11

Homework 11.pdf

1. (20 points) By considering appropriate Taylor polynomials, evaluate the limits below:

(a)\(\lim_{x \to 0} \frac{\ln(1+x^2)}{x \sin(x)}\)

\(\lim_{x\to0}\frac{\ln(1+x^2)}{x\sin(x)}=\lim_{x\to0}\frac{\ln(1+x^2)}{x\cdot\left(T_{1,0}(x)+R_{1,0}\right)}\)​\(=\lim_{x\to0}\frac{\ln(1+x^2)}{x\cdot\left(x+R_{1,0}\right)}\)​\(=\lim_{x\to0}\frac{\ln(1+x^2)}{x^2+Ax^3}=\lim_{x\to0}\frac{\frac{2x}{1+x^2}}{2x+3Ax^2}=\lim_{x\to0}\frac{\frac{2+2x^2+4x^2}{\left(1+x^2\right)^2}}{2+6Ax}=\frac22=1\)

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(b)\(\lim_{x \to 0} \frac{\sin^3(x)}{x(1-\cos(x))}\)

\(\lim_{x\to0}\frac{\left(T_{1,\sin x,0}+R_{\sin x,0}\right)^3}{x\left(1-\left(T_{2,\cos x,0}+R_{2,\cos x,0}\right)\right)}\)​\(= \lim_{x \to 0} \frac{(x + A x^2)^3}{x \left(1 - \left(1 - \frac{x^2}{2} + B x^3\right)\right)}\)

\(= \lim_{x \to 0} \frac{(x + A x^2)^3}{x \left(\frac{x^2}{2} - B x^3\right)}\)​\(=\lim_{x\to0}\frac{x^3(1+Ax)^3}{\frac{x^3}{2}-Bx^4}\)​\(= 2 \lim_{x \to 0} \frac{(1 + A x)^3}{1 - 2 B x}\)​\(= 2\)

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(c)\(\lim_{x \to \infty} \sqrt{1+x+x^2} - x\)

Hint: Consider a change of variable.

We want to expand \(\sqrt{1+x+x^2}\), but \(x \to \infty\).

Then consider \(x = \frac{1}{t}\) where \(t \to 0\).

Thus \(\lim_{x\to\infty}\sqrt{1+x+x^2}-x=\lim_{t\to0}\left(\sqrt{1+\frac{1}{t}+\frac{1}{t^2}}-\frac{1}{t}\right)\) \(=\lim_{t\to0}\left(\sqrt{\frac{t^2+t+1}{t^2}}-\frac{1}{t}\right)=\lim_{t\to0}\left(\frac{\sqrt{t^2+t+1}-1}{t}\right)\)

Then we expand \(\sqrt{t^2 + t + 1} = f(t)\). Since \(f(t)-T_{2,0}=R_{2,0}\), and \(T_{2,0}=1+\frac12t+\frac38t^2\)

Then \(f(t)=1+\frac12t+\frac38t^2+\frac{A}{6}t^3\)​\(,\quad\lim_{t\to0}\frac{\frac12t+\frac38t^2+\frac{A}{6}t^3}{t}=\lim_{t\to0}\frac{\frac12+\frac34t+\frac{A}{2}t^2}{1}=\frac12\)

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2. (20 points) Suppose \(a_i, b_i\) are coefficients of Taylor polynomials at \(a\) of \(f\) and \(g\) respectively. Find the coefficients \(c_i\) of Taylor polynomials at \(a\) of the function \(h(x) = f(x)g(x)\).

We have \(T_{n, f, a}(x) = a_0 + a_1(x - a) + a_2(x - a)^2 + \dots + a_n(x - a)^n\)

\(T_{n, g, a}(x) = b_0 + b_1(x - a) + b_2(x - a)^2 + \dots + b_n(x - a)^n\) where \(a_{i}=\frac{f^{(i)}(a)}{i!}\) and \(b_{i}=\frac{g^{(i)}(a)}{i!}\)

Thus \(T_{n,f\cdot g,a}(x)=c_0+c_1(x-a)+c_2(x-a)^2+\dots+c_{n}(x-a)^{n}\) where \(c_{i}=\frac{(f\cdot g)^{(i)}(a)}{i!}\)​

Since we know \((f\cdot g)^{(i)}=\sum_{k=0}^{i}\binom{i}{k}f^{\left(i-k\right)}g^{(k)}\), then \(c_{i}=\frac{(f\cdot g)^{(i)}(a)}{i!}=\frac{\sum_{k=0}^{i}\binom{i}{k}f^{\left(i-k\right)}g^{(k)}}{i!}=\frac{\sum_{k=0}^{i}\binom{i}{k}\cdot\left(i-k\right)!\cdot a_{i-k}\cdot k!\cdot b_{k}}{i!}\)

Thus \(c_{i}=\sum_{k=0}^{i}\frac{\binom{i}{k}\cdot\left(i-k\right)!\cdot k!}{i!}\cdot a_{i-k}\cdot b_{k}=\sum_{k=0}^{i}a_{i-k}\cdot b_{k}\)​

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3. (20 points) Show that for all \(x > 0\), the following inequality holds:

\(1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2}\)

We need to expand \(\sqrt{1+x}=f(x)\), since \(f(x)-T_{2,0}=R_{2,0}\)

And \(T_{2,0}=1+\frac12x-\frac18x^2,\quad R_{2,0}=Ax^3\) where \(A=\frac{f'''(c)}{6},c\in[0,x]\)​

Then \(f(x)=1+\frac12x-\frac18x^2+Ax^3\)

Then clearly \(f(x)>1+\frac x2-\frac{x^2}{8}\)

For another part, consider \(1+x\leq (1+\frac{x}{2})^2\Rightarrow 1+x\leq 1+\frac{x^2}{4}+x\) clearly, this is true.

Since \(x>0\), then \(\sqrt{1+x}<1+\frac{x}{2}\)

Thus \(1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2}\)

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4. (20 points) Compute the Taylor polynomial of order \(N\) at \(x = 0\) for the function:

\(f(x) = \frac{1}{1+x}.\)

\(T_{N,0} = f(0) + \frac{f^{(1)}(0)}{1!} x + \frac{f^{(2)}(0)}{2!} x^2 + \cdots + \frac{f^{(N)}(0)}{N!} x^N\)

\(f(x) = \frac{1}{1+x}, \quad f'(x) = -1(1+x)^{-2}, \quad f''(x) = 2(1+x)^{-3}\)

\(f^{(3)}(x) = -6(1+x)^{-4}, \quad f^{(4)}(x) = 24(1+x)^{-5}, \, \dots\)

\(f^{(N)}(x) = (-1)^N \cdot N! \cdot (1+x)^{-(N+1)}\)

Then \(T_{N,0} = 1 + \frac{-1}{1!} x + \frac{2}{2!} x^2 + \cdots + \frac{(-1)^N \cdot N!}{N!} x^N\)

\(= 1 - x + x^2 - \cdots + (-1)^N x^N\)

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5. (20 points) Using exercise 4, compute the Taylor polynomials of order \(N\) at \(x = 0\) of the following functions:

(a)\(\frac{1}{x-1}\)

Since \(\frac{1}{x-1}=-\frac{1}{1+\left(-x\right)}\), then we use exercise4. Since \(T_{N,\frac{1}{1+x},0}=1-x+x^2-\cdots+(-1)^{N}x^{N}\)

Then \(T_{N,\frac{1}{1+\left(-x\right)},0}=1+x+x^2+\cdots+x^{N}\)

Then \(T_{N,-\frac{1}{1+\left(-x\right)},0}=-\left(1+x+x^2+\cdots+x^{N}\right)=-1-x-x^2-\cdots-x^{N}\)

(b)\(\frac{1}{(x-1)^2}\)

Since \(\frac{1}{(x-1)^2}=\frac{1}{x-1}\cdot\frac{1}{x-1}\), let \(\frac{1}{(x-1)^2}=f\cdot g\) where \(f=g=\frac{1}{x-1}\).

We know the coefficients of \(f\cdot g\) are \(\sum_{k=0}^{i}a_{i-k}\cdot b_{k}=\sum_{k=0}^{i}a_{i-k}\cdot a_{k}=\sum_{k=0}^{i}\left(-1\right)\cdot\left(-1\right)=\sum_{k=0}^{i}1=i\)

Thus \(T_{N,\frac{1}{(x-1)^2},0}=1+2x+3x^2+\cdots+\left(N+1\right)x^{N}\)

(c)\(\frac{2x+3}{(x+1)(x+3)}\)

Hint: Write \(\frac{2x+3}{(x+1)(x+3)}\) as a sum of two rational expressions.

Since \(\frac{2x+3}{(x+1)(x+3)}=\frac{\frac32}{x+3}+\frac{\frac12}{x+1}\), then for the first one \(\frac{\frac32}{x+3}=\frac32\cdot\frac{1}{x+3}=\frac32\cdot\frac13\cdot\frac{1}{\frac{x}{3}+1}=\frac12\cdot\frac{1}{\frac{x}{3}+1}\)​

\(T_{N,\frac{1}{\frac{x}{3}+1},0}=1-\frac{x}{3}+(\frac{x}{3})^2-\cdots+(-1)^{N}(\frac{x}{3})^{N}=1-\frac{x}{3}+\frac19x^2-\cdots+(-\frac13)^{N}x^{N}\)

Then \(T_{N,\frac{2x+3}{(x+1)(x+3)},0}=\frac12T_{N,\frac{1}{\frac{x}{3}+1},0}+\frac12T_{N,\frac{1}{x+1},0}=\frac12\left\lbrack1-\frac{x}{3}+\frac19x^2-\cdots+(-\frac13)^{N}x^{N}+1-x+x^2-\cdots+(-1)^{N}x^{N}\right\rbrack\)

\(=\frac12\left(2-\frac43x+\frac{10}{9}x^2-\cdots+\left(-1\right)^{N}\cdot\frac{1+3^{N}}{3^{N}}x^{N}\right)=1-\frac23x+\frac59x^2-\cdots+\left(-1\right)^{N}\cdot\frac{1+3^{N}}{2\cdot3^{N}}x^{N}\)​

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