Question by Student 201527121 1. wanted Prove $$\frac{1}{1+y}=1-y+y^2-y^3+...$$      2. Given condition $$|y|<<1$$     3. Solution $$\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=1}^\infty (x)^n$$ Let's substitute $$x=-y$$ $$\frac{1}{1+y}=1-y+y^2-y^3+...$$      4. Let's prove $$\frac{1}{1-x}=\sum_{n=1}^\infty (x)^n$$ $|x|<<1$ (It should be for convergence) $$S_n=\sum_{k=1}^n (x)^k$$ $$S_n=x+x^2+x^3+...+x^n$$ $$(1-x)S_n=(1-x)(x+x^2+x^3+...+x^n)$$ $$=x+x^2+...+x^n-x^2-x^3-...-x^{n+1}=x-x^{n+1}$$ $$S_n=\frac{x}{1-x}-\frac{x^{n+1}}{1-x}$$ Because of $|x|<<1$, If $n\rightarrow\infty$ then $S_n\rightarrow\frac{x}{1-x}$ $$\sum_{k=N}^n (x)^k=x^N+...+x^n=x^{N-1}\sum_{k=1}^{n-N+1} (x)^k$$ $$\sum_{k=N}^\infty x^k=\varinjlim({n}\rightarrow{\infty})\sum_{k=N}^n x^k=\varinjlim({n}\rightarrow{\infty})\cdot x^{N-1}\sum_{k=1}^{n-N+1} x^k=\frac{x^N}{1-x}$$ If $$N=0$$ then $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$      $$5. Conclusion$$ $$\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=1}^\infty (x)^n$$ $$x=-y$$ $$\frac{1}{1-x}=\sum_{n=1}^\infty (x)^n$$ $$\frac{1}{1+y}=1-y+y^2-y^3+...$$
 10.01.18
This is great. I'll give you 3 points bonus. It could be made a bit shorter thus, and you shouldn't separate the proof into sections: this makes it harder to read. Also, I was looking for a simpler proof using Taylor series. If someone else can do it, another 3 bonus points are up for grabs.
 Question by Student 201427113 Professor, I'll prove $\frac{1}{1+x}=1-x+x^2-x^3...$ Taylor series is one of the power series. So $f(x)=C_0+C_1(x-a)+C_2(x-a)^2+C_3(x-a)^3+.... → A-equation$ And we have to decide what is $C_0, C_1, C_2, C_3, .....$ 1. Substitute x=a on A-equation So $f(a)=C_0$ 2. Apply differential on A-equation So $f^\prime(x)=C_1+2C_2(x-a)+3C_3(x-a)^2+....→ B-equation$ And substitute x=a on B-equation So $f^\prime(a)=C_1$ 3. Apply differential on B-equation So $f^{\prime\prime}(x)=2C_2+6C_3(x-a)+12C_4(x-a)^2....→ C-equation$ And substitute x=a on C-equation So $f^{\prime\prime}(a)=2C_2 → C_2=\frac{f^{\prime\prime}(a)}{2}$ 4. Apply differential on C-equation So $f^{\prime\prime\prime}(x)=6C_3+24C_4(x-a)....→ D-equation$ And substitute x=a on D-equation So $f^{\prime\prime\prime}(a)=6C_3 → C_3=\frac{f^{\prime\prime\prime}(a)}{6}$ 5. Apply this sequence infinitely and generalized the eqs. $C_n=\frac{f^n(a)}{n!}$ Finally substitute $C_n=\frac{f^n(a)}{n!}$ on A-equation So $f(x)=f(a)+\frac{f^\prime(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+...\\= \sum_{n=0}^\infty\frac{f^n(a)}{n!}(x-a)^n$ $f(x)=\frac{1}{1+x}=(1+x)^{-1} → P-equation$ $f^\prime(x)=-(1+x)^{-2} → Q-equation$ $f^{\prime\prime}(x)=2(1+x)^{-3} → R-equation$ $f^{\prime\prime\prime}(x)=-6(1+x)^{-4} → S-equation$ Apply x=a=0 on eqs. P,Q,R,S $f(0)=1\\ f^\prime(0)=-1\\ f^{\prime\prime}(0)=2\\ f^{\prime\prime\prime}(0)=-6$ So $f(x)= 1-1\times x+\frac{2}{2!}x^2+\frac{-6}{3!}x^3+...$ Prove is finished.!!
 10.02.18
This is fine, I'll give you 2 points bonus. I would have given 3 if you would have typeset correctly your equations with proper equation numbers as specified in the mini HOWTO in the Skylounge. But, this proof is still not what I was looking for.. There's a much easier way to prove it using Taylor series (without power series). Three bonus points are still up for grabs.
 Question by Student 201527119 Proof. $$\frac{1}{1+x}=1-x+x^2-x^3+...$$ $$f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2!}f^”(a) + \frac{(x-a)^3}{3!}f”'(a) + ...$$ In $$\frac{1}{1+x}$$ a=0 So, $$f(x) = f(0) + (x-0)f'(0) + \frac{(x-0)^2}{2!}f”(0) + \frac{(x-0)^3}{3!}f”'(0) + ...$$ $$f'(x)=\frac{-1}{(1+x)^2}, f”(x)=\frac{2}{(1+x)^3}, f”'(x)=\frac{-6}{(1+x)^4}$$ $$f(0)=1,f'(0)=-1,f”(0)=2,f”'(0)=-6$$ Put them in equation. $$f(x)=\frac{1}{1+x}=1-x+x^2-x^3+...$$ End.
This is what I was looking for. I'll give you 2.5 points bonus. The only problem here is that you don't define $f(x)$ prior to doing the expansion with the Taylor series.
 Question by Student 201427144 Dear professor, I have question about A2#Q6(a). According to the question, iterative method is $x^{n+1}=x^{n}-0.05*\frac{f(x^{n})}{f{\prime}(x^{n})}-0.95*\frac{f(x^{n})(x^{n-1}-x^{n-2})}{f(x^{n-1})-f(x^{n-2})}$However, in order to use this as an iterative method, isn`t this should be like this;$x_{n+1}=x_{n}-0.05*\frac{f(x_{n})}{f{\prime}(x_{n})}-0.95*\frac{f(x_{n})(x_{n-1}-x_{n-2})}{f(x_{n-1})-f(x_{n-2})}$
 10.04.18
Well, in this case, it's clear what the notation means. It doesn't matter if it's a superscript or subscript..
 Question by Student 201527119 Proof. $$\frac{\epsilon_{(n+1)(Newton)}}{\epsilon_{(n+1)(secant)}}=|\epsilon_{0}*k|^{{2^n}-{1.618^n}}$$ $$k=|\frac{f”(r)}{2f'(r)}|$$ $$\epsilon_{(n+1)(Newton)}=k\epsilon^2$$ $$\epsilon_1=k*\epsilon_{0}^2$$ $$\epsilon_2=k*\epsilon_{1}^2=k^3*\epsilon_{0}^4$$ $$\epsilon_3=k*\epsilon_{2}^2=k^7*\epsilon_{0}^8$$ ... $$\epsilon_n=k^{2^n-1}*\epsilon_{0}^{2^n}$$ $$\epsilon_{(n+1)(secant)}=k^{\frac{1}{1.168}}\epsilon^{1.168}$$(1.168=p) $$\epsilon_1=k^{p^{-1}}*\epsilon_{0}^p$$ $$\epsilon_2=k^{p^{-1}}*\epsilon_{1}^p=k^{1+p^{-1}}*\epsilon_{0}^{p^2}$$ $$\epsilon_3=k^{p^{-1}}*\epsilon_{2}^p=k^{p+1+p^{-1}}*\epsilon_{0}^{p^3}$$ ... $$\epsilon_n=k^{\sum_{m=0}^{n-1} p^{m-1}}*\epsilon_{0}^{p^n}$$ Consequently, $$\frac{\epsilon_{n(newton)}}{\epsilon_{n(secant)}}=\frac{k^{2^n-1}*\epsilon_{0}^{2^n}}{k^{\sum_{m=0}^{n-1} p^{m-1}}*\epsilon_{0}^{p^n}}=\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k*k^{\sum_{m=0}^{n-1} p^{m-1}}*\epsilon_{0}^{p^n}}$$ $$k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}}$$ $$1+\sum_{m=0}^{n-1} p^{m-1}=1+p^{-1}+1+p+...+p^{n-2}$$ Using Geometric sequence sum, $$p^{-1}+1+p+...+p^{n-2}=\frac{p^n-1}{p^2-p}$$ $$p^2=p+1$$ $$So, \frac{p^n-1}{p^2-p}=p^n-1$$ =>$$\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k*k^{p^n-1}*\epsilon_{0}^{p^n}}=\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k^{p^n}*\epsilon_{0}^{p^n}}$$ $$Consequently, |\epsilon_{0}*k|^{{2^n}-{1.618^n}}$$ $$So, \frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=|\epsilon_{0}*k|^{{2^n}-{1.618^n}}$$ $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=|\epsilon_{0}*\frac{f”(r)}{2f'(r)}|^{{2^n}-{1.618^n}}$$
 10.08.18
I don't understand this part: $$k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}}$$ This doesn't make sense to me. Is this a typo or a misunderstanding? Please fix this and retype your proof below.
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