ax2+bx+c=0x1=−b+b2−4ac2ax2=−b−b2−4ac2aax^{2}+bx+c=0\\ x_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}\\ax2+bx+c=0x1 =2a−b+b2−4ac x2 =2a−b−b2−4ac
已知三角形abc设p=(a+b+c)2S=p(p−a)(p−b)(p−c) 已知三角形abc\\ 设p=\frac{(a+b+c)}{2}\\ S=\sqrt{p(p-a)(p-b)(p-c)}\\已知三角形abc设p=2(a+b+c) S=p(p−a)(p−b)(p−c)
π=16arctan15−4arctan1239π=∑k=0∞(−1)k12k+1π=229801∑k=0∞(4k)!(1103+26390k)(k!)4⋅(396)4k \pi=16\arctan{\frac{1}{5}}-4\arctan{\frac{1}{239}}\\ \pi=\sum_{k=0}^{∞} (-1)^{k} \frac{1}{2k+1}\\ \pi=\frac{2\sqrt{2}}{9801}\sum_{k=0}^{∞}\frac{(4k)!(1103+26390k)} {(k!)^{4} \cdot(396)^{4k}} π=16arctan51
−4arctan2391 π=∑k=0∞ (−1)k2k+11 π=980122 ∑k=0∞ (k!)4⋅(396)4k(4k)!(1103+26390k)
a0=1b0=12t0=14p0=1an+1=an+bn2bn+1=an⋅bntn+1=tn−pn(an−an+1)2pn+1=2pnπ≈(an+1+bn+1)24tn+1 a_{0}=1\\ b_{0}=\frac{1}{\sqrt{2}}\\ t_{0}=\frac{1}{4}\\ p_{0}=1\\ a_{n+1}=\frac{a_{n}+b_{n}}{2}\\ b_{n+1}=\sqrt{a_{n}\cdot b_{n}}\\ t_{n+1}=t_{n}-p_{n}(a_{n}-a{n+1})^{2}\\ p_{n+1}=2p_{n}\\
\pi\approx\frac{(a_{n+1}+b_{n+1})^{2}}{4t_{n+1}} a0 =1b0 =2 1 t0 =41 p0 =1an+1 =2an +bn bn+1 =an ⋅bn tn+1 =tn −pn (an −an+1)2pn+1 =2pn π≈4tn+1 (an+1 +bn+1 )2
e=∑k=0∞1k! e=\sum_{k=0}^{∞}\frac{1}{k!} e=∑k=0∞ k!1
eix=cosx+isinx将x=π带入，则有上帝公式：eiπ=−1 e^{ix}=\cos{x}+i\sin{x}\\ 将x=\pi带入，则有上帝公式：\\ e^{i\pi}=-1 eix=cosx+isinx将x=π带入，则有上帝公式：eiπ=−1
