{"id":342,"date":"2024-03-15T17:38:06","date_gmt":"2024-03-15T08:38:06","guid":{"rendered":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=342"},"modified":"2024-06-03T09:13:18","modified_gmt":"2024-06-03T00:13:18","slug":"top-math2-1q-4h-2","status":"publish","type":"post","link":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=342","title":{"rendered":"\u6570\u5b66II_1Q4H_2"},"content":{"rendered":"\n\n\n\n\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nBack<\/span><\/a><\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u518d\u751f\u30ea\u30b9\u30c8<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n\n\n\n
\u30c6\u30ad\u30b9\u30c8<\/span><\/td>\n\u6f14\u7fd2<\/span><\/td>\n\u6f14\u7fd2\u89e3\u7b54<\/span><\/td>\n\u8ab2\u984c<\/span><\/td>\n\u89e3\u8aac<\/span><\/td>\n<\/tr>\n
1Q4H_7<\/span><\/a><\/td>\n1Q4H_E7<\/span><\/a><\/td>\n1<\/span>Q4H_ES7<\/span><\/a><\/td>\n1Q4H_K7<\/span><\/a><\/td>\n1Q4H_V7<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/span> \u3079\u304d\u95a2\u6570\u306e\u6975\u9650 lim$r^n=?$\u3001\u521d\u9805 $a$ \u516c\u6bd4 $r$ \u306e\u7b49\u6bd4\u7d1a\u6570 $S=?$ <\/div>
$ \\displaystyle{\\lim_{n\\to \\infty}r^n}=\\begin{cases}0 \\: (-1<r<1)\\\\ 1 \\:\u00a0 (r=1) \\\\ \\text{\u767a\u6563} \\begin{cases} \\infty & (r>1)\u00a0 \\\\ \\text{\u632f\u52d5} &(r<-1) \u3000\\end{cases} \\end{cases}$
$S=a+ar+ar^2+\\cdots=\\begin{cases}\\dfrac{a}{1-r} & (-1<r<1)\\\\ \\text{\u767a\u6563} & (\u305d\u306e\u4ed6)\\end{cases}$<\/div><\/div>
<\/span>Targets<\/div>
1. \u3079\u304d\u95a2\u6570\u306e\u6975\u9650 $\\displaystyle{\\lim_{n\\to \\infty}r^n}$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
2. 1\u306e\u5fdc\u7528\u3068\u3057\u3066\u6975\u9650 $\\displaystyle{\\lim_{n\\to \\infty}(a^n \\pm b^n \\pm c^n)}$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
3. 1\u306e\u5fdc\u7528\u3068\u3057\u3066\u6975\u9650 $\\displaystyle{\\lim_{n\\to \\infty}\\frac{a^n \\pm b^n}{c^n \\pm d^n}}$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
4. \u7b49\u6bd4\u7d1a\u6570\u306e\u53ce\u675f\u30fb\u767a\u6563\u3092\u5224\u5b9a\u3067\u304d\u308b\u00a0<\/div><\/div> <\/div>\n\n\n\n\n\n
1Q4H_8<\/span><\/a><\/td>\n1Q4H_E8<\/span><\/a><\/td>\n1Q4H_ES8<\/span><\/a><\/td>\n1Q4H_K8<\/span><\/a><\/td>\n1Q4H_V8<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/span> $a_k$$_+$$_1-a_k=d$ \u306e\u3068\u304d\u3001\u7121\u9650\u548c \u03a3$1\/a_ka_k$$_+$$_1=?$
\u56e0\u6570\u5b9a\u7406\uff1a$f(a)=0$ \u306e\u3068\u304d$f(x)=?$ <\/div>
$a_{k+1}-a_k=d\\neq 0$ \u306e\u3068\u304d
$\\displaystyle{\\sum_{k=1}^{\\infty}\\frac{1}{a_ka_{k+1}}}=\\displaystyle{\\lim_{n\\to \\infty}\\sum_{k=1}^{n}\\frac{1}{a_ka_{k+1}}}=\\displaystyle{\\lim_{n\\to \\infty}\\dfrac{1}{d}\\Big( \\dfrac{1}{a_1}-\\dfrac{1}{a_{n+1}} \\Big)}=\\dfrac{1}{da_1}$
$f(a)=0$<\/span> \u306e\u3068\u304d $f(x)=$$(x-a)$<\/span>$g(x)$\u3000\u56e0\u6570\u5b9a\u7406<\/div><\/div>
<\/span>Targets<\/div>
1.\u7121\u9650\u548c $\\displaystyle{\\sum_{k=1}^{\\infty}\\frac{1}{a_ka_{k+1}}}\\; (a_{k+1}-a_k=d)$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
2. \u5206\u6bcd\u30fb\u5206\u5b50\u3092\u56e0\u6570\u5206\u89e3\u3057\u3066\u3001\u4e0d\u5b9a\u5f62 $\\dfrac{0}{0}$ \u306e\u6975\u9650\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b<\/div><\/div> <\/div>\n\n\n\n\n\n
1Q4H_9<\/span><\/a><\/td>\n1Q4H_E9<\/span><\/a><\/td>\n1Q4H_ES9<\/span><\/a><\/td>\n1Q4H_K9<\/span><\/a><\/td>\n1Q4H_V9<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/span> lim$1\/x^n=?$\u3001\u5206\u5b50\u306e\u6709\u7406\u5316 $($\u221aA-\u221aB$)=?$ <\/div>
$\\displaystyle{\\lim_{x\\to \\infty}\\dfrac{1}{x^n}}=0$\u3001$\\sqrt{A}-\\sqrt{B}=\\dfrac{\\sqrt{A}-\\sqrt{B}}{1}=\\dfrac{A-B}{\\sqrt{A}+\\sqrt{B}}$<\/div><\/div>
<\/span>Targets<\/div>
1. \u6975\u9650 $\\displaystyle{\\lim_{n\\to \\infty}f(x)}\\; (f(x)$\u306f\u591a\u9805\u5f0f$)$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
2. \u6975\u9650 $\\displaystyle{\\lim_{n\\to \\infty}\\frac{f(x)}{g(x)}}\\; (f(x), g(x)$\u306f\u591a\u9805\u5f0f$)$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
3. \u6975\u9650 $\\displaystyle{\\lim_{n\\to \\infty}\\big(\\sqrt{ax^2+bx+c}-\\sqrt{ax^2+dx+e}\\big)}$ \u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b <\/div><\/div> <\/div>\n\n\n\n\n\n
1Q4H_10<\/span><\/a><\/td>\n1Q4H_E10<\/span><\/a><\/td>\n1Q4H_ES10<\/span><\/a><\/td>\n1<\/span>Q4H_K10<\/span><\/a><\/td>\n1Q4H_V10<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/span> \u95a2\u6570 $f(x)$ \u306e\u5e73\u5747\u5909\u5316\u7387 \u0394$f\/$\u0394$x=?$\u3001\u5fae\u5206\u4fc2\u6570 $f'(a)=?$\u3001\u5c0e\u95a2\u6570 $f'(x)=?$ <\/div>
$\\dfrac{\\Delta f}{\\Delta x}=\\dfrac{f(b)-f(a)}{b-a}$
$f'(a)=\\displaystyle{\\lim_{h\\to 0}\\frac{f(a+h)-f(a)}{h}}$\u3001$f'(x)=\\displaystyle{\\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h}}$ <\/div><\/div>
<\/span>Targets<\/div>
1. \u95a2\u6570 $f(x)$ \u306e\u533a\u9593 $[a,b]$ \u306b\u304a\u3051\u308b\u5e73\u5747\u5909\u5316\u7387\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
2. \u5b9a\u7fa9\u304b\u3089 $f(x)$ \u306e $x=a$ \u306b\u304a\u3051\u308b\u5fae\u5206\u4fc2\u6570\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b
3. \u5b9a\u7fa9\u304b\u3089 $f(x)=$(3\u6b21\u4ee5\u4e0b\u306e\u591a\u9805\u5f0f)\u306e\u5c0e\u95a2\u6570\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b\u00a0<\/div><\/div> <\/div>\n\n\n\n\n\n
1Q4H_11<\/span><\/a><\/td>\n1Q4H_E11<\/span><\/a><\/td>\n1Q4H_ES11<\/span><\/a><\/td>\n1Q4H_K11<\/span><\/a><\/td>\n1Q4H_V11<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/span>\u5fae\u5206\u516c\u5f0f\u3000$(x^n)'=?$\u3001$(1\/x^n)'=?$\u3001$(fg)'=?$\u3001$(f\/g)'=?$\u3001$(f\/g^n)'=?$ <\/div>
$(c)’=0$\u3001$(x)’=1$\u3001$(x^n)’=nx^{n-1}$\u3001$(f\\cdot g)’=f’\\cdot g + f\\cdot g’$
$\\left( \\dfrac{1}{x^n} \\right)^{\\prime}=\\dfrac{-n}{x^{n+1}}$\u3001$\\left( \\dfrac{f}{g} \\right)^{\\prime}=\\dfrac{f’\\cdot g-f \\cdot g’}{g^2}$ \u8a3c\u660e\uff1a1Q4H_V11_pf<\/a><\/span><\/div><\/div>
<\/span>Targets<\/div>
1. \u591a\u9805\u5f0f\u95a2\u6570\u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b
2. (\u591a\u9805\u5f0f)\u00d7(\u591a\u9805\u5f0f) \u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b
3.$\\dfrac{(\u591a\u9805\u5f0f)}{(\u591a\u9805\u5f0f)}$ \u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b
4. $\\dfrac{1}{x^n}=x^{-n}$ \u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b <\/div><\/div> <\/div>\n\n\n\n\n\n
1Q4H_12<\/span><\/a><\/td>\n1Q4H_E12<\/span><\/a><\/td>\n1Q4H_ES12<\/span><\/a><\/td>\n1Q4H_K12<\/span><\/a><\/td>\n1Q4H_V12<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n
<\/span>\u5fae\u5206\u516c\u5f0f\u3000$(f^n)'=?$\u3001$(1\/g^n)'=?$\u3001$(x^a)'=?$ <\/div>
$(f^n)’=nf’\\cdot f^{n-1}$\u3001$\\left( \\dfrac{f}{g^n} \\right)^{\\prime}=\\dfrac{f’\\cdot g-nf \\cdot g’}{g^{n+1}}$\u3001$\\left( \\dfrac{1}{g^n} \\right)^{\\prime}=\\dfrac{-ng’}{g^{n+1}}$
$\\left(\\sqrt[m]{x^n} \\right)’=\\left(x^{\\frac{n}{m}} \\right)’=\\frac{n}{m}x^{\\frac{n}{m}-1}$\u3001$\\left(\\dfrac{1}{\\sqrt[m]{x^n}} \\right)’=\\left(x^{-\\frac{n}{m}} \\right)’=\\frac{-n}{m}x^{-\\frac{n}{m}-1}$
\u8a3c\u660e\uff1a1Q4H_V12_pf<\/a><\/span><\/div><\/div>
<\/span>Targets<\/div>
1. $(\u591a\u9805\u5f0f)^n$ \u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b
2.$\\dfrac{1}{(\u591a\u9805\u5f0f)^n}$ \u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b
3.$\\sqrt[n]{x^m}=x^{\\frac{m}{n}},\\;\\dfrac{1}{\\sqrt[n]{x^m}}=x^{-\\frac{m}{n}}$ \u306e\u5fae\u5206\u304c\u8a08\u7b97\u3067\u304d\u308b <\/div><\/div> <\/div>\n","protected":false},"excerpt":{"rendered":"

\u00a0 \u00a0 \u00a0 \u00a0 Back \u00a0 \u00a0 \u00a0 \u00a0 \u518d\u751f\u30ea\u30b9\u30c8 \u30c6\u30ad\u30b9\u30c8 \u6f14\u7fd2 \u6f14\u7fd2\u89e3\u7b54 \u8ab2\u984c \u89e3\u8aac 1Q4H_7 1Q4H_E7 1Q4H_ES7 1Q4H_K7 1Q4H_V7 1Q4H_8 1Q4H_E8 1Q4H_ES8 […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[42,14,4,7,32,33],"tags":[],"class_list":["post-342","post","type-post","status-publish","format-standard","hentry","category-si","category-top","category-math2","category-math2-1q","category-math2-1q4h-1","category-math2-1q4h-2"],"_links":{"self":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/342","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=342"}],"version-history":[{"count":162,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/342\/revisions"}],"predecessor-version":[{"id":1956,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/342\/revisions\/1956"}],"wp:attachment":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}