{"product_id":"mathematical-recreations-containing-solutions-of-ai-ja","title":"Mathematical recreations, containing solutions of many very difficult and important equations, and of several useful problems in geometry, surveying and astronomy, together with a method of finding roots and equations by projection — 数学遊戯","description":"\u003cp style=\"font-weight:bold;margin:0 0 .6em;\"\u003e全355ページ\u003c\/p\u003e\u003cp\u003eMathematical recreations, containing solutions of many very difficult and important equations, and of several useful problems in geometry, surveying and astronomy, together with a method of finding roots and equations by projection（数学遊戯）の日本語版です。分野は「科学遊戯・パズル」です。AI翻訳をもとに校閲・編集しました（全355ページ）。\u003c\/p\u003e\u003cp\u003e会員登録とメルマガ配信登録（どちらも無料）をすると、ページ下部のビューアでオンライン閲覧できます。この商品をご購入いただくと、文字情報入りのPDF（355ページ）をダウンロードできます。\u003c\/p\u003e\u003cp style=\"margin:1em 0 0;\"\u003e\u003ca href=\"\/en-intl\/customer_authentication\/login?return_to=%2Fproducts%2Fmathematical-recreations-containing-solutions-of-ai-ja\" style=\"display:inline-block;padding:.7rem 1.4rem;background:#334FB4;color:#fff;text-decoration:none;border-radius:4px;margin:0 .5rem .5rem 0;font-weight:bold;\"\u003e会員登録はこちら\u003c\/a\u003e\u003ca href=\"#cld-gate\" style=\"display:inline-block;padding:.7rem 1.4rem;background:#334FB4;color:#fff;text-decoration:none;border-radius:4px;margin:0 .5rem .5rem 0;font-weight:bold;\"\u003eメルマガ配信登録はこちら\u003c\/a\u003e\u003c\/p\u003e\u003cdetails style=\"margin-top:1.2em;border:1px solid #ddd;border-radius:6px;padding:.6em 1em;\"\u003e\u003csummary style=\"cursor:pointer;font-weight:bold;\"\u003e内容の要約を読む\u003c\/summary\u003e\u003cdiv style=\"margin-top:1em;font-size:.95em;line-height:1.9;\"\u003e\n\u003cp\u003e※全文は文字数が多いため、各項目の冒頭部分を抜粋した要約を掲載しています。全文はページ下部のビューアでお読みいただけます。\u003c\/p\u003e\n\u003ch4\u003e1\u003c\/h4\u003e\n\u003cp\u003eMathematical recreations, containing solutions of many verydifficult and important equations, and…\u003c\/p\u003e\n\u003ch4\u003e数式・計算（OCR転写）：From [1], we find f-Cf = (i-l)i\u003c\/h4\u003e\n\u003cp\u003etherefore i-l = (r-Q)i,原文 From [1], we find f-Cf = (i-l)i therefore i-l = (r-Q)i, B0021｜原版スキャンp.8…\u003c\/p\u003e\n\u003cp\u003e数式・計算（OCR転写）：Hence ^ = 5 or --, 3\/ = 3 or -^g原文 Hence ^ = 5 or --, 3\/ = 3 or -^gB0086｜原版スキャンp.1…\u003c\/p\u003e\n\u003ch4\u003e数式・計算（OCR転写）：But -^x = I\/; -^x = 4 or 1, x = 16\u003c\/h4\u003e\n\u003cp\u003eor 1.原文 But -^x = I\/; -^x = 4 or 1, x = 16 or 1. B0107｜原版スキャンp.13／原版本文p.6｜原文順0107｜ロビンソン本文原版記号・OCR…\u003c\/p\u003e\n\u003ch4\u003e数式・計算（OCR転写）：10— lOiz+lOy^-lOy'+lO)\/'' =\u003c\/h4\u003e\n\u003cp\u003eUy+Uy'; 10(l+y^+\/) = 2l{y+f). Divide troth members by y^;then we have\u003c\/p\u003e\n\u003ch4\u003e原文 This is as far as we can reduce this equation, without the aid of the other\u003c\/h4\u003e\n\u003cp\u003eequation; therefore clear the second equation of fractions, and we haveB0176｜原版スキャンp.16／原版本文p.9｜原…\u003c\/p\u003e\n\u003ch4\u003e原文 Equation [a] shosvs us that Q^^^J = 24^\/3+5-30. Suhstituting this value of\u003c\/h4\u003e\n\u003cp\u003eG^\/^nj in [b], we haveB0220｜原版スキャンp.19／原版本文p.12｜原文順0220｜ロビンソン本文原版記号・OCR転写を要照合：数式、単位、数表、天文値は巻末の原版…\u003c\/p\u003e\n\u003ch4\u003e原文 --JThrough the three points A, B and H, describe a circle; the line FH will touch\u003c\/h4\u003e\n\u003cp\u003ethat circle at the point H. Join AH, BH, and the angle AHB is the angle required; andit is tbe gr…\u003c\/p\u003e\n\u003ch4\u003e原文 4. If, ffovn any two points in the circumference of a circle, there be dravm two\u003c\/h4\u003e\n\u003cp\u003estraight lines to a -point in a tangent to that circle, they will make the greatestangle when dra…\u003c\/p\u003e\n\u003ch4\u003e原文 S. if, from the tJvree angles of any triangle, lines be drawn to the points of\u003c\/h4\u003e\n\u003cp\u003ebisection of the opposite sides, these lines intersect each other in the same point.B0301｜原版スキャンp…\u003c\/p\u003e\n\u003ch4\u003e原文 The angle ABQ is a right angle, and equal to PBV. From each of these equals,\u003c\/h4\u003e\n\u003cp\u003etake away the common angkB0326｜原版スキャンp.29／原版本文p.22｜原文順0326｜ロビンソン本文『数学大辞典』p.102。\u003c\/p\u003e\n\u003ch4\u003e原文 Let ABC be the equilateral tjiangle in the circle ABC, and designate each side by\u003c\/h4\u003e\n\u003cp\u003et. Take D any point in the arc BC, and join AD, BD and BC.B0351｜原版スキャンp.31／原版本文p.24｜原文順0351｜ロビンソ…\u003c\/p\u003e\n\u003ch4\u003e原文 Scholium. By factoring this eqiiatioii, we have {y+ztn-z) = (i+.)(6-a), or (y+z):\u003c\/h4\u003e\n\u003cp\u003e(b+a) - (6-.): (y-z) __.[p]B0393｜原版スキャンp.33／原版本文p.26｜原文順0393｜ロビンソン本文この比率は三角法に用いられることもあり、次のような一般的…\u003c\/p\u003e\n\u003ch4\u003e原文 Let AB be the given side, A the gif en angle, and AD the given difference of [he\u003c\/h4\u003e\n\u003cp\u003eother tivo sides.B0414｜原版スキャンp.35／原版本文p.28｜原文順0414｜ロビンソン本文DBに接続し、HでBDを二分する。\u003c\/p\u003e\n\u003ch4\u003e原文 But ' this manner of presenting scientific truth is not only unsatisfactory, but\u003c\/h4\u003e\n\u003cp\u003eperplexing. We must therefore elucidate.B0438｜原版スキャンp.37／原版本文p.30｜原文順0438｜ロビンソン本文直角に交わる二つの線を描きます…\u003c\/p\u003e\n\u003ch4\u003e原文 x^—px = q, or (^x—p'jx = q. Construct the figure as here represented, and it is\u003c\/h4\u003e\n\u003cp\u003eobvious thatB0462｜原版スキャンp.39／原版本文p.32｜原文順0462｜ロビンソン本文原版記号・OCR転写を要照合：数式、単位、数表、天文値は巻末の原版ページ画像を確認してく…\u003c\/p\u003e\n\u003ch4\u003e原文 The most diificult form of a quadratic is the following; x^—px = —q. In\u003c\/h4\u003e\n\u003cp\u003erespect to the signs, it is proper to remark, that in numbers and magnitudes, thereis no such thi…\u003c\/p\u003e\n\u003ch4\u003e原文 To iind the other root, take the logarithm of q already found, md from it\u003c\/h4\u003e\n\u003cp\u003esubtract hgx last found, thus: 6,256273 2,698969 log.... 3,556304.. N. —3600 bychanging signs. B0…\u003c\/p\u003e\n\u003ch4\u003e数式・計算（OCR転写）：y^~{4ap—4f)y^-~8bpy+{a^-i-'l^~\u003c\/h4\u003e\n\u003cp\u003er^)4f = [G]原文 y^~{4ap—4f)y^-~8bpy+{a^-i-'l^~r^)4f = [G] B0604｜原版スキャンp.46／原版本文p.39｜原文順0604｜ロビンソン本…\u003c\/p\u003e\n\u003ch4\u003e原文 An equation with large coefficients is inconvenient for projection, as the\u003c\/h4\u003e\n\u003cp\u003erequired circle would be correspondingly large; but we can readily transform anyequation into ano…\u003c\/p\u003e\n\u003cp\u003e原文 44 MATHEMATICAl. KECKKAXIOUS. B0671｜原版スキャンp.51／原版本文p.44｜原文順0671｜ロビンソン本文ボラはMにあり、VPから2.8ユニット離れて…\u003c\/p\u003e\n\u003ch4\u003e原文 To show that we can determine the cuhe roof of any number (at least\u003c\/h4\u003e\n\u003cp\u003eapproximately) by projection, we require the cube root of 4096; that is, find a rootof the equati…\u003c\/p\u003e\n\u003cp\u003e数式・計算（OCR転写）：73 54原文 73 54 B0752｜原版スキャンp.54／原版本文p.47｜原文順0752｜ロビンソン本文原版記号・OCR転写を要照合：数式、単位、数表、天文値は巻…\u003c\/p\u003e\n\u003ch4\u003e158\u003c\/h4\u003e\n\u003cp\u003eproperty of the parabola, we find the angle nop = 109° 39' 45\"; to which add Cop(34° 6' 30\"), and…\u003c\/p\u003e\n\u003ch4\u003e原文 expression shows that the first member of equation [3] is a complete square,\u003c\/h4\u003e\n\u003cp\u003eon condition that we makeB0839｜原版スキャンp.58／原版本文p.51｜原文順0839｜ロビンソン本文原版記号・OCR転写を要照合：数式、単位、数表、天文値は巻末の…\u003c\/p\u003e\n\u003ch4\u003e原文 an equation showing that a and r are incommensurable with each other in this\u003c\/h4\u003e\n\u003cp\u003ecase.B0867｜原版スキャンp.60／原版本文p.53｜原文順0867｜ロビンソン本文与えられた円の中で互いに触れ合うときcは与えられている。\u003c\/p\u003e\n\u003ch4\u003e原文 When any angle of a triangle:d, the base is cut proportionally to the sides of\u003c\/h4\u003e\n\u003cp\u003ethe triangle. Put AB = X. Consider AE as the base of the triangle ABE; and as BDbisects the angle…\u003c\/p\u003e\n\u003ch4\u003e原文 Because the triangle BDC is rightangled at D, (he square on BD is equal to\u003c\/h4\u003e\n\u003cp\u003eBC.Bm, that is, equal to the rectangle Bn; but the square on BD is equal to cos^ \\k,that is, B094…\u003c\/p\u003e\n\u003ch4\u003e原文 The angle BCD, being the exterior angle of the triangle ACB, is equal to the sum\u003c\/h4\u003e\n\u003cp\u003eof the angles at the base; and as DEB is at the circumference, and stands on thesame arc, DEB is …\u003c\/p\u003e\n\u003ch4\u003e原文 On the longer side as radius, describe a circle. Divide the diameter into two\u003c\/h4\u003e\n\u003cp\u003eparts AE, EG, so that AE shall he the sum of the sides; then EG will be thedifference of the side…\u003c\/p\u003e\n\u003ch4\u003e原文 gitude. Let SE represent the distance between the earth and sun at this\u003c\/h4\u003e\n\u003cp\u003eparticular time; and to correspond with ourB1031｜原版スキャンp.71／原版本文p.64｜原文順1031｜ロビンソン本文例えば、地球の位置Eはa図…\u003c\/p\u003e\n\u003ch4\u003e原文 But AH is the sine of the arc PA, and CE is the sine of the arc PC. Let X = the\u003c\/h4\u003e\n\u003cp\u003earc from P to the midway point between A and C, and put a = half the arc AC; then\u003c\/p\u003e\n\u003ch4\u003eB1101｜原版スキャンp.74／原版本文p.67｜原文順1101｜ロビンソン本文\u003c\/h4\u003e\n\u003cp\u003e(a;-6sinA)'a;-6sinA'。原文 (a;— 6sinA)' a;— 6sinA ' B1102｜原版スキャンp.74／原版本文p.67｜原文順1102｜ロビンソン本文原版記号・O…\u003c\/p\u003e\n\u003ch4\u003e原文 4. Given the latitude, the sun's dedinaiion, or the declination of any planet or\u003c\/h4\u003e\n\u003cp\u003estar, to find the semidiurnal arc for the svn, planet or star.B1135｜原版スキャンp.77／原版本文p.70｜原文順1135｜…\u003c\/p\u003e\n\u003ch4\u003e15, we obtain time; when we multiply time by 15, we obtain arc; hence 1 hour\u003c\/h4\u003e\n\u003cp\u003emultiplied by 15, gives 15°. Therefore the sum of the arcsB1154｜原版スキャンp.78／原版本文p.71｜原文順1154｜ロビンソン…\u003c\/p\u003e\n\u003ch4\u003e数式・計算（OCR転写）：1. N 14=00' W, Distance 15,20\u003c\/h4\u003e\n\u003cp\u003echains = a;原文 1. N 14=00' W, Distance 15,20 chains = a; B1198｜原版スキャンp.81／原版本文p.74｜原文順1198｜Gummere…\u003c\/p\u003e\n\u003ch4\u003e38,72, and we have AH the distance required = 4,646, which ditfers from the given\u003c\/h4\u003e\n\u003cp\u003eanswer about one livJc of the chain.B1228｜原版スキャンp.83／原版本文p.76｜原文順1228｜Gummere由来の問題・ロビンソンによる解説どんな…\u003c\/p\u003e\n\u003ch4\u003e原文 The area of the triangle GAB = 270,5 square chains; GA = 24,05, GB = 36,40:\u003c\/h4\u003e\n\u003cp\u003ehence GD = 73,69, GC=66,90. Let PQ be drawn parallel to DC, and let it divide thefield into tvi'O…\u003c\/p\u003e\n\u003ch4\u003e原文 The distance from O to the middle point of QD is one half of 41,19, or 20,59;\u003c\/h4\u003e\n\u003cp\u003ebut this is greater by 3,89 than the distance from Q to the perpendicular from 0;that is, from Q …\u003c\/p\u003e\n\u003ch4\u003e原文 If the plane ACBD revolve on its axis CD, it will generate two equal cones; and\u003c\/h4\u003e\n\u003cp\u003eif from these we take the cylinder QR and the two small coaes on its ends, we shallhave the solid…\u003c\/p\u003e\n\u003ch4\u003e原文 Solving equation [1], we find three roots; one near 56, another 33,19 nearly,\u003c\/h4\u003e\n\u003cp\u003eand a third near the locality of mi««s 21, As neither the first nor the third canapply to our pro…\u003c\/p\u003e\n\u003ch4\u003eDIAG-P036｜原版スキャンp.36／本文p.29｜Geometrical Problems\u003c\/h4\u003e\n\u003cp\u003eDIAG-P038｜原版スキャンp.38／本文p.31｜Construction of Quadratic EquationsDIAG-P039｜原版スキャンp.39／本文p.32｜Constr…\u003c\/p\u003e\n\u003c\/div\u003e\u003c\/details\u003e","brand":"Straight Magic Shop","offers":[{"title":"Default Title","offer_id":68165717885165,"sku":"W-c41a168d","price":4.5,"currency_code":"USD","in_stock":true}],"thumbnail_url":"\/\/cdn.shopify.com\/s\/files\/1\/0813\/3027\/4541\/files\/nb_wc41a168d_p001.jpg?v=1793760319","url":"https:\/\/sp.straightmagicshop.com\/en-intl\/products\/mathematical-recreations-containing-solutions-of-ai-ja","provider":"Straight Magic Shop","version":"1.0","type":"link"}