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Tutor profile: Mark B.

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Mark B.
Tutor and mentor of three years
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Questions

Subject: Trigonometry

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Question:

Solve: 2$$\theta cos(\theta)+\theta=0$$

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Mark B.
Answer:

Subtracting $$\theta$$ to both sides we obtain: $$2\theta cos(\theta)=-\theta $$. From this we divide $$2\theta$$ to both sides and obtain: $$cos(\theta)=\frac{-1}{2}$$. Therefore, $$\theta= cos^{-1}(\frac{-1}{2})$$ which is equivalent to $$\theta= \frac{2\pi}{3}+2\pi k$$ and $$\theta=\frac{4\pi}{3}+2\pi k$$ for some integer k.

Subject: Calculus

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Question:

Prove the series $$\sum{a_n} $$ converges if $$\limsup_{n->\infty}|\frac{a_{n+1}}{a_n}|<1$$

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Mark B.
Answer:

Let $$c<1$$. For $$n\geq N$$ we can find a $$c$$ such that $$|\frac{a_{n+1}}{a_n}|<c$$. In particular: $$|a_{N+1}|<c|a_N|$$.This implies: $$|a_{N+2}|<c|a_{N+1}|<c^2|a_N|$$. Constructing an inequality in this fashion we have: $$|a_{N+p}|<c^p|a_N|$$ where $$p\in N$$ and $$N+p=n$$. This implies $$|a_n|<|a_N|c^{n-N}=|a_N|c^{-N}c^{n}$$ Taking the sum of both sides we have: $$\sum{|a_n|}<|a_N|c^{-N}\sum{c^{n}}$$. Since $$\sum{c^{n}}$$ is a geometric series we obtain an upper bound on $$\sum{|a_n|}$$ and we conclude $$\sum{|a_n|}$$ converges. From this we determine $$a_n$$ converges absolutely which implies $$\sum{a_n}$$ converges.

Subject: Differential Equations

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Question:

Consider the class of equation in the form of: $$y'+p(x)y= q(x)y^n$$. This is commonly referred to as a Bernoulli Equation. Show that if $$n\neq 0, 1$$ then the substitution $$v=y^{1-n} $$reduces this class of equations to a linear differential equation.

Inactive
Mark B.
Answer:

Let $$v=y^{1-n}$$ where $$n\neq 0,1$$. Taking the derivative with respect to x we have: $$\frac{dv}{dx}=(1-n)y^{-n}y'=v'$$. Dividing our Bernoulli Equation by $$y^n$$ and multiplying by $$(1-n)$$ yields: $$(1-n)y^{-n}y'+(1-n)p(x)y^{1-n}=q(x)(1-n)$$. Since $$\frac{dv}{dx}=(1-n)y^{-n}y'$$ and $$v=y^{1-n}$$ our equation reduces to: $$\frac{dv}{dx}+(1-n)p(x)v=q(x)(1-n)$$. Letting $$(1-n)p(x)=a(x)$$ and $$(1-n)q(x)= g(x)$$ our final expression is $$v'+a(x)v=g(x)$$ , which is a linear differential equation as desired.

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