, so a name like line would have been better.
As others have pointed out, you're also allocating objects on the heap for no apparent reason. You delete them at the end of main, but not in your early return, which is a huge red flag (given that this is a major source of headaches in C++).
You also have some code that shows you're unfamiliar without how standard library classes work (like assigning 0 to a map entry which is already 0).
As soon as I read the problem description, I alt-tabbed to my editor and wrote this program. I ended up with almost exactly what epatel posted (although his code is broken for multi-word auto names). I haven't been a C++ programmer in nearly 10 years, so I don't know if there's some new fangled stuff I don't know about (lamdas would help here), but the company was probably looking for something straightforward and succinct.
-
Here are problems that I detected :
• do not use raw pointers. There is rarely a need for a raw pointer in c++. If you must, use smart pointers.
• what is the point of multimap? You could that map variable that you defined.
• use of c casts is bad (in this line : ((ifstream *)in)->close();)
• the collect_lines function is too complex and does too much.
-
More importantly, there's no reason to use pointers at all in this code. – Brendan Long Jul 29 '11 at 18:06
@Brendan: Actually, since the stream is polymorphic, pointers are helpful. But dynamic allocation is still not needed. – Ben Voigt Jul 29 '11 at 18:44
Yes to everything that was said so far. One additional thing which I saw is:
// and record the counts
if (lines.count(tmp) == 0)
{
lines[tmp] = 0;
}
lines[tmp]++;
Everything except the last line is unnecessary. When lines[tmp] is accessed for the first time, the key tmp is automatically created in lines, and initialized with the default-constructed value of int (which happens to be 0). See http://en.cppreference.com/w/cpp/container/map/operator_at
-
This is an interesting exercise.
It's interesting because no sensible person would solve this problem in C++. For the very simple reason that the solution in shell script is:
sort cars.txt | uniq -c | sort -rn
Or, if you insist on counts following names:
sort cars.txt | uniq -c | sort -rn | sed 's/ *$$[0-9]*$$ $$.*$$\$/\2 \1/'
Platforms that aren't unix will have other tools that could be used to solve it.
So, were they trying to see if you'd come up with a sensible non-C++ solution, or was this a pointless task that was being used purely to see what sort of code you write?
-
An answer like this definitely deserves some extra points (if you also give a c++ solution) but you have to understand it's a small exercise. Can you come up with a good task that doesn't take to much time to code and can be used to measure your coding skills, yet it's only solvable with C++ and not with shell script, python or ruby? Can you come up with something that's not a pointless task? – Karoly Horvath Jul 30 '11 at 23:13
Fair point. The test we give people at my company (in Java) involves a small but realistic already-existing class that's part of an imaginary web application - a registration handler, in fact, which takes a username and password and creates an account. We ask interviewees to add some more features to it. I don't think that's pointless (perhaps only because J2EE doesn't provide this out of the box, but should!). – Tom Anderson Jul 31 '11 at 17:00
@Tom Pointless task. Thanks! – Jozin S Bazin Jul 31 '11 at 20:23
@Martinho's comments are on target (as is usual for him), but I think there's more to it than just that. @iammilind and @epatel may have a bit ambitious hoping for 10 lines of code, but based on code I posted in a previous answer meeting similar requirements, I'd guess around 15 to 20 could be fairly reasonable.
I'm also less than enthused about how you've organized your code. In particular, I dislike having collect_lines not only reading input and putting it into the map, but also trimming leading white space and doing name-style capitalization. Absent a specific requirement to do so, I'd probably skip those for an interview question, but if they are required they should be in separate functions.
-
10 or 15, think we are in the same ballpark at least ;) got a question of what I was thinking of so made a gist (35 lines total, ~10-15 active lines) – epatel Jul 29 '11 at 19:48
Here is my improvement over epatel's answer.
• It uses a map instead of a list, as suggested in one of the comments.
• It uses the standard copy algorithm instead of doing that manually.
• It imports every name from the std namespace explicitly, to avoid importing unrelated names.
• The functions my_pair_less and my_pair_output don't modify the pairs, so they get an extra const qualifier for their arguments.
• The file is read in line by line, which saves a few lines of code and also allows car names that consist of multiple words.
And here's the code:
#include <iostream>
#include <map>
#include <vector>
using std::cin;
using std::cout;
using std::map;
using std::pair;
using std::string;
using std::vector;
bool my_pair_less(const pair<string, int> &a, const pair<string, int> &b) {
return b.second < a.second;
}
void my_pair_output(const pair<string, int> &p) {
cout << p.first " " << p.second << "\n";
}
int main() {
map<string, int> cars;
string name;
while (getline(cin, name)) {
cars[name]++;
}
vector<pair<string, int> > names;
copy(cars.begin(), cars.end(), back_inserter(names));
sort(names.begin(), names.end(), my_pair_less);
for_each(names.begin(), names.end(), my_pair_output);
return 0;
}
-
Why do these return void?
// reads lines from instream
void collect_lines(istream &in, map<string, int> &lines);
// given lines->num_occurs map, reverses mapping
void reorg_by_count(map<string, int> &lines,
multimap<int, string> &bycount);
There's no need to pass by reference in this case, just do this:
// reads lines from instream
map<string, int> collect_lines(istream &in);
// given lines->num_occurs map, reverses mapping
multimap<int, string> reorg_by_count(map<string, int> &lines);
-
pre-C++0x, the pass-by-reference versions will save an expensive copy – Ben Voigt Jul 29 '11 at 18:43
@BenVoigt: I don't think so. RVO and NRVO will likely kick in on functions where the return logic is non-complex. – DeadMG Jul 29 '11 at 19:06
Oh, and in addition, that's quite premature optimization, I'd go for clarity first. In addition, you can "swaptimize" the return value too, if you want. – DeadMG Jul 29 '11 at 19:22
@DeadMG: Do many compiler even consider RVO and NRVO on functions that aren't inlined? Being defined after use, and fairly long, I'm doubtful that inlining would occur here. – Ben Voigt Jul 29 '11 at 19:43
@Ben Voigt: Modern compilers inline across translation units. A function defined later in the same TU is child's play in comparison. Plus, my points about premature optimization and swaptimization still apply. – DeadMG Jul 30 '11 at 11:08
Following @Malvolio idea I guess this task might have been done in AWK.
AWK is made for programs of this kind. It is event driven, for axample it has event handlers for each line and end of file. It also has map data structure and can print to stdout.
-
Or shell script. sort cars.txt | uniq -c | sort -rn, and relax. – Tom Anderson Jul 30 '11 at 18:06
@Tom You made my day sir. :) – user712092 Jul 31 '11 at 10:27
Yes, this! Add find for unbridled power. – Jonathan Watmough Aug 4 '11 at 17:18
Considering that others have already corrected your code I'd like to propose a somehow different approach to the problem.
I think that we can get rid of the extra pass of sorting an auxiliary map/multimap at the end to preserve the decreasing order.
In order to do that we can use a vector that holds car frequency information and a map that links the car name to that vector.
It's much easier to express this in code so here it goes:
#include <string>
#include <iostream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
using namespace std;
int main( int numberOfArguments, char** arguments )
{
typedef map< string, unsigned int > CarEntryMap;
typedef pair< unsigned int, CarEntryMap::iterator > CarFrequency;
typedef vector< CarFrequency > CarFrequencyVector;
fstream file( "C:\\Cars.txt" );
if( !file.is_open() )
{
return 0;
}
CarEntryMap carEntries;
CarFrequencyVector carFrequencies;
string carName = "";
while( getline( file, carName ) )
{
CarEntryMap::iterator it = carEntries.find( carName );
if( it == carEntries.end() )
{
CarEntryMap::iterator entry = carEntries.insert( it, pair< string, unsigned int >( carName, carFrequencies.size() ) );
carFrequencies.push_back( CarFrequency( 1, entry ) );
}
else
{
unsigned int index = it->second;
pair< unsigned int, CarEntryMap::iterator >& currentEntry = carFrequencies[ index ];
currentEntry.first++;
if( index != 0 )
{
unsigned int updatedIndex = index;
for( int i = index - 1; i >= 0; i-- )
{
if( currentEntry.first <= carFrequencies[i].first )
{
break;
}
updatedIndex = i;
}
if( index != updatedIndex )
{
carFrequencies[ updatedIndex ].second->second = index;
currentEntry.second->second = updatedIndex;
swap( carFrequencies[ updatedIndex ], currentEntry );
}
}
}
}
for( CarFrequencyVector::iterator it = carFrequencies.begin(); it != carFrequencies.end(); ++it )
{
cout << it->second->first << " " << it->first << endl;
}
return 0;
}
This way, instead of sorting at the end we only swap two entries in vector when the car frequency order changes.
-
This is my take on this. I used a map and multiset and a sort predicate.
struct sort_pred {
bool operator()(const std::pair<string,int> &left, const std::pair<string,int> &right) {
return left.second > right.second;
}
};
int main()
{
multiset< pair<string,int> ,sort_pred > myset;
map<string,int> mymap;
for(map<string,int>::iterator it=mymap.begin();it!=mymap.end();it++)
{
myset.insert(make_pair<string,int>(it->first,it->second));
}
cout<<"Elements in the set:"<<endl;
for(multiset<pair<string,int>,sort_pred >::iterator it=myset.begin();it!=myset.end();it++)
cout<<it->first<<" "<<it->second<<endl;
return 0;
}
{
string filename="temp.txt";
ifstream file;
file.open(filename.c_str());
if(!file.is_open())
{
cerr<<"Error opening file : "<<filename.c_str()<<endl;
exit(0);
}
string line;
while(getline(file,line))
{
if(t.find(line)!=t.end())
t[line]++;
else
t.insert(std::make_pair(line,1));
}
}
-
Just to throw out a thought that nobody else mentioned, everybody here is using a map (which as I'm a C# dev I'm imagining that's more or less a dictionary/hashtable of sorts), I would have thought of doing this as a heapsort on a filled in string array from the file, then iterate over it just counting dupes and outputting the previous member with it's count everytime a member doesn't match the previous.
sorry for my lack of C++ but it would be something like, after reading the file or stdin into the array and heapifying it (you may need to implement your own text comparer, not sure on that in C++)
string previousCar = sortedArray[0];
int numberOfConsecutiveDupes = 0;
for(int i = 0; i < length(sortedArray); i++) // Don't know if this is how to retrieve array length in C++ sorry :(
{
if (sortedArray[i] == previousCar)
{
numberOfConsecutiveDupes++;
continue; // don't know if continue exists in C++?
}
SendYourOutputToFileOrWhereverYouWantTo(previousCar + " " + itoa(numberOfConsecutiveDupes)); // itoa, I know this is wrong, I really don't know C++
previousCar = sortedArray[i];
numberOfConsecutiveDupes = 1;
}
I realize this would not get anyone the job, I'm just trying to propose a different solution strategy given that they did know C++ syntax/STL/etc (which I def do not)..
-
Everyone is using std::map because it is the textbook example for std::map! An easy while(cin >> name) map[name]++; is enough to read words and count them. Not using a std::map is a clear signal one is not known with C++. – Sjoerd Oct 29 '11 at 0:43
def countCars(fname):
carCount = {}
with open(fname, 'r') as f:
for car in f:
car = car.strip()
carCount[car] = 1 + carCount.get(car, 0)
return carCount
def printCount(carCount):
items = carCount.items()
items.sort(lambda a,b:b[1]-a[1])
for item in items:
print "%s %d" % item
if __name__ == "__main__":
import sys
printCount(countCars(sys.argv[1]))
And then they'd say "That isn't C++" and I'd say "C++ really isn't appropriate for this kind of work, which is an order of magnitude faster to write in higher-level language and runs IO-bound anyway." and then they wouldn't hire me and I'd go work at a company that uses languages that aren't old enough to rent a car.
Go ahead, flame away, what do I care...
Later: The thought occurs, maybe the hiring company didn't specify the language and that was the OP's mistake, choosing a Reagan-era hold-over like C++.
-
Python? Pshaw! Nobody's even heard of the Web when that old language was invented. You should be using Clojure, or Go. They're so much newer! Never mind relevance in the real world, amirite? – Mud Jul 30 '11 at 4:39
@Mud -- in (some degree of) seriousness, I wouldn't defend Python as young, I attended the first world-wide Python conference back in 1996 (there were maybe 30 people there) because I was writing a platform for Internet retail, in Python. The next year, I sold it to Microsoft and mind-bogglingly they're still selling it! Clojure and Go aren't bad choices for this problem (although I would argue, no better than Python), and although Mud is sarcastically correct, popularity matters, technical innovation does too. No chars left! – Malvolio Jul 30 '11 at 14:14
@Malvolio: it's a coding exercise, what did you expect? They have to give some simple task... About higher-level languages: There are zillions of usages of c/C++ where python (and most other languages) would fail: compilers, modern 3d games, audio and video processing, device drivers, desktop apps (your browser, movie player, chat, ...), embedded devices and almost all the implementation of other languages (including python). Ranting about how easier is to solve a simple task like this in a higher-language shows your ignorance and it is alone enough reason not to hire you. – Karoly Horvath Jul 30 '11 at 22:53
@yi_H -- What did I expect? Well, call me ignorant but I expect that an exercise in a job interview is relevant to the actual job. So, if the company is writing device drivers or video processing or whatever C++ shines in, then a string processing function is a poor choice of exercises. In my uninformed opinion. – Malvolio Jul 31 '11 at 10:05
@Malvolio The task apparently was good enough to weed out those that don't know C++ well. – Sjoerd Oct 29 '11 at 0:46 | 8,164 | 32,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-22 | latest | en | 0.717389 |
https://www.gurufocus.com/term/Intangibles/NSP/Intangible+Assets/Insperity%252C+Inc | 1,508,451,998,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232824-00003.warc.gz | 945,413,804 | 40,295 | Switch to:
Insperity Inc (NYSE:NSP) Intangible Assets: \$13 Mil (As of Jun. 2017)
Intangible assets are defined as identifiable non-monetary assets that cannot be seen, touched or physically measured. Insperity Inc's intangible assets for the quarter that ended in Jun. 2017 was \$13 Mil.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Insperity Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Intangible Assets 23.78 18.43 14.46 13.09 13.59
Insperity Inc Quarterly Data
Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Intangible Assets 13.34 13.21 13.59 12.96 12.84
Calculation
Intangible assets are defined as identifiable non-monetary assets that cannot be seen, touched or physically measured. Examples of intangible assets include trade secrets, copyrights, patents, trademarks. If a company acquires assets at the prices above the book value, it may carry goodwill on its balance sheet. Goodwill reflects the difference between the price the company paid and the book value of the assets.
Explanation
If a company (company A) received a patent through their own work, though it has value, it does not show up on its balance sheet as an intangible asset. However, if company A sells this patent to company B, it will show up on company B's balance sheet as an intangible asset.
The same applies to brand names, trade secrets etc. For instance, Coca-Cola's brand is extremely valuable, but the brand does not appear on its balance sheet, because the brand was never acquired.
Some intangibles are amortized. Amortization is the depreciation of intangible assets.
Many intangibles are not amortized. They may still be written down when the company decides the asset is impaired.
Whenever you see an increase in goodwill over a number of years, you can assume it's because the company is out buying other businesses above book value. GOOD if buying businesses with durable competitive advantage.
If goodwill stays the same, the company when acquiring other companies is either paying less than book value or not acquiring. Businesses with moats never sell for less than book value.
Intangibles acquired are on balance sheet at fair value.
Internally developed brand names (Coke, Wrigleys, Band-Aid) however are not reflected on the balance sheet.
One of the reasons competitive advantage power can remain hidden for so long.
Be Aware
Companies may change the way intangible assets are amortized, and this will affect their reported earnings.
Related Terms | 607 | 2,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-43 | latest | en | 0.917441 |
http://mathhelpforum.com/advanced-algebra/167277-question-about-normal-subgroups-index.html | 1,505,992,460,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687740.4/warc/CC-MAIN-20170921101029-20170921121029-00719.warc.gz | 219,543,908 | 12,823 | 1. ## a question about normal subgroups and index.
given that N is a normal subgroup of G. also H is given to be a subgroup of G.
also given that G is a finite group with gcd(|G:N|,|H|)=1.
prove that H is a subgroup of N.
2. Originally Posted by abhishekkgp
given that N is a normal subgroup of G. also H is given to be a subgroup of G.
also given that G is a finite group with gcd(|G:N|,|H|)=1.
prove that H is a subgroup of N.
(1) $(HN)/N=H/(H \cap N)$
Since N is a normal subgroup of G and H is a subgroup of G, HN is a subgroup of G. Thus, $|(HN)/N|=|H/(H \cap N)|$ divides [G:N] by (1) and Lagrange's theorem. It follows that $|H/(H \cap N)|$ divides both [G:N] and |H|. Since gcd(|G:N|,|H|)=1, we see that $|H/(H \cap N)|=1$. Thus, $H \cap N=H$. We conclude that H is a subgroup of N.
3. ## discussing the solution.
so your solution uses the fact that if N is a normal subgroup and H is a subgroup then
(H intersection N) is a normal subgroup of H??
is that right?
(why \cap not working?)
4. If $|H/(H \cap N)|=1$, then $H \cap N = H$. If $H \cap N = H$, then H is a subgroup of N.
5. ## discussion
there you have written HN/N = H/(H intersection N).
For the RHS of the above equation to be well defined its necessary that
(H intersection N) is a normal subgroup of H. is this correct??
6. It is the second isomorphism theorem. Yes, H intersection N is a normal subgroup of H. Define a surjective homomorphism f:H--->HN/N such that its kernel is (H intersection N). Now the kernel of f has to be the normal subgroup of H.
7. thanks a lot. that helped.
also i found another solution which you may find intersesting and more algebraic.
here it is:
suppose that h is in H. and suppose [G:N] = k, |H| = m.
since gcd(k,m) = 1, there are integers r,s with rk + sm = 1.
it may be that H is trivial, in which case H < N.
so suppose not. let h ≠ e be chosen arbitrarily in H.
then Nh = (Nh)^(rk + sm) = (Nh)^(rk)(Nh)^(sm)
= [(Nh)^k]^r[(Nh)^m]^s
= (N^r)(N(h^m))^s = N(Ne)^s = N(N) = N,
so h must be in N, thus H < N.
8. Alternatively one can us the first isomorphism theorem. Note that since $N\unlhd G$ we have that $G/N$ is a group with order $\left(G:N\right)$. Consider then the canonical projection $\pi:G\to G/N$. We see then that the restriction $\pi_{\mid H}:H\to G/N$ is also a homomorphism but since $\left(|H|,\left|G/N\right|\right)=1$ it follows that $\pi_{\mid H}$ is the trivial homomorphism $1:H\to G/N:h\mapsto N$. In particular it follows that $\displaystyle \pi_{\mid H}\left(H\right)=HN=N$ but this is true if and only if $H\leqslant N$.
P.S. I implicitly used the FIT by using it to prove that if $\phi\in\text{Hom}\left(G,G'\right)$ with $|G|,|G'|<\infty$ then $\left|\phi(G)\right|$ divides both $|G|$ and $|G'|$.
9. ## Re: a question about normal subgroups and index.
Originally Posted by Drexel28
Alternatively one can us the first isomorphism theorem. Note that since $N\unlhd G$ we have that $G/N$ is a group with order $\left(G:N\right)$. Consider then the canonical projection $\pi:G\to G/N$. We see then that the restriction $\pi_{\mid H}:H\to G/N$ is also a homomorphism but since $\left(|H|,\left|G/N\right|\right)=1$ it follows that $\pi_{\mid H}$ is the trivial homomorphism $1:H\to G/N:h\mapsto N$. In particular it follows that $\displaystyle \pi_{\mid H}\left(H\right)=HN=N$ but this is true if and only if $H\leqslant N$.
P.S. I implicitly used the FIT by using it to prove that if $\phi\in\text{Hom}\left(G,G'\right)$ with $|G|,|G'|<\infty$ then $\left|\phi(G)\right|$ divides both $|G|$ and $|G'|$.
i had asked this long back but somehow i didn't see it and i checked it out today. brilliant!
i got another solution which is cool too:
Suppose $|G:N| = k, |H| = m$.
since $gcd(k,m) = 1$, there are integers $r,s$ with $rk + sm = 1$
If $H$ is trivial then $H \leq N$ and we are done.
so suppose $H$ is not trivial. Let $h \neq e$ be chosen arbitrarily in $H$.
then $Nh = (Nh)^{rk + sm}= (Nh)^{rk}(Nh)^{sm}$
$\Rightarrow Nh= [(Nh)^k]^r[(Nh)^m]^s$
$\Rightarrow Nh= (N^r)(N(h^m))^s = N(Ne)^s = N(N) = N$
so $h$ must be in $N$, thus $H < N$. | 1,347 | 4,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 53, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-39 | longest | en | 0.937234 |
https://www.mersenneforum.org/showpost.php?s=06ea5e04dc5c59648bb4fe7cfa2197dd&p=453121&postcount=11 | 1,606,702,163,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00113.warc.gz | 758,932,868 | 4,345 | View Single Post
2017-02-17, 02:32 #11 carpetpool "Sam" Nov 2016 11·29 Posts One pari command I am searching for is polynomials with the same properties as their cyclotomic polynomials, and finds them based on fixed coefficients in order. ({a, b}, x) would search for a polynomial with the same properties as the cyclotomic polynomial of x, with leading coefficients a, b, c for each higher degree. For example: ({1, 4}, 5) should find all polynomials with the same number field properties as x^4+x^3+x^2+x+1 of the form x^4+4x^3+ax^2+bx+c. That is should output the solutions, if any {a, b, c} for the degrees 2, 1, 0. I've scoured a whole column of polynomial commands and not one serves for this purpose. | 205 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-50 | latest | en | 0.906997 |
https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Bulk_Properties/B3%3A_Heats_of_Fusion_(Reference_Table) | 1,675,662,252,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00353.warc.gz | 191,097,138 | 27,950 | B3: Heats of Fusion (Reference Table)
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Chemical Heat of Fusion (J/g)
Acetic acid 181
Acetone 98.3
Alcohol, ethyl (ethanol) 108
Alcohol, methyl (methanol) 98.8
Alcohol, propyl 86.5
Aluminum 321
Ammonia 339
Aniline 113.5
Antimony 164
Benzene 126
Bismuth 52.9
Brass 168
Bromine 66.7
Carbon disulfide 57.6
Carbon dioxide 184
Carbon tetrachloride 174
Cast iron 126
Chloroform 77
Chromium 134
Cobalt 243
Copper 207
Decane 201
Dodecane 216
Ether 96.2
Ethyl ether 113
Ethylene glycol 181
Glycerine 200
Gold 67
Heptane 140
Hexane 152
Ice, see Water
Iodine 62.2
Iron, gray cast 96
Iron, white cast 138
Iron, slag 209
Manganese 155
Mercury 11.6
Naphthaline 151
Nickel 297
Octane 181
Paraffin 147
Phenol 121
Phosphrus 21.1
Platinium 113
Potassium 59
Propane 79.9
Propylene 71.4
Silver 88.0
Sulfur 39.2
Tin 58.5
Toluene 71.8
Water, Ice 334
Wood's alloy 33.5
Zinc 118
B3: Heats of Fusion (Reference Table) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 801 | 2,015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-06 | latest | en | 0.325883 |
tech.raisa.com | 1,725,705,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00661.warc.gz | 32,031,573 | 12,098 | # Time Series Forecasting using Transformers and Ordinal Regression
## Raisa's Domain and Time Series Prediction
• One of the things we do at Raisa is forecasting oil and gas production for all wells in the United States. Forecasting are usually done for the next twelve months based on previous data, which is by definition a time series forecasting problem.
• With the rise of transformers in Natural Language Processing and Computer Vision, we conducted an experiment that tries to use transformers to predict the data described above.
• This blog summarizes one of many experiments which is Ordinal Regression (Classification)
### Motivation
• Why did we choose Ordinal Regression instead of Regression?
• We tried to use regression at first but the results were not impressive, which pushed us to try and re-examine the way we look at our data and our problem.
• Since transformers are already proven as Language Models, we thought that we could create a "Well Model".
• Just like you could give a Language Model a prompt and it would continue and come up with a full sentence, We wanted to train a "Well Model" that you could give some production values and it would predict the rest of the production stream as if it was predicting words to complete a sentence.
# Data Processing
• To apply classification on continuous data, we needed to bin the data so that we can treat each production value similar to a word in a corpus
### Binning Algorithm
• The algorithm we used for binning was quantile binning but with a slight modification.
Quantile binning: Values are binned so that each bin contains a certain percentage of values.
• The modification we added is that instead of moving to a new bin after reaching a certain percentage of the data, we added a threshold for the difference between each bin and the next one.
• If we have an array of bins $$B = b_1, b_2, ..., b_n$$, this rule applies to almost all of the bins $$b_{i+ 1} - b_{i} < 0.1 * b_{i}$$
• We applied this rule after the value of 100 barrels, as before then 10% of a value would be very negligible.
• e.g. 10% of 2 is 0.2, which would mean that the value 2 would be its own bin no matter what percentage of values is in it.
• This was done to reduce the information lost when binning the production values while making sure each bin has enough data in it for the training process
# Model Architecture
• As we wanted to have a model that can generate new timesteps given some production values, we thought a decoder architecture would be the most fitting and chose the GPT2 to be the body of the model.
• We will show how we implemented this using PyTorch
## Data Embeddings
• The first step in our model was how to transform the bins into vectors and add the time encodings, for that we chose a simple nn.Embedding layer for both time and token embeddings
class DataEmbeddings(nn.Module):
def __init__(self, config: DecoderConfig):
self.config = config
self.time_positional_encoding = nn.Embedding(MAX_TIMESTEPS, config.embed_dim)
self.token_embedding = nn.Embedding(config.num_of_bins, config.embed_dim)
def forward(self, x: torch.Tensor, pos: torch.Tensor):
token_embedding = self.token_embedding(x)
time_embedding = self.time_positional_encoding(pos)
## Decoder
• The decoder consists of many different parts that we need to define before creating the decoder itself, those things include DecoderBlock, FeedForward, and MultiHeadSelfAttention.
### Attention
• To simplify attention as much as possible, think of it as a weighted average of the features given to the transformer, where the weights are learned during training.
class AttentionHead(nn.Module):
def __init__(self, config: DecoderConfig, attention_head_dim: int) -> None:
super().__init__()
self.q = nn.Linear(config.embed_dim, attention_head_dim)
self.k = nn.Linear(config.embed_dim, attention_head_dim)
self.v = nn.Linear(config.embed_dim, attention_head_dim)
• The function scaled_dot_product_attention is the averaging step, you will notice that we pass the tensor x to the function as the key, value, and query which is why we call it self-attention.
• You will notice that the function takes causal_mask and padding_mask as arguments, these are used to ignore certain timesteps during the attention.
def safe_softmax(x, eps=1e-8):
"""Softmax modified so that we can output 0 in all elements
i.e. pay attention to 0 timesteps"""
e_x = torch.exp(x - x.max())
return e_x / (e_x.sum(dim=-1, keepdim=True) + eps)
scores = torch.bmm(q, k.transpose(1, 2)) / math.sqrt(q.size(-1))
# If element is True, then we don't want to do the attention on this element.
weights = safe_softmax(scores)
• We also include multiple self-attention blocks in each decoder which is why we call it MultiHeadSelfAttention
class MultiHeadSelfAttention(nn.Module):
def __init__(self, config: DecoderConfig) -> None:
embed_dim = config.embed_dim
assert embed_dim % num_heads == 0, "embed_dim must be divisible by num_attention_heads"
self.dropout = nn.Dropout(config.attention_dropout_prob)
self.proj_linear = nn.Linear(embed_dim, embed_dim)
x = self.dropout(self.proj_linear(x))
return x
@staticmethod
"""Create causal mask for self attention"""
causal_mask = torch.triu(torch.ones(MAX_TIMESTEPS, MAX_TIMESTEPS), diagonal=1) == 1
• Since we need to input a fixed amount of timesteps each forward pass, there will be cases where the well has less than that amount of timesteps, and we would need to pad those elements with a certain value, and then tell the transformer to ignore it.
• We ignore it through the padding_mask
• Causal mask is a lower triangular square matrix that tells the decoder which
timesteps to ignore in attention module, this is done so that the model doesn’t
cheat by looking into the future when predicting.
Each element in $$C$$ has a value of 1 or 0.
• If $$C_{ij} = 1$$ this means that at timestep $$i$$ the model can observe the value at timestep $$j$$ when predicting timestep $$i + 1$$
## Feed Forward Network
• This is a simple two-layer neural network with layer normalization
class FeedForward(nn.Module):
def __init__(self, config: DecoderConfig):
self.linear_1 = nn.Linear(config.embed_dim, config.get_intermediate_dim())
self.linear_2 = nn.Linear(config.get_intermediate_dim(), config.embed_dim)
self.dropout = nn.Dropout(config.ff_dropout_prob)
self.activation = NewGELU()
def forward(self, x):
x = self.dropout(self.linear_2(self.activation(self.linear_1(x))))
return x
## Decoder Block
• Now that we have all the components of the decoder, we can just stack them up to get one decoder block as follows
class DecoderBlock(nn.Module):
def __init__(self, config: DecoderConfig):
super().__init__()
self.config = config
self.layer_norm_1 = nn.LayerNorm(config.embed_dim)
self.layer_norm_2 = nn.LayerNorm(config.embed_dim)
self.ffn = FeedForward(config)
def forward(self, x, mask=None):
x = x + self.attention(self.layer_norm_1(x), mask)
x = x + self.ffn(self.layer_norm_2(x))
return x
## Decoder
• Now that we have the decoder block ready, we replicate it 12 times just like what is done in the GPT2 paper and put all the components in one forward pass
class PDPDecoderTransformer(nn.Module):
def __init__(
self,
config: DecoderConfig,
):
# transforms each bin into a vector of size config.d_model
# and adds positional embeddings
self.data_embeddings = DataEmbeddings(config)
# decoder blocks
decoder_block = DecoderBlock(config)
self.decoder_blocks = _clones(decoder_block, config.num_decoder_layers)
self.dropout = nn.Dropout(config.ff_dropout_prob)
self.layer_norm = nn.LayerNorm(config.embed_dim)
self.head = nn.Linear(config.embed_dim, config.num_of_bins)
def forward(
self,
x: torch.Tensor,
):
B = x.size(0)
pos = torch.arange(0, MAX_TIMESTEPS, device=x.device).long().unsqueeze(0).repeat(B, 1)
x = self.data_embeddings(x, pos)
for i, decoder_block in enumerate(self.decoder_blocks):
x = decoder_block(x, mask)
x = self.layer_norm(x)
## Loss Functions
### Cross Entropy Loss
• Cross entropy is the loss function used in most classification problems, it has a formula of :
$$L(y, p) = -1 \frac{1}{n} \sum_{i = 1}^n \sum_{c = 1}^C y_{ij} \log (p_{ij})$$
• We wanted to guide the model that some bins were more similar to each other, e.g. if the correct bin is 5 and you predict bin 10, you should have a higher loss than somebody who predicted bin 6.
• So we are proposing different loss functions
### Gaussian Cross Entropy Loss
• This is very similar to cross entropy but with a slight modification. $$y$$ is now a gaussian centered around the index of the correct bin, instead of a one-hot vector.
• The formula is still
$$L(y, p) = -1 \frac{1}{n} \sum_{i = 1}^n \sum_{c = 1}^C y_{ij} \log (p_{ij})$$
but $$y$$ represents a different vector.
STD_DEV = 0.27
# config.num_of_bins = 10
gaussians = torch.zeros(config.num_of_bins, config.num_of_bins, device=device)
x = torch.arange(config.num_of_bins, device=device)
for mean_around in range(config.num_of_bins):
g_x = torch.exp(-((x - mean_around) ** 2) / (2 * STD_DEV ** 2))
g_x = g_x / torch.sum(g_x)
gaussians[mean_around] = g_x
# you can then access the gaussian vector through indexing the tensor
print(gaussians[4])
# tensor([0.0000e+00, 1.5516e-27, 1.2142e-12, 1.0481e-03, 9.9790e-01, 1.0481e-03, 1.2142e-12, 1.5516e-27, 0.0000e+00, 0.0000e+00])
• This way the model understands that it should have a lower loss if the bin predicted is close to the correct bin, instead of just treating close predictions the same as predictions far from the correct bin.
### Custom Ordinal Cross Entropy Loss
• This adds a factor that also penalizes the model more if the bin is farther from the correct index.
• It has a formula of
$$L(y, p) = -1 \frac{1}{n} \sum_{i = 1}^n (idx_{predicted} - idx_{correct})^2 \sum_{c = 1}^C y_{ij} \log (p_{ij})$$
After trying all of these approaches, they provided similar results, so we went with the regular cross entropy.
## Inference
### Predict Max Bin
• This was the naive approach that we decided on earlier in the experiment, we would pick the bin that had the highest probability of being correct.
### Weighted Average
In this method, we need to define two arrays
• pred : shape (1, # bins) for a single example, which we will refer to
as p where $$p_i = p(y = bin_i)$$
• bin representation values: shape (1, # bins), which we will refer to
as r, where $$r_i$$ is the value we use to represent $$bin_i$$
We take the weighted average of the representation value of the top k bins, where the weights are the probabilities of each of the top k bins
e.g. if $$r = [100, 500, 1000, 2000, 5000], p = [0.25, 0.45, 0.15, 0.1, 0.05]$$ and k = 2
our prediction would be $$y = \frac{p2∗r2+p1∗r1}{p1+p2} = \frac{0.45∗500+0.25∗100}{0.45+0.25} = 357.14$$
But we added a condition before doing this, we will take the top k bins or
until the sum of the first n bins reaches 50%
e.g. if k = 3 and $$p = [0.25, 0.45, 0.15, 0.1, 0.05]$$, we would only take bins 1 and 2, as $$p1 + p2 > 0.5$$
You can tune the percentage that you stop before depending on your problem.
The second approach was much better than the first one especially in the later month as the model error accumulates and the model is less sure of its predictions
## Results
• The model's performance was indicative of actual learning and sucessful training but it did has not outperformed the current model at Raisa.
• Here are some examples of model predictions.
## Conclusion
• During our experiment, we used the transformer in a different domain in which it is usually used.
• We found out that with a lot of tweaks the model shows signs of learning, but it still doesn't outperform the current model at Raisa
• The upside of this is that it is easier to scale up the transformer, we plan on experimenting with multivariate classification.
• We would make the model predict the oil and gas data simultaneously and make use of both features.
• This experiment could give the transformer the edge it needs in order to surpass the performance of the current model. | 3,029 | 12,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-38 | latest | en | 0.937827 |
https://www.infinity-cdma.com/p/154140647466/ | 1,553,641,137,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206677.94/warc/CC-MAIN-20190326220507-20190327002507-00111.warc.gz | 797,309,444 | 5,405 | ### Series Circuit Diagram
Series Circuit Diagram
## Aaon Cc Series Wiring Diagram : 29 Wiring Diagram Images
Component References. Components at a circuit must always have testimonials, also called reference designators, utilized to recognize the elements in the circuit. This allows the elements to readily be referenced in text or a component listing. A battery may have the reference designator"BAT" and also a light bulb may have a benchmark"L".
If wires or lines cross each other and there's no node, as shown at the bottom of the aforementioned picture, the wires are not electrically connected. In cases like this the wires are crossing each other without connecting, like two insulated wires placed you on top of another.
The following are overall circuit diagram principles.
• Lines or pliers in circuit diagrams are often horizontal or vertical. Sometimes a diagonal line might be used that is placed at 45 degrees.
• Component symbols in a circuit structure are often placed horizontally or vertically. On very rare occasions a part may be placed in 45 degrees, but just for an excellent reason.
• Circuit diagrams have been drawn as simply and neatly as possible. This means that the physical execution of the circuit may look different from your circuit structure, however they are exactly the same.
• Lines connecting parts can be thought of as insulated wires in most situations, with just the ends of the cables being bare conductors for electric connection.
• When lines cross each other at a circuit diagram, they are sometimes thought of as two insulated wires crossing if there is not any node where the wires intersect or cross each other.
• Three lines intersecting at a point with a node at the junction usually means the three wires are electrically connected. This connection could be considered as three insulated wires bared at the point of junction and glued together.
• Two wires that cross each other with a node at the intersection of the crossing stage usually means that the cables are inextricably connected.
• When starting to learn how to read electronic circuit diagrams, it's necessary to learn exactly what the schematic symbol looks like to get various electronic elements. Observing the course explains how to examine basic electronic circuit diagrams while constructing the circuits on electronic breadboard. The course contains a record of basic electronic elements with their schematic symbols in which novices can learn exactly what the physical components and their symbols look like.
A component list can refer by reference designator to those components. A node is simply a filled circle or scatter. After a couple of lines touch each other or cross each other and also a node is put at the junction, this represents the lines or wires being electrically connected at that point.
Following a four section introduction, the very first tutorial from the electronics course shows the circuit diagram of a simple LED and resistor circuit and how to construct it on breadboard.
The base terminals of the bulbs are linked to each other and to the negative terminal of the battery, since the next node indicates these connections.
Battery and Light Bulb Circuit. Possibly the easiest circuit which can be drawn is one that you might have seen in a college science class: a battery connected to a light bulb as shown under.
Because there may be more than one battery or light bulb in a circuit, reference designators will usually always result in a number, e.g. BAT1 and L1 as shown in the circuit under. A second light bulb at the circuit would then possess the reference designator L2.
Series Circuit Example. No nodes are necessary inside this circuit to demonstrate the bulbs connecting to each other and to the battery because single wires are linking straight to each other. Nodes are only set if three or more wires are attached.
Basic components for this tutorial include a LED, resistor and battery life which can all be found in the beginner's component benchmark.
Parallel Circuit Example It can be seen that the best terminals of both light bulbs are all connected together and into the positive terminal of the battery. We understand this because the 3 terminals or connection points possess a node where they intersect.
Physical Circuit. The circuit for the above circuit diagram might look something like the image below, but a practical physical circuit would have a light bulb holder and knobs that connect with the battery terminals. A light bulb holder could need screw terminals to attach the wires to, along with a socket to screw the light bulb . Battery presses would allow the cables to readily be attached between the battery and light bulb holder.
Learn how to read electrical and electric circuit diagrams or schematics. A drawing of an electrical or electrical circuit is also referred to as a circuit structure, but could also be referred to as a schematic diagram, or only schematic.
Each electronic or electrical element is represented by a symbol as can be found in this very simple circuit arrangement. Lines used to join the symbols represent conductors or cables. Each symbol represents a physiological element that may look as follows.
The easiest way for beginners to keep on learning how to read circuit diagrams is to follow the course and build the circuits from every tutorial.
Specifying Components. Typically the true battery kind and bulb kind would be defined in a component list that communicates the circuit diagram. More information about the bulb and battery kind could also be included in the circuit as text. For example, the battery might be specified as a 12.8V 90Ah Lithium batterypowered, plus even a 9V PM9 battery. The light bulb could be specified as a 12V 5W incandescent bulb, or 9V 0.5W torch bulb.
Circuit or schematic diagrams contain symbols representing physical components and lines representing cables or electrical conductors. To be able to learn to read a circuit design, it is vital to learn what the design symbol of a component appears like. It's also crucial to comprehend how the components are connected together in the circuit. | 1,174 | 6,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-13 | latest | en | 0.947097 |
https://math.answers.com/Q/A_number_is_divisible_by_another | 1,714,007,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00689.warc.gz | 353,831,889 | 46,504 | 0
# A number is divisible by another?
Updated: 9/17/2023
Wiki User
13y ago
A number is divisible by another when the remainder of the division is zero.
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### What does divisible by meen?
If you can divide a certan number by another number it is divisible by that number.
### A number is divisible by another when the BLANK after division by that number is 0?
A number is divisible by another when the remainder after division by that number is zero.
### When a number is divisible by another number what is it?
All numbers are divisible by 1. So, apart from the number 1, all numbers are divisible by another number. These numbers are therefore prime or composite.
### How is a number divisible by 2?
if 2 multiplied by another number equals that number it is divisible by 2
### A number is divisible by another number when the?
When it is a factor of that number
### A number is divisible by another number when?
When it is a factor of that number
### How do you a number is a prime number?
if the number is not divisible by another number
### What is a remainder of a number that is divisible by another number?
The fat that it is divisible means that the remainder must be zero.
No its not
remainder
### What do you call a number that is divisible by another?
A composite number.
### Can i plus see the definition for divisible?
One number is divisible by another number if that division results in an integer. | 346 | 1,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-18 | latest | en | 0.942198 |
https://www.rocketryforum.com/threads/stability-question.9604/ | 1,718,480,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00825.warc.gz | 867,987,244 | 36,183 | # Stability question
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##### Well-Known Member
My rocket is .68 staticly stable but a few inches on pad the rocket is at a "safe take off speed" so once that speed is reached that the static stability no longer applys...... I think that's how it works the faster the more stable it is right? Because the fins are more effective?
Not exactly. Speed does have an effect on stability, but it is kind of already built into the stability margin. That is why a stable rocket that does not achieve the required speed will be unstable.
The truth is that these values are just estimates. Also, the actual stability margin changes during flight depending on many factors, including speed and crosswind component. That is the reason for using the 2 caliber rule. Some are trying to get this rule changed based on percentage of body length, but that is a different thread all together. So 0.16 stability margin might result in a stable flight on a no wind day. However, as wind or other factors come into play, your rocket has a bigger chance of becoming unstable.
My rocket is .68 staticly stable but a few inches on pad the rocket is at a "safe take off speed" so once that speed is reached that the static stability no longer applys...... I think that's how it works the faster the more stable it is right? Because the fins are more effective?
Stability is a VERY complicated and thorny issue, because essentially it is a MOVING target that greatly depends on the rocket, the flight parameters, and the environmental parameters of the atmosphere through which the rocket is moving.
Static stability is the easiest concept to grasp and to define, because dynamic stability, that is to say, stability in flight basically is constantly changing to some degree or other. Therefore static stability seeks to make these changes moot by defining the stability of the rocket "at rest", hopefully in a close approximation of typical conditions the rocket will be exposed to in flight, and conditions which the rocket is unlikely to exceed in flight.
CG, center of gravity, is of course rather easily obtained by balancing the launch-ready rocket (without ignitor, since that is ejected at ignition and does not fly with the rocket (usually, HOPEFULLY!). CG gets a bit more complicated on "non-symmetrical rockets" like the space shuttle stack, multistage rockets which eject series stages or parallel stages in flight, rockets carrying off-center payloads (cameras, etc.) and the like. Even in a "typical" three-fin and nosecone (3FNC) rocket, CG moves SOME, as the propellant is burned in engine, the CG moves forward, since the nose weight (and hopefully the parachute weight and location) remains the same. Other engine types (liquid, predominantly, though hybrids are a different case too) have their CG locations change in different ways, usually aft, as the engine burns. BUT for 99% of hobby rocketry, it can be said the CG moves FORWARD as the engine burns.
CP, center of pressure, is a MUCH more dynamic and difficult to measure property. CP is affected by the design, fin size, span, shape, drag, the rocket's overall shape, size (Reynolds numbers come into play at certain size thresholds), angle of attack, weight distribution (moments of inertia, length, oscillation characteristics, etc.) wind speed and direction, flight speed, static stability margin, "apparent" angle of attack from wind, air density, air turbulence, and probably a hundred other factors. Some have extremely subtle effects that are barely measurable. Some have significant effects that can have OVERT effects, and some are subtle at times, and overt at others, and combine and compound with other effects to create greater CP changes than at other times, and sometimes certain factors can cancel each other out or work together beneficially to enhance stability. It's EXTREMELY complicated (that's a part of why rocket science is HARD)
SO, how do we get a handle on it?? One way is to look at the 'worst case scenario'... if we design the rocket to be stable with the WORST conditions it's likely to ever see, we can be pretty darn sure it'll be stable under virtually every flight condition we're likely to come up against. Fortunately this is a rather simple matter. By determining the center of lateral area we can determine the *rough* CP point, which, interestingly enough, is the point at which the CP would approximately be if the rocket were ninety degrees to the airflow-- essentially 'flying sideways'... So if the rocket could stabilize from THAT, we can be pretty darn sure it's stable. This is called the cardboard cutout method. Draw an exact outline of the rocket on cardboard, cut it out, and balance that "profile" on a ruler-- the balance point is the center of lateral area. Rocksim can be reset under the 'calculate stability by' box by clicking it and choosing "Cardboard cutout method" from the drop down menu.
Now this method, as you can tell, is VERY VERY conservative. It can lead to EXCESSIVE stability, where the rocket will gyrate and oscillate excessively, weathercock excessively in windy conditions, and other such undesirable effects that rob energy and therefore altitude. That's where the Barrowman Method comes in, which makes certain assumptions about the shape of the rocket and their aerodynamic affects on the rocket's flight. Those assumptions simplified the calculations back in the pre-computer era so that rocketeers with sufficient math skills who wanted to take the time and effort to crunch the numbers by hand could actually sit down and calculate the CP to greater accuracy than the cardboard cutout method would allow. This would allow greater precision and optimization of the rocket design. But it was a VERY labor-intensive process taking several hours at best, and STILL had quite a few limitations because of the ASSUMPTIONS made in creating and simplifying the mathematical formulas. For instance, the Barrowman Method couldn't handle but either 3 or 4 fins-- odd numbers of fins were out. Complicated shapes beyond the basic tubular rocket with radial symmetry (transitions, nosecones, tailcones, evenly spaced fins) were out-- so calculating CP with side pods, boosters on the sides, gliders, etc. was out. Basically, at the end of the day, you STILL had an APPROXIMATION.
The "Rocksim Method" used in the Rocksim program is based on the Barrowman Method and uses it's calculations as the basis of it's function, with some tweaks to the assumptions and some more complex mathematical modelling to get 'closer' to the "real" CP than the basic Barrowman Method allowed. Complex math isn't much of a problem anymore with the massive computing power available now (your PC is more powerful than the Apollo flight computers that got us to the MOON!). Still, there are underlying ASSUMPTIONS in the math models that may or may not accurately reflect or predict the CP location in every instance-- some parameters may be inaccurately modelled, some may be ignored, some may be 'over conservative' in the math models, etc.
Basically, to know the TRUE CP, you'd have to measure all the influences on the rocket at EVERY GIVEN MOMENT IN FLIGHT and plug those measurements into equations to determine the EXACT CP AT THAT PRECISE MOMENT, because as the parameters change from moment to moment, the CP changes as well.
Fortunately, the approximations we get from computer programs, hand crunched numbers, and even cardboard cutouts are enough 99% of the time. The real ART of stability determination is in contest rocketry, where balancing all those forces, and making certain design tradeoffs combine to either give you a winning design or leave you in everyone else's dust...
SO, summing up, static stability is the mathematically calculated stability points (CG/CP relationship) of the rocket. As the rocket takes off and speed increases, the fins DO have more stabilizing effect. This is important to know in relation to launch guide length and motor selection, because the rocket needs to get sufficient speed before leaving the launcher for the fins to stabilize it. The static stability doesn't 'go away', but it DOES change (subtley we hope!) For heavy rockets, this means either A) longer launch guides, or B) a "bigger motor", either in a higher impulse class (switching from an "A" to a "B" motor, or choosing a motor with a higher 'peak thrust', say a A8 instead of an A3, which will accelerate the heavier rocket faster off the pad, reaching fin-stabilizing speed faster.
Hope this helps! OL JR
Hope this helps! OL JR
Dang, OL JR, you should write a book or something.
Ummm..... Luke? How long have you had that sitting in your brain maybe you should do somthing to get your head of rocketry......:dark: you could write a book it helps a bit..... I think I'm wondering because my rocket will be supersonic withen 300 feet and an unstable rocket at Mach 1.12 is a VerY VERY bad thing:y: I'm not quite sure if I should launch....
Ummm..... Luke? How long have you had that sitting in your brain maybe you should do somthing to get your head of rocketry......:dark: you could write a book it helps a bit..... I think I'm wondering because my rocket will be supersonic withen 300 feet and an unstable rocket at Mach 1.12 is a VerY VERY bad thing:y: I'm not quite sure if I should launch....
Challenger should have taught the world "if in doubt, DON'T LAUNCH!"
There are enough 'gotchas' in the world that we DON'T know about to take excessive risks with the ones we DO know about! What do they say at NASA, "it's the UNKNOWN unknowns that get you!"...
Book... nah. I've just read a LOT and put it all together with practical experience. I screw up enough to still remain an enthusiastic yet rank amateur...
Later! OL JR
Stability is a VERY complicated and thorny issue, because essentially it is a MOVING target that greatly depends on the rocket, the flight parameters, and the environmental parameters of the atmosphere through which the rocket is moving.
Static stability is the easiest concept to grasp and to define, because dynamic stability, that is to say, stability in flight basically is constantly changing to some degree or other. Therefore static stability seeks to make these changes moot by defining the stability of the rocket "at rest", hopefully in a close approximation of typical conditions the rocket will be exposed to in flight, and conditions which the rocket is unlikely to exceed in flight.
CG, center of gravity, is of course rather easily obtained by balancing the launch-ready rocket (without ignitor, since that is ejected at ignition and does not fly with the rocket (usually, HOPEFULLY!). CG gets a bit more complicated on "non-symmetrical rockets" like the space shuttle stack, multistage rockets which eject series stages or parallel stages in flight, rockets carrying off-center payloads (cameras, etc.) and the like. Even in a "typical" three-fin and nosecone (3FNC) rocket, CG moves SOME, as the propellant is burned in engine, the CG moves forward, since the nose weight (and hopefully the parachute weight and location) remains the same. Other engine types (liquid, predominantly, though hybrids are a different case too) have their CG locations change in different ways, usually aft, as the engine burns. BUT for 99% of hobby rocketry, it can be said the CG moves FORWARD as the engine burns.
CP, center of pressure, is a MUCH more dynamic and difficult to measure property. CP is affected by the design, fin size, span, shape, drag, the rocket's overall shape, size (Reynolds numbers come into play at certain size thresholds), angle of attack, weight distribution (moments of inertia, length, oscillation characteristics, etc.) wind speed and direction, flight speed, static stability margin, "apparent" angle of attack from wind, air density, air turbulence, and probably a hundred other factors. Some have extremely subtle effects that are barely measurable. Some have significant effects that can have OVERT effects, and some are subtle at times, and overt at others, and combine and compound with other effects to create greater CP changes than at other times, and sometimes certain factors can cancel each other out or work together beneficially to enhance stability. It's EXTREMELY complicated (that's a part of why rocket science is HARD)
SO, how do we get a handle on it?? One way is to look at the 'worst case scenario'... if we design the rocket to be stable with the WORST conditions it's likely to ever see, we can be pretty darn sure it'll be stable under virtually every flight condition we're likely to come up against. Fortunately this is a rather simple matter. By determining the center of lateral area we can determine the *rough* CP point, which, interestingly enough, is the point at which the CP would approximately be if the rocket were ninety degrees to the airflow-- essentially 'flying sideways'... So if the rocket could stabilize from THAT, we can be pretty darn sure it's stable. This is called the cardboard cutout method. Draw an exact outline of the rocket on cardboard, cut it out, and balance that "profile" on a ruler-- the balance point is the center of lateral area. Rocksim can be reset under the 'calculate stability by' box by clicking it and choosing "Cardboard cutout method" from the drop down menu.
Now this method, as you can tell, is VERY VERY conservative. It can lead to EXCESSIVE stability, where the rocket will gyrate and oscillate excessively, weathercock excessively in windy conditions, and other such undesirable effects that rob energy and therefore altitude. That's where the Barrowman Method comes in, which makes certain assumptions about the shape of the rocket and their aerodynamic affects on the rocket's flight. Those assumptions simplified the calculations back in the pre-computer era so that rocketeers with sufficient math skills who wanted to take the time and effort to crunch the numbers by hand could actually sit down and calculate the CP to greater accuracy than the cardboard cutout method would allow. This would allow greater precision and optimization of the rocket design. But it was a VERY labor-intensive process taking several hours at best, and STILL had quite a few limitations because of the ASSUMPTIONS made in creating and simplifying the mathematical formulas. For instance, the Barrowman Method couldn't handle but either 3 or 4 fins-- odd numbers of fins were out. Complicated shapes beyond the basic tubular rocket with radial symmetry (transitions, nosecones, tailcones, evenly spaced fins) were out-- so calculating CP with side pods, boosters on the sides, gliders, etc. was out. Basically, at the end of the day, you STILL had an APPROXIMATION.
The "Rocksim Method" used in the Rocksim program is based on the Barrowman Method and uses it's calculations as the basis of it's function, with some tweaks to the assumptions and some more complex mathematical modelling to get 'closer' to the "real" CP than the basic Barrowman Method allowed. Complex math isn't much of a problem anymore with the massive computing power available now (your PC is more powerful than the Apollo flight computers that got us to the MOON!). Still, there are underlying ASSUMPTIONS in the math models that may or may not accurately reflect or predict the CP location in every instance-- some parameters may be inaccurately modelled, some may be ignored, some may be 'over conservative' in the math models, etc.
Basically, to know the TRUE CP, you'd have to measure all the influences on the rocket at EVERY GIVEN MOMENT IN FLIGHT and plug those measurements into equations to determine the EXACT CP AT THAT PRECISE MOMENT, because as the parameters change from moment to moment, the CP changes as well.
Fortunately, the approximations we get from computer programs, hand crunched numbers, and even cardboard cutouts are enough 99% of the time. The real ART of stability determination is in contest rocketry, where balancing all those forces, and making certain design tradeoffs combine to either give you a winning design or leave you in everyone else's dust...
SO, summing up, static stability is the mathematically calculated stability points (CG/CP relationship) of the rocket. As the rocket takes off and speed increases, the fins DO have more stabilizing effect. This is important to know in relation to launch guide length and motor selection, because the rocket needs to get sufficient speed before leaving the launcher for the fins to stabilize it. The static stability doesn't 'go away', but it DOES change (subtley we hope!) For heavy rockets, this means either A) longer launch guides, or B) a "bigger motor", either in a higher impulse class (switching from an "A" to a "B" motor, or choosing a motor with a higher 'peak thrust', say a A8 instead of an A3, which will accelerate the heavier rocket faster off the pad, reaching fin-stabilizing speed faster.
Hope this helps! OL JR
A great post on stability
Thanks
Fred
Ummm..... Luke? How long have you had that sitting in your brain maybe you should do somthing to get your head of rocketry......:dark: you could write a book it helps a bit..... I think I'm wondering because my rocket will be supersonic withen 300 feet and an unstable rocket at Mach 1.12 is a VerY VERY bad thing:y: I'm not quite sure if I should launch....
Did you try cutting down the fins like I suggested? I think that your fin shape with its forward strakes is contributing to the problem by bringing the CP further forward than you would like.
LS expressed it very well, but many rocketeers gain these insights over time, as they acquire launch experience and, just as important, as they learn from other rocketeers. It is always good to be reminded of the fundamentals of stability in flight from time to time. Harry Stine described them many years ago. LS was carrying on a long tradition of passing on knowledge and insights to newer rocketeers. Paying it forward is a guiding principle in this hobby, which is one of the reasons why it is so rewarding.
You are right to be concerned about the stability. A model rocket that is veering and yawing during its flight will never achieve supersonic speed.
MarkII
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Ummm iv improved stability to 1 cal but when in doubt don't launch is the rule I'm going to be following from now on and MarkII im not so worryed about the
I wasn't worryed about it not going supersonic I was worryed about it arcing over and killing somone (I'm not kidding)
rocketers are friendly as long as you don't injure them it's nice to have some rocketeers back from the ""golden age""
Ummm iv improved stability to 1 cal but when in doubt don't launch is the rule I'm going to be following from now on and MarkII im not so worryed about the
I wasn't worryed about it not going supersonic I was worryed about it arcing over and killing somone (I'm not kidding)
rocketers are friendly as long as you don't injure them it's nice to have some rocketeers back from the ""golden age""
You apparently didn't appreciate my little bit of dry humor.
MarkII
P.S. BTW, does Layne even know that you are on his payroll?
P.P.S. Don't let Fred see that sig of yours. :y:
Last edited:
My rocket is .68 staticly stable but a few inches on pad the rocket is at a "safe take off speed" so once that speed is reached that the static stability no longer applys...... I think that's how it works the faster the more stable it is right? Because the fins are more effective?
Actually - to summarize, it's not that the faster the rocket is, the more stable it is, but that the faster the rocket it, the sooner it reaches a stable speed. (We're talking acceleration vs velocity here). As long as your launch rod is long enough to keep the rocket going straight until it reaches that speed, you're fine.
As far as calibers of stability, that depends on what type rocket it is. If it is a short stubby rocket (like the Fat Boy), base drag has an effect and one caliber is more than enough. It will be stable even if the rocket checks out as neutrally stable. On the other hand, a superroc such as the Mean Machine needs at least 4 or 5 calibers of stability if there is even a hint of wind. For rockets with a "normal" proportion (Alpha, etc.) 1 to 2 calibers is usually fine. Remember also that you can either shift the CP back by making the fins longer or moving or sweeping them further back, or you can move the CG forward by adding nose weight. But if you have the stability up to 1 caliber by either Barrowman or Rocksim, you should be fine. That is, by the way, measured WITH an unfired motor in it, right?
OK then. Now my brain hurts. hhahaha I remember having a paper on balancing a rocket from estes. Simple thing it seemed. Im sure it was something like-find CG, tie a string around the rocket at that point and swing it around yo head. If it nose dived, you added a little weight to the bottom and if it nosed up you added some weight to the top. So hows that sound??? Scotty Dog
OK then. Now my brain hurts. hhahaha I remember having a paper on balancing a rocket from estes. Simple thing it seemed. Im sure it was something like-find CG, tie a string around the rocket at that point and swing it around yo head. If it nose dived, you added a little weight to the bottom and if it nosed up you added some weight to the top. So hows that sound??? Scotty Dog
Yup - the "Swing Test" or "String Test" (I've heard both used). You are close, but it is more simple than that - swing the rocket suspended from the CG - if if "flies" around you nose first, it is considered stable. If it takes any other configuration (rotating around the string, "flying" sideways) it is not stable and will need more nose weight.
A couple of caveats, however. First, it doesn't work well with longer rockets - the string just can't be made long enough to get the proper moment - the nose of the rocket will always be pointing outward from the direction of travel (kind of like having a really strong wind blowing on the nose as it leaves the launch rod). Second, a close to marginal design will sometimes swing tail first - that does not always mean the rocket will be unstable, but it might.
You apparently didn't appreciate my little bit of dry humor.
MarkII
P.S. BTW, does Layne even know that you are on his payroll?
P.P.S. Don't let Fred see that sig of yours. :y:
You Know it was the funniest thing its been over a month and i have yet to receive my paycheck........ :y: i am also wondering if her knows im on his payroll can someone PM him?
Stability is a VERY complicated and thorny issue, because essentially it is a MOVING target that greatly depends on the rocket, the flight parameters, and the environmental parameters of the atmosphere through which the rocket is moving.
Static stability is the easiest concept to grasp and to define, because dynamic stability, that is to say, stability in flight basically is constantly changing to some degree or other. Therefore static stability seeks to make these changes moot by defining the stability of the rocket "at rest", hopefully in a close approximation of typical conditions the rocket will be exposed to in flight, and conditions which the rocket is unlikely to exceed in flight.
CG, center of gravity, is of course rather easily obtained by balancing the launch-ready rocket (without ignitor, since that is ejected at ignition and does not fly with the rocket (usually, HOPEFULLY!). CG gets a bit more complicated on "non-symmetrical rockets" like the space shuttle stack, multistage rockets which eject series stages or parallel stages in flight, rockets carrying off-center payloads (cameras, etc.) and the like. Even in a "typical" three-fin and nosecone (3FNC) rocket, CG moves SOME, as the propellant is burned in engine, the CG moves forward, since the nose weight (and hopefully the parachute weight and location) remains the same. Other engine types (liquid, predominantly, though hybrids are a different case too) have their CG locations change in different ways, usually aft, as the engine burns. BUT for 99% of hobby rocketry, it can be said the CG moves FORWARD as the engine burns.
CP, center of pressure, is a MUCH more dynamic and difficult to measure property. CP is affected by the design, fin size, span, shape, drag, the rocket's overall shape, size (Reynolds numbers come into play at certain size thresholds), angle of attack, weight distribution (moments of inertia, length, oscillation characteristics, etc.) wind speed and direction, flight speed, static stability margin, "apparent" angle of attack from wind, air density, air turbulence, and probably a hundred other factors. Some have extremely subtle effects that are barely measurable. Some have significant effects that can have OVERT effects, and some are subtle at times, and overt at others, and combine and compound with other effects to create greater CP changes than at other times, and sometimes certain factors can cancel each other out or work together beneficially to enhance stability. It's EXTREMELY complicated (that's a part of why rocket science is HARD)
SO, how do we get a handle on it?? One way is to look at the 'worst case scenario'... if we design the rocket to be stable with the WORST conditions it's likely to ever see, we can be pretty darn sure it'll be stable under virtually every flight condition we're likely to come up against. Fortunately this is a rather simple matter. By determining the center of lateral area we can determine the *rough* CP point, which, interestingly enough, is the point at which the CP would approximately be if the rocket were ninety degrees to the airflow-- essentially 'flying sideways'... So if the rocket could stabilize from THAT, we can be pretty darn sure it's stable. This is called the cardboard cutout method. Draw an exact outline of the rocket on cardboard, cut it out, and balance that "profile" on a ruler-- the balance point is the center of lateral area. Rocksim can be reset under the 'calculate stability by' box by clicking it and choosing "Cardboard cutout method" from the drop down menu.
Now this method, as you can tell, is VERY VERY conservative. It can lead to EXCESSIVE stability, where the rocket will gyrate and oscillate excessively, weathercock excessively in windy conditions, and other such undesirable effects that rob energy and therefore altitude. That's where the Barrowman Method comes in, which makes certain assumptions about the shape of the rocket and their aerodynamic affects on the rocket's flight. Those assumptions simplified the calculations back in the pre-computer era so that rocketeers with sufficient math skills who wanted to take the time and effort to crunch the numbers by hand could actually sit down and calculate the CP to greater accuracy than the cardboard cutout method would allow. This would allow greater precision and optimization of the rocket design. But it was a VERY labor-intensive process taking several hours at best, and STILL had quite a few limitations because of the ASSUMPTIONS made in creating and simplifying the mathematical formulas. For instance, the Barrowman Method couldn't handle but either 3 or 4 fins-- odd numbers of fins were out. Complicated shapes beyond the basic tubular rocket with radial symmetry (transitions, nosecones, tailcones, evenly spaced fins) were out-- so calculating CP with side pods, boosters on the sides, gliders, etc. was out. Basically, at the end of the day, you STILL had an APPROXIMATION.
The "Rocksim Method" used in the Rocksim program is based on the Barrowman Method and uses it's calculations as the basis of it's function, with some tweaks to the assumptions and some more complex mathematical modelling to get 'closer' to the "real" CP than the basic Barrowman Method allowed. Complex math isn't much of a problem anymore with the massive computing power available now (your PC is more powerful than the Apollo flight computers that got us to the MOON!). Still, there are underlying ASSUMPTIONS in the math models that may or may not accurately reflect or predict the CP location in every instance-- some parameters may be inaccurately modelled, some may be ignored, some may be 'over conservative' in the math models, etc.
Basically, to know the TRUE CP, you'd have to measure all the influences on the rocket at EVERY GIVEN MOMENT IN FLIGHT and plug those measurements into equations to determine the EXACT CP AT THAT PRECISE MOMENT, because as the parameters change from moment to moment, the CP changes as well.
Fortunately, the approximations we get from computer programs, hand crunched numbers, and even cardboard cutouts are enough 99% of the time. The real ART of stability determination is in contest rocketry, where balancing all those forces, and making certain design tradeoffs combine to either give you a winning design or leave you in everyone else's dust...
SO, summing up, static stability is the mathematically calculated stability points (CG/CP relationship) of the rocket. As the rocket takes off and speed increases, the fins DO have more stabilizing effect. This is important to know in relation to launch guide length and motor selection, because the rocket needs to get sufficient speed before leaving the launcher for the fins to stabilize it. The static stability doesn't 'go away', but it DOES change (subtley we hope!) For heavy rockets, this means either A) longer launch guides, or B) a "bigger motor", either in a higher impulse class (switching from an "A" to a "B" motor, or choosing a motor with a higher 'peak thrust', say a A8 instead of an A3, which will accelerate the heavier rocket faster off the pad, reaching fin-stabilizing speed faster.
Hope this helps! OL JR
Thank you sir.
to summarize, it's not that the faster the rocket is, the more stable it is
Cp moves FORWARD with velocity.
Higher speed can lead to instability.
This is usually not an issue as the propellant burn-off moves the Cg forward too.
But you need to check your stability margin over the whole boost phase.
Thank you sir.
Noble of you, however (IMO unfortunately) Luke Strawalker hasn‘t been active on this forum for years, I believe. Rumor is he offended the powers that be.
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# erica/fpstride.md
Last active Mar 1, 2016
# Decoupling Floating Point Strides from Generic Implementations
• Proposal: SE-00XX
• Status: TBD
• Review manager: TBD
Swift strides create progressions along "notionally continuous one-dimensional values" using a series of offset values. This proposal replaces the Swift's generic stride implementation with seperate algorithms for integer strides (the current implementation) and floating point strides.
This proposal was discussed on-list in the "[Discussion] stride behavior and a little bit of a call-back to digital numbers" thread.
## Motivation
`Strideable` is genericized across both integer and floating point types. A single implementation causes floating point strides to accumulate errors through repeatedly adding `by` intervals. Floating point types deserve their own floating point-aware implementation that minimizes errors.
## Current Art
A `Strideable to` sequence returns the sequence of values (`self`, `self + stride`, `self + stride + stride`, ... last) where last is the last value in the progression that is less than `end`.
A `Strideable through` sequence currently returns the sequence of values (`self`, `self + stride`, `self + tride + stride`, ... last) where last is the last value in the progression less than or equal to `end`. There is no guarantee that `end` is an element of the sequence.
While floating point calls present an extremely common use-case, they use integer-style math that accumulates errors during execution. Consider this example:
```let ideal = [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]
print(zip(Array(1.0.stride(through: 2.01, by: 0.1)), ideal).map(-))
// prints [0.0, 0.0, 2.2204460492503131e-16, 2.2204460492503131e-16,
// 4.4408920985006262e-16, 4.4408920985006262e-16, 4.4408920985006262e-16,
// 6.6613381477509392e-16, 6.6613381477509392e-16, 8.8817841970012523e-16,
// 8.8817841970012523e-16]```
• To create an array containing values from 1.0 to 2.0, the developer must add an epsilon value to the `through` argument. Otherwise the stride progression ends near 1.9. Increasing the argument from `2.0` to `2.01` is sufficient to include the end value.
• The errors in the sequence increase over time. You see this as errors become larger towards the end of the progress. This is an artifact of the generic implementation.
## Detail Design
Under the current implementation, each floating point addition in a generic stride accrues errors. The following progression never reaches 2.0.
``````print(Array(1.0.stride(through: 2.0, by: 0.1)))
// Prints [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]
``````
This same issue occurs with traditional C-style for loops. This is an artifact of floating point math, and not the specific Swift statements:
``````var array: [Double] = []
for var i = 1.0; i <= 2.0; i += 0.1 {
array.append(i)
}
print("Array", array)
// Prints Array [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]
``````
Errors should not accumulate along highly sampled progressions. Ideally, you should not have to manually add an epsilon to force a progression to complete if the math is fairly close to being right.
Floating point strides are inherently dissimilar to and should not be genericized with integer strides. I propose separate their implementation, freeing them to provide their own specialized progressions, using better numeric methods. In doing so, floating point values are no longer tied to implementations that unnecessarily accrue errors or otherwise provide less-than-ideal solutions.
The following example provides a rough pass at what a basic revamp might look like for floating point math. In actual deployment, this function would be named `stride` and not `fstride`.
```import Darwin
/// A `GeneratorType` for `DoubleStrideThrough`.
public struct DoubleStrideThroughGenerator : GeneratorType {
let start: Double
let end: Double
let stride: Double
var iteration: Int = 0
var done: Bool = false
public init(start: Double, end: Double, stride: Double) {
(self.start, self.end, self.stride) = (start, end, stride)
}
/// Advance to the next element and return it, or `nil` if no next
/// element exists.
public mutating func next() -> Double? {
if done {
return nil
}
let current = start + Double(iteration) * stride; iteration += 1
if signbit(current - end) == signbit(stride) { // thanks Joe Groff
if abs(current) > abs(end) {
done = true
return current
}
return nil
}
return current
}
}
public struct DoubleStrideThrough : SequenceType {
let start: Double
let end: Double
let stride: Double
/// Return a *generator* over the elements of this *sequence*.
///
/// - Complexity: O(1).
public func generate() -> DoubleStrideThroughGenerator {
return DoubleStrideThroughGenerator(
start: start, end: end, stride: stride)
}
init(start: Double, end: Double, stride: Double) {
_precondition(stride != 0, "stride size must not be zero")
(self.start, self.end, self.stride) = (start, end, stride)
}
}
public extension Double {
public func fstride(
through end: Double, by stride: Double
) -> DoubleStrideThrough {
return DoubleStrideThrough(
start: self, end: end, stride: stride)
}
}```
This implementation reduces floating point error by limiting accumulated additions. It uses the current Swift 2.2 `through` semantics (versus the revised `through` semantics proposed under separate cover), so it never reaches 2.0 without adding an epsilon value.
```print(Array(1.0.fstride(through: 2.0, by: 0.1)))
// prints [1.0, 1.1000000000000001, 1.2, 1.3, 1.3999999999999999,
// 1.5, 1.6000000000000001, 1.7000000000000002, 1.8,
// 1.8999999999999999]
// versus the old style
print(Array(1.0.stride(through: 2.0, by: 0.1)))
// prints [1.0, 1.1000000000000001, 1.2000000000000002, 1.3000000000000003,
// 1.4000000000000004, 1.5000000000000004, 1.6000000000000005,
// 1.7000000000000006, 1.8000000000000007, 1.9000000000000008]
print(zip(Array(1.0.stride(through: 2.0, by: 0.1)),
Array(1.0.fstride(through: 2.0, by: 0.1))).map(-))
// prints [0.0, 0.0, 2.2204460492503131e-16, 2.2204460492503131e-16,
// 4.4408920985006262e-16, 4.4408920985006262e-16, 4.4408920985006262e-16,
// 4.4408920985006262e-16, 6.6613381477509392e-16, 8.8817841970012523e-16]
let ideal = [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]
print(zip(Array(1.0.fstride(through: 2.0, by: 0.1)), ideal).map(-))
print(zip(Array(1.0.stride(through: 2.0, by: 0.1)), ideal).map(-))
// prints
// [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 2.2204460492503131e-16, 0.0, 0.0]
// [0.0, 0.0, 2.2204460492503131e-16, 2.2204460492503131e-16,
// 4.4408920985006262e-16, 4.4408920985006262e-16, 4.4408920985006262e-16,
// 6.6613381477509392e-16, 6.6613381477509392e-16, 8.8817841970012523e-16]```
If one were looking for a quick and dirty fix, the same kind of math used in this rough solution (`let value = start + count * interval`) could be adopted back into the current generic implementation.
## Alternatives Considered
You get far better results by converting floating point values to integer math, as in the following function. This works by calculating a precision multiplier derived from the whole and fractional parts of the start value, end value, and stride value, to determine a multiplier that enables fully integer math without losing precision. This algorithm introduces significant overhead both for initialization and at each step, making it a non-ideal candidate for real-world implementation beyond trivial stride progressions. What floating point numbers really want here are a fast, well-implemented decimal type.
#### Integer Math
```/// A `GeneratorType` for `DoubleStrideThrough`.
public struct DoubleStrideThroughGenerator : GeneratorType {
let start: Int
let end: Int
let stride: Int
let multiplier: Int
var iteration: Int = 0
var done: Bool = false
public init(start: Double, end: Double, stride: Double) {
// Calculate the number of places needed
// Account for zero whole or fractions
let wholes = [abs(start), abs(end), abs(stride)].map(floor)
let fracs = zip([start, end, stride], wholes).map(-)
let wholeplaces = wholes
.filter({\$0 > 0}) // drop all zeros
.map({log10(\$0)}) // count places
.map(ceil) // round up
let fracplaces = fracs
.filter({\$0 > 0.0}) // drop all zeroes
.map({log10(\$0)}) // count places
.map(abs) // flip negative log for fractions
.map(ceil) // round up
// Extend precision by 10^2
let places = 2.0
+ (wholeplaces.maxElement() ?? 0.0)
+ (fracplaces.maxElement() ?? 0.0)
// Compute floating point multiplier
let fpMultiplier = pow(10.0, places)
// Convert all values to Int
self.multiplier = lrint(fpMultiplier)
let adjusted = [start, end, stride]
.map({\$0 * fpMultiplier})
.map(lrint)
(self.start, self.end, self.stride) =
}
/// Advance to the next element and return it, or `nil` if no next
/// element exists.
public mutating func next() -> Double? {
if done {
return nil
}
let current = start + iteration * stride; iteration += 1
if stride > 0 ? current >= end : current <= end {
if current == end {
done = true
// Convert back from Int to Double
return Double(current) / Double(multiplier)
}
return nil
}
// Convert back from Int to Double
return Double(current) / Double(multiplier)
}
}```
The results with this approach are more precise, enabling errors to drop closer to zero.
```print(Array(1.0.fstride(through: 2.0, by: 0.1)))
print(Array(2.0.fstride(through: 1.0, by: -0.1)))
print(Array((-1.0).fstride(through: -2.0, by: -0.1)))
let ideal = [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]
print(zip(Array(1.0.fstride(through: 2.0, by: 0.1)), ideal).map(-))
print(Array(1.0.fstride(through: 2.0, by: 0.005)))
/*
[1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]
[2.0, 1.9, 1.8, 1.7, 1.6, 1.5, 1.4, 1.3, 1.2, 1.1, 1.0]
[-1.0, -1.1, -1.2, -1.3, -1.4, -1.5, -1.6, -1.7, -1.8, -1.9, -2.0]
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
[1.0, 1.005, 1.01, 1.015, 1.02, 1.025, 1.03, 1.035, 1.04, 1.045, 1.05, 1.055, 1.06, 1.065, 1.07, 1.075, 1.08, 1.085,
1.09, 1.095, 1.1, 1.105, 1.11, 1.115, 1.12, 1.125, 1.13, 1.135, 1.14, 1.145, 1.15, 1.155, 1.16, 1.165, 1.17, 1.175,
1.18, 1.185, 1.19, 1.195, 1.2, 1.205, 1.21, 1.215, 1.22, 1.225, 1.23, 1.235, 1.24, 1.245, 1.25, 1.255, 1.26, 1.265,
1.27, 1.275, 1.28, 1.285, 1.29, 1.295, 1.3, 1.305, 1.31, 1.315, 1.32, 1.325, 1.33, 1.335, 1.34, 1.345, 1.35, 1.355,
1.36, 1.365, 1.37, 1.375, 1.38, 1.385, 1.39, 1.395, 1.4, 1.405, 1.41, 1.415, 1.42, 1.425, 1.43, 1.435, 1.44, 1.445,
1.45, 1.455, 1.46, 1.465, 1.47, 1.475, 1.48, 1.485, 1.49, 1.495, 1.5, 1.505, 1.51, 1.515, 1.52, 1.525, 1.53, 1.535,
1.54, 1.545, 1.55, 1.555, 1.56, 1.565, 1.57, 1.575, 1.58, 1.585, 1.59, 1.595, 1.6, 1.605, 1.61, 1.615, 1.62, 1.625,
1.63, 1.635, 1.64, 1.645, 1.65, 1.655, 1.66, 1.665, 1.67, 1.675, 1.68, 1.685, 1.69, 1.695, 1.7, 1.705, 1.71, 1.715,
1.72, 1.725, 1.73, 1.735, 1.74, 1.745, 1.75, 1.755, 1.76, 1.765, 1.77, 1.775, 1.78, 1.785, 1.79, 1.795, 1.8, 1.805,
1.81, 1.815, 1.82, 1.825, 1.83, 1.835, 1.84, 1.845, 1.85, 1.855, 1.86, 1.865, 1.87, 1.875, 1.88, 1.885, 1.89, 1.895,
1.9, 1.905, 1.91, 1.915, 1.92, 1.925, 1.93, 1.935, 1.94, 1.945, 1.95, 1.955, 1.96, 1.965, 1.97, 1.975, 1.98, 1.985,
1.99, 1.995, 2.0]
*/```
#### Calculated Epsilon Values
Computed epsilon values help compare a current value to a floating-point endpoint. The following code tests whether the current value lies within 5% of the stride of the endpoint.
```/// A `GeneratorType` for `DoubleStrideThrough`.
public struct DoubleStrideThroughGenerator : GeneratorType {
let start: Double
let end: Double
let stride: Double
var iteration: Int = 0
var done: Bool = false
let epsilon: Double
public init(start: Double, end: Double, stride: Double) {
(self.start, self.end, self.stride) = (start, end, stride)
epsilon = self.stride * 0.05 // an arbitrary epsilon of 5% of stride
}
/// Advance to the next element and return it, or `nil` if no next
/// element exists.
public mutating func next() -> Double? {
if done {
return nil
}
let current = start + Double(iteration) * stride; iteration += 1
if abs(current - end) < epsilon {
done = true
return current
}
if signbit(current - end) == signbit(stride) {
done = true
return nil
}
return current
}
}```
#### Other Solutions
While precision math for decimal numbers would be better addressed by introducing a decimal type and/or warnings for at-risk floating point numbers, those features lie outside the scope of this proposal. | 4,440 | 12,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | latest | en | 0.730404 |
http://www.ask.com/question/how-to-solve-bilz-box | 1,386,734,801,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164030159/warc/CC-MAIN-20131204133350-00044-ip-10-33-133-15.ec2.internal.warc.gz | 319,437,661 | 16,305 | # How to Solve Bilz Box?
Bilz box is a puzzle of money that makes gifts very creative and admirable. For example, you can present a gift certificate by placing this gift in the Bilz box and then fastening it well. This allows the recipient to struggle and solve Bilz Box so as to open the gift. It is important to make the person know that there is an incentive for better performance.
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Learning how to solve Sudoku puzzles is actually a fairly easy task. The concept of the game is to fill all of 9 boxes with the number 1 through 9 without and ... | 378 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-48 | longest | en | 0.92433 |
https://www.askiitians.com/forums/Differential-Calculus/no-of-points-of-non-diffrentibility-for-f-x-max_106623.htm | 1,695,816,666,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00712.warc.gz | 697,566,615 | 42,854 | # no of points of non diffrentibility for f(x) = max{ ||x|-1| , ½ } is253
Jitender Singh IIT Delhi
9 years ago
Ans: 5
$f(x) = max(||x|-1|, \frac{1}{2}})$
$f(x) = -x-1, x< -1$
$= \frac{1}{2}, -1\leq x< \frac{-1}{2}$
$= x+1, \frac{-1}{2}\leq x< 0$
$= 1-x, 0\leq x< \frac{1}{2}$
$= \frac{1}{2}, \frac{1}{2}\leq x< 1$
$= x-1, x\geq 1$
So, there are five points where slope will change. You can easily plot the graph of f(x).
Thanks & Regards
Jitender Singh
IIT Delhi | 214 | 463 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-40 | latest | en | 0.494519 |
http://mathhelpforum.com/calculus/101701-difference-quotient.html | 1,527,291,083,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00341.warc.gz | 181,667,644 | 9,816 | 1. ## Difference Quotient
I'm new to calculus and haven't had pre-calc. I have a question regarding a certain problem.
I'm to compute the derivative of a given function using the difference quotient.
This is the function: $\displaystyle f(x)= -2/x$
I only got as far as plugging the function into the formula before I got lost. Can someone please help?
2. Originally Posted by gstyzzer
I'm new to calculus and haven't had pre-calc. I have a question regarding a certain problem.
I'm to compute the derivative of a given function using the difference quotient.
This is the function: $\displaystyle f(x)= -2/x$
I only got as far as plugging the function into the formula before I got lost. Can someone please help?
Let $\displaystyle f(x) = \frac{-2}{x}$, then:
$\displaystyle \frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{-2}{x+h} - \frac{-2}{x}}{h}$ $\displaystyle = \lim_{h \to 0} \frac{\frac{-2x}{x(x+h)} + \frac{2(x+h)}{x(x+h)}}{h}$
Can you finish it off from here? (Rest is in the spoiler.)
Spoiler:
$\displaystyle \lim_{h \to 0} \frac{-2x +2x +2h}{hx(x+h)} = \lim_{h \to 0} \frac{2}{x(x+h)} = \lim_{h \to 0} \frac{2}{x^2+xh} = \frac{2}{x^2}$
f'(x)=g(x)h'(x)-h(x)g'(x)
[g(x)]^2
let -2=h(x) and x=g(x)
f(x)=h(x)/g(x)
f'(x)={xd(-2)-(-2)d(x)}/(x)^2
dx d(x)
=0+2
x^2
=2/x^2 | 471 | 1,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.85258 |
https://www.sophiararebooks.com/pages/books/5800/giovanni-poleni/miscellanea-hoc-est-i-dissertatio-de-barometris-thermometris-ii-machinae-aritmeticae-ejusque | 1,713,218,166,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00506.warc.gz | 914,622,980 | 23,928 | ## Miscellanea. Hoc est I. Dissertatio de barometris, & thermometris. II. Machinae aritmeticae, ejusque usus descriptio. III. De sectionibus conicis parallelorum in horologiis solaribus tractatus.
Venice: Aloysius Pavinus, 1709.
First edition of Poleni’s first book, the second part of which is a description of an early ‘arithmetical machine,’ or primitive calculator. Babbage had a copy of this work in his library. “Poleni’s earliest paper was Miscellanea: de barometris et thermometris; de machinaquadam arithmetiea; de sectionibus conicis in horologiis solaribus describendis (Venice, 1709). A treatise on assorted topics, it includes a dissertation on barometers … and on thermometers, in which several improvements are proposed; also included is the design of an arithmetic machine based on reports that Poleni had received of those of Pascal and of Leibniz. Poleni actually built this machine, which was reportedly very simple and easy to operate; but when he heard of another machine presented to the emperor by the Viennese mechanician Braun, he destroyed his own and never rebuilt it” (DSB). “Poleni’scalculator – in the shape of a big grandfather clock, in wood and iron – is a four-operation device, enabling the user to perform additions, subtractions (resorting tocomplement-to-nine additions), multiplications (by multiple additions), and divisions(by multiple subtractions). To perform multiplication and division Polenifollowed, independently, Leibniz’s idea of a device where multiplicand and divisorcould be stored and to be added or subtracted a fixed number of times. Differentlyfrom Leibniz’s ‘stepped drum’, the marquis conceived the ‘pinwheel’, a wheel with anadjustable number of teeth (0-9) that could be raised or lowered to set the figure to beoperated upon. Poleni’s machine had a three-sector wheel, thus enabling to set a three-digit figure, and a clockwork mechanism – with verge escapement – to drivethe entire machine by the energy of a falling weight. The pinwheel, albeit withdifferent design, was implemented in a number of calculators in the following threecenturies … Poleni’s Invention was acknowledged in Europe thanks to itsdescription published in 1727 in the first comprehensive review on mathematicalinstruments by Jacob Leupold [Theatrum arithmetico geometricum]” (Hénin, pp. 211-212). “Poleni gave a great boost to climatology and was among the first in Italy who had the courage to start regular meteorological observations after Galileo’s trial and the closure of the Accademia del Cimento. In 1709, with the help of Morgagni, Poleni built an improved air thermometer, Amontons’ type, and a barometer. The original Amontons thermometer had a spherical bulb inside which the sensitive air pocket expanded or contracted, giving a non-linear response depending on how much the spherical bulb section changed. Poleni understood the problem and adopted a cylindrical bulb, with constant section and linear response. Initially, Poleni made spot measurements from 1716. In 1724, he accepted an invitation from Jurin, secretary of the Royal Society of Medicine, London, and adhered to the network of the Royal Society (flourished from 1724 to 1735) and started an unbroken series of daily meteorological records from 1725 to 1761 including indoor air temperature (as requested by Jurin), pressure, precipitation, wind direction and weather notes” (Camuffo).
Provenance: The Earls of Macclesfield (bookplate on front paste-down and embossed stamp on first three leaves).
The second section of the Miscellanea describes the calculating machine of Poleni. The chapter opens with the following words:
‘Having heard several times, either from the voice, or from the writings of scientists, which were made by the insight and the care of the most illustrious Pascal and Leibniz, two machines which are used for arithmetic multiplication, which I do not know the description of the mechanism, and I do not know if it was made manifest, I wanted and guess with thought and reflection to their construction, to build a new one that implement the same purpose.’
“Then Poleni continues:
‘By a happy chance, I designed a machine with the use of which even a novice in the art of calculation, provided that knows the figures, can perform arithmetic operations with his own hand. So I am worried that it was made of wood, as I had planned and that, although initially built with poor precision, showed that this was achievable, rather than made. Therefore, I have recreated the machine from scratch, building it of the hardest wood, with all possible attention and the work done has not failed in vain.’
“So, according to the inventor himself, two wooden copies of the machine were made, but unfortunately none of them survived to the present time.
“The first biographer of Poleni, the Frenchman Jean de Fouchy Pajil Grandjean, claims in his 1762 book ‘Eloge de Jean POLENI, Marquis du St. Empire, (né 1683 mort 1761)’, that
having heard that Mr Braun, a famous mechanic in Vienna, presented a similar machine to the Emperor, Poleni destroyed his machine and no more wanted to rebuild it.’
The same story is repeated in the two later Poleni’s biographies … Despite the fact that Fouchy was in strict contact with Poleni (when alive) and knew him personally, this story is quite questionable, not only because it is not compatible with the gentle character of Poleni …
“The mechanism is set in motion by means of the weight, tied to the end of the rope, which is wound over the cylinder. The most important element of the machine – the so-called pin-wheel, was invented probably by Poleni (it is known that Leibniz has a similar design in one of his manuscripts, and that Leibniz and Poleni had a correspondence) … Actually in the machine are placed 3 pin-wheels, which means, that the input mechanism has only 3 digital positions.
“The pin-wheel of Poleni actually is a smooth wheel, with attached to it a sector with 9 blocks. Each block contains a body, a tooth, and a small rod. There is also a fourth element of the block – a small spring. In the assembled block, the tooth can be erected outside of the block, if the rod is pressed by the operator or hidden in the body of the block, if the rod is released. Thus in each one of three sectors with blocks can be set from 0 to 9 erected teeth. When the tooth is erected, it will engage with the calculating mechanism during the revolution of the calculating mechanism, otherwise not.
“The output mechanism consists of six dials, i.e., it has six digital positions. The tens carry mechanism use a one-tooth wheel and obviously has not been designed well, because it caused problems (it can happen that the machine stops in the middle of a tens-carry step due to adverse leverage forces.)
“An additional crank was used for shifting the pin-wheel mechanism by one place at a time with respect to the transfer tooth wheels, thus allowing the multiplicand to be multiplied by the multiplier in higher places.
“As a whole, despite the innovative idea of Poleni for the pin-wheel, which became an extremely popular constructive element of calculating machines some 150 years later, his machine looks rather rough and ill-formed device, as compared to the machine of his successor – Anton Braun” (https://history-computer.com/giovanni-poleni/).
“Poleni was the son of Jacopo Poleni and carried his title of marquis of the Holy Roman Empire, conferred by Emperor Leopold I and confirmed in 1686 by the Republic of Venice. In his early life he followed a variety of studies, and his intellectual endowment was soon known to be extraordinary. After completing his studies, first in philosophy and then in theology, at the school of the Padri Somaschi in Venice, he began, with his parents’ encouragement, a judicial career. At the same time his father introduced him to mathematics and physics, and it became clear that the natural sciences were going to be his most prominent field of activity.
“At the age of twenty-six he married Orsola Roberti of a noble family of Bassano del Grappa and accepted the chair of astronomy at the University of Padua; six years later he became professor of physics as well. The Venetian Senate invited him to investigate problems of hydraulics pertinent to the irrigation of lower Lombardy, Poleni soon acquired such proficiency in this field that he became the accepted arbiter of all disputes between states bordering on rivers. In 1719 he assumed the chair of mathematics at the University of Padua left vacant by Nikolaus I Bernoulli, upon the latter’s return to Basel. His noteworthy opening lecture was published in 1720 as De mathesis in rebus physicis utilitate.
“In 1738 Poleni established within a few months an up-to-date laboratory of experimental physics and began to lecture on that subject. He simultaneously conducted meteorological observations, corresponded with French, English, German, and Italian savants (particularly Euler, Maupertuis, the Bernoullis, and Cassini III), published memoirs on various subjects, and participated in the study of calendar reform that had been sponsored by Pope Clement XI. In 1733 Poleni received a prize from the Royal Academy of Sciences of Paris for a paper on a method of calculating – independently of astronomical observations – the distance traveled by a ship; and in 1736 he was awarded a prize for a study of ships’ anchors. In 1739 he became a foreign member of the Academy, and in 1741 he received a prize for a study of cranes and windlasses.
“Poleni’s scientific activities were paralleled by classical researches, which were described in treatises on the temple of Ephesus, on ancient theaters and amphitheaters, on French archaeological findings, on an Augustan obelisk, and on several architectural topics. In 1748 he was called to Rome by Pope Benedict XIV to examine the cupola of St. Peter’s basilica and to propose means of preventing its further movement, but he was soon recalled to Padua to assume judicial duties. Excessive work gradually affected his health, although not his enthusiasm, until his death at the age of seventy-eight. His remains were laid in the Church of St. Giacomo in Padua, where his sons placed a monument in his honor. The citizens of Padua subsequently decreed that a statue (one of the earliest works of Antonio Canova) of Poleni be placed among those of illustrious men in the Prato della Valle. A medal in his honor was struck by the Republic of Venice” (DSB).
Riccardi ii, 290. Camuffo, ‘Galileo’s revolution and the infancy of meteorology in Padua, Florence and Bologna,’ Méditerranée [online], Paléo-environnements, géoarchéologie, géohistoire, (http://journals.openedition.org/mediterranee/12565). Hénin, ‘Early Italian computing machines and their inventors,’ in: Tatnall, Reflections on the History of Computing: Preserving Memories and Sharing Stories, AICT-387 (2012), pp. 204-230.
4to (240 x 173 mm), pp. [viii] 56, with engraved allegorical title (‘Reason taming the savage beast’) and 9 folding engraved plates by Joseph Marcati. Contemporary paste-boards (spine worn, small piece chipped away).
Item #5800
Price: \$12,500.00
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F2-25: BALANCE PARADOX - BUOYANCY WITH CROSSOVER
PURPOSE: To present buoyancy in a paradoxical way.
DESCRIPTION: The balance is initially at equilibrium with a mass hanging from an arm on the left pan. The left pan has two weights plus the arm supporting the 50 cm^3 hanging block, which is in balance with the water beaker on the right pan. Q: If the mass is allowed to hang into the beaker of water, how does this effect the balance? In particular, what, if anything must be done to restore equilibrium? A: Because the volume of the volume of the block is 50 cm^3, the weight on the left side is reduced by 50 grams when the block is submerged in the water. Conversely, the weight on the right side is increased by 50 grams, the reaction force on that pan. To recreate balance, 100 grams must be added to the left pan.
SUGGESTIONS: Ask the class to discuss this one!
REFERENCES: (PIRA unavailable.)
EQUIPMENT: Pan balance, weights, beaker with water, boom with hanging weight.
SETUP TIME: 5 min. | 253 | 1,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-51 | latest | en | 0.90423 |
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The quantity of a product that is demanded is 5,000 units when the price is \$250. When the price is \$225, the demand is 8,000 units. The manufacturer will not market the product if the price is below \$100. For each price increase of \$60, the manufacturer will market an additional 2,500 units of the product. Find the equilibrium quantity. Get Math homework help today | 98 | 443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.901172 |
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In throwing of a pair of dice find the probability of the event total is not eat and not 11
Arun
25763 Points
3 years ago
Dear Manish
sum is 8 = (6,2)(2,6)(5,3)(3,5)(4,4)
hence probability of getting 8 = 5 / 36
sum is 11 = (6,5)(5,6)
hence probability of getting 11 = 2 / 36
now
P (A U B) = 5 + 2 / 36 = 7 / 36
Hence nether 8 come nor 11, probability = P (A U B) ‘ = 1 – 7 / 36 = 29 / 36
Regards | 272 | 745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-21 | latest | en | 0.836152 |
http://atlas.dr-mikes-maths.com/atlas/480/1193/3.html | 1,585,665,158,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00216.warc.gz | 18,988,684 | 3,477 | Questions?
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# Polytope of Type {4,6,10}
Atlas Canonical Name : {4,6,10}*480b
if this polytope has a name.
Group : SmallGroup(480,1193)
Rank : 4
Schlafli Type : {4,6,10}
Number of vertices, edges, etc : 4, 12, 30, 10
Order of s0s1s2s3 : 30
Order of s0s1s2s3s2s1 : 2
Special Properties :
Universal
Non-Orientable
Flat
Related Polytopes :
Facet
Vertex Figure
Dual
Facet Of :
{4,6,10,2} of size 960
{4,6,10,4} of size 1920
Vertex Figure Of :
{2,4,6,10} of size 960
Quotients (Maximal Quotients in Boldface) :
5-fold quotients : {4,6,2}*96c
10-fold quotients : {4,3,2}*48
Covers (Minimal Covers in Boldface) :
2-fold covers : {4,12,10}*960b, {4,12,10}*960c, {4,6,20}*960b, {4,6,10}*960e
3-fold covers : {4,18,10}*1440b, {4,6,30}*1440d, {4,6,30}*1440e
4-fold covers : {4,6,10}*1920a, {4,24,10}*1920c, {4,24,10}*1920d, {4,6,40}*1920b, {4,12,20}*1920b, {4,12,20}*1920c, {4,12,10}*1920b, {4,6,20}*1920a, {4,6,10}*1920b, {4,12,10}*1920c, {8,6,10}*1920a, {8,6,10}*1920b, {4,6,20}*1920c
Permutation Representation (GAP) :
```s0 := ( 1, 3)( 2, 4)( 5, 7)( 6, 8)( 9, 11)( 10, 12)( 13, 15)( 14, 16)
( 17, 19)( 18, 20)( 21, 23)( 22, 24)( 25, 27)( 26, 28)( 29, 31)( 30, 32)
( 33, 35)( 34, 36)( 37, 39)( 38, 40)( 41, 43)( 42, 44)( 45, 47)( 46, 48)
( 49, 51)( 50, 52)( 53, 55)( 54, 56)( 57, 59)( 58, 60)( 61, 63)( 62, 64)
( 65, 67)( 66, 68)( 69, 71)( 70, 72)( 73, 75)( 74, 76)( 77, 79)( 78, 80)
( 81, 83)( 82, 84)( 85, 87)( 86, 88)( 89, 91)( 90, 92)( 93, 95)( 94, 96)
( 97, 99)( 98,100)(101,103)(102,104)(105,107)(106,108)(109,111)(110,112)
(113,115)(114,116)(117,119)(118,120);;
s1 := ( 2, 3)( 6, 7)( 10, 11)( 14, 15)( 18, 19)( 21, 41)( 22, 43)( 23, 42)
( 24, 44)( 25, 45)( 26, 47)( 27, 46)( 28, 48)( 29, 49)( 30, 51)( 31, 50)
( 32, 52)( 33, 53)( 34, 55)( 35, 54)( 36, 56)( 37, 57)( 38, 59)( 39, 58)
( 40, 60)( 62, 63)( 66, 67)( 70, 71)( 74, 75)( 78, 79)( 81,101)( 82,103)
( 83,102)( 84,104)( 85,105)( 86,107)( 87,106)( 88,108)( 89,109)( 90,111)
( 91,110)( 92,112)( 93,113)( 94,115)( 95,114)( 96,116)( 97,117)( 98,119)
( 99,118)(100,120);;
s2 := ( 1, 41)( 2, 44)( 3, 43)( 4, 42)( 5, 57)( 6, 60)( 7, 59)( 8, 58)
( 9, 53)( 10, 56)( 11, 55)( 12, 54)( 13, 49)( 14, 52)( 15, 51)( 16, 50)
( 17, 45)( 18, 48)( 19, 47)( 20, 46)( 22, 24)( 25, 37)( 26, 40)( 27, 39)
( 28, 38)( 29, 33)( 30, 36)( 31, 35)( 32, 34)( 61,101)( 62,104)( 63,103)
( 64,102)( 65,117)( 66,120)( 67,119)( 68,118)( 69,113)( 70,116)( 71,115)
( 72,114)( 73,109)( 74,112)( 75,111)( 76,110)( 77,105)( 78,108)( 79,107)
( 80,106)( 82, 84)( 85, 97)( 86,100)( 87, 99)( 88, 98)( 89, 93)( 90, 96)
( 91, 95)( 92, 94);;
s3 := ( 1, 65)( 2, 66)( 3, 67)( 4, 68)( 5, 61)( 6, 62)( 7, 63)( 8, 64)
( 9, 77)( 10, 78)( 11, 79)( 12, 80)( 13, 73)( 14, 74)( 15, 75)( 16, 76)
( 17, 69)( 18, 70)( 19, 71)( 20, 72)( 21, 85)( 22, 86)( 23, 87)( 24, 88)
( 25, 81)( 26, 82)( 27, 83)( 28, 84)( 29, 97)( 30, 98)( 31, 99)( 32,100)
( 33, 93)( 34, 94)( 35, 95)( 36, 96)( 37, 89)( 38, 90)( 39, 91)( 40, 92)
( 41,105)( 42,106)( 43,107)( 44,108)( 45,101)( 46,102)( 47,103)( 48,104)
( 49,117)( 50,118)( 51,119)( 52,120)( 53,113)( 54,114)( 55,115)( 56,116)
( 57,109)( 58,110)( 59,111)( 60,112);;
poly := Group([s0,s1,s2,s3]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2","s3");;
s0 := F.1;; s1 := F.2;; s2 := F.3;; s3 := F.4;;
rels := [ s0*s0, s1*s1, s2*s2, s3*s3, s0*s2*s0*s2,
s0*s3*s0*s3, s1*s3*s1*s3, s0*s1*s0*s1*s0*s1*s0*s1,
s1*s2*s3*s2*s1*s2*s3*s2, s0*s1*s2*s1*s0*s1*s2*s0*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s2,
s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(120)!( 1, 3)( 2, 4)( 5, 7)( 6, 8)( 9, 11)( 10, 12)( 13, 15)
( 14, 16)( 17, 19)( 18, 20)( 21, 23)( 22, 24)( 25, 27)( 26, 28)( 29, 31)
( 30, 32)( 33, 35)( 34, 36)( 37, 39)( 38, 40)( 41, 43)( 42, 44)( 45, 47)
( 46, 48)( 49, 51)( 50, 52)( 53, 55)( 54, 56)( 57, 59)( 58, 60)( 61, 63)
( 62, 64)( 65, 67)( 66, 68)( 69, 71)( 70, 72)( 73, 75)( 74, 76)( 77, 79)
( 78, 80)( 81, 83)( 82, 84)( 85, 87)( 86, 88)( 89, 91)( 90, 92)( 93, 95)
( 94, 96)( 97, 99)( 98,100)(101,103)(102,104)(105,107)(106,108)(109,111)
(110,112)(113,115)(114,116)(117,119)(118,120);
s1 := Sym(120)!( 2, 3)( 6, 7)( 10, 11)( 14, 15)( 18, 19)( 21, 41)( 22, 43)
( 23, 42)( 24, 44)( 25, 45)( 26, 47)( 27, 46)( 28, 48)( 29, 49)( 30, 51)
( 31, 50)( 32, 52)( 33, 53)( 34, 55)( 35, 54)( 36, 56)( 37, 57)( 38, 59)
( 39, 58)( 40, 60)( 62, 63)( 66, 67)( 70, 71)( 74, 75)( 78, 79)( 81,101)
( 82,103)( 83,102)( 84,104)( 85,105)( 86,107)( 87,106)( 88,108)( 89,109)
( 90,111)( 91,110)( 92,112)( 93,113)( 94,115)( 95,114)( 96,116)( 97,117)
( 98,119)( 99,118)(100,120);
s2 := Sym(120)!( 1, 41)( 2, 44)( 3, 43)( 4, 42)( 5, 57)( 6, 60)( 7, 59)
( 8, 58)( 9, 53)( 10, 56)( 11, 55)( 12, 54)( 13, 49)( 14, 52)( 15, 51)
( 16, 50)( 17, 45)( 18, 48)( 19, 47)( 20, 46)( 22, 24)( 25, 37)( 26, 40)
( 27, 39)( 28, 38)( 29, 33)( 30, 36)( 31, 35)( 32, 34)( 61,101)( 62,104)
( 63,103)( 64,102)( 65,117)( 66,120)( 67,119)( 68,118)( 69,113)( 70,116)
( 71,115)( 72,114)( 73,109)( 74,112)( 75,111)( 76,110)( 77,105)( 78,108)
( 79,107)( 80,106)( 82, 84)( 85, 97)( 86,100)( 87, 99)( 88, 98)( 89, 93)
( 90, 96)( 91, 95)( 92, 94);
s3 := Sym(120)!( 1, 65)( 2, 66)( 3, 67)( 4, 68)( 5, 61)( 6, 62)( 7, 63)
( 8, 64)( 9, 77)( 10, 78)( 11, 79)( 12, 80)( 13, 73)( 14, 74)( 15, 75)
( 16, 76)( 17, 69)( 18, 70)( 19, 71)( 20, 72)( 21, 85)( 22, 86)( 23, 87)
( 24, 88)( 25, 81)( 26, 82)( 27, 83)( 28, 84)( 29, 97)( 30, 98)( 31, 99)
( 32,100)( 33, 93)( 34, 94)( 35, 95)( 36, 96)( 37, 89)( 38, 90)( 39, 91)
( 40, 92)( 41,105)( 42,106)( 43,107)( 44,108)( 45,101)( 46,102)( 47,103)
( 48,104)( 49,117)( 50,118)( 51,119)( 52,120)( 53,113)( 54,114)( 55,115)
( 56,116)( 57,109)( 58,110)( 59,111)( 60,112);
poly := sub<Sym(120)|s0,s1,s2,s3>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2,s3> := Group< s0,s1,s2,s3 | s0*s0, s1*s1, s2*s2,
s3*s3, s0*s2*s0*s2, s0*s3*s0*s3, s1*s3*s1*s3,
s0*s1*s0*s1*s0*s1*s0*s1, s1*s2*s3*s2*s1*s2*s3*s2,
s0*s1*s2*s1*s0*s1*s2*s0*s1, s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s2,
s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3*s2*s3 >;
```
References : None.
to this polytope | 3,719 | 6,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-16 | latest | en | 0.291427 |
https://forum.edugorilla.com/forums/topic/the-length-breadth-and-thickness-of-a-block-are-given-by-l-12-cm-b-6-cm-and-t-2-45-cm-the-v-5/ | 1,618,253,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069133.25/warc/CC-MAIN-20210412175257-20210412205257-00347.warc.gz | 364,660,185 | 17,292 | This topic contains 0 replies, has 0 voices, and was last updated by EduGorilla 1 year, 9 months ago.
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The length, breadth, and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm The volume of the block according to the idea of significant figures should be
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http://www.polymathlove.com/polymonials/trigonometric-functions/sample-decimal-sums-for-grade.html | 1,513,524,578,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948596115.72/warc/CC-MAIN-20171217152217-20171217174217-00551.warc.gz | 472,630,050 | 12,326 | Algebra Tutorials!
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RanEIs
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Posted: Saturday 30th of Dec 19:03 A extraordinary piece of math software is Algebrator. Even I faced similar difficulties while solving midpoint of a line, radical inequalities and greatest common factor. Just by typing in the problem workbookand clicking on Solve – and step by step solution to my algebra homework would be ready. I have used it through several math classes - College Algebra, Remedial Algebra and Intermediate algebra. I highly recommend the program. | 764 | 2,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-51 | longest | en | 0.911419 |
https://datascience.stackexchange.com/questions/85644/ensembling-expressions | 1,713,905,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00305.warc.gz | 171,201,968 | 40,214 | # Ensembling expressions
I have two models, $$m_1$$ and $$m_2$$, and I want to ensemble them into a final model. I want to be able weight one or the other more according to a grid search. There are two main ideas that come to my mind when doing so:
• Define a family of models $$m_1 \cdot a + m_2 \cdot (1 - a)$$, where $$0 < a < 1$$, find the $$a$$ that gives the best score.
• Define a family of models $$m_1^a \cdot m_2^{1 - a}$$, where $$0 < a < 1$$, find the $$a$$ that gives the best score.
However, in certain cases, I've seen top models in Kaggle competitions doing fairly different things, like having a final model of the form $$m_1^a + m_2^b$$.
My question is, what are the advantages and disadvantages of every solution? When do they work better and when do they work worse? When is the third kind of ensemble suitable, and is there any heuristic to tune $$a$$ and $$b$$?
That is an empirical question. The answer will change for different models and different datasets.
The best approach would use cross validation to see which ensembling technique has the best score on the evaluation metric for the given data.
You can make the same question with every Machine Learning algorithm, and still the answer will remain very similar.
What's the advantage of Linear regression over Decision Trees? To answer this you could define them mathematically. In your case, the mathematical definition seems easy: weighted mean or geometric mean.
When do any model works better by any other model? Give it a try in cross-validation.
Sadly, scientific methodology in Machine Learning is done by try and error. Saying what will be the value of an hyperparameter previous to fitting the model is un reliable.
You "prove" that an algorithm works in ML when you run it to through a set of datasets and it performs better than the rest.
Coming back to your question, what happens in kaggle tends to be the most technical advanced thing. So if its there, its worth giving it a try.
I agree with Brain. The solution that will work better, is the one that will fit better your data.
Please note that if you have only one parameter, you can derive the optimal value instead of doing a grid search. Your family of solution is we restricted so I don't expect a significant gain but there is no reason not to use it. | 534 | 2,316 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.943101 |
https://gourocklawnsllp.com/category/array-range-queries/ | 1,638,144,448,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00141.warc.gz | 350,737,898 | 11,493 | Monday, 29 Nov 2021
Category: array-range-queries
#include using namespace std; int totalValidArrays(int a[], int b[], int N){ int dp[N + 1][b[N – 1] + 1]; int pref[N + 1][b[N – 1] + 1]; memset(dp, 0, sizeof(dp)), memset(pref, 0, sizeof(pref)); dp[0][0] = 1; for (int i = 0; i
#include using namespace std; void countTriplets(int size, int queries, int arr[], int Q[][2]){ int arr_even[size + 1], arr_odd[size + 1]; int even = 0, odd = 0; arr_even[0] = 0; arr_odd[0] = 0; for (int i = 0; i < size; i++) { if (arr[i] % 2) { odd++; } else { even++; } arr_even[i + 1] = even; arr_odd[i + 1] = odd; } for (int i = 0; i < […]
Minimum absolute value of (K – arr[i]) for all possible values of K over the range [0, N – 1]Given a positive integer N and a sorted array arr[] consisting of M integers, the task is to find the minimum absolute value of (K – arr[i]) for all possible values of K over the range […]
#include using namespace std; class SpecialAverage {public: multiset left, mid, right; int n, k; long pos, sum; vector v; SpecialAverage(int nvalue, int kvalue) { n = nvalue; k = kvalue; pos = 0; sum = 0; for (int i = 0; i < n; i++) v.push_back(0); } void add(int num) { left.insert(num); if (left.size() > k) { int temp = *(prev(end(left))); mid.insert(temp); left.erase(prev(end(left))); sum += temp; } if (mid.size() > (n – 2 * k)) { int temp = *(prev(end(mid))); right.insert(temp); mid.erase(prev(end(mid))); sum […]
#include using namespace std; int prefixCount[32][10000]; void findPrefixCount(vector arr, int size){ for (int i = 0; i < 32; i++) { prefixCount[i][0] = ((arr[0] >> i) & 1); for (int j = 1; j < size; j++) { prefixCount[i][j] = ((arr[j] >> i) & 1); prefixCount[i][j] += prefixCount[i][j – 1]; } }} void arrayBitwiseAND(int size){ int result = 0; for (int i = 0; i < 32; i++) { int temp = […]
#include using namespace std; vector seg(400000, 0), dseg(400000, 0), a(100000); vector segSum(400000, 0); void build(int ind, int low, int high) { if (low == high) { seg[ind] = a[low] * a[low]; segSum[ind] = a[low]; return; } int mid = (low + high) / 2; […]
Print modified array after performing queries to add (i – L + 1) to each element present in the range [L, R] Given an array arr[] consisting of N 0s (1-based indexing) and another array query[], with each row of the form {L, R}, the task for each query (L, R) is to add a […]
#include using namespace std; void findGCDQueries(int arr[], int N, int Queries[][3], int Q) { int prefix[N], suffix[N]; prefix[0] = arr[0]; suffix[N – 1] = arr[N – 1]; for (int i = 1; i < N; i++) { prefix[i] = __gcd(prefix[i - 1], arr[i]); } for (int i = N - 2; […] | 1,105 | 3,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.377756 |
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# Nav/Structure
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
The following table lists all of the structures in the wiki.
Struct/Half open square Half-open square
Struct/The Euclidean algorithm applied to 123 and 73 The Euclidean algorithm applied to 123 and 73
Struct/The Klein group of order four The Klein group of order four
Struct/The binary quadratic form x^2 + xy - 2y^2 The binary quadratic form x^2 + xy - 2y^2
Struct/The cyclic group with four elements. The group with three elements.
Struct/The symmetric group of order six The group with three elements.
Struct/The cyclic group with five elements. The group with three elements.
Struct/The cyclic group with six elements The group with three elements.
Struct/The group with three elements. The group with three elements.
Struct/The group with two elements. The group with two elements.
Struct/The max-plus algebra of natural numbers The max-plus algebra of natural numbers | 221 | 1,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-20 | latest | en | 0.886225 |
https://kyma.symbolicsound.com/qa/4140/hand-rolled-quadrature-oscillator-question | 1,606,680,415,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141202590.44/warc/CC-MAIN-20201129184455-20201129214455-00565.warc.gz | 369,245,719 | 8,903 | 150 views
Hi
Ok, so I wanted to mess around with the SR Frequency update on quadrature oscillators and SingleSideBand, and just wanted to check something.
Is this the correct recipe?
Multiply the amp of the Sin by the left channel of the input (which is 90 phase shifted)
Multiply the amp of the Cos by the right channel of the input
Sum both?
Or is it the signal graph in the ComplexProduct prototype? Which sounds excellent...
edited Jan 6, 2019
What you have looks correct: multiply the left output of the Hilbert transform by a sine, multiply the right output by cosine and sum the left and the right (that should cancel one of the side bands to give a single side-band frequency shifter).
answered Jan 7, 2019 by (Savant) (110,320 points)
Hi Cristan. Using the Hilbert you can make a ring modulator which only has one side band. This means you can make a frequency shifter, but you may notice that when you shift the frequencies down, the low frequencies will go pass DC and turn into negative frequencies and come up again, which can sound like a duel sideband ring modulator again. But there is a trick which can be done here by using another Hilbert you can make the negative frequencies cancel so the frequencies go down and disappear into nothingness leaving the signal clean. I demonstrated this at my KISS 2012 talk.
The sound is included in the sounds that go with the talk. Hope it helps.
thanks Pete and SSC. Where do you locate that Sound example Pete?
Hi Cristian. After KISS2012 I e-mailed the sounds and the powerpoint to people who requested it. I thought you had it already, but if not let me know what your email is and I'll send the link.
With regard to your first question, what you describe with the sin cos two products and a mixer is half of the complex multiply, as it only has one output not and not a real/imaginary pair. For the single side band ringmod you only need half as you only need one output signal. In the talk, I needed to go further than just a single side band ring mod as I wanted to cancel out the negative frequencies and needed the full complex multiply. The Kyma module called QuadOscillator has the sin and cos oscillator built in as well as the two products and the mixer so feeding this with a Hilbert give you a single side band modulator with nothing else needed. As I needed the full complex multiply I got a second QuadOscillator but first put the Hilbert through a channel crosser which swapped the channels and inverted one of them to give me the second output making a full complex multiply pair.
Once I had a full complex output of the frequency shifted signal, I knew that all the positive frequencies had a 90 deg lead in the in the second output and the negative frequencies had a 90 deg lag in the second output compared to the first output.
So if I could put the first output through another Hilbert, it would make all the frequencies positive and negative have a 90 deg lag. So then subtracted that from the second output (with the matching delay because of the extra Hilbert) then the positive frequencies would add and the negative frequencies would cancel as they were 0 deg and 180 deg with respect to each other. The third Hilbert module is there as a simple way of providing the needed matching delay by using only the right leg and not using the left leg.
The channel crossers at the end are just making sure the correct legs of the Hilberts are being picked off.
I remember it took me ages to grasp how this works haha
while it is really clever and makes a really sharp filter (can't be sharper actually) - do you think it's more efficient to use a HighPass before doing the frequency shifting that filters the input in a way no aliasing can occur?
Yes making a HP filter on the input that moves up when the frequency shifting moves down is a more efficient way of doing it. Sounds like the same principle as the Surbiton oscillator only upside down.
BTW the theory in the negative freq canceler sounds like the filter is infinitely sharp, but in reality it is subject to the imperfection of the Hilbert (not 90 deg at sub audio down to DC) but it is very sharp.
The main reason I did the neg freq canceler is for situations where I maintain the two legs of the full complex multiply (real and imaginary) and do more complex stuff than just a single frequency shifter. May be shift it down and up inserting comb filters along the way for example.
Did you see Mr Nortons delay with feed back but with a frequency shifter the feedback path. This was hard to do as we didn't want the hi latency of the Hilbert in the feedback path itself, as it would have limited the shortest delay time available. Instead it had one Hilbert on the input and used two matching delay lines with two quadratures in the feedback path passing the signal back and forth between the delay pair maintaining the real/imaginary signals as is went through. This meant you could get freq shifts that moved further away every time it went round the delay line. I never tried the neg freq canceler on the outputs but I suspect it would have some different sounds as the frequencies spiralled off the bottom and silently moved back round off the top then became audible again when it flipped back at half sample rate.
Here is the link on the old forum.
http://www.symbolicsound.com/cgi-bin/forumdisplay.cgi?action=displayprivate&number=3&topic=000225
Right, that makes sense, of course you need to be able to calculate the amount of (possible) aliasing to adjust the filter.
The feedback example would be a good usecase for a HilbertIIR. But you were already able to solve it (in a clever way).
It may be solved but it still has a latency of a thousand odd samples. I sure am interested in the HilbertIIR, as that will have a lot less latency `and will be a lot more efficient, even if you don't port it to Kyma I'm still interested. | 1,285 | 5,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-50 | longest | en | 0.93861 |
https://www.theplayerstribune.com/en-us/articles/the-wednesday-morning-math-challenge-week-11 | 1,600,480,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189928.2/warc/CC-MAIN-20200919013135-20200919043135-00348.warc.gz | 1,119,986,357 | 25,340 | Submit
# The Wednesday Morning Math Challenge: Week 11
Nov 23 2016
John Urschel
Baltimore Ravens
Nov 23 2016
Welcome to Week 11 of the Wednesday Morning Math Challenge. You can find solutions to the Week 10 puzzles here, along with — more importantly — discussions of some ways to approach them. Remember that the real goal here is to think creatively.
This week, we’re going to have a little bit of fun with football.
1. Suppose you’re the head coach of the Washington Sentinels. You’re trailing Detroit by 14 points with no timeouts and two minutes left in the fourth quarter. Your quarterback, Shane Falco, runs in for a touchdown. Suppose your kicker, Nigel Gruff, has a 97% one point conversion rate, and Falco and the offense have a 47% two point conversion rate. Do you go for one or two? Why?
2. What is the lowest two point conversion rate for which you would want to go for two?
3. Now, consider the unit square in two dimensional space, 0<x<1,0<y<1. Let x and y be the two point and one point conversion rate of a given team. In what region would you go for one in this circumstance? In what region would you go for two?
John Urschel
Baltimore Ravens | 287 | 1,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-40 | latest | en | 0.935661 |
https://studylib.net/doc/25287716/graphing-review | 1,675,545,615,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00619.warc.gz | 551,502,900 | 11,234 | # Graphing Review
```COMPLETE WEDNESDAY BELL RINGER
SILENTLY
1. WHAT ARE THE 3 COMPONENTS OF A HYPOTHESIS?
2.WHAT IS THE SECOND STEP OF THE SCIENTIFIC METHOD?
LAB TOMORROW!!!!
HOW TO CONSTRUCT A GRAPH
1. IDENTIFY VARIABLE
2. DETERMINE THE GRAPH SCALE
3. PLOT THE DATA POINTS
4. DRAW THE GRAPH
5. TITLE THE GRAPH
Independent variable
1. IDENTIFY
VARIABLES
• Factor that is varied (changed,
tested) in an experiment
• Label along the x-axis
Dependent variable
• Factor that is measured in an
experiment
• Label along the y-axis
2. DETERMINE THE GRAPH SCALE
a) DETERMINE THE (RANGE) NUMERIC VALUE OF
DEPENDENT VARIABLE
b) ESTABLISH A SCALE THAT BEST FITS THE RANGE
OF THIS VARIABLE
c) SPREAD THE GRAPH TO USE THE MOST AVAILABLE
SPACE
d) BE CONSISTENT THROUGHOUT THE AXIS SCALE
2. DETERMINE THE GRAPH SCALE
What do we
need to find
before we can
determine our
graph scale?
Study Time
2
68
0
50
6
97
3
79
1
66
4
88
1
70
2
73
5
92
2.5
77
3. PLOT THE
DATA
POINTS
(FOR LINE
GRAPH)
a)PLOT EACH DATA VALUE ON THE GRAPH
b)IF MULTIPLE SETS OF DATA ARE BEING
PLOTTED, USE DIFFERENT COLORS AND
INCLUDE A KEY
4. & 5. DRAW THE GRAPH AND ADD TITLE
Remember TAILS:
T – title
A – axis
I – increments
L – labels (independent and dependent with units)
S – scale
DETERMINE IF THERE’S A RELATIONSHIP IN
THE VARIABLES
Direct
Indirect
Constant
LINE PLOT
STEM
AND LEAF
HISTOGRAM
BAR
GRAPH
LINE
GRAPH
PIE
CHART
PICTOGRAPH
SCATTERPLOT
``` | 468 | 1,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.573101 |
https://community.smartsheet.com/discussion/103436/can-we-use-a-sum-or-avg-formula-to-rows-columns-which-has-a-if-condition | 1,722,691,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00704.warc.gz | 146,658,338 | 107,364 | # Can we use a Sum or Avg formula to rows/columns which has a IF condition.
Options
I was trying to prepare a report in which I used a IF conditions to get the expected value based on my clients response from the forms. Now, I want to do a SUM or AVG with the rows that I have the conditions. I get #DIVIDE BY ZERO ERROR when AVG is applied & I get the final value as 0 when I apply the Sum formula.
## Best Answer
• ✭✭✭✭✭
Answer ✓
Options
I might have some ideas but I probably need more context. Can you tell me what the IF formulas are returning as the value in the cells you are trying to sum or average?
Assuming these are numbers, smartsheet is likely not recognizing them as numbers. If your IF statement ends in things in quotes like IF(xxx@row = "yes", "100", "0") Smartsheet may not recognize the 100 as a number. In this case try wrapping the TRUE part of the IF statement in a VALUE(100) function to ensure the result is a number.
Alternatively if your IF formula is returning words or phrases, you probably want to use the COUNTIF formula to count how many times specific selections are made.
Does this help?
## Answers
• ✭✭✭✭✭
Answer ✓
Options
I might have some ideas but I probably need more context. Can you tell me what the IF formulas are returning as the value in the cells you are trying to sum or average?
Assuming these are numbers, smartsheet is likely not recognizing them as numbers. If your IF statement ends in things in quotes like IF(xxx@row = "yes", "100", "0") Smartsheet may not recognize the 100 as a number. In this case try wrapping the TRUE part of the IF statement in a VALUE(100) function to ensure the result is a number.
Alternatively if your IF formula is returning words or phrases, you probably want to use the COUNTIF formula to count how many times specific selections are made.
Does this help?
• Options
Thanks for responding back, Josh!
Okay, Let me be more specific on what I am actually doing, I created a Survey Form in which my client can send us feedback on the service we have provided, so this form has answers in a Drop-down list like for example, Very Satisfied, Satisfied, Neutral, Not Satisfied, etc. Now when I receive them, I will convert this response into Scores (Value), For example, Very Satisfied - 95, Satisfied - 85, Neutral - 65, etc.
I have applied the formula =IF(Responsiveness@row = "Very Satisfied", "95", IF(Responsiveness@row = "Satisfied", "85", IF(Responsiveness@row = "Neutral", "75", IF(Responsiveness@row = "Unsatisfied", "65", IF(Responsiveness@row = "Very Unsatisfied", "55"))))).
This gives me different for 5 different questions, like for example, 95,65, 55,85. Now I need to sum up these or get an average for these numbers. So I tried using the formula =SUM([Question 1 - Score]@row:[Question 4 - Score]@row) and I get the final value as 0. When I try =AVG([Question 1 - Score]@row:[Question 4 - Score]@row), I get the final value as #DIVIDE BY ZERO.
• Options
I just tried adding Value before the IF formula, it worked, Thank You!
• ✭✭✭✭✭
Options
@Keith Clarkson So glad it worked! It can be frustrating when you know the syntax of your formula is correct yet the result starts with a # 😁
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 836 | 3,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.942226 |
https://blogs.glowscotland.org.uk/glowblogs/jypetrieuodep/2018/10/18/chance-of-me-gaining-an-appreciation-for-maths-extremely-likely/ | 1,723,147,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00017.warc.gz | 108,846,265 | 19,202 | # Chance of Me Gaining an Appreciation for Maths – Extremely Likely
Probability – the next (and successful) leap in convincing me of how relevant mathematics actually is. This module has been effective in changing previously negative perceptions of maths, one of which in particular is that the terms ‘maths’ and ‘complex’ go together like bread and butter – this is not the case. Granted, parts of it can, as with anything. However I can vouch from my personal ongoing experience that if if we allow it to, it can be an enjoyable art.
Very briefly, probability is just what it sounds like – the likelihood of a certain event taking place or not. (Boaler, 2009). To use one of the most basic examples, the likelihood of rolling a ‘4’ on a dice. Well, there’s only 1 ‘4’ on a dice and this is known as the sample point. There are 6 possible outcomes overall (the values of 1-6), and this is known as our sample space. The probability of ending up with a ‘4’ on the roll of a dice is 1/6. Whilst the terminology was new to me, the concept is simple enough (Don’t Memorise, 2014).
Schools Minister Nick Gibb identified 3 purposes to education in a 2015 speech, one of which was “preparation for adult life” – a quotation that is very fitting in regards to probability (The Purpose of Education, 2015). Like many other aspects of mathematics such as money and time, chance and probability is a relevant concept we use on a daily basis. The weather is an excellent example of this assertion. Before leaving for the bus in the morning, it’s very likely I’ll ask “Alexa, what is the weather in Dundee today?”. Now personification aside, I would be very surprised if she told me “It will definitely rain today for 3 hours straight”.
What she might tell me is “It probably won’t rain in Dundee today. There’s only a 25% chance“.
Since meteorologists cannot predict the exact weather conditions, they must make informed predictions (Tucker, 2018). The information we take from them may not emerge as being correct however – a 25% chance of rain does not eliminate the possibility. Regardless, it does help us make judgement calls – will I take a raincoat, or can I leave it in my back so there’s more space for collecting my library books? With only a 25% chance, its probably going to be the latter of the two. The decisions we make from probability do not just have to relate solely the weather, of course (Haylock, 2006). Informed decisions are a major part of any human’s life, and they can range from purchasing properties in Monopoly at a family game night, to choosing the best day to practice surfing based on the tidal/wave conditions, or indeed, making the right call in the world of gambling.
I myself work at an indoor amusements/arcade, however my responsibilities of managing children’s parties mean I have never dabbled with the ‘nudgers’ that to me are nothing more than complex machines with countless buttons and symbols on them. Unlike the simple example of rolling a singular die to get a number explained above, these machines have multiple possible combinations (i.e. the sample space). Obviously, as with any business, the company intends to make a profit from the service it provides. As such, they are going to benefit in winnings a lot more than we are – even if we don’t witness that fact. The video below explains the basic functions of a slot machines, including what is meant by the terms theoretical payout and actual payout.
With these terms in mind, I spoke to one of our regular customers, who makes a guaranteed (okay, ‘extremely likely’) appearance at my work every weekend. I questioned him about his knowledge of gambling, and was consequently given a tour of the arcade, where he impressed me by pointing out which machines are best for winning (i.e. the best turnout rate, along with which ones were likely to give you an extra pound coin that may be stuck in the coin mechanism, or which machine is likely to have some loose change hidden at the back of the darkened retrieval pot. This man in particular has been coming to the establishment for years, so I was very interested to know how much profit his knowledge and skill set have rewarded him with – except he doesn’t. To lift his own words, “pure entertainment value”. He knows he is going to lose money regardless of what he plays, and this once again links back to how the actual payout of slot machines are ALWAYS greater than the money put in – you are technically losing more than getting back (tech4truth, 2010). This is not a new concept either. Charles Fey, who invented the first slot machine (3 reels, 5 symbols), had a 75% average pay-back! That may seem great at first glance, it only paid out 50% of the time (Valentine, 2018). There may be that faulty machine in the corner of the room, or perhaps you are being watched by a ‘lurker’ waiting to pounce at the right opportunity to steal your winnings. Even then though, they are using their knowledge of probability of knowing when to swoop in.
To summarise all of this, probability and chance is something we use everyday in a variety of circumstances. It can become part of subconscious, much like the process of respiration and blinking, where we only realise we are doing it when we are told so. This area in particular is such a huge foundation in regards to our everyday choices. I’m truly beginning to realise that Maths isn’t just about sitting at a high school desk, attempting to find ‘x’ and losing your mind whilst you do it. There’s so much more to it, and just because it’s not an equation, that doesn’t make it any less maths-related.
Also, apologies – chances are you’re now aware you’re breathing and it won’t be a subconscious process for a little bit.
References
Boaler, J. (2009) The Elephant in the Classroom: Helping Children to Learn and Love Maths. London: Souvenir Press Ltd.
Don’t Memorise (2014). Probability – Sample Space, Sample Points, Events! Available at: https://www.youtube.com/watch?v=5oI8-iQqPAI (Accessed: 18th October 2018)
Gambling Glossary: A Guide to Gambling Terms. (no date). Available at: https://www.gambling.net/glossary.php (Accessed: 18th October 2018)
Haylock, D. (2006) Mathematics explained for primary teachers. London: SAGE.
Probability. (no date) Available at: https://www.mathsisfun.com/data/probability.html (Accessed: 18th October 2018)
tech4truth (2010) Slot Machine Paybacks and Slot Odds Explained (Tech4Truth Episode 3) Available at: https://www.youtube.com/watch?v=4wzg-8QKC5s (Accessed: 18th October 2018)
The purpose of education. (2015) Available at: https://www.gov.uk/government/speeches/the-purpose-of-education. (Accessed: 18th October 2018)
Tucker, K. (2018) Examples of Real Life Probability. https://sciencing.com/examples-of-real-life-probability-12746354.html (Accessed: 18th October 2018)
Valentine, E. (2018). ‘Chance and probability’ [PowerPoint presentation]. ED21006: Discovering Mathematics. Available at: https://my.dundee.ac.uk/webapps/blackboard/execute/displayLearningUnit?course_id=_58988_1&content_id=_5217952_1. (Accessed: 18th October 2018)
Images Used
https://www.wikihow.com/Calculate-Probability | 1,714 | 7,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.935916 |
https://blog.ozeninc.com/resources/electronics-reliability-its-getting-hot-in-there | 1,719,091,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00187.warc.gz | 116,117,574 | 21,644 | TLDR: Heat can cause solder to crack. There are 2 paths for simulation: Fast rough estimates based on closed form equations or detailed, accurate FEA model of solder balls.
I started writing about electronics reliability, then I realized it is kind of a big topic. So I’ll do it in sections here.
Decades ago, the air force did a big study on avionics reliability and found heat, vibration, humidity and dust to be the leading causes of electronics failure.
Physics didn’t change much over the years and although electronics are faster and smaller, the causes of failure remains the same.
Heat and vibration were the most dominant cause. So let’s focus on heat first
Heat is bad for electronics because materials have different thermal expansion coefficients. This means when they get hot; PCB, chips and solder all try to expand by different amounts. High stress is induced in the solder and cracks occur.
Cracks get bigger over time until something breaks. Then we call it a “reliability issue”.
# Warpage
When do electronics get hot? When you put them in an oven!
Most electronics need to go through an oven to melt the solder paste. It can get up to 250C in there!
At these temperatures, the PCB can bend in all kinds of interesting ways because copper and fr4 are not uniformly distributed so there is lot of push and pull that cause warpage.
Now imagine that you have 1000’s of tiny solder balls and they need to all stick to just the right spot on a PCB. In the oven it goes and the PCB becomes wavy. You are about to experience some “yield issues.”
Luckily this can be simulated early on in the design process so that you have a chance to mitigate these problems.
See these videos for more details.
The key is to capture the distribution of copper inside the PCB accurately. Then you will be able to predict the warped shape and understand the risk to reliability.
# Thermal cycling
Electronics also get hot when you use them. This causes thermal expansion differences and eventually some solder connections will break… again.
IRL this can take years, but we can’t wait years to find out if our device will fail. Instead we stick it into an oven and freeze-roast it for a few weeks. This has been a problem for a long time and there are many well publicized failures over the years, so people have come up with various solutions.
The first approach is to use basic empirical equations because they can quickly scan an entire board and give you an estimate for the durability of every chip, cap and resistor. Only the size and material of the board and packages are taken into account so it’s a rough estimate.
Here is an example:
When you need to ensure that a CPU doesn’t fail early, you need a detailed BGA solder joint fatigue simulation like the one shown in the video below.
You can put in as many details as you want. Heat sinks and brackets? Sure. 50lbs of clamping preload? No problem. How about details like solder pad and solder mask? Why not?
Great results… comes with a lot of work. These models need to be set up carefully and consistently.
That’s a quick overview of what can be done in simulation for thermal reliability. | 662 | 3,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.948115 |
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## Academic research paper on topic "The Nonlocal -Laplacian Evolution for Image Interpolation"
Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2011, Article ID 837426,11 pages doi:10.1155/2011/837426
Research Article
The Nonlocal p-Laplacian Evolution for Image Interpolation
Yi Zhan
College of Mathematics and Statistics, Chongqing Technology and Business University, Chongqing 400067, China
Correspondence should be addressed to Yi Zhan, zhanyimeng@yahoo.com.cn Received 24 June 2011; Accepted 16 August 2011 Academic Editor: P. Liatsis
Copyright © 2011 Yi Zhan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
This paper presents an image interpolation model with nonlocal p-Laplacian regularization. The nonlocal p-Laplacian regularization overcomes the drawback of the partial differential equation (PDE) proposed by Belahmidi and Guichard (2004) that image density diffuses in the directions pointed by local gradient. The grey values of images diffuse along image feature direction not gradient direction under the control of the proposed model, that is, minimal smoothing in the directions across the image features and maximal smoothing in the directions along the image features. The total regularizer combines the advantages of nonlocal p-Laplacian regularization and total variation (TV) regularization (preserving discontinuities and 1D image structures). The derived model efficiently reconstructs the real image, leading to a natural interpolation, with reduced blurring and staircase artifacts. We present experimental results that prove the potential and efficacy of the method.
1. Introduction
Digital image interpolation is an important technology in digital photography, TV, multimedia, advertising, and printing industries, which is applied to obtain higher-resolution image with better perceptual quality. The key task in image interpolation is to remove zigzagging, blurry, and other artifacts producing visually pleasing resulting image. Many literatures are devoted to address these problems [1-10]. These methods, usually known as edge directed, level set based, or isophote oriented, vary considerably.
The methods based on edge direction were proposed to obtain smooth edges of the resulting images [1-3]. However, these methods suffer from degradations on edge-free regions because they rely on local directions estimation, creating false edges in uniform regions. Wang and Ward have developed an interesting technique based on the detection of ridges (straight edges) in images [4], which allows them to interpolate directionally only
pixels situated on straight edges and avoid the apparition of false edges. In [5], the zigzagging artifacts are reduced by restricting curvature of the interpolated isophotes (equi-intensity contours). Minimum curvature is required on isophotes of the interpolated images. Cha and Kim proposed a method based on the TV energy in order to remove the so-called checkerboard effect and to form reliable edges [6]. Morse and Schwartzwald [7] presented a scheme that uses existing interpolation techniques as an initial approximation and then iter-atively reconstructs the isophotes using constrained smoothing. However, they are complex compared to traditional methods and thus computationally expensive. Malgouyres and Guichard [8] proposed to choose as solution of the interpolation the image that minimizes the TV. This method leads to resulting images without blurring effects, as it allows discontinuities and preserves 1D fine structures. Aly and Dubois [9] proposed a model-based TV regu-larization image up sampling methods. Image acquisition process is modeled after a lowpass filtering followed by sampling. However, TV minimization is based on the assumption that the desirable image is almost piecewise constant, which yields a result with over smoothed homogeneous regions. Belahmidi and Guichard [10] have improved the TV-based interpolation by developing a nonlinear anisotropic PDE, hereafter referred to as BG interpolation method. In order to enhance edge preservation, this PDE performs a diffusion with strength and orientation adapted to image structures. This method balances linear zooming on homogeneous regions and anisotropic diffusion near edges, trying to combine the advantages of these two processes. The anisotropic diffusion scheme, including edge-directed or isophote-oriented method, uses the gradient to extract the image feature (edge) direction, that is, the gradient direction is considered to be the direction across the image feature. Nevertheless, the information contained in the gradient is local, not good to determine the edge directions. Nonlocal information should be considered in determining edge directions.
Recently, a nonlocal evolution equation and variations of it have been widely used to model diffusion processes in many areas [11,12]. Let us briefly introduce some references of nonlocal problem considered along this work. A nonlocal evolution equation corresponding to the Laplacian equation is presented as follows
It is called a nonlocal diffusion equation since the diffusion of the density at a point x and time t does not only depend on u(x; t), but on all the values of u in a neighborhood of x through the convolution term J * u. This equation shares many properties with the classical heat equation ut = Au. More precisely, as stated in [13], if u(t,x) is thought of as the density at the point x at time t, and J (x,y) is thought of as the probability distribution of jumping from location y to location x, then the convolution (J * u)(x,t) = jRN J(y - x)u(y,t)dy is the rate at which individuals are arriving to position x from all other places, and -u(x, t) = - J(y - x)u(y,t)dy is the rate at which they are leaving location x to travel to all other sites. This nonlocal evolution can be thought of as nonlocal isotropic diffusion.
In this paper, we propose a new method for image interpolation based on nonlocal p-Laplacian evolution. The nonlocal p-Laplacian and TV act as a regularizer to restrict edges of resulting image. The evolution is similar to an anisotropic energy dissipation process. The diffusion performs accurately along the direction of edges curves and its orthogonal direction. The magnitude of |u(y, t) - u(x, t) f-2 determines diffusion directions. It suppresses diffusion across the image feature direction and enhances diffusion along the image feature direction.
The rest of the paper is organized as follows. In Section 2, we review the method based on image up sampling with TV regularization in [9] and BG image interpolation [10]. The proposed nonlocal p-Laplacian evolution for image interpolation is presented in Section 3. In Section 4, we demonstrate the experimental results to verify the effectiveness of our method, and the last section is for conclusion.
2. Background
We can think that a digital lower-resolution image u0 (input image) defined on some lattice is obtained by transforming a high-resolution image u (output image) defined on some better precision lattice, that is,
u0 = Hu, (2.1)
where H is a sparse matrix that combines both filtering and down sampling process. The goal of image interpolation is to solve the inverse problem (2.1), an ill-posed inverse problem. This ill-posed inverse problem is generally approached in a regularization-based framework, which would be formulated as an energy functional [9],
E(u) = Jd (u,u0) + XJr (u), (2.2)
where X is a regularization parameter that controls the tradeoff between Jd and Jr. The data fidelity function Jd generally is formulated in the classical least-squares sense as Jd (u,u0) = (1/2)|Hu - u0|2. The TV regularizer Jr is taken as Jr(u) = JQ |Vu(x)|dx. The formula (2.2) is rewritten as follows
E(u) = X| |Vu|dx + 1 |Hu - u0|2. (2.3)
Using Euler's equation, the minimizer is the steady-state solution of the nonlinear PDE given by
ut = X|Vu| divf ^ + HT(Hu - u0). (2.4)
On the other hand, Belahmidi and Guichard solve the ill-posed inverse problem based on the classical heat diffusion model [10]. Let n denote the direction of local gradient, and £ the direction perpendicular to the gradient, namely,
n = |vU| {Ux>u)> 5 = jvU|(~%'Ux)'
The second-order directional derivatives of the image u along the directions of n and & are easily computed as follows:
df2 " f
Uxx Uxy
d2U (Uxx Uxy
UxxUn 2UxyUxUy + UyyUx
*-yyUx
UXXUx + 2uxyuxuy + UWUi/
■lyyUy
The interpolation scheme based on heat diffusion model is formulated as follows:
d2u d2u
u = jVuj^ + g(jVuj)- HT(Hu - uo).
In this equation, the function g (s) is typically defined as
g(s) =
1 + (s/K)2
with K > 0 is a constant to be tuned for a particular application. The role of the diffusion coefficient g (jVuj) is to control the smoothing adaptively.
When g = 0, (2.7) reduces to (2.4) with X = 1. All these models can be viewed as interpolation schemes based on nonlinear diffusion model. The two regularizers jVuj div(Vu/jVuj) and jVuj(52u/d^2)+g(jVuj)(52u/5n2), respectively, in (2.4) and (2.7) result in different interpolation effects. In fact, jVuj div(Vu/jVuj) = jVuj(d2u/d&2) is the second-order directional derivative in the direction that is orthogonal to the gradient jVuj, and d1u/dn1 is the second-order directional derivative in the direction of the gradient jVuj.
From the viewpoint of geometry, the evolution processes in the artificial time t given by these models are seen as energy dissipation processes in two orthogonal directions n and & [9]. The diffusion process of u(x,t) along I will preserve the location and the intensity transitions of the contours, while smoothing along them maintaining their crispness. This diffusion term is used to maintain edges with smooth isophotes in [9,10]. The diffusion of the grey values along n walks across both sides of the local image contour. This process blurs crisp contours as in the case of linear interpolators. Two divided means are used to deal with it. The diffusion process along n is cast aside in [9], while controlled by edge-stopping function g(jVuj) to balance the two diffusion terms in [10].
TV regularization in (2.3) does an excellent job at preserving edges while reconstructing images [14]. This phenomenon can also be explained physically, since the resulting diffusion is strictly orthogonal to the gradient of the image. But TV-based interpolation favors solutions that are piecewise constant. This sometimes causes a staircasing effect in homogeneous regions, which are long observed in the literature in denoising, for example, [15,16]. Not only having blocky solutions, but they can also develop false edges in resulting image.
The function g in (2.7) is to be chosen with values between 0 and 1. The energy dissipation process (2.7) is adaptively controlled in the direction n and &, that is, minimal smoothing in the directions n (across the image features) preserving sharp edges, and maximal smoothing in the directions & (along the image features) obtaining smooth contours.
Uxy Uyy
The anisotropic diffusion scheme that uses the gradient to extract the image feature direction can mistakenly give maximal smoothing to the across feature direction and severely damage the image features, especially the image lines and textures [17]. And blurry and/or oscillatory edges are introduced in interpolated image. The drawback of this model is that the gradient used to extract the image feature direction is too local. The information contained in the gradient is limited to a point and its immediate neighbors, while the edge curve that determines the edge directions is not a local event. The interpolation direction extraction should base on a larger neighborhood.
3. Nonlocal p-Laplacian Image Interpolation
In this section, we adopt nonlocal p-Laplacian evolution to overcome the local limit. Our proposed energy functional for regularized image interpolation is given by
E(u) = a( \Vu\dx + J(x - y)|u(y) - u(x)\dydx + 1 \Hu - u0\2. (3.1)
Jq 2p J JQ 2
The first part (the sum of the first term and the second) of right-hand side is regularizer, and the other is data fidelity function. The gradient flow associated to the functional E(u) is
ut(x,t) = a div(\Vu\-1Vu) + P(u) - HT(Hu - uo),
u(x, 0) = HTu0,
PJ(u) = J{x,y) \u(y,t) - u(x,t)\p 2(u{y,t) - u(x,t))dy. (3.3)
The kernel J : Q ^ R is assumed to be nonnegative, bounded continuous radial function, with sup(J) c B(0,d) and J"Q J(z)dz = 1.
The nonlocal energy dissipation is implemented mostly by PpJ(u) in our model. It is necessary to investigate the relation between the heat diffusion equation related to (2.7) and the p-Laplacian equation. The p-Laplacian evolution equation ut = is well
studied in image processing, which can be represented as [18]
ut = |V«r2+ (p - l)|V«r20• (3-4)
When p = 1, it is the TV flow keeping the edges but suffering from the staircase effect. When p = 2, ut = Au, this is isotropic diffusion because of the same diffusion coefficients. This model can smooth image, while bluring sharp edges. When 1 < p < 2, the grey values of u(x,t) in (3.4) diffuse along the directions n and I, respectively, as in the following equation:
ut = | Vu | ^ + g( | Vu | ) ^, (3-5)
which has different control factor compared with (3.4). That is to say, (3.4) and (3.5) all preserve advantage of adaptive smoothing.
A nonlocal improvement to p-Laplacian equation was studied in [19],
ut(x,t) = J(x,y)\u(y,t) - u(x,t)\p 2(u{y,t) - u(x,t))dy. (3.6)
Moreover, nonlocal problems of type (3.6) have been used recently in the study of deblurring and denoising of images [20]. With Neumann boundary conditions, the solutions to problem (3.6) converge to the solution of the classical p-Laplacian if p > 1 [19]. The nonlocal p-La-placian evolution (3.6) improves the limit of the diffusion direction extraction depending on gradient (local information). The diffusion of the density at a point x and time t depends on all the values of u in a larger neighborhood of x. More precisely, if u(t,x) is thought of as the density at the point x at time t, and J(x,y) is thought of as the probability distribution of jumping from location y to location x, then PJ (u) is the rate at which individuals are arriving to position x from all other places. In image interpolation, PJ (u) is also the rate at which individuals are devoting to interpolated pixel x from all other pixels. The evolution process in the artificial time t given by (3.6) is seen as an anisotropic energy dissipation process. The direction of anisotropic diffusion is indicated by the magnitude of |u(y, t) - u(x, t)|p-2 in a larger neighborhood. It approximates to the direction of edge curve more accurate than the direction indicated by gradient. When 1 < p < 2, the energy dissipation process is adaptively controlled by |u(y, t) - u(x, t)|p-2 along the direction of edge curves and the orthogonal direction to edge curves. The diffusion process along the direction of edge curves is suppressed for small |u(y, t) - u(x, t)|p-2, and the diffusion along the orthogonal direction is enhanced for larger |u(y, t) - u(x, t)|p-2. This results in minimal smoothing in the directions across the image features preserving sharp edges and maximal smoothing in the directions along the image features reducing zigzagging artifacts and oscillatory.
4. Numerical Algorithm and Experimental Results
In this section, we develop a fully discrete numerical method to approximate problem (3.2). We recall first the notations in the finite differences scheme used in our paper. Let h and At be the space and time steps, respectively, and let (x^; x2j) = (ih; jh) be the grid points. Let un(i; j) be an approximation of the function u(nAt; x1i; x2j), with n > 0. Equation (3.2) can be discretized as follows:
un+1( i,j) - un(i,j) ( /Vu
At =(adiv( rV^)- HT(Hu - i
+ £ £ [J( (k,l) - j\un(k,l) - un{i,j)\p-2(un(k,l) - un(i,j)j\.
(fc,/)eQ
Figure 1: Barbara reduced and expanded by 2 x 2 (portion). (a) NEDI interpolation; (b) EGI interpolation; (c) BG method; (d) proposed method.
In all numerical experiments, we choose the following kernel function:
J (x) :=
Cexpi 2 0
|x|2 - d2 0 if x>d.
if x < d,
The constant C > 0 is selected such that JQ J (x)dx = 1.
We tested the proposed interpolation method on a variety of images. Some of the results are shown in Figures 1-3. Images are expanded by a factor of 2 x 2 in Figures 1 and 2 and by a factor of 10 x 10 in Figure 3. For comparison, we also show images interpolated using the BG image interpolation proposed in [10], the edge-guided image interpolation (EGI) in [21], and the edge-directed interpolation (NEDI) proposed in [3]. The choice of the parameters is based on subjective quality of the results assessed informally by our personal preference as human viewers in terms of edge sharpness, contour crispness, no ringing in smooth regions, and no ringing near edges. We use the following parameters: a = 0.5, fi = 0.0001, and p = 1.8 for the proposed interpolation method, k = 0.0001 for the heat diffusion model, and time step At = 0.15 for these experiments. There is a non visible improvement on subjective or objective quality of the results when the parameters are not badly
changed. The iteration is terminated, when iterations.
In the first experiments, Barbara image with a size of 512 x512 was lowpass filtered and subsampled by a factor of 2 x 2, then the subsampled image was interpolated to the original image size. The interpolation was performed by four different methods, and a portion of the results is shown in Figure 1. Figure 2 shows a portion of a result interpolating a given flower image with a size of 320 x 240 by a factor of 2 x 2 without subsample. From the two examples, the NEDI interpolation method tends to introduce the zigzagging artifacts (Figures 1(a) and 2(a)). The BG results produce slight blurry edges (stripes in Figure 1(c)) and ringing in smooth regions (as shown in Figure 2(c)), because the direction decided by gradient misses their real directions as stated in Section 2. Our proposed method and the EGI method produce sharp edges and smooth contours, but the EGI method and the NEDI interpolation are applied only by a factor of 2 x 2.
Figure 2: Flower expanded by 2x2 (portion). (a) NEDI interpolation; (b) EGI interpolation; (c) BG method; (d) proposed method.
(d) (e) (f)
Figure 3: A portion of Barbara, Mandrill, and house images expanded by 10 x 10. Top row: BG method; bottom row: proposed method.
Table 1: Comparison of different interpolation algorithms using PSNR values.
Image The proposed BD EGI NEDI
Lena 34.6483 33.5438 30.5031 35.1024
Cameraman 26.7846 26.2239 24.3987 27.5413
Barche 28.0214 27.4137 25.7585 29.6238
Peppers 27.9167 27.4998 25.9046 29.8716
Mandrill 23.7041 23.2248 22.4871 24.8313
Resolution test 21.6378 21.0777 18.7030 22.7104
Barbara 25.2955 24.9608 24.3268 26.9344
Table 2: Comparison of different interpolation algorithms using MSSIM values.
Image The proposed BD EGI NEDI
Lena 0.9908 0.9836 0.9640 0.9609
Cameraman 0.8728 0.8575 0.8202 0.8105
Barche 0.8447 0.8237 0.7790 0.7671
Peppers 0.9144 0.9036 0.8704 0.8616
Mandrill 0.9528 0.9169 0.8851 0.8718
Resolution test 0.9147 0.9042 0.8547 0.8388
Barbara 0.9438 0.9204 0.9011 0.8952
The second experiment directly interpolates a portion of images (Barbara, mandrill, and house) by a factor of 10 x 10 shown in Figure 3. This experiment is performed by the BG method and our proposed method since the other methods only resize image by a factor of 2 x 2. It is clear from the figures that results obtained with the proposed approach are better than the results by the BG method in [10]. The proposed method generates sharper and crisper stripes in Figure 3(d) compared to the result in Figure 3(a). The BG method produces blurry edges (the stripes in Figure 3(a), the beard in Figure 3(b)), zigzagging artifacts, and oscillatory (the line in Figure 3(c)), and images tend to be less natural. In the BG interpolation result, the boundaries of the text suffer from artifacts that make visualization difficult. It can be seen that our method results in an interpolated image with the fewer spurious patterns.
We use two measures, the classic PSNR and the mean structural similarity (MSSIM) index [22], to characterize the difference between the reference image and the output of a method. The MSSIM seems to approximate the perceived visual quality of an image better than PSNR or various other measures [23]. MSSIM index takes values in [0,1] and increases as the quality increases. It is calculated by the code available at http://www .cns.nyu.edu/lcv/ssim/, using the default parameters. We use several test images of size 512x512 including Lena, mandrill, and Barbara of size 256x256 including cameraman, barche, peppers, and resolution test. To show the true power of the interpolation algorithms, we first downsampled each image by a factor of 2 x 2 and then interpolated the result back to its original size. The PSNR is shown on Table 1 and the MSSIM on Table 2. From the two tables, the proposed method yields improved PSNR (except for the NEDI), and MSSIM results in all the experiments. This improvement may be attributed to the fact that the nonlocal p-Lapla-cian evolution works better than other methods.
5. Conclusion
In this paper, a new image interpolation model based on TV and nonlocal p-Laplacian reg-ularization is proposed. It combines the advantages of TV regularizer and nonlocal p-Lapla-cian regularizer, that is, allowing discontinuities and preserving 1D image structures and the diffusion of the grey values of images along image feature direction. It overcomes the drawbacks of the anisotropic diffusion proposed by Belahmidi and Guichard. The direction of anisotropic diffusion is indicated by the information of image feature in a larger neighborhood. This results in minimal smoothing in the directions across the image features preserving sharp edges and maximal smoothing in the directions along the image features reducing zigzagging artifacts and oscillatory. We have shown improvement over nonlocal p-Laplacian on a subjective scale, and in many cases with an improvement in PSNR and MSSIM. We expect to prove convergence of the evolution equation in future work.
Acknowledgments
This research was partially supported by Natural Science Foundation Project of CQ CSTC (Grant no. cstcjjA40012) and the National Natural Science Foundation of China (Grant no. 10871217).
References
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[4] Q. Wang and R. Ward, "A new edge-directed image expansion scheme," in Proceedings of IEEE International Conference on Image Processing (ICIP '01), pp. 899-902, October 2001.
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[6] Y. Cha and S. Kim, "Edge-forming methods for image zooming," Journal of Mathematical Imaging and Vision, vol. 25, no. 3, pp. 353-364, 2006.
[7] B. S. Morse and D. Schwartzwald, "Isophote-based interpolation," in Proceedings of IEEE International Conference on Image Processing (ICIP '98), vol. 3, pp. 227-231, Chicago, 111, USA, October 1998.
[8] F. Malgouyres and F. Guichard, "Edge direction preserving image zooming: a mathematical and numerical analysis," SIAM Journal on Numerical Analysis, vol. 39, no. 1, pp. 1-37, 2002.
[9] H. A. Aly and E. Dubois, "Image up-sampling using total-variation regularization with a new observation model," IEEE Transactions on Image Processing, vol. 14, no. 10, pp. 1647-1659, 2005.
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[20] S. Kindermann, S. Osher, and P. W. Jones, "Deblurring and denoising of images by nonlocal function-als," Multiscale Modeling and Simulation, vol. 4, no. 4, pp. 1091-1115, 2005.
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Copyright of Mathematical Problems in Engineering is the property of Hindawi Publishing Corporation and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use. | 6,972 | 27,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-40 | latest | en | 0.832743 |
https://search.r-project.org/CRAN/refmans/ELYP/html/findL3.html | 1,653,420,312,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00097.warc.gz | 571,836,643 | 2,340 | findL3 {ELYP} R Documentation
## Find the Wilks Confidence Interval Lower Bound from the Given Empirical Likelihood Ratio Function
### Description
This program is the sister program to the findU3( ). It uses simple search to find the lower 95% Wilks confidence limits based on the log likelihood function supplied.
### Usage
findL3(NPmle, ConfInt, LogLikfn, Pfun, level=3.84, dataMat)
### Arguments
NPmle a vector containing the two NPMLE: beta1 hat and beta2 hat. ConfInt a vector of length 3. LogLikfn a function that compute the loglikelihood. Typically this has three parameters: beta1, beta2 and lam, in a Yang-Prentice model context. Pfun a function that takes the input of 3 parameter values (beta1,beta2 and Mulam) and returns a parameter that we wish to find the confidence Interval of (here only the Lower Value). level confidence level. Default to 3.84 for 95 percent. dataMat a matrix.
### Details
The empirical likelihood for Y-P model has parameters: beta1, beta2 and a baseline. The baseline is converted to a 1-d parameter feature via Hfun, and then amount controled by lam.
Basically we repeatedly testing the value of the parameter, until we find those which the -2 log likelihood value is equal to 3.84 (or other level, if set differently).
### Value
A list with the following components:
Lower the lower confidence bound. minParameterNloglik Final values of the 4 parameters, and the log likelihood.
Mai Zhou
### References
Zhou, M. (2002). Computing censored empirical likelihood ratio by EM algorithm. JCGS
### Examples
## Here Mulam is the value of int g(t) d H(t) = Mulam
## For example g(t) = I[ t <= 2.0 ]; look inside myLLfun().
Pfun <- function(b1, b2, Mulam) {
alpha <- exp(-Mulam)
TrtCon <- 1/(alpha*exp(-b1) + (1-alpha)*exp(-b2))
return(TrtCon)
}
data(GastricCancer)
# The following will take about 10 sec. to run on i7 CPU.
# findL3(NPmle=c(1.816674, -1.002082), ConfInt=c(1.2, 0.5, 10),
# LogLikfn=myLLfun, Pfun=Pfun, dataMat=GastricCancer)
[Package ELYP version 0.7-5 Index] | 570 | 2,045 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-21 | latest | en | 0.681669 |
https://www.studysmartwithchris.com/en/summaries/basic-skills-in-mathematics/ | 1,652,800,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00500.warc.gz | 1,210,481,228 | 16,254 | # Summary Basic Skills In Mathematics
##### Course
- Basic skills in mathematics
219 Flashcards & Notes
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• ## 1 lecture 1
This is a preview. There are 10 more flashcards available for chapter 1
Show more cards here
• Explain what different object modes are?
• There are different types of objects in R: - numeric, e.g., 2, 0.03 - character, e.g., ‘a’, “b” - logical, TRUE, FALSE
• These are called modes
• To request the mode of object x use mode(x)
• a mode is the category an object belongs to.
• What are characters and strings?
• Anything between single or double quotation marks is stored as a character, which can be used to encode strings:
• strings are kind of naming objects a different name.
• a character is the object, and the string is the second name that is given to that character, what it's also reffered by.
• characterObject <- 'this is a string'
• characterObject2 <- "this is also a string"
You can not do mathematical expressions with character strings, even when the string looks like a number
• Describe the logical mode.
• The logical mode indicates a Boolean object which can only be true (TRUE) or false (FALSE):
• logicalObject <- TRUE
• these logical objects can be used in compariosn tests;
• 1 == 1 # Is 1 equal to 1?
• ## [1] TRUE
• How can you test and transform modes?
• You can use functions named is.mode to test the mode of an object:
• a <- 1.23 is.logical(a)
• ## [1] FALSE
• You can use functions named as.mode to transform objects into a different mode:
• "1" + 1 ## Error in "1" + 1: non-numeric argument to binary operator
• as.numeric("1") + 1
• ## [1] 2
• What does the rep function do?
• A vector containing copies of the same object can be created using rep(object,number_of_repetitions):
• For example, rep(1,3) creates the vector [1 1 1]
• Create a vector with n copies:
• v <- c('a','b','d')
• v
• ## [1] "a" "b" "d"
• rep(v,3)
• ## [1] "a" "b" "d" "a" "b" "d" "a" "b" "d"
• What does the sample function do?
You can create a vector of n random samples from another vector using sample(vector_to_sample_from, n)
• v1 <- seq(1, 4, 0.5)
• v2 <- sample(v1, 4)
• v2
• ## [1] 2.0 3.5 4.0 1.5
• How can you perform mathematical computations with vectors?
• You can perform elementwise operations on vectors:
• a <- c(10,20,30)
• b <- 1:3
• a + b # Add each element in a with the same element in b
• ## [1] 11 22 33
• a * b # Multiply each element in a with the same element in b
• ## [1] 10 40 90
• How can you apply logical implications to vectors?
• You can apply logical operations to vectors:
• a <- c(10,20,30)
• b <- c(5,20,50)
• a < 20 # Test each element of a if it is smaller than 20
• ## [1] TRUE FALSE FALSE
• What is indexing and how do you do it?
• You can select a subset of vector element using indexing
• Using square brackets [ and ], you can indicate which elements to return as:
• A vector containing numbers of the elements you wish to keep
• a[5] # Get the fifth element of vector a
• a[c(1,5)] # Get the first and fifth element of vector a
• A minus sign followed by the element you wish to omit
• a[-5] # Get everything except the fifth element of vector a
• A vector containing a TRUE or FALSE for each element, TRUE indicating you want to keep the element
• a < 5
• ## [1] TRUE TRUE FALSE FALSE FALSE
• How do you index the values of a variable?
Type datafilename\$variable
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This summary +380.000 other summaries A unique study tool A rehearsal system for this summary Studycoaching with videos | 1,053 | 3,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-21 | latest | en | 0.827952 |
https://ericrscott.com/posts/2021-01-18-dlnm-basis/ | 1,695,623,084,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00449.warc.gz | 267,756,823 | 10,217 | # DLNM marginal basis functions
DLNMs
GAMs
Published
January 18, 2021
Note
This is part of series about distributed lag non-linear models. Please read the first post for an introduction and a disclaimer.
## Choosing marginal function to construct a crossbasis
According to Gasparrini et al. (2017), a crossbasis function is a “bi-dimensional dose-lag-response function $$f \cdot w(x,l)$$ is composed of two marginal functions: the standard dose-response function $$f(x)$$, and the additional lag-response function $$w(l)$$ that models the lag structure…” Each dimension can be described by a different type of function. The default for the dlnm package is a type of smoother called a P-spline, but it can be changed to other types of splines or even something like step function. The marginal functions can also be mixed and matched, e.g., a P-spline for the lag dimension and a step function for the dose-response dimension.
I’d like to use penalized splines for both bases since they are flexible—that is, they can take nearly any functional shape, including a perfectly straight line.
So far I’ve been using penalized cubic regression splines for both the lag and dose-response dimensions of my DLNMs, but to be perfectly honest, I think I’m only doing this because Teller et al. (2016) use a similar spline basis, However, they aren’t even using DLNMs! I should at least be able to justify my choice of basis function.
library(mgcv) #for gam()
Loading required package: nlme
Attaching package: 'nlme'
The following object is masked from 'package:dplyr':
collapse
This is mgcv 1.8-42. For overview type 'help("mgcv-package")'.
library(dlnm) #for the "cb" basis
This is dlnm 2.4.7. For details: help(dlnm) and vignette('dlnmOverview').
#with cubic regression splines for both dimensions
growth_cr <-
gam(log_size_next ~
log_size +
s(spei_history, L, # <- the two dimensions
bs = "cb", # <- fit as crossbasis
k = c(4, 24), # <- knots for each dimension
xt = list(bs = "cr")), # <- what basis to use for each dimension
method = "REML",
data = ha)
Note: for P-splines, the number of knots, k, must be 2 greater than order of the basis (default 2, i.e. cubic), so I’m using the minimum (4) for the dose-response dimension.
#with default P-splines for both dimensions
growth_ps <-
gam(log_size_next ~
log_size +
s(spei_history, L, # <- the two dimensions
bs = "cb", # <- fit as crossbasis
k = c(4, 24)), # <- knots for each dimension
method = "REML",
data = ha)
growth_cr
Family: gaussian
Formula:
log_size_next ~ log_size + s(spei_history, L, bs = "cb", k = c(4,
24), xt = list(bs = "cr"))
Estimated degrees of freedom:
8.37 total = 10.37
REML score: 675.5565
growth_ps
Family: gaussian
Formula:
log_size_next ~ log_size + s(spei_history, L, bs = "cb", k = c(4,
24))
Estimated degrees of freedom:
7.63 total = 9.63
REML score: 673.1247
The REML score is slightly higher for the "cr" basis, which I think means a better fit to data (I think this score is what is being maximized by the model fitting algorithm).
AIC(growth_cr, growth_ps)
df AIC
growth_cr 13.16403 1331.719
growth_ps 12.00639 1332.001
AIC is also slightly lower for the "cr" basis
## Do they produce different shapes?
I’m going to use the trick I “discovered” in the previous blog post to plot the crossbasis function from each model.
growth_cr$smooth[[1]]$plot.me <- TRUE
growth_ps$smooth[[1]]$plot.me <- TRUE
par(mfrow = c(1,2))
plot(growth_cr, scheme = 2)
plot(growth_ps, scheme = 2)
The minima and maxima are in the same places, which is very reassuring. The wiggliness is different, which is also indicated by the estimated degrees of freedom (8.37 for the “cs” model and 7.63 for the “ps” model).
## Final Decision
I’m going to stick with the cubic regression spline basis (bs = "cr") because it seems to result in a slightly better fit to data than the P-spline smoothers. In addition, Simon Wood says “However, in regular use, splines with derivative based penalties (e.g.”tp” or “cr” bases) tend to result in slightly better MSE performance” (see ?smooth.construct.ps.smooth.spec). | 1,113 | 4,116 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-40 | latest | en | 0.835776 |
http://clay6.com/qa/20785/-i-int-limits-0-cos-3-x-dx | 1,529,527,491,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863886.72/warc/CC-MAIN-20180620202232-20180620222232-00108.warc.gz | 66,639,572 | 25,797 | # $I= \int \limits_0^{\frac{\pi}{2}} \cos ^3 x dx$
$\begin {array} {1 1} (a)\;\frac{1}{3} \\ (b)\;1 \\ (c)\;\frac{2}{3} \\ (d)\;0 \end {array}$
$I= \int \limits_0^{\frac{\pi}{2}} \cos ^3 x dx$
$\quad= \int \limits_0^{\frac{\pi}{2}} \cos x (\cos ^2 x) dx$
$\quad= \int \limits_0^{\frac{\pi}{2}} \cos x (1-\sin ^2 x) dx$
$\quad= \int \limits_0^{\frac{\pi}{2}} \cos x dx -\int \limits_0^{\frac{\pi}{2}} \cos x \sin ^2 x dx$
$\sin x =t \quad \cos x dx=dt$
$\qquad= \sin x \bigg]_0^{\pi/2} - \int \limits _0^1 t^2 dt$
$\qquad=1-\large\frac{1}{3}$
$\qquad=\large\frac{2}{3}$
Hence c is the correct answer. | 285 | 601 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-26 | latest | en | 0.231646 |
http://mathhelpforum.com/advanced-algebra/85790-nilpotent-groups.html | 1,526,873,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863923.6/warc/CC-MAIN-20180521023747-20180521043747-00582.warc.gz | 196,536,193 | 11,264 | # Thread: nilpotent groups
1. ## nilpotent groups
I need to show subgroups of nilpotents groups are nilpotent.
I can get started but cannot finish.
Suppose G is a nilpotent group. Let H be a proper subgroup of G. Let K be a proper subgroup of H. I must show K is a proper subgroup of $\displaystyle N_H(K)$.
I assume $\displaystyle N_H(K)=K$. I know I must get a contradiction here but I can't figure it out.
Any hints?
2. Originally Posted by ziggychick
I need to show subgroups of nilpotents groups are nilpotent.
I can get started but cannot finish.
Suppose G is a nilpotent group. Let H be a proper subgroup of G. Let K be a proper subgroup of H. I must show K is a proper subgroup of $\displaystyle N_H(K)$.
I assume $\displaystyle N_H(K)=K$. I know I must get a contradiction here but I can't figure it out.
Any hints?
what you're trying to prove doesn't prove that H is nilpotent unless G is finite. is G a finite group?
3. in this case, yes G is finite. didn't realize that was an issue.. :-/
4. Originally Posted by ziggychick
in this case, yes G is finite. didn't realize that was an issue.. :-/
yes, the "normalizer condition" that you're trying to prove is equivalent to "nilpotent" only for finite groups. but the fact is that a subgroup of any nilpotent group is nilpotent and we prove
results about nilpotent groups using central series not the normalizer condition!
5. do you only know the definition of "finite nilpotent groups" or you're also familiar with lower and upper central series, which give the definition of "nilpotent" for general (finite or infinite) groups?
6. my text is focused on finite nilpotent groups but it mentions that most of the results hold for infinite ones. i'm familiar with upper and lower central series, but we haven't really done any work with them.
7. Originally Posted by ziggychick
my text is focused on finite nilpotent groups but it mentions that most of the results hold for infinite ones. i'm familiar with upper and lower central series, but we haven't really done any work with them.
ok! can we use this theorem that a finite group G is nilpotent iff every Sylow subgroup of G is normal?
8. yes we can definitely use that.
i tried to go that route as well, but maybe i didn't think hard enough about it.
Suppose H is a proper subgroup of finite nilpotent group G. Let Q be a Sylow p-subgroup of H. Then we get Q is a subgroup of the Sylow p-subgroup P of G. We know P is normal in G, and is unique.
If $\displaystyle |Q|=p^k$ can P have another subgroup of the same order?
9. Originally Posted by ziggychick
yes we can definitely use that.
i tried to go that route as well, but maybe i didn't think hard enough about it.
Suppose H is a proper subgroup of finite nilpotent group G. Let Q be a Sylow p-subgroup of H. Then we get Q is a subgroup of the Sylow p-subgroup P of G. We know P is normal in G, and is unique.
If $\displaystyle |Q|=p^k$ can P have another subgroup of the same order?
well, $\displaystyle P \cap H$ is a p subgroup of $\displaystyle H$ and it contains $\displaystyle Q.$ thus $\displaystyle P \cap H=Q,$ because $\displaystyle Q$ is a Sylow p subgroup of $\displaystyle H.$ can you finish the proof now?
10. yes, i believe i can. thanks so much!
11. Originally Posted by NonCommAlg
the "normalizer condition" that you're trying to prove is equivalent to "nilpotent" only for finite groups.
just to correct myself (for the sake of completeness): the "normalizer condition" is equivalent to "nilpotent" in "finitely generated" groups and not only in finite groups. | 915 | 3,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-22 | latest | en | 0.956459 |
http://mathhelpforum.com/calculus/36977-solved-logarithmic-derivation.html | 1,529,790,814,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865250.0/warc/CC-MAIN-20180623210406-20180623230406-00339.warc.gz | 210,781,883 | 10,041 | 1. ## [SOLVED] Logarithmic derivation
Good Morning everybody,
I'd like to show the equality
$\displaystyle \frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^{ms}}$
By logarithmic differenciation(using Euler product for zeta), I get
$\displaystyle \frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}$
But I can't figure how to reach the last step, to the double sum.
Could someone help me ?
Thank you
2. Originally Posted by Klaus
Good Morning everybody,
I'd like to show the equality
$\displaystyle \frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^ms}$
By logarithmic differenciation(using Euler product for zeta), I get
$\displaystyle \frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}$
But I can't figure how to reach the last step, to the double sum.
Could someone help me ?
Thank you
Shouldnt it read $\displaystyle \frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^{ms}}$ ??
You are nearly there
Can you see the geometric progression below?
$\displaystyle -\sum_p \frac{p^{-s}}{1-p^{-s}}=-\sum_p \sum_m (p^{-s})^m$
3. Oh yeah, that's it !
I hadn't thought of geometric progessions
Thank you very much Isomorphism(what a cool name!) | 424 | 1,248 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-26 | latest | en | 0.763144 |
http://mathhelpforum.com/advanced-statistics/145290-two-means-one-sample-print.html | 1,524,629,172,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00168.warc.gz | 194,508,939 | 3,077 | # Two means in one sample
• May 18th 2010, 03:34 AM
Solphist
Two means in one sample
Hi (Hi),
I'm not sure if this belongs in the basic bit but the question is pretty simple...
I have some biological orientation data in a histogram that looks like it has two normally dsitributed means in it, they are spread out with one mean a good distance from the other.
Does anyone know how to test for each of the peaks in the data?
Is there a mathematical test for this,
a) Number of peaks in the histogram. (I could do this in code)
b) It's normally distributed.(Based soley on 1 image)
I could just separate them in code and test each peak within it's range, but I'm not sure if that is mathematically sound? Or whether that's a statistical no no :eek:...
Apparently this is called a bi-polar distribution, and I've been told it's ok to separate them, I'm still not 100% convinced though as it seems arbitrary.
Any help would be appreicated (Happy)
• May 22nd 2010, 08:59 PM
matheagle
Clearly, there can't be two means.
I'm guessing that you have two modes.
http://en.wikipedia.org/wiki/Bimodal_distribution
It's called bimodal and not bipolar.
I know too many people that can best be described as bipolar.
• May 24th 2010, 12:44 AM
Solphist
Thanks for the reply, I know what to search for now...
Thanks for clarifying my mistakes, mode not mean, and bimodal not polar.
hehe it's no wonder I couldn't find it looking for bipolar (Giggle)
If you happen to know a good book for this area I'd appreciate it (Bow) | 394 | 1,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.97192 |
http://www.weegy.com/?ConversationId=RCBTI8KP | 1,526,962,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864624.7/warc/CC-MAIN-20180522034402-20180522054402-00178.warc.gz | 478,364,576 | 8,618 | You have new items in your feed. Click to view.
Q: the accompanying graph shows all of the possibilities for the number of refrigerators and the number of TVs that will fit into an 18-wheeler.a)write an inequality to describe this region. b) will
the truck hold 71 refrigerators and 118 TVs? c) will the truck hold 51 refrigerators and 176 TVs? The Burbank buy more stores is going to make an order which will include at most 60 refrigerators. What is the maximum number of TVs which could also be delivered on the same 18-wheeler? Describe the restrictions this would add to the existing graph. The next day, the Burbank buy more decides they will have a television sale so they change their order to include at least 200 TVs. what is the maximum number of refrigerators which could also be delivered in the same truck? Describe the restrictions this would add to the original graph.
A: Answer is: Since you identify "The graph shows (0,330) and (110,0) ", we will assume this is a linear relationship and therefore the graph shows a straight line connecting those two points... [ otherwise, this answer is probably completely wrong we let 'r' represent refrigerators and 't' represent tvs and we assume that the point (0,330) indicates "330 tvs, zero refrigerators" and therefore the
point (110,0) indicates "zero tvs, 110 refrigerators we then have this linear relationship: t = -3r + 330 if we let r=71: -3(71) + 330 = 117 this means that if we have 71 refrigerators, then 117 tvs would be the maximum number allowed if we let r=51: -3(51) + 330 = 177 this means there would be room for 176 tvs if we have 51 refrigerators ]
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# 1. About Line Following and Maze Solving
Line following is a great introduction to robot programming, and it makes a great contest: it’s easy to build a line-following course, the rules are simple to understand, and it’s not hard to program a robot to follow a line. Optimizing your program to make your robot zoom down the line at the highest speed possible, however, is a challenge that can introduce you to some advanced programming concepts.
When you’re ready for a more difficult task, build a line maze for your robot to solve. You can start by solving the maze with a simple “left hand on the wall” strategy and move up to more and more advanced techniques to make your robot faster.
In this tutorial, we’ll show you how to build a good-looking line following course and a line maze, step by step. The materials you’ll need (posterboard and electrical tape) are easy to find and should cost you under \$10.
### Suggested line following contest rules:
The contest is a single-elimination tournament in which two robots race against each other at a time, on two separate courses that are mirror images of each other. In each match of the contest, a robot races once on each course, and the robot with the best time (out of both races) is declared the winner of the match. A starting line must be marked on each course, in a way that will not interfere with the robots’ sensors. For example, the boundary between two segments of the course could be used. Two timers are used to time each robot for three laps.
A race proceeds as follows:
1. The two robots are prepared for the race, including battery testing, calibration, etc.
2. Each robot is placed approximately 1’ behind its starting line.
3. The contestants wait for a starting signal, then start their robots simultaneously.
4. When the robots cross the starting line, their timers are started. The starting time might be slightly different for the two robots.
5. The time for three laps is measured, and the contestants stop their robots.
The robots must follow the line for the entire race. If there is ever a point in time when some part of the robot is not over the line, or if the robot skips a section of the course, it is disqualified for that race.
### Suggested line maze contest rules:
Each robot makes four timed attempts to solve the maze, and the best time out of all attempts wins. The idea is that the first attempt is for learning the maze, and the robot remembers the path that it learns for subsequent attempts.
The robot may not be given advance knowledge about the details of the maze (such as whether it is better to follow a left-turning or right-turning strategy). However, the operator of a robot may clear its memory if he suspects the course was learned incorrectly, or adjust parameters like the speed to try to get the best possible time.
As in the line following contest, the robot is required to stay on the line at all times and may not take shortcuts.
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https://datascience.stackexchange.com/questions/77921/test-for-feature-dependencies-in-time-series-modelling | 1,653,455,004,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00477.warc.gz | 239,817,344 | 65,934 | # Test for feature dependencies in time series modelling
I have time-series data that track event occurrence in 3 locations. Here's a sample:
Count Total
Location A B C
Date
2018-06-22 0 1 1 2
2018-06-23 2 1 0 3
2018-06-24 0 0 1 1
2018-06-25 2 2 1 5
2018-06-26 0 3 1 4
I would like to use the data to predict the total number of event occurrences at a given date in the future. How do I test if an event happening in one location has an impact on events happening in another location (dependency)? I believe that if an event happening in locations B and C are dependant, I should sum the 2 columns together as 1 feature in my model.
• A $$\chi$$-square test would tell you whether there is a significant difference between an observed variable (e.g. count in one location) and an expected variable (count in the other location). In other words, it can tell you whether the variables are independent or not.
• The conditional probability $$p(A|B)$$ of a variable A given the other variable B tells you how likely the event A is assuming the event B happens. $$A$$ and $$B$$ are independent if $$p(A|B)=p(A)$$ (note that it's unlikely to be exactly equal in the case of a real sample). | 348 | 1,265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.906722 |
https://www.remodelormove.com/do-you-center-pictures-to-the-wall-or-couch/ | 1,718,510,201,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00163.warc.gz | 850,274,908 | 21,798 | # Do you center pictures to the wall or couch?
When deciding how to center pictures to the wall or couch, it ultimately depends on the size and shape of the pictures, as well as the size and shape of the wall or couch. If you have a medium-sized piece of artwork that you would like to hang, then it is best to center it to the wall by using a picture frame and measuring the distance from both sides of the frame to ensure it is even.
If you have a large piece, then you may have to offset it a bit from the center to make it look balanced. For a couch, centering a piece of artwork or a decorative item should be based on the size of the piece and the width of the couch.
If it is a small item, then you could place it centered on a pillow or even near the center of the back of the couch, making sure it is balanced on all sides. Ultimately, the answer to centering pictures to the wall or couch depends on the size and shape of the piece and the wall/couch it will be placed on.
## How do you center art over a couch?
The best way to center art over a couch is to use the Rule of Thirds. This idea states that when you divide an image into nine equal parts, the four points of intersection are the best placements for the main point of interest.
To do this, measure the wall where you want your art to hang, and then divide that distance in three. Once you have determined the three spots, you can hang your art in the centermost spot. Alternatively, you can also use a tape measure or a level to draw an imaginary line across the couch, and then hang your art directly above the center of the couch.
This will help to ensure that the art is centered in the room. Additionally, it’s always important to consider the size of the artwork and the size of the couch. Make sure the artwork isn’t too overwhelming or too small in comparison to the couch.
## How far above a couch should a picture be hung?
The general rule of thumb is to hang artwork at eye-level, which is typically 56-58 inches from the floor to the center of the piece. However, this can vary depending on where the piece will be hung in relation to the couch and the size of the artwork.
If the artwork is large and will be hung above the couch, it should be placed at least six to twelve inches above the top of the back of the couch to give it the most impact and to prevent furniture from blocking the artwork.
Conversely, if the artwork is small, it can be hung a bit higher, around eight to ten inches above the back of the couch. Additionally, if the artwork is going to be hung above a loveseat or a pair of chairs, it should be hung about four to six inches higher than the back of the furniture to allow for visual balance and separation.
Lastly, if you want to create a grouping of artwork or decorative pieces, hang them so that the center is at eye-level and the groups fan out below it.
## What size picture do I need over a couch?
When it comes to size, it all depends on your own preference and the size of the couch you are hanging the picture over. Generally speaking, a good guideline to follow is to pick a picture that is no larger than two-thirds of the width of the couch itself.
For example, if your couch is 6 feet in width, then you should select a picture that is 4 feet in width or smaller. Additionally, make sure that the distance from the top of the couch to the bottom of the picture is less than 7 feet.
Too large of a picture can make your room appear smaller than it actually is, and too small of a picture may not be the proper accent for the size of your couch.
## What is the rule for hanging pictures?
There are some important rules you should follow when hanging pictures in your home.
1. Measure: Make sure to always measure twice, mark once! Measure the wall and where you plan on hanging the picture. Make sure to bring the picture and any mounting hardware to compare with the measurements.
2. Hang at Eye Level: Artwork should be hung at eye level. This means that an adult’s eye level should be 57″ from the floor or around 5 feet. To make sure you get the right placement, measure the center of the art and mark it with a pencil on the wall, ensuring your artwork is hung in the right height.
3. Groupings: If hanging several pieces, consider creating a group of artwork like a mini-gallery wall. Most standard homes will have walls 8 to 9 feet high and 12 to 14 feet across. Determine a desired shape of the grouping (vertical, horizontal or freeform), then determine the size of the overall group.
Create that same size on the floor prior to beginning work on the wall.
4. Consider Proximity: Generally, if you are hanging art in a room with furniture, it should be at least 6 to 8 inches away from the piece of furniture and 4 to 6 inches away from other pieces of artwork.
5. Level: Make sure your artwork is level! As a general rule, it is better to hang art too high than too low. Even though it is important to hang artwork at eye level, it is still important to consider the special aesthetic of the artwork.
Following these rules carefully will help to ensure that you hang your artwork correctly and safely!
## Where should pictures be placed on the wall?
The placement of pictures on a wall is largely a matter of personal preference. However, it’s generally a good idea to keep things balanced, with the same amount of space between the pictures.
In terms of height, the center of the picture should be at eye level, around 5 to 6 feet off the floor. It’s best to hang pictures in a slightly staggered arrangement with the center of each slightly higher than the last.
This will make the arrangement look more balanced.
Additionally, consider the shape of the pictures when arranging your wall art. You can create a pleasing visual effect by combining different shapes, such as squares and rectangles. Mixing in some circular or oval pictures can make the arrangement seem more interesting.
You can also use accent pieces such as clocks or mirrors to create depth and texture.
Creating a gallery wall is also very popular. This involves creating a cluster of pictures in the same area, arranged in groupings to create a cohesive look. When creating a gallery wall, you’ll want to make sure there is balance between the sizes, shapes, and colors of the pieces.
Ultimately, when deciding how to create your wall art arrangement, it’s important to focus on how the pieces look together as a whole. With some patience and creativity, you can create a beautiful wall art display that’s unique to your home.
## How much space should be between two pictures on a wall?
The ideal amount of space between two pictures on a wall depends on the size and style of the walls, as well as the size and style of the pictures. Generally, for standard size walls and similarly sized pictures, 4-6 inches of space is recommended.
This creates a balanced and visually pleasing display. If you are hanging large pictures on the wall, more space may be needed. A good rule of thumb is to use a distance of roughly one-third to one-half the width of the picture.
Alternatively, if you are hanging smaller pictures on the wall, such as 4 inches by 6 inches, you can reduce the spacing distance to 2-3 inches. With any spacing, aim for a resolution that looks good to the eye and creates a cohesive appearance of the pictures on the wall.
## Are pictures supposed to be hung at eye level?
The general rule of thumb when hanging pictures is to hang them at eye level. This ensures that the artwork can be comfortably viewed from both a standing and seated position, and that it will have the most impact on a viewer.
However, this rule does not need to be strictly adhered to; it is perfectly acceptable to hang pictures at different heights depending on the space and the desired effect. For example, some artwork can look particularly stunning when hung higher on a wall, especially if it is a large piece.
On the other hand, small pieces often look better when hung a bit lower. Ultimately, there is no definitive answer as to where artwork should be hung; it is up to the preferences and personal style of the individual.
## Should you put pictures on every wall?
No, you should not always put pictures on every wall. While pictures can add aesthetic appeal to your home and can be a stylish way to display art, it may not always be appropriate to have pictures on every wall.
Depending on the size and style of your room, a wall with no pictures may look more appealing. Additionally, placing pictures in strategic places can make the room look more interesting and balanced, rather than having pictures everywhere.
Ultimately, it is important to consider the overall feel of the room when deciding whether or not to put pictures on every wall. | 1,855 | 8,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.944659 |
http://forums.4aynrandfans.com/index.php?/topic/13953-self-funded-asteroid-mapping-project/ | 1,610,967,163,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00037.warc.gz | 41,163,411 | 13,475 | Followers 0
Self-funded asteroid mapping project
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Towing asteroids. That is way cool and so very rational. Hollywood well never make a move about how a small vessel towed an asteroid and saved the earth.
ruveyn
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I'm not sure I understand how you're using "towed", ruveyn. I may have missed something, but it seems the idea is to spin a satellite of some sort around an incoming body, using gravity to change the body's direction. Of course if its the right asteroid we can mine it, either reducing it to nothing or altering its path so we never have to deal with it again . . . (I'm sure I just typed out a Green nightmare, but what can you do.)
http://www.popularmechanics.com/science/space/news/how-to-mine-an-asteroid-11644811
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I'm not sure I understand how you're using "towed", ruveyn. I may have missed something, but it seems the idea is to spin a satellite of some sort around an incoming body, using gravity to change the body's direction. Of course if its the right asteroid we can mine it, either reducing it to nothing or altering its path so we never have to deal with it again . . . (I'm sure I just typed out a Green nightmare, but what can you do.)
http://www.popularmechanics.com/science/space/news/how-to-mine-an-asteroid-11644811
The t.v. bit I saw has an ion powered vessel of small mass sail beside the asteroid to be diverted. If the vessel can "herd" the asteroid over say a period of 15 to 20 years in a particular direction then the asteroid will miss its mark (us). At least that was the theory. From a purely gravitational point of view it makes sense. The small vessel exerts a force on the asteroid. It does not have to be a big force but it has to be exerted over an extended period of time which implies an extended distance over with the force is exerted F x distance is energy and that is how you put a lot of energy (aka. work) on the asteroid to pull it into a better trajectory.
ruveyn
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Towing asteroids. That is way cool and so very rational. Hollywood well never make a move about how a small vessel towed an asteroid and saved the earth.
ruveyn
I'm not sure I understand how you're using "towed", ruveyn. I may have missed something, but it seems the idea is to spin a satellite of some sort around an incoming body, using gravity to change the body's direction. Of course if its the right asteroid we can mine it, either reducing it to nothing or altering its path so we never have to deal with it again . . . (I'm sure I just typed out a Green nightmare, but what can you do.)
http://www.popularmechanics.com/science/space/news/how-to-mine-an-asteroid-11644811
My understanding is that one pulls up alongside the rock, and uses enough power to keep from 'sticking' to it. As you slowly pull away, you drag the rock with you.
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Jules Verne had an interesting idea about what we could get from a meteor in The Hunt for the Meteor: precious metals.
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Jules Verne had an interesting idea about what we could get from a meteor in The Hunt for the Meteor: precious metals.
If we had decent propulsion systems for space vehicles, mining the Asteroids could eventually be done. Unfortunately our space vehicles are propelled by rockets producing hot gases by way of oxidation reactions. This technology is no more advanced -- in principle -- than the solid rockets invented by the Chinese during the Tang Dynasty. The best we can do with this kind of propulsion is burn then coast (i.e. move in free fall in some gravitational field). That means trips to Mars, even at close conjunction will take 9 months in free fall (sometimes called zero-g). That is plenty long enough for the crew'sw bones to become brittle, their muscles atrophy and their bodies bombarded with cosmic rays. A long distance trip outside of our magnetosphere with current and even likely technology is a suicide mission.
What we need is propulsion system than can emit reactants over a very long period of time, (or use light pressure from the sun as a propulsion) and thus achieve speed 2 to 3 orders of magnitude greater than we can achieve now. This would make a trip to the Asteroids a matter of weeks rather than the best part of a year. Also we need designs that produce pseudo-gravity by means of centrifugal force and vessels that are well insulated against cosmic rays. Until that happens we are NOT going to become a real space faring culture. With current technology long trips to beyond the orbit of Mars is more difficult than paddling the width of the Pacific Ocean in a dugout log canoe.
ruveyn
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I'm not sure I understand how you're using "towed", ruveyn...
... From a purely gravitational point of view it makes sense. The small vessel exerts a force on the asteroid. It does not have to be a big force but it has to be exerted over an extended period of time which implies an extended distance over with the force is exerted F x distance is energy and that is how you put a lot of energy (aka. work) on the asteroid to pull it into a better trajectory...
Orbiting masses are 'towed' by what you acknowledge is a gravitational force rather than falling into a gravitational mathematical ditch? :-)
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I'm not sure I understand how you're using "towed", ruveyn...
... From a purely gravitational point of view it makes sense. The small vessel exerts a force on the asteroid. It does not have to be a big force but it has to be exerted over an extended period of time which implies an extended distance over with the force is exerted F x distance is energy and that is how you put a lot of energy (aka. work) on the asteroid to pull it into a better trajectory...
Orbiting masses are 'towed' by what you acknowledge is a gravitational force rather than falling into a gravitational mathematical ditch? :-)
In this instance using the Newtonion locution is easier. The General Theory of Relativity will also explain a "tow" but the mathematics are horrendous.
Sometimes it make more sense to regard gravitation as a force somewhat akin to electro-magnetic forces.
ruveyn | 1,394 | 6,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-04 | latest | en | 0.947582 |
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The average of a set a) mode b) median c) mean d) range the number(s0 that occur most often a) mode b) median c) mean d) range The middle number in a set of numbers a) mode b) median c) mean d) range The largest number - the smallest number (distance frome ach other on a number line) a) mode b) median c) mean d) range This displays countable data with symbols or pictures a) line graph b) bar graph c) pictograph d) circle graph This uses horizontal or vertical bars to display countable data a) line graph b) bar graph c) pictograph d) circle graph Shows how parts of the data are related to the whold and each other a) line graph b) bar graph c) pictograph d) circle graph This is a pattern over time a) herringbone b) plaid c) trend d) style Part of a circle graph a) section b) line c) picture d) trend To place a point on a graph is called a) drawing points b) plotting points c) sketching points d) scattering points
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Question
# A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. The wavelength of the fourth harmonic of the fundamental will be
A
80 cm
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B
60 cm
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C
40 cm
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D
20 cm
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Solution
## The correct option is D 40 cmm= mass per unit length=4×10−380×10−2=0.005Kg/mgiven T=50NL=80cm=0.8m∴v=√Tm=√500.005=100m/secFundamental frequency f0=12L√Tm=12×0.8√500.005=625Hz∴f4 Frequency of fourth harmonic 4f0=4×62.5=250HzAs we knowv4=f4λ4∴λ4=v4f4=100250=0.4m=40cm
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Join BYJU'S Learning Program | 348 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.752628 |
https://denisegaskins.com/2020/09/02/prime-factor-art-on-a-hundred-chart/ | 1,685,432,696,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00716.warc.gz | 243,984,278 | 31,039 | # Prime Factor Art on a Hundred Chart
The best way to practice math is to play with it — to use the patterns and connections between math concepts in your pursuit of something fun or beautiful.
So this art project is a great way to practice multiplication. Use the prime factors of numbers from one to one hundred to create a colorful design.
First, download this printable file of hundred charts in non-photo blue (or light gray, if you’re printing in grayscale). The file includes:
• Line-by-line traditional chart, counting from top to bottom.
• Line-by-line bottom’s-up chart, counting from bottom to top.
• Ulam’s Spiral chart, spiraling out from the center.
• Blank grids for making your own patterns.
Decide whether you want to work in straight lines or in a spiral — or in any random pattern you like. Print the hundred chart that fits your artistic vision.
Or, if you feel especially creative, use a long ruler or drafting square to draw a grid on art paper. Make it as big as you like — you’ll never run out of numbers!
### Prime Factor Coloring Rules
Choose a neutral color for the 1 square. Then choose brighter, distinct colors for 2 and 3, the first two prime numbers. Pick your favorite color for 2 because it will show up the most in your design.
Color the composite numbers according to their prime factors. For example:
• 4 = 2 × 2, so divide the square into two parts and color both parts with your color for the number 2.
• 5 is prime, so it gets a new color of its own.
• 6 = 2 × 3, so divide the square into two parts and color one part with your color for 2 and the other for 3.
• 7 is prime, so it gets a new color of its own.
• 8 = 2 × 2 × 8, so divide the square into three parts and color each of them with your color for 2.
• 9 = 3 × 3, so divide the square into two parts and color both parts with your color for 3.
• 10 = 2 × 5, so divide the square into two parts and color one part with your color for 2 and the other for 5.
• 11 is prime, so it gets a new color of its own.
• 12 = 2 × 6 = 2 × 2 × 3, so divide the square into three parts and color two of the parts for 2 and the other for 3.
Continue along the chart, figuring out the prime factors of each number (the smallest numbers you can multiply to get it, not counting 1). Split that number’s square so you can color one section for each factor.
### More Ways to Play
For students who haven’t studied prime factors before, here is a fun way to teach them:
Not feeling artistic? Try these math games:
And check out all the wonderful ways to play with a hundred chart in this reader-favorite post:
### Credits
This math art project was inspired by Natalie Wolchover’s beautiful spiral creation:
Feature photo (top) by Washington Oliveira 🇧🇷 via Unsplash.com.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 689 | 2,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-23 | latest | en | 0.909259 |
http://docs.php.net/manual/uk/function.gmp-invert.php | 1,670,352,047,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711111.35/warc/CC-MAIN-20221206161009-20221206191009-00383.warc.gz | 14,150,508 | 7,109 | # gmp_invert
(PHP 4 >= 4.0.4, PHP 5, PHP 7)
gmp_invertInverse by modulo
### Опис
GMP gmp_invert ( GMP `\$a` , GMP `\$b` )
Computes the inverse of `a` modulo `b`.
### Параметри
`a`
Either a GMP number resource in PHP 5.5 and earlier, a GMP object in PHP 5.6 and later, or a numeric string provided that it is possible to convert the latter to a number.
`b`
Either a GMP number resource in PHP 5.5 and earlier, a GMP object in PHP 5.6 and later, or a numeric string provided that it is possible to convert the latter to a number.
### Значення, що повертаються
A GMP number on success or `FALSE` if an inverse does not exist.
### Приклади
Приклад #1 gmp_invert() example
``` <?phpecho gmp_invert("5", "10"); // no inverse, outputs nothing, result is FALSE\$invert = gmp_invert("5", "11");echo gmp_strval(\$invert) . "\n";?> ```
Наведений вище приклад виведе:
```9
```
``` Example #2 gmp_invert() example<?phpecho gmp_invert("5", "10"); // no inverse, outputs nothing, result is FALSEIt means (5 * x ) mod 10 = 1. And with this function we want a value for x, because of mod 10, x should be {1.. 10-1(9)}, so :5 * 1 mod 10 = 5 5 * 2 mod 10 = 05 * 3 mod 10 = 55 * 4 mod 10 = 05 * 5 mod 10 = 55 * 6 mod 10 = 05 * 7 mod 10 = 55 * 8 mod 10 = 05 * 9 mod 10 = 5We don't have any 1 in the results. so it will be False.\$invert = gmp_invert("5", "11");echo gmp_strval(\$invert) . "\n";?>The above example will output:95 * 9 mod 11 = 1 ``` | 525 | 1,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | latest | en | 0.266003 |
https://www.convertunits.com/from/gram/millilitre/to/milligram/cubic+meter | 1,695,550,384,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00807.warc.gz | 799,970,272 | 12,549 | ## Convert gram/millilitre to milligram/cubic metre
gram/millilitre milligram/cubic meter
How many gram/millilitre in 1 milligram/cubic meter? The answer is 1.0E-9. We assume you are converting between gram/millilitre and milligram/cubic metre. You can view more details on each measurement unit: gram/millilitre or milligram/cubic meter The SI derived unit for density is the kilogram/cubic meter. 1 kilogram/cubic meter is equal to 0.001 gram/millilitre, or 1000000 milligram/cubic meter. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between grams/milliliter and milligrams/cubic meter. Type in your own numbers in the form to convert the units!
## Quick conversion chart of gram/millilitre to milligram/cubic meter
1 gram/millilitre to milligram/cubic meter = 1000000000 milligram/cubic meter
2 gram/millilitre to milligram/cubic meter = 2000000000 milligram/cubic meter
3 gram/millilitre to milligram/cubic meter = 3000000000 milligram/cubic meter
4 gram/millilitre to milligram/cubic meter = 4000000000 milligram/cubic meter
5 gram/millilitre to milligram/cubic meter = 5000000000 milligram/cubic meter
6 gram/millilitre to milligram/cubic meter = 6000000000 milligram/cubic meter
7 gram/millilitre to milligram/cubic meter = 7000000000 milligram/cubic meter
8 gram/millilitre to milligram/cubic meter = 8000000000 milligram/cubic meter
9 gram/millilitre to milligram/cubic meter = 9000000000 milligram/cubic meter
10 gram/millilitre to milligram/cubic meter = 10000000000 milligram/cubic meter
## Want other units?
You can do the reverse unit conversion from milligram/cubic meter to gram/millilitre, or enter any two units below:
## Enter two units to convert
From: To:
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 601 | 2,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-40 | latest | en | 0.583962 |
http://stackoverflow.com/questions/7715406/how-can-i-efficiently-transform-a-numpy-int8-array-in-place-to-a-value-shifted-n?answertab=votes | 1,406,947,094,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510276250.57/warc/CC-MAIN-20140728011756-00460-ip-10-146-231-18.ec2.internal.warc.gz | 272,145,073 | 15,896 | # How can I efficiently transform a numpy.int8 array in-place to a value-shifted numpy.uint8 array?
I have a large numpy array of signed bytes (`dtype int8`). It contains values in the full range -128 to +127. I'd like to convert the efficiently to an array of unsigned bytes (`dtype uint8`) by adding 128 to each element, such that -128 → 0, 0 → 128, +127 → 255, etc. so of course the results still fit into an unsigned byte.
Simple elementwise addition given the correct numerical result, but creates a result array using twice the memory (`dtype int16`) in addition to the source array, even though only the low bytes of the result elements are needed.
``````>>> import numpy
>>> a = numpy.array( [-128, -1, 0, 1, 127 ], dtype=numpy.int8)
>>> b = a + 128
>>> b
array([ 0, 127, 128, 129, 255], dtype=int16)
``````
Is there a way to control the `dtype` of the result array to be `uint8`?
The alternative approach of modifying the values in-place and "casting" the data to a new type, like this:
``````>>> for i in xrange(0, 5):
... if a[i] < 0:
... a[i] -= 128
... elif a[i] >= 0:
... a[i] += 128
...
>>> a
array([ 0, 127, -128, -127, -1], dtype=int8)
>>> a.view(dtype=numpy.uint8)
array([ 0, 127, 128, 129, 255], dtype=uint8)
``````
is much more space efficient but very costly in time for large arrays with the transformation in Python.
How can I do this transformation in-place and quickly?
-
```import numpy as np
a = np.array([-128, -1, 0, 1, 127], dtype=np.int8)
a = a.view(np.uint8)
a += 128
print a
# -> array([ 0, 127, 128, 129, 255], dtype=uint8)
```
This creates no copies, and all operations are in-place.
EDIT: safer to cast first to uint --- unsigned wrap-around is defined. EDIT2: s/numpy/np/g;
-
Almost perfect. Just change `numpy.int8` to `np.int8` in the second line. – Petr Viktorin Oct 10 '11 at 16:02
Thank you. It didn't occur to me to exploit the unsigned wrap around combined with the add-assign operator to have some control over the type of the result of the sum. – Rob Smallshire Oct 10 '11 at 18:15
``````In [18]: a = numpy.array( [-128, -1, 0, 1, 127 ], dtype=numpy.int8)
In [19]: z = a.view(dtype=numpy.uint8)
In [20]: z += 128
In [21]: z
Out[21]: array([ 0, 127, 128, 129, 255], dtype=uint8)
``````
I hope I haven't misunderstood the requirements.
- | 720 | 2,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2014-23 | latest | en | 0.775793 |
https://www.wikibacklink.com/search/538 | 1,639,004,307,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00279.warc.gz | 1,185,118,336 | 18,167 | # Keyword Analysis & Research: 538
## Keyword Research: People who searched 538 also searched
What does 538 stand for?
- Definition of 538 WI - 538 WI stands for Madison.
What does the number 538 mean?
Angel number 538 is a sign from the guardian spirits advising you to begin living your life consciously aware that others are watching. Constantly learn to lead by example . Do not be afraid to make mistakes but learn to do this by growing and reminding yourself that you are not perfect.
What are the factors of 538?
Factors of 538 are 1, 2, 269, 538. So, 538 can be derived from smaller number in 2 possible ways using multiplication. Factors of 538 are 1, 2, 269, 538. | 167 | 680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.935698 |
http://listserv.uga.edu/cgi-bin/wa?A2=ind9604d&L=sas-l&D=0&P=28042 | 1,369,408,340,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704666482/warc/CC-MAIN-20130516114426-00031-ip-10-60-113-184.ec2.internal.warc.gz | 154,482,484 | 4,272 | ```Date: Fri, 26 Apr 1996 16:14:37 GMT Reply-To: Michael Friendly Sender: "SAS(r) Discussion" From: Michael Friendly Organization: York University, Ontario, Canada Subject: Lags with BY variables: Merge seeks help, possibly from SQL Many thanks to all those who answered my query re constructing lags with BY variables. Here's a new problem for anyone willing to help, but first I'll explain the old problem and the context: The simplest solutions to make the lags restart when the BY variable(s) change were variations of the code below to produce the dataset lags data codes; input subj @; do i=1 to 16; input code \$ @; output; end; cards; 1 c a a b a c a c b b a b a a b c 2 c c b b a c a c c a c b c b c c ; data lags; set codes; by subj; drop cnt; lag1 = lag(code); lag2 = lag2(code); if first.subj then cnt=0; cnt+1; if cnt<=1 then lag1 = ' '; if cnt<=2 then lag2 = ' '; The goal here (for a macro I'm working on) is to produce a complete n-way frequency table giving joint frequencies of events for an arbitrary number of lags. By 'complete' I mean (for 2 subjects, 3 codes, lag1 x lag0), a dataset with 2 x 3 x 3 observations, including those which occur with zero frequencies. Unfortunately, it does not appear that I can convince PROC FREQ to include the zero counts in the output dataset. proc freq data=lags; tables lag1 * code / noprint out=freq; by subj; proc print; id subj; by subj; which gives only counts >=1: SUBJ LAG1 CODE COUNT PERCENT 1 c 1 . a a 2 13.3333 a b 3 20.0000 a c 2 13.3333 b a 3 20.0000 b b 1 6.6667 b c 1 6.6667 c a 2 13.3333 c b 1 6.6667 2 c 1 . a c 3 20.0000 b a 1 6.6667 b b 1 6.6667 b c 2 13.3333 c a 2 13.3333 c b 3 20.0000 c c 3 20.0000 My original idea to fill in the zero entries was to construct a dataset of all zeros and merge it with the freq datset for each subject. That is, I want all combinations of a,b,c in pairs filled in with observations of COUNT=0 for those combinations which do not occur. This worked when there were no BY variables for the lag dataset, but does not work now. Here's the simplified verion of the code: *-- Sort, putting subject last; proc sort data=freq; by lag1 code subj; *-- Create a table of 0 counts to fill in missing entries; data zeros; do lag1='a', 'b', 'c'; do code='a', 'b', 'c'; count = 0; end; end; data freq; merge freq(drop=percent) zeros; by lag1 code; if lag1 = ' ' then delete; *-- Resort, putting subject first; proc sort data=freq; by subj lag1 code; proc print; id subj; by subj; and the output: SUBJ LAG1 CODE COUNT 1 a a 2 a b 3 a c 2 b a 3 b b 1 b c 1 c a 2 c b 1 2 a c 3 b a 1 b b 1 b c 2 c a 2 c b 3 c c 0 Using SET instead of MERGE above gives a different, but incorrect result. The whole job of filling in the zero entries BY subject seems like something that could be much more easily accomplished with PROC SQL, but I'm not familiar enough with it to see how. Can anyone help? -- Michael Friendly Internet: friendly@hotspur.psych.yorku.ca (NeXTmail OK) Psychology Department York University Voice: 416 736-5118 4700 Keele Street http://www.math.yorku.ca/SCS/friendly.html Toronto, ONT M3J 1P3 CANADA ```
Back to: Top of message | Previous page | Main SAS-L page | 969 | 3,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2013-20 | latest | en | 0.811684 |
http://www.worldplenty.com/g71n_mx1.htm | 1,669,949,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00495.warc.gz | 101,917,542 | 3,010 | Multiplying Scientific Notation Numbers How to multiply scientific notation numbers: Multiply the base numbers Add the exponents of the tens Adjust the base number to have one digit before the decimal point by raising or lowering the exponent of the tens Return to Top
#### Multiplying Scientific Notation Numbers
x10 * x10 = x10 | 66 | 332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.692953 |
https://mathematica.stackexchange.com/questions/256109/integral-in-ndsolve-boundary-conditions-and-initial-conditions-cant-both-satis?noredirect=1 | 1,713,507,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00833.warc.gz | 351,048,375 | 41,104 | # Integral in NDSolve, boundary conditions and initial conditions can't both satisfied
I have serval reaction-diffusion equations to be solved. Which are equivalent to the following problem:
First is a simple diffusion equation of $$Z(t,r)$$ like below: $$\dfrac{\partial Z}{\partial t}=A(t,r)\dfrac{\partial^2 Z}{\partial r^2}$$
Then $$Z(t,r)$$ is actually defined from: $$Z(t,r) \equiv \dfrac{\partial V(t,r)}{\partial r}$$ In my problem, I must have the value of $$V(t,r)$$ at every $$(t,r)$$ to determine the value of, for example, $$A(t,r)=F(V(t,r))$$. Bondary condtions are $$V(t,0)=0,V(t,1)=0,\\ Z_r(t,0)=0,Z_r(t,1)=0$$
For testing, I just set $$A$$ a constant, and tried something like below:
A = 0.01; L = 1; tmax = 10;
eqns = {V[t, r] == Integrate[Z[t, r], r],
D[Z[t, r], t] - A D[Z[t, r], {r, 2}] == NeumannValue[0, r == 0 || r == 1]};
INIs = {Z[0, r] == 2 Pi Cos[2 Pi r]};
BCs = {DirichletCondition[V[t, r] == 0, r == 0 || r == 1 ]};
sys = {eqns, BCs, INIs};
sol = NDSolve[sys, {V, Z}, {r, 0, L}, {t, 0, tmax}]
The codes run successfully without warning, but the results are quite strange: In the result $$Z(t, r)$$ is correctly calculated, while $$V(t, r)$$ is not correct. Does anyone know why it behaviors like this? Or any other way to do deal with this problem? Thanks a lot.
# Update
I found that this system is solved by NDSovle as differential algebraic equations. I tried index reduction of DAE as described in this. Here are the codes:
A = 0.01; L = 1; tmax = 10;
eqns = {V[t, r] == Integrate[Z[t, r], r],
D[Z[t, r], t] - A D[Z[t, r], {r, 2}] ==
NeumannValue[0, r == 0 || r == L]};
INIs = {Z[0, r] == 2 Pi Cos[2 Pi r],
V[0, r] == Sin[2 Pi r]};
BCs = {DirichletCondition[V[t, r] == 0, r == 0 || r == L]
};
sys = {eqns, BCs, INIs};
sol = NDSolve[sys, {V, Z}, {r, 0, L}, {t, 0, tmax},
Method -> {"IndexReduction" -> {"StructuralMatrix"}}]
And the results are different.
• In the previous codes, all BCs are satisfied, but initial value of $$V(t,r)$$ is not I wanted.
• In the updated codes, with index reduction. I can add the initial value of $$V(t,r)$$, which is satisfied by the result. But one of the boundary conditions of $$Z(t,r)$$ is violated.
Why can't it satisfy both ICs and BCs as I gave?
• > the results are quite strange... Can you clarify why you think the results are strange? At least I don't have physical intuition on this problem and the plot seems rather ok. What is the issue exactly? Sep 25, 2021 at 12:20
• @HansOlo, the two equations above are decoupled since I set A as a constant. First, the equation for $Z(t, r)$ is simply a diffusion equation, in which the initial value of a Cos function will be smoothed and finally decays to 0. So the first figure is correct. But then the integral of a Cos[r] in space should be Sin[r]. Therefore the second plot is not correct. You can found those strange bump up around $t=0$ and $r\approx 1$ in the second plot. Sep 25, 2021 at 13:12
• @YanQH It is not clear why NDSolve getting solution for Z since there is no any special algorithm to solve mixture integral and differential equations, see, for example, my post about it on mathematica.stackexchange.com/questions/217201/… Sep 26, 2021 at 7:16
• @AlexTrounev NDSolve seems like solved the system as DAEs, and satisfy those boundary condition strictly. But the initial value of $V(t, r)$ isn't what I expected. As for your post you refereed, I think there might be an error in the integral. Integrate[W[x, y, t], {y, 0, y}] doesn't make sense to me. I would write it as Integrate[W[x, yy, t], {yy, 0, y}] or Integrate[W[x, y, t], {y, 0, L}] depending on the problem. Sep 26, 2021 at 12:29
• @YanQH There is no special algorithm to solve mixture from integral and differential equations in NDSolve. But we can try to implement some method like FDM or Method of Lines for this case. See my answer with FDM algorithm. Sep 29, 2021 at 16:41
We can recommend FDM algorithm of 8 order with using DAE solver as follows
Clear["Global*"]
n = 137; grid = Range[0, n]/n;
fd1 = NDSolveFiniteDifferenceDerivative[Derivative[1], grid,
DifferenceOrder -> 8]; m1 = fd1["DifferentiationMatrix"]; fd2 =
NDSolveFiniteDifferenceDerivative[Derivative[2], grid,
DifferenceOrder -> 8]; m2 = fd2["DifferentiationMatrix"]; varz =
Table[z[i][t], {i, Length[grid]}]; varz1 =
Table[z[i]'[t], {i, Length[grid]}]; zxx = m2 . varz; varZ =
Table[z[i], {i, Length[grid]}]; varv =
Table[v[i][t], {i, Length[grid]}]; vx = m1 . varv; varV =
Table[v[i], {i, 2, n}];
A = .01; v[1][t_] := 0;
v[n + 1][t_] := 0; eq1 =
Table[z[i]'[t] - A zxx[[i]] == 0, {i, 2, n, 1}]; eq2 =
Table[z[i][t] - vx[[i]] == 0, {i, 2, n, 1}]; ic =
Table[z[i][0] == 2 Pi Cos[2 Pi grid[[i]]], {i, 2, n}]; ic1 =
Table[v[i][0] == Sin[2 Pi grid[[i]]], {i, 2, n}]; bc = {z[1][t] ==
z[2][t], z[n + 1][t] == z[n][t]};
eqn = Join[eq1, eq2, ic, ic1, bc]; var = Join[varV, varZ];
sol = NDSolve[eqn, var, {t, 0, 10},
Method -> {"EquationSimplification" -> "Residual"}];
Visualization
lst = Flatten[
Table[{t, grid[[i]], z[i][t] /. sol[[1]]}, {t, 0, 10, .1}, {i,
Length[grid]}], 1];
ListPlot3D[lst, Mesh -> None, ColorFunction -> Hue, PlotRange -> All,
AxesLabel -> {"t", "x", "Z"}]
lst1 = Flatten[
Table[{t, grid[[i]], v[i][t] /. sol[[1]]}, {t, 0, 10, .1}, {i,
Length[grid]}], 1];
ListPlot3D[lst1, Mesh -> None, ColorFunction -> Hue, PlotRange -> All,
AxesLabel -> {"t", "x", "V"}]
`
• Thank you very much for such detailed answer. Since I only started to use MMA 1 year ago, I still have many things to learn. Sometimes, procedures of NDSolve work like black boxes. So I am curious and think there might be a chance that it will work for this special case. Anyway, a FDM algorithm like yours could be used more universally, and also more suitable for my real problem. It's very helpful for me to see a workable example with MMA's FDM packages. I can learn to build my algorithm from it. Thank you again! Sep 30, 2021 at 16:07
• @YanQH You are welcome! Oct 1, 2021 at 6:21 | 2,000 | 5,966 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-18 | latest | en | 0.861414 |
https://towardsunlight.com/nova-scotia/first-law-of-thermodynamics-and-their-applications.php | 1,618,711,802,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464146.56/warc/CC-MAIN-20210418013444-20210418043444-00618.warc.gz | 649,763,006 | 8,404 | # And thermodynamics applications their law of first
12.2 the first law of thermodynamics and some simple. Application of first law of thermodynamics. process at constant volume; according to the first law of thermodynamics. list of laboratory apparatus and their uses..
## Biochemical Thermodynamics Jones & Bartlett Learning
Thermodynamics Questions and Answers Sanfoundry. Under its more formal name of the first law of thermodynamics, it governs all aspects of energy in science and engineering applications. if their values, first law of thermodynamics: applications. they can then be cooled further by their own expansion the general expression of the first law of thermodynamics.
### Biochemical Thermodynamics Jones & Bartlett Learning
Conceptual questions The first law of thermodynamics and. 21/01/2018в в· thermodynamics is the study of heat & energy transfer. learn about the laws of thermodynamics, their applications & the different state functions on which the, but their difference does not. their difference does not. the first law can be construed to be a вѕso we must know how the first law of thermodynamics can be.
Although q and w are not state functions on their own, their sum the first law of thermodynamics states that the energy of the universe is constant. applications of thermodynamics: the first law of thermodynamics and some simple processes of the first law of thermodynamics is that machines can be
Under its more formal name of the first law of thermodynamics, it governs all aspects of energy in science and engineering applications. if their values thermodynamics - fundamentals and its application in application of thermodynamics and kinetics in registered as vat taxable entities in their own eu
Chapter 3 - the first law of thermodynamics. pages and planetary scientists seeking a review of thermodynamic principles and their application to a specific engineering thermodynamics/applications. the liquid and vapor states merge together and all their thermodynamic properties become the from the first law,
Although q and w are not state functions on their own, their sum the first law of thermodynamics states that the energy of the universe is constant. thermodynamics is the branch of physics concerned with heat and temperature and their the initial application of thermodynamics the first law of
Engineering thermodynamics/applications. the liquid and vapor states merge together and all their thermodynamic properties become the from the first law, first law of thermodynamics and its applications problems from iit jee. problem (iit jee 2014): a thermodynamic system is taken from an initial state i with internal
Describe thermodynamic concepts and their applications. o extend the first law of thermodynamics to various daily life activities. o identify the... applications of thermodynamics. 1) all types of vehicles that we use, cars, motorcycles, trucks, ships, aeroplanes, and many other types work on the basis of second
First law of thermodynamics and its applications problems from iit jee. problem (iit jee 2014): a thermodynamic system is taken from an initial state i with internal the first law. the 1st law of thermodynamics states that internal energy the first law of thermodynamics is a version of both applications of the first law
First Law of Thermodynamics- Internal Energy (in Telugu. The first law of thermodynamics is a version of the law of an introductory treatise dealing mainly with first principles and their direct applications, b. g, вђ“ anyone wishing to sharpen their knowledge of thermodynamics subject questions & answers on first law of thermodynamics . entropy applications and.
## (Solved) Describe thermodynamic concepts and their
Biochemical Thermodynamics Jones & Bartlett Learning. First law of thermodynamics and internal energy discussed briefly. first law of thermodynamics and its applications. 13 this is a statement of the first law, 2. 1 first law of thermodynamics [vw, s & b: 2.6] we will spend most of the course dealing with various applications of the first law -- in one form or another..
## What are the implications of the First Law of Thermodynamics?
Practical Chemical Thermodynamics for Geoscientists. First law of thermodynamics and internal energy discussed briefly. first law of thermodynamics and its applications. 13 this is a statement of the first law https://en.m.wikipedia.org/wiki/Thermodynamic_equilibrium 4:04 application of the first law; referred to as the first law of thermodynamics or the law of energy helped over half a million teachers engage their.
• 12.2 The first law of thermodynamics and some simple
• (Solved) Describe thermodynamic concepts and their
• Ucl wiki. ucl wiki. spaces; the first law of thermodynamics was derived the streams of material crossing the control surface must not change their state or first law of thermodynamics and its applications problems from iit jee. problem (iit jee 2014): a thermodynamic system is taken from an initial state i with internal
4:04 application of the first law; referred to as the first law of thermodynamics or the law of energy helped over half a million teachers engage their application of first law of thermodynamics. process at constant volume; according to the first law of thermodynamics. list of laboratory apparatus and their uses.
The first significance is related with law of conservation of energy, heat and work both are different forms of same entity known as energy which is conserved this вђ“ anyone wishing to sharpen their knowledge of thermodynamics subject questions & answers on first law of thermodynamics . entropy applications and
Chapter 3 - the first law of thermodynamics. pages and planetary scientists seeking a review of thermodynamic principles and their application to a specific under its more formal name of the first law of thermodynamics, it governs all aspects of energy in science and engineering applications. if their values
Engineering thermodynamics/applications. the liquid and vapor states merge together and all their thermodynamic properties become the from the first law, thermodynamics and its applications вђ“ an overview by abstract: the laws of thermodynamics provide an elegant mathematical their values can be calculated
Examples of the first law of thermodynamics energy flow in a diesel engine when an engine burns fuel it converts the energy stored in the fuel's chemical bonds into application of first law of thermodynamics. process at constant volume; according to the first law of thermodynamics. list of laboratory apparatus and their uses.
The first law of thermodynamics, the application of thermodynamic principles begins by defining a system the molecules can rotate about their centre first law of thermodynamics and its applications problems from iit jee. problem (iit jee 2014): a thermodynamic system is taken from an initial state i with internal
The first law of thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. the first law of thermodynamics . the first law may be stated as and leaving the system is a very common phenomenon in most of the engineering applications.
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