Number & Algebra

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For this kind of task, you only need to represent things(sum, a certain term, etc.) using r(common ratio). Then solve the equation of r.

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It is after n years, not the n-th year. In that way, the formula should be \(x^n\), not \(x^{n-1}\).

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![[RV prac/Maths/_resources/Pasted image 20240920141504.png]]

\(log\) or \(ln\) on both side to get the indices.

Trig

for \(a \times sin(x)+b \times cos(x)\) to be written as \(R \times sin(x+c)\), \(R\) can be found by \(\sqrt{a^2+b^2}\)

An example for solving \(arctan\) related equation

The equation given involves an expression of the form \[\tan(\arcsin x).\] In order to solve for \(x\), one of the key steps shown in the solution is to rewrite \[\tan(\arcsin x)\] as \[\frac{\sin(\arcsin x)}{\cos(\arcsin x)}.\] From there, we use \(\sin(\arcsin x) = x\) and \(\cos(\arcsin x) = \pm \sqrt{1 - x^2}\) to express everything purely in terms of \(x\). This approach makes it straightforward to isolate \(x\) and solve the resulting algebraic equation.

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Why We Write \(\tan(\arcsin x)\) as \(\frac{\sin(\arcsin x)}{\cos(\arcsin x)}\) Instead of Using \(\arctan\) Directly

1. Directly Relating to \(x\)

   • By definition, \(\arcsin x\) is the angle \(\theta\) in the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) for which \(\sin \theta = x\).
   • Hence, \(\sin(\arcsin x) = x\) and \(\cos(\arcsin x) = \pm \sqrt{1 - x^2}\).
   • Rewriting \[\tan(\arcsin x) = \frac{\sin(\arcsin x)}{\cos(\arcsin x)}\] gives an expression directly in terms of \(x\). This makes the equation purely algebraic in \(x\), which is much more straightforward to solve.

2. Avoiding Domain/Range Complications

   • If we tried to “apply \(\arctan\)” on both sides or use an identity like \(\arctan(\tan(\arcsin x))\), we would run into potential domain-range issues.
   • \(\arcsin x\) takes values in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), while \(\arctan\) has its own principal range \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Even though these intervals overlap, confusion can arise if one tries to cancel \(\arctan\) and \(\tan\) without carefully tracking signs and quadrants.
   • Writing \[\tan(\arcsin x) = \frac{x}{\sqrt{1 - x^2}}\] sidesteps these complications by keeping the equation entirely in real-algebraic form.

3. Clear Path to an Algebraic Equation

   • Once \[\tan(\arcsin x) = \frac{x}{\sqrt{1 - x^2}},\] the rest of the steps (isolating \(x\), squaring, etc.) become a straightforward exercise in algebra.
   • This method also allows us to keep track of the necessary domain restrictions (like \(|x|\le 1\) from the definition of \(\arcsin\), and sign choices for \(\cos(\arcsin x)\)).

In short, rewriting \[\tan(\arcsin x)\] in terms of sine and cosine (and then in terms of \(x\)) is the cleanest, most direct way to solve the equation without running into tricky domain/range problems that arise from trying to apply inverse functions in the wrong order.

Calculus

条件洞察：know what information and situation is provided, especially when only one integral and differentiation is provided 符号抽象

concave-up: \(f''(x)>0\) concave-down: \(f''(x)<0\)

易错

积分内的系数不要忘记: \(\int a \times f(x) dx\) 不要看错带指数的括号：\((2x)^3 \ne 2x^3\) M series的公式divide by factorio 不要忘记

Integration

reverse chain rule

\[ \int f'(x)g'(f(x)) dx = g(f(x))+c \]

M series

Just expand. If composite function, expand outer layer, then expand inner layer. When see function capable of transformation, use transformation first, and then expand.

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use \(\frac{\text{M series}}{x^3}\), see it is undetermined form. Use L’Hopital. Ignore all other terms except the one with the same power as \(x^3\), apply L’Hopital rule 指数 times, you get result very fast

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Brutal Force, first expand the \(\cos(...)\), make \(\ln(1+x)\) as \(x\), and then expand the \(\ln(1+x)\) term into M series, then expand. Fuck.

Inverse Trig Integration

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enhance this flow and read method again

Complex Number

\[ z_1=r_1cis(\theta_1) \]

\[ z_1z_2=r_1r_2\times cis(\theta_1+\theta_2) \]

\[ \frac{z_1}{z_2}=\frac{r_1}{r_2} \times cis(\theta_1-\theta_2) \]

\[ cis(\theta_1)\times cis(\theta_2)=cis(\theta_1+\theta_2) \] if the equation has complex number as 系数, then the complex solutions do not come in conjugate pairs