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can any body solve the equation $x^7 + 2x +(61/20)=0 \tag{A}$ here is my method $\cdots x^7 + 2x +3=0$ hence $x=-1$. Then I assume (A) as $x^7+2x+3+q=0$ where $q=1/20$ writing $x=x(0)+qx(1)+((q^2)x(2))\tag{B}$ (I ignore powers of $x$ which are more than $2$) then plugging (B) in (A) we may get the answer through approximation method There's my problem : I can't plug (B) in (A) and get the answer!!please help me.... |
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You have $$x_0^7 + 2 x_0+3 = 0 \implies x_0=-1$$ Now you want to solve $$x^7+2 x+(3+\delta)=0$$ where $\delta = 1/20$. Assume $x=x_0+\epsilon$, where $\epsilon$ is small compared with $x_0$. Then $$(x_0+\epsilon)^7 + 2 (x_0+\epsilon) + (3+\delta)=0 $$ Taylor expand the lead term to get $$x_0^7 + 7 x_0^6 \epsilon + 2 x_0 + 2 \epsilon + 3 + \delta = O(\epsilon^2)$$ The $O$ term on the right-hand side means that we are ignoring any powers of $\epsilon$ beyond linear. Now use the original equation and get a simple equation in $\epsilon$: $$7 x_0^6 \epsilon+ 2 \epsilon+ \delta = 0$$ or $$\epsilon = -\frac{\delta}{7 x_0^6+2} = -\frac{1}{180}$$ Thus, the approximate solution is $x_0+\epsilon= -181/180 = -1.055\bar{5}$. WA gives the respective solution as $\approx -1.00548$; not bad but clearly some room for improvement. |
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A straightforward and efficient method is Newton--Raphson: Let $f(x)=x^7+2x+61/20\;(x\in \Bbb R)$. Start with $x_0=-1$, and iterate with $$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\quad(n=0,1,\dots),$$where $f'(x)=7x^6+2\;(x\in \Bbb R)$. Two or three iterations should suffice for good accuracy. |
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Since it seems that a root $x_0 \approx -1$, let $x = -1+c$ and see what happens (historically, I think this is akin to Horner's rule). If $f(x) = x^7+2x+(61/20)$, and $c$ is small, $\begin{align} f(-1+c) &=(-1+c)^7+2(-1+c)+61/20\\ &\approx (-1)^7 + 7(-1)^6 c -2+2c+61/20\\ &=-3+7c+2c+61/20\\ &=1/20+9c\\ \end{align} $ or, to make $f(-1+c)=0$, $c = -1/180$. So, a better root is $1-1/180$. |
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